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allsm [11]
3 years ago
11

What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro

priate unit?
Chemistry
1 answer:
monitta3 years ago
3 0
Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C° 
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n =  1.45/mass weight of argon
   = 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm

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Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

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where,

P_1 = initial pressure of gas at STP = 10^5Pa

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V_1 = initial volume of gas = 700.0 ml

V_2 = final volume of gas = 200.0 ml

T_1 = initial temperature of gas = 273 K

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Now put all the given values in the above equation, we get:

\frac{10^5\times 700.0ml}{273K}=\frac{P_2\times 200.0ml}{303K}

P_2=388462Pa

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