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anonymous
 one year ago
If you have 10.0 g of a substance that decays with a halflife of 14 days, then how much will you have after 42 days?
A. 0.10 g
B. 0.31 g
C. 1.25 g
D. 2.50 g
anonymous
 one year ago
If you have 10.0 g of a substance that decays with a halflife of 14 days, then how much will you have after 42 days? A. 0.10 g B. 0.31 g C. 1.25 g D. 2.50 g

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we have to apply this formula: \[\Large m\left( t \right) = {m_0}{e^{  t/\tau }}\] where m_0 is the initial mass of our sample, m(t) is the mass of our sample at time t, and \tau is the half life

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! what do we plug in?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is the next step: \[\Large m\left( {42} \right) = 10 \times {e^{  42/14}} = \frac{{10}}{{{e^3}}} = ...grams\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait sorry haha i totally blanked out :P

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1e is the Neperus constant, namely: \[e = 2.71828\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1my formula is the correct one!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay!! what would the solution be though? 0.49 is not a choice :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please wait, I'm checking my computation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that \[\tau \] is the average life, and it is given by the subsequent formula: \[\tau = \frac{{14}}{{0.693}} = 20.2\;days\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay! i am confused, though... which choice would be the accurate solution? :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so the right answer is given by the subsequent computation: \[\Large m\left( {42} \right) = 10 \times {e^{  42/20.2}} = \frac{{10}}{{{e^{2.079}}}} = ...grams\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh! ok! so we get 1.25! so choice C is the solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! that's right!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay! thank you!!:) okay! onto the next:)
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