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Charge in water

  1. Aug 28, 2006 #1
    The problem states: Calculate the number of coulombs of positive charge in 250 cm3 of (neutral) water. (Hint: A hydrogen atom contains one proton and an oxygen atom contains eight protons.)

    Ok, first I converted to moles. Using the density of water.

    [tex] 250mL * \frac{1g}{1mL} * \frac{1 mole H_2O}{18.02g}=13.8735moles H_2O [/tex]

    [tex]13.8735molesH_2O*6.022*10^{23}=8.3546*10^{24} atoms H_2O [/tex]

    [tex]8.3546*10^{24}*(2*(1 proton H)+8 proton O)=8.3546*10^{25} protons [/tex]

    [tex]8.3546*10^{25} protons * 1.6*10^{-19} Coulombs=1336736Coulombs[/tex]

    Does this seem right? That seems really large. However, the electrons calculate up to exactly the same amount and cancel it out.
    Last edited: Aug 28, 2006
  2. jcsd
  3. Aug 28, 2006 #2


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  4. Aug 28, 2006 #3
    oh ok, I didn't check the second page. Thanks!
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