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:
B. CH3Br
Explanation:
Dipole -Dipole interactions take place in polar molecules.
CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.
While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.
Hence, the correct option is B. CH3Br.
Answer:
3224 kJ/mol
Explanation:
The combustion of benzoic acid occurs as follows:
C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE
The change in temperature in the reaction is the change due the energy released, that is:
3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts
To find the heat released per mole we have to find the moles of benzoic acid:
<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>
1.250g * (1mol / 122.12g) = 0.0102 moles
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The dE combustion per mole of benzoic acid is:
33.00kJ / 0.0102moles =
<em>3224 kJ/mol </em>