What might happen if water molecules did not have a slight negative charge on one end and a slight positive charge on the other?

Physics

1 answer:

Katena32 [7]3 years ago

5 0

Answer:

It would not be possible the cohesion among water molecules by the polar covalent bonding.

Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.

In this sense, one of the main properties of water is cohesion (molecular cohesion), which is the attraction of molecules to others of the same type. So, water molecule ( $H_{2}O$ ) has 2 hydrogen atoms attached to 1 oxygen atom and can stick to itself through hydrogen bonds.

How is this possible?

By the polar covalent bonding, a process in which electrons are shared unequally between atoms, due to the unequal distribution of electrons between atoms of different elements. In other words: slightly positive and slightly negative charges appear in different parts of the molecule.

Now, it can be said that a water molecule has a negative side (oxygen) and a positive side (hydrogen). This is how the oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Thanks to this polarity, water molecules can stick together.

You might be interested in

a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec

MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

$\frac{1}{2} mu_1^2+0=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $\frac{1}{2} m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2$

∴ $\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $mv$ ₁ $v$ ₂ = 0

It is impossible for the mass to be zero.
Because the second ball moves, velocity v2 cannot be zero.
As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.

<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0

1 year ago

BEST ANSWER GET BRAINLEIST!!!!!!!!!!!!!!! PLEASE HELP ASAP!!!!!!!!!!!!!!

ozzi

No person shall be judged by their financial status in the process of applying for a job or position of power in the government.

6 0

3 years ago

A jetliner has a velocity of 95 m/s. What is the displacement of the jetliner at t=3.0 seconds?

charle [14.2K]

Distance = speed / time

speed = 95 m/s
time = 3 s

distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)