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Math Help - Chemistry Question

  1. #1
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    Chemistry Question

    Assume that an atom of neon is spherical with a volume V of 4.85 x 10^-24 m^3. If the equation relating radius r and volume V of a sphere is

    V = 4/3 pie r^3 (sorry, i don't know how to do the pie symbol on the computer)

    determine the atomic radius of a neon atom.



    Not sure if I'm right, but I have:

    V = 4.85 x 10^-24m^3 = 4/3 pie r^3

    0.0485m^3 = 4/3 pie r^3


    Is this right? And if so, do I solve it any further? If I'm supposed to, how?

    Cheers
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  2. #2
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    V = \frac{4}{3} \pi r^3

    \frac{3V}{4\pi} = r^3

    r = \sqrt[3]{\frac{3V}{4\pi}}


    btw, this is pie ...



    ... and this is pi.

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  3. #3
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    Chemistry Question

    lol, I it was late and I had food on the brain .

    So, where it says 3V, would you have

    3 x 4.85x10^-24
    _________________________________
    4pi

    ?

    So radius would equal

    r = 1.050068691x10^-8

    Don't know if I'm right, or if I'm completely confused?
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by AJShaw View Post
    lol, I it was late and I had food on the brain .

    So, where it says 3V, would you have

    3 x 4.85x10^-24
    _________________________________
    4pi

    ?

    So radius would equal

    r = 1.050068691x10^-8

    Don't know if I'm right, or if I'm completely confused?
    You would never round it off that much anyway. As many digits are given in the question will do - in this case 3sf.

    Where skeeter put 3V you put in the value of V given then evaluate the expression.

    \sqrt[3]{\frac{3 \times 4.85 \times 10^{-24}}{4\pi}} = 1.05 \times 10^{-8} \text { m } = 10.5 \text{ nm }


    According to Data this is about 2 orders of magnitude too large but the calculated answer holds given the volume named
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  5. #5
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    I get it now. Thanks heaps
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