- Earth Science
- Sheet that identifies rocks, minerals, and fossils in the kit
Yes. You can find it here:
- Will I be updating or redoing my upper grade Science courses?
The answer to this question depends on how Apologia ends up updating the courses. When I sold Apologia, I sold the copyrights of all my upper-grade courses. As a result, Apologia has the right to do whatever they want with them now. Thus, they will be producing new editions for those courses. In fact, they have already put out a new edition of The Human Body, and I find it to be, on balance, better than the older edition. I posted a review of that new edition here:
If the other editions are of the same quality, there will be no reason for me to do anything, as the courses will still be excellent for college-bound high school and junior high school students. However, if a new edition ends up being of lower quality (from a Christian, college-prep point of view), then I will most likely write a brand new one. I really pray that doesn't happen.
The answer above was written before the third edition of chemistry was published. Unfortunately, the third edition of chemistry is riddled with errors (even the second printing), and Apologia refuses to make the second edition available. Thus, I had to write a new chemistry course so that homeschoolers would still have a reliable source for high school chemistry. You can learn more about that course at Berean Builders.
- What type of microscope should a family buy?
Although there are inexpensive microscopes designed for elementary-age students, I don't recommend them. I suggest that you buy just one micrscope that you can use through high school. The younger children will need to be taught how to respect and care for it, but a good microscope should last more than 20 years if handled properly, and that way, you don't waste your money buying several different ones over the years.
For high school, you need a microscope that has three different magnifications, with the maximum magnifaction at 400x or higher. In addition, it needs both coarse and fine focus. This will allow you to see the cellular world, including yeast, pond water organisms, and the like. I think this is one of the best deals out there:
You don't need that specific one, of course. It just needs to have at least those specifications. If you don't have a specific curriculum that gives you instruction, I also strongly recommend a book to go with it, such as:
- Books I Recommend on Evolution
There are several good books about the inadequacies of the theory of evolution. Here are a few with some of my thoughts about them:
The New Creationism by Paul Garner. This book covers both evolution and the age of the earth. It is one of the best overall books in the creation/evolution controversy.
Refuting Evolution 2 by Jonathan Sarfati. This is more focused specifically on evolution.
Creation V. Evolution: What They Won’t Tell You in Biology Class. This has several authors and deals with specific questions, such as, "Can We Trust the Bible?" and "Are Humans and Chimps Really 99% Similar?"
A new one has just come out that I haven't read but has an impressive list of authors:
- Why does a pool noodle hold me up in the water?
In order to sink, the weight of any object must be larger than the weight of an equal amount of water. Corks float, for example, because a gallon of cork weighs significantly less than a gallon of water. A steel ball sinks because a gallon of steel weighs significantly more than a gallon of water.
People weigh a bit more than an equal volume of water, so they tend to sink. However, the Styrofoam out of which a noodle is made weighs a lot less than an equal volume of water. When you get on a noodle, it increases your weight by a small amount, but it adds a lot to your volume. The combined weight of you and the noodle is now less than the wieght of an equal volume of water, so you float.
- Why does the face turn blueish when there is a lack of oxygen?
When your blood is full of oxygen, it is bright red. When it lacks oxygen, it is dark red. Your skin distorts these colors, so when your blood is bright red, your skin looks pink. When it is dark red, your skin looks blue. If you look at your wrist on the palm side of your hand, for example, you will see blue lines. Those are veins that are close to the skin. The blood inside them is dark red, because it lacks oxygen. The way your skin distorts the color, however, makes them look blue.
Usually, your arteries have blood that is full of oxygen, and that bright red color (distorted by your skin to look pink) is more visible than the dark red color (distorted by your skin to look blue) that comes from the veins, which carry blood that has little oxygen. Thus, your skin looks pink. If the blood in your arteries starts to lose oxygen, it becomes dark red, like the blood in your veins. At that point, the dark red becomes more visible, and your skin distorts it to make it look blue.
Technically, this is called cyanosis.
- The Second Law and Evolution
Question: In Discovering Design with Chemistry and your blog, you say that the Second Law of Thermodynamics does not contradict biological evolution. Would you say that the evolutionary explanation for the formation of matter DOES contradict the Second Law of Thermodynamics (since the evolutionary model requires matter, or energy, [that is, the entire universe] to be formed from nothing)?
Answer: The Second Law of Thermodynamics says something very specific. It says that the entropy of the UNIVERSE can only increase or stay the same; it cannot decrease. This is why biological evolution doesn't contradict the Second Law. After all, an evolving animal is only one part of the universe. If an evolutionary event that causes a decrease in entropy for an organism also requires an increase in the entropy of the surroundings, then the total entropy of the UNIVERSE could stay the same or go up, allowing the evolutionary event to happen.
For this same reason, the formation of matter would, indeed, violate the Second Law. Even if the universe was a sea of matter and antimatter (as most evolutionary models require), at some point, the distribution of energy would have to shift so that there was more matter than antimatter. That would decrease the entropy of the entire UNIVERSE, which is not allowed.
- Does boiling water freeze faster than cooler water? If so, why?
Hot water can freeze faster than cooler water, depending on the specific situation. There are several factors involved. One of the important ones is mass. The more mass there is, the longer it takes for water to cool to the point where it can freeze. Hot water evaporates faster than cooler water, so it gets rid of some mass. As a result, if you put equal amounts of hot and cold water in the freezer, the hot water might freeze faster, but there will be less ice from the hot water than what will form from the cooler water.
There is also an effect related to the cooling of the water itself. Most of the cooling takes place at the surface. In hot water, this causes the difference between the surface temperature and the rest of the water to be greater, which encourages mixing, which allows more efficient cooling. Also, when you put the container in the freezer, it's possible for frost to build up on the surface, and frost is an insulator. Frost usually forms quickly, and if the water is hot, the frost will melt after it forms, getting rid of the insulation effect of the frost.
Finally (and I personally think this is the most important one), hot water has fewer gases dissolved in it, because heating drives gases out of solution. Dissolved gases lower the freezing point, so water with a lot of gas dissolved in it will have to cool to a lower temperature than water with less gas dissolved in it.
These factors are all fairly small, however, so you have to have just the right set of conditions to see the effect, which is now called the Mpemba effect. It is named after Erasto Batholomeo Mpemba, who did some experiments as a high school student to show that if the conditions are just right, hot water will freeze faster that cooler water. However, it has been noticed throughout history. Aristotle (384 BC - 322 BC) even mentions it.
- Physics first?
The reason a student usually does biology first, then chemistry,then physics is because of the mathematics needed. High school biology uses only simple arithmetic, high school chemistry uses algebra 1, and high school physics uses algebra 1, geometry, and basic trigonometry. Thus, it makes no sense to do physics first. The student has all the math skills needed for biology in 9th grade, so that's when biology can be done. Most students take algebra 1 in 9th grade, so by 10th grade, they are ready for chemistry. Most students take geometry in 10th grade, and most geometry courses have basic trignometry, so by 11th grade, most students are ready to take physics.
You can certainly do chemistry before biology, as long as the student has finished algebra 1, but your student should be taking biology while he or she is taking algebra 1.
- Was the water created before the light?
It's not clear. God's spirit was hovering over "the deep," and some have interpreted that to mean water. However, it could also mean "darkness." If "the deep" means "water," it's nothing like the water we see on earth now, as the earth was "formless and void" at that time.
- How do you use the measuring scoop?
I know it seems odd, but you do want to use the flat, rectangular spatula-like edge. When that reactangle is full, you have a "measure."
- At what stage does a hypothesis become a theory?
That question depends on several factors, including how detailed the hypothesis is. If the hypothesis covers one small aspect of science, it can become a theory pretty quickly, as long as there is confirming evidence. However, if a hypothesis is detailed, it can take a long time to test it well enough to determine whether or not it is a theory.
- Why can matter not be destroyed?
Honestly, we don't know. We just know that based on all the experiments we have done, chemical processes must have the same amount of matter after they complete as they did before they started.
- Why don'
- Are there enough labs in my courses for university preparation?
There are many, many homeschooled students who have used my materials and gone on to university, majoring in many scientific fields, including medicine. In this article, for example, I discuss a mother whose eldest son is in the fifth year of his MD/PhD program. My courses not only inspired him to study medicine, they allowed him to excel at university. Another student wrote me to tell me that he won the outstanding chemsitry student of the year award at his university, and he credits my course. Another student wrote to tell me that she originally went to university to study music and nutrition, but my books had prepared her so well for university-level science that she ended up getting a double-major in biology and chemistry. She then went to Harvard to get a Ph.D. in biomedical sciences.
I could go on and on. My courses clearly prepare students to study science at the univeristy level. If you are worried about labs, please understand that students do more labs at home using my courses than most high school students do at public school.
- Should Christians use evolution-based science materials?
I think it is very important to teach students critical thinking. The only way that can be done is to expose them to different views, even ones that you don't agree with. At the same time, however, it is your job as a parent to raise your children with the spiritual values that you want them to have. As a result, I think that in the early years, a Christian parent should use science curriculum that is written from a Christian point of view. As the children get older, you can use your parental judgement to decide when they are ready to be exposed to ideas that go counter to the spiritual values you want them to have. When they reach that point, you can start exposing them to those ideas.
How do you start exposing them? I think you have to make that decision for yourself, but let me tell you how I did it with my teenage daughter. All her science textbooks were from a Christian worldview. However, we read books written by scientists that argued for evolution. For example, we read Richard Dawkins's The Blind Watchmaker. That way, she was exposed to what were thought to be the best arguments for evolution at that time, but it wasn't the focus of her science education. Even in the teen years, I would keep the focus on a Christian view of science, but I would definitely incorporate good materials that give other views.
- Is general science necessary for those who finish the elementary
If you do all of my elementary books, general science is not necessary. You could go straight to physical science. If you plan to do that, I strongly encourage you to start using the tests that come with the elementary books as the student nears the end of the series. One of the goals of the general science course is to get them developing their study skills. If you are not doing general science, you will want to do that in the later parts of the elementary series.
- Is evolution not in essence contradictory to itself?
I wouldn't say evolution is contradictory to itself, since most evolutionists believe there are ways for new organs and structures to form without reducing fitness, and eventually, the new organs result in gains in fitness. Consider, for example, the way an evolutionist would imagine a fish evolving a lung. All fishes have swim bladders. They hold air, and the fish inflates the swim bladder to rise in the water and deflates it to sink.
From an evolutionist's point of view, some of the genes that produce and maintain that swim bladder got duplicated, and the duplicates started to mutate. Since they were duplicates, if the mutations caused the genes to stop working, that's no problem, because the original genes still work. Eventually, however, those mutated genes might produce more blood vessels around the air bladder. At that point, the air bladder inflates and deflates more easily, so that gives the fish an advantage. Eventually, tubes might be constructed from the mouth to the swim bladder. Until they are all connected, they might not provide an advantage, but they probably wouldn't cause a disadvantage, so the fish survives and passes those genes on. Eventually, the tubes connect the mouth to the swim bladder, and now the fish can rise to the surface and take in air. Because of the extra blood vessels around the swim bladder, this gets oxygen into the blood when the water is oxygen-poor. This offers an advantage, so the fish survives and passes that structure on. When the fish develops the other things it needs to venture onto land, it already has a primitive lung that allows it to breathe outside the water for a while. That produces even more advantage.
So I don't think evolutionists contradict themselves. However, there are two major problems. First, we know what mutations do - they either do nothing or damage the information content of the genome. They don't add information (like making tubes from the air bladder to the mouth) to the genome. However, evolutionists fervently believe that despite the fact that experiments involving tens of thousands of generations of fruit flies, bacteria, etc., have not seen mutations that add information to the genome, there must have been such mutations in the past. Second, there is no fossil record of such a transition happening. Thus, there is no evidence that this ever happened, and there is no way we can imagine it happening based on the experimental evidence we have now. That's why evolution is unscientific.
- When do you use H+ and when H3O+ in balancing Redox?
You can use either, because they are really the same thing. There are no free H+ ions in solution. They are always added to a water molecule. Thus, you can use either one. The only thing that changes is the number of water molecules on the other side. So a reaction like
ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+
ClO3¯ + 9H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H3O+
Both equatons are equivalent.
- Rounding up
When I say "round up," it means add one to the last digit you are keeping. So if you need to report 3.45 to two significant figures, you round up to 3.5. Since you are dropping a "5," the last digit you keep is one higher. If that's what you mean, round up when you are dropping a 5, 6, 7, 8, or 9.
- Am I going to put out any lectures for the advanced chemistry?
I don't have any plans at this time. You can always use this question/answer service to ask me questions about it, however.
- What math does Dr. Wile recommend for middle and high school?1. For most students, I like Math-U-See more than Saxon, but if the student prefers Saxon, it's a fine program. However, if the parent knows math and has time to help the student, you should look at Singapore Math, as it is a bit more rigorous than either of the others.2. Students who are gifted in math should consider The Art of Problem Solving.3. If the student prefers video, I think Videotext interactive is best.
- Ammonia Substitutes
Lye will work for most experiments that call for ammonia. Just use a weak solution of it. Also, many (not all) drain cleaners will work as well (once again, make a weak solution if it is not liquid already). The active ingredient should be sodium hydroxide or potassium hydroxide. This will not work in experiments where we are using ammonia to make a gas. For those experiments there is no substitute.
- High School Science
- Discovering Design with Chemistry
- Periodic Table of the Elements
If you would like a copy of the periodic table to print out, you can find it here:
- Sample Laboratory Notebook Entries
- Printable Worksheets
There is a document on the course website that contains all the "Comprehension Check" and Review questions, with space between them so you can write in your answers. You can get that document here:
- Can I use a weight scale instead of a mass scale?
Yes, you can use a weight scale instead of a mass scale for the experiments. However, you will need to convert from the weight unit to grams. If your scale reads the weight in ounces, multiply by 28.35 to convert to grams. If it reads in pounds, multiply by 453.6 to convert to grams. If it reads in Newtons, multiply by 101.97 to get grams.
Weight and mass are two different things, but on the surface of the earth, the factors given above allow you to convert between them.
Please note that the scale needs to be fairly precise. It needs to have an precision of 0.004 ounces (which is the same as 0.0002 pounds or 0.0009 Newtons).
- Do students need advanced chemistry after Discovering Design?
It depends on what you are trying to cover. If you want to cover a full year's worth of university-level chemistry in high school (this is typically referred to as an "AP course"), then yes, the student needs to take Advanced Chemistry in Creation after he or she takes Discovering Design with Chemistry. However, if your goal is a solid, university-prep chemistry course, then Discovering Design with Chemistry is all the student needs.
- How much weight should be given to each semester final?
Each semester final should be given twice the weight of an individual chapter test. Since there are eight chapters for each semester, that means you should add the eight chapter test grades together and then add the semester test grade multiplied by 2. Once you have that sum, divide by 10, and that will be the test average for the semester.
- How do you grade lab reports?
The grades on the lab reports shouldn't be based on the results, but instead on three criteria:
(1) How well did the student follow instructions?
(2) If there are calculations involved, did the student do the calculations properly (including with the proper number of significant figures)?
(3) If you had never read the student text, could you understand what the student did and what was learned by simply reading the student’s notebook?
The third one is the most important. If you want to assign point totals, I would say:
(1) How well did the student follow instructions? (30 points)
(2) If there are calculations involved, did the student do the calculations properly (including with the proper number of significant figures)? (20 points)
(3) If you had never read the student text, could you understand what the student did and what was learned by simply reading the student’s notebook? (50 points)
- Would this course work in a 32-33 week co-op?
This course will work well in a 32-week co-op. Two weeks per chapter is a good pace. The experiments are also well-suited to groups. The pace of this book is very similar to the pace of the second edition of Exploring Creation with Chemistry, so if you have used that book in your co-op, this one will work nicely as well.
- Do you need the Microchem kit for this course?
There really is no need for the Microchem kit with Discovering Design with Chemistry. There are 45 experiments in the course, and many of them are more detailed than what is in the Microchem kit. That was a great kit for supplementing Exploring Creation with Chemistry, but it just isn't necessary for Discovering Design with Chemistry.
- Solubility of Copper Iodide
Copper ions have similar solubility properties as silver ions. Thus, copper iodide (and all copper + Group 7A ions) are only slightly soluble in water. As a result, any significant concentration of it will result in a precipitate. Here is an expanded list of solubility rules that includes copper:
The actual solubility of copper iodide is 0.004 grams of copper iodide per 100 grams of water. Thus, any reaction between copper nitrate and potassium iodide that makes more than 0.004 grams of copper iodide per 100 grams of water will produce a precipitate.
- Can this be done in a 24-26 week co-op schedule?
Yes, you could do this in a 24-26 week schedule. There are two options. I prefer the first one:
1) Encourage the students to work through the book at their own pace, not worrying about getting done when the co-op is done. That way, they will be doing some experients ahead of the content, but that's fine. If you do that, here are the experiments I would do: 1.3, 2.1, 2.3, 3.2, 4.2, 4.3, 5.2, 6.1, 6.3, 7.1, 7.3, 8.1, 9.1, 9.3, 10.3, 10.4, 11.2, 11.3, 12.2, 13.2, 13.3, 14.1, 15.1, 16.2, 16.3
2) Try to cover the content at the same pace as the co-op. This will be tough for many students. I would do the same experiments, but I would monitor the students carefully, because that is a short period over which to cover a dense course!
- Extra practice problems and solutions
The course website, which is discussed in the introduction to the book is at:
On the website, you will find a link for each chapter. When you click on the link, there will be several links that appear, all related to that chapter. At the very bottom is a link to extra practice problems, with another link to the solutions.
- Is there a audio CD (book-on-CD) for the chemistry course
Yes, there is. The recording was completed in the middle of June (2016), and it is in the final editing stage. It should be realeased in the third week of July, 2016. It will be available from Berean Builders:
- What if my student's answer is a bit different from the book's?
There is always error in the last significant figure. As a result, when numbers differ by 1 or 2 in the last significant figure, they are treated by scientists as being the same. If your student's answer differs from the book's answer by 1 or 2 in the last significant figure, you shouldn't take any points off.
For example, 18.99 and 19.00 differ by only 1 in the last significant figure. Thus, from a scientific point of view, they are the same. Either is correct. Answers such as 18.98, 19.01, and 19.02 would also receive full credit.
- Are there alternate tests for the course?
There are alternate tests for the course. To get them, email me at:
Just let me know that you are a teacher needing alternate tests, and I will send them to you.
- A list of the elements, ordered by their chemical symbols
You can get the list of elements that is on the inside cover of the book here:
- Even more practice than the extra problems on the website
The only problems I have for the course are the ones in the book and the extra practice problems on the course website. However, there are a few resources online that might help. You can look at these websites and try to find the topics that you need more problems for:
- What calculator should I buy for use with chemistry?
Any calculator that has the word "scientific" in its description will work for the chemistry course. I personally prefer no-frills models, like this one:
- Description of the purpose of each experiment
The purpose of each experiment is discussed in the book itself, because I am using the experiment to illustrate something that the student is learning at the time. For example, in Experiment 1.1, the student is determining the conversion factor between inches and centimeters. Right before that experiment, I tell the student:
"Centimeters comes from a completely different unit system (the metric system), but since they also measure length, there should be some relationship between centimeters and inches. In the following experiment, you are going to figure out that relationship."
After the experiment, I discuss the result and what it means. This is true for each experiment.
- Strontium chloride for the experiments
Strontium chloride comes with the kit that is included in the course. However, if you are putting something together yourself, you can get strontium chloride from:
You can also substitute lithium chloride for strontium chloride in Expeirment 3.4 and calcium chloride for strontium chloride in Experiment 9.4.
- How to get denatured alcohol when it is unavailable
You can find denatured alcohol at most hardware stores as well as home-improvement stores. You can also find it in stores that sell catering materials (fondue and chaffing fuel), camping equipment (stove fuel), and marine supplies (marine stove fuel). It is sometimes called "denatured ethanol," "denatured ethyl alcohol," or "methylated spirits." Methanol and methyl hydrate are acceptable substitutes. Here are some brand names:
Kleen Strip Green
Blue Flame Fondue Fuel
Crown Alcohol Stove Fuel
Heet (NOT ISO Heet)
You may also use 70% isopropyl alcohol found in many stores. Make sure you don't use rubbing alcohol.
There are a few different consumer products in Canada that you can use for the experiments that require denatured alcohol. Marine stores sell "Marine alcohol," which is denatured alcohol:
Hardware stores sell "bioflame," which is also denatured alcohol:
You can also use methanol, which is sold as Methyl Hydrate
Finally, you may also use "Blue Flame Fondue Fuel".
- Full lab report or summary with each experiment?
It is important for a homeschooled student to document his or her experiments, because some universities require proof that a homeschooled student actually had a laboratory component for the course. However, documentation doesn't mean a lab report. In the introduction to the book (p. iv), I discuss how students should keep a lab notebook. I think it would be good to have a lab notebook entry for each experiment. That way, an evaluator can see exactly how much experimentation was done in the course. The notebook entry isn't a report, however. It is simply a record of the data (observations and measurments made in the expeirment) and a brief (NOT step-by-step) summary of what was done and what was learned. There are samples of notebook entries on the course website, which is also discussed in the introduction to the book (pp. ix-x).
- Is there an advanced chemistry book?
I do have an advanced chemistry book. It is published by Apologia. It's called Advanced Chemistry in Creation, and when added to Discovering Design with Chemistry, it covers all that is covered in the Advanced Placement Chemistry syllabus.
- How much Sodium Hydroxide and Strontium Chloride do I need?
Each student will use about 15 g of sodium hydroxide and 10 g of strontium chloride throughout the course. If you are doing lab groups, each lab group will use those amounts.
- How much Cupric Sulfate is needed for the course?
Each student needs 30 grams of cupric sulfate to do all of the labs in the course. If you are doing groups, each group needs 30 grams.
- How to report doing this course over a year and a half
If you do your transcript by semester, just call the three semesters "Chemistry 1," Chemistry 2," and "Chemistry 3." If this course is done in a year, it is an honors-level course. Thus, it can be done over three semesters and count as a normal high-school-level course.
If you do your transcript by subject, list "Chemistry" with 1.5 years' worth of credit. This is commonly done with honors courses, so it won't look odd for chemistry to have more than 1 year's worth of credit.
- Disposal of chemicals used in the experiments
You can pour all of the chemicals used in this course down the drain. It is best to use lots of water, just to make sure there isn't any residue left in the sink.
- Minimum number of labs for high school chemistry
Technically, it is done by hours. Students need to spend 30 hours in lab. That would translate to about 30-35 experiments, depending on the experiments you choose. If you complete all 46 experiments and all the chapters, it is the equivalent of an high school honors chemistry course.
- Experiments to skip so you have 13 in the first semester.
Skip the following:
Experiment 1.1: Determining the Relationship Between in and cm
Experiment 2.2: The Conservation of Mass
Experiment 3.1: The Wavelength of Microwaves
Experiment 4.2: Comparing a Metal and a Nonmetal
Experiment 6.2: In Between and All Around
Experiment 6.3: Copper-Plated Nails
Experiment 6.4: Burning Iron
Experiment 7.2: The Limiting Reactant
- SPARKlab links on the Berean website
The SPARKlab links on the course website are for use with digital probes that have been made by PASCO. These are for people who want to use computer-integrated technology in some of their labs. The program that opens them is the program that controls and reads the probes:
This is a supplement that is not a part of the standard course.
- What is a chemistry word that makes you smile?
That is probably the most interesting question I have answered on this site! The word "amphoteric" (can act as either an acid or a base) always makes me smile.
- When to do semester tests
As I say on page v of the Answer Key and Tests book, I don't think semester tests are necessary. Thus, if you are following the schedule in the Appendix, don't do them. I just provide them in case parents want their students to have cumulative tests.
- How to get denatured alcohol in Australia and New Zealand
A product called methylated spirits in Australia and New Zealand is an acceptable denatured alcohol for use in the burner found in the lab kit.
- Formal lab reports
I do not require students to write formal lab reports. I do not see any serious benefit to that. They need to learn how to document their labs, and that's what the notebook is for. There are sample notebook entries on the course website that help them see how a lab should be documented, but that's all they should be required to do.
- Can isopropyl alcohol be used in the alcohol burner?
