Question

for each problem below, write the equation and show your work. always use units and box in your final answer.

1. the density of silver (ag) is 10.5 g/cm³. find the mass of ag that occupies 965 cm³ of space.

m = d·v m = 10.5·965 = 10,132.5
10,132.5 g

1. a 2.75 kg sample of a substance occupies a volume of 250.0 cm³. find its density in g/cm³.

d = m/v d = 2.75/250 = 0.011
11 g/cm³

1. under certain conditions, oxygen gas (o₂) has a density of 0.00134 g/ml. find the volume occupied by 250.0 g of o₂ under the same conditions.

v = m/d v = 250/0.00134 = 186,567.2
186,567.2 ml

1. find the volume that 35.2 g of carbon tetrachloride (ccl₄) will occupy if it has a density of 1.60 g/ml.

v = m/d v = 35.2/1.60 = 22
22 ml

1. the density of ethanol is 0.789 g/ml at 20°c. find the mass of a sample of ethanol that has a volume of 150.0 ml at this temperature.

m = d·v m = 0.789·150.0 = 118.4
118.4 g

1. 30.0 g of each of the following acids are needed. find the volume of each that must be measured out in a graduated cylinder.

a. hydrochloric acid (hcl), density = 1.164 g/ml
v = m/d v = 30.0/1.164 = 25.8
25.8 ml
b. sulfuric acid (h₂so₄), density = 1.834 g/ml
v = m/d v = 30.0/1.834 = 16.4
16.4 ml
c. nitric acid (hno₃), density = 1.251 g/ml
v = m/d v = 30.0/1.251 = 24
24 ml

Explanation:

Step1: Recall density - mass - volume formula

The formula is $M = D\times V$ (where $M$ is mass, $D$ is density, and $V$ is volume) or $V=\frac{M}{D}$ when solving for volume.

Step2: Solve problem 1

Given $D = 10.5\ g/cm^{3}$ and $V = 965\ cm^{3}$, using $M = D\times V$, we have $M=10.5\times965 = 10132.5\ g$.

Step3: Solve problem 2

First convert mass to grams: $2.75\ kg=2750\ g$. Given $M = 2750\ g$ and $V = 250.0\ cm^{3}$, using $D=\frac{M}{V}$, we get $D=\frac{2750}{250}=11\ g/cm^{3}$.

Step4: Solve problem 3

Given $M = 250.0\ g$ and $D = 0.00134\ g/mL$, using $V=\frac{M}{D}$, we have $V=\frac{250}{0.00134}\approx186567.2\ mL$.

Step5: Solve problem 4

Given $M = 35.2\ g$ and $D = 1.60\ g/mL$, using $V=\frac{M}{D}$, we get $V=\frac{35.2}{1.60}=22\ mL$.

Step6: Solve problem 5

Given $D = 0.789\ g/mL$ and $V = 150.0\ mL$, using $M = D\times V$, we have $M=0.789\times150.0 = 118.35\ g\approx118.4\ g$.

Step7: Solve problem 6A

Given $M = 30.0\ g$ and $D = 1.164\ g/mL$, using $V=\frac{M}{D}$, we get $V=\frac{30.0}{1.164}\approx25.8\ mL$.

Step8: Solve problem 6B

Given $M = 30.0\ g$ and $D = 1.834\ g/mL$, using $V=\frac{M}{D}$, we have $V=\frac{30.0}{1.834}\approx16.4\ mL$.

Step9: Solve problem 6C

Given $M = 30.0\ g$ and $D = 1.251\ g/mL$, using $V=\frac{M}{D}$, we get $V=\frac{30.0}{1.251}\approx24\ mL$.

Answer:

1. $10132.5\ g$
2. $11\ g/cm^{3}$
3. $186567.2\ mL$
4. $22\ mL$
5. $118.4\ g$

6A. $25.8\ mL$
6B. $16.4\ mL$
6C. $24\ mL$