The atomic number is also the number of protons in the nucleus of an atom of a specific element, in this case Xenon. Xenon is represented on the periodic table as Xe. Find that and look at the number in the top center of the square for Xenon. In this case, the atomic number is 54.
<span>On the periodic table, the majority of elements are classified as "Metals"
In short, Your Answer would be Option A
Hope this helps!</span>
Answer:
22:
Formular:

substitute:

23:
<em>Same</em><em> </em><em>element</em><em> </em><em>is</em><em> </em><em>represented</em><em> </em><em>by</em><em> </em><em>same</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>protons</em><em>.</em><em> </em>
Answer:
6 protons. 6 protons
7 neutrons. 8 neutrons
6 electrons. 6 electrons
Note: <u>Atoms</u><u> </u><u>with</u><u> </u><u>same</u><u> </u><u>proton</u><u> </u><u>number</u><u> </u><u>but</u><u> </u><u>different</u><u> </u><u>mass</u><u> </u><u>number</u><u> </u><u>are</u><u> </u><u>called</u><u> </u><u>isotopes</u>
Explanation:
The given data is as follows.


Now, according to Michaelis-Menten kinetics,
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
where, S = substrate concentration =
M
Now, putting the given values into the above formula as follows.
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%20%5B%5Cfrac%7B10.4%20%5Ctimes%2010%5E%7B-6%7D%20M%7D%7B%2810.4%20%5Ctimes%2010%5E%7B-6%7DM%20%2B%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M%29%7D%5D)

= 
This means that
would approache
.
Answer:
i and ii
Explanation:
In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD