Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
C
experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons
Answer : The pH of the solution is, 3.37
Explanation : Given,
Concentration of = 0.3 M
Concentration of = 0.8 M
Now we have to calculate the pH of solution.
Using Henderson Hesselbach equation :
Now put all the given values in this expression, we get:
Thus, the pH of the solution is, 3.37
Tuesday if your hypnosis is correct and has been examined than it will be come a theory
Wednesday because without it your experiment would fail
Thursday volume
Friday I don't know
This problem is giving information about the proton concentrations of three solutions at 25 °C. Despite they are not numerically given, we can propose three scenarios to see how to approach the question.
Let the following solutions to come up:
[H⁺] = 2.63x10⁻³ M
[H⁺] = 1.00x10⁻⁷ M
[H⁺] = 4.511x10⁻⁹ M
The first step, will be the calculation of the pH for each solution via:
pH = -log([H⁺])
So that they turn out to be:
pH = -log(2.63x10⁻³ M) = 2.580
pH = -log(1.00x10⁻⁷ M) = 7.000
pH = -log(4.511x10⁻⁹ M) = 8.3457
In such a way, since acidic solutions have a pH below 7, neutral have a pH equal to 7 and basic have it above 7, we infer the first one is acidic, second one is neutral and third one is basic.
Thus, you can reproduce this methodology with the proton concentrations you are given.
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