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Answers to Critical Thinking Questions
for
CHEMISTRY
A Guided Inquiry
Fifth Edition, 2011
Richard S. Moog
Franklin & Marshall College
John J. Farrell
Franklin & Marshall College
Latest Update: May 12, 2011
John Wiley & Sons, Inc.
, 2
Answers to Critical Thinking Questions
Please do not give these answers to students.
Quite often, the answers given here are less
detailed than what is expected for a student’s
answer.
ChemActivity 1
1. 6, 6, 6
2. 6, 7, 7
3. 6, 6, 7
4. All carbon atoms and ions have six protons in the nucleus.
5. All hydrogen atoms and ions have one proton in the nucleus.
6. Z is the number of protons in the nucleus of that atom.
7. Twentyeight protons in the nucleus.
8. a) Because there are six protons and 7 electrons, so there is a net charge of 1. b) In an
ion the number of protons and electrons are not equal. c) The charge on an ion = # of
protons – # of electrons.
9. Because every H atom and ion must have one proton, the number of protons is 1. The
number of neutrons is zero, analogous to 1H and 1H. The number of electrons is zero
because the charge is 1+ and there has to be one proton.
10. Different isotopes on a particular element have differing numbers of neutrons in the
nucleus (but the same number of protons).
11. The mass number (A) is the sum of the number of protons and the number of neutrons in
the nucleus.
12. The O atom has 8 protons and 8 neutrons in its nucleus; hence, the mass number is 16. The
O atom has 8 protons and 10 electrons; hence the charge on the ion is 2–. The Na atom has
11 protons and 12 neutrons; hence, the mass number is 23. The Na atom has 11 protons
and 10 electrons; hence the charge on the ion is 1+.
13. Most of the mass is in the nucleus—where the protons and neutrons are. Note that the
difference in mass between a 13C and a 13C– atom (which differ by only one electron) is
only 0.0005 amu.
ChemActivity 2
1. 3
2. All isotopes of magnesium have twelve protons in the nucleus. The three isotopes of
magnesium have 12, 13, and 14 neutrons in the nucleus.
3. 12.0000 amu
4. a) 1200.00 amu; b) 1300.34 amu
, 3
5. Slightly more than 1200.00 amu because 1200 amu would be the minimum and there
13
should be, on average, 1.11 C atoms among the 100 total atoms.
6. None
7. You must know the total number of marbles to use the first method. Therefore, the second
method must be used in this case.
8. a) 0.7577 34.9689 amu + 0.2423 36.9659 amu = 35.45 amu. b) Zero.
–23
9. a) 35.45 amu. b) 5.887 x 10 g
10. 6.022 x 10 atoms x 5.887 x 1023 g/atom = 35.45 g
23
23
11. a) 24.31 amu. b) 4.037 x 10 g
12. 6.022 x 1023 atoms x 4.037 x 1023 g/atom = 24.31 g
13. a) It is the same number. b) It is the same number.
14. a) It is the same number. b) It is the same number.
15. 12.001 is a) the average mass in amu of one C atom and b) the mass in grams of 6.022 x
1023 C atoms.
16. Zero %.
17. a) 12. b) They have the same number—12. c) 6.022 1023. d) They have the same
number, 6.022 1023. e) They have the same number, 12. f) They have the same
number, 6.022 1023.
18. a) Two dozen elephants. b) One mole of sodium atoms.
19. There is one mole of H atoms and one mole of argon atoms; they both have the same
number of atoms.
ChemActivity 3
1. The magnitude of V decreases.
2. V = 0.
3. k and d must be positive. q1q2 must be positive.Therefore, V must be positive. That is, V > 0
4. a) q = +1 for a proton. b) q = 0 for a neutron. c) q = +6 for the nucleus of a carbon atom.
5. V is a negative number because k(1)(–1)/d is negative.
6. V is negative because the electron is negative and the proton is positive.
7. I would expect V to be become more negative as d becomes smaller.
8. V (10–18 J) = 0, –.0462, –0.231, –0.462, –1.16, –2.31.
9. V = – IE
10. a) The electron that is closer to the nucleus would have the larger ionization energy because
it is at a lower (more negative) potential energy according to Coulomb’s potential energy
equation.
11. a) The electron that is at d1 from the +2 nucleus would have the larger ionization energy
because it is at a lower (more negative) potential energy according to Coulomb’s potential
energy equation.
12. The ionization energy is larger by a factor of 2.
13. He+ would have a larger ionization energy than H because q = +2 for He and it would have
a lower (more negative) potential energy according to Coulomb’s potential energy
equation.
, 4
ChemActivity 4
1. 1.31 MJ/mole
2. The electron that is farthest from the nucleus will have the least negative potential energy
(d is larger) and the lowest ionization energy.
