Answer:
Cl⁻, Na⁺, OH⁻
Explanation:
The titration is:
CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)
In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.
Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.
molaity= moles/L
molarity= 5.0 moles/ 10 L
molarity= 0.5
a)
A: Copper
B: CuO
C: 
D: $\mathrm{CuCO_3}$
E: $\mathrm{CO_2}$
F: $\mathrm{Cu(NO_3)_2}$
b)
$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$
c)
$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$
