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2 years ago

What is thought to cause the dispersion forces?

Chemistry

1 answer:

Ilya [14]2 years ago

4 0

a) attraction between ions causes the dispersion forces.

Attraction between ions is a temporary attractive force which occurs when electrons in two consecutive adjacently placed atoms occupies positional confined areas that make the atoms acting like Temporary dipoles.

Dispersion forces or London Dispersion forces is the weakest force caused due to motion of electron.

On increasing the number of electrons the magnitude to Dispersion forces increases.

Dispersion forces also depends upon the atomic or the molecular weight of the given substance.

It is caused by an unequal and uneven distribution of the electrons inside an atom.

This causes a slight positive and slight negative charge to be formed inside an atom by establishing a temporary dipole. This temporary dipole induces a nearby atom by inducing another temporary dipole in the other one.

Learn more about Dispersion force here, brainly.com/question/14958417

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A proton is fired toward a lead nucleus from very far away. How much initial kinetic energy does the proton need to reach a turn

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4 years ago

A scientist is studying rock formations. She notices some unusual rocks like those pictured here. What can she conclude about th

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3 years ago

Read 2 more answers

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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4 years ago

The green light emitted by a stoplight has a wavelength of 525 nm. What is the frequency of this photon? (c = 3.00 × 10⁸ m/s).

Reil [10]

The frequency of this photon is 5.94 x $10^{14} H_{Z}$ .

Solution:

The speed of light has a value of approximately 3.00×108 m/s

The wavelength should have units of meters.

The frequency should have units of Hz or s−1. Hz is hertz or reciprocal seconds.

We know the value of the speed of light and we are given the wavelength. All we have to do is rearrange the equation to solve for the frequency:

v = c/λ

= 3.00× $10^{8}$ m/s/505× $10^{-9}$ m

= 5.94X $10^{14}$ s-1

= 5.94X $10^{14}$ Hz

Frequency describes the number of waves passing through a particular location at a particular time. A class interval frequency is the number of observations that occur in a particular predefined interval. So, for example, if 20 of her ages 5-9 appear in the survey's data, the interval 5-9 has a frequency of 20. The class interval endpoints are the lowest and highest possible values of the variable.

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1 year ago