AP Chemistry Practice Test 5

Correct: (C)

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

Correct: (A)

The peaks give the configuration 1s^2 2s^2 2p^4, a total of 8 electrons, which is oxygen.

Correct: (D)

Ionization energy generally increases across a period, and nitrogen, with its stable half-filled 2p subshell, has the highest first ionization energy of these elements.

Correct: (B)

(0.5069)(79) + (0.4931)(81) = 40.05 + 39.94 = 79.99, approximately 80.0.

Correct: (A)

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

Correct: (D)

Chromium has the configuration [Ar] 4s^1 3d^5, with one unpaired 4s electron and five unpaired 3d electrons, for a total of 6.

Correct: (C)

After losing one electron, Na+ has a stable noble-gas configuration, so removing a second (core) electron requires an exceptionally large amount of energy.

Correct: (A)

Nitrogen has 5 valence electrons and is assigned half of the 8 bonding electrons (4) with no lone pair electrons: formal charge = 5 – 4 = +1.

Correct: (D)

Sulfur has five electron domains (four bonds and one lone pair). The lone pair occupies an equatorial position of a trigonal bipyramid, giving a seesaw molecular geometry.

Correct: (B)

XeF4 has six electron domains (four bonding pairs and two lone pairs), which requires sp^3d^2 hybridization and gives a square planar geometry.

Correct: (C)

Lattice energy increases with ionic charge and decreases with ion size. MgO combines +2 and −2 charges with a relatively small cation, giving the greatest lattice energy.

Correct: (A)

The C=O double bond can be placed on any of the three oxygen atoms, giving three equivalent resonance structures.

Correct: (B)

Bond length decreases as bond order increases, so the triple bond in C2H2 is shorter than the double bond in C2H4 and the single bond in C2H6.

Correct: (D)

CHCl3 is tetrahedral with different atoms attached to carbon, so its bond dipoles do not cancel. CCl4, CO2, and BF3 are symmetric and nonpolar.

Correct: (B)

Each water molecule can form an extensive three-dimensional hydrogen-bonding network (two O-H bonds and two lone pairs), giving water the highest boiling point, 100°C.

Correct: (D)

Methanol is polar and hydrogen bonds with water, making it miscible in all proportions. The other choices are nonpolar or only slightly soluble.

Correct: (A)

36.0 g / 18.0 g/mol = 2.00 mol of ice; 2.00 mol × 6.01 kJ/mol = 12.0 kJ.

Correct: (C)

Dimethyl ether has no O-H bond, so its molecules cannot hydrogen bond with each other. It has the weakest intermolecular forces and therefore the highest vapor pressure.

Correct: (A)

Quartz (SiO2) is held together by covalent bonds extending throughout the entire structure. NaCl is ionic, solid CO2 is molecular, and Fe is metallic.

Correct: (C)

The O-H group in ethanol allows intermolecular hydrogen bonding, which takes much more energy to overcome than the dipole-dipole forces between dimethyl ether molecules.

Correct: (D)

The balanced equation is C3H8 + 5 O2 → 3 CO2 + 4 H2O, which balances the C, H, and O atoms.

Correct: (B)

BaSO4 is insoluble, so Ba^2+ and SO4^2- combine to form the precipitate; Na+ and Cl- are spectator ions and are omitted.

Correct: (C)

O2 is the limiting reactant: 0.80 mol O2 × (2 mol H2O / 1 mol O2) = 1.60 mol H2O.

Correct: (A)

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

Correct: (D)

Zn is oxidized (its oxidation number increases from 0 to +2), so Zn is the species that causes reduction and is itself the reducing agent.

Correct: (B)

8.80 g C3H8 / 44.1 g/mol = 0.200 mol C3H8. Each mole of C3H8 produces 3 mol of CO2, so 0.600 mol × 44.0 g/mol = 26.4 g of CO2.

Correct: (C)

Moles of HCl = 0.02500 L × 0.100 M = 0.00250 mol. The volume of NaOH = 0.00250 mol / 0.125 M = 0.0200 L = 20.0 mL.

Correct: (D)

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

Correct: (B)

For rate = k[A]^2, k = rate / [A]^2 = (M/s) / M^2 = M^-1 s^-1.

Correct: (A)

For a first-order reaction, t1/2 = ln 2 / k = 0.693 / 0.0693 s^-1 = 10.0 s.

Correct: (C)

ln([A]0/[A]) = kt, so ln(0.80/0.20) = ln 4 = 1.39 = k(30. min), giving k = 0.046 min^-1.

