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aleksandr82 [10.1K]
3 years ago
14

Which of the following bond types is the strongest?

Chemistry
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

b) Ion-dipole

Explanation:

Intermolecular forces are the forces of attraction or repulsion between molecules, they are significantly weaker than intramolecular forces like covalent or ionic bonds.

  • <em>Hydrogen bonds</em> happen between a partially positively charged hydrogen and another partially negatively charged, it's a type of dipole-dipole interaction, one of the strongest among intermolecular forces.
  • <em>Ion-dipole</em> involves an ion and polar molecule, its strength is proportional to the charge of the ion. It's stronger than hydrogen bonds because the ion and the polar molecule align so positive and negative charges are next to another allowing maximum attraction.
  • <em>Dipole-dipole </em>is an interaction between two molecules that have permanent dipoles, aligning to increase attraction.
  • <em>Ion-dipole</em> induced usually happens when a non-polar molecule interacts with an ion causing the molecule to be temporary partially charged. It's a weaker interaction.
  • <em>Dipole- Induced Dipole</em>, like ion-dipole induced this interaction causes one of the two involved molecules to be temporary partially charged.

Considering this information we can conclude that Ion-Dipole interaction is the strongest force among intermolecular forces.

I hope this information is useful to you!

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Answer:

6.82 kg

Explanation:

Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,

mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.

The sublimation enthalpy of dry ice is 571 KJ/kg.

Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.

Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.

This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.

Now, Q' =m'L' = heat lost by water = 3892.98KJ.

And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)

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