AAMC 7R: Discuss

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NUKid

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Just took this test today and made 9,8,9. Any thoughts on this test as compared to the real thing (for those who took in April) and AAMC 8? I had trouble finishing all sections (which is usual for me) and got a ton wrong at the end.

I thought the PS section was definately fair and would have liked it more had I finished it on time.

Verbal I did not like as I had crappy timing and was caught up in the beginning.

Bio seemed okay but was rushed on the last passages/stand alones and therefore crapped out at the end.

What do people think about doing stand alones first??


Thanks and good luck to all in these remaining weeks.

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i took it today, and damn that verbal was the toughest ive ever taken. These 10 question passages are messing with my timing pretty bad, i wish kaplan would do the 10 question passages too so i can get it down. I didnt have time for the last passage rushed through it made a 9, but still pretty disappointed im hoping for an 11 on V its suppose to be my strong section lol. I didn't understand the atonality passage at all, and the high low road stunk but not as bad as the atonality. I reread after the test was done, i still have no clue what its trying to say lol. I just suck at music passages.
PS wasnt bad at all, and BS was usual not too hard not too easy.
 
can someone explain #24 seemed like a basic question it is, but i missed it of course, i put 1.91, i thought that E cell=Ecathode+ E anode, shouldn't both E cathode and E anode be set as reduction potentials than add them? I know that Cu2+ is being reduced and H20 is being oxidized. So E cathode has to be a reduction potential and E anode has to be an oxidation potential for the formula to work? I always get confused by this damn equation.
 
Abe said:
can someone explain #24 seemed like a basic question it is, but i missed it of course, i put 1.91, i thought that E cell=Ecathode+ E anode, shouldn't both E cathode and E anode be set as reduction potentials than add them? I know that Cu2+ is being reduced and H20 is being oxidized. So E cathode has to be a reduction potential and E anode has to be an oxidation potential for the formula to work? I always get confused by this damn equation.


what you do is add the two electrode potentials together +.34 is the reduction potential -1.23 is the oxidation potential and the answer is -.89, you neeever multiply the potential by coeffecients, the potential is intrinsic to the species
 
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The correct formula here is Ecell=Ered +E ox. If you are given only reduction potentials, then you need to reverse the sign of one of them (the less positive one) to obtain the oxidation potential for the oxidized species and then add the two. If you are given a reduction and an oxidation potential, then simply add them, but always make sure that the oxidation potential is for the species that is actually oxidized and vice versa. Hope that makes sense.
 
Hey guys, I just got out of taking the test today. I scored a 14P 12V and 11B (sorry to be an ass). I'm really happy with my score but would still like to see some improvement as I did better than I usually do in P but worse in bio. I'm going to have to spend the next week memorizing a lot of junk if this test is indicative of the real thing--I keep missing easy knowledge questions in bio.

Anyways, I don't understand some of the questions. I've looked on the Princeton explanations, and I'm still a bit confused. I'd appreciate any help you can give me.

1) Why do you choose C over D? Both have complete octets, and the triple bonded nitrogen would probably be more stable.

45) Can you explain why choice D is incorrect? "HOCl allows the oxygen transfer to take place more quickly because the hydrogen atom reduces the charge on the oxygen atom, facilitating the breaking of the O-Cl bond." Reducing the attraction between Cl and O seems to be a restatement from the passage.

161) Why is an alcohol more polar than a ketone? I can't think of a resonance structure putting a negative charge on the phenol thing. I really need to review my basic chemistry.

191) I can see C as being the correct answer, but thet term "lipases" implies that more than one enzyme is doing the exciting hydrolysis action. I picked B as I thought that was more reasonable. Should I just have known this due to the immense variety of fats and stuff that we eat?

211) That chick is such a slut.

Abe: I dunno if you still need help, but if you just look at the equations and stuff in #22, the Cu is clearly getting reduced and the water is getting oxidized. The tricky part is that they give you those E0 values, but we're used to getting the the standard reduction potentials, not a mix of the oxidation and reduction ones.

In this particular problem, we're given the proper ones ahead of time. They're trying to trick you into subtracting the 2nd one thinking you gotta reverse the one that's getting oxidized. Also, remember that you never multiply the junk in the redox things. Hopefully this makes sense.
 
dsh said:
1) Why do you choose C over D? Both have complete octets, and the triple bonded nitrogen would probably be more stable.

