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Write a molecular equation for the precipitation reaction

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

Solution for problem 77E 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 77E

Problem 77E

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. If no reaction occurs, write “NO REACTION.”

a. potassium carbonate and lead(II) nitrate

b. lithium sulfate and lead(II) acetate

c. copper(II) nitrate and magnesium sulfide

d. strontium nitrate and potassium iodide

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the molarity of a solution of acetic acid made by dissolving 20.00 mL of glacial acetic acid in enough water to make 250 mL of solution. Step1: Molarity is defined as the number of moles of the solute(Here acetic acid is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: In order to calculate molarity of acetic acid solution, we have to first calculate the number of moles of acetic acid dissolved in the solution. Number of moles can be calculated by the following mathematical formula: Mass of the solute Number of moles(n) = ------------------------ Molar mass of the solute. Step3: Given, density of glacial acetic acid = 1.049 g/mL Volume of the glacial acetic acid used = 20.00 mL Now, density is defined as mass per unit volume, i.e., Density = mass / volume Therefore, mass = density X volume = 1.049 x 20.00 [(g/mL) X mL] = 20.98 g Thus, mass of the glacial acetic acid used = 20.98 g Molar mass of glacial acetic acid(CH C3H) = 60.04 g/mol Therefore, number of moles of acetic acid = 20.98 / 60.04 = 0.349 mol Volume of the solution in litres = 250 mL = 250/1000 L = 0.250 L Putting these values in equation (1), we get, Molarity = 0.349 mol / 0.250 L = 1.396 M Thus , the molarity of the acetic acid solution is 1.396 M. ----------------

Step 2 of 3

Chapter 4, Problem 77E is Solved
Step 3 of 3

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

The full step-by-step solution to problem: 77E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. This full solution covers the following key subjects: Acetic, acid, solution, glacial, liquid. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Since the solution to 77E from 4 chapter was answered, more than 578 students have viewed the full step-by-step answer. The answer to “Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. If no reaction occurs, write “NO REACTION.”a. potassium carbonate and lead(II) nitrateb. lithium sulfate and lead(II) acetatec. copper(II) nitrate and magnesium sulfided. strontium nitrate and potassium iodide” is broken down into a number of easy to follow steps, and 47 words.

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