- A$N{H_3} + NaN{O_2}$
- ✓$N{H_4}Cl + NaN{O_2}$
- C${N_2}O + Cu$
- D${(N{H_4})_2}C{r_2}{O_7}$
$N{{H}_{4}}N{{O}_{2}}\,\xrightarrow{\text{heat}}\,\underset{(g)}{\mathop{{{N}_{2}}}}\,+\underset{(l)}{\mathop{2{{H}_{2}}O}}\,$
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Sucrose $\xrightarrow[{Cleavage\,\,(Hydrolysis)}]{{Gly\cos idic\,bond}}A + B\xrightarrow[{{\text{reagent}}}]{{{\text{Seliwanoff 's}}}}?$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_2} = CH - C{H_2} - C - H}
\end{array} \to C{H_3} - C{H_2} - C{H_2} - C{H_2}OH$

Given : $K _{ sp } Cu ( OH )_2=1 \times 10^{-20}$
$\operatorname{Take} \frac{2.303 RT }{ F }=0.059 \,V$
The reduction potential at $pH =14$ for the above couple is $(-) x \times 10^{-2}\,V$. The value of $x$ is $........$.