AP Chemistry Practice Test 15

Correct: (B)

Chromium is an exception to the expected filling order; a half-filled 3d subshell (3d5) with a half-filled 4s gives extra stability, so the configuration is [Ar] 4s1 3d5.

Correct: (C)

Average mass = (0.692)(63) + (0.308)(65) = 43.6 + 20.0 = 63.6 u, consistent with copper.

Correct: (A)

Atomic radius decreases across a period as effective nuclear charge increases, so potassium, farthest left in Period 4, is the largest.

Correct: (C)

The very large jump between the fourth and fifth ionization energies shows that four electrons are removed before a stable noble-gas core is reached, so the element has 4 valence electrons (Group 14, silicon).

Correct: (D)

For isoelectronic species, radius decreases as nuclear charge increases. O2− has the fewest protons holding the same number of electrons, so it is the largest.

Correct: (B)

The peaks correspond to 1s2 2s2 2p1, a total of 5 electrons, which is boron.

Correct: (D)

Ionization energy generally increases across a period; chlorine is farthest right among the choices and holds its valence electrons most tightly.

Correct: (B)

Formal charge = 6 valence electrons − (6 nonbonding electrons + 1 bonding electron) = −1.

Correct: (A)

Sulfur has five electron domains (four bonds and one lone pair) in a trigonal-bipyramidal arrangement; the lone pair occupies an equatorial position, giving a seesaw shape.

Correct: (C)

Nitrogen in ammonia has four electron domains (three bonds and one lone pair), which corresponds to sp3 hybridization and a trigonal-pyramidal geometry.

Correct: (A)

Lattice energy increases with ion charge and decreases with ion size. MgO combines 2+ and 2− charges with small ions, giving the largest lattice energy.

Correct: (D)

The Lewis structure of CN− contains a carbon-nitrogen triple bond, so the bond order is 3.

Correct: (B)

Each nitrogen atom in N2 shares three pairs of electrons to complete its octet, forming a triple bond.

Correct: (C)

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

Correct: (B)

Water forms an extensive three-dimensional hydrogen-bonding network (two O−H bonds and two lone pairs per molecule), giving it a higher boiling point than HF or NH3.

Correct: (C)

I2 is nonpolar, so the only attractions between its molecules are London dispersion forces arising from temporary dipoles.

Correct: (D)

"Like dissolves like": nonpolar I2 dissolves best in a nonpolar solvent; the ionic and strongly polar substances do not.

Correct: (C)

Moles of ice = 36.0 g / 18.0 g/mol = 2.00 mol; q = (2.00 mol)(6.01 kJ/mol) = 12.0 kJ.

Correct: (B)

At higher temperature the distribution of molecular energies shifts, so more molecules can overcome intermolecular attractions and enter the vapor phase.

Correct: (D)

In quartz, silicon and oxygen atoms are joined by covalent bonds extending throughout the entire crystal, making it a covalent network solid.

Correct: (C)

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

Correct: (A)

Only Ag+ and Cl− change form, producing solid AgCl; Na+ and NO3− are spectator ions, and option (D) is the molecular equation, not the net ionic equation.

Correct: (B)

Zinc loses electrons (its oxidation number rises from 0 to +2), so Zn is oxidized; Cu2+ is reduced.

Correct: (C)

The mole ratio is 2 mol H2O per 1 mol O2, so 3.0 mol O2 produces 6.0 mol H2O.

Correct: (B)

H2 is the limiting reactant (4.0 mol H2 would require only 1.33 mol N2, but 2.0 mol N2 would require 6.0 mol H2). NH3 formed = 4.0 mol H2 × (2 mol NH3 / 3 mol H2) = 2.67 mol.

Correct: (D)

Percent yield = (actual / theoretical) × 100% = (20.0 / 25.0) × 100% = 80.0%.

Correct: (C)

At equivalence, moles HCl = moles NaOH = (0.100 mol/L)(0.02000 L) = 0.00200 mol; [HCl] = 0.00200 mol / 0.02500 L = 0.0800 M.

Correct: (D)

Doubling [A] (Exp. 1→2) doubles the rate, so the reaction is first order in A; doubling [B] (Exp. 1→3) quadruples the rate, so it is second order in B.

Correct: (A)

Falling to 1/8 requires three half-lives (1 → 1/2 → 1/4 → 1/8), so t = 3 × 15 min = 45 min.

