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Given that heat of neutralisation of strong acid and strong base is -57.1 kJ. Calculate the heat produced when 0.25 mole of HCl is neutralised with 0.25 mole of NaOH in aqueous solution.
  • a)
    22.5 kJ
  • b)
    57.1 kJ
  • c)
    14.275 kJ
  • d)
    28.6 kJ
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Given that heat of neutralisation of strong acid and strong base is -5...
57.1 is the heat of neutralization of 1 mole. In this question we asked about 0.25 mole hence it is equal to 0.25*57.1=14. 2
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Given that heat of neutralisation of strong acid and strong base is -57.1 kJ. Calculate the heat produced when 0.25 mole of HCl is neutralised with 0.25 mole of NaOH in aqueous solution.a)22.5 kJb)57.1 kJc)14.275 kJd)28.6 kJCorrect answer is option 'C'. Can you explain this answer?
Question Description
Given that heat of neutralisation of strong acid and strong base is -57.1 kJ. Calculate the heat produced when 0.25 mole of HCl is neutralised with 0.25 mole of NaOH in aqueous solution.a)22.5 kJb)57.1 kJc)14.275 kJd)28.6 kJCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Given that heat of neutralisation of strong acid and strong base is -57.1 kJ. Calculate the heat produced when 0.25 mole of HCl is neutralised with 0.25 mole of NaOH in aqueous solution.a)22.5 kJb)57.1 kJc)14.275 kJd)28.6 kJCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given that heat of neutralisation of strong acid and strong base is -57.1 kJ. Calculate the heat produced when 0.25 mole of HCl is neutralised with 0.25 mole of NaOH in aqueous solution.a)22.5 kJb)57.1 kJc)14.275 kJd)28.6 kJCorrect answer is option 'C'. Can you explain this answer?.
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