Class 12th

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

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

$\underset{x\to \frac{\pi }{2}}{lim}\frac{si{n}^{2}x\left(si{n}^{2}x-3sinx+2\right)}{co{s}^{2}x.\left(\sqrt[]{2si{n}^{2}x+3sinx+4}+\sqrt[]{si{n}^{2}x+6sinx+2}\right)}$  

$\downarrow$                                                       $\downarrow$                            

3                                                       3

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{9}.\frac{\left(sinx-1\right)\left(sinx-2\right)}{\left(1-sinx\right)\left(1+sinx\right)}=\frac{1}{6}\left(-1\right).\frac{-1}{2}=\frac{1}{12}$

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

$\Delta =\left|\begin{array}{ccc}\hfill -k\hfill & \hfill 3\hfill & \hfill -14\hfill \\ \hfill -15\hfill & \hfill 4\hfill & \hfill -k\hfill \\ \hfill -4\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|$

$=\left(-k\right)\left(12-k\right)+3\left(4k+45\right)-14\left(-15+16\right)$

$\Delta =0\Rightarrow k=\pm 11$             

 For k = -11,

->11x + 3y – 14z = 25

-4x + y + 3z = 4

$\left\{\begin{array}{l}-11x+3y-14z=25\\ -15x+4y-11z=3\\ -4x+y+3z=4\end{array}\right\}\to$ No solution for k = $\pm$ 11

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