Answer:
D because i did this last week and got it right.
The solubility equilibrium of
:
[tex] CaCrO_{4}(aq)<===>Ca^{2+}(aq) + CrO_{4}^{2-}(aq)\\
Q_{sp}=[Ca^{2+}][CrO_{4}^{2-}]\\
= (0.0200 M)(0.0300 M) \\
= 0.0006
Ksp (0.00071) > Qsp (0.0006). So, <u>no precipitate would form</u>.