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TEST BANK FOR CHEMISTRY 8TH EDITION BY ZUMDAHL
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56 Teacher’s Resource Guide
VI. Answers to AP Questions
A. Multiple Choice
Chapter 1
1. a) Quantitative means a specific amount. Only “measurement” meets this requirement.
2. e) Only “hypothesis” is an explanation.
3. d) A law is defined as a statement of fact deduced from observation.
4. b) By definition, Kg is the SI unit of mass.
5. a) Since the actual density is not known, the accuracy cannot be determined.
6. d) Ice melting is a physical change.
7. c) III and IV are correct. Answer I would not apply to addition and subtraction. Answer II is
incorrect because leading zeros are not significant.
8. c) Answer c) is correct because leading zeros are not significant.
9. a) Answer a) is correct because multiplication conserves the least number of significant digits.
10. d) Answer d) is correct because when adding, significant digits are based on the least significant
decimal place.
11. b) Bisecting the paper 30 times results in a strip approximately 3 x 10–8 cm wide.
12. d) The Celsius and Kelvin scales have the same size degree. Each is 9/5 the size of 1 ºF.
13. e) 10.00 g ÷ 1.40 mL = 7.14 g/mL
14. b) Answer b) is correct because carbon dioxide is a pure compound.
15. b) A solution (on the molecular level) is the definition of a homogeneous mixture.
Chapter 2
1. b) An example of this would be carbon reacting with oxygen and forming either CO or CO2.
The small whole-number mass ratio would be 1.33g ÷ 2.66 g, or 1 to 2.
2. d) Isotopes were not known during the time when Dalton proposed his theory.
3. a)
4. c)
5. d)
6. b)
7. b) The mass number is the sum of the protons and neutrons within the nucleus. The atomic
symbol is defined by the number of protons.
8. c) Isotopes have the same number of protons.
9. c)
10. e)
11. a)
12. e) Chloride has a 1– charge and M will have a 2+ charge within the compound. The neutral
metal will have 28 electrons, and therefore 28 protons, which is nickel.
,Teacher’s Resource Guide 57
13. e) Barium is in Group 2A on the periodic table and would have a 2+ charge, losing its 2 valence
electrons.
14. a) Per IUPAC naming conventions, Type I metal cations have the same name as the parent
atom.
15. d) Per IUPAC naming conventions; iron is a Type II metal.
Chapter 3
1. c) The “mystery” metal has an atomic mass ¼ the mass of silver (108 ÷ 4 = 27).
2. d) This is the only example that represents 6.022 x 1023.
3. d) The mass of nitrogen is 14 g/mol and iodine is 127 g/mol, which explains a molar mass of
395 g/mol for nitrogen triiodide: 14 + 3(127) = 395.
4. b) (0.7 x 62.93 amu) + (0.3 x 64.93 amu) = 63.53 amu
5. a) The fraction of the total mass is 14/17 = 0.824 x 100 = 82.4%.
6. e) Substance AB2 is 40% B, assuming a 100-g sample. AB2 going to AB would lose half the
initial mass of B, or 20 g. Therefore, AB would consist of 80 g A and 20 g B, or 75% A, by
mass.
7. d) Any total mass can be assumed.
8. e) The balanced equation is 4 FeCr2 O4 + 8 K2CO3 + 7 O2 ! 8 K2 CrO4 + 2 Fe2 O3 + 8 CO2 .
9. e) The balanced equation is Fe2 O3 + 3 C ! 2 Fe + 3 CO.
10. b) The mass ratio for the reaction is 84.01 g/mol÷188.18 g/mol = 0.4464; 0.4464 x 8.0 g = 3.6 g.
11. a) 16 g of methane is one mole. As per the balanced chemical equation, one mole of methane
produces two moles of water: 18 g x 2 = 36 g.
12. e) The limiting reactant in each mixture limits the amount of product (NH3) to 2 moles.
13. b) 2 H2 + O2 ! 2 H2 O. The mass ratio of hydrogen to oxygen is 4:32, or 1:8.
14. b) The methanol is the limiting reactant.
15. c) The reaction requires 7 moles of oxygen per 2 moles of ethane. The molar mass of ethane is
30 g/mol and the molar mass of oxygen is 32 g/mol.
Chapter 4
1. d) Water is polar; “like dissolves like.”
2. a) Strong electrolytes dissociate completely.
3. b) The formula for solution dilution is M1 i V1 = M2 i V2 . 100 i 1 = X i 1.33; X = 75, therefore
evaporation must account for 25.
4. d) Determine the molar mass for each salt; the one with the lowest molar mass would produce
the highest molarity.
5. d) The number of moles of Cl– in the NaCl solution is 0.200. The number of moles of Cl– in the
MgCl2 solution is 0.600. Adding the total moles of Cl– and dividing by the total volume of
solution (0.500 L) gives 1.60 M.
6. d) II: PbCl2(s); III: CaCO3(s); IV: AgOH(s).
7. b) K+ and NO3– are spectator ions.
,58 Teacher’s Resource Guide
8. a) Pb2+ + 2 Cl– ! PbCl2(s). Given 0.100 mol Pb2+ and 0.150 mol Cl–, Cl– is limiting and 0.0250
mol Pb2+ will be remaining in a total volume of 0.250 L. 0.0250 mol ÷ 0.250 L = 0.100 M .
9. c) None of the other options are acid-base reactions; none involves proton transfer.
10. a) 0.00750 mol H+ converts 0.0075 mol OH–, out of an initial 0.0100 mol, to water, leaving
0.0250 mol OH– in solution. All of the 0.0100 mol of Na+ is still present.
