1
Answers to Exercises and Problems
I I I I
for
CHEMISTRY
A Guided Inquiry
I I
Seventh Edition, 2017
I I
Richard S. Moog I I
Franklin & Marshall College
I I I
John J. Farrell I I
Franklin & Marshall College
I I I
Latest Update: December 20, 2017
I I I I
John Wiley & Sons, Inc.
I I I I
, 2
Answers to Exercises and Problems
I I I I
These Ianswers Imay Ibe Iprovided Ito Istudents Iat Ithe Iinstructor’s Idiscretion.
Do Inot Ipost Ion Ipublicly Iavailable Iwebsite.
These Ianswers Ido Inot Icontain Ifull Iexplanations Ior Isolutions.
ChemActivity I1
Exercises
1.
Isotope Atomic Mass Number Iof
INumber INumbe IElectrons
Z r
A
31P 15 31 15
18O 8 18 8
39K+ 19 39 18
58Ni2+ 28 58 26
2. 1.674 I I10–24 Ig. I 1.993 I I10–23 Ig.
3. 8.67 I I10–17 Ig.
4. 12.00 Ig.
5. 2.6621 I I10–23 Ig.
6. a) IThe Imass Inumber Iis Ithe Isum Iof Ithe Inumber Iof Iprotons Iand Ithe Inumber Iof Ineutrons Iin
ItheInucleus. I b) I The Iatomic Inumber Iis Ithe Inumber Iof Iprotons Iin Ithe Inucleus.
7. False. I 18O Ihas I8 Iprotons Iand I10
Ineutrons. I8.
Species Electrons Protons Neutrons
24
Mg 12 12 12
23
Na + 10 11 12
35
Cl 17 17 18
35 - 18 17 18
Cl
56 3+ 23 26 30
Fe
15
N 7 7 8
16 2- 10 8 8
O
27 3+ 10 13 14
Al
9.
Isotope Atomic Mass Number Iof
INumber INumbe IElectrons
Z r
A
59Co2+ 27 59 25
14N 7 14 7
7Li 3 7 3
6Li 3 6 3
58Zn2+ 30 58 56
19F– 9 19 10
, 3
10. I I IAll Iisotopes Iof Ian Ielement Ihave Ithe Isame Inumber Iof Iprotons Iin Ithe Inucleus. IOne Iisotope Iof
Ian Ielement Iis Idifferentiated Ifrom Ianother Iisotope Iof Ithe Isame Ielement Iby Ithe Inumber Iof
Ineutrons Iin Ithe Inucleus.
Problems
1. Using I carbon-13 I and I carbon-12, I approx I mass I of I neutron I = I 13.0034 I – I 12 I = I 1.0034
I amu.Iapprox Imass Iof I
14C I= I13.0034 I+ I1.0034 I= I14.0068 Iamu.
2. a) IUsing I14C Iand I14C–, I mass Iof Ielectron I is I approximately I13.0039 I amu I – I 13.0034 I amu I=
I0.0005 Iamu
b) Using I1H, Imass Iof Iproton Iis Iapproximately I1.0078 Iamu I– I0.0005 Iamu I= I1.0073 Iamu
c) Using I 1H and I 2H, I mass I of I neutron I is I approximately I 2.0140 I amu I – I 1.0078
I amu =I1.0062 Iamu
(Note Ithat Islightly Idifferent I values I for Ithe I masses I of Iprotons Iand I neutrons Iwill I be Iobtained
if Idifferent Ielements/isotopes Iare Iused Ito Icalculate Ithese Imasses.)
3. The Icalculated Imass Iof I12C Ibased Ion Ithe I masses Iof Ithe Iconstituent Iparticles Iis I12.099
I amu;Iand Ithe Iactual Imass Iof I12C Iis Iexactly I12 Iamu.
ChemActivity I2
Exercises
1. a) I1.008 Ig. b) I 39.10 Ig.
2. a) I45.98 Ig. b) I 57.27 Ig.
3. Isotopes I are I versions I of I atoms I of I an I element I that I have I the I same I number I of I protons I in
I theInucleus Ibut Idiffer Iin Ithe Inumber Iof Ineutrons Ithat Ithey Ihave.
4. 37Cl Ihas Itwo Imore Ineutrons Iin Iits Inucleus.
3
5. average Imass Iof Ia Imarble I = I (1 I I 5.00 Ig I+ I3 I I 7.00 Ig)/4 I I I= I 5.00 Ig I+ I7.00 Ig I =
I 1
4 I4
0.2500 I I5.00 Ig I + I 0.7500 I I7.00 Ig I = I 6.50 Ig I I(this Iis Ieqn I(2))
6. 10.44 Iamu
7. 35Cl, I75.76%. I 37Cl, I24.24%.
