$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
$\frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]}=\frac{\mathrm{Ka}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}$
$=\frac{3.8 \times 10^{-7}}{10^{-6}}=3.8 \times 10^{-1}$
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|
$A$ $(mol/l)$ |
$B$ $(mol/l)$ |
Rate |
| $0.05$ | $0.05$ | $1.2\times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4\times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2\times 10^{-3}$ |
$(i)\,\, Cl^-$ can give up an electron more easily than $F^-$
$(ii) \,\,Cl^-$ is a better reducing agent than $F^-$
$(iii)\,\, Cl^-$ is smaller in size than $F^-$
$(iv)\,\, F^-$ can be oxidized more readily than $Cl^-$
$\mathrm{E}_{\mathrm{n}}=$ Total energy, $\mathrm{K}_{\mathrm{n}}=$ Kinetic energy, $\mathrm{V}_{\mathrm{n}}=$ Potential energy , $\mathrm{r}_{\mathrm{n}}=$ Radius of $\mathrm{n}^{\text {th }}$ orbit
Match the following:
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{V}_{\mathrm{n}} / \mathrm{K}_{\mathrm{n}}=$ ? | $(P)$ $0$ |
| $(B)$ If radius of $n^{\text {th }}$ orbit $\propto E_n^x, x=$ ? | $(Q)$ $-1$ |
| $(C)$ Angular momentum in lowest orbital | $(R)$ $-2$ |
| $(D)$ $\frac{1}{\mathrm{r}^{\mathrm{n}}} \propto \mathrm{Z}^{\mathrm{y}}, \mathrm{y}=$ ? | $(S)$ $1$ |