- A${C_6}{H_5}OH$
- B${C_6}{H_5}C{H_2}OH$
- C$C{H_3}C \equiv CH$
- ✓$C{H_3}NH_3^ + C{l^ - }$
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$\begin{array}{*{20}{c}}
{HO - C{H_2} - CH - CH = C - C{H_2} - C - C - OH} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||\,\,\,\,\,\,\,||} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O\,\,\,\,\,\,O\,}
\end{array}$
$H _2 O ( g ) \rightarrow H _2( g )+\frac{1}{2} O _2( g )$
The percent of water decomposing at $2300\,K$ and $1\,bar$ is $...........$ (Nearest integer).
Equilibrium constant for the reaction is $2 \times 10^{-3}$ at $2300\,K$