45) Can you explain why choice D is incorrect? "HOCl allows the oxygen transfer to take place more quickly because the hydrogen atom reduces the charge on the oxygen atom, facilitating the breaking of the O-Cl bond." Reducing the attraction between Cl and O seems to be a restatement from the passage.

161) Why is an alcohol more polar than a ketone? I can't think of a resonance structure putting a negative charge on the phenol thing. I really need to review my basic chemistry.

191) I can see C as being the correct answer, but thet term "lipases" implies that more than one enzyme is doing the exciting hydrolysis action. I picked B as I thought that was more reasonable. Should I just have known this due to the immense variety of fats and stuff that we eat?

211) That chick is such a slut.

1) Actually only one of them has a full octet. The other one has 10 electrons (three from triple bonds, two from hydrogens). Don't worry, I made the same exact careless mistake.

45) (A) was the only answer that made any sense to me. Sorry, can't help.

161) An alcohol can hydrogen bond, whereas a ketone cannot.

For some reason, the next two questions you ask do not seem to correspond with my AAMC7.
 
Mister Pie said:
1) Actually only one of them has a full octet. The other one has 10 electrons (three from triple bonds, two from hydrogens). Don't worry, I made the same exact careless mistake.

161) An alcohol can hydrogen bond, whereas a ketone cannot.

For some reason, the next two questions you ask do not seem to correspond with my AAMC7.

Thanks for the quick response. I hate making dumb mistakes like that first one haha. I still don't get why hydrogen bonding makes you more polar, but I'll go ask a prof or someone at school if no one can explain it better. Sometimes I just don't get things even if they're blatantly obvious, but I don't get how you relatively gauge the charge distributions on a ketone/phenol using some basic rules.

Is it okay to post a full question/answer on the board? It's the 2nd to last problem on the discrete section between passages 8 and 9.

The last problem is the one with the really snanky female with the 20 kids or whatever from 2 men.
 
If you take 7R the week before the MCAT, with as much studying as you're going to have, it's a really good predictor of your real score, IMO. I took the 7R as part of my PR course one week before the real thing, and it predicted my real score exactly (38 both times). Interestingly, the difficulty of the 7R is nothing like the real thing, IMO. Some of the 7R questions are laughably easy (I recall some of them were just definitions...). Still though, it's a good test!
 
dsh said:
Thanks for the quick response. I hate making dumb mistakes like that first one haha. I still don't get why hydrogen bonding makes you more polar, but I'll go ask a prof or someone at school if no one can explain it better. Sometimes I just don't get things even if they're blatantly obvious, but I don't get how you relatively gauge the charge distributions on a ketone/phenol using some basic rules.

Is it okay to post a full question/answer on the board? It's the 2nd to last problem on the discrete section between passages 8 and 9.

The last problem is the one with the really snanky female with the 20 kids or whatever from 2 men.

Hydrogen bonds are more polar because the electronegativity difference between oxygen and hydrogen is greater than the difference between oxygen and carbon. In a sense, you have almost a "bare" proton exposed because the oxygen is able to pull off a substantial amount of the hydrogen's electron density. Since the charge separation is greater, the O-H bond is more polar than a C=O bond, therefore the alcohol is more polar than the ketone. Hope that helped a little!
 
11-11-11

Very happy...i was just under a 12 for BS which i found pretty difficult in comparison to the others.
 
12P 11V 12B

I attribute my good BS score to the fact that they asked the EXACT same question 3 times (what's the IR spectra of an OH group-->3500cm-1).

Do we need to know any other IR Spectra other than OH and Carbonyl group?
 
ctv1116 said:
12P 11V 12B

I attribute my good BS score to the fact that they asked the EXACT same question 3 times (what's the IR spectra of an OH group-->3500cm-1).

Do we need to know any other IR Spectra other than OH and Carbonyl group?


i love IR questions, i find them pretty simple and you are right they asked that question 3 times
 
Anyone know where I can get the solutions to this test. I wrote it as part of a prep course but all I got was the answer grid, not the detailed solutions?
 
god, i am finally feeling burned out.

lol. Felt like breaking the chair in the middle of the passage 9 of BS.

score was abit better than yesterday's 8R but still in that lousy zone where i seemed to be stuck in.

