The greatest aqueous freezing point is (D) 0.10 KCI
Answer:
Option A. 9.4 L
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 8 L
Initial temperature (T₁) = 293 K
Final temperature (T₂) = 343 K
Final volume (V₂) =?
V₁ / T₁ = V₂ / T₂
8 / 293 = V₂ / 343
Cross multiply
293 × V₂ = 8 × 343
293 × V₂ = 2744
Divide both side by 293
V₂ = 2744 / 293
V₂ = 9.4 L
Therefore, the final volume of the gas is 9.4 L
Hey there!
Values Ka1 and Ka2 :
Ka1 => 8.0*10⁻⁵
Ka2 => 1.6*10⁻¹²
H2A + H2O -------> H3O⁺ + HA⁻
Ka2 is very less so I am not considering that dissociation.
Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]
lets concentration of H3O⁺ = X then above equation will be
8.0*10−5 = [x] [x] / [0.28 -x
8.0*10−5 = x² / [0.28 -x ]
x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵
solve the quardratic equation
X =0.004693 M
pH = -log[H⁺]
pH = - log [ 0.004693 ]
pH = 2.3285
Hope that helps!
Answer:
i pretty sure it decreses
Explanation: