AAMC CBT3 and 3R OFFICIAL Q&A

[Vihsadas]

This is the official Q&A thread for AAMC CBT3 and 3R.

Please post ONLY questions pertaining to AAMC CBT3 and 3R.
Out of respect for people who may not have completed the other exams, do not post questions or material from any other AAMC exam.

Please see this thread for the rules of order before you post.

Good luck on your MCAT!

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[Snowy]

Yeah, I don't get why intensity (rather than wavelength) indicates probability. I don't think it says that in the passage. When I was doing the question originally, I saw that the intensity is lower on the right side than the left, so I figured peak 2 would be less probable because the intensity surrounding it is lower. I just don't get why you choose D over B and C. thanks again.

 

[andafoo]

Yeah, I don't get why intensity (rather than wavelength) indicates probability. I don't think it says that in the passage. When I was doing the question originally, I saw that the intensity is lower on the right side than the left, so I figured peak 2 would be less probable because the intensity surrounding it is lower. I just don't get why you choose D over B and C. thanks again.

K.

Remember that for X-ray spectra there are 2 different features. The part that you may have been looking at is the continuous part.

The continuous portion as mentioned in the beginning of the reading is produced by bremsstrahlung radiation.

The events that are mentioned in the question are different from the continuous nature of bremsstrahlung radiation because it is an actual collision between an X-ray photon and an electron.

So, lets look at each answer choice:

B) P1 b/c peak 1 has lower intensity
although peak 1 having a lower intensity may be true, P1 is not the most probable.

C) P2, b/c the peak has the longer wavelength
again, it is true that P2 has a longer wavelength, but that is not the reason why it is more probable.

Note: The graph is designed specifically for the question. The wavelength is the dependent variable and is located on the x-axis. So if you really have NO idea at all, the graph may hint at intensity being the indicator.

D) The reason is given in the solution.


But going back to why the intensity (rather than wavelength) indicates probability, I guess this requires a bit more thinking.

The intensity is proportional to the # of electrons emitted at a given wavelength. We know that the x-ray will emit at a multitude of wavelengths. So the x-ray photons are at many different wavelengths and have many different energies. Your task is to determine which event has a higher probability. The event the refer to is the collision of the x-ray and the electron. So if more electrons are ejected for P2, then it goes to show that the event occurs more.

 

[Snowy]

OH...your last sentence helps a lot, that makes sense.

 

[andafoo]

Glad it helped.

 

[HippocratesX]

Yeah, I don't get why intensity (rather than wavelength) indicates probability. I don't think it says that in the passage. When I was doing the question originally, I saw that the intensity is lower on the right side than the left, so I figured peak 2 would be less probable because the intensity surrounding it is lower. I just don't get why you choose D over B and C. thanks again.

It says in the passage (last paragraph) that the intensity measures probability. You dont have to understand the actual science of it, ....mostly everything comes from the passage. That passage especially has answers that come directly from what they've told u in the paragraphs.

 

[RySerr21]

question 4:

i dont understand how they determined that PbCO3 is less soluble than PbI2 which is less soluble than PbSO4.

They describe the order that it was precipitated in the reactions completed, which was PbSO4, PBI2 and then PbCO3. So if PbSO4 was precipiated first, wouldnt that mean it was the least soluble??

 

[tncekm]

question 4:

i dont understand how they determined that PbCO3 is less soluble than PbI2 which is less soluble than PbSO4.

They describe the order that it was precipitated in the reactions completed, which was PbSO4, PBI2 and then PbCO3. So if PbSO4 was precipiated first, wouldnt that mean it was the least soluble??

Here is the answer; hope it helps.

Yeah, I originally got that one wrong, too. So, here is my best shot at explaining it.

In short, in the first step Pb(SO4) (Compound A) precipitated after Na2SO4 was added to aqueous Pb(NO3)2. This tells us Pb(SO4) is less soluble then Pb(NO3)2.

The second step involved adding KI and agitating the solution until some Pb(SO4) (Compound A) was dissolved. The result was that PbI precipitated, so PbI2 is less soluble than Pb(SO4).

In the third step PbI2 was then mixed with Na2CO3 and all of the PbI2 was converted to PbCO3, which precipitated out of solution. So, This tells us that PbCO3 is less soluble than PbI2.

