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
It is going to be <span>Molar Volume
</span><span>3H2 + N2 --> 2NH3
</span><span> 54.1L*22.4 L/mol H2 , you can find mol of H2, then mol of NH3, and then L of NH3</span>
<span>Answer:
Correct answers are- Electron affinity decreases; Cl has 7 valence electrons but Na has only 1. So Na is going to lose its e, Cl is going to gain an e.</span>
Explanation:
The given data is as follows.
Pressure (P) = 760 torr = 1 atm
Volume (V) = 
= 0.720 L
Temperature (T) = 
= (25 + 273) K = 298 K
Using ideal gas equation, we will calculate the number of moles as follows.
PV = nRT
Total atoms present (n) = 
= 
= 0.0294 mol
Let us assume that there are x mol of Ar and y mol of Xe.
Hence, total number of moles will be as follows.
x + y = 0.0294
Also, 40x + 131y = 2.966
x = 0.0097 mol
y = (0.0294 - 0.0097)
= 0.0197 mol
Therefore, mole fraction will be calculated as follows.
Mol fraction of Xe = 
= 
= 0.67
Therefore, the mole fraction of Xe is 0.67.
