
- ✓$H^+$
- B${H^\circleddash }$
- C${H^ \bullet }$
- D${H^{ - 2}}$


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$\mathrm{HF}, \mathrm{H}_2 \mathrm{O}, \mathrm{SO}_2, \mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{NH}_3, \mathrm{HCl}, \mathrm{CHCl}_3, \mathrm{BF}_3$
$Ca(OH)_2 + H_3PO_4\to CaHPO_4 + 2H_2O$ is
| | A | B |
| (a) | $\text{H}_2\text{C}=\text{CH}_2$ | $\text{CH}\equiv\text{CH}$ |
| (b) | $\text{CH}_2=\text{CHBr}$ | $\text{CH}\equiv\text{CH}$ |
| (c) | $\text{CH}_2=\text{CHBr}$ | $\text{CH}_2=\text{CH}_2$ |
| (d) | $\text{CH}_2=\text{CH}_2$ | $\text{CH}\equiv\text{CBr}$ |
$A.$ $KF > KI ; LiF > KF$
$B.$ $KF < KI ; LiF > KF$
$C.$ $SnCl _4 > SnCl _2 ; CuCl > NaCl$
$D.$ $LiF > KF ; \quad CuCl < NaCl$
$E.$ $KF < KI ; CuCl > NaCl$
$2A{B_3}(g) \rightleftharpoons {A_2}(g) + 3{B_2}(g)$. At equilibrium, $2\, mol$ of $A_2$ are found to be present. The equilibrium constant of this reaction is