AAMC CBT8 and 8R OFFICIAL Q&A

[Vihsadas]

This is the official Q&A thread for AAMC CBT8 and 8R.

Please post ONLY questions pertaining to AAMC CBT8 and 8R.
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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[G1SG2]

If the valve is opened to drain the tank, where is the speed of the flowing water the greatest?

A) At the top of the tank
B) At the bottom of the tank
C) At the wide end of the pipe
D) At the narrow end of the pipe

I picked B (v=sqrt2gh) but the answer is D. I understand the reasoning (flow rate Q=AV) but why can't it be B?

 

[G1SG2]

Bump, anyone?

 

[Greenbayslacker]

.

 

[wanderer]

Bump, anyone?

Post a sketch or picture of the problem.

 

[G1SG2]

Post a sketch or picture of the problem.

Hey, it was just a tank full of water with a hole at the bottom right side.

 

[supras2kracer]

Well, it's not stated (at least I dont think it is) but according to the EK book we should assume ideal flow. That means Q will be constant. In that case the area at the bottom of the tank will be much greater than in the little pipe so velocity would be smaller compared to in the little pipe.

The velocity is inversely proportional to the area. Greater area, lower velocity.

If the valve is opened to drain the tank, where is the speed of the flowing water the greatest?

A) At the top of the tank
B) At the bottom of the tank
C) At the wide end of the pipe
D) At the narrow end of the pipe

I picked B (v=sqrt2gh) but the answer is D. I understand the reasoning (flow rate Q=AV) but why can't it be B?

 

[kingko]

Doesn't the pancreas also produce lipase which is an emulsifier????



The pancreas produces which of the following substances for the digestive system? A
) Bile salts B
) Emulsifier C
) Gastric juices D
) Proteolytic enzymes
The pancreas produces several proteolytic enzymes, which are released into the small intestine where they are converted to their active forms of trypsin, chymotrypsin, and carboxypeptidase. Thus, D is the best answer.

 

[unique135]

Doesn't the pancreas also produce lipase which is an emulsifier????



The pancreas produces which of the following substances for the digestive system? A
) Bile salts B
) Emulsifier C
) Gastric juices D
) Proteolytic enzymes
The pancreas produces several proteolytic enzymes, which are released into the small intestine where they are converted to their active forms of trypsin, chymotrypsin, and carboxypeptidase. Thus, D is the best answer.

A) Bile salts emulsify fats - produced by liver
B) Emulsifier ...Bile salts do this work
C) Gastric juices...stomach
D) Proteolytic enzymes....such trypsin,chymotrypsin and other enzymes such as lipase from pancreas

 

[traitorman]

this test owned me. im scared

 

[UCB05]

Doesn't the pancreas also produce lipase which is an emulsifier????



The pancreas produces which of the following substances for the digestive system? A
) Bile salts B
) Emulsifier C
) Gastric juices D
) Proteolytic enzymes
The pancreas produces several proteolytic enzymes, which are released into the small intestine where they are converted to their active forms of trypsin, chymotrypsin, and carboxypeptidase. Thus, D is the best answer.

Lipase is not an emulsifier. Emulsifiers only solubilize the fats. Lipase chemically degrades them.

 

[kingko]

my mistake.. good thing i have that cleared up.

 

[joshto]

A) Bile salts emulsify fats - produced by gall bladder
B) Emulsifier ...Bile salts do this work
C) Gastric juices...stomach
D) Proteolytic enzymes....such trypsin,chymotrypsin and other enzymes such as lipase from pancreas

Just to correct the above:

**bile salts are produced in the liver and stored in the gall bladder (they are not produced by gall bladder)

 

[befuddle]

Quote:
Originally Posted by phEight
#102 in Bio..

Q: Which semicarbazone is the product of thermodynamic control?

A: 2-Furaldeyhde's because it is produced under equilibrium conditions and is more stable than the other product.

I understand the reasoning of the answer, saying that the higher the melting point, the greater the stability. What I don't understand is what the "under equilibrium conditions" has anything to do with anything. Not really seeing the connection with equilibrium conditions vs. non equilibrium conditions I guess..

i think the keypoint is that it's "more stable than the other product"

melting point is a function of intermolecular bonding, not intra...so why would higher melting point indicate a more stable product?

 

[GASPER20]

If the valve is opened to drain the tank, where is the speed of the flowing water the greatest?

