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Topic: Rate constant  (Read 2212 times)

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Offline islandsssol

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Rate constant
« on: September 13, 2012, 10:13:21 AM »
The rate constant for bacterial growth is 0.1 day-1. If the likely number of a given bacteria needed to cause an infection is 1000000 how long would you store a food with an initial load of a) 1 bacteria and b) 1000 bacteria?

I think I'm supposed to use this equation

Ln(Qa)=ln(Qa)0-kt

Can you help me with this?

Offline curiouscat

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Re: Rate constant
« Reply #1 on: September 13, 2012, 10:32:43 AM »
Your eq. (signs?) seems wrong.

This will cause decimation not growth.

In any case, you know k, Q and Q0.

Find t.


Offline islandsssol

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Re: Rate constant
« Reply #2 on: September 13, 2012, 10:47:07 AM »
   ln(1000000) = ln(1) - 0.1t

not like this?

Offline curiouscat

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Re: Rate constant
« Reply #3 on: September 13, 2012, 10:59:33 AM »
   ln(1000000) = ln(1) - 0.1t

not like this?

Not unless you like working with  negative times  :)

But close....

Offline islandsssol

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Re: Rate constant
« Reply #4 on: September 13, 2012, 11:02:14 AM »
I know what you are meaning, but the teacher put this equation like this, and I get negative days from this, as expected, but what do you suggest is the right way?

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