Identify $(A)$ to $(E)$ and also explain the reactions involved.
$CuCO_3 \to CuO$ $\xrightarrow{\text{Heat with } CuS} (A)$ $\xrightarrow{\text{conc. } HNO_3} (B)$ $\xrightarrow{NH_{3(aq)}} (C) \text{ (Blue solution)}$
$CuCO_3 \to (D)$ $\xrightarrow{Ca(OH)_2} (E) \text{ (Milky)}$ $\xrightarrow{CO_2} Ca(HCO_3)_2 \text{ (Clear solution)}$

(N/A) The reactions are as follows:
$1. \text{ Decomposition of } CuCO_3: CuCO_3 \xrightarrow{\Delta} CuO + CO_2 \uparrow (D)$
$2. \text{ Self-reduction of copper: } 2CuO + CuS \xrightarrow{\Delta} 3Cu (A) + SO_2 \uparrow$
$3. \text{ Reaction with conc. } HNO_3: Cu (A) + 4HNO_3 \text{ (conc.)} \to Cu(NO_3)_2 (B) + 2NO_2 + 2H_2O$
$4. \text{ Formation of blue complex: } Cu^{2+} + 4NH_3 \to [Cu(NH_3)_4]^{2+} (C)$
$5. \text{ Test for } CO_2: CO_2 (D) + Ca(OH)_2 \to CaCO_3 (E) \text{ (Milky)} + H_2O$
$6. \text{ Formation of clear solution: } CaCO_3 (E) + CO_2 + H_2O \to Ca(HCO_3)_2 \text{ (Clear solution)}$
Therefore,the identities are:
$(A) = Cu$
$(B) = Cu(NO_3)_2$
$(C) = [Cu(NH_3)_4]^{2+}$
$(D) = CO_2$
$(E) = CaCO_3$