Explanation:
Xenon is a noble gas element that is highly unreactive. It has completely filled outermost shell and very stable on its own.
Xenon is an element with an atomic number of 54; it has an electronic configuration below:
[Kr] 4d¹⁰ 5s² 5p⁶
As we can see, the outermost energy shell, level 5; has a configuration of:
5s² 5p⁶
S is a sublevel with 2 electrons made up of 1 orbital
P is a sublevel with 6 electrons made up of 3 orbitals
These orbitals are filled in xenon and so it is stable and will not want to react with any other kind of atom.
This is why it is unreactive.
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I believe it doesn't as Xenon is a noble gas and is un-reactive.
Answer:
A. K = 59.5
Explanation:
Hello there!
In this case, since this reaction seems to start moving leftwards due to the fact that neither A nor Y are present at equilibrium, we should rewrite the equation:
3C (g) + D (g) <-- --> 2A (g) + Y (g)
Thus, the equilibrium expression is:
Next, according to an ICE table for this reaction, we find that:
Whereas x is calculated by knowing that the [C] at equilibrium is 0.456M; thus:
Next, we compute the rest of the concentrations:
Thus, the equilibrium constant for the leftwards reaction is:
Nonetheless, we need the equilibrium reaction for the rightwards reaction; thus, we take the inverse to get:
Therefore, the answer would be A. K = 59.5.
Regards!
Answer:
26.6
Explanation:
Step 1: Calculate the molar concentrations
We will use the following expression.
M = mass solute / molar mass solute × liters of solution
[CO]i = 26.6 g / (28.01 g/mol) × 5.15 L = 0.184 M
[H₂]i = 2.36 g / (2.02 g/mol) × 5.15 L = 0.227 M
[CH₃OH]e = 8.63 g / (32.04 g/mol) × 5.15 L = 0.0523 M
Step 2: Make an ICE chart
CO(g) + 2 H₂(g) ⇄ CH₃OH(g)
I 0.184 0.227 0
C -x -2x +x
E 0.184-x 0.227-2x x
Since [CH₃OH]e = x, x = 0.0523
Step 3: Calculate all the concentrations at equilibrium
[CO]e = 0.184-x = 0.132 M
[H₂]e = 0.227-2x = 0.122 M
[CH₃OH]e = 0.0523 M
Step 4: Calculate the equilibrium constant (Kc)
Kc = [CH₃OH] / [CO] [H₂]²
Kc = 0.0523 / 0.132 × 0.122² = 26.6
Answer:
The final dilution is 1:400
Explanation:
Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.
So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400
This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.