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UWW CHEM 104 - Review for Chapters 10-11

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Chem 104 1st Edition Lecture 6 Outline of Last Lecture II. Liquids and their properties.A. Definition of surface tensionB. Definition of viscosityC. Definition of meniscusD. Definition of vaporizationIII. Introduction to Vaporization and Condensation.A. The factors that affect vaporization and condensationOutline of Current Lecture II. Review of in class quiz on 02/02/2015A. Example questionsIII. Introduction to heat of vaporizationCurrent LectureI. On Monday, 02/02/2015 there will be an in class quiz on Chapters 10-11. This will include: intermolecular forces, strong vs. weak attractions; dipole-dipole attractions, hydrogen bonds, dispersion forces, and ion-dipole attractions.A. Example 1:a. Place these in order of increasing intermolecular forces NH2CH3 CO2 F2The answer you should have gotten is: F2 < CO2 < NH2CH3This is because F2 only has dispersion forces, CO2 has only dispersion forces but has a larger molar mass than F2, and NH2CH3 has a hydrogen bond, dipole-dipole attraction, and dispersion forces.B. Example 2:a. Which of these has the smallest dipole-dipole forces?CH3Cl HBr O2 NOO2. This is because O2 has the ability to obtain temporary dipole-dipole attractions.C. Example 3:a. Which does not have dipole-dipole as their strongest forces? CH2Cl2 CH3OCH3 CH3Br HCCl3 CO2CO2. This contains only dispersion forces.D. Example 4:a. Which exhibits hydrogen bonding? CH3Cl HI CH3OCH3 NH3NH3. Only hydrogen bonds are possible through O, N, and F.E. Example 5:a. Give the change in condition to go from a liquid to a gas.1. Increase heat or reduce pressure2. Increase heat or increase pressure3. Cool or reduce pressure4. Cool or increase pressure5. None of the above.It’s #2 because of the molecules need to be “excited” through increased pressure or temperature which causes more interactions between molecules that will rise to the surface and vaporize as a gas.F. Example 6:a. What is the strongest force present in CHF3?1. Ion-dipole2. Dispersion3. Hydrogen bond4. Dipole-dipole5. None of the aboveIt’s #4. By process of elimination we’re figured out that ion-dipole doesn’t apply in this particular problem, there are no hydrogen bonds,and the molecule is polar. Thus, dipole-dipole is the answer.II. Heat of vaporization: the amount of heat energy required to vaporize one mole of liquid. This is also known as enthalpy of vaporization. It is always endothermic, therefore ∆Hvap is positive. This is somewhat temperature dependent.∆Hcondensation = -∆Hvaporization∆ = change = Hfinal - HintialA reminder from Chem 102 is that ∆E = q + w. q = heat, w = work. Work is dependent on the path traveled. This does not make it a state function. A state function is a property of something that only depends on its current state, not the journey to get there. E and H are state functions.a. Example: Hypothetically, I am asked to meet a friend down the street for dinner. Ican choose to fly because I’m also superman. It does not matter which path I flew to my dinner date, the only thing that matters is my position after my journey. My position is a state


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UWW CHEM 104 - Review for Chapters 10-11

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