5 Steps to a 5: AP Chemistry 2017 (2016)

STEP 2

Determine Your Test Readiness

CHAPTER 3   Take a Diagnostic Exam

CHAPTER 3

Take a Diagnostic Exam

IN THIS CHAPTER

Summary: The diagnostic exam is for your benefit. It will let you know where you need to spend the majority of your study time. Do not make the mistake of studying only those parts you missed; you should always review all topics. It may be to your advantage to take the diagnostic exam again just before you begin your final review for the exam. This exam has only multiple-choice questions. It will give you an idea of where you stand with your chemistry preparation. The questions have been written to approximate the coverage of material that you will see on the AP exam and are similar to the review questions that you will see at the end of each chapter. However, there will be a few questions on content that will not be directly tested on the AP exam; these questions refer to basic chemistry knowledge that your teacher will expect you to know and that you will need to know before taking the AP Chemistry Exam. Once you are done with the exam, check your work against the given answers, which also indicate where you can find the corresponding material in the book. We also provide you with a way to convert your score to a rough AP score.

Key Ideas

  Practice the kind of multiple-choice questions you will be asked on the real exam.

  Answer questions that approximate the coverage of topics on the real exam.

  Check your work against the given answers.

  Determine your areas of strength and weakness.

  Highlight the pages that you must give special attention to.

Getting Started: The Diagnostic Exam

The following problems refer to different chapters in the book. The important thing is not whether you get the correct answer, but whether you have difficulty with one or more questions from a chapter. If so, then review the material in that chapter. You may use a calculator and periodic table. For each question, circle the letter of your choice.

Chapter 5

  1. In most of its compounds, this element exists as a monatomic cation.

(A) F

(B) S

(C) N

(D) Ca

  1. Which of the following groups has the species correctly listed in order of decreasing radius?

(A) Cu2+ , Cu+ , Cu

(B) V, V2+ , V3+

(C) F , Br , I

(D) B, Be, Li

  1. Which of the following elements has the lowest electronegativity?

(A) F

(B) I

(C) Ba

(D) Al

  1. Which of the following represents the correct formula for hexaamminecobalt(III) nitrate?

(A) [Co(NH3 )6 ](NO3 )3

(B) [Co(NH3 )6 ](NO2 )3

(C) Am6 Co(NO3 )3

(D) (NH3 )6 Co3 (NO3 )

  1. Which of the following represents the correct formula for hexaamminechromium(III) chloride?

(A) [Cr(NH3 )6 ](ClO3 )3

(B) (NH3 )6 Cr3 Cl

(C) Am6 CrCl3

(D) [Cr(NH3 )6 ]Cl3

  1. The discovery that atoms have small, dense nuclei is credited to which of the following?

(A) Einstein

(B) Dalton

(C) Bohr

(D) Rutherford

Chapter 6

  1. ____ Mn(OH)2(s) + ____ H3 AsO4 (aq) → ____ Mn3 (AsO4 )2 (s) + ____ H2 O(l)

After the above chemical equation is balanced, the lowest whole-number coefficient for water is:

(A) 6

(B) 2

(C) 12

(D) 3

  1. Which of the following best represents the net ionic equation for the reaction of calcium hydroxide with an aqueous sodium carbonate solution to form a precipitate?

(A) Ca2+ + Na2 CO3 → CaCO3 + 2 Na+

(B) 2 Ca(OH) + Na2 CO3 → Ca2 CO3 + 2 NaOH

(C) Ca(OH)2 + CO3 2− → CaCO3 + 2 OH

(D) Ca2+ + CO3 2− → CaCO3

  1. A student mixes 50.0 mL of 0.10M Fe(NO3 )2 solution with 50.0 mL of 0.10 M KOH. A green precipitate forms, and the concentration of the hydroxide ion becomes very small. Which of the following correctly places the concentrations of the remaining ions in order of decreasing concentration?

(A) [Fe2+ ] > [NO3  ] > [K+ ]

(B) [Fe2+ ] > [K+ ] > [NO3  ]

(C) [NO3  ] > [K+ ] > [Fe2+ ]

(D) [K+ ] > [Fe2+ ] > [NO3  ]

  1. Solutions containing this ion give a reddish-brown precipitate upon standing.

(A) Cu2+

(B) CO3 2–

(C) Fe3+

(D) Al3+

Chapter 7

  1. 14 H+(aq) + 6 Fe2+ (aq) + Cr2 O7 2− (aq) → 2 Cr3+ (aq) + 6 Fe3+ (aq) + 7 H2 O(l)

The above reaction is used in the titration of an iron solution. What is the concentration of the iron solution if it took 45.20 mL of 0.1000 M Cr2 O7 2− solution to titrate 75.00 mL of an acidified iron solution?

(A) 0.1000 M

(B) 0.4520 M

(C) 0.3616 M

(D) 0.7232 M

  1. Manganese, Mn, forms a number of oxides. A particular oxide is 49.5% by mass Mn. What is the simplest formula for this oxide?

(A) MnO

(B) Mn2 O3

(C) Mn2 O7

(D) MnO2

  1. 2 KMnO4(aq) + 5 H2 C2 O4 (aq) + 3 H2 SO4 (aq) → K2 SO4 (aq) + 2 MnSO4 (aq) + 10 CO2 (g) + 8 H2 O(l)

How many moles of MnSO4 are produced when 2.0 mol of KMnO4 , 2.5 mol of H2 C2 O4 , and 3.0 mol of H2 SO4 are mixed?

