Chemistry

According to Faraday law

  $\mathrm{W}=\mathrm{Z}\mathrm{I}\mathrm{t}*\eta \mathrm{}$ Where $\eta =$ efficiency

or $W=\frac{E}{96500}*1*t*\eta$

Putting values

$.104=\frac{(52/3)*2*8*60*\eta }{96500}\Rightarrow \eta =0.6$

So percentage efficiency $=60\mathrm{\%}$

For isotonic solution

${\mathrm{i}}_1{\mathrm{C}}_1={\mathrm{i}}_2{\mathrm{C}}_2\mathrm{}$ ffor protein $\left.\mathrm{i}=1\right\}$

${\mathrm{C}}_1={\mathrm{C}}_2$

$\frac{0.73*1000}{M_A*250}=\frac{1.65}{M_B*1}$

$\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}=\frac{0.73*4}{1.65}=1.77=177*{10}^{-2}$

Asp - Glu - Lys pripeptide is:

No. of $\mathrm{C}\mathrm{O}$ group $=5$

Phosphinic acid is hypo phosphorous acid $\left({\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2\right)$ .

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+{\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2?{\mathrm{N}\mathrm{a}\mathrm{H}}_2{\mathrm{P}\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

For neutrization

${\left({\mathrm{N}}_1{\mathrm{V}}_1\right)}_{\text{acid}}={\left({\mathrm{N}}_2{\mathrm{V}}_2\right)}_{\text{base}}$

$0.1*10=0.1*{\left({\mathrm{V}}_{\mathrm{m}\mathrm{L}}\right)}_{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}$

For a given reaction, rate $=-\frac{1}{2}\frac{d{n}_A}{dt}=-\frac{1}{3}\frac{d{n}_B}{dt}=-\frac{2}{3}\frac{d{n}_C}{dt}$

$\text{rate}=\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{A}}}{\mathrm{d}\mathrm{t}}=\frac{2}{3}\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{B}}}{\mathrm{d}\mathrm{t}}=\frac{4}{3}\frac{{\mathrm{d}\mathrm{n}}_{\mathrm{C}}}{\mathrm{d}\mathrm{t}}$

Due to inter molecular $\mathrm{H}$ -Bonding in $\mathrm{B}$ , than $\mathrm{A},\mathrm{B}$  is more soluble and having more B.P point than A.

Charge / radius ratio of $\mathrm{B}\mathrm{e}$ and $\mathrm{A}\mathrm{l}$ is same because of diagonal relationship. Remaining statements are correct.

$P_{\text{gas}}=\frac{{\mathbf{n}}_{\text{gas}}RT}{V}$

as $n,T$ & $V$  constant So

${\mathrm{P}}_{{\mathrm{H}}_2}={\mathrm{P}}_{{\mathrm{O}}_2}={\mathrm{P}}_{{\mathrm{H}}_{\theta }}=2\mathrm{a}\mathrm{t}\mathrm{m}$

So, ${\mathrm{P}}_{\text{Total}}={\mathrm{P}}_{{\mathrm{H}}_2}+{\mathrm{P}}_{{\mathrm{O}}_2}+{\mathrm{P}}_{\mathrm{H}\mathrm{e}}=6\mathrm{a}\mathrm{t}\mathrm{m}$