214 - Chemistry Questions Answers

Steric hindrance is a relative word, so is compared between two carbons.

Tertiary carbon is always more sterically hindered than primary,

$so \beta carbon is more hindered than \alpha in this compound$

In first case reactions involved are$$\begin{gathered} \mathrm{Reduction} \mathrm{of} {\mathrm{Fe}}^{+++}, {\mathrm{Fe}}^{+++}\to {\mathrm{Fe}}^{++} \\ \mathrm{Titration} \mathrm{step}, 6{\mathrm{Fe}}^{++}+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to 6{\mathrm{Fe}}^{+++}+2{\mathrm{Cr}}^{+++} \\ \mathrm{By} \mathrm{using} \mathrm{titration} \mathrm{reaction}, \mathrm{solve} \mathrm{the} \mathrm{question} \end{gathered}$$

 In second case reaction involved is${\mathrm{As}}_4{\mathrm{O}}_6+4{{\mathrm{I}}_3}^{-}\to {\mathrm{As}}_4{\mathrm{O}}_{10}+12{\mathrm{I}}^{-}$Now solve

inform if any issue

1. It is cis-trans pair (no chiral carbon atom so no optical isomerism)

2. has chiral carbon atom so case of optical isomerism, mirror images so dextro and laevo

Zn                  ----------->   Zn⁺⁺   +   2e^-

Cu⁺⁺ + 2e^-   ----------->   Cu

H₂  -------->  2H⁺  +   2e^- 

Cl₂ + 2e^-      -------->  2Cl^-

3/2 times of the previous number of elements

by formula PV/T = constant

(752-28)*150/301 = 760*V/273

solve for V, this will be the equivalent volume at STP

now convert this volume in gm by the formula 2V/22400

then finally m = zit, where 

z = 1/96500

solve it for i

molarity = 1 

and N = nM = 2(1) = 2

hence it is clear 

by gas eq.  PV/nT = constant

so 0.5*2V/0.7*300 = PV/n₁400 + PV/n₂300        (1)

no. of moles are also constant

so 0.7 = n₁ + n₂

similarly by gas eq. 

PV = n₁R400                      for first flask

PV = n₂R300                      for second flask

now solve

b = 4V, where V is volume of 1 mole particles

so 24*10^-6 cm³/mole = 4*[4πr³/3] cm³/mole 

now solve