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STP gchem problem

Discussion in 'DAT Discussions' started by keibee82, Aug 6, 2006.

  1. keibee82

    keibee82 Blue_tooth...
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    For which one of the following mixture is the ideal gas law
    most likely to produce an accurate prediction of volume at STP?

    a. HCl and HBr
    b. HCl and NH3
    c. HF and HCl
    d. HF and CH4
    e. HCl and CH4



    How many seconds will it take to produce 11.2 liters of Cl2 (g),
    measured at STP, by electrolysis of molten NaCl with a 12amp
    current? (F=96500 C/mol)

    a. 96500/12
    b. 12/96500
    c. (12)(96500)
    d. (2)(12)/(96500)
    e. 96500/(2)(12)


    Can you guys help me with these problems? thx!
     
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  2. dat_student

    dat_student Junior Member
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    you can easily eliminate three of the answer choices: b,c and d. NH3 & HF form hydrogen bonds (i.e. intermolecular interaction). It's either A or E. Since HBr is larger & heavier than CH4 I'd pick E. HCl and CH4. That's my guess. What's the answer?


    This one is easy:

    1 mole -> 96500 c
    11.2 liters at STP = 1/2 moles or 1/2(96500) C

    I (amp) = Q (charge) / t (time)

    12 = [(1/2)(96500)] / t

    t = 96500 / [(2)(12)]
     
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  3. OP
    OP
    keibee82

    keibee82 Blue_tooth...
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    hmm answer is E for the first one and A for the second one.
     
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  4. Envision

    Envision Envisioning...
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    How many seconds will it take to produce 11.2 liters of Cl2 (g),
    measured at STP, by electrolysis of molten NaCl with a 12amp
    current? (F=96500 C/mol)

    a. 96500/12
    b. 12/96500
    c. (12)(96500)
    d. (2)(12)/(96500)
    e. 96500/(2)(12)

    This is a very long and time consuming problem that requires several steps of dimensional analysis. i don't think you have to worry about anything this hard on the actual exam.

    first, figure out the rxn: 2Cl- <-> Cl2 + 2e-

    note: amp= C/seconds
    note: 12 columbs=1 sec
    note: a mol e-= 96500C
    note: 11.2 L in stp is half a mole
    note: use stoichiometric coeficients from rxn above

    (11.2L) (1 mol Cl2/22.4L) (2 mol e-/1 mol Cl2)(96500C/mol e-)(1 sec/12 columbs) = 96500/12
    (E)

    for the first problem, i'd pick e over a.
     
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  5. dat_student

    dat_student Junior Member
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    Envision is right. I forgot to multiply by 2 :)

    I narrowed it down to a or e. I chose e because HBr is more polar, larger and heavier. Like I said, it's either a or e. I'm NOT sure why Kaplan says a. Kaplan probably has an explanation for it. I'd very much like to know their reasonings behind picking a over e. They may be right. I guessed b/w a and e.
     
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  6. OP
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    keibee82

    keibee82 Blue_tooth...
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    answer for the first one is E. you guys are right :)
    thanks for the help !!
     
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  7. Sorry boys,....correct answer. wrong reasoning! question one is a physical chemistry question. The molecules are in the GAS phase. That's important. Hydrogen bonding comes into play more in SOLUTIONS!
    The reasoning is different. A gas is ideal when its collisions are inelastic! That's a fundamental assumption of the ideal gas law. Here, you're looking to minimize the intermolecular forces. Specifically, you're looking to minimize the DIPOLE-DIPOLE interactions between the molecules! H-X has a strong dipole. NH3 has a small dipole. Methane has no dipole! Thus, D and E are the better answers. However, F is more electronegative. Thus, out of D and E, E has less dipole-dipole interactions.

    ....again, this is a physical chemistry question!
     
  8. DrGeek

    DrGeek Member
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    http://hyperphysics.phy-astr.gsu.edu/HBASE/chemical/bond.html

    "HYDROGEN BONDING" is a form of DIPOLE-DIPOLE interactions. See below:

    Hydrogen Bonding

    "Hydrogen bonding differs from other uses of the word "bond" since it is a force of attraction between a hydrogen atom in one molecule and a small atom of high electronegativity in another molecule. That is, it is an intermolecular force, not an intramolecular force as in the common use of the word bond.

    When hydrogen atoms are joined in a polar covalent bondwith a small atom of high electronegativity such as O, F or N, the partial positive charge on the hydrogen is highly concentrated because of its small size. If the hydrogen is close to another oxygen, fluorine or nitrogen in another molecule, then there is a force of attraction termed a dipole-dipole interaction. This attraction or "hydrogen bond" can have about 5% to 10% of the strength of a covalent bond."
     
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  9. um, alright.... btw, for the DAT, I completely DIDN'T follow strict definitions. All I used was common sense. That itself got me a 24TS while studying for one month,...and on the phone with the gf for at least 3-5 hrs/day.

    but anyway....you're still going to think in terms of dipole-dipole interaction. in fact, the WORST choice for this question is A. somehow, ...people were choosing between A and E.


    ...btw, the def of h-bond is standardized using a linear molecule. it's not incorrect. it just not intended to be a stead-fast rule. The molecule NH3 has 3 potential h-bond sites, but the molecule only has ONE dipole moment! but enough semantics. think dipole moment!
     
  10. DrGeek

    DrGeek Member
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    Dude, did you read the definition of hydrogen bonds? A & E don't form hydrogen bonds.
     
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  11. Envision

    Envision Envisioning...
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    true story. Only F, N, and O can form hydrogen bonds. Thus, A is not the worst answer...i would say C is the worse answer.
     
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