AP Chemistry Practice Test 3

Correct: (A)

Fe is [Ar] 4s^2 3d^6, and the two 4s electrons are removed first when Fe^2+ forms, leaving [Ar] 3d^6.

Correct: (D)

All four are in period 3; atomic radius decreases across a period as effective nuclear charge increases, so S, farthest to the right, is smallest.

Correct: (C)

The huge jump between the first and second ionization energies shows that only one electron is easily removed; the second electron comes from a stable noble-gas core, so the atom has 1 valence electron.

Correct: (B)

(34.969)(0.7577) + (36.966)(0.2423) = 26.50 + 8.96 = 35.45 amu.

Correct: (A)

The peaks total 11 electrons with configuration 1s^2 2s^2 2p^6 3s^1, which is sodium.

Correct: (D)

All four ions have 10 electrons, but O^2- has the fewest protons (8), so its electron cloud is held least tightly and the ion is largest.

Correct: (C)

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

Correct: (B)

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

Correct: (A)

Sulfur has five electron domains (four bonding pairs and one lone pair); the lone pair occupies an equatorial position and the atoms form a seesaw shape.

Correct: (C)

Nitrogen has four electron domains (three N-H bonds and one lone pair), which corresponds to sp3 hybridization.

Correct: (D)

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

Correct: (A)

Lattice energy increases with ion charge and decreases with ion size; Mg^2+ and O^2- combine the largest charge product (+2/-2) with small ionic radii.

Correct: (B)

CH3OH contains three C-H bonds, one C-O bond, and one O-H bond, all single bonds, giving 5 sigma bonds and 0 pi bonds.

Correct: (C)

CO2 is linear and symmetrical, so the dipoles of the two C=O bonds point in opposite directions and cancel.

Correct: (C)

CH3OH molecules form hydrogen bonds, the strongest intermolecular force among the choices, so the most thermal energy is needed to separate its molecules.

Correct: (D)

I2 is a nonpolar molecule, so the only attractions between its molecules are London dispersion forces.

Correct: (B)

CH3OH is polar and can hydrogen-bond with water, while CCl4, C6H14, and CH4 are nonpolar and dissolve poorly in water.

Correct: (A)

Diethyl ether has an oxygen atom but no O-H bond, so it cannot hydrogen-bond with itself; its weak intermolecular forces let its molecules escape into the vapor most easily.

Correct: (D)

36.0 g / 18.0 g/mol = 2.00 mol, so q = (2.00 mol)(40.7 kJ/mol) = 81.4 kJ.

Correct: (C)

Diamond is a network covalent solid, and melting it requires breaking strong covalent bonds that extend throughout the entire lattice.

Correct: (A)

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

Correct: (B)

Ag+ and Cl- combine to form the insoluble precipitate AgCl; Na+ and NO3- are spectator ions and do not appear in the net ionic equation.

Correct: (D)

44.0 g / 44.1 g/mol ≈ 1.00 mol of C3H8 produces 3.00 mol of CO2, and 3.00 mol × 44.0 g/mol = 132 g.

Correct: (C)

4.0 mol of H2 requires only 2.0 mol of O2, so H2 is the limiting reactant and forms 4.0 mol of H2O, with 1.0 mol of O2 left over.

Correct: (B)

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

Correct: (A)

Using H = +1 and O = -2: 2(+1) + S + 4(-2) = 0, so S = +6.

Correct: (D)

Moles of NaOH = (0.100)(0.02250) = 2.25 × 10^-3 mol = moles of HCl, so [HCl] = 2.25 × 10^-3 / 0.02500 = 0.0900 M.

Correct: (C)

Doubling [A] quadruples the rate, so the reaction is second order in A; doubling [B] leaves the rate unchanged, so it is zero order in B.

Correct: (A)

For a first-order reaction, t1/2 = 0.693/k = 0.693/0.0347 = 20.0 s.

Correct: (D)

Since rate (M/s) = k[A]^2 (M^2), k must carry units of M^-1·s^-1 for the units to balance.