(A) 1.0 mol

(B) 3.5 mol

(C) 2.0 mol

(D) 2.5 mol

  1. ____ KClO3→ ____ KCl + ____ O2

After the above equation is balanced, how many moles of O2 can be produced from 1.0 mol of KClO3 ?

(A) 1.5 mol

(B) 3.0 mol

(C) 1.0 mol

(D) 3.0 mol

Chapter 8

  1. Ba(s) + 2 H2O(l) → Ba(OH)2 (aq) + H2 (g)

Barium reacts with water according to the above reaction. What volume of hydrogen gas, at standard temperature and pressure, is produced from 0.400 mol of barium?

(A) 8.96 L

(B) 5.60 L

(C) 4.48 L

(D) 3.36 L

  1. A sample of chlorine gas is placed in a container at constant pressure. The sample is heated until the absolute temperature is doubled. This will also double which of the following?

(A) potential energy

(B) moles

(C) density

(D) volume

  1. A balloon contains 2.0 g of hydrogen gas. A second balloon contains 4.0 g of helium gas. Both balloons are at the same temperature and pressure. Which of the following statements is true?

(A) The number of hydrogen molecules is less than the number of helium atoms in each balloon.

(B) The density of the helium in its balloon is less than the density of the hydrogen in its balloon.

(C) The volume of the hydrogen balloon is less than that of the helium balloon.

(D) The average kinetic energy of the molecules/atoms in each balloon is the same.

  1. The volume and pressure of a real gas are NOT the same as those calculated from the ideal gas equation, because the ideal gas equation does NOT take into account:

(A) the volume of the molecules and the attraction between the molecules.

(B) the volume of the molecules and the mass of the molecules.

(C) the attraction between the molecules and the mass of the molecules.

(D) the volume of the molecules and variations in the absolute temperature.

  1. Aluminum metal reacts with gaseous HCl to produce aluminum chloride and hydrogen gas. What volume of hydrogen gas, at STP, is produced when 13.5 g of aluminum is mixed with an excess of HCl?

(A) 22.4 L

(B) 33.6 L

(C) 11.2 L

(D) 16.8 L

  1. A sample containing the gases carbon dioxide, carbon monoxide, and water vapor was analyzed and found to contain 4.5 mol of carbon dioxide, 4.0 mol of carbon monoxide, and 1.5 mol of water vapor. The mixture had a total pressure of 1.2 atm. What was the partial pressure of the carbon monoxide?

(A) 0.48 atm

(B) 0.18 atm

(C) 5.4 atm

(D) 0.54 atm

  1. An ideal gas sample weighing 0.548 g at 100°C and 0.993 atm has a volume of 0.237 L. Determine the molar mass of the gas.

(A) 71.3 g/mol

(B) 143 g/mol

(C) 19.1 g/mol

(D) 0.0140 g/mol

  1. If a sample of He effuses at a rate of 30 mol per hour at 45°C, which of the gases below will effuse at approximately one-half the rate under the same conditions?

(A) CH4

(B) O3

(C) N2

(D) H2

  1. The average kinetic energy of nitrogen molecules changes by what factor when the temperature is increased from 30°C to 60°C?

(A) (333 − 303)

(B) 2

(C) 1/2

(D) 

Chapter 9

  1. Which of the following is the required energy to produce a gaseous cation from a gaseous atom in the ground state?

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) ionization energy

  1. The average ____________ is the same for any ideal gas at a given temperature.

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) ionization energy

  1. Which of the following is the maximum energy available for useful work from a spontaneous reaction?

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) ionization energy

  1. The energy required to completely separate the ions from a solid is which of the following?

(A) free energy

(B) lattice energy

(C) kinetic energy

(D) ionization energy

  1. Oxidation of ClF by F2yields ClF3 , an important fluorinating agent formerly used to produce the uranium compounds in nuclear fuels:

ClF(g) + F2 (g) → 2 ClF3 (l)

Use the following thermochemical equations to calculate ΔH °rxn for the production of ClF3 :

(A) +270.2 kJ

(B) −135.1 kJ

(C) 0.0 kJ

(D) −270.2 kJ

  1. Choose the reaction expected to have the greatest increase in entropy.

(A) N2 (g) + O2 (g) → 2 NO(g)

(B) CO2 (g) → CO2 (s)

(C) 2 XeO3 (s) → 2 Xe(g) + 3 O2 (g)

(D) 2 K(s) + F2 (g) → 2 KF(s)

  1. A certain reaction is nonspontaneous under standard conditions but becomes spontaneous at lower temperatures. What conclusions may be drawn under standard conditions?

