Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
15th Edition
ISBN: 9781119231318
Author: Morris Hein
Publisher: Wiley (WileyPLUS Products)
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Chapter 7, Problem 1PE

(a)

Interpretation Introduction

Interpretation:

The molar mass of KBr has to be given.

(a)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of KBr is 119g.

Explanation of Solution

The molecular formula has one potassium atom, and one bromine atoms.  The molecular mass of KBr can be calculated using the atomic masses of each atom in the molecular formula.

The atomic mass of potassium =39g

The atomic mass of bromine =80g

Adding the values of atomic masses of each atom the molar mass of KBr is found as 119g.

(b)

Interpretation Introduction

Interpretation:

The molar mass of Na2SO4 has to be given.

(b)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of Na2SO4 is 142g.

Explanation of Solution

Given,

The atomic mass of Sodium atom =23g

The atomic mass of sulfur atom =32g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of Na2SO4=2×23+32+4×16

Molecular formula of Na2SO4=142g

The molar mass of Na2SO4 is 142g.

(c)

Interpretation Introduction

Interpretation:

The molar mass of Pb(NO3)2 has to be given.

(c)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of Pb(NO3)2 is 331.2g.

Explanation of Solution

Given,

The atomic mass of lead atom =207.2g

The atomic mass of nitrogen atom =14g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of Pb(NO3)2=207.2+2×(14+3×16)

Molecular formula of Pb(NO3)2=331.2g

The molar mass of Pb(NO3)2 is 331.2g.

(d)

Interpretation Introduction

Interpretation:

The molar mass of C2H5OH has to be given.

(d)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of C2H5OH is 46g.

Explanation of Solution

Given,

The atomic mass of carbon atom =12g

The atomic mass of Hydrogen atom =1g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of C2H5OH=2×12+6×1+16

Molecular formula of C2H5OH=46g

The molar mass of C2H5OH is 46g.

(e)

Interpretation Introduction

Interpretation:

The molar mass of HC2H3O2 has to be given.

(e)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of HC2H3O2 is 60g.

Explanation of Solution

Given,

The atomic mass of carbon atom =12g

The atomic mass of Hydrogen atom =1g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of HC2H3O2=2×12+4×1+2×16

Molecular formula of HC2H3O2=60g

The molar mass of HC2H3O2 is 60g.

(f)

Interpretation Introduction

Interpretation:

The molar mass of Fe3O4 has to be given.

(f)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of Fe3O4 is 231.53g.

Explanation of Solution

Given,

The atomic mass of iron atom =55.845g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of Fe3O4=3×55.845+4×16

Molecular formula of Fe3O4=231.53g

The molar mass of Fe3O4 is 231.53g.

(g)

Interpretation Introduction

Interpretation:

The molar mass of C12H22O11 has to be given.

(g)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of C12H22O11 is 342g.

Explanation of Solution

Given,

The atomic mass of carbon atom =12g

The atomic mass of hydrogen atom =1g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of C12H22O11=12×12+22×1+11×16

Molecular formula of C12H22O11=342g

The molar mass of C12H22O11 is 342g.

(h)

Interpretation Introduction

Interpretation:

The molar mass of Al2(SO4)3 has to be given.

(h)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of Al2(SO4)3 is 342g.

Explanation of Solution

Given,

The atomic mass of luminium atom =27g

The atomic mass of sulfur atom =32g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of Al2(SO4)3=2×27+3×(32+4×16)

Molecular formula of Al2(SO4)3=342g

The molar mass of Al2(SO4)3 is 342g.

(i)

Interpretation Introduction

Interpretation:

The molar mass of (NH4)2HPO4 has to be given.

(i)

Expert Solution
Check Mark

Answer to Problem 1PE

The molar mass of (NH4)2HPO4 is 132g.

Explanation of Solution

Given,

The atomic mass of nitrogen atom =14g

The atomic mass of phosphorus atom =31g

The atomic mass of oxygen atom =16g

The atomic mass of hydrogen atom =1g

The molecular formula can be calculated as,

Molecular formula of (NH4)2HPO4=2×(14+4×1)+1+31×1+4×16

Molecular formula of (NH4)2HPO4=132g

The molar mass of (NH4)2HPO4 is 132g.

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Chapter 7 Solutions

Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card

Ch. 7.4 - Prob. 7.11PCh. 7.5 - Prob. 7.12PCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQCh. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 1PECh. 7 - Prob. 2PECh. 7 - Prob. 3PECh. 7 - Prob. 4PECh. 7 - Prob. 5PECh. 7 - Prob. 6PECh. 7 - Prob. 7PECh. 7 - Prob. 8PECh. 7 - Prob. 9PECh. 7 - Prob. 10PECh. 7 - Prob. 11PECh. 7 - Prob. 12PECh. 7 - Prob. 13PECh. 7 - Prob. 14PECh. 7 - Prob. 15PECh. 7 - Prob. 16PECh. 7 - Prob. 17PECh. 7 - Prob. 18PECh. 7 - Prob. 19PECh. 7 - Prob. 20PECh. 7 - Prob. 21PECh. 7 - Prob. 22PECh. 7 - Prob. 25PECh. 7 - Prob. 26PECh. 7 - Prob. 27PECh. 7 - Prob. 28PECh. 7 - Prob. 29PECh. 7 - Prob. 30PECh. 7 - Prob. 31PECh. 7 - Prob. 32PECh. 7 - Prob. 33PECh. 7 - Prob. 34PECh. 7 - Prob. 35PECh. 7 - Prob. 36PECh. 7 - Prob. 37PECh. 7 - Prob. 38PECh. 7 - Prob. 39PECh. 7 - Prob. 40PECh. 7 - Prob. 41PECh. 7 - Prob. 42PECh. 7 - Prob. 43PECh. 7 - Prob. 44PECh. 7 - Prob. 45PECh. 7 - Prob. 46PECh. 7 - Prob. 47PECh. 7 - Prob. 48PECh. 7 - Prob. 49PECh. 7 - Prob. 50PECh. 7 - Prob. 51PECh. 7 - Prob. 52PECh. 7 - Prob. 53AECh. 7 - Prob. 54AECh. 7 - Prob. 55AECh. 7 - Prob. 56AECh. 7 - Prob. 57AECh. 7 - Prob. 58AECh. 7 - Prob. 59AECh. 7 - Prob. 60AECh. 7 - Prob. 61AECh. 7 - Prob. 62AECh. 7 - Prob. 63AECh. 7 - Prob. 64AECh. 7 - Prob. 65AECh. 7 - Prob. 66AECh. 7 - Prob. 67AECh. 7 - Prob. 68AECh. 7 - Prob. 69AECh. 7 - Prob. 70AECh. 7 - Prob. 71AECh. 7 - Prob. 72AECh. 7 - Prob. 73AECh. 7 - Prob. 74AECh. 7 - Prob. 75AECh. 7 - Prob. 76AECh. 7 - Prob. 77AECh. 7 - Prob. 78AECh. 7 - Prob. 79AECh. 7 - Prob. 80AECh. 7 - Prob. 81AECh. 7 - Prob. 82AECh. 7 - Prob. 83AECh. 7 - Prob. 84AECh. 7 - Prob. 88AECh. 7 - Prob. 89CECh. 7 - Prob. 90CE
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