
Spring, 2000
ConcepTests for Exam 4 (ConcepTests for earlier exams can be found by scrolling down)
44. The oxidation number for carbon in CO2 is
a. -4
b. -2
c. 0
d. +2
e. +4
ANS: The oxidation number for the oxygen atoms is -2 and the sum of all oxidation numbers is zero for a molecule. So:
(oxidation number for C) + 2 (oxidation number for O) = 0
(oxidation number for C) + 2 (-2) = 0
(oxidation number for C) = +4
45. The oxidation number for carbon in CH4 is
a. -4
b. -2
c. 0
d. +2
e. +4
ANS: The oxidation number for the hydrogen atoms is +1 and the sum of all oxidation numbers is zero for a molecule. So:
(oxidation number for C) + 4 (oxidation number for H) = 0
(oxidation number for C) + 4 (+1) = 0
(oxidation number for C) = -4
46. For the spontaneous process
MnO4- (aq) + 5 Fe2+ (aq) + 8 H+ (aq) ----------> Mn2+ (aq) + 5 Fe3+ (aq) + 4 H2O (l)
The reducing agent is
|
a. MnO4- (aq) |
|
b. Fe2+ (aq) |
|
c. H+ (aq) |
The oxidation number of Mn decreases from +7 to +2, so the MnO4- (aq) is reduced by the Fe2+ (aq).
47. For the spontaneous process
MnO4- (aq) + 5 Fe2+ (aq) + 8 H+ (aq) ----------> Mn2+ (aq) + 5 Fe3+ (aq) + 4 H2O (l)
The stronger of the two potential reducing agents is
|
a. Fe2+ (aq) |
|
b. Mn2+ (aq) |
In a spontaneous process, a stronger reducing agent, Fe2+ (aq), reacts and a weaker reducing agent, Mn2+ (aq), is formed.
48. For the spontaneous process
MnO4- (aq) + 5 Fe2+ (aq) + 8 H+ (aq) ----------> Mn2+ (aq) + 5 Fe3+ (aq) + 4 H2O (l)
Eocell must be
|
a. negative |
|
b. positive |
For a spontaneous process Eocell is positive (D Go = -nFEocell).
49. Table 18.1 tells us
Fe2+ (aq) ---------> Fe3+ (aq) + e- Eo = -0.771 V
If the stoichiometric coefficients of the half reaction are all multiplied by 5 so that it can be combined with the half reaction MnO4- (aq) + 8 H+ (aq) + 5 e- -----> Mn2+ (aq) + 4H2O (l), the value of Eo used in the calculation of Ecell is
a. -0.386
b. -0.771 V
c. -1.542 V
The standard reduction potentials are intensive. The value does not depend on the amount reacting, so it is the same whether 1 or 5 moles of Fe2+ are reacting.
50. Exposure to small amounts of ionizing radiation is:
|
a) good for you |
|
b) neither good nor bad. |
|
c) bad for you. |
This is an active area of scientific, social, and political discussion. We don't know the answer and we probably cannot know it, given the complexity of the problem and the ethical restraints on doing controlled experiments. While some may find the conclusion unacceptable, an article in Science vol. 271, 29 Mar 1996, p. 1821-1822 by Marvin Goldman, "Cancer Risk of Low Level Exposure", makes an argument for accepting the argument that low level exposure is not demonstrably harmful.
51. The process 714N + 24He -------> 817O + 11H
|
a) is a balanced nuclear process |
|
b) is not a balanced nuclear process. |
|
c) hey, that's alchemy |
This transmutation (alchemy) show conservation of charge and conservation of nucleon number.
52. Complete conversion of 1 g of matter to energy is equivalent to burning (in Oa.)
a. 2 kg C
b. 2000 kg C
c. 2 x 106 kg C
d. 2 x 109 kg C
D
E = D (m) c2D
E = (-0.001 kg) (3 x 108 m/s)2 (1 J / kg m2 / s2)D
E = - 9 x 1013 J = - 9 x 1010 kJD
E for combustion of 1 mol (12 g) C is -394 kJ soD
E for combustion of 1 g C is -394/12 = -32.8 kJ = soyou would need - 9 x 1010 kJ / (-32.8 kJ / g C) = 2.7 x 109 g C = 2.7 x 106 kg C.
