FOUNDATIONS OF CHEMISTRY

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SoluteFoundations of Chemistry

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Extra Practice

Additional practice for every module — matter and measurement, atoms and formulas, the mole, equations, stoichiometry, solutions, acid–base and redox, energy — plus a cumulative mixed set, each with fully worked answers.

Course document · about 35 min read · updated 2026-09-13

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Additional practice for every module, with fully worked answers. Each set adds problems beyond the ones in Course Lessons — same skills, more repetitions, and a few that combine ideas from earlier modules. Work a set after you have read the module and tried its own practice; check each answer only after a genuine attempt.

Notation matches the rest of the course: ×, →, Δ, and ^ for exponents (3.0 × 10^8). Give answers to the correct number of significant figures.

Set 0 — Chemistry math and reasoning

Problems

  1. Express each in scientific notation with the stated significant figures: (a) 0.00104060 (six), (b) 47,900 (three), (c) 1,000,000 (one).
  2. Evaluate (6.022 × 10^23)(1.60 × 10^-19) and report with correct significant figures.
  3. Solve PV = nRT for T, then for n.
  4. Convert 0.85 g/cm³ to kg/m³.
  5. A data table shows y values of 2.0, 4.1, 5.9, 8.0 for x values of 1, 2, 3, 4. What relationship is suggested, and what would confirm it?
  6. Round 0.0834519 to three significant figures, then to one.
  7. How many significant figures are in each: 0.00500, 3400, 3400., 6.02 × 10^23?
  8. A rectangular block measures 2.54 cm × 1.20 cm × 0.75 cm. Give its volume with the correct significant figures.

Answers

  1. (a) 1.04060 × 10^-3; (b) 4.79 × 10^4; (c) 1 × 10^6.
  2. (6.022)(1.60) = 9.6352, exponents 23 + (-19) = 4, so 9.64 × 10^4 (three significant figures, limited by 1.60).
  3. T = PV/(nR); n = PV/(RT).
  4. 0.85 g/cm³ × (1 kg / 1000 g) × (10^6 cm³ / 1 m³) = 850 kg/m³.
  5. A direct proportion (y ≈ 2x). Confirm with more points, a line of best fit through the origin, and a near-zero intercept within uncertainty.
  6. 0.0835, then 0.08.
  7. 0.00500: three; 3400: two (ambiguous, treat as two); 3400.: four; 6.02 × 10^23: three.
  8. 2.54 × 1.20 × 0.75 = 2.286, round to two significant figures (0.75 limits): 2.3 cm³.

Set 1 — Matter, measurement, and density

Problems

  1. Classify each as element, compound, homogeneous mixture, or heterogeneous mixture: helium, table salt, seawater, sand, air, stainless steel, chicken noodle soup.
  2. Label each change physical or chemical: dry ice subliming, iron rusting, milk souring, dissolving sugar in tea, cutting aluminum foil, a glow stick lighting.
  3. Which properties are intensive: mass, temperature, density, volume, melting point, length, boiling point?
  4. A 25.0 mL sample of a liquid has a mass of 22.15 g. Find its density and identify it from this list: ethanol (0.789 g/mL), water (1.00 g/mL), glycerol (1.26 g/mL).
  5. A metal cube 2.00 cm on each side has a mass of 62.4 g. Find its density. Could it be aluminum (2.70 g/mL), iron (7.87 g/mL), or lead (11.3 g/mL)?
  6. A graduated cylinder holds 18.0 mL of water. A stone is added; the level rises to 27.6 mL. The stone's mass is 25.9 g. Find its density.
  7. Four measurements of a 50.00 g standard read 50.42, 50.41, 50.43, 50.42 g. Describe the precision and accuracy.
  8. Convert: 2.5 kg to g; 0.045 L to mL; 350 nm to m; 98.6 °F to °C.

Answers

  1. Helium: element; table salt: compound; seawater: homogeneous mixture; sand: heterogeneous mixture; air: homogeneous mixture; stainless steel: homogeneous mixture; chicken noodle soup: heterogeneous mixture.
  2. Physical: subliming, dissolving sugar, cutting foil. Chemical: rusting, souring, glow stick.
  3. Intensive: temperature, density, melting point, boiling point. Extensive: mass, volume, length.
  4. d = 22.15 g / 25.0 mL = 0.886 g/mL. None match exactly; closest is ethanol, but the value suggests an ethanol–water mixture rather than a pure substance.
  5. V = 2.00³ = 8.00 cm³; d = 62.4 / 8.00 = 7.80 g/mL. Consistent with iron.
  6. V = 27.6 − 18.0 = 9.6 mL; d = 25.9 / 9.6 = 2.7 g/mL (two significant figures).
  7. Precise (values agree closely) but inaccurate (all high by about 0.42 g) — a systematic bias, likely a miscalibrated balance.
  8. 2500 g; 45 mL; 3.50 × 10^-7 m; (98.6 − 32)/1.8 = 37.0 °C.

Set 2 — Atoms, isotopes, ions, formulas, names

Problems

  1. Give the number of protons, neutrons, and electrons in each: ¹⁹F⁻, ²⁴Mg²⁺, ³¹P, ⁵⁶Fe³⁺.
  2. Chlorine has isotopes ³⁵Cl (34.969 u, 75.77%) and ³⁷Cl (36.966 u, 24.23%). Calculate the atomic weight.
  3. Write formulas: potassium sulfide, aluminum oxide, iron(III) chloride, ammonium phosphate, calcium hydrogen carbonate, magnesium nitride.
  4. Name: Na₂CO₃, Cu(NO₃)₂, FeS, (NH₄)₂SO₄, PCl₅, N₂O₄, HClO₃(aq).
  5. How many electrons does a neutral atom of each have, and what ion does each most commonly form: Na, O, Al, S, Ca, Br?
  6. An element X forms an oxide X₂O₃ and a chloride XCl₃. Which group is X most likely in?
  7. Write the formula and name for the compound of Sr²⁺ with the phosphate ion.
  8. Identify the error and correct it: "calcium chloride is CaCl."

