Question

1. an irregularly shaped stone was lowered into a graduated... volume of water equal to 2.0 ml. the height of the water rose to 7.0 ml. if the mass of the stone was 25 g, what was its density?

... of copper has a mass of 89.6 g. what is the density of copper?

Explanation:

For the stone's density problem:

Step1: Find the volume of the stone

The volume of the stone is the change in water volume, so we subtract the initial volume from the final volume. Initial volume \( V_1 = 2.0\space mL \), final volume \( V_2 = 7.0\space mL \). The volume of the stone \( V=V_2 - V_1 \).
\( V = 7.0\space mL - 2.0\space mL= 5.0\space mL \) (since \( 1\space mL = 1\space cm^3 \), the volume is also \( 5.0\space cm^3 \))

Step2: Calculate the density of the stone

Density formula is \(
ho=\frac{m}{V} \), where \( m = 25\space g \) and \( V = 5.0\space cm^3 \)
\(
ho=\frac{25\space g}{5.0\space cm^3} \)

Step3: Simplify the density

\(
ho = 5.0\space g/cm^3 \) (or \( 5.0\space g/mL \) since \( 1\space mL = 1\space cm^3 \))

Step1: Recall density formula

\(
ho=\frac{m}{V} \), \( m = 89.6\space g \), assume \( V = 10.0\space cm^3 \) (since the problem likely has a volume of \( 10.0\space cm^3 \) or \( 10.0\space mL \) which is missing)

Step2: Calculate density

\(
ho=\frac{89.6\space g}{10.0\space cm^3}= 8.96\space g/cm^3 \)

Answer:

The density of the stone is \( \boldsymbol{5.0\space g/cm^3} \) (or \( 5.0\space g/mL \))

For the copper's density problem (assuming the volume of copper is missing, but if we assume a common volume for copper sample, for example, if the volume of copper is \( 10.0\space cm^3 \) (a typical value for a simple problem, but since it's not given, maybe there was a typo and the volume is \( 10.0\space cm^3 \) or maybe the original problem had a volume like \( 10.0\space mL \) or \( 10.0\space cm^3 \)):

Wait, the original problem for copper: "a mass of 89.6 g. What is the density of copper?" seems to be missing the volume. If we assume the volume of the copper sample is \( 10.0\space cm^3 \) (a common example), but actually, the density of copper is a known value, but if we calculate it from mass and volume, let's assume the volume is \( 10.0\space cm^3 \) (but this is an assumption as the problem is incomplete). However, the standard density of copper is \( 8.96\space g/cm^3 \). If we use \( m = 89.6\space g \) and assume \( V = 10.0\space cm^3 \) (since \( 89.6\div10 = 8.96 \)):