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Two charge problems

  1. Mar 7, 2006 #1
    I have been trying to figure out these two problems and for some reason my logic or my math is wrong...please help.

    1) What is the total charge (in C) on all the electrons in 7.0kg of water.
    --what I did. 7000g/18g * 6.022E23 to give me the # of electrons (2.3419E26) then I plugged that into the N=q/e and got q= 37,470,222C. But according to our problem site it is wrong, so where did I go wrong?

    2) A 2.0g Cu penny has a positive charge of 0.27millicoulombs. What decimal fraction of its electrons has it lost?
    --what I did. 2g/63.546g 8 6.022E23 to get # of electrons it started with (1.895E22) then I plugged used the N=q/e (N= 0.27E-6C/1.6E-19 which came out to 1.6875E12) THen I divided 1.895E22/1.6875E12 to get 1.122E10, but the answer is incorrect.

    I would truly apprecate ANY help.
     
  2. jcsd
  3. Mar 7, 2006 #2

    Hootenanny

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    (1)For question one as far as I can see your right. I got a similar answer: 37 357 777C. Just make sure you are using the given constants.
     
  4. Mar 7, 2006 #3

    Hootenanny

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    (2) This is what I did:
    Number of electrons lost = [itex]\frac{q}{e} = \frac{0.00027}{1.6\times 10^{-19}} = 1.6875\times 10^{15}[/itex]

    Number of atoms of Copper in penny = [itex]\frac{2}{63.5} \times 6.02\times 10^{23} = 1.896\times 10^{22}[/itex]

    Number of electrons (Each atom has 29 electrons) = [itex]\left(1.896\times 10^{22} \right) \times 29 = 5.50 \times 10^{23}[/itex]

    Ratio = [itex]\frac{lost}{total} = \frac{1.6875 \times 10^{15}}{5.50 \times 10^{23}} = 3.07 \times 10^{-9}[/itex]
     
  5. Mar 7, 2006 #4
     
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