Answers

  1. ¹⁹F⁻: 9 p, 10 n, 10 e⁻. ²⁴Mg²⁺: 12 p, 12 n, 10 e⁻. ³¹P: 15 p, 16 n, 15 e⁻. ⁵⁶Fe³⁺: 26 p, 30 n, 23 e⁻.
  2. (34.969)(0.7577) + (36.966)(0.2423) = 26.50 + 8.957 = 35.45 u.
  3. K₂S, Al₂O₃, FeCl₃, (NH₄)₃PO₄, Ca(HCO₃)₂, Mg₃N₂.
  4. Sodium carbonate; copper(II) nitrate; iron(II) sulfide; ammonium sulfate; phosphorus pentachloride; dinitrogen tetroxide; chloric acid.
  5. Na: 11 e⁻ → Na⁺; O: 8 → O²⁻; Al: 13 → Al³⁺; S: 16 → S²⁻; Ca: 20 → Ca²⁺; Br: 35 → Br⁻.
  6. Group 13 (the boron/aluminum group): a +3 ion is consistent with X₂O₃ and XCl₃.
  7. Sr₃(PO₄)₂, strontium phosphate.
  8. Calcium is Ca²⁺ and chloride is Cl⁻, so charge balance gives CaCl₂.

Set 3 — The mole and chemical composition

Problems

  1. Calculate the molar mass of: CO₂, Ca(OH)₂, (NH₄)₂SO₄, C₆H₁₂O₆, MgSO₄·7H₂O.
  2. How many moles are in 12.0 g of NaCl? How many formula units? How many chloride ions?
  3. What is the mass of 2.50 × 10^22 molecules of water?
  4. Calculate the percent by mass of nitrogen in NH₄NO₃.
  5. A compound is 40.00% C, 6.71% H, and 53.29% O by mass. Find its empirical formula. If its molar mass is about 180 g/mol, find its molecular formula.
  6. A 0.500 mol sample of a metal has a mass of 32.7 g. Identify the metal.
  7. How many grams of oxygen are in 25.0 g of Fe₂O₃?
  8. Which has more atoms: 10.0 g of He or 10.0 g of Ne? Show the reasoning.

Answers

  1. CO₂: 44.01; Ca(OH)₂: 74.10; (NH₄)₂SO₄: 132.15; C₆H₁₂O₆: 180.16; MgSO₄·7H₂O: 246.48 g/mol.
  2. 12.0 / 58.44 = 0.205 mol; 0.205 × 6.022 × 10^23 = 1.24 × 10^23 formula units; the same number of Cl⁻ ions, 1.24 × 10^23.
  3. 2.50 × 10^22 / 6.022 × 10^23 = 0.04152 mol; × 18.02 = 0.748 g.
  4. N mass = 2 × 14.01 = 28.02; molar mass of NH₄NO₃ = 80.05; 28.02 / 80.05 × 100 = 35.00%.
  5. Moles: C 40.00/12.01 = 3.331, H 6.71/1.008 = 6.657, O 53.29/16.00 = 3.331. Divide by 3.331: C₁H₂O₁ → empirical CH₂O (30.03 g/mol). 180 / 30.03 ≈ 6, so molecular formula C₆H₁₂O₆.
  6. 32.7 g / 0.500 mol = 65.4 g/mol; that is zinc.
  7. Molar mass Fe₂O₃ = 159.69; O fraction = 48.00 / 159.69 = 0.3006; 25.0 × 0.3006 = 7.51 g of oxygen.
  8. He: 10.0/4.003 = 2.50 mol; Ne: 10.0/20.18 = 0.4955 mol. Helium has about five times as many atoms because its molar mass is smaller.

Set 4 — Chemical equations and reaction patterns

Problems

  1. Balance: Al + O₂ → Al₂O₃; C₃H₈ + O₂ → CO₂ + H₂O; Fe + H₂O → Fe₃O₄ + H₂; KClO₃ → KCl + O₂.
  2. Classify each as combination, decomposition, single replacement, double replacement, or combustion: 2H₂ + O₂ → 2H₂O; CaCO₃ → CaO + CO₂; Zn + CuSO₄ → ZnSO₄ + Cu; AgNO₃ + NaCl → AgCl + NaNO₃; CH₄ + 2O₂ → CO₂ + 2H₂O.
  3. Predict the products and balance: aqueous barium chloride + aqueous sodium sulfate.
  4. Predict whether a reaction occurs when solid copper is placed in aqueous zinc nitrate. Explain using the activity series.
  5. Assign oxidation numbers to every atom in KMnO₄, Cr₂O₇²⁻, H₂O₂, and NaH.
  6. In 2Na + Cl₂ → 2NaCl, identify what is oxidized, what is reduced, the oxidizing agent, and the reducing agent.
  7. Write the balanced molecular, complete ionic, and net ionic equations for aqueous lead(II) nitrate + aqueous potassium iodide.
  8. Balance and classify: NH₄NO₃ → N₂O + H₂O.