(A) ΔH > 0, ΔS > 0 and ΔG > 0

(B) ΔH < 0, ΔS < 0 and ΔG = 0

(C) ΔH < 0, ΔS > 0 and ΔG > 0

(D) ΔH < 0, ΔS < 0 and ΔG > 0

Chapter 10

  1. Which of the following groups contains only atoms that are paramagnetic in their ground state?

(A) Be, O, and N

(B) Mg, He, and Rb

(C) K, C, and Fe

(D) Br, Sb, and Kr

  1. Which of the following could be the electron configuration of a transition metal ion?

(A) 1s2 1p6 2s2 2p3

(B) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

(C) 1s2 2s2 2p6 3s2 3p6 3d3

(D) 1s2 2s2 2p5

  1. Which of the following is the configuration of a noble gas?

(A) 1s2 1p6 2s2 2p3

(B) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

(C) 1s2 2s2 2p6 3s2 3p6 3d3

(D) 1s2 2s2 2p5

  1. Which of the following is the electron configuration of a halogen?

(A) 1s2 1p6 2s2 2p3

(B) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

(C) 1s2 2s2 2p6 3s2 3p6 3d3

(D) 1s2 2s2 2p5

  1. Which of the following is an impossible electron configuration?

(A) 1s2 1p6 2s2 2p3

(B) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

(C) 1s2 2s2 2p6 3s2 3p6 3d3

(D) 1s2 2s2 2p5

  1. This explains why the exact position of an electron is not known.

(A) Pauli exclusion principle

(B) Electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

  1. This is why nitrogen atoms, in their ground state, are paramagnetic.

(A) Pauli exclusion principle

(B) Electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

  1. This means that an atomic orbital can hold no more than two electrons.

(A) Pauli exclusion principle

(B) Electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

  1. Which of the following explains why the 4s orbital fills before the 3d?

(A) Pauli exclusion principle

(B) Electron shielding

(C) Hund’s rule

(D) Heisenberg uncertainty principle

  1. Magnesium reacts with elementX to form an ionic compound. If the ground-state electron configuration of X is 1s2 2s2 2p3 , what is the simplest formula for this compound?

(A) MgX 2

(B) Mg2 X 3

(C) Mg3 X 2

(D) MgX

Chapter 11

  1. VSEPR predicts that an IF5molecule will have which of the following shapes?

(A) tetrahedral

(B) trigonal bipyramidal

(C) square pyramid

(D) trigonal planar

  1. Which of the following does NOT have one or moreπ bonds?

(A) SO2

(B) SF6

(C) O2

(D) SO3

  1. Which of the following is nonpolar?

(A) IF5

(B) BrF3

(C) CF4

(D) SF4

  1. The only substance listed below that contains ionic, σ, and π bonds is:

(A) Na3 N

(B) NO2

(C) NaNO3

(D) NH3

  1. Which molecule or ion in the following list has the greatest number of unshared electron pairs around the central atom?

(A) SO2

(B) CO3 2–

(C) XeF2

(D) CF4

  1. What types of hybridization of carbon are in the compound acetic acid, CH3COOH?

(A) sp3 , sp2 , and sp

(B) sp3 only

(C) sp3 and sp2

(D) sp2 and sp

Chapter 12

  1. Which of the following is the best description of the structure of graphite?

(A) composed of atoms held together by delocalized electrons

(B) composed of molecules held together by intermolecular dipole-dipole interactions

(C) composed of positive and negative ions held together by electrostatic attractions

(D) composed of macromolecules held together by strong bonds

  1. Which of the following best describes Ca(s)?

(A) composed of atoms held together by delocalized electrons

(B) composed of molecules held together by intermolecular dipole-dipole interactions

(C) composed of positive and negative ions held together by electrostatic attractions

(D) composed of macromolecules held together by strong bonds

  1. Which of the following categories best describes CaCO3(s)?

(A) composed of atoms held together by delocalized electrons

(B) composed of molecules held together by intermolecular dipole-dipole interactions

(C) composed of positive and negative ions held together by electrostatic attractions

(D) composed of macromolecules held together by strong bonds

  1. Which of the following is applicable to SO2(s)?

(A) composed of atoms held together by delocalized electrons

(B) composed of molecules held together by intermolecular dipole-dipole interactions

(C) composed of positive and negative ions held together by electrostatic attractions

(D) composed of macromolecules held together by strong bonds

  1. The critical point on a phase diagram represents:

(A) the highest temperature and pressure where a substance can sublime

(B) the highest temperature and pressure where the substance may exist as discrete solid and gas phases

(C) the temperature and pressure where the substance exists in equilibrium as solid, liquid, and gas phases

(D) the highest temperature and pressure where the substance may exist as discrete liquid and gas phases

  1. This explains why copper is ductile.

(A) London dispersion forces

(B) covalent bonding

(C) hydrogen bonding

(D) metallic bonding

  1. This is why acetic acid, CH3COOH, molecules exist as dimers in the gaseous phase.

(A) London dispersion forces

(B) covalent bonding

(C) hydrogen bonding

(D) metallic bonding

  1. For the following, pick the answer that most likely represents their relative solubilities in water.

(A) CH3 CH2 CH2 OH < HOCH2 CH2 OH < CH3 CH2 CH2 CH3

(B) CH3 CH2 CH2 CH3 < HOCH2 CH2 OH < CH3 CH2 CH2 OH

(C) CH3 CH2 CH2 CH3 < CH3 CH2 CH2 OH < HOCH2 CH2 OH

(D) CH3 CH2 CH2 OH < CH3 CH2 CH2 CH3 < HOCH2 CH2 OH

  1. The above diagram represents the heating curve for a pure crystalline substance. The solid is the only phase present up to point:

(A) C

(B) B

(C) E

(D) A

  1. For all one-component phase diagrams, choose the correct statement from the following list.

(A) The line separating the gas from the liquid phase may have a positive or negative slope.

