Chemistry

$BaC{l}_2+S{O}_3^{2-}\stackrel{}{\to }\underset{White}{BaS{O}_3}\downarrow \stackrel{dil.HCl}{\to }\underset{\begin{array}{l}burningsulphur\\ likesmell\end{array}}{S{O}_2\uparrow }$

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

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

Antiseptic Dettol is mixture of chloroxylenol and terpineol

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

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

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

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

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

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