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timo86m

  • one year ago

for nin And using calc to derive this formula

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  1. timo86m
    • one year ago
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  2. timo86m
    • one year ago
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    @nincompoop

  3. nincompoop
    • one year ago
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  4. timo86m
    • one year ago
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    Yes i remember integrating with time.

  5. timo86m
    • one year ago
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    The central science book is not that advanced then since it just gives you formulas but doesn't explain how to reach that conclusion

  6. timo86m
    • one year ago
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    calc 2 in this case. There is also some advanced wave function kinda stuff.

  7. timo86m
    • one year ago
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    in the electronic structure of matter chapter luckily it dont go too far. It just gives you equations you will use.

  8. abb0t
    • one year ago
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    Is this for thermodynamics? D:

  9. abb0t
    • one year ago
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    For a reaction of the form: A --> product a first order rate law takes the form: \[r = k_{1^{st}}[A]^1\] writing it in differentials, you get: \[\frac{ d[A] }{ dt } = -k_{1^{st}}[A]\] The minus sign as A is a reactant and so its concentration decreases with time (whici is the slope of a graph of [A] against time will be negativ when you graph it) The variables [A] and t can be separated to give: \[\frac{ 1 }{ [A] }d[A] = -k_{1^{st}}dt\] which CAN be integrated easily! \[\int\limits \frac{ 1 }{ [A] }d[A] = \int\limits -k_{1^{st}}dt\] this gives: \[\ln[A] = -k_{1^{st}}t + C\] The constant can be removed by supposing that at t = 0 since \[[A] = [A]_0\] which is the initial concentration of A giving us: \[\ln[A] = -k_{1^{st}}t + \ln[A]_0\] to simplify it a bit more to a form that might be more familiar to you is: \[[A] = [A]_0 e^{(-k_{1^{st}}t)} \]

  10. nincompoop
    • one year ago
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    @abb0t I think what you did is the same to what I attached

  11. abb0t
    • one year ago
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    Oh. Lol. My bad. haha.

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