(B) The line separating the solid from the liquid phase may have a positive or negative slope.

(C) The line separating the solid from the liquid phase has a positive slope.

(D) The temperature at the triple point is the same as at the freezing point.

Chapter 13

  1. A solution is prepared by dissolving 0.500 mol of NaCl in 500.0 g of water. Which of the following would be the best procedure to determine the molarity of the solution?

(A) Measure the volume of the solution.

(B) Titrate the solution with standard silver nitrate solution.

(C) Determine the freezing point of the solution.

(D) Determine the osmotic pressure of the solution.

  1. A chemist needs 800.0 mL of a 0.50M bromide ion, Br , solution. She has 800.0 mL of a 0.20 M KBr solution. How many moles of solid MgBr2 will she need to add to increase the concentration to the desired value?

(A) 0.24

(B) 0.50

(C) 0.30

(D) 0.12

  1. How many grams of HNO3(molecular weight 63.0) are in 500.0 mL of a 5.00 M solution?

(A) 31.5 g

(B) 63.0 g

(C) 5.00 g

(D) 158 g

  1. If a solution of ethyl ether, (C2H5 )2 O, in ethanol, C2 H5 OH, is treated as an ideal solution, what is the mole fraction of ethyl ether in the vapor over an equimolar solution of these two liquids? The vapor pressure of ethyl ether is 480 mm Hg at 20°C, and the vapor pressure of ethanol is 50 mm Hg at this temperature.

(A) 0.50

(B) 0.76

(C) 0.91

(D) 0.27

Chapter 14

  1. Step 1: 2 NO2(g) → N2 (g) + 2 O2 (g)
    Step 2: 2 CO(g) + O2 (g) → 2 CO2 (g)
    Step 3: N2 (g) + O2 (g) → 2 NO(g)

The above represents a proposed mechanism for the reaction of NO2 with CO. What are the overall products of the reaction?

(A) NO and CO2

(B) O2 and CO2

(C) N2 and NO

(D) NO only

  1. The difference in energy between the transition state and the reactants is:

(A) the kinetic energy

(B) the activation energy

(C) the free energy

(D) the reaction energy

  1. The table below gives the initial concentrations and rates for three experiments.

The reaction is 2ClO2 (aq) + 2OH (aq) → ClO2  (aq) + ClO3  (aq) + H2 O(l). What is the rate law for this reaction?

(A) Rate = k [ClO2 ]2 [OH ]2

(B) Rate = k [ClO2 ]

(C) Rate = k [ClO2 ]2 [OH ]

(D) Rate = k [OH ]2

Chapter 15

  1. Which of the following CANNOT behave as both a Brønsted base and a Brønsted acid?

(A) HCO3 

(B) HPO4 2−

(C) HSO4 

(D) CO3 2–

Using the information from the preceding table, which of the following is the best choice for preparing a pH = 8.5 buffer?

(A) K2 HPO4 + K3 PO4

(B) K2 HPO4 + KH2 PO4

(C) K3 PO4

(D) K2 HPO4

  1. What is the ionization constant,K a , for a weak monoprotic acid if a 0.5 molar solution has a pH of 5.0?

(A) 3 × 10−4

(B) 2 × 10−10

(C) 7 × 10−8

(D) 1 × 10−6

  1. Assuming all concentrations are 1M , which of the following is the most acidic solution (lowest pH)?

(A) KBr (potassium bromide) and HBr (hydrobromic acid)

(B) H2 C2 O4 (oxalic acid) and KHC2 O4 (potassium hydrogen oxalate)

(C) NH3 (ammonia) and NH4 NO3 (ammonium nitrate)

(D) (CH3 )2 NH (dimethylamine) and HC2 H3 O2 (acetic acid)

  1. Assuming all concentrations are equal, which of the following solutions has pH nearest 7?

(A) KBr (potassium bromide) and HBr (hydrobromic acid)

(B) H2 C2 O4 (oxalic acid) and KHC2 O4 (potassium hydrogen oxalate)

(C) NH3 (ammonia) and NH4 NO3 (ammonium nitrate)

(D) (CH3 )2 NH (dimethylamine) and HC2 H3 O2 (acetic acid)

  1. Which of the following yields a buffer with a pH > 7 upon mixing equal volumes of 1M solutions?

(A) KBr (potassium bromide) and HBr (hydrobromic acid)

(B) H2 C2 O4 (oxalic acid) and KHC2 O4 (potassium hydrogen oxalate)

(C) NH3 (ammonia) and NH4 NO3 (ammonium nitrate)

(D) (CH3 )2 NH (dimethylamine) and HC2 H3 O2 (acetic acid)

  1. Which of the following will give a buffer with a pH < 7 when equal volumes of 1M solutions of each of the components are mixed?

