Chemistry

Wt of Cl^- in 100 ml = 1.8 * 10^-3 gm

Mol. of Cl^- in 100 ml = $\frac{1.8*10^{-3}}{35.5}=0.0507*10^{-3}mole$  

$\therefore \left[C{l}^{-}\right]=\frac{0.0507*10^{-3}*1000}{100}=5.07*10^{-4}M$   

i.e. 0.507 milli mole in one lit required in one hr.

Coagulation value = (millimole/lit) required in one hr = 0.507

= 1 (the nearest integer)

Here, $\lambda =\frac{h}{\sqrt[]{2*m*eV}}=\frac{6.63*10^{-34}}{\sqrt[]{2*1.6*10^{-19}*40000*9.1*10^{-31}}}m$  

$\therefore \lambda =0.614*10^{-11}m$  

$\lambda =6.14*10^{-12}m$ [the nearest integer = 6]

From   $\Delta H=\Delta G+T\Delta S\therefore \Delta S=\frac{\Delta H-\Delta G}{T}$

$\therefore$ $\Delta S=\left[\frac{51.4-\left(-49.4\right)}{300}\right]*1000J{K}^{-1}mo{l}^{-1}=336J{K}^{-1}mo{l}^{-1}$

0.15 gm (organic compounds) $\underset{}{\overset{AgN{O}_3}{\to }}AgBr\left(s\right)$

$\downarrow$                                                                    

0.2397 gm

Weight of Br in AgBr = $\left(\frac{80}{188}*0.2397\right)=0.102gm$

% Br in compound = $\frac{0.102*100}{0.15}=68\mathrm{\%}$

Total wt = 4gm = W_NaOH + $W_{N{a}_{2}C{O}_3}$  

Let us suppose moles are 'm' for each.

4 = 40m + 106m

$\therefore m=\left(\frac{4}{146}\right)$ moles of NaOH and Na₂CO₃ each .

$\therefore$ Mass of NaOH (x gm) = $\frac{4}{146}*40=1.095\approx 1.0$

Due to the larger acid dissociation constant (Ka) sulphuric acid acts as an acid and HNO₃ acts as a base.

$\left[C{l}^{-}\right]=10^{-1}M$

$\left[Cr{O_4}^{--}\right]=10^{-3}M$

for AgCl ppt¹, ${\left[A{g}^{+}\right]}_{req}=\frac{1.7*10^{-10}}{10^{-1}}M$  

For Ag₂CrO₄ppt², ${\left[A{g}^{+}\right]}_{req}=\sqrt[]{\frac{1.9*10^{-12}}{10^{-3}}}M$

${\left[A{g}^{+}\right]}_{req}=4.3*10^{-5}M$

Being lower concentration of [Ag⁺] in case of AgCl, it will precipitate first.

Kindly consider the following figure

Regarding industrial consumption of H₂ manufacturing of NH₃ is the largest applications.