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lbvjy [14]
3 years ago
8

List and define three types of intermolecular forces and identify which types of molecules each forces affects.

Physics
1 answer:
Zigmanuir [339]3 years ago
3 0

Van der waals forces

Hydrogen bonding

Crystal lattice forces

Explanation:

Intermolecular forces exists between molecules.

  • Van der waals forces are weak attractions that joins non-polar and polar molecules together. London dispersion forces occurs between non-polar molecules(polar) and noble gases. Dipole -dipole attraction occurs between polar molecules. Van der waal forces occurs in graphite layers, HCl e.t.c
  • Hydrogen bonding is force of attraction between polar molecules in which a hydrogen atom is directly joined to a highly electronegative atom. Examples occur in water.
  • Ionic crystal lattice forces are strong electrostatic forces of attraction between oppositely charged ions arranged into a crystal lattice of ionic compounds. For example in NaCl

Learn more:

Intermolecular forces brainly.com/question/3622116

#learnwithBrainly

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An experiment is carried out to measure the extension of a rubber band for different loads.
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Complete question is;

An experiment is carried out to measure the extension of a rubber band for different loads.

The results are shown in the image attached.

What figure is missing from the table?

Answer:

17.3 cm

Explanation:

The image attached showed values for load, extension and initial length.

Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.

The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.

This gives: 16.2 - 15.2 = 1 cm

This corresponds to what is given in the table.

For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.

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A field measuring 12 meters by 16 meters is to have a brick paver walkway installed all around it, increasing the total area to
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Answer:

1.5 m

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New Area, A' = L' x W' = (12 + 2d)(16 + 2d) = 192 + 56 d + 4d^2

Difference in area = A' - A

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4d^2 + 56 d - 93 = 0

d = \frac{-56\pm \sqrt{56^{2}+4\times 4\times 93}}{8}

d=\frac{-56\pm 87.72}{8}\

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