Ncert Solutions Chemistry Class 12th

Compound react with S_N¹ mechanism in presence of polar protic solvent, which follows carbocation forming path.

Hence correct order is.

In electrolytic reduction of Al₂O₃, cryolite (Na₃AlF₆) is used to increase conductivity & decrease melting point. Oxidation state of Al in cryolite (Na₃AlF₆) is (+3).

By increases in temperature absorption decreases, T₁ > T₂ means higher absorption at T₂ temperature.

Kindly consider the following Image.

One water molecule is associated with hydrogen bond.

Ans. = 1

Cl, Br & I form halic (V) acid i.e. HClO₃, HBrO₃ & HlO₃.

HgS, PbS, CuS, Sb₂S₃, As₂S₃, & CdS are given sulphides

              CdS, PbS, As₂S₃ & CuS are soluble in 50% HNO₃ but Sb₂S₃ & HgS are not soluble.

              Ans. = 4

Energy of emitted photon in 0.1 sec = 10^-4 J

$n*\frac{hc}{\lambda }=10^{-4}$

$\frac{n*6.63*10^{-34}*3*10^8}{1000*10^{-9}}=10^{-4}$

$n=5.02*10^{14}=50.2*10^{13}\approx 50*10^{13}$

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)*8.31*\frac{284}{3*10^{-3}}*10^{-5}bar=0.84287bar$

$P=84.28*10^{-2}bar$

$\approx 84*10^{-2}bar$