Class 12th

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$  

3.3 * 10^-4 =   $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$

$t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$  

so; the nearest integer is 26.

Coordination no. of an atom in BCC is 8.

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$

&     $x^2+y^2=\frac{31}{4}...........\left(ii\right)$      

Equation of any tangent to (i) be y = mx +  $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$

OR 36m² + 16 = 31 + 31m²

=>m² = 3

Using $\lambda =\frac{h}{\sqrt[]{2qVm}}$  

${\lambda }_{Li}=\frac{h}{\sqrt[]{2*3e*V*8.3m}}$                 m = mass of proton

${\lambda }_p=\frac{h}{\sqrt[]{2*e*V*m}}$  

$\frac{{\lambda }_{Li}}{{\lambda }_p}=\sqrt[]{\frac{2eVm}{2*24.9eVm}}$            

= 0.2 = 2 * 10^-1

So; x = 2

The balanced equation is :

$S_8\left(s\right)+12O{H}^{-}\left(aq\right)\to 4S^{2-}\left(aq\right)+2S_2{O}_3^{2-}\left(aq\right)+6H_{2}O\left(\mathcal{l}\right)$           

Here ; a = 12

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

  $\Delta T_f=i{k}_f.m$           

  $0.5=i*1.86*\frac{9.45}{94.5*500}*1000$         

i = 1.344

Now, using $\alpha =\frac{i-1}{n-1}$  

n for ClCH₂COOH = 2

a =  $\frac{1.344-1}{2-1}=0.344$  

Using

$K_a=\frac{C{\alpha }^2}{1-\alpha }$  

$K_a=\frac{0.2*{\left(0.344\right)}^3}{0.66}=36*10^{-3}$       

So; x = 36

  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :      $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$

Cathode :  2Ag⁺(aq) + 2e^- ® 2Ag(s)

Zn(s) + 2Ag⁺(aq) ® Zn²⁺ (aq) + 2Ag(s)

$E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$    

 x = 147

Kindly consider the following figure

Fraction of molecules having enough energy to form product = $e^{-Ea/RT}$ ${}^{}$

Fraction of molecules having enough energy to form product = $e^{-\frac{80.9*10^3}{8.314*700}}$ ${}^{\frac{{}^{}}{}}$

$=e^{-13.8}\cong e^{-14}$

So, x = 14

$\Delta T_f=K_f*m*i$

0.93 = 1.86 * 1 * i

i = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ $i=1+\left(\frac{1}{n}-1\right)$ $\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$\therefore n=2$