Practice Exams

Four assembled, timed practice exams — three unit exams and a cumulative final — each built to the assessment blueprints and each with a fully worked solution. The Assessment Package has the blueprints and loose item banks; this page is the part students ask for: practice tests that look like the real thing, with the working shown.

How to use these. Do one under exam conditions — a timer, a calculator, one sheet of allowed notes, no looking anything up — then grade yourself against the solution. A problem you get right but slowly is a problem you do not really own yet. Where you get stuck, the solution names which module, playbook step, or misconception is involved.

Constants and masses used here. Use a periodic table for molar masses; approximate values used in the keys: H 1.01, C 12.01, N 14.01, O 16.00, Na 22.99, Mg 24.31, S 32.07, Cl 35.45, K 39.10, Ca 40.08, Fe 55.85, Cu 63.55, Zn 65.38. Avogadro's number 6.022 × 10²³ mol⁻¹. R = 8.314 J·mol⁻¹·K⁻¹ or 0.08206 L·atm·mol⁻¹·K⁻¹. Specific heat of water 4.18 J·g⁻¹·°C⁻¹. 0 °C = 273.15 K.

Practice Unit Exam 1 — Quantifying Chemical Composition and Change

75 minutes · 100 points · modules 1–7

1. (6 pts) A student measures a length four times: 12.41, 12.44, 12.42, 12.43 cm. The true value is 12.67 cm. Is the data set precise? Accurate? Explain in one sentence each.

2. (6 pts) Convert 4.50 g/cm³ to kg/m³. Show the setup.

3. (5 pts) Give the number of significant figures in each: (a) 0.003080, (b) 45 000, (c) 2.50 × 10⁴, (d) 100.0.

4. (8 pts) Chlorine has two stable isotopes: Cl-35 (mass 34.969 u, 75.77 %) and Cl-37 (mass 36.966 u, 24.23 %). Calculate the atomic weight and state which isotope the periodic-table value sits closer to.

5. (6 pts) An ion has 20 protons, 20 neutrons, and 18 electrons. Give its symbol with mass number and charge, and name it.

6. (9 pts) Name or give the formula: (a) Fe₂(SO₄)₃, (b) N₂O₅, (c) HClO₂(aq), (d) copper(I) oxide, (e) ammonium phosphate, (f) diphosphorus pentachloride.

7. (8 pts) How many oxygen atoms are in 25.0 g of Ca(NO₃)₂?

8. (12 pts) A 4.60 g sample of a compound contains 1.20 g C, 0.30 g H, and the rest O. Find the empirical formula. If the molar mass is 92 g/mol, find the molecular formula.

9. (10 pts) Balance and classify: ___ C₃H₈ + ___ O₂ → ___ CO₂ + ___ H₂O. In this reaction, what is oxidized and what is reduced?

10. (14 pts) 2 Al + 3 Cl₂ → 2 AlCl₃. You start with 5.40 g Al and 21.3 g Cl₂. (a) Which is limiting? (b) What mass of AlCl₃ forms in theory? (c) If 22.0 g is actually collected, what is the percent yield?

11. (16 pts) Ammonia is made by N₂ + 3 H₂ → 2 NH₃. A reactor is fed 14.0 kg N₂ and 3.20 kg H₂. (a) Limiting reactant? (b) Theoretical mass of NH₃? (c) Mass of the excess reactant left over?

Solutions — Unit Exam 1

1. Precise: yes — the four readings agree to within 0.03 cm, so the measurement is repeatable. Accurate: no — every reading is about 0.24 cm below the true value, a consistent offset that points to a systematic error (a mis-zeroed or mis-calibrated instrument). Precision without accuracy is the classic calibration problem; see Lab Math.

2. Set it up as 4.50 g/cm³ × (1 kg / 1000 g) × (100 cm / 1 m)³. The factor is (1/1000) × (100)³ = (1/1000) × 10⁶ = 1000, so the answer is 4.50 × 10³ kg/m³ (4500 kg/m³). The trap is cubing the length conversion: (100)³ = 10⁶, not 100.

