Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 17, Problem 85QRT

(a)

Interpretation Introduction

Interpretation:

The structure formula of CH3SO2F has to be drawn.  The geometry around the S atom has to be stated.  The bond angles OSO and OSF has to be stated.

Concept Introduction:

The molecular shape is determined primarily by the repulsions between pairs of electrons in the molecule or molecular ion, be they bonding pairs or lone pairs by the theory of the valence-shell-electron-pair repulsion (VSEPR).  The specific sets of bond pairs and lone pairs of electrons give different molecular shape to molecules.

(a)

Expert Solution
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Explanation of Solution

The given compound is CH3SO2F.  In CH3SO2F, the sulfur is the center atom.  The sulfur atom is attached to the two oxygen, a fluorine atom, and methyl group.

Therefore, the structure formula of CH3SO2F is shown below.

Chemistry: The Molecular Science, Chapter 17, Problem 85QRT

Figure 1

The number of ligands around sulfur atom of the molecule is 4.

The number of lone pairs on sulfur atom of the molecule is 0.

The formula to calculate number of sets of electron pairs is shown below.

    N=Nligand+Nlonepair

Where,

  • N is number of sets of electron pairs.
  • Nligand is the number of ligands around the chromium atom of the molecule.
  • Nlonepair is number of lone pairs on the chromium atom of the molecule.

Substitute the values of Nligand and Nlonepair in the above equation.

    N=4+0=4

When number of sets of electron pairs is 4, then the four orbital are required.  Therefore, the hybridization of inner atom of the molecule is sp3.  The corresponding geometery will be tetrahedral.

The bond angle for tetrahedral geometry is 109.5°.  Therefore, the bond angles OSO and OSF are apporximatly 109.5°_.

(b)

Interpretation Introduction

Interpretation:

The mass of HF required to electrolyze 150gCH3SO2F has to be calculated.  The mass of each product has to be calculated.

Concept Introduction:

The stoichiometry of a chemical species involved in a chemical reaction represents the number of chemical species involved in the chemical reaction.  The stoichiometry of a chemical species helps in calculating the expected mass of reactant and product.  The stoichiometry of a chemical species is also represented in a number of moles.

(b)

Expert Solution
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Explanation of Solution

The given mass of CH3SO2F is 150g.

The molar mass of CH3SO2F is 98.09g/mol.

The molar mass of HF is 20.01g/mol.

The number of moles a substance is given by the formula as shown below.

  n=mM

Where,

  • m is the mass of the substance.
  • M is the molar mass of the substance.

Substitute the value of mass and molar mass of CH3SO2F in the above equation.

    n=150g98.09g/mol=1.53mol

The given reaction is shown below.

    CH3SO2F+3HFCF3SO2F+3H2

Three moles of HF reacts with one mole of CH3SO2F.  Therefore, the relation between the number of moles of HF and CH3SO2F is given by the expression as shown below.

  nHF=3nCH3SO2F

Where,

  • nHF is the number of moles of HF.
  • nCH3SO2F is the number of moles of CH3SO2F.

Substitute the value of nCH3SO2F in the above equation.

  nHF=3(1.53mol)=4.59mol

The relation between number of moles and mass of a substance is given by the expression as shown below.

  m=n×M        (1)

Where,

  • m is the mass of the substance.
  • n is the number of moles of the substance.

Substitute the value of number of moles and molar mass of HF in the above equation.

  m=(4.59mol)×(20.01g/mol)=91.84g_

Therefore, the mass of HF required to electrolyze 150gCH3SO2F is 91.84g_.

The molar mass of H2 is 2.016g/mol.

Three moles of H2 produced by one mole of CH3SO2F.   Therefore, the relation between the number of moles of H2 and CH3SO2F is given by the expression as shown below.

  nH2=3nCH3SO2F

Where,

  • nH2 is the number of moles of H2.
  • nCH3SO2F is the number of moles of CH3SO2F.

Substitute the value of nCH3SO2F in the above equation.

  nHF=3(1.53mol)=4.59mol

Substitute the value of number of moles and molar mass of H2 in the equation (1).

  m=(4.59mol)×(2.016g/mol)=9.25g_

Therefore, the mass of H2 produced by 150gCH3SO2F is 9.25g_.

One

The molar mass of CF3SO2F is 282.13g/mol.

One mole of CF3SO2F produced by one mole of CH3SO2F.  Therefore, 1.53mol of CF3SO2F produced by 1.53mol of CH3SO2F.

