Answer:
- From octane:
- From ethanol:
Explanation:
Hello,
At first, for the combustion of octane, the following chemical reaction is carried out:
Thus, the produced mass of carbon dioxide is:
Now, for ethanol:
Best regards.
Answer: C. The Number of Molecules
Explanation: I just took the test, it's correct.
Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
Answer: 8.691 mols of CO₂
Explanation:
To find the number of moles in a given grams, you want to use the molar mass.
Let's first find the molar mass of CO₂.
Carbon's molar mass is 12.011 g/mol
Oxygen's molar mass is 15.999 g/mol
To find molar mass of CO₂, we want to add up the molar mass of carbon and oxygen. Remember, there are 2 Oxygens so we need to mulitply that by 2.
12.011+2(15.999)=44.009 g/mol
Now that we have molar mass, we can convert 382.5 g to mols.
There are about 8.691 mols of CO₂.