8. a) I4.003 Ig b) I39.10 Ig
9. a) I1 IHe Iatom I I I I4.003 Ig IHe/mole IHe Iatoms
= I6.647 I I 10–24 Ig Iof IHe.
b) 1 IK Iatom I I I I39.098 Ig IK/mole IK Iatoms I=
6.493 I I10–23 Ig. Iof IK
10. 60.06 Ig
11. 3.613 I I1024 I atoms.
12. 2.619 I I1024 Iatoms.
13. a) I 3.029 I I 1025 I atoms. I b) I 1.022 I I 1019 I atoms. I c) I 1.878 I I 1025 I atoms. I d) I 9.782 I I 1027
atoms.
I
14. Phosphorus
I15. 89.5 Ig II
, 4
Problems
1. Assume I that I mass I of I 22Ne I is I 22 I amu. I Calculated I avg I mass I of I Ne I is I 20.18—close I to
I theIe xperimental Ivalue Iof I20.179.
2. a) IFalse. I 6.941 Ig Iper Imole Iof ILi Iatoms. I b) I False. INo IH Iatoms Iweighs I1.008 Iamu.
c) True. INa Iatoms Iare Imore Imassive. Id) IFalse. I1H Iis Ithe Ilightest Iatom Iand Ihas Ia Imass Iof I1
Ig/mole. IThis Iis I(essentially) Ithe Imass Iof Ione Iproton. ITherefore, Ithere Icannot Ibe Ian Iatom Ithat
Ihas Ia Ilower Iaverage Iatomic Imass Ithan I1 Ig/mole.
3. 17: Iprotons Iin Inucleus Iand Ielectrons Iin Ithe Ineutral Iatom. I35.453: Iavg Iamu Iof Ia ICl Iatom Iand
Igrams Iof Ione Imole Iof ICl Iatoms.
4. I 187Re
ChemActivity I3
Exercises
1. 5.47 I I 10–18 IJ.
2. i) IIEa I = I –(2)(–1)/d1 I = I I2/d1 I I I ii) I IEb I = I –(1)(–1)/2d1 I = I 1/2d1 IEi I> I IEii
3. The Iionization Ienergy Iof Icase Ii) Iis Ilarger, I1.20 Ik/d1, Ithan Ithat Iof Icase Iii), I1.17 Ik/d1.
Problems
1. large Iand Inegative
2. V I= I
ChemActivity I4
Exercises
1. No. IThe Iionization Ienergy Iof IHe Iwould Ibe Iabout I4 I(twice Ithe Icharge Iand Ihalf Ithe Idistance)
It he Iionization Ienergy Iof IH Iif Ithis Iwere Ithe Icase. IThe Idata Iin ITable I1 Iindicate Ithat Ithe
Iionization Ienergy Iof IHe Iis Ionly Iabout Itwice Ias Igreat.
2. Several Ipossible Ianswers. IOne Ipossibility: IAll Ithree Ielectrons Iat Ia Ifarther Idistance I(about
Ithree Itimes Ias Ifar Ias Iin IH) Ifrom Ithe Inucleus.
Problem
1. a) IV I= I
b) IThe IIE Iof IHe Iis Islightly Iless Ithan Itwice Ithe IIE Iof IH Ibecause Ithe Ielectron-electron
Irepulsion Imakes Ithe Ipotential Ienergy Imore Ipositive. INote Ithat Ithe Ifirst Itwo Iterms Iin I1a) Iare
Inegative Iand Ithe Ithird Iterm Iis Ipositive.
ChemActivity I5
Exercises
1. a) I6. b) I 5. c) I I8.
2. a) I+6. b) I+5. c) I+8.
3. The I IE I of I Br I should I be I less I than I the I IE I of I Cl. I There I is I about I a I 0.4 I MJ/mole
I differenceIbetween Ithe IIEs Iof IF Iand ICl. I Prediction: IBr, I0.8 IMJ/mole.
4. The IIE Iof ILi+ Ishould Ibe Ilarger Ithan Ithe IIE Iof IHe Ibecause Iboth Iatoms Ihave I2 Ielectrons Iin Ithe
I1
st Ishell Iand ILi+ I has Ia Icore Icharge Iof I+3 Iwhereas IHe Ionly Ihas Ia Icore Icharge Iof I+2.
5. The I IE I of I F– I should I be I less I than I the I IE I of I Ne I because I both I atoms I have I eight I electrons
I inIt he I2
nd Ishell Iand IF– I has Ia Icore Icharge Iof I+7 Iwhereas INe Ihas Ia Icore Icharge Iof I+8.
