Class 11th

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V
Vishal Baghel

Contributor-Level 10

Sm ⇒ +2, +3 (4f? , 6 s²); Gd ⇒ +3 (4f?5 d¹6 s²)
Tm ⇒ +2 + 3 (4f¹³,6 s²); Nd ⇒ +2, +3, +4 (4f? , 6 s²)

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Raj Pandey

Contributor-Level 9

η 1 = 1 - T 2 T 1 , η 2 = 1 - T 3 T 2

as η 1 = η 2 T 2 T 1 = T 3 T 2 T 2 = T 1 T 3 1 2 = 400 * 900 = 600 K

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A
alok kumar singh

Contributor-Level 10

Kindly go therough the solution

Ans. (122.00)

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A
alok kumar singh

Contributor-Level 10

N 2 O , C 2 H 2 , C O 2 , C 3 O 2 , B e F 2

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Raj Pandey

Contributor-Level 9

 Newton's law of cooling, - d T d t = k T - T 0 - d T T - T 0 = k d t - l n ? T - T 0 T T + T 0 / 2 = k t ?   Heat   T - T 0 - l n ? T - T 0 / 2 T - T 0 = k t - l n ? 1 2 = k t t = l n ? 2 k = l n ? 4 2 k n = 4

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Vishal Baghel

Contributor-Level 10

Factual:
Lanthanoid ⇒ 57 to 71 and actinoids ⇒ 89 to 103

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Raj Pandey

Contributor-Level 9

Kindly consider the following Image 

 

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A
alok kumar singh

Contributor-Level 10

Molecular formula from calculation comes to be C3H6O i.e., it stands for the following compounds.

No addition reaction with Br2

 

VD * 2 = MM         = 29 * 2 = 58

?  58 g of ether combines with 2 * 80g Br2

? 2.9 g of ether will combines with   2 * 8 0 5 8 * 2 . 9 = 8 g o f B r o m i n e

Y = 9 and X = 8

Y - X = 9 - 8 = 1

 

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alok kumar singh

Contributor-Level 10

  Z n + 2 N a O H N a 2 Z n O 2 + H 2

2 A I + 2 N a O H + H 2 O 2 N a A l O 2 + 3 H 2 S n + 2 N a O H + H 2 O N a S n O 3 + 2 H 2 P 4 + 3 N a O H + 3 H 2 O P H 3 + 3 N a H 2 P O 2

            

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