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Aleks [24]
3 years ago
8

What is the pressure in a 27.0-L cylinder filled with 44.9 g of oxygen gas at a temperature of 315 KK

Chemistry
1 answer:
steposvetlana [31]3 years ago
6 0

Answer:

1.343 atm

Explanation:

We are given the following;

Pressure, p = ?

Volume v = 27 L

Mass of oxygen = 44.9 g

Temperature, T = 315 K

The formular relating all these variables is the equation;

PV = nRT

where R = gas constant = 0.08206 L atm / mole K

To obtain n, we use;

number of moles, n = Mass /  molar mass =  44.9 / 32 = 1.403 moles

From the ideal gas equation;

P = nRT / V

P = 1.403 * 0.08206 * 315 / 27

P = 36.27 / 27 = 1.343 atm

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<h3>Answer:</h3>

2 L Ne

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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<u>Chemistry</u>

<u>Atomic Structure</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

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

STP - 22.4 L per mole

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})
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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

1.568 L Ne ≈ 2 L Ne

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