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.

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$        

$A^4\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$          

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=   $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$        

  $2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$          

So, b = 2

Hence b - a = 4

Freundlich adsorption Isotherm

$\left(\frac{x}{m}\right)=k{\left(P\right)}^{\frac{1}{n}}$

At moderate pressure ; $\frac{x}{m}$ $\frac{}{}$ varies non-linearly with P.

So, $x=\frac{1}{n}$

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

$Y-y=-\frac{dx}{dy}\left(X-x\right)$           

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$  

->(a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  $\left(4,-2\sqrt[]{2}\right)$  then

3a – 3b – c = 9       .(ii)

& 4a -  $2\sqrt[]{2}b$ - c = 12           .(iii)

also given   $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9

$l^{-}+H_2{O}_2+2H^{+}\to l_2+2H_{2}O$

(-1) oxidation (0)

Here, I is reducing agent

${}_{}$ $\therefore H_2{O}_2$  behaves like oxidizing agent

In Ce & Eu (+3) oxidation state is more stable

$\therefore C{e}^{+4}+e^{-}\to C{e}^{+3}\left(Reduction\right)$ so CeO₂ is oxidising agent

$E{u}^{+2}\to E{u}^{+3}+e-\left(oxidation\right)$ so EuSO₄ is reducing agent

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$  

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$  

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$                              

(i)   $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$  

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$  

Putting   $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence  $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = a lm, n

-> a = 1

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$           

So at least one root will lie in (-2, -1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$  

$=10\left[x^4+2x^3+3x^2+2x+1\right]$            

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$           

So, f(x) be purely increasing function so exactly one root of f(x) that will lie in (-2, 1). Hence |a| = 2

Kindly consider the following figure