3. Openended question. Students might predict that IE1 increases with atomic number
(higher nuclear charge). Or, they might predict that IE 1 decreases if they put additional
electrons farther away from the nucleus.
4. The answer here depends on the students’ answer to CTQ 3.
5. a) The IE1 for He is greater than the IE1 for H because the nuclear charge on a He atom is
+2 whereas the nuclear charge on H is +1. b) The IE 1 for Li is less than the IE1 for He
because at least one of the electrons of Li must be farther away from the nucleus than any
electron of He. Otherwise, the +3 nuclear charge of Li would hold the electron more
tightly.
6. Each electron is held by a +2 charge, rather than a +1 as in H. Therefore, if the electrons is
He and H are at the same distance the Coulombic Potential Energy for He ought twice the
Coulombic Potential for He and IE1 for He ought to be about 2 IE1 for H.
7. If the 3rd electron was at the same distance as H, the IE 1 would be about 3 IE1 for H—
3.93 MJ/mole. However, it would be a bit less because of the repulsion between the
electrons; the best answer is 3.6 MJ/mole.
8. For this model the IE1 of Li would be about 3.93 MJ/mole (probably a bit less due to the
repulsion between the electrons—see the previous CTQ). The IE 1 of Li is 0.52 MJ/mole—
much smaller than 3.93 MJ/mole (see the previous CTQ). This low value is inconsistent
with the least tightly held electron being this close to the nucleus.
9. If the 3rd electron is farther away from the nucleus than the electron in H the IE 1 would be
much lower that the IE1 for H. Because the IE 1 of Li is 0.52 MJ/mole and the IE 1 of H is
0.52 MJ/mole, this model is consistent with the experimental data.
10. The electrons in the first shell of Li are closer to the nucleus and should be harder to ionize
(have a greater ionization energy) than the electron in the second shell.
ChemActivity 5
1. a) H, one electron b) Li, one electron c) He, two electrons
2. a) H, zero b) Li, two c) He, zero
3. The core charge of Li is +1.
4. Electron “b” experiences more electronelectron repulsion than electron “a” because
electron “a” is farther from the other electrons than electron “b”. Therefore the IE 1 of
electron “b” would be less than the IE1 of electron “a”.
5. a) Be has 4 protons in its nucleus. b) two c) two d) 4 – 2 = 2
e) The core charge is equal to the number of valence electrons for a neutral atom.
6. Be has a higher core charge (+2) than Li (+1) and a higher IE1 than Li.
7. Ne: 10–2 = 8
Answers to Critical Thinking Questions
for
CHEMISTRY
A Guided Inquiry
Fifth Edition, 2011
Richard S. Moog
Franklin & Marshall College
John J. Farrell
Franklin & Marshall College
Latest Update: May 12, 2011
John Wiley & Sons, Inc.
, 2
Answers to Critical Thinking Questions
Please do not give these answers to students.
Quite often, the answers given here are less
detailed than what is expected for a student’s
answer.
ChemActivity 1
1. 6, 6, 6
2. 6, 7, 7
3. 6, 6, 7
4. All carbon atoms and ions have six protons in the nucleus.
5. All hydrogen atoms and ions have one proton in the nucleus.
6. Z is the number of protons in the nucleus of that atom.
7. Twentyeight protons in the nucleus.
8. a) Because there are six protons and 7 electrons, so there is a net charge of 1. b) In an
ion the number of protons and electrons are not equal. c) The charge on an ion = # of
protons – # of electrons.
9. Because every H atom and ion must have one proton, the number of protons is 1. The
number of neutrons is zero, analogous to 1H and 1H. The number of electrons is zero
because the charge is 1+ and there has to be one proton.
10. Different isotopes on a particular element have differing numbers of neutrons in the
nucleus (but the same number of protons).
11. The mass number (A) is the sum of the number of protons and the number of neutrons in
the nucleus.
12. The O atom has 8 protons and 8 neutrons in its nucleus; hence, the mass number is 16. The
O atom has 8 protons and 10 electrons; hence the charge on the ion is 2–. The Na atom has
11 protons and 12 neutrons; hence, the mass number is 23. The Na atom has 11 protons
and 10 electrons; hence the charge on the ion is 1+.
13. Most of the mass is in the nucleus—where the protons and neutrons are. Note that the
difference in mass between a 13C and a 13C– atom (which differ by only one electron) is
only 0.0005 amu.
ChemActivity 2
1. 3
2. All isotopes of magnesium have twelve protons in the nucleus. The three isotopes of
magnesium have 12, 13, and 14 neutrons in the nucleus.
3. 12.0000 amu
4. a) 1200.00 amu; b) 1300.34 amu
, 3
5. Slightly more than 1200.00 amu because 1200 amu would be the minimum and there
13
should be, on average, 1.11 C atoms among the 100 total atoms.
6. None
7. You must know the total number of marbles to use the first method. Therefore, the second
method must be used in this case.
