
- A$CH_3-CH_2-CH_2-Cl \xrightarrow{{LiAl{H_4}}}$
- B$CH_3-CH_2-COOH \xrightarrow{{P + HI}}$
- ✓$C{H_3} - C \equiv CH\xrightarrow[{{H_2}O}]{{H{g^{ + 2}}}}$
- DAll

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. . . . . (Given that the value of solubility product of $A B \left( K _{ sp }\right)=2 \times 10^{-10}$ and the value of ionization constant of $H B \left( K _{ a }\right)=1 \times 10^{-8}$ )
$\Delta {H^o}\, = \, +234.1\,\,kJ$
$C(s)\, + {O_2}(g)\, \to \,\,C{O_2}(g)$
Use these equation and $\Delta H^o$ value to calculate $\Delta H^o$ for this reaction:
$4Fe(s) + 3O_2(g) \to 2Fe_2O_3(s)$
.....$kJ$
Reason : Pure water is weak electrolyte.
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,||} \\
{C{H_3}C{H_2} - O - C - C{H_2} - C - OH} \\
{||\,\,\,\,\,} \\
{O\,\,\,\,\,\,}
\end{array}$
$PCl_5 \longrightarrow PCl_4^+ + Cl^-$

(image) $\xrightarrow[{N{H_4}Cl}]{{Zn}}.....?$