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NEET Class neet Answered

  CH3COOH(aq) → CH3COO- (aq) + H+ (aq), ΔrH° = 0.005 kcal g-1 Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution is a) -27.4 kcal mol-1 b) -13.4 kcal mol-1 c) -26.8 kcal mol-1 d) -27.1 kcal mol-1 Correct answer is option 'C'. Can you explain this answer?
Asked by ntg432000 | 06 Mar, 2019, 11:26: AM
answered-by-expert Expert Answer

the change in enthalpy is calculated as follows:

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Answered by Ramandeep | 06 Mar, 2019, 06:35: PM
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