Particulate Nature (S1.1)
State the difference between an element, compound, and mixture at the particulate level.
Element: contains only one type of atom. Compound: contains two or more different types of atoms chemically bonded in fixed ratios. Mixture: contains two or more substances that are not chemically bonded and retain individual properties.
Define the mole and state the value of Avogadro's constant.
A mole is the SI unit for amount of substance, containing exactly 6.02 × 10²³ particles (atoms, molecules, ions, etc.). Avogadro's constant = 6.02 × 10²³ mol⁻¹.
State three assumptions of the ideal gas model.
1) Gas particles have negligible volume compared to container volume, 2) No intermolecular forces between particles, 3) Collisions between particles and container walls are elastic.
Balance the following equation: ___Al + ___O₂ → ___Al₂O₃
4Al + 3O₂ → 2Al₂O₃
How many grams of KCl are needed to prepare 250 cm³ of 0.15 mol dm⁻³ solution?
Moles needed = concentration × volume = 0.15 mol dm⁻³ × 0.250 dm³ = 0.0375 mol
Molar mass KCl = 39.10 + 35.45 = 74.55 g/mol
Mass = 0.0375 mol × 74.55 g/mol = 2.8 g
Explain what happens to particle motion and kinetic energy when a solid is heated and melts into a liquid.
As temperature increases, particles gain kinetic energy and vibrate more vigorously. At the melting point, particles have enough energy to overcome some intermolecular forces, allowing them to move more freely while remaining close together.
Calculate the mass of 0.75 mol of calcium carbonate (CaCO₃).
Molar mass of CaCO₃ = 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol
Mass = moles × molar mass = 0.75 mol × 100.09 g/mol = 75.1 g
State the ideal gas equation and define each variable.
PV = nRT
P = pressure (Pa or atm), V = volume (m³ or dm³), n = amount in moles, R = gas constant (8.314 J mol⁻¹ K⁻¹), T = temperature (K)
In the reaction N₂ + 3H₂ → 2NH₃, how many moles of NH₃ can be produced from 1.5 mol of N₂?
From the balanced equation, mole ratio N₂ : NH₃ = 1 : 2
Moles of NH₃ = 1.5 mol N₂ × (2 mol NH₃/1 mol N₂) = 3.0 mol NH₃
How many moles are in 50.0 g of iron(III) oxide (Fe₂O₃)?
Molar mass = (2×55.85) + (3×16.00) = 159.70 g/mol
Moles = 50.0 g ÷ 159.70 g/mol = 0.313 mol
Convert 25°C to Kelvin and explain why the Kelvin scale is used in gas law calculations.
5°C + 273.15 = 298.15 K (or 298 K). Kelvin is used because it's an absolute temperature scale starting at absolute zero, where particle motion theoretically stops. This allows for direct proportional relationships in gas laws
Calculate the molar concentration of a solution made by dissolving 8.5 g of ammonia (NH₃) in enough water to make 500 cm³ of solution.
Molar mass of NH₃ = 14.01 + (3 × 1.01) = 17.04 g/mol
Moles of NH₃ = 8.5 g ÷ 17.04 g/mol = 0.499 mol
Volume = 500 cm³ = 0.500 dm³
Concentration = 0.499 mol ÷ 0.500 dm³ = 0.998 mol dm⁻³
Calculate the volume occupied by 0.50 mol of gas at 25°C and 1.0 atm pressure.
Using PV = nRT
V = nRT/P = (0.50 mol)(0.0821 L atm mol⁻¹ K⁻¹)(298 K)/(1.0 atm)
V = 12.2 L
Calculate the mass of CO₂ produced when 10.0 g of C₂H₆ undergoes complete combustion.
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Molar mass C₂H₆ = 30.08 g/mol, CO₂ = 44.01 g/mol
Moles C₂H₆ = 10.0 g ÷ 30.08 g/mol = 0.332 mol
Moles CO₂ = 0.332 mol C₂H₆ × (4 mol CO₂/2 mol C₂H₆) = 0.665 mol
Mass CO₂ = 0.665 mol × 44.01 g/mol = 29.3 g
For the reaction 4Fe + 3O₂ → 2Fe₂O₃, determine the limiting reactant when 10.0 g Fe reacts with 5.0 g O₂.
