If 200mL of 0.60 M MgCl2 (aq) is added to 400mL of distilled water, what is the concentration of Mg2+(aq) in the resulting solution with the total volume of 600mL? A.0.20M. B.0.30M. C.0.60M D.1.2M

A.0.20M

Explanation:

c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

Final Volume, V2 = 200 + 400 = 600 mL

Initial Concentration, c1 = 0.60 M

Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

c2 = 0.60 * 200 / 600

c2 = 0.20M

mark him as !

i would say b

explanation:

but i think i would need more information than that