Question

Check my work Be sure to answer all parts. Earth's oceans have an average depth of 3800 m, a total area of 3.63 x 103 km', and an average concentration of dissolved gold of 5.8 x 10L. If a recent price of gold was $757.00/troy oz, what is the value of gold in the oceans? (I troy oz-31.1 g) Enter your answer in scientific notation.

Answer 1

--Average depth of an ocean = 3800 m = 3.8 x 10³ m

--Total surface area of ocean surface = 3.63 X 10³ km² [1km = 10² m,1km² = (10³m)² = 10⁶ m²]

= 3.63 X 10³ X 10⁶ m²

= 3.63 X 10⁹m²

--Therefore, Volume of the ocean water = Surface area of ocean X depth of the ocean

= (3.63 X 10⁹m² ) x (3.8 x 10³ m)

= 13.794 X 10¹²m³    [1L=0.001 m³ = 10^-3m³]

= 13.794 X 10¹⁵ L

= 1.3794 x 10¹⁶ L

-- Concentration of dissolved gold in ocean water = 5.8 X 10^-9 gL^-1

Amount of gold dissolved in 1.3794 x 10¹⁶ L of ocean water = (5.8 X 10^-9 gL^-1) X 1.3794 x 10¹⁶ L

= 8.00052 x 10⁷ g

-- Amount of gold interms of troy oz =

   ^{$= \frac{8.00052\times 10^{7}g}{31.1g/troy oz} = 0.2573 \times 10^{7} troy oz$}

= 2.573 X 10⁶ troy oz

--Recent prize of the gold is = $ 757.0/troy oz

The value of the gold present in ocean = (2.573 X 10⁶ troy oz) X ($ 757.0/troy oz)

= $ 1947.3934 X 10⁶

= $ 1.9474 X 10⁹

The value of the gold present in the ocean of depth 3.8 x 10³ m and surface area 3.63 X 10³ km² is

= $ 1.9474 X 10⁹