Answer: [tex]6.3\times 10^7m^3[/tex]
Explanation:
Required amount of magnesium = [tex]1.00\times 10^5 tons[/tex]
Given : 1 ton = 2000 lb
[tex]1.00\times 10^5 tons=\frac{2000}{1}\times 1.00\times 10^5=2\times 10^8lb[/tex]
1 lb = 453.592 g
[tex]2\times 10^8 lb=\frac{453.592 }{1}\times 2\times 10^8 lb=907.184\times 10^8g[/tex]
Given : 1.4 g of magnesium is produced by 1000 g of sea water
[tex]907.184\times 10^8g[/tex] of magnesium is produced by =[tex]\frac{1000}{1.4}\times 907.184\times 10^8g=6.5\times 10^{13}[/tex] g of sea water
Density of sea water = 1.025 g/ml
Volume of sea water =[tex]\frac{\text {mass of sea water}}{\text {density of sea water}}=\frac{6.5\times 10^{11}g}{ 1.025g/ml}=6.3\times 10^{13}ml[/tex]
1 ml = [tex]10^{-6}m^3[/tex]
[tex]6.3\times 10^{13}ml=\frac{10^{-6}}{1}\times 6.3\times 10^{13}=6.3\times 10^7m^3[/tex]
Volume of seawater, in cubic meters is [tex]6.3\times 10^7[/tex]
