1) If k = 0.022 sec-1 for a first order reaction, what will be the final concentration after 30. seconds if the initial concentration is 0.10 M? 2) How long will it take for the [ ] to drop from 1.00 mol/L to 0.040 mol/L?
3) What is the half-life of the reaction above?

1. [A] = 0.0517M

2. t = 146.3s

3. t (1/2) = 31.5s

Explanation:

The general kinetic law for a first-order reaction is:

ln[A] = -kt + ln [A]₀

Where [A] is actual concentration after time t, k is rate constant and [A]₀ is initial concentration of the reactant.

1. Replacing with the values of the first point:

ln[A] = -kt + ln [A]₀

ln[A] = -0.022sec⁻¹*30s + ln [0.10]

ln[A] = -2.9626

[A] = 0.0517M

2. Now, with the values of the second point:

ln[A] = -kt + ln [A]₀

ln[0.04] = -0.022s⁻¹t + ln [1]

t = 146.3s

3. Half-life is the time the reaction need to decrease the initial concentration to the half:

ln[A] = -kt + ln [A]₀

ln[1/2] = -0.022s⁻¹t + ln [1]

t (1/2) = 31.5s

It will appear as o2(g) as a reactant 

First up write the aim
then write the materials reqd.
then the procedure which you follow
then observation
then conclusion
then you can write lab safety rules or precautions(you can avoid if u want).
lemme' give an example
aim: to observe the double displacement reaction
materials reqd: bacl2 and na2so4, test tubes and a dropper
procedure:
1. take bacl2 in a test tube and na2so4 in the other.
2. then pour bacl2 into na2so4 solution.
3. observe
observation:
you will see that a white precipitate is formed called baso4. along with that common salt or nacl is formed.
the equation is
bacl2+na2so4=baso4+2nacl
conclusion:
a double displacement reaction is observed.
hope it