The concentration of carbon monoxide at equilibrium is 0.209 M.
<h3>Concentration of each gas</h3>
CO2 = 2 mol/5 L = 0.4
H2 = 1.5 mol/5L = 0.3
<h3>ICE table</h3>
Create ICE table as shown below
CO2(g) + H2(g) ↔ CO(g) + H2O(g)
I 0.4 0.3 0 0
C - x - x x x
E 0.4 - x 0.3 - x x x
Kc = [CO][H₂O] / [CO₂][H₂]
2.5 = (x²)/(0.4 - x)(0.3 - x)
x² = 2.5(0.4 - x)(0.3 - x)
x² = 2.5(0.12 - 0.7x + x²)
x² = 0.3 - 1.75x + 2.5x²
0 = 1.5x² - 1.75x + 0.3
solve the quadratic equation using formula method;
x = 0.209
Thus, the concentration of carbon monoxide at equilibrium is 0.209 M.
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Answer:
It will take 77.8 seconds for 95% of sample to react
Explanation:
As the given reaction obeys 1st order therefore-
Where is the concentration of after "t" time, is the initial concentration of and is half life
Here, and
So,
or,
So, it will take 77.8 seconds for 95% of sample to react
One electron has a mass of approximately 1 amu
Answer:
(a) 133.33nm
(b) 600nm
(c) 7,600nm
Explanation:
The concentration of Y can be determined by using the formula:
where;
[L] = concentration of the binding ligand.
kd = 400 nm
Thus:
When Y = 0.25; we get :
0.25 (400 + [L]) = [L]
100 + 0.25[L] = [L]
100 = [L] - 0.25 [L]
100 = 0.75 [L]
[L] = 100/0.75
[L] = 133.33 nm
At, Y = 0.6
0.6 (400 + [L]) = [L]
240 + 0.6[L] = [L]
240 = [L] - 0.6 [L]
240 = 0.4 [L]
[L] = 240/0.4
[L] = 600 nm
At, Y = 0.95
0.95 (400 + [L]) = [L]
380 + 0.95[L] = [L]
380 = [L] - 0.95 [L]
380 = 0.05 [L]
[L] = 380/0.05
[L] = 7600 nm
Answer
im not quite sure but I think the answer is <em>D atom</em><em> </em>
Explaination