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:
It should be 1. 1.2 X 10^24
Explanation:
D. Hydrogen chemical bonds are found within water molecules.
Answer:
4.2 Liters (2 sig-figs)
Explanation:
Apply Boyles Law ... P₁V₁ = P₂V₂
P₁ = 712 Torr P₂ = 3560 Torr
V₁ = 21.0L V₂ = ?
P₁V₁ = P₂V₂ => V₂ = P₁V₁/P₂ = (712Torr)(21.0L)/(3560Torr) = 4.2 Liters (2 sig-figs)
Here are 3 main types of models are ( Physical, Mathematical, and Conceptual. Models have limitations but are useful and can be changed based on a new evidence.