CH3COOH
Explanation:
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The mole fraction of KBr in the solution is 0.0001
<h3>How to determine the mole of water</h3>
We'll begin by calculating the mass of the water. This can be obtained as follow:
- Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
- Density of water = 1 g/mL
- Mass of water =?
Density = mass / volume
1 = Mass of water / 400
Croiss multiply
Mass of water = 1 × 400
Mass of water = 400 g
Finally, we shall determine the mole of the water
- Mass of water = 400 g
- Molar mass of water = 18.02 g/mol
- Mole of water = ?
Mole = mass / molar mass
Mole of water = 400 / 18.02
Mole of water = 22.2 moles
<h3>How to de terminethe mole of KBr</h3>
- Mass of KBr = 0.3 g
- Molar mass of KBr = 119 g/mol
- Mole of KBr = ?
Mole = mass / molar mass
Mole of KBr = 0.3 / 119
Mole of KBr = 0.0025 mole
<h3>How to determine the mole fraction of KBr</h3>
- Mole of KBr = 0.0025 mole
- Mole of water = 22.2 moles
- Total mole = 0.0025 + 22.2 = 22.2025 moles
- Mole fraction of KBr =?
Mole fraction = mole / total mole
Mole fraction of KBr = 0.0025 / 22.2025
Mole fraction of KBr = 0.0001
Learn more about mole fraction:
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A - 1 CH4+ 1 O2 = 1 CO2+2 H2
b - 2 Al+3 Cl2 = 2 AlCl3
c - 1 CH2O+ 1 H2 = 1 CH3OH
*The ones that have the coefficient of 1 you can leave blank but if you have to put a number just put 1
Answer:
3.62x10⁻⁷ = Kb
Explanation:
The acid equilibrium of a weak acid, HX, is:
HX + H₂O ⇄ X⁻ + H₃O⁺
Where Ka = [X⁻] [H₃O⁺] / [HX]
And basic equilibrium of the conjugate base, is:
X⁻ + H₂O ⇄ OH⁻ + HX
Where Kb = [OH⁻] [HX] / [X⁻]
To convert Ka to Kb we must use water equilibrium:
2H₂O ⇄ H₃O⁺ + OH⁻
Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]
Thus, we can obtain:
Kw = Ka*Kb
Solving for Kb:
Kw / Ka = Kb
1x10⁻¹⁴ / 2.76x10⁻⁸ =
3.62x10⁻⁷ = Kb
Ruler is the answer of this question
