Because they frequently have a long half-lives, therefore his stay in the middle is long.
Number 4
If you notice any mistake in my english, please let me know, because i am not native.
The atom has only one isotope which means 100 % of same atom is present in nature. The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. Mass of one atom of that isotope is 9.123 ✕ 10⁻²³ g, so mass of one mole of atom that is Avogadro's number of atom is 6.023 X 10²³ X 9.123 X 10⁻²³ g=54.94 g = 55 g (approximate).
So, the atom having atomic mass 55 will be Cesium (Cs). Only one isotope of Cesium is stable in nature.
Nitrogen lone pair will act as a base,removing H+ from water leaving behind OH- ion.
Why ?
Because N is a better donor than O.
Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.
We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y .......> equation I
We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) ............> equation II
Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml
Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml
Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml
Hope this helps :)
<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.
<u>Explanation:</u>
The chemical formula of carbonic acid is 
. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water
The chemical equation for the first dissociation of carbonic acid follows:
The expression of first equilibrium constant equation follows:
![Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}](https://tex.z-dn.net/?f=Ka_1%3D%5Cfrac%7B%5BH%5E%2B%5D%5BHCO_3%5E%7B-%7D%5D%7D%7B%5BH_2CO_3%5D%7D)
The chemical equation for the second dissociation of carbonic acid follows:
The expression of second equilibrium constant equation follows:
![Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}](https://tex.z-dn.net/?f=Ka_2%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BHCO_3%5E-%5D%7D)
Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.
