Chemistry

$C_v=\frac{3R}{2}$  

AC  polytropic P $\propto$  V

$\Rightarrow P.V^{-1}=constant$            

$C_{AC}=C_V+\frac{R}{1-\left(-1\right)}=2R$

AB isobaric

$C_{AB}=C_P=\frac{5R}{2}$       

  $\therefore \frac{Q_{AC}}{Q_{AB}}=\frac{C_{AC}}{C_{AB}}\left[?\Delta Tsameforboth\right]$

$=\frac{2R}{\left(\frac{5R}{2}\right)}=\frac{4}{5}=0.8$           

Factual:
Lanthanoid ⇒ 57 to 71 and actinoids ⇒ 89 to 103

(i)  give NO₂   (ii) NH₄NO₃         (iii) gives NO₂     (iv) Give NO₂      (v) give NO₂       (vi) Given N₂O

(vii) Gives NO₂       (viii) Gives NO₂  (ix) Gives NO

σ bonded organometallic compound ⇒ M – C
σ-bond
and in π – bonded organo metallic compound
M – C
π bond
In ferrocene, there is π-bond

Molecular formula from calculation comes to be C₃H₆O i.e., it stands for the following compounds.

No addition reaction with Br₂

VD * 2 = MM         = 29 * 2 = 58

$?$  58 g of ether combines with 2 * 80g Br₂

$?$ 2.9 g of ether will combines with   $\frac{2*80}{58}*2.9=8gofBromine$

Y = 9 and X = 8

Y - X = 9 - 8 = 1

Let V ml is taken

  $\frac{5*V+1*150}{\left(V+150\right)+0.9}=2.5$            

5V + 150 = 2.25 V + 337.5

$V=\frac{187.5}{2.75}=68.18ml$