Report your answer with the proper number of significant figures. magnesium is used in lightweight alloys for airplane bodies and other structures. the metal is obtained from seawater in a process that includes precipitation, neutralization, evaporation, and electrolysis. how many kilograms of magnesium can be obtained from 8.01 km3 of seawater if the initial mg2+concentration is 0.12% by mass? (d of seawater = 1.04 g/ml)

The density given is 1.04 g/mL, or also equivalent to:

density = 1040 kg/m^3

The total mass of seawater is calculated by multiplying density and volume. conversion factor is 1 km = 1000 m:

total mass = (1040 kg/m^3) * (8.01 km^3) * (1000 m / 1 km)^3

total mass = 8.3304 x 10^12 kg

Since Mg2+ is 0.12% by weight, in fraction this is 0.0012 by weight. So:

mass Mg2+ = (8.3304 x 10^12 kg) * 0.0012

mass Mg2+ = 9.996 x 10^9 kg

your answer would be a. 12.5 kg

the dependant variable is spilled water

the independent is the heat