Chapter 18 Practice Test 3: Answers and Explanations

2. A Allotropes are different forms or molecular arrangements of the same element. Carbon has three common allotrophic forms at standard conditions (25°C and 1 atm): amorphous carbon (charcoal), graphite, and diamond. Graphite is unique among nonmetals in that it conducts electricity.

3. B Two elements that are essential to plant growth but are depleted in most soils are nitrogen and phosphorous. Phosphorous is not given as a choice, but nitrogen is. Plants cannot utilize atmospheric nitrogen gas because the strong triple bond in N₂ makes it virtually inert to biological processing.

4. C Again, allotropes are different forms or molecular arrangements of the pure element. Oxygen has two allotropic forms at standard conditions: molecular oxygen, O₂, and ozone, O₃. Ozone (and to a lesser extent, molecular oxygen) is the primary absorber of UV light in Earth’s atmosphere.

5. D A catalyst decreases the activation energy, or energy barrier, that must be overcome for reactants to become products. In this way, catalysts increase the rate of chemical reactions.

6. B Entropy, S, is the measure of the amount of disorder in a molecular system. When a gas condenses into a liquid, the molecules become more organized, so the entropy of the system decreases.

7. A It may be tempting to choose (E), but remember that by definition, the change in Gibbs free energy, ∆G, must be negative for a reaction to be spontaneous. Standard voltaic potential, Eº, is related to ∆Gº by the equation

∆Gº = –nFEº

where n is the number of transferred electrons and F is Faraday’s constant. Since n and F are always positive numbers, you can see that for a spontaneous reaction—where ∆Gº is negative—Eº will always be positive.

8. B By definition, entropy for a pure element in crystalline form at absolute zero (0 K) is zero.

9. C The atoms of noble gas elements have filled valence shells and, therefore, are extremely unreactive— more so than any other family.

10. D To form a negative ion, an atom needs to acquire electrons. This sounds like a nonmetal, not a metal. Eliminate (A), (B), and (E). Noble gases are essentially inert, so that leaves the halogens. Halogens need 1 valence electron to complete their valence shell and so will readily gain an electron and form an anion.

11. E When the test writers start talking about the “d” subshell, think “transition metals.”

12. A Ionization energy is needed to remove an electron from an atom. Which kind of elements tend to give up electrons? Metals, of course. Of the metals, alkali metals, having only 1 valence electron per atom, will lose an electron most easily because this allows an alkali metal atom to assume a stable noble gas electron configuration.

13. B Given that the strong acids are HCl, HBr, HI, HNO₃, H₂SO₄, and HClO₄ and the strong bases are LiOH, NaOH, KOH, RbOH, and CsOH, the only compound that could result from a neutralization between a strong acid and strong base is KI.

14. C Compounds composed of only nonmetal elements tend to form covalent bonds, while compounds composed of metals and nonmetals tend to form ionic bonds. Therefore, N₂ and CCl₄ are expected to be covalent compounds. However, N₂ is a gas at standard conditions; only CCl₄ is a liquid.

15. E Carbonates, CO₃^2–, and bicarbonates, HCO₃^–, form CO₂ gas when mixed with acid. That’s why baking soda (NaHCO₃) fizzes when added to vinegar (acetic acid).

16. A Oxygen gas is able to effectively take electrons from most metals, oxidizing them. This process is also colloquially known as rusting.

17. B The primary source of solar energy is energy released by the nuclear fusion of hydrogen at high pressure and temperature to form helium. 4¹H → ⁴He + 2 electrons + lots of energy.

18. A The only nuclear process that does not change the number of protons and neutrons of a nucleus is gamma decay. Gamma decay involves stabilization of a nucleus by loss of energy in the form of a gamma ray photon.

19. C Given the nuclear reaction

1by4

When conserving mass and charge, the missing particle must be He. Therefore, the nuclear process responsible for this transmutation is alpha decay.

