C2H4 is oxidized and O2 is reduced in both reactions.
<h3>What is oxidation/reduction?</h3>
Oxidation is defined in several ways. Some of the definitions are:
- The addition of oxygen or removal of hydrogen
- Increase in the oxidation number of atoms
- Addition of electronegative or the removal of electropositive elements
Reduction, on the other hand, is defined as:
- Removal of oxygen or addition of hydrogen
- Decrease in the oxidation number of atoms
- Addition of electropositive elements or the removal of electronegative elements.
In the two reactions, oxygen is being added to C2H4. Thus, C2H4 is being oxidized.
The oxidizing agent is O2. In oxidation reactions, the oxidizing agents usually get reduced. Thus, O2 is reduced in both reactions.
More on oxidation and reduction can be found here: brainly.com/question/3867774
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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:
I'm gonna say igneous
Explanation:
The texture looks rough and the rock looks like it's composed of different types of minerals making it an igneous rock
I hope this helps :)
Chemical reactions are a change in substance while nuclear is destruction