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:
0.125 moles
Explanation:
2.8 litres is equivalent to 2.8dm³
At STP,
1 mole = 22.4 dm³
x mole = 2.8 dm³
Cross multiply
22.4x = 2.8
Divide both sides by 22.4
x = 2.8/22.4
x = 0.125
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Thomson suggested the model of atom which was a sphere of positive matter within which electronic forces determined the positioning of the corpuscles. The corpuscles were distributed in a uniform sea of positive charge. This was so-called "plum pudding" model.
Answer: C ) Thomson
