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FinnZ [79.3K]
3 years ago
5

If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?

Chemistry
1 answer:
meriva3 years ago
5 0
Use the clapeyron equation:

T in kelvin : 6.80 + 273 => 279.8 K

R = 0.082 

n = 71.5 moles

P = 5.03 atm

Therefore:

P x V = n x R x T

5.03 x V = 71.5 x 0.082 x 279.8

5.03 x V = 1640.4674

V = 1640.4674 / 5.03

V = 326.13 L

hope ths helps!
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A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.
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Answer:

  • <em>Hydration number:</em> 4

Explanation:

<u>1) Mass of water in the hydrated compound</u>

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

<u>2) Number of moles of water</u>

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  • molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

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  • Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol =  0.108 mol

<u>4) Ratio</u>

  • 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈  4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

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Answer:

0.0303 Liters

Explanation:

Given:

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The number of moles of 0.2352 g of potassium hydrogen phosphate

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= 0.0027 mol of HNO₃

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for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

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