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Situation:
A 11 gram sample of a substance that's
used to treat thyroid disorders has a k-
value of 0.125.
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

Respuesta :

semsee45

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

Given formula:

[tex]\sf N=N_0e^{-kt}[/tex]

  • [tex]\sf N_0[/tex] = initial mass (at time t=0)
  • N = mass (at time t)
  • k = a positive constant
  • t = time (in days)

Given values:

  • [tex]\sf N_0[/tex] = 11 g
  • k = 0.125

Half-life:  The time required for a quantity to reduce to half of its initial value.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

[tex]\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}[/tex]

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

https://brainly.com/question/28016999

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MisterBrainly

[tex]\\ \rm\Rrightarrow N=N_oe^{-kt}[/tex]

  • We need half-life
  • put N=N_o/2

[tex]\\ \rm\Rrightarrow \dfrac{N_o}{2}=N_oe^{-kt}[/tex]

  • Cancel N_o

[tex]\\ \rm\Rrightarrow \dfrac{1}{2}=e^{-kt}[/tex]

[tex]\\ \rm\Rrightarrow 2^{-1}=e^{-kt}[/tex]

  • Cancel negative

[tex]\\ \rm\Rrightarrow 2=e^{kt}[/tex]

  • Apply natural log

[tex]\\ \rm\Rrightarrow ln2=lne^{kt}[/tex]

[tex]\\ \rm\Rrightarrow ktlne=ln2[/tex]

[tex]\\ \rm\Rrightarrow kt=ln2[/tex]

[tex]\\ \rm\Rrightarrow kt=0.693[/tex]

[tex]\\ \rm\Rrightarrow T=\dfrac{0.693}{k}[/tex]

  • Where T denotes half-life

Put k from the question

[tex]\\ \rm\Rrightarrow T=\dfrac{0.693}{0.125}[/tex]

[tex]\\ \rm\Rrightarrow T=8(0.693)[/tex]

[tex]\\ \rm\Rrightarrow T=5.54days[/tex]

Note:-

  • This equation is popularly known as Arrhenius equation
  • The decay constant for first order reactions and half-life don't depend upon the initial concentration so 11g we haven't used

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