(A) KBr (potassium bromide) and HBr (hydrobromic acid)

(B) H2 C2 O4 (oxalic acid) and KHC2 O4 (potassium hydrogen oxalate)

(C) NH3 (ammonia) and NH4 NO3 (ammonium nitrate)

(D) (CH3 )2 NH (dimethylamine) and HC2 H3 O2 (acetic acid)

  1. Determine the OH(aq) concentration in 0.0010 M pyridine (C5 H5 N) solution. (The K b for pyridine is 9 × 10−9 .)

(A) 5 × 10−1 M

(B) 1 × 10−3 M

(C) 3 × 10−6 M

(D) 9 × 10−9 M

  1. SnS(s) + 2H+(aq)  Sn2+ (aq) + H2 S(aq)

The successive acid dissociation constants for H2 S are 9.5 × 10−8 (K a1 ) and 1 × 10−19 (K a2 ). The K sp , the solubility product constant, for SnS equals 1.0 × 10−25 . What is the equilibrium constant for the above reaction?

(A) 9.5 × 10−8 /1.0 × 10−25

(B) 9.5 × 10−27 /1.0 × 10−25

(C) 1.0 × 10−25 /9.5 × 10−27

(D) 1 × 10−19 /1.0 × 10−25

  1. N2O4 (g)  2NO2 (g) endothermic

An equilibrium mixture of the compounds is placed in a sealed container at 150°C. Which of the following changes may increase the amount of the product?

(A) raising the temperature of the container

(B) increasing the volume of the container and raising the temperature of the container

(C) adding 1 mole of Ar(g) to the container

(D) adding 1 mole of Ar(g) to the container and raising the temperature of the container

  1. TheK sp for LaF3 is 2 × 10−9 . What is the molar solubility of this compound in water?

(A) 2 × 10–9 /27

(B) 

(C) 

(D) 

  1. TheK sp for Mn(OH)2 is 1.6 × 10−13 . What is the molar solubility of this compound in water?

(A) 

(B) 1.6 × 10–13

(C) 

(D) 

  1. FeS(s) + 2 H+(aq)  Fe2+ (aq) + H2 S(aq)

What is the equilibrium constant for the above reaction? The successive acid dissociation constants for H2 S are 9.5 × 10−8 (K a1 ) and 1 × 10−19 (K a2 ). The K sp , the solubility product constant, for FeS equals 5.0 × 10−18 .

(A) 9.5 × 10−27 /5.0 × 10−18

(B) 5.0 × 10−18 /9.5 × 10−27

(C) 5.0 × 10−18 /9.5 × 10−8

(D) 9.5 × 10−8 /5.0 × 10−18

  1. TheK sp for Cr(OH)3 is 1.6 × 10−30 . What is the molar solubility of this compound in water?

(A) 

(B) 

(C) 1.6 × 10–30

(D) 1.6 × 10–30 /27

Chapter 16

  1. I(aq) + H+ (aq) + MnO4  (aq) → Mn2+ (aq) + H2 O(l) + I2 (s)

What is the coefficient of H+ when the above reaction is balanced?

(A) 12

(B) 32

(C) 16

(D) 8

  1. How many moles of Au will deposit on the cathode when 0.60 Faradays of electricity is passed through a 1.0M solution of Au3+ ?

(A) 0.60 mol

(B) 0.30 mol

(C) 0.40 mol

(D) 0.20 mol

  1. Sn2+(aq) + 2 Fe3+ (aq) → Sn4+ (aq) + 2 Fe2+ (aq)

The reaction shown above was used in an electrolytic cell. The voltage measured for the cell was not equal to the calculated E° for the cell. Which of the following could explain this discrepancy?

(A) Both of the solutions were at 25°C instead of 0°C.

(B) The anode and cathode were different sizes.

(C) The anion in the anode compartment was chloride instead of nitrate, as in the cathode compartment.

(D) One or more of the ion concentrations was not 1 M .

Questions 81–82 refer to the following half-reaction in an electrolytic cell:

2 SO4 2− (aq) + 10 H+ (aq) + 8 e → S2 O3 2− (aq) + 5 H2 O(l)

  1. Choose the correct statement from the following list.

(A) The sulfur is oxidized.

(B) This is the cathode reaction.

(C) The oxidation state of sulfur does not change.

(D) The H+ serves as a catalyst.

  1. If a current of 0.60 amperes is passed through the electrolytic cell for 0.75 h, how should you calculate the grams of S2O3 2− (aq) formed?

(A) (0.60) (0.75) (3,600) (112)/(96,500) (8)

(B) (0.60) (0.75) (3,600) (112)/(96,500) (10)

(C) (0.60) (0.75) (60) (32)/(96,500) (8)

(D) (0.60) (0.75) (3,600) (112)/(10)

  1. 2 BrO3 (aq) + 12 H+ (aq) + 10 e → Br2 (aq) + 6 H2 O(l)

If a current of 5.0 A is passed through the electrolytic cell for 0.50 h, how should you calculate the number of grams of Br2 that will form?

(A) (5.0) (0.50) (3,600) (159.8)/(10)

(B) (5.0) (0.50) (3,600) (159.8)/(96,500) (10)

(C) (5.0) (0.50) (60) (159.8)/(96,500) (10)

(D) (5.0) (0.50) (3,600) (79.9)/(96,500) (10)

  1. 2 IO3 (aq) + 6 H2 O(l) + 10 e → I2 (s) + 12 OH (aq)

Using the above reaction, if a current of 7.50 A is passed through the electrolytic cell for 0.45 h, how should you calculate the grams of I2 that will form?

