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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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Answer:

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General Formulas and Concepts:

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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Explanation:

<u>Step 1: Define</u>

0.200 mol H₂S

<u>Step 2: Identify Conversions</u>

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<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

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