Answers to Exercises and Problems
I I I I
for
CHEMISTRY
A Guided Inquiry
I I
Seventh Edition, 2017
I I
Richard S. Moog I I
Franklin & Marshall College
I I I
John J. Farrell I I
Franklin & Marshall College
I I I
Latest Update: December 20, 2017
I I I I
John Wiley & Sons, Inc.
I I I I
, 2
Answers to Exercises and Problems
I I I I
These Ianswers Imay Ibe Iprovided Ito Istudents Iat Ithe Iinstructor’s Idiscretion.
Do Inot Ipost Ion Ipublicly Iavailable Iwebsite.
These Ianswers Ido Inot Icontain Ifull Iexplanations Ior Isolutions.
ChemActivity I1
Exercises
1.
Isotope Atomic Mass Number Iof
INumber INumbe IElectrons
Z r
A
31P 15 31 15
18O 8 18 8
39K+ 19 39 18
58Ni2+ 28 58 26
2. 1.674 I I10–24 Ig. I 1.993 I I10–23 Ig.
3. 8.67 I I10–17 Ig.
4. 12.00 Ig.
5. 2.6621 I I10–23 Ig.
6. a) IThe Imass Inumber Iis Ithe Isum Iof Ithe Inumber Iof Iprotons Iand Ithe Inumber Iof Ineutrons Iin
ItheInucleus. I b) I The Iatomic Inumber Iis Ithe Inumber Iof Iprotons Iin Ithe Inucleus.
7. False. I 18O Ihas I8 Iprotons Iand I10
Ineutrons. I8.
Species Electrons Protons Neutrons
24
Mg 12 12 12
23
Na + 10 11 12
35
Cl 17 17 18
35 - 18 17 18
Cl
56 3+ 23 26 30
Fe
15
N 7 7 8
16 2- 10 8 8
O
27 3+ 10 13 14
Al
9.
Isotope Atomic Mass Number Iof
INumber INumbe IElectrons
Z r
A
59Co2+ 27 59 25
14N 7 14 7
7Li 3 7 3
6Li 3 6 3
58Zn2+ 30 58 56
19F– 9 19 10
, 3
10. I I IAll Iisotopes Iof Ian Ielement Ihave Ithe Isame Inumber Iof Iprotons Iin Ithe Inucleus. IOne Iisotope Iof
Ian Ielement Iis Idifferentiated Ifrom Ianother Iisotope Iof Ithe Isame Ielement Iby Ithe Inumber Iof
Ineutrons Iin Ithe Inucleus.
Problems
1. Using I carbon-13 I and I carbon-12, I approx I mass I of I neutron I = I 13.0034 I – I 12 I = I 1.0034
I amu.Iapprox Imass Iof I
14C I= I13.0034 I+ I1.0034 I= I14.0068 Iamu.
2. a) IUsing I14C Iand I14C–, I mass Iof Ielectron I is I approximately I13.0039 I amu I – I 13.0034 I amu I=
I0.0005 Iamu
b) Using I1H, Imass Iof Iproton Iis Iapproximately I1.0078 Iamu I– I0.0005 Iamu I= I1.0073 Iamu
c) Using I 1H and I 2H, I mass I of I neutron I is I approximately I 2.0140 I amu I – I 1.0078
I amu =I1.0062 Iamu
(Note Ithat Islightly Idifferent I values I for Ithe I masses I of Iprotons Iand I neutrons Iwill I be Iobtained
if Idifferent Ielements/isotopes Iare Iused Ito Icalculate Ithese Imasses.)
3. The Icalculated Imass Iof I12C Ibased Ion Ithe I masses Iof Ithe Iconstituent Iparticles Iis I12.099
I amu;Iand Ithe Iactual Imass Iof I12C Iis Iexactly I12 Iamu.
ChemActivity I2
Exercises
1. a) I1.008 Ig. b) I 39.10 Ig.
2. a) I45.98 Ig. b) I 57.27 Ig.
3. Isotopes I are I versions I of I atoms I of I an I element I that I have I the I same I number I of I protons I in
I theInucleus Ibut Idiffer Iin Ithe Inumber Iof Ineutrons Ithat Ithey Ihave.
4. 37Cl Ihas Itwo Imore Ineutrons Iin Iits Inucleus.
3
5. average Imass Iof Ia Imarble I = I (1 I I 5.00 Ig I+ I3 I I 7.00 Ig)/4 I I I= I 5.00 Ig I+ I7.00 Ig I =
I 1
4 I4
0.2500 I I5.00 Ig I + I 0.7500 I I7.00 Ig I = I 6.50 Ig I I(this Iis Ieqn I(2))
6. 10.44 Iamu
7. 35Cl, I75.76%. I 37Cl, I24.24%.
