Calabasas HS
posted by Michael .
A 46.2mL, 0.568 M calcium nitrate [Ca(NO3)2] solution is mixed with 80.5 mL of1.396 M calcium nitrate solution. calculate the concentration of the final solution

.

Not 100% sure if this is right, but here's what I think it is.
M = mol/L
.568 = mol/.0462 L
= .0262416 mol
1.396 = mol/.0805 L
= .112378 mol
(.0262416 mol + .112378 mol) / (.0462 L + .0805 L) = M
M = 1.0094077348 
Thanks, the answer's 1.09M

oops!
(.0262416 mol + .112378 mol) / (.0462 L + .0805 L) = M
M = 1.0941 and not 1.0094077