MCQ
Compound $A$ and $B$, both were treated with $NaOH,$ producing a single compound $C$.


- ✓

- B

- C

- D








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${K_p} = 8 \times {10^{ - 2}}$$C{O_{2(g)}} + {C_{(s)}} \to 2C{O_{(g)}}$ ; ${K_p} = 2$
$CaC{{O}_{3(s)}}\xrightarrow{\Delta }Ca{{O}_{(s)}}+C{{O}_{2}}\uparrow $; ${{K}_{p}}=8\times {{10}^{-2}}$

${{C}_{2}}{{H}_{5}}N{{H}_{2}} \xrightarrow{HN{{O}_{2}}}A \xrightarrow{PC{{l}_{5}}}B \xrightarrow{H.N{{H}_{2}}}C$