JAMB Chemistry 2014 Objective — Question 48
Question 48 of 50 from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2014 Objective paper, with the correct answer and a full explanation.
Advertisement
Acids (both organic and inorganic) react with carbonates to liberate carbon(IV) oxide. Since CH₃COOH (ethanoic acid) is an acid, the same must be expected of it: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. What type of substance is CH₃COOH?
- A. an organic acidCorrect
- B. an inorganic acid
- C. a base
- D. a neutral salt
Explanation
CH3COOH (ethanoic acid) is an organic acid, and like inorganic acids, it reacts with carbonates to liberate carbon(IV) oxide.
Advertisement
Sign up free to unlock
- Score tracking
- Practice history
- Saved questions
- Progress dashboard
- Personalized sessions
- Weak-topic breakdown
…and/or go further with premium services and No Ads.