A) At the top of the tank
B) At the bottom of the tank
C) At the wide end of the pipe
D) At the narrow end of the pipe

I picked B (v=sqrt2gh) but the answer is D. I understand the reasoning (flow rate Q=AV) but why can't it be B?

i also went by your logic but chose A because height would be greatest at the top of the tank no? does A1V1=A2V2 trump V=(2gh)^1/2?

 

[puravida85]

Item 19)

For some odd reason I chose A

A) Final velocity of the sphere is smaller.

I was thinking of the work function. W = F * d. If we increased the distance, the force would decrease. Because force decreases the acceleration also decreases making final velocity smaller. Also, down an incline an object accelerates at a fraction of gravity due to mgsintheta. Therefore if acceleration is decreased, I figure final velocity is decreased.

But what I maybe didn't think about is that time increases, therefore the force has greater time to act on the object making it reach the same final velocity but just in a greater amount of time?

Is it reasonably to set up potential energy = kinetic energy and solving for v= 2gh ^1/2 ? Thus, final velocity is only dependent on height and gravity? Any inputs appreciated. Thanks

 

[luctoretemergo]

On question 49, wouldn't alpha be greater than theta if the refractice index is greater inside the medium, thus it bends away from the normal? whereas theta coming in would bend towards the normal thus theta would be smaller?

 

[surag]

Item 19)

For some odd reason I chose A

A) Final velocity of the sphere is smaller.

I was thinking of the work function. W = F * d. If we increased the distance, the force would decrease. Because force decreases the acceleration also decreases making final velocity smaller. Also, down an incline an object accelerates at a fraction of gravity due to mgsintheta. Therefore if acceleration is decreased, I figure final velocity is decreased.

But what I maybe didn't think about is that time increases, therefore the force has greater time to act on the object making it reach the same final velocity but just in a greater amount of time?

Is it reasonably to set up potential energy = kinetic energy and solving for v= 2gh ^1/2 ? Thus, final velocity is only dependent on height and gravity? Any inputs appreciated. Thanks

you are making it too complicated. Think about the falling object in terms of how much kinetic energy represents the object at any position along its path. I use kinetic energy because this is a way to measure the velocity of the object (at any point)-whereas work and potential energy cannot measure the speed.

We know that both objects dropped from same height(and assuming an frictionless plane) will reach the bottom at the same time. however, as you said the object will be doing more work along the plane(which really just translates to potential energy-again remember, the potential energy of the object along the plane is still mgh-it is independent of the plane itself!!!-if you dont beleive me reread your physics text)

now, as I said before, since the object on the plane has more potential energy (the path is longer and thus more work which in this case equals potential energy) then it has less time to convert itself to kinetic energy-thus by the time it gets to the bottom it has converted less of its energy to kinetic energy which reperesents speed-since a lower amount of speed is present in the ball moving through the plane than being just dropped straight down the object takes longer (even though the speed at the bottom for both is same).

 

[sangria1986]

yea can someone explain BS #106? I don't get it at all:

Assuming Hypothesis B to be correct, which of the following endocrine disorders would cause hypertension that could NOT be rectified by physiologically normal kidneys?

A.) An excess of aldosterone
B.) An excess of glucagon
C.) A shortage of thyroxine
D.) A shortage of insulin

What's the thought process for this one

I know this was a year ago but I'll try. Both d and b could technically cause hypertension because of increased solute-glucose-In blood. However since the kidney functions it will remove these. On the other hand with excess aldosterone even if kidney works right too much aldosterone causes an increase in blood solute by removing said solute from kidney-which increases bp. Thus the functionality of the kidney does not matter. Aldosterone is messing it up. Hence a

 

[FarMD2012]

VR CBT Q67, the question about environmentalists advise congress to reduce the problem of red tide. Answer was D "plant nutrients be removed from wastewater before it is released into waterways."

I put B to drive marine life away from affected areas, answer explains its wrong because it is inpracticable, but my reasoning was I eliminated the other choices. Which environmentalist would actually support releasing waste water into the coastal waters regardless if there is nutrients in it or not?

 

[Bali]

Can anyone explain #38 on the PS to me?

If red litmus paper is dipped into the Na2CO3 solution, it will:
A) remain red, because carbonate is an acidic salt.
B) remain red, because sodium carbonate is neutral.
C) turn blue, because carbonate reacts with water to produce OH–.

In water, carbonate will undergo the following reaction: CO32–(aq) + H2O(l) ® HCO3–(aq) + OH–(aq). Red litmus paper will turn blue in a base. Thus, C is the best answer.

D) turn blue, because sodium ions form sodium hydroxide in water.