It can only be used if it is 70% pure. Most isopropyl alcohol has a lot of water in it, so it won't burn hot enough to boil water. It is best to go to the hardware store and buy denatured alcohol.
- Are there reviews for the semester exams?
There aren't reviews for the semester exams. The best review would be to go over each chapter test in the semester.
- PDF of the reading schedule (Appendix B)
There is a PDF of the reading schedule on the course website. It is the second link on this page:
- How to make this an honors-level courseIf the student completes all the chapters and completes a lab notebook entry for every lab in the book (as discussed in the introduction to the course), the course is an honors course.
- Where are the answers to the comprehension checks
The answers to the comprehension checks are near the end of the chapter, right before the sample experiment calculations (if the chapter has them) and the chapter review. For Chapter 1, they start on page 27.
- How long should the tests take?
I wrote the tests to take an hour, but I don't necessarily think students should be required to get it done in that amount of time. It is a good benchmark, however.
- In a co-op setting, when should we do the experiments?
I am a strong proponent of doing the experiments before the students do the reading. That's because I designed the course for the students to do the experiment and then read about what happened. That way, they have the experience in mind as they read the explanation. In a co-op setting, this is best acheived by having the students do the experiment before they do the reading. Of course, this means they will not be able to write up the experiment until later, but that's fine. You can always help them with questions they have about the write up the next time you meet for class.
- I have the class recordings. Where are the videos you refer to?
The best way to watch the videos is to go here:
"Syllabus" tells you what each class covers, and when you click on the class, you see the video that you rented at the top, and all the videos to which I refer directly below.
- What is the difference between my new and old chemistry courses?
I wrote a blog article that details the answer to this question. You can find it here:
- Answer for question 29 on the second semester exam
The 2.0x10^3 g has only two significant figures, so the answer can have only two. The number is, indeed 44,400J, but you can only report that to two significant figures, which means 44,000J
- Sample Laboratory Notebook Entries
- Does it matter how many zeroes are between two significant figs?
It doesn't matter. In the number 300.0002, all five zeroes are significant. Remember, a significant figure represents a digit that has actually been measured. If the instrument used was precise enough to measure the "2," it had to be able to measure all the decimal places in between as well.
- Scale overload in Experiment 1.3
If you are getting a "scale overload" error on your scale when you measure the filled graduated cylinder in Experiment 1.3, the scale is probably bad. It is supposed to go from 0 to 500 g. The graduated cylinder doesn't even weigh 100 g, and the water would only weigh 50, so there is no reason the scale should overload. I guess one thing is to make sure you hit the "tare" button when there is nothing on the scale, to make sure it is reading 0 grams when nothing is on the scale. If you have done that and it still reads an overload, your scale is not working.
If you bought it from Nature's Workshop Plus or Berean Builders, call 317-745-0443 to get it replaced. If you bought it from another vendor call that vendor. Make sure the vendor understands that the scale is supposed to have a range of at least 0 to 500 g.
- Test Question #14
The student is expected to recognize that the ruler is marked off in centimeters, since there are 10 marks between the numbers. If the student reports the answer in inches, take off 1/4 of a point.
- When to round significant figures
As a general rule of thumb, round at the end of every equation, unless there is a change of rules. So, if the equation is: (223.1)x (13.7)/12 Do all the multiplication and division and then round to two significant figures. However, if the equation is: 12.3x(4.5 - 4.22) You need to do the subtraction first, round to the tenths place, then do the multiplication and then round to one significant figure. If a problem's solution has multiple steps involving multiple equations, follow this rule for each equation.
- Why doesn't the answer to the first part of Ex 1.3 need a unit?
The first answer in Example 1.3 doesn't need a unit because there isn't one. When the centimeters cancel in the math, you are left with no unit. Thus, you can't list one. The answer isn't 0.151 inches or 0.151 centimeters, because there isn't any kind of unit. The answer is just 0.151.
There is actually a physical reason for this. Suppose I measure the length of something to be 15.423 cm (the first number in the problem). Then, suppose I measure its width to be 102 cm (the second number in the problem). If I divide the length by the width, I get 0.151. Now...suppose I take the same object and measure its length and width in inches instead of centimeters. When I divide the two, I will still get 0.151. This is why 0.151 has no units. Regardless of the units you use, as long as you use the same unit in each measurement, you will always get 0.151 when you divide the two measurements. Thus, 0.151 is independent of units. That's why it has no units.
How do you know whether or not to list units? You always list the unit, if there is one. In the second part of Example 1.3, 7.0 in is mulitplied by 4.209 in. The units don't cancel there. They end up becoming in2. Since the units don't cancel, you must list the result: in2.
- Do significant figures decrease accuracy?
Significant figures do not reduce accuracy. They ensure that your answer is reported to the proper precision.
In a calculation like 70 g x 4 g, the initial answer is 280 g2, but that is too precise for the data you have. A measurement of 70 means the measurement could be something like 68 g or 73 g. If your measuring device can only be read to ten grams, then it cannot determine the ones position, so masses of 68 g and 73 g, for example, would both read as 70. In the same way, a measurement like 4 g could be something like 3.8 g or 4.3 g. Both of those masses would read as 4 g in a scale that can only read to the ones place. So, when you multiply the two, the answer could something like:
68 g x 3.8 g = 258.4 g2
or it could be something like:
73 g x 4.3 g = 313.9 g2
Based on the precision to which you know each mass, those results would be possible as well. Note how both of them round to 300 g2.
By properly reporting the answer to one significant figure (300 g2), you are indicating just how well you know the result. If you reported your answer as 280 g2, it would indicate that you know the answer to a higher precision than you actually know the answer.
- Why is the answer to Comp. Check #2 12.1 instead of 12?
Remember that addition and subtraction use different rules from multiplication and division. In addition and subtraction, you do not count significant figures. Instead you report your answer to the same precision (decimal place) as the least precise number in the problem (see the pink box on page 6).
In this problem, the numbers are 3.1 and 8.991. The first has its last significant figure in the tenths place. The second has its last significant figure in the thousandths place. The tenths place is less precise, so you must report your answer to the tenths place. That's why the answer is 12.1 inches.
You might want to relook at Example 1.2, which shows you how to use significant figures in addition and subtraction.
- Why is a graduated cylinder's precision not to the hundredths?
On page 3, when explaining a ruler, I tell you, "As a general rule of thumb, when you are reading a measuring device that has a scale to read, you can read the scale to a division below what is marked off." This is because you can estimate between the lines that are marked off.
If you look at the graduated cylinder on page 16, there are 9 little marks between the numbers, and each number marks off 10's of mL. That means each little mark represents 1 mL. After all, if there are 9 marks between 90 and 100, the marks are 91, 92, 93,...99.
You can estimate where the meniscus is between those little marks, so that allows you to read to the tenths place. So, as the discussion on that page tells you, you can see that the meniscus is a bit above 87, but still below 88. That means it is something like 87.2. The "87" is marked, and the "0.2" is your estimate between the lines. That's the best you can do. There is no way you can estimate hundreths, so a graduated cylinder can't be read to the hundredths place.
- How many sig figs are in 76000.0 g and 76000 g?
76000.0 g has 6 significant figures, because the last zero is significant, since it is at the end of the number and to the right of the decimal point. That makes all the other zeroes significant as well.
76000 g has only 2 significant figures, because there is no decimal, so none of the zeroes at the end of the number are signficant.
The idea here is that the zeroes in 76000 g are not measured. They are there because they have to be, to give the right value to the number. However, the measuring device wasn't precise enough to measure them. If you wanted to report 76000 g to a precision that indicated all those zeroes are significant, the only way you could do that is to put it in scientific notation:
With the decimal point there, now all the zeroes are significant. So if you had measured the mass to a precision of 1 gram, that's how you would have to report it.
- Why isn't 7 13/32 EXACTLY 7.40625 on page 2?
Think about the ruler for a moment. You are determining the length of what you are measuring using the lines that are drawn on the ruler. How thick is one of the lines that represents each 16th of an inch? It is probably more than 1/100 of an inch wide. Thus, if something ends up on the mark, it could be as much as 1/100 less than 1/16 or 1/100 more than 1/16. There is simply no way to know what part of the line actually represents that exact 16th of an inch. As a result, there is no way to know what is EXACTLY 1/16 of an inch, or 1/32 of an inch. Because of this, there is no way to read the ruler to that kind of precision. The width of the lines themselves limit how precise the ruler is.
- How to convert from liters to cubic centimeters
Your experiment showed you that:
1mL = 1 cubic centimter
So...if you have liters, you need to convert to milliliters. After that, whatever you have in mL is also the volume in cubic centimers.
For example, if you have 0.456 liters, you first convert to mL:
(0.456 L/1) x (1 mL/0.001 L) = 456 mL
That means the volume is also 456 cubic centimeters.
- In comp. check #7, how can 53000000000 have 5 significant figs?According to the rules of significant figures, 53000000000 has only two significant figures. However, suppose you were doing the following problem:5300.0x10000000.0The answer would be 53000000000. How many signficant figures does it need to have? According to the rules, 5300.0 has five significant figures, and 10000000.0 has 8. So, the rules tell us that we must report our answer to five significant figures. Since 53000000000 has only two significant figures, that can't be the answer. We have to report a number that has that VALUE, but has five significant figures. We do that with scientific notation. 5.3000x1010 has the same value, but it has five significant figures.That's one of the main reasons we use scientific notation. Sometimes, we get a decimal answer that we can't report, because there is no way to write it with the correct number of significant figures. Thus, we convert the answer into scientific notation to get the significant figures correct. This problem is just practice in how to do that.
- Ch. 1, comprehension check #4, Why isn't the answer 410?
Remember the rules of significant figures. When multiplying, the answer has to have the same number of significant figures as the measurement with the least. 45.5 cm has three significant figures, but 9 cm has only one. So the answer can have only one significant figure. 410 has two significant figures. 400 has only one significant figure, so that's the correct answer.
- In Ch. 1 rev #22, how is 1.0x10 g/mL the same as 10.49 g/mL?
Since you can only have two significant figures in the answer, you have to drop th "4" and "9." The problem is, that leaves you with "10," which has only one significant figure. As discussed in the text, this is a situation in which you need to use scientific notation. By putting the answer in scientific notation, you put in a decimal. That makes the zero significant. Thus, the answer has to be 1.0x10^1 g/mL, or 1.0x10 g/mL. that way, the zero is at the end of the number and to the right of the decimal, making it significant. As a result, that gives you two significant figures, which is what the rules say you need.
- How does mass relate to density??
Mass tells you the total amount of matter in a substance, while density tells you how tightly-packed that matter is.
Consider two objects, each of which has a mass of 1,000.0 grams. The first object is the size of a golf ball, and the second object is the size of a basketball. They each have the same total amount of matter in them (1,000.0 grams). However, the first object has that matter packed in a small amount of space, while the second object has it packed in a much larger amount of space. So while both have the same mass, the density of the first object is much higher than the density of the second object.
- Please explain conversion relationships.
A conversion relationship tells you how two things relate to one another. For example, the prefix "centi" means 0.01. That means a centimeter is 0.01 meters. We can write that as an equation:
1 cm = 0.01 m
That's a conversion relationship. Now...since the to sides equal each other, multiplying by a fraction that has one side on the top and the other side on the bottom is like multiplying by 1. It doesn't change the value, but it can change the unit. So....if I have something that measured 15.1 cm, I could multiply it with a fraction made up by the conversion relationship. If I take the "1 cm" side of the equation and put it on the bottom of the fraction that will allow the "cm" unit to cancel. Thus:
(15.1 cm/1)x(0.01 m/1 cm)
will cancel the cm and leave m. In other words, this is how we can convert between cm and m. 15.1 times 0.01 is 0.151, so 15.1 cm is the same as 0.151 m. On the other hand, if I have something that measured 4.3 m, I could convert it to cm by multiplying with the fraction the other way around:
(4.3 m/1)x(1 cm / 0.01 m)
Now m cancels and we are left with cm. Since 4.3 divided by 0.01 is 430, that means 4.3 m is the same as 430 cm.
- Why is the answer to #21 of review 1 20,600 instead of 20,618.5
Remember that you have to worry about significant figures. The mass (55,670 g) has four. That last zero is not significant, because it is not to the right of the decimal. It needs to be BOTH at the end of the number AND to the right of the decimal. The density (2.70 g/mL) has three. That last zero is significant, because it is both at the end of the number and to the right of the decimal. That means the lowest number of significant figures you have is three, so your answer can have only three. 20,600 mL has three. The last two zeroes are not significant, because they are not to the right of the decimal. They need to be BOTH at the end of the number AND to the right of the decimal. The zero after the "2" is significant, since it is between two significant figures.
- Stopping in between conversions
I stop between conversions in order to make the steps more clear. However, students need not do that. Their answers might not exactly match mine, but that's okay, because there is always error in the last significant figure.
The only time you should stop and worry about significant figures is when you change from addition/subtraction to multiplication/division or vice-versa. When that happens, you also change the rules for significant figures, so you have to round with the rules of the process you are currently using and then continue with the next process and round that process with the new rules. So in a situation like this:
(12.12)x(1.2) - (4.56)x(92.111)
You would do each multiplication and round based on the number of significant figures. Then you would do the subtraction and round based on the precision of the numbers that you just determined.
- With 3 sig.figs, what is 5996.25?
When you round, you look at the first number you drop. If it's 0-4, you do nothing, if it's 5-9, you round up. Since you are dropping a "6" to make it three significant figures, you have to round up, so the final answer is 6.00x103. This is an example of an answer that MUST be in scientific notation, since that's the only way to give it three significant figures.
- Comp. Check #3: Why isn't the answer 12 because of sig figs?
When you add and subtract, the errors affect each other in a specific way. When you multiply and divide, they affect each other in a completely different way. That's why there are two completely different rules. One rule deals with how error is affected by adding and subtracting, while the other rule deals with how error is affected my multiplying and dividing.
When you add 3.1 and 8.991, the second number is so very precise that there is no question about the value of the first three digits (8.99). Only the last digit (the 1) has some error in it. However, for 3.1, the 1 has error in it. Thus, the 3 is the only number that is certain. Both measurements are certain in the ones place, so your answer can be certain in the ones place. However, the first number (3.1) has uncertainty in the tenths place, so adding it to anything will also produce uncertainty in the tenths place. Thus, your answer must have uncertainty in the tenths place. As a result, the answer is 12.1, because it is certain in all 12 of the ones, but it is not certain in the tenths place. It doesn't matter that 12.1 has more significant figures than 3.1 and less than 8.991. It only matters which decimal place is uncertain.
For multiplying and dividing, the decimal place doesn't play a role anymore, because you are making a much bigger change than adding or subtracting. Counting significant figures takes that into account.
- Conversions: Where do I put the "1"
Remember that the "1" always goes with the unit that has the prefix. The definition of the prefix always goes with the base. So:
1 km = 1,0000 m
1 cL = 0.01 L
1 mg = 0.001 g
Notice that in all cases, the 1 is with the unit that has the prefix (km, cL, mg). The definition of the prefix is then with the base unit (m, L, g). When you multiply, you put the unit you want to cancel on the bottom of the fraction, along with its number. So to convert from km to m, you want km to cancel, so you would multiply by (1,000 m/ 1 km). If you want to convert from L to cL, you need L to cancel, so you would multiply by (1 cL / 0.01 L).
- How many significant figures in 800.0?
There are four. The last zero is significant, because it is at the end of the number and to the right of the decimal. That makes the other two zeroes in between significant figures. 800 has only one significant figure, but 800.0 has four.
- How do I know when to use scientific notation for an answer?
You can always use it if you want to use it, but the only time you MUST use it is when significant figures demand it. For example, if you are solving this:
The answer is 100.2, but you can have only two significant figures because of the 3.0. Thus, you have to report 100 to two significant figures. The only way to do that is with scientific notation: 1.0x102. However, if you have this equation:
The answer is 100.2 and you can keep all those significant figures. Thus, you can call it 100.2 or 1.002x102.
- Why is the answer for C C #11 not in scientific notation?
It doesn't need to be. The answer only requires two significant figures, and 740,000 cg has two significant figures. You could put it in scientific notation if you want (7.4x105 g), but you don't need to. The only time you must use scientific notation is when it is required by the significant figures rules. Any other time, its use is optional.
- Sample Laboratory Notebook Entry
- Chapter 2, Review Question 6b: Can student use a proportion?
Using a proportion is fine. If the student sets it up as:
11.2/75.0 = amount copper for new/29.86 g
11.2/75.0 = amount sulfur for new/15.06 g
11.2/75.0 = amount oxygen for new/30.1 g
he or she is using the same reasoning that I am using. The answers the student gets might be a bit different from the ones in the book, but only by 1 or 2 digits in the last significant figure. Since there is always error in the last significant figure, however, that is not a problem.
- Why is the answer to Comp. Check #9 12.8 instead of 13?
Multiplying by an exact number is the same as adding. Thus, when "2" is exact, 2x6.4 is really 6.4 + 6.4. Since that is an addition problem, we use the rules of addition and subtraction, which go by the decimal place. Since 6.4 goes out to the tenths place, the answer must go out to the tenths place as well. That's why the answer is 12.8 g.
However, the student might not understand that yet, so if the student gives the answer as 13 g, that's fine. When the student starts adding atomic masses together, the fact that multiplying by an exact number is actually addition will be discussed explicitly.
- Why aren't alpha particles deflected in the plum pudding model?
In the plum pudding model, the electrons are in the pudding. The alpha particles are repelled by the pudding, but they are attracted to the plums. On average, then, the attraction and repulsion cancel, and the alpha particles travel pretty much straight. This is different from the Rutherford model, because there, the electrons are away from the positive charges. Thus, there is nothing to cancel the positive charge's repulsion, and the alpha particles are deflected.
- Adding numbers and significant figures
When you have a situation where you add masses, like:
10.2 g + 98.1 g + 45.6 g = 153.9 g
You are gaining significant figures, but that's okay, because you are using the rule of addition and subtraction, which doesn't count significant figures.
Remember, the rules are different because an operation like adding or subtracting is different from an operation like multiplying or dividing. Addition and subtraction doesn't change the measurements nearly as much as multiplication and division, so the operations affect the precision of the measurements in different ways. The different rules take that into account.
- Review #6:Why is the factor 11.2g/75.0g instead of 75g/11.2g?
In this case, we want to make LESS than the known recipe, so we have to scale down. That means the multiplication factor is less than one. The way to remember it is ALWAYS divide the amount you want to make by the amount you know how to make. That will always give you the right multiplication factor.
- We get the wrong wavelength in Experiment 3.1
The frequency of a microwave is generally about 2,500 megaHertz, which is 2.5x109 Hz. If you see "60 Hz" on a label, that's for the power coming in from the power cord, not the microwaves. Look for a frequency labeled in "MHz," which is megaHertz. That stands for a million Hertz.
- How do I do the wavelength and frequency math on the calculator?
Let's suppose you have a wavelength of 3.95x10^-7 m. To turn that into frequency, you would need to do:
f = (c) / (wavelength)
So you need to divide c (3.0 x 10^8 m/s) by 3.95x10^-7 m.
When you divide 3.0 x 10^8 by 3.95 x 10^-7, you are dividing 3 by 3.95 and 10^8 by 10^-7. 3.0 divided by 3.95 is 0.76. What's 10^8 divided by 10^-7? When you divide numbers with exponents, you SUBTRACT the exponents. So 10^8 divided by 10^-7 is 10^(8 - -7), which is 10^15. Thus, the answer is 0.76 x 10^15, which is the same as 7.6 x 10 ^14.
Of course, it is much easier to do this on your calculator. Most calculators have an "EE" or "EXP" key. That key abbreviates "x 10^". So, let's assume your calculator has an "EXP" key. The calculator that comes on Windows computers is like that. Here is how you would input the division:
3.0 EXP 8 divided by 3.95 EXP 7 +/- =
That should give you the proper answer, 7.6 x 10^14.
- The egg whites only cook on the edges of the pan in Exp. 3.1
Try doing it without the juice glasses, sitting the pan on the bottom of the microwave oven. Different microwaves are set up differently. For most, the upper part of the oven is sensitive to the crests and troughs of the waves. For some, it is the lower part. If nothing else, you can watch this video:
- Review # 11, how do you do this on a calculator?
The important thing is to learn how to put scientific notation in your calculator. Most calculators have an "EE" or an "EXP" button. That abbreviates "x10^" Thus, when you put in
3.00 x 10^8
You must input it as 3.00 EE (or EXP) 8. Your calculator will then show 3.00 with a little "08" on the right. That means 3.00 x 10^8. To continue, you would hit the divide button, and then to input
2.31 x 10^-7
you would hit 2.31 EE (or EXP) 7 +/-. Your calculator will then show 2.31 with a little "-07" on the right. That means 2.31 x 10^-7. When you hit the equal sign, you should see a 1.2987 with a little "15" on the right. That means 1.2987x10^15.
If your calculator doesn't have an EE or EXP button, you need to look at the manual and learn the proper way to input scientific notation.
- Strontium chloride in Experiment 3.2
Strontium chloride comes with the kit that is included in the course. However, if you are putting something together yourself, you can get strontium chloride from:
You can also substitute lithium chloride for strontium chloride in this experiment.
- Why don't we see color from the other ions in Exp. 3.2?
You don't see color from the chlorine ion in NaCl and SrCl2 or the sulfate ion in cupric sulfate because those substances do not emit light in the visible spectrum. The electron energy levels of negative ions tend to be very far apart, so the light they emit when excited is very high energy (ultraviolet). As a result, the light is there, but you don't see it.
- How to do the math on Example 3.3
It is best to use a scientific calculator in these problems. Most calculators have an "EE" or an "EXP" key, which abbreviates "x10^" Thus, for the first equation:
you would enter:
1 EXP 9 +/-
The calculator should display 4.75 with a small -7 at the top right, indicating the answer is 4.75x10-7
For the next equation:
(3.00 x 108) / (4.75x10-7)
you would enter
3.00 EXP 8
4.75 EXP 7 +/-
That will give you 6.32 with a little 14 on the top right, indicating 6.32x1014
You can do the equations longhand, remebering that when you multiply numbers with exponents, the exponents add and when you divide, they subtract. Thus:
(3.00 x 108) / (4.75x10-7)
3.00/4.75 x 108/10-7
3.00 divide by 4.75 is 0.632
108 divided by 10-7 is 10 raised to the 8 - -7, which is 15. Thus, you have:
0.632 x 1015
To put this in scientific notation, you have to move the decimal to the right one place. That makes the number bigger, so to compensate for that, you make the exponent smaller:
6.32 x 1014
- What happens to energy when electrons move away from the nucleus
When an electron moves to a lower orbital (closer to the nucleus), it has to emit light because it has too much energy to be in the lower orbital. So, it gets rid of that excess energy by emitting light. When an electron moves to a higher orbital (away from the nucleus), it needs to gain energy, because it doesn't have enough energy to get to the higher orbital. It gains that energy by absorbing light.
Remember, energy can't be created or destroyed. So, when an electron does something that requires more energy than it has, it needs to gain that energy. The energy can't be created, so it must be taken from something. The electron absorbs light, taking the light's energy. When it does something that requires less energy than what it has, it needs to lose that energy. It does that by emitting light, and the light takes that energy away.
- Saucer for Experiment 3.2
Any "oven safe" saucer will work great. If it shouldn't be placed under a broiler, it probably shouldn't be used. You also don't have to use a saucer. If you have a metal pie pan or something like that which will fit in your sink, that will work as well. It just needs to be something that can get hot without melting or catching on fire and can also fit in a sink.
- The Math for Review problem 3.
If you are getting a number that is way off for your answer, you are not entering the scientific notation properly in your calculator. If you do the math longhand, you can come up with the proper answer:
(2.49 x 10-23) / (1.66 x 10-24)
You can do the numbers and exponents separately:
2.49 divided by 1.66 is 1.50
When dividing numbers with exponents, you subtract, so:
(10-23/10-24) = 10-23 - -24 = 101
So the answer is 1.50x101, or 15.0.
You need to find out how to properly put scientific notation into your calculator. It depends on the calculator, but you should be able to google instructions, if you use your calculator's make and modle.
- Why do groups 1B and 2B come after 3-8B on the periodic table?
