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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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Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

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The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

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The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

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Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

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