I think A.weathering, is the best possible answer
PV = nRT —> n = PV/RT
P = 2.90 atm
V = 4.80 L
R = 0.08206 L atm / mol K
T = 62.0 + 273 = 335 K (make sure you convert from celsius to kelvin)
n = (2.90 • 4.80) / (0.08206 • 335) = 0.506 moles of gas
Answer:
Oxygen is oxidized and hydrogen is reduced.
Explanation:
Let's consider the redox reaction during the electrolysis of water in an electrolytic cell.
2 H₂O ⇒ 2 H₂ + O₂
The corresponding half-reactions are:
2 e⁻ + 2 H₂O ⇒ H₂ + 2 OH⁻
2 H₂O ⇒ O₂ + 4 H⁺ + 4 e⁻
Oxygen is oxidized since its oxidation number increases from -2 to 0.
Hydrogen is reduced since its oxidation number decreases from +1 to 0.
Answer:
Mass percent N₂ = 89%
Mass percent H₂ = 11%
Explanation:
First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:
- 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
So now we know that
- MolH₂ + MolN₂ = 0.307 mol
and
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g
So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:
Express MolH₂ in terms of MolN₂:
- MolH₂ + MolN₂ = 0.307 mol
Replace that value in the second equation:
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
- (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
- 0.614 - 2MolN₂ + 28molN₂ = 3.49
Now we calculate MolH₂:
- MolH₂ + MolN₂ = 0.307 mol
Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:
- N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
- H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂
Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%
Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%
