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Chem 105 |
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Spring 2001 |
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ConcepTests |
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ConceptTests for Exam 3
ConcepTests for exams 1 and 2 can be found by paging down
25. The hybrid orbitals used to explain the bonds formed form C to H, Cl and S in CHSCl (carbon in center)
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a. sp |
1 |
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b. sp2 |
8 |
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c. sp3 |
12 |
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d. dsp |
5 |
The Lewis electron dot structure shows C=S, C-H, and C-Cl bonds from the central C. Three electron groups (two single bonds, one double bond) around the C requires sp2 hybridization.
26.
The hybrid orbitals used to explain the bonds formed from P to H in PH3 are
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a. sp |
0 |
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b. sp2 |
6 |
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c. sp3 |
n-6 |
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d. dsp |
0 |
The Lewis electron dot structure shows three P-H bonds and a nonbonding pair of electrons (P:) from the central P. Four electron groups (three single bonds, one nonbonding pair) around the P requires sp3 hybridization.
27. Formaldehyde ( CH2O) is a molecule with shape best described as
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a. pyramidal & trigonal |
1 |
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b. linear |
0 |
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c. planar & trigonal |
12 |
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d. planar & T-shaped |
1 |
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e. bent |
0 |
The Lewis electron dot structure shows a C=O and two C-H bonds from the central C. Three electron groups (two single bonds, one double bond) around the C requires trigonal planar array of electron groups to minimize electron pair repulsion. Since all of the electron groups are involved in bonding, the molecular shape is also trigonal planar.
28. The amide ion, NH2-, is a molecule with shape best described as
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a. pyramidal & trigonal |
0 |
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b. linear |
7 |
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c. planar & trigonal |
0 |
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d. planar & T-shaped |
0 |
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e. bent |
6 |
The Lewis electron dot structure shows a two N-H bonds and two nonbonding pairs (:N:) around the central N. Four electron groups (two single bonds, two nonbonding pairs) around the N requires a tetrahedral array of electron groups to minimize electron pair repulsion. Since only two of the electron groups are involved in bonding, the molecular shape is bent.
29. Electronegativities: H = 2.2 C = 2.5 F = 4.0
i. The polarity of CH4 is
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a. zero |
1 |
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b. a bit larger than zero |
11 |
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c. large |
4 |
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|
|
The individual bonds have very little polarity (for practical purposes, zero), and the symmetry of the molecule guarantee zero molecular polarity.
ii. The polarity of CF4 is
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a. zero |
1 |
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b. a bit larger than zero |
0 |
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c. large |
n-1 |
Although the individual bonds have large polarity, the tetrahedral symmetry of the molecule guarantee zero molecular polarity.
The polarity of CH2F2 is
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a. zero |
0 |
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b. a bit larger than zero |
7 |
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c. large |
10 or so |
The individual C-F bonds have large polarity, and the geometry of the molecule results in a large nonzero molecular dipole moment.
30. When you numerically interpret the observations made in a calorimetry experiment, the key relationship is
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a. q reaction = 1/q calorimeter |
0 |
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b. q reaction = q calorimeter |
0 |
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c. q reaction = -1/q calorimeter |
0 |
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d. q reaction = -q calorimeter |
n |
For an exothermic process, the energy gained by the calorimeter is equal to the energy released from the chemical reaction. For an endothermic process, the energy gained by the reaction is equal to the energy released from the calorimeter.
31. Fe2O3(s) + 2Al(s) ------------> 6 Al2O3(s) + 2Fe(s)
DH ~ q ~ -849 kJ
When iron (III) oxide and elemental aluminum react, the temperature of the reaction mixture should
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a. decrease |
4 |
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b. be constant |
2 |
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c. increase |
n-6 |
For an exothermic process, the temperature of the reaction mixture increases as heat energy is released as the reaction proceeds.
