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Nataly [62]
2 years ago
7

What number and type of hybrid orbital(s) form(s) when one p and one s atomic orbital mix?

Physics
1 answer:
Harrizon [31]2 years ago
3 0

Two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

<h3>What are orbitals?</h3>

Orbital is the place around nucleus where mostly the electrons are present. There are four types of orbitals are present, s, p, d, and f.

The orbitals that are formed by the mixing of these orbitals are called hybrid orbitals.

Thus, two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

Learn more about orbitals

brainly.com/question/18914648

#SPJ4

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Explanation:

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The correct option is <u><em>(d) Velocity of a moving body is its speed in a given direction.</em></u>

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Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
kirill [66]

Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

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Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

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Answer:

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