What is the density (g/mL) of an object that has a mass of 42.1 grams and, when placed into a graduated cylinder, causes the water level to rise from 33.2 mL to 59.1 mL?

Sure! Let's break down the problem and solve it step-by-step.

1. Identify the Given Values:
- Mass of the object [tex]\( m = 42.1 \)[/tex] grams
- Initial water level in the graduated cylinder [tex]\( V_i = 33.2 \)[/tex] mL
- Final water level in the graduated cylinder [tex]\( V_f = 59.1 \)[/tex] mL

2. Calculate the Volume Displaced:
- When the object is placed in the water, the water level rises from [tex]\( V_i \)[/tex] to [tex]\( V_f \)[/tex].
- The volume displaced by the object is given by the difference between the final and initial water levels.
Therefore,
[tex]\[
\text{Volume Displaced} = V_f - V_i = 59.1 \, \text{mL} - 33.2 \, \text{mL} = 25.9 \, \text{mL}
\][/tex]

3. Calculate the Density:
- Density [tex]\( \rho \)[/tex] is defined as mass divided by volume. The formula for density is:
[tex]\[
\rho = \frac{m}{V}
\][/tex]
- Here, [tex]\( m \)[/tex] is the mass of the object, and [tex]\( V \)[/tex] is the volume displaced calculated previously.
Therefore,
[tex]\[
\rho = \frac{42.1 \, \text{grams}}{25.9 \, \text{mL}}
\][/tex]

4. Compute the Density:
- Performing the division:
[tex]\[
\rho = \frac{42.1}{25.9} \approx 1.625 \, \text{g/mL}
\][/tex]

Thus, the density of the object is approximately [tex]\( 1.625 \, \text{g/mL} \)[/tex].