Question

1. the density of a copper pendant is 20 g/cm³ and volume is 5 cm³. what is the mass of the copper pendant?
2. a pencil has a mass of 3 grams and a volume of 1.5 cm³. what is the density of the pencil?
3. the mass of a nickel is 5 grams and the volume is 1 ml. what is the density of the nickel?
4. a wooden box has a length of 3 cm, a width of 2 cm, and a height of 4 cm. if the mass of the wooden box is 96 grams, what is the density of the wooden box?
5. water has a density of 1g/ml. what is the mass of the water in a 60,000 ml pool?
6. what is the density of a bowling ball that has a volume of 15.8 cm³ and a mass of 72 grams?
7. a cube has a side length of 4 cm. if the density of the cube is 12 g/cm³. what is the mass of the cube?
8. a brick has a length of 8 cm, a width of 3 cm and a height of 2.5 cm. if the mass of the brick is 360 grams, what is the density of the brick?

Explanation:

Step1: Recall density formula

The formula for density $
ho=\frac{m}{V}$, where $m$ is mass and $V$ is volume. We can also re - arrange it as $m =
ho V$ when we need to find mass.

Step2: Solve problem 8

Given $
ho = 20\ g/cm^{3}$ and $V=5\ cm^{3}$. Using $m=
ho V$, we have $m = 20\times5=100$ g.

Step3: Solve problem 9

Given $m = 3$ g and $V = 1.5\ cm^{3}$. Using $
ho=\frac{m}{V}$, we get $
ho=\frac{3}{1.5}=2$ g/cm³.

Step4: Solve problem 10

Given $m = 5$ g and $V = 1$ mL. Since 1 mL = 1 cm³, using $
ho=\frac{m}{V}$, we have $
ho=\frac{5}{1}=5$ g/mL.

Step5: Solve problem 11

First, find the volume of the wooden box. The volume of a rectangular - box $V=l\times w\times h$, where $l = 3$ cm, $w = 2$ cm and $h = 4$ cm. So $V=3\times2\times4 = 24$ cm³. Given $m = 96$ g. Using $
ho=\frac{m}{V}$, we get $
ho=\frac{96}{24}=4$ g/cm³.

Step6: Solve problem 12

Given $
ho = 1$ g/mL and $V = 60000$ mL. Using $m=
ho V$, we have $m=1\times60000 = 60000$ g.

Step7: Solve problem 13

Given $m = 72$ g and $V = 15.8$ cm³. Using $
ho=\frac{m}{V}$, we get $
ho=\frac{72}{15.8}\approx4.56$ g/cm³ (the answer 0.2194 in the picture is incorrect).

Step8: Solve problem 14

First, find the volume of the cube. The volume of a cube $V=a^{3}$, where $a = 4$ cm. So $V = 4^{3}=64$ cm³. Given $
ho = 12$ g/cm³. Using $m=
ho V$, we have $m=12\times64 = 768$ g (the answer 48 in the picture is incorrect).

Step9: Solve problem 15

First, find the volume of the brick. The volume of a rectangular - box $V=l\times w\times h$, where $l = 8$ cm, $w = 3$ cm and $h = 2.5$ cm. So $V=8\times3\times2.5=60$ cm³. Given $m = 360$ g. Using $
ho=\frac{m}{V}$, we get $
ho=\frac{360}{60}=6$ g/cm³.

Answer:

1. 100 g
2. 2 g/cm³
3. 5 g/mL
4. 4 g/cm³
5. 60000 g
6. Approximately 4.56 g/cm³
7. 768 g
8. 6 g/cm³