Question

part 5: irregular object
mass (g):\tmeasurement
63.34
initial volume (ml):\t30ml
final volume (ml):\t40ml
volume (cm³):\t(show all work!)
\t\t\t\tvolume: _____________
density (g/cm³):\t(show all work!)

Explanation:

Volume Calculation

Step1: Recall Volume Displacement Formula

The volume of an irregular object is found by the displacement of water, so \( \text{Volume} = \text{Final Volume} - \text{Initial Volume} \). Also, \( 1 \, \text{mL} = 1 \, \text{cm}^3 \).

Step2: Substitute Values

Initial Volume \( = 30 \, \text{mL} \), Final Volume \( = 40 \, \text{mL} \).
\( \text{Volume} = 40 - 30 = 10 \, \text{mL} = 10 \, \text{cm}^3 \).

Density Calculation

Step3: Recall Density Formula

Density \(
ho = \frac{\text{Mass}}{\text{Volume}} \).

Step4: Substitute Values

Mass \( = 63.34 \, \text{g} \), Volume \( = 10 \, \text{cm}^3 \).
\(
ho = \frac{63.34}{10} = 6.334 \, \text{g/cm}^3 \).

Answer:

• Volume: \( 10 \, \text{cm}^3 \)
• Density: \( 6.334 \, \text{g/cm}^3 \)