Question

A 46.2-mL, 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

Answer 1

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

Answer 2

Thanks, the answer's 1.09M

Answer 3

oops!

(.0262416 mol + .112378 mol) / (.0462 L + .0805 L) = M

M = 1.0941 and not 1.0094077