5 wrongs in PS, 10 wrongs in VR, and 3 wrongs in BS.

I am not sure whether it was the my lousy mood or that tacky ***** behind me annoyingly tapping her antiquated laptop in the kaplan center, but BS section felt abit more difficult than 8R, especially the orgo passages.

the quake passage in PS section also threw me off a bit because whenever i dont get the passage i just get confused.

anyhoo, that BS passage concerning IBD was really odd one for me because i am suffering from ulcerative colitis and took my medications right before starting that section.
 
dsh said:
45) Can you explain why choice D is incorrect? "HOCl allows the oxygen transfer to take place more quickly because the hydrogen atom reduces the charge on the oxygen atom, facilitating the breaking of the O-Cl bond." Reducing the attraction between Cl and O seems to be a restatement from the passage.

THe others have been answered. For this, the question begins:

"In addition to the explanation in the passage, the rate..."

I actually didn't notice that till now, but luckily I thought A was more straightforward.

191) I can see C as being the correct answer, but thet term "lipases" implies that more than one enzyme is doing the exciting hydrolysis action. I picked B as I thought that was more reasonable. Should I just have known this due to the immense variety of fats and stuff that we eat?

I dunno dude, it's just interpretation of the question...though I can't see how the question means that there are multiple enzymes rather than a class of enzymes doing the ester hydrolysis.

211) That chick is such a slut.

hahaha, yep.
 
I have a couple questions -

Physics #35: The correct answer is D, which says the transmissions would not interfere if two different frequencies are used. Isn't this wrong? I thought any two transverse waves could interfere, regardless of their frequencies. This is how a beat frequency is created right?

Physics #63: Why would fuel efficiency remain unchanged if you change the frictional coefficients? With more sliding, you must lose energy right?

Thanks!
 
totalcommand said:
I have a couple questions -

Physics #35: The correct answer is D, which says the transmissions would not interfere if two different frequencies are used. Isn't this wrong? I thought any two transverse waves could interfere, regardless of their frequencies. This is how a beat frequency is created right?

Physics #63: Why would fuel efficiency remain unchanged if you change the frictional coefficients? With more sliding, you must lose energy right?

Thanks!


p35) the passage mentions about the bandwidth(30k Hz or something) thus eliminating possiblity of interference, if i recall correctly. Your thinking process about the interference of two transverse wave seems correct. But if the source of a particular EMW is dispersing the wave in a spherical manner(like a shining light bulb), the field density will be decreasing rapidly as it moves away from the source making the inteference pattern negligible. Also we are assuming that the distance between the two sources is far apart unlike most of the interference pattern experiment where the wave sources are close together, and this contributes to the condition where we can ignore the interference pattern. Furthermore, if you look at the answer choices for that question, none of them makes sense except the correct answer key.

p63) again i took this test yesterday in the kaplan center so i dont remember the exact question, but i believe one of the condition was an increase in static frictional coefficient. The wheels provides the thrust for the car because of the static friction and the static friction is the actual force that actually moving the car, and rolling wheel simply provides the new advancing contact sites for static frictional force to work on. This is why if the road has low static frictional coefficient the wheels simply rotates without actually moving the car. The kinetic frictional force plays little role on movement of a car unless the movement of car is not parallel to rotational velocity. It also plays a role when the wheel is not rolling such as when one applys break, for example. So when you consider this simple fact it should increase the fuel efficiency of the car because the fuel is only used to generate the rotational acceleration of the wheels and you get more frictional thrust for the same distance(thus more work done on the car by the friction) the car travels by rolling, if the static frictional coeffienct increases. However, since there is force associated with breaking the temporary bond(electrical) between the tire and the road(which is essentially cause for the static frictional force), more energy will be spent to roll the wheel. Subsequently, the energy expenditure for movement car by rolling mech. will remain more or less(abit on the "more" side) the same, as the static frictional coeffienct increases.
 
10p 10v 11b

those verbal passages were practially all one full page! and like most others I got a lot(4) wrong in the high road/low road passage. hopefully this is my actual score.
 
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