Answer:
C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]
Explanation:
In this question our options are:
A. [tex]Ca^+^2_(_a_q_)+NO_3^-_(_a_q_)->Ca(NO_3)_2_(_a_q_)[/tex]
B. [tex]Ag^2^+_(_a_q_)+2Cl^-_(_a_q_)->AgCl_2_(_s_)[/tex]
C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]
D. None of the above because no reaction occurs
We have to remember that the ions produced by [tex]AgNO_3[/tex] are:
[tex]Ag^+[/tex] and [tex]NO_3^-[/tex]
And the ions produced by [tex]CaCl_2[/tex] are:
[tex]Ca^+^2[/tex] and [tex]Cl^-[/tex]
Additionally, we will have a double displacement reaction so the compounds produce are:
[tex]AgCl[/tex] and [tex]Ca(NO_3)_2[/tex]
If we remember the solubility rules, all the nitrate salts are soluble and the salts made with silver are not soluble. With this in mind, we will have a solid-state for [tex]AgCl_(_s_)[/tex] and an aqueous state for [tex]Ca(NO_3)_2_(_a_q_)[/tex].
If this is true, the final answer can be B or C. The charge of Ag is +1 so the final answer is C.
I hope it helps!