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Triss [41]
2 years ago
3

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in chemical synthesis. It is obtained mostly from seawater,

which contains about 1.3 g of Mg for every kilogram of seawater. Calculate the volume of seawater (in liters) needed to extract 8.0 x 104 tons of Mg, which is roughly the annual production in the United States. (rho seawater = 1.03 g/ml).
Chemistry
1 answer:
alexira [117]2 years ago
3 0

Answer:

The volume of seawater needed to extract 8.0\times 10^4 tons of magnesium is 5.420\times 10^{10} L.

Explanation:

Concentration magnesium in sea water = 1.3 g /kg of seawater

Extracted mass of magnesium ,= 8.0\times 10^4 tons

(1 ton = 907185 g)

8.0\times 10^4 tons=8.0\times 10^4\times 907185 g

=7.257\times 10^{10} g

If 1 kg of sea water contains 1.3 grams of magnesium. Then 7.257\times 10^{10} g of magnesium will be contained by:

\frac{1}{1.3} kg\times 7.257\times 10^{10}=5.582\times 10^{10} kg sea water.

Mass of sea water,m =5.582\times 10^{10} kg=5.582\times 10^{13} g

Volume of sea water = v

Density of sea water ,d= 1.03 g/mL

v=\frac{m}{d}=\frac{5.582\times 10^{13} g}{1.03 g/mL}=5.420\times 10^{13} mL

1 mL = 0.001 L

v = 5.420\times 10^{10} L

The volume of seawater needed to extract 8.0\times 10^4 tons of magnesium is 5.420\times 10^{10} L.

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Answer:

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Explanation:

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