<span>Filtration, if its a precipitate that means its insoluble. </span>
Question:
Which of the following statements correctly describe(s) the driving forces for diffusion of Na+ and K+ ions through their respective channels? Select all that apply.
A)The diffusion of Na+ ions into the cell is facilitated by the Na+ concentration gradient across the plasma membrane.
B)The diffusion of Na+ ions into the cell is impeded by the electrical gradient across the plasma membrane.
C)The diffusion of K+ ions out of the cell is impeded by the K+ concentration gradient across the plasma membrane.
D)The diffusion of K+ ions out of the cell is impeded by the electrical gradient across the plasma membrane. The electrochemical gradient is larger for Na+ than for K+.
Answer:
"The concentration gradient and the electro-chemical gradient" describes the driving forces for diffusion of Na+ and K+ ions through their respective channels
Explanation:
The Na ions diffusion inside the cell is facilitated by the concentration gradient of the Na ions which is present across the plasma membrane. Hence, the diffusion of the K ions which is present outside the cell and will be impeded due to the electrical gradient which is present near the plasma membrane. Thus, the electro-chemical gradient is greater as compared to the Na ion than that of the K ion.
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.
<span>sudden increase of Nitrate and Phosphate in March to April that occurred, it causes the fish tank to produce algae</span>
4/325 = 2/unknown temperature
unknown temperature= 2/(4/325)=162.5k
