Respuesta :
If the lead can be extracted with 92.5 efficiency, the mass of ore is required to make a lead sphere with a 5.00 cm radius is 6.45 kg.
Here's how,
Given,
- The radius of the lead sphere is r = 5.0 cm,
- The lead can be extracted with, η = 92.5% = 0.925 efficiency.
We know the density of lead as,
⇒ p = 11.4 [tex]g/cm^{3}[/tex],
The volume of the lead sphere is,
⇒ V = [tex]\frac{4}{3} \pi r^{3}[/tex]
⇒ V = [tex]\frac{4}{3} \pi 5^{3}[/tex]
⇒ V = 523.3 [tex]cm^{3}[/tex],
Mass of the lead present in the lead sphere is,
⇒ m = p . V
⇒ m = 11.4 [tex]\frac{g}{cm^3}[/tex] × 523.3 [tex]cm^{3}[/tex]
⇒ m = 5966 g
⇒ m = 5.97 kg
From the expression of efficiency, we calculate the mass of the ore,
⇒ η = Mass of lead obtained / Mass of the ore to be taken
⇒ 0.925 = 5.97 kg / Mass of the ore to be taken
⇒ Mass of the ore to be taken = 6.45 kg
To learn more about Lead extraction, here
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