This question requires the knowledge of density.
The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³
Density = Mass / Volume
By applying ethyl alcohol,
789 kg m⁻³ = Mass / 0.9 m³
Mass = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.
Then by applying water,
1000 kg m⁻³ = 710.1 kg / Volume
Volume = 0.7101 m³
= 0.7 m³
hence the equal water volume is 0.7 m³
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
Answer is: Kb for methylamine is 4.37·10⁻⁴.<span>
Chemical reaction: CH</span>₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻.
c(CH₃NH₂) = 0.253 M.
α = 4.07% ÷ 100% = 0.0407.
[CH₃NH₃⁺] = [OH⁻] = c(CH₃NH₂) · α.
[CH₃NH₃⁺] = [OH⁻] = 0.253 M · 0.0407.
[CH₃NH₃⁺] = [OH⁻] = 0.0103 M.
[CH₃NH₂] = 0.253 M - 0.0103 M.
[CH₃NH₂] = 0.2427 M.
Kb = [CH₃NH₃⁺] · [OH⁻] / [CH₃NH₂].
Kb = (0.0103 M)² / 0.2427 M.
Kb = 4.37·10⁻⁴.
