AP Chemistry Practice Test 16

Correct: (B)

Copper is an exception to the simple filling order: a completely filled 3d subshell (3d10) is especially stable, giving [Ar] 4s1 3d10.

Correct: (C)

Average mass = (0.790)(24) + (0.100)(25) + (0.110)(26) = 18.96 + 2.50 + 2.86 = 24.3 u.

Correct: (A)

Radius increases down a group and decreases across a period; fluorine is in the top row and farthest right among these choices, so it is the smallest.

Correct: (D)

Removing a second electron from Na+ means breaking into a stable noble-gas (neon) core, which requires a very large amount of energy.

Correct: (D)

The peaks correspond to 1s2 2s2 2p4, a total of 8 electrons, which is oxygen.

Correct: (A)

All have 10 electrons; Al3+ has the most protons (13) pulling on them, so its radius is smallest.

Correct: (B)

Across a period, protons are added while the added electrons enter the same shell and shield one another poorly, so Zeff felt by valence electrons increases.

Correct: (A)

Formal charge = 5 valence electrons − (0 nonbonding electrons + 4 bonding electrons) = +1.

Correct: (B)

Xenon has six electron domains (four bonds and two lone pairs); the lone pairs sit opposite each other, leaving the four fluorines in a square planar arrangement.

Correct: (C)

Boron in BF3 has three electron domains and no lone pairs (trigonal planar), which corresponds to sp2 hybridization.

Correct: (B)

All contain singly charged ions, so size decides: Li+ and F− are the smallest pair, giving LiF the largest lattice energy.

Correct: (C)

O2 has a bond order of 2; removing one electron from an antibonding pi* orbital to form O2+ raises the bond order to (10 − 5)/2 = 2.5.

Correct: (D)

Bond length decreases as bond order increases: single (C2H6) > benzene (1.5) > double (C2H4) > triple (C2H2), so the triple bond in C2H2 is shortest.

Correct: (B)

The double bond can be drawn to either of the two equivalent oxygen atoms, giving two resonance structures.

Correct: (A)

Although HF has the lowest molar mass, its strong hydrogen bonding gives it a much higher boiling point than the other hydrogen halides, which interact mainly through dipole and dispersion forces.

Correct: (B)

CH3OCH3 is polar but has no H bonded to O, N, or F, so it cannot hydrogen bond; dipole-dipole attractions are its strongest intermolecular force.

Correct: (B)

Nonpolar I2 dissolves readily in a nonpolar solvent ("like dissolves like"), whereas the ionic and polar substances listed do not.

Correct: (D)

Moles of water = 18.0 g / 18.0 g/mol = 1.00 mol; q = (1.00 mol)(40.7 kJ/mol) = 40.7 kJ.

Correct: (C)

Glycerol has three −OH groups per molecule and forms an extensive hydrogen-bonding network, making it far more viscous than the other liquids.

Correct: (D)

Dry ice consists of discrete CO2 molecules held together by weak intermolecular forces, the definition of a molecular solid.

Correct: (D)

The balanced equation is C2H5OH + 3 O2 → 2 CO2 + 3 H2O, so the coefficient of O2 is 3.

Correct: (C)

Only Pb2+ and I− combine to form the solid (in the correct 1:2 ratio); K+ and NO3− are spectators, and option (D) is the molecular equation.

Correct: (B)

With H at +1 and O at −2: 2(+1) + S + 4(−2) = 0 gives S = +6.

Correct: (C)

Moles CaCO3 = 50.0 g / 100 g/mol = 0.500 mol, and the 1:1 mole ratio gives 0.500 mol CO2.

Correct: (D)

Cl2 is limiting: 0.400 mol Al would require 0.600 mol Cl2, but only 0.300 mol is present. AlCl3 formed = 0.300 mol Cl2 × (2/3) = 0.200 mol; mass = (0.200)(133.3) = 26.7 g.

Correct: (C)

Percent yield = (18.0 / 24.0) × 100% = 75.0%.

Correct: (B)

Moles KOH = (0.150)(0.02000) = 0.00300 mol; since H2SO4 + 2 KOH → products, moles H2SO4 = 0.00150 mol; [H2SO4] = 0.00150 / 0.02500 = 0.0600 M.

Correct: (C)

Doubling [B] (Exp. 1→2) does not change the rate (zero order in B); doubling [A] (Exp. 1→3) quadruples the rate (second order in A).

Correct: (B)

For second order, 1/[A] = kt + 1/[A]0: 1/0.10 − 1/0.40 = 10 − 2.5 = 7.5 = (0.050)t, so t = 150 s.

Correct: (A)

For first order, rate (M/s) = k[A] (M), so k has units of s^-1.

Correct: (C)

An intermediate is produced in one step and consumed in a later step; C is formed in Step 1 and used up in Step 2.

Correct: (B)

ln(k2/k1) = (Ea/R)(1/300 − 1/310) = (75,000/8.314)(1.08 × 10^-4) = 0.97, so k2/k1 = e^0.97 ≈ 2.6.

