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
After the water is boiled off, the mass is 9.7 grams, so find the weight of the water.
12.98 - 9.7 = 3.28g H2O
Find how many moles of CuSO4 and H2O you have.
If there are about 18g per mole of H2O
3.28/18 = .1822 moles of H2O
If there are about 160g per mole of CuSO4
9.7/160 = .0606 moles of CuSO4
.1822/.0606 = 3
Ratio of H2O to CuSO4 = 1:3
Formula
<span>CuSO4.3H2O
</span>
Full Name
<span>Copper (II) Sulphate Trihydrate</span>
First blank- space
second blank- time
Hope this is extra helpful but also check your notes! have a wonderful day and stay safe is you are up north in the snow!
Answer:
2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF
Explanation:
AlF₃ + Li₂O —> Al₂O₃ + LiF
The above equation can be balanced as follow:
AlF₃ + Li₂O —> Al₂O₃ + LiF
There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by writing 2 before AlF₃ as shown below:
2AlF₃ + Li₂O —> Al₂O₃ + LiF
There are 6 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 6 before LiF as shown below:
2AlF₃ + Li₂O —> Al₂O₃ + 6LiF
There are 2 atoms of Li on the left side and 6 atoms on the right side. It can be balance by writing 3 before Li₂O as shown below:
2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF
Thus, the equation is balanced..!
Answer:
The metals are located on the left and the non-metals are located on the right.
Explanation:
Hope this helps :)
