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#. How many grams of gold are in the oceans?

Answer 1

Over 8 million kilograms; i.e. #8xx10^9*g#

We need to find the volume of the ocean in #m^3#, and then multiply this volume by the average concentration in #g*L^-1# knowing that there are #1000 *L# in a #m^3#.
#"Volume of the oceans"# #=# #3.63xx10^11*m^2xx3800*m# #=# #1.38xx10^15*m^3#.
#"Mass of gold"# #=# #"Volume "xx" concentration"#
#=# #1.38xx10^15*cancel(m^3)xx5.8xx10^-9*g*cancel(L^-1)xx1000*cancelL*cancel(m^-3)#
#=# #8004000000# #g#
#=# #8004000*kg#

Nevertheless, carefully review my figures; a lot of math is involved.

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Answer 2

The Earth's oceans contain approximately 1.05 x 10^14 grams of gold.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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