Answer:
494.1 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question,
P1 = 294 kPa
P2 = ?
V1 = 42.9 liters
V2 = 22.8 liters
T1 = 76.0°C = 76 + 273 = 349K
T2 = 38.7°C = 38.7 + 273 = 311.7K
294 × 42.9/349 = P2 × 22.8/311.7
12612.6/349 = 22.8 P2/311.7
36.14 = 22.8P2/311.7
Cross multiply
36.14 × 311.7 = 22.8P2
11264.605 = 22.8P2
P2 = 11264.605 ÷ 22.8
P2 = 494.1 kPa
Your answer is sublimation.
Answer:
9.8 × 10²⁴ molecules H₂O
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Organic</u>
<u>Stoichiometry</u>
- Analyzing reaction rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 130 g CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[RxN] 1 mol CH₄ → 2 mol H₂O
[PT] Molar Mass of C: 12.01 g/mol
[PT] Molar Mass of H: 1.01 g/mol
Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