Answer:
See below.
Explanation:
The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.
The balanced equation is:
2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)
From the equation, using atomic masses:
228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O
So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29
= 42.43 g of CO2.
Amount of water = 324.3 * 13.759) / 228.29
= 19.55 g of H2O.
Answer:
4.82 g
Explanation:
To solve this problem we'll use the <em>definition of molarity</em>:
- Molarity = moles / liters
We are given the volume and concentration (keep in mind that 100mL=0.100L):
- 0.825 M = moles / 0.100 L
Now we <u>calculate the number of NaCl moles required</u>:
Then we convert 0.0825 NaCl moles into grams, using its molar mass:
- 0.0825 mol * 58.44 g/mol = 4.82 g
The equation is as follows:
Cyclohexane (C6H12) ⇔ Methyl cyclopentane (C6H12)
The equilibrium constant Kc = 0.143 >>> (1)
Qc is the reaction quotient
where; If Q = K >>>> No shift left or right >>> (2)
Q > K >>>> Reaction shifts left >>> (3)
Q < K >>>> Reaction shifts right >>> (4)
And in our equation; Q = 0.3 / 0.4 = 0.75 >>> (5)
From (1), (3) and (5), Q > Kc
∴ the reaction shifts left
