Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 10, Problem 112A

Convert each to mass in grams.

a. 4.22   ×  10 15 atoms c

b. 8 .65  ×  10 25 atoms H

c. 1 .25  ×  10 22 atoms O

d. 4 .44  ×  10 23 atoms Pb

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

4.22×1015 atoms of U should be converted into mass of U in grams.

Concept introduction:

The number of moles of a compound is defined as the ratio of the given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = given mass of the compoundmolar mass of the compound

The one mole of any substance is equal to the Avogadro number, , i.e., , 6.022×1023 atoms.

Answer to Problem 112A

Mass of U in grams = 166.80×108 g

Explanation of Solution

Since, one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Number of moles = number of atoms ×1 mole6.022×10 23atoms

Put the values,

Number of moles = 4.22×1015 atoms ×1 mole6.022×10 23atoms

=0.70076×108 mole

Number of moles =  mass of the compoundmolar mass of the compound

Molar mass of U = 238.029 g/mole

Put the values,

0.70076×108 mole = mass238.029 g/mole

Mass = 0.70076×108 mole ×238.029 g/mole

Mass of U in grams = 166.80×108 g

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

8.65×1025 atoms of H should be converted into mass of U in grams.

Concept introduction:

The number of moles of a compound is defined as the ratio of the given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = given mass of the compoundmolar mass of the compound

The one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Answer to Problem 112A

Mass of H in grams = 144.76 g

Explanation of Solution

Since, one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Number of moles = number of atoms ×1 mole6.022×10 23atoms

Put the values,

Number of moles = 8.65×1025 atoms ×1 mole6.022×10 23atoms

=1.43639×102 mole

Number of moles =  mass of the compoundmolar mass of the compound

Molar mass of H = 1.0078 g/mole

Put the values,

1.43639×102 mole= mass1.0078 g/mole

Mass = 1.43639×102 mole×1.0078 g/mole

Mass of H in grams = 144.76 g

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

1.25×1022 atoms of O should be converted into mass of U in grams.

Concept introduction:

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = given mass of the compoundmolar mass of the compound

The one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Answer to Problem 112A

Mass of O in grams = 0.332 g

Explanation of Solution

Since, one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Number of moles = number of atoms×1 mole6.022×10 23atoms

Put the values,

Number of moles = 1.25×1022 atoms ×1 mole6.022×10 23atoms

=0.2076×101 mole

Number of moles =  mass of the compoundmolar mass of the compound

Molar mass of O = 15.999 g/mole

Put the values,

0.2076×101 mole = mass15.999 g/mole

Mass = 1.43639×102 mole×1.0078 g/mole

Mass of O in grams = 0.332 g

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

4.44×1023 atoms of Pb should be converted into mass of U in grams.

Concept introduction:

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = given mass of the compoundmolar mass of the compound

The one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Answer to Problem 112A

Mass of Pb in grams = 152.8 g

Explanation of Solution

Since, one mole of any substance is equal to the Avogadro number, i.e., 6.022×1023 atoms.

