Answer:
The correct order of increasing reactivity toward nucleophilic acyl substitution is E < D < C < A < F < B.
Explanation:
The stability of the leaving group best determines the manner of reactivity of carboxylates to nucleophilic substitution after the substitution of the nucleophile to the leaving group. The leaving group should, therefore, be protonated with hydrogen ion in the solution to form a stable molecule. From the given list: The leaving group for A, Ethyl thioacetate will be ethanethiol. For B, Acetyl chloride will be Hydrochloric acid. For C, Sodium acetate will be Sodium Hydroxide. For D, Ethyl acetate will be Ethanol. For E, Acetamide will be Ammonia, and for F, Acetic anhydride will be Ethanoic acid. The reactivity of the substitution reaction is dependent on the stability of these leaving groups. The stability of these leaving groups depends on their pKa, and the more the pKa, the lesser the acidity of the leaving group, and the lower the reactivity. Therefore, considering their pKa: A is 8.5, B is -7, C is 13.8, D is 15.9, E is 36, and F is 4.8. When we rearrange this pKa in descending order, we have E, D. C, A, F, B. Which is also the increased reactivity of the nucleophilic acyl substitution.
Answer:
Explanation:
Pair 2.50g of O₂ and 2.50g of N₂
The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:
Number of moles = 
Atomic mass of O = 16g and N = 14g
Molar mass of O₂ = 16 x 2 = 32gmol⁻¹
Molar mass of N₂ = 14 x 2 = 28gmol⁻¹
Number of moles of O₂ = 
= 0.078mole
Number of moles of N₂ = 
= 0.089mole
We see that N₂ has the largest number of moles
The position of equilibrium lies far to the right, with products being favored.
Answer:
I A hole
Explanation:
I must be in a museum, because you truly are a work of art
