Answer:
Explanation:
Hello,
In this case, for the given reaction at equilibrium:
We can write the law of mass action as:
![Keq=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
That in terms of the change 
due to the reaction extent we can write:
![Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7Bx%7D%7B%28%5BCO%5D_0-x%29%28%5BH_2%5D_0-2x%29%5E2%7D)
Nevertheless, for the carbon monoxide, we can directly compute as shown below:
![[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\](https://tex.z-dn.net/?f=%5BCO%5D_0%3D%5Cfrac%7B0.45mol%7D%7B1.00L%7D%3D0.45M%5C%5C)
![[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\](https://tex.z-dn.net/?f=%5BH_2%5D_0%3D%5Cfrac%7B0.57mol%7D%7B1.00L%7D%3D0.57M%5C%5C)
![[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\](https://tex.z-dn.net/?f=%5BCO%5D_%7Beq%7D%3D%5Cfrac%7B0.28mol%7D%7B1.00L%7D%3D0.28M%5C%5C)
![x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M](https://tex.z-dn.net/?f=x%3D%5BCO%5D_0-%5BCO%5D_%7Beq%7D%3D0.45M-0.28M%3D0.17M)
Finally, we can compute the equilibrium constant:
Best regards.
I'm going to assume you are speaking about the characteristics of weather.
Here they are:
Humidity
Air Temperature and Pressure
Wind Speed/Direction
Cloud Cover and what kind of clouds.
Also the amount of precipitation as well as what effects the different kind of weather phenomenon to change into e.g. Freezing rain, Hail, Straight Line winds, etc.
mass percent concentration = 15.7 %
molar concentration of glucose solution 1.03 M
Explanation:
To calculate the mass percent concentration of the solution we use the following formula:
concentration = (solute mass / solution mass) × 100
solute mass = 60.5 g
solution mass = solute mass + water mass
solution mass = 60.5 + 325 = 385.5 g (I used the assumption that the solution have a density of 1 g/mL)
concentration = (60.5 / 385.5) × 100 = 15.7 %
Now to calculate the molar concentration (molarity) of the solution we use the following formula:
molar concentration = number of moles / volume (L)
number of moles = mass / molecular weight
number of moles of glucose = 60.5 / 180 = 0.336 moles
molar concentration of glucose solution = 0.336 / 0.325 = 1.03 M
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molarity
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The number of atoms of fluorine that are present in the molecules of CF4 is calculated as follows
1 x4 = 4 atoms of fluorine
in CF4 they are four atoms of fluorine since the subscribe 4 indicate that they are four atoms of fluorine
The components : a solvent which is the component of the solution and the solute.