Answers

  1. 4Al + 3O₂ → 2Al₂O₃; C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; 3Fe + 4H₂O → Fe₃O₄ + 4H₂; 2KClO₃ → 2KCl + 3O₂.
  2. Combination; decomposition; single replacement; double replacement; combustion.
  3. BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq) — barium sulfate is insoluble and precipitates.
  4. No reaction. Copper is below zinc in the activity series, so copper cannot displace zinc from solution.
  5. KMnO₄: K +1, Mn +7, O −2. Cr₂O₇²⁻: Cr +6, O −2. H₂O₂: H +1, O −1. NaH: Na +1, H −1.
  6. Sodium is oxidized (0 → +1); chlorine is reduced (0 → −1); Cl₂ is the oxidizing agent; Na is the reducing agent.
  7. Molecular: Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq). Complete ionic: Pb²⁺ + 2NO₃⁻ + 2K⁺ + 2I⁻ → PbI₂(s) + 2K⁺ + 2NO₃⁻. Net ionic: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
  8. NH₄NO₃ → N₂O + 2H₂O — a decomposition reaction.

Set 5 — Stoichiometry, limiting reactant, yield

Problems

  1. For N₂ + 3H₂ → 2NH₃, how many moles of NH₃ form from 4.00 mol of H₂ (excess N₂)?
  2. How many grams of CO₂ are produced by burning 10.0 g of CH₄ in excess oxygen?
  3. 2Al + 3Cl₂ → 2AlCl₃. Starting with 5.40 g Al and 12.0 g Cl₂, find the limiting reactant and the mass of AlCl₃ formed.
  4. In problem 3, how many grams of the excess reactant remain?
  5. A reaction has a theoretical yield of 8.50 g and an actual yield of 7.14 g. Find the percent yield.
  6. How many grams of oxygen are needed to react completely with 24.0 g of Mg to form MgO?
  7. CaCO₃ → CaO + CO₂. What mass of CaCO₃ is needed to produce 5.00 L of CO₂ measured at STP (22.4 L/mol)?
  8. Fe₂O₃ + 3CO → 2Fe + 3CO₂. If 100.0 g of Fe₂O₃ gives 62.5 g of iron, what is the percent yield?

Answers

  1. 4.00 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 2.67 mol NH₃.
  2. 10.0 / 16.04 = 0.6234 mol CH₄ → 0.6234 mol CO₂ → × 44.01 = 27.4 g.
  3. Al: 5.40 / 26.98 = 0.2001 mol. Cl₂: 12.0 / 70.90 = 0.1693 mol. Ratio needed is 2 Al : 3 Cl₂. From Cl₂: 0.1693 × 2/3 = 0.1129 mol Al needed — less than available, so Cl₂ is limiting. AlCl₃: 0.1693 × 2/3 = 0.1129 mol × 133.33 = 15.0 g.
  4. Al used = 0.1129 mol × 26.98 = 3.05 g; remaining = 5.40 − 3.05 = 2.35 g Al.
  5. 7.14 / 8.50 × 100 = 84.0%.
  6. 2Mg + O₂ → 2MgO. 24.0 / 24.31 = 0.9873 mol Mg → 0.4936 mol O₂ → × 32.00 = 15.8 g.
  7. 5.00 / 22.4 = 0.2232 mol CO₂ → 0.2232 mol CaCO₃ → × 100.09 = 22.3 g.
  8. 100.0 / 159.69 = 0.6262 mol Fe₂O₃ → 1.2524 mol Fe → × 55.85 = 69.95 g theoretical. 62.5 / 69.95 × 100 = 89.3%.

Set 6 — Solutions and aqueous reactions

Problems

  1. What is the molarity of a solution made by dissolving 14.6 g of NaCl in enough water to make 500.0 mL?
  2. How many grams of KNO₃ are needed to prepare 250.0 mL of 0.400 M solution?
  3. You dilute 25.0 mL of 6.00 M HCl to 500.0 mL. Find the new concentration.
  4. How many milliliters of 12.0 M stock are needed to make 2.00 L of 0.150 M solution?
  5. Classify each in water as strong electrolyte, weak electrolyte, or nonelectrolyte: NaCl, CH₃COOH, C₆H₁₂O₆ (glucose), HCl, NH₃.
  6. Write the net ionic equation for mixing aqueous Na₂CO₃ and aqueous CaCl₂.
  7. Using the compact solubility guide, predict whether each is soluble: AgCl, KNO₃, BaSO₄, PbI₂, Na₂S.
  8. A 0.150 M solution of CaCl₂ — what is the concentration of chloride ions?

Answers

  1. 14.6 / 58.44 = 0.2498 mol; 0.2498 / 0.5000 L = 0.500 M.
  2. 0.2500 L × 0.400 M = 0.1000 mol × 101.11 = 10.1 g.
  3. M₁V₁ = M₂V₂: (6.00)(25.0) = M₂(500.0); M₂ = 0.300 M.
  4. (12.0)V₁ = (0.150)(2000); V₁ = 25.0 mL.
  5. NaCl: strong; CH₃COOH: weak; glucose: nonelectrolyte; HCl: strong; NH₃: weak.
  6. Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s).
  7. Soluble: KNO₃, Na₂S. Insoluble: AgCl, BaSO₄, PbI₂.
  8. CaCl₂ → Ca²⁺ + 2Cl⁻, so [Cl⁻] = 2 × 0.150 = 0.300 M.

Set 7 — Acid–base and redox stoichiometry

Problems

  1. Write balanced neutralization equations: HCl + NaOH; H₂SO₄ + KOH; HNO₃ + Ca(OH)₂; H₃PO₄ + NaOH (complete neutralization).
  2. A 25.00 mL sample of HCl is titrated with 0.1050 M NaOH; the endpoint is at 22.40 mL. Find the concentration of the HCl.
  3. It takes 18.6 mL of 0.200 M NaOH to neutralize 20.0 mL of H₂SO₄. Find the H₂SO₄ concentration.
  4. How many milliliters of 0.150 M Ca(OH)₂ neutralize 30.0 mL of 0.250 M HNO₃?
  5. Identify the oxidizing agent and reducing agent in MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
  6. In an acidic titration, 24.10 mL of 0.02000 M KMnO₄ reacts completely with an iron(II) solution using the equation in problem 5. How many moles of Fe²⁺ were present?
  7. Is the reaction Zn + 2H⁺ → Zn²⁺ + H₂ an oxidation–reduction reaction? Identify the electron transfer.
  8. What volume of 0.500 M HCl contains exactly 0.0250 mol of acid?

