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, no precipitate would form.
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