94.7k views
0 votes
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?

2 Answers

6 votes

Final answer:

To recover 500 mg of bromine from the ocean, one must process approximately 7.69231 liters or 7.69231 x 10^-3 cubic meters of sea water, considering the bromine content and the density of sea water.

Step-by-step explanation:

The student asked how many cubic meters of ocean must be processed to recover 500 mg of bromine, given that the ocean contains about 65 g of bromine per million g of sea water and the density of sea water is 1.0x 103 kg/m3.

To find the volume of ocean water needed to extract 500 mg (0.5 g) of bromine, we use the proportion:

  • 65 g bromine/1,000,000 g sea water = 0.5 g bromine/x g sea water

Solving for x gives us:

  • x = (0.5 g bromine * 1,000,000 g sea water) / 65 g bromine
  • x ≈ 7692.31 g of sea water

To convert grams of sea water to cubic meters, we use the density of sea water:

  • density = mass/volume, which rearranges to volume = mass/density
  • volume = 7692.31 g sea water / (1.0 x 103 kg/m3)
  • Since 1 kg = 1000 g, we have volume ≈ 7692.31 g / (1.0 x 106 g/m3)
  • volume ≈ 7.69231 x 10-3 m3 or 7.69231 liters

Therefore, to recover 500 mg of bromine, approximately 7.69231 liters or 7.69231 x 10-3 cubic meters of ocean water would need to be processed.

User Sergey Teplyakov
by
8.7k points
3 votes

Answer:

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 answer:

1,000,000 g __________ 1 m3

7.69g __________ X

X= (7.69* 1)/ 1,000,000

X= 7,69 x 10^-6 m3

User Ellison
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories