Answer:
Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ
2. Mass of O₂
Answer is: K <span>be for the reaction at 375 K is 326.
</span>Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
</span><span>Δ<span>H = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K</span></span></span>·mol.<span>
K</span>₁ = 6,8·10⁵.<span>
K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.
Answer:
C
Explanation:
For an organism scientific names the first part is the genus and the second is the species
I believe the atomic level increases hope this helps.