Explanation Given: Concentration of CaCO 3 in hard water = 75 ppm Concentration of MgCO 3 in hard water = 55 ppm Volume of hard water = 10.0 L Formula: Parts per million by mass (ppm) is expressed as: 1ppm = 1mg solute 1kg solvent ------(1) Calculation: Step 1: Calculate the moles of CaCO 3 and MgCO 3 in 10.0 L of hard water Based on equation (1) we have, Mass of CaCO 3 (75 ppm) = 75 mg of CaCO 3 in 1000 g water Mass of MgCO 3 (55 ppm) = 55 mg of MgCO 3 in 1000 g water The total volume of hard water = 10.0 L = 10 , 000 m l The density of water = 1 g/ml; The mass of water = 10,000 g The total amount of CaCO 3 present in 10,000 g of hard water = 750 m g = 0.75 g MgCO 3 present in 10,000 g of hard water = 750 m g = 0...

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Chapter 17, Problem 137E

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To determine:

The mass of Na₂CO₃ required to remove 90% of Ca²⁺ and Mg²⁺ ions from 10.0 L of hard water.

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Chapter 17, Problem 136E

Chapter 17, Problem 138E

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Chemistry: Structure and Properties, Books a la Carte PACKAGE W/MasteringChemistry, 2nd Edition

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The mass of Na 2 CO 3 required to remove 90% of Ca 2+ and Mg 2+ ions from 10.0 L of hard water.

Solution Summary: The author calculates the mass of Na 2 CO 3 required to remove 90% of the total ions from Step 1.

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To determine:

The mass of Na₂CO₃ required to remove 90% of Ca²⁺ and Mg²⁺ ions from 10.0 L of hard water.

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Chapter 17 Solutions

The mass of Na 2 CO 3 required to remove 90% of Ca 2+ and Mg 2+ ions from 10.0 L of hard water.

Solution Summary: The author calculates the mass of Na 2 CO 3 required to remove 90% of the total ions from Step 1.

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To determine: The mass of Na2CO3 required to remove 90% of Ca2+ and Mg2+ ions from 10.0 L of hard water.

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Chapter 17 Solutions

To determine:

The mass of Na₂CO₃ required to remove 90% of Ca²⁺ and Mg²⁺ ions from 10.0 L of hard water.