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The correct answer is 2.73.
HF is a weak acid which partially dissociates to release H+ and F-
HF → H⁺ + F⁻
Initial 0.0050 0 0
Change -x +x +x
Equilibrium 0.0050–x +x +x
Solve by using the equilibrium expression: = [H⁺] [F⁻]/ [HF]
6 .8 x 10⁻⁴= x. x / 0.0050 –x
6 .8 x 10⁻⁴= x² /0.0050
x² = 6 .8 x 10⁻⁴ x 0.0050
x² = 3.4 x 10⁻⁶
x = 3.4 x 10⁻⁶
[H⁺] = 1.84 x 10⁻³
pH = - log [H⁺] = - log (1.84 x 10⁻³)
pH = 2.73
Answer:
The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Explanation:
- To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
- The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
- The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
- ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
- Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
- ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
- ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
- (760 torr /P₂) = 0.01075
- Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.
So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Answer:
condensation:The act or process of condensing or of being condensed pressureA pressing; a force applied to a surface.
temperature The state or condition of being tempered or moderated.
n:
