- A${H^ + }$
- B$O{H^ - }$
- COnly $NH_4^ + $
- ✓$O{H^ - },$ and $N{H_4}OH$ molecules
$N{H_3} + {H_2}O\, \to \,NH_4^ + + O{H^ - }$ $\rightleftharpoons$ $N{H_4}OH$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
hence $A$ $\xrightarrow{{conc.\,{H_2}S{O_4}}}$ $B,$ $B$ is :
$1.$ The compound ${X}$ is
$(A)$ $\mathrm{NaNO}_3$ $(B)$ $\mathrm{NaCl}$ $(C)$ $\mathrm{Na}_2 \mathrm{SO}_4$ $(D)$ $\mathrm{Na}_2 \mathrm{~S}$
$2.$ The compound ${Y}$ is
$(A)$ $\mathrm{MgCl}_2$ $(B)$ $\mathrm{FeCl}_2$ $(C)$ $\mathrm{FeCl}_3$ $(D)$ $\mathrm{ZnCl}_2$
$3.$ The compound ${Z}$ is
$(A)$ $\left.\mathrm{Mg}_2 \mid \mathrm{Fe}(\mathrm{CN})_6\right]$ $(B)$ $\operatorname{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$
$(C)$ $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$ $(D)$ $\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$
Give the answer question $1,2$ and $3.$