@Tobster18 If all else is equal (# of electrons) than D-D & LDF is stronger than just LDF… But the data showed that the CCl4 was stronger… CollegeBoard has loved these types of questions for the last few years… You can’t just look at the IMF’s present if they are different sized molecules.
If your teacher has taught you that D-D is always stronger than something with just LDF, I’m sorry, but I’d then ask you to explain to me why iodine is a solid at room temperature when water (which has hydrogen bonding) is a liquid.
No the question reads that the CCl4 condenses while cooling at 70, which implies that the HCl is still a gas below 70.
I think a right explanation is that although in both cases .55 grams were lost, if there is water, that constitutes part of the .55 grams but does not combust and instead evaporates, which releases no heat, and then the alcohol constitutes less than .55 grams, so less alcohol is combusted, releasing less heat.
@Soccer1235 That would be an acceptable explaination… Could also talk about how some of the energy of the combustion would have to go towards vaporizing the water that was drawn up and not into the water in the container.
@superbunny320 I don’t predict the cut lines since I only get to see half of the exam, and we don’t ever get to know the cut lines… However with that being said, I think it is unlikely it will go that low.
@sjman1 I think I did about the same as you. Depending on the cutoff, you’re probably a borderline 5 or a solid 4. I was checking ap pass calculators and if the curve is like the 2015 test, you’l def get a 5. if it’s like the 2016, then it would be a 4. im praying the curve is generous this year tho lol
a. (i) n=PV/RT=(0.40)(25.0)/[(0.08206)(120+273)]=0.31 mol Cl2
(ii) 0.31mol Cl2/3 *(76.13g/mol)=7.87 g CS2
b. (i) Higher temperature means higher average kinetic energy and velocity of reactants. Reactants collide more forcefully (i.e. with more energy) and more frequently, thereby resulting in increased number of effective collisions and increasing the rate of reaction.
(ii) Graph should be more skewed right and decreased fraction of collisions at activation energy.
c. (i) Lewis structure should have single bonds between all the chemicals with 6 and 4 electrons around Cl and S respectively.
(ii) Both S have a sp3 hybridization with two lone pairs. Therefore, the molecular geometry wrt S is bent, so the bond angle of both should be around 109.
d. (i.) HCl experiences London dispersion forces and dipole-dipole, but NOT hydrogen bonding because the Cl- ion is too big and not electronegative enough.
(ii) Since CCl4 condenses first (higher boiling point), it has greater IMF than HCl because the particles interact more with each other, thereby forming a liquid.
a. The first diagram assigns a formal charge of -1 to oxygen, which is correct since it is the most electronegative atom in the structure. The second diagram is incorrect in that it assigns a formal charge of -1 to carbon and 0 to oxygen, even though oxygen is more electronegative.
b. ΔH=Σbonds broken-Σbonds formed=(413+891+201)-(391+615+745)=-246 kJ/mol
c. Since the reaction features one mole of gas on both sides of the equation, there is not an increase in number of macrostates, so entropy is zero.
d. ΔG=ΔH-TΔS=ΔH since ΔS=0; since ΔH is negative, ΔG is negative, so this reaction is thermodynamically favored at 298 K and will proceed in the forward direction. Therefore, isocyanic acid will be in greater concentration. Also, ΔG=-RTlnK; for ΔG to be negative, K has to be positive, meaning [product]>[reactant].
e. (i) A graph of ln[urea] vs. times is linear, so the reaction is first order.
(ii) Using the integrated rate law ln[A]t=ln[A]-kt, solve for k=0.693
f. Set initial concentration as constant, run the reaction in various pH (i.e. varying levels of [OH]), graph them, determine k and rate.
a. Kp=[NO]^2/([N2][O2])
b. N2 + O2 =2NO
I 6.01 1.61 0
C -0.61 -0.61 +0.122
E 5.40 1.00 0.122
Kp=(0.122)^2/(5.4)=0.00278
c. (i) Addition of acid will be neutralized by NO2- and addition of base will be neutralized by HNO2.
HNO2+OH-=H2O+NO2-
(ii) pKa=pH at half equivalence point, where half of the acid has been neutralized.
