- A

- B

- C

- ✓







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$(A)$ $q=0$ $(B)$ $T_2=T_1$ $(C)$ $P_2 V_2=P_1 V_1$ $(D)$ $P_2 V_2^\gamma=P_1 V_1^\gamma$
$C_2H_4(g) + H_2(g) \to C_2H_6(g)$
| Bond | Bond energy $(kJ)$ |
| $C-H$ | $413$ |
| $C-C$ | $348$ |
| $C=C$ | $610$ |
| $H-H$ | $436$ |
(Round off to the Nearest Integer)
$\mathrm{NaCl}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{Conc} .) \rightarrow(\mathrm{A})+$ Side products
$(\mathrm{A})+\mathrm{NaOH} \rightarrow(\mathrm{B})+$ side product
$(\mathrm{B})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{dilute})+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow(\mathrm{C})+$ Side product
The sum of the total number of atoms in one molecule each of $(A), (B)$ and $(C)$ is
$3 s^{2}$; $3 s^{2} 3 p^{1}$; $3 s ^{2} 3 p ^{3}$; $3 s^{2} 3 p^{4}$
The correct order of first ionization enthalpy for them is.