3. (a) 4 — leading zeros never count; the trailing zero after a decimal does. (b) 2 — trailing zeros with no decimal point are ambiguous and taken as not significant (write 4.5 × 10⁴ to mean 2, 4.5000 × 10⁴ to mean 5). (c) 3. (d) 4 — the decimal point makes all trailing zeros count.

4. Atomic weight = (34.969)(0.7577) + (36.966)(0.2423) = 26.496 + 8.957 = 35.45 u. It sits much closer to Cl-35, because Cl-35 is roughly three-quarters of every chlorine sample — a weighted average is pulled toward the more abundant isotope, not the midpoint.

5. Protons = atomic number = 20 → calcium. Mass number = 20 + 20 = 40. Charge = protons − electrons = 20 − 18 = +2. Symbol: ⁴⁰Ca²⁺, the calcium ion (calcium-40). Losing electrons, not protons, is what makes an ion — the element identity never changes.

6. (a) iron(III) sulfate — three SO₄²⁻ (−6) balanced by two Fe, so Fe is +3. (b) dinitrogen pentoxide — two nonmetals, use prefixes. (c) chlorous acid — the anion is chlorite (ClO₂⁻, -ite → -ous). (d) Cu₂O — Cu⁺ and O²⁻ cross to Cu₂O. (e) (NH₄)₃PO₄ — NH₄⁺ and PO₄³⁻ cross. (f) P₂Cl₅ — di- and penta- are the subscripts. See Naming & Reactions.

7. Molar mass Ca(NO₃)₂ = 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol. Moles of compound = 25.0 / 164.10 = 0.15234 mol. Each formula unit has 6 O atoms, so O atoms = 0.15234 mol × 6 × 6.022 × 10²³ = 5.50 × 10²³ oxygen atoms. The two places students lose this: forgetting the ×6 inside the parentheses, and stopping at moles instead of converting to atoms.

8. Mass of O = 4.60 − 1.20 − 0.30 = 3.10 g. Moles: C 1.20/12.01 = 0.0999; H 0.30/1.01 = 0.297; O 3.10/16.00 = 0.194. Divide by the smallest (0.0999): C 1.00, H 2.97 ≈ 3, O 1.94 ≈ 2 → empirical formula CH₃O₂ (empirical mass 12.01 + 3.03 + 32.00 = 47.0 g/mol). 92 / 47.0 ≈ 2, so the molecular formula is 2 × CH₃O₂ = C₂H₆O₄.

9. C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Balance C (3), then H (8 → coefficient 4 on water), then O last (right side has 6 + 4 = 10 O, so 5 O₂). It is a combustion (and therefore a redox) reaction. Carbon in C₃H₈ goes from roughly −8/3 to +4 in CO₂ — carbon is oxidized. Oxygen goes from 0 in O₂ to −2 in the products — oxygen is reduced. The fuel is always the thing oxidized.

10. Molar masses: Al 26.98, Cl₂ 70.90, AlCl₃ 133.33. Moles: Al 5.40/26.98 = 0.2001; Cl₂ 21.3/70.90 = 0.3004. Ratio needed is 2 Al : 3 Cl₂. Cl₂ needed for all the Al = 0.2001 × 3/2 = 0.300 mol — you have exactly 0.3004, so it is a near-perfect match; Al is limiting by a hair (0.3004 available vs 0.300 required — round the other way in an exam and note it is essentially stoichiometric). (b) AlCl₃ = 0.2001 mol Al × (2 AlCl₃ / 2 Al) × 133.33 g/mol = 26.7 g. (c) Percent yield = 22.0 / 26.7 × 100 = 82.4 %.