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ConcepTests for Exam 3 (ConcepTests for earlier exams can be found by scrolling down)
17. A typical covalent bond energy is about
a. 20 kJ / mol bonds
b. 200 kJ / mol bonds
c. 2000 kJ / mol bonds
18. In simple, small molecules, "hydrogen bonding" typically occurs
a. within one molecule
b. between two molecules
ANS. "Hydrogen bonding" is an intermolecular force.
19. The shape of the methane (CH4) molecule is
a. flat, square planar
b. tetrahedral
20. The shape of the ammonia (:NH3) molecule is
a. flat
b. pyramidal
21. The shape of the formaldehyde (H2C=O) molecule is
a. flat
b. pyramidal
22. Vitamin A is predominantly

a. nonpolar
b. polar
23. Vitamin C is predominantly

a. nonpolar
b. polar
24. Vitamin C is more soluble in
a. hexane
b. water
25. I2 is
a. nonpolar
b. moderately polar
c. very polar
26. I2 is
a. more soluble in water than in hexane.
b. less soluble in water than in hexane.
c. equally soluble in water and in hexane.
d. insoluble in water and in hexane.
ANS. Nonpolar solutes are more soluble in nonpolar solvents than in polar solvents. (Like dissolves like.)
27. When liquid water boils at 100o C, covalent bonds in the water molecules are broken.
a. true
b. false
28. Consider two systems at equilibrium: {CaCO3 (s) <=====> Ca2+ (aq) + CO32- (aq)}


Which statement is true?
a. [Ca2+] in the left system is less than [Ca2+] in the right system.
b. [Ca2+] in the left system is equal to [Ca2+] in the right system.
c. [Ca2+] in the left system is greater than [Ca2+] in the right system.
ANS. The amount of solid does not affect the degree of reaction (the position of equilibrium).
29. Northern pike (a fish) does not live in warm water because
a. it is a True Alaskan fish
b. it requires a lot of O2 to thrive
c. many predators live in warm water
30. When a gas dissolves in a liquid, we can approximate DG = -TDS. Considering this, we would then predict that the solubility of a gas in a liquid
a. decreases as the temperature increases
b. does not change with temperature
c. increases as the temperature increases
31. In a 0.010 M HNO3 solution the [H3O+] = 0.010 M. The [OH-] is
|
a. 1.00 M |
|
b. 0.010 M |
|
c. 1.0 x 10-7 M |
|
d. 1.0 x 10-12 M |
|
e. 0 M |
For an aqueous solution near room temperature, [H3O+] [OH-] = Kw = 1.0 x 10-14
Thus, since [H3O+] = 0.010 M,
[OH-] = 1.0 x 10-14 / [H3O+] = 1.0 x 10-14 / 0.010 = 1.0 x 10-12 M.
32. A 0.001 M HBr solutions has a pH of about
|
a) 1.0 |
|
b) 3.0 |
|
c) 5.0 |
|
d) 7.0 |
The strong acid is completely dissociated. The hydronium ion concentration is 0.001 M.
33. A 0.001 M HOCl (hypochlorous acid) solutions has a pH of about
|
a) 1.0 |
|
b) 3.0 |
|
c) 5.0 |
|
d) 7.0 |
The weak acid is less than 100% dissociated. This the hydronium ion concentration must be less than 0.001 M. Since this is an acid the hydronium ion concentration must be greater than 10-7 M.
34. When NaHCO3(s) is placed in water, you get mostly
|
a) Na+(aq), H+(aq), CO32-(aq) |
|
b) Na+(aq), HCO3-(aq) |
|
c) NaHCO3(aq) |
|
d) NaHCO3(s) - not soluble |
|
e) Na+(aq), CO2(g), H2O(l) |
NaHCO3 is a strong electrolyte.
35. A solutions with pH = 4.00 is more acidic than a solution with pH 6.0 by a factor of
|
a) 6/4 = 1.5 |
|
b) 2 |
|
c) 10 |
|
d) 100 |
The pH scale is logarithmic. A difference in 2 pH units is a difference in 102 in hydronium ion concentration.
