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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

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 · 18/11/2009

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 · 26/10/2008

a level teaspoon is generally around 5ml. .

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

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

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

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

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

It is a 1:1 reaction. So 1 mL of HCl will react with 1.2/0.516 mL of NaOH. Since we are estimating this is somewhat less than 2.4 mL, so it will take somewhat less than 24 mL of the NaOH to react with 10...

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

2.96 : 100 = x : 677 x = 20.0 mL

1 Answers · Science & Mathematics · 11/12/2007

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