- A(2, 6, 6)
- B(4, 6, 4)
- C(2, 4, 8)
- D(2, 8, 4)
Explanation:
Electronic configuration of element with atomic number with 14 is 2, 8, 4
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${A_2}\left( g \right) + {B_2}\left( g \right) \rightleftharpoons 2AB\left( g \right)$
${\Delta _r}{G^o}$ and ${\Delta _r}{S^o}$ are $20\, kJ/mol$ and $-20\, JK^{-1}\, mol^{-1}$ respectively at $200\, K$.
If ${\Delta _r}{C_P}$ is $20\, JK^{-1}\, mol^{-1}$ then ${\Delta _r}{H^o}$ at $400\, K$ is.....$kJ/mol$

a $20\ L$ box containing $X_6$ and $X_3$ at equilibrium at $1500\ K$ . $K_p = 4 \times 10^{18}\ atm$ . Assuming ${P_{{x_3}}} > > {P_{{x_6}}}$ and total pressure is $10\ atm$ .The partial pressure of $X_6$ will be
$\begin{array}{*{20}{c}}
{C{H_3} - C = CH - C{H_2} - COOH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$