Correct: (B)

NO3 is produced in step 1 and consumed in step 2, so it is an intermediate. It does not appear in the overall equation, NO2 + CO → NO + CO2.

Correct: (D)

Using ln(k2/k1) = (Ea/R)(1/T1 − 1/T2): ln 2 = (Ea/8.314)(1/300 − 1/310), which gives Ea ≈ 5.4 × 10^4 J/mol = 54 kJ/mol.

Correct: (A)

A catalyst works by providing a different mechanism with a lower activation energy. It does not change ΔH or the position of equilibrium.

Correct: (C)

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

Correct: (A)

q = mcΔT = (50.0 g)(4.184 J/g·°C)(10.0°C) = 2092 J = 2.09 kJ.

Correct: (D)

ΔH = [−393.5 + 2(−285.8)] − [−74.8] = −965.1 + 74.8 = −890.3 kJ.

Correct: (B)

Bonds broken: 436 + 242 = 678 kJ. Bonds formed: 2(431) = 862 kJ. ΔH = 678 − 862 = −184 kJ.

Correct: (C)

Dissolving the solid salt produces mobile aqueous ions, which greatly increases the disorder of the system. The other processes all decrease disorder.

Correct: (B)

c = q / (mΔT) = 650 J / [(25.0 g)(26.0°C)] = 650 / 650 = 1.00 J/g·°C.

Correct: (D)

34.0 g NH3 / 17.0 g/mol = 2.00 mol of NH3, which is exactly the amount in the thermochemical equation, so 92.2 kJ of heat is released.

Correct: (A)

Kc equals the product concentrations raised to their stoichiometric coefficients divided by the reactant concentrations raised to theirs.

Correct: (C)

Let x be the amount of H2 consumed. Then [HI] = 2x and (2x)^2 / (0.50 − x)^2 = 50, so 2x / (0.50 − x) = 7.07, giving x = 0.39 and [HI] = 2x = 0.78 M.

Correct: (B)

Kp = Kc(RT)^Δn, and Δn = 2 − 1 = +1, so Kp = (4.6 × 10^-3)(0.0821)(298) = 0.11.

Correct: (D)

The reaction converts 3 mol of gas into 1 mol of gas, so increasing the pressure shifts the equilibrium toward products. A catalyst does not change the yield, and for an exothermic reaction heating shifts the equilibrium toward reactants.

Correct: (B)

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

Correct: (A)

The Cl- already present is a common ion and suppresses the solubility: Ksp = s(0.10), so s = 1.8 × 10^-10 / 0.10 = 1.8 × 10^-9 M.

Correct: (D)

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

Correct: (C)

HCl is a strong acid and dissociates completely, so [H+] = 0.0025 M and pH = −log(2.5 × 10^-3) = 2.60.

Correct: (A)

[H+] = sqrt(Ka × C) = sqrt(1.8 × 10^-5 × 0.10) = 1.34 × 10^-3 M, so pH = 2.87.

Correct: (D)

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

Correct: (B)

At the equivalence point all of the NH3 has been converted to NH4+ at a concentration of 0.050 M. NH4+ is a weak acid with Ka = Kw/Kb = 5.6 × 10^-10, so [H+] = sqrt(5.6 × 10^-10 × 0.050) = 5.3 × 10^-6 M and pH = 5.28.

Correct: (C)

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

Correct: (B)

Each formula unit provides two OH- ions, so [OH-] = 0.020 M. pOH = −log(0.020) = 1.70, and pH = 14.00 − 1.70 = 12.30.

Correct: (A)

ΔG = ΔH − TΔS = −100 kJ − (298 K)(−0.200 kJ/K) = −100 + 59.6 = −40.4 kJ, so the reaction is spontaneous at 298 K.

Correct: (D)

ΔG = ΔH − TΔS becomes negative only when T is greater than ΔH/ΔS = 50,000 J / 100 J/K = 500 K.

Correct: (B)

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

Correct: (C)

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

Correct: (A)

6.35 g of Cu is 0.100 mol, which requires 0.200 mol of electrons. Q = (0.200)(96,485) = 1.93 × 10^4 C, and t = Q / I = 1.93 × 10^4 / 2.00 = 9.65 × 10^3 s.

Correct: (D)

Oxidation always occurs at the anode. In a galvanic cell the anode supplies electrons to the external circuit and is the negative terminal; reduction occurs at the positive cathode.