- A$294$
- B$298$
- ✓$49$
- D$50$
No. of electrons lossed $= 12 -6 = 6$
Equivalent weight $=\frac{M}{6} = \frac{{294}}{6} = 49$ .
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$2NOCl\left( g \right) \rightleftharpoons 2NO\left( g \right) + C{l_2}\left( g \right)$
Given : $K_P = 8 \times 10^{12}\, atm$ at $500\, K$ use $R = 0.08\, L\, atm\, mol^{-1}\, K^{-1}$
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{{HBr}}A$
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{{HBr + ROOR}}B$
$\begin{array}{*{20}{c}}
{C{H_3} - CH - CH = C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\xrightarrow{{NBS}}C$


(Assume: Momentum is conserved when photon is absorbed. Use: Planck constant $=6.6 \times 10^{-34} J s$, Avogadro number $=6 \times 10^{23} mol ^{-1}$, Molar mass of $He =4 g mol ^{-1}$ )