Explanation:
Alkali metals forms the group 1 of the modern periodic table. They include Lithium, sodium, potassium, rubidium, cesium, and francium.
The general electronic configuration of the alkali metals -
. They have 1 valence electron.
Ionization energy is the minimum amount of energy which is required to knock out the loosely bound valence electron from the isolated gaseous atom.
<u>Thus, removal of the valence electron leads to the alkali metals having noble gas electronic configuration and which is stable. Thus, the removal of the second electron is very tough and thus, there a large increase between the first and second ionization energies of the alkali metals.</u>
Answer:
I'm pretty sure they both relate with matter
Answer:
Eight
Explanation:
The L shell only holds eight electrons.
Answer:
The right solution according to the question is provided below.
Explanation:
According to the question,
(a)
The initial conditions will be:
DO = 
= 
= 
The initial oxygen defict will be:
Do = 
= 
The initial BOD will be:
Lo = 
= 
= 
(b)
The time reach minimum DO:
tc = ![\frac{1}{(kr-kd)} ln{(\frac{0.76}{0.61} )[1-\frac{1.674(0.76-0.61)}{0.61\times 6.453} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28kr-kd%29%7D%20ln%7B%28%5Cfrac%7B0.76%7D%7B0.61%7D%20%29%5B1-%5Cfrac%7B1.674%280.76-0.61%29%7D%7B0.61%5Ctimes%206.453%7D%20%5D%7D)
= 
By putting the values of log, we get
= 
The distance to reach minimum DO will be:
Xc = 
= 
= 