11. e) H2SO4 + 2 KOH ! K2SO4 + 2 H2 O. There are 0.04000 mol KOH, therefore it will require
0.0200 mol H2SO4 for neutralization. This amount is delivered in 200.0 mL.
12. c) This is a precipitation reaction. None of the reactants’ oxidation states changes.
13. e) Nitrogen’s oxidation state is 5+ in NO3–; it is 3– in NH3 and NH4+, 4+ in NO2, and 2+ in NO.
14. b) Na ! Na+ + e–; Cl + e– ! Cl–.
15. d) The balanced equation is 16 H+ + 2 Cr2 O72– + C2H5 OH ! 4 Cr3+ + 2 CO2 + 11 H2 O
Chapter 5
1. c) P1V1 = P2 V2; 2.00 atm x 9.00 L = P2(Ne) x 12.00 L, therefore P2 = relative pressure of Ne,
which is 1.50 atm. 4.00 atm x 3.00 L = P2(He) x 12.00 L, therefore P2(He) = 1.00 atm.
2. a) All gases occupy the same volume/mole at the same temperature and pressure. The gas with
the lightest molar mass will occupy the greatest volume.
3. b) Assume 1 mol air; then 0.79 mol N2 is present. 0.79 mol N2 x 28 g/mol = 22 g N2. 0.21 mol
O2 x 32 g/mol = 6.7 g O2. Total mass = 28.7 g; 22 g/28.7 g = 0.77 x 100 = 77% (w/w).
4. d) 300 g = a little less than 10 mol of oxygen gas. The molar volume at standard conditions is
about 24 L/mol; 9 x 24 = 216 L; 220 L is the closest.
5. c) All gases have the same molar volume at the same temperature and pressure. The molar mass
of the mystery gas is approximately ½ the mass of argon, which fits Ne.
6. c) There are approximately 30 g of air in one mole of air. This occupies ~25 L, therefore 30/25
= 1.2, or approximately 1 g/L.
7. d) All gases exert the same pressure/mol at a given temperature. The number of moles of He is
approximately 5 times the original number of moles of Ne in the container, therefore the
pressure will be more than doubled.
8. b) All gases exert the same pressure/mol at a given temperature. The gas with the smallest molar
mass will have the greatest number of moles present, therefore the greatest partial pressure.
9. d) The kinetic molecular theory for gases is based on the KE of the gas molecules. This KE is
linearly dependent on the mass of the gas molecules. (KE = ½ mv2.)
10. a) The model of an ideal gas assumes gas molecules occupy zero space and are non-associating
(not attracted to each other). These assumptions are not explicitly true for real gases.
11. c) Gases behave most ideally at low pressures and high temperatures.
12. e) The smallest non-polar molecule will be the least associating, therefore most ideal.
13. b) PV = nRT; at constant T, PV = k.
14. a) PV = nRT; at constant T, as n increases, so must the PV product.
15. e) PV = nRT; at constant P, as T increases, V increases, so d must decrease for a fixed mass of
gas.
, Teacher’s Resource Guide 59
Chapter 6
1. c) w = – P∆V; – (1 atm x 9 L) = – 9 L•atm, which is the largest of the four.
2. b) Since energy is released from the system, the products must be at a lower energy state.
3. a) ∆E = q + w = – 28 J + 63 J = 35 J
4. d) Pb has the lowest specific heat capacity, so will have the largest ∆T.
5. d) The calorimeter is in thermal contact with the system. It has its own heat capacity.
6. d) The other choices are pathway-dependent.
7. e) Ice cream must absorb heat from its surroundings in order for melting to occur.
8. b) The definition of enthalpy is heat at constant pressure: ∆H = qp.
9. a) The first reaction is reversed and multiplied by ½, the second is multiplied by 2, and the third
is multiplied by ½, so ∆Hreaction = – ½ (–2600 kJ) + 2(– 393 kJ) + ½ (– 574 kJ) = 225 kJ.
10. c) Using Hess’s Law: reaction 1 + (– reaction 3) = the overall reaction (reaction 2 is not needed)
and so –1411 kJ + (+ 44 kJ) = – 1367 kJ.
11. a) The sum of ∆Hºf (CO2) – ∆Hºf (CO) – ∆Hrxn (PbO + CO ! Pb + CO2) = ∆Hºf (PbO). Thus
– 394 kJ/mol + 111 kJ/mol + 131 kJ = – 152 kJ/mol.
12. d) ∆Hºrxn = – 2 ∆Hºf (H2 O ) + 2 ∆Hºf(NaOH) = 572 kJ + (– 940 kJ) = – 368 kJ
13. b)
14. e) Each of I and III represents the enthalpy of formation of the elements in their most stable
form.
15. b)
Chapter 7
1. e) III: E = hν; IV: n = 4 is a higher energy state than n = 2, so the transition will release energy;
V: hν = hc/λ.
2. c) E = hν = hc/λ.
3. a) As n increases, the energy separation between principal levels decreases.
4. e) Atomic radii decrease moving to the right and up on the periodic table.
5. c) Due to Hund’s rule of maximum multiplicity, a d8 configuration will have two unpaired
electrons.
6. c) Hund’s rule of maximum multiplicity states that electrons fill all degenerate states before
pairing up. Thus, 2p4 will contain two unpaired electrons.
7. a) The allowable values of l are 0 to (n – 1), therefore n ≠ 1.
8. e) For the hydrogen atom, all n = 2 orbitals are energetically degenerate.
9. d) Nitrogen requires sharing 3 electrons to complete its octet and element X requires one.
10. b) In the progression of the periodic table, element 118 falls under radon, filling the 7p6.
11. d) The second ionization energy requires more energy to remove an electron than the first.
12. d)
13. a) ∆Eionization is proportional to 1/n2, therefore 1/12 = 1, 1/22 = ¼.
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