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.
Answer:
Boyle's Law
Explanation:
- This law states volume and pressure have an inverse relationship when temperature is constant.
- Hope this helped! If you would like me to further explain gas laws please let me know.
3 C6H12O6,
C=3*6=18 Carbon atoms
H=3*12=36 Hydrogen atoms
O=6*3=18 Oxygen atoms
PbH4 will be formed as a result of a polar covalent bond between the H and the Pb.
Since H is more electronegative than the Pb, it is, thus, expected that the H would be able to pull the electron charge towards itself. This will result in the H being negative.
Based on this:
PbH4 would be expected to <span>have polar covalent bonds with a partial negative charges on the H atoms. </span>
