In teh hypothetical case of aliquid that had such strong intermolecular forces it would never evaporate, what temperature change would you expect to see

A wide temperature change.

Explanation:

The intermolecular force is the force that puts the molecules together in a substance. If the substance is ionic (formed by ions) than the force is called ion-ion, which is very strong. If it's a covalent compound (the elements share electron pairs), then it can be a polar or a nonpolar substance.

The polar substance has a huge difference in electronegativity of its components, so partial charges are presented. In this case, the intermolecular force is called dipole-dipole, because charged poles are formed. In the other case, of nonpolar, the dipoles are induced, so it's called dipole induced-dipole induced force.

When a polar compound has a hydrogen-bonded to a high electronegative element (F, O, or N), the dipole-dipole is extremely strong and it's called a hydrogen bond.

Thus, as strong is the force (Ionic > hydrogen bond > dipole-dipole > dipole induced - dipole induced), as difficult is to broken these bonds. In a phase change, those bonds must be broken. So if the liquid has so strong forces, it would be necessary a large amount of energy to evaporate it.

The temperature is a measure of the kinetic energy of the molecules, so if the energy applied is too high, the temperature change must be extremely high too!

This situation is not real, because if the forces are so strong than the material would be a solid and not a liquid. Besides, even an extremely strong bond may be broken with the right temperature increase, so the liquid would evaporate!

i look it up online and i'll tell you

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.