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dedylja [7]
1 year ago
13

Compare the relative strength of the two forces B and C. Explain how you determined this comparison by identifying the forces.

Chemistry
1 answer:
Mars2501 [29]1 year ago
4 0

The intermolecular forces or the strength of C is more than that of B.

What are intermolecular forces?

  • Intermolecular forces are the forces that hold the molecules of any object in a particular state of matter.
  • It exists between molecules.
  • These forces are weak when compared with intramolecular forces.
  • Hydrogen bonding, dipole-dipole, Vander Walls, and ion-dipole are the types of intermolecular forces.
  • At a short distance, these forces are repulsive but at large distances they are attractive.

Here, we can see that C has dipole-dipole interaction while B has none.

Therefore, it states that the intermolecular forces present in C are much more than that in B.

To learn more about intermolecular forces, visit: brainly.com/question/9007693

#SPJ9

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Magnesium has three naturally occurring isotopes with masses of 23.99 amu, 24.99 amu, and 25.98 amu and natural abundances of 78
mihalych1998 [28]

Answer:24.31

Explanation:Contribution made by isotope of mass 23.99= 23.99×78.99=1894.97

Contribution made by isotope of mass 24.99=24.99×10.00=249.9

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3 0
2 years ago
Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. N 3−, O 2
love history [14]

Answer:

<u>Option 1</u>:  N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²    

Explanation:

<u>Ionic radius is the radius of an atom´s ion in ionic crystal structure</u>.<u> </u><u><em>In an ion that lose an electron, to form a cation, the radius of the ion gets smaller</em></u><em>, </em>because the repulsion between electrons decrease because fewer electrons are present. Conversely, <u><em>adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.</em></u>                  

<u><em>Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers</em></u>.                        

<u><em>In a isolelectronic series (same number of electrons),</em></u> <u><em>the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius </em></u>beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.  

<u>We will use this principle to solve our problem</u>.  

In our case, the given ions are:  

  • N⁻³ :    Z = 7,  e⁻ = 10
  • O⁻²:     Z= 8,   e⁻ =10
  • F⁻:       Z = 9,  e⁻ = 10
  • Na⁺:    Z= 11,   e⁻ = 10
  • Mg⁺²:  Z=12,   e⁻ =10

where Z= number of protons, and e⁻ = number of electrons.

<em><u>Hence the decreasing order of ionic radius is:</u></em>

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²  

Have a nice day!

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