- A$\sqrt {{K_{sp}}} $
- B$^3\sqrt {{K_{sp}}} $
- ✓$^3\sqrt {\frac{{{K_{sp}}}}{4}} $
- D$\sqrt {8{K_{sp}}} $
${K_{sp}}$ of $PbC{l_2} = [P{b^{2 + }}] \times {[C{l^ - }]^2}$ ;${K_{sp}} = S \times {(2S)^2}$
${K_{sp}} = S \times 4{S^2}$$ = 4{S^3}$; ${S^3} = \frac{{{K_{sp}}}}{4}$; $S = \sqrt[3]{{\frac{{{K_{sp}}}}{4}}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)\,\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,C{H_3}}
\end{array}}\limits_{(Neo - pen\tan e)\,(i)} $
$(ii\mathop {)\,\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}}\limits_{(Iso - pen\tan e)\,\,(ii)} $
$(iii)\,\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}}\limits_{(n - pen\tan e)} $
$B.E(N -N) = 159\, kJ\, mol^{-1}$,
$B.E(H -H) = 436\, kJ\, mol^{-1}$
$B.E.(N \equiv N) = 941\, kJ\, mol^{-1}$,
$B.E(N -H) = 398\, kJ\, mol^{-1}\, .......\,kJ\, mol^{-1}$
$CH_3-CH = CH -CH_2 -CH_2 -CH(CH_3)_2$
$(A)$ $(B)\,\,\,(C)$ $(D)$ $(E)\,\,\,(F)$
arrange them in decreasing order of reactivity towards free radical substitution
$k=A\,e-^{E/RT}$
In this equation, $E$ represents
$U\,\,\, 1s^2 \,2s^2 \,2p^3$
$V\,\,\, 1s^2\,2s^2 \,2p^6 \,3s^1$
$W\,\,\, 1s^2\,2s^2\, 2p^6\,3s^2\,3p^2$
$X\,\,\, 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^5\, 4s^2$
$Y\,\,\, 1s^2\, 2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\, 4s^2\, 4p^6$
Determine which sequence of elements satisfy the following statements :
$(i)$ Element forms a carbonate which is not decomposed by heating
$(ii)$ Element is most likely to form coloured ionic compounds
$(iii)$ Element has largest atomic radius
$(iv)$ Element forms only acidic oxide