Molar Solutions - NaOH

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• Solutions, Precipitations, and Balancing Equations

  NaOH+HNO3NaNO3+H2O Since in the balanced equation, the molar amount of NaOH is equal to molar amount of HNO3. The volume of HNO3: 25*0.103/0.150=17.17 ml The molecules which are soluble are determined.

• Titration of an Antacid

  The volumes of these two solutions are then used to find the neutralizing power of the antacid.

• Organic Chemistry - Calculating Molar Solubility

  d) Solution D is made by adding 0.0150 mil solid NaOH into Solution A e) Calculate the pH for each of the solutions, A,B,C, and D 6.

• Calculation of buffer solutions

  57552 Calculation of buffer solutions Please show me how to solve this problem, step by step. 0.50 molar solutions of the following weak acids HOCL and NaOCL are available.

• 3 acid base titrations solved and explained

  contains 1.35g of NaOH per liter. 0.105M HCl means 0.105 moles of hydrochloric acid per liter of solution Other useful numbers: molar mass of HCl = 36.46 g/mol molar mass of NaOH = 39.9971 g/mol molar mass of NH3 = 17.0306 (though

• Molar solubility of Fe(OH)2 in NaOH

  17200 Molar solubility Calculate molar solubility of Fe(OH)2 in 0.10 Molar NaOH. we know that Fe(OH)2 undergoes the following equilibrium:
  Fe(OH)2 ---> Fe+2 + 2OH- ( we also know that it has a cetrain Ksp=8e-16 where e-16 is ten to the power of -

• Titration, estimation of carbon dioxide in carbonated beverage

  (g/l) Molar mass(g/mol) Molar mass of NaOH = (23 + 1 + 16) g/mol = 40g/mol Molar conc. (mol/l) = 5.67g/l 40g/mol = 0.1475 mol/l No. of mole = Conc.

• Converting molarity to mL

  Now, moles of NaOH = Mass of NaOH dissolved/Molar mass = y/40 Here, y is not given in the problem.

• A 100 ml sample of .10 molar NAOH solution was added to 100 ml of .10 molar H3C6H5O7

  63728 A 100 ml sample of .10 molar NAOH solution was added A 100 ml sample of .10 molar NAOH solution was added to 100 ml of .10 molar H3C6H5O7.