32. In a calorimetry experiment, you determine DHrxn by using
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a. DHrxn = DHcalorimeter |
0 |
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b. DHrxn = -DHcalorimeter |
n |
33. Standard enthalpies of formation (DHof) are used to calculate enthalpy of reaction,
D
Horxn, by
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a. DHorxn = 3mDHof, products + 3nDHof, reactants |
0 |
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b. DHorxn = 3mDHof, products + 3nDHof, reactants |
n |
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c. DHorxn = 3mDHof, products + 3nDHo, reactants |
0 |
34. The "formation reaction", for benzene, for which DHof, C6H6(l) = + 49.0 kJ/mol applies is
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a. 3C2H2(q) -------> C6H6(l) |
0 |
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b. 6C(q) + 6H(q) -------> C6H6(l) |
0 |
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c. 6C(s) + 3H2(q) -------> C6H6(l) |
25 |
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d. C60(s) + 30 H2(q) -------> 10 C6H6(l) |
0 |
35. A liquid boils when
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a. covalent bonds are broken |
0 |
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b. the temperature of the liquid reaches 100oC (373 K) |
2 |
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c. The vapor pressure of the liquid equals the pressure above its surface. |
27 |
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d. The vapor pressure of the liquid equals 1.00 atm (760 mm Hg) |
0 |
36. Dogs pant to
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a. annoy people |
1 |
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b. adjust the pH of their blood |
0 |
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c. cool off |
n-1 |
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d. warm up |
0 |
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e. smell their surroundings |
0 |
37. Snow, H2O (s)
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a. must turn to liquid water, H2O (l), before it turns to gaseous water, H2O (g) . |
10 |
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b. may turn into gaseous water without turning into liquid water. |
16 |
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c. never goes away |
0 |
Snow may sublime, which is the process of solid converting directly to gas.
38. It takes 3 minutes to make a "soft boiled egg" in Fairbanks. To make an egg of the same sort at 8000 feet in the Alaska Range might take
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a. 1 minute |
11 |
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b. 3 minutes |
0 |
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c. 5 minutes |
n-11 |
At 8000 feet altitude, the pressure on the water is less than in Fairbanks (500 feet). The water will boil at a lower temperature in the mountains, and the egg will cook at a slower rate.
39. I2 is
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a. nonpolar |
n-5 |
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b. moderately polar |
4 |
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c. very polar |
1 |
Diatomic molecules containing a single element have no net dipole moment because the electronegativities of the two atoms are identical.
40. I2 is
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a. more soluble in water than in hexane |
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b. less soluble in water than in hexane |
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c. equally soluble in water and hexane |
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d. insoluble in water and hexane |
Like dissolves like. The nonpolar iodine is more soluble in the nonpolar hexane than in the polar water.
41. Vitamin A
is predominantly
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a. nonpolar |
1 |
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b. polar |
n-1 |
Despite, the OH functional group, the molecule as a whole is very nonpolar. Vitamin A is a "fat soluble vitamin".
42. Vitamin C (ascorbic acid)
is predominantly
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a. nonpolar |
2 |
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b. polar |
n-2 |
The many OH groups in the relatively small molecule make for a nonpolar structure. Vitamin C is a "water soluble vitamin".
43. Northern Pike (a fish) do not live in warm water because:
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a. it is a true Alaskan fish |
0 |
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b. it requires a lot of oxygen to thrive |
n-1 |
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c. many predators live in warm water |
1 |
The solubility of oxygen decreases as the temperature of the water increases. Active predators like the northern pike nedd a high concentration of oxygen to live, and they can only get that if the water temperature is sufficiently low.
ConceptTests for Exam 2
ConcepTests for exam 1 can be found by paging down
9. When 10.0g of Na2CO3 is dissolved in water and diluted to 350 mL, we might say "The concentration of sodium carbonate is 0.270 molar." What is the actual concentration of " Na2CO3 " molecules?
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a. 0 M |
3 |
19 |
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b. 0.135 M |
2 |
0 |
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c. 0.270 M |
12 |
0 |
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0.540 M |
2 |
1 |
The sodium carbonate is a strong electrolyte. The salt dissociates completely to Na+ (aq) and CO32- (aq) when it dissolves, leaving no " Na2CO3 " molecules.
10. What is the actual concentration of sodium cations, Na+ (aq)?
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a. 0 M |
0 |
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b. 0.135 M |
10 |
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c. 0.270 M |
1 |
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d. 0.540 M |
17 |
The sodium carbonate dissolves and dissociates to form two sodium cations per unit of sodium carbonate:
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H2O |
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Na2CO3 (s) |
-------> |
2 Na+ (aq) + CO32- (aq) |
The "0.270 M sodium carbonate" solutions has 2 x 0.270 = 0.540 moles of sodium cation per liter of solution.
11. If I mix a solution of 1 M CaCl2 (aq) and another solution of 1 M Na2CO3 (aq), what will happen?
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a. Nothing |
1 |
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b. CO2 (g) will form |
1 |
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c. NaCl (s) will form |
5 |
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d. CaCO3 (s) will form |
8 |
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e. Cl2 (g) and Na (s) will form |
4 |
The reaction forms insoluble calcium carbonate, based on solubility rules.
CaCl2 (aq) + Na2CO3 (aq) -----> 2 NaCl (aq) + CaCO3 (s)
12. The oxidation number for carbon in CO2 is
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a. -4 |
3 |
1 |
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b. -2 |
1 |
0 |
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c. 0 |
6 |
0 |
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d. +2 |
7 |
0 |
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e. +4 |
7 |
25 |
The oxidation number of oxygen is -2. The sum of the oxidation numbers equals zero for a molecule.