Answers

  1. HCl + NaOH → NaCl + H₂O; H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O; 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O; H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O.
  2. mol NaOH = 0.02240 L × 0.1050 M = 2.352 × 10^-3; equal moles of HCl; 2.352 × 10^-3 / 0.02500 L = 0.09408 M.
  3. mol NaOH = 0.0186 × 0.200 = 3.72 × 10^-3; H₂SO₄ needs 2 NaOH per acid, so mol H₂SO₄ = 1.86 × 10^-3; / 0.0200 L = 0.0930 M.
  4. mol HNO₃ = 0.0300 × 0.250 = 7.50 × 10^-3; Ca(OH)₂ supplies 2 OH⁻, so mol Ca(OH)₂ = 3.75 × 10^-3; / 0.150 M = 0.0250 L = 25.0 mL.
  5. MnO₄⁻ is the oxidizing agent (Mn goes +7 → +2); Fe²⁺ is the reducing agent (Fe goes +2 → +3).
  6. mol MnO₄⁻ = 0.02410 L × 0.02000 M = 4.820 × 10^-4; the ratio is 1 MnO₄⁻ : 5 Fe²⁺, so mol Fe²⁺ = 2.410 × 10^-3.
  7. Yes. Zn is oxidized (0 → +2, loses two electrons); H⁺ is reduced (+1 → 0, gains electrons) to form H₂.
  8. 0.0250 mol / 0.500 M = 0.0500 L = 50.0 mL.

Set 8 — Energy, heat, and calorimetry

Problems

  1. How much heat is required to raise 250.0 g of water from 22.0 °C to 95.0 °C? (c of water = 4.184 J/g·°C)
  2. A 55.0 g piece of copper (c = 0.385 J/g·°C) cools from 99.0 °C to 25.0 °C. How much heat does it release?
  3. A 45.0 g metal at 100.0 °C is dropped into 100.0 g of water at 23.0 °C in an insulated cup; the final temperature is 26.8 °C. Find the specific heat of the metal.
  4. Combustion of 0.500 g of a snack in a bomb calorimeter (heat capacity 6.20 kJ/°C) raises the temperature by 2.15 °C. Find the energy released per gram.
  5. Convert: 500.0 cal to J; 2.50 kJ to J; 1.00 Cal (food calorie) to J.
  6. Is a reaction that raises the temperature of its surroundings exothermic or endothermic? What is the sign of ΔH?
  7. How much heat is absorbed when 20.0 g of ice at 0 °C melts? (heat of fusion of water = 334 J/g)
  8. If 1.00 kJ of heat is added to 100.0 g of aluminum (c = 0.897 J/g·°C) at 20.0 °C, what is the final temperature?

Answers

  1. q = (250.0)(4.184)(95.0 − 22.0) = 250.0 × 4.184 × 73.0 = 7.64 × 10^4 J (76.4 kJ).
  2. q = (55.0)(0.385)(25.0 − 99.0) = −1.57 × 10^3 J; it releases about 1.57 kJ.
  3. Heat lost by metal = heat gained by water. (100.0)(4.184)(26.8 − 23.0) = 1590 J. 1590 = (45.0)(c)(100.0 − 26.8); c = 1590 / (45.0 × 73.2) = 0.483 J/g·°C.
  4. q = (6.20 kJ/°C)(2.15 °C) = 13.3 kJ; per gram: 13.3 / 0.500 = 26.7 kJ/g.
  5. 500.0 cal × 4.184 = 2092 J; 2.50 kJ = 2500 J; 1.00 Cal = 1000 cal = 4184 J.
  6. Exothermic; ΔH is negative.
  7. q = (20.0 g)(334 J/g) = 6.68 × 10^3 J (6.68 kJ).
  8. ΔT = 1000 J / (100.0 × 0.897) = 11.1 °C; final temperature = 20.0 + 11.1 = 31.1 °C.

Set 9 — Enthalpy and Hess's law

Problems

  1. Given C(s) + O2(g) → CO2(g), ΔH₁ = −393.5 kJ, and CO(g) + 1/2 O2(g) → CO2(g), ΔH₂ = −283.0 kJ, use Hess's law to find ΔH for C(s) + 1/2 O2(g) → CO(g).
  2. Using ΔH°f[C2H5OH(l)] = −277.6 kJ/mol, ΔH°f[CO2(g)] = −393.5 kJ/mol, ΔH°f[H2O(l)] = −285.8 kJ/mol, and ΔH°f[O2(g)] = 0, calculate ΔH°rxn for C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l).
  3. If A → B has ΔH = +25 kJ, what is ΔH for 2 B → 2 A?
  4. A reaction run in a bomb calorimeter releases 50 kJ to its surroundings. Is the reaction exothermic or endothermic, and what is the sign of ΔH for the reaction as written?
  5. A thermochemical equation is multiplied through by 1/2 to balance it against another step in a Hess's law problem. What happens to its ΔH, and why is that the correct move rather than leaving ΔH unchanged?
  6. Using ΔH°f[NH3(g)] = −46.1 kJ/mol, ΔH°f[N2(g)] = 0, and ΔH°f[H2(g)] = 0, find ΔH°rxn for N2(g) + 3 H2(g) → 2 NH3(g).
  7. A reaction has ΔH = −150 kJ. What is ΔH for the reverse reaction? Does the "reverse a step, flip the sign" rule still apply once a step has also been scaled by some factor?
  8. Given A → B, ΔH = +40 kJ, and B → C, ΔH = −90 kJ, find ΔH for A → C and state whether the overall process is exothermic or endothermic.

