Answer:
Scientists use models to study atoms because atoms are extremely small and can't even be seen. If atoms can't be seen, it's easier to study and.
Explanation:
hope it helps you
<u>Answer:</u> The rate constant for the given reaction is 
<u>Explanation:</u>
For the given chemical equation:

We are given that the above equation is undergoing first order kinetics.
The equation used to calculate rate constant from given half life for first order kinetics:

The rate constant is independent of the initial concentration for first order kinetics.
We are given:
= half life of the reaction = 
Putting values in above equation, we get:

Hence, the rate constant for the given reaction is 