Earth\'s oceans have an average depth of 3800 m, a total areaof 3.63E8 km^2, and

Question

Earth's oceans have an average depth of 3800 m, a total areaof 3.63E8 km^2, and an average concentration of dissolved gold of5.8E-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) Assuming the price of is $370.00/troy oz, what is thevalue of gold in the oceans (1 troy oz = 31.1g; d of gold = 19.3g/cm^3)? Earth's oceans have an average depth of 3800 m, a total areaof 3.63E8 km^2, and an average concentration of dissolved gold of5.8E-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) Assuming the price of is $370.00/troy oz, what is thevalue of gold in the oceans (1 troy oz = 31.1g; d of gold = 19.3g/cm^3)?

Explanation / Answer

(a) How many grams of gold are in the oceans?
Volume of water = area * depth
                         = 3.63E8 km2 * 3.8 km
                         = 1.3794E9 km3 * (10003m3/km3)* (1003cm3/m3) * (1 mL/cm3)
                         = 1.3794E24 mL (1.3794E24 L)
Mass of gold = concentration of gold * volume
                         = 5.8E-9 g/L * 1.3794E24 L
                         = 8.00052E12 g (8E9 kg)

(b) How many m^3 of gold are in the oceans?
Density of gold is 19.3 g/cm3 (from part (c))
Volume of gold = mass / density
                       = 8.00052E12 g / 19.3 g/cm3
                       = 4.145E11 cm3 * (1 m3/1003cm3)
                       = 4.145E5 m3 of gold

(c) Assuming the price of is $370.00/troy oz, what is the value ofgold in the oceans (1 troy oz = 31.1g; d of gold = 19.3g/cm^3)?
Price = ($370 / troy oz) * (1 troy oz / 31.1 g) * 8.00052E12 g
         = $9.5 x1013

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