20. D A pH of 13 indicates a basic solution; therefore, there must be a base in solution in the first place. NH₄OH and KOH are bases. However, a pH of 13 means that the pOH is 1, i.e., [OH^–] = 0.1 M. For the [OH^–] to be the same as the base, that base must completely dissociate and be strong. KOH is the only strong base given.

21. A This is a question about the colligative property freezing point depression. Remember that for all colligative properties, the greater the number of dissolved particles, the greater the effect. Therefore, this question is really asking which solution has the greatest number of dissolved particles. Since all of them have the same molar concentration, this is really just a contest of which compound breaks up into the most individual particles.

MgCl₂ → Mg²⁺ + Cl^– + Cl^–	(3 particles)
HClO₄ → H⁺ + ClO₄^–	(2 particles)
NH₄OH → NH₄⁺ + OH^–	(2 particles)
KOH → K⁺ + OH^–	(2 particles)
LiNO₃ → Li⁺ + NO₃^–	(2 particles)

Therefore, the winner is MgCl₂.

22. A This is a question about the colligative property boiling point elevation. As in question 21, recall that for all colligative properties, the greater the number of dissolved particles, the greater the effect. Therefore, this question is really asking which solution has the greatest number of dissolved particles. Since all of them have the same molar concentration, this too is a contest of which compound breaks up into the most individual particles.

MgCl₂ → Mg²⁺ + Cl^– + Cl^–	(3 particles)
HClO₄ → H⁺ + ClO₄^–	(2 particles)
NH₄OH → NH₄⁺ + OH^–	(2 particles)
KOH → K⁺ + OH^–	(2 particles)
LiNO₃ → Li⁺ + NO₃^–	(2 particles)

Again, the winner is MgCl₂.

23. E Certain metal ions produce characteristic colors when ionized in a flame—that’s how fireworks are made to have different colors. Here are the most common ions, and the color they produce.

Red:	Lithium, strontium
Orange:	Calcium
Yellow:	Sodium
Green:	Barium, copper
Violet:	Potassium

Therefore, lithium ions, Li, will produce a red flame—(E).

Part B

101. T, T, CE

Divide and conquer. Both statements are true. Nearly all colored compounds fall into two categories: 1) those that are colored because they are organic molecules that have conjugation, and 2) those that are colored because they have transition metal atoms with partially filled d subshells. That’s why sodium oxide is colorless, but iron(II) oxide is orange.

102. T, F Divide and conquer. The first statement is true. It is a fundamental law of chemical kinetics that all chemical processes slow down at lower temperatures—that’s why refrigerating food retards the growth of microbes. At lower temperatures, reactant molecules have less kinetic energy to use to overcome the energy barrier for the formation of products, called the activation energy. The second statement is false. The only way to lower activation energy is to add a catalyst.

103. F, T Divide and conquer. Exothermic reactions release heat energy (notice exo- looks like exit), endothermic reactions absorb energy (notice endo- looks like enter), so the first statement is false. The second statement is true and is an important law in chemistry.

104. F, T Divide and conquer. The first statement is false. The solubility of a gas in a liquid is very sensitive to pressure, such that the solubility of gases in liquids increases with increasing pressure. That’s why when we release the pressure trapped in a bottle of soda by opening it, a sudden surge of carbon dioxide bubbles races to get out of the container. The second statement is true but has no relevance to the solubility of gases.

105. T, T, CE

Divide and conquer. All ionic compounds have relatively high melting points (all are solids at room temperature) because ionic forces between ions are very strong. In the case of MgO, the +2 and –2 charges on Mg and O, respectively, result in very strong intermolecular forces. Not surprisingly, MgO has a melting point—it’s about 2,000°C.

