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sertanlavr [38]
3 years ago
13

A mixture of gases contains 4.46 moles of neon (Ne), 0.74 mole of argon (Ar), and 2.15 moles of xenon (Xe). Calculate the partia

l pressures of the gases if the total pressure is 2.00 atm at a certain temperature.
Chemistry
1 answer:
viva [34]3 years ago
5 0

Explanation:

The partial pressure of an individual gas is equal to the total pressure of the mixture multiplied by the mole fraction of the gas.

Total pressure = 2atm

Mole Fraction = number of moles / total number of moles

Neon

Mole Fraction = 4.46 / 7.35 = 0.607

Partial Pressure = 0.607 * 2 = 1.214 atm

Argon

Mole Fraction = 0.74 / 7.35 = 0.101

Partial Pressure = 0.101 * 2 = 0.202 atm

Xenon

Mole Fraction = 2.15 / 7.35 = 0.293

Partial Pressure = 0.293 * 2 = 0.586 atm

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Explanation:

The relation between K_a\&K_b is given by :

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3 0
3 years ago
Help in chem please!!!!!!
IgorLugansk [536]

This is a one-step unit analysis problem.  Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}

<h3>Answer:</h3>

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3 0
3 years ago
A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres
tatiyna

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 m^{3}

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Gas constant for Argon R = 208.13 \frac{J}{kg k}

From ideal gas equation P V = m RT

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