Areaction known to be first-order has a half-life of 23.6 min. if the initial concentration of the reactant is 2.50 m, what will the concentration of the reactant be after 2.00 hr?

First i wanted to say you look preety LOL

1) Determine the rate constant for the reaction: 

k = (ln 2) / t_1/2 <--- I'll leave you to figure out how that came to be. Hint: use the integrate form of the first-order rate law 

k = (ln 2) / 23.6 min 

k = 0.02937 min^-1 <--- keep a few extra digits 

1) use the integrated form of the first-order rate law: 

ln A = -kt + ln A_o 

ln A = - (0.02937 min^-1) (120 min) + ln 2.50 

ln A = -3.5244 + 0.91629 

ln A = -2.60811 

A = 0.07367 M <--- round off more as you see fit 

Here's another way: 

120 min / 23.6 min = 5.106383 half-lives 

(0.5)^5.106383 = 0.02902856 <--- the decimal amount remaining after 5.106383 half-lives 

0.02902856 x 2.50 M = the answer

Hi. where is the graph?

Solubility would be your answer.