Substitute the value of number of moles and molar mass of CF3SO2F in the equation (1).

  m=(1.53mol)×(282.13g/mol)=431.66g_

Therefore, the mass of CF3SO2F produced by 150gCH3SO2F is 431.66g_.

(c)

Interpretation Introduction

Interpretation:

Whether the hydrogen gas is produced at anode or cathode of the electrolysis cell has to be stated.

Concept Introduction:

A reducing agent is a reacting species in the reaction that reduces another reacting species.  A reducing agent itself gets oxidized when reacts with another species.  The reducing agent donates electrons.  An oxidizing agent is a reacting species in the reaction that oxidize another reacting species.  An oxidizing agent itself gets reduced when it reacts with another species.  The oxidizing agent accepts electrons.

(c)

Expert Solution
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Explanation of Solution

The given reaction is shown below.

    CH3SO2F+3HFCF3SO2F+3H2

The reduction of hydrogen ion is shown below.

    H+(aq)+eH2(g)

Hydrogen has been reduced in the reaction.  The reduction takes place at cathode.  Therefore, H2 is produced at the cathode.

(d)

Interpretation Introduction

Interpretation:

The amount of energy transfer in the cell at 250A and 8V for 24h has to be calculated.

Concept Introduction:

During a redox reaction, the amount of electrons transferred is determined by measuring the current flowing in an external electric circuit for a particular period of time.  The electric charge flowing through an electric circuit is calculated by multiplying the current and the time interval.

(d)

Expert Solution
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Explanation of Solution

The cell potential is 8V.

The unit conversion of cell potential from V to J/C is shown below.

  Cell potential=(8V)(1J/C1V)=8J/C

The current pass through the cell is 250A.

The unit conversion of current from A to C/s is shown below.

  Current=(250A)(1C/s1A)=250C/s

The time of current flow is 24h.

The unit conversion of time from h to s is shown below.

  Time=(24h)(60s1h)=1440s

The energy of the cell is calculated by the formula shown below.

    Energy=Cell potential×Current×Time

Substitute the value of cell potential, current, and time in the above equation.

    Energy=(8J/C)×(250C/s)×(1440s)=2.88×106J

The unit conversion of energy from J to kWh is shown below.

  Energy=(2.88×106J)(1kWh3.60×106J)=0.8kWh_

Therefore, the amount of energy transfer in the cell at 250A and 8V for 24h is 0.8kWh_.