Answers

  1. Target = (equation 1) − (equation 2): ΔH = ΔH₁ − ΔH₂ = (−393.5) − (−283.0) = −110.5 kJ.
  2. ΔH°rxn = [2(−393.5) + 3(−285.8)] − [(−277.6) + 3(0)] = [−787.0 − 857.4] − [−277.6] = −1644.4 + 277.6 = −1366.8 kJ.
  3. Reverse A → B to get B → A, ΔH = −25 kJ; scale by 2 for 2B → 2A: ΔH = −50 kJ.
  4. Exothermic (energy leaves the system into the surroundings); ΔH is negative for the reaction.
  5. ΔH is also multiplied by 1/2 — enthalpy scales with the amount of reaction, so an equation and its ΔH must always be scaled together. Leaving ΔH unchanged would report the energy for the original (larger) amount while describing the smaller, halved reaction.
  6. ΔH°rxn = [2(−46.1)] − [0 + 3(0)] = −92.2 kJ.
  7. ΔH = +150 kJ for the reverse reaction. Yes — reversing always flips the sign regardless of whatever scaling has also been applied; scaling and reversing are independent operations that both have to be tracked and applied consistently.
  8. Add the steps: ΔH = 40 + (−90) = −50 kJ. Exothermic — the net ΔH for the overall process is negative, so energy is released overall even though the first step absorbs energy.

Set 10 — Light, quantization, and the atomic model

Problems

  1. Find the frequency and energy of a photon with wavelength 486 nm (a line in hydrogen's visible emission spectrum). (c = 2.998 × 10^8 m/s, h = 6.626 × 10^-34 J·s)
  2. A photon carries 3.37 × 10^-19 J of energy. What is its wavelength, in nm?
  3. An electron drops from a higher energy level to a lower one and emits a photon. Does the photon's energy equal the energy of the upper level, the energy of the lower level, or something else? State the relationship.
  4. Without calculating exact values, rank photons of wavelength 400 nm, 550 nm, and 700 nm from highest energy to lowest.
  5. In one or two sentences, explain why the quantum mechanical model describes an orbital as a probability region rather than a traceable path.
  6. Using Eₙ = −2.18 × 10^-18 J / n², find the energy of hydrogen's n = 2 level.
  7. A metal's photoelectric threshold requires at least 3.5 × 10^-19 J per photon to eject an electron. Does a 600 nm photon of visible red light carry enough energy? Show the comparison.
  8. Explain why atomic emission spectra consist of discrete lines rather than a continuous rainbow of color.

Answers

  1. ν = c/λ = (2.998 × 10^8) / (486 × 10^-9) = 6.17 × 10^14 Hz; E = hν = (6.626 × 10^-34)(6.17 × 10^14) = 4.09 × 10^-19 J.
  2. ν = E/h = (3.37 × 10^-19) / (6.626 × 10^-34) = 5.09 × 10^14 Hz; λ = c/ν = (2.998 × 10^8) / (5.09 × 10^14) = 5.89 × 10^-7 m = 589 nm.
  3. The photon's energy equals the difference between the two levels: E_photon = E_upper − E_lower. It is not the energy of either level by itself.
  4. Highest to lowest energy: 400 nm > 550 nm > 700 nm. Energy is inversely proportional to wavelength (E = hc/λ), so the shortest wavelength carries the most energy.
  5. The uncertainty principle means an electron's exact position and exact momentum cannot both be known at once, so quantum mechanics can only describe where an electron is likely to be found — a probability distribution — rather than a specific path it travels along.
  6. E₂ = −2.18 × 10^-18 / 2² = −2.18 × 10^-18 / 4 = −5.45 × 10^-19 J.
  7. E_photon = hc/λ = (6.626 × 10^-34)(2.998 × 10^8) / (600 × 10^-9) = 3.31 × 10^-19 J. That is less than the 3.5 × 10^-19 J threshold, so no — this photon cannot eject an electron from this metal, regardless of how many of these photons arrive.
  8. Electrons occupy only specific, quantized energy levels, so an emitted photon's energy must equal one specific allowed difference between two levels. Only certain photon energies — and therefore only certain wavelengths, seen as specific colors — are possible, producing distinct lines rather than every wavelength blending into a continuum.

Problems

  1. Write the full electron configuration for Fe (Z = 26) and for the ion Fe²⁺.
  2. Write the electron configuration for Cu (Z = 29). Why doesn't it follow the "expected" fill order?
  3. Actual first ionization energies: N ≈ 1402 kJ/mol, O ≈ 1314 kJ/mol, F ≈ 1681 kJ/mol. Rank the three, and explain why O doesn't fall between N and F the way a strict left-to-right trend would predict.
  4. Is the oxidation number of sulfur in SO2 the same as an ionic charge sulfur actually carries? Explain.
  5. Which has the larger radius, Na or Na⁺? Explain in terms of what changes when the electron is removed.
  6. Write the condensed electron configuration for Br⁻ (Z = 35, one extra electron). What familiar noble-gas configuration does it match?
  7. Rank atomic radius for P, S, and Cl (all period 3) from smallest to largest, and explain the trend in terms of effective nuclear charge.
  8. Explain why sodium's second ionization energy is dramatically higher than its first, rather than just somewhat higher.

