Chem 161-2013 exam I + solutions
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Chemistry 161
Exam I
October 2, 2013
Student Name (Print): _____TAVSS______________
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Chem161f13e1v1
1
Your EXAM FORM is
①
1
Chem 161-2013 Exam I
B&O Chapter 1 – Chemistry: Matter and Measurement
Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement
Significant Figures/Precision/Accuracy
A homogeneous mixture was prepared by mixing 0.15g A, 112.3g B and 24.55g C. The
volume is found to be 85.80 cm
3
. What is the density of the mixture to correct number of
significant digits?
A. 1.5967 g/cm
3
B. 1.59674 g/cm
3
C. 1.6 g/cm
3
D. 1.60 g/cm
3
E
. 1.597 g/cm
3
0.15g
112.3g
24.55g
137.00g
The rule for addition-subtraction is to consider the smallest number of digits to the right of
the decimal point. Since 112.3 has only one digit to the right of the decimal point, then the
sum, 137.00, can only have one digit to the right of the decimal point, i.e., 137.0, which
has four significant figures. Nevertheless, don’t round off in the middle, so the next
process should use the full 137.00, but we should keep in mind that this number really has
only four significant figures.
D = g/cm
3
D = 137.00g/85.80cm
3
= 1.59674g/cm
3
Since 137.00 has four significant figures and 85.80 has four significant figures then the
final answer must have only four significant figures. Rounding off gives 1.597g/cm
3
.
E
Chem161f13e1v1
2
2
Chem 161-2013 Exam I
B&O Chapter 1 – Chemistry: Matter and Measurement
Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement
Units/unit conversions/dimensional analysis
Which of the following is
not
an SI base unit?
A. kilogram
B. mole
C
. liter
D. Kelvin
E. second
The SI base units are: meter, kilogram, second, Kelvin, ampere, mol and candela. Liters
are derived SI units. meter → cm → cm
3
→ mL → L
C
Chem161f13e1v1
3
3
Chem 161-2013 Exam I
B&O Chapter 4 – Periodic Trends of the Elements
Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends
Periodic table trends
Which of the following would have the highest 2nd ionization energy?
A. Ba
B. Ga
C. Sr
D
. K
E. Ca
D
Chem161f13e1v1
4
The first ionization energy is M → M
+
+ 1e
-
.
The second ionization energy is M
+
→ M
2+
+ 1e
-
Therefore, plot M
+
, and mentally consider the relative energy required to remove another electron to form M
2+
.
20
Ca
+
= 1s
2
2s
2
2p
6
3s
2
3p
6
4s
1
56
Ba
+
= 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
4p
6
5s
2
4d
10
5p
6
6s
1
19
K
+
= 1s
2
2s
2
2p
6
3s
2
3p
6
38
Sr
+
= 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
4p
6
5s
1
56
Ba
+
31
Ga
+
= 1s
2
2s
2
2p
6
3s
2
3p
6
4s
2
3d
10
20
Ca
+
19
K
+
38
Sr
+
31
Ga+
K
+
has its valence electrons in the third principal shell. The other four ions have their valence electrons in
higher principal shells. This means that the K
+
ion has the highest ionization energy, in that the valence
electrons are held relatively strongly by the protons in the nucleus because the valence shell is close to the
nucleus. Now look at the effective nuclear charge of the ions. The effective nuclear charge on K
+
is far
greater than the effective nuclear charge on any of the other ions, adding further strength to the conclusion
that K
+
has the highest second ionization energy. This conclusion is different than what one would expect
Chem161f13e1v1
5
regarding the generalization that atoms and ions in the lower left-hand corner of the periodic table have
low ionization energies.
Chem161f13e1v1
6
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Arial
11
BIUA
E = E = EE E - E E X
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AIV-20: Getting Connected (Writing lonic Compounds)
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CI
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Be
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