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marshall27 [118]
3 years ago
9

A 0.5 mol sample of N2 is in a 6L container at 2 atm. what is the temperature of the gas in K

Chemistry
1 answer:
Andreas93 [3]3 years ago
8 0

Answer:

300 K

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Diatomic Elements
  • Moles

<u>Gas Laws</u>

Ideal Gas Law: PV = nRT

  • <em>P</em> is pressure
  • <em>V</em> is volume
  • <em>n</em> is moles
  • <em>R</em> is gas constant
  • <em>T</em> is temperature

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>n</em> = 0.5 mol N₂

[Given] <em>V</em> = 6 L

[Given] <em>P</em> = 2 atm

[Given] <em>R</em> = 0.0821 L · atm · mol⁻¹ · K⁻¹

[Solve] <em>T</em>

<em />

<u>Step 2: Solve for </u><em><u>T</u></em>

  1. Substitute in variables [Ideal Gas Law]:                                                          (2 atm)(6 L) = (0.5 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)T
  2. Multiply [Cancel out units]:                                                                               12 atm · L = (0.04105 L · atm · K⁻¹)T
  3. Isolate <em>T</em> [Cancel out units]:                                                                             292.326 K = T
  4. Rewrite:                                                                                                             T = 292.326 K

<u>Step 3: Check</u>

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

292.326 K ≈ 300 K

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Answer:

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Explanation:

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\frac{0.04~mol~X}{1}=~0.04

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The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

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0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

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