Class 12th

2x + 3y + 2z = 9 . (i)

3x + 2y + 2z = 9     . (ii)

x – y + 4z = 8          . (iii)

(ii) – (i) ⇒ x = y

Then (iii) ⇒ z = 2

(i) ⇒ 5x + 4 = 9 ⇒ (x = 1, y = 1, z = 2) unique solution

(1) $C{r}^{+1}=4s^{0}3d^{5}M{n}^{+2}=4s^{0}3d^5$  

(2)    $N{i}^{+2}=4s^{0}3d^{8}C{u}^{+}=4s^{0}3d^{10}$

(3) $F{e}^{+2}=4s^{0}3d^{6}C{o}^{+}=4s^{1}3d^7$  

(4)  $V^{+2}=4s^{0}3d^{3}C{r}^{+}=4s^{o}3d^5$

Freundlich adsorption Isotherm

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

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

So, $x=\frac{1}{n}$  

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

  (-1)                 oxidation       (0)

Here, I^- is reducing agent

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

Cu⁺ = 3d⁹ 4s⁰

n = 1

$\mu =\sqrt[]{1*\left(1+2\right)}=\sqrt[]{3}=1.73BM$

the nearest integer is 2.

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

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

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

$AB?A^{+}+B^{-}$

1-x x x   $\therefore i=1+x$

$\begin{array}{l}\Delta T_b=i*k_b*m\\ 2.5=\left(1+x\right)*0.52m\end{array}$

$\therefore molality=\frac{2.5}{1.75*0.52}=2.74$

the nearest integer is 3.

Density = $\frac{M*z}{N_A{a}^3}$

$d=\frac{63.5*4}{6.02*10^{23}*{\left(3.596*10^{-10}\right)}^3*1000}=9076kg/m^3$

$log\left(\frac{k_2}{k_1}\right)=\frac{Ea}{2.303*8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

log (5) = $\frac{Ea}{2.303*8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

$\therefore Ea=\frac{0.7*2.303*8.314*300*325}{25}=52271.7$

Ea in kJ/mole = $\frac{52271.7}{1000}=52.2kJ/mol$

the nearest integer is 52.