ACS General Chemistry Practice Test 14

Correct: (B)

Atomic mass on the periodic table is the abundance-weighted average of isotopic masses: (0.692)(63.0) + (0.308)(65.0) = 63.55.

Correct: (D)

Cations are smaller than their neutral atoms because electron loss reduces electron-electron repulsion, while anions are larger than their neutral atoms. Mg2+ < Mg < S < S2-.

Correct: (A)

X is chlorine and Y is argon. Chlorine readily accepts one electron to achieve the noble-gas configuration (high electron affinity), whereas argon has essentially zero electron affinity since its shell is already filled.

Correct: (C)

The large jump between the third (2745) and fourth (11577) ionization energies indicates removal of a core electron after three valence electrons are removed. Aluminum ([Ne]3s23p1) has three valence electrons.

Correct: (B)

Effective nuclear charge increases across a period as protons are added with little additional shielding from same-shell electrons. Among these period 3 elements, phosphorus (highest Z) has the greatest Zeff.

Correct: (A)

Atomic radius decreases across a period because increasing nuclear charge pulls same-shell electrons closer, while electronegativity, ionization energy, and electron affinity magnitude generally increase.

Correct: (D)

Cl contributes 7, four O atoms contribute 4(6) = 24, and the 1- charge adds 1: 7 + 24 + 1 = 32 valence electrons.

Correct: (C)

XeF4 has six electron domains around Xe (4 bonding pairs + 2 lone pairs). The lone pairs occupy opposite positions in an octahedral electron geometry, leaving the four F atoms in a square planar arrangement.

Correct: (B)

Bond polarity increases with electronegativity difference. Fluorine is the most electronegative element, so the F-H bond has the largest dipole of those listed.

Correct: (D)

SF6 has six bonding domains around sulfur, requiring six equivalent hybrid orbitals: one s + three p + two d = sp3d2.

Correct: (A)

O2+ has 11 valence electrons; its MO configuration gives (8 bonding – 3 antibonding)/2 = 2.5. O2 has bond order 2, O2- has 1.5, and N2 has 3.

Correct: (C)

Intramolecular forces are the covalent bonds within a molecule; intermolecular forces act between molecules. Boiling breaks intermolecular attractions, not covalent bonds.

Correct: (D)

H2O is bent (two lone pairs on O), so its bond dipoles do not cancel. CO2 (linear), CCl4 (tetrahedral), and BF3 (trigonal planar) are symmetric, so their bond dipoles cancel.

Correct: (B)

Each formula unit of Al2(SO4)3 contains 3 x 4 = 12 oxygen atoms, so 3.0 mol x 12 = 36 mol of O atoms.

Correct: (A)

8.0 mol Al would require 12 mol Cl2, but only 9.0 mol Cl2 is available, so Cl2 is the limiting reactant: 9.0 mol Cl2 x (2 mol AlCl3/3 mol Cl2) = 6.0 mol AlCl3.

Correct: (C)

Moles CaCO3 = 25.0/100.1 = 0.2498 mol; theoretical CaCl2 = 0.2498 mol x 111.0 g/mol = 27.7 g; percent yield = (10.5/27.7) x 100 = 37.9%.

Correct: (D)

M1V1 = M2V2 gives V1 = (0.100 M)(500. mL)/(0.250 M) = 200. mL.

Correct: (A)

PbI2 is the insoluble precipitate; K+ and NO3- are spectator ions and are omitted from the net ionic equation.

Correct: (B)

Half-reactions: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O and C2O42- → 2CO2 + 2e-. Balancing 10 electrons: 2MnO4- + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O. The H+ coefficient is 16.

Correct: (C)

Molarity = moles solute / liters solution = 0.125 mol / 0.500 L = 0.250 M.

Correct: (B)

In a fixed-volume container at constant temperature, the partial pressure of O2 depends only on n(O2), which is unchanged. Adding N2 raises total pressure but not the O2 partial pressure (Dalton's law).

Correct: (A)

By Gay-Lussac's law (P1/T1 = P2/T2 at constant V, n): P2 = 1.00 atm × (228 K / 298 K) = 0.77 atm.

Correct: (B)

At equal T and V, P is proportional to moles. Moles: He 1.00 mol, CH4 0.25 mol, SO2 0.0625 mol, so pressure follows the same order (i) > (ii) > (iii).