8. a) 0.7577 34.9689 amu + 0.2423 36.9659 amu = 35.45 amu. b) Zero.
–23
9. a) 35.45 amu. b) 5.887 x 10 g
10. 6.022 x 10 atoms x 5.887 x 1023 g/atom = 35.45 g
23
23
11. a) 24.31 amu. b) 4.037 x 10 g
12. 6.022 x 1023 atoms x 4.037 x 1023 g/atom = 24.31 g
13. a) It is the same number. b) It is the same number.
14. a) It is the same number. b) It is the same number.
15. 12.001 is a) the average mass in amu of one C atom and b) the mass in grams of 6.022 x
1023 C atoms.
16. Zero %.
17. a) 12. b) They have the same number—12. c) 6.022 1023. d) They have the same
number, 6.022 1023. e) They have the same number, 12. f) They have the same
number, 6.022 1023.
18. a) Two dozen elephants. b) One mole of sodium atoms.
19. There is one mole of H atoms and one mole of argon atoms; they both have the same
number of atoms.
ChemActivity 3
1. The magnitude of V decreases.
2. V = 0.
3. k and d must be positive. q1q2 must be positive.Therefore, V must be positive. That is, V > 0
4. a) q = +1 for a proton. b) q = 0 for a neutron. c) q = +6 for the nucleus of a carbon atom.
5. V is a negative number because k(1)(–1)/d is negative.
6. V is negative because the electron is negative and the proton is positive.
7. I would expect V to be become more negative as d becomes smaller.
8. V (10–18 J) = 0, –.0462, –0.231, –0.462, –1.16, –2.31.
9. V = – IE
10. a) The electron that is closer to the nucleus would have the larger ionization energy because
it is at a lower (more negative) potential energy according to Coulomb’s potential energy
equation.
11. a) The electron that is at d1 from the +2 nucleus would have the larger ionization energy
because it is at a lower (more negative) potential energy according to Coulomb’s potential
energy equation.
12. The ionization energy is larger by a factor of 2.
13. He+ would have a larger ionization energy than H because q = +2 for He and it would have
a lower (more negative) potential energy according to Coulomb’s potential energy
equation.
, 4
ChemActivity 4
1. 1.31 MJ/mole
2. The electron that is farthest from the nucleus will have the least negative potential energy
(d is larger) and the lowest ionization energy.
3. Openended question. Students might predict that IE1 increases with atomic number
(higher nuclear charge). Or, they might predict that IE 1 decreases if they put additional
electrons farther away from the nucleus.
4. The answer here depends on the students’ answer to CTQ 3.
5. a) The IE1 for He is greater than the IE1 for H because the nuclear charge on a He atom is
+2 whereas the nuclear charge on H is +1. b) The IE 1 for Li is less than the IE1 for He
because at least one of the electrons of Li must be farther away from the nucleus than any
electron of He. Otherwise, the +3 nuclear charge of Li would hold the electron more
tightly.
6. Each electron is held by a +2 charge, rather than a +1 as in H. Therefore, if the electrons is
He and H are at the same distance the Coulombic Potential Energy for He ought twice the
Coulombic Potential for He and IE1 for He ought to be about 2 IE1 for H.
7. If the 3rd electron was at the same distance as H, the IE 1 would be about 3 IE1 for H—
3.93 MJ/mole. However, it would be a bit less because of the repulsion between the
electrons; the best answer is 3.6 MJ/mole.
8. For this model the IE1 of Li would be about 3.93 MJ/mole (probably a bit less due to the
repulsion between the electrons—see the previous CTQ). The IE 1 of Li is 0.52 MJ/mole—
much smaller than 3.93 MJ/mole (see the previous CTQ). This low value is inconsistent
with the least tightly held electron being this close to the nucleus.
9. If the 3rd electron is farther away from the nucleus than the electron in H the IE 1 would be
much lower that the IE1 for H. Because the IE 1 of Li is 0.52 MJ/mole and the IE 1 of H is
0.52 MJ/mole, this model is consistent with the experimental data.
10. The electrons in the first shell of Li are closer to the nucleus and should be harder to ionize
(have a greater ionization energy) than the electron in the second shell.
ChemActivity 5
1. a) H, one electron b) Li, one electron c) He, two electrons
2. a) H, zero b) Li, two c) He, zero
3. The core charge of Li is +1.
4. Electron “b” experiences more electronelectron repulsion than electron “a” because
electron “a” is farther from the other electrons than electron “b”. Therefore the IE 1 of
electron “b” would be less than the IE1 of electron “a”.
5. a) Be has 4 protons in its nucleus. b) two c) two d) 4 – 2 = 2
e) The core charge is equal to the number of valence electrons for a neutral atom.
6. Be has a higher core charge (+2) than Li (+1) and a higher IE1 than Li.
7. Ne: 10–2 = 8
