AP Chemistry Practice Test 11

Correct: (B)

Iron is [Ar] 4s^2 3d^6. When Fe forms Fe^3+, the two 4s electrons are removed first, followed by one 3d electron, leaving [Ar] 3d^5.

Correct: (A)

The total number of electrons is 2 + 2 + 6 + 2 + 4 = 16, which is the atomic number of sulfur.

Correct: (D)

First ionization energy generally increases across a period because of increasing effective nuclear charge. Si (787 kJ/mol) is greater than Mg (738), Al (578), and Na (496).

Correct: (C)

(0.199)(10.013) + (0.801)(11.009) = 1.99 + 8.82 = 10.81 u, closest to 10.8 u.

Correct: (B)

(0.692)(62.93) + (0.308)(64.93) = 43.55 + 20.00 = 63.55 u, closest to 63.5 u (the element is copper).

Correct: (D)

All species have 18 electrons. S^2- has the fewest protons (16), so its electron cloud is held least tightly and it has the largest radius.

Correct: (A)

Copper is an exception to the expected filling order; a half-filled or completely filled d subshell is especially stable, so Cu is [Ar] 4s^1 3d^10 (29 electrons).

Correct: (C)

The double bond can be placed between carbon and any one of the three equivalent oxygen atoms, giving three equivalent resonance structures.

Correct: (D)

CO2 is linear, so the two equal and opposite bond dipoles cancel, giving a nonpolar molecule. The others have asymmetric geometries with net dipoles.

Correct: (B)

XeF4 has four bonding pairs and two lone pairs on xenon (AX4E2), giving an octahedral electron geometry and a square planar molecular geometry.

Correct: (C)

Nitrogen in NH3 has three bonding pairs and one lone pair (four electron domains), which corresponds to sp3 hybridization.

Correct: (A)

Lattice energy increases with greater ionic charges and smaller ionic radii. MgO has 2+ and 2- ions that are both relatively small, giving it by far the largest lattice energy.

Correct: (B)

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: (B)

With two S=O double bonds and two S-O single bonds, sulfur has 12 bonding electrons and no lone pairs: formal charge = 6 – 0 – 12/2 = 0.

Correct: (A)

CH3OH can form hydrogen bonds, the strongest intermolecular force among these substances, giving it the highest boiling point.

Correct: (C)

Ethanol is polar and hydrogen bonds with water, making it fully miscible. The others are nonpolar hydrocarbons or nonpolar molecules that are nearly insoluble in water.

Correct: (D)

36.0 g of H2O is 2.00 mol. Heat required = (2.00 mol)(40.7 kJ/mol) = 81.4 kJ.

Correct: (B)

HCl is a polar molecule, so dipole-dipole forces operate. It does not exhibit significant hydrogen bonding, which requires H bonded to N, O, or F.

Correct: (A)

SiO2 (quartz) consists of silicon and oxygen atoms joined by covalent bonds in a continuous three-dimensional network. NaCl is ionic, I2 is molecular, and Fe is metallic.

Correct: (C)

Diethyl ether has the weakest intermolecular forces of the four liquids (no hydrogen bonding between its molecules), so its molecules escape most easily and it has the highest vapor pressure.

Correct: (D)

Producing NH3 from 2.0 mol N2 would require 6.0 mol H2, but only 4.0 mol is available, so H2 is limiting. NH3 formed = 4.0 mol H2 × (2 mol NH3 / 3 mol H2) = 2.67 mol.

Correct: (B)

Percent yield = (36.0 g / 45.0 g) × 100% = 80.0%.

Correct: (A)

Ba^2+ and SO4^2- combine to form the insoluble solid BaSO4. Na+ and Cl- are spectator ions and do not appear in the net ionic equation.

Correct: (C)

With K at +1 and O at -2: 2(+1) + 2(x) + 7(-2) = 0, so 2x = +12 and x = +6.

Correct: (D)

16.0 g of CH4 is 1.00 mol. It requires 2.00 mol of O2, and 2.00 mol × 32.0 g/mol = 64.0 g.

Correct: (B)

Moles of HCl = (0.100 M)(0.02500 L) = 0.00250 mol, which equals the moles of NaOH. Molarity = 0.00250 mol / 0.02000 L = 0.125 M.

Correct: (A)

The balanced equation is C3H8 + 5 O2 → 3 CO2 + 4 H2O, so the coefficient of O2 is 5.

Correct: (C)

Doubling [A] doubles the rate (first order in A), and doubling [B] quadruples the rate (second order in B), so Rate = k[A][B]^2.

Correct: (D)

For a first-order reaction, k = 0.693 / t(1/2) = 0.693 / 20.0 s = 0.0347 s^-1.

Correct: (A)

The rate law is determined by the slow, rate-determining step (Step 1), which involves two NO2 molecules: Rate = k[NO2]^2.

