Class 12th

$\left|adj\left(24A\right)\right|=\left|adj\left(3adj\left(2A\right)\right)\right|$

$\Rightarrow {\left|24A\right|}^2={\left|3adj\left(2A\right)\right|}^2$

$24^6{\left|A\right|}^2=3^6.{\left(2^3\right)}^4{\left|A\right|}^4$

${\left|A\right|}^2=\frac{24^6}{3^6.2^{12}}=\frac{2^{18}.3^6}{3^6.2^{12}}=2^6$

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

  $\Rightarrow \left(x^2+1+x\right)\left(x^2-x+1\right)=0$

  $x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$  

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$  

= 1 + 1 – 1 = 1

CoCl₃.NH₃ + AgNO₃ $\stackrel{}{\to }AgC\mathcal{l}\downarrow \left(2mol\right)$  

$\left[Co{\left(N{H}_3\right)}_{5}C\mathcal{l}\right]C{\mathcal{l}}_2+AgN{O}_3\stackrel{}{\to }AgC\mathcal{l}\downarrow \left(2mol\right)$  

x = 5

$M{n}^{2+}\stackrel{}{\to }{t}_{2g^{111}}{e_g}^{11}$

(Number of unpaired electron = 5)

${\mu }_S=\sqrt[]{35}=5.91\cong 6$

$K=A.e^{-Ea/RT}=\left(6.5*10^2\right)e^{-26000K/T}$

$\frac{Ea}{8.314}=26000$

Ea = 216.164kJ/mol $\cong$ $$ 216

H₂ (g) + Cu²⁺ (aq) $\stackrel{}{\to }$  2H⁺ (aq) + Cu (s)

  $\therefore E_{cell}=E_{cell}^0-\frac{2.303RT}{nf}logQ$

0.31 = 0.34 - $\frac{0.06}{2}log\frac{\left[H^{+}\right]}{C{u}^{2+}}$  

  $\left[C{u}^{2+}\right]=10^{-7}\left(?\left[H^{+}\right]=10^{-3}\right)$

x = 7

$f\left(x\right)=\frac{x-1}{x+1}\Rightarrow f\left(f\left(x\right)\right)=\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=-\frac{1}{x}$

$f^3\left(x\right)=-\frac{x+1}{x-1}\Rightarrow f^4\left(x\right)=\frac{\frac{x-1}{x+1}+1}{\frac{x-1}{x+1}-1}=x$

$So,f^6\left(6\right)+f^7\left(7\right)=f^2\left(6\right)+f^3\left(7\right)$

= $-\frac{1}{6}-\frac{7+1}{7-1}=-\frac{9}{6}=-\frac{3}{2}$

$\Delta T_b=i{K}_{b}m$

$\Delta T_f=i{K}_{f}m$

$\therefore \frac{4}{4}=\frac{K_b*1.5}{K_f*4.5}$

$\frac{K_b}{K_f}=3$

Kindly go through the image