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MakcuM [25]
3 years ago
7

if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius

Chemistry
1 answer:
Lady_Fox [76]3 years ago
6 0
The mass of ore required is 21 700 t.

r = 750 cm

V = \frac{4}{3}  \pi  r^{3} = \frac{4}{3}  \pi  (750 cm)^{3} = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ × \frac{11.34 g}{1 cm^{3} } = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g × \frac{1 kg}{1000 g} = 2.004 × 10⁷ kg

2.004 × 10⁷ kg × \frac{1 t}{1000 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 ×\frac{100 tore}{92.5 t Pb} = 2.17 × 10⁴ t ore = 21 700 t ore

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SO the reaction after add the NaOH:
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initial                  0.00525                0                         0
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equilibrium      0.002625             0.002625        0.002625
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