## Photon336 one year ago Question

1. Photon336

Questions attached

2. Photon336

Another here

3. Photon336

@rushwr @cuanchi @rvc

4. Rushwr

for the 1st one i'll go with C

5. Rushwr

156. it is heat of sublimation right?

6. Rushwr

@Empty U too give it a try

7. anonymous

D. heat of sublimation,it is the transition from solid to gas

8. Photon336

The first one i was thinking intermolecular forces.

9. Photon336

I chose C for the first one

10. Photon336

heat of fusion (s)-->(l) heat of vaporization (l)-->(g) heat of sublimation (s)--->(g) that makes most sense to me

11. Photon336

$K = C + 273$ 273+56 = 329K $0.5 \frac{ kj }{ g } x 100g$ = 50 kj we have already reached the boiling point, so vapor pressure = external pressure. don't know why i'm skeptical of that answer of 50 kj.

12. Rushwr

So what are the answers given / ????? @Photon336

13. Photon336

You guys are correct for all three

14. Photon336

First one was C IMF second one was D sublimation the other one was B

15. Ciarán95

172. When we boil something and convert it from a liquid to a gas, we lose any intermolecular interactions between the individual molecules present, with each now becoming independent species, free to move with respect to one another. So, the amount of energy to convert a substance into the gaseous phase depends on the strength of these intermolecular interactions (i.e. the degrees of attraction between molecules) |dw:1438191745228:dw| Whilst both molecules contain a dipole due to the uneven distribution of electrons in one or more covalent bonds (thus leading to electrostatic dipole-dipole interactions between oppositely charged ends of identical molecules), you have to consider whether there is an especially strong interaction in one over the other, that would take more energy to overcome. This arises from Hydrogen Bonding, where we have H directly convalently bonded to one of the most electronegative elements (Oxygen, Flourine, Nitrogen), leading to very large partial charges forming and much stronger intermolecular interactions that 'ordinary' dipole-dipole interactions (large delta + and delta -). Once considering this, you should be able to decipher the answer. 173. I'm not 100% sure about this one, but from looking at it, here's my proposed answer: The heat of vaporisation of acetone is 0.500 kJ/g. That is, if we had 1 gram of liquid acetone (or about 1.26 cm3, given the density of acetone is 0.791 g/cm3), it would take 0.500 kJ of heat/energy input to overcome the interactions of the molecules as the slide past one another and convert them into independent gas molecules. The boiling point of acetone is 56 degrees Celsius. On the Kelvin scale, this is: 273.15 + 56 = 329.15 K When rounded down to 329 K, this is also the temperature at which we're hoping our 100 g (~126 cm3) of acetone will vaporise at. So, both the vaporisations mentioned in the question are taking place under the same conditions (i.e. bringing them to the boil at the same temperature value). Obviously, the higher the mass, the more molecules that are present and the greater the degree of interactions available between these in the sample. So, to get it to boil at the same temperature, we need to compensate by providing it with more energy, or heat. $1~g = 0.500~kJ$ implying that: $100~g = (100)(0.500~kJ)$ $= 50~kJ$

16. Photon336

Ciaran wow great explanation