questions 8-12: calculations. use the information below to answer the following questions. show all work for full credit. all answers should include significant figures and the appropriate units. note: 1 cm³ = 1 cc = 1 ml
density ($\frac{g}{ml}$ or $\frac{g}{cm^{3}}$) = $\frac{mass (g)}{volume (ml or cm^{3})}$
8. calculate the density of a substance with a mass of 24.756 grams and occupies a volume of 10.5 millimeters. (3)
9. a liquid has a density of 1.25 g/ml and occupies a volume of 35.6 ml. calculate the mass of the liquid.(3)
10. an object has a mass of 85.2 grams. when it is submerged in water, the initial water level rises from 50.0 ml to 68.5 ml. what is the density of the object? (3)
11. what volume of copper metal will weigh exactly 250.0 g? the density of copper is 8.90 g/cm³.(3)
12. an object with a mass of 45.6 g and a volume of 12.0 cm³ will float in which of the following: (3)
| substance | density at 20 °c (g/cm³) |
| ---- | ---- |
| wood | 0.70 |
| corn oil | 0.92 |
| water | 1.00 |
| rubber | 1.34 |
| corn syrup | 1.38 |
| copper | 8.80 |

Question

Accepted Answer

### Question 8
# Explanation:
## Step1: Identify formula
The density formula is $	ext{Density}=\frac{	ext{mass}}{	ext{volume}}$.
## Step2: Substitute values
We have mass $m = 24.756\ g$ and volume $V=10.5\ mL$. So, $	ext{Density}=\frac{24.756\ g}{10.5\ mL}$.
## Step3: Calculate density
$	ext{Density}=\frac{24.756}{10.5}\ g/mL\approx2.36\ g/mL$ (3 significant - figures)

# Answer:
$2.36\ g/mL$

### Question 9
# Explanation:
## Step1: Rearrange density formula
From $	ext{Density}=\frac{	ext{mass}}{	ext{volume}}$, we can get $	ext{mass}=	ext{Density}	imes	ext{volume}$.
## Step2: Substitute values
Given $	ext{Density}=1.25\ g/mL$ and volume $V = 35.6\ mL$. Then $m=1.25\ g/mL	imes35.6\ mL$.
## Step3: Calculate mass
$m = 1.25	imes35.6\ g=44.5\ g$

# Answer:
$44.5\ g$

### Question 10
# Explanation:
## Step1: Calculate volume of the object
The volume of the object is equal to the volume of water displaced. $V=68.5\ mL - 50.0\ mL=18.5\ mL$.
## Step2: Use density formula
We know mass $m = 85.2\ g$ and volume $V = 18.5\ mL$. So, $	ext{Density}=\frac{85.2\ g}{18.5\ mL}$.
## Step3: Calculate density
$	ext{Density}=\frac{85.2}{18.5}\ g/mL\approx4.60\ g/mL$ (3 significant - figures)

# Answer:
$4.60\ g/mL$

### Question 11
# Explanation:
## Step1: Rearrange density formula
From $	ext{Density}=\frac{	ext{mass}}{	ext{volume}}$, we get $	ext{volume}=\frac{	ext{mass}}{	ext{Density}}$.
## Step2: Substitute values
Given mass $m = 250.0\ g$ and density of copper $
ho=8.90\ g/cm^{3}$. Then $V=\frac{250.0\ g}{8.90\ g/cm^{3}}$.
## Step3: Calculate volume
$V=\frac{250.0}{8.90}\ cm^{3}\approx28.1\ cm^{3}$ (3 significant - figures)

# Answer:
$28.1\ cm^{3}$

### Question 12
# Explanation:
## Step1: Calculate density of the object
Using the density formula $	ext{Density}=\frac{	ext{mass}}{	ext{volume}}$, with $m = 45.6\ g$ and $V = 12.0\ cm^{3}$. So, $	ext{Density}=\frac{45.6\ g}{12.0\ cm^{3}} = 3.80\ g/cm^{3}$.
## Step2: Compare densities
An object will float in a substance if its density is less than the density of the substance. The object will not float in any of the given substances (wood: $0.70\ g/cm^{3}$, corn - oil: $0.92\ g/cm^{3}$, water: $1.00\ g/cm^{3}$, rubber: $1.34\ g/cm^{3}$, corn - syrup: $1.38\ g/cm^{3}$, copper: $8.80\ g/cm^{3}$) since $3.80\ g/cm^{3}$ is greater than all of them.

# Answer:
The object will not float in any of the given substances.

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QUESTION IMAGE

Question

Explanation:

Question 8

Step1: Identify formula

Step2: Substitute values

Step3: Calculate density

Step1: Rearrange density formula

Step2: Substitute values

Step3: Calculate mass

Step1: Calculate volume of the object

Step2: Use density formula

Step3: Calculate density

Answer:

Question 9