Question

what is the density of an object that has a mass of 15.1 g and, when placed into a graduated cylinder, causes the water level to rise from 25.2 ml to 33.6 ml?
1.7 g/ml
1.8 g/ml
0.60 g/ml
2.4 g/ml
question 7 (0.0862 points)
the initial volume of a gas cylinder is 750.0 ml. if the pressure of a gas inside the cylinder changes from 360.0 mm hg to 840.0 mm hg, what is the final volume the gas occupies?
3.151 l
1.750 l
321.4 ml
630.0 ml

Explanation:

First Question (Density Calculation)

Step1: Find the volume of the object

The volume of the object is the change in water level, so we calculate the difference between the final and initial volumes.
Final volume - Initial volume = \( 33.6 \, \text{mL} - 25.2 \, \text{mL} = 8.4 \, \text{mL} \)

Step2: Use the density formula

The formula for density \(
ho \) is \(
ho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume.
Substitute \( m = 15.1 \, \text{g} \) and \( V = 8.4 \, \text{mL} \) into the formula:
\(
ho = \frac{15.1 \, \text{g}}{8.4 \, \text{mL}} \approx 1.8 \, \text{g/mL} \)

Step1: Recall Boyle's Law

Boyle's Law states that \( P_1V_1 = P_2V_2 \), where \( P_1 \) and \( V_1 \) are the initial pressure and volume, and \( P_2 \) and \( V_2 \) are the final pressure and volume.

Step2: Rearrange the formula to solve for \( V_2 \)

\( V_2 = \frac{P_1V_1}{P_2} \)

Step3: Substitute the values

\( P_1 = 360.0 \, \text{mm Hg} \), \( V_1 = 750.0 \, \text{mL} \), \( P_2 = 840.0 \, \text{mm Hg} \)
\( V_2 = \frac{360.0 \, \text{mm Hg} \times 750.0 \, \text{mL}}{840.0 \, \text{mm Hg}} \)
First, calculate the numerator: \( 360.0 \times 750.0 = 270000 \)
Then divide by the denominator: \( \frac{270000}{840.0} \approx 321.4 \, \text{mL} \)

Answer:

1.8 g/mL

Second Question (Gas Law - Boyle's Law)