(A) (7.50) (0.45) (3,600) (253.8)/(10)

(B) (7.50) (0.45) (3,600) (126.9)/(96,500) (10)

(C) (7.50) (0.45) (60) (253.8)/(96,500) (10)

(D) (7.50) (0.45) (3,600) (253.8)/(96,500) (10)

Chapter 17

  1. When decays, it emits two α particles, then a β ‚ particle, followed by an α particle. The resulting nucleus is:

(A) 

(B) 

(C) 

(D) 

  1. Which of the following lists the types of radiation in the correct order of increasing penetrating power?

(A) α , γ , β

(B) β , α , γ

(C) α , β , γ

(D) β , γ , α

  1. What is the missing product in the following nuclear reaction?

(A) 

(B) 

(C) 

(D) 

  1. If 75% of a sample of pure decays in 24.6 yr, what is the half-life of  ?

(A) 24.6 yr

(B) 18.4 yr

(C) 12.3 yr

(D) 6.15 yr

Chapter 18

  1. Alkenes are hydrocarbons with the general formula Cn H2n . If a 1.40 g sample of any alkene is combusted in excess oxygen, how many moles of water will form?

(A) 0.2

(B) 0.1

(C) 1.5

(D) 0.7

  1. What type of compound is shown?

(A) an alcohol

(B) an aldehyde

(C) a ketone

(D) an ester

Chapter 19

Questions on this chapter are incorporated into the chapters concerning the specific experiments.

 Answers and Explanations

Chapter 5

  1. D—The others (nonmetals) form anions.
  2. B—Decreasing radii is related to increasing charges, or for going up a column (with equal charges), or moving toward the right in a period of the periodic table. This explanation will not be sufficient for the free-response portion of the test, where it is necessary to address such factors as the effective nuclear charge.
  3. C—The element that is farthest away from F on the periodic table.
  4. A—Hexaammine = (NH3 )6 ; cobalt(III) = Co3+ ; and nitrate = NO3  .
  5. D—Hexaammine = (NH3 )6 ; chromium(III) = Cr3+ ; chloride = Cl
  6. D—Rutherford, and students, determined this by bombarding gold foil with alpha particles and detecting the deflection of some of the particles.

Chapter 6

  1. A—The balanced chemical equation is:

3 Mn(OH)2 (s) + 2 H3 AsO4 (aq) → Mn3 (AsO4 )2 (s) + 6 H2 O(l)

  1. D—Ca(OH)2 , NaOH, and Na2 CO3 are strong electrolytes (strong bases or soluble salts) and should be separated. You should know all the strong bases and that sodium compounds are soluble. Cancel all spectator ions (Na+ and OH ).
  2. C—The hydroxide is low because it combined with some of the iron, so Fe2+ will be low. There is no other ion that the hydroxide ion could combine with to form a precipitate. The nitrate is double the potassium because there are two moles of nitrate per mole of iron(II) nitrate instead of one ion per mole, as in potassium hydroxide.
  3. C—Copper is blue, not red, and carbonate and aluminum are colorless. Iron slowly hydrolyzes (reacts with water) to form solid Fe(OH)3 (rust).

Chapter 7

  1. C—(0.1000 mol Cr2 O7 2− /1,000 mL)(45.20 mL) (6 mol Fe2+ /1 mol Cr2 O7 2− )(1/75.00 mL) (1,000 mL/L)
  2. C—Either calculate the percent Mn in each oxide: (A) 77.4%; (B) 69.6%; (C) 49.5%; (D) 63.2% or determine the empirical formula from the percent manganese and the percent oxygen ( = 100.0 − 49.5).
  3. A—H2 C2 O4 is the limiting reagent as the amount is less than the stoichiometric ratio indicates. The calculation is 
  4. A—The coefficients in the balanced equation are 2, 2, and 3. Therefore, (1.0 mol KClO3 ) (3 mol O2 /2 mol KClO3 ) = 1.5 mol.

Chapter 8

  1. A—(0.400 mol Ba)(1 mol H2 /1 mol Ba) (22.4 L/mol). Note that the 22.4 L/mol only works at STP.
  2. D—This is an application of Charles’s law, which relates volume to temperature. There is a direct relationship between volume and the absolute temperature. Doubling either volume or temperature, with moles and pressure remaining constant, doubles the other.
  3. D—The average kinetic energy of a gas depends upon the temperature. Since the temperature of the two gases is the same, the average kinetic energy of the gases is the same. The moles are the same, so the number of particles and the volumes must be the same. Density is mass over volume, and since the balloons have the same volume, the one with more mass will have the higher density.
  4. A—These are the basic differences between ideal and real gases.
  5. D—It is necessary to first write the balanced chemical equation:

2 Al(s) + 6 HCl(g) → 2 AlCl3 (s) + 3 H2 (g)

(13.5 g Al)(1 mol Al/27.0 g Al)(3 mol H2 /2 mol Al)(22.4 L/mol H2 )

  1. A—The mole fraction of CO times the total pressure yields the partial pressure. The mole fraction of CO is the moles of CO (4.0) divided by the total moles (10.0).