Correct: (B)

Initial P_A V_A = 0.60 RT and P_B V_B = 0.40 RT. After opening, total moles = 1.00 mol in volume 2V_A, so P_final = 1.00 RT/(2V_A) = (P_A + P_B)/2 = 1.5 atm.

Correct: (C)

Ethanol forms intermolecular hydrogen bonds (highest bp); dimethyl ether is polar but lacks H-bond donors (middle); butane has only London dispersion forces (lowest bp).

Correct: (B)

Heat lost by metal = heat gained by water: 65.0 × c × 71.5 = 150.0 × 4.18 × 4.5 = 2822 J. c = 2822/(65.0 × 71.5) = 0.607 J/g·°C ≈ 0.61 J/g·°C.

Correct: (B)

33.0 g C3H8 = 0.750 mol. Heat = 0.750 × 2,220 kJ = 1,665 kJ released.

Correct: (A)

ΔH° = [4(90) + 6(-242)] – [4(-46) + 0] = [360 – 1,452] + 184 = -908 kJ ≈ -906 kJ (rounding differences give -906 kJ).

Correct: (B)

Hess's law: reverse the second equation and add: -394 + (+283) = -111 kJ for formation of 1 mol CO from graphite and ½ O2.

Correct: (C)

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

Correct: (C)

Total mass = 97.0 g; q = 97.0 × 4.18 × 6.8 = 2,757 J absorbed. Moles NH4NO3 = 0.150 mol. ΔH = +2,757/0.150 = +18.4 kJ/mol ≈ +19 kJ/mol (endothermic, temperature dropped).

Correct: (C)

Photon energy = KE + work function = 5.68 × 10⁻¹⁸ J. Threshold wavelength = hc/φ = (6.626 × 10⁻³⁴)(3.00 × 10⁸)/(4.2 × 10⁻¹⁸) = 4.73 × 10⁻⁷ m = 473 nm.

Correct: (B)

The n = 4 → n = 2 transition releases more energy (larger ΔE), so the photon has higher frequency and shorter wavelength.

Correct: (B)

Ground-state P has configuration [Ne] 3s² 3p³; the 3d subshell (l = 2, n = 3) is empty in the ground state, so no electron can have l = 2.

Correct: (C)

2p⁴ would require pairing two electrons in the same 2p orbital before the third 2p orbital is singly occupied, violating Hund's rule of maximum multiplicity.

Correct: (A)

Rate scales as (3)²(1/2) = 9/2 = 4.5 times the original rate.

Correct: (A)

7.20 h / 2.40 h = 3 half-lives. [N2O5] = 0.800/2³ = 0.100 M.

Correct: (D)

ln(k2/k1) = (Ea/R)(1/T1 – 1/T2) = (54,000/8.314)(1/300 – 1/330) = 6,495 × 3.030 × 10⁻⁴ = 1.968. k2/k1 = e^1.968 ≈ 7.2 (with rounding ≈ 7.4).

Correct: (C)

In a proposed mechanism, the rate-determining (slow) step determines the rate law. Only choice C has a slow step involving exactly one ClO2 and one OH⁻, matching rate = k[ClO2][OH⁻]. The fast second step consumes the remaining reactants without affecting the rate law.

Correct: (B)

Slope = k = 0.025 M⁻¹s⁻¹. 1/[A]t = kt + 1/[A]0 → 1/0.125 = 0.025t + 1/0.500 → 8.00 = 0.025t + 2.00 → t = 6.00/0.025 = 240 s.

Correct: (A)

First-order kinetics (units of k are min⁻¹): ln([A]/[A]₀) = −kt → ln(0.25) = −0.0256t → t = 1.386/0.0256 = 54.1 min, which is exactly two half-lives.

Correct: (C)

Kc = Kp(RT)^(−Δn) with Δn = 2 − 4 = −2, so Kc = Kp(RT)² = 6.0×10⁻³ × (0.0821 × 773)² ≈ 6.0×10⁻³ × 4.02×10³ ≈ 24.

Correct: (D)

[SO₂] = 0.100 M, [O₂] = 0.0250 M, [SO₃] = 0.0500 M, so Qc = (0.100)²(0.0250)/(0.0500)² = 0.100. Since Qc > Kc (0.0420), the system shifts left to re-establish equilibrium.