It's an attempt to label the most common charges for the last two transition metal groups. In most ionic compounds, copper, silver, and gold usually take on the 1+ charge, while zinc, cadmium, and mercury usually take on the 2+ charge. However, that doesn't really apply to the numbers of the other B groups, so it's not clear why it's still in use.
- How do you do the conversion in comprehension check 5?
We have the equation
f = (c) / (wavelength)
Doing algebra to solve for wavelength tells us:
wavelength = (c) / (f)
So you need to divide c (3.00 x 108 m/s) by 7.89x1012 1/s.
It is best to do this on your calculator. Most calculators have an "EE" or "EXP" key. That key abbreviates "x 10^". So, let's assume your calculator has an "EXP" key. The calculator that comes on Windows computers is like that. Here is how you would input the division:
3.00 EXP 8 divided by 7.89 EXP 12 =
That should give you the proper answer, 3.80 x 10-7 m.
To get to nanometers, you use the fact that 1 nm = 10-9 m.
(3.80 x 10-5 m / 1) * (1 nm / 10-9 m)
Once again, it is best to do this on your calculator. Remember that 10-9 can also be expressed as 1x10-9. In the calculator, then, you would put:
3.80 EXP 5 +/- divided by 1 EXP 9 +/- =
That should give you 380 nm. You need three significant figures, however, so the only proper way to express the answer is 3.80x102 nm.
- Are there really 98 naturally occurring elements?
Yes, there really are 98 naturally-occurring elements. Most websites and texts list 92, but that's just not true. Elements 1-92 are stable (with the exception of technetium), so some texts and websites say there are 91 naturally-occurring elements. However, both of those numbers are wrong. Some of the unstable elements are made by natural processes and exist in trace quantities in nature. In uranium-rich pitchblende, you can find traces of technetium, neptunium, plutonium, americium, curium, berkelium, and californium. They are all produced in small quantities by the breakdown of uranium-235 and uranium-238. Now, it was once thought that all seven of those elements can only be artificially produced (indeed, that's how they were discovered), but we now know differently.
Thus, the current number of naturally-occurring elements is 98.
- Comprehension check questions 9 and 10
For #9, remember that the energy of light that is emitted is equal to the difference in energy between the orbits. Orbit 2 is farther from the nucleus than orbit 1, so it has a higher energy than orbit 1. However, orbit 5 is even farther away, so it has an even higher energy. When the electron moves from 5 to 1, then, it must emit more energy than when it moves from 2 to 1. Remember that energy is determined by frequency, so the higher the energy, the higher the frequency. However, frequency are wavelength are inversely related, so the higher the frequency, the SHORTER the wavelenth.
For #10, you have to think about every possible way the electron can move from n=4 to n=1 (which is always the ground state). It can jump directly from 4 to 1. That will result in one wavelength of light. However, it can go from 4 to 3 and then 3 to 1. That will result in 2 more wavelengths. Continue that reasoning to figure out the answer.
- Comprehension Check #10
Remember, an electron has many choices as to how it releases its energy, but it must make it to the n=1 orbit. Each jump from one orbit to another requires a specific energy and thus a specific wavelength.
It starts in n=4. It could jump straight down to n=1. That would be one amount of energy to release, and therefore one wavelength of light that must be emitted. However, it could also go first to n=3 and then to n=1. The first jump requires a specific energy, as does the second. That's two more wavelengths of light. It could also go first to n=2 and then n=1. That's two more energies and thus two more wavelengths. It could also go to n=3, then n=2, then n=1. However, we already considered it going from n=2 to n=1, so that doesn't count as a new energy and thus it is not a new wavelength. We also already considered it going first to n=3, so that is also not a new energy and not a new wavelength. We have not considered it going from n=3 to n=2, so that IS a new energy and thus a new wavelength. That makes a total of six.
- How does hydrogen get its ideal electron configuration
Hydrogen's ideal electron configuration exists when it has two electrons, because then it is like the nearest noble gas, helium. As a nonmetal, it will gain an electron to get its ideal configuration.
- No bubbles on one tack in Experiment 4.3
If bubbles appear on one tack and not the other, your battery is fine. The only way bubbles could form on either tack is if electricity is being conducted.
The tack touching the positive terminal of the battery might not have looked like it was bubbling, but it probably was. The oxygen that is forming there tends to make tiny bubbles that can be hard to see through the water. One thing you might try to do is add significantly less salt (1/2 of a teaspoon instead of a tablespoon). When the concentration of salt is high, other reactions can occur, which might obscure the oxygen being formed.
- Did the salt or sugar change in Experiment 4.3?
On page 119, just below the definition of electrolysis, the text says, "So the only chemical in the experiment that was actually changing was the water. The salt didn’t change, nor did the sugar." However, they both dissolved in the water. Isn't that a change?
As you will learn in an upcoming chapter, when a chemical dissolves, the chemical itself does not change. Salt water tastes salty because the dissolved salt is still salt. Sugar water tastes sweet because the dissolved sugar is still sugar. So the only chemical that changed in the experiment was water. It stopped being water and became hydrogen and oxygen instead. The salt stayed salt, and the sugar stayed sugar.
- Succinct definition of the quantum mechanical model of the atom
The quantum mechanical model of the atom is one in which the electrons are treated as waves rather than particles. As the text discusses, electrons can act as either particles or waves. The Bohr model treats the electrons as particles that orbit the nucleus. The quantum mechanical model treats them as waves that occupy orbitals. As the book says:
"[Schrodinger] thought that perhaps the problem with the Bohr model was that it forced electrons to behave like particles. Of course, it makes sense that they are particles, but it also made sense that light is a wave. If light (which should be a wave) can act like a particle, perhaps electrons (which should be particles) can act like waves. As a result, he analyzed the electrons in the atom as if they behaved like waves instead of particles...because he analyzed the electrons as waves instead of particles, they weren’t constrained to circular orbits. Instead, they were constrained to specific shapes called orbitals."
- How do I determine charge from the periodic table?
To determine the charge from the periodic table, look at the column the element is in, and think about what it needs to do to reach group 8A. Those in group 1A (with the exception of hydrogen), for example, must lose one electron. That makes them have the same valence electron configuration as the elements in group 8A. Since they lose one electron, they are 1+. For those in group 2A, they must lose two electrons, so they become 2+. The elements in Group 7A need to gain one electron to become like the elements in Group 8A, so they become 1-. The elements in Group 6A need to gain two electrons to become like the elements in Group 8A, so they become 2-.
- Light bulb for Experiment 4.1
A compact flourescent light bulb will work for this experiment. A white LED bulb will also work.
- Ex 4.2 What is the ground state electron configuration of Pd?
Pd is an exception to the rules we have been discussing. It probably wasn't the best example to give because of that. Based on our rules, the ground state configuration is what I give in the book:1s22s22p63s23p64s23d104p65s24d8As far as a general chemistry student goes, that's correct. However, there is another effect that our rules don't take into account.A full orbital is actually lower in energy than a partially-filled orbital. Thus, atoms will often shift their electrons around to get the most possible decrease in energy by filling an orbital. Well, it turns out that going from 4d8 to 4d10 reduces the energy substantially, because the 4d orbital loses a lot of energy when filled. In fact, the energy reduction is more than the difference between the 4d and 5s orbitals. So even though the 5s orbital (when fully filled) is higher in energy than the 4d orbital (when fully filled), a configuration of 4d10 is a lower energy state than a configuration of 5s24d8.There are other exceptions like that. For general chemistry students, however, the exceptions aren't worth learning.
- How do we determine the valence electrons in the B columns?
The number of valence electrons in an element in one of the B columns is the number of s and d electrons, which is equal to the number of boxes from the left side of the periodic table. For example, zircon (Zr) has four valence electrons. Its electron configuration is [Kr]5s24d2, so it has two "s" electrons and two "d" electrons for a total of four. It is also the fourth box from the left. Silver has 11 because it is the eleventh from the left and its electron configuration is [Kr]5s24d9.
It's important to note, however, that the valence electrons in the B-column elements doesn't tell you much about their charges in ionic compounds. Silver, for example, has 11 valence electron, but its main charge in ionic compounds is 1+. These are the "rule-breakers," so it is hard to determine their charge from their valence electrons.
- Why aren't the 1s orbitals included in the second energy level?
We don't include the 1s orbital in the number of orbitals in the 2nd energy level, because the number right before the letter indicates the energy level. The 1s orbital is in the first energy level, while the 2s and 2p orbitals are in the second energy level. It is true that the 1s orbital must fill up before the 2s orbital can start filling, but the energy of the 2s orbital is much higher than the energy of the 1s orbital, because it is on a different energy level. You might want to review page 101, right under the illustration of the various s orbitals.
- Why does a 2p orbital fill before a 3s orbital?
Electrons fill orbitals based on the energy of the orbital. 1s has the lowest energy, so it fills first. 2s has the next lowest energy, so it fills next. 2p has lower energy than 3s but higher energy than 2s. So it fills after the 2s but before the 3s. In the same way, 3p is higher in energy than 4s, so after 2p and 3s fill, then the 3p orbitals fill. However, 3d orbitals are higher in energy than 4s, so after 3p fills, the 4s fills and then the 3d fills. This can be hard to remember, so that's why I have you walk through the periodic table. It is arranged so that you know the order in which the orbitals fill.
- Which group 8A atom do you use to abbreviate electron configs?
You use the one that has the closest atomic number that is LOWER than the atomic number of the element whose electron configuration you want to abbreviate. So, if you want to do Se (atomic number 34), you would use Ar, because its atomic number is lower than 34 but closer than the other two Group 8A elements with lower atomic numbers. In the same way, for iodine (atomic number 53), you would use Kr (atomic number 36). It the group 8A element with the closest atomic number that is lower than iodine's.
- Can any metal chemically bond with any other metal?It is very rare for metals to bond with other metals. They generally lose electrons, so they like to bond to nonmetals. You can form mixtures of metals, called alloys. However alloys are mixtures. They are not compounds. Brass, for example is a mixture of copper and zinc.
- Lanthanide and Actinide electron configurations
The Lanthanides and Actinides have pretty complex electron configurations, due to their access to f orbitals. This page discusses them.
- Experiment 4.1 - How far away are we supposed to stand?
You should start as far from the light as you can get and still be in the same room. You will eventually walk closer and then back again, so you will end up exploring a wide range of distances.
- Gold-colored coating coming off one thumbtack in Experiment 4.3
Most thumbtacks are made of steel with a nickel or bronze coating. It sounds like yours had a bronze coating. The bronze is not very reactive, but the steel likes to react with the oxygen made from the breakdown of water. The coating does not make a perfect seal, so some of the oxygen got under the coating and reacted with the steel. That made rust, which does not adhere to the bronze coating, so the coating came off.
This happens only at the postive side of the battery, because that's where oxygen is made. In order to make oxygen from water, electrons have to be pulled from the oxygen. That can only happen at the positive terminal. The hydrogen is made by forcing electrons only hydrogen, and therefore the hydrogen is made at the negative terminal. Hydrogen doesn't react with eigther bronze or steel, so nothing happened there.
- Experiment 4.3: Water turned yellow and black solid produced
In this experiment, there are other reactions happening besides the breakdown of water. The metal touching the positive terminal is reacting with the oxygen produced there, and that tend to make a metal oxide, which is the black solid. Most thumbtacks have a lot of nickel, so it is nickel (II) oxide. The yellow is caused by the small amount of metal oxide that dissolves.
- Answers to Review 21
I am sorry that the names were left off. They are:
a. Rubidium oxide
b. Strontium sulfide
c. Calcium nitride
d. Magnesium iodide
- Experiment 4.3: Can any color plastic cup be used?
Any color will work, but if you can't see through the side of the cup, you will need to look into the cup from the top, and that might make it harder to see what is going on.
- How many electrons can a single orbital hold?
A single orbital can hold only two electrons. The reason is that each electron in an orbital must have its own spin state, which reduces the repulsion that the electrons feel. As a result, each orbital can hold only two. However, remember that there are multiple p and d orbitals for every energy level. There are three p orbitals in every energy level, and each of them can hold two electrons, so there can be up to six electrons in the p orbitals of a given energy level. There are five d orbitals, and each can hold two electrons, so there can be up to ten electrons in the d orbitals of a given energy level.
- Do you need an incandescent light bulb for experiment 4.1?
You do not need an incandescent light bulb. You will get better rainbows with an incandescent light bulb, but a fluorescent or LCD light bulb will work as well.
- Why isn't Astatine listed as a homonuclear diatomic?
In the second edition of Exploring Creation with Chemistry (which I wrote), Astatine (At) is listed as a homonuclear diatomic. However, in Discovering Design with Chemistry, it is not. The reason is that astatine is only theoretically a homonuclear diatomic. All its isotopes are radioactive, and they decay fairly quickly. Thus, it is hard for two astatine atoms to live long enough to encounter one another and form a bond. If they did, however, they would form a bond and become a homonuclear diatomic molecule. That means you can think of it as a homonuclear diatomic, even though you will probably never encounter it in that form.
- Why doesn't CF4 form purely covalent bonds?
CF4 is a purely covalent molecule, but it does not have purely covalent bonds. A purely covalent bond is one in which the electrons are shared equally. CF4 doesn't have those bonds, because F is much more electronegative than C. It is a purely covalent molecule because its polar covalent bonds cancel each other out. So the question is asking specifically about the bonds,not the molecule as a whole.
- How do you determine if a molecule is polar or nonpolar?
This is a three-step process:
1. Determine if the molecule is covalent. It must have only nonmetals in it to be covalent. If there are metals in the molecule, it is ionic.
2. If the molecule is covalent, draw its Lewis structure. If there are bonds between different elements, then the electrons are shared unevenly, and the molecule has polar BONDS. It might not be a polar molecule, however. If the bonds are between idental atoms (or atoms with the same electronegativity), then the molecule is nonpolar, which is the same as purely covalent.
3. If there are polar bonds in the molecule, determine its geometry. If the molecule is tetrahedral with the same elements surrounding the central atom (like CH4), then the polar bonds cancel each other out, and the molecule is nonpolar, which is the same as purely covalent. If the molecule is linear with the same elements on either side of the central atom (like CO2), then the polar bonds cancel out, and the molecule is once again nonpolar, which is the same as purely covalent. If the geometry is pyramidal (like NH3), bent (like H2O), or linear with only two atoms in the molecule (like HCl), then it is polar.
You might want to check out the videos on the course website (which is given in the introduction to the course). There are several vidoes that discusse drawing Lewis structures, determing shapes, and determining polarity.
- Why are commas used in the electronegativity chart?
There are some countries where a comma has been traditionally used instead of a decimal point. While most countries now accept that the decimal point is an international standard, some still use the traditions of their country. The layout person who put together the chart on page 138 is from France, where the comma was traditional. In case you are interested, here is a map that shows you the places in the world where a comma has been traditionally used in place of a decimal point:
- More explanation for Example 5.4
Have you taken a look at the course website, which is discussed in the introduction to the book?
The section on Chapter 5 has two very good videos that might help:
The links are labelled "VSPER theory" and "More on VSEPR Theory." They may help.
The real key in figuring the shapes out is to figure out how many GROUPS of electrons are around the central atom and how many of those groups are bonds. In the first molecule of Example 5.4, there is a triple bond and a single bond on the carbon, which is the central atom. Each bond is a group, so there are two groups of electrons around it. How can two groups get as far apart as possible? By forming a line, as discussion on the bottom of page 145. Thus, this molecule is linear.
In the second molecule, there are three bonds on the N and one lone pair of electrons. That's 4 groups. How can 4 groups get as far away from each other as possible? By forming a tetrahedron, as discussed on page 143. A real tetrahedron, however, has 4 bonds, with each bond being a leg on the tetrahedron. This molecule has only 3. So this molecule is a tetrahedron with one leg removed. As discussed on page 144, that's pyramidal.
For the last molecule in the example, there are three bonds. One is a double bond, and the other two are single bonds. Each bond is a group, so there are three groups. How can three groups get as far away from each other as possible? By forming a triangle, as discussed on page 146. Thus, this is triangular.
Notice that the last two molecules each have three BONDS around the central atom, but they have different shapes. Why? Because the NH3 has four GROUPS of electrons, and the number of GROUPS form the basic shape. 4 groups form the basic shape of a tetrahedron. For each group that is not a bond, you remove a leg from the tetrahedron. So when one group isn't a bond, it's a tetrahedron with one leg removed, which is pyramidal. When 2 groups aren't bonds, it is a tetrahedron with two legs removed, which is bent. 3 groups make the basic shape of a triangle. If one of the groups isn't a bond, it is a triangle with one leg removed, which is bent. 2 groups form the linear shape.
- How do you determine the number of moveable groups of electrons?
A moveable group of electrons is either a bond or a pair of electrons that don't form a bond. Remember that a bond counts as a single group, regardless of whether it is a single, double, or triple bond. So, in a molecule like CH2O (shown on the bottom of p. 133, there are two single bonds and a double bond on the C. That means three moveable groups. In HCN (shown on the bottom of p. 134), there is a single bond and a triple bond on the C. That means two moveable groups. In H2O (shown in the middle of page 140), there are two single bonds and two pairs of electrons that don't form a bond on the O. That means there are four moveable groups.
- Why is water bent? Shouldn't the bonds be 180 degrees apart.
Remember that there are a total of four pairs of electrons around the central atom. All of those electron pairs repel each other, so they all want to get as far apart from one another as possible. If the bonds were 180 degrees apart, they would be farther from each other than in the bent configuration, but the lone pairs would be only 90 degrees away from the bonds. So the bonds would be farther from one another, but the lone pairs would be closer to the bonds. Those lone pairs repel as well, so they won't want to get that close. So...all four pairs of electrons get as far apart as possible. Thus, they form a tetrahedron. However, two of the pairs are not bonds, so they don't make "legs" on the tetrahedron. That means you have a tetrahedron with two legs missing, which is a bent structure.
- Why commas instead of decimals in electronegativity table?
There are many countries in which a comma is used in numbers instead of a decimal point. One of those countries is Sweden. The chemist who first came up with the idea of electronegativity was Jöns Jacob Berzelius, who was from Sweden. In addition, the man who came up with today's scale (Wolfgang Pauli) was Austrian. Austria is another country that uses commas instead of decimal points. Since the two main people associated with electronegativity come from countries that use commas instead of decimal points, most electronegativity tables do that.
- How do we know the angles between the bonds of CH4 are 109.5
Here is a page that goes through the geometric proof to show that a tetrahedral molecule has bond angles of 109.5 degrees.
- Do electrons move in nonpolar molecules?
The electrons definitely are moving in a polar compound. They move in orbitals around both atoms that share them. However, there is no NET motion. In a polar molecule, the electrons move around one of the atoms more than they move around the other. Thus, there is net motion that favors one atom. That atom is the one with the partial negative charge. The atom that is not favored in the net motion has the partial positive charge. In nonpolar molecules, the atoms are moving, but they move around both atoms equally, so there is no net motion.
- Can the Cl and F positions in the SiF2Cl2 diagram be switched?
Yes. As log as you have all Fs and Cls attached to the Si with single bonds. The order is not important.
- Step 28 in Experiment 6.2
You do need to repeat steps 24 and 25 once you have poured the two liquids together. You measured the mass of the first graduated cylinder plus its 25 mL of liquid. You then measured the mass of the second graduated cylinder plus its 25 mL of liquid. When you add the two, you get the total mass of the system, which includes both graduated cylinders and both samples of liquid. To see that the mass doesn't change, then, you need the compare it to the final mass of both graduated cylinders and both samples of liquid.
Both samples of liquid are in only one graduated cylinder in step 28, so when you put that on the scale, you are measuring the mass of one graduated cylinder and the mass of both samples together. However, to get the total mass of the system at step 28, you still have to measure the mass of the empty graduated cylinder. That way, the total mass still includes both graduated cylinders and both samples of liquid.
- How do you get the subscripts in review questions 15a and 15b?
15a asks for an equation for the formation of CaCO3. As discussed on pp. 182-183 in the book, a formation reaction has reactants that are all elements, and the only product is what is being formed (CaCO3, in this case).
Now, if you look at the elements, you have Ca, C, and O. From chapter 5 (p. 128), students were told to memorize seven homonuclear diatomic elements. O is on that list. This means the element O exists as O2. Since Ca and C are not on that list, those elements exist just as Ca and C. Thus, the unbalanced equation becomes:
Ca + C + O2 --> CaCO3
15b asks for a decomposition equation. Those reactions start with the chemical given (K2CrO4), but the products are all elements. Thus, the products involve K, Cr, and O. Once again, O is a homonuclear diatomic, so it is O2. K and Cr are not, so they are just K and Cr. That means the unbalanced equation is:
K2CrO4 --> K + Cr + O2
- Ch 6, comprehension check #2: how to get -60.0 from 9/5x(-108)
I strongly recommend using a scientific calculator in this course. They are very inexpensive, and they help you do this kind of math more efficiently. This is the calculator I recommend:
However, if you want to do it longhand, you need to remember that a fraction means multiply by the number in the numerator and divide by the number in the denominator. Thus, you first multiply by the 5:
5x-108 = -540
Then, you divide by the 9:
-540 ÷ 9 = -60
Since 108 has three significant figures and 9/5 is exact, the answer should have three. That's why it is -60.0 degrees C.
- Experiment 6.1, opposite results from what's expected
What was the temperature of the water and alcohol compared to the temperature of the room? If the water and alcohol were a lot colder than the room, the thermometer might be increasing in temperature because the room is warming it more than the evaporation is cooling it. Even if that's the case, the thermometer should never reach room temperature again, because the evaporation should keep it cooler than room temperature.
Let the water and alcohol sit for a while so they come to room temperature, then try it again.
- Exp. 6.3: why did the galvanized nail turn black not coppery?
While it is copper that is formed on the nail, it might not look "coppery" in color because of contaminants in the water and on the nail. Those contaminants get incorporated into the copper and can turn it black.
- Substitute for liquid food coloring in Experiment 6.2
If you can't find liquid food coloring, you could use something like flavored drink mix that has a definite color. Add very little water to a lot of the drink mix to make a concentrated liquid that has a deep color. You could also go the other way. Take something like coffee or tea and then boil lots of the water away to make a concentrated, dark liquid.
- Comprehension Check #12
Number 12 asks, "Give the balanced chemical equation for the formation of HNO3." Remember that formation equations start with elements and make the compound of interest, as given in the definition on page 182. Thus, we need to start with the elements in the molecule: hydrogen, nitrogen, and oxygen. All three of those are homonuclear diatomics, so they exist as H2, N2, and O2. That means the unbalanced equation is:
H2 + N2 + O2 --> HNO3
Notice that on the reactants side, we have only elements, and on the products side, we have the compound being made. That's the definition of a formation reaction.
- Comprehension check 10: why is iron (III) oxide Fe2O3?
Remember from Chapter 4 that the Roman numeral after the metal in an ionic compound indicates the metal's positive charge. Metals give up electrons, so they are always positive. The name iron (III) oxide tells you that in this ionic compound, the iron has a charge of 3+. Since oxygen is in group 6A, it has a charge of 2-. To get the chemical formula, you drop the signs and switch the numbers, meaning the "2" in "2-" goes with the iron, and the "3" in "3+" goes with the oxygen, making it Fe2O3.
If you don't like the "switch the numbers" shortcut, think about the charges. They have to cancel one another out. Since Fe is 3+ and iron is 2-, you will need two iron atoms (which total 6+) to cancel out the charge of three oxygens (which total 6-). That makes Fe2O3.
- Charges in ionic compounds
Remember from chapter 4 that atoms want to get to the ideal electron configuration: that of a Noble Gas. Nonmetals do that by gaining electrons. As indicated on the periodic table in the book, all elements in Group 7A are nonmetals, so they all need to gain electrons to get the electron configuration of a Noble Gas. Well, any element in 7A can gain one electron, and they will end up having the electron configuration of the Noble Gas that is right next to them. Thus, Group 7A elements gain one electron and therefore become 1- in ionic compounds. Elements in Group 6A can gain two electrons to get the Noble Gas electron configuration, so they are 2- in ionic compounds. Nonmetals in group 5A gain three electrons and become 3-. Nonmetals in group 4A gain four electrons and become 4-.
