# Solutions for Homework 1 Practice Problems.

The solutions manual for the exercises are primarily given in four parts: Question, Answer, Hint, and Solution. Not all problems have all four of these components.

#### 1.1

Calculate the result of each of the following and record answer with the proper number of significant figures.

1. 13.4 + 9.8 + 55.99
2. 340.5 * 419.23
3. 19.45 / 14.23
4. 7 - 2.8

1. 79.2
2. 1.427 x 105
3. 1.367
4. 4

Hint: Significant Digits

Solution: When adding and subtracting numbers, the answer may not contain more decimal places than the least accurate number. When multiplying or dividing numbers, record the answer to the number of significant figures as the least accurate number.

#### 1.2

Calculate and express the result in the correct number of significant figures.

1. (1.602 x 10-19 ) x (2.15 x 103)
2. (3.1667 x 1023 ) x (9.018 x 10-9) / (7.52 x 104)
3. 1.714 x 10-1 + 1.96 x 10-3 + 1.255 x 10-2
4. (7.00743 - 7.00521) / (5.9831 x 1019 )
5. (4.832 x 10-5 - 4.760 x 10-5) / (4.832 x 10-5) x 100 (100 is exact)
6. (9.02 x 103 + 8.99 x 10+ 1.065 x 104) / 3 (3 is exact)

1. 3.44 x 10-16
2. 3.80 x 1010
3. 1.859 x 10-1
4. 3.71 x 10-23
5. 1.5
6. 9.55 x 103

Solution:

1. 3.4443 x 10-16 = 3.44 x 10-16 (3 sig. figs)
2. 3.797513378 x 1010 = 3.80 x 1010 (3 sig. figs)
3. 1.714 x 10-1 = 0.1714 (4 dec. places)

1.96 x 10-3 = 0.00196 (5 dec. places)

1.255 x 10-2 = 0.01255 (5 dec. places)

the answer should have 4 decimal places (the least amount of decimal places)

1.714 x 10-1 + 1.96 x 10-3 + 1.255 x 10-2 = 0.18591 = 0.1859 = 1.859 x 10-1

1. (7.00743 - 7.00521) / (5.9831 x 1019 ) = 0.00222 / (5.9831 x 1019) = 3.7104511 x 10-23 = 3.71 x 10-23 (3 sig. figs)
2. (4.832 x 10-5 - 4.760 x 10-5) / (4.832 x 10-5) x 100 = (7.2 x 10-7) / (4.832 x 10-5) x 100 = 1.4906623 = 1.5 (2 sig.figs)
3. (9.02 x 103 + 8.99 x 10+ 1.065 x 104) / 3 = (2.8660 x 104 ) / 3 = 9.553333333 x 103 = 9.55 x 10(3 sig. figs; do not round up until the end of calculations)

#### 1.3

Convert the following temperatures to Kelvin and Fahrenheit degrees.

1. room temperature: 20.0 oC
2. the boiling point of water: 100.oC
3. the temperature of human: 37.0 oC

1. 293 K; 68.0 oF
2. 373 K; 212oF
3. 310.K; 98.6oF

Solution:

1. 20.0+273=293K; 20.0*9/5+32=68.0oF
2. 100+273=373K; 100*9/5+32=212oF
3. 37.0+273=310K; 37.0*9/5+32=98.6oF

Hint: Temperature

#### 1.4

Using the following table, calculate the volume of 50.0 g of each of the following substances at 20.0o C. What does the volume of oxygen gas tell you about the properties of gasses?

 Substance Physical State Density (g/cm3) Ethanol Liquid 0.789 Iron Solid 7.87 Oxygen Gas 0.00133
1. Oxygen gas
2. Iron
3. Ethanol

1. 3.76x104 cm3
2. 6.35 cm3
3. 63.4 cm3

The volume of 50.0 g of O2 (g) is much larger than the other volumes, indicating that gasses consist of mostly empty space.

Solution: Converting from mass to volume requires that you use an inverse density value as well as dimensional analysis.

