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Mathematics 16 Online
OpenStudy (anonymous):

A 10 gram sample of a substance that's used to detect explosives has a k-value of 0.1356. Find the subtance's half-life, in days.round to the nearest tenth

OpenStudy (anonymous):

please help

OpenStudy (anonymous):

I will try my best ><. This is chemistry ?

OpenStudy (anonymous):

no algebra

OpenStudy (anonymous):

okay give me a minute and let me work it out

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

are you still working it out?

OpenStudy (anonymous):

lol yesh, I am questionable about my answer considering I am given no other variables but the constant an mass

OpenStudy (anonymous):

I divided 10 by the k-value constant since this seem like rate of decay

OpenStudy (anonymous):

10/0.1356

OpenStudy (anonymous):

N=Noe^-kt No=initial mass( at time t=0) N=mass at time t k=a positive constant that depends on the substance itself and on the units used to measure time t=time, in days

OpenStudy (anonymous):

@LiteLegacy

OpenStudy (anonymous):

Sorry can't swing it I'm to unsure ~-~

OpenStudy (anonymous):

But insert the value you have Like No=10 for the intial and and K-0.1356

OpenStudy (anonymous):

can you tell me what you think it is?

OpenStudy (anonymous):

@LiteLegacy

OpenStudy (anonymous):

73.7 but that is highly wrong to me ><

OpenStudy (anonymous):

@rational

OpenStudy (anonymous):

@Loser66

OpenStudy (anonymous):

@coolman500

OpenStudy (irishboy123):

it works as follows: \[\frac{N}{N_o} = \frac {1}{2} = e^{- k t_{half}} \\ \ln ( \frac {1}{2} ) = -\ln ( 2 ) = - k t_{half} \\ t_{half} = \frac{\ln(2)}{k}\] in summary half life = ln(2)/k AND k = ln(2)/half life here you get 5.1 days the mass is a red herring, the half life for a kilo of the stuff should be the same as the half life for a gram.

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