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:
kilograms is the unit for mass
To find food. echolocation is when a sound bounces off of something and they can tell what direction it came from.
Answer:
Kindly check the explanation section.
Explanation:
From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.
In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;
∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).
That is the heat of formation of products minus the heat of formation of the reaction g specie(s).
Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.
−382.185 = x - 472.435kj/mol = 90.25 kJ/mol
I would say C but I’m not entirely sure