11. Moles: N₂ 14 000 g / 28.02 = 499.6; H₂ 3 200 g / 2.016 = 1587. Ratio needed 1 N₂ : 3 H₂. H₂ needed for all the N₂ = 499.6 × 3 = 1499 mol; you have 1587 mol, so N₂ is limiting (H₂ is in excess). (b) NH₃ = 499.6 mol N₂ × (2 NH₃ / 1 N₂) × 17.03 g/mol = 17 016 g ≈ 17.0 kg. (c) H₂ consumed = 499.6 × 3 = 1499 mol; H₂ left = 1587 − 1499 = 88 mol → 88 × 2.016 = 178 g (≈ 0.18 kg) H₂ remaining. Check by mass: 14.0 kg + 3.20 kg in; 17.0 kg + 0.18 kg out ≈ 17.2 kg — mass is conserved.

Practice Unit Exam 2 — Solutions and Energy

90 minutes · 100 points · modules 8–10

1. (8 pts) How many grams of NaCl are needed to prepare 500.0 mL of 0.250 M solution?

2. (8 pts) You dilute 25.0 mL of 6.00 M HCl to a final volume of 250.0 mL. What is the new molarity? How would you actually do this safely?

3. (8 pts) In 40.0 mL of 0.150 M Ca(NO₃)₂, how many moles of nitrate ion are present?

4. (12 pts) Mixing Pb(NO₃)₂(aq) and KI(aq) gives a bright yellow solid. Write (a) the balanced molecular equation with states, (b) the total ionic equation, (c) the net ionic equation, and (d) name the spectator ions.

5. (10 pts) 30.00 mL of HCl is titrated to the endpoint by 24.65 mL of 0.1050 M NaOH. What is the molarity of the HCl?

6. (14 pts) A 55.0 g piece of metal at 99.5 °C is dropped into 120.0 g of water at 21.2 °C in an insulated cup. The final temperature is 24.8 °C. Find the specific heat of the metal. (Assume no heat lost to the cup.)

7. (10 pts) When 1.75 g of NH₄NO₃ dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 22.7 °C to 21.1 °C. Is dissolving NH₄NO₃ endothermic or exothermic? Calculate q for the water and state the sign of the enthalpy of solution.

8. (14 pts) Given: C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ H₂(g) + ½ O₂(g) → H₂O(l), ΔH = −285.8 kJ CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), ΔH = −890.3 kJ Use Hess's law to find ΔH for C(s) + 2 H₂(g) → CH₄(g).

9. (16 pts) Calculate ΔH°rxn for 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l) using standard enthalpies of formation: ΔH°f: C₂H₆(g) = −84.7, CO₂(g) = −393.5, H₂O(l) = −285.8 kJ/mol (O₂ is 0).

Solutions — Unit Exam 2

1. mol = M × V = 0.250 mol/L × 0.5000 L = 0.125 mol. Mass = 0.125 mol × 58.44 g/mol = 7.31 g. Weigh 7.31 g NaCl, dissolve in less than 500 mL of water, then add water to the 500.0 mL mark (a volumetric flask) — you cannot add 500 mL of water, because the salt takes up volume too.

2. M₁V₁ = M₂V₂ → M₂ = (6.00)(25.0)/(250.0) = 0.600 M. Safe technique: add the concentrated acid to water (never water to acid), stirring, then dilute to the mark. "Do as you oughta, add acid to water."

3. mol Ca(NO₃)₂ = 0.150 mol/L × 0.0400 L = 0.00600 mol. Each formula unit releases 2 NO₃⁻, so mol NO₃⁻ = 2 × 0.00600 = 0.0120 mol. The subscript-2 is the whole question; molarity is per formula unit, and one formula unit is not one nitrate.

4. (a) Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq). (b) Pb²⁺(aq) + 2 NO₃⁻(aq) + 2 K⁺(aq) + 2 I⁻(aq) → PbI₂(s) + 2 K⁺(aq) + 2 NO₃⁻(aq). (c) Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s) — charge balances (+2 −2 = 0 → 0). (d) Spectators: K⁺ and NO₃⁻. See Naming & Reactions.

