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: 
Explanation:
According to the Arrhenius equation,

or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= 
= rate constant at
= 
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = 
Now put all the given values in this formula, we get
![\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)


Therefore, the value of the rate constant at 775.0 K is 
Hey there!:
V1 = 3.05 L
V2 = 3.00 L
P1 = 724 mmHg
P2 = to be calculated
T1 = 298 K
T2 = 273 K
Therefore:
P1*V1 / T1 = P2*V2 / T2
P2 = ( P1*V1 / T1 ) * T2 / V2
P2 = 724 * 3.05 * 273 / 298 * 3.00
P2 = 602838.6 / 894
P2 = 674.31 mmHg
1 atm ----------- 760 mmHg
atm ------------- 674.31 mHg
= 674.31 * 1 / 760
= 0.887 atm
Hope this helps!
When an object enters the Earth's atmosphere, it experiences a few forces, including gravity and drag. Gravity will naturally pull an object back to earth.
-Hope this helped, have a great day. ;)