Chemistry

  ${E^o}_{Cell}={E^o}_{A{g}^{+}/Ag}-{E^o}_{Z{n}^{+2}/Zn}=0.8+0.76=1.56V$

Anode :  $Zn\left(s\right)\to Z{n}^{2+}\left(aq\right)+2e^{-}$    

Cathode :  2Ag⁺(aq) + 2e^- -> 2Ag(s)

Zn(s) + 2Ag⁺(aq) -> Zn²⁺ (aq) + 2Ag(s)

  $E_{cell}={E^o}_{Cell}-\frac{0.0591}{n}log\frac{\left[Z{n}^{2+}\right]}{{\left[A{g}^{+}\right]}^2}=1.56-\frac{0.0591}{2}log\left(\frac{0.1}{10^{-4}}\right)$          

x = 147

Cis (optically active)                          Trans (optically inactive)

d & l form (2)                                                    (1)

So, total 3 isomerism.

pH =   $\frac{1}{2}\left[p{K}_w+p{K}_a-p{K}_b\right]$

$=\frac{1}{2}\left[14+4.75-5.23\right]$

$=6.76\cong 7$          

Uncertainty in speed = 90   $*\frac{5}{100}=4.5m/s=\Delta V$

  $\Delta x=\frac{h}{4\pi m\Delta V}=\frac{6.63*10^{-34}}{4*3.14*10^{-2}*4.5}=1.173*10^{-33}$            

Fraction of molecules having enough energy to form product  =   $e^{-Ea/RT}$

 Fraction of molecules having enough energy to form product =   $e^{-\frac{80.9*10^3}{8.314*700}}$

$=e^{-13.8}\cong e^{-14}$

So, x = 14

$\Delta T_f=K_f*m*i$  

0.93 = 1.86 * 1 * i

i =  $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  

$i=1+\left(\frac{1}{n}-1\right)$

$\therefore n=2$            

$Mn{O}_4^{-}+S_2{O}_3^{2-}\stackrel{O{H}^{-}}{\to }Mn{O}_2+S{O}_4^{2-}$

O.S of S in    $S{O}_4^{2-}=+6$

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

Ratio = n/n = 1

$S\left(g\right)+6F\left(g\right)\to S{F}_6\left(g\right)$  

$\Delta {H^0}_R=\Delta {H^0}_f\left(S{F}_6\right)-\Delta {H^0}_f\left(S\right)-6*\Delta {H^0}_f\left(F\right)$            

= (-1100) - (275) -6 (80) = -1855

$\Delta {H^0}_{S-F}=\frac{1855}{6}=309.16KJ/mol\cong 309$            

Seliwanoff's test is used to distinguish carbohydrates while xanthoproteic test is used to distinguish proteins