[기출문제] 울산과학기술원(UNIST) 일반화학 2020-1 중간 기출문제 (정답 포함)

울산과학기술원(UNIST) 일반화학 2020-1 중간 기출문제 (정답 포함)

 

 

 

1. 시험 정보

 

학교/과목	울산과학기술원(UNIST)/일반화학
시험명	2020-1 중간
교수명	-
문항수/형식	풀이형 14문제
정답/해설	✅ 있음
파일형식	-

 

 

 

2. 출제 범위 & 키워드

 

일반화학 전반의 단위와 SI, 화학결합·원자모형, 화학량론, 산화환원, 열화학, 기체 및 반응 에너지 계산을 종합적으로 다루는 문제

 

 

📚 키워드

 

 

 

3. 기출 미리보기

 

 

Ch1. 1. Derived SI units (in base SI units)

 

 

 

 

4. 자료 보기

 

[기출 문제 및 정답] 

 

 

Ch1. 1. Derived SI units (in base SI units) (a) Acceleration meter per second squared (m/s²) (b) Force kilogram meter per second squared (kg·m/s²) (c) Work kilogram meter squared per second squared (kg·m²/s²) (d) Pressure kilogram per meter per second squared (kg/m·s²) (e) Power kilogram meter squared per second cubed (kg·m²/s³) (f) Velocity meter per second (m/s) (g) Energy kilogram meter squared per second squared (kg·m²/s²) 2. Tanker truck capacity Given: 3.4 × 10⁴ liters Conversion: 1 gallon = 3.785 liters (a) Capacity in gallons 3.4 × 10⁴ L ÷ 3.785 L/gal ≈ 8.98 × 10³ gallons Answer: 8.98 × 10³ gallons (b) Value of full load 8.98 × 10³ gal × 3.00 USD/gal = 2.69 × 10⁴ USD Answer: 2.69 × 10⁴ USD Ch2. 1. Law of Multiple Proportions Assume 100 g of each compound. Compound 1 Carbon: 42.9 g Oxygen: 57.1 g Mass ratio (O/C) = 57.1 / 42.9 ≈ 1.33 Given that the formula is CO, this corresponds to 1 oxygen atom per carbon atom. Compound 2 Carbon: 27.3 g Oxygen: 72.7 g Mass ratio (O/C) = 72.7 / 27.3 ≈ 2.66 Comparison 2.66 ÷ 1.33 ≈ 2 Thus, for the same mass of carbon, Compound 2 contains twice as much oxygen as Compound 1. This shows that the law of multiple proportions is followed. Formula of Compound 2: CO₂ 2. Rutherford’s Gold Foil Experiment If the plum pudding model were correct, the positive charge would be spread uniformly throughout the atom. As a result: Alpha particles would pass through the gold foil with only slight deflections No large-angle deflections or backscattering would be observed This differs from Rutherford’s actual results, which showed occasional large deflections, indicating a small, dense nucleus. 3. Mass Spectrum of H₂ Hydrogen isotopes: ¹H: mass = 1.00783 u (99.9885%) ²H: mass = 2.01410 u (0.0115%) (a) Number of peaks Possible molecular combinations: ¹H–¹H ¹H–²H ²H–²H Answer: 3 peaks (b) Relative molecular masses ¹H–¹H: 1.00783 + 1.00783 = 2.01566 u ¹H–²H: 1.00783 + 2.01410 = 3.02193 u ²H–²H: 2.01410 + 2.01410 = 4.02820 u (c) Largest and smallest peaks Largest peak: ¹H–¹H (highest natural abundance) Smallest peak: ²H–²H (lowest natural abundance) Ch3 1. Reaction of sulfuric acid and lead(II) acetate Balanced reaction: Sulfuric acid + lead(II) acetate → lead(II) sulfate + acetic acid (1 : 1 → 1 : 2) Molar masses: Sulfuric acid (H₂SO₄): 98.08 g/mol Lead(II) acetate [Pb(C₂H₃O₂)₂]: 325.29 g/mol Lead(II) sulfate (PbSO₄): 303.26 g/mol Acetic acid (HC₂H₃O₂): 60.05 g/mol Moles initially present: H₂SO₄: 5.00 g ÷ 98.08 g/mol = 0.0510 mol Pb(C₂H₃O₂)₂: 5.00 g ÷ 325.29 g/mol = 0.0154 mol Limiting reactant: lead(II) acetate Moles reacted: H₂SO₄ reacted = 0.0154 mol Pb(C₂H₃O₂)₂ reacted = 0.0154 mol Moles remaining / formed: H₂SO₄ remaining = 0.0510 − 0.0154 = 0.0356 mol Pb(C₂H₃O₂)₂ remaining = 0 mol PbSO₄ formed = 0.0154 mol Acetic acid formed = 0.0308 mol Final masses after reaction: Sulfuric acid: 3.49 g Lead(II) acetate: 0.00 g Lead(II) sulfate: 4.66 g Acetic acid: 1.85 g 2. Formation of ammonia Balanced equation: Nitrogen + hydrogen → ammonia N₂ + 3 H₂ → 2 NH₃ Given final amounts: N₂ = 3.0 mol H₂ = 3.0 mol NH₃ = 3.0 mol Extent of reaction: 2 mol NH₃ produced per reaction Reaction extent = 3.0 ÷ 2 = 1.5 mol Reactants consumed: N₂ consumed = 1.5 mol H₂ consumed = 4.5 mol Original amounts: N₂ = 3.0 + 1.5 = 4.5 mol H₂ = 3.0 + 4.5 = 7.5 mol 3. Thermite reaction (a) Balanced equation with states Iron(III) oxide (s) + aluminum (s) → aluminum