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
Eta Carinae could be as large as 180 times the radius of the Sun, and its surface temperature is 36,000-40,000 Kelvin. Just for comparison, 40,000 Kelvin is about 72,000 degrees F. So it's the blue hypergiants, like Eta Carinae, which are probably the hottest stars in the Universe.
There are 34 g of oxygen in the container.
We can use the<em> Ideal Gas Law</em> to solve this problem.
But 
, so
and
STP is 0 °C and 1 bar, so
About 8.0 moles of methane.Number of moles = MassMolar mass.
And thus we get the quotient:
128.3⋅g16.04⋅g⋅mol−1=8.0⋅moles of methane.
Note that the expression is dimensionally consistent, we wanted an answer in moles, and the quotients gives, 1mol−1=11mol=mol as required.
Answer:
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Explanation:
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