To find the density of the rock, we use the formula:

$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

Given:

• Mass of the rock, $m = 240 \, \text{g}$
• Volume of the rock, $V = 87.0 \, \text{cm}^3$ (based on water displacement)

Plugging in the values:

$\text{Density} = \frac{240 \, \text{g}}{87.0 \, \text{cm}^3} \approx 2.76 \, \text{g/cm}^3$

Answer: The density of the rock is approximately $2.76 \, \text{g/cm}^3$.

Would you like more details or have any questions?

Here are 5 related questions to further explore the topic:

1. How would the density calculation change if the rock displaced 90 cm³ of water?
2. What are some limitations of using water displacement to measure volume, especially for porous objects?
3. How does Archimedes' principle relate to measuring the density of irregularly shaped objects?
4. Can you find the density if only the weight in newtons and volume in liters are provided?
5. How would temperature variations affect the density of the water and, consequently, the density calculation?

Tip: When using water displacement, ensure there are no air bubbles on the submerged object, as they can affect the volume reading.