The balanced chemical reaction is expressed as:
M + F2 = MF2
To determine the moles of the element fluorine present in the product, we need to determine the moles of the product formed from the reaction and relate this value to the ratio of the elements in MF2. We do as follows:
moles MF2 produced = 0.600 mol M ( 1 mol MF2 / 1 mol M ) = 0.600 mol MF2
molar mass MF2 = 46.8 g MF2 / 0.6 mol MF2 = 78 g/mol
moles MF2 = 46.8 g ( 1 mol / 78 g ) = 0.6 mol
moles F = 0.6 mol MF2 ( 2 mol F / 1 mol MF2 ) = 1.2 moles F
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)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
Compounds with Carbonate ions are less soluble than the compounds with Chlorate ions.
Explanation:
The reason is that the compounds with carbonate ions have less attraction between positive and negative charges. (Solvation)
<span>Like the previous post states, the number of valence electrons (typically the group number for nonmetals)relates directly to the number of covalent bonds available. Additionally, when you exceed this number you create a cation and when you are less the number of potential bonds, you have created a anion.
</span>
