Chemistry

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

 = 23 * 10^-2 V

$\therefore$  x = 23

 x = 23

In S?2 reaction, inversion takes place, so configuration get reversed.

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

As Zn (OH)? , BeO, Al? O? are amphoteric, so they can react with both HCl and NaOH.

When Ferric chloride reacts with potassium thiocyanate a blood red colour of Ferric thiocyanate is formed
FeCl? + 3KSCN → Fe (SCN)? + 3KCl

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

                             = 0.00002 * 10³ mm

Formation of passive layer of Fe? O? on the surface of Fe and NO? gas is evolved
[Co (NH? )? ]³? is diamagnetic (t? g? eg? ) and is inner orbital complex

$\Delta H=-165kJ/mole$

T =?

$\Delta S=-550J{K}^{-1}$

At equilibrium $\Delta G=0$

$\therefore T=\frac{\Delta H}{\Delta S}=\frac{-165*1000}{-550}K$

$\begin{array}{l}=3*100K\\ =300K\end{array}$

i. Zn + 2NaOH → Na? ZnO? + H?
ii. 4Au + 8NaCN + O? + 2H? O → 4Na [Au (CN)? ]+ 4NaOH
iii. Cu + 4HNO? → Cu (NO? )? + 2NO? + 2H? O
(conc.)

K = 3.3 * 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$  

3.3 * 10^-4 =   $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$

  $t=\frac{2.303}{3.3}*10^4*0.22\Rightarrow t=25.58mins$            

so; the nearest integer is 26.