Chemistry

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New answer posted

10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

The reaction is: FeCl? + 3H? C? O? + 3KOH → K? [Fe (C? O? )? ] (A) + 3HCl + 3H? O

In the complex [Fe (C? O? )? ]³? , the oxalate ion (C? O? )²? is a bidentate ligand.

There are three bidentate ligands, so the coordination number, or secondary valency, of Fe is 3 * 2 = 6.

New answer posted

10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

The cell constant (G) is given by G = κ * R, where κ is conductivity and R is resistance.

Since the cell constant is constant: (κ * R)KCl = (κ * R)HCl

0.14 Sm? ¹ * 4.19 Ω = κ_HCl * 1.03 Ω

κ_HCl = (0.14 * 4.19) / 1.03 = 0.569 Sm? ¹

This is equivalent to 56.9 * 10? ² Sm? ¹.

The answer, rounded off, is 57.

 

New answer posted

10 months ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

-SO? H acts as a cation exchanger.

-NH? acts as an anion exchanger.

New answer posted

10 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

3CaO + 2Al → 3Ca + Al? O?
ΔH? = ΣΔH? (Products) - ΣΔH? (Reactants)
ΔH? = [ (0) + (-1675) ] - [ (3 * -635) + (0) ]
ΔH? = -1675 - (-1905) = -1675 + 1905 = 230 kJ
Ans = 230

New answer posted

10 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Moles of benzene = 3.9 g / 78 g/mol . This would produce (3.9 / 78) moles of nitrobenzene in 100% conversion.
Produced moles of nitrobenzene = 4.92 g / 123 g/mol .
% yield = [ (4.92 / 123) / (3.9 / 78) ] * 100 = [ (4.92 * 100 * 78) / (123 * 3.9) ] = 80.0%
Ans = 80

New answer posted

10 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

Molality = (mole of solute * 1000) / wt of solvent (gm)
100 = (n_solute * 1000) / [ (1 - n_solute) * 18]
(1 - n_solute) / n_solute = 1000 / (100 * 18) = 10/18
18 (1 - n_solute) = 10 n_solute
18 - 18 n_solute = 10 n_solute
18 = 28 n_solute
n_solute = 18 / 28? 0.6428 = 64.28 * 10? ²
Ans = 64 (Rounded off)

New answer posted

10 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

For n = 4, the possible values of l are 0, 1, 2, 3.
For l = 3, and m = -3.
Radial nodes = (n - l - 1) = (4 - 3 - 1) = 0
Ans = 0

New answer posted

10 months ago

0 Follower 16 Views

V
Vishal Baghel

Contributor-Level 10

Cr? O? ²? + Fe²? - (H? )-> Cr³? + Fe³?
(n=6) (n=1)
Meq Cr? O? ²? = Meq Fe²?
20 * 0.03 * 6 = 15 * M * 1
M = (20 * 0.03 * 6) / 15 = 0.24 = 24 * 10? ²
Ans = 24

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