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The assignment involves solving six chemistry problems, including calculations related to chemical reactions, percent yield, molar masses, partial pressures, reaction stoichiometry, and kinetic energy of molecules. Each problem requires detailed explanations, organized work, and accurate application of chemical principles.

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Introduction

This paper addresses six distinct chemistry problems, each rooted in fundamental principles such as redox reactions, stoichiometry, kinetics, and gas laws. A comprehensive approach involving detailed calculations, explanations, and conceptual understanding is employed to provide accurate solutions. The problems examine a variety of topics including reaction mechanisms, yields, gas mixtures, and molecular energies, demonstrating the breadth of applied chemical knowledge.

Problem 1: Reacting Chlorine Gas with Sodium Hydroxide

Part a: Oxidation Number of Chlorine in Bleach (NaOCl)

The chemical formula of bleach is NaOCl. To determine the oxidation number of chlorine in NaOCl, we consider the known oxidation states of sodium (Na) and oxygen (O). Sodium generally exhibits an oxidation state of +1, and oxygen in oxides usually has an oxidation state of -2.

• Let the oxidation number of chlorine in NaOCl be x.
• Summing the oxidation states: (+1) + (x) + (-2) = 0 (since the molecule is neutral).
• Thus, +1 + x - 2 = 0 => x = +1.

Therefore, the oxidation number of chlorine in bleach (NaOCl) is +1.

Part b: Is the Reaction a Redox Reaction? Explain.

In the unbalanced chemical equation:

NaOH (aq) + Cl₂ (g) → NaOCl (aq) + NaCl (aq) + H₂O (l)

chlorine undergoes a change in oxidation state: it starts as 0 in Cl₂ and ends as +1 in NaOCl, indicating oxidation. Simultaneously, NaOH provides hydroxide ions, but sodium remains +1, and oxygen remains -2; there is no change in these oxidation states.

The key is that Cl₂ is being oxidized from 0 to +1 and likely reduced in some capacity, possibly involving water or hydroxide. Given chlorine's change from 0 to +1, the process involves oxidation of Cl, which is characteristic of a redox reaction (one substance oxidized and another potentially reduced).

Hence, this chemical process is indeed a redox reaction because chlorine gas is oxidized from 0 to +1. The reaction involves changes in oxidation states, confirming its redox nature.

Part c: Percent Yield

Given data:

• Chlorine gas bubbled: 83.0 g
• Bleach produced: 22.0 g

First, calculate moles of Cl₂:

molar mass of Cl₂ = 70.90 g/mol

n_Cl2 = 83.0 g / 70.90 g/mol ≈ 1.172 mol

The balanced chemical reaction (assuming simplified):

NaOH + Cl₂ → NaOCl + NaCl + H₂O

For each mole of Cl₂, 1 mole of NaOCl is produced.

Calculating the theoretical maximum mass of NaOCl:

molar mass of NaOCl = 74.44 g/mol

theoretical yield = 1.172 mol × 74.44 g/mol ≈ 87.3 g

Actual yield = 22.0 g

Percent yield = (actual / theoretical) × 100 = (22.0 / 87.3) × 100 ≈ 25.2%

The percent yield of the reaction is approximately 25.2%, indicating a significant loss or incomplete reaction.

Problem 2: Precipitate Mass from Ba(OH)₂ and Na₂SO₄

Given Data:

• Volume of Ba(OH)₂: 2.27 L
• Concentration of Ba(OH)₂: 0.0820 M
• Volume of Na₂SO₄: 3.06 L
• Concentration of Na₂SO₄: 0.0664 M

Calculations:

Calculate moles of each reactant:

n_Ba(OH)2 = 0.0820 mol/L × 2.27 L ≈ 0.186 mol

n_Na2SO4 = 0.0664 mol/L × 3.06 L ≈ 0.204 mol

The balanced reaction for precipitation:

Ba(OH)₂ + Na₂SO₄ → BaSO₄ (s) + 2 NaOH

1 mol of Ba(OH)₂ reacts with 1 mol of Na₂SO₄ to produce 1 mol of BaSO₄.

The limiting reagent is determined by comparing moles: since 0.186 mol of Ba(OH)₂ is less than 0.204 mol of Na₂SO₄, Ba(OH)₂ is limiting.

Mass of BaSO₄ formed:

molar mass of BaSO₄ = 233.39 g/mol

Mass = 0.186 mol × 233.39 g/mol ≈ 43.4 g

Hence, approximately 43.4 grams of barium sulfate precipitate forms when the solutions react.

