Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 3.120P

(a)

Interpretation Introduction

Interpretation:

The larger quantity between 0.4 mol of O3 molecules and 0.4 mol of O atoms is to be identified.

Concept introduction:

1 mol of any substance contains Avogadro’s number of entities that is 6.022×1023.

The formula to calculate the number of moles is:

  Number of moles=given mass  of sample(g)molar mass (g/mol)        (1)

The formula to calculate mass is as follows:

  Mass of a given sample(g)=(molar mass g/mol)(number of moles)        (2)

(a)

Expert Solution
Check Mark

Answer to Problem 3.120P

Both the quantities have an equal number of entities.

Explanation of Solution

Since an equal number of moles of various substances contains an equal number of entities. Both O3 molecules and O atoms have an equal number of moles being 0.4 mol hence the number of entities is equal for both the samples of O3 molecules and O atoms.

Conclusion

Ozone molecules present in 0.4 mol is equal to the number of oxygen atoms present in 0.4 mol.

(b)

Interpretation Introduction

Interpretation:

The larger of the two given quantities in terms of grams is to be identified.

Concept introduction:

1 mol of any substance contains Avogadro’s number of entities that is 6.022×1023.

The formula to calculate the number of moles is:

  Number of moles=given mass  of sample(g)molar mass (g/mol)        (1)

The formula to calculate mass is as follows:

  Mass of a given sample(g)=(molar mass g/mol)(number of moles)        (2)

(b)

Expert Solution
Check Mark

Answer to Problem 3.120P

0.4 mol O3 molecules has a larger mass than 0.4 mol O atoms.

Explanation of Solution

The molar mass of O is 16.00 g/mol.so the molar mass of O3 would be calculated as follows:

  Molar mass of O3=3(16.00 g/mol)=48.00 g/mol

The molar mass of O would be calculated as follows:

  Molar mass of O=1(16.00 g/mol)=16.00 g/mol

Since O3 has larger molar mass (48.00 g/mol) compared to the molar mass of O hence the former has greater mass than the latter. This can be verified as follows:

Substitute 48.00 g/mol for molar mass of O3 and 0.4 mol for the number of moles in equation (2).

  Mass of O3 in given sample(g)=(48.00 g/mol)(0.4 mol)=19.2 g

Substitute 16.00 g/mol for molar mass of O and 0.4 mol for the number of moles in equation (2).

  Mass of O3 in given sample(g)=(16.00 g/mol)(0.4 mol)=6.4 g.

Conclusion

For the same number of moles, the quantity with larger molar mass will have a larger mass.

(c)

Interpretation Introduction

Interpretation:

Whether the number of moles will be larger in 4.0 g N2O4 or 3.2 g SO2 is to be determined.

Concept introduction:

The formula to calculate the number of moles is:

  Number of moles=given mass  of sample(g)molar mass (g/mol)        (1)

The formula to calculate mass is as follows:

  Mass of a given sample(g)=(molar mass g/mol)(number of moles)        (2)

(c)

Expert Solution
Check Mark

Answer to Problem 3.120P

In terms of moles SO2 is larger compared to N2O4.

Explanation of Solution

The molar mass of N2O4 is 92.02 g/mol. Substitute 92.02 g/mol for molar mass and 4.0 g for given mass in equation (1) to calculate the moles of N2O4.

  Moles of N2O4(mol)= 4.0 g  92.02 g/mol = 0.043 mol

Substitute 3.3 g for the mass of SO2 and 64.06 g/mol for molar mass of SO2 in equation (1) to calculate the moles of SO2 as follows:

  Moles of SO2(mol)= 3.3 g  64.06 g/mol = 0.052 mol.

Hence SO2 is a larger quantity in terms of moles.

Conclusion

The number of moles in 3.2 g SO2  is greater than the number of moles in 4.0 g N2O4.

(d)

Interpretation Introduction

Interpretation:

Whether the mass of 0.6 mol C2H4 or 0.6 mol F2 is larger is to be determined.

Concept introduction:

The formula to calculate the number of moles is:

  Number of moles=given mass  of sample(g)molar mass (g/mol)        (1)

The formula to calculate mass is as follows:

  Mass of a given sample(g)=(molar mass g/mol)(number of moles)        (2)

(d)

Expert Solution
Check Mark

Answer to Problem 3.120P

In terms of mass in grams F2 is larger compared to C2H4.

Explanation of Solution

Substitute 28.05 g/mol for molar mass of C2H4 and 0.6 mol for moles of C2H4 in equation (2).

