This problem consists of three parts based on the concept of density. Let's work through each part of the question:

A. Express $V$ in terms of the other variables in the formula.

The formula for density is: $D = \frac{m}{V}$ Where:

• $D$ = density,
• $m$ = mass, and
• $V$ = volume.

To express $V$ in terms of $m$ and $D$, solve the equation for $V$: $V = \frac{m}{D}$

B. Osmium is the densest metal known. Use the information shown to find the volume of a sample of osmium that has a mass of 200 grams.

The density of osmium is approximately $D = 22.59 \, \text{g/cm}^3$ (this value is well-known for osmium). Using the formula for volume: $V = \frac{m}{D}$ Substitute $m = 200 \, \text{g}$ and $D = 22.59 \, \text{g/cm}^3$: $V = \frac{200}{22.59} \approx 8.86 \, \text{cm}^3$ So, the volume of the osmium sample is approximately $8.86 \, \text{cm}^3$.

C. Suppose the sample of osmium in Part B is shaped like a cube. What is the length of each edge of the cube?

For a cube, the volume $V$ is related to the edge length $s$ by the formula: $V = s^3$ We know $V \approx 8.86 \, \text{cm}^3$, so solve for $s$: $s = \sqrt[3]{8.86} \approx 2.06 \, \text{cm}$

Therefore, the length of each edge of the cube is approximately $2.06 \, \text{cm}$.

Would you like further details or have any other questions?
Here are some related questions to deepen your understanding:

1. What is the relationship between mass, volume, and density?
2. How would the volume change if the mass of osmium was doubled?
3. How does the shape of an object (e.g., cube, sphere) affect how we calculate its volume from density?
4. Can you find the surface area of the cube-shaped osmium sample?
5. What would happen to the density of an object if its volume is halved, but its mass remains constant?

Tip: Always check the units of density and ensure the mass and volume are in compatible units when performing these calculations.