Chemistry

Please find the below image

Angle of H_t – B – H_t > angle of H_b – B - H_b

Bond angle $\propto \mathrm{\%}S-Character\propto \frac{1}{\mathrm{\%}ofP-character}$  

$\angle H_t$ B H_t is more so % P-Character is less.

M_eq of NaOH = M_eq H₂C₂O₄

4.44 * N = 1.25 * 2 * 10

$\therefore N=\frac{1.25*2*10}{4.44}=5.63$

$\therefore N=M=5.63$

$AB?A^{+}+B^{-}$

1-x x x   $\therefore i=1+x$

$\begin{array}{l}\Delta T_b=i*k_b*m\\ 2.5=\left(1+x\right)*0.52m\end{array}$

$\therefore molality=\frac{2.5}{1.75*0.52}=2.74$

the nearest integer is 3.

$\underset{SS}{AgC{N}_{\left(S\right)}?A{g^{+}}_{\left(aq\right)}+C{N^{-}}_{\left(aq\right)}}$            

$C{N}^{-}+H^{+}?HCN$

Before reaction     S            10^-3       0

After reaction       0            10^-3       S

$\therefore HCN?H^{+}+C{N}^{-}$

$S^2=\frac{2.2*10^{-16}}{6.2*10^{-7}}=\frac{2.2}{6.2}*10^{-9}$

$S=\sqrt[]{\frac{2.2}{6.2}*10^{-9}}=\sqrt[]{\frac{22}{6.2}*10^{-10}}$

$=\sqrt[]{3.54*10^{-10}}=1.88*10^{-5}=1.9*10^{-5}$           

If $\Delta E=0$

Q = W

W = -P_ext $\left(\Delta V\right)=-4.3nRT\left[\frac{1}{P_2}-\frac{1}{P_1}\right]$

= $-4.3*5*8.314*293\left[\frac{1}{1.3}-\frac{1}{2.1}\right]$

= 15347.70 K = 15.3 kJ

Q = 15 kJ

Density = $\frac{M*z}{N_A{a}^3}$

$d=\frac{63.5*4}{6.02*10^{23}*{\left(3.596*10^{-10}\right)}^3*1000}=9076kg/m^3$

$log\left(\frac{k_2}{k_1}\right)=\frac{Ea}{2.303*8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

log (5) = $\frac{Ea}{2.303*8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

$\therefore Ea=\frac{0.7*2.303*8.314*300*325}{25}=52271.7$

Ea in kJ/mole = $\frac{52271.7}{1000}=52.2kJ/mol$

the nearest integer is 52.