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
By moles.
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).
To find the concentration of the resulting solution, you need to calculate the amount of copper (II) sulfate in the solution.
First, convert the volume of copper (II) sulfate given in milliliters (mL) to liters (L):
2.0 mL = 2.0 * 10^(-3) L
Next, calculate the amount of copper (II) sulfate in moles:
moles = concentration * volume
moles = 0.80 M * 2.0 * 10^(-3) L
Now, add the amount of copper (II) sulfate to the amount of water in the solution:
total volume = 2.0 * 10^(-3) L + 8.0 * 10^(-3) L
Finally, calculate the concentration of the resulting solution by dividing the moles of copper (II) sulfate by the total volume of the solution:
concentration = moles / total volume
By plugging in the values, you can calculate the concentration of the resulting solution.