Answer: I think it’s B. Because as you can see from the picture there are layers
Explanation:
I think but just in case ask for a second opinion
Octane just prevents the air-fuel mixture from igniting before the spark plug does it. Firing the air-fuel mixture at the proper time gives you the maximum power your engine was designed to get. Using higher-octane gasoline than your engine is designed to utilize is only wasting your money.
Answer:
All Are correct responses
Explanation:
We will use the following equation,
(a) Now, we will calculate the value of as follows.
= 0.377
As, the is 0.21 (which is smaller than the value) this means that equilibrium will shift to the left
.
(b)
= 0.1155
Here, is smaller than which means that the equilibrium will shift to the right.
(c)
= 0.119
Here, is smaller than which means that the equilibrium will shift to the right
.
(d)
= 0.21
Since, here the value of comes out to be equal to the given value. Therefore, this reaction is at equilibrium.
(e)
= 0.36
Here, is higher than the reaction will shift to the left.
Answer:
Mass of butane = 1.87 g
Explanation:
Form a balanced chemical equation for the reaction stated:
C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O
Number of moles of CO₂
From the equation,
1 mol C₄H₁₀ : 4 mol CO₂
0.085 mol C₄H₁₀ : 0.34 mol CO₂
Mass of butane