106. T, T, CE

Divide and conquer. According to the Aufbau principle, subshells need to be completely filled before moving up to the next higher one. However, completely half-filled and filled d subshells bestow extra stabilization to an atom. Therefore, Cr and Cu actually violate the Aufbau principle and promote a 4s electron to become [Ar] 4s¹3d ⁵ and [Ar] 4s¹3d ¹⁰, respectively. Remember this important exception.

107. T, T, CE

Divide and conquer. Isotopes are atoms of the same element that have differing numbers of neutrons. They have nearly identical chemical behavior because the number of protons and electrons in an atom (two quantities that are identical between isotopes) govern an atom’s chemical properties.

108. T, T, CE

Divide and conquer. No matter how complicated acid/base chemistry can appear, never forget that adding acid to any solution, buffered or not, always lowers the pH; adding base to any solution always raises the pH. A buffer does not prevent the pH from changing in these cases; it simply lessens by how much the pH changes.

109. T, F Divide and conquer. The first statement is true. Remember the colligative properties: Adding any solute to a liquid always raises the boiling point temperature of the resulting solution—that’s called boiling point elevation. This occurs because adding a solute to a liquid always lowers the vapor pressure of the solution—vapor pressure depression. Recall two more things: 1) The vapor pressure of a liquid always gets higher with higher temperature, and 2) a liquid boils when its vapor pressure is equal to the atmospheric pressure. So if the vapor pressure of a solution is lowered by the addition of a solute, you have to heat the solution to a higher temperature before the vapor pressure equals the atmospheric pressure and the solution will boil again.

110. F, T Divide and conquer. The Lewis dot structure for BF₃ is

Now, count the groups of electrons around the central atom (B), keeping in mind that every pair of nonbonding electrons, every single bond, every double bond, and every triple bond counts as one group. So here, boron is surrounded by three groups of electrons. Any atom that is surrounded by three groups of electrons has an sp² hybridization and a trigonal planar geometry. Of course, the second statement is true because this B doesn’t have a stable octet—it has only 6 electrons.

111. T, T Divide and conquer. Like all peroxides, hydrogen peroxide is a good oxidizing agent. It is also true that the hydrogen atoms in H₂O₂ have a +1 oxidation number. However, the oxidizing tendency of this molecule is not due to the H, but rather to the fact that each O has a –1 oxidation number, instead of the usual –2.

112. F, F Divide and conquer. Both statements are false. There are no ideal gases—end of story. Furthermore, H’s in H₂ are bonded together via a covalent bond. Hydrogen bonding refers to a specific dipole interaction between two or more different molecules where an H covalently bonded to an F, O, or N is electrostatically attracted to an F, O, or N on another molecule.

113. T, F As the temperature increases, the overall energy level of the gas increases. However, due to the randomized distribution of velocities, it’s impossible say that ALL of the particles will have more energy (and thus greater velocities). Yes, the average velocity of the gas particles will be greater, but there will still be some particles at the same, or even lower, velocities.

114. T, F Divide and conquer. Over of the mass of the average human is oxygen. However, most oxygen atoms in the bloodstream, over 99.99 percent of them, are in the form of H₂O, not as O₂. The first statement is true, but the second is false.

115. T, T Divide and conquer. Strong acids and strong bases are those that undergo 100 percent dissociation in water.

The strong acids are HCl, HBr, HI, HNO₃, H₂SO₄, and HClO₄.

The strong bases are LiOH, NaOH, KOH, RbOH, and CsOH.

Mixing any acid with any base will produce a neutralization reaction, regardless of whether the acid and base are strong or weak.

Part C

24. C Electrons don’t actually orbit the nucleus at fixed distances, although many textbooks use the graphical representation of the Bohr atom because it is difficult to draw the true electron orbitals (first proposed by Heisenberg) as a two-dimensional representation.

