Answer:
After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.
Explanation:
The first order kinetics reaction is:
ln [A] = ln [A]₀ - kt
<em>Where [A] is concentration after t time, [A]₀ is intial concentration and k is reaction constant.</em>
To convert half-life to k you must use:
t(1/2) = ln 2 / K
221s = ln 2 / K
K = ln 2 / 221s
<h3>K = 3.1364x10⁻³s⁻¹</h3>
If [A] = 1/64, [A]₀ = 1:
ln [A] = ln [A]₀ - kt
ln (1/64) = ln 1 - 3.1364x10⁻³t
4.1588 = 3.1364x10⁻³s⁻¹t
1326s = t
<h3>After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.</h3>
<em />
Given what we know, we can confirm that if further increases in substrate concentration do not result in further increases in reaction rate, then an enzyme is likely saturated.
<h3>What does it mean for an enzyme to be saturated?</h3>
Enzymes work by binding to the substrate in specific zones of the enzyme. The zones are known as the active sites on enzymes. Since enzymes have a limited amount of these zones, once they are all bonded to a substrate, we can say that it is saturated.
Therefore, the saturation of enzymes allows us to explain how further increases in substrate concentration do not result in further increases in reaction rate.
To learn more about enzymes visit:
brainly.com/question/24811456?referrer=searchResults
Answer:
See explanation
Explanation:
The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.
As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.
Answer: Concentration of 
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of 
= 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture = 
Thus concentration of in the equilibrium mixture is 0.31 M
