Chemistry

Kindly consider the following figure

Acid = M (OH)2 Salt + H2O

M.E of Acid = me of Base

$\therefore 10*\left(0.1*nf\right)=\left(0.05*2\right)*30$

$\therefore n_f=3$

Thus basicity of acid = 3

Biuret test is given by proteins and dipeptide & compounds having following groups

Hence positive test given by tripeptide & biuret.

Monomer of novolac is

Molecular mass of monomer unit

              = 12 * 7 + 1 * 8 + 16 * 2

              = 84 + 8 + 32

              = 124 g

Molecular mass of polymer = 963 g

$\therefore$ n * mass of monomer = mass of polymer + (n – 1) * 18

(n – 1) unit of H₂O is removed during formation

$\therefore 124*n=963+18\left(n-1\right)$

106n = 945

 $\therefore n=\frac{945}{106}=8.91\approx 9$

FeCl₃. 3H₂O i.e. [Fe (H₂O)₃], K₃ [Fe (CN]₆ and [Co (NH₃)₆]Cl₃ the last two complex are inner-orbital complex due to presence of strong ligand.

$\therefore {\Delta }_{0}of{K}_3\left[Fe{\left(CN\right)}_6\right]>{\Delta }_{0}of\left[Co{\left(N{H}_3\right)}_6\right]C{l}_3$

Because CN^- is strong ligand

$\therefore {\Delta }_0\propto \frac{1}{\lambda }$

More ${\Delta }_0$ smaller value of absorbed $\lambda$

$K_3\left[Fe{\left(CN\right)}_6\right]\to F{e}^{+3}=4s^{0}3d^{5}4p^0$

$\therefore u.e=1$

$\therefore \mu =\sqrt[]{1*\left(1+2\right)}B.M=\sqrt[]{3}B.M$

= 1.732 B.M

$\approx$ 2

Number of replaceable H is 2.

$F{e}^{+3}+3e^{-}\to Fe$

3 Faraday is required to deposit 1 mole Fe

 $?56g$  deposited by   $3*96500$ C charge

$\therefore 0.3482g$ deposited by $\frac{3*96500}{56}*0.3482C=\frac{100803.9}{56}C=1800.7C$

$\therefore Q=It$

1800.07 = 1.5 t

$\therefore t=1200sec=\frac{120\cancel{0}}{6\cancel{0}}$ min = 20 min 

t1/2 = 340 secP0 = 55.5 kPa

t1/2 = 170 sec P0 = 27.8 kPa

$\therefore t_{1/2}\propto {\left(P_0\right)}^{1-n}$

$\therefore 1-n=1n=0$

$\therefore$ zero order reaction

  $F-Be-F$   $\mu =0$ $$

BF3 $\mu =0$

H2O $\mu \ne 0$ $$

NH3 $\mu \ne 0$ $$

CCl4 $\mu =0$ $$

HCl $\mu \ne 0$ $$

$T_1=27°C=300K$ T2 = 45° C = 45 + 273 = 318 K

P1 = 30 atmP2 =?

V is constant due to rigid tank\

$\therefore \frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{30}{300}=\frac{P_2}{318}\therefore P_2=\frac{1}{10}*318=31.8atm\approx 32atm$