oxide (s) + iron (l) Fe₂O₃(s) + 2 Al(s) → Al₂O₃(s) + 2 Fe(l) (b) Aluminum needed for 500.0 g Fe₂O₃ Molar mass of Fe₂O₃ = 159.69 g/mol Moles of Fe₂O₃: 500.0 g ÷ 159.69 g/mol = 3.13 mol Stoichiometry: 1 mol Fe₂O₃ requires 2 mol Al Moles of Al needed: 3.13 × 2 = 6.26 mol Mass of Al: 6.26 mol × 26.98 g/mol = 169 g (c) Fe₂O₃ needed to produce 1.00 × 10⁴ kJ Heat released per mol Fe₂O₃ = 852 kJ Moles required: 1.00 × 10⁴ kJ ÷ 852 kJ/mol = 11.7 mol Mass required: 11.7 mol × 159.69 g/mol = 1.88 × 10³ g (d) Reverse reaction The reverse reaction would be endothermic. Heat would be a reactant. Ch4 1. Oxidation ability and activity series (a) Order of decreasing oxidation ability From the given reactions: C + B⁺ → C⁺ + B → C is more easily oxidized than B C⁺ + A → no reaction → C is more easily oxidized than A A⁺ + D → no reaction → A is more easily oxidized than D D + B⁺ → D⁺ + B → D is more easily oxidized than B Combining these results: C > A > D > B So, 1 > 2 > 3 > 4 = C > A > D > B (b) Predicted reactions (1) A⁺ + C → A + C⁺ The reaction will occur because C is higher than A in the activity series and can reduce A⁺. (2) A⁺ + B → A + B⁺ The reaction will not occur because B is lower than A in the activity series and cannot reduce A⁺. 2. Determination of iron content in ore Balanced equation: 2 Fe³⁺(aq) + Sn²⁺(aq) → 2 Fe²⁺(aq) + Sn⁴⁺(aq) Moles of Sn²⁺ used: 13.28 mL × 0.1015 mol/L = 0.001348 mol Stoichiometry: 1 mol Sn²⁺ reacts with 2 mol Fe³⁺ Moles of Fe³⁺: 2 × 0.001348 = 0.002696 mol Mass of Fe: 0.002696 mol × 55.85 g/mol = 0.150 g Mass percent Fe: (0.150 g ÷ 0.1875 g) × 100 = 80.3 % Answer: 80.3 % Fe 3. Compound X (C, H, O, S only) (a) Empirical formula From combustion data: CO₂: 4.83 g → 0.1097 mol C H₂O: 1.48 g → 0.164 mol H From titration of H₂SO₄: NaOH used = 109.8 mL × 1.00 M = 0.1098 mol H₂SO₄ formed = 0.0549 mol → 0.0549 mol S Oxygen by mass difference gives 0.1098 mol O. Mole ratios (dividing by 0.0549): C : 2.00 H : 3.00 O : 2.00 S : 1.00 Empirical formula: C₂H₃O₂S (b) Molecular formula NaOH required to neutralize X: 54.9 mL × 1.00 M = 0.0549 mol NaOH Two acidic H per molecule → moles of X = 0.02745 mol Molar mass of X: 5.00 g ÷ 0.02745 mol = 182 g/mol Empirical formula mass (C₂H₃O₂S) ≈ 91 g/mol Factor = 2 Molecular formula: C₄H₆O₄S₂ Ch5 1. Formation of ethyl chloride Reaction: C₂H₄(g) + HCl(g) → C₂H₅Cl(g) ΔH° = −72.3 kJ (per mole of reaction) Moles initially present: C₂H₄: 89.5 g ÷ 28.05 g/mol = 3.19 mol HCl: 125 g ÷ 36.46 g/mol = 3.43 mol Limiting reactant: C₂H₄ Moles reacted = 3.19 mol Total enthalpy change: ΔH° = 3.19 × (−72.3 kJ) = −231 kJ PV work: w = −PΔV = −(1 atm)(−71.5 L) = +71.5 L·atm 71.5 L·atm = +7.25 kJ Internal energy change: ΔE° = ΔH° − w = −231 − 7.25 = −238 kJ Answers: PV work = +7.25 kJ ΔE° = −238 kJ 2. Combustion of ethanol (a) Balanced equation C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g) (b) Standard enthalpy change Standard enthalpies of formation (kJ/mol): C₂H₅OH(l): −277.7 CO₂(g): −393.5 H₂O(g): −241.8 O₂(g): 0 ΔH° = [2(−393.5) + 3(−241.8)] − [−277.7] ΔH° = −1.23 × 10³ kJ/mol (c) Heat produced per liter of ethanol Density of ethanol = 0.789 g/mL Mass per liter = 789 g Moles of ethanol: 789 g ÷ 46.07 g/mol = 17.1 mol Heat produced: 17.1 × 1.23 × 10³ kJ = 2.11 × 10⁴ kJ (d) Mass of CO₂ produced per kJ From the balanced equation: 1 mol ethanol → 2 mol CO₂ = 88.0 g CO₂ 88.0 g ÷ 1.23 × 10³ kJ = 0.071 g CO₂ per kJ 3. Hess’s Law Given: A → B  ΔH = +60 kJ B → C  ΔH = −90 kJ (a) Enthalpy change for A → C ΔH = (+60) + (−90) = −30 kJ (b) Enthalpy diagram (description) A is at the highest enthalpy level B is 60 kJ lower than A C is 90 kJ lower than B Net drop from A to C is 30 kJ This illustrates Hess’s law, where the total enthalpy change depends only on the initial and final states.

 

 

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