(Oxidation number for carbon) + 2 (oxidation number for oxygen) = 0
(Oxidation number for carbon) + 2 (-2) = 0
(Oxidation number for carbon) = +4
12. The oxidation number for carbon in CH4 is
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a. -4 |
17 |
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b. -2 |
0 |
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c. 0 |
0 |
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d. +2 |
1 |
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e. +4 |
0 |
The oxidation number of hydrogen is +1. The sum of the oxidation numbers equals zero for a molecule.
(Oxidation number for carbon) + 4 (oxidation number for hydrogen) = 0
(Oxidation number for carbon) + 4 (+1) = 0
(Oxidation number for carbon) = -4
13. The gas pressure in a rigid container is 1.20 atm at room temperature (20oC). When the temperature increases to 40oC, the resulting gas pressure is
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a. 0.60 atm |
0 |
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b. a bit less than 1.20 atm. |
1 |
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c. 1.20 atm |
0 |
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d. a bit more than 1.20 atm |
15 |
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e. 2.40 atm |
13 |
ANS: Increasing the temperature increases the pressure in the rigid container (constant volume). The pressure does not double, however! To make the correct calculation, absolute temperature must be used, not Celsius temperature, and the pressure increase is a factor of (313K / 293 K =) 1.07, so the final pressure is 1.28 atm.
14. A balloon has a volume of 0.30 L at atmospheric pressure (760 torr). When it is placed in a vacuum chamber with pressure of 1/3 atm (253 torr), its volume will be
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a. 0.10 L |
4 |
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b. 0.30 L |
1 |
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c. 0.90 L |
n-5 |
ANS: Decreasing the pressure by a factor of three increases the volume by a factor of three.
15. If the dotted curve in the plot below represents the velocity distribution of N2 at 293K, the curve that represents the velocity distribution of N2 at 400K is
a. A
b. B
For a given type of gas molecule, the average gas velocity increases as temperature increases.
16. If the dotted curve in the plot below represents the velocity distribution of N2 at 293K, the curve that represents the velocity distribution of CH4 at 293K is
a. A
b. B
The molar mass of nitrogen is 28 g/mol and the molar mass of methane is 16 g/mole. The average gas velocity increases as the molar mass of the molecule decreases, at constant temperature.
17. The behavior of real gases will deviate most from that predicted by the ideal gas law when
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a. Both P and T are low. |
2 |
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b. P is low and T is high. |
3 |
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c. P is high and T is low. |
6 |
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d. Both P and T are high. |
4 |
18. Consider the following plotted data set of five measurements for carbon dioxide gas at P = 1 atm. If experiments were done at lower temperatures, would the volume decrease as predicted by the extrapolated dotted line?
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a. Yes |
2 |
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b. No |
9 |
The experimental data points would show negative deviations from the extrapolated straight line as the temperature decreased and the carbon dioxide behaves less ideally.
19. Electrons in atoms
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a. follow well-defined circular orbits. |
0 |
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b. follow well-defined elliptical orbits. |
13 |
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c. do not follow simple, closed paths that are easily described. |
17 |
20. The Bohr model of the atom was deficient because it
|
a. neglected relativistic effects in the mass of the electron. |
0 |
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b. neglected the wave nature of matter. |
9 |
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c. used electrostatic arguments. |
5 |
21. Bohr's theory of the atom is important because it was
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a. revolutionary and correct. |
8 |
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b. revolutionary but incorrect. |
19 |
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c. correct but mundane. |
1 |
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d. mundane and incorrect. |
0 |
22. Hydrogen has one electron and an electron configuration of 1s1.
Helium has two electrons and an electron configuration of 1s2.
Lithium has three electrons. Its electron configuration is
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a. 1s3. |
0 |
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b. 1s2 1p1. |
5 |
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c. 1s2 2s1. |
9 |
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d. 1s1 1s1 2p1. |
0 |
23. Consider Mg and S. Periodic trends allow the following prediction:
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a. Magnesium is smaller than sulfur. |
17 |
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b. Magnesium is larger than sulfur. |
9 |
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c. Magnesium is the same size as sulfur. |
4 |
The general trend is that atoms size decreases across the periodic table (from left to right). Sulfur is further to the right in the periodic table than magnesium (but in the same row). The valence electrons in sulfur experience a larger effective nuclear charge than those in magnesium, resulting in a smaller equilibrium distance for the valence electrons in sulfur. The sulfur atom is the smaller of the two.
