What type of hybridization is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen?: and
1 answer:
<u></u><u> hybridization and 1 lone pair</u> is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen.
b. hybridization and 1 lone pair
<u>Explanation:</u>
The Nitrogen particle is hybridized with one crossover orbital involved by the solitary pair. Likewise, nitrogen is
hybridized which implies that it has four
half and half orbitals. The sub-atomic structure of water is predictable with a tetrahedral game plan of two solitary sets and two holding sets of electrons. Two of the
hybridized orbitals cover with s orbitals from hydrogens to frame the two N-H sigma bonds.
Nitrogen utilizes orbitals to accomplish this geometry. Three of the mixtures are utilized to frame bonds to hydrogen and the fourth contains the solitary pair. Whereas lone pairs are the pairs of electron on atoms that don't take an interest in the holding bonding of two atoms. To distinguish solitary matches in a particle, make sense of the number of valence electrons of the molecule and subtract the number of electrons that have partaken in the holding.
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Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force, = m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F = = m·v²/r = 16 N
(1/2) × r × = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r × = (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.