AP Chemistry Practice Test 12

Correct: (C)

Chromium is an exception to the expected filling order; the half-filled 3d subshell is especially stable, so Cr (24 electrons) is [Ar] 4s^1 3d^5.

Correct: (D)

The total number of electrons is 2 + 2 + 6 + 2 = 12, which is the atomic number of magnesium.

Correct: (A)

K is farthest down and to the left in the periodic table among these elements; its single 4s electron is well shielded, so it has the smallest first ionization energy (419 kJ/mol).

Correct: (B)

(0.7899)(23.985) + (0.1000)(24.986) + (0.1101)(25.983) = 18.95 + 2.50 + 2.86 = 24.31 u, closest to 24.3 u.

Correct: (D)

(0.75)(35) + (0.25)(37) = 35.5 u, which matches chlorine. (Bromine's isotopes are near 79 and 81 u, giving an average near 80 u.)

Correct: (C)

All have 10 electrons. Mg^2+ has the most protons (12), which pull the electron cloud in most strongly, giving it the smallest radius.

Correct: (B)

Across a period, electrons are added to the same shell while the nuclear charge increases, so the effective nuclear charge rises and the electron cloud is drawn inward.

Correct: (A)

Nitrogen has 5 valence electrons; three are used in N-H bonds, leaving one lone pair on nitrogen.

Correct: (C)

The nitrate ion has a double bond that can be drawn to any of three equivalent oxygen atoms, giving three resonance structures. The others have only one valid Lewis structure.

Correct: (D)

PCl5 has five bonding pairs and no lone pairs on phosphorus (AX5), giving a trigonal bipyramidal geometry.

Correct: (B)

Carbon in CO2 has two electron domains (two double bonds) arranged linearly, which corresponds to sp hybridization.

Correct: (C)

All four contain singly charged ions, so lattice energy is governed by ionic size. NaF has the smallest cation-anion pair, giving it the largest lattice energy.

Correct: (A)

Bond length decreases as bond order increases because more shared electrons pull the nuclei closer together. A triple bond is shorter than a double bond, which is shorter than a single bond.

Correct: (D)

Oxygen in H3O+ has one lone pair (2 nonbonding electrons) and three bonds (6 bonding electrons): formal charge = 6 – 2 – 6/2 = +1.

Correct: (B)

Based on molar mass alone H2O would have the lowest boiling point of the group, but hydrogen bonding between water molecules makes its boiling point by far the highest.

Correct: (A)

F2 is a nonpolar molecule, so only London dispersion forces operate. HF and NH3 hydrogen bond, and CH3Cl is polar (dipole-dipole forces).

Correct: (C)

18.0 g of H2O is 1.00 mol. Condensation is the reverse of vaporization, so heat released = (1.00 mol)(40.7 kJ/mol) = 40.7 kJ.

Correct: (D)

I2 is nonpolar, so by the "like dissolves like" principle it dissolves best in the nonpolar solvent CCl4 rather than in polar or aqueous solvents.

Correct: (B)

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

Correct: (A)

Ion-dipole forces are the attractions between an ion and the partial charges of a polar molecule, such as Na+ and the oxygen ends of surrounding water molecules.

Correct: (C)

Reacting 0.30 mol of Cl2 consumes only 0.20 mol of Al (ratio 3 Cl2 : 2 Al), so Cl2 is limiting and 0.20 mol of AlCl3 is produced.

Correct: (D)

Percent yield = (27.0 g / 36.0 g) × 100% = 75.0%.

Correct: (B)

Ag+ and Cl- combine to form the insoluble solid AgCl. Na+ and NO3- remain in solution as spectator ions.

Correct: (A)

With H at +1 and O at -2: 2(+1) + x + 4(-2) = 0, so x = +6.

Correct: (C)

44.0 g of C3H8 is 1.00 mol, which produces 3.00 mol of CO2. Mass = 3.00 mol × 44.0 g/mol = 132 g.

Correct: (D)

Moles of NaOH = (0.150 M)(0.03000 L) = 0.00450 mol, which equals the moles of HCl. Molarity = 0.00450 mol / 0.02500 L = 0.180 M.

Correct: (B)

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

Correct: (A)

Doubling [X] leaves the rate unchanged (zero order in X), while doubling [Y] quadruples the rate (second order in Y), so Rate = k[Y]^2.

Correct: (C)

Each half-life halves the amount: after three half-lives the fraction remaining is (1/2)^3 = 1/8 = 12.5%.

Correct: (D)

The rate law comes from the slow, rate-determining step (Step 1): Rate = k[A2][B]. The intermediate A2B cannot appear in the final rate law.

