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erma4kov [3.2K]
3 years ago
6

In general, which attractions are stronger: intermolecular or intramolecular? Research and explain a quantitative comparison bet

ween the strength of a chemical bond and the strength of intermolecular forces of attraction.
- I understand that intramolecular forces are stronger but I don’t know what to put for the quantitative data
Chemistry
1 answer:
murzikaleks [220]3 years ago
8 0

Answer is: intramolecular attractions are stronger.

Intramolecular attractions are the forces between atoms in molecule.

There are several types of intramolecular forces: covalent bonds, ionic bonds.

Intermolecular forces are the forces between molecules. The stronger are intermolecular forces, the higher is boiling point of compound, because more energy is needed to break interaction between molecules.

There are several types of intermolecular forces: hydrogen bonding, ion-induced dipole forces, ion-dipole forces andvan der Waals forces.

Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.

Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.

20 kJ/350 kJ = 0.057 = 5.7 %.


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15 POINTS PLEASE HELP What volume of water must be added to 35mL of 2.6m KCl to reduce its concentration to 1.2m? Please explain
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First, find the volume the solution needs to be diluted to in order to have the desired molarity:
You have to use the equation M₁V₁=M₂V₂ when ever dealing with dilutions.

M₁=the starting concentration of the solution (in this case 2.6M)
V₁=the starting volume of the solution (in this case 0.035L)
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V₂=the volume of solution needed for the dilution (not given)

Explaining the reasoning behind the above equation:
MV=moles of solute (in this case KCl) because molarity is the moles of solute per Liter of solution so by multiplying the molarity by the volume you are left with the moles of solute.  The moles of solute is a constant since by adding solvent (in this case water) the amount of solute does not change.  That means that M₁V₁=moles of solute=M₂V₂ and that relationship will always be true in any dilution.

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 Second, Finding the amount of water needed to be added:
Since we know that the volume of the solution was originally 35mL and needed to be diluted to 75.8mL to reach the desired molarity, to find the amount of solvent needed to be added all you do is V₂-V₁ since the difference in the starting volume and final volume is equal to the volume of solvent added.
75.8mL-35mL=40.8mL
40.8mL of water needs to be added

I hope this helps.  Let me know if anything is unclear.
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