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Molarity = # Moles/ # Liters 0.400 M HCl = X Moles/.0600 Liters X Moles = .400 M HCl x .0600 Liters X Moles = 0.024 Moles

3 Answers · Science & Mathematics · 14/12/2013

Initial Volume X initial concentration = Final Volume X Final Concentration therefore: XL X 5.0M = 0.3L X 0.04M solve for X: X = (0.3L)(0.04M)/(5...

2 Answers · Science & Mathematics · 10/09/2007

You have : ------------------ ( M HCl ) ( V HCl ) = ( M NaOH ) ( V NaOH ) M HCl = ( M NaOH ) ( V NaOH / V HCl ) M HCl = ( 0.02387M ) ( 42.10...

1 Answers · Science & Mathematics · 08/10/2015

moles of MgCl2 ---> (3.67 mol/L) (0.0250 L) = 0.09175 mol For the equation, we will consume 2 moles of AgNO3 for every mole of MgCl2 consumed. Therefore: 0.09175 mol times 2 = 0.1835 moles of AgNO3...

2 Answers · Science & Mathematics · 05/05/2013

More than one way to do this. Here is what I think is the easiest. pOH = 14 - pH pOH goes from 5.9 to 2.4 Work out initial, final values of [OH-] From the equation, for every two [OH...

1 Answers · Science & Mathematics · 14/09/2011

M1V1 = M2V2 (1.75 mol/L) (x) = (0.183 mol/L) (100 mL) x = 10.457 mL Round off as you see fit.

1 Answers · Science & Mathematics · 20/04/2014

You want 0.001 mol in 1 liter. because that is what means 1.0mM solution. So as 180g is 1 mole you need 180/1000 = 0.180g (180mg) for a liter. So devide that by 10 to get how much you need for 100ml. That is 18mg or 0.0180g. Hope that helps.

1 Answers · Science & Mathematics · 27/01/2010

which buffer ?

1 Answers · Science & Mathematics · 05/03/2009

Qsp = [Ag+]²[CrO4+2] Qsp = [(0.0050 M)(200 ml) / (200 + 300 ml)]² [(0.0020 M)(300 ml) / (200 + 300 ml)] Qsp = 4.8e-9 Qsp > Ksp (precipitate forms)

1 Answers · Science & Mathematics · 05/03/2012

Moles LiI = 16.93 g / 133.845 g/mol = 0.126 2 LiI + Ag2SO4 >> 2 AgI + Li2SO4 moles Ag2SO4 needed = 0.126 / 2 =0.0630 V = 0.0630 / 1.63 = 0.0387 L => 38.7 mL

1 Answers · Science & Mathematics · 05/10/2008

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