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:
If
0.0357
mol
CaCl
2
is dissolved in water to make a
0.420
M
solution, what is the volume of the solution?
Answer:
Evaporated via combustion.
Explanation:
The loss on ignition (LOI) and Ash content are inorganic analytical techniques used to determine the percentage by mass of hydrocarbon in compounds. This process allows volatile matter and possibly impurities found in a material to evaporate, leaving behind the true chemical constituents of the material.
The 10kg lost could be moisture content of the tree, or other volatile matter which has escaped on combustion of the tree.
Answer:
Hello I didi this quiz and i got 100% and the answer is (a) hope this helps <3
Explanation: