Final_Exam_Fall_08

Final_Exam_Fall_08 - lastname rstname signature McCord 1pm...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
lastname frstname signature McCord 1pm class : #53745 Please read the following!! You MUST be in McCord’s 1PM class in or- der to take this exam. You will have to turn in ALL oF your exam materials at the end oF the exam. This includes your exam copy, your bubblesheet, and your scratch paper. Be sure and sign your exam copy (see above) as well as your bubblesheet. Your exam copy can be picked up at the window oF WEL 2.212 beginning Tuesday, 12/16 aFter 1PM. Your scores For the fnal should be avail- able on Quest by 7:00 PM this evening.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version 148 – Final – McCord – (53745) 2 This print-out should have 55 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points Which o± the ±ollowing statements concerning particle in a box is/are true? I) The energy o± the particle can be equal to zero. II) The wavelength λ is proportional to the length L o± the box. III) As the length L o± the box increases, the energy levels become closer in value. 1. III only 2. IandII 3. Ionly 4. II and III correct 5. II only 6. I, II and III 7. IandIII Explanation: The energy o± the particle in the box can never be zero; this is a consequence o± the ±act that it can never be stationary, and as a massive particle in motion, thus has to have some energy. The length L o± the box and the principle energy level n determine the wavelength λ o± the system, and because n can only have integer values, λ can only have certain discrete values. The energy E o± the particle is inversely proportional to L ,soas L increases, the di²erent energy levels are closer to total degeneracy. 002 10.0 points Bond Energies per mole o± bonds C–H 412 kJ C=O 743 kJ O–H 463 kJ O=O 496 kJ Using the provided bond enthalpy data, calculate the change in enthalpy ±or the ±ol- lowing reaction: CH 4 (g) + O 2 (g) ←→ CH 2 O(g) + H 2 O(g) 1. 710 kJ · mol - 1 2. 349 kJ · mol - 1 3. 577 kJ · mol - 1 4. - 577 kJ · mol - 1 5. - 710 kJ · mol - 1 6. - 349 kJ · mol - 1 correct Explanation: Δ H rxn =ΣBE reactants - ΣBE products =4 · 412 kJ · mol - 1 +496kJ · mol - 1 - 2 · 412 kJ · mol - 1 - 743 kJ · mol - 1 - 2 · 463 kJ · mol - 1 = - 349 kJ · mol - 1 003 10.0 points Sur±ace tension describes 1. the ±orces o± attraction between the sur- ±ace o± a liquid and the air above it. 2. the inward ±orces that must be overcome in order to expand the sur±ace area o± a liquid. correct 3. the ±orces o± attraction between sur- ±ace molecules o± a solvent and the solute molecules. 4. adhesive ±orces between molecules. 5. capillary action. 6. the resistance to ³ow o± a liquid. Explanation: Molecules in the interior o± a liquid inter- act with molecules all around them, whereas
Background image of page 2
Version 148 – Final – McCord – (53745) 3 molecules at the surface of a liquid can only be a±ected by those beneath the surface layer.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 17

Final_Exam_Fall_08 - lastname rstname signature McCord 1pm...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online