Correct: (B)

The rate law is determined by the slow, rate-determining step, whose reactants are two NO2 molecules, giving Rate = k[NO2]^2.

Correct: (C)

Using the second-order integrated law, 1/[A] = kt + 1/[A]0 = (0.10)(100) + 1/0.50 = 12, so [A] = 1/12 = 0.083 M.

Correct: (A)

A catalyst works by providing a different reaction pathway whose activation energy is lower, so a larger fraction of collisions can succeed; ΔH and the energies of reactants and products are unchanged.

Correct: (D)

From the Arrhenius relation, ln 2 = (Ea/8.314)(1/300 – 1/310), which gives Ea ≈ 5.4 × 10^4 J/mol ≈ 54 kJ/mol.

Correct: (B)

q = mcΔT = (50.0)(4.18)(8.0) = 1670 J = 1.67 kJ.

Correct: (D)

By Hess's law, subtracting the second equation from the first gives the target reaction: ΔH = -394 – (-283) = -111 kJ.

Correct: (A)

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

Correct: (C)

ΔH° = [-393.5 + 2(-285.8)] – [-74.8] = -965.1 + 74.8 = -890 kJ per mole of CH4.

Correct: (B)

Heat gained by the water = (200.0)(4.18)(3.5) = 2926 J, so (100.0)(c)(71.5) = 2926 and c = 0.409 J/g·°C.

Correct: (D)

Sublimation converts an ordered solid directly into a highly dispersed gas, which greatly increases the number of accessible microstates.

Correct: (A)

By definition, the standard enthalpy of formation of any element in its most stable form under standard conditions, such as O2(g), is zero.

Correct: (C)

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, so 2x/(0.10 – x) = 7 and x = 0.078 M; therefore [HI] = 2x = 0.156 M.

Correct: (A)

Kp = Kc(RT)^Δn with Δn = 2 – 3 = -1, so Kp = 100/(0.0821 × 1000) = 100/82.1 = 1.22.

Correct: (D)

For an exothermic reaction, heat can be treated as a product; adding heat drives the reaction in the reverse (endothermic) direction, favoring reactants and making K smaller.

Correct: (C)

AgCl dissolves to give equal concentrations of Ag+ and Cl-, so Ksp = s^2 and s = sqrt(1.8 × 10^-10) = 1.3 × 10^-5 M.

Correct: (B)

When Q > K, products are present in excess relative to equilibrium, so the net reaction runs in reverse, toward the reactants, until Q = K.

Correct: (A)

With the common ion F- already at 0.10 M, Ksp = [Ca^2+][F-]^2 = s(0.10)^2, so s = 4.0 × 10^-11/0.010 = 4.0 × 10^-9 M.

Correct: (D)

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

Correct: (A)

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

Correct: (C)

pH = pKa + log([base]/[acid]) = 4.74 + log(0.20/0.30) = 4.74 – 0.18 = 4.56.

Correct: (B)

A conjugate base is formed by removing one H+ from the acid; removing H+ from H2PO4- gives HPO4^2-.

Correct: (D)

At the equivalence point only the conjugate base of the weak acid remains, and it reacts with water to produce OH-, so the solution is basic with pH > 7.

Correct: (A)

Because the second ionization is so much weaker than the first, it barely changes any concentrations, and the equilibrium analysis of the second step gives [A^2-] ≈ Ka2.

Correct: (B)

ΔG° = ΔH° – TΔS° = -100 – (298)(-0.200) = -100 + 59.6 = -40.4 ≈ -40 kJ/mol, so the reaction is spontaneous under standard conditions.

Correct: (C)

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

Correct: (A)

6.35 g / 63.5 g/mol = 0.100 mol of Cu requires 0.200 mol of electrons = 1.93 × 10^4 C; time = Q/I = 1.93 × 10^4/2.00 = 9.65 × 10^3 s.

Correct: (D)

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

Correct: (C)

With ΔH negative and ΔS positive, ΔG = ΔH – TΔS is negative at every temperature.

Correct: (B)

Four moles of gas are converted into two moles of gas, so the system becomes more ordered and ΔS° is negative.