Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
Answer:
e- 7.25 x 10³.
Explanation:
∵ ΔG = -RTlnK,
where, ΔG is the free energy change.
R is the general gas constant (R = 8.324 J/mol.K).
K is the equilibrium constant of the reaction.
- For the reaction: <em>N₂(g) + 3H₂(g) → 2NH₃(g),</em>
K = (PNH₃)²/(PN₂)(PH₂)³ = (0.65)²/(1.9)(1.6)³ = 5.43 x 10⁻².
∵ ΔG = -RTlnK.
∴ ΔG = -(8.314 J/mol.K)(298 K) ln(5.43 x 10⁻²) = 7.218 x 10³ J/mol.
Answer:
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Answer:
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<h2>Answer:</h2>
2 hydrogen atoms.
<h2>Explanations:</h2>
Given the chemical formula H2SO4,
The compound shows that the formula has 4 atoms of oxygen, one atom of sulfur and 2 hydrogen atoms.
Therefore, the number of hydrogen atom in the molecule H2SO4 is 2 hydrogen atoms.
