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enot [183]
2 years ago
9

Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges.

Be sure to include all appropriate nonbonding electrons and charges. A three carbon chain with a double bond between carbons 1 and 2. Carbon 1 is bonded to two hydrogen atoms, carbon two is bonded to one hydrogen atom, and carbon three is bonded to two hydrogen atoms. There is a plus one charge on the third carbon atom. Draw the resonance structure of the allyl ion. A carbon atom is single bonded to a C H 3 group, double bonded to an oxygen atom, and single bonded to an N atom. The N atom is bonded to a hydrogen atom. It has two lone pairs and a minus one charge. The O atom has two lone pairs. Draw the resonance structure of the amidate ion.
Chemistry
1 answer:
pshichka [43]2 years ago
7 0

Three resonance structures  can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

Learn more: brainly.com/question/4933048

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- 1273.02 kJ.

Explanation:

This problem can be solved using Hess's Law.

Hess's Law states that <em>regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.</em>

  • We should modify the given 3 equations to obtain the proposed reaction:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • We should multiply the first equation by (6) and also multiply its ΔH by (6):

6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,

  • Also, we should multiply the second equation and its ΔH by (6):

6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.

  • Finally, we should reverse the first equation and multiply its ΔH by (- 1):

6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.

  • By summing the three equations, we cam get the proposed reaction:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:

<em>∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ =</em> (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = <em>- 1273.02 kJ.</em>

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