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The heat of neutralisation, when 0.5 mole of HNO₃ solution is added to 0.2 moles of NaOH solution, is (Heat of neutralisation = 57 KJ-mol-1)
  • a)
    11.4 KJ
  • b)
    23.5 KJ
  • c)
    34.7 KJ
  • d)
    58.8 KJ
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The heat of neutralisation, when 0.5 mole of HNO₃ solution is ad...
HNO3=0.5 mol then ΔH =57KJ-mol-1we have already 0.2 mol of NaOHwhen 1 mol of NaoH are neutralized heat released =57 KJ-mol-10.2 mol of NaOH are neutralized heat released = 57 multiply 0.2=11.4 KJ-mol-1
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The heat of neutralisation, when 0.5 mole of HNO₃ solution is ad...
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The heat of neutralisation, when 0.5 mole of HNO₃ solution is added to 0.2 moles of NaOH solution, is (Heat of neutralisation = 57 KJ-mol-1)a)11.4 KJb)23.5 KJc)34.7 KJd)58.8 KJCorrect answer is option 'A'. Can you explain this answer?
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The heat of neutralisation, when 0.5 mole of HNO₃ solution is added to 0.2 moles of NaOH solution, is (Heat of neutralisation = 57 KJ-mol-1)a)11.4 KJb)23.5 KJc)34.7 KJd)58.8 KJCorrect answer is option 'A'. 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 The heat of neutralisation, when 0.5 mole of HNO₃ solution is added to 0.2 moles of NaOH solution, is (Heat of neutralisation = 57 KJ-mol-1)a)11.4 KJb)23.5 KJc)34.7 KJd)58.8 KJCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The heat of neutralisation, when 0.5 mole of HNO₃ solution is added to 0.2 moles of NaOH solution, is (Heat of neutralisation = 57 KJ-mol-1)a)11.4 KJb)23.5 KJc)34.7 KJd)58.8 KJCorrect answer is option 'A'. Can you explain this answer?.
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