Chemistry

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

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

Change in oxidation number = 2

$\Delta G°=-n{E}_{cell}^{o}F$

$\Delta G°=-n{E}_{cell}^{o}F$

=-2 * 4.315 * 96487 Jmol^-1

^{$\therefore \Delta G°=\Delta H°-T\Delta S°$}

^{$\Delta S°=\frac{\Delta H°-\Delta G°}{T}$}

$=\frac{-825.2*1000-\left(-2*4.315*96487\right)}{298.15}J{K}^{-1}$

$=\frac{7482.81}{298.15}=25.09J{K}^{-1}$

$\approx$ 25.1 JK^-1

Ans. = 25

Molarity (M) = $\frac{w*1000}{molecularmass*volumeofsolution\left(ml\right)}$

= $\frac{6.3*1000}{126*250}=\frac{4}{20}=0.2M$

Molecular mass of oxalic acid $\left(H_2{C}_2{O}_4.2H_{2}O\right)$

= 1 * 2 + 12 * 2 + 16 * 4 + 2 * 18

              = 26 + 64 + 36 = 126

              M = 2 * 10^-1 M

              = 20 * 10^-2M

$\therefore x*10^{-2}=20*10^{-2}$

$\therefore x=20$

Ans. = 20

$A_3{B}_{2\left(s\right)}?3A_{\left(aq\right)}^{+2}+2B_{\left(aq\right)}^{3-}$

Solubility $\frac{x}{M}$ $\frac{3x}{M}$ $\frac{x}{M}$

$\therefore Ksp={\left[A^{+2}\right]}^3{\left[B^{-3}\right]}^2$

$={\left(\frac{3x}{M}\right)}^3{\left(\frac{2x}{M}\right)}^2$

$=108{\left(\frac{x}{M}\right)}^5$

$Ksp=a{\left(\frac{x}{M}\right)}^5=108{\left(\frac{x}{M}\right)}^5$

Ans. $\therefore$ a = 108

Molality (m) = $\frac{\left(40/180\right)}{0.2}=\left(\frac{10}{9}\right)molal$

$\Delta T_f=T_f-T_f^{\text{'}}=1.86*\frac{10}{9}K$

$T_f^{\text{'}}=273.15-1.86*\frac{10}{9}K$

$=271.08K\approx 271K$

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

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

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

$\approx 84*10^{-2}bar$  

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

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

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

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