Answer:
In the Lewis structure for IF5 you'll need to put a total of 12 valence electrons on the Iodine atom in order to draw the Lewis structure. Remember that Iodine (I) can hold more than eight valence electrons. For the IF5 Lewis structure, calculate the total number of valence electrons for the IF5 molecule
Explanation:
<em> I be knowing hope this helps</em>
The temperature at which the process be spontaneous is calculated as follows
delta G = delta H -T delta S
let delta G be =0
therefore delta H- T delta s =0
therefore T= delta H/ delta S
convert 31 Kj to J = 31 x1000= 31000 j/mol
T=31000j/mol /93 j/mol.k =333.33K
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
Answer:
it is a nice question....my mind tells me that the first is it use me as a good vibes and can use to anything the second i will do my best too absorbe it.
Explanation:
Hope this help...