Chemistry

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

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

Here 'X' is the Gypsum (CaSO₄.2H₂O) which is used to enhance the setting of time.

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

Stannane is the tetrahedral shape of molecule.

Hybridisation is- sp³, shape and structure are tetrahedral

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

t_1/2 = 340 sec      P₀ = 55.5 kPa

t_1/2 = 170 sec      P₀ = 27.8 kPa

$?t_{1/2}?{\left(P_0\right)}^{1?n}$

$?1?n=1n=0$

$?$ zero order reaction

$F-Be-F$ $\mu =0$      

BF₃             $\mu =0$            

H₂O         $\mu =0$

NH₃           $\mu =0$                          

CCl₄           $\mu =0$             

HCl             $\mu =0$            

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

P₁ = 30 atm        P₂ =?

V is constant due to rigid tank

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

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