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The mass of a 147 cm³ block of uranium metal is 2.18 lb.

A. What is this density expressed in g/cm³?

B. What would be the mass (in grams) for a 2.391 cm³ sample of uranium? ___g

C. What volume (in liters) would a uranium sample with a mass of 5.887 kg occupy?

Answer :

Final Answer:

a. The density of the uranium metal is 6.72 g/cm³.

b. The volume that a uranium sample with a mass of 5.887 kg would occupy is approximately 0.87614 liters.

Explanation:

the density (D) of a substance is defined as mass (m) divided by volume (V):

D = m / V

Let's solve each part of the question:

Part A: Density in g/cm³

Given:

Volume (V) = 147 cm³

Mass (m) = 2.18 lb

First, convert the mass from pounds to grams (1 lb = 453.592 g):

Mass (m) = 2.18 lb × 453.592 g/lb = 988.64 g

Now, plug the values into the density formula:

Density (D) = 988.64 g / 147 cm³ = 6.72 g/cm³

The density of the uranium metal is 6.72 g/cm³.

Part B: Mass of a 2.391 cm³ Sample

Given:

Volume (V) = 2.391 cm³

Using the previously calculated density (6.72 g/cm³), we can rearrange the density formula to solve for mass:

Mass (m) = Density (D) × Volume (V)

Mass (m) = 6.72 g/cm³ × 2.391 cm³ ≈ 16.01 g

The mass of the 2.391 cm³ sample of uranium is approximately 16.01 grams.

Part C: Volume for a Mass of 5.887 kg

Given:

Mass (m) = 5.887 kg

First, convert the mass from kilograms to grams (1 kg = 1000 g):

Mass (m) = 5.887 kg × 1000 g/kg = 5887 g

Using the previously calculated density (6.72 g/cm³), we can rearrange the density formula to solve for volume:

Volume (V) = Mass (m) / Density (D)

Volume (V) = 5887 g / 6.72 g/cm³ ≈ 876.14 cm³

Now, convert the volume from cm³ to liters (1 L = 1000 cm³):

Volume (V) = 876.14 cm³ / 1000 cm³/L = 0.87614 L

The volume that a uranium sample with a mass of 5.887 kg would occupy is approximately 0.87614 liters.

Learn more about volume at https://brainly.com/question/24788314

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