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aev [14]
2 years ago
11

Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You

will probably find some useful data in the ALEKS Data resource. acetic acid
Chemistry
1 answer:
Sliva [168]2 years ago
7 0

Answer:

0.11%

Explanation:

Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.

CH3COOH <=======================================> CH3COO⁻ + H⁺

Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻  and H⁺ is 0 respectively.

At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x  and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.

1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.

The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:

percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%

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PLS HELP-CHEMISTRY: What is the molar concentration of an ammonia solution in which there are 0.1557 grams
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a compound is 21.20% nitrogen,6.06% hydrogen, 24,30% sulfur, and 48.45% oxygen. write the empirical formula and name the compoun
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3 years ago
Determine if each statement is True or False. [ Select ] Central atoms with four electron groups will be sp3 hybridized. [ Selec
meriva

Answer:

Central atoms with four electron groups will be sp3 hybridized. True

Hybrid orbitals are delocalized over the entire molecule. False

The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. True

Atoms with a single pi bond and an octet are sp2 hybridized. True

Sometimes oxygen atoms will be sp3d hybridized in organic molecules. False

All resonance structures must be considered when assigning hybridization. False

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When a central atom has four electron groups attached to it, then it must be sp3 hybridized. This is the case in ammonia, water, hydrogen sulphide, methane etc.

Hybridization is a valence bond concept while delocalization is a molecular orbital theory concept. Hybridized orbitals are localized on central atoms in a molecule while delocalized orbitals spread across the entire molecule.

According to valence bond theory, the number of atomic orbitals that combined to give hybrid orbitals must be equal to the number of hybrid orbitals formed.

When an atom has a single pi bond and on octet of electrons, then it must be sp2 hybridized. Remember that in ethene for instance, carbon has one pi bond and an octet of electrons.

Oxygen has an empty n=3 level hence it can not have d-orbitals involved in hybridization.

The electron domain geometry of the molecule is considered when assigning hybridization and not the resonance structures. All the resonance structures must have the central atom in the same hybridization state.

8 0
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