Answer is: 10 gallons of 3% and 15 gallons of 8% solution.
V₁ + V₂ = 25 gallons.
0,03V₁ + 0,08V₂ = 0,06 · 25 gallons.
V₁ = 25 gallons - V₂.
0,03 · (25 gallons - V₂) + 0,08V₂ = 1,5 gallons.
0,75 gallons - 0,03V₂ + 0,08V₂ = 1,5 gallons.
0,05V₂ = 0,75 gallons.
V₂ = 0,75 gallons / 0,05.
V₂ = 15 gallons.
V₁ = 25 gallons - 15 gallons.
V₁ = 10 gallons.
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.
We can calculate the total number of moles using the ideal gas equation.
At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.
We can calculate the moles of nitrogen using the ideal gas equation.
The mole fraction of nitrogen in the mixture is:
A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.
Learn more: brainly.com/question/2060778
Answer:
pH = 4.164
Explanation:
The first process is to find the initial moles for the base (B) & the acid (HA)
i.e.
The acid with base reaction is expressed as;
HA + B → A⁻ + HB⁺
to 1.493 × 10⁻³ 2.047 × 10⁻³ - -
- 1.493 × 10⁻³ 1.493 × 10⁻³ 1.493 × 10⁻³ 1.493 × 10⁻³
0 5.54 × 10⁻⁴ 1.493 × 10⁻³ 1.493 × 10⁻³
From observation; both the acid & base weak
Given that:
The pKa for base = 4.594
The pKa for acid = 3.235
Recall that;
pKa = -log Ka
So; Ka =
By applying this:
For Base; Ka = = 2.5468 × 10⁻⁵
For Acid: Ka = = 5.821 × 10⁻⁴
After the reaction; we have the base with its conjugate acid & conjugate base of acid; Thus, since the conjugate acid of the base possesses a higher value of K, it is likely it would be the one to define the pH of the solution.
By analyzing the system, we have:
HB⁺ + H₂O ↔ B + H₃O⁺
to 0.01493 M 0.00554 M
- x x x
0.01493 - x 0.00554 - x x
Thus;
Using the common ion effect;
0.00554 - x 0.00554 &
0.01493 - x 0.01493
∴
x = [H₃O⁺] = 6.8635 × 10⁻⁵
∴
pH = -log(6.8635 × 10⁻⁵)
pH = 4.164
A neutralization reaction is when an acid and a base react to form water and a salt and involves the combination of H+ ions and OH- ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7.