Correct: (B)

Solids are excluded from the equilibrium expression, so adding or removing them does not change Q. Removing CO₂ lowers Q below Kp, so the reaction shifts right to restore equilibrium.

Correct: (B)

The forward reaction is endothermic and produces more moles of gas. Increasing the volume lowers the pressure, shifting equilibrium toward the side with more gas moles (right), consuming PCl₅. Decreasing T favors the reverse reaction; raising pressure favors the left; a catalyst does not shift equilibrium.

Correct: (B)

Let x = [Cl₂]; then [NO] = 2x and [NOCl] ≈ 0.0400 M (K is tiny). Kc = (2x)²(x)/(0.0400)² = 1.60×10⁻⁵ → 4x³ = 2.56×10⁻⁸ → x³ = 6.40×10⁻⁹ → x = 1.86×10⁻³ M.

Correct: (D)

Adding the two equations gives the target reaction, so K = Ksp × Kf = (1.8×10⁻¹⁰)(1.6×10⁷) = 2.9×10⁻³.

Correct: (D)

[H⁺] = √(Ka·C) = √(1.34×10⁻⁵ × 0.150) = √(2.01×10⁻⁶) = 1.42×10⁻³ M → pH = 2.85.

Correct: (C)

A buffer needs a weak acid and its conjugate base. HNO₂ is a weak acid and NO₂⁻ its conjugate base. (A) and (D) pair strong acids with spectator salts; (B) pairs a strong base with a spectator salt.

Correct: (C)

Mol NH₃ = 0.0150, mol NH₄⁺ = 0.00750; ratio base/acid = 2.00. pKa = 14 − 4.75 = 9.25. pH = pKa + log(2.00) = 9.25 + 0.30 = 9.56.

Correct: (D)

Remaining H⁺ = (0.0250 × 0.100) − (0.0125 × 0.100) = 1.25×10⁻³ mol in 0.0375 L → [H⁺] = 0.0333 M → pH = −log(0.0333) = 1.48.

Correct: (B)

At equivalence only benzoate remains ([C₆H₅COO⁻] = 0.0600 M after the volume doubles). Kb = Kw/Ka = 1.59×10⁻¹⁰. [OH⁻] = √(Kb·C) = 3.09×10⁻⁶ M → pOH = 5.51 → pH = 8.49 (basic, as expected for a weak-acid titration).

Correct: (C)

CN⁻ is the conjugate base of the weak acid HCN, so KCN is basic. NH₄⁺ is acidic (conjugate acid of weak base NH₃), HSO₄⁻ is an appreciably strong weak acid, and KCl is neutral; the basic solution has the highest pH.

Correct: (A)

Gases have far greater molar entropy than solids. Converting 1 mol of O₂(g) into solid product decreases the moles of gas and the overall disorder, so ΔS° < 0.

Correct: (D)

ΔG° = ΔH° − TΔS° = 1.9 kJ/mol − (298 K)(−3.4×10⁻³ kJ/mol·K) = +2.9 kJ/mol > 0, so the process is nonspontaneous (with ΔH > 0 and ΔS < 0 it is nonspontaneous at all temperatures).

Correct: (C)

E = E° − (0.0592/2)log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.0296·log(0.010) = 1.10 + 0.0592 = 1.16 V; lowering the product-ion concentration moves the cell away from equilibrium and raises the potential.

Correct: (A)

Q = It = 4.50 × 7200 s = 3.24×10⁴ C; mol e⁻ = 3.24×10⁴/96485 = 0.336 mol. Cr³⁺ + 3e⁻ → Cr gives 0.336/3 = 0.112 mol Cr; mass = 0.112 × 52.0 = 5.82 g.

Correct: (A)

E°cell = E°cathode − E°anode = 0.80 − 0.77 = +0.03 V. ΔG° = −nFE° = −(1)(96485)(0.03) = −2.9×10³ J/mol = −2.9 kJ/mol (spontaneous).

Correct: (B)

36.0 days is 3 half-lives; mass remaining = 2.50 × (1/2)³ = 2.50/8 = 0.313 g.

Correct: (C)

Alpha decay emits a ⁴₂He nucleus, so Z decreases by 2 and A decreases by 4. (A) describes beta decay; (B) describes alpha decay, not positron emission; (D) describes positron emission, not beta decay.