Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Answer:
El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.
Explanation:
Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.
HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−
Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M
Writing the information from the ICE Table in Equation form yields
6.6×10−4=x20.3−x6.6×10−4=x20.3−x
Manipulating the equation to get everything on one side yields
0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4
Now this information is plugged into the quadratic formula to give
x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2
The quadratic formula yields that x=0.013745 and x=-0.014405
However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86
Answer:
The molality of the solution is 0.3716 mol/kg
The number of moles of solute is 0.0157 mol
The molecular weight of the solute is 129.30 g/mol
The molar mass of the solute is 129.32 g/mol
Explanation:
m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg
Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol
Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol
When Kf = 7.66 °C.kg/mol
Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol
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