We can determine **[A] _{t}**, but lack the value of k.

k can be obtained from the equation of half-life for 1^{st} order reactions:

**Step 1**: We rearrange the half-life equation to solve for k :

$\overline{){\mathbf{k}}{\mathbf{=}}\frac{\mathbf{ln}\left(2\right)}{{\mathbf{t}}_{{\displaystyle \raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}}}$

${\mathbf{k}}{\mathbf{=}}\frac{\mathbf{ln}\mathbf{\left(}\mathbf{2}\mathbf{\right)}}{\mathbf{109}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}\mathit{i}\mathit{n}}{\mathbf{=}}$**6.32×10**^{-3}** min ^{-1}**

**Step 2**: We convert the time elapsed to minutes:

$\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{59}\mathbf{}\overline{)\mathbf{h}}\left(\frac{60\mathrm{min}}{1\overline{)h}}\right)\mathbf{=}$**335.4 min**

**Step 3**: We determine

$\mathbf{ln}\left[A\right]\mathbf{t}\mathbf{=}\mathbf{ln}{\left[A\right]}_{\mathbf{o}}\mathbf{-}\mathbf{kt}\phantom{\rule{0ex}{0ex}}\mathbf{ln}\mathbf{\left[}\mathbf{A}\mathbf{\right]}\mathbf{t}\mathbf{=}\mathbf{ln}\left(1\right)\mathbf{-}(6.32\times {10}^{-3}\overline{){\mathrm{min}}^{-1}})(335.4\overline{)\mathrm{min}})\phantom{\rule{0ex}{0ex}}$

**ln [A] _{t} = -2.119728**

Ln has a base of e. When you divide a number by ln you will take e to the power of the number.

Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is _{9} ^{18}F⟶ _{8} ^{18}O + _{+1}^{0}e)

Physicians use ^{18}F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.

(b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?

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