Example $$\PageIndex{1}$$

Detailed calculations for each of the three substances

Solution

$(50.0\; g\; Ethanol) \times \dfrac{1\;cm^3}{0.789\;g} = 63.4\; cm^3.$

$(50.0\; g\; Iron) \times \dfrac{1\;cm^3}{7.87\;g} = 6.35\; cm^3.$

$(50.0\; g\; Oxygen) \times \dfrac{1\;cm^3}{0.00133\;g} = 3.76\; \times 10^4 cm^3.$

Gasses tend to have much larger volumes than solids and liquids because they fill the container that they are placed in. Gasses exhibit weak intermolecular forces, causing them to expand readily and occupy a large volume.

#### 1.5

A student gets three thermometers from the lab. He notices that these thermometers use different units for measurements. They are oF, oC, and K. If he puts these thermometers into three water baths and observe a 10 degree change for each thermometer, which water bath has the smallest change in temperature ?

Answer: The water bath using oF thermometer

Hint: Consider the difference between two known temperature values whose difference is 10 degrees.

Solution: For convenience, choose to convert zero and 10 degrees for each unit to degrees C, then calculate the difference (final T - initial T).

Fahrenheit: For example, 0oF is about -18oC, which can be solving using the formula oF = 9/5oC + 32, solving for oC to get oC=(5/9)(oF-32) , and then plugging in 0 for oF. 10oF is about -12oC, using the same method. Thus, the change in temperature in oC, is -12oC-(-18​oC) = 5.6oC for a 10 degree shift in the Fahrenheit scale.

Kelvin: To convert Kelvin to oC, the formula K = 273.15 + oC is used, and for 0K, it results in -273oC; for 10K, it results in -263oC. Thus, the change in temperature in oC, is -263oC-(-273oC) = 10oC for a 10 degree shift in the Kelvin scale.

Celsius: Finally, the difference in temperature between 0oC and 10oC is 10oC in the Celsius scale.

Result: Comparing the 10 degree differences in the three units, 5.6oC is the smallest, which originated from a 10 degree difference in the Fahrenheit scale.

#### 1.6

What is the main difference between each of the following pairs?

1. An elemental substance (element) and a compound
2. Physical and chemical changes
3. Homogeneous and heterogeneous mixtures

1. An element (or elemental substance) is composed entirely of a single element. A compound has atoms bonded that are different elements.
2. A physical change is a change in the state of the matter without changing the substance itself, like ice changing into water or a foil pulled into a wire or a metal ball hammered into a sheet. A chemical change is a change where the original matter changes into different matter. An example of these is making and baking cookies. Mixing flour and sugar is a physical change because the individual grains are still made of flour and sugar. Baking cookies is a chemical change because the heat alters the bonds between atoms to create something new .
3. A homogeneous mixture is one in which, given any part of the mixture, it has the same concentration (amount of matter per volume) as any other part of the whole, like a brewed cup of coffee or a round of cheese. This is not the case for a heterogeneous mixture, which can be represented by a bowl of mixed jelly beans or gravel.

Hint:

1. One contains same elements, the other contains different elements.
2. Melting compared to burning.
3. Potting soil compared to tea.

#### 1.7

Define whether each of the following is a pure substance or a mixture.

2. copper
3. bronze
4. silk
5. ice
6. sand
7. magnesium
8. solid potassium iodide (KI)
9. honey

a) mixture

b) pure

c) mixture

d) mixture

e) pure

f) mixture

g) pure

h) pure

i) mixture

#### 1.8

Are these chemical or physical changes?

1. A nail rusting
2. Toast burning
3. Water boiling
4. Ice cream melting

1. Chemical change
2. Chemical change
3. Physical change
4. Physical change

Solution:

1. A nail rusting is a chemical change because a new substance,rust, has formed.
2. Burnt toast is a chemical change because a new substance, burned toast, was created.
3. Water boiling is a physical change because a new substance has not formed. The molecules of water are still composed of two hydrogen atoms and one oxygen atoms. Simply removing the heat source will cause the water to condense.
4. Melting of ice cream is a physical change because no new substance formed when the ice cream melted. Instead, it changed state from a solid to a liquid.

#### 1.9

The actual amount of magnesium in a sample is 21.14%. A student analyzed the percent composition of magnesium in the sample and obtained the results: 20.36%, 20.37%, 20.34%, 20.37%. Explain the accuracy and precision of these results.

Answer: The data the student collected about the amount of magnesium in the sample is precise, but not accurate.

Solution: The data that the student collected is precise because the four percentages are close to one another. The data is not accurate because the numbers collected are significantly lower than the actual percentage of magnesium in the sample.