posted by roger .
If 2.0 mL of 0.80M copper (II) sulfate and 8.0 mL of water are mixed, what is the concentration of the resulting solution?
0.80M x (2.0 mL/10.0 mL) = ?M assuming volumes are additive.
i think it might be
0.002 L CuSO4 (0.80 mol/L) = 0.0016 mol CuSO4/0.008 L = 0.02 mols/L CuSO4
someone should double check that thou
i am goining crazy please me out part 1 of 2 determine for further reading
Rose Bud is in error. But we can calculate it by moles. First my answer of
0.8m x (2/10) = 0.16M
0.002L x 0.80M = 0.0016 moles.
M = moles/L = 0.0016/0.01 = 0.16M
The total volume is 10 mL(0.010L) and not 8 mL(0.008L).