Answer:
silver nitrate decomposes when heated.
Answer:
The pH of the solution is 4.69
Explanation:
Given that,
Mass of potassium = 2.643 grams
Weight of water = 50.00 mL
Weight of HCl=100.00 ml
Mole = 0.120 M
We know that,
is a basic salt.
Let's write it as KY.
The acid
would become HY.
We need to calculate the moles of KY
Using formula of moles
![moles\ of\ KY=\dfrac{m}{M}\times1000](https://tex.z-dn.net/?f=moles%5C%20of%5C%20KY%3D%5Cdfrac%7Bm%7D%7BM%7D%5Ctimes1000)
![moles\ of\ KY=\dfrac{2.643}{126}\times1000](https://tex.z-dn.net/?f=moles%5C%20of%5C%20KY%3D%5Cdfrac%7B2.643%7D%7B126%7D%5Ctimes1000)
![moles\ of\ KY=20.97\ m\ mole](https://tex.z-dn.net/?f=moles%5C%20of%5C%20KY%3D20.97%5C%20m%5C%20mole)
The reaction is
![KY+HCl\Rightarrow HY+ KCl](https://tex.z-dn.net/?f=KY%2BHCl%5CRightarrow%20HY%2B%20KCl)
The number of moles of KY is 20.98 m
initial moles = 20.98
Final moles ![m=20.98-0.120\times100= 8.98](https://tex.z-dn.net/?f=m%3D20.98-0.120%5Ctimes100%3D%208.98)
We need to calculate the value of pKa(HY)
Using formula for pKa(HY)
![pKa_{HY}=-log Ka](https://tex.z-dn.net/?f=pKa_%7BHY%7D%3D-log%20Ka)
![pKa_{HY}=-log(1.5\times10^{-5})](https://tex.z-dn.net/?f=pKa_%7BHY%7D%3D-log%281.5%5Ctimes10%5E%7B-5%7D%29)
![pKa_{HY}=4.82](https://tex.z-dn.net/?f=pKa_%7BHY%7D%3D4.82)
We need to calculate the pH of the solution
Using formula of pH
![pH=pKa+log(\dfrac{[KY]}{[KH]})](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cdfrac%7B%5BKY%5D%7D%7B%5BKH%5D%7D%29)
Put the value into the formula
![pH=4.82+log(\dfrac{8.98}{12})](https://tex.z-dn.net/?f=pH%3D4.82%2Blog%28%5Cdfrac%7B8.98%7D%7B12%7D%29)
![pH=4,69](https://tex.z-dn.net/?f=pH%3D4%2C69)
Hence, The pH of the solution is 4.69
Answer: The partial pressure of
is 0.35 atm.
Explanation:
Given: Total pressure = 0.98 atm
Partial pressure of
= 0.48 atm
Partial pressure of Ar = 0.15 atm
Partial pressure of
= ?
Total pressure is the sum of partial pressure of each component present in a mixture of gases.
Hence, partial pressure of
is calculated as follows.
Total pressure = ![P_{N_{2}O} + P_{O_{2}} + P_{Ar}](https://tex.z-dn.net/?f=P_%7BN_%7B2%7DO%7D%20%2B%20P_%7BO_%7B2%7D%7D%20%2B%20P_%7BAr%7D)
Substitute the values into above formula as follows.
![Total pressure = P_{N_{2}O} + P_{O_{2}} + P_{Ar}\\0.98 atm = P_{N_{2}O} + 0.48 atm + 0.15 atm\\P_{N_{2}O} = 0.35 atm](https://tex.z-dn.net/?f=Total%20pressure%20%3D%20P_%7BN_%7B2%7DO%7D%20%2B%20P_%7BO_%7B2%7D%7D%20%2B%20P_%7BAr%7D%5C%5C0.98%20atm%20%3D%20P_%7BN_%7B2%7DO%7D%20%2B%200.48%20atm%20%2B%200.15%20atm%5C%5CP_%7BN_%7B2%7DO%7D%20%3D%200.35%20atm)
Thus, we can conclude that the partial pressure of
is 0.35 atm.
A lot ...................