MCQ
Carbylamine reaction is given by
- ✓${1^o}$ amine
- B${3^o}$ amine
- C${2^o}$ amine
- DQuarternary salts
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$(A)$ $P$ $(B)$ $Q$ $(C)$ $R$ $(D)$ $S$
The Compound $B$ is
$C_{(graphite)} + CO_{2(g)} \rightarrow 2CO _{(g)}$
are $170\, kJ$ and $170\, J K^{-1},$ respectively. This reaction will be spontaneous at ............ $\mathrm{K}$