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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%
Answer:
0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂
Explanation:
In first place, the balanced reaction between Mg and O₂ is:
2 Mg + O₂ ⇒ 2 MgO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:
- Mg: 2 moles
- O₂: 1 mole
- MgO: 2 moles
Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?
moles of MgO= 0.250
<u><em>0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂</em></u>
Answer:
mi juego favorito probablemente sería Mine craft cuando era más joven, aunque sigue siendo uno de mis juegos favoritos
Explanation:
Answer:
the other variable is also doubled
Explanation:
direct proportion, same thing has to happen to both variables
