Hi!
The radical bromination reaction of C₆H₅CH₂CH₃ is performed through a mechanism in which radical reactions are involved. This compound is an alkylbenzene compound, and the carbon that is more reactive towards radical bromination is the carbon bonded to the aromatic ring because in the reaction mechanism the intermediaries are stabilized by resonance in the aromatic ring.
A terminal substitution will not occur because substitution there will not be stabilized by resonance. The compound that will be formed in this reaction would be:
C₆H₅CH₂CH₃ + Br₂ → C₆H₅CH₂(Br)CH₃ + HBr
The ionization energy of an element is the amount of energy required to remove one mole of electrons from the element in its gaseous state. The equations for the first three are:
Fe(g) → Fe⁺(g) + e⁻
Fe⁺(g) → Fe⁺²(g) + e⁻
Fe⁺²(g) → Fe⁺³(g) + e
Answer:
ΔH°c = -2219.9 kJ
Explanation:
Let's consider the combustion of propane.
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)
We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.
ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]
ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]
ΔH°c = -2219.9 kJ
Answer:
D
Explanation:
The fact is that the both elements belong to different groups in the periodic table. Mg is in group 12 while Al is in group 13. The outer most electron configuration for Mg2+=[Ne]3s0
Al+= [Ne]3s13p0
It is evident that in Al+, the second electron is to be removed from a filled 3s subshell which is energetically unfavourable. Unlike In Mg2+ where the two electrons are removed. There is always a tendency towards a quick loss of all the valence electrons in the valence shell. It will be more difficult to remove an electron from a filled shell.
