Answer:
[tex]CuCl_{2}(aq.)[/tex] will form a precipitate of insoluble [tex]CuCO_{3}[/tex] when aqueous [tex]Na_{2}CO_{3}[/tex] is added.
Explanation:
According to solubility rule-
all carbonates are insoluble except group IA compounds and [tex]NH_{4}^{+}[/tex]
all salts of sodium are soluble
When [tex]Na_{2}CO_{3}[/tex] is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.
So, in accordance with solubility rule, addition of [tex]Na_{2}CO_{3}[/tex] into [tex]CuCl_{2}(aq.)[/tex] will result precipitation of insoluble [tex]CuCO_{3}[/tex]
Reaction: [tex]Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)[/tex]