Class 12th

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New answer posted

8 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Here, C3H6 is propene

   

Second reaction is iodoform reaction.

New answer posted

8 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Impurities present in electrolytic refining of blister Cu, removed as anode mud

= Sb, Se, Te, Ag, Au, Pt

Ans. = 6

New answer posted

8 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Given are the oxide of alkali and alkaline earth metals which are ionic in nature.

Simple oxide are Li2O, CaO, MgO and K2O.

Peroxide is Na2O2 and superoxide is KO2.

All simple oxides are diamagnetic as it has no unpaired electron.

New answer posted

8 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

log k = 20.35 - ( 2 . 4 7 * 1 0 3 ) T  

Comparing with,

l o g k = l o g A E a 2 . 3 0 3 R T

E a 2 . 3 0 3 R = 2 . 4 7 * 1 0 3

= 47.29 kJ/mole

Ans. = 47

New answer posted

8 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

8 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

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

A m a x = A c + A m

A m i n = A c A m

A m i n A m a x = A c A m A c + A m = 2 5 0 1 5 0 2 5 0 + 1 5 0

1 0 0 4 0 0 = 5 0 2 0 0  

New answer posted

8 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

P (E1)= 0.9 P ( E ¯ t ) = 0 . 1 a n d P ( E 2 ) = 0 . 8 , P ( E ¯ 2 ) = 0 . 2

R e q u i r e d p r o b a b i l i t y P = 0 . 8 * 0 . 1 0 . 8 * 0 . 1 + 0 . 2 * 0 . 1 + 0 . 9 * 0 . 2 = 0 . 8 2 . 8 = 2 7

9 8 P = 2 8

New answer posted

8 months ago

Answer the questions on the basis of the information given below.

In a school when a survey is conducted on liking of subjects out of three subjects physics, chemistry mathematics then outcome is as follow.

Total number of students who like physics, chemistry and mathematics is 30, 40, 50 respectively. 10 students shown interest in all three subjects, Number of students who like physics and mathematics both is equal to number of students who like chemistry and mathematics both and is equal to 3/4 times number of students who like physics and chemistry both 15 students like only chemistry.

When class teacher saw this survey report he to

...more
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

From the given condition g = 10,

Number of students who like both Physics and Maths = 10 + f

Number of students who like both chemistry and Maths = 10 + e

From given condition 10 + f = 10 + e or e = f

Similarly from 3rd condition ¾ (10 + e) = 10 + d

Or 30 + 3e = 40 + 4d

Or

 Since 15 students like only chemistry hence b = 15

Number of students who like only chemistry is 40

Hence e + g + d + b = 40

or e + 10 + (3e – 10)/4 + 15 = 40

or (7e – 10)/4 = 15 or 7e = 70 or e = 10

Hence a = 5, c = 20, f = e = 10, d = 5

Hence the distribution is as follows

As per the observation of  class teacher actual numbers of students who like Physics in

...more

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