  Mass of C2H4 in given sample(g)=( 28.05 g/mol)(0.6 mol)=17 g

Substitute 38.00 g/mol for molar mass of F2 and 0.6 mol for moles of F2 in equation (2).

  Mass of F2 in given sample(g)=( 38.00 g/mol)(0.6 mol)=23 g.

Hence F2 has larger mass compared to C2H4.

Conclusion

For the same number of moles, the quantity with larger molar mass will have a larger mass.

(e)

Interpretation Introduction

Interpretation:

Whether the number of ions in 2.3 mol NaClO3 or 2.2 mol MgCl2 is larger is to be determined.

Concept introduction:

The electrolyte is the substance that produces ions when it is dissolved in a polar solvent. It breaks into positively and negatively charged ions that spread uniformly through the solvent. The electrolytic solution, as a whole, is electrically neutral. Sodium chloride, potassium chloride, calcium phosphate are some of the examples of electrolytes.

Strong electrolytes are those electrolytes that completely dissociates into its ions. These have a very high value of electrical conductance. Sodium chloride and potassium chloride are strong electrolytes.

Weak electrolytes are those electrolytes that partially dissociates into its ions. They are poor conductors of electricity. Acetic acid and carbonic acid are weak electrolytes.

(e)

Expert Solution
Check Mark

Answer to Problem 3.120P

The number of ions is larger in 2.2 mol MgCl2 compared to 2.3 mol NaClO3.

Explanation of Solution

The chemical formulas for sodium chlorate and magnesium chloride are NaClO3 and MgCl2 respectively.

On dissociation NaClO3 gives Na1+ and ClO32 that is 2 mol of ions while MgCl2 on dissociation gives Mg2+ and 2Cl ions that is 3 mol of ions.

Hence total moles of ions present in 2.3 mol of NaClO3 is calculated as follows:

  Total moles of ions in 2.3 mol NaClO3=(2.3)(2 mol 1 mol )=4.6 mol

Total moles of ions present in 2.2 mol of MgCl2 is calculated as follows:

  Total moles of ions in 2.2 mol MgCl2=(2.2)(3 mol 1 mol )=6.6 mol.

Hence MgCl2 is a larger quantity in terms of the total number of ions compared to NaClO3.

Conclusion

Magnesium chloride and sodium chlorate are strong electrolyte and give corresponding ions in the solution. Magnesium chloride gives more number of ions per mole compared to sodium chlorate.

(f)

Interpretation Introduction

Interpretation:

Whether the number of molecules in 1.0 g H2O or 1.0 g H2O2 is larger is to be determined.

Concept introduction:

The formula to calculate the number of molecules from the number of moles is as follows:

  Numberof  molecules =[(number of moles )(6.022×1023molecules)]        (3)

(f)

Expert Solution
Check Mark

Answer to Problem 3.120P

H2O has a greater number of molecules compared to H2O2.

Explanation of Solution

Substitute 18.02 g/mol for molar mass of H2O and 1.0 g for the mass of H2O in equation (1).

  Moles of H2O(mol)= 1.0 g  18.02 g/mol =0.05549 mol

Substitute 0.05549 mol for the number of moles in equation (3) to calculate the number of molecules of H2O.

  Numberof H2O molecules=(0.05549 mol)(6.022×1023molecules)=0.33416×1023 molecules0.3342×1023 molecules

Substitute 34.02 g/mol for molar mass of H2O2 and 1.0 g for the mass of H2O2 in equation (1).

  Moles of H2O2(mol)= 1.0 g  34.02 g/mol =0.02939 mol

Substitute 0.02939 mol for the number of moles in equation (3) to calculate the number of molecules of H2O2.

  Numberof H2O2 molecules =(0.05549 mol)(6.022×1023molecules)=0.11701×1023 molecules0.1170×1023 molecules

Hence H2O has a greater number of molecules compared to H2O2.

Conclusion

Greater the number of molecules present in the given mass of the sample larger will be the number of molecules present.

(g)

Interpretation Introduction

Interpretation:

Whether the mass of 6.022×1023atoms 235U or 6.022×1023atoms 238U is larger is to be determined.

Concept introduction:

1 mol of any substance contains Avogadro’s number of entities that is 6.022×1023.

