Chemistry

Rate constant, k = 5.5 * 10^-14 s^-1.

$t=\frac{2.303}{k}log\frac{{\left[R\right]}_0}{\left[R\right]}$                

= $\frac{2.303}{k}log3-(i)$  

$t_{50\mathrm{\%}}=\frac{2.303}{k}log2-(ii)$                

From (i) & (ii)

$\frac{t_{67\mathrm{\%}}}{t_{50\mathrm{\%}}}=\frac{log3}{log2}$                

= 1.58 t_50%

So; t_67% is 15.8 * 10^-1 times half life.

X = 16 (the nearest integer)

$C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$  

Here, 6F electricity is required to reduce 1 mol  $C{r}_2{O}_7^{2-}$ to Cr³⁺.

Using ; k_sp = 108 S⁵

1.1 * 10^-23 = 108 S⁵

S = ${\left(\frac{110}{108}*10^{-25}\right)}^{1/5}M\approx 10^{-5}M$  

Specific conductance,

${\lambda }_m^0=\frac{\kappa }{5*1000}S{m}^{2}mo{l}^{-1}$                              

= 3 * 10^-3 Sm² mol^-1

So; x = 3

$P_A^0=50torr$

$P_B^0=100torr$

Mole fraction of A in liquid phase, x_A = 0.3

Mole fraction of B in liquid phase, x_B = 0.7

Now; $P_A=P_A^0{x}_A$

$P_B=P_B^0{x}_B$

= 100 * 0.7 = 70 torr

Mole fraction of B in vapour phase,  $y_B=\frac{P_B}{P_A+P_B}=\frac{70}{85}$

$\frac{70}{85}=\frac{x}{17}$

X = 14

For expansion in vacuum, workdone, w = 0

For isothermal process,  $\Delta U=0$

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$

Hybridization of P in PF₅ is sp³d, so value of y = 1.

Work function,  ${?}_0=6.63*10^{-19}J$

Threshold wavelength ${\lambda }_0=?$

Using ; ${?}_0=\frac{hc}{{\lambda }_0}$

${\lambda }_0=\frac{hc}{{?}_0}$

= 300 * 109 m = 300 nm

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$  

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$               

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}*1000=M*n-factor*V$                              

$\frac{w}{87/2}*1000=0.1*1*60$                              

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample =   $\frac{0.261}{2}*100$

= 13.05%

$\underset{\begin{array}{l}Sodiumextract\\ ofsulphur\end{array}}{N{a}_{2}S}+N{a}_2\underset{Sodiumnitroprusside}{\left[Fe{\left(CN\right)}_{5}NO\right]}\to N{a}_4\underset{violet/purple}{\left[Fe{\left(CN\right)}_{5}NOS\right]}$

$\alpha -$ Helix (2°) structure of protein is stabilized by intermolecular H-bonding.