Ncert Solutions Chemistry Class 12th

Anhydride of HNO₃ produced by the treatment of P₄O₁₀ (anhydring reagent) in ratio of 4 : 1 is N₂O₅ which is acidic in nature. $4HN{O}_3+P_4{O}_{10}\to \underset{Acidicnature}{\underset{?}{2N_2{O}_5}}+{\left(HP{O}_3\right)}_4$

Calcinations is the decomposition in absence of air, $ZnC{O}_3\underset{\Delta }{\overset{}{\to }}ZnO+C{O}_2\uparrow$ ${}_{}\underset{}{\overset{}{}}{}_{}$

while roasting is the oxidation process, $2ZnS+3O_2\to 2ZnO+2S{O}_2\uparrow$

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