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
Answer:
Second order
Explanation:
We could obtain the order of reaction by looking at the table very closely.
Now notice that in experiment 1 and 2, the concentration of [OH^-] was held constant while the concentration of [S8] was varied. So we have;
a situation in which the rate of reaction was tripled;
0.3/0.1 = 2.10/0.699
3^1 = 3^1
Therefore the order of reaction with respect to [S8] is 1.
For [OH^-], we have to look at experiment 2 and 3 where the concentration of [S8] was held constant;
x/0.01 = 4.19/2.10
x/0.01 = 2
x = 2 * 0.01
x = 0.02
So we have;
0.02/0.01 = 2^1
2^1 = 2^1
The order of reaction with respect to [OH^-] = 1
So we have the overall rate law as;
Rate = k[S8]^1 [OH^-] ^1
Overall order of reaction = 1 + 1 = 2
Therefore the reaction is second order.
Answer:
change in the total mass of substances