Metals, on the other hand, do not gain electrons. They lose electrons. They get the Noble Gas configuration by losing all their valence electrons. That gives them the electron configuration of the Noble Gas that is ABOVE them in the periodic table. Aluminum is in group 3A, so it has three valence electrons. It must lose all of them, which makes it 3+. Elements in group 2A lose their two valence electrons to become 2+, and elements in group 1A lose their only valance electron to become 1+
You might want to review pages 114-117, where this is covered.
- Can steel wool scrubbing pads be used in Exp 6.4?The strands need to be very thin for the experiment to work as written. However, you can probably take a single strip, tape it down to a ceramic plate so it is flat, and then put the two terminals on the strip. That should make it red hot and then make it burn.
- Review question 10: Why is the answer 22 and not 23?
Remember that when you add liquids, the volumes do not add. As Experiment 6.2 showed, the final volume is LESS than the sum of the individual volumes, because the molecules get in between each other. So, the sum of the volumes is 23, but we know the actual volume is less than that. 22 is the only number on the list that is less than 23.
- Determining the Limiting Reactant
You don't need to do this for the course, but if you are interested in figuring out which reactant is the limiting reactant, here's how:
You just have to start the stoichiometry with each reactant in the equation. Convert each reactant to moles (if it isn't in moles already), and then use it to calculate the moles of one of the products. When you have the moles of product for each separate reactant, choose the SMALLEST result. After all, the limiting reactant runs out first, so it will make the smallest amount of product. Use that result in the rest of the stoichiometry.
Here's an example:
Determine the number of grams of hydrogen gas made when 50.0 grams of magnesium and 50.0 grams of HCl react according to the following equation:
Mg(s) + 2HCl (aq) --> H2 (g) + MgCl2 (aq)
First, use Mg:
(50.0 grams Mg/1)x(1 mole Mg/24.31 g Mg) = 2.06 moles Mg
(2.06 moles Mg/1)x(1 mole H2/1 mole Mg) = 2.06 moles H2
Next, use HCl:
(50.0 grams HCl)x(1 mole HCl/36.46 g HCl) = 1.37 moles HCl
(1.37 moles HCl)x(1 mole H2/2 moles HCl) = 0.685 moles H2
So, 50.0 g of Mg could make 2.06 moles of H2, but 50.0 g of HCl can make only 0.685 moles H2. That means HCl is the limiting reactant, since it produces the smaller amount of product. That means we use the number of moles of H2 calculated from HCl to finish the problem.
(0.685 moles H2/1)x(2.02 grams H2/1 mole H2) = 1.38 grams H2
- When do the numbers in equations represent moles?
The numbers in front of molecules in a chemical equation represent BOTH moles and molecules. So, in a chemical equation like:
2Al + 6HBr --> 2AlBr3 + 3H2
You can say:
Two atoms of Al react with six molecules of HBr to make two molecules of AlBr3 and three molecules of hydrogen
or you can say:
Two moles of Al react with six moles of HBr to make two moles of AlBr3 and three moles of hydrogen
Remember, a mole is simply a certain number of molecules. So any relationship that works with molecules will work with moles as well. If I run the reaction just one time, two atoms of Al react with 6 molecules of HBr to make two molecules of AlBr3 and three molecules of hydrogen. If I run it a dozen times, two dozen Al react with six dozen HBr to make two dozen AlBr3 and three dozen hydrogen. If I run it 6.02x1023 times, two moles of Al react with six moles of HBr to make two moles of AlBr3 and three moles of hydrogen.
- Incorrect Results for Experiment 7.1
If you got more than 5 molecules of water for every molecule of CuSO4, that's not really a problem. Most likely, some of the solid popped out of the beaker during your experiment (or you lost some in another way), causing you to read too much mass loss between the hydrated and non-hydrated form. Just check your calculations against the ones at the end of the chapter to make sure you didn't make a mistake.
- Chapter 7, Comprehension Check 6
Remember, the chemical equation gives you the relationship in moles. Thus, since the chemical equation is:
2Al (s) + 3Br2 → 2AlBr3
You know that
2 moles of Al = 2 moles of AlBr3
You can use that conversion relationship to convert 0.407 moles of AlBr3 into moles of Al. In the same way, the equation tells you:
3 moles Br2 = 2 moles of AlBr3
Once again, that's a conversion relationship which allows you to convert 0.407 moles AlBr3 into moles of Br2.
- Review problem #10, why multiply molecules H2O by moles MgSO4?
In the end, we only know how many molecules of water can be absorbed for each molecule of magnesium sulfate. We know that every molecule of MgSO4 can absorb 7 molecules of water. Thus, if we have 10 molecules of MgSO4, they can absorb 7x10 = 70 molecules of water.
Now remember, since a mole is just a bunch of molecules we can also say that for if we have 10 moles of MgSO4, they can absorb 7x10 = 70 moles of water. Of course, we don't have 10 moles of MgS04. Instead, we have 0.8308 moles. That means they can absorb 7x0.8308 moles of water.
We keep the answer in moles (rather than molecules) because the question asks for the mass of water. Thus, we need to convert to grams. Well, the mass of the molecule allows us to easily convert from moles of water into grams of water, so it is best to keep the answer in moles.
- Experiment 7.2: Calculating moles of acetic acid
If you are trying to calculate the moles of acetic acid in the experiment to see how close they are to the moles of sodium bicarbonate, you need to take into account the fact that vinegar is only about 5% (by mass) acetic acid. Thus, you must multiply the mass of the vinegar by 0.05 to get the mass of acetic acid. If you then convert that to moles, you will have the moles of acetic acid.
- Sig figs in extra practice problem 6 (from website)
Remember that when you are determining the mass of a molecule, you are really adding, not multiplying, because the numbers of atoms are exact. Thus, it's not really:
2×22.99 amu + 12.01 amu + 3×16.00 amu
22.99 amu + 22.99 amu + 12.01 amu + 16.00 amu + 16.00 amu + 16.00 amu
When you add, you don't count significant figures. Instead, you report your answer to the same decimal place as the least precise number in the problem. Since all the numbers go out to the hundreths place, you report your answer to the hundredths place. That's why it is 105.99.
This is discussed in Example 7.1 on pp 199-200.
- Review #8: Are the significant figures in the solution correct?
The significant figures are correct as given in the solution. In the first conversion, 1.0x1022 has two significant figures, 6.02x1022 has three significant figures, and the 1's are exact. Thus, the lowest number of significant figures is two, so the answer should have two. That's why it is 0.017. Remember that a zero is significant only if it is between two significant figures or at the end of the number AND to the right of the decimal. Thus, neither zero in 0.017 is significant.
It's important to stress that "significant" doesn't mean "important." It means "measured." The second zero in 0.017 is very important, but it is not significant, because there is no way to measure it. That's what the rules for zeroes tell you. If a zero is between two significant figures, it was measured. If it is at the end of the number and to the right of the decimal, it was measured.
- Experiment 7.2 balloon shape
As long as all the balloons are the same, the shape doesn't matter. You will just be looking at how much they inflate.
- For test #7, why isn't the answer in scientific notation?
You can certainly put the answer in scientific notation if you wish. However, you don't have to, since the significant figures don't require it. Remember, you can usually choose whether or not to use scientific notation (see page 15, right below the Comprehension Check box). However, if the significant figures require it, you have no choice. Since the "169" in this problem limits you to three significant figures, 14,900 grams is a good answer. So is 1.49x104 grams. They are identical, since they both refer to the same value, and they both have three significant figures.
- Why does anhydrous CuSO4 heat up when water is added?
The water is attracted to the CuSO4, which is why it becomes a part of the solid. When things move towards the things they are attracted to, they reach a lower-energy state. So when the water joins the CuSO4, it goes to a lower-energy state than when it was not part of the solid. Energy cannot be created or destroyed, so it has to go somewhere. The energy that the water lost gets released into the surroundings, heating the surroundings.
- How do I know when to convert moles or grams in stoichiometry?
If you are converting from one chemical to another, you must use the chemical equation, which can only be used with moles. Thus, before you do that, the amount must be in moles. In Example 7.7, you are given grams of PCl5 and are asked for grams of Cl2. To go from PCl5 to Cl2, you must use the chemical equation. Thus, your amount of PCl5 must be in moles. It is not, so you have to convert it to moles before you can use the chemical equation. Since the chemical equation uses only moles, the answer will be in moles of Cl2. Since the problem asked for grams of Cl2, you must convert to grams.
Compare that with Example 7.6. There, you are given moles of Fe and are asked for moles of Fe2O3. To go from Fe to Fe2O3, you must use the chemical equation, which uses moles. However, you already have moles of Fe, so you can just go straight to the chemical equation and convert from Fe to Fe2O3. That will give you moles of Fe2O3, which is what the problem asks for, so there is no reason to convert back to grams.
So it depends on what you are given and what you are asked for. If you are given moles, you go straight to the chemical equation and use it to calculate the moles of what is being asked for. If you are given grams, then you must first convert to moles before using the chemical equation. Once you have used the chemical equation, your answer will be in moles. If that's what the problem asks for, you are done. If not, it must be asking for grams, so you must convert to grams.
- How to translate stoichiometry problems
Remember that the chemical equation gives you the conversion relationship between two chemicals, but only in moles. So:
1) You will be given the amount of one chemical. Is it in moles? If not, you must convert it to moles. If it already is in moles, go to step 2.
2) Use the chemical equation to convert the moles of the chemical you were given to the moles of the chemical you need to find out.
3) Does the question ask for the moles of the chemical you need to find out? If so, you have the answer. If not, convert it to the units the question asks for (usually grams).
- Review Question 13
This question asks for the molecular formula, but in the first printing of the book, the solution doesn't supply it. I am sorry that was overlooked. In the end, to get the correct molar mass, you need to multiply the molar mass of the empirical formula by 3. Thus, the molecular formula is the empirical formula times 3:
- Example 8.5: Where did the "3" go?
The final answer is incorrect. Unfortunately, the "3" was accidentally dropped. The proper answer is
I am very sorry for the mistake!
- Example 8.8: What numbers are being switched?
In Chapter 4 (p. 116, Example 4.4) you learned how to determine the chemical formula of an ionic compound. First, you determined the charge of each ion. Then, you dropped the signs on the charges, switched the numbers on the charges, and then turned those numbers into subscripts. In this example, you are using that same method again, but this time, with polyatomic ions.
For sodium carbonate, then, you know that sodium becomes Na+ because it is in group 1A. That means its charge is +1. Carbonate is CO32-, so its charge is -2. Dropping the signs means you have a 1 with Na and a 2 with CO3. Now you switch the numbers, so the 2 goes with Na and the 1 goes with CO3. That's how you get the formula Na2CO3.
It might be helpful to review Example 4.4 on page 116 to refamiliarize yourself with the process.
- Example 8.4 - How do moles of CO2 determine moles of carbon?
Think about where the carbon dioxide comes from. There are only two reactants: the fuel and oxygen. Oxygen has no carbon in it, so every carbon atom in CO2 must come from the fuel. So, supose you burned the fuel completely and got 5 molecules of CO2. How many atoms of carbon are in 5 molecules of CO2? Each molecule has one carbon atom, so if there are 5 CO2 molecules, there are 5 carbon atoms. Where did those atoms come from? They all came from the fuel. So, how many atoms of carbon must have been in the fuel? If there are 5 carbon atoms, and they came from the fuel, that must means there were 5 carbon atoms in the fuel. Since the fuel is the only source of carbon atoms, every carbon atom in CO2 must have been in the fuel.
Now remember, moles are just a way of counting atoms and molecules, so anything that works for atoms and molecules also works for moles. So, supose you burned the fuel completely and got 5 moles of CO2. How many moles of carbon are in 5 moles of CO2? Each molecule has one carbon atom, so each mole of CO2 has one mole of C. If there are 5 moles of CO2, there are 5 moles of C. Where did those moles come from? They came from the fuel. So, how many moles of carbon must be in the fuel? If there are 5 moles of carbon, and they came from the fuel, that must means there were 5 moles of carbon in the fuel. Since the fuel is the only source of carbon atoms, every mole of carbon in CO2 must have been in the fuel.
The same reasoning applies to hydrogen. The only source of hydrogen is the fuel. So any hydrogen in the water produced must come from the fuel. Since every mole of water has 2 moles of hydrogen, then the number of moles of hydrogen in the fuel is twice the number of moles of water produced.
- Example 8.3, why O instead of O2?
Oxygen is a homonuclear diatomic, but that only applies when it is by itself, not part of a compound. So you never see O as the product in a chemical reaction. You see O2. However, when you put O with any other element, it doesn't have to be O2. In carbon monoxide, for example, it is CO. In calcium carbonate, it is CaCO3. So when O is not with any other element, it is O2. When it is with any other element, it can be anything.
- Review Question 6: Isn't C5H10O a carbohydrate?
C5H10O is not a carbohydrate. There must be twice as many H's as O's. C6H12O6 is a carbohydrate, because there are 12 H's and 6 O's. Thus, it reduces to CH2O. C5H10O cannot be reduced, because there is only one O, so you can't divide it by anything. If you divided by 5, you would have CH2O1/5.
- Review question # 5 significant figures
The mass of CaCO3 (100.09) has 5 significant figures, 500.0 g has 4, and the mass of Ca3(PO4) (310.18) has 5. Thus, the answer should have 4.
- What does "switching the numbers and dropping the charges" mean?
This goes back to something you learned in Chapter 4.
When you figure out the charge of an ion, there is a number and a sign. Mg ions, for example, are Mg2+, while nitride ions are N3-. The number is what comes in front of the charge, so for Mg2+ the number is 2 and the charge is +. For N3-, the number is 3 and the charge is -. If Mg and N make an ionic compound it is Mg3N2, because you dropped the charges (+ and -), and you switched the numbers (Mg's 2 went to N, while N's 3 went to Mg).
- Determining the value for "i"
You determine the value for "i" by figuring out how many particles the solute splits into when dissolving.
For covalent compounds (compounds made of only nonmetals), i = 1, because they never split up into multiple particles, since all their elements are chemically bonded to one another. Thus, even though there are a lot of atoms in C6H12O6, i = 1, because all the atoms are nonmetals, so the compound is covalent and doesn't split up into multiple particles when it dissolves.
For ionic compounds, "i" equals the number of ions in the compound. If the ionic compound contains only single-atom ions, it is fairly easy, because each atom in the chemical formula becomes an ion. Thus, for Na2S, i = 3, because the compound has two Na+ ions and one S2- ions.
If the ionic compound contains polyatomic ions, you need to be able to recognize the polyatomic ions, which is why it is so important to keep the bold polyatomic ions in the table on page 245 memorized. For example, if you look at Na2CO3 and have those polyatomic ions memorized, you should recognize CO3 as representing the carbonate ion. Thus, this compound has two Na+ ions and one CO32- ion, so i = 3.
- What makes SrCl2 an ionic compound?
You might have forgotten a rule that was mentioned on the top of page 115 (in the pink box). It says that ionic compounds are formed between metals and nonmetals. If you look at the Periodic Table, Sr is in a box whose color indicates it is a standard metal. Cl is in a box whose color indicates it is a nonmetal. Thus, SrCl2 is a compound between a metal and a nonmetal. As a result, it is ionic.
Remember, metals like to give away electrons, and nonmetals like to gain them. Thus, when there is a metal and a nonmetal in a compound, the metal will give its electrons away, forming a positive ion, and the nonmetal will take them, forming a negative ion. Thus, all compounds that contain at least one metal and one nonmetal are ionic.
- Strontium chloride in Experiment 9.4
You can also substitute calcium chloride for strontium chloride in this experiment.
- How do I determine the charges on elements?
You might want to go back to Chapter 4 (pp. 114-117), where you learned how to determine the charge of ions in ionic compounds and how to use charge to determine the chemical formual of an ionic compound. As you learned there, the column in the periodic table tells you the charge of the element. Na is 1+ in ionic compounds because it is in group 1A and therefore must lose an electron to become like the nearest noble gas. As another example, O takes on a charge of 2- in ionic compounds, because it is in group 6A and must gain two electrons to become like the nearest noble gas.
In Chapter 8 on page 245, there is a table of polyatomic ions. The ones in bold need to be memorized. Notice that OH- is one of the bolded ones. Thus, when you see OH in an ionic compound, you should immediately recognize that it is the OH- ion.
As you learned in Chapter 4, once you have the charges, you drop the signs and switch the numbers to get the chemical formula. So, since Sr is in group 2A of the periodic table, it must lose two electrons to become like the nearest noble gas, so it is 2+. You are supposed to have OH- memorized. You drop the signs and switch the numbers, so the "2" in "2+" goes with the OH, and the "1" in "1-" goes with the Sr. Since the 2 goes with OH, that means you need to put the OH in parentheses and have the 2 on the outside, which is why it is Sr(OH)2.
- Chapter 9, Comprehension Check #13 signficant figures
In this problem, 102.0 degrees C has four significant figures, and 100 degrees C is exact. However, you are subtracting them to get the change in boiling point (Delta T)
Delta T = solution boiling point - solvent boiling point
Delta T = 102.0 degrees C - 100 degrees C
Remember that when subtracting (or adding), you don't count significant figures. As explained on page 6 (that was a long time ago!), you report your answer to the same precision (decimal place) as the least precise number in the problem. 100 is exact, so it is infinitely precise. 102.0 has its last significant figure in the tenths place, so your answer must have its last significant figure in the tenths place. That means
Delta T = 2.0 degrees C
- In extra problem #11 on the website, what is the solvent?
In this problem, the solvent is ethanol, and the solute is water. That's why ethanol's boiling point and Kb are given. In addition, the solution determines i for water, because it is the solute. I know we usually think of water as a solvent, but it is often a solute in alcohols. For example, distilled ethanol is about 95% ethanol and 5% water. Thus, it is a solution in which ethanol is the solvent and water is the solute. The rubbing alcohol you get at the drugstore is usually 70% isopropyl alcohol and 30% water. Once again, then, alcohol is the sovent, and water is the solute.
- In Experiment 9.3, where does the oxygen come from
The oxygen that the alcohol is reacting with comes from the air. The oxygen atoms in the water molecules will not react with alcohol, since they are in a lower-energy state inside the water molecule. If the oxygen in the water molecules did react with the alcohol, water could not be used to put out fires!
- Ch.9 review #21: How do you know that N2H4 is a covalent solute?
You know that N2H2 is a covalent solute because it is composed of only nonmetals. Remember, ionic compounds have at least one nonmetal and one metal. Covalent compounds are made of only nonmetals. As shown on the periodic table, H and N are both nonmetals (the legend on the periodic table tells you that all blue boxes are nonmetals), so N2H2 is covalent. All covalent molecules have i = 1, because they do not split up into particle when they dissolve. They dissolve as an entire molecule.
- How can carbon dioxide dissolve in water, it's nonpolar.
Carbon dioxide doesn't really dissolve in water. It reacts with water to make carbonic acid:
H2O + CO2 --> H2CO3
This reaction is reversible, however, so it's like carbon dioxide has dissolved. This is covered in Chapter 11.
- Why are salt ions attracted more to the slight charges in water?
The ions in salt have full electrical charges, and the water has only partial electrical charges. However, there are A LOT of water molecules, and they can surround the salt ions. Lots of slight charges add up to more charge than the full charge on the oppositely-charged ion, so the ions are more attracted to the water molecules than they are to one another.
- Significant figures in Comprehension Check 5
The volume is given as 200.0 mL. That's four significant figures. The zero at the end is significant, since it is both at the end of the number and right of the decimal. That makes the other two zeroes significant, since they are between two significant figures. The metric conversion factors are exact, so the 0.001 has an infinite number of signficant figures. Thus, the answer needs to have four significant figures. 0.2000 has four. The zero at the end is significant, since it is both at the end of the number and right of the decimal. That makes the two zeroes between it and the 2 significant, since they are between two significant figures. The first zero is not significant. It is not at the end of the number and right of the decimal, and it is not between two significant figures.
The same reasoning can be used for moles. 15.67 has four, 96.11 has four, so the answer must have four. That's why it is 0.1630. Then, in the final step, you have four significant figures in both 0.2000 and 0.1630, so the answer must have four, which is why it is 0.8150 M.
- Are the phase symbols necessary in review question 11?
In this case, they are. For many reactions, it doesn't matter. However, this is a reaction that forms a precipitate. You need the phase symbols so it is clear what the precipitate is.
- In Comprehension Check #3, why don't you warm the solution?
The problem tells you the solution is not saturated. Thus, you could dissolve more sodium bromide in it. However, you don't have any more sodium bromide to add. A saturated solution would not allow you to dissolve more sodium bromide. So how do you turn this into a solution that cannot dissolve any more sodium bromide?
Well, since sodium bromide is a solid, its solubility decreases with decreasing temperature. If you decrease the temperature, then, less sodium bromide can dissolve. That means the amount which is already dissolved in solution is closer to the maximum amount that can be dissolved. If you lower the temperature enough, the amount that is already in solution will be the maximum amount that can be dissolved. At that point, no more sodium bromide can be dissolved, so the solution is saturated.
Warming the solution will increase the solubiluty of sodium bromide, which means even more can be dissolved. In that case, you are making the solution even less saturated.
- Why does food take longer to cook at low pressure in review #22?
Think about putting something in the oven to cook. If the recipe says that the meal will take 30 minutes to cook at 400 degrees, how long will it take to cook at 300 degrees? It will take longer than 30 minutes. After all, the hotter the oven, the more quickly the food cooks. Conversely, the cooler the oven, the longer it takes to cook. Crock pots take a long time to cook specifically because they cook at a low temperature.
If you are boiling something in water, you are using the temperature of the water to cook the food. Sure, the water might reach boiling more quickly because it doesn't have to be heated to as high a temperature. However, that lower temperature will take longer to cook the food.
- Test for Chapter 10, Question #9
Remember what Avagadro's Law tells us. At the bottom of p. 305, the pink box has the following:
"When all substances of interest are gases, as long as pressure and temperature remain the same, the chemical equation gives you the relationship between the volumes of those gases."
The only two substances of interest in this equation are H2 and CH4. Carbon is there as a solid, but it is not mentioned at all. You are given the volume of CH4 and asked to determine the voume of H2. Thus, the only two substances of interest are gases. That means the chemical equation tells you the relationship between the two gases: 2 liters of H2 make 1 liter of CH4. Thus, you can use that as a relationship to convert between the two.
- Comprehension Check #4
If you don't get exactly the answer that the book has, that's not necessarily a problem. Remember, the last signficnt figure has error in it (see the box on p. 9). Thus, when you solve the combined gas law equation and get T2 = 286 K, that's fine. Even though the book's answer is 285 K, that "5" has error in it. It could be 4, 6, 3, even 7. From a scientific point of view, 286 K and 285 K are the same (see the discussion under the box on p. 9). That's why it is so important to report your answer with the proper number of significant figures. That way, everyone knows which digit has error in it.
However, if you want to get exactly what the book gets, you should only round for significant figures at the end of the equation or when you are forced to change the rules. In this problem, for example, I multiply all the top numbers, then divide by all the bottom numbers, and then round. That gives me 285 K as the answer for T2 in the combined gas law.
If the problem had been something like this one:
q = (1.00g) (1.0000 J/gC)(14.6 C - 14.3 C)
You would have to do the subtraction and then round to 0.3 C, since subtraction uses different rules from multiplication. Thus, the equation becomes:
q = (1.00g) (1.0000 J/gC)(0.3 C) = 0.3 J
The only time I round in the middle of an equation, then, is when it changes from addition/subtraction to multiplication/division.
- How do you get that 1 mole of gas occupies 22.4 liters at STP?
You can use the Ideal Gas Law to determine this. After all, standard temperature is 0 C, which is 273.15 K. Standard pressure is 1 atm. R = 0.0821 (l atm)/(mole K), and if there is 1 mole, n = 1. So, the Ideal Gas Law says:
PV = nRT
(1 atm)(V) = (1 mole)(0.0821 [l atm]/[mole k])(273.15 K)
V = (1 mole)(0.0821 [l atm]/[mole k])(273.15 K) / 1 atm = 22.4 liters
The 1's are exact, since they are coming from definitions, so you are limited to three significant figures by 0.0821.
- Temperatures and Pressures Ideal Gas Law
Remember what I say on page 301:
"In this course, you can assume gases will behave ideally if their temperature is near or above 0 ºC and their pressure is near or below 1 atm."
