Ncert Solutions Chemistry Class 12th

In fcc structure of diamond four C present in fcc lattice and other four C present in tetrahedral voids where 50% of tetrahedral voids are occupied. Hence number of carbon atoms present per unit cell of diamond is 8.

$NiC{l}_2+NaCN\underset{O.A.}{\overset{strong}{\to }}{\left[Ni{\left(CN\right)}_6\right]}^{2-}$

Complex has Ni⁴⁺ and strong ligand, hence following are the metal ion electronic configuration

Change of unpaired electron = 2

Wt of Cl^- in 100 ml = 1.8 * 10^-3 gm

Mol. of Cl^- in 100 ml = $\frac{1.8*10^{-3}}{35.5}=0.0507*10^{-3}mole$  

$\therefore \left[C{l}^{-}\right]=\frac{0.0507*10^{-3}*1000}{100}=5.07*10^{-4}M$   

i.e. 0.507 milli mole in one lit required in one hr.

Coagulation value = (millimole/lit) required in one hr = 0.507

= 1 (the nearest integer)

Here, $\lambda =\frac{h}{\sqrt[]{2*m*eV}}=\frac{6.63*10^{-34}}{\sqrt[]{2*1.6*10^{-19}*40000*9.1*10^{-31}}}m$  

$\therefore \lambda =0.614*10^{-11}m$  

$\lambda =6.14*10^{-12}m$ [the nearest integer = 6]

0.15 gm (organic compounds) $\underset{}{\overset{AgN{O}_3}{\to }}AgBr\left(s\right)$

$\downarrow$                                                                    

0.2397 gm

Weight of Br in AgBr = $\left(\frac{80}{188}*0.2397\right)=0.102gm$

% Br in compound = $\frac{0.102*100}{0.15}=68\mathrm{\%}$

Due to the larger acid dissociation constant (Ka) sulphuric acid acts as an acid and HNO₃ acts as a base.

Kindly consider the following figure

Due to the absence of acidic hydrogen in But – 2-yne, metallic sodium does not react with this.

Orlon fibres are the polymer of acrylonitrile