5. At the endpoint mol H⁺ = mol OH⁻ (1:1 for HCl + NaOH). mol NaOH = 0.1050 mol/L × 0.02465 L = 0.0025883 mol = mol HCl. M(HCl) = 0.0025883 mol / 0.03000 L = 0.08628 M (0.0863 M). Use the titrant volume and molarity to get moles; divide by the analyte volume.

6. Heat lost by metal = heat gained by water (opposite signs, equal magnitude): −m_metal · c_metal · ΔT_metal = m_water · c_water · ΔT_water Water: q = 120.0 g × 4.18 J/g°C × (24.8 − 21.2) = 120.0 × 4.18 × 3.6 = 1805.8 J gained. Metal: ΔT = 24.8 − 99.5 = −74.7 °C. So −(55.0)(c)(−74.7) = 1805.8 → (55.0)(74.7) c = 1805.8 → 4108.5 c = 1805.8 → c = 0.440 J/g·°C (consistent with a metal like iron, 0.449). Sign check: the metal cooled, so it released energy; the water warmed, so it absorbed the same amount.

7. The temperature of the solution fell, so the water lost heat to the dissolving salt: the process is endothermic. q_water = m·c·ΔT = 60.0 g × 4.18 J/g°C × (21.1 − 22.7) = 60.0 × 4.18 × (−1.6) = −401 J (the water lost 401 J). By energy balance, the dissolving absorbed +401 J, so q_solution for the process is +401 J and ΔH_solution is positive (endothermic). This is why NH₄NO₃ is used in instant cold packs.

8. Target: C(s) + 2 H₂(g) → CH₄(g). Take equation 1 as written: C + O₂ → CO₂, ΔH = −393.5. Take equation 2 ×2: 2 H₂ + O₂ → 2 H₂O(l), ΔH = −571.6. Reverse equation 3: CO₂ + 2 H₂O(l) → CH₄ + 2 O₂, ΔH = +890.3. Add: C + O₂ + 2 H₂ + O₂ + CO₂ + 2 H₂O → CO₂ + 2 H₂O + CH₄ + 2 O₂. Cancel CO₂, 2 H₂O, and 2 O₂ from both sides → C + 2 H₂ → CH₄. ΔH = −393.5 + (−571.6) + 890.3 = −74.8 kJ.

9. ΔH°rxn = Σ nΔH°f(products) − Σ nΔH°f(reactants). Products: 4(−393.5) + 6(−285.8) = −1574.0 + (−1714.8) = −3288.8. Reactants: 2(−84.7) + 7(0) = −169.4. ΔH°rxn = −3288.8 − (−169.4) = −3119.4 kJ (for 2 mol of ethane, so −1559.7 kJ per mole burned). Elements in their standard state have ΔH°f = 0 — that is why O₂ drops out.

Practice Unit Exam 3 — Atomic Structure and Bonding

90 minutes · 100 points · modules 11–14

1. (8 pts) A photon has a wavelength of 486 nm. Find its frequency and energy. (c = 3.00 × 10⁸ m/s, h = 6.626 × 10⁻³⁴ J·s.)

2. (6 pts) Which transition in a hydrogen atom emits a photon of higher energy: n = 4 → n = 2, or n = 3 → n = 2? Explain without a calculation.

3. (8 pts) Give a valid set of four quantum numbers for one electron in a 3p orbital.

4. (12 pts) Write the full and the noble-gas (condensed) electron configuration for (a) S, (b) Fe, (c) Fe³⁺. Draw the orbital diagram for the valence subshells of neutral Fe.

5. (10 pts) For each pair, choose the atom with the larger value and give the reason in one clause: (a) atomic radius — Na or Cl; (b) first ionization energy — Mg or Al; (c) electronegativity — N or P.

6. (10 pts) Rank NaF, MgO, and NaCl by lattice energy (magnitude), largest first, and justify with Coulomb's law.

