It is the same like the aldaric acid of L-Allose because the top and bottom groups are the same we can rotate it without changing the configuration
check the attached picture:
- The structure is in Meso form so both compound and its mirror image have the same structure (optically inactive)
Answer:
I think its biosphere nd atmosphere
Answer:
165 ml
Explanation:
We are given;
Initial volume; V_a = 55 ml
Initial molarity; M_a = 3 M
Molarity of desired solution; M_b = 0.75 M
Volume of desired solution; V_b = (55 + x) ml
Where x is the volume of water to be added.
To solve for V_b, we will use the equation ;
M_a•V_a = M_b•V_b
V_b = (M_a•V_a)/M_b
V_b = (3 × 55)/0.75
V_b = 220 mL
Thus;
(55 + x) = 220
x = 220 - 55
x = 165 mL
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
