The bromine content of average ocean water is 65 parts by weight per million. Assuming 100% recovery, how many cubic metres of ocean water must be processed to produce 0.61 kg of bromine? Assume sea water is 1.0 * 10 3 kg/m 3.
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Bromine content is 65 parts per million parts of water. That means for every 65 parts of bromine is 1000000 parts of ocean water.
therefore if 65parts= 0.61kg
then 100000parts = 1000000 x 0.61kg
= 610000kg
density of of ocean water = 1000kg/m3
that means for every 1000kg is 1m3
therefore 610000kg will be 610000kg/1000kg/m3
= 610m3
Therefore the volume of ocean water used = 610 cubic meters
therefore if 65parts= 0.61kg
then 100000parts = 1000000 x 0.61kg
= 610000kg
density of of ocean water = 1000kg/m3
that means for every 1000kg is 1m3
therefore 610000kg will be 610000kg/1000kg/m3
= 610m3
Therefore the volume of ocean water used = 610 cubic meters
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