Given that, if we now know the following in terms of solubility:

PbCO3 < PbI2 < Pb(SO4) < Pb(NO3)2

So, on to the answers:

(a) SO4 then I --- No, from our trend above you can see that PbI is less soluble than Pb(SO4), so I would precipitate as PbI2 BEFORE SO4 would precipitated as PbSO4
(b) CO3 then I -- Correct order
(c) SO4 then CO3 -- No, see explanation to (a)
(d) I then CO3 -- No, see explanation to (a)

 

[RySerr21]

Here is the answer; hope it helps.

thanks! that makes sense! sorry if that question was alreayd asked earlier in this thread,i was too lazy to search, haha.

 

[tncekm]

thanks! that makes sense! sorry if that question was alreayd asked earlier in this thread,i was too lazy to search, haha.

No prob, glad it helped.

 

[Lizzie Bartlet]

This might be a really basic question, but can someone explain number 48 on 3R to me? I got it right only because I knew you'd get ZnCl2 but how are you supposed ot know from the problem?

 

[Lizzie Bartlet]

Anyone?

 

[Kaustikos]

This might be a really basic question, but can someone explain number 48 on 3R to me? I got it right only because I knew you'd get ZnCl2 but how are you supposed ot know from the problem?

The idea I used was process of elimination. Cross out A and C. A because ZnCl2 is formed, not ZnCl. C because Zn is definitely more reactive than H2 (g). Now it's between B and D. B worked for me because of your mentioned response. I knew a reaction would occur between HCl and Zn, which eliminated C and D from the getgo. The negative E and the solution involved definitely points to a reaction. The Zn has a high potential of oxidizing and becoming an ion and would react with the Chlorine atoms to produce ZnCl2. H+ is less reactive than Zn+2.

 

[Revilla]

I used process of elimination as well and got it right. If anyone can explain it, I'd be interested.

 

[wooki]

for passage 7..can anyone show me the dehydration process?
..........................O........OH
..........................||....... |
dehydration of CH3-C-CH2-C(CH3)2

 

[tncekm]

If its a simple dehydration, then likely heat + a little acid will cause that alcohol to pop off, leaving a tertiary carbocation. The hydrogen next to the tertiary carbocation will lose its electrons to the carbocation and it will pop off as a proton, get picked up by OH- and become water.

......O........OH + H+ <-- protonation
......||....... |
CH3-C-CH2-C(CH3)2

......O...H...
......||...|... +
CH3-C-CH-C(CH3)2 + HOH

......O.......
......||.......
CH3-C-CH=C(CH3)2 + H2O

 

[Kaustikos]

If its a simple dehydration, then likely heat + a little acid will cause that alcohol to pop off, leaving a tertiary carbocation. The hydrogen next to the tertiary carbocation will lose its electrons to the carbocation and it will pop off as a proton, get picked up by OH- and become water.

......O........OH
......||....... |
CH3-C-CH2-C(CH3)2

......O...H...
......||...|... +
CH3-C-CH-C(CH3)2 + -OH

......O.......
......||.......
CH3-C-CH=C(CH3)2 + H2O

Hey, LOOK! Aldol condensation!

- edit -

Actually, I'm pretty sure the acid protonates the OH group and then the OH2 displaces from the compound and leaves the cation. The compound then loses H+ and forms a double bond, the H+ goes into solution giving back the proton to the acidic solution. But this could be a difference in text?


double edit - seems there isn't any real clarity in how the proton is lost. Some say a water molecule comes and steals the proton from the compound, forming the double bond?

But basically, the OH is protonated by the acidic solution and then the H is lost which forms the double bond.

 

[tncekm]

Yeah, it would be protonated. that's generally the purpose of the acid catalyst, to protonate an alcohol and make a better leaving group. I guess I should have expanded on that rather than saying "heat + a little acid will cause the alcohol to pop off" Edit: looking at my "drawing" yestarday, looks like it totally skipped my mind.... -OH leaving? Meh, w/e, you cleared it up!

And, I'm not "quite" sure that it would be called an aldol condensation unless the condensation was preceded by an aldol addition, but that's definitely what it would look like if it were preceded by an aldol addition!

