Which calculation set-up would be correct to determine the volume of concentrated hydrochloric acid (12.0 M HCl) that is required to make 500.0 milliliters of a 6.0 M HCl?
12.0 M 0.500 l = 6.0 M V2
12.0 M 6.0 M = 0.500 l V2
12.0 M V1 = 6.0 M 0.500 l
6.0 M V1 = 12.0 M 0.500 l
(12.0 M + 6.0 M) = 0.500 l * V2
12.0 M 0.500 l = 6.0 M V2
12.0 M 6.0 M = 0.500 l V2
12.0 M V1 = 6.0 M 0.500 l
6.0 M V1 = 12.0 M 0.500 l
(12.0 M + 6.0 M) = 0.500 l * V2