This example illustrates the importance of rounding in calculations where no calculator is available. The answers are not close together; therefore, a rough calculation will lead to the correct answer. Also, you should notice the answer D is impossible for any substance.

  1. A—For the rate to be one-half, the molar mass of the other gas must be the square of the molar mass of helium (42 = 16).
  2. A—The average kinetic energy of the molecules depends on the temperature. The correct answer involves a temperature difference (333 K − 303 K). Do not forget that ALL gas law calculations require Kelvin temperatures.

Chapter 9

  1. D—This is the definition of the ionization energy.
  2. C—This is a basic postulate of kinetic molecular theory.
  3. A—This is one of the properties of free energy.
  4. B—This is the definition of the lattice energy.
  5. B—This is an application of Hess’s law.

As always, rounding and estimating will save time.

  1. C—The one with the greatest increase in the moles of gas.
  2. D—Nonspontaneous means ΔG > 0. For a reaction to become spontaneous at lower temperature (ΔG < 0) means ΔH < 0 and ΔS < 0.

Chapter 10

  1. C—Atoms with only completely filled shells or subshells are not paramagnetic, they are diamagnetic. From the choices in this problem, these are: Be, Mg, He, Kr, and Zn; therefore, any answer containing one of these cannot be the correct choice. It is not necessary to work through a possible solution until encountering a diamagnetic species. Also, it might be helpful to look on the periodic table.
  2. C—Transition metal ions are, in general, s0 and p0 or p6 with the possibility of having one or more electrons in the d orbitals. C could be Cr3+ .
  3. B—The noble gases, except helium, are n s2 n p6 . In this case, n = 4, and the gas is krypton.
  4. D—Halogens are n s2 n p5 . In this case, n = 2, and the halogen is F.
  5. A—The 1p orbital does not exist.
  6. D—This is a statement of the uncertainty principle.
  7. C—According to Hund’s rule, the nitrogen 2p electrons enter the 2p orbitals individually (with spins parallel).
  8. A—The Pauli exclusion principle states this limitation for all orbitals.
  9. B—The d orbitals are less effectively shielded than the s orbitals. Due to this difference, the s orbitals have lower energy.
  10. C—Mg becomes Mg2+ . The element is N, which can become N3− .

Chapter 11

  1. C—The iodine has five bonding pairs and one lone pair.
  2. B—This is the only one with only single bonds. The other molecules have double or triple bonds. All double and triple bonds are a combination of σ and π bonds.
  3. C—Use VSEPR; only the tetrahedral CF4 is nonpolar. The other materials form a square pyramidal (IF5 ), T-shaped (BrF3 ), and irregular tetrahedral (SF4 ), and, therefore, are polar.
  4. C—The only ionic bonds are present in the sodium compounds (eliminating B and D). The nitride ion has no internal bonding (eliminating A), but the nitrate ion has both σ and π bonds.
  5. C—Draw the Lewis structures. The number of unshared pairs: (A) 1; (B) 0; (C) 3; (D) 0.
  6. C—Draw the Lewis structure; the carbon on the left in the formula is sp3 , and the other is sp2 .

Chapter 12

  1. D—Both graphite and diamond are covalent network solids.
  2. A—Calcium is a metal, and answer A applies to metallic bonding.
  3. C—Calcium carbonate is an ionic compound.
  4. B—Sulfur dioxide consists of polar molecules.
  5. D—This is the definition of the critical point.
  6. D—This is a consequence of metallic bonding as the atoms can easily move past each other without breaking any bonds.
  7. C—The carbonyl, C═O, and —OH groups are capable of participating in hydrogen bonds.
  8. C—The more —OH groups, the more hydrogen bonding, and the more soluble in water (where hydrogen bonding also occurs).
  9. D—The solid begins to melt at A and finishes melting at B.
  10. B—The gas-liquid line always has a positive slope, which eliminates A. Answer B negates C; therefore, both cannot be correct. The triple point is not the same as the freezing point.

Chapter 13

  1. A—Molarity is moles per liter, and the moles are already known; therefore, only the volume is necessary to complete the determination.
  2. D—(0.800 L)(0.50 mol Br /L) = 0.40 mol needed.

(0.800 L)(0.20 mol Br /L) = 0.16 mol present.

[(0.40 − 0.16) mol Br to be added] (1 mol MgBr2 /2 mol Br )

  1. D—(0.5000 L)(5.00 mol/L)(63.0 g/mol) = 158 g As always, estimate the answer by rounding the values.
  2. C—Equimolar gives a mole fraction of 0.5. 0.5 × 480 mm Hg + 0.5 × 50 mm Hg = 265 mm Hg (total vapor pressure) mole fraction ethyl ether = (0.5 × 480 mm Hg)/265 mm Hg.

Chapter 14

  1. A—Add the equations and cancel anything that appears on both sides of the reaction arrows.
  2. B—This is the definition of the activation energy.
  3. C—The table shows second order in chlorine dioxide (comparing experiments 1 and 2), because doubling the ClO2 concentration quadruples (22 ) the rate. The reaction is first order in the hydroxide ion (comparing experiments 2 and 3), because doubling the OH concentration doubles (21 ) the rate. When making this determination, make sure there is only one concentration changing, i.e., do not compare experiments 1 and 3.

