Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 2, Problem 54A

(a)

Interpretation Introduction

Interpretation: The relationship between hydrogen, oxygen, and water needs to be identified. Each item needs to be identified as an element, compound, or mixture.

Concept Introduction: An element is a substance that cannot be decomposed further into simpler substances by any chemical process. On the other hand, compounds are composed of two or more elements. They can be decomposed into their elemental components. A mixture in solid, liquid, or gaseous form is said to be homogenous if it has the same proportions of components throughout the sample. Thus, in any homogenous mixture, only one phase of matter is observed. In heterogeneous mixtures, the components of the mixture can be easily identified. There is more than one phase that can be seen by the naked eye. It has a non-uniform composition.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given items are hydrogen, oxygen, and water. Here, water is made up of hydrogen and oxygen combined in a 2:1 ratio. In all the samples of water, hydrogen and oxygen are present in a fixed ratio by mass which is 2 mol of hydrogen and 1 mol of oxygen. Here, hydrogen and oxygen are elements, but water is a compound.

(b)

Interpretation Introduction

Interpretation: The relationship between nitrogen, oxygen, and air needs to be identified. Each item needs to be identified as an element, compound, or mixture.

Concept Introduction: An element is a substance that cannot be decomposed further into simpler substances by any chemical process. On the other hand, compounds are composed of two or more elements. They can be decomposed into their elemental components. A mixture in solid, liquid, or gaseous form is said to be homogenous if it has the same proportions of components throughout the sample. Thus, in any homogenous mixture, only one phase of matter is observed. In heterogeneous mixtures, the components of the mixture can be easily identified. There is more than one phase and can be seen by the naked eye. It has a non-uniform composition.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given items are nitrogen, oxygen, and air. Air is composed of gases which include nitrogen and oxygen gas. Air is a mixture of different gases but the major percentage is nitrogen and oxygen gas. The approximate percentages of nitrogen and oxygen gas in the air are 78% and 21% respectively. Here, nitrogen and oxygen are elements, but the air is a homogenous mixture. It is homogenous because all the gases are uniformly distributed.

(c)

Interpretation Introduction

Interpretation: The relationship between sodium, chlorine, and table salt needs to be identified. Each item needs to be identified as an element, compound, or mixture.

Concept Introduction: An element is a substance that cannot be decomposed further into simpler substances by any chemical process. On the other hand, compounds are composed of two or more elements. They can be decomposed into their elemental components. A mixture in solid, liquid, or gaseous form is said to be homogenous if it has the same proportions of components throughout the sample. Thus, in any homogenous mixture, only one phase of matter is observed. In heterogeneous mixtures, the components of the mixture can be easily identified. There is more than one phase and can be seen by the naked eye. It has a non-uniform composition.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given items are sodium, chlorine, and table salt. Here, the name and formula for table salt are sodium chloride and NaCl respectively. It is made up of sodium and chlorine combined in a 1:1 ratio. In all the samples of table salt, sodium and chlorine elements are present in a fixed ratio by mass which is 1 mol of sodium and 1 mol of chlorine. Here, sodium and chlorine are elements, but table salt or NaCl is a compound.

(d)

Interpretation Introduction

Interpretation: The relationship between carbon, water, and table sugar needs to be identified. Each item needs to be identified as an element, compound, or mixture.

Concept Introduction: An element is a substance that cannot be decomposed further into simpler substances by any chemical process. On the other hand, compounds are composed of two or more elements. They can be decomposed into their elemental components. A mixture in solid, liquid, or gaseous form is said to be homogenous if it has the same proportions of components throughout the sample. Thus, in any homogenous mixture, only one phase of matter is observed. In heterogeneous mixtures, the components of the mixture can be easily identified. There is more than one phase that can be seen by the naked eye. It has a non-uniform composition.