I picked D... why is this not the case?

d) is technically true, because this will happen. but c) has to happen first before d) happens....you have to make the -OH before Na can form NaOH.

so while both occur, the REASON the paper turns blue is because the solution turns basic, which happens because carbonate reacts with water to make -OH

 

[conebone69]

correct symbols for #49 are there (with thetas) in the pdf version.

 

[Handinhand]

Just took this test. 13,10,14(48/52, 30/40, 49/52). God damn verbal! I was shocked when I finished, I felt like I was getting owned by this test from start to finish.

Anybody from years past who has already taken the MCAT want to weigh in on the the compared difficulty of AAMC 8 compared to the real thing? Or better yet, anybody who has very recently taken the MCAT? I take it on 8/24, and have yet to do AAMC9 and 10, but I'm getting damn nervous!

 

[Rabolisk]

All the AAMC tests were easier than the real MCAT, especially the PS section. VR was more or less the same. BS was heavily skewed towards genetics.

 

[phltz]

Just took this test. 13,10,14(48/52, 30/40, 49/52). God damn verbal! I was shocked when I finished, I felt like I was getting owned by this test from start to finish.

Anybody from years past who has already taken the MCAT want to weigh in on the the compared difficulty of AAMC 8 compared to the real thing? Or better yet, anybody who has very recently taken the MCAT? I take it on 8/24, and have yet to do AAMC9 and 10, but I'm getting damn nervous!

I took my MCAT 5/27 this year. My actual MCAT score was +1 better in PS and BS as compared to AAMC8, and -1 worse in VR. So for me at least, it was pretty comparable.

 

[Stephen2009]

BS # 102
Passage 2: Semicarbazide and SemiCarbazones:


Which semicarbozone is the product of thermodynamic control?


Answer: 2-furaldehyde b/c it is produced under equilibrium conditions and is more stable than the other product

Solution says: The melting point data in Table 1 indicate that the therm. controlled product is the semicarbazone of 2-furaldehyde. The thermodynamically controlled product is the one that is formed under equilibrium conditions and is more stable.


OK. First off, how in the world do we make sense of the melting point data in relevance to this question?

AND, how do we know that it was produced under equilibrium conditions?


I don't fully comprehend the experiment that was performed; googling didn't help either. so as an extra, if anyone can explain it that would be great, thanks!



I have been on this question for 30+ minutes alone so please help with as much info as possible... thanks!

 

[Stephen2009]

1. BS Passage 3: Question 108 : If restriction of blood flow to the kidneys (by placing clamps on the renal arteries ) resulted in an immediate but small increase in blood pressure, followed by the gradual development of severe hypertension, which hypothesis would these results best support?


Answer: Hypothesis B, b/c the kidneys were responding to decreased glomerular blood pressure


So the reduced flow of blood through the renal arteries due to the clamps would cause a decrease in glomerular blood pressure.

So let me get this straight: "clamping" the renal arteries means just clamping it enough that blood still comes out of the artery?
I mean I interpreted clamping as COMPLETELY STOPPING FLOW at the point of clamping so 0% of blood passes through.

Per the exact wording of the AAMC question above, why am I wrong? Do we have to ASSUME that clamping only stops some or most of the blood from passing through and NOT ALL of the blood?

 

[Handinhand]

For 102, if I remember correctly, the correct answer was really the only one that took thermodynamic properties into account. You should know that equilibrium is a thermodynamic property of the MCAT. Stability can also be interpreted as how badly something would want to form this product from reactants, also known as G (spontaneity), another thermo property.

As for 108, I thought the same thing as you. I assumed clamping caused a complete occlusion and that it was just a poorly worded question.

 

[Aiu]

1. BS Passage 3: Question 108 : If restriction of blood flow to the kidneys (by placing clamps on the renal arteries ) resulted in an immediate but small increase in blood pressure, followed by the gradual development of severe hypertension, which hypothesis would these results best support?


Answer: Hypothesis B, b/c the kidneys were responding to decreased glomerular blood pressure


So the reduced flow of blood through the renal arteries due to the clamps would cause a decrease in glomerular blood pressure.

So let me get this straight: "clamping" the renal arteries means just clamping it enough that blood still comes out of the artery?
I mean I interpreted clamping as COMPLETELY STOPPING FLOW at the point of clamping so 0% of blood passes through.

Per the exact wording of the AAMC question above, why am I wrong? Do we have to ASSUME that clamping only stops some or most of the blood from passing through and NOT ALL of the blood?

Does the degree of clamping matter? Any decrease in blood flow is detected by the kidney?