 

[Kaustikos]

Yeah, it would be protonated. that's generally the purpose of the acid catalyst, to protonate an alcohol and make a better leaving group. I guess I should have expanded on that rather than saying "heat + a little acid will cause the alcohol to pop off" Edit: looking at my "drawing" yestarday, looks like it totally skipped my mind.... -OH leaving? Meh, w/e, you cleared it up!

And, I'm not "quite" sure that it would be called an aldol condensation unless the condensation was preceded by an aldol addition, but that's definitely what it would look like if it were preceded by an aldol addition!

No, you're right. But that reaction is basically the end-point of an aldol condensation and I thought I'd mention it.

 

[tco]

K.

Remember that for X-ray spectra there are 2 different features. The part that you may have been looking at is the continuous part.

The continuous portion as mentioned in the beginning of the reading is produced by bremsstrahlung radiation.

The events that are mentioned in the question are different from the continuous nature of bremsstrahlung radiation because it is an actual collision between an X-ray photon and an electron.

So, lets look at each answer choice:

B) P1 b/c peak 1 has lower intensity
although peak 1 having a lower intensity may be true, P1 is not the most probable.

C) P2, b/c the peak has the longer wavelength
again, it is true that P2 has a longer wavelength, but that is not the reason why it is more probable.

Note: The graph is designed specifically for the question. The wavelength is the dependent variable and is located on the x-axis. So if you really have NO idea at all, the graph may hint at intensity being the indicator.

D) The reason is given in the solution.


But going back to why the intensity (rather than wavelength) indicates probability, I guess this requires a bit more thinking.

The intensity is proportional to the # of electrons emitted at a given wavelength. We know that the x-ray will emit at a multitude of wavelengths. So the x-ray photons are at many different wavelengths and have many different energies. Your task is to determine which event has a higher probability. The event the refer to is the collision of the x-ray and the electron. So if more electrons are ejected for P2, then it goes to show that the event occurs more.

What subject is this passage concerning? (meaning for example, electrochemistry, optics, etc)

 

[dofith]

It's a verbal question:

Could anybody tell me why the third choice is wrong?

Thanks a lot!

 

[MichaelRowle]

I am confused as to how PbCO3 is less soluble because it precipitates last, my intuition would seem to think that things that would precipitate last would be more soluble because they would be so dissolved that they would want to come out of solution last. Any clarification would be appreciated. Thanks.

 

[tncekm]

I am confused as to how PbCO3 is less soluble because it precipitates last, my intuition would seem to think that things that would precipitate last would be more soluble because they would be so dissolved that they would want to come out of solution last. Any clarification would be appreciated. Thanks.

Here is the answer. Hope that helps.

Yeah, I originally got that one wrong, too. So, here is my best shot at explaining it.

In short, in the first step Pb(SO4) (Compound A) precipitated after Na2SO4 was added to aqueous Pb(NO3)2. This tells us Pb(SO4) is less soluble then Pb(NO3)2.

The second step involved adding KI and agitating the solution until some Pb(SO4) (Compound A) was dissolved. The result was that PbI precipitated, so PbI2 is less soluble than Pb(SO4).

In the third step PbI2 was then mixed with Na2CO3 and all of the PbI2 was converted to PbCO3, which precipitated out of solution. So, This tells us that PbCO3 is less soluble than PbI2.

Given that, if we now know the following in terms of solubility:

PbCO3 < PbI2 < Pb(SO4) < Pb(NO3)2

So, on to the answers:

(a) SO4 then I --- No, from our trend above you can see that PbI is less soluble than Pb(SO4), so I would precipitate as PbI2 BEFORE SO4 would precipitated as PbSO4
(b) CO3 then I -- Correct order
(c) SO4 then CO3 -- No, see explanation to (a)
(d) I then CO3 -- No, see explanation to (a)

 

[MichaelRowle]

Thanks. I think I understand how to get the answer but not the complete concept. Though in working towards it I gain some better understanding in general of these concepts so working it out is good for me and thanks for your help. It is still hard for me to understand that the relationship between precipitation and solubility is that things that precipitate first are the more soluble compound though it would seem that the less soluble compound would precipitate first due to it's insolubility (I am picturing it as a chunk or separate layer). I guess it is what it is. And that means that when you go in and put the Pb2+ ion in later to precipitate the anion the reverse is true, the less soluble is precipitated first.