25. D First, draw the Lewis dot structure for XeOF₄.

Now, count the groups of electrons around the central atom (Xe), keeping in mind that every pair of nonbonding electrons, every single bond, every double bond, and every triple bond counts as one group. So here, Xe is surrounded by six groups of electrons. Any atom that is surrounded by six groups of electrons has an sp³d² hybridization and an octahedral geometry. However, because the question asks about the molecular shape (as opposed to the geometry), look at the arrangement of the surrounding atoms. In this case, the F would make a flat square around the Xe with the oxygen atom lying directly above. This traces out a square pyramid, (D).

26. A After examining the periodic table, realize that the full atomic symbol for this isotope of technicium is

Remembering that the superscript (the mass number) represents the total number of protons and neutrons, and the subscript (the atomic number) represents just the number of protons, has 43 protons and 99 – 43 = 56 neutrons—(A).

27. D Using the coefficients, you can determine that 4.0 moles of Fe would only require 3.0 moles of oxygen to react fully, so there is excess oxygen, making iron the limiting reactant. The theoretical yield of the Fe₂O₃ for the reaction would then be 2.0 moles, and 1.5/2.0 × 100% = 75%.

28. D Recall that the six strong acids are HCl, HBr, HI, HNO₃, HClO₄, and H₂SO₄. Phosphoric acid, H₃PO₄, is a weak acid.

29. B First, the correct abbreviation for heat is Q. So you can eliminate (C), (D), and (E). During a phase change, the temperature of a substance remains constant. Therefore, (A) must be wrong because it has a term for changing temperature, ΔT. In fact, (A) is the equation used to determine the change in temperature when heat is added or removed from a substance that isn’t undergoing a phase change. Choice (B) is the correct choice, where Q is the heat added or removed, n is the number of moles of substance, and ΔH is the heat of phase change, a quantity that is unique for every substance.

30. A The first quantum number refers to the energy level, so it would be 3. A d-orbital is represented by a 2, giving us the second number. For the third number, the five possible numerical representations for a d-orbital orientation are –2, –1, 0, 1, or 2. The final quantum number, representing the spin, could be either or –.

31. E First, according to the periodic table, Cr has 24 electrons. You can eliminate (B) and (C) because those configurations have 25 electrons. Choice (A) is wrong because the 3s subshell is already accounted for in the [Ar] core—i.e., [Ar] stands for 1s²2s²2p⁶3s²3p⁶. Now, according to the Aufbau principle, completely fill subshells before moving up to the next higher one. So the best answer would appear to be (D). However, remember that completely half-filled and filled d subshells bestow extra stabilization to an atom. Therefore, Cr and Cu actually violate the Aufbau principle and promote a 4s electron to become [Ar] 4s¹3d ⁵ and [Ar] 4s¹3d¹⁰, respectively. Choice (E) is the correct ground state configuration for Cr.

32. D At 25°C, pH + pOH = 14 for any solution. Therefore, if the pH is 10 for this solution, the pOH is 14 – 10 = 4. Taking the negative antilog of 4 gives 10⁻⁴, (D).

33. C Chromium is a transition metal, so the charge on the cation in this compound must be specified. Carbonate has a charge of negative two, and to balance it, the chromium must have a charge of positive two.

34. D First, find the number of moles of Br⁻.

molarity = moles/volume or moles = molarity × volume

= 0.1 M × 0.5 L

= 0.05 moles

Then figure out how much 0.05 moles of Br⁻ weighs.

grams = molecular weight × moles

= 80 g/mole × 0.05 moles

= 4 grams or 4,000 mg (D)

35. B This is another math problem, but this one involves the ideal gas law. A couple of things first: 1) Since you don’t have a calculator, round off the value of R to 0.1 L•atm/mol•K, and 2) remember that T must be in K, not °C, for PV = nRT to work correctly. So

PV = nRT or V = nRT/P

= (0.5 moles × 0.1 L•atm/mol•K × 300 K)/3 atm

= (15 L•atm)/3 atm

= 5 (B)

36. E For covalent compounds, a di- prefix signifies two, and the tetra- prefix signifies four.

37. E Recall that when ΔG is negative, a reaction is spontaneous, and when ΔG is positive, a reaction is nonspontaneous. The question indicates that the reaction in question is endothermic—i.e., ΔH is positive. Looking at the equation provided, the effect that decreasing T has will depend on the sign of ΔS. But the question doesn’t tell you the sign of ΔS, nor can you figure it out on your own. Therefore, there is no way to make any conclusions—(E).