The formula to calculate the number of moles is:

  Number of moles=given mass  of sample(g)molar mass (g/mol)        (1)

The formula to calculate mass is as follows:

  Mass of a given sample(g)=(molar mass g/mol)(number of moles)        (2)

The formula to calculate the number of molecules is as follows:

  Numberof  molecules =[(number of moles )(6.022×1023molecules)]        (3)

(g)

Expert Solution
Check Mark

Answer to Problem 3.120P

238U has a greater mass number (238) would have a greater total mass in grams compared to 235U.

Explanation of Solution

For an equal number of atoms of each isotope of uranium that is 6.022×1023, there would an equal number of moles too and since 6.022×1023 atoms are equivalent to 1 mol hence each of the given sample of 235U and 238U are 1 mol.

Equation (2) implies that for an equal number of moles the one with greater molar mass or whichever isotope is heavier would have greater mass in grams.

Hence 238U that has greater mass number 238 would have a greater total mass in grams compared to 235U.

Conclusion

For an equal number of particles (6.022×1023) the one having greater mass number should have greater mass. Hence 238U having greater mass number 238 would have a greater total mass in grams compared to 235U.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 3 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 3.1 - Prob. 3.6AFPCh. 3.1 - For many years, compounds known as...Ch. 3.1 - Use the information in Follow-up Problem 3.6A to...Ch. 3.1 - Prob. 3.7BFPCh. 3.2 - Prob. 3.8AFPCh. 3.2 - A sample of an unknown compound contains 6.80 mol...Ch. 3.2 - A sample of an unknown compound is found to...Ch. 3.2 - Prob. 3.9BFPCh. 3.2 - Prob. 3.10AFPCh. 3.2 - Prob. 3.10BFPCh. 3.2 - A dry-cleaning solvent (ℳ = 146.99 g/mol) that...Ch. 3.2 - Prob. 3.11BFPCh. 3.3 - Prob. 3.12AFPCh. 3.3 - Prob. 3.12BFPCh. 3.3 - Prob. 3.13AFPCh. 3.3 - Prob. 3.13BFPCh. 3.4 - Prob. 3.14AFPCh. 3.4 - The tarnish that forms on objects made of silver...Ch. 3.4 - Prob. 3.15AFPCh. 3.4 - In the reaction that removes silver tarnish (see...Ch. 3.4 - Prob. 3.16AFPCh. 3.4 - Prob. 3.16BFPCh. 3.4 - Prob. 3.17AFPCh. 3.4 - Prob. 3.17BFPCh. 3.4 - Prob. 3.18AFPCh. 3.4 - Prob. 3.18BFPCh. 3.4 - In the reaction in Follow-up Problem 3.18A, how...Ch. 3.4 - Prob. 3.19BFPCh. 3.4 - Prob. 3.20AFPCh. 3.4 - Prob. 3.20BFPCh. 3.4 - Marble (calcium carbonate) reacts with...Ch. 3.4 - Prob. 3.21BFPCh. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Why might the expression “1 mol of chlorine” be...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Calculate the molar mass of each of the...Ch. 3 - Prob. 3.10PCh. 3 - Prob. 3.11PCh. 3 - Calculate each of the following quantities: Mass...Ch. 3 - Calculate each of the following quantities: Amount...Ch. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Calculate each of the following: Mass % of H in...Ch. 3 - Calculate each of the following: Mass % of I in...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Calculate each of the following: Mass fraction of...Ch. 3 - Oxygen is required for the metabolic combustion of...Ch. 3 - Cisplatin (right), or Platinol, is used in the...Ch. 3 - Allyl sulfide (below) gives garlic its...Ch. 3 - Iron reacts slowly with oxygen and water to form a...Ch. 3 - Prob. 3.28PCh. 3 - Prob. 3.29PCh. 3 - The mineral galena is composed of lead(II) sulfide...Ch. 3 - Prob. 3.31PCh. 3 - Prob. 3.32PCh. 3 - List three ways compositional data may be given in...Ch. 3 - Prob. 3.34PCh. 3 - Prob. 3.35PCh. