Answer : The pH will be, 3.2
Explanation :
As we known that the value of solubility constant of ferric hydroxide at
is, 
Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L
The given solubility of iron convert from mg/L to mol/L.

The chemical reaction will be:

The expression of solubility constant will be:
![K_{sp}=[Fe^{3+}]\times [3OH^-]^3](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B3%2B%7D%5D%5Ctimes%20%5B3OH%5E-%5D%5E3)
Now put all the given values in this expression, we get the concentration of hydroxide ion.
![2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3](https://tex.z-dn.net/?f=2.79%5Ctimes%2010%5E%7B-39%7D%3D%282.7%5Ctimes%2010%5E%7B-7%7D%29%5Ctimes%20%5B3OH%5E-%5D%5E3)
![[OH^-]=1.5\times 10^{-11}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-11%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)


Now we have to calculate the pH.

Therefore, the pH will be, 3.2
Answer:
The correct answer is due to the difference in pressure inside and outside the bottle.
Explanation:
Liquids have melting and boiling points that depend on pressure and temperature. The pressure inside the bottle is higher than the pressure outside. This causes the melting point to drop, making the liquid freeze at a lower temperature than if it were at atmospheric pressure, and therefore has a lower temperature than it would freeze at atmospheric pressure. When the bottle is uncovered, the liquid becomes an atmospheric pressure, and due to the temperature acquired when the bottle was closed the liquid freezes.
Have a nice day!
Answer:
This is and ADDITION REACTION
Explanation:
Because your putting a compound and an element together
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.