MCQ
The compound $HOC{H_2} - C{H_2}OH$ is
- AEthane glycol
- ✓Ethylene glycol
- CEthylidene alcohol
- DDimethyl alcohol
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$2N{O_2}\left( g \right) \rightleftharpoons {N_2}{O_4}\left( g \right)$ is
$\left( {R = \frac{{25}}{3}J/K.mol,\ln 2 = 0.7,\ln 3 = 1.1} \right)$
${C_6}{H_5} - COO - C{H_3}\mathop {\xrightarrow{{1.\,LiAl{H_4}}}}\limits_{2.\,{H_2}O} $
Reason : Water is polar and benzene is nonpolar.