<span>1. D, 2. A, 3. B, 4. C, 5. D, 6. A, 7. B, 8.B....I think</span>
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.