Chapter 15

  1. D—To be an acid, the species must have an H+ to donate, and to be a base, the species must be able to accept an H+ . The carbonate ion has no H+ to donate to be an acid.
  2. B—Start with the acid with a pK a as near 8.5 (K = 10−8.5 ) as possible (H2 PO4  ). To go to a higher pH, add the acid (conjugate base) with the smaller K a (higher pK a ).
  3. B—This is an approximation. At pH = 5, [H+ ] = 10−5 M ; therefore, K a = (10−5 )2 /0.5.
  4. A—HBr is a strong acid, and with equal concentrations and no base present, it will give the lowest pH.
  5. D—The weak acid and weak base give a nearly neutral solution, as they will tend to neutralize each other.
  6. C—Only B and C are buffers. B is acidic (pH < 7) and C is basic (pH > 7).
  7. B—Only B and C are buffers. B is acidic (pH < 7) and C is basic (pH > 7).
  8. C—[OH ] = (0.0010 × 9 × 10−9 )1/2 = (9 × 10−12 )l/2 Estimate—the square root of 10−12 will be 10−6 .
  9. CK = K sp /K a1 K a2

In this case, the key is setting up the calculation but not doing the calculation.

  1. B—Adding Ar yields no change, as it is not part of the equilibrium. Increasing the temperature of an endothermic equilibrium will increase the amount of product.
  2. DK sp = [La3+ ][F ]3 = [x ][3x ]3 = 27x 4 . Solve for x . It is only necessary to set up the problem. This requires a knowledge of what the equilibrium is (LaF3 (s)  La3+ (aq) + 3 F (aq)) and how to write the equilibrium expression (K sp = [La3+ ][F ]3 ).
  3. A—The equilibrium constant expression for the dissolving of manganese(II) hydroxide is:

K sp = [Mn2+ ][OH ]2 = 1.6 × 10−13

If s is used to indicate the molar solubility, the equilibrium expression becomes:

  1. BK = K sp /K a1 K a2 = 5.0 × 10−18 /(9.5 × 10−8 ) (1 × 10−19 )
  2. BK sp = [Cr3+ ][OH ]3 = [x ][3x ]3 = 27x 4 = 1.6 × 10−30 . Solve for x .

Chapter 16

  1. C—The balanced equation is

10 I (aq) + 16 H+ (aq) + 2 MnO4  (aq) → 2 Mn2+ (aq) + 8 H2 O(l) + 5 I2 (s)

  1. D—It is only necessary to know the mole ratio for the reaction (Au3+ (aq) + 3 e → Au(s)), which gives (0.60 F)(1 mol Au/3 F) = 0.20 moles.
  2. D—The cell must be nonstandard. This could be due to variations in temperature (not 25°C) or concentrations (1 M) that are not standard.
  3. B—A reduction is shown. Reductions take place at the cathode.
  4. A—Use dimensional analysis:

(0.60 coul/s)(0.75 h)(3,600 s/h)(112 g S2 O3 2− /mol S2 O3 2− )/(96,500 coul/F)(8 F/mol S2 O3 2− )

  1. B—Recall that 5.0 amp is 5.0 C/s. The calculation would be:
  2. D—Dimensional analysis:

(7.50 coul/s)(0.45 h)(3,600 s/h)(253.8 g I2 /mol I2 )/(96,500 coul/F)(10 F)

Chapter 17

  1. D—The mass of an alpha particle is 4 and the mass of a beta particle is negligible. The mass number (superscript) should be 226 − (4 + 4 + 0 + 4) = 214. The charge on an alpha particle is +2 and the charge on the beta particle is −1; therefore, the atomic number (subscript) should be 88 − (2 + 2 − 1 + 2) = 83.
  2. C—Alpha particles are the least penetrating, and gamma rays are the most penetrating.
  3. C—Mass difference = 236 − 4(1) − 136 = 96.

Atomic number difference = 92 − 4(0) − 53 = 39.

  1. C—After one half-life, 50% would remain. After another half-life this would be reduced by 1/2 to 25%. The total amount decayed is 75%. Thus, 24.6 years must be two half-lives of 12.3 years each.

Chapter 18

  1. B—The general formula simplifies to CH2 , which has a molar mass of 14 g/mol. This leads to (1.40 g) (1 mol/14 g).
  2. B.

Chapter 19

Questions on this chapter are incorporated into the chapters concerning the specific experiments.

Scoring and Interpretation

Now that you have finished and scored the diagnostic exam, it is time for you to learn what it all means. First, note any area where you had difficulty. This should not be limited to unfamiliar material. You should do this even if you got the correct answer. Determine where this material is covered in the book. Plan to spend additional time on the chapter in question. There is material you may not recognize because you have not gotten that far in class.

There are no free-response questions on this diagnostic exam; such questions are not useful at this point. There will be examples of free-response questions later in this book. We will use the multiple-choice questions to provide an estimate of your preparation. This is a simplified approach based on these questions. Do not try to do more than use these results as a general guide.

If you did better than you expected—great! Be careful not to become overconfident. Much more will need to be done before you take the AP Chemistry exam.

If you did not do as well as you would have liked, don’t panic. There is plenty of time for you to prepare for the exam. This is a guide to allow you to know which path you need to follow.

No matter what your results were, you are about to begin your 5 steps to a 5.

Good Luck!