Answer:
0.295cm thick wall
Explanation:
In the complete question, the diameter of the gold sphere is said to be 6cm.
Step 1: Determine the apparent volume of 6cm diameter sphere (including wall and hollow inside)
[tex]V=\frac{4}{3}pir^3[/tex]
[tex]V=\frac{4}{3}pi(3)^3[/tex]
[tex]V=113.04 cm^3[/tex]
Step 2: Determine mass of gold required for sphere to have density of iron ore if the apparent volume is as calculated in Step 1
[tex]mass=density*volume[/tex]
[tex]mass=5.15*113.04[/tex]
[tex]mass=582.16g[/tex]
Step 3: Determine volume of gold based on mass calculated in Step 2
[tex]volume=\frac{mass}{density}[/tex]
[tex]volume=\frac{582.16}{19.3}[/tex]
[tex]volume=30.16cm^3[/tex]
Step 4: Determine thickness of wall of a hollow sphere for volume calculated in Step 3
[tex]V=\frac{4}{3}pi(r_{outer}^3-r_{inner}^3)[/tex]
[tex]30.16=\frac{4}{3}pi(3^3-(3-t)^3)[/tex]
[tex](3-t)^3=19.796[/tex]
[tex](3-t)=2.705[/tex]
[tex]t=0.295cm[/tex]