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luck♥♦♣â™
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follow you!? Hi--I am struggling with this problem. Could someone please help me?
Earth’s oceans have an average depth of 3800 m, a total area of 3.63 x 10^8 km^2, and an average concentration of dissolved gold of 5.8 x 10^-9 g/L. (a) How many grams of gold are in the oceans? (b) How many m^3 of gold are in the oceans? (c) If a recent price of gold was $370.00/troy oz, what is the value of gold in the oceans (1 troy oz=31.1 g ; d of gold = 19.3 g/cm^3)?
I know A, but b & c are confusing....
3.63 x 10(8)km(2) = 3.63 x 10(14)m(2)
3.63 x 10(14)m(2) x 3.8 x 10(3)m [depth] = 1.38 x 10(18)m(3) of water in the ocean. 1L = 10(-3)m(3), so 1.38 x 10(21) liters of water in ocean.
1.38 x 10(21)L x 5.8 x 10(-9)g/L = 8.04 x 10(12) grams of gold in the ocean.
Earth’s oceans have an average depth of 3800 m, a total area of 3.63 x 10^8 km^2, and an average concentration of dissolved gold of 5.8 x 10^-9 g/L. (a) How many grams of gold are in the oceans? (b) How many m^3 of gold are in the oceans? (c) If a recent price of gold was $370.00/troy oz, what is the value of gold in the oceans (1 troy oz=31.1 g ; d of gold = 19.3 g/cm^3)?
I know A, but b & c are confusing....
3.63 x 10(8)km(2) = 3.63 x 10(14)m(2)
3.63 x 10(14)m(2) x 3.8 x 10(3)m [depth] = 1.38 x 10(18)m(3) of water in the ocean. 1L = 10(-3)m(3), so 1.38 x 10(21) liters of water in ocean.
1.38 x 10(21)L x 5.8 x 10(-9)g/L = 8.04 x 10(12) grams of gold in the ocean.