Question

a 112 - g piece of metal has a density of 9.20 g/ml. a 50 - ml graduated cylinder contains 20.0 ml of water. what is the final volume after the metal is added to the graduated cylinder? (1 point)
solve for the volume of the object, then add that volume to the volume of water in the beaker.
20 ml
28 ml
22 ml
32 ml

Explanation:

Step1: Calculate the volume of the metal

Using the density formula \(d=\frac{m}{V}\), we can rearrange it to find \(V = \frac{m}{d}\). Given \(m = 112\space g\) and \(d=9.20\space g/mL\), then \(V=\frac{112\space g}{9.20\space g/mL}\approx12.2\space mL\)

Step2: Calculate the final volume

The initial volume of water \(V_{water}=20.0\space mL\). The final volume \(V_{final}=V_{water}+V_{metal}\). Substitute the values: \(V_{final}=20.0\space mL + 12.2\space mL=32.2\space mL\approx32\space mL\) (rounded to the nearest whole number as per the options)

Answer:

32 mL