I can say this because the lower the pressure and the higher the temperature, the more ideally a gas behaves. Thus, I don't have to be at STP for the Ideal Gas Law to work. I simply need to be near or below 1 atm and near or above 0 oC. So consider Example 10.4. The pressure is 1.02 atm, which is very near 1 atm. The temperature is not near 0 oC, but it is ABOVE 0 oC. That means the gas behaves even more ideally than it would at 0 oC. Thus, I can use the Ideal Gas Law. In the end, the Ideal Gas Law works well in most situations with which we are familiar, as the pressures we usually encounter are right around 1 atm, and the temperatures are usually higher than 0 0 oC.
- How does dividing by mass in Experiment 10.4 give us percent?
The hydrogen peroxide you use in the lab is a solution made by dissolving hydrogen peroxide in water. It is actually a pretty weak solution, because concentrated hydrogen peroxide can be very dangerous. The mass that you measured at the beginning was the total mass of the solution, which is the mass of hydrogen peroxide plus the mass of water. The mass you got from the moles of oxygen is the mass of only the hydrogen peroxide. So dividing the two gives you the fraction of hydrogen peroxide in the solution. Multiplying by 100 turns that into a percent.
- Example 10.8, why don't we consider the pressure of NaC2H3O2
The Na2C2H3O2 produced in the reaction is a solid that dissolves in the water. Since it is a dissolved solid, it doesn't exert pressure. Only liquids can evaporate, so a substance must be in its liquid phase to exert a vapor pressure. A dissolved solid is not a liquid.
- Comprehension check 5A - Why not (1)?
Gases become more ideal the warmer the temperature. After all, the warmer they are, the more space they occupy, so the less important the volume of their molecules is. So STP is a benchmark. 0 oC is a good estimate for the minimum temperature to consider it an ideal gas, but the higher the temperature, the more ideal the gas. In the same way, 1 atm is also a benchmark. It is a good estimate for the maximum pressure. The lower the pressure, the more ideal the gas.
- Comprehension Check #4 - why is the answer 12, not 11.85
Remember that when you add and subtract, you don't count significant figures. You look at the decimal place, and you report your answer to the same decimal place as the least precise number in the problem. 285 has its last significant figure in the ones place, while 273.15 has its last significant figure in the hundredths place. The ones place is less precise, so the answer can only be reported to the ones place. Thus, 11.85 has to become 12.
- What is a round balloon?
A round balloon is what most people think of when they think of balloons. However, there are tublular balloons, which are much longer than they are wide. There are also heart-shaped balloons, etc. You just want one that is fairly round when it is inflated, like these:
- What is the pressure in the jar for Experiment 10.3?
When you boil the water in the microwave, it is so hot that its vapor pressure is equal to atmospheric pressure. Thus, it boils. As a result, the empty space above the water in the jar is filled with water vapor at atmospheric pressure. When you put the lid on and turn it upside down, that space is still filled with water vapor at atmospheric pressure. That's what I mean by the "pressure in the jar." However, when you put the ice on the jar, that water vapor in the empty space gets cooled. So some of it condenses. When the water vapor that was in the empty space condences, the pressure in that empty space (the pressure in the jar) reduces.
The vapor pressure is the pressure of water vapor that evaporates from the water. That is determined only by the temperature of the water. Since the water cooled a bit while you were closing the lid and turning the jar upside down, the vapor pressure was lower than atmospheric pressure. That' why it stopped boiling as you were pulling the jar out of the microwave and putting the lid on. At that point, the pressure in the empty space (the pressure in the jar) was greater than the water's vapor pressure. As the pressure in the empty space (the pressure in the jar) reduced, it eventually reached the vapor pressure of the water, which is why the water started to boil again.
- Review Question #19
In this problem, you are being asked to determine the liters of one chemical in a reaction using the liters of another chemical. Remember that the numbers in a chemical equation always refer to the MOLES of each chemical. However, when the chemicals you are analyzing are all gases, they also refer to the VOLUMES of each chemical. So the chemical equation tells you that:
1 mole of C3H8 = 3 moles CO2
However, since both are gases, you also know that:
1 liter of C3H8 = 3 liters of CO2
This is now a conversion relationship between liters of C3H8 , which you were given, and liters of CO2, which you want to know. Thus, it is much like Example 10.5, except that the volume unit is liters, not mL.
- Could you explain the definition of vapor pressure?
When you have a sample of liquid at any temperature, some portion of it evaporates. Thus, there is always some gas right over the surface of the liquid. The warmer it is, the more gas there is above the liquid. Vapor pressure is a measure of how much of the gas phase exists right above the liquid phase. The higher the vapor pressure, the more the lquid has evaporated, so the more gas there is above the liquid.
- When litmus paper doesn't change color or turns purple
When both blue and red litmus papers retain their color for the same substance, that substance is neutral. Purple is actually an ambiguous color. It could mean the substance is slightly basic or slightly acidic. You can sometimes figure out which it is by comparing the color change it causes on both red and blue litmus paper. If the red litmus paper doesn't change much but the blue does, then it is probably slightly acidic. If the blue litmus paper doesn't change much but the red does, then it is probably slightly basic.
- Why is the endpoint in Experiment 11.3 clear?
While you would think the endpoint should be purple to indicate a neutral solution, the high base content of the sodium hydroxide solution actually causes a small structural change in the anthocyanins. The change is small, but it affects the colors in the red region and blue regions of the spectrum, making them much paler. As a result, the solution ends up being clear (not purple) when it first turns neutral.
- Comp check #6 - why do the products have the + and - signs?
Remember, there are two different kinds of bases - covalent bases and ionic bases. As you already learned, if a compound has a metal and a nonmetal, it is ionic. If it has only nonmetals it is covalent. So, when a base has a metal in it, like Al(OH)3, it is an ionic base. When it has no metals in it, like NH3, it is a covalent base.
When you are doing an acid/base reaction with an ionic base, you use the fact that an acid reacts with an ionic base to make water and a salt. Since the salt's ions come together, the opposite charges cancel one another out. That's why there are no positive and negatuve signs in Comprehension Check #7. It is a reaction between an acid and an ionic base.
For #6, however, the base is covalent. At that point, you don't make water and a salt. You have to rely on the definition of an acid and the definition of a base. An acid donates an H+, and a base accepts an H+. So, whatever is left after you pull the "H" of the acid has lost a positive charge. HClO3, then, becomes ClO3-. Since NH3 gains an H+, it gains an H and a positive charge. That makes it NH4+. The charges have to be left in there, because there is no such thing as ClO3 or NH4. Those chemicals can't exist. However, ClO3- and NH4+ do exist, because the charges make them stable.
In general, then, when you are dealing with covalent bases, you will have charges in your products. When you are dealing with ionic bases, there will be no charges in your products. There are actually charged things produced, because a salt is made of ions. However, the chemical formula of the salt cancels out those charges.
- Experiment 11.3 - What if you use 50 mL or more?
If you use more than 50 mL in your titration, it could be that your vinegar is stronger than most vinegar. However, you don't need to worry. If you use the entire graduated cylinder and reached the endpoint, then your volume was 50.0 mL. If you need more, just refill the graduated cylinder and use vinegar again until you reach the endpoint. Then add 50.0 to the volume that you took from the second filling, and you have the volume of vinegar used.
- Why isn't HNO3 broken into ions in Example 11.1 or H2SO4 in CC 4
Remember, an ionic compound needs at least one metal (to give away electrons) and one nonmetal (to take electrons). H, N, S, and O are all nonmetals. You can see that by looking at the periodic table and following the color coding. Because there are no metals among them, HNO3 and H2SO4 are not ionic.
- Touching sodium hydroxide in Exp 11.2
Solid sodium hydroxide is caustic, because you have only a little moisture on your hands, and there is plenty of sodium hydroxide, so it forms a concentrated solution. In Experiment 11.2, there is a lot more water, and so the concentration of the solution is low. As a result, it is not caustic. However, if you don't want to touch the solution, just put some plastic wrap over the top of the test tube before covering it with your thumb.
- Chapter 12, review question 17. Why is the metal at the anode?
Electroplating is the opposite of a Galvanic cell. In a Galvanic cell, you are using the chemicals to make a battery. In electroplating, you are using a battery to make a chemical. So, in a Galvanic cell, the reduced metal is at the cathode, because it wants electrons. Thus, it is attracting electrons towards it. In electroplating, you are forcing the reduced metal to take electrons. Thus, you have to shove the electrons onto it, and that requires a battery (or other power source). Since the anode is where electrons are forced off, that's where the reduced metal goes in electroplating. This is what you saw in the experiment. The copper plated on only one side of the battery: the negative side (anode).
- Chapter 12, review question 18. Why do you add acid to gold?
You don't actually have to know why in order to answer the question. At the bottom of page 379, the student is told how gold is treated in order to be used for electroplating. The student is specifically told that a strong acid is used.
To see why you add an acid, look at the chemical equation at the bottom of page 379. The gold is getting oxidized, because it is going from an oxidation state of 0 to an oxidation state of 3+. Thus, the acid oxidizes the gold. Why is that important? In electroplating, the metal ion is reduced. If I have gold that I want to plate onto something, then, I will first have to oxidize it so that it can then be reduced in the electroplating process.
- Experiment 12.2: Why were there holes in the aluminum foil?
As discussed on page 369, the reaction between the copper ions in solution and the aluminum in the foil is:
2Al (s) + 3Cu2+ (aq) --> 3Cu (s) + 2Al3+ (aq)
Notice what is happening. The dissolved copper ions are becoming solid copper, and the solid aluminum is becoming aluminum ions, which are dissolved in solution. So the holes that you saw in the aluminum were the result of aluminum changing to Al3+ and dissolving in solution. There still should have been "dirt" in the solution and on the foil. That's the copper.
- Which rule do you use if you have both Cl and O in a compound?
On page 361, I discuss the rule about oxygen and say, "The sixth “rule” is that in most compounds, oxygen has the oxidation state of 2-. This is true in both ionic compounds (like MgO) and in most covalent compounds (like CO2). However, there are a few covalent compounds in which oxygen’s oxidation state is 1-, such as in H2O2. You won’t have to worry about those kinds of compounds, however."
Since you don't have to worry about the compounds where O has an oxidation state of 1-, you can always assume that the rule for oxygen is true. It is a last resort, but it is more likely to be true than the rule for all Group 7A elements except for flourine. If you are ever presented with a molecule that has F and O, you know that F is 1-. However, if you have any other Group 7A element in a molecule along with O, assume that O is 2-.
- In Experiment 12.1, where did the H+ come from?
In Chapter 11, you learned that acids donate H+ ions. In the experiment, you added vinegar, which contains acetic acid. The acetic acid donated those H+ ions.
- Anodes and cathodes in Galvanic cells
Think about what happens in a Galvanic cell. Electrons are lost by the chemicals at the anode. What do those electrons do? They travel to the cathode, where the chemicals gain them. So in a Galvanic cell, electrons from from the anode to the cathode.
Think about how electrons react to charge. They are repelled by a negative charge, and they are attracted to a positive charge. If I look at how electrons are traveling in a Galvanic cell, then, it looks like the anode is negative, because electrons are traveling away from it, as if they are repelled by it. In the same way, it looks like the cathode is positive, because electrons are traveling towards it, as if they are attracted to it.
In electroplating, the positive ions in the solution absorb electrons. Where are those electrons coming from? They are coming from the anode. That's where the positive ions in solution first "see" them. so they are attracted to the anode, so they can absorb the electrons coming from it.
- For Experiment 12.2, does the penny have to be shiny?
It does not have to be shiny. In fact, the difference you are supposed to see will probably be easier to see if the penny is not shiny.
- Extra Practice Problem 2a
Remember that the always true rules beat the usually true rules. This is a covalent compound (no metals), so H must be +1. Thus, you have to choose between rule #6 and rule #7 to get the rest of the molecule. When I discuss Rule #6, I tell the student he won't have to worry about compounds in which O is not 2-. Thus, for the student, rule #6 is more reliable. Also, in discussing rule #7, I give an example of ClO2, and I tell the student that O is more electronegative than Cl. That's another reason to choose rule #6 over rule #7. So O is 2-, which makes Cl 5+.
- Mass of the metal object in Experiment 13.2
The mass of the metal object can be greater than 15 grams. Even something like 50 grams will work. My main concern is that the metal piece can't be so big that it is difficult to stir the water and evenly distribute the heat. So as long as the object isn't too big, then, the larger the mass, the better.
- Why are food calories really kilocalories?
I can't verify this, but I think the reason a food Calorie is 1,000 calories is really just for convenience. If nutritionists used regular calories, there would be a lot more zeroes to write.
- Chapter 14, Review #13: I Don't understand the answer
The question asks, "A student says that because a chemical reaction has a negative ∆S, it cannot be spontaneous. Is he correct? Why or why not?"
Students often think that the Second Law of Thermodynamics says that Delta S can never be negative, since entropy cannot decrease. Thus, a chemical reaction with a negative Delta S can never be spontaneous. However, the Second Law doesn't say that. It says that the TOTAL entropy of the UNIVERSE can never decrease. As a result, the Delta S of a chemical reaciton CAN be negative, as long as the resulting Delta S of the surroundings is so positive that the total entropy change of the universe (Delta S of the reaction + Delta S of the surroundings) is positive or zero.
- What happens to entropy when electrons return to ground state?
When electrons return to their ground state, the entropy of the atom itself decreases. However, in order to do that, the electrons must release energy, usually in the form of light. That energy increases the entropy of the surroundings. The entropy increase in the surroundings is always greater than or equal to the entropy decrease of the atom, so that the total entropy of the universe always increases or stays the same, in accordance with the Second Law of Thermodynamics.
- Rev. Quest. #6: How is 1.0x10^3 precise to the hundreds place?
When you are determining Delta_H, in this problem, you end up with the following result:
Delta_H = 1000 kJ + (1 mole)(triple CN bond energy)
However, because the bond energy of the CN single bond is 300 kJ/mole, you need to report that "1000 kJ" to the hundreds place. In other words, the first zero after the 1 needs to be significant, but the other cannot be. Since the rules say that none of the zeros in 1000 are significant, we have to make one of them significant using scientific notation. Thus, we convert it to:
Delta_H = 1.0x103 kJ + (1 mole)(triple CN bond energy)
Think about what this means. The number 1.0x103 has two significant figures: the 1, and the zero right after the 1 (because it is at the end of the number and to the right of the decimal). Now...the numerical value of 1.0x103 is 1000, and since only the first zero after the 1 is significant, that means the 0 in the hundreds place is significant. Thus, 1.0x103 has its last significant figure in the hundreds place. If it were 1.00x103, the second zero after the 1 would be significant. That would be the zero in the tens place of 1000, so the number has its last significant figure in the tens place. If it were 1.000x103, the third zero after the 1 would be significant. That would be the zero in the ones place of 1000, so the number has its last significant figure in the ones place.
- Is gaseous carbon possible at 25 C?
Gaseous carbon isn't stable at 25 C. The main reason it is on the table is that we often assume that the absolute entropy of a substance doesn't change a lot with temperature. Thus, even though it is an approximation, we can use it for higher temperatures, where gaseous carbon is stable.
- Determining the Lewis Structure for CO in CC problem 2
In Chapter 5, Comprehension Check 3a asks for that Lewis structure. The solution in on page 155. Look at the drawings there as you read this explanation. You put carbon's 4 valence electrons around it and oxygen's 6 valence electrons around it. When you link one of the lone electrons on carbon with one of the lone electrons on oxygen, the carbon has 5 and the oxygen has 7. That doesn't give either atom 8. However, if you put the other lone electron from oxygen in between the two atoms as well as one of the lone electrons on carbon between the two atoms, you now have a double bond between carbon and oxygen. The oxygen has 8, but the carbon only has six. To fix this, you take a pair of electrons on oxygen and put them in between the two atoms. That makes a triple bond, and now both carbon and oxygen have 8. To make things tidy, it is best to pair up the two lone electrons left on carbon.
- Specific heat of a paper cup (Experiment 14.1)
If you used a paper cup, the specific heat is just a bit higher. It's 1.4 J/gC.
- Negative and positive signs when using Hess's Law
Whenever you use Hess's Law for delta H, delta S, or delta G, you always add each product and subtract each reactant. For determining the delta S in Comprehension Check #11, for example, you would add 2 times the absolute entropy of NH3 (g), then you would subtract the absolute energy of N2 (g), and then you would subtract three times the absolute entropy of H2 (g).
- Experiment 14.1, calculating the heat gained by the calorimeter
In step 2, you were supposed to measure the mass of the top cup in the calorimeter. That's the part of the calorimeter that absorbed heat. In step 13, the lab tells you to multiply the mass of the top cup by 1.3 J/gC to get the heat capacity of your calorimeter. When you multiply mass by 1.3 J/gC, the unit will be J/C, which is the proper unit for heat capacity. That's what you use when calculating the heat absorbed by the calorimeter.
- Experiment 14.1: Should the two cups be put back together?
Yes. Once you measure the mass of the cup, then put it back into the other one. The idea here is that only the cup you measured the mass of will be in contact with the water, so it's the only thing that will heat up. However, putting it back into the other cup will help make the system more insulated so that less heat leaaks out.
- How can H2 have a zero delta Hf if its bond energy is 436 kJ/mol
Remember that Delta_Hf is Delta_H of the formation reaction for any chemical. So the Delta_Hf of CH4 (g) is the Delta_H of the formation reaction:
C (s) + 2H2 (g) --> CH4 (g)
What is the formation reaction for H2 (g)? A formation reaction is the reaction that takes elements and makes the molecule of interest, so it's
H2 (g) --> H2 (g)
The Delta_H of that reaction is obviously zero. Now, the Delta_Hf of H2 (l) is not zero, because the elemental form of hydrogen is H2 (g), so you have to take energy away from the elemental form to make it a liquid. That means the Delta_Hf of H2 (l) is negative. However, you don't have to make H2 (g). It is the simplest form of H2. That means there is no energy involved in making it.
- Why is the H-H bond energy different in different sources?
These are measured quantities, and remember that in measured quantities, there is always error in the last significant figure. The book says the H-H bond energy is 436 kJ/mole. However, others say 432 kJ/mole. The variance in the last significant figure reflects the error associated in measuring the bond energy.
- Why does adding -137 to 1.0 x 10^3 give you 900?
Remember that when you add or subtract, you report your answer to the same DECIMAL PLACE as the least precise number in the problem. When you just do the math, you get:
-137 + 1.0x103= 863
However, you now have to take significant figures into account. 137's last significant figure is the 7, and it's in the ones place. 1.0x103's last significant figure is in the hundreds place, because 1.0x103 is 1,000 with the first zero significant. Well, that first zero is in the hundreds place. The hundreds place is less precise than the ones place, so the answer must have its last significant figure in the hundreds place. Thus, you must drop the 6 (which is in the tens place) and the 3, which is in the ones place. However, since you are dropping a 6, you round up, so the answer is 900.
- Why does equilibrium shift the way it does with pressure?
There are two ways to think about this. I will give you the easy way first: An equilibrium will shift so as to reduce the stress. If you add more reactant, it shifts to the product so as to reduce the amount of reactant that you just added. In other words, an equilibrium shifts to counteract the way it was stressed. If I increase pressure, then, it will shift so that the total pressure is lowered. Thus, it will shift to the side with the fewest gas molecules. On the other hand, if I lower the pressure, it will shift so the pressure increases. That can be accomplished by shifting to make more gas molecules.
However, it really is all about reaction rates, and that brings us to the harder way to think about this: Think about how reaction rates change with changing pressure. If I lower the pressure, BOTH the forward AND reverse reaction rates will decrease, because the molecules aren't being pushed together as hard. As a result, there won't be as many effective collisions.
Now...think about which side will experience the most drastic reduction in rate. If I have only two molecules on one side of the equation, and they are suddently not forced together as hard, they wll have fewer effective collisions. So their reaction rate will be lower. However, if I have four molecules on the other side, and they all have to collide to make their reaction work, they will experience a GREATER DECREASE in rate, because their four-molecule collision is much less likely than a two-molecule collision. As a result, their successful collision is more dependent on pressure. When you reduce the pressure, then, the reaction that has to have more molecules colliding will have a greater reduction in its rate.
So BOTH sides of the equilibrium experience a decrease in rate, but the one that has the most gas molecules experiences the largest decrease in rate. Thus, it slows down more than the other reaction. That means the reaction that uses fewer gas molecules slows down less. As a result, the reaction that uses fewer gas molecules becomes faster, so the equilibrium will shift away from it and towards the side with more gas molecules.
- Equilibrium and the Second Law
Remember that for the Second Law to be followed, the total disorder in the universe must increase. Well, when a reaction is at equilibrium, there is more disorder than when it is all products or all reactants. After all, if it is a complete reaction that becomes nothing but products, all the mass is concentrated in a few chemicals (the products). If it is a reaction that never starts, all the mass is also concentrated in a few chemical (the reactants). When both the reactants and products are present, there is more disorder, as the mass is spread out over more chemicals (both the reactants and the products).
At the same time, however, some chemicals are more disordered than others. So, if I have all the mass concentrated in a few chemicals that are really disordered, that may be a more disordered situation than spreading the mass over the really disordered chemicals and some really ordered chemicals. The value for Delta_G takes all this into account. When Delta_G is zero, the reactants and products have similar disorder, so the universe is most disordered when things are spread evenly over reactants and products. When Delta_G is small and positive, the reactants are a bit more disordered than the products, but you still gain a bit of disorder by having some of the mass spread out over all the chemicals. However, that only gains you a bit of disorder, so the reaction only makes a few products. When the value of Delta_G is large and positive, the reactants are much more disordered than the products, so you don't gain any disorder by making products.
In the end, then, the state of the equilbirium contributes to the disorder of the universe, and that's why Delta_G really determines the equilibrium constant.
- Discovering Design with Biology
- Chapter 1
- Where do respiration and movement fit in life's characteristics
Respiration is a part of the metabolism characteristic. Organisms use oxygen to burn the molecules they get from food, while others (anaerobic organisms) use a different process to get energy. Thus, respiration is a part of metabolism. Movement is one of the ways organisms respond to stimulus, so it is a part of that characteristic.
- Can I use other fruits in Experiment 1.1?
Other fruits will work, but the amount of DNA extracted will be different, because of the chemical nature of the fruits themselves and the amount of DNA in them. Strawberries and bananas give the best results for this particular procedure, but other fruits will work. It just might be harder to see the DNA.
- What does "complexity that is not specified or specific" mean?
Think about a rock like this one:
It's got a complex shape (lots of bumps, valleys, and points) but the complexity doesn't serve any purpose. It's not specified or specific. It's just complex. Now consider this rock:
It also has a complex shape (pointed at one end, round at the other, sides sharpened). The difference is that this shape serves a purpose - it can be used to cut and scrape. In many ways, this rock is less complex than the first one, but the complexity of this rock is specific, so you know it was designed to be a tool.
Complexity can occur via random processes, but complexity that is specified or specific always originates in design.
- What is the difference between Irreducible Complexity and Design
Specific complexity (design) is the recognition that the system is complicated, but it is complicated in order to serve a function. So...if you see this rock:
You can recognize it is a complex shape, but that complexity doesn't serve any purpose. Thus, it is complex, but it does not have specified complexity. If you see this statue:
You see that it is complex, but you know the complexity has a purpose - to show the viewer what an elephant looks like. The first rock was not designed, because its complexity is not specified. The statue was designed, because its complexity is specified.
Irreducible complexity is complexity that comes from a bunch of components that must already exist in their current form and cannot come about by modifying something else. In both cases above, the complexity is not irreducible, because both come from modifying rocks, and many different kinds of rocks could be used in the process.
Compare both of those to this elephant robot:
Some of its electronic components might work in other robots, but some of them only work in this model of elephant robot. Those components cannot be the result of modifying other components. That means the elephant robot has irreducible complexity.
- Chapter 3
- Can you give more explanation for Comp. Check 3.14?
On page 85, we tell you that in aerobic respiration the final electron receptor molecule is oxygen. It receives the electrons at the end of the electron transport chain. Then, on page 87, we tell you that this kind of respiration can also happen without oxygen, if the organism has another final electron receptor molecule. For example, methanogens use carbon dioxide as the final electron receptor.
If the organism has no final electron receptor molecule, all it can do is glycolysis, so it must do glycolysis and fermentation, as discussed on pp. 80-83.
