milli mole of $\mathrm{HCl}=100 \times 0.1=10$ milli mole
milli mole of $\mathrm{NH}_{4} \mathrm{OH}=200 \times 0.1=20 \mathrm{mill}$ imole
$\mathrm{HCl}+\mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{NH}_{4} \mathrm{Cl}+\mathrm{H}_{2} \mathrm{O}^{-}$
$10 \quad \quad \quad 20\quad \quad \quad \quad -\quad \quad -$
$-\quad \quad \quad 10\quad \quad \quad \quad 10\quad \quad \quad -$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(a)$ Diborane is prepared by the oxidation of $\mathrm{NaBH}_{4}$ with $\mathrm{I}_{2}$.
$(b)$ Each boron atom is in sp $^{2}$ hybridized state.
$(c)$ Diborane has one bridged $3$ centre$-2-$electron bond.
$(d)$ Diborane is a planar molecule.
The option with correct statement(s) is :
$\mathop {\begin{array}{*{20}{c}}
{{\mkern 1mu} {\mkern 1mu} Ph}\\
|\\
{C{H_3} - C - OH}\\
|\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_2} - C{H_3}{\mkern 1mu} {\mkern 1mu} }
\end{array}}\limits_{{\rm{'D'}}} $ $\xrightarrow{{HI}}$ products
If the configuration of substrate is $D$, then configuration of products will be
$(I)\,\,Zn\,+\,Conc. HNO_3\,(hot) \to Zn(NO_3)_2\,+\,X\,+\,H_2O$
$(II)\,\,Zn\,+\,dil. HNO_3\,(Cold) \to Zn(NO_3)_2\,+\,Y\,+\,H_2O$
Compound $X$ and $Y$ are respectively