Problem 3: Calorimeter Heat Capacity from Aniline Combustion

Given Data:

• Mass of aniline: 6.55 g
• Molar mass of C₆H₅NH₂: 93.13 g/mol
• Temperature rise: 32.9°C
• Enthalpy change (ΔH°rxn): -1.28 kJ (for the balanced reaction)

Calculations:

Number of mols of aniline burned:

n = 6.55 g / 93.13 g/mol ≈ 0.0703 mol

Total heat released:

Q = n × ΔH°rxn = 0.0703 mol × (-1.28 kJ) ≈ -0.0900 kJ

The heat absorbed by the calorimeter is equal to the magnitude of heat released:

Q = C_cal × ΔT

Rearranged to find the calorimeter's heat capacity: C_cal = Q / ΔT = 0.0900 kJ / 32.9°C ≈ 0.00274 kJ/°C or 2.74 J/°C

The heat capacity of the calorimeter is approximately 2.74 Joules per degree Celsius.

Problem 4: Partial Pressures in Interconnected Gas Flasks

Given Data:

• Oxygen volume: 200 mL, pressure: 200 mmHg
• Nitrogen volume: 300 mL, pressure: 100 mmHg

Part a: Final Partial Pressures

Initial moles of gases, using ideal gas law (PV=nRT), are proportional to pressure because temperature and R are constant.

Final total volume: 200 mL + 300 mL = 500 mL

Initial moles of O₂: n_O2 ∝ 200 mmHg

Initial moles of N₂: n_N2 ∝ 100 mmHg

After connection, the gases distribute in the combined volume, maintaining their partial pressures proportionally due to the ideal gas law.

Final partial pressure of O₂:

P_{O2_final} = (initial P_O2 × initial volume) / final volume = (200 mmHg × 200 mL) / 500 mL = 80 mmHg

Similarly, for N₂:

P_{N2_final} = (100 mmHg × 300 mL) / 500 mL = 60 mmHg

Part b: Total Pressure

The total pressure is the sum of the partial pressures:

P_total = P_{O2_final} + P_{N2_final} = 80 mmHg + 60 mmHg = 140 mmHg

Thus, the final mixture has partial pressures of 80 mmHg for oxygen and 60 mmHg for nitrogen, with a total pressure of 140 mmHg.

Problem 5: Chlorine Gas Volume Needed for Aluminum Reaction

Given Data:

• Mass of aluminum: 7.85 g
• Reaction: 2Al + 3Cl₂ → 2AlCl₃
• Temperature: 298 K
• Pressure: 225 mmHg

Calculations:

Moles of aluminum:

n_Al = 7.85 g / 26.98 g/mol ≈ 0.291 mol

From the balanced equation:

2 mol Al react with 3 mol Cl₂

Thus, moles of Cl₂ needed: (0.291 mol Al) × (3/2) ≈ 0.436 mol

Using ideal gas law: PV=nRT, solve for V:

V = nRT / P

Convert pressure to atm:

225 mmHg / 760 mmHg = 0.295 atm

V = 0.436 mol × 0.082057 L·atm/(mol·K) × 298 K / 0.295 atm ≈ 34.2 L

Therefore, approximately 34.2 liters of chlorine gas are required to react completely with 7.85 grams of aluminum under the specified conditions.

Problem 6: Molecular Kinetic Energy of Fluorine Molecules

Given Data:

• Temperature: 25°C = 298 K
• Molecular mass of fluorine (F₂): 38.00 g/mol

Calculations:

Convert molar mass to kg:

m = 38.00 g/mol / 1000 = 0.038 kg/mol

Using RMS velocity formula:

u_rms = √(3RT / M)

where R = 8.314 J/(mol·K), M = molar mass in kg

u_rms = √(3 × 8.314 × 298 / 0.038) ≈ √(3 × 8.314 × 298 / 0.038)

Calculate numerator:

3 × 8.314 × 298 ≈ 7420.3

Then:

u_rms ≈ √(7420.3 / 0.038) ≈ √195,268.4 ≈ 441.8 m/s

Average kinetic energy per molecule:

KE_avg = (3/2)k_BT

where k_B = Boltzmann constant = 1.38 × 10^-23 J/K

KE_avg = (3/2) × 1.38 × 10^-23 × 298 ≈ 6.16 × 10^-21 J

Thus, the root-mean-square velocity of fluorine molecules at 25°C is approximately 442 m/s, and their average kinetic energy is about 6.16 × 10^-21 Joules.

Conclusion

In this analysis, each chemistry problem was approached with detailed calculations and relevant chemical principles, resulting in accurate and comprehensive solutions. The application of stoichiometry, gas laws, thermodynamics, and redox concepts demonstrates a solid understanding of fundamental chemistry topics.

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