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aev [14]
3 years ago
5

Select the true statements regarding these resonance structures of formate.? Each carbon-oyxgen bond is somewhere between a sing

le and double bond. The actual structure of formate switches back and forth between the two resonance forms. Each oxygen atom has a double bond 50% of the time.

Chemistry
2 answers:
Law Incorporation [45]3 years ago
8 0

True statement regarding the resonance structures of formate is that the .

\boxed{{\text{carbon - oxygen bond lies in between the single and double bond}}}

Further explanation:

The bonding between the different atoms in covalent molecules is shown by some diagrams known as the Lewis structures. These also show the presence of lone pairs in the molecule. These are also known as Lewis dot diagrams, electron dot diagrams, Lewis dot structures or Lewis dot formula. In covalent compounds, the geometry, polarity, and reactivity are predicted by these structures.

When more than one Lewis structures are possible for a single molecule but no single structure is able to explain all the properties of the molecule then resonance is used. All the structures thus formed are called resonating structures and the phenomenon is known as resonance. The resonance structures have the same placement of atoms but different locations of bond pairs and lone pairs of electrons. Moreover, various resonating structures can be converted to each other by moving lone pairs to bonding positions, and vice-versa.

The general rules that we follow to draw the resonance structures are as follows:

a. The \pi electrons or a lone pair of electrons can change their positions while the position of atoms remains fixed.

b. The count of the valence electrons in all the resonating structures should be same.

c. The transfer of electrons is shown by the curved arrows.

d. The resonating structure must follow octet rule that is all atoms should have 8 electrons

The resonating structures and resonance hybrid of the formate is represented in the attached image.

A single canonical structure such as (I) cannot describe all the experimentally observed properties of the formate ion. The non-bonding electrons belonging to the p orbitals of the oxygen undergo delocalization so that nature of the carbon-oxygen bond is neither purely single nor double bond but has an intermediate bond length that is 0.133{\text{ nm}}. This bond length is greater than the double bond 0.123{\text{ nm}} but less than the carbon-oxygen single bond that is  0.143{\text{ nm}}.

Learn more:

1. When titrating a strong acid with a strong base what will be pH? brainly.com/question/3168961

2. How many covalent bond nitrogen forms: brainly.com/question/5974553

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

Keywords: Lewis structure, valence electrons, formal charge, HCO2-, oxygen, p orbitals, double bonds, a single bond, bonding electrons and non-bonding electrons.

son4ous [18]3 years ago
7 0

Answer : Option 1) The true statement is each carbon-oxygen bond is somewhere between a single and double bond and the actual structure of format is an average of the two resonance forms.

Explanation : The actual structure of formate is found to be a resonance hybrid of the two resonating forms. The actual structure for formate do not switches back and forth between two resonance forms.

The O atom in the formate molecule with one bond and three lone pairs, in the resonance form left with reference to the attached image, gets changed into O atom with two bonds and two lone pairs.

Again, the O atom with two bonds and two lone pairs on the resonance form left, changed into O atom with one bond and three lone pairs. It concludes that each carbon-oxygen bond is neither a single bond nor a double bond; each carbon-oxygen bond is somewhere between a single and double bond.

Also, it is seen that each oxygen atom does not have neither a double bond nor a single bond 50% of the time.

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Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

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             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

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Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

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Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

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3 years ago
Chemical formula for Potassium Hydrogen Nitrate
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Explanation:

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In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.

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3 0
2 years ago
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume
MissTica

Answer:

4.56 X 10^ -4 g/mL

Explanation:

A solution is  prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.

(7.6 X10^-4  gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g

this is dissolved )in 10 m L=45.6 X 10^-4  g/ 10

4.56 X 10^ -4 g/mL

check

6/10 =0.6

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8 0
3 years ago
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