Not sure if I am 100% right
i messaged the guy on twitter and more people are replying with the same complaint but the AP guy has yet to respond.
For 3d, did anyone else get 9.23… cuz that was the only way that my titration curve looked slightly normal
I think you had do 14-9.23 bc 9.23 was the pOH
Which choice did you choose for 5b 
I’m concerned about my answers now
The bond angle is 109º. The hybridization is sp3
I got 4.76 bc I didn’t round before I subtracted but the answer should be 4.77
Anyone get 1c
@teamsparia11 I got that too, but it was for the pOH, which is incorrect I believe.
pt e and f were definitely on there. the questions just weren’t all on the same page.
if it’s a buffer solution tho, it doesn’t make sense for the pH to go from 9.4ish to 4.76 at half equivalence point… it’s supposed to be relatively unchanged so i changed my answer from 4.76 to 9.23 but idk
also, what was the answer to 5b? I put an increase in bleach concentration
@notreadyforlife I completely checked my frq’s to make sure I wasn’t missing any parts when I received them and those parts were not there. On twitter, people are messaging collegeboard that they are missing those parts as well so if a whole group of people didn’t see those questions, then we all need stronger glasses obviously.
I put an increase in bleach concentration too but I thought I messed that part up
Ok here it goes:
- (a)
(i) 1.35V
(ii) Arrow was like this: ←
(b)
(i) Remains the same
(ii) The overall cell reaction does not lose any mass. If you add up the molar mass of both sides it is equal.(c) (i) Lower (ii) Less O2 means less reaction occurring (d) Ca since it has 2+ charge instead of one 1.0g/ 40g/mol Ca x 2 e per mol = .05mol 1.0g/23g/mol Na = .0435mol .05>.0435mol (e) (i) 1s22s22p63s23p64s23d10 (ii) 4s2 - (a)
(i) PV=nrT
n = PV/rT
n = .822atm(.0854L)/(.0821)(305) = .00280mol / 2 =.0014mol
(ii) .200g / 46.1g/mol C2H5OH = .00434mol
(b) .0014mol / .00434 = 32.3%
© deltaH = 45.5kJ/mol
deltaS = .126kJ/Kmol
deltaG = deltaH - TdeltaS
deltaG = 45.5kJ/mol - 298K(.126kJ/Kmol)
deltaG = 7.952
7.952 = -RT ln K
7.952 = -(.0821)(298)lnK
K = .723
deltaG is positive and K<1 so thermodynamic data for the reaction supports the student’s claim(d) insert lewis dot structure here
(e) 109.5
(f) The IMFs in ethanol consisted of Hydrogen bonds which are much stronger than IMFS of C2H4 (a) C6H7O2- + H+ → HC6H7O2 (b) I suck at this rofl (c) Thymol blue since the approximate pKa range falls within 2.54pH (d) -log(1.7 x 10^-5) = 4.77pH (e) weak acid weak base curve point of inflection at second point (29.95mL, 2.54pH) (f) I suck at this rofl- (a) Ca(OH)2 → Ca2+ + 2OH- (b) nope (c) four molecules with the O part (large circle) touching the Ca2+ molecule
- (a) first order since ln vs t
(b) bleach is present in large excess so that the concentration of OCl− is essentially constant throughout the
reaction
(c) Change wavelength of spectrometer - (a) LiI and NaF: LiI has smaller cations and larger anions than NaF and therefore has a lower melting point (think electronegativity and radius) (b) LiF basic F- + H2O → HF + OH-
- (a) Q=mcdeltaT Q = 1mol(24J/molK)(933-298) = 15240J + 10700J = 25940J = 25.94kJ (b) Since 1675kJ > 25.94kJ, the recycling process requires less energy.
@whatisthepurpose Doesn’t the cell gain mass for 1b since it is taking in oxygen from the surroundings?
I just spouted the fact that cathode gains mass and the anode doesn’t.
I also gave Le Chatelier as my reasoning for the pressure at the mountain top question.
In 2a did you subtract the vapor pressure of the water ?
@Kitsyxoxo Yeah cuz I subtracted vapor pressure of water and got a different answer.
@whatisthepurpose I got 5 a and 5 c the same as you but what exactly is your answer for 5b…?