- Chapter 4
- Comprehension Check 4.7: Why 22 chromosomes?
Remember that mitosis starts with a diploid cell and ends with two duplicates of it. If the cell has a haploid number of 11, then its diploid number is 22. Thus, mitosis would start with a cell that has 22 chromosomes and will end with two duplicates, so each daughter cell will have 22.
When mitosis starts, the DNA is duplicated. So after the duplication, the cell will have 22 chromosomes, each of which is attached to its duplicate. Those 22 chromosomes with their duplicates attached line up at metaphase, and then anaphase separates the chromosomes from their duplicates. The original 22 chromosomes go into one daughter, and the duplicates go into the other daughter. Thus, each daughter has 22 chromosomes.
- Chapter 1
- Microscope needed for Biology
To do the microscope experiments in the biology course, you need a microscope with two main features:
1) At least three magnifications, from 40x to at least 400x. Note that for most scopes, the eyepiece is x10, so whatever the lens says, you multiply it by 10 to get the total magnficiation. Most student scopes have three lenses that are 4x, 10x, and 40x, which results in magnifications of 40x, 100x, and 400x.
2) Separate coarse and fine focus knobs. When you turn the coarse focus knob, you see the microscope move. When you turn the fine focus knob, you don't see it move, because it is moving on a microscopic scale.
- How to weight the grades
To weight the grades the way I suggest, calculate a percentage for each test, and give each lab report grade as a percent as well. Then, add all the regular tests' percents together, then add each quarterly test percent twice, and then divide all that by the total number of points, which should be 2,400 (1600 from the 16 module tests and 800 from the four quarterly tests added twice). That will give you a decimal number that represents the tests' average. Separately, add all the lab report percents and divide by the total possible points (100 for each lab), and that will give you a decimal for the lab reports' average. The final grade is then:
(0.65)x(decimal from the tests) + (0.35)x(decimal from the labs)
Here is an example:
Test 1: 87
Test 2: 91
Test 3: 93
Test 4: 88
Qtst 1: 95
Qtst 1: 95
Test 5: 87
Test 6: 89
Test 7: 96
Test 8: 95
Qtst 2: 89
Qtst 2: 89
Test 9: 91
Test 10: 99
Test 11: 100
Test 12: 97
Qtst 3: 92
Qtst 3: 92
Test 13: 99
Test 14: 88
Test 15 : 85
Test 16: 87
Qtst 4: 91
Qtst 4: 91
decimal: 2,206/2,400 = 0.919167
Do the same things for labs. Add up the percentages and divide by the number of labs times 100. You will get another decimal. Let's say it is 0.965143
The final grade is:
(0.65)x(0.919167) + (0.35)x(0.965143) = 0.933959
So the student's final grade is 93.4%
- Microscope light too bright
You can diffuse the light by placing a piece of tissue (Kleenex) over the LED aperature. If that isn’t sufficient try a strip of paper towel.
- Module 1
- What's the difference between exoskeletons and shells?
Exoskeletons are not shells. Think of a clam, snail, or oyster. They have shells. Shells are typically very hard, thick, and jointed in at most one place. A snail shell, for example, doesn't open, so it has no joints. A clam shell has one joint between the two halves. Now think of ants, lobsters, and crabs. They have exoskeletons. Exoskeletons are typically thinner (you can crush an ant easily, you can't crust a clam easily) and jointed in many places so the occupant can move its legs, arms, etc. Insects all have exoskeletons, as do spiders, millipedes, lobsters, crabs, etc.
- Module 2
- Plasma membrane when there is a cell wall or capsule
The cell wall and capsule don't interfere with the plasma membrane's job of bringing chemicals into the cell, because both the cell wall and capsule are permeable. The cell wall has large holes in it so that chemicals can easily pass through. The capsule isn't solid. It is more like a gel. So chemicals can easily pass through it as well. Thus, even when there is a cell wall and/or capsule, the plasma membrane has access to all the chemicals in the surroundings.
- Module 4
- Can you preserve fungi for later use in an experiment?
You can preserve fungi that you find early in the school year, but drying them out is the key. Ideally, you would use a food dehydrator. However, if you don't have one, there are some alternative methods:
As long as you get them dry, you should not refrigerate or freeze them. Just store them in a Ziploc bag and keep the bag in a cool, dry place.
- Module 6
- Should I memorize all 45 definitions for the Module 6 test?
Yes. It's a lot, but it is necessary. Biology is a vocabularly-driven science.
- Module 7
- Module 9
- Test Question 10
The key difference between neo-Darwinism and Darwinism is the recognization that information must be added to the genome. Nevertheless, the way that is assumed to be accomplished is through mutation. If the student mentions information being added but not mutatons (or vice-versa), I would give the student 1/2 of a point.
- convergent evolution and homologous structures
Convergent evolution is the idea that macroevolution produced similar structures in two organisms that are not closely related. As a result, the similar structures were not inherited from a common ancestor. The similarity is simply a coincidence.
Homologous structures are similar structures in two organisms that are closely related. As a result, they were inherited from a common ancestor.
Convergent evolution is a purely macroevolutionist idea made up to explain the problem that some creatures have similar structures even though they are distantly related to one another. In other words, it is simply an attempt to explain around evidence so as to preserve macroevolutionary theory.
Homologous structures apply to both macroevolution and microevolution. A macroevolutionist would believe that a penguin flipper and human arm are homologous. However, the term can apply to microevolution as well. All dogs have essentially the same set of teeth, so their teeth are homologous structures, because they are all related by microevolution.
- Module 12
- How do we kill insects without destroying them?
You can kill insects without destroying them by using a kill jar. Here are instructions on how to make one:
- Module 14
- Is the shape of a dandelion leaf elliptical or lobed?
It is lobed. Technically, we say it is "irregularly lobed," but lobed is good enough.
- General Questions
- Should my freshman skip physical science if she is behind?
My answer to your question depends on why your freshman is behind. If she find science challenging, then I would not skip physical science. Only about 20% of high school students complete biology, chemistry, and physics by the time they graduate, so your freshman isn't really behind. As long as she completes physical science, biology, and chemistry before she graduates, she will be on par with the average college-bound student. That will give her three high school sciences (physical science counts as a high school science, as it is used in ninth grade in most of the private high schools that use my books). Most colleges look for three high school sciences, one of which is from the physical sciences, one of which is from the life sciences, and the third can be from anything.
Now, of course, if she is not being challenged, I would skip the physical science. What she gains from the more challenging courses will be much more beneficial than what she loses by skipping physical science.
- Calculus-based physics book
I don't know of any physics text that is designed for homeschoolers and is calculus-based. However, Physics for Scientists and Engineers (3rd Edition) by Douglas C. Giancoli is an excellent traditional school text. Students often benefit from having a helpful book that has more example problems, like 100 Instructive Calculus-based Physics Examples. by Chris McMullen. There are two volumes to that, one for mechaninics and one for electricity and magnetism.
- How many points is a lab worth?
Typically, a lab is worth 35% of a student's grade. You can assign any number of points you want to a lab. Just make sure to record the grade as a percentage. For example, if a lab is worth 10 points and the student gets 8, he or she got an (8÷10)x100 = 80% on the lab.
When it comes to assigning a final grade, just do a weighted average between the tests and the labs. So, let's suppose the student averages 83% on all his tests and 91% on all his labs. The weighted average would be:
(0.65)x83 + (0.35)x91
The "0.65" tells you that the tests are worth 65% of the grade, and the "0.35" tells you that the labs are worth 35% of the grade. When you multiply 0.65 times 83, you get 53.95, and when you multiply 0.35 times 91, you get 31.85. They add together to make 85.8, which is the students final percentage, including labs.
- Calculus based physics
Unfortunately, I don't know of any calculus-based physics course designed for homeschoolers. There are a lot of calculus-based physics texts, such as the one I use when I teach it at the university level:
However, they are expensive and not designed for independent study.
- Advanced Physics in Creation
- Syllabus for Advanced Physics in Creation
There is a set of daily lesson plans available here:
- Review Question 1
Remember how we graphically add vectors. We put the tail of the second at the head of the first, and then we complete the triangle by drawing an arrow from the tail of the first to the head of the second. That's the sum of the two vectors. This is covered in Module 3 of Exploring Creation with Physics 2nd Edition, pp. 74-78.
In this question, we know the initial velocity vector and the final velocity vector. Well, v_final = v_initial + at. Since t = 1, we can say v_final = v_initial + a. If we had the acceleration vector, then, we would put the tail of the acceleration vector at the head of the initial velocity vector, and then we would draw the final velocity vector by starting at the tail of the initial velocity vector and ending at the head of the acceleration vector.
We don't have the acceleration vector. Instead, we have the final velocity vector. Well, it starts at the tail of the initial velocity vector, so we put it there. It's head tells us where the head of the acceleration vector would be. The tail of the acceleration vector would be at the head of the initial velocity vector, so we now know where both the head and tail of the acceleraiton vector is, so we can draw it.
In the end, the sum of two vectors is a triangle. You need two of the sides of that triangle to get the third side. Usually, you are given two vectors to add. The first vector is the first side of the triangle, the second vector is the second side of the triangle, and the sum is the third side. In this case, you were given the first side of the triangle (initial velocity) and third side of the triangle (final velocity), and you had to come up with the second side (acceleration).
- Module 3
- Advanced Physics n Your Own 3.9
For P1, the answer given in the book is 221 N. It should be 2.20x102 N. However, you shouldn't worry about answers that are off by a few in the last significant figure. Remember, there is always error in the last significant figure. Thus, from a scientific point of view, 221 N and 2.20x102 N are really equivalent, along with 223 N and 219 N.
- Module 5
- Module #5 Practice Problem 4b solution
The solution to this part of the problem should have made it clear that BOTH things were reversed. So, the piano is on the inner radius, and the rope that goes through the small pulley is on the outer radius. That's when the force needed would be 980 N.
- Module #5 Practice Problem 4b solution
The solution to this part of the problem should have made it clear that BOTH things were reversed. So, the piano is on the inner radius, and the rope that goes through the small pulley is on the outer radius. That's when the force needed would be 980 N.
- On Your Own 5.3: Angle of Fh
Remember that we are explicitly putting in the positives and negatives, so our forces are going to come out positive no matter what. I have to reason through their directions. The humerus is pushing down, and as the drawing says, its force vector is down and to the right. Thus, the x-component is positive, but the y-component is negative. Thus, it is in the 4th quadrant. The angle given by tan-1(148/43.7) is the reference angle, which is from the nearest horizontal axis. To convert that to an angle from the positive x-axis, we have to subtract it from 360.
- Module 7
- Module 8
- Module 8 Review problem #5
For all planets, T2/a3 has to be the same. If we do things in terms of earth, we know that T = 365.25 days, because it takes 365.25 days for the earth to orbit the sun. Since we are doing things in terms of earth's orbit, we can say that in some unit, a = 1 for earth. That means:
(365.25 days)2/13 = 133410 days2
Well, that's the same for all planets. Since a = 1 for earth, then a = 0.387 x1 for Mercury. So that means:
T2/(0.387)3 = 133410 days2
T2 = 7732.5 days2
T = 88 days
Remember, it takes 365 and 1/4 days for the earth to complete one orbit around the sun, which is why a leap year occurs every four years. That's why the orbital period is 365.25 days.
- Module 8 Review question #3
Think about the setup. Object 1 and 2 are separated by a distance x:
The "1" is object 1, the "2" is object 2, and the distance from 1 to 2 is x. Now, in order for the forces to be the same so they cancel each other out, it has to be farther from the more massive object. Since the more massive object is 16 times more massive, the distance has to offset the force by 16, so it needs to be 4 time farther away. So you want to place it like this:
The "1" is the more massive object, the "2" is the less massive object. The "o" is the new object. Now, the TOTAL distance from 1 to 2 is x. Thus, you have put "o" in between 1 and 2 so that "o" is 4 times farther from 1 than it is from 2. That means you have split the distance between the objects (which equals x) into two segments, one of which is 4 times longer than the other. But the SUM of those two distance must be x. One segment is 4 times the other segment:
distance 1 = 4x(distance 2)
But they must sum to x:
distance 1 + distance 2 = x
Substituting in 4x(distance 2) for distance 1 in the second equation:
4x(distance 2) + distance 2 = x
distance 2 = (1/5)x
Well, distance 1 = 4x(distance 2), so
distance 1 = 4x(1/5)x = (4/5)x
- Exploring Creation with Physics
- What Math is Necessary for Physics?
The student needs to have completed Algebra 1 as well as a course that teaches the three basic trig functions (sine, cosine, and tangent), how they are defined on a right triangle, and their inverses. Most of the major math curricual cover this in geometry. Math-U-See covers it in the last two lessons of the course, Teaching Textbooks covers it in Chapter 11, Bob Jones covers it in Chapter 12, and Abeka covers it in Chapter 7.
- Is it necessary to use a standard unit in my answer?
It is not necessary to report your answer in standard units if you are not asked to do so. You can certainly solve for a force and use the units (g·m)/sec2. Because I tend to give the answers in standard units, however, that means you will have to convert either my answer or your answer to see whether or not you are correct.
- Significant figures and operations
In general, I try to deal with significant figures only at the end of an equation or when the rules change. So if a single equation has multiple multplications and/or divisions, I wait until the end to round. However, if there is a mix of multiplication/division and addition/subtraction, then I round before I have to change between multiplication/division and addition/subtraction.
It really doesn't matter, however. Remember that there is error in the last signficiant figure. So as long as your answer has the same number significant figures as mine, if there is a slight difference in the last significant figure, your answer is equivalent to mine. So an answer of 43.1 N would be counted equally correct as 43.4 N, since there is always error in the last significant figure.
- Changes in sign
Which signs you use don't matter, as long as you use them consistently within the problem. In some problems, we define down the slope as positive, and in other problems, we define down the slope as positive. I want the student seen it done both ways to see there is no difference in the solutions.
Had you defined down the slope as positive in Practice problem 9, that wouldn't have affected the answer. In the end, you would get the same value for acceleration, but it would be positive. However, all that means is the acceleration is down the slope. Either way, you get the same number for acceleration, and you get the sign that indicates the acceleration is down the slope.
- Module 1
- How to get acceleration from a velocity-versus-time graph
Remember that acceleration is the change in velocity over time. So in a velocity-versus-time graph, the SLOPE of the graph gives you the acceleration. Notice that the graph is falling from about 12 seconds all the way to 20 seconds. When a graph is falling, its slope is negative. So, for the entire time from 12 seconds to 20 seconds, the acceleration is negative. Compare that to the range from 6 seconds to 12 seconds. the graph is rising during those times. When a graph rises, its slope is positive, so the acceleration is positive from 6 seconds to 12 seconds.
- The algebra for OYO 1.3
We start with:
15 = (3.41x104)/delta_t
We need to solve for delta_t, which means we need to get it out of the denominator and into the numerator. We do this by multiplying both sides by it:
delta_T x 15 = 15 = (3.41x104)/delta_t x delta_t
It cancels on the right, leaving
15 x delta_t = 3.41x104
Now we need to solve for delta_t. We do that by dividing both sides by 15
15 x delta_t / 15 = 3.41x104 /15
15 cancels on the left, leaving
delta_t = 3.41x104 /15
- Module 2
- Mod 2 Test Qest #14: Why is the left side of the equation zero?
In this question, we are trying to figure out the maximum height of something that travels straight up in the air. Now remember, when something travels straight up into the air, gravity is pulling it down, so as it travels upward, its upward velocity decreases. Once that upward velocity hits zero, it can no longer travel upwards, so it starts to fall. Thus, the maximum height occurs where the VELOCITY (not the displacement) equals zero. So that's something we know about the maximum height: at that point: the velocity is zero. Since we are working with free fall, we also know the acceleration: 9.8 m/sec2 down. Finally, the initial velocity is given: 2.0x102 m/sec upward. So if we look at the one-dimensional motion equations we have, there is only one that contains displacement (what we want to know), acceleration, initial velocity, and velocity:
v2 = vo2 + 2(a)(delta_x)
We know v = 0, because maximum height occurs when velocity is zero, we were given vo (2.0x102 m/sec), and we know acceleration (since we are using vo as positive, that means upward motion is positive, since gravity accelerates things downward, it is -9.8 m/sec2). That's how I get the equation in the answer key:
(0)2 = (2.0x102 m/sec)2 + 2(-9.8 m/sec2)(delta_x)
- Is the acceleration due to gravity infinitely precise?
The acceleration due to gravity is not infinitely precise. It is a measured quantity. Since the value given in the book has two significant figures, that limits a lot of the answers to two significant figures.
- Module 3
- Graphical subtraction of vectors
Here is a video explanation of the graphical subtraction of vectors:
- Graphical addition of vectors
- Module 5
- Why is the weight 730 on Practice Problem 6?
It is a significant figures thing. The weight is the mass times the acceleration, and the gravitational acceleration is 9.8 m/s2. So that means:
w = (74.9 kg)(9.8 m/s2)
74.9 has three significant figures, and 9.8 has only two, so the answer can have only two. So even though the answer is 734.02, it must be rounded to two significant figures (730 N).
- Module 7
- On Your Own problem 7.7, how do you get 4.8x10^3?
It is really best to use a calculator in this course, as math with scientific notation can be very confusing. The square root of 2.26x107 is 4.8x103. If you want to do it by hand, you need to remember how the decimal place moves in scientific notation. To take the square root, you need the exponent to be divisible by 2. Thus, you need to change it so the exponent is x106. When moving the decimal place, you need to think about what your exponent change did to the number. By changing the exponent from 107 to 106, you LOWERED the value of the number. You have to make up for that by RAISING the value of the decimal. Thus, 2.26 must become 22.6. That's why 2.26x107 is 22.6x106.
This can be confusing, because when changing a number like 3,450,000 to scientific notation, you are moving the decimal to the left, not to the right. Once again, however, this is because of the change you are making in the number. When you change 3,450,000 to 3.45, you are LOWERING the value of the number. As a result, your exponent must RAISE the value, so it must be a positive exponent, 3.45x106.
In scientific notation, you can't just memorize what do to when you move the decimal left and right. You have to think about what you are doing to the value of a number when you make the change. Of course, a scientific calculator takes this into account, and I strongly encourage you to use one.
- Module 8
- On Your Own 8.4 significant figures Remember that the rules of a
Remember that the rules of addition and subtraction go by precision, not number of significant figures. So..the equation is:
9.8*21 - 9.8*12
When you multiply, you count significant figures, so each term has two significant figures:
210 - 120
But now you are subtracting, so you have to report your answer to the same precision as the least precise number in the problem. Both numbers have their last significant figure in the tens place, so the answer must have its last significant figure in the tens place. That's why the answer is 90.
Typically, you don't bother with significant figures until the end of an equation, but when an equation mixed multiplication/division with addition/subtraction, you have to round before you switch the rules.
- Module 9
- Review Question #6, why are the momenta not necessarily equal?
Remember, kinetic energy is equal to (1/2)(m)(v2). So we know:
(1/2)(m_cat)(v_cat2) = (1/2)(m_mouse)(v_mouse2)
The 1/2 cancels, so:
(m_cat)(v_cat2) = (m_mouse)(v_mouse2)
Now, for the momenta to be equal, the following must be true:
(m_cat)(v_cat) = (m_mouse)(v_mouse)
The easiest way both equations could be true would be if
v_cat2 = v_cat and v_mouse2 = v_mouse
However, that would only happen if the v's were 0 or 1, because 02 = 0 and 12 = 1. There could be a few combinations of masses and velocities where both equations would be true, but they would be rare.
- OYO 9.8, why is 9.0 x 10^1 last sig fig in the ones place?
When you have a number in scientific notation, whether or not a figure is significant depends on the decimal place. So, in 9.0x102, the zero is significant. However, you have to think about where that zero actually is. It's not in the tenths place, because the 9.0 is mutiplied by 10. So that zero is actually in the ones place. That means 9.0x101 has its last significant figure in the ones place.
Think about it this way: 9.0x101 is 90, but it was put in scientific notation specifically to indicate that the zero is significant. 90 is its value, but 9.0x101 tells you what digits were measured. Both the 9 and the 0 were measured, so the last significant figure in 9.0x101 is the 0. Since the zero is in the ones place, 9.0x101 has its last significant figure in the ones place.
The same kind of reasoning applies to -1.0x102. The number's value is 100, but it was put in scientific notation specifically to indicate that the first zero is significant. Where is that first zero? It's in the tens place. Thus, -1.0x102 has its last significant figure in the tens place.
- Module 12
- The radius ray of a curved mirror.
If you have an object between the mirror and the radius, you draw the ray as if it came from the radius. When it reflects back, it can be extended to the radius and beyond, and it is not considered an extrapolation. Even though the ray did not come from the radius, it comes from the object at the same angle as if it came from the radius. Thus, from the object on, it is a real light ray. Once it reflects, it is still a real light ray and will continue traveling through the radius and beyond.
- Module 13
- Significant figures in practice problem 6Remember that addition and subtraction go by precision. To see the actual precision of a number in scientific notation, however, you need to think about the number's real value. The number 7.7x104 has a value of 77,000. The significant figures come from the scientific notation (both 7's are significant), but where is that second 7? It is in the thousands place. In the number 1.7x105, the value is 170,000. Once again, the 7 is significant, but it is in the ten thousands place. The ten thousands place is less precise than the thousands place, so the answer must be written to the ten thousands place. In decimal notation, that would be 90,000. In scientific notation, it would be 9x104.So you use the decimal place in the scientific notation to determine what is the last significant figure, but you must use the actual decimal place of the last significant figure when you are determining significant figures. In the second one, you have 2.6x104, and the 6 is in the thousands place. For 6.3x105, the last significant figure is in the ten thousands place, so once again, the answer must be limited to the ten thousands place, which is -660,000, or -6.6x105.
- In Extra Practice Problem 6, what are the significant figures?
Remember that when you add and subtract, you DO NOT count significant figures. Instead, you look at precision, and you report your answer to the same precision as the least precise number in the problem. To get Cx, we have this:
-6.8x106 N + -3.5x106 N
Both of these numbers have their last significant figure in the hundred thousands place (6.8x10^6, for example, is 6,800,000). So the answer must have its last significant figure in the hundred thousands place. When you add them, you get:
That's -10,300,000. So that is the right answer, since the last significant figure is in the hundred thousands place.
For Cy, you have:
-1.8x106 N + 6.3x106 N
Once again, both of them have their last significant figure in the hundred thousands place, so the answer should as well. -1.2x106 N has its last significant figure in the hundred thousands place, so it is correct.
- Extra practice problem 6: why isn't the angle 190?
This was a while ago, but in Module 3, you were told that the 180 degrees you use to adjust the angle is exact (see p. 82, for example). It has infinite precision. Thus, it's like 180.000000000000000... That means when you add 180 to something to adjust the angle, you look at the original angle's precision. So since 6.6 goes to the tenths place, the answer goes to the tenths place.
- Module 14
- Question 9 in the review questions
As you learned in Chapter 13, opposite charges attract and like charges repel. Since electrons are negative, they will be attracted to the positive side of the capacitor. Since they already have some speed, however, they will only be deflected from their straight path. This is like the deflection system discussed in the last section of the module.
- Module 15
- Significant figures in the parallel circuit problems
To do the significant figures, in these problems, you need to take the inverses, round to the same significant figures as the resistances, then add based on precision, and then take the inverse again, but count significant figures. So, a problem with two parallel resistors that are 15 Ohms and 12 Ohms, you would do this:
1/R = 1/r1 + 1/r2
1/r1 = 1/15 = 0.067 (round to two significant figures)
1/r2 = 1/12 = 0.098 (round to two significant figures)
1/R = 0.067 + 0.098 = 0.181 (each of the numbers has its last signifcant figure in the thousandths place, so the answer must have its last signficant figure in the thousandths place)
R = 5.52 Ohms (round to three significant figures)
- Advanced Physics in Creation
- Advanced Chemistry
- Other sources for extra problems
There are no other problems than what you see in the book. There are a couple of good websites that do a lot of things like that, however:
- Module 1
- Enthaplies and Significant Figures
You may have forgotten this from your first-year course, but enthalpies follow the rules for addition. That's because even though you are multiplying sometimes, the numbers you are multiplying by are exact. When you multiply a Delta_H by 3, for example, it is really just adding the Delta_H three times, since the three is exact. So you are really just adding Delta_H's and should therefore follow the addition rules, reporting your answer to the same precision as the least precise Delta_H.
