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  1. 1.2g/204.22 (MW of KHP) = .00588 moles KHP = .00588 moles NaOH at endpoint M=mol/L 0.5 M = 0.00588 moles NaOH/x L = .01175 L = 11.75 mL

    1 Answers · Science & Mathematics · 19/11/2009

  2. ppm is equal to 0.001g/1,000mL so you will need 0.0001g of each for 100mL of ethanol. 1 mL=10^3 µL 0.0001g/density= volume of each to add Methanol=0.0001g/0.791g/mL=1.26 x 10^-4mL=1 x 10^-4 mL=0.1 µL Acetone=0.0001g/0...

    1 Answers · Science & Mathematics · 15/12/2013

  3. a. n = m/M = 0.4935/(40.08+12.011+3*15.9994) n = 4.9306E-3 moles were weighed out b. v = 250 mL = 0.25 L c = n/v = 4.9306E-3 mol / 0.25 L = 1.9722E-2 mol/L is molarity c. 25 mL = 0.1*250 mL so aliquot...

    1 Answers · Science & Mathematics · 27/10/2008

  4. a level teaspoon is generally around 5ml. .

    1 Answers · Science & Mathematics · 13/11/2013

  5. 1420 g/L x 69.5 / 100 =986.9 g/L of HNO3 986.9 g/L / 63.012 g/mol =15.7 M is the concentration of the acid = concentration H+ we need pH = 2.70 => [H+] = 10^-2.70 = 0.00200 M M1V1 = M2V2 15.7 V1 = 0.00200 x 5.50 L...

    1 Answers · Science & Mathematics · 16/03/2008

  6. NaOH is 39.99 g/mol In order to make 500mL of a 0.450M solution, you need 0.225mol NaOH 0.225*39.99 =8.99775 grams NaOH

    4 Answers · Science & Mathematics · 25/04/2011

  7. since the mmoles never change ( mass ) only the concentration will so M1V1= M2V2 M1= 6 M Vol 1 = ? vol 2 =100 M2 = 1 M so Vol1 = (100 x 1 )/6M = 16.667 mL take 16.667 mL of the 6 M dilute to exactly 100...

    1 Answers · Science & Mathematics · 29/11/2010

  8. The pH of the required acid is 2.70, that is, [H+] = 10^-pH = 10^-2.70 = 1.99 x 10^-3 M ≈ 2x10^-3 M Since HNO3 is a strong acid, it ionizes completely; HNO3(aq) ------> H^+(aq) + NO3...

    1 Answers · Science & Mathematics · 15/03/2008

  9. 36.5% by weight means there are 36.5 g HCl for every 100 g of solution. 36.5 g is 1 mole, while 100 g /1.200 g/ml = 83.33 mL 1 mole / 0.08333 L = 12 M concentrated solution To make 100 mL of 2 M HCl, use this equation...

    1 Answers · Science & Mathematics · 13/08/2010

  10. NaBr = 102.9 g/mol 3.8 g / 102.9g/mol = 0.0369 mol NaBr 0.29 = 0.0369 / L 0.29(L) = 0.0369 L = 0.127 L 0.127 x 1000 = 127 mL

    2 Answers · Science & Mathematics · 28/05/2009

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