Let me go on, I don't get where this leap comes from here:

The order in which the anions precipitate Pb2+ is: CO32- then I- then SO42-.
The reactions described in the passage show that lead(II) is successively precipitated as PbSO4, PbI2, and PbCO3. This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is less soluble than PbI2, and PbI2 is less soluble than PbSO4. The order in which the anions precipitate Pb2+ is: CO32- then I- then SO42-. When this sequence is applied to the question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.

 

[MichaelRowle]

Item 41

Solution

Mark An object is placed upright on the axis of a thin convex lens at a distance of four focal lengths (4 f) from the center of the lens. An inverted image appears at a distance of 4/3 f on the other side of the lens. What is the ratio of the height of the image to the height of the object? A
) 1/3
The ratio of object to image distance equals the ratio of object to image height. The ratio of image to object height is found by rearranging the ratios to give 4f /(4/3)f = 1/3. The image is demagnified by a factor of 3. Thus, answer choice A is the best answer.


Anyone have a link to problems like this or explanations? Thanks.

 

[tncekm]

Magnification = -(i/o) = -(4f/3)/(-4f) = 1/3

In other words, the magnification of the image is always the ratio of the image to the object.

 

[MichaelRowle]

Which of the following species has the smallest concentration in 98% H2SO4?

I don't understand how they got the answer to this, even with the lengthy explanation?

 

[Chemist0157]

Which of the following species has the smallest concentration in 98% H2SO4?

I don't understand how they got the answer to this, even with the lengthy explanation?

I had missed this one too. Basically, there's not enough water for the acid to completely ionize. It's kind of a process of elimination. The first Ka is greater than the second Ka so we know there is more HSO4 than SO4. And since, like I said before, ionization is hindered, there must be an unusual amount of H2SO4 in the solution. That makes SO4 have the lowest concentration.

To sum it up, the lack of water hinders ionization. Since SO4 is the product of maximum ionization, there should be very little of it in solution.

 

[unsung]

Dang, I don't know what happened. My BS fell to a 9 but my verbal jumped 2 pts from last test to a 12?? Wth.

Did anyone find the bio to be really hard on this one? I guess everyone says #3 is "easy" :-/

Also, what does an "ebulliator" however you spell that do in the passage about the vacuum distillation? Can someone explain that? I couldn't figure out how the setup was supposed to work.

 

[unsung]

It's a verbal question:

Could anybody tell me why the third choice is wrong?

Thanks a lot!

How I understood this question is, people might respond to force, but whether or not they respond to force does NOT impact the author's argument that "people bind themselves more through ceremony" than through force. The key word here is *more*. The author's argument is that A has more of an effect on people than B. The fact that people also respond to B has no relationship to the original statement. So even if it were true, it does not weaken the author's argument.

 

[unsung]

Okay, can someone explain passage 4 #73?

Solution

According to the passage, magnetic fields are primarily by-products of:
A) fluid motion.
B) the rotation of the earth.
C) mechanical energy.
D) electrical currents.

The passage says:

"The only theory that can explain the persistence of the field and its propensity for reversing itself is the dynamo theory, which holds that the magnetic field results from fluid motions in the outer core. What sustains the motions, ultimately, is heat loss from the core."

But then the answer is supposed to be D. ??

 

[travelbug73]

Okay, can someone explain passage 4 #73?


According to the passage, magnetic fields are primarily by-products of:
A) fluid motion.
B) the rotation of the earth.
C) mechanical energy.
D) electrical currents.

The passage says:

"The only theory that can explain the persistence of the field and its propensity for reversing itself is the dynamo theory, which holds that the magnetic field results from fluid motions in the outer core. What sustains the motions, ultimately, is heat loss from the core."

But then the answer is supposed to be D. ??

Look at the beginning of paragraph 4, at least in the CBT version:

"The existence of the earth's magnetic field is the clearest evidence that the core is cooling. The earth's magnetic field, like any other, is a by-product of electric currents"

 

[unsung]

Look at the beginning of paragraph 4, at least in the CBT version:

"The existence of the earth's magnetic field is the clearest evidence that the core is cooling. The earth's magnetic field, like any other, is a by-product of electric currents"

Okay, I see that now. Thanks! Still seems kind of screwy, as the passage says at another pt that "fluid motions cause the magnetic field", so I'm not sure what that's about. But anyway, I get it now.

