Question

density
n has the greater mass, air or lead? most would answer lead, but this question
lly does not have an answer. to compare these two things, you need to know how
of each you have. a large amount of air could have a greater mass than a small
nt of lead. to compare different things, we have to compare the masses of each
occupy the same space, or volume. this is called density. it is measured in units of
or g/cm³.
density = \\(\frac{\text{mass}}{\text{volume}}\\) or \\( d = \frac{m}{v} \\)
ach problem.
what is the density of carbon dioxide gas if 0.196 g occupies a volume of 100 ml?
the density formula is d=m/v and volume is 2.0 ml
a block of wood that measures 3.0 cm on each side has a mass of 27 g. what is
the density of the block?
an irregularly shaped stone was lowered into a graduated cylinder holding a
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?
a 10.0 cm³ sample of copper has a mass of 89.6 g. what is the density of copper?
silver has a density of 10.5 g/cm³, and gold has a density of 19.3 g/cm³. which
would have a greater mass, 5 cm³ of silver or 5 cm³ of gold?
five ml of ethanol has a mass of 3.9 g , and 5.0 ml of benzene has a mass of
4.4 g. which liquid is denser?
a sample of iron in the shape of a rectangular prism has the dimensions of
3 cm × 2 cm. if the mass of this object is 94 g, what is the density of iron?
104442
13

Explanation:

Problem 6: Comparing density of ethanol and benzene

Given: For ethanol: \( M_{\text{ethanol}} = 3.9 \, \text{g} \), \( V_{\text{ethanol}} = 5 \, \text{mL} \); For benzene: \( M_{\text{benzene}} = 4.4 \, \text{g} \), \( V_{\text{benzene}} = 5.0 \, \text{mL} \)

• Density of ethanol: \( D_{\text{ethanol}} = \frac{M_{\text{ethanol}}}{V_{\text{ethanol}}} = \frac{3.9 \, \text{g}}{5 \, \text{mL}} = 0.78 \, \text{g/mL} \)
• Density of benzene: \( D_{\text{benzene}} = \frac{M_{\text{benzene}}}{V_{\text{benzene}}} = \frac{4.4 \, \text{g}}{5.0 \, \text{mL}} = 0.88 \, \text{g/mL} \)

Since \( 0.88 \, \text{g/mL} > 0.78 \, \text{g/mL} \), benzene is denser.

Problem 7: Density of iron (assuming the third dimension is missing, let's assume it’s a typo and the dimensions are \( 3 \, \text{cm} \times 2 \, \text{cm} \times h \, \text{cm} \), but wait—the mass is 94 g? Wait, maybe a typo. Wait, standard iron density is ~7.87 g/cm³. Wait, maybe the dimensions are \( 3 \, \text{cm} \times 2 \, \text{cm} \times 5 \, \text{cm} \)? No, let's check the original problem: "a sample of iron in the shape of a rectangular prism has the dimensions of 3 cm × 2 cm. If the mass of this object is 94 g, what is the density of iron?" Wait, the third dimension is missing? Maybe it's a typo (e.g., 3 cm × 2 cm × 5 cm? Or maybe 3 cm × 2 cm × 4 cm? Wait, no—maybe the original problem has a third dimension. Wait, perhaps it's a mistake, but let's assume the volume is \( 3 \times 2 \times l \), but since the mass is 94 g, let's see:

Wait, maybe the dimensions are \( 3 \, \text{cm} \times 2 \, \text{cm} \times 5 \, \text{cm} \) (volume \( 30 \, \text{cm}^3 \)), but 94 g / 30 cm³ ≈ 3.13 g/cm³ (not iron). Alternatively, maybe the mass is 94 g and the volume is \( 3 \times 2 \times 6 \, \text{cm} = 36 \, \text{cm}^3 \), 94 / 36 ≈ 2.61 (no). Wait, maybe the original problem has a typo (e.g., mass is 94 g, dimensions 5 cm × 3 cm × 2 cm: volume 30 cm³, density 94/30 ≈ 3.13—still not iron. Alternatively, maybe the mass is 940 g? Then 940 / 30 ≈ 31.3 (no). Wait, standard iron density is ~7.87 g/cm³. Let's check: if density is 7.87, then mass = 7.87 × volume. If volume is 3×2×h, then 7.87×6h = 47.22h. If h=2, volume 12, mass 94.64 g (close to 94 g). So maybe the dimensions are 3 cm × 2 cm × 2 cm (volume 12 cm³). Then:

Given: \( V = 3 \times 2 \times 2 = 12 \, \text{cm}^3 \), \( M = 94 \, \text{g} \) (approx 94.64, so maybe 94 g is a typo for 94.64 g)

Step 1: Apply the density formula

\( D = \frac{M}{V} = \frac{94 \, \text{g}}{12 \, \text{cm}^3} \approx 7.83 \, \text{g/cm}^3 \) (close to iron’s density)

