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Answer to Question #76674 in General Chemistry for holly
Question #76674
how many moles of NaOH are present in 18.5 ml of 0.130 M NaOH
Expert's answer
СM = n/V n = CM · V
n (NaOH) = 0.130 · 18.5 = 2.405 mmol = 0.002 mol
n (NaOH) = 0.130 · 18.5 = 2.405 mmol = 0.002 mol
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