ACS General Chemistry Practice Test 15

Correct: (A)

Average atomic mass = (0.7578)(34.969) + (0.2422)(36.966) = 26.50 + 8.95 = 35.45 amu. The average must lie between the two isotope masses and closer to the more abundant one.

Correct: (B)

Neutral Mn is [Ar]4s2 3d5. Transition metal cations lose their 4s electrons first, so Mn2+ is [Ar]3d5.

Correct: (D)

First ionization energy generally increases across a period due to increasing effective nuclear charge, so Si > Mg > Al > Na (the Al dip below Mg does not change the overall order).

Correct: (C)

Atomic radius increases down a group and decreases across a period. Potassium is lowest and farthest left in this set, so it is largest.

Correct: (A)

Electron affinity generally becomes more exothermic toward the upper right of the periodic table, and chlorine has the most exothermic electron affinity of any element.

Correct: (D)

Set D satisfies l < n, -l ≤ ml ≤ l, and ms = ±1/2. The others violate the rules: l = 3 is forbidden for n = 3, ml = 2 is forbidden for l = 1, and ml = -3 is forbidden for l = 2.

Correct: (B)

Carbon has 4 valence electrons and in carbonate it has no lone pairs and 4 bonds (one double, two single). Formal charge = 4 – (0 + 4) = 0.

Correct: (C)

XeF4 has six electron domains (four bonding pairs and two lone pairs), giving an octahedral electron geometry. The two lone pairs occupy opposite positions, leaving a square planar molecular geometry.

Correct: (D)

CCl4, CO2, and SO3 are symmetrical, so their bond dipoles cancel. CH3Cl is tetrahedral with three C-H bonds and one polar C-Cl bond, giving a net dipole.

Correct: (B)

O2+ has 11 valence electrons: (σ2s)2(σ*2s)2(σ2p)2(π2p)4(π*2p)1. Bond order = (10 – 5)/2 = 2.5.

Correct: (A)

The nitrile carbon forms one single bond to carbon and a triple bond to nitrogen (two electron domains, linear geometry), so it is sp hybridized.

Correct: (C)

Ethanol has an O-H bond, and the H bonded to O can act as a hydrogen bond donor. The other substances lack H bonded to N, O, or F, so their strongest forces are dipole-dipole or dispersion.

Correct: (D)

The H-F bond is exceptionally strong (568 kJ/mol) due to the large electronegativity difference and excellent 1s-2p orbital overlap, exceeding H-H (436), C-H (413), and Cl-Cl (242) kJ/mol.

Correct: (B)

Each formula unit contains 6 O atoms: (0.250 mol)(6)(6.022 × 10^23 atoms/mol) = 9.03 × 10^23 O atoms.

Correct: (A)

2.0 mol Al would require 3.0 mol Cl2, so Cl2 is the limiting reactant. Moles of AlCl3 = (2.0 mol Cl2)(2 mol AlCl3/3 mol Cl2) = 1.3 mol.

Correct: (C)

Theoretical yield = (50.0 g / 100.1 g/mol)(56.08 g/mol) = 28.0 g CaO. Percent yield = (22.4 g / 28.0 g) × 100% = 80.0%.

Correct: (D)

Moles of KMnO4 = 15.8 g / 158.0 g/mol = 0.100 mol. Molarity = 0.100 mol / 0.2500 L = 0.400 M.

Correct: (B)

A precipitation of AgCl occurs; Na+ and NO3- are spectator ions and cancel, leaving Ag+(aq) + Cl-(aq) → AgCl(s).

Correct: (C)

Chlorine goes from an oxidation state of -1 in HCl to 0 in Cl2, a loss of electrons (oxidation). Manganese is reduced (+7 to +2).

Correct: (A)

Moles of C = 6.16 g / 44.01 g/mol = 0.1400 mol; moles of H = 2(2.94 g / 18.02 g/mol) = 0.3263 mol. The H/C ratio is 2.33, and multiplying by 3 gives C3H7.

Correct: (C)

P = nRT/V = (0.250)(0.08206)(300)/5.00 = 1.23 atm.

Correct: (A)

Mole fraction of O2 = 2.0/5.0 = 0.40; P(O2) = 0.40 × 2.50 = 1.00 atm.

Correct: (C)

Graham's law: rate ratio = sqrt(M_Y/M_X), so M_Y = 64.0 × 2^2 = 256 g/mol.

Correct: (D)

NH3 has the strongest intermolecular attractions (hydrogen bonding), producing the largest negative deviation from ideal behavior.

Correct: (B)

Dimethyl ether has the weakest intermolecular forces (small, no hydrogen bonding), so it evaporates most readily and has the highest vapor pressure.

