Part A: Part B: Ore from gold mines is about 0.02% gold by mass. Seawater contai

Question

Part A:

Part B:

Ore from gold mines is about 0.02% gold by mass. Seawater contains gold, too, at about 1 part per trillion by volume. (Gold has a specific gravity of 19.3, while seawater has a specific gravity of 1.03.) Calculate the tons of gold ore and of seawater required to produce one ounce of gold. If you assume earth's diameter is 25.000 miles and the average depth of ocean water across the entire planet is 2.4 miles, how many tons of gold do you estimate are dissolved in the oceans, worldwide?

Explanation / Answer

1)one ounce of gold = 31 grams of gold

=> ore required for 1 ounce = 31/(0.02/100) = 15.5 kg = 15.5/900 = 00.017 tons

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1 ounce of gold has volume of 31/19.3 = 1.61 litres

=> volume of water = 1.61*10^12 litres

mass of water = 1.61*10^12*1000 = 1.61*10^15 grams = 1.8*10^12 tons

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25000 miles = 25000*1.6 km = 40000000 m

radius = 20000000m

2.4 miles = 3840 m

volume of water = 4pi(20000000^3 - (20000000 - 3840)^3)/3 = 7.72*10^19 litres

=> amount of gold = 7.72*10^19 /10^12 = 77200369 litres

=> mass = 77200369*19.3 g = 1505.017 tons

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