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prohojiy [21]
3 years ago
6

Why is a ylid a stable nucleophile? Select all that apply. Group of answer choices The proximity of the positively charged phosp

horous and negatively charged carbon stabilizes the charges. It is a stable Arrhenius base. Inductive effects and resonance stabilize the negative charge. It isn't stable, it only forms transiently.
Chemistry
1 answer:
pogonyaev3 years ago
6 0

Answer:

1)The proximity of the positively charged phosphorous and negatively charged carbon stabilizes the charges.

2) Inductive effects and resonance stabilize the negative charge

Explanation:

both atoms have full octets of electrons( I.e Carbon and say phosphorus). The result can be viewed as a structure in which two adjacent atoms are connected by both a covalent and an ionic bond; normally written X+–Y−. Ylides are thus 1,2-dipolar compounds, and a subclass of zwitterions

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Answer:

ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given) (exothermic)

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           3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g)  +  17H₂O(g)   +  10CO₂(g)

ΔH⁰(f):  3(-365.6)Kj     1(-301)Kj    14(0)Kj       3(0)Kj    17(-241.8)Kj    10(-393.5)Kj

            = -1096.8Kj     = -301Kj     = 0Kj         = 0Kj      = -4110.6Kj    = -3930.5Kj

ΔHₙ°(rxn) = ∑ (ΔH˚(f)products) - ∑(ΔH˚(f)reactants)

= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]

= [-(8041.1) - (-1397.8)]Kj

= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)  

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ΔH°(rxn for 14.11g  NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of  NH₄NO₃(s) given)

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