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.
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Answer:
1 mole of platinum
Explanation:
To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.
The empirical formula for the compound can be obtained as follow:
Platinum (Pt) = 117.4 g
Carbon (C) = 28.91 g
Nitrogen (N) = 33.71 g
Divide by their molar mass
Pt = 117.4 / 195 = 0.602
C = 28.91 / 12 = 2.409
N = 33.71 / 14 = 2.408
Divide by the smallest
Pt = 0.602 / 0.602 = 1
C = 2.409 / 0.602 = 4
N = 2.408 / 0.602 = 4
The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄
From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.
Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
Answer:
33300J
Explanation:
Given parameters:
Mass of ice = 100g
Unknown:
Amount of energy = ?
Solution:
This is a phase change process from solid to liquid. In this case, the latent heat of melting of ice is 3.33 x 10⁵ J/kg.
So;
H = mL
m is the mass
L is the latent heat of melting ice
Now, insert the parameters and solve;
H = mL
mass from gram to kilogram;
100g gives 0.1kg
H = 0.1 x 3.33 x 10⁵ = 33300J