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 10, Problem 57A

(a)

Interpretation Introduction

Interpretation: The representative particle needs to be determined in 3.00 mol Sn.

Concept Introduction: In a substance, a representative particle is the smallest unit of the substance in which it naturally exists. For example, for elements, the representative particle is an atom.

The four types of representative particles are as follows:

  1. Atom is the smallest particle of an element.
  2. Ions are atoms with positive or negative charges.
  3. Molecules are formed when two or more atoms are covalently bonded.
  4. Formula units are the simplest ratio of ions to form an ionic compound.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given substance is 3.0 mol of Sn. The smallest unit of the substance in which it naturally exists is an element. Thus, the representative element is the atom. According to Avogadro’s law, 1 mol of a substance contains 6.022×1023 atoms. Thus, the number of atoms in 3.0 mol of Sn is calculated as follows:

  N=3.0 mol6.022×1023 atoms1.0 mol=1.806×1024 atoms

Therefore, the number of atoms in 3.0 mol of Sn is 1.806×1024 atoms .

(b)

Interpretation Introduction

Interpretation: The representative particle needs to be determined in 0.400 mol KCl.

Concept Introduction: In a substance, a representative particle is the smallest unit of the substance in which it naturally exists. For example, for elements, the representative particle is an atom.

The four types of representative particles are as follows:

  1. Atom is the smallest particle of an element.
  2. Ions are atoms with positive or negative charges.
  3. Molecules are formed when two or more atoms are covalently bonded.
  4. Formula units are the simplest ratio of ions to form an ionic compound.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given substance is 0.40 mol of KCl. The smallest unit of the substance in which it naturally exists is the formula unit because it is an ionic compound. Thus, the representative element is the formula unit. According to Avogadro’s law, 1 mol of a substance contains 6.022×1023 formula unit. Thus, the number of formula unit in 0.40 mol of KCl is calculated as follows:

  N=0.40 mol6.022×1023 Formula units1.0 mol=2.41×1023 Formula units

Therefore, the number of formula units in 0.40 mol of KCl is 2.41×1023 Formula units .

(c)

Interpretation Introduction

Interpretation: The representative particle needs to be determined in 7.50 mol of SO2

Concept Introduction: In a substance, a representative particle is the smallest unit of the substance in which it naturally exists. For example, for elements, the representative particle is an atom.

The four types of representative particles are as follows:

  1. Atom is the smallest particle of an element.
  2. Ions are atoms with positive or negative charges.
  3. Molecules are formed when two or more atoms are covalently bonded.
  4. Formula units are the simplest ratio of ions to form an ionic compound.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given substance is 7.50 mol of SO2 . The smallest unit of the substance in which it naturally exists is a molecule because it is a covalent compound. Thus, the representative element is a molecule. According to Avogadro’s law, 1 mol of a substance contains 6.022×1023 molecule. Thus, the number of molecules in 7.50 mol of SO2 is calculated as follows:

  N=7.50 mol6.022×1023 molecules1.0 mol=4.52×1024 molecules

Therefore, the number of molecules in 7.50 mol of SO2 is 4.52×1024 molecules .

(d)

Interpretation Introduction

Interpretation: The representative particle needs to be determined in 4.80×103 mol of NaI.

Concept Introduction: In a substance, a representative particle is the smallest unit of the substance in which it naturally exists. For example, for elements, the representative particle is an atom.

(d)

Expert Solution
Check Mark

Explanation of Solution

The given substance is 4.80×103 mol of NaI. The smallest unit of the substance in which it naturally exists is the formula unit because it is an ionic compound. Thus, the representative element is the formula unit. According to Avogadro’s law, 1 mol of a substance contains 6.022×1023 formula unit. Thus, the number of formula unit in 4.80×103 mol of NaI is calculated as follows:

  N=4.80×103 mol6.022×1023 Formula units1.0 mol=2.90×1021 Formula units

Therefore, the number of formula units in 4.80×103 mol of NaI is 2.90×1021 Formula units .

