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andriy [413]
2 years ago
7

What Period would this element be in?

Chemistry
1 answer:
AlladinOne [14]2 years ago
5 0

Answer:

Period 2

Explanation:

From the picture provided, I am guessing this is Fluorine. Normal Fluorine has 9 protons, electrons, and atomic number. Thus, it can be found in the second period.

***Period # = # of electron shells

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A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

Answer :  The % of (+) limonene isomer = 79%


                The % of (-) limonene isomer = 0%


                The % of enantiomeric excess = 58%


Explanation :   Enantiomeric excess (ee) is the measurement of purity used for chiral substances.


Given,


% of pure limonene enantiomer = The % of (+) limonene isomer = 79%


Therefore, The % of (-) limonene isomer = 0%


Formula used :  

\%(+)\text{ isomer}=\frac{ee}{2}+50\%


Where,         ee → enantiomeric excess


Now, put all the values in above formula, we get the value of enantiomeric excess (ee).


     {ee}=\frac{\%(+)-50\%}{2}


            =\frac{79\%-50\%}{2}


              = 58%



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Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.

Along the period, the size of elements decreases. Down the group the size of elements increases. The atomic radius in decreasing order is Bi>Sb>As>N>O.

Therefore, atomic radius in decreasing order is Bi>Sb>As>N>O.

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