2PbO(s) + O₂(g) ⇄ 2PbO₂(s)
Then Δngas = -1
<h3>
What is Δngas?</h3>
The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.
Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.
<h3>
Definition of equilibrium</h3>
When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.
Learn more about Equilibrium
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Answer:
Hey there!
This is already rounded to four significant figures!
Zeroes after the decimal but before the 7 don't count, and 7, 0, 6, and 2 count as significant figures.
So, the answer would be 0.007062.
Let me know if this helps :)
Answer:
Uranium must be purified before it is used as a fuel source
Explanation:
The purer the uranium sample, the more the concentration of uranium in the fuel is.
Whenever uranium is extracted from nature, it contains a lot of impurities. Only a few special nuclear reactors can utilize uranium in this raw state. most of the others have to get uranium to become about 3% pure before they begin using it.
To do this, uranium has to be passed through a series of chemical reactions all with the aim of extracting the other compounds that may be present in the fuel.
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)
= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant can be calculated as:
0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
Answer:
awnser is true took the test
Explanation:
i took the test