Number of moles = number of atoms ×1 mole6.022×10 23atoms

Put the values,

Number of moles = 4.44×1023 atoms ×1 mole6.022×10 23atoms

=0.7373 mole

Number of moles =  mass of the compoundmolar mass of the compound

Molar mass of Pb in = 207.2 g/mole

Put the values,

0.7373 mole = mass207.2 g/mole

Mass = 0.7373 mole×207.2 g/mole

Mass of Pb in grams = 152.8 g

Chapter 10 Solutions

Chemistry: Matter and Change

Ch. 10.1 - Prob. 11SSCCh. 10.1 - Prob. 12SSCCh. 10.1 - Prob. 13SSCCh. 10.1 - Prob. 14SSCCh. 10.2 - Prob. 15PPCh. 10.2 - Prob. 16PPCh. 10.2 - Prob. 17PPCh. 10.2 - Prob. 18PPCh. 10.2 - Prob. 19PPCh. 10.2 - Prob. 20PPCh. 10.2 - Prob. 21PPCh. 10.2 - Prob. 22SSCCh. 10.2 - Prob. 23SSCCh. 10.2 - Prob. 24SSCCh. 10.2 - Prob. 25SSCCh. 10.2 - Prob. 26SSCCh. 10.2 - Prob. 27SSCCh. 10.2 - Prob. 28SSCCh. 10.3 - Prob. 29PPCh. 10.3 - Prob. 30PPCh. 10.3 - Prob. 31PPCh. 10.3 - Prob. 32PPCh. 10.3 - Prob. 33PPCh. 10.3 - Prob. 34PPCh. 10.3 - Prob. 35PPCh. 10.3 - Prob. 36PPCh. 10.3 - Prob. 37PPCh. 10.3 - Prob. 38PPCh. 10.3 - Prob. 39PPCh. 10.3 - Prob. 40PPCh. 10.3 - Prob. 41PPCh. 10.3 - Prob. 42PPCh. 10.3 - Prob. 43PPCh. 10.3 - Prob. 44PPCh. 10.3 - Prob. 45PPCh. 10.3 - Prob. 46PPCh. 10.3 - Prob. 47SSCCh. 10.3 - Prob. 48SSCCh. 10.3 - Prob. 49SSCCh. 10.3 - Prob. 50SSCCh. 10.3 - Prob. 51SSCCh. 10.3 - Prob. 52SSCCh. 10.3 - Prob. 53SSCCh. 10.4 - Prob. 54PPCh. 10.4 - Prob. 55PPCh. 10.4 - Prob. 56PPCh. 10.4 - Prob. 57PPCh. 10.4 - Prob. 58PPCh. 10.4 - Prob. 59PPCh. 10.4 - Prob. 60PPCh. 10.4 - Prob. 61PPCh. 10.4 - Prob. 62PPCh. 10.4 - Prob. 63PPCh. 10.4 - Prob. 64PPCh. 10.4 - Prob. 65PPCh. 10.4 - Prob. 66PPCh. 10.4 - Prob. 67SSCCh. 10.4 - Prob. 68SSCCh. 10.4 - Prob. 69SSCCh. 10.4 - Prob. 70SSCCh. 10.4 - Prob. 71SSCCh. 10.4 - Prob. 72SSCCh. 10.4 - Prob. 73SSCCh. 10.5 - Prob. 74PPCh. 10.5 - Prob. 75PPCh. 10.5 - Prob. 76SSCCh. 10.5 - Prob. 77SSCCh. 10.5 - Prob. 78SSCCh. 10.5 - Prob. 79SSCCh. 10.5 - Prob. 80SSCCh. 10.5 - Prob. 81SSCCh. 10.5 - Prob. 82SSCCh. 10 - Prob. 83ACh. 10 - Prob. 84ACh. 10 - Prob. 85ACh. 10 - Prob. 86ACh. 10 - Prob. 87ACh. 10 - Prob. 88ACh. 10 - Prob. 89ACh. 10 - Determine the number of representative particles...Ch. 10 - Determine the number of representative particles...Ch. 10 - Prob. 92ACh. 10 - Determine the number of moles in each substance....Ch. 10 - Prob. 94ACh. 10 - Prob. 95ACh. 10 - RDA of Selenium The recommended daily allowance...Ch. 10 - Prob. 97ACh. 10 - Prob. 98ACh. 10 - Prob. 99ACh. 10 - Prob. 100ACh. 10 - Prob. 101ACh. 10 - Prob. 102ACh. 10 - Prob. 103ACh. 10 - Prob. 104ACh. 10 - Prob. 105ACh. 10 - Prob. 106ACh. 10 - Prob. 107ACh. 10 - Calculate the mass of each element. a. 5.22 mol of...Ch. 10 - Perform the following conversions. a. 3.50 mol of...Ch. 10 - Determine the mass in grams of each element....Ch. 10 - Complete Table 10.3.Ch. 10 - Convert each to mass in grams. a. 4.221015 