Class 12th

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New answer posted

5 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

log (k2k1)=Ea2.303*8.314 [13001325]

log (5) = Ea2.303*8.314 [13001325]

Ea=0.7*2.303*8.314*300*32525=52271.7

Ea in kJ/mole = 52271.71000=52.2kJ/mol

the nearest integer is 52.

New answer posted

5 months ago

0 Follower 11 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Glycosidic linkage is present in lactose between C1 of galactose & C4 of glucose.

 

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5 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

5 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Stable carbocation formation is the important factor.

New answer posted

5 months ago

0 Follower 10 Views

P
Payal Gupta

Contributor-Level 10

Statement – I True

Statement – II False

Dimethyl glyoxime is the bidentate anionic ligand.

N i 2 + + 2 D M G [ N i ( D M G ) 2 ] R e d p p t

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

? 1 = 3 5 E 0 ( 0 . 2 ) N m 2 C 1 , a n d ? 2 = 4 5 E 0 ( 0 . 3 ) N m 2 C 1

? 1 ? 2 = 3 * 0 . 2 * 5 5 * 0 . 3 * 4 = 1 2

New answer posted

5 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

α-anomer of maltose is 1,4 combination. (C1-C4) glycosidic linkage

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