38. B First, ionization energy is a periodic trend that increases up and to the right on the periodic table. After looking at the position of these elements on the periodic table, carbon is clearly the best answer—(B).

39. D Hydrogen bonding is the strongest type of intermolecular force. Other types of permanent dipoles would be next strongest, and the temporary dipoles created through London dispersion forces would be the weakest type of intermolecular force.

40. B H₂S is a covalent substance, and when covalent substances undergo phase changes the intermolecular forces between various molecules must be neutralized. The strongest type of IMF present in a sample of H₂S is permanent dipoles. Note that H₂S does not have hydrogen bonding because the hydrogen is not bonded to N, O, or F.

41. D According to the reaction, 6 Ca atoms are present in the reactants. Therefore, 6 CaSiO₃ must be present as products, (D).

42. B Percent by mass is calculated by dividing the mass of the solute by the total mass of the solution. 40.0 mL of water would have a mass of 40.0 g (density of water = 1.0 g/mL), so the total mass of the solution would be 10.0 g + 40.0 g = 50.0 g. Then, (10.0 g/50.0 g) × 100% = 20%.

43. E Given that the molecular weight of AgCl is 143.4 g/mol, the number of moles of AgCl precipitated is

Moles AgCl = 5.70 g/143.4 g/mol

= 0.040 moles

Since the molar ratio of Ag+ in AgCl is 1, the number of moles of Ag⁺ in the original solution was also 0.040. Therefore, the concentration of the original solution was

Molarity Ag⁺ = moles Ag⁺/volume (L)

= 0.040 moles/0.100 L

= 0.40 M (E)

44. A Choices (B), (C), (D), and (E) are all true statements. Choice (A) is a false statement—the vapor pressure of a substance depends only upon 1) the substance’s temperature, and 2) its mole fraction when it’s in solution (see the colligative property of vapor pressure depression).

45. B This is a bit of a trick question; when using any gas law that involves temperature the Kelvin scale must be used. However, Boyle’s Law (P₂V₂ = P₂V₂) is only used when temperature is held constant, and thus, the temperature never enters into the calculations.

46. B Since these are all ionic compounds, electrostatic forces can be assumed to be entirely responsible for the cohesive forces on the lattice. According to Coulomb’s Law

F = Kq₁q₂/r² where Energy = Kq₁q₂/r

Since the charges for all of the ion pairs given in the choices (q₁ and q₂) are ±1, it is the internuclear distance, r, of each ion pair that is the determinant factor. According to the equations above, the smaller the r, the greater the energy. So using the periodic trend in atomic/ion size, LiF (B) is the ion pair with the smallest internuclear distance.

47. C Positron emission is a type of beta decay. During beta decay, nuclear mass remains constant. Therefore, (C) is the only possible answer.

Conserving mass (superscript) and charge (subscript) gives (C).

48. E A reducing agent describes a species that will lose electrons easily in a redox reaction. This in turn causes another species to gain electrons and thus be reduced. Out of the choices, sodium loses its one valence electron to form Na⁺ ions very readily. This also means that sodium would be very high on the activity series.

49. C Recall the solubility rules. Choices (A) and (B) are incorrect because neither NaCl nor BaOH are insoluble and therefore would not precipitate out of the aqueous solution. Because lead salts are insoluble except for their nitrates and perchlorates, lead iodide would precipitate out of solution as (C) depicts. Choices (D) and (E) are examples of oxidation-reduction reactions.

