Chemistry

$A_2{B}_3?2A^{+3}+3B^{-2}$

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$\therefore i=1+4\alpha =1+4*0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4*0.52*1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$

$\Delta U=-742.24kJ/mole\Delta H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $-742.24+\frac{1}{2}*\frac{8.314}{1000}*298$

$-742.24+1.238=-741.002\approx -741kJ/mol$

So, magnitude of $\Delta H$ = 741 kJ/mol.

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2*9}{\frac{10}{3}*2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24*10^{-2}=81*10^{-2}$

Gabriel phthalimide synthesis is used for 1° Aliphatic /alicyclic amine

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$

3.3 * 10^-4 = $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$  

  $t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$

so; the nearest integer is 26.

Coordination no. of an atom in BCC is 8.

(1) $C{r}^{+1}=4s^{0}3d^{5}M{n}^{+2}=4s^{0}3d^5$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(2) $N{i}^{+2}=4s^{0}3d^{8}C{u}^{+}=4s^{0}3d^{10}$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(3) $F{e}^{+2}=4s^{0}3d^{6}C{o}^{+}=4s^{1}3d^7$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(4) $V^{+2}=4s^{0}3d^{3}C{r}^{+}=4s^{o}3d^5$

Freundlich adsorption Isotherm

$\left(\frac{x}{m}\right)=k{\left(P\right)}^{\frac{1}{n}}$

At moderate pressure ; $\frac{x}{m}$ $\frac{}{}$ varies non-linearly with P.

So, $x=\frac{1}{n}$

$l^{-}+H_2{O}_2+2H^{+}\to l_2+2H_{2}O$

(-1) oxidation (0)

Here, I is reducing agent

${}_{}$ $\therefore H_2{O}_2$  behaves like oxidizing agent

In Ce & Eu (+3) oxidation state is more stable

$\therefore C{e}^{+4}+e^{-}\to C{e}^{+3}\left(Reduction\right)$ so CeO₂ is oxidising agent

$E{u}^{+2}\to E{u}^{+3}+e-\left(oxidation\right)$ so EuSO₄ is reducing agent