The following two compounds are constitutional isomers. Identify which of these is expected to be more acidic, and explain your
choice. a. The compound below is more acidic because its conjugate base is more resonance stabilized.b. The conjugate base of the other compound is not as much resonance stabilized.c. The compound below is more acidic because its conjugate base is more resonance stabilized.d. The conjugate base of the other compound is not as much resonance stabilized.e. The compound below is more acidic because its conjugate base is resonance stabilized.f. The conjugate base of the other compound is not resonance stabilized.Tg. he compound below is more acidic because its conjugate base is resonance stabilized.h. The conjugate base of the other compound is not resonance stabilized.
1 answer:
Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
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Answer: The final molarity of the diluted oxalic acid solution is 0.0032 M
Explanation:
Molarity: It is defined as the number of moles of solute present per liter of the solution.
Formula used :
where,
n= moles of solute
Moles=
= volume of solution in ml
To calculate the final molarity of the diluted oxalic acid solution
where,
are the molarity and volume of concentrated oxalic acid solution.
are the molarity and volume of diluted oxalic acid solution.
We are given:
Putting values in above equation, we get:
Thus the final molarity of the diluted oxalic acid solution is 0.0032 M