Solve both for me 2. A solution of 0.255 N KOH is needed to analyze acidic mine water.K-39, O=16, H=1a. Calculate the mass of KOH needed to prepare 500mL of the solutionmol kott- 10.255 more) (0.5x) = 0.1275 mol kot9 KOH = 0.127556 g/molxmolI mol- 7.14 кон93. 0.24 N HCI is needed to analyze a sample of soda ashCalculate the volume of conc. HCI to prepare 750 mL of the solution, Conc. HCI = 12a.C₁V₁ = C₂ V ₂CO-24 MC 750 m)V₂=15ml of HC112 1M(1215) (V₂)12M=x=39= 16i56 g/mol

2. A solution of 0.255 N KOH is needed to analyze acidic mine water. K=39, O=16, H=1 a. Calculate the mass of KOH needed to prepare 500mL of the solution mol kott- co.255 ml) (0.5) -0.1275 mol kot 9 кон - о- 1275 56 g/mol mol x I mol - 7.14, кон 3. 0.24 N HCI is needed to analyze a sample of soda ash a. Calculate the volume of conc. HCI to prepare 750 mL of the solution, Conc. HCI = 12 CiV₁ =C₂ V ₂ CO-24 MJC 750m) 12 M (1215) (V₂) 12M K=39 0 - 16 =-1 56 g/mol √₂-15 ml of HC1

2. A solution of 0.255 N KOH is needed to analyze acidic mine water. K=39, O=16, H=1 a. Calculate the mass of KOH needed to prepare 500mL of the solution mol kott- co.255 ml) (0.5) -0.1275 mol kot 9 кон - о- 1275 56 g/mol mol x I mol - 7.14, кон 3. 0.24 N HCI is needed to analyze a sample of soda ash a. Calculate the volume of conc. HCI to prepare 750 mL of the solution, Conc. HCI = 12 CiV₁ =C₂ V ₂ CO-24 MJC 750m) 12 M (1215) (V₂) 12M K=39 0 - 16 =-1 56 g/mol √₂-15 ml of HC1

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 121AP: Calcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion...

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