Question
A spherical ball of lead has a diameter of 5.0 cm . What is the mass of the sphere if lead has a density of 11.34 g/em'? (The volume of a sphere is r" where r is t Express your answer using two significant figures.
Answers
GPT-4o mini
To find the mass of the spherical ball of lead, we can use the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
First, we need to calculate the volume of the sphere. The formula for the volume \( V \) of a sphere is:
\[ V = \frac{4}{3} \pi r^3 \]
Given that the diameter of the sphere is 5.0 cm, we can find the radius \( r \):
\[ r = \frac{\text{diameter}}{2} = \frac{5.0 \, \text{cm}}{2} = 2.5 \, \text{cm} \]
Now, we can calculate the volume:
\[
V = \frac{4}{3} \pi (2.5 \, \text{cm})^3
\]
Calculating \( (2.5 \, \text{cm})^3 \):
\[
(2.5)^3 = 15.625 \, \text{cm}^3
\]
Now substituting back into the volume formula:
\[
V = \frac{4}{3} \pi (15.625 \, \text{cm}^3)
\]
Using \( \pi \approx 3.14159 \):
\[
V \approx \frac{4}{3} \times 3.14159 \times 15.625 \approx 65.45 \, \text{cm}^3
\]
Next, we will calculate the mass using the density of lead, which is given as 11.34 g/cm³:
\[
\text{Mass} = \text{Density} \times \text{Volume} = 11.34 \, \text{g/cm}^3 \times 65.45 \, \text{cm}^3
\]
Calculating the mass:
\[
\text{Mass} \approx 11.34 \times 65.45 \approx 742.78 \, \text{g}
\]
Finally, rounding this to two significant figures, we get:
\[
\text{Mass} \approx 740 \, \text{g}
\]
Therefore, the mass of the sphere is approximately **740 g**.
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
First, we need to calculate the volume of the sphere. The formula for the volume \( V \) of a sphere is:
\[ V = \frac{4}{3} \pi r^3 \]
Given that the diameter of the sphere is 5.0 cm, we can find the radius \( r \):
\[ r = \frac{\text{diameter}}{2} = \frac{5.0 \, \text{cm}}{2} = 2.5 \, \text{cm} \]
Now, we can calculate the volume:
\[
V = \frac{4}{3} \pi (2.5 \, \text{cm})^3
\]
Calculating \( (2.5 \, \text{cm})^3 \):
\[
(2.5)^3 = 15.625 \, \text{cm}^3
\]
Now substituting back into the volume formula:
\[
V = \frac{4}{3} \pi (15.625 \, \text{cm}^3)
\]
Using \( \pi \approx 3.14159 \):
\[
V \approx \frac{4}{3} \times 3.14159 \times 15.625 \approx 65.45 \, \text{cm}^3
\]
Next, we will calculate the mass using the density of lead, which is given as 11.34 g/cm³:
\[
\text{Mass} = \text{Density} \times \text{Volume} = 11.34 \, \text{g/cm}^3 \times 65.45 \, \text{cm}^3
\]
Calculating the mass:
\[
\text{Mass} \approx 11.34 \times 65.45 \approx 742.78 \, \text{g}
\]
Finally, rounding this to two significant figures, we get:
\[
\text{Mass} \approx 740 \, \text{g}
\]
Therefore, the mass of the sphere is approximately **740 g**.