Question

chemistry review & practice worksheet
density & significant figures - solve the following problems, reporting answers with the correct sig. figs.

1. a sample has a mass of 56.37 g and displaces 7.4 ml of water. what is its density?
2. a block has a mass of 125.4 g and a measured volume of 15.00 cm³. calculate the density.
3. a metal cylinder weighs 342.5 g and has a volume of 31.2 ml. what is its density?

si unit conversion - convert the following masses to kilograms (kg). (1kg = 1000 g; 1000 mg = 1 g)

1. 425 mg
2. 0.35 g
3. 1.25×10³ mg

dimensional analysis - set up and solve using dimensional analysis.

1. copper wire sells for $0.82 per meter. how much does 5.25×10³ cm cost?
2. a metal rod is 250 cm long and costs $1.50 per meter. what is the total cost?
3. string costs $2.40 per meter. what is the cost of 75 cm of string?

sds (safety data sheet) - answer the following based on the sds sample diagram below.

1. which section of the sds would tell you how to dispose of a chemical?
2. which section contains first aid measures?
3. where would you find information on chemical hazards?

Explanation:

1.

Step1: Recall density formula

Density ($
ho$) is given by $
ho=\frac{m}{V}$, where $m$ is mass and $V$ is volume. For the first problem, $m = 56.37$ g and $V=7.4$ mL.
$
ho=\frac{56.37\text{ g}}{7.4\text{ mL}}\approx7.6\text{ g/mL}$ (rounded to 2 significant - figures as 7.4 has 2 significant figures)

Step2: For the second problem

Using $
ho=\frac{m}{V}$, with $m = 125.4$ g and $V = 15.00$ cm³.
$
ho=\frac{125.4\text{ g}}{15.00\text{ cm}^3}=8.36\text{ g/cm}^3$ (4 significant - figures as all values in the calculation have 4 significant figures)

Step3: For the third problem

Using $
ho=\frac{m}{V}$, with $m = 342.5$ g and $V = 31.2$ mL.
$
ho=\frac{342.5\text{ g}}{31.2\text{ mL}}\approx10.98\text{ g/mL}\approx11.0\text{ g/mL}$ (rounded to 3 significant - figures as 31.2 has 3 significant figures)

Step4: For unit conversion (425 mg to kg)

Use the conversion factors: $1\text{ kg}=1000\text{ g}$ and $1\text{ g}=1000\text{ mg}$. So, $425\text{ mg}=425\times\frac{1\text{ g}}{1000}\times\frac{1\text{ kg}}{1000}=4.25\times10^{-4}\text{ kg}$

Step5: For 0.35 g to kg

$0.35\text{ g}=0.35\times\frac{1\text{ kg}}{1000}=3.5\times10^{-4}\text{ kg}$

Step6: For $1.25\times10^{3}$ mg to kg

$1.25\times10^{3}\text{ mg}=1.25\times10^{3}\times\frac{1\text{ g}}{1000}\times\frac{1\text{ kg}}{1000}=1.25\times10^{-3}\text{ kg}$

Step7: For dimensional - analysis (copper wire problem)

First, convert $5.25\times10^{3}$ cm to m. Since $1\text{ m}=100\text{ cm}$, $5.25\times10^{3}\text{ cm}=5.25\times10^{3}\times\frac{1\text{ m}}{100}=52.5\text{ m}$.
The cost is $0.82$ dollars per meter, so the cost $C = 0.82\frac{\text{ dollars}}{\text{ m}}\times52.5\text{ m}=43.05$ dollars

Step8: For the metal - rod problem

Convert 250 cm to m: $250\text{ cm}=250\times\frac{1\text{ m}}{100}=2.5\text{ m}$
The cost per meter is $1.50$ dollars, so the total cost $C = 1.50\frac{\text{ dollars}}{\text{ m}}\times2.5\text{ m}=3.75$ dollars

Step9: For the string problem

Convert 75 cm to m: $75\text{ cm}=75\times\frac{1\text{ m}}{100}=0.75\text{ m}$
The cost per meter is $2.40$ dollars, so the cost $C = 2.40\frac{\text{ dollars}}{\text{ m}}\times0.75\text{ m}=1.80$ dollars

Step10: For SDS questions

The section of the SDS that tells you how to dispose of a chemical is Section 13 (not shown in the provided diagram but generally).

Step11: The section that contains first - aid measures is Section 4.

Step12: Information on chemical hazards is found in Section 3.

Answer:

1. $7.6$ g/mL
2. $8.36$ g/cm³
3. $11.0$ g/mL
4. $4.25\times10^{-4}$ kg
5. $3.5\times10^{-4}$ kg
6. $1.25\times10^{-3}$ kg
7. $\$43.05$
8. $\$3.75$
9. $\$1.80$
10. Section 13
11. Section 4
12. Section 3