Correct: (D)

A higher concentration means more molecules per unit volume, so collisions occur more frequently and the rate increases.

Correct: (C)

A catalyst lowers the activation energy for both directions equally, so equilibrium is reached faster but neither K nor the equilibrium composition changes.

Correct: (A)

q = mcΔT = (200.0 g)(4.18 J/g·°C)(35.0°C) = 29,260 J ≈ 29.3 kJ released.

Correct: (D)

Adding the two reactions (the NO cancels) gives the target reaction, so ΔH = +180.5 + (−114.1) = +66.4 kJ.

Correct: (C)

Bonds broken: 941 + 3(436) = 2249 kJ; bonds formed: 6(391) = 2346 kJ (each NH3 has three N−H bonds); ΔH = 2249 − 2346 = −97 kJ.

Correct: (B)

ΔH = ΔHf°(N2O4) − 2(ΔHf° NO2) = 9.2 − 2(33.2) = −57.2 kJ.

Correct: (D)

Three moles of gas are converted into a liquid, a large decrease in disorder, so ΔS < 0; the other choices increase disorder.

Correct: (A)

Heat gained by water = (100.0)(4.18)(2.0) = 836 J. Heat lost by metal = (100.0)(c)(78.0) = 836 J, so c = 836 / 7800 = 0.107 J/g·°C.

Correct: (C)

ΔHf° = 0 for an element in its standard state; bromine's standard state is the liquid, whereas O3, diamond, and atomic H are not the standard forms of their elements.

Correct: (A)

Δn = 2 − 2 = 0 for this reaction, so Kp = Kc(RT)^0 = Kc = 50.

Correct: (B)

Solve x^2 / (0.20 − x) = 0.040. The quadratic x^2 + 0.040x − 0.0080 = 0 gives x = 0.072 M (the small-x approximation, 0.089 M, is not valid here since x is a large fraction of 0.20).

Correct: (C)

Decreasing the volume raises the pressure, so the equilibrium shifts toward the side with fewer moles of gas; SO3 (2 mol) is favored over SO2 + O2 (3 mol). Kc does not change with volume.

Correct: (D)

For PbI2, Ksp = [Pb2+][I−]^2 = (s)(2s)^2 = 4s^3 = 7.1 × 10^-9, so s = (1.78 × 10^-9)^(1/3) ≈ 1.2 × 10^-3 M.

Correct: (A)

When Q < K, reactants are in excess relative to equilibrium, so the net reaction proceeds in the forward direction (toward products) until Q = K.

Correct: (D)

With [Ca2+] ≈ 0.10 M from the common ion, Ksp = (0.10)(2s)^2 = 0.40s^2 = 3.9 × 10^-11, so s = sqrt(9.75 × 10^-11) ≈ 9.9 × 10^-6 M.

Correct: (C)

Reversing the reaction takes the reciprocal (1/4) and doubling it squares the value: (1/4)^2 = 1/16 = 0.0625.

Correct: (A)

NaOH is a strong base, so [OH−] = 0.010 M, pOH = 2.00, and pH = 14.00 − 2.00 = 12.00.

Correct: (D)

[OH−] ≈ sqrt(Kb × C) = sqrt((1.8 × 10^-5)(0.10)) = 1.3 × 10^-3 M; pOH = 2.87, so pH = 14.00 − 2.87 = 11.13.

Correct: (A)

Henderson-Hasselbalch: pH = 3.74 + log(0.10/0.15) = 3.74 − 0.18 = 3.56.

Correct: (C)

Adding a proton to HPO4^2− gives its conjugate acid, H2PO4−.

Correct: (B)

At the half-equivalence point, [HA] = [A−], so pH = pKa = 4.74, giving Ka = 10^-4.74 ≈ 1.8 × 10^-5.

Correct: (A)

HCO3− can donate a proton (forming CO3^2−) or accept a proton (forming H2CO3), so it is amphoteric.

Correct: (A)

ΔG° = −nFE° = −(2)(96,485)(0.80) = −154,000 J ≈ −154 kJ.

Correct: (D)

Charge = (1.50 A)(1932 s) = 2898 C = 0.0300 mol e−. Since Ag+ + e− → Ag, 0.0300 mol Ag forms; mass = (0.0300)(107.87) = 3.24 g.

Correct: (B)

Zn is oxidized at the anode, releasing electrons that travel through the external circuit to the cathode, where Cu2+ is reduced; salt-bridge anions migrate toward the anode compartment.

Correct: (A)

ΔG = ΔH − TΔS; with both ΔH and ΔS negative, the unfavorable −TΔS term is small only at low T, so the process is spontaneous at low temperatures.

Correct: (D)

A negative E° means ΔG° = −nFE° is positive, so the reaction is nonspontaneous as written and its equilibrium constant is less than 1.

Correct: (C)

ΔS° = 2(192.5) − [191.5 + 3(130.7)] = 385.0 − 583.6 = −198.6 J/K, consistent with a decrease from 4 mol of gas to 2 mol of gas.