8. a) I4.003 Ig b) I39.10 Ig
9. a) I1 IHe Iatom I I I I4.003 Ig IHe/mole IHe Iatoms
= I6.647 I I 10–24 Ig Iof IHe.
b) 1 IK Iatom I I I I39.098 Ig IK/mole IK Iatoms I=
6.493 I I10–23 Ig. Iof IK
10. 60.06 Ig
11. 3.613 I I1024 I atoms.
12. 2.619 I I1024 Iatoms.
13. a) I 3.029 I I 1025 I atoms. I b) I 1.022 I I 1019 I atoms. I c) I 1.878 I I 1025 I atoms. I d) I 9.782 I I 1027
atoms.
I
14. Phosphorus
I15. 89.5 Ig II
, 4
Problems
1. Assume I that I mass I of I 22Ne I is I 22 I amu. I Calculated I avg I mass I of I Ne I is I 20.18—close I to
I theIe xperimental Ivalue Iof I20.179.
2. a) IFalse. I 6.941 Ig Iper Imole Iof ILi Iatoms. I b) I False. INo IH Iatoms Iweighs I1.008 Iamu.
c) True. INa Iatoms Iare Imore Imassive. Id) IFalse. I1H Iis Ithe Ilightest Iatom Iand Ihas Ia Imass Iof I1
Ig/mole. IThis Iis I(essentially) Ithe Imass Iof Ione Iproton. ITherefore, Ithere Icannot Ibe Ian Iatom Ithat
Ihas Ia Ilower Iaverage Iatomic Imass Ithan I1 Ig/mole.
3. 17: Iprotons Iin Inucleus Iand Ielectrons Iin Ithe Ineutral Iatom. I35.453: Iavg Iamu Iof Ia ICl Iatom Iand
Igrams Iof Ione Imole Iof ICl Iatoms.
4. I 187Re
ChemActivity I3
Exercises
1. 5.47 I I 10–18 IJ.
2. i) IIEa I = I –(2)(–1)/d1 I = I I2/d1 I I I ii) I IEb I = I –(1)(–1)/2d1 I = I 1/2d1 IEi I> I IEii
3. The Iionization Ienergy Iof Icase Ii) Iis Ilarger, I1.20 Ik/d1, Ithan Ithat Iof Icase Iii), I1.17 Ik/d1.
Problems
1. large Iand Inegative
2. V I= I
ChemActivity I4
Exercises
1. No. IThe Iionization Ienergy Iof IHe Iwould Ibe Iabout I4 I(twice Ithe Icharge Iand Ihalf Ithe Idistance)
It he Iionization Ienergy Iof IH Iif Ithis Iwere Ithe Icase. IThe Idata Iin ITable I1 Iindicate Ithat Ithe
Iionization Ienergy Iof IHe Iis Ionly Iabout Itwice Ias Igreat.
2. Several Ipossible Ianswers. IOne Ipossibility: IAll Ithree Ielectrons Iat Ia Ifarther Idistance I(about
Ithree Itimes Ias Ifar Ias Iin IH) Ifrom Ithe Inucleus.
Problem
1. a) IV I= I
b) IThe IIE Iof IHe Iis Islightly Iless Ithan Itwice Ithe IIE Iof IH Ibecause Ithe Ielectron-electron
Irepulsion Imakes Ithe Ipotential Ienergy Imore Ipositive. INote Ithat Ithe Ifirst Itwo Iterms Iin I1a) Iare
Inegative Iand Ithe Ithird Iterm Iis Ipositive.
ChemActivity I5
Exercises
1. a) I6. b) I 5. c) I I8.
2. a) I+6. b) I+5. c) I+8.
3. The I IE I of I Br I should I be I less I than I the I IE I of I Cl. I There I is I about I a I 0.4 I MJ/mole
I differenceIbetween Ithe IIEs Iof IF Iand ICl. I Prediction: IBr, I0.8 IMJ/mole.
4. The IIE Iof ILi+ Ishould Ibe Ilarger Ithan Ithe IIE Iof IHe Ibecause Iboth Iatoms Ihave I2 Ielectrons Iin Ithe
I1
st Ishell Iand ILi+ I has Ia Icore Icharge Iof I+3 Iwhereas IHe Ionly Ihas Ia Icore Icharge Iof I+2.
5. The I IE I of I F– I should I be I less I than I the I IE I of I Ne I because I both I atoms I have I eight I electrons
I inIt he I2
nd Ishell Iand IF– I has Ia Icore Icharge Iof I+7 Iwhereas INe Ihas Ia Icore Icharge Iof I+8.
