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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Potassium (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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can someone tell me if my answer is correct, please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hey, I'm back, I was away from my computer for a while. Thanks! :)
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