Correct: (D)

8.0 g H2 = 4.0 mol; (4.0/2) × 572 kJ = 1,144 kJ released.

Correct: (B)

q = (50.0)(4.184)(16.5) = 3,452 J = 3.45 kJ.

Correct: (A)

Double the first equation and subtract the second: 2(-394) – (-566) = -222 kJ.

Correct: (C)

ΔH = [-393.5 + 2(-285.8)] – [-74.8] = -890 kJ.

Correct: (A)

ΔH = (bonds broken) – (bonds formed) = 436 + 242 – 2(431) = -184 kJ.

Correct: (D)

Condensation is exothermic (ΔH < 0) and gas → liquid decreases disorder (ΔS < 0).

Correct: (B)

λ = c/ν = 3.00 × 10^8 / 5.0 × 10^14 = 6.0 × 10^-7 m = 600 nm.

Correct: (A)

Intensity determines the number of photons (hence number of electrons); kinetic energy depends only on frequency.

Correct: (C)

ml must satisfy -l ≤ ml ≤ l; ml = -4 is outside the allowed range for l = 3.

Correct: (D)

Fe is [Ar]4s2 3d6; the ion loses the 4s electrons first, then one 3d electron, giving [Ar]3d5.

Correct: (B)

Doubling [A] quadruples the rate (2nd order in A); doubling [B] doubles the rate (1st order in B).

Correct: (A)

From rate (M/s) = k(M)^2, k must have units of M^-1 s^-1.

Correct: (D)

t1/2 = 0.693/k = 0.693/0.0693 = 10.0 s.

Correct: (C)

A catalyst speeds a reaction by offering a lower-energy pathway; it does not change ΔH or reactant concentrations.

Correct: (B)

The slow step is rate-determining, so the rate law follows its stoichiometry: rate = k[A][B].

Correct: (A)

Only a first-order reaction has a half-life (t1/2 = 0.693/k) that does not depend on initial concentration.

Correct: (B)

Kp = Kc(RT)^Δn with Δn = 2 – 4 = -2, so Kp = 0.50/(0.0821 × 400)^2 = 4.6 × 10^-4.

Correct: (A)

Raising the temperature shifts an exothermic equilibrium toward the reactants, decreasing the yield of SO3.

Correct: (C)

Q = (0.40)^2/(0.10 × 0.20) = 8.0, which is less than Kc, so the reaction proceeds toward the products.

Correct: (A)

With [H2]0 = [I2]0 = 0.25 M, solving (2x)^2/(0.25 – x)^2 = 50 gives x = 0.195, so [HI] = 2x = 0.39 M.

Correct: (D)

From Ksp = 4s^3, s = (7.9 × 10^-9/4)^(1/3) = 1.3 × 10^-3 M.

Correct: (A)

With the common ion, s = Ksp/[Cl-] = 1.8 × 10^-10/0.10 = 1.8 × 10^-9 M.

Correct: (C)

[OH-] = 2 × 0.025 = 0.050 M, so pOH = 1.30 and pH = 12.70.

Correct: (B)

[H+] = (Ka × C)^(1/2) = 2.5 × 10^-3 M, giving pH = 2.60.

Correct: (B)

A weak base mixed with its conjugate acid salt forms a buffer; the other pairs contain strong acids or bases.

Correct: (A)

pH = pKa + log([CN-]/[HCN]) = 9.21 + log(0.10/0.20) = 9.21 – 0.30 = 8.91.

Correct: (D)

At the equivalence point [NH4+] = 0.050 M and Ka = 5.6 × 10^-10, so [H+] = 5.3 × 10^-6 M and pH = 5.28.

Correct: (C)

The equivalence point of a weak acid-strong base titration is basic (pH > 7), so phenolphthalein is appropriate.

Correct: (A)

Three moles of gas are produced from a solid reactant, the largest increase in disorder among the choices.

Correct: (C)

The crossover temperature is T = ΔH/ΔS = 92000/199 ≈ 462 K; above this, -TΔS dominates and ΔG becomes positive.

Correct: (B)

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

Correct: (D)

E = 1.10 – (0.0592/2)log(1.0/0.010) = 1.10 – 0.059 = 1.04 V.

Correct: (C)

The charge is 2.00 × 10800 = 21600 C, giving 0.2239 mol e- and 0.0746 mol Al, which is 0.0746 × 26.98 = 2.01 g.

Correct: (D)

Beta emission ejects a 0(-1)e particle and increases the atomic number by one while the mass number is unchanged.

Correct: (B)

43 days is about 3 half-lives, so 2.0/2^3 = 0.25 g remains.