Can anyone explain #115 on the PS?

"Assuming defective Hormone X response is X-linked, which children of an affected male will most likely be unable to respond to Hormone X?"

A) All of the males and all of the females
B) Half of the males and half of the females
C) None of the males and half of the females
D) None of the males and none of the females

I get that for an x-linked disease, none of the male children would have the disease. What I don't get is how we're supposed to know that X^dX females are unaffected. Is that a characteristic of x-linked diseases that we can assume those females are never going to exhibit the disease?

Thanks!!

 

[bluemonkey]

Okay, I see that now. Thanks! Still seems kind of screwy, as the passage says at another pt that "fluid motions cause the magnetic field", so I'm not sure what that's about. But anyway, I get it now.

Can anyone explain #115 on the PS?

"Assuming defective Hormone X response is X-linked, which children of an affected male will most likely be unable to respond to Hormone X?"

A) All of the males and all of the females
B) Half of the males and half of the females
C) None of the males and half of the females
D) None of the males and none of the females

I get that for an x-linked disease, none of the male children would have the disease. What I don't get is how we're supposed to know that X^dX females are unaffected. Is that a characteristic of x-linked diseases that we can assume those females are never going to exhibit the disease?

Thanks!!

This problem is assuming that it is not only x-linked, but recessive as well. If a woman is heterozygous, XrX, and the trait is recessive, she will not express the trait, but rather will be a carrier. If, on the other hand, the trait were dominant, women would express it if they have one or two copies of the allele.

Is there a passage that came with this question? You should be able to figure out whether or not this is a recessive trait.

 

[travelbug73]

Okay, I see that now. Thanks! Still seems kind of screwy, as the passage says at another pt that "fluid motions cause the magnetic field", so I'm not sure what that's about. But anyway, I get it now.

Can anyone explain #115 on the PS?

"Assuming defective Hormone X response is X-linked, which children of an affected male will most likely be unable to respond to Hormone X?"

A) All of the males and all of the females
B) Half of the males and half of the females
C) None of the males and half of the females
D) None of the males and none of the females

I get that for an x-linked disease, none of the male children would have the disease. What I don't get is how we're supposed to know that X^dX females are unaffected. Is that a characteristic of x-linked diseases that we can assume those females are never going to exhibit the disease?

Thanks!!

You will have to figure out if the disease is recessive or dominant. Since thousands of families were screened and only two affected families were identified coupled with the fact that only the males were affected, you have to assume that it is a recessive condition.
You have to be very careful when you say that none of the male children will be affected for an X-linked disease. That is true only when it is an X-linked recessive disease. You will find, albeit rarely, X-linked dominant disorders also eg. Rett's syndrome. In these cases, females are affected more than males given that females have an extra X-chromosome (dosage effect presumably).

 

[Hopefullpremed]

Why is there no page for AAMC Test 1. I just took the test which I purchased a while back but there are no explanations for the test. AAMC website does not offer the test either.

 

[nfg05]

Nvm, answered my own question. Mod, please delete if possible.

 

[HippieChick]

I don't understand why the components are squared. Can someone please explain?

 

[spyderracing32]

I don't understand why the components are squared. Can someone please explain?

It's the Pythagorean theorem. You need to counteract the 3N force pointing down and the 4N force pointing to the left. They tell you arctan(.75) = theta which really doesn't matter because tan being O/A will be .75 if you simply cancel the force vectors. So you break the force vector into component vectors:
4N to the right
3N up
The resultant force is the sum of these two vectors: 4^2+3^2 = 25, take the square root (a^2 + b^2 = c^2), and your answer is 5 (you could also just know the 3,4,5 triangle relation too).

 

[landerman]

can someone please explain how i figure out the answer from the passage/graph?

How is the period of the waveform of 1c the same as the first harmonic?

thank you!

 

[engineeredout]

can someone please explain how i figure out the answer from the passage/graph?

How is the period of the waveform of 1c the same as the first harmonic?

thank you!

Intersection of the X-axis occurs at the same places and the same times as the first harmonic. Even if its not shaped like a traditional sine wave, this still holds true.

 

[landerman]

aha! thank you!

 

[mastamind518]

for 145 on the cbt
Ignoring stereochemistry, how many different tripeptides may exist that contain the same three amino acids as the molecule shown below?