- Module 2
- The signs of the magnetic and spin quantum numbers
The signs are really arbitrary. They depend on how you define the three-dimensional space. Thus, whether you use positive or negative magnetic and spin quantum numbers first doesn't matter. If you have l = 2 (for p orbitals), then you can use ml = -1 first or ml = 1 first. If you have l=3 (d orbitals), you can use ml = -2 first or ml = 2 first. In the same way, if there is only one electron in an orbital, you can give it an ms of -1/2 or an ms of 1/2. It doesn't matter.
- Module 6
- Significant figures, Example Problem 6.2a and b
There should be four significant figures in the exponent of "e" in both of those problems, which means the answers should have four as well.
- The significant figure rule when raising a number to a power
The phrase, "When raising a number to a power that has error in it..." refers to a situation when the exponent isn't an exact number. Think about a rate equation like:
R = k[A]2
The exponent (the square) is exact, because rate equations have integer or half-integer exponent. Thus, it can't be
R = k[A]1.9 or R = k[A]2.1
It is exactly R = k[A]2
In these thermodynamics problems, the values for Delta_G are not exact. They have error in them. Thus, in Example 6.2a, the equation:
could actually be
e0.483 or e0.487
In other words, the exponent itself has error in it. You need to know how that affects the error in your answer. That's where the significant figure rule comes into play. It tells you that in this case, your answer needs to have three significant figures.
- Example 6.2: How do you know where to round your exponent?
You use the standard rules for multiplication and division to get the exponent. So you are dividing a number with three significant figures by two numbers, each with four significant figures. That means you can have only three significant figures in your answer, which is why the exponent is 67.4.
- Module 7
- Why does the method work in Example 8.10?
The method works because of the nature of an equilibrium reaction. Remember, for a given set of reactants at a given temperature, there is only one equilibrium position. If I start with a certain concentration of NO and a certain concentration of Cl2, the reaction will reach an equilibrium position. If I somehow then force it out of equilibrium without changing the amounts of anything, the reaction will strive to get back to that same equilibrium point. So, in this method, I am starting with a certain amount of NO and a certain concentration of Cl2, and then I am artificially forcing the reaction to go as far to the right as possible. That's okay, though, because the reaction will strive to get back to equilibrium. So all I have to do is figure out what the concentrations of all the chemicals are once the reaction is artificially forced to the right, and then I can track how it gets back to equilibrium.
Why do I use the K as it's set up for the forward reaction if I am tracking the reverse reaction? Because it doesn't matter. If you wanted to use the K for the reverse reaction, you could. What would it be? It would be [NO]2[Cl2]/[NOCl]2. What's the value of that equilibrium constant? Since it is the inverse of the K as given, the value is the inverse of the given K's value, or 1/(2.5x103). In the end, if you use that K to set up the equilibrium, you just get the inverse of the first equation on page 251. That will end up giving you the same value for x, which will lead to the same answer.
Now there is an error in the solution, which I didn't notice until this question was asked. In the solution, I failed to square the concentration of NOCl. Thus, the top term on the right-hand side of the equation on page 251 should be (3.80-x)2. That leads to a MUCH longer set of approximations, and x = 0.11 instead of 0.049.
- In Ex 8.8, why isn't water ignored in the equilibrium constant?
You ignore water when you know it is a liquid or solid. Since the phases aren’t given in the problem, you don’t know that it is a liquid or solid, so you can’t ignore it.
Also, on a more technical note, you ignore liquid water only in aqueous solutions. When water is the solvent, you have enough of it to keep the concentration constant. However, if it is not a solvent and is either being used or made, the concentration does change under certain conditions. As a result, it needs to be kept in the equilibrium constant.
- Module 10
- When do you use H+ and when H3O+ in balancing Redox?
You can use either, because they are really the same thing. There are no free H+ ions in solution. They are always added to a water molecule. Thus, you can use either one. The only thing that changes is the number of water molecules on the other side. So a reaction like
ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+
ClO3¯ + 9H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H3O+
Both equatons are equivalent.
- Advanced Physics
- Chapter 6
- Why is the distance 4A in Review Question 3?
Remember that A is the maximum distance from equilibrium. So....let's assume it is pulled all the way to the right and let go. When it travels left to the equilibrium position, it has traveled A. Then it travels left through the equilibrium position to the left side, traveling another A. That's only half the trip back, however. It must now travel A to the right so it gets back to the equilibrium position and another A to get back to the A on the right. That's a total round trip, and it is 4A.
- Chapter 6
- Discovering Design with Chemistry
- Junior High Science
- Science in the Atomic Age
- Chapter 2
- Comprehension Check 2.15
Isotope is a relative term. It means in relation to something else. So you have to compare two atoms. It's like the term "brother." A single male child cannot be a brother. There must be at least two siblings. If so, the male siblings are the brothers. In the same way, you must compare two atoms to determine if there are isotopes. The isotopes will be the atoms that BOTH have the same number of protons but a different number of neutrons. So A and C are isotopes since they both have six protons (they have the same number of protons) but they have different numbers of neutrons (one has six neutrons, and the other has eight).
- Chapter 6
- Comprehension Check 1: Why do larger cells move things faster?
Speed and time are not the same thing. Let's suppose a substance moves at the same speed in all cells, no matter the size. In a small cell, that substance would get to its destination in less time, because it had less surface area to cover. However, if a substance is supposed to get to its destination in a set time, it must move faster in a larger cell, because it has more surface area to cover.
Think about it this way - It's 7:00 AM, and you have to get to an appointment by 7:30. If the appointment is 5 miles away, you wouldn't have to travel very fast to get there on time. If it is 25 miles away, you would have to travel faster to get there on time. So...if a substance in a cell must make it to a destination, it must travel faster the farther away that destination is.
- Chapter 2
- General Science
- Is General Science too difficult for a bright 5th grader?
I have heard from parents whose 5th graders have done the course successfully. In my opinion, there are two things to consider:
1. What is the child's math level? If he or she will be doing 7th-grade math at the same time (that means the student will be one year before pre-algebra), then he or she is also at the 7th-grade level when it comes to science.
2. Does the student have experience with studying for and then taking tests? That's a big adjustment for some students. If the student has already learned how to study for and take tests, that will help a lot.
- Module 2
- Module 3
- Question 3.2
A student could easily get tired after scrubbing 2 or 3 tiles. It all depends on how much force he or she uses and how fast he or she scrubs. I can tire after just a minute of exerting a lot of force quickly.
The problem doesn't say that there are different amounts of heel marks, but it also doesn't say that there are the same amounts of heel marks. Thus, that could be a variable. Think about heel marks on a floor. Do they happen uniformly everywhere? No. Most likely, then, the amount of heel marks varies.
- Module 5
- Getting Alum in the UK
If you are having trouble finding alum in the UK, try Amazon:
You can also try Oxford ChemServe
- Physical Science
- What math is needed for Physical Science?
I don't have a scope and sequence of the math in Exploring Creation with Physical Science, but I can tell you what the student needs to know to understand what's in the course:
1) The student needs to be able to multiply fractions, recognizing that like terms in the numerator and denominator cancel.
2) The student needs to be able to square numbers, like 32 = 9.
3) The student needs to be able to plug numbers into equations and evaluate them. For example, if distance = (1/2)(a)(t)2, and a = 9.8 m/sec and t = 1.0 sec, what is the distance?
4) The student needs to understand proportional and inversely proportional, including squares. For example, the force of gravity is proportional to the mass. So...if mass is multiplied by 2, the force of gravity is multiplied by 2. However, it is inversely proportional to the square of the distance. So, if distance is multiplied by 2, the force of gravity is divided by 4.
- Module 1
- Module 4
- What do the Epsom salts do in Experiment 4.1?
In order to get the water to break down into hydrogen and oxygen, you have to add energy. In Experiment 4.1, you are adding that energy with electricity. However, for electricity to add energy to the water, it must flow through the water. Pure water cannot conduct electricity. Thus, this experiment wouldn't work in pure water, because electricity could not flow through it. As a result, you must dissove something in the water that will allow it to conduct electricity.
There are lots of things you could have added, including normal table salt. However, normal table salt will cause a side reaction to occur that will take oxygen away. You don't want that, so you need to add something that both makes the water conduct electricity and doesn't promote a lot of side reactions. Epsom salts fit that bill, so that's why they are used.
- Module 10
- Velocity, Force, and acceleration
The idea is that the equation
F = ma
only works when F is the total of all the forces acting in the problem. Friction always acts against motion, so if you have a man pushing something:
F = F_man - F-friction
For example, if a man applues 100 N of force, but there is a 35 N frictional force:
F = 100 N - 35 N = 65 N
65 N = ma
Alternatively, lets say a man is using a 50 N force, and a 10-kg object is accelerating at 3 m/sec2. We can calculate the total force:
F = ma
F = 10 kg x 3 m/sec2 = 30 N
However, that's the TOTAL force, which include both the man pushing and friction. So:
F = F_man - F_friction
Since F is 30 N and the man is pushing with a 50 N force:
30 N = 50 N - F_friction
The only way this works is for the frictional force to be 20 N.
- Exp. 10.3: Why the ice cube slid faster down the rough board
It is quite possible that the ice cube will slide down the rough board faster than the smooth board. That's because the ice cube is melting, and therefore there is water coming off it. That water can fill up the pits in the rough board, making a thin layer of water that the ice cube can slide on. This can't happen as well on the smooth board, because the water just rolls down it. In this case, then, the WATER is sliding down the rough board more slowly, but that allows the ice cube to slide down more quickly,
- Module 11
- Mass Bends Spacetime
In the General Theory of Relativity, mass bends space and time together. In this view of physics, space and time are not separate. They are connected together in a four-dimensional matrix called "spacetime." Mass bends spacetime because it interacts with the matrix. We have no idea exactly how this happens. However, if we assume it does, we can explain several observations that otherwise could not be explained.
Because of this effect, mass affects the passage of time. In the presence of a lot of mass, time moves slowly. In the presence of less mass, time moves more quickly. This sounds odd, but it is has been experimentally confirmed. For example, time moves more quickly on the global positioning system satellites than it does on the surface of the earth, since the satellites are farther away from the mass of the earth. If this difference in the way time behaves were not taken into account, the global positioning system would not work. Since the moon has less mass than the earth, time moves more quickly on the surface of the moon than it does here on earth. As a result, a person would age more quickly on the moon than he would on earth.
- Module 14
- Is it possible for sound waves to have negative decibels?
Yes, but you won't hear them. However, some animals can. A level of 0 decibels means you are at a volume where human ears can barely hear. So, if a sound has such a low volume that a human can't hear it, the sound would have a negative decibel measurement. Dogs, for example, can hear certain sounds that have volumes of up to -4 decibels. However, this is frequency dependent. Typically, the higher the frequency, the more decibels of volume it needs to have for the dog to hear it.
- Science in the Atomic Age
- Elementary Science
- The Series
- How my series differs from the Young Explorer Series
The Young Explorers Series published by Apologia is an excellent series that has been used successfully by many, many homeschooled students. My elementary series is quite different from the Young Explorers Series in many ways, because I am trying to help those for whom the Young Explorers Series is not a good fit. Here are the main differences:
1. The Young Explorers Series is topical. You study Astronomy for a year, Botany for a year, etc. My series presents science in a historical framework. The first book deals with the six creative days in Genesis 1 as the beginning of history, and the student learns about what was created each day. So when the student studies the first day of creation, the student learns about light. When the student studies the second day of creation, the student learns about air and water, etc. The rest of the series will then present science topics as we learned about them in history. So rather than being topical, my series will be more chronological.
2. The Young Explorers Series has about 13-15 lessons per book, so each lesson must be spread over several days. My series will have 90 lessons per book, and each lesson is designed to be done in one sitting. So my books are a bit more "scheduled" than the Young Explorers Series. You know exactly where to start and stop. Start at the beginning of a lesson, and science is over for the day when you reach the end of the lesson.
3. The Young Explorers Series does not have as many hands-on activities as my series. In the Young Explorers Series, there are many days where you do nothing but reading. In my series, every day that you do science has a hands-on activity.
4. The Young Explorers Series is based on notebooking and narration so that it can be multigrade. My series is also multigrade, but it does so by having three different levels of review for each lesson. The youngest students orally answer two questions about the lesson, and that's all they do. The older students have a notebooking assignment for their review, and the oldest students have a more complicated notebooking assignment for their review.
5. The Young Explorers Series does not have tests. I am not a fan of giving tests at the elementary level in science, but I understand that there are many parents who want to give tests to their students. Thus, my series will have 12 tests per book, so you could give the student a test every 3 weeks, if you choose.
- When will the elementary books be available in non-US countries?
The answer to this depends a lot on what the Lord is leading. However, here is the plan. If the field test of the first book in the series (Science in the Creation Week) indicates that homeschoolers find it useful, I plan to make it available in book form starting in the summer of 2012. Currently, I do not plan to publish it myself, but instead will try to get a several non-US publishers (each in a different country) to publish it. That way, its cost will depend totally on the economy of the country you are in. So for Australia, for example, I would like to get a company like Adnil Press to publish it. They would then set the price, based on what their costs are. Please note that Adnil Press has not agreed to this. I am just using them as an example (and as one I am truly interested in).
- What ages are my elementary courses geared to?
In general, the courses are multigrade, K-6 courses. I do think that it is best done when the oldest student is at least in third grade. Once that happens, however, the younger ones can be a part of the course, too. Since there is an experiment or activity for every lesson, even the youngest will get something out of the course. Also, the review part of each lesson has three levels of difficulty, so you can find a review that works for nearly every student.
The big thing, however, is that you need to judge for yourself. For example, if you have students that are 12 and 14, at first glance, I would say that the 14-year-old is too old. However, it all depends on where he is in his development and interaction. If he has not really learned much science, the courses might work well for him, although they are geared (on average) to younger students.
I would encourage you to read through some of the lessons to see what you think. You will be able to judge whether or not the course will be too basic for your children. If so, you might consider looking at Exploring Creation with General Science, which is a book I wrote that targets seventh grade. It is available here:
- Fulbright Zoology to My Elementary Courses
I think that going from Jeannie Fulbright's Zoology series to my series would be a great transition, since Science in the Beginning does a lot of chemistry and physics, which the students haven't had yet. There will be some overlaps, but not a lot. Science in the Acient World doesn't do much biology at all, so it will definitely expose the students to a lot of things they haven't seen before.
You might consider having the 5th grader do the tests to get him or her ready for the testing that will happen in General and Physical Science.
- Grading in the Elementary Series
If you want to assign grades in this course, I would break it down into two segments:
(1) Notebooking journal (which includes the review questions and the labs)
The notebooking journal should be 40% of the grade, while the tests should be 60%. So if the student averages 75% on the tests and 90% on the notebooking journal, the total grade would be:
0.40x90 + 0.60x75 = 81%
- Teachning multiple students with this series
You only need one book to teach however many elementary students you have. They should all be doing the lesson at the same time. They can all refer to the same book when doing the experiments, and when it comes to the reading, either you read aloud to everyone or you have the good readers take turns reading aloud to everyone.
- What would your suggestion be for a notebook for journaling?
The journal will definitely need a place to draw pictures. I personally like notebooks with blank pages so the student can do anything he or she wants with it. You could also go here:
and click on "downloads." One of the options is a "Printable Notebook." If you click on that, you can get a template that two mothers made for Science in the Beginning. Even if you are not using that specific book, it gives you an idea of how you might make one or two template sheets that contain both places to draw pictures and lines for writing.
- Using Science in the Beginning in a co-op setting
When using any book of this series in a co-op setting, I suggest that you focus on doing the experiments as a group. That way, the student can do the reading and the notebooking exercises on his or her own. I suggest that you do the experiments that pertain to the NEXT WEEK'S reading. That way, the students have already experienced the experiments before learning what the experiments show.
- Coloring pages to go along with the course?
Unfortunately, we don't have any coloring pages to go with the elementary courses. However, Here are two free sites that might be worth looking at:
They each have search boxes, so if you search for "light" you will find a light bulb to color, while searching for "rainbow" gives you rainbow coloring pages.
- Is Bernoulli's principle taught in your elementary science?
Yes, it is. I discuss Daniel Bernoulli in Lesson 12 of Science in the Age of Reason. I focus on the fact that the pressure in a moving fluid gets lower the faster the fluid moves. The experiment involves blowing over a straw that is in a glass of water. The reduced air pressure causes the water to rise up and out of the straw, spraying water in the direction that the student is blowing. I then tie that into how an airplane flies.
- Which of my elementary books should a 6th-grader use?
I would recommend that a 6th-grader who has taken a lot of science should do my "Science in the Ancient World" book, including the tests that come with it. If the student does the "Oldest student" reviews, it will be an appropriate level for a 6th-grader, and if the student takes the tests, he will get some testing experience, which will prepare him for junior high school science.
- Time periods and history curricula
Berean Builders has a website that has six popular history curricula and how my courses line up with each:
One of those is Story of the World. Here are the time periods covered in each book:
Science in the Beginning: The six creative days in the Genesis Account
Science in the Ancient World: 600 BC to AD 1540
Science in the Scientific Revolution: 1540 to 1700
Science in the Age of Reason: 1700 to 1800
Science in the Industrial Age: 1800 to 1900
- How long does a book take?
The amount of time the course takes depends on which lessons you do. There are two types of lessons: normal and challenge. The titles of the challenge lessons are in red. If you skip the challenge lessons, there are a total of 72 lessons in the course. If you do science twice a week, that means the course will take 36 weeks. If you do the challenge lessons, there are a total of 90 lessons in the course. That means it would take 45 weeks at two lessons per week. Of course, if you do more lessons per week, the course takes less time. Doing a lesson every other day, for example, would cover 90 lessons in 36 weeks.
- Is this series a cycle? Do the books get harder, or the review?
The series is designed to be a cycle. You start with the first one with all your K-6 children, and you continue through the series. As your family grows, you just keep bringing children in to join the group, wherever you are in the series. When you are done with the series and the oldest children start doing junior high school work, you start over with the rest of the K-6 children.
I tried to write the books all at the same difficulty level, so that the only thing that changes for the student is the level of review. The older the student is, the more detailed the review becomes.
- Timeline to go with the series
I have considered producing a timeline for the series. However, I doubt that I will get around to it any time soon. There always seems to be something else that needs to be done.
- Science In The Beginning
- What about when the season is wrong for outdoor experiments?
In the situation where the season is wrong for outdoor experiments, I would just skip the experiment. I would go ahead and do the lesson, since some lessons build on previous lessons. However, I would mark the lesson in some way so that you can do the experiment later when the season changes. One other thing to consider is that you could do a YouTube search for a video of a similar experiment. For example, if it is just too overcast and cloudy to do the experiment where you burn paper with a magnifying glass, just go on YouTube and search "burning paper with a magnifying glass," and you can watch a video of it.
- Type of Notebook for Science in the Beginning
I think the ideal notebook for Science in the Beginning would be a spiral-bound notebook that has blank pages. There is a lot of drawing in the course, and lined paper makes that just a little harder. Also, I like spiral-bound for elementary because it lays flat. In high school, the student will need a bound notebook that makes it difficult to tear paper out. However, for elementary science, that's not necessary. If you can't find a spiral-bound notebook with blank pages, use a 3-ring binder and punch holes in blank, white paper.
- Is order important in Science in the Beginning
I did design the course assuming that the student starts at day 1 and goes through the days in order. For example, in Day 1, I define certain kinds of energy (mechanical energy, chemical energy, radiant energy, etc.). In later days I simply refer to those kinds of energy, assuming the student will know what they are. Of course, they are defined in the glossary, so by looking at the glossary more often, etc., you probably could do the days out of order.
- Notebook for Science in the Beginning
Two mothers who used Science in the Beginning have made a notebook that is free to download:
- Errata for the first printing of the course.
- Can this be taught as a class one hour per week
I think it would be hard to teach this as a one-hour-per-week class. If you skip the challenge lessons, you still need to do two lessons per week to cover the book in a year. In a class setting, that means two experiments per class. I would think doing two experments and explaining their meaning would be difficult in one hour.
You could do one lesson in class and then suggest that they do the other lesson at home. Alternatively, you might be able to do two lessons if you had 90 minutes instead of just an hour.
One other thought, which I am not a big fan of but could work, is to do the expeirments as demonstrations. I want the students to do the experiments themselves, but if you did them as demonstrations, they would go faster, and you could probably do two in an hour.
- Ages for the review exercises
I would say that the "youngest" students are those who aren't able to write much. They have two questions to answer orally. Once the students are comfortable writing, they should do the "older" student exercises. The "oldest" student exercises should be done by fifth and sixth grade students.
- What are the "hot pads" needed for one of Day 2's lessons?
They are the pads you use to hold on to hot things. When you pull a casserole dish out of the oven, you use hot pads to protect your hands from being burnt.
- Lesson 1
- Is the moon a mirror? Do stars make their own light?
The moon is like a mirror in the sense that it reflects light, but a mirror reflects light differently than the moon. You will learn the difference between the way a mirror reflects light and the way things like the moon or a sheet of white paper reflect light in Lesson 5.
Stars do make their own light. They are too far away to reflect light from the sun.
- How do we know that the moon reflects the sun?
We know that the moon reflects light from the sun for at least three reasons. The most important one is that it goes through phases. At times, some parts of the moon are dark, while other parts are light. When you compare where the moon is to where the sun is, you see that the light parts of the moon are the parts that can reflect light from the sun to the earth. The dark parts are not in the right place to reflect the sun's light back to earth. Thus, the light from the moon is the sun's reflected light.
Another reason is what we call "earth shine," and it was figured out by Da Vinci in the 1500s. When the moon is a thin crescent, you can often see the rest of the moon, but it is dim. That's because the earth is reflecting the sun's light to the moon, and then the moon is reflecting that light back to the earth. It is dimmer than the sun's light, so those parts of the moon are dimmer.
Nowadays, we can actually measure precisely how much light is coming from the moon. We can do the same thing for the sun. We see that when the sun gets a bit brighter, the moon does as well. When the sun gets a bit dimmer, the moon does as well. Thus, the moon's light is actually the sun's light.
- Lesson 2
- Why Didn't God Put Pink in the Rainbow?
Actually, there is pink in a rainbow. However, it is difficult for us to see, since our eyes see the red more vividly. If you are fortunate enough to see a wide rainbow where the colors are more streched out, you might be able to find pink in it.
- Lesson 3
- What makes objects reflect certain colors
All objects are made up of either atoms or molecules. These molecules are able to absorb certain colors, and they reflect other colors. Two shirts made from the same material can be different colors because dyes are added to the fabric. Those dyes are chosen so that they absorb the colors you don't want to see and reflect the colors you do want to see. A red shirt, for example, is dyed with a chemical that absorbs all colors except red. A blue shirt made of the same material is dyed with a chemical that absorbs all colors except blue.
- Why can we see black?
Black is the absence of all color. We see it because we are comparing it to other things that send light to our eyes. You see a black box sitting on a table, for example, because the table is reflecting light, sending it to your eyes. The black box is not reflecting any light, so it is sending nothing to your eyes. As a result, your eyes see light coming from the table, and then this box-shaped area from which no light is coming. That lack of light surrounded by light from the table defines the box.
Suppose you had a box that was open on the top but completely black inside. Now suppose you put a red ball in it. You would see the ball because light comes in through the open top, hits the ball, and the ball reflects red light into your eyes. Now suppose you replaced the red ball with a black ball. The inside of the box isn't sending any light to your eyes, and neither is the ball. Thus, you wouldn't see the ball.
- Lesson 4
- Can artificial light be used in Lesson 4?
Artificial light can be used in lesson 4, but it has to be from an incandescent light bulb (the kind with a filament - not a compact flourescent light bulb or an LED light bulb) that is around 100 Watts or more. You need to get the light bulb directly above the plastic in some way, either by using a lamp with a movable head or by laying the lamp on its side on books so the bare light bulb is hanging above the plastic. Also, you have to play a bit with the distance between the light and the plastic. If you get the light too close, it will heat both of them up so much that you won't notice a difference. If you have it too far away, it won't heat either of them enough to notice.
