The bromine content of the ocean is about 65 g of bromine per million g of sea water. how many cubic meters of ocean must be processed to recover 500 mg of bromine, if the density of sea water is 1.0x 103 kg/m3?

7.69 X 10^-6 cubic meters of sea water must be processed to recover 500 mg of bromine.

Step-by-step explanation:

First of all We need to work with consistent units, so let's convert mg of bromine to grams:

500 mg of bromine (divided by 1000) = 0,5g of bromine

So, to find the right answer to this problem we have to apply a direct rule of three:

If there are 65 g of bromine in 1,000,000 grams of sea water, How many grams of sea water is needed to recover 0,5 grams

65g (bromine)   1,000,000 g (sea water)

0,5g (bromine)  X

X = (0,5 * 1,000,000)/(65)

X = 7.69 g of sea water

But now we need to convert these 7.69 g of sea water to cubic meters.

Sea water density is 1.0 x 10^3 kg/m3 = 1,000 kg/m3

If we multiply this by 1000 to convert kg to grams, We will find that

the equivalent sea water density is: 1.0 x 10^6 g/m3

It means that each cubic meter of sea water weights 1,000,000 g.

So, If  one cubic meter of sea water equals to 1,000,000 g  of sea water

We need to apply another rule of three to find the final

1,000,000 g 1 m3

          7.69g  X

X=  (7.69* 1)/ 1,000,000

X= 7,69 x 10^-6 m3    

step-by-step explanation:

the answer to this question is 3/19

step-by-step explanation: