Chemistry

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having  manufacturing of N-based fertizers)

% of C in organic compound

$=\frac{12}{14}*\frac{WC{O}_2}{W.O.C.}*100$  

$=\frac{952.56}{21.648}=44\mathrm{\%}$  

$\mathrm{\%}ofH=\frac{2}{18}*\frac{W{H}_{2}O}{W.O.C.}*100$  

= $\frac{20}{18}*\frac{0.4428}{0.492}*100$  

=  $\frac{88.56}{8.856}=10\mathrm{\%}$  

= 100 – 54 = 46%

$\left[Cu{\left(en\right)}_2{\left(SCN\right)}_2\right]$

More stable isomers = 3 (trans isomers)

Invertase → Cane sugar to glucose and fructose

Zymase → Glucose to ethanol

Diastase → Starch to Maltose

Maltase → Maltose to glucose

In 4d orbital, n = 4 and $\mathcal{l}=2$

Radial nodes = $n-\mathcal{l}-1$

Radial nodes = 4 – 2 – 1 = 1

And angular nodes,  $\mathcal{l}=2$

$Mn{O}_4^{-2}\to A+B$  

Oxidation state of Mn in B < A 

$Mn{O}_4^{-2}+H^{+}\to Mn{O}_4^{-}+Mn{O}_2$  

B is MnO₂

$\therefore$  Oxidation state of Mn = +4

${\therefore }_{25}M{n}^{+4}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^3$  

  $\therefore$ unpaired electron = 3

Spin only magnetic moment  $\left(\mu \right)$  

$=\sqrt[]{n\left(n+2\right)}=\sqrt[]{3\left(3+2\right)}=\sqrt[]{15}$  

$=3.87\approx 4$  

$C\mathcal{l}{F}_3$ ->T-shaped (sp³d)

IF₇ ->Pentagonal bipyramidal (sp³d³)

BrF₅ -> Square pyramidal (sp³d²)

BrF₃ ->T-Shaped (sp³d)

I₂Cl₆ -> Triangular bipyramidal (sp³d)

$I{F}_5\to$  Square Pyramidal (sp³d²)

ClF -> (sp³)

ClF₅ -> Square Pyramidal (sp³d²)

Br₅, IF₅ & ClF₅ ® Square Pyramidal

t_1/2 = 0.301 min

t = 2 min

$K=\frac{2.303}{t}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{0.693}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{2.303*0.301}{0.301}=\frac{2.303}{2}\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\therefore 2=\mathcal{l}og\left(\frac{Co}{Ct}\right)$  

$\frac{Co}{Ct}=10^2=100$  

              Ans. 100

Ka for C₃H₇COOH = 2 * 10^-5

$pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$  

=5 – 0.3 = 4.7

pH of 0.2 (M) solution = 

$pH=\frac{pKa-\mathcal{l}ogC}{2}$  

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27*10^{-1}$     

Ans 27

$m=\frac{w*1000}{molecularwt*w_{solvent}}=\frac{10.2*1000}{176*150}$  

$=\frac{1020\cancel{0}}{176*15\cancel{0}}=0.386$  

  $\Delta T_f=K_f*m$

3.9 * 0.386

$\therefore x*10^{-1}=15.05*10^{-1}$  

Ans = 15