World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6.2, Problem 1RQ

(a)

Interpretation Introduction

Interpretation: The number of moles of Ba and Cl in BaCl2 needs to be determined.

Concept Introduction: The number of moles of an element can be determined from the molecular formula of the compound. The subscript for an element is equal to the number of moles of that element present in 1 mol of the compound.

(a)

Expert Solution
Check Mark

Answer to Problem 1RQ

The number of moles of Ba is 0.15 mol and that of Cl is 0.30 mol.

Explanation of Solution

The number of moles of BaCl2 is 0.15 mol.

From the formula, it can be seen that 1 mol of the compound contains 2 mol of Cl thus, number of moles of Cl in 0.15 mol of the compound can be calculated as follows:

  n=0.15×2=0.30 mol

Thus, the number of moles of Cl is 0.30 mol.

Similarly, 1 mol of compound has 1 mol of Ba thus, number of moles of Ba in 0.15 mol will be 0.15 mol.

(b)

Interpretation Introduction

Interpretation: The number of moles of C and O in 2.0 mol of CO2 needs to be determined.

Concept Introduction: The number of moles of an element can be determined from the molecular formula of the compound. The subscript for an element is equal to the number of moles of that element present in 1 mol of the compound.

(b)

Expert Solution
Check Mark

Answer to Problem 1RQ

The number of moles of C and O is 2 mol and 4 mol respectively.

Explanation of Solution

The given formula of compound is CO2. There is 1 mol of C and 2 mol of O in 1.0 mol of CO2.

Thus, the number of moles of C and O in 2 mol of CO2 will be:

  nC=2 mol nO=4 mol

(c)

Interpretation Introduction

Interpretation: The number of moles of H, N and O in 3.50 moles needs to be determined.

Concept Introduction: The number of moles of an element can be determined from the molecular formula of the compound. The subscript for an element is equal to the number of moles of that element present in 1 mol of the compound.

(c)

Expert Solution
Check Mark

Answer to Problem 1RQ

The number of moles of H, N and O is 3.5 mol, 3.5 mol and 10.5 moles respectively.

Explanation of Solution

The given compound is HNO3.

Thus, 1 mol of compound has 1 mol of H, 1 mol of N and 3 mol of O.

The number of moles of compound is 3.50 mol thus, number of moles of H and N will also be 3.50 mol.

The number of moles of O will be:

  nO=3×3.50=10.5 mol

(d)

Interpretation Introduction

Interpretation: The number of moles of N, H, S and O in 0.65 mol of compound needs to be determined.

Concept Introduction: The molar mass is the mass of a given chemical element or chemical compound divided by the amount of substance. The molar mass of a compound can be calculated by adding the standard atomic masses (in g/mol) of the constituent atoms.

(d)

Expert Solution
Check Mark

Answer to Problem 1RQ

The number of moles of N, H, S and O is 1.3 mol, 5.2 mol, 0.65 mol and 2.6 mol respectively.

Explanation of Solution

The given compound is (NH4)2SO4 . In 1 mol there are 2 mol of N, 8 mol of H, 1 mol of S and 4 mol of O.

The given number of moles of compound is 0.65 mol thus, number of moles of each element can be calculated as follows:

  nN=2×0.65=1.3 molnH=8×0.65=5.2 molnS=1×0.65=0.65 molnO=4×0.65=2.6 mol

Chapter 6 Solutions

World of Chemistry, 3rd edition

Ch. 6.2 - Prob. 5RQCh. 6.2 - Prob. 6RQCh. 6.3 - Prob. 1RQCh. 6.3 - Prob. 2RQCh. 6.3 - Prob. 3RQCh. 6.3 - Prob. 4RQCh. 6.3 - Prob. 5RQCh. 6.3 - Prob. 6RQCh. 6 - Prob. 1ACh. 6 - Prob. 2ACh. 6 - Prob. 3ACh. 6 - Prob. 4ACh. 6 - Prob. 5ACh. 6 - Prob. 6ACh. 6 - Prob. 7ACh. 6 - Prob. 8ACh. 6 - Prob. 9ACh. 6 - Prob. 10ACh. 6 - Prob. 11ACh. 6 - Prob. 12ACh. 6 - Prob. 13ACh. 6 - Prob. 14ACh. 6 - Prob. 15ACh. 6 - Prob. 16ACh. 6 - Prob. 17ACh. 6 - Prob. 18ACh. 6 - Prob. 19ACh. 6 - Prob. 20ACh. 6 - Prob. 21ACh. 6 - Prob. 22ACh. 6 - Prob. 23ACh. 6 - Prob. 24ACh. 6 - Prob. 25ACh. 6 - Prob. 26ACh. 6 - Prob. 27ACh. 6 - Prob. 28ACh. 6 - Prob. 29ACh. 6 - Prob. 30ACh. 6 - Prob. 31ACh. 6 - Prob. 32ACh. 6 - Prob. 33ACh. 6 - Prob. 34ACh. 6 - Prob. 35ACh. 6 - Prob. 36ACh. 6 - Prob. 37ACh. 6 - Prob. 38ACh. 6 - Prob. 39ACh. 6 - Prob. 40ACh. 6 - Prob. 41ACh. 6 - Prob. 42ACh. 6 - Prob. 43ACh. 6 - Prob. 44ACh. 6 - Prob. 45ACh. 6 - Prob. 46ACh. 6 - Prob. 47ACh. 6 - Prob. 48ACh. 6 - Prob. 49ACh. 6 - Prob. 50ACh. 6 - Prob. 51ACh. 6 - Prob. 52ACh. 6 - Prob. 53ACh. 6 - Prob. 54ACh. 6 - Prob. 55ACh. 6 - Prob. 56ACh. 6 - Prob. 57ACh. 6 - Prob. 58ACh. 6 - Prob. 59ACh. 6 - Prob. 60ACh. 6 - Prob. 61ACh. 6 - Prob. 62ACh. 6 - Prob. 63ACh. 6 - Prob. 64ACh. 6 - Prob. 65ACh. 6 - Prob. 66ACh. 6 - Prob. 67ACh. 6 - Prob. 68ACh. 6 - Prob. 69ACh. 6 - Prob. 70ACh. 6 - Prob. 1STPCh. 6 - Prob. 2STPCh. 6 - Prob. 3STPCh. 6 - Prob. 4STPCh. 6 - Prob. 5STPCh. 6 - Prob. 6STPCh. 6 - Prob. 7STPCh. 6 - Prob. 8STPCh. 6 - Prob. 9STPCh. 6 - Prob. 10STPCh. 6 - Prob. 11STPCh. 6 - Prob. 12STP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY