Ncert Solutions Chemistry Class 12th

Data is not sufficient but official answer is (D)

This is the formation of free radical mechanism, hence free radical will be formed

$P_4+3NaOH+3H_{2}O\to P{H}_3+3Na{H}_{2}P{O}_2$

$RedP+alkali\to \underset{\left(NoP-Hbond\right)}{H_4{P}_2{O}_6}$

According to Ellingham diagram, the metal oxide with lower $\Delta G^o$ is more stable than metal oxide of higher $\Delta G^o$ at same temperature.

$\frac{dE}{dT}$ is the temperature Co-efficient of cell. The cell having less variation of EMF, with respect to temperature have high efficiency.

Organic compound → AgBr (s)

  0.5 g                    0.40 g

Here; Moles of Br in organic compound = Moles of Br in Ag Br

= Moles of AgBr

=  $\frac{0.40}{188}mol$

Mass of Br in organic compound =  $\frac{0.4}{188}*80g$  

$\begin{array}{l}\mathrm{\%}ofBr=\frac{0.4*80}{188*0.5}*100\mathrm{\%}\\ =34\mathrm{\%}\end{array}$

$P_A^0=50torr$  

$P_B^0=100torr$  

Mole fraction of A in liquid phase,           x_A = 0.3

Mole fraction of B in liquid phase,            x_B = 0.7

Now;     $P_A=P_A^0{x}_A$  

$P_B=P_B^0{x}_B$

= 100 * 0.7 = 70 torr

Mole fraction of B in vapour phase,   $y_B=\frac{P_B}{P_A+P_B}=\frac{70}{85}$  

$\frac{70}{85}=\frac{x}{17}$  

                             X = 14

$\underset{\begin{array}{l}Sodiumextract\\ ofsulphur\end{array}}{N{a}_{2}S}+N{a}_2\underset{Sodiumnitroprusside}{\left[Fe{\left(CN\right)}_{5}NO\right]}\to N{a}_4\underset{violet/purple}{\left[Fe{\left(CN\right)}_{5}NOS\right]}$

Copolymer is the polymer formed by two or more monomers having multiple bonds.

Buna – S, PHBV and butadiene – styrene are copolymers. Neoprene is homopolymer.

$R-N{H}_2\underset{273K}{\overset{HN{O}_2\left(aq.\right)}{\to }}R-\stackrel{\oplus }{N_2}\underset{298K}{\overset{H_{2}O/-N_2\uparrow }{\to }}R-OH$

1° aliphatic amine           Diazonium salt                Alcohol