AP Chemistry Practice Test 19

Correct: (C)

Neutral Fe is [Ar] 4s^2 3d^6. When iron ionizes, the 4s electrons are removed first, so Fe^3+ loses both 4s electrons and one 3d electron, giving [Ar] 3d^5.

Correct: (B)

The peaks account for 2 + 2 + 6 + 2 = 12 electrons, so the element has atomic number 12, which is magnesium (1s^2 2s^2 2p^6 3s^2).

Correct: (A)

Atomic radius decreases across a period because increasing effective nuclear charge pulls the electron cloud inward. Na is farthest left in Period 3, so it has the largest radius.

Correct: (D)

(0.6915)(62.93) + (0.3085)(64.93) = 43.52 + 20.03 = 63.55 amu, identifying the element as copper.

Correct: (C)

The very large jump between the third and fourth ionization energies shows that three valence electrons are removed easily before a core electron must be removed, indicating three valence electrons and Group 13.

Correct: (B)

Coulombic attraction increases with the product of the ionic charges. The +2 and −2 charges of Mg^2+ and O^2- give a charge product of 4, greater than for any of the singly charged pairs.

Correct: (D)

E = hc/λ = (6.626 × 10^-34)(3.00 × 10^8)/(4.00 × 10^-7) = 4.97 × 10^-19 J.

Correct: (A)

Xe has 8 valence electrons. Four are used in the four Xe-F bonds, leaving 4 electrons, which form 2 lone pairs on xenon (an AX4E2, square planar arrangement).

Correct: (C)

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

Correct: (B)

SF4 has five electron domains around sulfur (four bonding pairs and one lone pair), giving a trigonal bipyramidal electron geometry and a seesaw molecular geometry.

Correct: (D)

Carbon in CO2 has two electron domains (the two double bonds), so it uses sp hybrid orbitals and the molecule is linear.

Correct: (A)

MgO combines small ions with +2 and −2 charges, giving the strongest electrostatic attractions and therefore the greatest lattice energy of the compounds listed.

Correct: (C)

The extra electron in O2^- enters a pi antibonding orbital. This reduces the bond order of O2 (2.0) by 0.5, giving a bond order of 1.5.

Correct: (D)

CO2 has two C=O double bonds, and each double bond contains one pi bond, for a total of 2 pi bonds (along with 2 sigma bonds).

Correct: (B)

CH3OH molecules form hydrogen bonds, which are much stronger than the dipole-dipole and dispersion forces in the other substances, so methanol boils at the highest temperature.

Correct: (A)

Octane is a nonpolar hydrocarbon that cannot form favorable interactions with water, so it is essentially insoluble. The other substances are polar or hydrogen-bonding and dissolve readily.

Correct: (D)

18.0 g H2O corresponds to 1.00 mol, so q = (1.00 mol)(6.01 kJ/mol) = 6.01 kJ.

Correct: (C)

By Raoult's law, P = (mole fraction of water)(P°) = (9.00/10.00)(23.8 torr) = 21.4 torr. The nonvolatile glucose lowers the vapor pressure.

Correct: (B)

Quartz, SiO2, consists of an extended three-dimensional network of covalent bonds. NaCl is ionic, and ice and dry ice are molecular solids.

Correct: (A)

The largest ΔHvap means the most energy is required to separate the molecules into the gas phase, so H2O, with strong hydrogen bonding, has the strongest intermolecular forces.

Correct: (D)

C3H8 + 5 O2 → 3 CO2 + 4 H2O. Balancing carbon, then hydrogen, then oxygen gives an O2 coefficient of 5.

Correct: (C)

Na^+ and NO3^- are spectator ions and remain in solution. The only chemical change is Ag^+ combining with Cl^- to form solid AgCl.

Correct: (B)

44.0 g C3H8 is 1.00 mol. Each mole of C3H8 produces 3 mol of CO2, so 3.00 mol × 44.0 g/mol = 132 g of CO2 is produced.

Correct: (A)

The 0.200 mol of Al would require 0.300 mol of Cl2, but only 0.150 mol is available, so Cl2 is limiting. Product = 0.150 × (2/3) = 0.100 mol AlCl3 = 0.100 × 133 = 13.3 g.

Correct: (D)

Percent yield = (actual/theoretical) × 100% = (42.5/50.0) × 100% = 85.0%.

Correct: (C)

Moles of HCl = (0.0250 L)(0.100 M) = 2.50 × 10^-3 mol, which equals the moles of NaOH. Concentration = 2.50 × 10^-3/0.0200 L = 0.125 M.

Correct: (B)

With hydrogen at +1 and oxygen at −2, the sum must be zero for a neutral compound: 2(+1) + S + 4(−2) = 0, so S = +6.

Correct: (A)

Doubling [A] (Expt 1 to 2) quadruples the rate, so the reaction is second order in A. Doubling [B] (Expt 1 to 3) doubles the rate, so it is first order in B. Rate = k[A]^2[B].

Correct: (D)

For a first-order reaction, t1/2 = 0.693/k = 0.693/0.0693 = 10.0 s.

Correct: (C)

Using the second-order integrated rate law: 1/[A] = 1/[A]0 + kt = 1/0.80 + (0.50)(3.0) = 1.25 + 1.50 = 2.75, so [A] = 1/2.75 = 0.36 M.

