EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 7, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

The molar mass of NaOH has to be given.

(a)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of NaOH is 40g.

Explanation of Solution

Given,

The atomic mass of sodium atom =23g

The atomic mass of oxygen atom =16g

The atomic mass of Hydrogen atom =1g

The molecular formula can be calculated as,

Molecular formula of NaOH=23+1+16

Molecular formula of NaOH=40g

The molar mass of NaOH is 40g.

(b)

Interpretation Introduction

Interpretation:

The molar mass of Ag2CO3 has to be given.

(b)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of Ag2CO3 is 276g.

Explanation of Solution

Given,

The atomic mass of carbon atom =12g

The atomic mass of silver atom =108g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of Ag2CO3=2×108+12+3×16

Molecular formula of Ag2CO3=276g

The molar mass of Ag2CO3 is 276g.

(c)

Interpretation Introduction

Interpretation:

The molar mass of Cr2O3 has to be given.

(c)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of Cr2O3 is 152g.

Explanation of Solution

Given,

The atomic mass of chromium =52g

The atomic mass of oxygen atom =16g

The molecular formula can be calculated as,

Molecular formula of Cr2O3=2×52+3×16

Molecular formula of Cr2O3=152g

The molar mass of Cr2O3 is 152g.

(d)

Interpretation Introduction

Interpretation:

The molar mass of (NH4)2CO3 has to be given.

(d)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of (NH4)2CO3 is 96g.

Explanation of Solution

Given,

The atomic mass of Nitrogen atom =14g

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 (NH4)2CO3=2×(14+4)+12+3×16

Molecular formula of (NH4)2CO3=96g

The molar mass of (NH4)2CO3 is 96g.

(e)

Interpretation Introduction

Interpretation:

The molar mass of Mg(HCO3)2 has to be given.

(e)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of Mg(HCO3)2 is 146.3g.

Explanation of Solution

Given,

The atomic mass of magnesium =24.3g

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 Mg(HCO3)2=24.3+2×(1+12+3×16)

Molecular formula of Mg(HCO3)2=146.3g

The molar mass of Mg(HCO3)2 is 146.3g.

(f)

Interpretation Introduction

Interpretation:

The molar mass of C6H5COOH has to be given.

(f)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of C6H5COOH is 122g.

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 C6H5COOH=7×12+6+2×16

Molecular formula of C6H5COOH=122g

The molar mass of C6H5COOH is 122g.

(g)

Interpretation Introduction

Interpretation:

The molar mass of C6H12O6 has to be given.

(g)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of C6H12O6 is 180g.

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 C6H12O6=6×12+12×1+6×16

Molecular formula of C6H12O6=180g

The molar mass of C6H12O6 is 180g.

(h)

Interpretation Introduction

Interpretation:

The molar mass of K4Fe(CN)6 has to be given.

(h)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of K4Fe(CN)6 is 211.845g.

Explanation of Solution

Given,

The atomic mass of potassium =39g

The atomic mass of iron atom =55.845g

The atomic mass of carbon atom =12g

The atomic mass of nitrogen atom =14g

The molecular formula can be calculated as,

Molecular formula of K4Fe(CN)6=4×39+55.845+6×(12+14)

Molecular formula of K4Fe(CN)6=211.845g

The molar mass of K4Fe(CN)6 is 211.845g.

(i)

Interpretation Introduction

Interpretation:

The molar mass of BaCl2.2H2O has to be given.

(i)

Expert Solution
Check Mark

Answer to Problem 2PE

The molar mass of BaCl2.2H2O is 244.2g.

Explanation of Solution

Given,

The atomic mass of barium atom =137.327g

The atomic mass of chlorine atom =35.45g

The atomic mass of oxygen atom =16g

The atomic mass of hydrogen atom =1g

The molecular formula can be calculated as,

Molecular formula of BaCl2.2H2O=137.327+2×35.45+2×(2+16)

Molecular formula of BaCl2.2H2O=244.2g

The molar mass of BaCl2.2H2O is 244.2g.

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Chapter 7 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

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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