SDN members see fewer ads and full resolution images. Join our non-profit community!

More Chem and Ochem ?s

Discussion in 'Pre-Dental' started by Beagle, Aug 8, 2002.

  1. Beagle

    Beagle Meet BEAGLE 7+ Year Member

    596
    0
    Jul 16, 2001
    California
    Does anyone have answers for these ?s

    1. Which of the following statementst about straight-chain alkanes is true?
    A. As the molecular weight increases, the symetry of the molecules decrease
    B. As the molecular weight increaes, the surface area per molecule decreases.
    C. As mol. weight increases, dispersion forces between molecules become weaker
    D. As the mol. weight increases, the boiling pt decreases
    E. As the MW increases, dispersion forces between the molecules become stronger

    2. What property do CH3CH2COOH, HCL, HCOOH, and HNO3 have in common?

    A. they are all strong acids
    B. they accept protons
    C. They can all act as bronsted acids
    D. They can all act as bronsted bases
    E. They are nonpolar molecules

    3. When a nonvolatile hydrocarbon is dissolved in benzene, the partial pressure of benzene over the resulting solution

    A. is greater than the vapor pressure of pure benzene
    B. Decreases as the mole fraction of benzene decreases
    C. Is lower than the vapor pressure of pure hydrocarbon
    D. Increases as the mole fraction of benzene decreases
    E. Does not change


    4. An excited electron drops from a high energy level, Ea, to ground level state, Eb. The wavelength of the photon emitted is:
    A. h/(Ea-Eb)
    B. (Ea-Eb)/h
    C. (Eb-Ea)/h
    D. hc/(Ea-Eb)
    E. c(Ea-Eb)/h

    5. Catalysts have which of the following effects on chemical reactions.

    A. They cause the reaction to proceed spontaneously
    B. they increase the available free energy
    C. They increase the rate at which products are formed
    D. They lower the energy of formation of the products

    I am stuck--- I read in barrons that B was correct. However, C seems to be a good answer?? Or is rate not acted on with catalysts?. And D sounds good as well?


    6. CaF2 is added to 0.1 M Ca(NO3)2 solution. at waht concentration of F- will CaF begin to precipitate. (Ksp of CaF2=
    4 x 10^-11

    A. 4 x 5 ^-5
    B. 2 x 5 ^-5
    C. 4 x 10 ^-10
    D. 2 x 10 ^ -10
    E. 4 x 10 ^-11

    I have no idea how to approach this problem

    7. The eq for this reaction
    N204 (g) -----> 2 NO2 (g) delta H= 13.1 kcal/mol can be shifted to the right by
    A. decreasing the volume
    B. decreasing the pressure
    C. decreasing the temp
    D. two of the above?


    Thank you
     
  2. SDN Members don't see this ad. About the ads.
  3. DesiDentist

    DesiDentist G. S. Khurana, DMD, MBA Moderator Emeritus Exhibitor 10+ Year Member

    1,573
    2
    Jul 21, 2001
    Washington, DC
    SDN Exhibitor
     
  4. vixen

    vixen I like members 10+ Year Member

    5,760
    1
    Oct 17, 2000
    upstate ny
    the last one is lechatliers principle....when you decrease pressure (or increase temp), you go to the side that has MORE moles.
     
  5. vixen

    vixen I like members 10+ Year Member

    5,760
    1
    Oct 17, 2000
    upstate ny
    I *think* 3 is B...not sure though....I think its talking about partial pressures....if you have benzene + X....then the partial pressure for benzene is benzene/benzene + X....so then it would now benzene would have a lower # (its being divided by a larger #)....


    This is just what I would think, I could be totally off :)
     
  6. portlander

    portlander Member 7+ Year Member

    93
    0
    Jun 18, 2002
    1. Dispersion forces become weaker (c). I'm not sure why, I got this wrong on the diagnostic exam because I put that they would become stronger, but the key said (c).

    2. Bronsted acids is correct.

    3. The answer is (b) as the mole fraction increases, the partial pressure increases also. Volitile means a liquid that will evaporate quickly.

