Complete Question
The complete question is shown on the first uploaded image
Answer:
At Equilibrium the number of moles of 
=0.0104
At Equilibrium the number of moles of 
=0.0052
Explanation:
The explanation is shown on the second uploaded image
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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:
La reacción de etano, CH₃CH₃ y cloro, Cl₂ en presencia de luz solar se conoce como reacción de halogenación.
Los radicales de cloro se forman cuando el cloro se expone a la luz solar a continuación;
Cl₂ → 2Cl *
Se forma un radical etano cuando el etano, C₂H₆ reacciona con un radical libre de cloro, Cl *, como sigue;
C₂H₆ + Cl → C₂H₅ * + HCl
Se forma un producto halogenado cuando un radical libre de etano reacciona con una molécula de cloro como sigue;
C₂H₅ * + Cl₂ → C₂H₅Cl + Cl *
Un producto halogenado puede reaccionar además con un radical libre para producir como sigue;
C₂H₅Cl + Cl * → C₂H₅Cl₂
Explanation:
Assuming ideal behavior of the gas for a fixed amount when temperature is held constant, the pressure and volume are inversely proportional as given by the expression
P1V1 = P2V2
where the terms with subscripts of one represent the initial conditions for pressure and volume of the gas while for terms with subscripts of two represent the final conditions.
Rearranging the Boyle's law equation to calculate for the final volume V2:
V2 = P1V1 / P2
V2 = (99.7 kPa)(150 mL) / 99.8 kPa
V2 = 149.85 mL
