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Solved: Complete and balance each acid–base equation.

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

Solution for problem 86E 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 86E

Complete and balance each acid–base equation.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the molar mass of the metal hydroxide used in the above titration. Also, we are going to identify the metal cation in the metal hydroxide used. Step1: a) Given, molarity of the hydrochloric acid used = 2.50 M Volume of the hydrochloric acid used = 56.9 mL = 56.9 / 1000 L = 0.0569 L Therefore, number of moles of hydrochloric acid used = volume of the solution in litres X molarity = 0.0569 L X 2.50 M = 0.142 mol Step2: Since, 2A group metal forms dihydroxide, 2 moles of HCl will be required to neutralize 1 mole of the metal hydroxide. Thus, to neutralize 0.142 mol of HCl, (0.142 / 2 = 0.071) moles of the metal hydroxide will be required. Now, Number of moles of the metal hydroxide = mass of...

Step 2 of 3

Chapter 4, Problem 86E is Solved
Step 3 of 3

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

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Solved: Complete and balance each acid–base equation.

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