Chemistry

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 =

              $pH=\frac{pKa-\mathcal{l}ogC}{2}$  

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

              $pH=27*10^{-1}$     

          $\therefore$    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

$B_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^1={\pi }_{2py}^1\to$ Paramagnetic

$\begin{array}{l}L{i}_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2\to Diamagnetic\\ C_2={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2\to Diamagnetic\\ C_2^{-}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\sigma }_{2pz}^1\to Paramagnetic\end{array}$

$O_2^{-2}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*2}\equiv {\pi }_{2py}^{*2}$ Diamagnetic

$O_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}{\sigma }_{2pz}^2{\pi }_{2px}^2\equiv {\pi }_{2py}^2{\pi }_{2px}^{*1}\equiv {\pi }_{2py}^0\to$ Paramagnetic

$H{e}_2^{+}={\sigma }_{1s}^2{\sigma }_{1s}^{*1}\to$ Paramagnetic

$\therefore$ Paramagnetic molecules are $=B_2,C_2^{-},O_2^{+},H{e}_2^{+}$

x      +      y    +   3z     =     xyz₃

1 mole 1 mole 0.05 mole

$\therefore n_x=1$

$n_y=1$

$n_z=\frac{0.05}{3}=0.0167$  here z is limiting reagent.

$?\frac{0.05}{3}$  mole z gives 1 mole xyz₃

mass of xyz₃ = n * molecular mass

=  $\frac{0.05}{3}*\left(10+20+3*30\right)a.m.u.$  

= 0.5 * 4 = 2g

$2B{F}_3+6NaH\to B_2{H}_6+6NaF$

${\mathrm{B}}_2{\mathrm{H}}_6+2\stackrel{?}{\mathrm{N}}{\left(\mathrm{M}\mathrm{e}\right)}_3\to 2{\mathrm{H}}_3\mathrm{B}\leftarrow :\mathrm{N}{\left(\mathrm{M}\mathrm{e}\right)}_3$

Density of metal a very specific and depends upon many factors.

Li ® 0.53 gm/cm³

Na ® 0.97 gm/cm³

K ® 0.86 gm/cm³

Rb ®1.53 gm/cm³

Cs ® 1.90 gm/cm³

$$ $2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$
 

$FeO+\underset{\left(Acidicflux\right)}{Si{O}_2}\to \underset{\left(Slag\right)}{FeSi{O}_3}$