Use your density triangle to solve for the missing variable. SHOW YOUR WORK! 22. An object has a volume of $$ 200 \mathrm{~cm}^{3} $$ and a density of $$ 14.2 \mathrm{~g} / \mathrm{cm}^{3} $$. Find its mass. 23. A liquid has a density of $$ 0.92 \mathrm{~g} / \mathrm{mL} $$. What volume is occupied by 50 g of the liquid?

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To solve these problems, we will use the density triangle, which relates density ($$d$$), mass ($$m$$), and volume ($$V$$). The relationships can be expressed as:

$$
d = \frac{m}{V}
$$

From this, we can derive the following formulas:

1. To find mass: $$ m = d \times V $$
2. To find volume: $$ V = \frac{m}{d} $$

Now, let's solve each problem step by step.

### Problem 22
**Given:**
- Volume ($$V$$) = $$200 \, \text{cm}^3$$
- Density ($$d$$) = $$14.2 \, \text{g/cm}^3$$

**To find:** Mass ($$m$$)

**Step 1:** Use the formula for mass:
$$
m = d \times V
$$

**Step 2:** Substitute the known values:
$$
m = 14.2 \, \text{g/cm}^3 \times 200 \, \text{cm}^3
$$

**Step 3:** Calculate the mass:
$$
m = 2840 \, \text{g}
$$

### Problem 23
**Given:**
- Density ($$d$$) = $$0.92 \, \text{g/mL}$$
- Mass ($$m$$) = $$50 \, \text{g}$$

**To find:** Volume ($$V$$)

**Step 1:** Use the formula for volume:
$$
V = \frac{m}{d}
$$

**Step 2:** Substitute the known values:
$$
V = \frac{50 \, \text{g}}{0.92 \, \text{g/mL}}
$$

**Step 3:** Calculate the volume:
$$
V \approx 54.35 \, \text{mL}
$$

### Summary of Results
- For Problem 22, the mass of the object is $$2840 \, \text{g}$$.
- For Problem 23, the volume occupied by the liquid is approximately $$54.35 \, \text{mL}$$.
- For Problem 22, the mass is 2840 grams. - For Problem 23, the volume is approximately 54.35 milliliters.

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