Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
Question
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Chapter 6, Problem 1CE
Interpretation Introduction

Interpretation:

An explanation as to how a binary molecular compound can be distinguished from a binary acid is to be stated.

Concept introduction:

Binary molecular compounds are the compounds formed by the combination of two nonmetals. Binary acid is the aqueous solution of the compounds containing one element as hydrogen and other one as a nonmetal.

Expert Solution & Answer
Check Mark

Answer to Problem 1CE

Binary acids are the aqueous solutions and are ultimately represented by the symbol (aq) in the end whereas binary molecules are not the aqueous solutions and are never represented by the symbol (aq).

Explanation of Solution

Hydrogen is a nonmetal and there is a possibility that a binary molecular compound contains hydrogen and one nonmetal but still, it is not a binary acid. For example, H2O is a binary molecular compound as it contains hydrogen. Binary acid is the aqueous solution of compounds containing hydrogen and one other nonmetal.

The special feature of binary acids is that they are represented by the symbol (aq) in the end of the formula of acid. For example, HCl(aq). The binary molecular compounds are never represented by the symbol (aq).

Conclusion

Binary acids are the aqueous solutions and are represented by the symbol (aq) in the end while binary molecules compounds are not the aqueous solutions and are never represented by the symbol (aq).

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Chapter 6 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 6 - Prob. 11CECh. 6 - Prob. 12CECh. 6 - Prob. 13CECh. 6 - Prob. 14CECh. 6 - Prob. 15CECh. 6 - Prob. 16CECh. 6 - Prob. 17CECh. 6 - Prob. 18CECh. 6 - Prob. 19CECh. 6 - Prob. 1KTCh. 6 - Prob. 2KTCh. 6 - Prob. 3KTCh. 6 - Prob. 4KTCh. 6 - Prob. 5KTCh. 6 - Prob. 6KTCh. 6 - Prob. 7KTCh. 6 - Prob. 8KTCh. 6 - Prob. 9KTCh. 6 - Prob. 10KTCh. 6 - Prob. 11KTCh. 6 - Prob. 12KTCh. 6 - Prob. 13KTCh. 6 - Prob. 14KTCh. 6 - Prob. 15KTCh. 6 - Prob. 16KTCh. 6 - Prob. 17KTCh. 6 - Prob. 1ECh. 6 - Prob. 2ECh. 6 - Prob. 3ECh. 6 - Prob. 4ECh. 6 - Prob. 5ECh. 6 - Prob. 6ECh. 6 - Prob. 7ECh. 6 - Prob. 8ECh. 6 - Prob. 9ECh. 6 - Prob. 10ECh. 6 - Prob. 11ECh. 6 - Prob. 12ECh. 6 - Prob. 13ECh. 6 - Prob. 14ECh. 6 - Prob. 15ECh. 6 - Prob. 16ECh. 6 - Prob. 17ECh. 6 - Prob. 18ECh. 6 - Prob. 19ECh. 6 - Prob. 20ECh. 6 - Prob. 21ECh. 6 - Prob. 22ECh. 6 - Prob. 23ECh. 6 - Prob. 24ECh. 6 - Prob. 25ECh. 6 - Prob. 26ECh. 6 - Prob. 27ECh. 6 - Prob. 28ECh. 6 - Prob. 29ECh. 6 - Prob. 30ECh. 6 - Prob. 31ECh. 6 - Prob. 32ECh. 6 - Prob. 33ECh. 6 - Prob. 34ECh. 6 - Prob. 35ECh. 6 - Prob. 36ECh. 6 - Prob. 37ECh. 6 - Prob. 38ECh. 6 - Prob. 39ECh. 6 - Prob. 40ECh. 6 - Prob. 41ECh. 6 - Prob. 42ECh. 6 - Prob. 43ECh. 6 - Prob. 44ECh. 6 - Prob. 45ECh. 6 - Prob. 46ECh. 6 - Prob. 47ECh. 6 - Prob. 48ECh. 6 - Prob. 49ECh. 6 - Prob. 50ECh. 6 - Prob. 51ECh. 6 - Prob. 52ECh. 6 - Prob. 53ECh. 6 - Prob. 54ECh. 6 - Prob. 55ECh. 6 - Prob. 56ECh. 6 - Prob. 57ECh. 6 - Prob. 58ECh. 6 - Prob. 59ECh. 6 - Prob. 60ECh. 6 - Prob. 61ECh. 6 - Prob. 62ECh. 6 - Prob. 63ECh. 6 - Prob. 64ECh. 6 - Prob. 65ECh. 6 - Prob. 66ECh. 6 - Prob. 67ECh. 6 - Prob. 68ECh. 6 - Prob. 69ECh. 6 - Prob. 70ECh. 6 - Prob. 71ECh. 6 - Prob. 72ECh. 6 - Prob. 73ECh. 6 - Prob. 74ECh. 6 - Prob. 75ECh. 6 - Prob. 76ECh. 6 - Prob. 77ECh. 6 - Prob. 78ECh. 6 - Prob. 79ECh. 6 - Prob. 80ECh. 6 - Prob. 1STCh. 6 - Prob. 2STCh. 6 - Prob. 3STCh. 6 - Prob. 4STCh. 6 - Prob. 5STCh. 6 - Prob. 6STCh. 6 - Prob. 7STCh. 6 - Prob. 8STCh. 6 - Prob. 9STCh. 6 - Prob. 10STCh. 6 - Prob. 11STCh. 6 - Prob. 12STCh. 6 - Prob. 13STCh. 6 - Prob. 14STCh. 6 - Prob. 15STCh. 6 - Prob. 16STCh. 6 - Prob. 17STCh. 6 - Prob. 18STCh. 6 - Prob. 19STCh. 6 - Prob. 20ST
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