 

[rls303]

Hi,
I am wondering why exactly for the hyocalcemia question it would increase the excitability? How does this affect the membrane potential, wouldn't there now be a gradient for Calcium ions to leave the cell? Thus becoming more negative in the cell?

 

[IamSuperDoctor]

the metathesis question was bs, i didnt even know what that was. at least, it isnt in my princeton review book.

 

[fizzgig]

i'mma get kicked out of this coffee shop, but the carbymazide whatever thermo/kinetics question:

my understanding is this. you have mp of hexothing and furaldehyde.
in COLD conditions, where you dont have lots of energy, the mp of the result is close to hexothing. so the activation E for hexothing is low and that product dominated.
in HOT conditions, where you have energy to spare, the mp of the result is close to furaldehyde. so given plenty of E the thermo product is furaldehyde.

this is confirmed in the second exp, where they did something i didn't understand but the point was they heated samples containing each of these things back up to the HOT temp, then let them cool. in this case, you'll notice the mp for both D and E end up close to the thermo product, melting point of furaldehyde again, so it confirms this is the thermo product.

 

[IamSuperDoctor]

yea fizz that passage was messed up, but the following questions were easy - the ones after the kinetic/thermo control. so im expecting something similar to this on the mcat, a messed up passage but easy/doable questions.

 

[fizzgig]

PS Question 16:

rxn is I2 +2S2O3^-2 -> 2I^-1 + S4O6^-2
starch is an indicator that reacts with excess I2 once the other reactant is used up. it is a fast reaction so we know iodine isn't removed til you run out of S2O3 i'm assuming.

question asks for the graph of moles of the product S4O6^-2 over time.
correct graph rises steadily then flatlines.

this answer would make sense to me if the reaction were isolated, but as starch removes iodine by binding with it, wouldn't the reaction move left for a while, resulting in a decrease in the S4O6?

i picked a graph that went up then back down...

their explanation says 'the data show the reaction is complete in 19s' so i should know it stops moving after that? thanks all.

 

[IamSuperDoctor]

i think that, because its a fast step the left side reacts fast and violently to produce the products, which are heavily favored. You have to consider that reaction as irreversible. My browser doesnt show the arrow, but i think its in one direction only --->. Although the starch may, in reality, decrease the amount of s4o6, that amount would be very small and completely negligible.

 

[fizzgig]

fair enough... i thought about the 'fast step' idea but then i said no wait, trickery, fast doesn't say anything about WHERE eqlb is, just that it gets there quickly... but if the arrow is one way then that does imply products are heavily favored so that argument could make sense...

and yeah the symbols that show up in my browser for some stuff is nuts... this reaction arrow was the R with a circle around it symbol for trademarks or whatever. super helpful but i assume that stuff won't be a problem on the real thing.

maybe if i'd seen the one way arrow i wouldn't have overcomplicated it, who knows...

thanks

 

[IamSuperDoctor]

glad to help good luck ! take things easy and dont overstudy ive been watching movies for the last 5 hours or so

 

[fizzgig]

new question. saponification of a TAG to glycerol and 3 fatty acids.

question is does it use hydroxide or is hydroxide a catalyst.

i get that an extra oxygen is thrown in there (the carbonyl attack), so the hydroxide IS used. but the glycerol O- it leaves behind: since water is a stronger acid than ROH, wouldn't water donate an H to the glycerol (especially as it would end up with 3 negs otherwise) and regenerate HO-??

thanks peoples.

 

[Rabolisk]

When the intermediate collapses, glycerol O- leaves. But water does not protonate glycerol. Rather, the carboxylic acids (weak acid) react with glyceroxide to produce carboxylate and glycerol. This is the step that drives the entire reaction forward. OH- is never regenerated.

http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch20/ch20-3-3-1.html

 

[fizzgig]

i wish stuff didn't make MORE sense in the middle of the night, but yeah, it looks like i dropped a hydrogen out of my picture somewhere and i left both new products deprotonated. thanks for the help!

 

[linusp]

I know this is passage has been brought up, but for Passage II of the BS (The infamous Semicarbazone one)

For the first two entries, does the table on the bottom show the melting point data for Cyclohexanone and 2-Furaldehydes as pure compounds, OR the semicarbazone crystals of them? I'm now assuming its the latter, in which case, the passage makes much, much more sense. Definitely didn't see this the first time.

Can somebody confirm? Thank you!

Edit: I'm also assuming that (lit) stands for the value as found in the literature?

 

[emericana]

This makes NO SENSE to me. Pls help.

How are the rates the same here?

