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 $CH<em>4$ and $CCl</em>4$:
  • $CCl_4$ 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:
  • $CH<em>4(g) + 2O</em>2(g) \rightarrow CO<em>2(g) + 2H</em>2O(l)$
• Enthalpy of Combustion ($\Delta H_{comb}$):
  • $\Delta H<em>{comb} = \Sigma n \cdot \Delta H</em>{f,prod} - \Sigma m \cdot \Delta H_{f,react}$
  • Given:
    • $\Delta H<em>f(H</em>2O, l) = -285.82 \frac{kJ}{mol}$
    • $\Delta H<em>f(CO</em>2, g) = -393.5 \frac{kJ}{mol}$
    • $\Delta H<em>f(CH</em>4, g) = -74.85 \frac{kJ}{mol}$
  • $\Delta H_{comb} = [1 \cdot (-393.5) + 2 \cdot (-285.82)] - [1 \cdot (-74.85) + 2 \cdot 0]$
  • $\Delta H_{comb} = -890.3 \frac{kJ}{mol}$
• Pressure Calculation:
  • Given:
    • Volume (V) = 5.0 L
    • Temperature (T) = 25 °C = 298.15 K
  • Ideal Gas Law: $PV = nRT$
  • $n = \frac{PV}{RT} = \frac{P \cdot 5.0}{0.08206 \cdot 298.15}$
  • $P = \frac{nRT}{V}$
  • $P = \frac{0.0478 \cdot 0.08206 \cdot 298.15}{5.0}$
  • $P = 0.23 atm$