7. (16 pts) Draw the Lewis structure of (a) CO₂, (b) SO₂, (c) NO₃⁻. For NO₃⁻, show the resonance structures and give the formal charge on each atom in one structure.

8. (14 pts) For SO₂ and CO₂: give the electron-domain geometry, the molecular shape, the approximate bond angle, and state whether the molecule is polar. Explain the difference.

9. (16 pts) A gas sample occupies 2.50 L at 1.20 atm and 25 °C. (a) How many moles? (b) It is compressed to 1.00 L and heated to 75 °C — what is the new pressure?

Solutions — Unit Exam 3

1. ν = c/λ = (3.00 × 10⁸ m/s) / (486 × 10⁻⁹ m) = 6.17 × 10¹⁴ Hz. E = hν = (6.626 × 10⁻³⁴)(6.17 × 10¹⁴) = 4.09 × 10⁻¹⁹ J (per photon). Convert nm to m first — the single most common slip here.

2. n = 4 → 2 releases more energy. The energy levels get closer together as n rises, so the gap from 4 to 2 is larger than the gap from 3 to 2. A bigger energy drop means a higher-energy (shorter-wavelength, bluer) photon.

3. 3p: n = 3, ℓ = 1 (p), mₗ = any of −1, 0, +1 (say mₗ = 0), mₛ = +½ (or −½). Example valid set: (3, 1, 0, +½). ℓ must be less than n; mₗ runs from −ℓ to +ℓ.

4. (a) S (Z = 16): 1s² 2s² 2p⁶ 3s² 3p⁴ = [Ne] 3s² 3p⁴. (b) Fe (Z = 26): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ = [Ar] 4s² 3d⁶. (c) Fe³⁺: remove 3 electrons — the 4s electrons go first, then one 3d: [Ar] 3d⁵. Orbital diagram, valence subshells of neutral Fe: 4s [↑↓] 3d [↑↓][↑][↑][↑][↑] — the 3d electrons singly occupy all five orbitals before pairing (Hund's rule), giving four unpaired electrons.

5. (a) Na — same period, and radius shrinks left-to-right as nuclear charge pulls the same shell in tighter. (b) Mg — removing Al's first electron empties a 3p orbital that is already higher in energy and less tightly held than Mg's filled 3s², so Al's IE₁ dips slightly below the trend. (c) N — electronegativity decreases down a group as the bonding electrons sit farther from the nucleus.

6. Lattice energy scales as |q₁q₂| / d. MgO has +2 and −2 charges (product 4) versus +1/−1 (product 1) for the others, so MgO ≫. Between NaF and NaCl, both are +1/−1, but F⁻ is smaller than Cl⁻, so the ions sit closer (d smaller) in NaF. Order: MgO > NaF > NaCl.

7. (a) CO₂: O=C=O, two double bonds, each O with two lone pairs, C with none; octet on all. (b) SO₂: bent, one S–O single and one S=O double (a resonance pair), one lone pair on S; S carries an expanded octet in the common depiction. (c) NO₃⁻: N centre, three N–O bonds, one double and two single in any single structure, three equivalent resonance structures with the double bond rotating among the three oxygens. Formal charges in one structure: N = +1 (5 − 0 − 4), the double-bonded O = 0 (6 − 4 − 2), each single-bonded O = −1 (6 − 6 − 1). Sum = +1 − 1 − 1 = −1, matching the ion charge.

8. CO₂: two electron domains on C, linear, 180°, and though each C=O bond is polar the two dipoles point exactly opposite and cancel — nonpolar. SO₂: three domains on S (two bonds + one lone pair), electron geometry trigonal planar, molecular shape bent, ≈ 119°, and the lone pair plus the bent shape mean the S–O dipoles do not cancel — polar. Same central atom group, different answer, entirely because of the lone pair on sulfur.

