Answer:
+1.76 V
Explanation:
Ecell = Ecathode - Eanode
= -0.19 - (-1.95) = + 1.76 V
Answer:
i. Keq=4157.99.
ii. More hydrogen sulfide will be produced.
Explanation:
Hello,
i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:
![Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5BS_2%5D%7D%20%3D%5Cfrac%7B%280.97M%29%5E2%7D%7B%280.051M%29%5E2%280.087%29%7D%20%3D4157.99)
ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.
Best regards.
Answer:b
Explanation:
I honestly don’t know if this I right but that would be my guess
Answer:
27 min
Explanation:
The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:
![v = \frac{vmax[S]}{Km + [S]}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bvmax%5BS%5D%7D%7BKm%20%2B%20%5BS%5D%7D)
Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.
So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min
If Km is a thousand times smaller then [S], then
v = vmax[S]/[S]
v = vmax
vmax = 1.33 μmol/min
For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:
vmax = 1.33/3 = 0.443 μmol/min
Km will still be much smaller then [S], so
v = vmax
v = 0.443 μmol/min
For 12 μmol formed:
0.443 = 12/t
t = 12/0.443
t = 27 min
Which type of solution are you talking about?