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Mkey [24]
2 years ago
13

How many electrons are present in the nonbonding π molecular orbital of the allyl anion? a. 2 b. 1 c. 3 d. 0

Chemistry
1 answer:
Nostrana [21]2 years ago
6 0

The number of electrons, which are present in the nonbonding π molecular orbital of the allyl anion is "0".

Anions, cations, but also allylic radicals have always been frequently mentioned as reaction intermediates. Each one has three adjacent sp^{2}-hybridized carbon centers, and they all rely on resonance for stability. Two resonance structures would be used to present each species, with the charge as well as unpaired electron scattered across both the 1,3 and 0 positions.

The Aufbau principle states that these orbitals would fill up based on the order of stability, therefore a typical pi bond, will have 2 electrons in the Pi orbital as well as zero in the Pi* orbital.

Therefore, the correct answer will be option (d)

To know more about allyl anion

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When an iron nail is ground into powder, its mass ____.
Gre4nikov [31]

Explanation:

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3 0
3 years ago
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Crazy boy [7]

1. The volume of ammonia consumed in the reaction is 23.2 L

2. The volume of oxygen consumed in the reaction is 29 L

<h3>Balanced equation</h3>

4NH₃(g) + 5O₂(g) --> 4NO(g) + 6H₂O(g)

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃ and 5 L of O₂

<h3>1. How to determine the volume of ammonia, NH₃ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 4) / 6 = 23.2 L of NH₃

Thus, 23.2 L of NH₃ were consumed

<h3>2. How to determine the volume of oxygen, O₂ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 5 L of O₂

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 5) / 6 = 29 L of O₂

Thus, 29 L of O₂ were consumed

Learn more about stoichiometry:

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7 0
2 years ago
How much heat is evolved in converting 1.00 mol of steam at 135.0 ∘c to ice at -45.0 ∘c? the heat capacity of steam is 2.01 j/(g
Viktor [21]
1 mole of steam (H2O) = 18g
Heat evolved = MCФ where m is the mass and c s specific heat capacity while Ф is change in temperature. Latent heat of fusion is 334 kj/kg while latent heat of vaporization is 2260 kj/kg and specific heat of water is 4.2 j/g/c
= 18 ×2.01 × (135-100) = 1266.3 J
    0.018 × 334000 =  6012 J (change of state from gas (steam) to liquid (water)
 18 × 4.186× (100 -0) =  7534.8 J
 0.018 × 2260000 = 40680 J (change of state from liquid to solid ice)
 18 × 2.09 × (0--45) = 1692.9 J
The total heat evolved is therefore 57186 J or 57.186 kJ
4 0
3 years ago
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