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
Answer:
A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.
What is the specific heat of the substance?
If it is one of the substances found in Table 8.1.1, what is its likely identity?
There are a lot of empty space between the particles
Newton first law of motion
Explanation:
This law by Newton states that an object will remain at rest, or a moving object will stay in motion in a straight line and with the same speed unless an external and unbalanced force acts on the object.
One application of this Newtonian law is in the launching of projectiles like ballistic missiles. A ballistic missile parabolic path is affected by its inertia and the unbalanced forces acting on the object such as air resistance/ drag and gravity.
Answer:
Explanation:
In an aqueous solution of potassium sulfate (K₂SO₄), the solute is K₂SO₄ and the solvent is water. The percentage by mass describes the grams of solute there are dissolved per 100 grams of solution. It can be calculated as:
mass percentage = (mass of solute/total mass of solution) x 100%
For example, in an aqueous solution which is 2% by mass of K₂SO₄, there are 2 grams of K₂SO₄ per 100 g of solution.
