Chemistry

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New answer posted

8 months ago

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V
Vishal Baghel

Contributor-Level 10

Due to formation of anilinium ioin in acidic medium meta product is also obtained.

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Vishal Baghel

Contributor-Level 10

ln (A) and (C) CH3 – group is activating group

ln (B) Br is weakly deactivating group

ln (D) NO2 is strongly deactivating group

Order of reactivity towards nitration.

D < B < E < A < C.

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8 months ago

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alok kumar singh

Contributor-Level 10

Z=101 belong to actinoids

104 belong to group 4

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Vishal Baghel

Contributor-Level 10

 N2+O2? VeryhighTemp2NO

Reaction is endothermic and required very high temperature for comobination

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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alok kumar singh

Contributor-Level 10

4Li+O2? 2Li2O oxide

2Na+O2? Na2O2 peroxide

K+O2? KO2 superoxide

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Vishal Baghel

Contributor-Level 10

 2MnO4+16HCl2MnCl2+2KCl+8H2O+Cl2

KMnO4 oxidizes HCl to Cl2, so HCl is not used in permanganate titration.

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alok kumar singh

Contributor-Level 10

U and H are temperature dependent

CP, m-CV, m=R  (for 1 mole of ideal gas)

dU=CVdt

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Vishal Baghel

Contributor-Level 10

 No. of electron = 62 – 2 = 60

68Er+3 No. of electron = 68 – 3 = 65

70Yb+2 No. of electron = 70 – 2 = 68

71Lu+3 No. of electron = 71 – 3 = 68

65Tb+4 No. of electron = 65 – 4 = 61

63Eu+2 No. of electron = 63 – 2 = 61

65Tb+2 No. of electron = 65 – 2 = 63

69Tm+4 No. of electron = 69 – 4 = 65

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8 months ago

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Vishal Baghel

Contributor-Level 10

I 2 + 1 0 H N O 3 ( c o n c ) 2 H l O 3 + 1 0 N O 2 + 4 H 2 O

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