Ch.11 - Liquids, Solids & Intermolecular ForcesWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Place these hydrocarbons in order of decreasing boiling point. Rank from highest to lowest boiling point. paraffin, octadecane , methane, hexane , 2,2-dimethylbutane

Solution: Place these hydrocarbons in order of decreasing boiling point. Rank from highest to lowest boiling point. paraffin, octadecane , methane, hexane , 2,2-dimethylbutane

Problem

Place these hydrocarbons in order of decreasing boiling point. Rank from highest to lowest boiling point. 

paraffin, octadecane , methane, hexane , 2,2-dimethylbutane

Solution

Recall: For hydrocarbons, a large surface area leads to a high boiling point, i.e. a large compound will have lots of bonds, therefore it will require more energy (and thus a higher temperature) for those bonds to break and vice–versa.

View the complete written solution...