An estuary is partly enclosed and forms a transition zone between rivers and oceans/ seas
Answer:
Part a: The rate of the equation for 1st order reaction is given as ![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b: The integrated Rate Law is given as ![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c: The value of rate constant is 
Part d: Concentration after 4000 s is 0.043 M.
Explanation:
By plotting the relation between the natural log of concentration of
, the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.
Part a
Rate Law
The rate of the equation for 1st order reaction is given as
![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b
Integrated Rate Law
The integrated Rate Law is given as
![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c
Value of the Rate Constant
Value of the rate constant is given by using the relation between 1st two observations i.e.
t1=0, M1=1.00
t2=120 s , M2=0.91
So k is calculated as

The value of rate constant is 
Part d
Concentration after 4000 s is given as

Concentration after 4000 s is 0.043 M.
<u>Answer:</u> The mass of potassium superoxide required is 142.2 grams
<u>Explanation:</u>
The chemical equation for the reaction of potassium superoxide with water follows:

Number of moles of potassium superoxide reacted = 2 moles for given amount of heat released
To calculate the mass for given number of moles, we use the equation:

Molar mass of potassium superoxide = 71.1 g/mol
Moles of potassium superoxide = 2 moles
Putting values in above equation, we get:

Hence, the mass of potassium superoxide required is 142.2 grams