9. (a) n = PV/RT = (1.20 atm)(2.50 L) / [(0.08206)(298 K)] = 3.00 / 24.45 = 0.1227 mol. (b) Combined gas law with fixed n: P₁V₁/T₁ = P₂V₂/T₂. T₁ = 298 K, T₂ = 348 K. P₂ = P₁ · (V₁/V₂) · (T₂/T₁) = 1.20 · (2.50/1.00) · (348/298) = 1.20 × 2.50 × 1.168 = 3.50 atm. Temperatures must be in kelvin; using °C here would give nonsense.

Practice Cumulative Final

120 minutes · 150 points · modules 1–14

Twelve problems that mix domains the way the real final does. Full solutions follow.

1. (10) A 15.00 mL sample of a 1.20 g/mL sulfuric acid solution that is 37.0 % H₂SO₄ by mass — how many moles of H₂SO₄ does it contain?

2. (12) A hydrocarbon is 82.66 % C and 17.34 % H by mass; its molar mass is 58.12 g/mol. Find the molecular formula and give a valid name or common use.

3. (14) 3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O. Starting with 12.7 g Cu and 60.0 mL of 6.00 M HNO₃: (a) limiting reactant, (b) grams of Cu(NO₃)₂ formed, (c) volume of NO gas produced at 1.00 atm and 25 °C.

4. (10) How many mL of 0.200 M NaOH neutralize 25.0 mL of 0.150 M H₃PO₄ completely (all three protons)?

5. (14) 2.00 g of solid NaOH dissolves in 100.0 g of water in a coffee-cup calorimeter; the temperature rises from 23.0 °C to 28.3 °C. Find the molar enthalpy of solution of NaOH in kJ/mol, with sign.

6. (12) Using ΔH°f: CH₃OH(l) = −238.6, CO₂(g) = −393.5, H₂O(l) = −285.8 kJ/mol, find ΔH° for the combustion of one mole of methanol, CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2 H₂O(l).

7. (10) Write the noble-gas electron configuration for Se and for Cu (note the anomaly), and give the number of unpaired electrons in each.

8. (10) Arrange in order of increasing first ionization energy: K, Ca, Br, Ga. One sentence of justification.

9. (14) Draw Lewis structures and give the molecular shape and polarity of (a) NH₃, (b) CH₄, (c) H₂O, (d) BF₃.

10. (12) Rank the boiling points of CH₄, CH₃OH, CH₃CH₃, and NaCl from lowest to highest, naming the dominant intermolecular (or interionic) force in each.

11. (16) A 3.20 L flask holds N₂ at 0.80 atm and O₂ at 1.60 atm, both at 300 K. (a) Total pressure? (b) Mole fraction of O₂? (c) If the mixture is cooled to 150 K at constant volume, the total pressure? (d) Grams of O₂ present?

12. (16) 4.00 g of a mixture of CaCO₃ and CaCl₂ is treated with excess HCl. Only CaCO₃ reacts: CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂. If 0.66 g of CO₂ is released, what is the mass percent of CaCO₃ in the original mixture?

Solutions — Cumulative Final

1. Mass of solution = 15.00 mL × 1.20 g/mL = 18.00 g. Mass of H₂SO₄ = 18.00 × 0.370 = 6.66 g. M(H₂SO₄) = 98.08 g/mol → mol = 6.66 / 98.08 = 0.0679 mol. Three conversions chained: volume → mass of solution → mass of solute → moles.

2. In 100 g: 82.66 / 12.01 = 6.883 mol C; 17.34 / 1.01 = 17.17 mol H. Divide by smallest: C 1.00, H 2.49 ≈ 2.5 → multiply by 2 → C₂H₅ empirical (mass 29.06). 58.12 / 29.06 = 2 → molecular formula C₄H₁₀ — butane, a common fuel gas.

