AP Chemistry Practice Test 6

Correct: (B)

The atom has 10 electrons, like Ne, but one electron occupies the 3s subshell instead of completing the 2p subshell, which is an excited-state arrangement.

Correct: (D)

(0.790)(24) + (0.100)(25) + (0.110)(26) = 18.96 + 2.50 + 2.86 = 24.32, approximately 24.3.

Correct: (A)

Atomic radius decreases across a period because the increasing nuclear charge pulls the electrons closer. Fluorine is farthest to the right in period 2 and has the smallest radius.

Correct: (C)

Copper is an exception to the usual filling order: a filled 3d subshell with a half-filled 4s subshell is especially stable, giving [Ar] 4s^1 3d^10 (29 electrons).

Correct: (B)

The 1s electrons are closest to the nucleus and are attracted most strongly, so they require the most energy to remove and appear at the highest binding energy.

Correct: (A)

Electronegativity increases toward the top right of the periodic table, and fluorine is the most electronegative element.

Correct: (D)

The very large jump between the third and fourth ionization energies shows that the first three electrons are valence electrons and the fourth is a core electron, so the element has 3 valence electrons and is in Group 13.

Correct: (C)

Sulfur has 6 valence electrons and is assigned half of the 12 bonding electrons (6) with no lone pairs: formal charge = 6 – 6 = 0.

Correct: (A)

Nitrogen has four electron domains (three bonds and one lone pair). The electron geometry is tetrahedral, but the lone pair makes the molecular geometry trigonal pyramidal.

Correct: (D)

The C-Cl bonds in CCl4 are polar, but the four bonds are arranged symmetrically in a tetrahedron, so the bond dipoles cancel and the molecule is nonpolar.

Correct: (B)

Lattice energy decreases as the distance between ion centers increases. KF combines a large cation with a relatively large overall ion separation, giving the smallest lattice energy of the choices.

Correct: (C)

O2 has a bond order of 2. Removing one electron to form O2+ removes an antibonding electron, increasing the bond order to 2.5.

Correct: (A)

BF3 has three electron domains around boron and is trigonal planar, so the bond angles are 120°.

Correct: (D)

Each carbon has three electron domains (two C-H bonds and one C=C bond region) in a trigonal planar arrangement, which corresponds to sp^2 hybridization.

Correct: (A)

Propane is nonpolar and experiences only weak London dispersion forces. The other substances are polar, and the alcohol and carboxylic acid also hydrogen bond, so all have higher boiling points.

Correct: (C)

Moles of NaCl = 11.7 / 58.44 = 0.200 mol. Molarity = 0.200 mol / 0.2500 L = 0.801 M.

Correct: (B)

9.00 g / 18.0 g/mol = 0.500 mol of water; 0.500 mol × 40.7 kJ/mol = 20.3 kJ.

Correct: (D)

Boiling point elevation depends on the total concentration of dissolved particles. CaCl2 produces three ions per formula unit, more than NaCl or KBr (two each) or glucose (one molecule), so it has the highest boiling point.

Correct: (B)

In metals, mobile delocalized electrons carry current, and the nondirectional bonding between the positive ion cores allows the solid to be deformed without breaking apart.

Correct: (A)

H2S is a bent, polar molecule, so dipole-dipole forces dominate. Hydrogen bonding requires hydrogen bonded to N, O, or F, and sulfur does not meet that requirement.

Correct: (C)

The balanced equation is 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O, so the coefficient of CO2 is 4.

Correct: (A)

AgCl is insoluble, so Ag+ and Cl- combine to form the precipitate; K+ and NO3- are spectator ions and are omitted.

Correct: (D)

28.0 g of N2 is 1.00 mol, which can produce 2.00 mol of NH3 = 34.0 g. The 9.00 g of H2 (4.46 mol) is more than the 3.00 mol needed, so N2 is the limiting reactant.

Correct: (B)

16.0 g of CH4 is 1.00 mol, which produces 2.00 mol of H2O; 2.00 mol × 18.0 g/mol = 36.0 g.

Correct: (C)

The oxidation number of Sn increases from +2 to +4, meaning Sn^2+ loses electrons and is oxidized. Fe^3+ is reduced.

Correct: (A)

12.0 g / 16.0 g/mol = 0.750 mol of CH4, which requires 2 × 0.750 = 1.50 mol of O2; 1.50 mol × 32.0 g/mol = 48.0 g.

Correct: (D)

Moles of NaOH = 0.03000 L × 0.200 M = 0.00600 mol. Because H2SO4 reacts with 2 NaOH, moles of H2SO4 = 0.00300 mol, and the concentration is 0.00300 mol / 0.02000 L = 0.150 M.

Correct: (B)

Doubling [Y] (experiments 1 and 2) quadruples the rate, so the reaction is second order in Y. Doubling [X] (experiments 1 and 3) doubles the rate, so it is first order in X.

Correct: (C)

A half-life that is independent of the starting concentration is the defining characteristic of first-order kinetics.

Correct: (D)

For a second-order reaction, 1/[A] = kt + 1/[A]0 = (0.050)(100.) + 1/0.40 = 5.0 + 2.5 = 7.5, so [A] = 1/7.5 = 0.13 M.

Correct: (A)

At a higher temperature the molecules move faster, increasing both the collision frequency and, most importantly, the fraction of collisions energetic enough to overcome the activation energy barrier.

Correct: (C)

The slow step gives rate = k[A][B]. From the fast equilibrium in step 1, [A] is proportional to [A2]^1/2, so substituting gives rate = k[A2]^1/2[B].

