One molecule of N2O4 contains
two nitrogen
four oxygen
The new temperature (in °C) of the gas, given the data is –148.20 °C
<h3>Data obtained from the question </h3>
- Initial temperature (T₁) = 149.05 °C = 149.05 + 273 = 422.05 K
- Initial pressure (P₁) = 349.84 KPa
- Volume = constant
- New pressure (P₂) = 103.45 KPa
- New temperature (T₂) =?
<h3>How to determine the new temperature </h3>
The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:
P₁V₁ / T₁ = P₂V₂ / T₂
Since the volume is constant, we have:
P₁ / T₁ = P₂ / T₂
349.84 / 422.05 = 103.45 / T₂
Cross multiply
349.84 × T₂ = 103.45 × 422.05
Divide both side by 349.84
T₂ = (103.45 × 422.05) / 349.84
T₂ = 124.80 K
Subtract 273 from 124.80 K to express in degree celsius
T₂ = 124.80 – 273
T₂ = –148.20 °C
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Liquids have fixed volume but not fixed shape, so B.
Answer:
1.55 × 10²⁴ atoms Xe
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 57.5 L Xe at STP
[Solve] atoms Xe
<u>Step 2: Identify Conversions</u>
[STP] 22.4 L = 1 mol
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.54583 × 10²⁴ atoms Xe ≈ 1.55 × 10²⁴ atoms Xe