0.5(0.1L*0.1M HNO2)=0.005 mol HNO2 at half equivalence point
0.005 mol/0.1M NaOH=0.05 L NaOH
d. Higher concentration of HNO2 and NaOH means buffer can neutralize more acid and base, increasing its buffer capacity. Besides, buffers are generally made with a weak acid and the salt of its conjugate base, which is stronger than adding a strong base (NaOH) to a weak acid.
e. Since there are more HNO2 than NO2-, the solution is not at half-equivalence point, so pH<3.40. pH=3.40=log([NO2-]/[HNO2])=3.4+log(0.5)=3.10
a. C would be the least polar since it is most attracted to the NONPOLAR solvent and would have traveled the farthest.
b. In the unknown, the dye traveled half the distance as that of the solvent front Rf=0.5. In teh standard, A traveled half the distance as that of the solvent, so A is most likely the unknown.
a. q=mcΔT=125.00(4.18)(51.1-22.0)=15205J=15.2 kJ
b. 0.55g C3H7OH/60.09g/mol=0.0092 mol 15.2kJ/0.0092 mol=1700 kJ
c. Since water has a higher specific heat than 2-propanol, to burn the same mass requires a greater energy, which is transferred to the beaker of water. The beaker of water would have a higher final temperature since mass and specific heat remain constant, so ΔT increases. Since initial temperature remained constant, then final temperature must have increased.
a. Mg(OH2)–>2OH-+Mg 2+ Ksp=[OH-]^2[Mg2+]=(2x)^2(x)=4x^3=1.8e-11 x=[Mg2+]=2.62e-4 mol/L
I - 0 0
C - 2x x
E - 2x 2x
2.62e-4M58.32g/mol0.1L=0.00153 g Mg(OH)2
b. Mg2+ and Sr2+ are in the same group. Atomic radius increases as one moves down a group due to increased shielding effect. Sr2+ has a greater atomic radius because it has more electrons to shield its valence electrons, which do not experience as great of an attraction towards the positive nucleus as Mg2+ does. Coulomb’s law states that energy is inversely related to distance between two charges, so Mg2+, with its smaller atomic radius, experiences a greater lattice energy.
a. (i) Chromate is reduced, H2O2 is oxidized: E=1.33-0.7=0.63 V
(ii) Co2+ is oxidized and H2O2 is reduced: E=1.77-1.84=-0.07 V
b. (i) The galvanic cell with chromate and H2O2 yields a positive cell potential, meaning that it’s thermodynamically favored since ΔG=-nFE. For ΔG to be negative, E must be positive.
(ii) ΔG=-nFE=-(6 mol e-/mol rxn)(96485 coulomb/mole e-)(0.63 joule/coulomb)/1000=-365 kJ/mol rxn
@APChemTeacher do you think this year’s test was harder than last years (in terms of the frq)? I’m just wondering because I did tests from previous years and this year’s frq seemed a lot harder.
Thanks for your input in this thread by the way. It was really helpful!
yeah cutoffs have been creeping up for everything since people are getting smarter
even though everyone thought exams were hard this year the cutoff is probably still similar to last year @Dewzeph that’s really good lol you’ll def get a 5
1b should only have 2 sig figs
3b, your ice chart is wrong… the change (x) is .061, not .61… That’s gonna change your calculation
3d, your 2nd explaination is flawed as by adding the NaOH, you are essentially making the salt… your 1st explaination is good though
5c I disagree with your answer… Since you are burning the 2-propanol, you would burn less of it if the water is also being drawn up, therefore there would be less temperature change… the alcohol burner is contaminated, not the water above
6a. also only 2 sig figs
7bii, also only 2 sig figs
I say the sig figs even though 5b will be the sig fig question this year, so for the others it wouldn’t hurt you.
@btsftw… It’s been a long time since I sat down and actually took a test in a full chunk… honestly, years tests blend together for me… I just have a feel that it wasn’t too difficult, although I do think it was probably a bit harder than last years operational, and certainly harder than last years international (which is what the 79 is based on).
@sjman1… My understanding is that it is scaled using some college students every 5 years or so, and that determines the cut lines the first year of a cycle. Then they use the 10 “field” questions to compare year to year performance on the various exams, and reset the cut lines to match the abilities of where they put that first group. So if each year, the average score on the “field” questions is 6/10, but lets say the first year the average student was a 55%, and this year the average was a 58%, that tells the statisticians that this years test was “easier” than the original, but that the ability level of the students based on those 10 questions were equal… Hopefully that makes sense
@APChemTeacher They aren’t too harsh with the sigfigs unless they specifically ask for it
Perhaps I was under the impression that the alcohol burner also evaporates water and that the container of the burner was contaminated with water. The burner’s mass decreases by the same amount (which doesn’t differentiate between mass of alcohol and water). The student burns both alcohol and water, which due to the latter’s higher specific heat, transfers more heat to the beaker of water above, and therefore result in a higher final temperature of the water beaker.
I’m curious as to how they’re going to score my explanation for 3d. I offered a correct solution first, but then offered another incorrect response. Will they score the correct one as it came first, or will they not give any credit?
