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New answer posted

5 months ago

0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

use sin1x=cos11x2

tan11x2=cot111x2

sin1x1x2=cos112x21x2

Sum of roots b = 1 + 2  (k21)k22

Product of roots 5 = 2  (k21k22)

b = 4, k213

New answer posted

5 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

Let initial moles of reactant taken = n. Total moles obtained for benzene sulphonic acid = 0.6 n (with % yield = 60%)

Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield 50%) = 0.6 * 0.5 n = 0.3 n

So, overall % age yield of complete reaction = 0.3nn*100=30%

New answer posted

5 months ago

0 Follower 2 Views

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Payal Gupta

Contributor-Level 10

 %Opticalpurity=Observedrotationofmixture*100rotationofpureenautiomer

=+12.6+30*100=42%

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

2KMnO4+5H2C+32O4+3H2SO4 (dil)K2SO4+2MnSO4+10CO+42+8H2O

Oxidation state of carbon changes from +3 to +4

 change in O. S is 1

New answer posted

5 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Complex is  [Co (NH3)4Cl2]Cl

Primary valency = No. of ionisable species = 1

New answer posted

5 months ago

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Payal Gupta

Contributor-Level 10

Solubility of CaF2 = S mol/L

S=2.34*1030.1*78=2.3478*102=3*104mol/L

Ksp (CaF2)=4S3=4 (3*104)3=108*1012

= 0.0108 * 10-8 (mol/L)3

New answer posted

5 months ago

0 Follower 16 Views

P
Payal Gupta

Contributor-Level 10

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

 Order of reaction w.r.t 'NO' = 2

New answer posted

5 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Meq of K2Cr2O7 = Meq of Fe2+

(Molarity * Volume * nf) of K2Cr2O7 = (molarity * volume * nf) of Fe2+

0.02 * 20 * 6 = M * 10 * 1

M = 0.24 M

Molarity = 24 * 10-2 M

New answer posted

5 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

x=2ε0ρ

E2πxl=ρπx2lε0

Ex=ρx22xε0=ρx2ε0

=ρ2ε0*2ε0ρ=1v/m

New answer posted

5 months ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

i = 1 A j=σE

A = 2mm2

ρ=1.7*108Ωm F=eE=ejσ=eiAσ=eiρA

1.6*1019*1.7*1082*106

= 136 * 10-23 N

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