Answers

  1. Fe: [Ar]4s²3d⁶. Fe²⁺: [Ar]3d⁶ — the two 4s electrons are removed first, not two 3d electrons.
  2. Cu: [Ar]4s¹3d¹⁰, not the "expected" [Ar]4s²3d⁹. A completely filled 3d¹⁰ subshell is extra stable, so the atom's actual lowest-energy configuration promotes one 4s electron into 3d to complete it.
  3. Order: F (1681) > N (1402) > O (1314). Nitrogen has a half-filled 2p³ subshell, which is extra stable and resists losing an electron; oxygen's fourth 2p electron must pair up with an existing one, and that electron-electron repulsion makes it easier to remove than one of nitrogen's unpaired electrons — so oxygen's ionization energy dips below nitrogen's even though oxygen is farther right.
  4. No. Oxidation number is a bookkeeping assignment applied to every atom in a formula (sulfur in SO2 is assigned +4), not a measurement of real, countable charge — SO2 is a covalent molecule, and sulfur does not carry an actual +4 ionic charge.
  5. Na is larger. Removing Na's single valence electron leaves Na⁺ with one fewer occupied shell and a full outer shell held more tightly by the same nuclear charge acting on fewer electrons, pulling the remaining electrons in closer.
  6. Br: [Ar]4s²3d¹⁰4p⁵; gaining one electron fills the 4p subshell: Br⁻: [Ar]4s²3d¹⁰4p⁶. That is the same electron count and arrangement as krypton — Br⁻ is isoelectronic with Kr.
  7. Smallest to largest: Cl < S < P. All three are in the same period, so they share the same outer shell (n = 3), but effective nuclear charge increases left to right as protons are added; Cl's higher effective nuclear charge pulls its outer electrons in tighter than S's or P's, making it the smallest of the three.
  8. Removing Na's single 3s valence electron leaves Na⁺ with a complete, neon-like inner shell. Removing a second electron means breaking into that already-complete, tightly held shell sitting much closer to the nucleus — a fundamentally harder task than removing one loosely held outer electron, which is why the jump is dramatic rather than incremental.

Set 12 — Bonding and Lewis structures

Problems

  1. CO3²⁻ has three equivalent resonance structures. What does that imply about the lengths of its three C–O bonds, and why isn't the ion better described as "two single bonds and one double bond, fixed in place"?
  2. In the best Lewis structure for NO3⁻ (one N=O double bond, two N–O single bonds each bearing a negative charge), find the formal charge on the nitrogen and on each type of oxygen, and confirm they sum to the ion's overall −1 charge.
  3. SF6 has sulfur surrounded by six bonding pairs — more than an octet. Why is this structure accepted despite "breaking" the octet rule?
  4. Two candidate Lewis structures for the same molecule are proposed: one has formal charges of 0 on every atom, the other has +2 and −2 on two atoms. Which is generally the better structure, and why does formal charge help decide?
  5. Does BF3 satisfy the octet rule at boron? What is unusual about boron's bonding here?
  6. CO's best Lewis structure has a carbon–oxygen triple bond with one lone pair on each atom. Find the formal charge on each atom, and explain why this structure — with a negative formal charge on carbon, the less electronegative atom — is still accepted as the best structure for CO.
  7. NO has 11 valence electrons total. Explain why it can never satisfy the octet rule for both atoms no matter how the structure is drawn.
  8. Expanded-octet species like PCl5 and SF6 only occur for elements in period 3 and below. Why can't a period-2 element like nitrogen or oxygen form an expanded octet the same way?

Answers

  1. All three C–O bonds are experimentally identical in length — each intermediate between a single and a double bond — because the true structure is a hybrid of all three resonance forms, not a molecule that is "really" one particular resonance structure or that flips between them over time.
  2. Nitrogen (one double bond, three single... here: one double + two single = 4 bond pairs, 0 lone pairs): FC = 5 − 0 − 8/2 = +1. The doubly bonded oxygen (2 lone pairs, one double bond): FC = 6 − 4 − 4/2 = 0. Each singly bonded oxygen (3 lone pairs, one single bond): FC = 6 − 6 − 2/2 = −1. Sum: +1 + 0 + (−1) + (−1) = −1, matching the ion's charge.
  3. Sulfur is in period 3, which has accessible d-character in bonding that period-2 elements (like carbon or nitrogen) lack, allowing it to accommodate more than eight electrons around the central atom. The octet rule is a strong guideline for period-2 elements, not a universal law.
  4. The all-zero structure is generally better. Lewis structures with formal charges as close to zero as possible (and any negative formal charge on the most electronegative atom present) are the most stable, most realistic candidates — large formal charges on adjacent or nearby atoms usually signal a less favorable electron arrangement.
  5. No — boron ends up with only 6 electrons (three bonding pairs, no lone pair) in BF3, not 8. Boron is a well-known, stable exception to the octet rule; its compounds are often described as "electron deficient," which is also why BF3 readily accepts a lone pair from another species (like NH3) to complete an octet.
  6. Carbon (triple bond, one lone pair): FC = 4 − 2 − 6/2 = −1. Oxygen (triple bond, one lone pair): FC = 6 − 2 − 6/2 = +1. Formal charge is only one factor in judging a Lewis structure — completing every atom's octet takes priority, and the triple-bonded structure is the only arrangement that gives both carbon and oxygen a full octet, so it is accepted as the best structure even though it puts a negative formal charge on the less electronegative atom.
  7. Eleven is an odd number of electrons, which can never be split entirely into bonding pairs and lone pairs — one electron is always left unpaired on some atom no matter how the structure is arranged, so at least one atom always falls short of (or exceeds) a filled octet. NO is a stable example of an odd-electron, radical molecule.
  8. Expanded octets require low-energy, accessible orbitals beyond the outer s and p subshells that period-3-and-below elements have available in the same valence shell. Period-2 elements have only 2s and 2p orbitals to work with — a hard ceiling of eight valence electrons — so nitrogen and oxygen have no equivalent extra orbital space to expand into.

