Chemistry

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New answer posted

10 months ago

0 Follower 10 Views

R
Raj Pandey

Contributor-Level 9

Fe? ³ + e? → Fe? ² E° = 0.77V
Zn (s) → Zn? ² + 2e? ; E° = 0.76V
Cell reaction: 2Fe? ³ + Zn → 2Fe? ² + Zn? ² E°cell = 1.53V
Ecell = E°cell - (0.059/2)log ( [Zn? ²] [Fe? ²]²/ [Fe? ³]²)
1.5 = 1.53 - (0.06/2)log (1 * [Fe? ²]²/ [Fe? ³]²)
-0.03 = -0.03 log ( [Fe? ²]/ [Fe? ³])²
1 = log ( [Fe? ²]/ [Fe? ³])² => [Fe? ²]/ [Fe? ³] = 10
Let total iron = T. [Fe? ³] + [Fe? ²] = T. [Fe? ³] + 10 [Fe? ³] = T. 11 [Fe? ³] = T.
fraction of Fe? ³ = [Fe? ³]/T = 1/11 ≈ 0.09
This solution seems to differ from the image. Let's follow the image's steps.
log ( [Fe? ²]/ [Fe? ³])² = 1 => ( [Fe? ²]/ [Fe? ³])² = 10
[Fe? ²]/ [Fe?

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New answer posted

10 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Cr? O? ²? + 6Fe²? + 14H? → 2Cr³? + 6Fe³? + 7H? O
n-factor for Cr? O? ²? = 6, for Fe²? = 1
M.E Cr? O? ²? = M.E Fe²?
M? V? n? = M? V? n?
0.02 * 15 * 6 = M? * 10 * 1
M? = (0.02 * 15 * 6)/10 = 0.18 M = 18 * 10? ² M

New answer posted

10 months ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 * 10³) = 0.5 * 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) * n (H? O) = 0.5 * 10? ³ * 50 = 25 * 10? ³ moles = 25 mmol

New answer posted

10 months ago

0 Follower 10 Views

R
Raj Pandey

Contributor-Level 9

C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) * 72g = 46.4 g
= 464 * 10? ¹g

New answer posted

10 months ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

ΔH? = ΔH? + ΔH?
= 2.8 + 98.2 = 101 kJ/mole

New answer posted

10 months ago

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R
Raj Pandey

Contributor-Level 9

Number of sigma bonds are 10.
[Structure showing 10 sigma bonds in the molecule]

 

New answer posted

10 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

Solubility of 2 group hydroxide increases down the group.


New answer posted

10 months ago

0 Follower 26 Views

R
Raj Pandey

Contributor-Level 9

[Reactions showing reduction of various functional groups]
(1) Nitrile reduced to primary amine with H? /Ni.
(2) Nitrile and aldehyde reduced to primary amine and alcohol with LAH.
(3) Nitro group reduced to primary amine with Na/Hg, C? H? OH.
(4) Nitrile reduced to aldehyde with SnCl? /HCl.
Products of reaction 1, 2 and 3 are 1° amines, so react with Hinsberg's reagent to form sulphonamide. Product of (4) is an aldehyde, which does not react.

New answer posted

10 months ago

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R
Raj Pandey

Contributor-Level 9

Cl?C=CCl? is not used for dry cleaning of clothes due to safety purpose.

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

H? O (l) → H? O (g)
ΔH° = ΔU° + ΔngRT
ΔH° - ΔU° = ΔngRT
= 1 * 8.31 * 373
= 3099.63 J/mol
= 30.9963 * 10² J/mol
≈ 31 * 10² J/mol

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