Correct: (B)

A catalyst speeds up a reaction by providing an alternative mechanism with a lower activation energy. It does not change ΔH or the equilibrium position.

Correct: (C)

Rate (M/s) = k[A]^2 (M^2), so k must have units of M^-1 s^-1 for the units to balance.

Correct: (A)

At higher temperature, the average kinetic energy increases, so a much larger fraction of collisions possess at least the activation energy needed to react.

Correct: (D)

Using the Arrhenius equation, ln(k2/k1) = (50000/8.314)(1/300 – 1/310) = 0.647, so k2/k1 = e^0.647 ≈ 1.91 and k2 ≈ 1.9 × 10^-3 s^-1.

Correct: (B)

q = mcΔT = (50.0 g)(4.184 J/g·°C)(20.0°C) = 4184 J = 4.18 kJ.

Correct: (C)

By Hess's law, subtract the second equation from the first: ΔH = -393.5 – (-283.0) = -110.5 kJ.

Correct: (A)

ΔH = (bonds broken) – (bonds formed) = (436 + 242) – 2(431) = 678 – 862 = -184 kJ.

Correct: (D)

A formation reaction produces one mole of the compound from its elements in their standard states. Carbon's standard state is graphite, not diamond.

Correct: (B)

Sublimation converts an ordered solid into a gas, greatly increasing the disorder (entropy) of the system. The other processes all decrease entropy.

Correct: (C)

ΔH = [(-393.5) + 2(-285.8)] – [(-74.8) + 0] = -965.1 + 74.8 = -890.3 kJ, approximately -890 kJ.

Correct: (A)

q = mcΔT = (200.0 g)(4.18 J/g·°C)(10.0°C) = 8360 J = 8.36 kJ.

Correct: (D)

Kc equals the product concentrations raised to their coefficients divided by the reactant concentrations raised to their coefficients: [NH3]^2 / ([N2][H2]^3).

Correct: (B)

Setting up an ICE table: (2x)^2/(0.10 – x)^2 = 49. Taking the square root gives 2x/(0.10 – x) = 7, so x = 0.078 M and [HI] = 2x = 0.16 M.

Correct: (C)

Because the forward reaction is endothermic, increasing the temperature shifts the equilibrium to the right, producing more NO2. Decreasing volume shifts toward the side with fewer gas moles (left).

Correct: (A)

For AgCl, Ksp = s^2, so s = sqrt(1.8 × 10^-10) = 1.3 × 10^-5 M.

Correct: (D)

If Q < K, the mixture has too little product relative to equilibrium, so the net reaction proceeds forward until Q = K.

Correct: (B)

The common ion Cl- suppresses the solubility. Ksp = s(0.10 + s) ≈ 0.10s, so s = 1.8 × 10^-10 / 0.10 = 1.8 × 10^-9 M.

Correct: (C)

Kp = Kc(RT)^Δn, where Δn = 2 – 3 = -1. Since RT (about 24.5) is greater than 1 and Δn is negative, Kp = Kc/24.5, which is less than Kc.

Correct: (A)

HCl is a strong acid that dissociates completely, so [H+] = 0.010 M and pH = -log(0.010) = 2.00.

Correct: (D)

[H+] = sqrt(Ka × C) = sqrt(6.8 × 10^-4 × 0.10) = sqrt(6.8 × 10^-5) = 8.2 × 10^-3 M, so pH = 2.08.

Correct: (B)

A buffer requires a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts. CH3COOH/CH3COO- is such a pair.

Correct: (C)

A conjugate base is formed by removing one proton (H+) from the acid: H2PO4- – H+ → HPO4^2-.

Correct: (A)

At the equivalence point, the solution contains NH4+, the conjugate acid of the weak base NH3, which hydrolyzes to give an acidic solution (pH < 7).

Correct: (D)

pH = pKa + log([base]/[acid]) = 4.74 + log(0.10/0.20) = 4.74 – 0.30 = 4.44.

Correct: (C)

ΔG = ΔH – TΔS = -100 kJ/mol – (298 K)(-0.200 kJ/mol·K) = -100 + 59.6 = -40.4 kJ/mol, so the reaction is spontaneous.

Correct: (A)

E°cell = E°(cathode) – E°(anode) = 0.34 – (-0.76) = +1.10 V. The positive value confirms the reaction is spontaneous as written.

Correct: (B)

Charge = (0.50 A)(1800 s) = 900 C. Moles of electrons = 900/96,485 = 0.00933 mol, giving 0.00933 mol Ag × 107.9 g/mol = 1.0 g.

Correct: (B)

With ΔH negative and ΔS positive, ΔG = ΔH – TΔS is negative at every temperature, so the reaction is always spontaneous.

Correct: (C)

A gas is produced from a solid, so the disorder of the system increases and ΔS is positive.

Correct: (A)

Reduction always occurs at the cathode and oxidation always occurs at the anode, regardless of the type of cell. (In a galvanic cell the anode is negative, and E°cell is positive.)