atoms c...Ch. 10 - Prob. 113ACh. 10 - Prob. 114ACh. 10 - Prob. 115ACh. 10 - Prob. 116ACh. 10 - Prob. 117ACh. 10 - Prob. 118ACh. 10 - Prob. 119ACh. 10 - Prob. 120ACh. 10 - Prob. 121ACh. 10 - Prob. 122ACh. 10 - Prob. 123ACh. 10 - Prob. 124ACh. 10 - Prob. 125ACh. 10 - Prob. 126ACh. 10 - Prob. 127ACh. 10 - Prob. 128ACh. 10 - Prob. 129ACh. 10 - Prob. 130ACh. 10 - Prob. 131ACh. 10 - Prob. 132ACh. 10 - Prob. 133ACh. 10 - Prob. 134ACh. 10 - Prob. 135ACh. 10 - Prob. 136ACh. 10 - Prob. 137ACh. 10 - Prob. 138ACh. 10 - Prob. 139ACh. 10 - Prob. 140ACh. 10 - Prob. 141ACh. 10 - Prob. 142ACh. 10 - Prob. 143ACh. 10 - Prob. 144ACh. 10 - Prob. 145ACh. 10 - Prob. 146ACh. 10 - Prob. 147ACh. 10 - Pain Relief Acetaminophen, a common aspirin...Ch. 10 - Prob. 149ACh. 10 - Prob. 150ACh. 10 - Prob. 151ACh. 10 - Prob. 152ACh. 10 - Prob. 153ACh. 10 - Prob. 154ACh. 10 - Prob. 155ACh. 10 - Prob. 156ACh. 10 - Prob. 157ACh. 10 - Prob. 158ACh. 10 - Prob. 159ACh. 10 - Prob. 160ACh. 10 - Prob. 161ACh. 10 - Prob. 162ACh. 10 - Express the composition of each compound as the...Ch. 10 - VitaminD3 Your body's ability to absorb calcium is...Ch. 10 - Prob. 165ACh. 10 - Cholesterol Heart disease is linked to high blood...Ch. 10 - Prob. 167ACh. 10 - Prob. 168ACh. 10 - Prob. 169ACh. 10 - Prob. 170ACh. 10 - Prob. 171ACh. 10 - Prob. 172ACh. 10 - Prob. 173ACh. 10 - Prob. 174ACh. 10 - Prob. 175ACh. 10 - Prob. 176ACh. 10 - Prob. 177ACh. 10 - Prob. 178ACh. 10 - Prob. 179ACh. 10 - Determine the mass percent of anhydrous sodium...Ch. 10 - Table 4 shows data from an experiment to determine...Ch. 10 - Chromium(lll) nitrate forms a hydrate that is...Ch. 10 - Determine the percent composition of MgCO35H2O and...Ch. 10 - What is the formula and name of a hydrate that is...Ch. 10 - Gypsum is hydrated calcium sulfate. A 4.89-g...Ch. 10 - A 1.628-g sample of a hydrate of magnesium iodide...Ch. 10 - Borax Hydrated sodiumtetraborate (Na2B4O7xH2O) is...Ch. 10 - Prob. 188ACh. 10 - Prob. 189ACh. 10 - Prob. 190ACh. 10 - Prob. 191ACh. 10 - Prob. 192ACh. 10 - Prob. 193ACh. 10 - Prob. 194ACh. 10 - Prob. 195ACh. 10 - Prob. 196ACh. 10 - Prob. 197ACh. 10 - Prob. 198ACh. 10 - Prob. 199ACh. 10 - Prob. 200ACh. 10 - Prob. 201ACh. 10 - Prob. 202ACh. 10 - Prob. 203ACh. 10 - Prob. 204ACh. 10 - Prob. 205ACh. 10 - Prob. 206ACh. 10 - Prob. 207ACh. 10 - Prob. 208ACh. 10 - Prob. 209ACh. 10 - Prob. 210ACh. 10 - Prob. 211ACh. 10 - Prob. 212ACh. 10 - Prob. 213ACh. 10 - Prob. 214ACh. 10 - Prob. 215ACh. 10 - Prob. 216ACh. 10 - Prob. 1STPCh. 10 - Prob. 2STPCh. 10 - Prob. 3STPCh. 10 - Prob. 4STPCh. 10 - Prob. 5STPCh. 10 - Prob. 6STPCh. 10 - Prob. 7STPCh. 10 - Prob. 8STPCh. 10 - Prob. 9STPCh. 10 - Prob. 10STPCh. 10 - Prob. 11STPCh. 10 - Prob. 12STPCh. 10 - Prob. 13STPCh. 10 - Prob. 14STPCh. 10 - Prob. 15STPCh. 10 - Prob. 16STP
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