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)

First, convert 8.01 km^3 to L 

(1) 1 km^3 = (10^3 m)^3 = 10^9 m^3 

1 m^3 = 10^3 L 

10^9 m^3 = (10^9 m^3)*(10^3 L/m^3) = 10^12 L 

So 8.01 km^3 = 8.01 x 10^12 L 

(2) calculate the mass of that volume of seawater (1.04 g/mL = 1.04 kg/L) 

(8.01 x 10^12 L)*( 1.04 kg/L) = 8.3304 x 10^12 kg 

(3) calculate the Mg in that mass of water 

(8.3304 x 10^12 kg)*(0.0012) = 9.99648 x 10^10 kg of Mg

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.