The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.
An ion can be cation (positively charged) or anion (negatively charged).
Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).
The remaining elements are completed as follows to attain octet structure;
<u>Element</u>--<u>valence electron</u>--<u>electrons to gain</u>--<u>electrons to lose</u>--<u>ion formed</u>
O ------------ 6 ---------------------- 2 ------------------------ none -------------- 
Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ 
Br ----------- 7 --------------------- 1 ------------------------ none --------------- 
S ------------ 6 ----------------------- 2 ------------------------ none --------------- 
Cl ------------ 7 ----------------------- 1 ------------------------ none ----------------
K -------------- 1 ----------------------- none ----------------------- 1 ------------------ 
Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- 
Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- 
Learn more here: brainly.com/question/21089350
Answer:
0.0010 mol·L⁻¹s⁻¹
Explanation:
Assume the rate law is
rate = k[A][B]²
If you are comparing two rates,
![\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7Brate%7D_%7B2%7D%7D%7B%5Ctext%7Brate%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bk_%7B2%7D%5Ctext%7B%5BA%5D%7D_2%5B%5Ctext%7BB%5D%7D_%7B2%7D%5E%7B2%7D%7D%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_1%5B%5Ctext%7BB%5D%7D_%7B1%7D%5E%7B2%7D%7D%3D%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%5Cright%20%29%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%5Cright%20%29%5E%7B2%7D)
You are cutting each concentration in half, so
![\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctext%7B%20and%20%7D%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Then,

Answer:
Beryllium has an electronic configuration of 1s2 2s2 in its natural state. When in an ion state of Be2+ it loses the electrons in a 2s shell and has a configuration of 1s2. This means that Be2+ has 2 electrons.
With the principle quantum number being 2, the maximum number that can share this is 8. You can use the general formula 2n^2 to calculate this number (n=quantum level), or you can use the concept of quantum numbers (n, l, m, s) to justify this answer.