Question

bronze is an alloy made of copper (cu) and tin (sn). calculate the mass of a bronze cylinder of radius 7.94 cm and length 38.87 cm. the composition of the bronze is 79.42 percent cu and 20.58 percent sn and the densities of cu and sn are 8.94 g/cm³ and 7.31 g/cm³, respectively.
part 1 of 2
what assumption should you make in this calculation?
assume that the density is exactly halfway between 8.94 g/cm³ and 7.31 g/cm³.
assume the volumes of the two components are additive.
part 2 of 2
round your answer to 3 significant digits.
the mass of the bronze cylinder is: □g

Explanation:

Step1: Calculate the volume of the bronze cylinder

The volume \( V \) of a cylinder is given by the formula \( V=\pi r^{2}h \).
Here, \( r = 7.94\,\text{cm} \) and \( h=38.87\,\text{cm} \).
So, \( V=\pi\times(7.94)^{2}\times38.87\)
\( V=\pi\times63.0436\times38.87\)
\( V = 3.1416\times63.0436\times38.87\)
\( V\approx7639.9\,\text{cm}^{3}\)

Step2: Calculate the density of bronze

The density of bronze is assumed to be exactly halfway between \( 8.94\,\text{g/cm}^{3} \) (density of \( Cu \)) and \( 7.31\,\text{g/cm}^{3} \) (density of \( Sn \)).
The formula for the average density \(
ho=\frac{8.94 + 7.31}{2}\)
\(
ho=\frac{16.25}{2}=8.125\,\text{g/cm}^{3}\)

Step3: Calculate the mass of the bronze cylinder

We know that mass \( m=
ho\times V \)
Substitute \(
ho = 8.125\,\text{g/cm}^{3} \) and \( V\approx7639.9\,\text{cm}^{3}\)
\( m=8.125\times7639.9\)
\( m\approx62174\,\text{g}\)

Answer:

\(62200\,\text{g}\) (rounded to 3 significant digits)