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Chapter 17 Solutions

Chemistry: The Molecular Science

Ch. 17.4 - Given this reaction, its standard potential, and...Ch. 17.5 - Prob. 17.5PSPCh. 17.5 - Prob. 17.8CECh. 17.5 - Prob. 17.9CECh. 17.5 - Prob. 17.10CECh. 17.6 - Prob. 17.6PSPCh. 17.6 - Prob. 17.11ECh. 17.6 - Prob. 17.7PSPCh. 17.7 - Calculate the cell potential for the Zn(s) +...Ch. 17.7 - Prob. 17.9PSPCh. 17.8 - Prob. 17.12ECh. 17.8 - Prob. 17.13ECh. 17.8 - Prob. 17.14ECh. 17.10 - Predict the results of passing a direct electrical...Ch. 17.10 - In 1886. Henri Moissan was the first to prepare...Ch. 17.11 - In the commercial production of sodium metal by...Ch. 17.11 - Prob. 17.16CECh. 17.11 - Prob. 17.17ECh. 17.11 - Prob. 17.18CECh. 17.11 - Prob. 17.19ECh. 17.12 - Prob. 17.20CECh. 17.12 - Prob. 17.21CECh. 17 - Prob. 2SPCh. 17 - Prob. 1QRTCh. 17 - Prob. 2QRTCh. 17 - Prob. 3QRTCh. 17 - Prob. 4QRTCh. 17 - Identify each statement as true or false. Rewrite...Ch. 17 - Prob. 6QRTCh. 17 - Prob. 7QRTCh. 17 - Prob. 8QRTCh. 17 - Answer Question 8 again, but this time find a...Ch. 17 - Prob. 10QRTCh. 17 - Prob. 11QRTCh. 17 - For the reaction in Question 6, write balanced...Ch. 17 - Prob. 13QRTCh. 17 - Prob. 14QRTCh. 17 - Prob. 15QRTCh. 17 - Prob. 16QRTCh. 17 - Prob. 17QRTCh. 17 - For the reaction Cu2+(aq) + Zn(s) → Cu(s) + Zn2+...Ch. 17 - Prob. 19QRTCh. 17 - Prob. 20QRTCh. 17 - Prob. 21QRTCh. 17 - Prob. 22QRTCh. 17 - Draw a diagram of each cell. Label the anode, the...Ch. 17 - Prob. 24QRTCh. 17 - Prob. 25QRTCh. 17 - Prob. 26QRTCh. 17 - Prob. 27QRTCh. 17 - Prob. 28QRTCh. 17 - Prob. 29QRTCh. 17 - Prob. 30QRTCh. 17 - Prob. 31QRTCh. 17 - Consider these half-reactions: (a) Which is the...Ch. 17 - Consider these half-reactions: (a) Which is the...Ch. 17 - In principle, a battery could be made from...Ch. 17 - Prob. 35QRTCh. 17 - Hydrazine, N2H4, can be used as the reducing agent...Ch. 17 - Prob. 37QRTCh. 17 - Prob. 38QRTCh. 17 - Prob. 39QRTCh. 17 - Prob. 40QRTCh. 17 - Prob. 41QRTCh. 17 - Prob. 42QRTCh. 17 - Prob. 43QRTCh. 17 - Prob. 44QRTCh. 17 - Prob. 45QRTCh. 17 - Prob. 46QRTCh. 17 - Consider the voltaic cell 2 Ag+(aq) + Cd(s) 2...Ch. 17 - Consider a voltaic cell with the reaction H2(g) +...Ch. 17 - Calculate the cell potential of a concentration...Ch. 17 - Prob. 50QRTCh. 17 - Prob. 51QRTCh. 17 - Prob. 52QRTCh. 17 - Prob. 53QRTCh. 17 - NiCad batteries are rechargeable and are commonly...Ch. 17 - Prob. 55QRTCh. 17 - Prob. 56QRTCh. 17 - Prob. 57QRTCh. 17 - Hydrazine, N2H4, has been proposed as the fuel in...Ch. 17 - Consider the electrolysis of water in the presence...Ch. 17 - Prob. 60QRTCh. 17 - Prob. 61QRTCh. 17 - Prob. 62QRTCh. 17 - Identify the products of the electrolysis of a 1-M...Ch. 17 - Prob. 64QRTCh. 17 - Prob. 65QRTCh. 17 - Prob. 66QRTCh. 17 - Prob. 67QRTCh. 17 - Prob. 68QRTCh. 17 - Prob. 69QRTCh. 17 - Prob. 70QRTCh. 17 - Prob. 71QRTCh. 17 - Prob. 72QRTCh. 17 - Prob. 73QRTCh. 17 - Prob. 74QRTCh. 17 - Calculate how long it would take to electroplate a...Ch. 17 - Prob. 76QRTCh. 17 - Prob. 77QRTCh. 17 - Prob. 78QRTCh. 17 - Prob. 79QRTCh. 17 - Prob. 80QRTCh. 17 - Prob. 81QRTCh. 17 - Prob. 82QRTCh. 17 - Prob. 83QRTCh. 17 - Prob. 84QRTCh. 17 - Prob. 85QRTCh. 17 - Prob. 86QRTCh. 17 - Prob. 87QRTCh. 17 - Prob. 88QRTCh. 17 - You wish to electroplate a copper surface having...Ch. 17 - Prob. 90QRTCh. 17 - Prob. 91QRTCh. 17 - Prob. 92QRTCh. 17 - Prob. 93QRTCh. 17 - An electrolytic cell is set up with Cd(s) in...Ch. 17 - Prob. 95QRTCh. 17 - Prob. 96QRTCh. 17 - Prob. 97QRTCh. 17 - Prob. 98QRTCh. 17 - Prob. 99QRTCh. 17 - Prob. 100QRTCh. 17 - Prob. 101QRTCh. 17 - Prob. 102QRTCh. 17 - Prob. 103QRTCh. 17 - Prob. 104QRTCh. 17 - Prob. 105QRTCh. 17 - Prob. 106QRTCh. 17 - Prob. 107QRTCh. 17 - Prob. 108QRTCh. 17 - Prob. 109QRTCh. 17 - Prob. 110QRTCh. 17 - Prob. 111QRTCh. 17 - Prob. 17.ACPCh. 17 - Prob. 17.BCP
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