when a solution of sodium chloride is added to a solution of copper (ii) nitrate, no precipitate is observed. Write the molexular equation that describes this behavior,

1.

Cu (NO3) 2 + 2NaCl (aq) - -> CuCl2 (aq) + 2NaNO3 (aq)

2.

Cu (NO3) 2 + 2NaOH (aq) - -> Cu (OH) 2 (s) + 2NaNO3 (aq)

A light blue precipitate of Cu (OH) 2 is formed and NaNO3 in solution.

3.

Cu (NO3) 2 (aq) - -> Cu2 + (aq) + 2NO3^-2 (aq)

2NaOH (aq) - -> 2Na + (aq) + 2OH - (aq)

Cu2 + (aq) + 2OH - (aq) - -> Cu (OH) 2 (aq)

2Na + (aq) + 2NO3^-2 (aq) - -> 2NaNO3 (aq)

4.

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.