36. What is the best "picture" of acetic acid (HAc) dissolved in water?

Acetic acid, a weak acid, is partially dissociated in water. Thus c is the best picture.
37. A solution of 0.026 M NH3 is
|
a) very acidic ([H+] ~ 0.026 M) |
|
b) moderately acidic (10-7<[H+]<0.026 M) |
|
c) neutral (pH ~ 7) |
|
d) moderately basic (10-7<[OH-]<0.026 M) |
|
e) very basic ([OH-] ~ 0.026 M) |
NH3 is a weak base. The hydroxide concentration must be greater than 1 x 10-7 M, but less than 0.026 M.
38. A solution of 1.1 M HNO2 is
|
a) very acidic ([H+] ~ 1.1 M) |
|
b) moderately acidic (10-7<[H+]<1.1 M) |
|
c) neutral (pH ~ 7) |
|
d) moderately basic (10-7<[OH-]<1.1 M) |
|
e) very basic ([OH-] ~ 1.1 M) |
HNO2 is a weak acid. The hydronium concentration must be greater than 1 x 10-7 M, but less than 1.1 M.
39. When dissolved in water, which salt will produce a basic solution?
|
a) LiNO3 |
|
b) KF |
|
c) NH4I |
For a salt to be basic, the anion must hydrolyze. F- reacts with water, while NO3- and I- do not.
F- (aq) + H2O (l) < ===== > HF (aq) + OH- (aq)
40. When 5 mmol of some strong acid, HX, is added to 4 mmol of some base, B, what happens?
a. It is not possible to say.
b. All of the base B (4 mmol) reacts with 4 mmol of the acid. The final mixture has 1 mmol unreacted HX, no B, 4 mmol of conjugate base X-, and 4 mmol conjugate acid HB+.
ANS: The strong acid will react stoichiometrically with the base, consuming the limiting reactant and forming stoichiometric amounts of product: HX + B -----> X- + HB+.
41. When 25 mL of 0.20 M NaOH and 40 mL of 0.10 M HCl mix and react
NaOH (aq) + HCl (aq) ------- > NaCl(aq) + H2O (l)
The resulting solution is
|
a) neutral |
|
b) acidic |
|
c) basic |
The amount of strong base NaOH is (25 mL) (0.20 mmol / mL) = 5 mmol.
The amount of strong acid HCl is (40 mL) (0.10 mmol / mL) = 4 mmol.
The HCl is the limiting reactant, and it is all consumed, along with 4 mmol of NaOH. There is 1 mmol unreacted NaOH dissolved in 65 mL of solution, giving a basic pH. [OH] = 1 mmol / 65 mL = 0.0154 M, thus pOH = -log (0.0154) = 1.81 and pH = 14.00 - pOH = 12.19.
42. When 25 mL 0.20 M NaCN and 40 mL 0.10 M HCl mix, the resulting solution contains as major protic (not NaCl) solutes
|
a) NaCN only |
|
b) HCl only |
|
c) NaCN and HCl |
|
d) NaCN and HCN |
|
e) HCN only |
Note the similarities to the problem directly above.
The amount of weak base NaCN is (25 mL) (0.20 mmol / mL) = 5 mmol.
The amount of strong acid HCl is (40 mL) (0.10 mmol / mL) = 4 mmol.
The HCl is the limiting reactant, and it is all consumed, along with 4 mmol of NaCN, which is converted to 4 mmol HCN. There is also 1 mmol unreacted NaCN.
NaCN (aq) + HCl (aq) ------- > NaCl(aq) + HCN (l)
43. Which solutes would give the best buffer at pH 10.6?
|
a) Acetic acid + (sodium) acetate (Ka = 1.8 x 10-5) |
|
b) H2CO3 + NaHCO3- (Ka = 4.4 x 10-7) |
|
c) NH4Cl + NH3 (Ka = 5.7 x 10-10) |
|
d) HOI + NaOI (Ka = 2.3 x 10-11) |
The weak acid portion of the conjugate pair should have pKa near the desired pH. pKa for HOI is -log(2.3 x 10-11) = 10.6, which makes it ideal.
