Ncert Solutions Chemistry Class 12th

Bakelite is cross linked polymer of formaldehyde and novolac. Novolac is linear polymer of phenol-formaldehyde.

Kindly consider the following figure

Mechanism path is free radical.

Conductivity;    $\kappa =1.07*10^{6}S{m}^{-1}$

Resistance ; R = 0.243    $\Omega$

cell constant ; G^* =?

Using ;

$\kappa =\frac{G^{*}}{R}$      

$G*=\kappa *R$          

$=1.07*10^6*0.243m^{-1}$                

$=26*10^4{m}^{-1}$              

$Cl{O}_2^{-},C{l}_2,M{n}^{3+}$ can show disproportionation reaction while $Mn{O}_4^{-}$ cannot show disproportionation reaction is Mn is in +7 oxidation state.

Kindly consider the following figure

Power = 50 watt = 50 J/sec

Energy emitted per second = 50 J

Wavelength ; $\lambda$  = 795 nm

Energy of one photon =   

$=0.025*10^{-17}J$             

Number of photons emitted per second = $\frac{50}{0.025*10^{-17}}$  

= 2 * 10²⁰

Kindly consider the following figure

$M{n}^{+2}\to groupIV\left(M{n}^{+2},C{o}^{+2},Z{n}^{+2},N{i}^{2+}\right)$

$A{s}^{+3}\to groupIIB\left(A{s}^{+3},A{s}^{+5},S{b}^{+3},S{b}^{+5},S{n}^{+2},S{n}^{+4}\right)$

$\begin{array}{l}C{u}^{+2}\to groupIIA\left(C{u}^{+2},P{b}^{+2},H{g}^{+2},C{d}^{+2},B{i}^{+3}\right)\\ A{\mathcal{l}}^{+3}\to groupIII\left(F{e}^{+3},A{\mathcal{l}}^{+3},C{r}^{+3}\right)\end{array}$