answer:
lets say we have a big lake with a lot of animals and you want to move the water to somewhere else. A lot of animals and fish, frogs will have to find a new home!
Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.
A rate law indicates the rate of a chemical response depends on reactant concentration. For a response inclusive of the price regulation commonly has the form rate = ok[A]ⁿ, in which okay is a proportionality constant known as the fee regular and n is the order.
The charge of a chemical response is, perhaps, its maximum crucial asset because it dictates whether or not a reaction can arise all throughout an entire life. knowing the charge regulation, an expression concerning the price to the concentrations of reactants can assist a chemist to modify the response conditions to get an extra suitable rate.
half-life is the time taken for the radioactivity of a substance to fall to 1/2 its authentic cost whereas implies existence is the common life of all the nuclei of a particular risky atomic species.
Learn more about rate law here:-brainly.com/question/7694417
#SPJ4
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M