Chemistry

The structure of an atom reflects how the atom is made, which includes the nucleus at the centre where protons and neutrons are present, and electrons revolving around the nucleus.

Mass of Na⁺ in 50 ml = 70 * 50 = 3500 mg

23000 mg of Na⁺ is present in 85000 mg of NaNO₃ (1 mole NaNO₃ contains 1 mole Na⁺)

$\therefore 3500$   mg Na⁺ will be present in   $\frac{85000}{23000}*3500$

= 12934.78 mg

= 12.93478 gm

$\cong 13$          

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

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