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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

Solve both for me

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- 10.255 more) (0.5x) = 0.1275 mol kot
9 KOH = 0.1275
56 g/mol
x
mol
I mol
- 7.14 кон
9
3. 0.24 N HCI is needed to analyze a sample of soda ash
Calculate the volume of conc. HCI to prepare 750 mL of the solution, Conc. HCI = 12
a.
C₁V₁ = C₂ V ₂
CO-24 MC 750 m)
V₂=15ml of HC1
12 1
M
(1215) (V₂)
12M
=
x=39
= 16
i
56 g/mol

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