SOLUTION: I a longer word problem that I have not been able to get the second part answered correctly. Reacting with water in an acidic solution at a particular temperature, compound A

Question 886121: I a longer word problem that I have not been able to get the second part answered correctly.
Reacting with water in an acidic solution at a particular temperature, compound A decomposes into compounds B and C according to the law of uninhibited decay. An initial amount of 0.60 M of compound A decomposes to 0.55 M in 30 minutes. How much of compound A will remain after 2 hours? .42 is the answer I came up for on this part.
How long will it take until 0.10 M of compound A remains?
I got to .10=0.60e^ln (.55/.60)/30 times t
I then divided both sides by .60 and
ln.10/.60=e^ln(.55/.60)/30 times t
If I'm on the right track I'm kind of stuck now.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Reacting with water in an acidic solution at a particular temperature, compound A decomposes into compounds B and C according to the law of uninhibited decay. An initial amount of 0.60 M of compound A decomposes to 0.55 M in 30 minutes. How much of compound A will remain after 2 hours? .42 is the answer I came up for on this part.
Model:: y = ab^x
0.55 = 0.6*b^(1/2)
sqrt(b) = 0.9167
b = 0.8403
----
Equation:
y = 0.6*0.8403^x
-------------------------
Your Problem
y = 0.6*(0.8403)^2
y = 0.4236
===========================

How long will it take until 0.10 M of compound A remains?
0.10 = 0.6*0.8403^x
0.8403^x = 0.167
x = ln(0.167)/ln(0.8403) = 10.29 hours
=================
Cheers,
Stan H.
=================

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