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Solved: A 35.0-mL sample of 1.00 M KBr and a 60.0-mL

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 99 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 99

A 35.0-mL sample of 1.00 M KBr and a 60.0-mL sample of 0.600 M KBr are mixed. The solution is then heated to evaporate water until the total volume is 50.0 ml. How many grams of silver nitrate are required to precipitate out silver bromide in the final solution?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the quantity of silver nitrate required to precipitate out silver bromide in the resulting solution of the given reaction. Step1: 35 Given, volume of the first solution of KBr = 35.0 mL = 1000 L = 0.035 L Molarity of the first KBr solution = 1.00 M Therefore, number of moles of first KBr solution = Volume in litres X Molarity = 0.035 X 1.00 =0.035 mol Volume of the second solution of KBr = 60.0 mL = 1000 L = 0.060 L Molarity of the second KBr solution = 0.600 M Therefore, number of moles of second KBr solution = Volume in litres X Molarity = 0.060 X 0.600 =0.036 mol Step2: Thus, total number of moles of KBr = 0.036 + 0.035 = 0.071 mol On heating the above mixture,...

Step 2 of 3

Chapter 4, Problem 99 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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Solved: A 35.0-mL sample of 1.00 M KBr and a 60.0-mL

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