Determining the Locations of Subatomic
1 answer:
You might be interested in
The mass of ore required is 21 700 t.
r = 750 cm
V =
= 
= 1.767 × 10⁹ cm³
The density of lead is 11.34 g/cm³.
So mass of lead sphere = 1.767 × 10⁹ cm³ ×
= 2.004 ×10¹⁰ g
2.004 ×10¹⁰ g ×
= 2.004 × 10⁷ kg
2.004 × 10⁷ kg ×
= 2.004 × 10⁴ t
92.5% efficiency means 92.5 t Pb per 100 t of ore.
Mass of ore = 2.004 × 10⁴ t Pb ×
= 2.17 × 10⁴ t ore = 21 700 t ore
r = 750 cm
V =
The density of lead is 11.34 g/cm³.
So mass of lead sphere = 1.767 × 10⁹ cm³ ×
2.004 ×10¹⁰ g ×
2.004 × 10⁷ kg ×
92.5% efficiency means 92.5 t Pb per 100 t of ore.
Mass of ore = 2.004 × 10⁴ t Pb ×