50. A H₂CO₃ is carbonic acid, and when carbonic acid is mixed with water it dissociates. One of the products of that dissociation is carbon dioxide gas, CO₂, which will then bubble up out of the solution.

51. E According to Le Châtelier’s principle, the direction in which an equilibrium is disturbed can be predicted if ΔH_rxn is known (eliminate (B)). A straightforward way of solving this is to write “HEAT” into the reaction either as a reactant for endothermic reactions or as a product for exothermic reactions. Here, the reaction is exothermic, so

A ⇋ B + C + “HEAT”

Then, since temperature is a measure of “HEAT,” increasing T, or “HEAT,” would be expected to shift the system to the left. Choice (E) is the answer.

52. D Recall some fundamental solubility rules.

• All group 1 metals and NH₄⁺ salts are soluble.

• All NO₃^– and ClO₄^– salts are soluble.

• All silver, lead, and mercury salts are insoluble.

Therefore, the addition of Li⁺, NH₄⁺, or Cs⁺ would not produce a precipitate with either acid (eliminate (A), (B), and (C)). Silver, Ag⁺, would form precipitates with both acids (eliminate (E)). Ba²⁺ is somewhat unique, even among other group 2 elements, because BaCl₂ is soluble, while BaSO₄ is not, (D).

53. A Writing equilibrium expressions is a three-step process.

1. First, ignore any molecule that is in the solid (s) or liquid (l) phases.

2. Second, write K = [products]/[reactants]

3. Third, all coefficients in front of molecules become exponents in the equilibrium expression.

Therefore, the correct choice is (A).

54. D A conjugate acid/base pair is a set of 2 molecules/ions that have identical molecular formulas, except that one of them has one more H⁺ than the other. Only (D) represents a true pair of conjugates.

55. B The starting solution has a pH of 3.5. Therefore, the starting solution must be acidic. You can eliminate (C) and (D) because they are bases. The subsequent titration experiment revealed a single equivalence point, so the acid in question must be monoprotic—eliminate (E). Since the equivalence point comes at pH 7, the unknown acid must be strong. HCl is the only monoprotic strong acid given, (B).

56. E Statements (A) through (D) can all be determined with data obtained from a titration experiment.

57. A The phase changes associated with the terms are

deposition (gas → solid)

sublimation (solid → gas)

liquefaction (gas→ liquid)

fusion (solid → liquid)

vaporization (liquid → gas)

58. D The molecular formulas associated with the IUPAC names for organic molecules are

methane:	CH₄
ethane:	CH₃CH₃
methylethane:	does not exist in the IUPAC system
propane:	CH₃CH₂CH₃
isopropane:	does not exist in the IUPAC system

59. E First, figure out the concentration of [H⁺] at pH 1 and pH 3.

pH = 1; [H⁺] = 0.1 M

pH = 3; [H⁺] = 0.001 M

Then realize that 0.001 M is 1 percent of (or 100 times smaller than) 0.1 M. Therefore, 99 percent of the H⁺ is neutralized when going from pH 1 to pH 3.

60. A Avogadro’s number is nothing more than the conversion faction among two measures of mass, atomic mass units, and grams. Just as you can say that there are 2.54 centimeters in an inch, so too can you say there are 6.02 × 10²³ amu in a gram.

61. C POE is the best tool to use here. First, we can eliminate mercury, (A), because mercury metal is a liquid at standard conditions, yet the table indicates that unknown metal #1 is a solid. Second, we can eliminate copper, (B), because copper metal is brownish, yet the table says unknown metal #1 is dull gray. Now, consider the presence of the white oxide coat. The oxide of iron, better known as rust, is orange brown, not white, so eliminate (D). Last, the oxide of silver is gray-black, better known as tarnish, so eliminate (E). Therefore, zinc is the best choice—(C).

