Answer:
The correct answer is A electron on the aniline nitrogen are somehow delocalized to the aromatic ring.
Explanation:
The structure of aniline contain double bonds and lone pair of electron in the nitrogen atom of -NH2 group that is attached to the benzene ring.
The electron pair present in the nitrogen atom of -NH3 group of aniline undergo delocalization with the aromatic ring of benzene resulting in the formation of resonance hybrid that increases the ability of nitrogen atom of -NH2 group of aniline to easily donate that lone pair of electron.
ON the other hand the resonance stabilization cannot be possible with the cyclohexylamine ring as it is saturated.
Hey there!
A ) PO₄³⁻
The element exhibits resonance (4 times) as there is a double bondwith 1 oxygen and a single with the other 3 oxygen. This switchesin each resonance structure. (I know its weird b/c I leaves Phosphorus with 10 electrons surrounding it)
b) SO₄²⁻
Sulfur is the central atom ,double bond 2 oxygens so that they can only have 4 valence electrons ( the little dots around the oxygen ) ,single bond 2 oxygens ( So that they have 6 valence electrons )
Formal charges are 0 for the two double bonded oxygens [ 6-6 = 0 ]and 0 for the sulfur [ 6-6=0 ]
But the oxygens are -1 [ 6-7 = -1 ]
Since there are two of them, there is a negative charge of -2
Bracket the whole structure and on the top right of the outside ofthe brackets write 2- and include a - (negative sign) next to eachof the single bonded oxygens
Don't forget there are resonance structures. The double bonds canbe placed around any of the oxygens
Hope that helps!
The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:
CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol
As observed O2 will not have enthalpy of formation as it is a pure substance.
To calculate for the enthalpy of reaction,
enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
= 294.5 kJ/mol
ANSWER: 294.5 kJ/mol