AP Chemistry Practice Test 4

Correct: (C)

Copper is an exception to the usual filling order: one 4s electron moves into the 3d subshell because a completely filled 3d^10 subshell is especially stable.

Correct: (A)

All four are in period 4, and radius decreases across a period as effective nuclear charge increases, so K, farthest to the left, has the largest radius.

Correct: (B)

Na+ has the stable neon configuration, so removing a second electron from sodium means breaking into a filled core shell, which requires an exceptionally large amount of energy.

Correct: (D)

(23.985)(0.7899) + (24.986)(0.1000) + (25.983)(0.1101) = 18.95 + 2.50 + 2.86 = 24.31 amu.

Correct: (C)

E = hc/λ = (6.626 × 10^-34)(3.00 × 10^8)/(6.00 × 10^-7) = 3.31 × 10^-19 J.

Correct: (A)

Each occupied subshell produces one peak. Sulfur (1s^2 2s^2 2p^6 3s^2 3p^4) has electrons in five subshells: 1s, 2s, 2p, 3s, and 3p.

Correct: (B)

First ionization energy increases up a group because the valence electrons are in shells closer to the nucleus; oxygen is at the top of group 16 among the choices.

Correct: (D)

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

Correct: (C)

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

Correct: (A)

The carbon atom in CO2 has two electron domains (two double bonds) in a linear arrangement, which corresponds to sp hybridization.

Correct: (B)

NO has 8 electrons in bonding orbitals and 3 in antibonding orbitals, so bond order = (8 – 3)/2 = 2.5.

Correct: (D)

All four compounds have +1 and -1 ions, so the smallest ions produce the strongest attraction; Li+ is the smallest cation, giving LiCl the greatest lattice energy.

Correct: (A)

C2H4 has four C-H sigma bonds and one C-C sigma bond (5 sigma total); the second component of the C=C double bond is 1 pi bond.

Correct: (C)

The double bond can be drawn to any one of the three equivalent oxygen atoms, giving 3 equivalent resonance structures.

Correct: (B)

HF forms strong hydrogen bonds between its molecules, giving it an anomalously high boiling point despite having the lowest molar mass of the group.

Correct: (C)

The O-H group in ethanol allows extensive hydrogen bonding between molecules, which is the strongest intermolecular force present.

Correct: (A)

I2 is nonpolar and dissolves best in a nonpolar solvent according to the "like dissolves like" principle; CCl4 is the only nonpolar solvent listed.

Correct: (D)

Acetone has no O-H or N-H bonds and cannot donate hydrogen bonds, so its intermolecular forces are relatively weak and its molecules escape into the vapor most readily.

Correct: (B)

90.0 g / 18.0 g/mol = 5.00 mol, so q = (5.00 mol)(6.01 kJ/mol) = 30.1 kJ.

Correct: (A)

In molten NaCl the ions are free to move and carry charge, but in the solid the ions are locked into fixed positions in the lattice. (Copper conducts in both states; diamond and I2 conduct in neither.)

Correct: (D)

The balanced equation is Fe2O3 + 3 CO → 2 Fe + 3 CO2, so the coefficient of CO is 3.

Correct: (A)

Zinc is more easily oxidized than copper, so Zn reduces Cu^2+ to Cu while forming Zn^2+; sulfate is a spectator ion.

Correct: (C)

32.0 g / 16.0 g/mol = 2.00 mol of CH4 produces 4.00 mol of H2O, and 4.00 mol × 18.0 g/mol = 72.0 g.

Correct: (B)

The 0.100 mol of CaCO3 would require 0.200 mol of HCl, but only 0.150 mol is available, so HCl is limiting and 0.150/2 = 0.0750 mol of CO2 forms.

Correct: (A)

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

Correct: (D)

Using K = +1 and O = -2: 2(+1) + 2(Cr) + 7(-2) = 0, so 2(Cr) = +12 and Cr = +6.

Correct: (C)

Moles of NaOH = (0.200)(0.03000) = 6.00 × 10^-3 mol; since each H2SO4 reacts with 2 NaOH, moles of H2SO4 = 3.00 × 10^-3, and [H2SO4] = 3.00 × 10^-3/0.02000 = 0.150 M.

Correct: (B)

Doubling [X] does not change the rate (zero order in X), while doubling [Y] quadruples the rate (second order in Y).

Correct: (D)

For a second-order reaction, t1/2 = 1/(k[A]0) = 1/((0.50)(0.40)) = 1/0.20 = 5.0 s.

Correct: (A)

Each half-life cuts the amount in half: (1/2)^3 = 1/8 = 12.5% remains.

Correct: (C)

The slow step gives Rate = k[O3][O], but O is an intermediate; from the fast equilibrium, [O] = K[O3]/[O2], so substituting gives Rate = k'[O3]^2/[O2].