Correct: (B)

The reverse reaction goes from products (50 kJ/mol) to the transition state (100 kJ/mol), so its activation energy is 100 – 50 = 50 kJ/mol.

Correct: (A)

Rate (M/s) = k[A] (M), so k must have units of s^-1.

Correct: (C)

A catalyst lowers the activation energy of both the forward and reverse reactions equally; it changes neither ΔH nor K, and equilibrium is simply reached faster.

Correct: (D)

Average rate = (0.80 – 0.20) M / 30.0 s = 0.60 M / 30.0 s = 0.020 M/s.

Correct: (A)

q = mcΔT = (200.0 g)(4.184 J/g·°C)(20.0°C) = 16,736 J ≈ 16.7 kJ released.

Correct: (B)

By Hess's law, subtract the first equation from the second: ΔH = 66.4 – 180.5 = -114.1 kJ.

Correct: (D)

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

Correct: (C)

A formation reaction produces exactly one mole of the compound from its elements in their standard states, which is what this equation shows.

Correct: (A)

Entropy increases from solid to liquid to gas because particles have more freedom of motion and more accessible arrangements. A gas such as H2O(g) has the greatest molar entropy.

Correct: (B)

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

Correct: (D)

Heat gained by water = (100.0)(4.184)(5.0) = 2092 J. Heat lost by metal = (100.0)(c)(70.0), so c = 2092 / 7000 = 0.30 J/g·°C.

Correct: (C)

Kc = [PCl3][Cl2] / [PCl5] = (0.40)(0.40) / 0.20 = 0.80.

Correct: (A)

From an ICE table, (2x)^2/(0.50 – x) = 0.36, which gives 4x^2 + 0.36x – 0.18 = 0 and x = 0.17 M. Therefore [NO2] = 2x = 0.34 M.

Correct: (B)

For an exothermic reaction, decreasing the temperature favors the heat-producing (forward) direction, shifting equilibrium toward products. Increasing volume would shift toward the side with more gas moles (reactants).

Correct: (D)

For CaF2, Ksp = [Ca^2+][F-]^2 = (s)(2s)^2 = 4s^3. So s = (3.9 × 10^-11 / 4)^(1/3) = (9.75 × 10^-12)^(1/3) ≈ 2.1 × 10^-4 M.

Correct: (C)

If Q > K, there is too much product relative to equilibrium, so the net reaction proceeds in the reverse direction (toward reactants) until Q = K.

Correct: (A)

Adding the common ion CrO4^2- shifts the dissolution equilibrium toward the solid, decreasing the solubility of Ag2CrO4. Ksp itself is unchanged at constant temperature.

Correct: (B)

Kp = Kc(RT)^Δn. Here Δn = 2 – 2 = 0, so (RT)^0 = 1 and Kp = Kc.

Correct: (D)

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

Correct: (C)

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

Correct: (A)

pOH = pKb + log([acid]/[base]) = 4.74 + log(0.20/0.30) = 4.74 – 0.18 = 4.56, so pH = 14.00 – 4.56 = 9.44.

Correct: (B)

A conjugate acid is formed by adding one proton (H+): CO3^2- + H+ → HCO3-.

Correct: (D)

At the half-equivalence point, [weak acid] = [conjugate base], so the Henderson-Hasselbalch equation gives pH = pKa + log(1) = pKa.

Correct: (C)

For a diprotic acid with Ka1 much greater than Ka2, the second ionization is so limited that [S^2-] is approximately equal to Ka2, about 1 × 10^-19 M.

Correct: (B)

ΔG = ΔH – TΔS = 50 kJ/mol – (298 K)(0.200 kJ/mol·K) = 50 – 59.6 = -9.6 kJ/mol, so the reaction is spontaneous at 298 K.

Correct: (D)

E°cell = E°(cathode) – E°(anode) = 0.80 – (-0.25) = +1.05 V. (The coefficient of 2 does not multiply the potential, since E° is an intensive property.)

Correct: (A)

Charge = (1.50 A)(7200 s) = 10,800 C. Moles of e- = 10,800/96,485 = 0.112 mol, so moles of Cu = 0.0560 mol and mass = 0.0560 × 63.55 = 3.56 g.

Correct: (C)

Three moles of gas are converted into a liquid, so the disorder of the system decreases greatly and ΔS is negative.

Correct: (D)

Oxidation always occurs at the anode. In a galvanic cell the anode is the negative electrode, and electrons flow from the anode to the cathode through the external circuit.

Correct: (B)

The reaction becomes spontaneous when TΔS exceeds ΔH. The crossover temperature is ΔH/ΔS = 120,000 J/mol / 400 J/mol·K = 300 K, so the reaction is spontaneous above 300 K.