A chemical reaction proceeds in such a way that after the first second, the amount of a
certain chemical involved in the reaction changes at a rate that’s inversely proportional to the product of the mass of the chemical present (in grams) and the time elapsed since the reaction began (in seconds).
C. Suppose that the amount of this chemical involved in the reaction is 40 grams at time t = 1 second and 30 grams at time t = 10 seconds. Find an explicit equation for the mass m of the chemical as a function of t, for t ≥ 1. Your equation should not involve any unknown constants or any calculator numbers.
D. According to your equation for m(t) in part C, at what time does the mass of the chemical involved in the reaction become zero? (You may use your calculator here.)
So far I have A as dm/dt=k/mt and B for m(t)=sqrt 2kln(t)+c
I really need help on C and D show your work and give a thourgh explaination
Okay, so you have
m = √(2k ln(t) + c)
Now (C) tells you that
m(1) = 40
m(10) = 30
So now you can use that to find k and c, since 1/2 m^2 = k ln(t) + c
k ln(1) + c = 800
k ln(10) + c = 450
since ln(1) = 0, c = 800, and
k ln(10) + 800 = 450
k = -350/ln10
Now you have
m(t) = √(-350/ln10 ln(t) + 800)
Now, for (D),
-350/ln10 ln(t) + 800 = 0
ln(t) = (800 ln10)/350 = 16/7 ln10
t = e^(16/7 ln10) = 10^(16/7) = 100*10^(2/7) ≈ 193.07