Moles Fe = 10.0 g ÷ 55.85 g/mol = 0.179 mol
Moles O₂ = 5.0 g ÷ 32.00 g/mol = 0.156 mol
From stoichiometry: 0.179 mol Fe needs 0.134 mol O₂
Have 0.156 mol O₂ → Fe is limiting reactant
Using kinetic molecular theory, explain why gases are compressible while liquids and solids are not.
In gases, particles are far apart with large spaces between them, so applying pressure can push particles closer together. In liquids and solids, particles are already close together with minimal empty space, making compression very difficult.
A student dilutes 25.0 cm³ of 2.0 mol dm⁻³ HCl solution to a final volume of 250 cm³. Calculate the final concentration.
Using C₁V₁ = C₂V₂
C₂ = C₁V₁/V₂ = (2.0 mol dm⁻³)(25.0 cm³)/(250 cm³)
C₂ = 0.20 mol dm⁻³
A gas occupies 2.0 L at 20°C and 1.5 atm. What volume will it occupy at STP (0°C, 1 atm)?
Using combined gas law: P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂/(P₂T₁) = (1.5 atm)(2.0 L)(273 K)/[(1 atm)(293 K)]
V₂ = 2.8 L
In the reaction 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu, identify the limiting reactant when 5.4 g Al reacts with 25.0 g CuSO₄.
Moles Al = 5.4 g ÷ 26.98 g/mol = 0.20 mol
Moles CuSO₄ = 25.0 g ÷ 159.61 g/mol = 0.157 mol
From stoichiometry: 0.20 mol Al needs 0.30 mol CuSO₄
Only 0.157 mol CuSO₄ available → CuSO₄ is limiting reactant
A student has 100 cm³ of 3.0 mol dm⁻³ H₂SO₄. What volume of water must be added to make a 0.50 mol dm⁻³ solution?
Using C₁V₁ = C₂V₂
V₂ = C₁V₁/C₂ = (3.0)(100)/(0.50) = 600 cm³
Water to add = 600 - 100 = 500 cm³
Describe the changes in particle arrangement, motion, and energy during sublimation. Give an example of a substance that undergoes sublimation.
Sublimation: solid → gas directly. Particles gain enough energy to completely overcome intermolecular forces and move from fixed positions to completely free motion. Energy increases significantly. Example: dry ice (CO₂)
What volume of 0.50 mol dm⁻³ NaOH solution contains 0.025 mol of NaOH? If this solution is diluted to 500 cm³, what is the new concentration?
Volume = moles / concentration = 0.025 mol / 0.50 mol dm⁻³ = 0.050 dm³ = 50 cm³
After dilution: C₂ = C₁V₁/V₂ = (0.50)(50)/(500) = 0.050 mol dm⁻³
Explain two conditions under which real gases deviate significantly from ideal behavior and why this occurs.
1) High pressure: particles are forced closer together, so particle volume becomes significant compared to container volume. 2) Low temperature: particles move slowly, so intermolecular forces become significant and affect particle behavior. Both violate ideal gas assumptions.
In the reaction Mg + 2HCl → MgCl₂ + H₂, calculate the mass of hydrogen gas produced when 12.0 g of Mg reacts with 15.0 g of HCl. Identify the limiting reactant.
Moles Mg = 12.0 g ÷ 24.31 g/mol = 0.494 mol
Moles HCl = 15.0 g ÷ 36.46 g/mol = 0.411 mol
From stoichiometry: 0.494 mol Mg needs 0.988 mol HCl
Only 0.411 mol HCl available → HCl is limiting
Moles H₂ produced = 0.411 mol HCl × (1 mol H₂/2 mol HCl) = 0.206 mol
Mass H₂ = 0.206 mol × 2.02 g/mol = 0.416 g
Calculate the number of oxygen atoms in 15.0 g of aluminum oxide (Al₂O₃).
Molar mass of Al₂O₃ = (2 × 26.98) + (3 × 16.00) = 101.96 g/mol
Moles of Al₂O₃ = 15.0 g ÷ 101.96 g/mol = 0.147 mol
Each Al₂O₃ has 3 oxygen atoms
Oxygen atoms = 0.147 mol × 3 × 6.022 × 10²³ = 2.66 × 10²³ atoms