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New answer posted

8 months ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

(a) C o C l 3 . 4 N H 3 i s [ C o C l 2 ( N H 3 ) 4 ] C l will show cis- trans isomerism as;

(b) C o C l 3 . 5 N H 3 i s [ C o ( N H 3 ) 5 C l ] C l 2 will not show cis- trans isomerism.

(c) C o C l 3 . 6 N H 3 i s [ C o ( N H 3 ) 6 ] C l 3 will not show cis- trans isomerism.

(d) C o C l ( N O 3 ) 2 . 5 N H 3 i s [ C o ( N H 3 ) 6 C l ] ( N O 3 ) 2  will not show cis-trans isomerism.

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

CN is a strong field ligand, so pairing occurs.

1. [ F e ( C N ) 6 ] 4

So; it is diamagnetic

2. [ F e ( C N ) 6 ] 3

So; it is paramagnetic.

3. [ T i ( C N ) 6 ] 3

Ti3+ = 4s03d1

 It is paramagnetic

4. [ N i ( C N ) 4 ] 2

It is diamagnetic

5. [ C o ( C n ) 6 ] 3

Hence,  [ F e ( C N ) 6 ] 3 a n d [ T i ( C N ) 6 ] 3  are paramagnetic.

New answer posted

8 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

t 1 / 2 = 2 0 0 d a y s = 0 . 6 9 3 k

t = 2 . 3 0 3 k l o g 1 0 N 0 N

83 =  2 . 3 0 3 0 . 6 9 3 * 2 0 0 l o g N 0 N

8 3 = 2 0 0 0 . 3 0 1 0 l o g N 0 N

0.125 = l o g N 0 N

N 0 N = a n t i l o g ( 0 . 1 2 5 )

= 1.333

Activating remaining = N N 0 * 1 0 0

= N 0 1 . 3 3 3 N 0 * 1 0 0

= 75%

 

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

π = n v R T

π = W M * v R T

5 . 0 3 * 1 0 3 = 2 . 5 M * 0 . 5 * 0 . 0 8 3 * 3 0 0

M = 2 . 5 * 0 . 0 8 3 * 3 0 0 5 . 0 3 * 0 . 5 * 1 0 3 g / m o l

Molar mass of protein, M = 24751 g/mol

 Molar mass of glycine = 75 g/mol

So; number of glycine units =   2 4 7 5 1 7 5 = 3 3 0

New answer posted

8 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

E S n 2 + / S n 0 = 0 . 1 4 0 V = E 2 0

E S n 4 + / S n 0 = + 0 . 0 1 0 V = E 3 0

Here, Δ G 3 0 = Δ G 1 0 + Δ G 2 0

4 F E 3 0 = 2 F E 1 0 2 F E 2 0

E 3 0 = E 1 0 + E 2 0 2

+ 0 . 0 1 0 = E 1 0 + ( 0 . 1 4 0 ) 2

E 1 0 = E S n 4 + / S n 2 + 0 = + 0 . 1 6

= 16 * 10-2 V

 

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

For combustion of Mg:

Mg (s) + 1 2 O 2 ( g ) M g O ( s )

Here,   Δ n g = 1 2

Now using

Δ H = Δ U + Δ n g R T

-601.7 =   Δ U + ( 1 2 ) * 8 . 3 * 1 0 3 * 3 0 0

Δ U = 6 0 0 . 4 5 k J

So; magnitude of   Δ U is 600 kJ (the nearest integer).

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Using : PV = nRT

1.5 * 416 = n * 0.083 * 300

n = 25mol =  W M

25 =  1 0 0 M

So, molar mass, M = 4 g/mol.

New answer posted

8 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

8 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Compound X is sucrose as,

Here, glucose is A which on oxidation with HNO3 gives saccharic acid as

B is b-D fructose which is laevarotatory.

New answer posted

8 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Natural rubber is a linear polymer of isoprene called cis-polyisoprene and has elastic properties.

Ci-polyisoprene molecule consists of various chains held together by weak vander Waals interactions and has a coiled structure.

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