62. E Again, POE is the best tool to use here. First, you can eliminate carbon, (A), because carbon is not a metal. Second, you can again eliminate copper, (B), because copper metal is brownish, yet the table says unknown metal #2 is silver-gray. Third, you can eliminate zinc, (C), because it is a fairly reactive metal that always has a whitish oxide coat in air. Fourth, you can definitely eliminate sodium because sodium metal explodes with yellow flame on contact with even plain water, let alone acidic solutions—the table doesn’t report any explosions! Finally, the best choice is silver, a relatively inert metal along with copper, gold, and platinum (it’s no accident that these metals are used for jewelry and coinage)—(E).

63. B There are several ways to approach this one. Again, POE is very useful here. Chlorine, (A), and nitrogen dioxide, (E), gases are colored—greenish and orange, respectively. Therefore, they cannot be the gas in question. Furthermore, it’s difficult to choose carbon dioxide, (D), because there are no carbon atoms anywhere in this experiment. Finally, as a rule of thumb, a colorless gas produced from reactions between metals and acids is hydrogen—(B).

64. E This requires a bit of general chemistry knowledge. The colors of the gases are

Cl₂—green

H₂—colorless

O₂—colorless

CO₂—colorless

NO₂—orange/brown

Choice (E) is the best answer.

65. B First, the neutralization reaction that occurs here is

KOH + HI → KI + H₂O

The trick is to realize that the number of K’s and I’s are not changing during the reaction; but since the solutions are being added, the volume is doubling. If the volume doubles, then the initial solution concentrations (0.2 M) are halved (0.1 M).

66. D An O^2– ion has gained 2 electrons to fill its outer shell. This gives an O^2– ion the same electron configuration as Ne. That is, the 2 are isoelectronic.

67. D The addition of a catalyst to a system at equilibrium has NO effect. That’s because all a catalyst does is increase the rate at which a nonequilibrated system reaches equilibrium.

68. A It’s important to know the phase of the elements at standard conditions (25°C at 1 atm).

Gases:	hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine, argon, krypton, xenon, and radon
Liquids:	mercury and bromine (A)
Solids:	the rest of the elements

69. C By definition, accuracy is the measure of how close experimental data are to true data, while precision is the measure of how similar experimental data are to one another. Clearly, the student’s data is very precise; the values of 1.65, 1.68, 1.71 vary by no more than 4 percent. However, the average value of the experimental data is nearly 50 percent off the true value, meaning these results are not very accurate—(C).

70. C As a weak acid, very little H₂SO₃ actually dissociates, meaning the H₂SO₃ molecule itself will be present in the highest concentration. Each ensuing dissociation is weaker than the previous one, and thus the ion that is only created after the final dissociation will be present in the lowest concentration.

HOW TO SCORE PRACTICE TEST 3

When you take the real exam, the proctors will collect your test booklet and bubble sheet and send your answer sheet to a processing center, where a computer looks at the pattern of filled-in ovals on your answer sheet and gives you a score. We couldn’t include even a small computer with this book, so we are providing this more primitive way of scoring your exam.

Determining Your Score

STEP 1
Using the answer key, determine how many questions you got right and how many you got wrong on the test. Remember: Questions that you do not answer don’t count as either right or wrong answers.

STEP 2
List the number of right answers here.
(A) ________

STEP 3
List the number of wrong answers here. Now divide that number by 4. (Use a calculator if you’re feeling particularly lazy.)
(B) ________ ÷ 4 = (C) _______

STEP 4
Subtract the number of wrong answers divided by 4 from the number of correct answers. Round this score to the nearest whole number. This is your raw score.
(A) ________ – (C) ________ = _________

STEP 5
To determine your real score, take the number from Step 4 above, and look it up in the left column of the Score Conversion Table on the next page; the corresponding score on the right is your score on the exam.

