O N-H || ||
H-C-H H-C-H
Méthanal Méthanimine

1. Recopier les schémas de Lewis incomplets, puis les
compléter en ajoutant un (ou des) doublet(s) non liant(s),
sachant que chaque atome vérifie la règle de stabilité,

2. L'énergie d'atomisation est l'énergie à fournir pour
rompre toutes les liaisons d'une molecule et obtenir des
atomes,
a. Calculer les énergies de liaison des liaisons C-O et C=N
b. En deduire, parmi les doubles liaisons C=O et C-N, celle
qui est la plus stable
Données:
Eliaison (C-H) - 413 USI.
Eliaison (N-H)= 391 USI
Eatomisation (methanal) = 1567 USL
Eatomisation (méthanimine) - 1564 USI
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Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge