The bromine content of average ocean water os 65 parts per million. Assuming 100 percent recovery, how many cubic meters of ocean water must be processed to produce .75 kg of bromine? Assume the density of seawater is 1.023x10^3 kg/m^3.
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Density of seawater = 1, 023 kg/m³
Mass of 1 m³ volume = 1, 023 kg
Composition by mass (weight) of Br in 1 m³ = 1, 023 * 65/1000,000 kg
= 0.066495 kg
Volume of ocean water needed to produce 0.75 kg of Br
= 0.75 / 0.066495 m³
= 11.279 m³
Mass of 1 m³ volume = 1, 023 kg
Composition by mass (weight) of Br in 1 m³ = 1, 023 * 65/1000,000 kg
= 0.066495 kg
Volume of ocean water needed to produce 0.75 kg of Br
= 0.75 / 0.066495 m³
= 11.279 m³
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