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Some troubling problems...

Discussion in 'DAT Discussions' started by Predentknight, Jun 12, 2008.

  1. Predentknight

    Predentknight 5+ Year Member

    Feb 12, 2008
    Some problems I am having trouble with:

    1. Which of the following descriptions relate to nuclear reactions involving one beta emission. I.Transformation of a nuclear proton into a nuclear neutron II.Transformation of a nuclear neutron into a nuclear protonIII.Emission of one electron

    Whats the ans. and why?

    2.Balance the following redox equation and determine the sum of coefficients for both the reactants and the products.
    Al(s) + NO3-(aq) + H+(aq) à N2O4(g) + H2O(l) + Al3+(aq)


    I will post answers once I get some responses
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  3. doc3232

    doc3232 7+ Year Member

    Feb 15, 2008
    For the second one

    I got 2 NO3 4 H 1 Al
    Right side: 1 N2O4 2 H2O and 1 Al
    The electrons should be set correctly.
  4. osimsDDS

    osimsDDS 5+ Year Member

    Jun 8, 2007
    for #1: is the answer II and III?

    so basically when you undergo B emission you are adding 1 prton and 0 neutrons. So you would have lets say C-12 with 6 are adding a proton so you would have 7 protons. Now take the difference to see how many neutrons you have: 12-7= 5, and before you have 21-6= 6. So you went from 6 neutrons to 5 neutrons so 1 of those neutrons had to be changed into a for answer III, when it is B positron emission it is called B + decay and when its B electron emission it is called B- just have to know that when you have beta emission it emitts an electron and when its B+ emission or positron emission it emitts a positron (e+).

    Hope thats right haha
  5. happyasaclam88

    happyasaclam88 7+ Year Member

    Oct 16, 2007
    when i did the redox, i got a lot messier numbers in order to get the electrons to even out. i got:
    2 Al + 12 H+ + 6 NO3- for reactants
    3 N2O4 + 2 Al3+ + 6H2O for products.

    what is the answer? thanks!
  6. osimsDDS

    osimsDDS 5+ Year Member

    Jun 8, 2007
    I got:

    two half reaction are:
    Al----> Al3+ + 3e-
    2e- + 2NO3- + 4H+ ----> N2O4 + 2H20

    then you multiply the bottom half reaction by 3 and the top by 2 to cancel out the electrons then add them up and you should get:

    Reactants: 2Al, 6 NO3-, 12H+
    Products: 3 N2O4, 6 H20, and 2 Al3+
    Last edited: Jun 12, 2008
  7. Predentknight

    Predentknight 5+ Year Member

    Feb 12, 2008
    This is the correct answer to #1, that really helps,thanks for the explanation

    And for #2 the ans. is 31. Happyasaclam88 has the right answer do you mind explaining how you come to that.

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