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anonymous

  • one year ago

How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all of the work used to solve this problem. 2 K + F2 yields 2 KF

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  1. anonymous
    • one year ago
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    Potassium (K) = 39.098 g/mol 23.5g / 39.098 g/mol = 0.601 mol ratio: 2:1 0.601 mol / 2 = 0.3005 mol V = nRT/P = 22.4 L - one mole of any gas occupies a volume of 22.4 L n = 0.3005 mol R = 0.0821 T = 273.15 K P = 1 atm (0.3005 * 0.0821 * 273.15) / 1 = 6.73 V = 6.73 L

  2. anonymous
    • one year ago
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    Can someone tell me if my answer is correct, please?

  3. anonymous
    • one year ago
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    @ganeshie8

  4. anonymous
    • one year ago
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    @Cuanchi

  5. cuanchi
    • one year ago
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    yes its correct!

  6. anonymous
    • one year ago
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    Hey, I'm back, I was away from my computer for a while. Thanks! :)

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