Enzymes catalyze the chemical reactions, they act upon the reaction substrates and speed up the reaction. Enzymes have active sites, the places where the reaction substrates interact with the enzyme bringing about the conversion of substrates to products. So, as the enzyme concentration increases the rate of reaction increases till a point where the rate is leveled off. The rate does not further increase, as the substrate might have become limiting at that point. All the available amount of substrate would have been associated with the active sites of the enzymes. So, at that point although there is enough catalyst, lack of substrate would limit the rate of reaction.
Answer:
v = 37.9 ml
Explanation:
Given data:
Mass of compound = 1.56 kg
Density = 41.2 g/ml
Volume of compound = ?
Solution:
First of all we will convert the mass into g.
1.56 ×1000 = 1560 g
Formula:
D=m/v
D= density
m=mass
V=volume
v = m/d
v = 1560 g / 41.2 g/ml
v = 37.9 ml
The question is incomplete, here is the complete question:
At elevated temperature, nitrogen dioxide decomposes to nitrogen oxide and oxygen gas
The reaction is second order for 
with a rate constant of 
at 300°C. If the initial [NO₂] is 0.260 M, it will take ________ s for the concentration to drop to 0.150 M
a) 1.01 b) 5.19 c) 0.299 d) 0.0880 e) 3.34
<u>Answer:</u> The time taken is 5.19 seconds
<u>Explanation:</u>
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant =
t = time taken = ?
[A] = concentration of substance after time 't' = 0.150 M
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.260 M
Putting values in above equation, we get:
Hence, the time taken is 5.19 seconds
When light travels from air into water,it slows down,causing it to change directions slightly
