MCQ
In $C{H_3}C{H_2}Br,\,\,$$\% $ of $Br$ is
- A$80$
- ✓$75$
- C$70$
- D$7$
$ = \frac{{80}}{{109}} \times 100 = 73.39\,\% $ or approx. $ 75$ $\%$
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Product obtained in the above reaction is
$A{l_{\left( s \right)}}|A{l^{ + 3}}\left( {0.1\,M} \right)||F{e^{ + 2}}\left( {0.001\,M} \right)|Fe\left( s \right)$ If $E_{A{l^{ + 3}}/Al}^o = - 1.66\,V$ and $E_{Fe/F{e^{ + 2}}}^o = + 0.44\,V$


