Answer: The given statement is false.
Explanation:
Precipitation reaction is defined as the chemical reaction in which two aqueous solution upon mixing together results in the formation of an insoluble solid.
For example, 
Here AgCl is present in solid state so, it is the precipitate.
But it is not necessarily true that two aqueous solutions will always result in the formation of a precipitate.
For example, 
Hence, we can conclude that the statement precipitation reactions always occur when two aqueous solutions are mixed, is false.
Answer:
A) The mass would be the same.
Explanation:
Since there is no loss of any particle to vapor during the phase change process from solid to liquid, the mass of the before and after the process will remain the same.
- In this way, the law of conservation of mass is obeyed.
- Mass is the amount of matter contained in a substance.
- Since there is no room for escape or matter loss, the mass will remain the same.
A buffer solution contains an equivalent amount of acid and base. The pH of the solution with an acid dissociation constant (pKa) value of 3.75 is 3.82.
<h3>What is pH?</h3>
The amount of hydrogen or the proton ion in the solution is expressed by the pH. It is given by the sum of pKa and the log of the concentration of acid and bases.
Given,
Concentration of salt [HCOO⁻] = 0.24 M
Concentration of acid [HCOOH] = 0.20 M
The acid dissociation constant (pKa) = 3.75
pH is calculated from the Hendersons equation as,
pH = pKa + log [salt] ÷ [acid]
pH = 3.75 + log [0.24] ÷ [0.20]
= 3.75 + log (1.2)
= 3.75 + 0.079
= 3.82
Therefore, 3.82 is the pH of the buffer.
Learn more about pH here:
brainly.com/question/27181245
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When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M
