ACS General Chemistry Practice Test 18

Correct: (C)

Cr3+ is [Ar]3d3, giving three unpaired electrons in the 3d subshell.

Correct: (A)

Kr is a noble gas with a complete shell and high effective nuclear charge, so it has the highest first ionization energy of the four.

Correct: (D)

For l = 1, ml can only be -1, 0, or +1, so ml = 2 is impossible.

Correct: (B)

Among the group 16 elements listed, S has the most negative electron affinity (O is anomalously low due to small size and electron-electron repulsion).

Correct: (C)

ΔE = 13.6 eV(1/4 – 1/16) = 2.55 eV; λ = 1240 eV·nm / 2.55 eV ≈ 486 nm (blue-green line of the Balmer series).

Correct: (D)

S2- and Cl- both have 18 electrons, while the other pairs differ in electron count.

Correct: (B)

ICl3 has 5 electron domains (3 bonding, 2 lone pairs), giving a T-shaped molecular geometry.

Correct: (C)

Si-Cl has the largest electronegativity difference (≈1.26), making it the most polar of the listed bonds.

Correct: (A)

XeF4 has 6 electron domains (4 bonding, 2 lone pairs), requiring sp3d2 hybridization for square planar geometry.

Correct: (D)

O2+ has bond order 2.5, higher than O2 (2.0), O2- (1.5), and O22- (1.0).

Correct: (B)

Formal charge = 5 – (0 nonbonding + 8/2 bonding) = +1 on nitrogen in the standard Lewis structure.

Correct: (C)

H2S has a bent geometry with two lone pairs on S, so its bond dipoles do not cancel; the others are symmetric and nonpolar.

Correct: (B)

C2 has no unpaired electrons in its molecular orbital configuration (bond order 2), so it is diamagnetic; B2, O2, and NO each have unpaired electrons.

Correct: (A)

Al is limiting (0.926 mol Al needs 0.695 mol O2, and 0.938 mol O2 is available); 0.926/2 mol Al2O3 × 102 g/mol = 47.2 g.

Correct: (C)

M2 = (2.50 M)(15.0 mL)/(250. mL) = 0.150 M.

Correct: (B)

1:1 stoichiometry gives M(HCl) = (0.100 M)(32.50 mL)/(20.00 mL) = 0.1625 M.

Correct: (D)

Ag+ and Cl- combine to form the insoluble precipitate AgCl; Na+ and NO3- are spectator ions.

Correct: (A)

Percent yield = (9.80 g / 12.5 g) × 100% = 78.4%.

Correct: (C)

Mole ratio: C 3.33 : H 6.7 : O 3.33 = 1 : 2 : 1, giving the empirical formula CH2O.

Correct: (B)

With K at +1 and O at -2: 2(+1) + 2x + 7(-2) = 0 gives x = +6.

Correct: (D)

P = nRT/V = (0.500)(0.08206)(300.)/10.0 = 1.23 atm.

Correct: (A)

By Graham's law, rate ratio = √(M(O2)/M(H2)) = √(32/2) = 4.00.

Correct: (C)

P(O2) = mole fraction × total = (3.0/5.0)(2.50 atm) = 1.50 atm.

Correct: (B)

NH3 has H bonded directly to the highly electronegative N atom, enabling hydrogen bonding; the others lack this feature.

Correct: (D)

Diethyl ether has the weakest intermolecular forces (dipole-dipole only, no hydrogen bonding), so it evaporates most readily.

Correct: (D)

q = mcΔT = (50.0 g)(4.184 J/g·°C)(25.0°C) = 5230 J = 5.23 kJ.

Correct: (B)

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

Correct: (A)

ΔHf° is the enthalpy change when 1 mol of compound forms from its elements in standard states; the negative sign means heat is released.

Correct: (C)

Exothermic means heat flows out of the system (ΔH < 0), warming the surroundings.

Correct: (D)

ΔH = (436 + 242) – 2(431) = 678 – 862 = -184 kJ.

Correct: (A)

Bonds broken: 4(413) + 614 + 3(498) = 3760 kJ; bonds formed: 4(799) + 4(463) = 5048 kJ; ΔH = 3760 − 5048 = −1288 kJ.

Correct: (C)

Removing a third electron from Mg strips a core electron from a neon-like Mg2+ ion, and its higher nuclear charge makes this harder than for Na, while Al and Si still lose valence electrons.

