Answer:
-O CH30
Explanation:
My Head Is About To Explode
B.
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Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
Answer:
climate is average of weather over time and place and weather can change from day to day.Weather is also affected by humidity, rain, cloudiness.
Explanation:
In lower temperatures, the molecules of real gases tend to slow down enough that the attractive forces between the individual molecules are no longer negligible. In high pressures, the molecules are forced closer together- as opposed to the further distances between molecules at lower pressures. This closer the distance between the gas molecules, the more likely that attractive forces will develop between the molecules. As such, the ideal gas behavior occurs best in high temperatures and low pressures. (Answer to your question: C) This is because the attraction between molecules are assumed to be negligible in ideal gases, no interactions and transfer of energy between the molecules occur, and as temperature decreases and pressure increases, the more the gas will act like an real gas.
