How are molecule shapes determined by electrons

2 answers:

4 0

The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction.

il63 [147K]3 years ago

3 0

Answer:

Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule. The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction.Explanation:

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Plz answer this! I am stumped

Gwar [14]

MgCl2
Mg = magnesium
Cl = chlorine

Magnesium + chlorine = magnesium chloride.

This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.

So ur answer would be magnesium chloride. :)

6 0

4 years ago

If the pendulum took longer to complete one oscillation, how would the graph change?

Sindrei [870]

Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

7 0

3 years ago

Read 2 more answers

A dog is walking at 2m/s and then begins to run at a speed of 6m/s. What is his acceleration if his total travel time is 2 secon

fiasKO [112]

The formula for velocity vf = vi + at

First list your given information

2m/s Is your initial velocity (vi)
6m/s is you final velocity (vf)
2 seconds is your time (t)

Since you want the a for acceleration get a by itself

a = (vf-vi)/t

So a= (6-2)/2

a= 4/2

a=2

Now units

the units for acceleration are m/s $x^{2}$

2m/s $x^{2}$

7 0

3 years ago

A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?

docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

$v=\sqrt{2gh}$