Chemistry

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New answer posted

7 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

  Δ T b = i K b m

m = w * 1 0 0 0 M * W s o l v e n t

For acetone solution,

0 . 1 7 = 1 * 1 . 7 * 1 . 2 2 * 1 0 0 0 M * 1 0 0

For Benzene solution,

2 A c i d ( A c i d ) 2 i = 1 / 2

Δ T b = i * K b * m

= 1 2 * 2 . 6 * 1 . 2 2 * 1 0 0 0 1 2 2 * 1 0 0 ° C

= 0 . 1 3 ° C = 1 3 * 1 0 2 ° C

x * 1 0 2 = 1 3 * 1 0 2

x = 1 3

New answer posted

7 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Effective number of atom in C.C.P

= 1 8 * 8 + 1 2 * 6     

= 4

Number of octahedral void = 4

Number of cations = 4

Number of anion = 4

Formula of compound = A4B4

Empirical formula = AB

Ans. x = 1

New answer posted

7 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H2O

30 * 1 mmol      50 * 1 mmol     

0 mmol               20 mmol

[ H + ] m i x = 2 0 5 0 + 3 0 M = 2 0 8 0 M = 1 4 M = 0 . 2 5 M

x * 1 0 4 = 6 0 2 1 * 1 0 4    

x = 6021

Ans. = 6021

New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

O 2 2 = 8 * 2 + 2 = 1 8 e

σ 1 s 2 σ 1 s * 2 σ 2 s 2 σ 2 s * 2 σ 2 p z 2 π 2 p x 2 = π 2 p y 2 π 2 p x * 2 = π 2 p y * 2      

Number of unpaired e- = 0

Ans. = 0

New answer posted

7 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

( x m ) = K P 1 / n           

1 0 1 = K ( 1 0 0 ) 1 / n . . . . . . . . . . . . . . . . ( 1 )     

1 5 1 = K ( 2 0 0 ) 1 / n . . . . . . . . . . . . . . . . ( 2 )

V 1 = K ( 3 0 0 ) 1 / n . . . . . . . . . . . . . . . ( 3 )

Dividing (2) by (1)

1 5 1 0 = ( 2 1 ) 1 / n = (2)1/n

V 1 0 = 1 0 0 . 2 7 9

V = 1 0 * 1 0 0 . 2 7 9 = 1 0 1 . 2 7 9 = 1 0 x

x = 1 . 2 7 9 = 1 2 7 . 9 1 * 1 0 2 1 2 8 * 1 0 2

Ans. = 128

New answer posted

7 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

C 2 H 7 N + ( 2 x + y 2 ) C u O x C O 2 + y 2 H 2 O + Z 2 N 2 + ( 2 x + y 2 ) C u

C 2 H 7 N 1 + ( 2 * 2 + 7 2 ) C u O 2 C O 2 + 7 2 H 2 O + 1 2 N 2 + ( 2 * 2 + 7 2 ) C u              

y = 7              

Ans. = 7

New answer posted

7 months ago

0 Follower 3 Views

J
Jaya Sinha

Beginner-Level 5

Students should prepare complete syllabus when they have time to prepare. However, you can use the list of highweightage chapters in last minute revision for scoring well.

  • The p-Block Elements: This chapter holds a high weightage of 8–10 marks. (in the latest syllabus this is deleted)
  • Aldehydes, Ketones, and Carboxylic Acids: This chapter contributes around 8–10 marks.

  • Biomolecules: This chapter accounts for around 8 marks.

  • Chemical Kinetics: This chapter holds a high weightage of 5-6 marks. 

  • The d- and f-Block Elements: This chapter contributes around 5-6 marks.

  • Amines: This chapter contributes around 5-6 marks.

New answer posted

7 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider following image

 

New answer posted

7 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

PV = nRT             PV = constant (at constant T)

Pressure increases & volume decreases, PV remains constant at constant T.

New answer posted

7 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Kindly Consider the following Image

 

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