The bromine content of the ocean is about 65 grams of bromine per million grams of sea water. How many cubic

meters of ocean must be processed to recover 1.0 pounds of bromine if the density of sea water is 1.0 x 103 kg/m3?

what the heck are you even doing

I got 6.99 m^3. It's

1 lb Br/1
1,000,000 lb Water/65 lb Br
1 kg Water/2.2 lb Water
m^3 Water/1000 kg Water
(Multiply (1)*(1,000,000)*(1)*(m^3))
&
(Multiply (1)*(65)*(2.2)*(1,000))
Take the first answer (1,000,000 m^3)
and divide by the second answer (143,000)
to get 6.99 m^3. Enjoy!

65 ppm is 65 grams/10^6 g sea water. We want 1 lb of Br2 which is about 454 grams. So how much sea water do we need to do that?

That's 1E6 g sea water x (454/65) = about 7E6 grams. I would change that to 7E3 kg and use the density.
volume = mass/density = 7E3/1E3 = aboaut 7 cubic meters. You need to go through and clean up the numbers because I've estimated here and there although I think the 7 is close. Check my thinking.

I got 67.75m cubed.

To solve this problem, we need to convert the given information into consistent units, and then use the density of seawater to determine the volume of ocean water that needs to be processed.

First, let's convert the given bromine content from grams per million grams of seawater to grams per cubic meter (g/m³).

Since there are 1 million grams in 1 metric ton (1000 kg), we can calculate the bromine content in g/m³ as follows:

(65 g / 1,000,000 g) * (1,000,000 g / 1 metric ton) = 65 / 1,000,000 metric tons

Next, let's convert the density of seawater from kilograms per cubic meter (kg/m³) to g/m³:

1.0 x 10³ kg/m³ = 1.0 x 10³ * 1000 g/m³ = 1.0 x 10⁶ g/m³

Now we have the bromine content in g/m³ and the density of seawater in g/m³. We can use these values to determine the volume of ocean water needed to recover 1.0 pound of bromine.

1 pound is equal to 453.592 grams. Therefore, 1.0 pound of bromine is equal to 453.592 grams.

Let V be the volume of ocean water needed in cubic meters.

(65 / 1,000,000) g/m³ * V m³ = 453.592 g

Solving for V:

V = 453.592 g / (65 / 1,000,000) g/m³

V ≈ 6,981.48 m³

Therefore, approximately 6,981.48 cubic meters of ocean water must be processed to recover 1.0 pound of bromine.