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  1. 30.0 mL * 0.750 M = 22.5 mmoles H3PO4. H3PO4 can releases up to 3 H+ ions, so 22.5 mmoles H3PO4 can give 67.5 mmoles H+. To neutralize 67.5 mmoles H+, you need an equivalent in OH...

    1 Answers · Science & Mathematics · 20/05/2010

  2. Sr(OH)2 + 2 HCl → SrCl2 + 2 H2O (11.2 mL) x (0.340 M HCl) x (1 mol Sr(OH)2 / 2 mol HCl) / (0.228 M Sr(OH)2) = 8.35 mL Sr(OH)2

    1 Answers · Science & Mathematics · 14/10/2012

  3. Do you know the MaVa=MbVb equation, as in the molarity of acid multiplied by its volume is equal to the molarity of the base multiplied by its volume? Well, judging by the question, I would look for key words and...

    1 Answers · Science & Mathematics · 02/10/2011

  4. Here's one I answered a bit before yours: https://answers.yahoo.com/question/index?qid=20161202102857AApz0z9 For your problem, we need the moles of NaOH: (0.03 mol/L) (0.6 L) = 0.018 mol...

    1 Answers · Science & Mathematics · 15/12/2016

  5. 24*725= 61*x. x= 17400/61. x= 285. 725-285= 440[mL] To evaporate.

    1 Answers · Science & Mathematics · 09/12/2012

  6. N1 x V1=N2 x V2, so 16 x187=2992. 2992/619=4.834M

    2 Answers · Science & Mathematics · 22/01/2009

  7. HCl(aq) + NaOH(aq) --> NaCl(aq) + HOH(l) 50.0mL ...90.0mL ?M ..........0.650M 0.0900 L x (0.650 mol NaOH / 1L) x (1 mol HCl / 1 mol NaOH) / 0.0500L = 1.17 mol HCl / 1L ...... or ..... 1...

    1 Answers · Science & Mathematics · 05/11/2014

  8. To answer this question, you need the Ka (or pKa) value for propionic acid. I looked up the Ka and it's 1.3 x 10^-5. So pKa = -log Ka = -log (1.3 x 10^-5) = 4.89. I'll abbreviate propionic acid as HPr. Buffer problems...

    1 Answers · Science & Mathematics · 11/02/2012

  9. moles NH3 = 1.5 M x 0.030 L= 0.045 moles HNO3 = 0.100 L x 0.50 = 0.050 moles HNO3 in excess = 0.050 - 0.045 =0.005 total volume = 0.130 L [H+]= 0.005/ 0.130 =0.385 M pH = 1.41

    1 Answers · Science & Mathematics · 06/05/2009

  10. Why would you add sodium azide to make a buffer with ammonia? It would seem more likely that you would form a buffer using ammonium nitrate or something like that. The buffer couple is NH3/NH4+. On the other hand, do you mean HN3 (hydrazoic acid)? It will form a...

    1 Answers · Science & Mathematics · 26/09/2010

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