Question

chemistry: density problems
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.
2. a 2.75 kg sample of a substance occupies a volume of 250.0 cm³. find its density in g/cm³.
3. 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.
4. find the volume that 35.2 g of carbon tetrachloride (ccl₄) will occupy if it has a density of 1.60 g/ml.

Explanation:

Step1: Recall density formula

The density formula is $d=\frac{m}{V}$, where $d$ is density, $m$ is mass and $V$ is volume. We can re - arrange it to solve for different variables. For example, $m = d\times V$ and $V=\frac{m}{d}$.

Step2: Solve problem 1

Given $d = 10.5\ g/cm^{3}$ and $V=965\ cm^{3}$. Using $m = d\times V$, we have $m=10.5\ g/cm^{3}\times965\ cm^{3}=10132.5\ g$.

Step3: Solve problem 2

First, convert mass from kg to g. Since $1\ kg = 1000\ g$, a $2.75\ kg$ sample has a mass $m = 2.75\times1000\ g=2750\ g$. Given $V = 250.0\ cm^{3}$. Using $d=\frac{m}{V}$, we get $d=\frac{2750\ g}{250.0\ cm^{3}}=11\ g/cm^{3}$.

Step4: Solve problem 3

Given $d = 0.00134\ g/mL$ and $m = 250.0\ g$. Using $V=\frac{m}{d}$, we have $V=\frac{250.0\ g}{0.00134\ g/mL}\approx186567.16\ mL$.

Step5: Solve problem 4

Given $m = 35.2\ g$ and $d = 1.60\ g/mL$. Using $V=\frac{m}{d}$, we get $V=\frac{35.2\ g}{1.60\ g/mL}=22\ mL$.

Answer:

1. Mass of Ag: $10132.5\ g$
2. Density of the substance: $11\ g/cm^{3}$
3. Volume of $O_2$: $186567.16\ mL$
4. Volume of $CCl_4$: $22\ mL$