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - Prob. 3.38PCh. 3 - Prob. 3.39PCh. 3 - What is the molecular formula of each...Ch. 3 - Prob. 3.41PCh. 3 - Prob. 3.42PCh. 3 - Find the empirical formula of each of the...Ch. 3 - An oxide of nitrogen contains 30.45 mass % N. (a)...Ch. 3 - Prob. 3.45PCh. 3 - A sample of 0.600 mol of a metal M reacts...Ch. 3 - Prob. 3.47PCh. 3 - Prob. 3.48PCh. 3 - Prob. 3.49PCh. 3 - Prob. 3.50PCh. 3 - Prob. 3.51PCh. 3 - Prob. 3.52PCh. 3 - Prob. 3.53PCh. 3 - Prob. 3.54PCh. 3 - Prob. 3.55PCh. 3 - Prob. 3.56PCh. 3 - Prob. 3.57PCh. 3 - Prob. 3.58PCh. 3 - Prob. 3.59PCh. 3 - Prob. 3.60PCh. 3 - Prob. 3.61PCh. 3 - Prob. 3.62PCh. 3 - Prob. 3.63PCh. 3 - Prob. 3.64PCh. 3 - Prob. 3.65PCh. 3 - Prob. 3.66PCh. 3 - Prob. 3.67PCh. 3 - Prob. 3.68PCh. 3 - Prob. 3.69PCh. 3 - Prob. 3.70PCh. 3 - Prob. 3.71PCh. 3 - Prob. 3.72PCh. 3 - Prob. 3.73PCh. 3 - Prob. 3.74PCh. 3 - Elemental phosphorus occurs as tetratomic...Ch. 3 - Prob. 3.76PCh. 3 - Solid iodine trichloride is prepared in two steps:...Ch. 3 - Prob. 3.78PCh. 3 - Prob. 3.79PCh. 3 - Prob. 3.80PCh. 3 - Prob. 3.81PCh. 3 - Prob. 3.82PCh. 3 - Prob. 3.83PCh. 3 - Prob. 3.84PCh. 3 - Prob. 3.85PCh. 3 - Prob. 3.86PCh. 3 - Prob. 3.87PCh. 3 - Prob. 3.88PCh. 3 - Prob. 3.89PCh. 3 - Prob. 3.90PCh. 3 - Prob. 3.91PCh. 3 - Prob. 3.92PCh. 3 - Prob. 3.93PCh. 3 - Prob. 3.94PCh. 3 - When 20.5 g of methane and 45.0 g of chlorine gas...Ch. 3 - Prob. 3.96PCh. 3 - Prob. 3.97PCh. 3 - Prob. 3.98PCh. 3 - Prob. 3.99PCh. 3 - Prob. 3.100PCh. 3 - Sodium borohydride (NaBH4) is used industrially in...Ch. 3 - Prob. 3.102PCh. 3 - The first sulfur-nitrogen compound was prepared in...Ch. 3 - Prob. 3.104PCh. 3 - Prob. 3.105PCh. 3 - Prob. 3.106PCh. 3 - Serotonin () transmits nerve impulses between...Ch. 3 - In 1961, scientists agreed that the atomic mass...Ch. 3 - Prob. 3.109PCh. 3 - Isobutylene is a hydrocarbon used in the...Ch. 3 - The multistep smelting of ferric oxide to form...Ch. 3 - Prob. 3.112PCh. 3 - Prob. 3.113PCh. 3 - Prob. 3.114PCh. 3 - Prob. 3.115PCh. 3 - Prob. 3.116PCh. 3 - Prob. 3.117PCh. 3 - Prob. 3.118PCh. 3 - Prob. 3.119PCh. 3 - Prob. 3.120PCh. 3 - For the reaction between solid tetraphosphorus...Ch. 3 - Prob. 3.122PCh. 3 - Prob. 3.123PCh. 3 - Prob. 3.124PCh. 3 - Prob. 3.125PCh. 3 - Prob. 3.126PCh. 3 - Prob. 3.127PCh. 3 - Prob. 3.128PCh. 3 - Ferrocene, synthesized in 1951, was the first...Ch. 3 - Prob. 3.130PCh. 3 - Prob. 3.131PCh. 3 - Citric acid (below) is concentrated in citrus...Ch. 3 - Prob. 3.133PCh. 3 - Nitrogen monoxide reacts with elemental oxygen to...Ch. 3 - Prob. 3.135PCh. 3 - Prob. 3.136PCh. 3 - Manganese is a key component of extremely hard...Ch. 3 - The human body excretes nitrogen in the form of...Ch. 3 - Aspirin (acetylsalicylic acid, C9H8O4) is made by...Ch. 3 - Prob. 3.140PCh. 3 - Prob. 3.141PCh. 3 - Prob. 3.142PCh. 3 - When powdered zinc is heated with sulfur, a...Ch. 3 - Cocaine (C17H21O4N) is a natural substance found...Ch. 3 - Prob. 3.145P
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
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY