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
<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.
<u>Explanation:</u>
Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:

Or,

We are given:
Mass of iron (III) chloride = 1.01 g
Molar mass of iron (III) chloride = 162.2 g/mol
Volume of the solution = 10 mL
Putting values in above equation, we get:

Hence, the molarity of Iron (III) chloride is 0.622 M.
Answer:
This part require data such as Avogadro's number and the molar mass of water. But first, let's find the mass of water in the specified volume by making use of the density formula:
Density = mass/volume
1 g/mL = Mass/70 mL
Mass = 70 g
Each water contains 18 grams per mole, and each mole contains 6.022×10²³ molecules of water. Thus,
70 g * 1mole/18 g * 6.022×10²³ molecules/mole = 2.342×10²⁴ molecules of water
Explanation:
329g Cl (You're welcome :D)