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Brilliant_brown [7]
3 years ago
15

The oxides of nitrogen are very important ingredients in determining urban air pollution. Name each of the following compounds.(

a) N2O(b) NO(c) NO2(d) N2O5(e) N2O4
Chemistry
1 answer:
deff fn [24]3 years ago
6 0

Answer:

(a) N₂O - Dinitrogen monoxide or Dinitrogen (I) oxide

(b) NO - Nitrogen monoxide or Nitrogen (II) oxide

(c) NO₂ - Nitrogen dioxide or Nitrogen (IV) oxide

(d) N₂O₅ - Dinitrogen pentoxide

(e) N₂O₄ - Dinitrogen tetraoxide.

Explanation:

To name each of the compounds,

(a) N2O; It is properly written as N₂O. N₂O is a colorless, sweet tasting gas also known as "laughing gas".

N₂O - Dinitrogen monoxide OR Dinitrogen (I) oxide.

(b) NO; NO is a colorless gas.

NO - Nitrogen monoxide OR Nitrogen (II) oxide.

(c) NO2; It is properly written as NO₂

NO₂ is a reddish-brown gas.

NO₂ - Nitrogen dioxide or Nitrogen (IV) oxide.

(d) N2O5; It is properly written as N₂O₅

N₂O₅ is a white solid.

N₂O₅ - Dinitrogen pentoxide.

(e) N2O4; It is properly written as N₂O₄;

N₂O₄ is a red-brown liquid with an unpleasant smell.

N₂O₄ - Dinitrogen tetraoxide.

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Answer:

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Explanation:

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ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

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<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

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Answer:

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