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.
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