Correct: (B)

The rate law is determined by the slow, rate-determining step. Step 1 involves two NO2 molecules, so Rate = k[NO2]^2. Species in the fast step do not appear.

Correct: (A)

From the Arrhenius equation, ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). With ln 2 = 0.693 and (1/300 − 1/310) = 1.08 × 10^-4 K^-1, Ea = (0.693)(8.314)/(1.08 × 10^-4) ≈ 5.4 × 10^4 J/mol = 54 kJ/mol.

Correct: (D)

A catalyst provides an alternative reaction pathway with a lower activation energy. It speeds up both the forward and reverse reactions equally, so ΔH, K, and the equilibrium position are unchanged, and it is regenerated rather than consumed.

Correct: (C)

Raising the temperature increases the average kinetic energy of the molecules, so a much larger fraction of collisions have enough energy to overcome the activation energy barrier and react.

Correct: (B)

q = mcΔT = (50.0 g)(4.18 J/g·°C)(8.0°C) = 1670 J = 1.67 kJ.

Correct: (A)

By Hess's law, subtract the second equation from the first: ΔH = −394 − (−283) = −111 kJ.

Correct: (D)

ΔH = (energy to break bonds) − (energy released forming bonds) = (436 + 242) − 2(431) = 678 − 862 = −184 kJ.

Correct: (C)

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

Correct: (B)

Entropy increases when the number of freely moving particles increases. Decomposing a solid to produce a gas greatly increases the disorder of the system, so ΔS > 0.

Correct: (A)

Heat gained by the water = (100.0)(4.18)(5.0) = 2090 J. This equals the heat lost by the metal: (100.0)(c)(70.0). Solving, c = 2090/7000 = 0.30 J/g·°C.

Correct: (D)

Moles of glucose = 2.00/180 = 0.0111 mol. Heat released = 0.0111 × 2808 = 31.2 kJ.

Correct: (C)

Kc is the ratio of product concentrations to reactant concentrations, each raised to the power of its stoichiometric coefficient.

Correct: (A)

Setting up an ICE table gives x^2/(1.00 − x) = 0.0211, where x = [Cl2] at equilibrium. Solving the quadratic (x^2 + 0.0211x − 0.0211 = 0) gives x = 0.135 M.

Correct: (B)

Decreasing the volume increases the pressure, and by Le Chatelier's principle the equilibrium shifts toward the side with fewer moles of gas. The product side has 2 mol of gas versus 4 mol on the reactant side, so more NH3 forms.

Correct: (D)

For AgCl, Ksp = s^2, so s = (1.8 × 10^-10)^1/2 = 1.3 × 10^-5 M.

Correct: (C)

Because Q > K, the mixture contains too much product relative to equilibrium, so the net reaction proceeds toward reactants until Q = K.

Correct: (B)

The common ion F^- from NaF makes [F^-] ≈ 0.10 M. Ksp = [Ca^2+][F^-]^2 = s(0.10)^2, so s = 3.9 × 10^-11/0.010 = 3.9 × 10^-9 M.

Correct: (A)

Kp = Kc(RT)^Δn = (2.0 × 10^3)(0.0821 × 500)^-2 = (2.0 × 10^3)/(41.05)^2 = 2000/1685 ≈ 1.19.

Correct: (D)

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

Correct: (C)

For a weak acid, [H^+] ≈ (Ka × C)^1/2 = (6.8 × 10^-4 × 0.10)^1/2 = 8.2 × 10^-3 M, so pH = −log(8.2 × 10^-3) = 2.08.

Correct: (A)

Using the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]) = 4.74 + log(0.10/0.20) = 4.74 + log(0.50) = 4.74 − 0.30 = 4.44.

Correct: (B)

A conjugate base is formed when an acid loses one proton. Removing one H^+ from H2PO4^- gives HPO4^2-.

Correct: (D)

At the equivalence point, all the NH3 has been converted to NH4^+, which is a weak acid. Its hydrolysis produces H^+, so the pH is below 7.00.

Correct: (C)

For a conjugate acid-base pair, Ka × Kb = Kw. Kb = (1.0 × 10^-14)/(4.9 × 10^-10) = 2.0 × 10^-5.

Correct: (A)

ΔG = ΔH − TΔS = −100 kJ − (298 K)(−0.200 kJ/K) = −100 + 59.6 = −40.4 kJ. A negative ΔG means the process is spontaneous.

Correct: (B)

E°cell = E°(cathode) − E°(anode) = 0.34 − (−0.76) = +1.10 V. Copper(II) is reduced at the cathode and zinc is oxidized at the anode.

Correct: (D)

ΔG° = −nFE° = −(2)(96,485)(1.10) = −2.12 × 10^5 J = −212 kJ.

Correct: (C)

Charge passed = It = (2.00)(965) = 1930 C, which is 1930/96,485 = 0.0200 mol of electrons. Since Cu^2+ + 2 e^- → Cu, this deposits 0.0100 mol of Cu = 0.0100 × 63.55 = 0.635 g.

Correct: (B)

Two moles of gas combine to form one mole of gas, a decrease in the number of freely moving particles, so entropy decreases and ΔS < 0.

Correct: (A)

In aqueous solution, water is reduced at the cathode in preference to Na^+, producing H2 gas. Sodium metal cannot be obtained from aqueous electrolysis; Cl2 forms at the anode instead.