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Topic: Rate constant (Read 2215 times)
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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?
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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.
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islandsssol
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Re: Rate constant
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Reply #2 on:
September 13, 2012, 10:47:07 AM »
ln(1000000) = ln(1) - 0.1t
not like this?
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curiouscat
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Re: Rate constant
«
Reply #3 on:
September 13, 2012, 10:59:33 AM »
Quote from: islandsssol on September 13, 2012, 10:47:07 AM
ln(1000000) = ln(1) - 0.1t
not like this?
Not unless you like working with negative times
But close....
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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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Rate constant