24. Consider Mg and Ca. Periodic trends allow the following prediction:
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a. Magnesium is smaller than calcium. |
n-2 |
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b. Magnesium is larger than calcium. |
1 |
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c. Magnesium is the same size as calcium. |
1 |
The general trend is that atoms size increases down the periodic table. Calcium is below magnesium in the periodic table. The valence electrons in calcium occupy the 3s orbital, which is larger than the 2s orbital that the valence electrons in magnesium occupy. The calcium atom is the larger of the two.
ConceptTests for Exam 1
1. The identity of an element is determined by the number of
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answer |
# student responses |
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a. electrons |
6 |
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b. protons |
n-6 |
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c. neutrons |
0 |
In each atom of the element.
2. Gasoline is a
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a. pure compound |
1 |
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b. mixture of a few compounds |
23 |
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c. mixture of dozens of compounds |
21 |
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d. mixture of many scores of compounds |
3 |
Rationale: The gas chromatogram of gasoline showed many scores of peaks, so we would infer that the gasoline is made up of many scores of compounds.
3. Some NaCl is weighed and used as limiting reactant in a chemical process.
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Data: |
NaCl plus beaker |
83.164 g |
|
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beaker alone |
83.008 g |
The amount of NaCl is
n = (mass NaCl) / (formula weight NaCl)
n = (mass NaCl) / (58.443 g NaCl / mole NaCl)
The most reasonable way to report the value of n is
|
a. 2.7 x 10-3 mol NaCl |
1 |
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b. 2.67 x 10-3 mol NaCl |
3 |
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c. 2.669 x 10-3 mol NaCl |
8 |
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d. 2.6693 x 10-3 mol NaCl |
14 |
Solution: n = (mass NaCl) / (58.443 g NaCl / mole NaCl)
mass NaCl = 83.164 g - 83.008 g = 0.156 g
Note that I haven't rounded, but the net weight of NaCl has 3 significant figures (not five!)
Thus the calculation is now n = 0.156 g NaCl / (58.443 g NaCl / mole NaCl)
n = 2.6693 x 10-3 mol NaCl.
The weak link is the mass of NaCl, which has three significant figures. The answer must be rounded to 3 significant figures in spite of all of the raw rata looking more impressive. n = 2.67 x 10-3 mol NaCl
4. Fact: Sulfur and oxygen can combine to form SO, SO2, SO3, and SO4.
Question: Can lithium and chlorine combine to form
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a. LiCl only? |
53 |
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b. LiCl2 only? |
0 |
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c. LiCl, LiCl2, LiCl3, and LiCl4? |
0 |
Rationale: Li is a metal, Cl is a nonmetal, and the resulting compound is ionic. The ionic compound is made of Li+ ions and Cl- ions. Since the compound must have zero net charge, there must be equal numbers of each mono-positive lithium ion and mono-negative chloride ion.
5. The only expected formula for a compound of aluminum and oxygen is
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a. AlO |
0 |
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b. Al2O |
10 |
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c. AlO3 |
7 |
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d. Al2O3 |
n-17 |
Rationale: Al is a metal, O is a nonmetal, and the resulting compound is ionic. The ionic compound is made of Al3+ and O2- ions. Since the compound must have zero net charge, the simplest ration of whole numbers that achieves that goal is two Al3+ cations to three O2- anions: Al2O3
6. The name for Na2SO3 is
|
a. sodium sulfide |
1 |
|
b. sodium sulfite |
n-6 |
|
c. sodium sulfate |
5 |
|
d. sodium persulfate |
0 |
Rationale: The ionic compound is made of sodium cations and sulfite anions.
7. The molar mass for C2H4O2 is approximately
(use atomic weights for H = 1 g/mol, C = 12 g/mol, O = 16 g/mol)
|
a. 29 g/mol |
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b. 44 g/mol |
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c. 60 g/mol |
(This is what I meant to have in class)
Calculation: molar mass =
(2 mol C)(12 g / mol C) + (4 mol H)(1 g / mol H) + (2 mol O)(16 g / mol 0) = 60 g / mol
8. Potassium metal reacts with water:
2 K (s) + 2 H2O (l) ----> 2 KOH (aq) + H2 (g)
When 3.9 g of K (s) is added to 500 g of water,
|
a. All of the K and all of the H2O react |
0 |
|
b. All of the K and some of the H2O react. Some of the H2O is left unreacted. |
15 |
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c. Some of the K and all of the H2O react. Some of the K is left unreacted. |
2 |
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d. Some of the K and some of the H2O react. Some of the K and some of the H2O are left unreacted. |
10 |
In this "obvious" case. The K is the limiting reactant. The K is consumed completely. Some of the excess reactant remains when the reaction is complete.
______________________________________________________________________
Email Prof. Stolzberg: ffrjs@uaf.edu
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Updated May 2, 2001