Chapter 2 - FAQ

Here are some questions that some very good students have asked me. I hope my answers are some help to all of you. To the students who asked: Thank you. ~~~~DWL

1. How would you figure out the purity of lets say "gold' if you only had the > mass and the measurement in cm3? Also, I have seem Milliliter written ml and mL. Which way is correct?

 

Compare the density you calculate to the density of pure gold.

 

ml is what I usually use. mL is also used, and but I'm not sure why one would capitalize liter. Capitals are used for things like Joule, which was named after somebody. I suppose it depends on whom you are asking. Since you're asking me, I'll take either one. Just be consistent. ml and mL are fine, Ml and ML mean something completely different.

 

2. This weekend I plan on spending quality time studying for next weeks test. However my weakness are in the word problems. From listening to your lecture on Sept. 1, the word problems do not seem that complicated, but when I try them on my own it becomes confusing. The algebra is the simplest part, I just have a hard time setting up the equation. What suggestion do you have?

 

I think that you share the difficulty with setting up word problems with many others in the class. There are a few suggestions in the book and on my study guides about how to set them up. Also, if you took detailed notes, you might be helped by the way I set up the "Plan" in many of the problems. I do realize that even all that might not be enough, so here are a few more ideas.

 

When you are first reading the problem, write down all the numbers WITH THEIR UNITS.

 

If two numbers or units are related (for example, with the word PER liter or, another example, FOR EVERY 2 feet), then see if you can write an equality that can be used as a conversion factor. That road marker problem in the book is a good one for that. They said that every 1.0 quarts of paint will cover 43 square feet of markers. That's an equality that relates one unit to another. So you can use it as a conversion factor.

 

Once you've written all that down, look for what the problem is asking for. Figure out what kind of quantity it will be and what the units for the quantity will be. This is important because you want to know the units so you can have a target in your plan.

 

The Plan: This is a sequence of units that have to be converted to reach the target. Sometimes you'll have to look up conversion factors to get from the units you have to your target units. The only way to make this easy is to practice, so you can recognize the types of conversions I'll be asking about, and so you'll be open-minded enough to recognize the weird ones.

 

Conversions: Once you have the plan of unit conversions, then set up the problem in dimenional analysis format and do the math.

 

Remember, sometimes you have to convert two different kinds of units, like if you're converting a density in g/ml to a density in pounds per cubic inch. Then you have two plans, really. One to convert the grams to pounds and the other to convert ml to cubic inches. Once you've got the plan(s) do the math all at once so you avoid rounding errors.

 

Good luck. And please email me again if you want to talk about a specific problem, from our book, or another book, or the practice problems on the web. Send your emails to both of my addresses at the same time, so to make sure I will see them at home and here at work.

 

3. I had a couple of questions re: sig figs and rounding on problems #6 & #7 in you 1st set of practice problems (online). On Problem #6, I'm a bit confused about which factor determines the number of sig figs in the answer. Is it the 28 miles? I'm still in the process of grasping defined & exact vs estimated numbers.

In problem #6, it is the 28 miles per gallon that determines the sig figs. The 3.785 liters per gallon is 4 s.f., and the others are all exact conversions. (For some reason, 2.54 cm per inch is exact, but this is the only cross-system conversion that I know to be an exact one.)

 

On Problem #7, my final answer came to 585,000 = 5.9 x 10 five. The answer posted was 5.8 x 10 five, which led to some rounding confusion.

 

For the first conversion (miles to meters) I got 3.86242560 x 10^8. I keep that in my calculator to convert the meters to seconds, using the speed, and get 5.7936384 x 10^5, which rounds off to 5.8 x 10^5.

 

4. I'm still muddling through your practice probs (part 2), getting more wrong than right, but I'm working on them. For now, I have some questions relating to the Ch. 2 study guide questions. First, you asked how temperature affects density, but our text does not exlain this in depth, using only examples of different densities at different temps. The text briefly discusses (in a separate section) heat energy and its association with the motion of molecules/matter, but it doesn't really state in depth what heat's relationship is to density. My question: Is the grouping or slow down of molecules at low temps and the increased activity of molecules at higher temps explain changes in density? This brings me to my next question.

The effect of temperature on density of a substance depends on the substance itself. Different substances behave differently. Most gases, for example, are less dense at higher temperatures. Solids and liquids sometimes change in odd ways. For example, most forms of ice is less dense than liquid water. The main thing I wanted you to think about is that density is not constant and will vary with temperature, especially if you have a phase change. Density has to do with the packing or grouping of molecules or atoms, not the speed, although the speed is sometimes related (e.g., gases at higher temperatures are less dense and the particles are moving faster).

 

Secondly, regarding "Absolute": Our text states that the Kelvin temp scale is the abolute temp scale, 0 K (absolute zero) being the lowest attainable temp. My dictionary's definition of absolute zero = "the hypothetical point at which all molecules cease activity." However, -73.16 C & -459.69 F, along with 0 K, are used as examples of absolute zero. Thoroughly confused, my question is: What exactly is "absolute" in terms of chemistry/temp? Is the Kelvin scale the only absolute scale for temp measurement?

 

As for absolute zero, it's the same temperature, no matter what units you have it in. Absolute zero is the point at which motion ceases, which has never actually been reached. (They've come quite close, but there's always some vibration or rotation, etc.) Anyway, an "absolute scale" is one which has absolute zero as zero. There is the Kelvin scale and there is something that is similar to the Fahrenheit scale that has absolute zero as zero. I forgot the name of it: something that starts with an "R", I think. It's not used much, which is probably why I can't remember it.

 

5. Mr. Lingner, I am having a little bit of difficulty with #1 and #2 of the practice problems which you gave us. With #1 I ended up with the answer of 19.04 degrees Fahrenheit using the formula degrees F=1.8(degreesC)+ 32. degreesF=1.8(-7.2) + 32 = 19.04 degrees F. My question is how would you even attempt the second part of the problem involving Bromine. The only information I could come up with is that Bromine has an Atomic mass of 79.904. I figured that whatever state Bromine is normally in, it would stay in that state right now since it its about 80degrees and the melting point is 19.04 degrees. ?????

 

Go back to the practice problems page on the web site and hit reload or refresh on your internet browser (e.g., Netscape). The answer links should appear there. You can click where it says "Answer" and see my hand-written answers to each problem. I forgot to add the answer to the second part, so try this. Compare the melting point for bromine to the temperature in the room where you are right now. Estimate if you have to. It's probably about 70 F.

Also #2. My plan is this: K to degrees C to degrees F using formula: 301.9K=degrees C + 273.15 I am having trouble isolating Kelvins. Could you possibly steer me in the right direction?????

 

Problem 2 is about the same, except that you have to do two conversions. Check my answers. For part 2 of #2, assume the temperature in Death Valley is about 120 F.