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Dominik [7]
2 years ago
10

What type of van der Waals interactions occur between molecules

Chemistry
1 answer:
BigorU [14]2 years ago
7 0

Given what we know, we can confirm that the type of van der Waals interactions that occur between molecules of O2, SCl2, and CH4 in liquids of these substances are the presence of <u>London dispersion forces</u>.

<h3>What are London dispersion forces?</h3>
  • They are a force of attraction between atoms.
  • They are generated by electrostatic attraction.
  • These forces are common between atoms in close proximity and occur often when compounds have a symmetrical distribution of atoms.
  • They are generated by the formation of temporary dipoles.

Therefore, given the symmetry of the atoms disposition in these compounds and the temporary dipoles generated by the atoms being in close proximity, we can confirm that the van der Waals forces present in each compound are London dispersion forces.

To learn more about van deer Waals forces visit:
brainly.com/question/13201335?referrer=searchResults

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The goal of Science is to expand knowledge.

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Nitric acid can be produced by the reaction of gaseous nitrogen dioxide with water. 3 no2(g + h2o(? ?? 2 hno3(? + no(g if 538 l
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The balanced chemical reaction is written as:

<span>3NO2 + H2O = 2HNO3 + NO

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Does CO2 contain molecules?
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3 years ago
A sample of a compound of xenon and fluorine contains molecules of a single type; XeFn, where n is a whole number. If 5.0 x 1020
kicyunya [14]

Answer:

4

Explanation:

Avogadro’s number represent the number of the constituent particles which are present in one mole of the substance. It is named after scientist Amedeo Avogadro and is denoted by N_0.

Also, it is the number of particles in exactly 12.000 g of isotope carbon 12.

Avogadro constant:-

N_a=6.023\times 10^{23}

Hence,

Mass of XeF_n = 131.293+ n18.998 g

So,

6.023\times 10^{23}  molecules have a mass of 131.293+ n18.998 g

Also,

1  molecules have a mass of \frac{(131.293+ n18.998)}{6.023\times 10^{23}} g

So,

5.0\times 10^{20}  molecules have a mass of \frac{(131.293+ n18.998)}{6.023\times 10^{23}}\times 5.0\times 10^{20} g

Also, given mass = 0.172 g

Thus,

\frac{(131.293+ n18.998)}{6.023\times 10^{23}}\times 5.0\times 10^{20}=0.172

\frac{18.998n+131.293}{1204.6}=0.172

18.998n+131.293=207.1912

n=\frac{75.8982}{18.998}=4

<u>Thus, value of n is 4.</u>

4 0
3 years ago
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