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A 31.5 mL aliquot of HNO3 (aq) of unknown concentration was titrated with 0.0134 M NaOH (aq). It took 23.9 mL of the base to reach the endpoint of the titration. The concentration (M) of the acid was __________.
Answer:
Moles NaOH = 23.9 x 0.0134 /1000 = 0.000320 = moles HNO3
M = moles / L = 0.000320 / 0.0315 L = 0.0101 M
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Answer:
Moles NaOH = 23.9 x 0.0134 /1000 = 0.000320 = moles HNO3
M = moles / L = 0.000320 / 0.0315 L = 0.0101 M
you need to do your own homework or your going to fail your exams cos you have no clue what is going on and FunQA.com won't be there. If you need this much help you should be asking your teacher for help after class.
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