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A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)

$\mathrm{XeF}_{2}$

01:57

Aadit S.

Chemistry 101

Chapter 9

Gases

Yanbing Z.

December 21, 2021

Is remained H2 is 18torr, shouldn’t reacted H2 be 36torr? It should be XeF4 then?

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so the pressure exerted by each individual gas in the mixture of the gas is has called its partial pressure. So the partial pressure of the gas in a mixture is equal to the mathematical product of total pressure of the gaseous mixture. It's more fraction in the mixture, so we can express this as P I is equal to X. I was supplied by Pete Total Excise the mole fraction of the gas. So we have a reaction to consider here on p of H two is equal to Pete Total. Take away p x e yeah, X. This is equal to 54 top than P X e is equal to 18. Tour multiplied by Juan Mall XY f x one mall XY. So partial pressure of the H two, which remains and reacted, is as follows. We have 36 take away 18. That gives us a value of 18 tour. And so the pressure of hydrogen gas, which reacts, is calculated by having 36 take away. 18. That's got 18 tall, so partial pressure of the number of moles of gas a proportional to each other. So with 18 tools of AC CF. X on 18 tools off H to have been used. So therefore, we have x divided by two t cost. One X is equal to two. So therefore the formula is XY F two. Yeah.

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