Correct: (D)

The azimuthal quantum number l must be less than n, so l = 2 is not allowed when n = 2.

Correct: (B)

Chromium is an exception to the filling order; a half-filled 3d subshell with one 4s electron is more stable than [Ar]4s2 3d4.

Correct: (C)

The number of angular nodes equals the azimuthal quantum number l, which is 2 for any d orbital.

Correct: (D)

Doubling [A] quadruples the rate (second order in A); doubling [B] doubles the rate (first order in B).

Correct: (A)

135 min is three half-lives, so the concentration falls by (1/2)3: 0.800 M / 8 = 0.100 M.

Correct: (B)

From ln(k2/k1) = (Ea/R)(1/T1 − 1/T2), Ea = (8.314 J/mol·K)(ln 2)/(1/300 − 1/310) ≈ 53.6 kJ/mol.

Correct: (C)

The rate-determining slow step is bimolecular, so the rate law is first order in each reactant of step 1.

Correct: (D)

A catalyst lowers the activation barrier via a new mechanism but leaves ΔH, K, and the overall stoichiometry unchanged.

Correct: (B)

The reaction is third order overall, so k has units of M(1−3)·s−1 = M−2·s−1.

Correct: (A)

Reversing the reaction inverts K and halving the coefficients takes the square root: K = (1/280)1/2 = 0.0598.

Correct: (D)

Forming 0.20 M NH3 consumes (3/2)(0.20) = 0.30 M H2, so [H2]eq = 0.80 − 0.30 = 0.50 M.

Correct: (B)

The reaction is endothermic, so higher temperature favors products; pure solids do not affect the equilibrium position, and compression favors the side with fewer gas moles.

Correct: (C)

Qc = 0.50/(0.30)2 = 5.56, which is less than Kc, so the reaction shifts right to reach equilibrium.

Correct: (A)

For CaF2(s) ⇌ Ca2+ + 2F−, Ksp = s(2s)2 = 4s3 = 4(2.0 × 10−4)3 = 3.2 × 10−11.

Correct: (D)

The added chloride is a common ion; by Le Châtelier's principle it suppresses the dissolution of AgCl even though Ksp itself is unchanged.

Correct: (B)

[H+] ≈ (Ka·C)1/2 = (6.8 × 10−5)1/2 = 8.2 × 10−3 M, so pH = −log(8.2 × 10−3) = 2.08.

Correct: (C)

Removing one proton (H+) from the acid H2PO4− gives its conjugate base, HPO42−.

Correct: (B)

A weak base (NH3) plus its conjugate acid (NH4+) resists pH changes; the other mixtures contain only strong acids/bases and spectator ions.

Correct: (A)

At the equivalence point only the conjugate base A− remains, and its hydrolysis (A− + H2O ⇌ HA + OH−) makes the solution basic.

Correct: (D)

By the Henderson–Hasselbalch equation, pH = pKa + log(10.0) = 4.76 + 1.00 = 5.76.

Correct: (C)

For a conjugate pair, Ka·Kb = Kw, so Kb = 1.0 × 10−14 / 6.8 × 10−4 = 1.5 × 10−11.

Correct: (C)

Four moles of gas combine to form two moles of gas, decreasing disorder; the other reactions increase disorder.

Correct: (D)

ΔG° = ΔH° − TΔS° = −120 − (500)(−0.150) = −45 kJ/mol; a negative ΔG° means the reaction is spontaneous.

Correct: (A)

Copper has the more positive reduction potential, so Cu2+ is reduced at the cathode while Zn is oxidized at the anode.

Correct: (C)

By the Nernst equation, E = 1.10 − (0.0592/2)log(0.10/2.0) = 1.10 + 0.039 = 1.14 V.

Correct: (B)

Charge = (2.00 A)(1800 s) = 3600 C = 0.0373 mol e−; moles Cu = 0.0373/2 = 0.0187 mol, giving (0.0187)(63.55) = 1.19 g.

Correct: (D)

The mass number decreases by 4 and the atomic number by 2, which identifies the emitted particle as an alpha particle (4He nucleus).

Correct: (A)

Strontium produces a characteristic crimson-red flame; sodium gives yellow, potassium gives lilac, and copper gives green.