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Roman55 [17]
4 years ago
13

What type of hybridization is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen?: and

how many lone
a. sp hybridization and 2 lone pairs
b. sp hybridization and 1 lone pair
c. sp hybridization and 2 lone pairs
d. sp hybridization and I lone pair
e. sp hybridization and 1 lone pair
f. sp hybridization and 2 lone pair
Physics
1 answer:
insens350 [35]4 years ago
7 0

<u></u>sp^3<u> hybridization and 1 lone pair</u> is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen.

b. sp^3 hybridization and 1 lone pair

<u>Explanation:</u>

The Nitrogen particle is sp^3 hybridized with one crossover orbital involved by the solitary pair. Likewise, nitrogen is sp^3 hybridized which implies that it has four sp^3 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 sp^3 hybridized orbitals cover with s orbitals from hydrogens to frame the two N-H sigma bonds.

Nitrogen utilizes sp^3 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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L = 1.11 x 10^{6} m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x 10^{5} m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x 10^{-2} m

Length = ?

So,

we know that,

U = 1/2 C Δv^{2}

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δv^{2}  

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And we also know that,

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K = 3.7

A = 0.20 x L

d = 2.6 x 10^{5} m

Plugging in the values into the formula, we get:

0.28 = \frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }

Solving for L, we get:

L = 1.11 x 10^{6} m,

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