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An aqueous solution containing $\ce{Ca^2+}$, $\ce{Cl^-}$, $\ce{CO3^2-}$, and $\ce{NO3^-}$ is allowed to evaporate. Which compound will precipitate first?
The correct answer is $\ce{CaCO3}$. I had guessed that, and I know nitrates are soluble, so that wouldn't precipitate, but why the $\ce{CO3^2-}$ and not the $\ce{Cl-}$? Is it because of the equal and opposite charges between them? Does electronegativity play a role?
This question is testing your knowledge of general solubility trends, which largely need to be memorized. Here is a fairly comprehensive list, and there are many others scattered through the internet.
You are correct that nearly all nitrates are soluble, so that would be a bad guess.
The halides are generally pretty soluble. There are more exceptions than for nitrates, but as a rule halide salts tend to be soluble.
Carbonates however are frequently insoluble. The Group I cations tend to be an exception, but not the Group II cations, like $\ce{Ca^2+}$.
So, the general trend in solubility for the three anions in your question is $\ce{NO3- > Cl- > CO3^2-}$, so you would expect $\ce{CaCO3}$ to precipitate first.
We can verify this against the known solubilities of each of the salts at room temperature:
$\begin{array}{rr} \pu{\ce{Ca(NO3)2}}: & \pu{1212 g/L }\\ \pu{\ce{CaCl2}}: & \pu{75g/L}\\ \pu{\ce{CaCO3}}: & \pu{0.01g/L} \end{array}$
So, the quantitative solubility data confirms the original prediction and $\ce{CaCO3}$ will clearly precipitate first.
