Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =6.46 x 10⁻¹²mol dm³
Hi there I believe it’s 18 please let me know if I’m wrong :)
Answer:
(c) The retention time would be higher (d) The retention time would be lower.
Explanation:
For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.
However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.
When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.
Learn more: brainly.com/question/4400908