I am not sure about this but I think it’s y
Answer:
9.21 kJ is the enrgy required to vaporize 11.0 grams of ethanol.
Explanation:
Energy of vaporization can be stated as 38.6 kj/mole for ethanol. [<u>"It takes 38. 6 kj of energy to vaporize 1. 00 mol of ethanol"</u>]
11.0 grams of ethanol is not 1 mole. Calculate the moles of ethanol by dividing t=it's mass by its molar mass:
(11.0 grams)/(46.07 g/mole) = 0.2388 moles ethanol
(0.2388 moles ethanol)*(38.6 kj/mole) = 9.21 kJ is the enrgy required to vaporize 11.0 grams of ethanol.
1) The answer is: 2 moles of oxygen molecules.
Balanced chemical reaction of methane combustion:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
Coefficients with the lowest ratio indicate the relative amounts of substances in a reaction.
In fronf of oxygen molecule is coefficient 2.
2) The answer is: 4 moles of oxygen atoms.
In one molecule of oxygen there are two oxygen atoms, so in two molecules there are four oxygen atoms.
Answer:
The answer is #1
Explanation:
oil does not come out of valcanos