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Lerok [7]
3 years ago
13

What are the trends and exceptions to the trends in electron affinity? Check all that apply.

Chemistry
2 answers:
Montano1993 [528]3 years ago
7 0

The trends and exceptions to the trends in electron affinity;

  • Elements in Group 14 have larger (more negative) electron affinities than elements in Group 15.
  • Elements in Group 2 have very large electron affinities.

Explanation:

Electron affinity can be defined as the tendency of a neutral atom to pick an additional electron to its orbital shell. The smaller the atomic radius of an atom, the higher the electron affinity and the less energy is expended in adding an electron to it. This is because the smaller the atomic radius,  the stronger the attraction by the nucleus even to its outermost shell. This means less energy is required to add the electron the smaller the atomic radius.

Therefore, electron affinity increases to the right of a period and decreases down a group in the periodic table.

Learn More:

For more on electron affinity check out;

brainly.com/question/1199346

brainly.com/question/2881335

#LearnWithBrainly

Readme [11.4K]3 years ago
5 0

Answer

it is A and C

Explanation:

The electron affinities of the elements in Group 17 are larger (more negative) than the elements in Group 1.

Elements in Group 14 have larger (more negative) electron affinities than elements in Group 15.

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The reaction between nitrogen and oxygen is given below: 2 N2(g) + O2(g) 2 N2O(g) We therefore know that which of the following
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Answer:

  • a)  2N₂O(g) → 2N₂(g)  + O₂(g)

Explanation:

Arrange the equations in the proper way for better understanding.

T<em>he reaction between nitrogen and oxygen is given below:</em>

<em />

  • <em>2N₂(g) + O₂(g) → 2N₂O(g)</em>

<em />

<em>We therefore know that which of the following reactions can also occur?</em>

<em />

  • <em>a)  2N₂O(g) → 2N₂(g)  + O₂(g)</em>
  • <em>b)  N₂(g) + 2O₂(g) → 2NO₂(g)</em>
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<h2>Solution</h2>

Notice that the first equation,  a) 2N₂O(g) → 2N₂(g)  + O₂(g), is the reverse of the original equation, 2N₂(g) + O₂(g) → 2N₂O(g).

The reactions in gaseous phase are reversible reactions that can be driven to one or other direction by modifying the conditions of temperature or pressure.

Thus, the equilibrium equation would be:

  • 2N₂(g) + O₂(g) ⇄ 2N₂O(g)

Which shows that both the forward and the reverse reactions occur.

Whether one or the other are favored would depend on the temperature and pressure: high temperatures would favor the reaction that consumes more heat (the endothermic reaction) and high pressures would favor the reaction that consumes more moles.

Thus, by knowing that one of the reactions can occur you can conclude that the reverse reaction can also occur.

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Answer:

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Explanation:

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