Answer:
2.00X10^5 x 20gNe/6.02x10^23=6.46x10^-18 but books answer is 797.
Explanation:
Balanced:
1. <span>Na2O + H2O ---> 2NaOH
2. </span><span>K2O + H2O ---> 2KOH
3. </span><span>MgO + H2O ---> Mg(OH)2
4. </span><span>CaO + H2O ---> Ca(OH)2
5. </span><span>SO2 + H2O ⇄ H2SO3
6. </span>SO3 + H2O ---> H2SO4
All except by 2 were balanced.
I believe that it is B chemicals containing carbon
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100