Final Answers:

1. \( \boldsymbol{0.00196 \, \text{g/mL}} \)
2. \( \boldsymbol{1.0 \, \text{g/cm}^3} \)
3. \( \boldsymbol{5.0 \, \text{g/cm}^3} \)
4. \( \boldsymbol{8.96 \, \text{g/cm}^3} \)
5. \( \boldsymbol{\text{Gold}} \)
6. \( \boldsymbol{\text{Benzene}} \)
7. \( \boldsymbol{\approx 7.83 \, \text{g/cm}^3} \) (assuming volume \( 12 \, \text{cm}^3 \))

Answer:

Problem 6: Comparing density of ethanol and benzene

Given: For ethanol: \( M_{\text{ethanol}} = 3.9 \, \text{g} \), \( V_{\text{ethanol}} = 5 \, \text{mL} \); For benzene: \( M_{\text{benzene}} = 4.4 \, \text{g} \), \( V_{\text{benzene}} = 5.0 \, \text{mL} \)

• Density of ethanol: \( D_{\text{ethanol}} = \frac{M_{\text{ethanol}}}{V_{\text{ethanol}}} = \frac{3.9 \, \text{g}}{5 \, \text{mL}} = 0.78 \, \text{g/mL} \)
• Density of benzene: \( D_{\text{benzene}} = \frac{M_{\text{benzene}}}{V_{\text{benzene}}} = \frac{4.4 \, \text{g}}{5.0 \, \text{mL}} = 0.88 \, \text{g/mL} \)

Since \( 0.88 \, \text{g/mL} > 0.78 \, \text{g/mL} \), benzene is denser.

Problem 7: Density of iron (assuming the third dimension is missing, let's assume it’s a typo and the dimensions are \( 3 \, \text{cm} \times 2 \, \text{cm} \times h \, \text{cm} \), but wait—the mass is 94 g? Wait, maybe a typo. Wait, standard iron density is ~7.87 g/cm³. Wait, maybe the dimensions are \( 3 \, \text{cm} \times 2 \, \text{cm} \times 5 \, \text{cm} \)? No, let's check the original problem: "a sample of iron in the shape of a rectangular prism has the dimensions of 3 cm × 2 cm. If the mass of this object is 94 g, what is the density of iron?" Wait, the third dimension is missing? Maybe it's a typo (e.g., 3 cm × 2 cm × 5 cm? Or maybe 3 cm × 2 cm × 4 cm? Wait, no—maybe the original problem has a third dimension. Wait, perhaps it's a mistake, but let's assume the volume is \( 3 \times 2 \times l \), but since the mass is 94 g, let's see:

Wait, maybe the dimensions are \( 3 \, \text{cm} \times 2 \, \text{cm} \times 5 \, \text{cm} \) (volume \( 30 \, \text{cm}^3 \)), but 94 g / 30 cm³ ≈ 3.13 g/cm³ (not iron). Alternatively, maybe the mass is 94 g and the volume is \( 3 \times 2 \times 6 \, \text{cm} = 36 \, \text{cm}^3 \), 94 / 36 ≈ 2.61 (no). Wait, maybe the original problem has a typo (e.g., mass is 94 g, dimensions 5 cm × 3 cm × 2 cm: volume 30 cm³, density 94/30 ≈ 3.13—still not iron. Alternatively, maybe the mass is 940 g? Then 940 / 30 ≈ 31.3 (no). Wait, standard iron density is ~7.87 g/cm³. Let's check: if density is 7.87, then mass = 7.87 × volume. If volume is 3×2×h, then 7.87×6h = 47.22h. If h=2, volume 12, mass 94.64 g (close to 94 g). So maybe the dimensions are 3 cm × 2 cm × 2 cm (volume 12 cm³). Then:

Given: \( V = 3 \times 2 \times 2 = 12 \, \text{cm}^3 \), \( M = 94 \, \text{g} \) (approx 94.64, so maybe 94 g is a typo for 94.64 g)

Step 1: Apply the density formula

\( D = \frac{M}{V} = \frac{94 \, \text{g}}{12 \, \text{cm}^3} \approx 7.83 \, \text{g/cm}^3 \) (close to iron’s density)

Final Answers:

1. \( \boldsymbol{0.00196 \, \text{g/mL}} \)
2. \( \boldsymbol{1.0 \, \text{g/cm}^3} \)
3. \( \boldsymbol{5.0 \, \text{g/cm}^3} \)
4. \( \boldsymbol{8.96 \, \text{g/cm}^3} \)
5. \( \boldsymbol{\text{Gold}} \)
6. \( \boldsymbol{\text{Benzene}} \)
7. \( \boldsymbol{\approx 7.83 \, \text{g/cm}^3} \) (assuming volume \( 12 \, \text{cm}^3 \))