Repulsion of electrons within two interacting molecules produces changes in electron distribution. This change in electron distribution creates temporary dipole moments. Which of the following does this explain? Choose one or more: A. This explains how two noble gases' molecules can have an attractive force between them. B. This explains why ammonia and nitrogen gas exhibit an attractive force between them. C. This explains why long hydrocarbon chains have relatively high boiling points. D. This explains how the molecules hydrogen fluoride and methanol can exhibit uncharacteristically strong intermolecular forces. E. This explains why the dipole-dipole attractive force between dimethyl ether and acetone does not entirely account for the attractive force between these molecules.

D

Explanation:

This explains how two noble gases molecules can have an attractive force between them.

This force is called as van dar Waals forces.

It plays a fundamental role in fields in as diverse as supramolecular chemistry structural biology .

If no other forces are present, the point at which the force becomes repulsive rather than attractive as two atoms near one another is called the van der Waals contact distance. This results from the electron clouds of two atoms unfavorably coming into contact.[1] It can be shown that van der Waals forces are of the same origin as the Casimir effect, arising from quantum interactions with the zero-point field.[2] The resulting van der Waals forces can be attractive or repulsive.[3] It is also sometimes used loosely as a synonym for the totality of intermolecular forces.[4] The term includes the force between permanent dipoles (Keesom force), the force between a permanent dipole and a corresponding induced dipole (Debye force), and the force between instantaneously induced dipoles

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

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hope i !