A particle of light is called a photon. I hope this helps! Please mark branliest!! ❤️
Answer:
The molal concentration of the solution is approximately 0.124 molal
Explanation:
The given parameters are;
The reduced freezing point of the water = -0.23°C
For freezing point depression of a solution, we have;
= · b·i
Where;
= The freezing point depression
= Cyroscopic constant (The freezing-point depression constant) = 1.86° Cm⁻¹ for water
b = The molality of the solution
i = van't Hoff factor = 1
Therefore;
=
Where;
= The freezing point of pure water = 0°C
= The freezing point of the solution = -0.23°C
∴ = = 0°C - (-0.23°C) = 0.23 °C
From, = · b·i, we have;
0.23°C = 1.86°C/m × b × 1
∴ b = (0.23°C/1.86°C/m) ≈ 0.124 molal
Therefore, the molal concentration of the solution, b ≈ 0.124 molal.
Answer:
a. CH3COO⁻.
Explanation:
Hello there!
In this case, according to the given ionization reaction of acetic acid:
CH₃COOH + H₂O --> H₃O⁺ + CH₃COO⁻
Whereas we can evidence how the CH₃COOH is the acid whereas H₂O the base, as the former donates H ions to the latter, which accepts them; we can infer that H₃O⁺ is the conjugate acid, resulting from the base and the a. CH3COO⁻ is the conjugate base, resulting from the acid.
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Using the ideal gas equation:
PV = nRT, we can form:
V/n = constant
Thus,
V₁/n₁ = V₂/n₂
n₂ = 2.3 x 85.5/50.5
= 3.89 moles