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

Answer:

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 10³ + 1.065 x 10⁴) / 3 (3 is exact)

Answer:

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

Hint: http://chemwiki.ucdavis.edu/Analytical_Chemistry/Quantifying_Nature/Significant_Digits

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

4. (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)
5. (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)
6. (9.02 x 10³ + 8.99 x 10³ + 1.065 x 10⁴) / 3 = (2.8660 x 10⁴ ) / 3 = 9.553333333 x 10³ = 9.55 x 10³ (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

Answer:

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

Solution:

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

Hint: Temperature

1.4 

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

Substance	Physical State	Density (g/cm^3)
Ethanol	Liquid	0.789
Iron	Solid	7.87
Oxygen	Gas	0.00133

1. Oxygen gas
2. Iron
3. Ethanol

Answer:

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

The volume of 50.0 g of O_{2 (g)} is much larger than the other volumes, indicating that gasses consist of mostly empty space.

Hint: http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Bulk_Properties/Density

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.

• Energy, Heat, and Temperature: An Introduction

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, 0^oF is about -18^oC, which can be solving using the formula ^oF = 9/5^oC + 32, solving for ^oC to get ^oC=(5/9)(^oF-32) , and then plugging in 0 for ^oF. 10^oF is about -12^oC, using the same method. Thus, the change in temperature in ^oC, is -12^oC-(-18^​oC) = 5.6^oC 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 -273^oC; for 10K, it results in -263^oC. Thus, the change in temperature in ^oC, is -263^oC-(-273^oC) = 10^oC for a 10 degree shift in the Kelvin scale.

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

Result: Comparing the 10 degree differences in the three units, 5.6^oC 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

Answer:

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.

• 1.3_A_Description_of_Matter
• http://chemwiki.ucdavis.edu/Analytical_Chemistry/Qualitative_Analysis/Chemical_Change_vs._Physical_Change

Solution: See answer.

1.7

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

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

Answer:

a) mixture

b) pure

c) mixture

d) mixture

e) pure

f) mixture

g) pure

h) pure

i) mixture

Hint: http://chemwiki.ucdavis.edu/Analytical_Chemistry/Qualitative_Analysis/Classification_of_Matter

1.8 

Are these chemical or physical changes?

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

Answer:

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

Hint: http://chemwiki.ucdavis.edu/Analytical_Chemistry/Qualitative_Analysis/Chemical_Change_vs._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.