JEE 2015 :: Advanced 2015 :: Chemistry 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

Under hydrolysis conditions, the compounds used for preparation  of linear polymer and for chain termination m respectively are

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Explanation:

The  number of hydroxyl group(s) in Q

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Explanation:

In dilute aqueous H₂SO_4, the complex diaquadioxalatoferrate (II) is oxidised by MnO₄^-. For this reaction, the ratio of the rate of change [H⁺] to the rate of change of [MnO₄^-]  is 

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Explanation:

The balanced redox reaction is

                  $MnO_{4}^{-}+[Fe(H_{2}O)_{2}(C_{2}O_{4})_{2}]^{2-}+8H^{+}\rightarrow Mn^{2+}+Fe^{3+}+4CO_{2}+6H_{2}O$

$\Rightarrow$                                $\frac{r[H^{+}]}{r[MnO_{4}^{-}]}=\frac{8}{1}=8$

A closed vessel with rigid walls contains  1 mole of  $^{238}_{92}U$  and 1 mole of air at 298 K. considering complete  decay of  $^{238}_{92}U$ to $^{206}_{82}Pb$, the ratio of the final pressure to the initial pressure of the system at 298 K  is

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Explanation:

$_{92}U^{238}\rightarrow _{82}Pb^{206}+8 _{2}He^{4}(g)+6 _{-1}\beta^{0}$

n(gas) [Initial]=1 (air)

n(gas) [final]=8(He)+1(air)=9

$\Rightarrow$    At constant temperature and volume:  $p \propto n$ . 

 So,        $\frac{P_{f}}{P_{i}}=\frac{n_{f}}{n_{i}}=\frac{9}{1}=9$

 The molar conductivity  of a solution of a weak acid HX (0.01 M)  is 10 times smaller than the molar conductivity  of a solution  of a weak  acid HY  (0.01 M), If  $\lambda^{0}_{X^{-}}\approx \lambda^{0}_{Y^{-}}$, the  difference  in their pK_a  values, $PK_{a}(HX)-PK_{a}(HY)$   is (consider  degree of ionisation  of both acids to be <<1)

Answer:

Explanation:

Degree of ionisation ($\alpha$)=  $\frac{^{\wedge}m}{\wedge^{\infty}}$

Let           $^{\wedge}m(HY)=x$

$\Rightarrow$                     $^{\wedge}m(HX)=\frac{X}{10}$

$\Rightarrow$       $\frac{^{\wedge}m(HX)}{^{\wedge}m(HY)}=\frac{1}{10}=\frac{\alpha(HX)}{\alpha(HY)}$

                                                                                                $[\because ^{\wedge \infty}(HX)=^{\wedge \infty} (HY)]$

         Also :   $K_{a}(HX)= (0.01)[\alpha(HX)]^{2}$

                        $K_{a}(HX)= (0.10)[\alpha(HX)]^{2}$

                                           $= (0.10)[10\alpha(HX)]^{2}$

                                $= 10[\alpha(HX)]^{2}$    ........(ii)

   $\Rightarrow$                  $\frac{K_{a}(HX)}{K_{a}(HY)}=\frac{0.01}{10}$

 $\Rightarrow$     $\log K_{a}(HX)-\log K_{a}(HY)=-3$

 $\Rightarrow$     $-\log K_{a}(HX)-[-\log K_{a}(HY)]=3$

 $\Rightarrow$   $pK_{a}(HX)-p K_{a}(HY)=3$

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015