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ConcepTests for Exam 2
1. Imagine mixing H2(g) and I2(g) so that [H2]i = 1.0 M, [I2] i = 2.0 M. Then cause them to react according to the rules of Chem 105.
H2 (g) + I2 (g) -----> 2 HI (g)
The concentration of HI when the reaction is done will be
|
a. 0.00 M |
|
b. 1.00 M |
|
c. 1.87 M |
|
d. 2.00 M |
|
e. 4.00 M |
ANS: H2 is the limiting reactant. If this were simply a matter of stoichiometry, then 1 mol/L of H2 should react completely to form 2 mol/L HI.
2. The equilibrium constant expression, Kc, for
CaCO3 (s) + 2H2O+ ------ > Ca2+ (aq) + CO2 (g) + 3 H2O (l) is:
|
a. |
|
|
|
b. |
|
|
|
c. |
|
ANS: This is it! Solids and liquids are excluded from the expressions. |
|
d. |
|
|
3. For COCl2 (g) < ==== > CO (g) + Cl2 (g) K1 = 8 x 10-4
If we turn the reaction around to CO (g) + Cl2 (g) < ======= > COCl2 (g), the value of K2 will be
|
a. 8 x 10-4 |
|
b. -8 x 10-4 |
|
c. 8 x 104 |
|
d. 1.2 x 103 |
ANS: Since the reaction is the reverse of the previous one, K2 is the reciprocal of K1. K2 = 1.2 x 103
4. For the reaction
CO (g) + Cl2 (g) < ======= > COCl2 (g), K2 = 1.2 x 103.
If we multiply all of the stoichiometric coefficients by 2, we get
2 CO (g) + 2 Cl2 (g) < ======= > 2 COCl2 (g).
The values of K3 for this new reaction is
|
a. 3.5 x 101 |
|
b. 0.6 x 103 |
|
c. 1.2 x 103 |
|
d. 2.4 x 103 |
|
e. 1.4 x 106 |
ANS: Since the new equilibrium constant expression ([COCl2]2 / [CO]2{Cl2]2) is the square of the original ([COCl2] / [CO] {Cl2]), the numerical value is squared.
5. Which picture best represents a solution of the strong acid, HCl?
|
a. |
HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) HCl(aq) |
(i.e., all HCl(aq)) |
|
b. |
HCl(aq) H3O+(aq) HCl(aq) Cl-(aq) HCl(aq) HCl(aq) HCl(aq) H3O+(aq) Cl-(aq) HCl(aq) Cl-(aq) HCl(aq) HCl(aq) HCl(aq) H3O+(aq) |
(i.e., some H3O+(aq), Cl-(aq), and HCl(aq)) |
|
c. |
H3O+(aq) Cl-(aq) H3O+(aq) H3O+(aq) Cl-(aq) Cl-(aq) H3O+(aq) Cl-(aq) Cl-(aq) H3O+(aq) Cl-(aq) H3O+(aq) Cl-(aq) |
(i.e., all H3O+(aq) and Cl-(aq)) |
ANS: The strong acid is completely dissociated in water. (The reaction
HCl (aq) + H2O (l) ------> Cl- (aq) + H3O+ (aq)
goes to completion.)
6.
|
For COCl2 (g) < ==== > CO (g) + Cl2 (g) |
K = 8 x 10-4 |
Assume you made an initial mixture of 0.01 M COCl2 (g), 0.01 M CO (g), and 0.01 M Cl2 (g). This is a nonequilibrium mixture. When the system reaches it equilibrium concentration, you should expect
|
a. more COCl2 (g) than CO (g) and Cl2 (g) |
|
b. more CO (s) and Cl2 (g) than COCl2 (g) |
|
c., similar concentrations of COCl2 (g), CO (g), and Cl2 (g) |
ANS: Since K is appreciably less than 1, this is a reactant favored process and we'd expect some COCl2 (g) to be formed from CO (g) and Cl2 (g). (The numerical result is 0.0163 M COCl2, 0.0037 M CO, 0.0037 M Cl2.
