1.) If the volume of the oceans is 1.35 x 10^18 m^3 and the abundance of deuteri

Question

1.) If the volume of the oceans is 1.35 x 10^18 m^3 and the abundance of deuterium is 0.0156 atom%, how many kilograms of deuterium are present in the oceans?  (Neglect the presence of dissolved substances, and assumer that the density of water is 1.00 g/cm^3).

2.) Do the same calculation for tritium, assuming that the abundance of tritium is approximately 10^-6 atome %.

I have no idea where to even begin on these. Please show me step by step with correct answers so I can try to understand.

Thank you!!

Explanation / Answer

Density of water is 1.00 g/cm^3 or 1.00 X 10^6 g/m^3 or 1.00 X 10^3 kg/m^3

Number of grams of water 1.35 X 10^18 m^3 * 1.00 X 10^3 kg/m^3 = 1.35 X 10^21 kg.

Now calculate the number of grams of hydrogen/deuterium/tritium.

Water is H2O, AW(H/D/T) ~ 1.0 AW(O) = 16. Therefore the mass of hydrogen/deuterium/tritium will be 16/18 times the mass of water.

Mass Hydrogen/Deuterium/Tritium = (16/18 ) * 1.35 X 10^21 kg = 1.20 * 10^21.

Since the abundance of deuterium is 0.0156% then the mass of deuterium will be
m(De) = 1.2 ^ 10^21 * 0.0156/100 = 1.87 ^ 10^17 kg

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