elohehl… Each calcuation on that problem will be worth 1 point (2 pts total for part A)
Hey guys, what did you say for 5c? (alcohol burning)
will it decrease?
also any clue with 3e? (buffer diagram)
thanks!
@APChemTeacher how many points do you think will be awarded if only the formal charges on each atom is calculated and no explanation was given?
Also 1d
which one has greater IMF’s???
lower condensation point= higher/lower Imf?
what do you think the curve will be based on the difficulty?
low 70’s? high 60’s?
superbunny… I’m thinking that 2a will only be worth one point… there’s 7 parts to that problem, so that’s 7 of the 10 points… I think that d, eii, and f will each be given a 2nd point in that problem (just a guess though).
@sjman1 I said HCl because the bonds are more polar so their dipole-dipole forces are stronger. I might be wrong tho
what would we need to get full points for 2f (the one about explain your own experiment thing)
So the E voltages are definitely +0.63 and -0.07 yeah? Cause you use the 1st and 4th half reactions for the first voltage and the 1st and 2nd for the second voltage.
@sjman1 For 3e, the proportion of conjugate base to acid is 5:10. This means that when using the hasselbalch equation, you would end up with 3.54 + log(5/10) since it deals with the proportion of base to acid. Log(1/2) is a negative number, so the buffer would have a lower pH than pka. For 5c, the final temp would be less since it water is mixed in with the alcohol. This means there wouldn’t be .55g of alcohol, but rather a mix of alcohol and water. Since there isn’t a pure sample of alcohol, less energy would be released since less alcohol is combusting, this means the final water temp would be lower than the experiment done with pure water
@sjman1 … yes it will decrease in 5c…
for 3e, it is below 3.40 (more acid form than base form, so not to half EP yet)
for 1d, lower condensation point also means lower boiling point… I think if you thought of it that way you’d realize lower boiling point means weaker total IMFs
Personally, I don’t think it was that bad… there really wasn’t that much math, everything in the buffer problem put it at the half EP, so as long as you knew about that you were fine… I don’t guess what the cut lines are because they never get released, but I think they’ll be lower than last years international version, just based on the FRQ, but from what it sounds like the MC was pretty easy, so it’s hard to say. Remember, I’m only getting to see the difficulty of one half of the test.
@sjman1 If it condenses at a higher temperature, it means it has the greater intermolecular forces because it doesn’t need to cool down so much to get a lower energy state, if that makes sense. So the CCl4 has the stronger intermolecular forces.
@btsftw CCl4 would have stronger imfs. The compound was hydrogen chloride, which has much lower LDFs due to its relatively small size when compared to Carbon tetrachloride. Since CCl4 is a larger molecule, its surface area would be greater which means a greater electron cloud is present. Thus allows for more instantaneous dipoles to occur, effectively allowing for higher a condensation temp than HCl.
@Dewzeph right lol i realized i didn’t even read the question on the test rip
@btsftw There are many ways to answer that question… But the easiest way would be to do the same experiment at a different pH… Again measure the concentration of the same species, and find the half-life or k… if they are the same at the new pH as the old pH, then OH doesn’t affect… If it does change k, then OH does affect…
I’ll tell you that graders hate that type of question because it is so open ended with so many possible experiments.
Would it be correct to say that the water can hydrogen bond with the 2-propanol, therefore hindering 2-propanol’s ability to combust to the fullest extent of 0.55 g of pure 2-propanol. I am talking about 5c.
@APChemTeacher Yeah, I was kinda shocked how little math there was on this year’s frqs. The buffer problem definitely saved me because my class covered buffers so in depth. As for the MC, there were a few questions that stumped me, not sure if they were field questions or not, but I can hope for the best.
@ImpossibleCube That allows it to be immisible (which they told you), but that would not be an acceptable reason for why it would produce less heat
Doesn’t HCL have stronger intermolecular forces as it has dipole-dipole and vanderwaals while CCl4 only has vanderwaals.
Plus HCL condenses at a higher temperature according to the question…