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Effectus [21]
4 years ago
14

2.What is the electronic geometry about a central atom which has the following number of regions of electron density? A.3 region

s of electron density B.4 regions of electron density C.5 regions of electron density
Physics
1 answer:
zloy xaker [14]4 years ago
7 0

Answer:

  • <u>The electronic geometry of the different compounds:</u>

The over all geometry or structure of the compound is greatly effected by the number of lone pair electrons on its central atom, as there are a number of free or non- bonding electrons present inside the structure of the compound that effects the over all structure of the compound.

Explanation:

A- SO₂ and BF₃ has 3 lone pair of electrons and it has a trigonal planer geometrical shape, due to the presence of the 3 different regions or points of electronic density on its central atom.

B-NH₃ and CH₄ have the four different zones or regions due to having the very concrete or strong electronic density on its central atom. As, the shape of the molecules itself tends to be more tetrahedral in its geometric form or shape.

C- PCL₅ and PF₅ are the two compounds having the 5 different regions of electronic density. As, the over all geometric shape or structure of the compound is more trigonal Bi-pyramidal in total, as there are the lone pair of electrons on the central atom which makes the whole structure of the compound.

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Answer:

When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion.Hope this answer helps.

7 0
3 years ago
Read 2 more answers
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
4 years ago
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nasty-shy [4]

Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

8 0
3 years ago
which statement best describes how changes behave a.opposites charges repel against each other and like charges attract to each
Anon25 [30]

I think it is B hope this helps

God Bless

7 0
3 years ago
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Help me with this physics math?
solong [7]
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes

(6.4)^2 x 10^12

= 40.96 x 10^12 .

Now it's just a matter of mashing out the fraction.

The 'mantissa' (the number part) is

6/40.96 = 0.1465

and the order of magnitude is

10^24 / 10^12 = 10^12 .

Put it all together and you've got

1.465 x 10^11 .
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3 years ago
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