7. Consider the reaction:
|
H2 (g) + I2 (g) -----> 2 HI (g) |
H = 50.5 |
A reaction mixture is examined and found to have this composition:
[H2] = 0.050 M, [I2] = 1.53 M, [HI] = 6.25 M.
You can state
a. The system is at equilibrium. There will be no net change in concnetrations.
b. The system is still reacting. HI is being formed.
c. The system is still reacting. H2 and I2 are being formed.
ANS: This is a nonequilibrium system. The value for Q = [HI]2 /{[H2] [I2]} = 6.252 / {(0.050)(1.53)} = 511 is greater than the value of K = 50.5. Because Q > K, the concentration of product HI is greater than that expected at equilibrium. The system is still reacting. H2 and I2 are being formed.
8. For the production of ammonia at 472 oC, Kc = 0.105.
N2 (g ) + 3 H2 (g) <=====> 2 NH3 (g)
A particular closed system contains 0.1207 M H2, 0.0402 M N2, and 0.00272 M NH3. This system is at equilibrium since
Qc = [NH3]2now/{[N2] now[NH3] 3now }= (0.00272)2/{(0.0402)(0.1207)3 }= 0.105
If some more H2 is added to the system so that the concentration of H2 has doubled to 0.2414 M, the system is still at equilibrium
|
a. true |
|
b. false |
Qc changes when more hydrogen is added:
Qc = (0.00272)2/{(0.0402)(0.2414)3 }= 0.0131
so that the system is no longer at equilibrium
9. Consider the following equilibrium process:
|
Co(H2O)62+(aq) |
+ |
4 Cl-(aq) |
< ======= > |
CoCl42-(aq) |
+ |
6 H2O(l) |
K<1 |
|
Red |
|
|
|
Blue |
|
|
|
A 0.1 M cobalt solution, at equilibrium, is red. What is the effect of adding HCl to this solution.
a. No change. The solution remains red.
b. The solution turns blue.
ANS: Addition of the HCl perturbs the system by increasing the concentration of the reactant chloride. The system responds by consuming chloride ion (and Co(H2O)62+). In doing this, the blue CoCl42-(aq) is formed.
10. Aqueous CoCl42-(aq) can be formed by mixing Co(H2O)62+(aq) with HCl(aq)
|
Co(H2O)62+(aq) |
+ |
4 Cl-(aq) |
< ======= > |
CoCl42-(aq) |
+ |
6 H2O(l) |
|
Red |
|
|
|
Blue |
|
|
Assume you have made a solution containing approximately equal amounts of Co(H2O)62+and CoCl42- .
The addition of AgNO3(aq) will
|
a) Cause no change |
|
b) Cause the solution to become red |
|
c) Cause the solution to become blue |
Addition of silver ions will consume chloride: Ag+ (aq) + Cl- (aq) < ==== > AgCl (s). This causes Q to increase. The system responds by producing more chloride ion and the red Co(H2O)62+(aq)
11. Aqueous CoCl42-(aq) can be formed by mixing Co(H2O)62+(aq) with HCl(aq)
|
Red |
Blue |
Co(H2O)62+(aq) + 4 Cl-(aq) < ======= > CoCl42-(aq) + 6 H2O(l)
Delta H for the above reaction is >0.
Assume you have made a solution containing approximately equal amounts of Co(H2O)62+and CoCl42-and have a murky purple solution. To get a pure blue color you should
|
a. heat the solution |
|
b. cool the solution |
To get a blue solution, the reaction must respond to the temperature perturbation by producing more CoCl42-(aq). When more CoCl42-(aq) is produced, the system will cool because the reaction is endothermic. Thus, the perturbation should be to heat the solution to a higher temperature, so that the system will respond by cooling.
12. 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
|
a. 0.60 atm |
|
b. a bit less than 1.20 atm. |
|
c. 1.20 atm |
|
d. a bit more than 1.20 atm |
|
e. 2.40 atm |
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.
13. 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
a. 0.10 L
b. 0.30 L
c. 0.90 L
ANS: Decreasing the pressure by a factor of three increases the volume by a factor of three.
14. 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 393K is
a. A
b. B
ANS. Gas velocity increases as temperature increases.

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Created February 11, 2000
Updated April 20, 2000