Answer:
(A) NaOH and Ca(OH)2: Ca(OH)2
(B) MgCl2 and MgF2: MgF2
(C) Agl and KI: AgI
(D) NH4Cl and PbCl2: PbCl2
Explanation:
We need to see the solubility in water, at similar temperatures, for each compound and see which one is less soluble than the other:
NaOH: 1000 g/L (25 °C)
Ca(OH)2: 1.73 g/L (20 °C)
MgCl2: 54.3 g/100 mL (20 °C)
MgF2: 0.013 g/100 mL (20 °C)
Agl: 3×10−7g/100mL (20 °C)
KI: 140 g/100mL (20 °C)
NH4Cl: 383.0 g/L (25 °C)
PbCl2: 10.8 g/L (20 °C)
After the comparison made we can conclude that the less soluble, after saturation of water, will precipitate first.
In Silver, the 4d orbitals will be completely filled. That
implies that it does not have two electrons in the 5s orbital. The electronic
configuration of Silver is :
Ag (47) = 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^1
4d^10
Answer : The value of 
of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.
Now we have to calculate the value of of the weak acid.
The equilibrium chemical reaction is:
Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
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{[HK]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHK%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the value of of the weak acid is, 4.72
Nitrogen tribromide is a chemical compound with the formula NBr3. It is extremely ... NBr3 was first prepared by reaction of bistrimethlysilylbromamine (bis(trimethylsilyl) amine bromide) ..
