AP Unit 4 Solutions

1. A temperature change establishes energy transfer, but mixing or dissolution can also cause cooling.
2. Evidence for a chemically distinct product, from a supplied spectrum or composition measurement, would distinguish formation of new species from unchanged species in a mixture.

1. During vaporization the H₂O units remain intact, with more units in the gas phase.
2. During decomposition the atoms regroup into H₂ and O₂; new species form. Both conserve atoms in a sealed boundary.

1. Molecular: CaCl₂ + Na₂CO₃ → CaCO₃(s) + 2NaCl, with (aq) on all other species.
2. Complete ionic: Ca²⁺ + 2Cl⁻ + 2Na⁺ + CO₃²⁻ → CaCO₃(s) + 2Na⁺ + 2Cl⁻, with (aq) on ions.
3. Cancel identical sodium and chloride terms; +2 and −2 sum to zero on the left and the solid is neutral.

1. Dissolution: K₂SO₄(s) → 2K⁺(aq) + SO₄²⁻(aq); charge is zero on each side.
2. NaOH supplies aqueous OH⁻; Na⁺ is spectator. HF is a weak acid and is retained.
3. HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l); both sides have charge −1.

1. Construct a before panel with seven separate bonded A–A pairs and three separate bonded B–B pairs. In the after panel, create six separate molecules each with two A atoms and one B atom connected within that molecular group; retain one intact A–A pair. Keep the original A/B symbols and scale. No B₂ molecule remains. The equation does not specify detailed molecular geometry, so do not grade a particular bond angle.
2. Three B₂ require six A₂, so B₂ is limiting. Before: 14 A atoms and 6 B atoms. After: 6 × 2 + 2 = 14 A atoms and 6 × 1 = 6 B atoms. All model species are neutral, so charge stays zero. Coefficients specify the 2:1 reacting ratio; the extra A₂ remains rather than being erased or split into atoms.

For a text coordinate construction, assign separate positions to seven A–A pairs and three B–B pairs in the before panel, then to six A–A–B groups and one A–A pair in the after panel. State that connections group atoms into molecules, not into one continuous network.

1. Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s). Two Mg²⁺ consume four OH⁻. The before panel shows all 18 supplied ions dispersed, with no bonded nitrate/sodium or magnesium/hydroxide molecules. The after panel has one solid array containing two Mg²⁺ and four intact OH⁻ units; draw these together in the labeled solid region rather than as two isolated Mg(OH)₂ molecules. Aqueous solution retains six Na⁺, four NO₃⁻ and two excess OH⁻. State that water is omitted.
2. The atom inventory is the same before/after: Mg 2, O 18 (12 in nitrate + 6 in hydroxide), H 6, N 4 and Na 6. In the final model, four hydroxide O/H units are in the solid and two remain aqueous. Before charge = 2(+2) + 4(−1) + 6(+1) + 6(−1) = 0. Solid charge = 2(+2) + 4(−1) = 0; final aqueous charge = 6(+1) + 4(−1) + 2(−1) = 0. Sodium and nitrate are spectators with unchanged counts and identities; cancelling them records the net chemical change, not physical removal.

A text coordinate answer should place all initial ion labels at separate aqueous positions, place two Mg²⁺ and four OH⁻ labels in one final solid region, and place six Na⁺, four NO₃⁻ and two OH⁻ at separate final aqueous positions. The layout is a schematic inventory, not a crystallographic structure or hydration simulation.

1. Evaporation leaves covalent bonds intact while intermolecular attractions are overcome.
2. Dissolution separates lattice ions and forms ion–dipole hydration interactions; water interactions reorganize.
3. The physical/chemical label alone does not specify every interaction; salt dissolution is a boundary case that needs a stated criterion.

1. Nitrogen could support 0.800 mol NH₃; hydrogen supports 0.900 ×2/3 =0.600 mol. Hydrogen limits.
2. N₂ used =0.900/3 =0.300 mol; 0.100 mol remains.
3. At 1.80 mol H₂, nitrogen now limits; maximum NH₃ is 0.800 mol, rather than 1.20 mol.