(d)

Expert Solution
Check Mark

Explanation of Solution

The given items are carbon, water, and table sugar. Sugar is a carbohydrate that mainly contains carbon, hydrogen, and oxygen atoms. Here, carbon is an element, water is a compound of hydrogen and oxygen and table sugar is a homogenous mixture as components are uniformly distributed.

Chapter 2 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 2.1 - Prob. 11SPCh. 2.2 - Prob. 12LCCh. 2.2 - Prob. 13LCCh. 2.2 - Prob. 14LCCh. 2.2 - Prob. 15LCCh. 2.2 - Prob. 16LCCh. 2.2 - Prob. 17LCCh. 2.2 - Prob. 18LCCh. 2.2 - Prob. 19LCCh. 2.3 - Prob. 20SPCh. 2.3 - Prob. 21SPCh. 2.3 - Prob. 22LCCh. 2.3 - Prob. 23LCCh. 2.3 - Prob. 24LCCh. 2.3 - Prob. 25LCCh. 2.3 - Prob. 26LCCh. 2.3 - Prob. 27LCCh. 2.3 - Prob. 28LCCh. 2.3 - Prob. 29LCCh. 2.3 - Prob. 30LCCh. 2.3 - Prob. 31LCCh. 2.4 - Prob. 32LCCh. 2.4 - Prob. 33LCCh. 2.4 - Prob. 34LCCh. 2.4 - Prob. 35LCCh. 2.4 - Prob. 36LCCh. 2.4 - Prob. 37LCCh. 2.4 - Prob. 38LCCh. 2 - Prob. 39ACh. 2 - Prob. 40ACh. 2 - Prob. 41ACh. 2 - Prob. 42ACh. 2 - Prob. 43ACh. 2 - Prob. 44ACh. 2 - Prob. 45ACh. 2 - Prob. 46ACh. 2 - Prob. 47ACh. 2 - Prob. 48ACh. 2 - Prob. 49ACh. 2 - Prob. 50ACh. 2 - Prob. 51ACh. 2 - Prob. 52ACh. 2 - Prob. 53ACh. 2 - Prob. 54ACh. 2 - Prob. 55ACh. 2 - Prob. 56ACh. 2 - Prob. 57ACh. 2 - Prob. 58ACh. 2 - Prob. 59ACh. 2 - Prob. 60ACh. 2 - Prob. 61ACh. 2 - Prob. 62ACh. 2 - Prob. 63ACh. 2 - Prob. 64ACh. 2 - Prob. 65ACh. 2 - Prob. 66ACh. 2 - Prob. 67ACh. 2 - Prob. 68ACh. 2 - Prob. 69ACh. 2 - Prob. 70ACh. 2 - Prob. 71ACh. 2 - Prob. 72ACh. 2 - Prob. 73ACh. 2 - Prob. 74ACh. 2 - Prob. 75ACh. 2 - Prob. 76ACh. 2 - Prob. 77ACh. 2 - Prob. 78ACh. 2 - Prob. 79ACh. 2 - Prob. 80ACh. 2 - Prob. 81ACh. 2 - Prob. 82ACh. 2 - Prob. 83ACh. 2 - Prob. 84ACh. 2 - Prob. 85ACh. 2 - Prob. 86ACh. 2 - Prob. 87ACh. 2 - Prob. 88ACh. 2 - Prob. 89ACh. 2 - Prob. 90ACh. 2 - Prob. 93ACh. 2 - Prob. 94ACh. 2 - Prob. 95ACh. 2 - Prob. 1STPCh. 2 - Prob. 2STPCh. 2 - Prob. 3STPCh. 2 - Prob. 4STPCh. 2 - Prob. 5STPCh. 2 - Prob. 6STPCh. 2 - Prob. 7STPCh. 2 - Prob. 8STPCh. 2 - Prob. 9STPCh. 2 - Prob. 10STPCh. 2 - Prob. 11STPCh. 2 - Prob. 12STPCh. 2 - Prob. 13STPCh. 2 - Prob. 14STP
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