    4. Oooh, I know this!! Well, E=hv (E is energy, v is wavelength, and h is planks constant) So wavelength equals energy over planks constant, so you can discount A and D. E is the change in energy, and since you are going from high to low, it should be Ea-Eb, so B is the correct answer.

    I have no idea how to approach this problem.

    7. The eq for this reaction
    N204 (g) -----> 2 NO2 (g) delta H= 13.1 kcal/mol can be shifted to the right by
    A. decreasing the volume
    B. decreasing the pressure
    C. decreasing the temp
    D. two of the above?

    Well, I think that decreasing the presure would push it to the right. I know that increasing pressure would push it the other way, (because there is 1 mole as compared to 2 on the right) so I assume that doing decreasing pressure would do the opposite. Anyone else??
     
  7. wasabi007

    wasabi007 Senior Member 7+ Year Member

    161
    0
    Sep 16, 2001
    San Francisco, CA
    4) i think the answer to #4 is C instead of B...

    portlander was right in that you should use planck's theory: E=hf or E=hV, where E is the energy of the electron, h is planck's constant, and f or V is FREQUENCY. Just remember that f=c/wavelength. so plug in this new value for f, and solve for wavelength. i believe you end up with
    C: hc/(Ea-Eb) = wavelength.

    7) answer is B....le chatelier's principle

    when you decrease P of a system, your rxn will move in the direction where there are more moles of gas (right). decreasing V is the same as increasing P, so that would move the rxn to the left...decreasing T would also shift the rxn to the left because the rxn is endothermic (delta H > 0).
     
  8. Viraj

    Viraj Senior Member 10+ Year Member

    156
    1
    Jan 18, 2002
    Boston
    For #4
    E= h*c/lambda
    where E= Energy
    h= Planck's const
    lamba= wavelength
    so according to the question :
    4. An excited electron drops from a high energy level, Ea, to ground level state, Eb. The wavelength of the photon emitted is:
    A. h/(Ea-Eb)
    B. (Ea-Eb)/h
    C. (Eb-Ea)/h
    D. hc/(Ea-Eb)
    E. c(Ea-Eb)/h


    Ea-Eb= hc/lambda
    this implies that Ea-Eb/hc=1/lambda
    this implies that lambda= hc/Ea-Eb
    Therefore answer to this question is gonna be choice D
    I hope this solves the problem for all of you guys
     
  9. Beagle

    Beagle Meet BEAGLE 7+ Year Member

    596
    0
    Jul 16, 2001
    California
    Thanks everyone for your help!
    I am still stuck on the catalyst question though.
    Does anyone have an answer?

    Thank you!!!!
     
  10. DesiDentist

    DesiDentist G. S. Khurana, DMD, MBA Moderator Emeritus Exhibitor 10+ Year Member

    1,573
    2
    Jul 21, 2001
    Washington, DC
    SDN Exhibitor
    i believe the answer to the "catalyst" question is C. Catalysts only increase the rate of the RXN, they don't affect anything else, hence my answer in red.

    DesiDentist
     
  11. Beagle

    Beagle Meet BEAGLE 7+ Year Member

    596
    0
    Jul 16, 2001
    California
    Desi
    Do you think increasing the available energy is wrong?
    I swear I saw that in barrons and wondered if it was a mistake??
     
  12. UBTom

    UBTom Class '04 official geezer 10+ Year Member

    1,459
    3
    Jul 24, 2002
    Queens, NY
    Well, all catalysts actually do is lower the activation energy required to get a reaction going... It does not increase available energy.

    If one looks at the Gibbs free energy change of a reaction, written as negative-delta-G (don't know how to type the delta symbol) of a catalyzed vs. an uncatalyzed reaction involving the same reactants, it would be absolutely the same.

    Desi is right-- a catalyst increases the rate of a reaction. The caveat is that it is still subject to equilibrium laws same as uncatalyzed reactions! :p

    HTH!
     
  13. wasabi007

    wasabi007 Senior Member 7+ Year Member

    161
    0
    Sep 16, 2001
    San Francisco, CA
    whoops...sorry...i put the wrong letter answer for #4...but my answer and process in coming to it was correct...i just correlated it with the wrong letter...
     

Share This Page