- Where Does Energy Go After It Is Used?
When you use mechanical energy to move, you are using your muscles. As you walk, for example, your muscles contract and relax to move the bones in your skeleton. As the muscles are doing that, the fibers that make them up are rubbing against each other, causing friction. Your muscles expend energy fighting that friction. When you want to stop, your muscles stop fighting the friction, and the friction between the muscle fibers quickly causes the muscles to stop contracting, which causes you to stop your motion.
Friction converts mechanical energy into heat energy. If you rub your palms together vigorously, for example, you will feel them heat up. That's because friction is converting some of that mechanical energy into heat energy.
Most mechanical energy, then, gets converted into heat energy via friction. That's where most of the energy in motion goes to when you stop moving. Some of it is also absorbed by the molecules in the air, and they end up moving more quickly as a result. If you could really account for it, you would find that all the energy you used for motion was converted into some other kind of energy (heat via friction, motion of air molecules, etc.)
[Names are never published with the answers. Only the answers are published.]
- Lesson 3
- Why Don't the Dark Colors Cover the Light Ones in White Light?
All colors are present in white light. However, when they hit an object, the object refects only certain colors, giving the object the color you see. The darker colors don't overpower the lighter colors because an object reflects light based on the ENERGY of the light, and each color of light has a different energy. Thus, the colors don't overlap. This is different from when you draw with a ight-colored crayon and then draw over it with a darker-colored crayon. When you do that, the light hits BOTH colors, and they each reflect the light that corresponds to their color. However, they both absord the OTHER COLOR'S light. In the end, the darker color absorbs a lot more light, so most of the lighter color's light is taken away. As a result, you see mostly the darker color. In white light, the colors don't overlap. They each have a different energy, and since an object reflects light based on its energy, the colors don't interfere with one another.
- Why Does a Colored Object Reflect Only Certain Colors of Light?
All objects are made of atoms or molecules. Even molecules are made up of atoms, so essentially, all objects are made of atoms. An atom has tiny particles called electrons that orbit its center. The center of the atom is made up of other tiny particles called protons and neutrons. When light hits an atom, if it has the right energy, the atom's electrons can absorb the light. This usually results in the light's energy being converted into thermal energy, which heats up the object. The key is that when this happens, the light is gone. Any light that has energy which cannot be absorbed is typically reflected, and it becomes the light you see coming off the object. Well, each color of light has its own energy, so when certain energies are absorbed, certain colors are taken away. Whatever is not abosorbed is reflected, and those colors mix to form what you see as the color of the object. So when I want to paint an object red, I will use a paint whose molecules have atoms that absorb energies corresponding to all colors except the shade of red that I want. That way, the paint gets rid of all the colors in the white light except the shade of red I want, and it reflects that color.
- Lesson 5
- Trouble burning paper with a magnifying glass
This experiment does work better when the sun is higher in the sky. If it is fall or winter, it doesn't work as well. Nevertheless, you should still be able to get it to work. First, try a single sheet of newspaper and wad it up very loosely. The paper is thin and will burn more easily. Second, play with the tilt of the magnifyinng glass so that the circle of light is as tight as possible. It's not just the distance between the magnifying glass and the paper that is important. It is also how the magnifying glass is tilted. Play with both the distance and the tilt to get as tight a circle of light as possible.
- Lesson 6
- If you can't create energy, how can you make it?
You can make energy by converting it from one form to another. For example, we make electricity by burning something (like coal). The heat is used to boil water, and the motion of the steam is used to turn turbines, which generate electrical energy. Throughout that process, no energy is created. Instead, the energy that is stored in the chemical bonds of the wood and the oxygen in the air gets released in the form of heat. So chemical energy has been converted into heat energy. That heat energy then boils the water, and that causes the steam to move. Boiling the water, then, converts the heat energy into mechanical energy. The motion of the steam is used to move the turbines that generate the electricity. This means the mechanical motion of the steam is converted into electrical energy.
- Why is it called the Law of Energy Conservation?
In science, when something is "conserved," it means that it is kept intact. The Law of Mass Conservation, for example, says that in chemical processes the total amount of mass cannot change. The chemicals can change, but the total mass of all the chemicals after a reaction have to add up to the total mass of the chemicals before the reaction. Since all the mass is kept intact, the mass is "conserved." In the same way, the Law of Energy Conservation says that even though the energy changes form, the total amount has to remain intact.
Now, of course, the Law of Energy Conservation is important when we do energy conversions. When a black sheet of paper absorbs light, the radiant energy is converted to thermal energy, and the Law of Energy Conservation says that the total amount of energy before has to equal the total amount of energy after. However, the law of Energy Conservation applies to things other than energy conversions. For example, when sunlight is split into its many colors by drops of water in the air (forming a rainbow), the total amount of energy in the white light must be the same as the total amount of energy in all the colors of the rainbow. In this case, energy isn't converted. It is radiant energy when it is white, and it is still radiant energy when it is in the rainbow. Even though energy isn't converted, it is still conserved.
That's the long answer to your question. The shorter answer is that The Law of Energy Conservation applies to situations other than just energy conversion.
- Will an LED light work for the experiment in Lesson 6?
Unfortunately, an LED light will not work. The only kind of light that works in this experiment is a fluorescent light.
- Lesson 9
- Microwave Ovens and Radio Waves
Microwave ovens use microwave radiation, which is light waves that have wavelengths between 0.001 meters and 1 meter. Most microwave ovens today use microwaves with wavelengths of around 0.12 meters.
Radio waves are also light waves. They comprise a broad range of light waves that have wavelengths between 0.001 meters and 100,000,000 meters. In other words, microwaves are a subset of radio waves.
So you can say that microwave ovens use radio waves - they just use the radio waves with the smallest wavelengths. However, radio waves are just another form of light. The light waves have longer wavelengths than the light waves we can see with our eyes, so your eyes can't detect the light. Nevertheless, they are just a form of light.
- Lesson 14
- Why Does Light Slow Down When It Travels Through a Substance?
Click here fore an excellent video that explains why light slows down when it travels through a substance like glass:
- Lesson 19
- Does the size of the soda can matter in the experiment?
Yes, it does. You need 12-ounce cans, because they have the right density.
- What size cans should be used in the experiment?
You need to use standard, 12-ounce cans. Smaller cans will be less dense than water, regardless of what is inside.
- Lesson 21
- Why does water form droplets instead of one big glob?
Water is more attracted to itself than many surfaces, which causes the water molecules to move more closely to one another than to those surfaces. You might think this would cause water to form big globs, because the water molecules would just keep getting closer and closer together. Remember, however, that gravity is working on everything as well. If the water molecules got as close as they wanted to get, they would end up forming a big sphere. However, gravity keeps that from happening, because it pulls down on the water molecules, "squishing" the sphere that would form. The result is a water droplet. This happens all over the surface, so there are lots of little water droplets instead of just one big glob of water.
- Lesson 23
- Do the same ions recombine when saltwater evaporates?
When salt dissolves in water, the sodium and chloride ions move about in the solution. As a result, they "loose track" of the ions that were right next to them. When the water evaporates, they recombine with the nearest ions, which probably aren't the ones they were originally next to. If you want to add this to the story, you could have the ions looking for each other as the water evaporates, wondering if they can find each other before it's "too late."
- Lesson 24
- Does the size of the bottle matter in the experiment?
Any size bottle will produce a fountain, but any bottle smaller than 2 liters will be much more likely to tip over, spraying soda everywhere!
- Lesson 26
- Why did my plastic bottle shrink when put in boiling water?
Plastic is made from long molecules that chemists call "polymers." In order to reduce the cost of making the bottle, it is common these days to use polymers that are not stable at high temperatures in bottles that aren't supposed to reach high temperatures. Since soda bottles are never supposed to reach the temperature of boiling water, many soda manufacturers use a cheaper polymer that breaks down at those temperatures. When the polymers break down, the structural integrity of the bottle is reduced, and this causes the bottle to collapse a bit. The only result is that the balloon in the experiment won't deflate as much, but it stil should have deflated noticeably.
- Lesson 32
- The size of pores in soil
In general, the size of the bits of rock that make up the soil determines the size of the pores in the soil. The larger the bits of rock, the larger the pores, and the smaller the bits of rock, the smaller the pores. Sand, for example, is made up of fairly large particles. As a result, it has very large pores and doesn't hold water very well. Clay, on the other hand, is made of very tiny particles, so it has very small pores and holds onto water really well. The ideal growing soil has a mixture of different sized particles so that the pores are small, but not too small.
- Lesson 33
- Can sedimentary rock change to magma and then igneous rock?
If sedimentary rock is subject to enough heat to become magma, it will first become metamorphic rock. So in the end, sedimentary rock cannot become magma directly. It must first pass through a stage where it is metamorphic rock.
- Lesson 36
- Molecules, atoms, ions, etc
Atoms are the basic building blocks of molecules. If you have two or more atoms linked together in some way, you have a molecule. Atoms are made of smaller particles called protons, neutrons, and electrons. So...molecules are bigger than atoms, which are bigger than protons, neutrons and electrons.
An ion is an atom that has an imbalance of protons or electrons and therefore has an electrical charge. Protons are positive, and electrons are negative. If an atom ends up with more electrons than protons, it is a negatively-charged ion. If it ends up with more protons than electrons, it is a positively-charged ion. Whereas ions have electrical charge, molecules do not.
So sugar is made of a bunch of atoms linked together in a molecule. There is no charge on a molecule of sugar. When it dissolves, the water simply pulls the molecules of sugar away from each other, spreading them out in the solution.
Salt, on the other hand, is made of two IONS. When it dissolves, it splits up into its positively-charged ion (sodium) and its negatively-charged ion (chloride). Because of its ions, salt can conduct electricity when dissolved in water. Sugar cannot, because it has no charge.
- Lesson 40
- Do roots have gateways shaped to fit only certain nutrients?
Roots, like all parts of a plant, are made of cells. Most of the absorption in roots come from the tiny "hairs" that grow off them. Those root hairs are very thin - only one cell or a few cells thick. Those cells are covered in a cell wall that has large holes in it. Under the cell wall is a plasma membrane that has channels in it. For something to get into the plant, it has to go through one of those channels. There are some channels that have specific shapes to allow in only specific molecules. However, many of the channels are less discriminating. They are typically the smaller channels, and they allow many different small molecules to get inside the plant. If a plant absorbs poison through its roots, for example, it usually comes through one of these smaller, less discriminating channels.
- Lesson 43
- Replacement for Iodine in Lesson 43?
Unfortunately, I don't know of anything you can use to replace the iodine in that experiment. Since that lesson is a challenge lesson, I think you should just skip it if you think your daughter will be allergic to the iodine.
- Lesson 48
- What happens to the ROY colors during the daytime?
That's a great question! When you look at the sun when it is higher in the sky, you see it is white. That's where the "ROY" colors are. They are adding to the other colors to make white light. However, the light isn't perfectly white, since some of the "BIV" colors have been removed because they are bouncing off the dust particles in the sky. The ROY colors don't bounce off the dust particles, so they travel straight to your eyes, along with the BIV parts. Because of the bouncing, though, there is slightly more ROY in the light coming straight from the sun, and all the light coming from the rest of the sky is BIV.
- How fast does the earth spin?
The earth spins so that it makes a complete rotation every 24 hours. The speed at which a person on the surface of the earth moves in order to make that rotation depends on where you are on the earth. If you are on the equator (the imaginary line that runs all the way around the center of the earth), you move at just over 1,000 miles per hour. In the U.S., you move at about 850 miles per hour.
Of course, you don't notice this, because the entire surface of the earth is spinning with you. Think about being in an airplane. It doesn't seem like you are moving at several hundred miles per hour when you are in an airplane, because the entire airplane and everyone in it is moving right along with you.
- Why does the earth revolve around the sun?
Think about tying a toy plane to a string and then twirling it around in the air above your head. The plane is revolving around you, because the string you are holding is keeping the plane from flying away. You are much heavier than the plane, so you can stay in one place while the the plane moves.
It's similar for the earth and sun. The sun holds the earth in orbit with its gravity, much like the string holds the plane in orbit around you. The sun is much, much heavier than the earth, so it stays in one place while the earth revolves around it.
- How far away is earth from the moon?
The distance between the moon and earth changes a bit. The closest the moon gets to the earth is 226,000 miles (363,000 kilometers). The farthest it is from the earth is 252,000 miles (406,000 km). The average distance between the two is about 239,000 miles (384,000 km).
- Why is the moon white?
The light we see coming from the moon is reflected light from the sun. The light comes from the sun, hits the moon, reflects off the moon, and hits our eyes. The surface of the moon reflects all colors of light. Remember that white light contains all colors of light, so since the moon reflects all colors, the light coming from it is white.
- Lesson 59
- Relationship between sun temperature and weather or climate?
The sun's temperature is definitely not responsible for the shift from summer to winter. Remember, when the Northern Hemisphere of the earth is experiencing summer, the Southern Hemisphere is experiencing winter, and the temperature of the sun is the same for each of them! The changing of the seasons is caused by the tilting of the earth. Because the earth is tilted, one hemisphere is pointed directly towards the sun, and the other is pointed away from the sun. The one pointed directly towards the sun is heated more efficiently, so it experiences summer. The other one is heated less efficiently, so it experiences winter.
The temperature of the sun doesn't affect weather on a day-to-day basis, because the changes in temperature due to weather patterns are much stronger than the changes in temperature due to the temperature fluctuations in the sun. However, a given seaon (winter or summer) might be a bit warmer when there are a lot of sunspots and a bit cooler when there are few sunspots. This is a small effect, however, and it isn't noticeable in any meaningful way.
- Lesson 61
- Color of plants in Venn Diagram
Most plants must have green in them, because green is the color of chlorophyll, which is the main pigment used for photosynthesis. Thus, while a plant can have other colors (especially a flowering plant), it must have green on it. That's why the diagram says, "Have a lot of green in them." It's not that they are exclusively green, but the green parts give them the ability to make food for themselves, so the gree parts are necessary.
When the diagram says that animals "Can be many different colors," it just means that animals don't need green in them.
- Lesson 63
- Osmosis or Diffusion in Lesson 63's experiment?
The dye in the experiment for Lesson 63 operates under diffusion, not osmosis. The dye is moving from the water (where there is a lot of dye) into the egg (where there is little dye). This happens because the dye can travel through the membrane, and it does so in order to spread out as evenly as possible. That's diffusion.
Water is being pulled in the egg because of osmosis. There are a lot of solutes inside the egg, but the membrane holds them in, so they can't travel into the water. As a result, they pull the water inside the egg. Thus, the solvent (water) gets pulled into the egg because of osmosis. However, the dye (a solute) moves into the egg because of diffusion.
- Lesson 65
- Can I use a round balloon for the experiment?
Yes. However, you need to make sure it goes in a consistent direction. To do that, cut the ends out of a long, narrow bottle (like a water bottle) and blow the balloon up inside the cut-up bottle. That way, the bottle will force the balloon to take on the shape of a tube.
- Lesson 66
- Having Trouble With The Experiment
I am sorry that you are having trouble with that experiment. The best thing to do is use a longer piece of straw. It sounds like you aren't losing enough volume when you squeeze the bottle. To get more volume variation, cut a 15-cm section of straw rather than an 10-cm section. You could also try a variation on the experiment, like this one:
- Lesson 80
- Does fat protect animals from heat?
Yes, fat does protect animals from the heat. It can prevent heat from coming into an animal's body, which is good if it's too hot. However, SOME animals want heat to come into their bodies. Reptiles, for example, love to live in hot places and soak up the warmth, because they can't keep themselves warm. So those animals that live in hot places only use fat to store excess energy. However, there are animals (especially mammals) that live in hot places that do use their fat to protect them from the heat.
- Science in the Ancient World
- Ages for Science in the Ancient World
You can use Science in the Ancient World for all students ages 6-12. It has three different levels of review at the end of each lesson, so you can determine how much each child should retain by choosing the level of review appropriate for each student. There are 90 lessons in the course. If you do science every other day, it will last the full year with 10 extra days so you can be a bit flexible with the schedule.
- Table of Contents for Science in the Ancient World
- List of Supplementary Reading for Science in the Ancient World
There are a few resources that would work really well as supplemental reading. An excellent web series can be found here:
It starts at AD 1000, so it relates to the later parts of the book, but it is very good.
There is a list of famous scientists here:
Clicking on the name will lead you to a biography.
There are also two books you might consider. The first is
Once again, it deals with scientists from the later parts of the course. For the more ancient natural philosophers, you could use
- Workbook for Science in the Ancient World
Currently, there is no workbook for Science in the Ancient World. There is, however, a lapbook for the course:
While we don't plan to produce one, it is quite possible an independent person will. If so, we will work to make it available for free or for a very low cost. Two such workbooks have been made for Science in the Beginning already.
(Click on "Downloads" and "Printable Notebook")
- Errata for the first and second printings
- Lesson 1
- In Experiment 1, what is the "middle" of the Tree?
When I say middle of the tree, I am talking about the middle of the tree's trunk. In other words, I am not talking about the middle of its height. I am talking about the middle of its width. Perhaps the better word is "center" of the tree.
- What if the shadow length is longer than my height?
Actually, depending on where you live and what time of day you did the experiment, it's possible that the shadow is longer. When the sun is low in the sky, the shadows increase in length, so if it's low enough, the shadows can be longer than the objects. However, the math still works. Just do the math as shown in the Helps&Hints book. Your factors will be less than one, but they should still give you the correct height for the tree or whatever you ended up measuring.
- Lesson 5
- Does the speed at which you speak affect pitch
The sounds that you make when you talk are produced by your vocal cords vibrating and making waves. That's what determines the number of clumps of air that hit your ear every second, and that's what determines the pitch. The speed at which you speak affects how quickly the waves change.
When you speak in a low tone, for example, you are using your vocal cords to produce waves that have low frequencies, and therefore not many crests hit your ears per second. Speaking quickly simply makes each sound wave you are making change quickly. It does not affect the pitch.
- Lesson 28
- Eratosthenes and the shape of the earth
Most people think that the ancients thought the earth was flat, but in fact, no one of any repute believed that, even in BC times. In fact, even uneducated people who lived by the sea knew the earth isn't flat, because they saw a departing ship's hull disappear before its mast, at least on a really clear day. As a result, they understood that the earth is curved. Here is a good discussion of this fact and why even Christopher Columbus's detractors didn't think the earth was flat.
- Lesson 39
- Why does your pulse rate change over the course of your life?
The answer to the "Oldest Student" review question says, "As you mature, the body's need for nutrients drops, so you don't pump as much blood." The phrase "As you mature" refers to children maturing into adults. The more you mature, the less nutrients you need, because your body isn't growing and developing as much. Thus, the answer is more over the course of a child's life, not the course of an entire life.
As an adult gets older, other things happen, including the blood vessels becoming more stiff. This reduces the pulse rate even more.
- Lesson 46
- Infinity and numbers
Numbers are, indeed, infinite. No matter how large a number you say, I can add one to it and make a larger number. Therefore, numbers are infinite. However, there is no way for a single number to be infinite, because I can always add one to it, making a larger number. So numbers go on without end (which means they are infinite), which is why there is no single number that can actually be infinite.
- Lesson 52
- How much gelatin mix should be used in the experiment?
Usually, this kind of gelatin comes in a box that has several packets in it. You need only one packet.
- Which recipe should I use from the gelatin container?
You should ignore any "extras" like sugar, fruit, etc. Those are to make the gelatin flavored in some way, and you want plain, unflavored gelatin. Just use the water and the gelatin mix. Nothing else.
- Lesson 75
- Can nonfat, dry milk be used in the experiment?
I have not tested the experiment with nonfat, dry milk, but it should work. The important component is the protein part, and once you mix it with water, the proteins in dry milk should be similar enough to get the job done.
- Lesson 88
- How do scientists know what's in a bacterium?
When scientists want to know what's inside an animal, they typically do a dissection, where they cut open a dead animal and see what's inside. However, that doesn't work for microscopic creatures. Instead, scientists use microscopes. Many details of microscopic creatures can be seen with really good microscopes that use light, but light microscopes aren't strong enough to allow us to see the details of what's inside a bacterium. For that, scientists use electron microscopes. They shoot beams of electrons at the bacteria, and the way the electrons bounce off the bacteria allow scientists to make an image of the bacteria. High-energy electrons can pass through the outer layers of the bacteria and bounce of the structures inside, so by varying the electron energy, scientists can see both the outside and the inside of the bacterium.
- Science in the Scientific Revolution
- Science in the Scientific Revolution in a 27-week co-op.
If you wanted to cover the book in 27 weeks, you would be skipping the challenge lessons and covering 3 lessons each week. I am not sure how you would do that in a co-op, since most co-ops concentrate on the experiments, which means doing three experiments each week. You could choose the two most interesting of the three lessons to do in co-op, and have the students do the other one at home.
Alternatively, you could just do two lessons per week, skipping the challenge lessons, and have the students finish the book at home when the co-op is over.
- Lesson 57
- Some more illustrations from Hooke's book
Here are some more illustrations from Micrographia:
- Lesson 69
- How do I make the notecard in Lesson 69?I have made a drawing of the card here:Folding the notecard gives the student a way to fix how far in front if his or her eyes the holes are. Notice in the picture that the young lady has one end of the notecard almost touching her nose, as discussed in step 7. That allows her to keep the holes, which are on the other end of the card, a specific distance from her eyes.
- Lesson 79
- Lesson 79
- Why isn't the older student answer to (a) Newton's First Law?
Newton's First Law does apply, since the coin was at rest and then started to move. However, what would have happened had the other coin been moving slowly when it hit the stack? The bottom coin wouldn't have slid out, right? So the force with which the bottom coin gets hit must produce enough acceleration to make the coin on the bottom of the stack move quickly enough to slide out. So...while both laws apply, the second law is more important, since it determines whether or not the coin slides out from under the stack.
- Science in the Age of Reason
- Lesson 15
- If you can't do the experiment
Here is a video of a man making paper from old jeans. It is similar to what you do in the experiment:
- Lesson 42
- In Lesson 42, the balloon inflated instead of being sucked in.
Rusting does release heat, so it's possible that the balloon inflated because the rusting heated up the interior, increasing the volume of the gas inside. However, I wouldn't expect that, because the heat releases pretty slowly, so I would expect that it wouldn't get significantly warmer. I suppose that the temperature of the room could change enough to inflate the baloon, but you would certainly have noticed that.
It's also possible that there was something contaminating the steel on the steel wool, and it reacted to produce a gas.
- Lesson 15
- Science in the Industrial Age
- Lesson 15
- Lesson 15 trouble with the experimentI am sorry you are having trouble with that one. You might try adding some paperclips on the hanging wire, as shown in this video:You can also try starting the motion with your hands and seeing if it will continue on its own.
- Lesson 15
- The Series
- Earth Science
- What Day is the Sabbath?
Exodus 31:16-17 says, "So the sons of Israel shall observe the sabbath, to celebrate the sabbath throughout their generations as a perpetual covenant.’ It is a sign between Me and the sons of Israel forever; for in six days the Lord made heaven and earth, but on the seventh day He ceased from labor, and was refreshed.” So the Sabbath was ordained to be on the seventh day of the week, which is Saturday.
Christians do not worship on the Sabbath. They worship on Sunday because that's when Christ rose from the dead. This is not a problem, as the New Testament says that keeping the Sabbath is not necessary for Christians. For example, Romans 14:5 says, "One person regards one day above another, another regards every day alike. Each person must be fully convinced in his own mind."
- RoundingThe only time you MUST round is when you are about to change the rules. In an equation like this, for example:2.1x3.45 - 1.1x5.81You must do the multiplication and round, because in multiplication you count significant figures. So you have:7.2 - 6.4Now because you are subtracting, you must look at precision. As a result, the final answer is0.8If you never change the rules during a calculation, you can round at the end if you like. It doesn't matter, however, since there is always error in the last significant figures. Thus, if someone who rounds at each step gets an answer of 4.51 and someone who rounds at the end gets 4.53, it is IMPOSSIBLE to know who is right, because that last significant figure has error in it.
- Math CoursesThere are different math curricula I recommend, depending on the situation:2. Students who are gifted in math should consider The Art of Problem Solving.3. If the student prefers video, I think Videotext interactive is best.