Chapter 10 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 10.1 - Prob. 11LCCh. 10.1 - Prob. 12LCCh. 10.1 - Prob. 13LCCh. 10.1 - Prob. 14LCCh. 10.1 - Prob. 15LCCh. 10.2 - Prob. 16SPCh. 10.2 - Prob. 17SPCh. 10.2 - Prob. 18SPCh. 10.2 - Prob. 19SPCh. 10.2 - Prob. 20SPCh. 10.2 - Prob. 21SPCh. 10.2 - Prob. 22SPCh. 10.2 - Prob. 23SPCh. 10.2 - Prob. 24LCCh. 10.2 - Prob. 25LCCh. 10.2 - Prob. 26LCCh. 10.2 - Prob. 27LCCh. 10.2 - Prob. 28LCCh. 10.2 - Prob. 29LCCh. 10.2 - Prob. 30LCCh. 10.2 - Prob. 31LCCh. 10.2 - Prob. 32LCCh. 10.3 - Prob. 33SPCh. 10.3 - Prob. 34SPCh. 10.3 - Prob. 35SPCh. 10.3 - Prob. 36SPCh. 10.3 - Prob. 37SPCh. 10.3 - Prob. 38SPCh. 10.3 - Prob. 39SPCh. 10.3 - Prob. 40SPCh. 10.3 - Prob. 41SPCh. 10.3 - Prob. 42SPCh. 10.3 - Prob. 43LCCh. 10.3 - Prob. 44LCCh. 10.3 - Prob. 45LCCh. 10.3 - Prob. 46LCCh. 10.3 - Prob. 47LCCh. 10.3 - Prob. 48LCCh. 10.3 - Prob. 49LCCh. 10 - Prob. 50ACh. 10 - Prob. 51ACh. 10 - Prob. 52ACh. 10 - Prob. 53ACh. 10 - Prob. 54ACh. 10 - Prob. 55ACh. 10 - Prob. 56ACh. 10 - Prob. 57ACh. 10 - Prob. 58ACh. 10 - Prob. 59ACh. 10 - Prob. 60ACh. 10 - Prob. 61ACh. 10 - Prob. 62ACh. 10 - Prob. 63ACh. 10 - Prob. 64ACh. 10 - Prob. 65ACh. 10 - Prob. 66ACh. 10 - Prob. 67ACh. 10 - Prob. 68ACh. 10 - Prob. 69ACh. 10 - Prob. 70ACh. 10 - Prob. 71ACh. 10 - Prob. 72ACh. 10 - Prob. 73ACh. 10 - Prob. 74ACh. 10 - Prob. 75ACh. 10 - Prob. 76ACh. 10 - Prob. 77ACh. 10 - Prob. 78ACh. 10 - Prob. 79ACh. 10 - Prob. 80ACh. 10 - Prob. 81ACh. 10 - Prob. 82ACh. 10 - Prob. 83ACh. 10 - Prob. 84ACh. 10 - Prob. 85ACh. 10 - Prob. 86ACh. 10 - Prob. 87ACh. 10 - Prob. 88ACh. 10 - Prob. 89ACh. 10 - Prob. 90ACh. 10 - Prob. 91ACh. 10 - Prob. 92ACh. 10 - Prob. 93ACh. 10 - Prob. 94ACh. 10 - Prob. 95ACh. 10 - Prob. 96ACh. 10 - Prob. 97ACh. 10 - Prob. 98ACh. 10 - Prob. 99ACh. 10 - Prob. 100ACh. 10 - Prob. 101ACh. 10 - Prob. 102ACh. 10 - Prob. 103ACh. 10 - Prob. 104ACh. 10 - Prob. 105ACh. 10 - Prob. 106ACh. 10 - Prob. 108ACh. 10 - Prob. 109ACh. 10 - Prob. 110ACh. 10 - Prob. 111ACh. 10 - Prob. 112ACh. 10 - Prob. 113ACh. 10 - Prob. 114ACh. 10 - Prob. 115ACh. 10 - Prob. 116ACh. 10 - Prob. 117ACh. 10 - Prob. 118ACh. 10 - Prob. 119ACh. 10 - Prob. 120ACh. 10 - Prob. 121ACh. 10 - Prob. 122ACh. 10 - Prob. 123ACh. 10 - Prob. 124ACh. 10 - Prob. 125ACh. 10 - Prob. 126ACh. 10 - Prob. 127ACh. 10 - Prob. 128ACh. 10 - Prob. 1STPCh. 10 - Prob. 2STPCh. 10 - Prob. 3STPCh. 10 - Prob. 4STPCh. 10 - Prob. 5STPCh. 10 - Prob. 6STPCh. 10 - Prob. 7STPCh. 10 - Prob. 8STPCh. 10 - Prob. 9STPCh. 10 - Prob. 10STPCh. 10 - Prob. 11STPCh. 10 - Prob. 12STPCh. 10 - Prob. 13STPCh. 10 - Prob. 14STP
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