MR

Sample Final Exam Part 2 Notes

Accuracy vs. Precision

  • Definition:
    • Accurate: Data is close to the true or accepted value.
    • Precise: Data is consistent and repeatable.
    • Student D's Data: Accurate but not precise.
      • Average: 122 °C (matches the known value).
      • Range: Large compared to other students.

Electrolytes

  • Strong Electrolytes:
    • Completely dissociate into ions in water.
    • Ions move freely, conducting electricity well.
    • HCl: A strong electrolyte.
      • Separates into H^+ and Cl^- ions in water.

Atomic Models

Nuclear Model

  • Claim:
    • The atom has a very small and very dense nucleus.
  • Evidence:
    • Most alpha particles directed at gold foil went through.
    • Occasionally, alpha particles deflected at large angles.
  • Reasoning:
    • Most particles going through indicates the nucleus is very small.
    • Deflection of alpha particles at large angles indicates the nucleus is very dense.

Plum Pudding Model

  • Claim: Atoms are divisible with subatomic particles that we now know as electrons.
  • Evidence:
    • When applying a potential difference between two electrodes, green rays were observed, which deflected by an electric field. These green rays were observed regardless of the material of the electrodes.
  • Reasoning:
    • Deflection of green ways towards + of the electric field indicates the particle carries - charge. The independence of material indicates it is present in all types of atoms.

Intermolecular Forces and Boiling Point

  • Comparison of CH4 and CCl4:
    • CCl_4 has a higher boiling point.
  • Reasoning:
    • CCl_4 has more electrons and is more polarizable.
    • Stronger dispersion forces require more energy to overcome, leading to a higher boiling point.

Ionization Energy Trends

  • First Ionization Energy (Li to Ne):
    • Trend: Increases from Li to Ne.
    • Cause: Effective nuclear charge increases, pulling valence electrons closer to the nucleus and increasing their attraction.
  • Large Increase in Ionization Energy:
    • Observation: Very large increase between the third and fourth ionization energy of B.
    • Explanation: After 3 ionizations, B has a noble gas electron configuration, requiring a large amount of energy to remove an electron.

Kinetic Theory of Gases

  • Pressure as Evidence:
    • Claim: Gas molecules are in constant, random motion.
    • Evidence: Pressure is created when gas particles randomly collide with the walls of their container.
    • Reasoning: Collisions generate a force. Summing all of these forces over a given area accounts for the pressure observed.

Lewis Structures and Molecular Geometry

  • Glycine in Aqueous Solution:
    • Includes drawing the most stable Lewis structure with lone pairs and bonding pairs.
    • Assigning formal charges to the structure.
  • O-C-O Bond Angle:
    • Angle: 120°
    • Reasoning: In O-C-O, the central C atom has 3 electron domains, and all 3 are bonding pairs. This indicates a trigonal planar molecular geometry, and the bond angle is 120°.

Hydrogen Bonding vs. Dipole-Dipole Forces

  • Hydrogen Bonding Strength:
    • Hydrogen bonding intermolecular forces are stronger than ordinary dipole-dipole intermolecular forces.
  • Justification using Coulomb's Law:
    • The covalent bond between H and N, O, or F is very polar, resulting in large partial charges (\delta^+/\delta^-).
    • The small size of H allows the molecules to get closer together.

Combustion of Methane

  • Balanced Chemical Equation:
    • CH4(g) + 2O2(g) \rightarrow CO2(g) + 2H2O(l)
  • Enthalpy of Combustion (\Delta H_{comb}):
    • \Delta H{comb} = \Sigma n \cdot \Delta H{f,prod} - \Sigma m \cdot \Delta H_{f,react}
    • Given:
      • \Delta Hf(H2O, l) = -285.82 \frac{kJ}{mol}
      • \Delta Hf(CO2, g) = -393.5 \frac{kJ}{mol}
      • \Delta Hf(CH4, g) = -74.85 \frac{kJ}{mol}
    • \Delta H_{comb} = [1 \cdot (-393.5) + 2 \cdot (-285.82)] - [1 \cdot (-74.85) + 2 \cdot 0]
    • \Delta H_{comb} = -890.3 \frac{kJ}{mol}
  • Pressure Calculation:
    • Given:
      • Volume (V) = 5.0 L
      • Temperature (T) = 25 °C = 298.15 K
    • Ideal Gas Law: PV = nRT
    • n = \frac{PV}{RT} = \frac{P \cdot 5.0}{0.08206 \cdot 298.15}
    • P = \frac{nRT}{V}
    • P = \frac{0.0478 \cdot 0.08206 \cdot 298.15}{5.0}
    • P = 0.23 atm