I do not know where to begin in part a with the information givenin the problem.
ID: 685170 • Letter: I
Question
I do not know where to begin in part a with the information givenin the problem.Here is the problem from the text:
Earth's oceans have an average depth of 3800 m, a total area of3.63 *10^8 km^2, and an average concentration of dissolved gold of5.8 * 10^-9 g/L.
(a). How many grams of gold are there in the oceans?
(b). How many m^3 of gold are in the oceans?
(c). Assuming the price of gold is $370.00/troy oz., what is thevalue of gold in the oceans ( 1 troy oz= 31.1g; density ofgold= 19.3 g/cm^3)?
Explanation / Answer
The important things about this problem are 1) unit conversions 2) concentration = dissolved mass/volume 3) density = mass/volume so total ocean volume = depth* area = (3800 m)(3.83 x108 km2)=(3.8 km)(3.83 x 108km2)=1.46 x 109 km3 now we convert cubic kilometers to meters there are 1000 L in 1 cubic meter there are 1000 meters in 1 km so 1.46 E9km3*(1000m/km)3*1000L/m3=1.46 E21L then we multiply the concentration of gold times the volume of theocean to find hte mass of gold 5.8 x 10-9*1.46 x 1021 = 8.5 x1012 g of gold now to find the volume of GOLD we divide mass found above by density (given as 19.3g/cm3) to get volume of gold. you'll get an answer incm^3 which you can divide by (100)3 to get in m^3 the last one required similar steps, conversion andmultiplication!
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