Correct: (B)

Completing 75% of a first-order reaction means two half-lives have elapsed (100% → 50% → 25% remaining), so the half-life is 60.0 / 2 = 30.0 min.

Correct: (D)

The activation energy of the reverse reaction is the energy difference between the products and the transition state: 90 − 60 = 30 kJ/mol. (The forward activation energy is 70 kJ/mol.)

Correct: (A)

Subtracting the first equation from the second (Hess's law) gives the target reaction, with ΔH = 66.4 − 180.5 = −114 kJ.

Correct: (D)

Heat gained by the water = (100.0)(4.184)(5.0) = 2092 J. This equals the heat lost by the metal: (100.0)(c)(70.0), so c = 2092 / 7000 = 0.299 J/g·°C.

Correct: (B)

ΔH = [2(−395.7)] − [2(−296.8) + 0] = −791.4 + 593.6 = −197.8 kJ.

Correct: (C)

Bonds broken: 941 + 3(436) = 2249 kJ. Bonds formed: 6(391) = 2346 kJ. ΔH = 2249 − 2346 = −97 kJ.

Correct: (A)

The reaction converts 3 mol of gas into 2 mol of gas, decreasing the number of freely moving particles and therefore the entropy. The other choices produce a gas from a condensed phase and increase entropy.

Correct: (D)

Melting 1.00 mol of ice requires 6.01 kJ. Heating the liquid from 0°C to 50.0°C requires (18.0)(4.184)(50.0) = 3766 J = 3.77 kJ. The total is 6.01 + 3.77 = 9.78 kJ.

Correct: (B)

The heat absorbed by the water is q = (200.0)(4.184)(4.5) = 3766 J = 3.77 kJ, so the reaction released approximately 3.77 kJ of heat.

Correct: (C)

Q = [N2O4] / [NO2]^2 = 0.50 / (0.10)^2 = 50. Because Q is less than Kc (170), the net reaction proceeds toward the products.

Correct: (D)

With x = [Cl2] at equilibrium, Kc = x^2 / (0.20 − x) = 0.040, so x^2 + 0.040x − 0.0080 = 0. Solving the quadratic gives x = 0.072 M.

Correct: (A)

Removing the product CO2 shifts the equilibrium toward the products, forming more CaO. Adding solids does not affect the equilibrium position, and decreasing the volume shifts the equilibrium toward fewer gas molecules (the reactants).

Correct: (B)

Kp = Kc(RT)^Δn, so Kp = Kc when Δn = 0. The reaction CO + H2O ⇌ CO2 + H2 has 2 mol of gas on each side, so Δn = 0.

Correct: (C)

For CaF2(s) ⇌ Ca^2+(aq) + 2 F-(aq), Ksp = (s)(2s)^2 = 4s^3. So s = (3.9 × 10^-11 / 4)^1/3 = 2.1 × 10^-4 M, and [Ca^2+] = s.

Correct: (D)

The Cl- in NaCl is a common ion. Adding it shifts the dissolution equilibrium PbCl2(s) ⇌ Pb^2+(aq) + 2 Cl-(aq) toward the solid, decreasing the solubility.

Correct: (A)

Pure solids do not appear in the equilibrium expression, so Kc = [H2O]^4 / [H2]^4, with each gas raised to its stoichiometric coefficient.

Correct: (D)

NaOH is a strong base, so [OH-] = 0.050 M. pOH = −log(0.050) = 1.30, and pH = 14.00 − 1.30 = 12.70.

Correct: (B)

[OH-] = sqrt(Kb × C) = sqrt(1.8 × 10^-5 × 0.10) = 1.34 × 10^-3 M. Then pOH = 2.87 and pH = 14.00 − 2.87 = 11.13.

Correct: (A)

pH = pKa + log([base]/[acid]) = 3.17 + log(0.15/0.25) = 3.17 − 0.22 = 2.95.

Correct: (C)

The first ionization produces HCO3-, while the second ionization is far weaker (Ka2 is much smaller than Ka1), so HCO3- is far more concentrated than CO3^2-.

Correct: (D)

Half of the acid has been neutralized, leaving 0.00125 mol of H+ in a total volume of 37.5 mL, so [H+] = 0.0333 M and pH = −log(0.0333) = 1.48.

Correct: (A)

A conjugate acid is formed by adding one H+ to the base. Adding H+ to HCO3- gives H2CO3.

Correct: (B)

ΔG° = −RT ln K = −(8.314)(298) ln(1.0 × 10^5) = −(2478)(11.51) ≈ −28,500 J = −28.5 kJ.

Correct: (C)

ΔG = ΔH − TΔS. With ΔH positive and −TΔS also positive, ΔG is positive at every temperature, so the reaction is never spontaneous.

Correct: (A)

The strongest oxidizing agent is the species most easily reduced, which is the one with the most positive reduction potential. Ag+, with E° = +0.80 V, is reduced more readily than Ni^2+.

Correct: (D)

Q = It = (5.00)(1800 s) = 9000 C, which is 9000 / 96,485 = 0.0933 mol of electrons. Since Na+ + e- → Na, 0.0933 mol of Na forms, with a mass of 0.0933 × 22.99 = 2.14 g.

Correct: (B)

A negative E°cell means ΔG° = −nFE° is positive, so the reaction is nonspontaneous as written; the reverse reaction has E° = +0.40 V and is spontaneous.

Correct: (C)

Electrolysis drives a nonspontaneous reaction, one with ΔG > 0 (equivalently E°cell < 0), by supplying electrical work from an external power source.