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Paraphin [41]
3 years ago
9

g When a species can be represented by two or more resonance structures, the actual representation of bonding in the species is

given by: A. the resonance structure with no multiple bonds B. the resonance hybrid of all structures C. the best resonance structure D. the resonance structure with no lone pairs on the central atom. E. the resonance structure with the most multiple bonds
Chemistry
2 answers:
AlekseyPX3 years ago
6 0

Answer:

B. the resonance hybrid of all structures

Explanation:

The idea of resonance is used to explain bonding in compounds where a single structure does not fully account for all the bonding interactions in a molecule.

A number of equivalent structures are then used to show the nature of bonding in such a molecule. Such structures are called resonance structures or canonical structures. None of these structures individually offer a holistic explanation to the bonding interactions in the molecule under study.

However, a hybrid of all the canonical structures does explain the nature of bonding in the molecule.

Luba_88 [7]3 years ago
4 0

Answer:

B. the resonance hybrid of all structures

Explanation:

When a species can be represented by two or more resonance structures, the actual representation of bonding in the species is given by: <u>the resonance hybrid of all structures</u>

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Calculate the volume of a 2.0 molar aqueous solution made from 14 miles of K2S
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Answer:

volume is 7.0 liters

Explanation:

We are given;

  • Molarity of the aqueous solution as 2.0 M
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We are required to determine the volume of the solution;

We need to know that;

Molarity = Moles ÷ volume

Therefore;

Volume = Moles ÷ Molarity

Thus;

Volume of the solution = 14 moles ÷ 2.0 M

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3 years ago
Calculate hydrochloric acid (umol)in 200 ul of a<br> 0.5173Msolution of acid?
strojnjashka [21]

<u>Answer:</u> The moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Or,

\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}

We are given:

Molarity of solution = 0.5173 M

Volume of solution = 200\mu L

Putting values in above equation, we get:

0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol

Hence, the moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

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Answer:

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