Answer:
-767,2kJ
Explanation:
It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ
The sum of (4) - (2) produce:
6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ
(6) + 4×(3):
7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ
(7) - 2×(1):
8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ
(8) - 2×(5):
9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>
I hope it helps!
Answer:
The number of moles of the gas is 9.295 moles or 9.30 moles
Explanation:
We use PV = nRT
Where P = 4.87 atm;
V = 67.54 L
R= 0.0821Latm/molK
T = 158 C = 158 +273 K = 431 K
the number of moles can be obtained by substituting the values in the respective columns and solve for n
n = PV / RT
n = 4.87 * 67.54 / 0.0821 * 431
n = 328.9198 / 35.3851
n = 9.295moles
The number of moles is approximately 9.30moles.
PH = -log [H3O+]
4.15 = -log [H3O+]
[H3O+] = 10^(-4.15)
[H3O+]= 7.08 × 10^-5
Both products will start to cancel the acidity and how strong the base is if they are mixed. If the acid is stronger than the base then it will be an acidic product and visa versa if the base is stronger than the acid.
Answer:
0.846 moles.
Explanation:
- This is a stichiometric problem.
- The balanced equation of complete combustion of butane is:
C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O
- It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.
<u><em>Using cross multiplication:</em></u>
- 1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂
- ??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂
- The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.
