Answer:
54.4 mol
Explanation:
the equation for complete combustion of butane is
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
molar ratio of butane to CO₂ is 2:8
this means that for every 2 mol of butane that reacts with excess oxygen, 8 mol of CO₂ is produced
when 2 mol of C₄H₁₀ reacts - 8 mol of CO₂ is produced
therefore when 13.6 mol of C₄H₁₀ reacts - 8/2 x 13.6 mol = 54.4 mol of CO₂ is produced
therefore 54.4 mol of CO₂ is produced
Answer:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
Explanation:
For the reaction:
H₂C₂O₄(g) → CO₂(g) + HCOOH(g)
At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:
H₂C₂O₄(g) = P₀ - x
CO₂(g) = x
HCOOH(g) = x
P at t=20000 is:
P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x
For 1st point:
x = 92,8-65,8 = 27
Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8
2nd point:
x = 130-92,1 = 37,9
H₂C₂O₄(g): 92,1 - 37,9 = 54,2
3rd point:
x = 157-111 = 46
H₂C₂O₄(g): 111-46 = 65
Now, as the rate law is :
v = k P[H₂C₂O₄]
Based on integrated rate law, k is:
(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k
1st point:
k = 2,64x10⁻⁵
2nd point:
k = 2,65x10⁻⁵
3rd point:
k = 2,68x10⁻⁵
The averrage of this values is:
k = 2,66x10⁻⁵
That means law is:
v = 2,66x10⁻⁵ P[H₂C₂O₄]
I hope it helps!
Answer:
The black snakes will survive and reproduce while the orange snakes die out.
Explanation:
The orange snakes will be easier to see by their prey and predators, and therefore will die out while the black snakes thrive because of their camouflage.
Definitely definite mass because liquids can take on any shape and can have different volumes.
Answer: The answer is Fossil A is younger than the index fossil.
Explanation: Because it’s the newest layer
