Class 12th

Primary structure of protein in unaffected by physical or chemical changes.

Most basic oxide V₂O₃

Here V has +3 O.S. Hence V⁺³ -> $\left[Ar\right]3d^2$  

two unpaired e^- in d- subshell

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

  $=\left(l_L+l_C\right)$  

Therefore current through R circuit at resonance will be zero

 

Kindly consider the following figure

Volume of H₂ adsorbed = $\frac{nRT}{P}=\frac{2*0.083*300}{2*1}=24.9lit=24900ml$  

Therefore volume of gas adsorbed per gram of the adsorbent =  $\frac{24900}{2.5}=9960$  

Aniline show acid-base reaction with AlCl₃

aniline is a Lewis base while AlCl₃ acts as lewis acid.

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100}*t=2.303lo{g}_{10}\frac{A_0}{A_t}$ ………….(i)

and       $\frac{0.693}{50}*t=2.303lo{g}_{10}\frac{B_0}{B_t}$              ………….(ii)

from equation (i) and (ii)

$0.693t\left[\frac{1}{50}-\frac{1}{100}\right]=\left[log\frac{B_0}{B_t}-log\frac{A_0}{A_t}\right]*2.303$  

Here; A₀ = B₀ and $A_t=4*B_t$  

Therefore  $0.693t\left[\frac{1}{100}\right]=2.303\left[log\left(\frac{B_0}{B_t}*\frac{A_t}{A_0}\right)\right]$  

$\therefore t=\frac{2.303*0.3010*2*100}{693}=200s$  

$\begin{array}{l}{D}_1\\ D_3\end{array}\}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I=\frac{v}{R}=\frac{10}{10}=1Amp$

0.5 % KCl solution has molality (m) = $\frac{0.5*1000}{74.5*99.5}$  

$KC{l}_{\left(aq\right)}?K_{\left(ag\right)}^{+}+C{l}_{\left(aq\right)}^{-}$  

1 - a        a          a

And I =  $\left(1-\cancel{\alpha }+\cancel{\alpha }+\alpha \right)=1+\alpha$  

$i=\frac{\Delta Tf}{kf*m}=\frac{0.24*74.5*99.5}{1.8*0.5*1000}=1+\alpha$  

1.976 = 1 + a

$\therefore \alpha =0.976$  

% = 97.6%

the nearest 98.

Complete reaction of acid rain is

$2S{O}_2+O_2+2H_{2}O\stackrel{}{\to }2H_{2}S{O}_4\left(acidrain\right)$