All categories

2 years ago

What intermolecular force is responsible for the attraction between an ion and a polar molecule?

Physics

1 answer:

Pani-rosa [81]2 years ago

8 0

Answer:

Dipole ion forces

Explanation:

Dipole ion forces are attractive forces between an ion (an atom that has lost or gained an electron, so it has a charge) and a polar molecule. A molecule is a dipole when there is an asymmetric distribution of electrons because the molecule is made up of atoms of different electronegativity. The ion then attaches to the part of the molecule that has its opposite charge: the positive end of the polar molecule faces the anion (negatively charged ion) and the negative end of the polar molecule faces the cation ( positively charged ion).

You might be interested in

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

3 0

3 years ago

a train moving west with an initial velocity of 20m/s accelerates 4m/s2 for 10 seconds . during this time , the train moves a di

Blababa [14]

X=1/2at^2+vt+x0
x=1/2(4)(100)+20(10) = 400 m

8 0

3 years ago

In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the

amid [387]

Answer:

dd=4.3 ly=4.3 •9.46•10°15 m=4.1 10°16m.

Explanation:

5 0

2 years ago

A certain string just breaks when it is under400N of tension.

OLEGan [10]

Answer:

The string will break with a speed of 20 m/s.

Explanation:

It is given that,

Tension at which the string just breaks, T = 400 N

Mass of the stone, m = 10 kg

Radius of the circle, r = 10 m

We need to find the speed at which the string will break. The boy continuously increases the speed of the stone. The tension acting on the stone is equal to the centripetal force. It is a force that acts towards the center of circle. It is given by :

$T=F=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{Tr}{m}}$

$v=\sqrt{\dfrac{400\ N\times 10\ m}{10\ kg}}$

v = 20 m/s

So, the string will break with a speed of 20 m/s. Hence, this is the required solution.

4 0

3 years ago

An arrow is shot horizontally from the top of a building and it lands 200m from the foot of the building after 10s.Assuming air

Alexandra [31]

The initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

u is the initial velocity of fall = ? m/sec

h is the distance of fall = 200 m

g is the acceleration of free fall = 9.81 m/sec²

H is the height of the building

t is the time period = 10 second

According to Newton's second equation of motion,

$\rm h= ut+\frac{1}{2} gt^2 \\\\\ h-\frac{1}{2} gt^2 =ut \\\\ 200 - 0.5 \times(9.81) \times 10^2 = 10 u \\\\ u = - 29.05 \ m/sec$

- ve shows the direction is downward.The magnitude of the initial velocity is found as;

u = 29.05 m/sec

The height of the building

$\rm H= ut+\frac{1}{2} gt^2 \\\\\ H = 29.05 \times 10 + 0.5 \times 9.81 \times 10^2 \\\\ H = 781 \ m$

Hence the initial velocity of the arrow and the height of the building will be 29.05 m/s² and 781 m respectively.

To learn more about the velocity, refer to the link: brainly.com/question/862972

SPJ1

4 0

1 year ago