3. mol Cu = 12.7 / 63.55 = 0.1999. mol HNO₃ = 6.00 × 0.0600 = 0.360. Required ratio 3 Cu : 8 HNO₃. HNO₃ needed for all the Cu = 0.1999 × 8/3 = 0.533 mol — you only have 0.360, so HNO₃ is limiting. (b) Cu(NO₃)₂ = 0.360 mol HNO₃ × (3 Cu(NO₃)₂ / 8 HNO₃) × 187.57 g/mol = 0.135 × 187.57 = 25.3 g. (c) NO = 0.360 × (2 NO / 8 HNO₃) = 0.0900 mol. V = nRT/P = (0.0900)(0.08206) (298) / 1.00 = 2.20 L.

4. mol H₃PO₄ = 0.150 × 0.0250 = 0.003750. Neutralizing all three protons needs 3 mol NaOH per mol acid → mol NaOH = 0.01125. V = 0.01125 / 0.200 = 0.05625 L = 56.3 mL.

5. q_water = 100.0 g × 4.18 J/g°C × (28.3 − 23.0) = 100.0 × 4.18 × 5.3 = 2215.4 J gained by the water → the dissolving released 2215.4 J, so it is exothermic. mol NaOH = 2.00 / 40.00 = 0.0500. Per mole: −2215.4 J / 0.0500 mol = −44 308 J/mol = −44.3 kJ/mol (negative: exothermic).

6. ΔH° = [(−393.5) + 2(−285.8)] − [(−238.6) + 0] = (−393.5 − 571.6) − (−238.6) = −965.1 + 238.6 = −726.5 kJ per mole of methanol.

7. Se (Z = 34): [Ar] 4s² 3d¹⁰ 4p⁴ — the 4p⁴ has two orbitals paired-and-two... actually p⁴ is [↑↓][↑][↑] → 2 unpaired electrons. Cu (Z = 29): the anomaly — [Ar] 4s¹ 3d¹⁰ (a filled 3d is extra stable, so one 4s electron drops in). The 3d is full and paired; the lone 4s¹ is unpaired → 1 unpaired electron.

8. Increasing IE₁: K < Ca < Ga < Br. K and Ca are period-4 metals with low IE (Ca a bit higher, greater nuclear charge, same shell); Ga is further right so higher still; Br is near the right end of the period, highest of the four. Down-a-group would lower IE, but all four are period 4, so the left-to-right trend dominates.

9. (a) NH₃ — N with three bonds and one lone pair; electron geometry tetrahedral, shape trigonal pyramidal, ≈ 107°, polar. (b) CH₄ — four bonds, no lone pairs; tetrahedral, 109.5°, bonds polar but symmetric → nonpolar. (c) H₂O — two bonds, two lone pairs; bent, ≈ 104.5°, polar. (d) BF₃ — three bonds, no lone pair (boron is an octet exception with 6 electrons); trigonal planar, 120°, symmetric → nonpolar.

10. Lowest → highest boiling point: CH₄ < CH₃CH₃ < CH₃OH < NaCl. CH₄ and CH₃CH₃ are nonpolar → only London dispersion forces, and ethane's are slightly stronger (more electrons). CH₃OH adds hydrogen bonding (O–H). NaCl is held by full ionic bonds, far stronger than any intermolecular force — it boils above 1400 °C.

11. (a) Partial pressures add: P_total = 0.80 + 1.60 = 2.40 atm. (b) X(O₂) = P(O₂)/P_total = 1.60/2.40 = 0.667. (c) At constant V and n, P ∝ T: P₂ = 2.40 × (150/300) = 1.20 atm. (d) mol O₂ = P(O₂)V/RT = (1.60)(3.20) / [(0.08206)(300)] = 5.12 / 24.62 = 0.2080 mol → × 32.00 g/mol = 6.66 g O₂.

12. mol CO₂ = 0.66 / 44.01 = 0.01500. From the equation, mol CaCO₃ = mol CO₂ = 0.01500. mass CaCO₃ = 0.01500 × 100.09 = 1.50 g. Mass percent = 1.50 / 4.00 × 100 = 37.5 % CaCO₃ (and 62.5 % CaCl₂). The CaCl₂ is a spectator here — the gas measured comes only from the carbonate.

Where to go from a wrong answer