Instructions are given for both the 11th Edition of the textbook and the 12th Edition.

Temperature:
11th Edition, Sec.2.11, p30-32
12th Edition, Sec. 2.8, p33-35

Read what the book says about temperature and the three scales used for temperature. Learn how to convert between the three scales. Notice that the book gives you names of the three scales, (Fahrenheit, Celsius, & Kelvin) but does not explain how these names came about or the how the temperature scales were created. I feel that if you don't know this background information, there will always be a hole in your learning. I will cover temperature in a separate tutorial, but if I didn't, you should do your own background check.

Density:

11th Edition, page 35:
12th Edition, page 37
Density is useful for many aspects of chemistry. It helps you identify unknown substances. It tells you whether a poisonous gas will rise or sink which helps you pick the right way to escape. It helps you purify mixtures and do many more things.

Density can be calculated with simple measurements-volume and mass.

Read Section on Density

The book shows both the algebraic approach and the Dimensional Analysis approach. Trust me, the Dimensional Analysis will be easier in the long run, so focus on that approach. They use parentheses to indicate multiplication, but as I've shown before, a faster way to write it is with vertical lines.

Book Example 2.23:
The density of gold is 19.3g/mL. What is the mass of 25.0 mL of gold?

Here is my suggested style. Write the 25mL as the starting amount and the grams as the end dimensions. The density is given as 19.3 g/mL. These dimensions are good because the grams stays and the mL cancels out the mL in 25.0 mL.

If this problem said, "You have 483 grams of gold. What is the volume?", then the 483 grams would be written first and the density would be written upside down (1 mL/19.3grams) so that the grams cancels out and the mL stays. The answer would be 25.0 mL.

Gold Density Problem 1:
11th Edition, Question 92, page 42
12th Edition: Question 81, page 45

The first part of this problem is to solve for the volume of the bar. They give you the dimensions of the bar. Realize the volume of a rectangular solid (a bar) is V=L x W x H (length x width x height). Use the density of gold mentioned above and dimensional analysis to convert the volume of the bar into grams. Compare the calculated grams to the 3300 grams that it actually weighed. (a) What should it have weighed if it were pure gold? (b) Was it pure gold and why?

Gold Density Problem 2:
11th Edition, Question 93, page 42
12th Edition: Question 83, page 45
Use dimensional analysis to start with 93.3 kg and end up with cubic centimeters. Realize that 1 milliliter (mL) is the same size as a cubic centimeter. Since the density of gold is 19.3 g/mL (grams per milliliter), you can keep the milliliter dimension. You don't have to convert it to cubic centimeters (they're the same size). Don't forget this is 93.3 kilograms not grams, so you will need to get rid of the "kilo" which you know is equivalent to 1,000.

The problem says to use the current price of $559 per oz (troy oz); however, look up the current price of gold on the Internet and then use that price. They give a conversion of 14.58 troy oz = 1 pound. You will also need to know that 2.2 lbs = 1 kg = 1,000 g.
(a) give me the volume of the nugget in milliliters (mL) or cubic centimeters (cm³ or cc)
(b) What is the current value of that nugget? Also, tell me what price per oz you are using.

Problem #3: In the textbook they show this graphic to explain density. What's nice they drew the boxes to scale. In other words, the sulfur box measured about 1.7 centimeters (cm) on each side. That gives the volume of 1.7cm x 1.7cm x 1.7cm = 4.9cm³, which is close to the volume they show for sulfur.
Now look at the large cube for water. It says it holds 10.00 mL of water. Compare that to the 4.83 mL cube of sulfur. Isn't 10mL about twice as big as 4.83mL? Does the blue cube look only twice as big? Not really. The blue cube is about 3.3cm on each side. #3a: In that case what is its volume? To draw a cube that is 10mL, the three sides have to multiply out to be 10mL. (Remember one cubic centimeter is the same as one mL). #3b. What should have been the size of the sides if they wanted to draw a 10mL cube?

Oil Slick Problem example:
11th Edition: Question 52, Page 42
12th Edition: Question 40, Page 44
How much area in m² (square meters) will 200cm³ of oil cover if it forms a layer 0.5nm thick?

This is a bit tricky problem. The slick is considered a very flat box with a volume equal to W x H x L. Volume is 200cm³(about a cup full). Height is 0.5nm (about 1/200 of human hair) and the width and length are the same size, let's call them M.
Algebra would set it up as
200cm3=0.5nm x M².
We would solve for "M" by dividing both sides by 0.5nm and taking the square root, but we still have problems with the different dimensions. So that doesn't work too well.

With dimensional analysis we put our starting amount on the left and what we want to find on the right. Realize that cm³ is cm x cm x cm. cm³ has "m" three times (cubic meter); however, our answer is square meters (m²). So that means we have to get rid of one of the meters (m). That's why the 0.5nm (nanometers) is on the bottom. The meter in "nm" will cancel one of the meters in the cubic centimeters. Then we end up with the meters squared (m²) that we want. The starting amount also has "centi" (c) three times, so we cancel them by multiplying by 0.01 over "c" three times. Remember c=0.01. Then "n" in 0.5nm needs to be canceled so we put "n" on top and its equivalent on the bottom (10^-9, which is one billionth).

If dimensions are in the right place, everything that needs to cancel will cancel and you are left with the correct final units (m²). Now we just multiply all of the numerators together and divide by each of the denominators. So the 200cm³, which is less than a cup of oil, can form an oil slick that covers 400,000 m², which is the area of about 100 football fields.

Problem #4: Here is a picture of the Exxon Valdez. It spilled 11 million gallons of oil. How many square miles of ocean could it cover if the oil also coated the surface also with 0.5nm thickness of oil?

Setup the problem the same as before except there will need to be more conversions. If you can do this problem, count yourself very smart.

Tip: find the gallon to cubic meter conversion and the square meters to square mile conversion.