Chemistry

The early 20th century experiments on electrical discharge through gases eventually led to the discovery of cathode rays (electrons). The major finding was that the characteristics of these cathode rays (electrons) were independent of the material used for the electrodes and the nature of the gas present in the cathode ray tube. This consistent behaviour across different substances led to the conclusion that electrons are a basic constituent of all atoms.

Before 20th century, atoms were widely considered the indivisible building blocks of matter. This view goes back to Indian and Greek philosophies, as old as 400 B.C. It also became one established thought on a scientific basis by John Dalton in 1808. With his theory, several fundamental laws of chemistry were established. But those laws failed against observations like static electricity. The experimental observations made towards the end of the 19th and beginning of the 20th century definitively proved that atoms are made of subatomic particles. 

$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{35}{300}=\frac{40}{T_2}$

$\therefore T_2=\frac{40*300}{35}=342.85K$

$T_2\left(°C\right)=69.707\approx 70°C$

$A_2{B}_3?2A^{+3}+3B^{-2}$

1-α        2α        3α

$\therefore i=1+4\alpha =1+4*0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4*0.52*1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$         

  $?U=?742.24\text{?}kJ/mole?H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

 = $?742.24+\frac{1}{2}*\frac{8.314}{1000}*298$  

$?742.24+1.238=?741.002??741kJ/mol$                  

So, magnitude of  $?H$ = 741 kJ/mol.

NaOH + Na₂CO₃

(1) When Hph is added

m.e NaOH + $\frac{1}{2}m.eN{a}_{2}C{O}_3=m.e.HCl=17.5*\frac{1}{10}=1.75$  (m.e = milli equivalents)

(2) When MeOH is added after Hph

$\frac{1}{2}meN{a}_{2}C{O}_3=m.eHCl=1.5*\frac{1}{10}=0.15$

$\begin{array}{l}\therefore m.eNaOH=1.75-0.15=1.6\\ m.eN{a}_{2}C{O}_3=0.15*2=0.3\end{array}$  

$\frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}*1000=0.3$   

$Weight\mathrm{\%}ofN{a}_{2}C{O}_3=\left(\frac{\frac{0.3*53}{1000}}{0.4}\right)*100=\frac{0.3*53}{10*0.4}=\frac{15.9}{4}=3.975\mathrm{\%}\approx 4\mathrm{\%}.$               

$\begin{array}{l}\left(+6\right)\left(+2\right)\left(+3\right)\left(+6\right)\\ Cr{O_4}^{-2}+\underset{}{\underset{?}{S_2{O_3}^{-2}\to Cr{{\left(OH\right)}_4}^{-}+S{O_4}^{-2}}}\end{array}$  

O.N.C = 4

n factor of S₂O₃^-- = 8

n factor of CrO₄^-- = 3

$m.eCr{O}_4^{-2}=m.e.S_2{O}_3^{-2}$        

$V*\left(0.154*3\right)=40*\left(0.25*8\right)\therefore v=\frac{80}{0.462}$       

= 173.16 ml $\approx$ 173 ml

$logK=logA-\frac{Ea}{2.303RT}$

$Slope=\frac{Ea}{2.303R}=-10,000K$  

$\frac{Ea}{2.303R}=10^4$            

$log\left(10^{-5}\right)=logA-10^4*\frac{1}{500}$   (at 500 K temperature)

$\therefore T=\frac{10^4}{19}K=\frac{100}{19}*10^{2}K=5.2631*10^{2}K=526.31K\approx 526K$

(A) Ethane                   one $\left(C-C\right)\sigma$ bond

(B) Ethene                         one $\left(C-C\right)\sigma$ and one $\left(C-C\right)\pi$ bond

(C) $C_2$                                                       two $\left(C-C\right)\pi$ bonds

(D) Ethyne $H-C\equiv C-H$                       two $\left(C-C\right)\pi$ bonds and one $\left(C-C\right)\sigma$ bond

$E_n=-R_H\left(\frac{Z^2}{n^2}\right)J$

For ${}^{+}\left(n=1\right)$ ,

$E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4R_H$

$\therefore R_H=\frac{x}{4}$

For ${}^{3+}\left(n=2\right)$ ,

$E_n=-R_H\left(\frac{z^2}{n^2}\right)J$

$=-\frac{x}{4}*\left(\frac{4*4}{2*2}\right)=-xJ$