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
This link might help you!!
https://www.eiu.edu/biology/bio1500/writing_a_lab_report.pdf
NaCI is the answer, in what my teacher taught me.
Hope I helped :)
Answer:
The answer to your question would be substance, but chocolate power mixed into milk would be more of a suspension.
Explanation:
Neither chocolate powder nor milk are elements. They are both complex molecules. Their mixture will not result in the formation of a compound since no chemical reaction will take place.
The molecules of the chocolate powder will simply intermingle with the fatty molecules of the milk to form the substance.
When thoroughly mixed the solution will become homogeneous so there will be no lumps of chocolate power visible. But after time, the chocolate will become visible at the bottom of the clear container in which we asked you to prepare the mixture.
HOPE THIS HELPS :)
