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 11PE

(a)

Interpretation Introduction

Interpretation:

The number of atoms in 25 molecules of P2O5 has to be given.

(a)

Expert Solution
Check Mark

Answer to Problem 11PE

The number of atoms in 25 molecules of P2O5 is 175atoms.

Explanation of Solution

Given,

The number of molecules of P2O5 is 25.

There are seven atoms in P2O5 (two phosphorus and five oxygen atoms).

The number of atoms in 25 molecules of P2O5 can be calculated as,

  (25molecules)×(7atoms1molcule)=175atoms

The number of atoms in 25 molecules of P2O5 is 175atoms.

(b)

Interpretation Introduction

Interpretation:

The number of atoms in 3.62mol of O2 has to be given.

(b)

Expert Solution
Check Mark

Answer to Problem 11PE

The number of atoms in 3.62mol of O2 is 43.59×1023atoms.

Explanation of Solution

Given,

The number of moles of O2 is 3.62mol.

The Avogadro’s number is 6.022×1023molecules.

There are two atoms in O2.

The number of atoms in 3.62mol of O2 can be calculated as,

  (3.62moles)×(6.022×1023molecules1mol)(2atoms1molcule)=43.59×1023atoms

The number of atoms in 3.62mol of O2 is 43.59×1023atoms.

(c)

Interpretation Introduction

Interpretation:

The number of atoms in 12.2mol of CS2 has to be given.

(c)

Expert Solution
Check Mark

Answer to Problem 11PE

The number of atoms in 12.2mol of CS2 is 2.20×1025atoms.

Explanation of Solution

Given,

The number of moles of CS2 is 12.2mol.

The Avogadro’s number is 6.022×1023molecules.

There are three atoms in CS2 (one carbon atom and two sulfur atom).

The number of atoms in 12.2mol of CS2 can be calculated as,

(12.2moles)×(6.022×1023molecules1mol)(3atoms1molcule)=220.40×1023atoms=2.20×1025atoms

The number of atoms in 12.2mol of CS2 is 2.20×1025atoms.

(d)

Interpretation Introduction

Interpretation:

The number of atoms in 1.25g of Na has to be given.

(d)

Expert Solution
Check Mark

Answer to Problem 11PE

The number of atoms in 1.25g of Na is 0.327×1023 atoms.

Explanation of Solution

Given,

The molar mass of Na is 23g/mol.

The mass of sodium is 1.25g.

The Avogadro’s number is 6.022×1023atoms.

The number of atoms in 1.25g of Na can be calculated as,

  1.25(6.022×1023atoms23g)=0.327×1023atoms

The number of atoms in 1.25g of Na is 0.327×1023 atoms.

(e)

Interpretation Introduction

Interpretation:

The number of atoms in 2.7g of CO2 has to be given.

(e)

Expert Solution
Check Mark

Answer to Problem 11PE

The number of atoms in 2.7g of CO2 is 1.1×1023 atoms.

Explanation of Solution

Given,

The molar mass of CO2 is 44.01g/mol.

The mass of CO2 is 2.7g.

The Avogadro’s number is 6.022×1023atoms.

The number of atoms in 2.7g of CO2 can be calculated as,

  2.7(6.022×1023atoms×344.01g)=1.1×1023atoms

The number of atoms in 2.7g of CO2 is 1.1×1023 atoms.

(f)

Interpretation Introduction

Interpretation:

The number of atoms in 0.25g of CH4 has to be given.

(f)

Expert Solution
Check Mark

Answer to Problem 11PE

The number of atoms in 0.25g of CH4 is 0.469×1023 atoms.

Explanation of Solution

Given,

The molar mass of CH4 is 16.04g/mol.

The mass of CH4 is 0.25g.

The Avogadro’s number is 6.022×1023atoms.

The number of atoms in 0.25g of CH4 can be calculated as,

  0.25(6.022×1023atoms×516.04g)=0.469×1023atoms

The number of atoms in 0.25g of CH4 is 0.469×1023 atoms.

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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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