This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,

where,
= molar solubility of
= ?
= partial pressure of
= 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:


Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is, 
M = n/V
.5M = n/.100 L
n = .1 L * .5M
n= .05 mols of MgCl2
mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2
mass of MgCl2 = 4.76 grams
4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused
Answer:
![AU^{3+} : [Rn] 5f^3](https://tex.z-dn.net/?f=AU%5E%7B3%2B%7D%20%3A%20%5BRn%5D%205f%5E3)
Explanation:
Writing electronic configuration of any element you should know atomic number of that element ,
and also electrons are filling according to their energy level and first electron is filled in the lower energy orbital
and it follows n+1 rule if n+1 is same for two orbital electron will go first in the lowest value of n.
writing electronic configuration of ion can be done like first for their neutral atom and then add or remove electron it will make things easy because there are also some eception case their you may do wrong.
![AU : [Rn] 5f^3 6d^1 7s^2](https://tex.z-dn.net/?f=AU%20%3A%20%5BRn%5D%205f%5E3%206d%5E1%207s%5E2)
remove three electron from outer most shell of AU
![AU^{3+} : [Rn] 5f^3](https://tex.z-dn.net/?f=AU%5E%7B3%2B%7D%20%3A%20%5BRn%5D%205f%5E3)
B) 6
one above one below and 2 on the left and right sides