What happens when two atoms form a chemical bond?A) Two atoms fuse together to form a chemical bondB) The atoms share or transfe
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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.
Answer:
The [SO₃²⁻]
Explanation:
From the first dissociation of sulfurous acid we have:
H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)
At equilibrium: 0.50M - x x x
The equilibrium constant (Ka₁) is:
With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:
Similarly, from the second dissociation of sulfurous acid we have:
HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)
At equilibrium: 7.94x10⁻²M - x x x
The equilibrium constant (Ka₂) is:
Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:
Therefore, the final concentrations are:
[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M
[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M
[SO₃²⁻] = 7.07x10⁻⁵M
[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M
So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.
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Answer:
0.159 \%
Explanation:
The acid will dissociate according to the reaction shown below:-
Given that, pH=3.8
The concentration of can be determined from the expression fo pH as:-
pH = - log
3.8 = - log
=
The initial concentration of RCOOH was 0.10 M, then the percent dissociation was- calculated as shown below:-