Set 13 — Molecular geometry and polarity

Problems

  1. For CO2: how many electron domains surround the central carbon? Name the electron-domain geometry, the molecular geometry, and state whether the molecule is polar.
  2. For NH3: how many electron domains surround nitrogen? Name the electron-domain geometry and the molecular geometry (they differ here), and state whether the molecule is polar.
  3. SO2 also has an AX2-style formula on paper, similar to CO2, but SO2 is polar while CO2 is not. What's different about the electron-domain count around the central atom, and how does that explain the difference in polarity?
  4. CH4 and CH3Cl are both roughly tetrahedral. Only one is polar. Which one, and why?
  5. Ozone, O3, has a central oxygen bonded to two other oxygens (with resonance) and one lone pair. How many electron domains surround the central oxygen, what is the molecular geometry, and is the molecule polar?
  6. SF4 has five electron domains around sulfur (four bonding, one lone pair). Name the electron-domain geometry and the molecular geometry, and state whether the molecule is polar.
  7. XeF4 has six electron domains around xenon (four bonding, two lone pairs). Name the electron-domain geometry and the molecular geometry, and state whether the molecule is polar.
  8. BF3 is nonpolar, but PF3 — which looks similar on paper — is polar. What structural difference explains this?

Answers

  1. Two electron domains (two C=O double-bond regions, each counted once). Electron-domain geometry: linear. Molecular geometry: linear. The two C=O bond dipoles point in exactly opposite directions and cancel, so CO2 is nonpolar overall despite having polar bonds.
  2. Four electron domains (three N–H bonds plus one lone pair). Electron-domain geometry: tetrahedral. Molecular geometry (naming only the atoms): trigonal pyramidal. The bond dipoles do not cancel because of the asymmetric, lone-pair-distorted shape, so NH3 is polar.
  3. SO2's central sulfur has three electron domains (two bonding regions to oxygen plus one lone pair), giving a bent molecular geometry — unlike CO2's two domains and linear shape. The bent shape means SO2's two S–O bond dipoles do not point in opposite directions, so they do not cancel, making SO2 polar; CO2's linear, symmetric shape lets its bond dipoles cancel exactly.
  4. CH3Cl is polar; CH4 is nonpolar. CH4's four identical C–H bonds are arranged symmetrically around carbon so their dipoles cancel exactly. Replacing one H with the more electronegative Cl breaks that symmetry — the C–Cl bond dipole no longer has an equal, opposite partner to cancel it, leaving CH3Cl with a net dipole.
  5. Three electron domains around the central oxygen (one lone pair plus two bonding regions to the outer oxygens, treating the resonance-delocalized double-bond character as one region each). Molecular geometry: bent. The bond dipoles do not cancel in a bent shape, so ozone is polar.
  6. Electron-domain geometry: trigonal bipyramidal. Molecular geometry: seesaw (the lone pair occupies an equatorial position, distorting the remaining four bonds out of a symmetric shape). Polar — the lone pair breaks the symmetry needed for the four S–F dipoles to cancel.
  7. Electron-domain geometry: octahedral. Molecular geometry: square planar (the two lone pairs sit opposite each other, on the axial positions, leaving the four Xe–F bonds in one symmetric plane). Nonpolar — the square-planar arrangement is symmetric enough that the four bond dipoles still cancel exactly, even with two lone pairs present.
  8. BF3 has only three electron domains around boron (three bonds, no lone pair — boron is electron-deficient), giving a perfectly symmetric trigonal planar shape whose three B–F dipoles cancel exactly. PF3 has four electron domains around phosphorus (three bonds plus one lone pair), giving a trigonal pyramidal shape; that lone pair breaks the symmetry the same way it does in NH3, so PF3's three P–F dipoles do not cancel and the molecule is polar.

Set 14 — Gases, intermolecular forces, and phases

Problems

  1. A 2.50 mol sample of an ideal gas is held at 1.15 atm and 298 K. Find its volume. (R = 0.08206 L·atm/mol·K)
  2. A gas occupies 5.00 L at 1.00 atm and 273 K. What volume does it occupy at 2.50 atm and 310 K?
  3. A gas mixture at a total pressure of 1.20 atm contains 0.40 mol of gas A and 0.60 mol of gas B. Find the partial pressure of gas A.
  4. Rank CH4, NH3, and H2O from lowest to highest boiling point, and explain the ranking using intermolecular forces.
  5. Real gases deviate from the ideal gas law at high pressure. Which assumption of the ideal gas model breaks down, and why does that matter specifically at high pressure rather than at low pressure?
  6. A gas sample at STP occupies 11.2 L. How many moles is this, and how many molecules? (STP molar volume ≈ 22.4 L/mol)
  7. Water has a much higher boiling point than CH4 or H2S, molecules of comparable or greater molar mass. Explain why, in terms of intermolecular forces.
  8. A rigid, sealed container of gas is heated, with volume and moles held fixed. Using the ideal gas law, explain what happens to the pressure and why.