Correct: (B)

In a zero-order reaction the concentration falls at a constant rate of 0.025 M/s, so t = 1.0 M / 0.025 M/s = 40 s.

Correct: (D)

A higher concentration means more particles per unit volume, so collisions occur more often and the number of effective collisions per second rises.

Correct: (A)

ln(k2/k1) = ln 4 = (Ea/8.314)(1/300 – 1/320) = (Ea)(2.08 × 10^-5), which gives Ea ≈ 5.5 × 10^4 J/mol = 55 kJ/mol.

Correct: (C)

q = mcΔT = (100.0)(4.18)(20.0) = 8360 J = 8.36 kJ of heat released.

Correct: (B)

Subtracting equation (2) from equation (1) gives the target reaction, so ΔH = 68 – 180 = -112 kJ.

Correct: (D)

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

Correct: (A)

ΔH° = 2(-285.8) – 2(-187.8) = -571.6 + 375.6 = -196 kJ.

Correct: (C)

Heat absorbed from the water = (100.0)(4.18)(5.5) = 2299 J for 5.00/80.0 = 0.0625 mol, so ΔH = 2299/0.0625 ≈ +36.8 kJ/mol; the positive sign reflects the endothermic dissolution.

Correct: (B)

Condensation changes a dispersed gas into a more ordered liquid, reducing the number of accessible microstates, so ΔS < 0.

Correct: (D)

A formation equation produces exactly one mole of the compound from its elements in their standard states: solid carbon (graphite), H2 gas, and O2 gas.

Correct: (A)

The concentrations of pure solids do not appear in equilibrium constant expressions, leaving only the gas: Kc = [CO2].

Correct: (C)

From an ICE table, x^2/(1.00 – x) = 0.50 gives x^2 + 0.50x – 0.50 = 0, whose positive root is x = 0.50 M, and [Cl2] = x = 0.50 M.

Correct: (B)

Kp = Kc(RT)^Δn with Δn = 2 – 1 = +1, so Kp = (0.36)(0.0821)(500) = 14.8.

Correct: (D)

Higher pressure favors the side with fewer moles of gas. The left side has 1 mol of gas and the right has 2 mol, so the equilibrium shifts toward N2O4; the value of K is unchanged because the temperature is constant.

Correct: (A)

PbCl2 dissolves to give [Pb^2+] = s and [Cl-] = 2s, so Ksp = (s)(2s)^2 = 4s^3 = 1.6 × 10^-5; thus s^3 = 4.0 × 10^-6 and s = 1.6 × 10^-2 M.

Correct: (C)

Q = [B]/[A] = 0.60/0.10 = 6.0, which is greater than K = 4.0, so the net reaction runs in reverse, toward the reactants, until Q = K.

Correct: (B)

The common ion Br- (already 0.010 M from NaBr) suppresses the solubility: s = Ksp/[Br-] = 5.0 × 10^-13/0.010 = 5.0 × 10^-11 M.

Correct: (C)

NaOH dissociates completely, so [OH-] = 0.0010 M and pOH = 3.00; therefore pH = 14.00 – 3.00 = 11.00.

Correct: (D)

[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)

pH = pKa + log([base]/[acid]) = 3.17 + log(0.40/0.50) = 3.17 – 0.10 = 3.07.

Correct: (B)

A conjugate acid is formed by adding one H+ to the base; adding H+ to CO3^2- gives HCO3-.

Correct: (C)

At the half-equivalence point exactly half of the acid has been converted to its conjugate base, so [HA] = [A-] and the Henderson-Hasselbalch equation reduces to pH = pKa.

Correct: (D)

HCO3- can donate a proton to form CO3^2- (acting as an acid) or accept a proton to form H2CO3 (acting as a base).

Correct: (A)

ΔG = ΔH – TΔS; with both terms positive, ΔG becomes negative only when TΔS is large enough to outweigh ΔH, which happens at high temperatures.

Correct: (B)

E°cell = E°(cathode) – E°(anode) = 0.80 – (-0.44) = 1.24 V.

Correct: (D)

Charge passed = (3.00)(9650) = 28,950 C = 0.300 mol of electrons; since each Cu^2+ requires 2 electrons, 0.150 mol of Cu is deposited, and 0.150 × 63.5 = 9.53 g.

Correct: (C)

ΔG° = -nFE° = -(2)(96,485)(0.50) = -9.65 × 10^4 J = -96.5 kJ.

Correct: (A)

ΔG° = -RT ln K = -(8.314)(298) ln(1.0 × 10^-5) = +2.85 × 10^4 J = +28.5 kJ, consistent with K < 1 meaning reactants are favored under standard conditions.

Correct: (B)

Oxidation at the anode releases electrons, which travel through the external circuit to the cathode, where reduction occurs; in a galvanic cell the anode is negative and E°cell is positive for a spontaneous reaction.