Answer NaBr :- Sodium bromide Pb(CH 3 COO) 2 :- Lead(II) acetate (NH 4 ) 2 CO 3 :- Ammonium carbonate Explanation NaBr :- Sodium bromide formed by sodium (Na + ) cation and bromide (Br - ) anion. Pb(CH 3 COO) 2 :- Lead(II) acetateformed by lead (Pb 2+ ) cation and acetate (CH 3 COO - ) anion. (NH 4 ) 2 CO 3 :- Ammonium carbonateformed by ammonium (NH 4 + ) cation and carbonate (CO 3 2- ) anion.

1st Edition

ISBN: 9780078746376

Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom

Publisher: Glencoe/McGraw-Hill School Pub Co

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Question

Chapter 14, Problem 123A

Interpretation Introduction

Interpretation:

The name of the compounds have to be written.

Concept introduction:

Cations and anions are combined together to form neutral compounds and the net charge of an ionic compound must be zero.

Expert Solution & Answer

Answer to Problem 123A

NaBr :- Sodium bromide
Pb(CH₃COO)₂ :- Lead(II) acetate
(NH₄)₂CO₃ :- Ammonium carbonate

Explanation of Solution

NaBr :- Sodium bromide formed by sodium (Na⁺) cation and bromide (Br^-) anion.
Pb(CH₃COO)₂ :- Lead(II) acetateformed by lead (Pb²⁺) cation and acetate (CH₃COO^-) anion.
(NH₄)₂CO₃ :- Ammonium carbonateformed by ammonium (NH₄⁺) cation and carbonate (CO₃^2-) anion.

Chapter 14, Problem 122A

Chapter 14, Problem 124A

Chapter 14 Solutions

Chemistry: Matter and Change

Ch. 14.1 - Prob. 1SSC Ch. 14.1 - Prob. 2SSC Ch. 14.1 - Prob. 3SSC Ch. 14.1 - Prob. 4SSC Ch. 14.1 - Prob. 5SSC Ch. 14.1 - Prob. 6SSC Ch. 14.1 - Prob. 7SSC Ch. 14.1 - Prob. 8SSC Ch. 14.2 - Prob. 9PP Ch. 14.2 - Prob. 10PP

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