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

For precipitation of two moles of AgCl

Two Cl^- will produce as a free anion

CoCl₃.4NH₃ -> complex will $\left[CO{\left(N{H}_3\right)}_{4}C{l}_2\right]$  Cl (will not give 2Cl^-)

$PtC{l}_4.2HCl\to$  complex will be H₂ [PtCl₆] will not any Cl^-

$NiC{l}_2.6H_{2}O\to \left[Ni{\left(H_{2}O\right)}_6\right]C{l}_2$  will produce two Cl^- ion.

${\left[Ni{\left(H_{2}O\right)}_6\right]}^{++}+2C{l}^{-}\underset{AgN{O}_3}{\overset{}{\to }}2AgCl\left(s\right)$  precipitate formation

$N{i}^{2+}\to \left[Ar\right]3d^{8}4s^0$  

Most basic oxide V₂O₃

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

two unpaired e^- in d- subshell

Kindly consider the following figure

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

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

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

 

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,

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

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

from equation (i) and (ii)

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

Here; A₀ = B₀ and $A_t=4\times B_t$  

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

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

Here, total meq of acetic acid = 50 × 0.1 = 5

And total meq of NaOH = 25 × 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$pH=pKa+lo{g}_{10}\frac{\left[S\right]}{\left[A\right]}$

$pH=4.76+lo{g}_{10}\frac{2.5}{2.5}=4.76=476\times 10^{-2}$  

With AlCl₃, alkyl halide will form cabocation which will show rearrangement.

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$\begin{array}{l}{D}_1\\ D_3\end{array}\}$ 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) = $\frac{0.5\times 1000}{74.5\times 99.5}$  

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

1 - a        a          a

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

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

1.976 = 1 + a

$\therefore \alpha =0.976$  

% = 97.6%

the nearest 98.

In this carbocation +M effect of -OCH₃ group stabilizes the carbocation.

While in option (A) and (B), +M of -OCH₃ will not work but in option (C), +M of -OCH₃ works so due to more delocalization in option (D), it is more stable.

Hence, x = 727 (the nearest integer)