Answer:
a. ![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. ![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Explanation:
Untuk semua jenis reaksi umum:

Konstanta kesetimbangan ![K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Dari pertanyaan yang diberikan:
a. 
Konstanta kesetimbangan:
![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. 
Konstanta kesetimbangan untuk tekanan parsial 
![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Karena Fe3O4 (s) hadir sebagai padatan.
Well, you don't need enzymes (biological catalysts) if you're willing to wait a century or two to digest a burger.
Without catalysts, complex reactions like digestion would take too long and the organism could not extract energy from the nutrients it eats in a practical time frame.
In addition, speed is everything in the biological world.
Some reactions and their speed relative to other organisms reactions determines who survives and who doesn't, among other aspects of life.
If a plant is slow to photosynthesize and grow in a habitat high in competition for sunlight real estate, other autotrophs will surely take over.
Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.
<h3>What is the equilibrium constant?</h3>
In a reaction, we can judge using the value of the equilibrium constant weather or not the reaction moves on to completion. If the reaction moves up to completion, it the follows that the value of the equilibrium constant ought to be large.
On the other hand, when we have a case that the equilibrium constant is small and is not so large, then the reaction does not proceed essentially to completion.
Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.
Learn more about equilibrium constant:brainly.com/question/10038290
#SPJ1
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
.
Note that water isn't part of this constant.
The value of
at 25 °C is
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
for pure water. - As a result,
at 25 °C.
Back to this question.
is given. 25 °C implies that
. As a result,
.
Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S = 
Number of moles of S =
= 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