Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 7, Problem 34E
Interpretation Introduction

(a)

Interpretation:

The number of atoms in 7.70g iodine atoms is to be calculated.

Concept introduction:

Mole is a basic unit used in the International system of units (SI). It is abbreviated as mol. It can be defined as 6.022×1023 units of some quantity which can be atoms, molecules or ions. The quantity 6.022×1023 is known as the Avogadro’s number.

Expert Solution
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Answer to Problem 34E

The number of atoms in 7.70g iodine is 3.65×1022atoms.

Explanation of Solution

The molar mass of iodine atom is 126.9gmol1.

Therefore, one mole of iodine atom is 126.9g.

The number of moles in 7.70g of iodine is given as shown below.

Molesofiodineatom=1moleofiodineatom126.9gofiodineatom×weightofiodineatoming=1moleofiodineatom126.9gofiodineatom×7.70gofiodineatom=6.06×102molesofiodineatom

The number of atoms in iodine atom is calculated as shown below.

Atomsofiodineatom=(6.022×1023atomsofiodineatom×molesofiodineatom1moleofiodineatom)

Substitute the number of atoms of iodine atom in above equation as shown below.

Atomsofiodineatom=(6.022×1023atomsofiodineatom×6.06×102molesofiodineatoms1moleofiodineatom)=3.65×1022atomsofiodineatom

Conclusion

The number of atoms in 7.70g iodine is calculated as 3.65×1022atoms.

Interpretation Introduction

(b)

Interpretation:

The number of molecules in 0.447gC9H20 is to be calculated.

Concept introduction:

Mole is a basic unit used in the International system of units (SI). It is abbreviated as mol. It can be defined as 6.022×1023 units of some quantity which can be atoms, molecules or ions. The quantity 6.022×1023 is known as the Avogadro’s number.

Expert Solution
Check Mark

Answer to Problem 34E

The number of molecules in 0.447gC9H20 is 2.1×1021molecules.

Explanation of Solution

The molecular formula is C9H20.

The molar mass of carbon is 12.01gmol1.

The molar mass of hydrogen is 1.008gmol1.

The molar mass of C9H20 is calculated below.

Totalmolarmass=(9×12.01)+(20×1.008)=128.25gmol1

Therefore, the molar mass of C9H20 is 128.25gmol1.

This means one mole of C9H20 is 128.25g.

The number of moles in 0.447gC9H20 is given as shown below.

MolesofC9H20=1moleofC9H20128.25gofC9H20×weightofC9H20ing=1moleofC9H20128.25gofC9H20×0.447gofC9H20=3.48×103molesofC9H20

The number of molecules in C9H20 is calculated as shown below.

MoleculesofC9H20=(6.022×1023moleculesof128.25g×molesofC9H201moleofC9H20)

Substitute the number of moles of C9H20 in above equation as shown below.

MoleculesofC9H20=(6.022×1023moleculesofC9H20×3.48×103molesofC9H201moleofC9H20)=2.1×1021moleculesofC9H20

Conclusion

The number of molecules in 0.447gC9H20 is calculated as 2.1×1021molecules.

Interpretation Introduction

(c)

Interpretation:

The number of formula units in 72.6g manganese (II) carbonate is to be calculated.

Concept introduction:

Mole is a basic unit used in the International system of units (SI). It is abbreviated as mol. It can be defined as 6.022×1023 units of some quantity which can be atoms, molecules or ions. The quantity 6.022×1023 is known as the Avogadro’s number.

Expert Solution
Check Mark

Answer to Problem 34E

The number of formula units in 72.6g manganese (II) carbonate is 3.80×1023formulaunits.

Explanation of Solution

The molecular formula for manganese (II) carbonate is Mn(CO3).

The molar mass of manganese is 54.94gmol1.

The molar mass of oxygen is 16.00gmol1.

The molar mass of carbon is 12.01gmol1.

The molar mass of Mn(CO3) is calculated below.

Totalmolarmass=54.94+12.01+(3×16.00)=114.95gmol1

Therefore, the molar mass of Mn(CO3) is 114.95gmol1.

This means one mole of Mn(CO3) is 114.95g.

The number of moles in Mn(CO3) is given as shown below.

MolesofMn(CO3)=1moleofMn(CO3)114.95gofMn(CO3)×weightofMn(CO3)ing=1moleofMn(CO3)114.95gofMn(CO3)×72.6gofMn(CO3)=0.632molesofMn(CO3)

The number of formula units in Mn(CO3) is calculated as shown below.

Formula unitsofMn(CO3)=6.022×1023formula unitsofMn(CO3)1moleofMn(CO3)×molesofMn(CO3) Substitute the number of moles of Mn(CO3) in above equation,

Formula unitsofMn(CO3)=(6.022×1023formula unitsofMn(CO3)×0.632molesofMn(CO3)1moleofMn(CO3))=3.80×1023formula unitsofMn(CO3)

Conclusion

The number of formula units in 72.6g manganese (II) carboante is calculated as 3.80×1023formulaunits.

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Chapter 7 Solutions

Introductory Chemistry: An Active Learning Approach

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