Answer: The oxidation state of selenium in SeO3 is +6
Explanation:
SeO3 is the chemical formula for selenium trioxide.
- The oxidation state of SeO3 = 0 (since it is stable and with no charge)
- the oxidation number of oxygen (O) IN SeO3 is -2
- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)
Hence, SeO3 = 0
Z + (-2 x 3) = 0
Z + (-6) = 0
Z - 6 = 0
Z = 0 + 6
Z = +6
Thus, the oxidation state of selenium in SeO3 is +6
<u>Answer:</u> The above reaction is non-spontaneous.
<u>Explanation:</u>
For the given chemical reaction:
Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.
We know that:
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Relationship between standard Gibbs free energy and standard electrode potential follows:
As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.
Hence, the above reaction is non-spontaneous.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law
Explanation:
Please, observe the solution in the attached Word document.
Answer:
See explanation
Explanation:
The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;
k= Ae^-Ea/RT
Where;
k= rate constant
A= pre-exponential factor
Ea=activation energy
R= gas constant
T= temperature
We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.
Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
