Oxygen gas expands in a container and when pressure is applied it liquifies and occupies a smaller volume
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The cell body is the spherical part of the neutron that contains the nucleus and connects to the dendrites
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:
V₁ = 0.025 Liters = 25 ml
Explanation:
P₁ = 6 Atm P₂ = 7800mm/760mm/Atm = 0.01 Atm
V₁ = ? V₂ = 29 Liters
T₁ = 115 K T₂ = 225 K
P₁V₁/T₁ = P₂V₂/T₂ => V₁ = P₂V₂T₁/P₁T₂
V₁ = (0.01 Atm)(29L)(115K)/(6 Atm)(225K) = 0.025 Liters = 25ml