​

shouldnt the answer be 3 since ABC/CBA ; BAC/CAB ; ACB/BCA are the same thing if u ignore the stereochemistry?

 

[spyderracing32]

for 145 on the cbt
Ignoring stereochemistry, how many different tripeptides may exist that contain the same three amino acids as the molecule shown below?

​

shouldnt the answer be 3 since ABC/CBA ; BAC/CAB ; ACB/BCA are the same thing if u ignore the stereochemistry?

I'm curious about this one as well because I thought the same thing.

 

[Lukkie]

for 145 on the cbt
Ignoring stereochemistry, how many different tripeptides may exist that contain the same three amino acids as the molecule shown below?

​

shouldnt the answer be 3 since ABC/CBA ; BAC/CAB ; ACB/BCA are the same thing if u ignore the stereochemistry?

no they're not. in ACB, the N terminus is on A, whereas in BCA the N terminus is on B. Vice versa for C terminus.

 

[mastamind518]

makes sense, thanks alot

 

[Deepa100]

Yeah, I originally got that one wrong, too. So, here is my best shot at explaining it.

In short, in the first step Pb(SO4) (Compound A) precipitated after Na2SO4 was added to aqueous Pb(NO3)2. This tells us Pb(SO4) is less soluble then Pb(NO3)2.

The second step involved adding KI and agitating the solution until some Pb(SO4) (Compound A) was dissolved. The result was that PbI precipitated, so PbI2 is less soluble than Pb(SO4).

In the third step PbI2 was then mixed with Na2CO3 and all of the PbI2 was converted to PbCO3, which precipitated out of solution. So, This tells us that PbCO3 is less soluble than PbI2.

Given that, if we now know the following in terms of solubility:

PbCO3 < PbI2 < Pb(SO4) < Pb(NO3)2

So, on to the answers:

(a) SO4 then I --- No, from our trend above you can see that PbI is less soluble than Pb(SO4), so I would precipitate as PbI2 BEFORE SO4 would precipitated as PbSO4
(b) CO3 then I -- Correct order
(c) SO4 then CO3 -- No, see explanation to (a)
(d) I then CO3 -- No, see explanation to (a)

I don't follow this explanation. If a precipitate is formed, isn't it considered to be insoluble? So, what determines various degrees of insolubility?

 

[junkct]

Are you serious?!?! I just took AAMC 3 and the curve is INSANE! This is the first AAMC i've taken (I've taken Kaplan FL 1-5 already... scored 40+ on FL5) and I got a 12/10/12 on this. Really happy about the VR, but my raw scores were 46/52 on both sciences, which would easily have been 14 or 15 on Kaplan (I've been scoring 13-14 consistently on kaplan sciences). but this curve is scaring the crap out of me! I mean, there's NO leeway for ANY error. Is this really like the curve on the actual MCAT? Everyone says that after they've taken the real thing (and before they know their scores), they all think they got dominated but then do pretty well (which is the reasoning behind why you shouldn't void if you feel like crap after the exam). After taking AAMC 3 I felt like I had demolished it... thought I would get AT LEAST 13's on the sciences. is the curve more lenient on real thing?

sorry, don't want to sound like a troll or anything, but I'm a little shocked about the fact that a few small errors led to a large reduction in points.... 50/52 is a 13 on PS (wtf??)

 

[LaCasta]

Did anyone else find the verbal for AAMC 3 pretty difficult? I got an 11 on the Kaplan diag. verbal, and I got a 10 on this one. I know its only one point, but I missed only 4 out of 32 on the diag. as opposed to 11 out of 40 on this aamc one. Is the real thing more like AAMC 3?

 

[Ibn Rushd]

For question #22 on the CBT (first one on the passage about waves, harmonics, etc) ... how am I supposed to know which of three superimposed waves in Figure 1a is first harmonic, which is second harmonic, or third harmonic? Is there a way to read a graph that has superimposed waves? A convention I just don't know about? I understand the question otherwise. I missed it because I thought the "first harmonic" was the "third harmonic."

Any help is appreciated.

 

[thatscorrect7]

i think that b/c the first harmonic had the biggest wave length and lowest frequency. the third harmonic by the fn= nv/2L has the higher frequency.