Question

1. given: d_{actinium}=10.1g/cm^{3},1.2g

wanted: volume of actinium in cm^{3}?

1. given: red brass,0.7700g,0.08800cm^{3}

wanted: density of red brass in g/cm^{3}?

1. given: d_{maple}=0.6900g/cm^{3},1.473g

wanted: volume of maple in cm^{3}?

1. given: krypton,1.661g,448.9ml

wanted: density of krypton at stp in g/ml?

1. given: d_{calcium}=1.5g/cm^{3},2.098g

wanted: volume of calcium in cm^{3}?

1. given: polonium,0.3670g,0.03938cm^{3}

wanted: density of polonium in g/cm^{3}?

1. given: phosphorus,2.0g,1.099cm^{3}

wanted: density of phosphorus in g/cm^{3}?

1. given: d_{cork}=0.2400g/cm^{3},0.538g

wanted: volume of cork in cm^{3}?

Explanation:

Step1: Recall the density formula

The formula for density is \(D=\frac{m}{V}\), where \(D\) is density, \(m\) is mass, and \(V\) is volume. We can rearrange it to solve for the unknown variable. For example, if we want to find volume \(V = \frac{m}{D}\), and if we want to find density \(D=\frac{m}{V}\).

Step2: Solve problem 1

Given \(D_{actinium}=10.1\ g/cm^{3}\), \(m = 1.2\ g\). Using \(V=\frac{m}{D}\), we substitute the values: \(V=\frac{1.2\ g}{10.1\ g/cm^{3}}\approx0.12\ cm^{3}\)

Step3: Solve problem 2

Given \(m = 0.7700\ g\), \(V=0.08800\ cm^{3}\). Using \(D=\frac{m}{V}\), we substitute the values: \(D=\frac{0.7700\ g}{0.08800\ cm^{3}} = 8.75\ g/cm^{3}\)

Step4: Solve problem 3

Given \(D_{maple}=0.6900\ g/cm^{3}\), \(m = 1.473\ g\). Using \(V=\frac{m}{D}\), we substitute the values: \(V=\frac{1.473\ g}{0.6900\ g/cm^{3}}\approx2.135\ cm^{3}\)

Step5: Solve problem 4

Given \(m = 448.9\ mL\) (Note: for gases at STP, we assume appropriate conditions. Given \(m = 1.661\ g\), \(V = 448.9\ mL\). Using \(D=\frac{m}{V}\), we substitute the values: \(D=\frac{1.661\ g}{448.9\ mL}\approx0.0037\ g/mL\)

Step6: Solve problem 5

Given \(D_{calcium}=1.5\ g/cm^{3}\), \(m = 2.098\ g\). Using \(V=\frac{m}{D}\), we substitute the values: \(V=\frac{2.098\ g}{1.5\ g/cm^{3}}\approx1.4\ cm^{3}\)

Step7: Solve problem 6

Given \(m = 0.3670\ g\), \(V = 0.03938\ cm^{3}\). Using \(D=\frac{m}{V}\), we substitute the values: \(D=\frac{0.3670\ g}{0.03938\ cm^{3}}\approx9.32\ g/cm^{3}\)

Step8: Solve problem 7

Given \(m = 2.0\ g\), \(V = 1.099\ cm^{3}\). Using \(D=\frac{m}{V}\), we substitute the values: \(D=\frac{2.0\ g}{1.099\ cm^{3}}\approx1.82\ g/cm^{3}\)

Step9: Solve problem 8

Given \(D_{cork}=0.2400\ g/cm^{3}\), \(m = 0.538\ g\). Using \(V=\frac{m}{D}\), we substitute the values: \(V=\frac{0.538\ g}{0.2400\ g/cm^{3}}\approx2.24\ cm^{3}\)

Answer:

1. \(0.12\ cm^{3}\)
2. \(8.75\ g/cm^{3}\)
3. \(2.135\ cm^{3}\)
4. \(0.0037\ g/mL\)
5. \(1.4\ cm^{3}\)
6. \(9.32\ g/cm^{3}\)
7. \(1.82\ g/cm^{3}\)
8. \(2.24\ cm^{3}\)