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
The correct answer is: [A]: "<span>points with the same elevation" .
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Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:
(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
Answer:
0.0187 M
Explanation:
Step 1: Write the balanced neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
18.7 mL of 0.01500 M HCl react.
0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol
Step 3: Calculate the reacting moles of NaOH
The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.
Step 4: Calculate the molarity of NaOH
2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.
[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M
Answer:
can u give us the options