PRACTICE TEST 3 SCORE CONVERSION TABLE

FORMULA REFERENCE GUIDE

•

Heat

• q = mcΔT

q = Heat

m = mass

c = specific heat

ΔT = Temperature change

Spontaneity

• ΔG = ΔH – TΔS

ΔG < 0 means that a reaction is spontaneous at this temperature

ΔG > 0	+ΔH, Low temp –ΔS, High temp –ΔH, +ΔS, Any temp
ΔG < 0	–ΔH, +ΔS, Any temp +ΔS, High temp

	• ΔH > 0 Endothermic
	• ΔH < 0 Exothermic

• ΔT = kmi

Freezing Point Depression
& Boiling Point Depression

i: whole number equaling the number of particles a substance dissolves into

k: solvent constant

m: molality (NOT molarity)

molarity =

molality =

mole fraction =

Ideal Gas Law

• PV = nRT

P = pressure in atm (or mmHg or torr)

V = volume in liters

n = number of moles of gas particles in the container

R = the ideal gas constant

T = temperature in Kelvin

To calculate the ideal gas constant, use the following formula:

R = .08

Conversions

• Pressure:

1 ATM = 760 torr = 760 mmHg

• Energy (Heat):

1 Calorie = 1000 calories = 4000 joules

• Temperature:

0°C = 273 Kelvin

25°C = 298 K

Facts to Know

• To find the mass of a mole of a substance, multiply the formula weight by the number of moles.

• 1 mol of gas at STP occupies 22.4 L of volume.

Solubility

• All alkali and ammonium (NH₄⁺) compounds are soluble. The alkali metals are included in group IA (Li⁺, NA⁺, K⁺, Rb⁺, Cs⁺).

• All acetates (CH₃COO^–), chlorates (ClO₃^–), nitrates (NO3^–), and perchlorates (ClO₄^–) are soluble.

• Pbⁿ⁺, Hgⁿ⁺, and Agⁿ⁺ salts are insoluble, unless they are paired with one of the anions from the previous rule.

Oxidation Rules

1. When oxygen is in a compound, its oxidation state is usually –2 (it has been reduced). One important exception is oxygen in a peroxide such as hydrogen peroxide (H₂O₂). In a peroxide, oxygen has an oxidation state of –1.

2. When an alkali metal (Li, Na, etc.) is involved in a compound, its oxidation state is always +1 (it’s been oxidized).

3. When an alkaline earth metal (Be, Mg, etc.) is involved in a compound, its oxidation state is +2.

4. When a halogen (F, Cl, etc.) is involved in a compound, its oxidation state is often –1. The oxidation state of fluorine in a compound is always –1.

5. When hydrogen is combined with a nonmetal, its oxidation state is +1. When hydrogen is combined with a metal, its oxidation state is –1.

6. In any compound, the sum of all oxidation states is zero.

Constants

• Avogadro’s number: 6 × 10²³ molecules/mole

Prefixes

Prefixes for Organics

Meth- 1

Eth- 2

Prop- 3

But- 4

Pent- 5

Hex- 6

Hep- 7

Oct- 8

Non- 9

Dec- 10

Prefix	Symbol	Notation
nano-	n	10^-9
micro-	μ	10^-6
milli-	m	10^-3
centi-	c	10^-2
kilo-	k	10^-3
mega-	M	10⁶

Electrochemistry

OIL

Oxidation Is Losing Electrons

RIG

Reduction Is Gaining Electrons

AN OX

Oxidation Occurs at Anode

RED CAT

Reduction Occurs at Cathode

• Galvanic Cell: E°_cell > 0 means reaction is spontaneous and energy is released (battery).

• Electrolytic Cell: E°_cell < 0 means the reaction is NOT spontaneous and requires a voltage source (electro plating).

Equilibrium

•

Periodic Trends

Acid-Base

pH = pK_a + log

pH = –log₁₀[H⁺]

pH > 7: Basic

pH = 7: Neutral

pH < 7: Acidic

Strong Acids:

HCI, HI, H₂SO₄, HBr, HClO₄, HNO₃

Strong Bases:

All Group I Hydroxides