Flat as more oxygen and water can react over it think of it like this would a cube rust faster than a sheet
Answer:
The two conditions that can limit the usefulness of the kinetic-molecular theory in describing gas behavior are "high pressure" and "low temperatures". At low temperatures or high pressures, real gases deviate significantly from ideal gas behavior.
Explanation:
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Answer:</h3>
4551.37 Pascals
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Explanation:</h3>
The pressure refers to the force exerted by a substance per unit area.
The pressure is liquid is calculated by;
P = height × density × gravitational acceleration
In this case;
Height = 636.2 mm
Density = 0.7300 g/cm³
g = 9.8 N/kg
We need to convert mm to m and g/cm³ to kg/m³
Therefore;
Height = 636.2 mm ÷1000
= 0.6362 m
Density = 0.73 g/cm³ × 1000 kg/m³
= 730 kg/m³
Then, we can calculate the pressure;
Pressure = 0.6362 m × 730 kg/m³ × 9.8 N/kg
= 4551.3748 pascals
= 4551.37 Pascals
Therefore, the pressure of the column of decane is 4551.37 Pascals
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
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