Respuesta :

znk
The mass of ore required is 21 700 t.

r = 750 cm

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

The density of lead is 11.34 g/cm³.

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

2.004 ×10¹⁰ g × [tex] \frac{1 kg}{1000 g} [/tex] = 2.004 × 10⁷ kg

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

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