Correct: (A)

For a second-order reaction, rate (M/s) = k[A]^2 (M^2), so k must have units of M^-1 s^-1.

Correct: (D)

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

Correct: (C)

From ln(k2/k1) = (Ea/R)(1/T1 − 1/T2): ln 4 = (Ea/8.314)(1/300 − 1/320) = (Ea/8.314)(2.08 × 10^-4), giving Ea ≈ 5.5 × 10^4 J/mol = 55 kJ/mol.

Correct: (B)

Raising the temperature shifts the energy distribution so that a much larger fraction of collisions can overcome the activation-energy barrier; the activation energy itself does not change.

Correct: (D)

A catalyst speeds up a reaction by providing a different pathway of lower activation energy; it is regenerated, and it changes neither ΔH nor the equilibrium yield.

Correct: (B)

q = mcΔT = (100.0 g)(4.18 J/g·°C)(5.0°C) = 2090 J = 2.09 kJ.

Correct: (A)

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

Correct: (B)

ΔH = bonds broken − bonds formed = (436 + 193) − 2(366) = 629 − 732 = −103 kJ.

Correct: (C)

ΔH = [−393.5 + 2(−285.8)] − [−74.8 + 0] = −965.1 + 74.8 = −890 kJ/mol.

Correct: (C)

Sublimation converts an ordered solid into a highly disordered gas, so entropy increases; the other processes decrease disorder or reduce the moles of gas.

Correct: (B)

c = q / (mΔT) = 1250 J / [(50.0 g)(25.0°C)] = 1.00 J/g·°C.

Correct: (D)

Heat is absorbed from the water (making it colder) to dissolve the solid, so the process is endothermic with a positive enthalpy of solution.

Correct: (A)

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

Correct: (D)

Let x be the amount reacted: (2x)^2 / (0.50 − x)^2 = 49, so 2x / (0.50 − x) = 7, giving x = 0.39 M and [HI] = 2x = 0.78 M.

Correct: (D)

For an endothermic reaction, added heat favors the forward direction (toward NO and O2), and Kc increases with temperature for an endothermic reaction; option (A) is incorrect because heating does not favor NO2 formation.

Correct: (D)

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

Correct: (A)

When 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: (D)

The common ion Cl− (0.10 M) suppresses solubility: Ksp = [Ag+][Cl−] = s(0.10), so s = 1.8 × 10^-10 / 0.10 = 1.8 × 10^-9 M.

Correct: (A)

Δn = 2 − 3 = −1, so Kp = Kc(RT)^−1 = 420 / [(0.0821)(298)] = 420 / 24.5 ≈ 17.

Correct: (B)

HCl is a strong acid, so [H+] = 0.050 M and pH = −log(0.050) = 1.30.

Correct: (C)

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

Correct: (A)

Henderson-Hasselbalch: pH = pKa + log([base]/[acid]) = 4.74 + log(0.10/0.20) = 4.74 − 0.30 = 4.44.

Correct: (D)

Removing a proton from HCO3− gives its conjugate base, CO3^2−.

Correct: (B)

At equivalence, all the weak acid has been converted to its conjugate base, which hydrolyzes (reacts with water) to generate OH−, making the solution basic.

Correct: (C)

Kb = Kw / Ka = (1.0 × 10^-14) / (4.9 × 10^-10) = 2.0 × 10^-5.

Correct: (B)

ΔG = ΔH − TΔS = −100 kJ − (298)(−0.200 kJ/K) = −100 + 59.6 = −40.4 kJ/mol; a negative ΔG means the reaction is spontaneous.

Correct: (A)

ΔG = ΔH − TΔS; with both terms positive, the −TΔS term dominates only at high T, making ΔG negative there.

Correct: (D)

E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V.

Correct: (C)

Charge = (2.00)(965) = 1930 C = 0.0200 mol e−. Cu2+ + 2e− → Cu, so 0.0100 mol Cu forms; mass = (0.0100)(63.55) = 0.635 g.

Correct: (A)

ΔG° = −nFE° = −(2)(96,485)(1.10) = −212,000 J ≈ −212 kJ.

Correct: (D)

Reduction always occurs at the cathode. In this cell Cu2+ is reduced to Cu; Zn is oxidized at the anode.