Answers

  1. V = nRT/P = (2.50)(0.08206)(298) / 1.15 = 53.2 L.
  2. Combined gas law: V2 = P1V1T2 / (T1P2) = (1.00)(5.00)(310) / [(273)(2.50)] = 2.27 L.
  3. P_A = P_total × (n_A / n_total) = 1.20 × (0.40 / 1.00) = 0.48 atm.
  4. Lowest to highest: CH4 < NH3 < H2O. CH4 has only weak dispersion forces (no polar bonds). NH3 and H2O both hydrogen-bond, but each water molecule can form more hydrogen bonds on average (two O–H donors and two lone pairs as acceptors, versus ammonia's one lone pair as acceptor), giving water the more extensive hydrogen-bonding network and the highest boiling point of the three.
  5. The ideal gas model assumes gas particles have negligible volume compared to the container and don't attract or repel each other. At high pressure, particles are forced close together, so their actual volume becomes a non-negligible fraction of the container's volume, and intermolecular attractions become significant — both assumptions break down, and real gas behavior diverges from the ideal gas law. At low pressure, particles are far apart on average, so both assumptions hold much better.
  6. n = V / 22.4 = 11.2 / 22.4 = 0.500 mol; molecules = 0.500 × 6.022 × 10^23 = 3.01 × 10^23 molecules.
  7. Despite a smaller molar mass than H2S and a similar one to CH4, water's O–H bonds let it hydrogen-bond extensively — each molecule can donate two hydrogen bonds (from its two O–H bonds) and accept two more (at its two lone pairs), building a demanding network that takes much more energy to break apart. CH4 has no polar bonds at all (only weak dispersion forces), and H2S — despite being polar — barely hydrogen-bonds because sulfur isn't electronegative enough for the S–H bond to qualify, leaving it with only dipole–dipole and dispersion forces.
  8. Pressure increases. With n and V fixed in PV = nRT, P is directly proportional to T — raising the temperature increases the average kinetic energy and collision force of the gas particles against the container walls, and with nowhere for the gas to expand into, that extra collision force shows up entirely as higher pressure.

Cumulative set — mixed skills

Problems

  1. A 0.750 g sample of an unknown group 2 carbonate MCO₃ is heated and loses 0.330 g as CO₂. Identify the metal M.
  2. How many milliliters of 0.200 M AgNO₃ are needed to precipitate all the chloride in 25.0 mL of 0.150 M CaCl₂ as AgCl?
  3. Burning 2.00 g of a hydrocarbon CₓHᵧ produces 6.29 g CO₂ and 3.44 g H₂O. Find the empirical formula.
  4. A 100.0 mL solution contains 0.0100 mol HCl. What volume of 0.0500 M NaOH neutralizes it, and what is the [Cl⁻] after neutralization (assume volumes add)?
  5. 2KClO₃ → 2KCl + 3O₂. If 4.90 g of KClO₃ decomposes completely, what volume of O₂ results at STP (22.4 L/mol)?
  6. A reaction releases 45.0 kJ and warms 1.50 kg of water in a jacket. By how many degrees does the water temperature rise? (c = 4.184 J/g·°C)
  7. Rank these by number of atoms, greatest first: 1.0 mol H₂O, 1.0 mol CO₂, 18 g H₂O, 6.022 × 10^23 molecules of O₂.
  8. A student reports the density of a liquid as 1.2456 g/mL from a mass of 12.4 g and a volume of 9.95 mL. Critique the reported significant figures and give the correct answer.

Answers

  1. mol CO₂ = 0.330 / 44.01 = 7.498 × 10^-3; same moles of MCO₃. Molar mass of MCO₃ = 0.750 / 7.498 × 10^-3 = 100.0 g/mol. Subtract CO₃²⁻ (60.01): M = 40.0 g/mol → calcium; the compound is CaCO₃.
  2. mol Cl⁻ = 2 × (0.0250 L × 0.150 M) = 7.50 × 10^-3; AgNO₃ needed 1:1, so 7.50 × 10^-3 / 0.200 M = 0.0375 L = 37.5 mL.
  3. mol C = 6.29 / 44.01 = 0.1429; mol H = 2 × (3.44 / 18.02) = 0.3818. Ratio H:C = 0.3818 / 0.1429 = 2.67 ≈ 8/3. Multiply by 3: C₃H₈ — the empirical formula is C₃H₈ (propane).
  4. mol NaOH needed = 0.0100; 0.0100 / 0.0500 M = 0.200 L = 200 mL. Total volume = 0.1000 + 0.200 = 0.300 L; mol Cl⁻ = 0.0100 (spectator, unchanged); [Cl⁻] = 0.0100 / 0.300 = 0.0333 M.
  5. mol KClO₃ = 4.90 / 122.55 = 0.03998; mol O₂ = 0.03998 × 3/2 = 0.05997; × 22.4 = 1.34 L.
  6. 45.0 × 10^3 J = (1500 g)(4.184)(ΔT); ΔT = 45000 / 6276 = 7.17 °C.
  7. 1.0 mol CO₂ = 3.0 mol atoms (greatest). 1.0 mol H₂O = 3.0 mol atoms and 18 g H₂O = 1.0 mol = 3.0 mol atoms (tie with it). 6.022 × 10^23 molecules O₂ = 1.0 mol × 2 = 2.0 mol atoms (least). Order: CO₂ ≈ 1.0 mol H₂O ≈ 18 g H₂O > O₂.
  8. The mass (12.4 g, three significant figures) limits the result to three significant figures, not five. 12.4 / 9.95 = 1.246, so the density should be reported as 1.25 g/mL.

How to use these sets

  • Do them after a real attempt at the module's own practice, not instead of it.
  • Write the full solution path — units, the conversion factor, the balanced equation — not just the final number. The Study Toolkit has the templates.
  • When an answer is wrong, find the exact step where your work and the solution diverge before rereading the module. That is where the misconception is.
  • The Assessment Package has cumulative practice and full unit and final blueprints when you are ready to test the whole span.