1. Ba moles =0.0300 ×0.120 =0.00360; sulfate moles =0.0200 ×0.150 =0.00300.
2. The ratio is 1:1; sulfate limits and gives 0.00300 mol BaSO₄.
3. Excess Ba moles =0.000600; final [Ba²⁺] =0.000600/0.0500 =0.0120 M.

1. Stoichiometry gives 0.0150 mol CO₂. Its partial pressure is 1.000 −0.0313 =0.9687 atm.
2. V =nRT/P =0.0150 ×0.08206 ×298/0.9687 =0.379 L.
3. If total pressure 1.000 atm is incorrectly treated as CO₂ pressure, the predicted volume is 0.0150 ×0.08206 ×298/1.000 =0.367 L. At fixed gas moles and temperature, an overestimated product pressure gives an underestimated predicted volume.
4. A different question—inferring CO₂ moles from a fixed measured volume—would instead overestimate moles. Here the requested quantity is volume, so the bias is low.

1. Delivered volume =19.75 −1.35 =18.40 mL.
2. HA + OH⁻ → A⁻ + H₂O; acid amount =0.01840 ×0.1250 =0.002300 mol.
3. [HA] =0.002300/0.02500 =0.09200 M. Label x-axis added NaOH volume (mL); mark transition/equivalence near 18.40 mL.

1. Excess base increases the recorded titrant moles. The calculation attributes them all to analyte consumption.
2. Calculated acid moles and concentration are too high.
3. Consistent overshoot is systematic; close repeats demonstrate precision, not correct equivalence. Use a more appropriate endpoint signal or analyze the transition around equivalence.

1. (i) NH₄⁺ transfers H⁺ to OH⁻; acid–base. N remains −3, O remains −2 and H +1.
2. (ii) Mg goes 0→+2; two Mg lose four electrons total. Two O atoms go 0→−2 and gain four total: redox.
3. (iii) Solvated ions make a sparingly soluble solid; Ba stays +2 and sulfate retains its internal oxidation numbers: precipitation.

1. Balance C and H first: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
2. Hydrogen is +1, so each equivalent carbon in ethane is −3; in CO₂ carbon is +4.
3. Each carbon increases by seven; four carbons give 28 electrons. Fourteen oxygen atoms each decrease by two, balancing 28.

1. Acetic acid donates to water. Pairs are CH₃COOH/CH₃COO⁻ and H₃O⁺/H₂O.
2. H₃O⁺ is the stronger acid; its conjugate base water is weaker than acetate.
3. The left side contains the weaker acid and weaker base, so reactants are favored.

1. I: HX + H₂O → X⁻ + H₃O⁺; water accepts a proton and is the base.
2. II: B + H₂O → BH⁺ + OH⁻; water donates a proton and is the acid.
3. Both start neutral and finish with +1 and −1, conserving total charge. Arrow direction describes the shown event rather than claiming completion for every acid/base.

1. Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
2. Oxidation: 2I⁻ → I₂ + 2e⁻; multiply by three.
3. Add and cancel six electrons. Reactant charge =−2 +14 −6 =+6; product charge =2×3 =+6.

1. Reduction: Cl₂ +2e⁻ →2Cl⁻. Each Cl atom is reduced from 0 to −1.
2. For oxidation, balance first in acidic bookkeeping form: Cl₂ +2H₂O →2ClO⁻ +4H⁺ +2e⁻. Waters balance O, protons balance H and electrons balance charge.
3. Add 4OH⁻ to both sides. The right-side H⁺ and OH⁻ make 4H₂O; cancel 2H₂O to obtain Cl₂ +4OH⁻ →2ClO⁻ +2H₂O +2e⁻.
4. Add the reduction half, cancel 2e⁻, and divide the entire equation by two: Cl₂ +2OH⁻ →Cl⁻ +ClO⁻ +H₂O. Atom counts and charge −2 match. Different Cl atoms are oxidized 0→+1 and reduced 0→−1.