If you look at the graph the slope (which is the rate?) is totally different for both at 2? How is the slope the same? i chose .25 because it looked like the slopes were the most similar here

 

[emericana]

One more question... I had this down first (B) but then I changed it.

I changed it because, following the logic I learned from EK 101, I thought that II was worded incorrectly.

As you can see in the picture above it states

"A mandate to increase the ratio of soft-to-hard energy sources by a specified amount within ten years" .

Why would an advocate ofthe soft-energy path support a proposition that increases the ratio of soft-to hard? wouldnt this reduce the amount of soft and increase the amount of hard? shouldnt the proper wording be hard to soft? A supporter of the soft would want to increase the soft and decrease the hard, not the other way around.

I had this down at first but changed it because i thought it was a trick question because of the wording.

 

[emericana]

bump... anyone know how to solve the above two?

 

[Rabolisk]

For the treadmill question, it's as the explanation has it. Apparently, treadmill means that the rate at + end is equal to the rate at - end. That is, you are adding at the + end at the same rate as you are losing from the - end. If you look at the graph, choice C is the concentration in which the rates are equal (equidistant from 0). You're either miscomprehending the passage or misreading the graph, or both. For one, rate has nothing to do with the slope. You can see that the y axis of the graph is explicitly labeled rate.

For the second question, ratio of soft-to-hard means soft:hard = soft/hard. That is the proper wording. You thought the opposite, which is just a problem with language/wording, not understanding (You have the reasoning right).

 

[plzNOCarribbean]

If the valve is opened to drain the tank, where is the speed of the flowing water the greatest?

A) At the top of the tank
B) At the bottom of the tank
C) At the wide end of the pipe
D) At the narrow end of the pipe

I picked B (v=sqrt2gh) but the answer is D. I understand the reasoning (flow rate Q=AV) but why can't it be B?

Yeah, PR says that d Vefflux=Sqrt 2gD where D is the distance from the surface of the liquid down to the hole, so escpae velocity should be highest at the bottom of the tank. I dont understand why D is the correct answer

 

[emericana]

Yeah, PR says that d Vefflux=Sqrt 2gD where D is the distance from the surface of the liquid down to the hole, so escpae velocity should be highest at the bottom of the tank. I dont understand why D is the correct answer

The reason why D is correct, or at least why I put D... which I learned from TPR, is the formula A1V1=A2V2. Essentially if the area is bigger on one side, the velocity is going to be bigger on the other side.

Bigger area, smaller velocity, smaller area, bigger velocity

 

[BenZq]

Can anyone explain to me why CuSO4 does not does not act as a common ion to reaction 1 and reduce the amount of I2 produced? I chose (c) inhibitor, because i thought that there would no longer be excess I2 to react with the starch indicator.

 

[Rabolisk]

Can anyone explain to me why CuSO4 does not does not act as a common ion to reaction 1 and reduce the amount of I2 produced? I chose (c) inhibitor, because i thought that there would no longer be excess I2 to react with the starch indicator.

But there is excess iodine, as shown by the table. It's plausible that copper sulfate reduces the equilibrium concentration of iodine in reaction 1, but that is irrelevant. Per reaction 2 S2O3 will react with any iodine formed from reaction 1 so the reaction will continuously happen until there is no more S2O3 to react with any iodine. At this point, apparently, there is enough excess iodine for the solution to turn blue, per the table. You know for a fact that the reaction reaches excess iodine faster with copper sulfate, hence it is a catalyst.

 

[BenZq]

But there is excess iodine, as shown by the table. It's plausible that copper sulfate reduces the equilibrium concentration of iodine in reaction 1, but that is irrelevant. Per reaction 2 S2O3 will react with any iodine formed from reaction 1 so the reaction will continuously happen until there is no more S2O3 to react with any iodine. At this point, apparently, there is enough excess iodine for the solution to turn blue, per the table. You know for a fact that the reaction reaches excess iodine faster with copper sulfate, hence it is a catalyst.

Thanks, I really think that I overthought this one. Just for clarity, is there a kinetic or thermodynamic inference that can be made from the passage for why rxn 2 will continue to completion, other than the passage stating that S2O3 2- was used up?

 

[leoni101]

PS 30 8R

The question says: If the pH of the water sample were high such that all the carbonate is present as CO3^2-, what would be the concentration of Ca^2+? (The Ksp of CaCO3 is 4.8x10^-9.)

Can someone please help with this? Maybe I've been studying too long and can't think straight, but I don't understand what the question is really asking nor what the explanation for this is.

Why would we assume that all of the carbonate has dissociated?