The answer to this question is yes i think
The mass would be 2 hydrogen electroydtyetes
<u>Answer:</u>
<em>54.9 mg of the radioisotope are still active after 110 min.</em>
<u>Explanation:</u>
Half-life of F-18 is found to be 109.7 minutes
Rate constant
![$k=\frac{0.693}{t_{\frac{1}{2}}}=0.00632$](https://tex.z-dn.net/?f=%24k%3D%5Cfrac%7B0.693%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%3D0.00632%24)
t = time taken = 110 minutes
![$\ln [A]=\ln [A]_{0}-k t$](https://tex.z-dn.net/?f=%24%5Cln%20%5BA%5D%3D%5Cln%20%5BA%5D_%7B0%7D-k%20t%24)
[A] is the final quantity
is the initial quantity
Plugging the values and solving for [A]
![\\$\ln [A]=\ln (110 m g)-\left(0.00632 \min ^{-1} \times 110 \min \right)$\\\\$\ln [A]=4.700-0.6952$\\\\$\ln [A]=4.0048$\\\\$[A]=e^{4.0048}$](https://tex.z-dn.net/?f=%5C%5C%24%5Cln%20%5BA%5D%3D%5Cln%20%28110%20m%20g%29-%5Cleft%280.00632%20%5Cmin%20%5E%7B-1%7D%20%5Ctimes%20110%20%5Cmin%20%5Cright%29%24%5C%5C%5C%5C%24%5Cln%20%5BA%5D%3D4.700-0.6952%24%5C%5C%5C%5C%24%5Cln%20%5BA%5D%3D4.0048%24%5C%5C%5C%5C%24%5BA%5D%3De%5E%7B4.0048%7D%24)
[A] = 54.9 mg is the Answer
Answer:
Atomic number
Explanation:
Atomic number is the number of protons, and therefore also the total positive charge, in the atomic nucleus. The Rutherford–Bohr model of the hydrogen atom (Z = 1) or a hydrogen-like ion (Z > 1).