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

supongo que te refieres al paralaje estelar (observable).

si es así, entonces se puede inferir que la estrella está a más de 200 parsecs.

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year