- ✓$7$
- B$20$
- C$49$
- D$21$
${K_c} = \frac{{[{A_2}]\,\,[{B_2}]}}{{{{[AB]}^2}}}$
For reaction $AB$ $ \rightleftharpoons $ $\frac{1}{2}{A_2} + \frac{1}{2}{B_2}$
${{K}_{c}}^{'}=\frac{{{[{{A}_{2}}]}^{{1}/{2}\;}}{{[{{B}_{2}}]}^{{1}/{2}\;}}}{[AB]}$; ${{K}_{c}}^{'} =\sqrt{{{K}_{c}}} =\sqrt{49} =7$
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$4 HNO _{3}(l)+3 KCl ( s ) \rightarrow Cl _{2}( g )+ NOCl ( g )+ 2 H _{2} O ( g )+3 KNO _{3}( s )$
The amount of $HNO _{3}$ required to produce $110.0 \;g$ of $KNO _{3}$ is $...... \;g$
(Given : Atomic masses of $H , O , N$ and $K$ are $1 , 16,14$ and $39$ respectively.)
$(A)$ $\Delta G$ is positive $(B)$ $\Delta S _{\text {system }}$ is positive
$(C)$ $\Delta S _{\text {surroundings }}=0$ $(D)$ $\Delta H =0$
$CH \equiv CH\xrightarrow[{(excess)}]{{NaN{H_2}/liq.N{H_3}}}A\xrightarrow{{2C{H_3} - I}}B\xrightarrow[{quinolene}]{{{H_2}/Pd - BaS{O_4}}}C$
$C\xrightarrow[{quinolene}]{{{H_2}/Pd - BaS{O_4}}}\xrightarrow[{liq.\,N{H_3}}]{{NaN{H_2}}}D$