Physical sciences - FL7

Hi Alex,
My question is about number 4 on the physical sciences section. I guess I just don't understand the concept behind setting the torque equal to the gravitational potential energy in that I can't visualize/understand why they would be equal.

Also, for number 49 on the biological science section of Kaplan Full Length 8, wouldn't the flow be reversed and be from the pulmonary artery to the aorta?

Thanks!
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For the first question, let's start by considering the forces acting on each of the small balls (m).  They each have a force of gravity created by Earth acting downwards (F = mg), and a force of gravity created by the large balls (F = GMm/r²).  Let’s then translate these forces into torques.  We have forces in two directions, so let’s consider rotation caused by torques in those two directions.  The two forces of gravity from Earth are acting in opposite directions by rotation, so the two torques cancel out.  That is to say that we have rotational equilibrium, looking only at the forces of gravity from Earth.

For the other direction, we have a torque on each ball pulling counterclockwise.  Since they’re pulling the same rotational direction, we can just add them up.  Torque = (lever arm)(force), so torque from each is (L/2)(GMm/r²).  Adding the torques together, we have 2(L/2)(GMm/r²) = GMm/r².  That’s answer (D).

Overall, we’re not setting equal to gravitational potential energy; we’re calculating the torque by multiplying the lever arm (L/2) by the force between the two masses – a gravitational force equal to GMm/r².

For the other question, remember that this question says after birth.  The ductus arteriosus connects the pulmonary artery and the aorta.  After birth, when an individual has the adult-style circulation, the pressure in the aorta is MUCH higher than the pulmonary artery.  Thus, we would expect blood to travel through the shunt – if it’s still open – from the area of higher pressure to lower pressure.  Therefore, there will be flow from the aorta to the pulmonary artery.

PS Test 2, question 30

Hi Alex, could you help explain question 30 from the physical sciences
section test 2. I found that question to be really challenging and I am
not quite understanding the explanation.
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This question reads:  “If Step 2 above were the rate-determining step of Reaction 1, which of the following equations would correctly define the rate?”

Remember that the rate-limiting step can be related to the rate by simply taking each of the reactants from that step and raising them to their stoichiometric coefficients.  For this question, that means that rate = k₂[B][D].  There’s a problem here, though.  When we write rate laws, all of the terms must be reactants of the overall reaction, meaning that we’ll have to have our rate dependent on A and D (based on the reaction above the mechanism).  So having [B] in our answer is a problem.  How do we fix this?

The key is then looking at Step 1, where [B] is manufactured.  To find out how to rewrite our concentration of B, let’s use the fact that it tells us this reaction is “slow.”  In reaction kinetics, “slow” means that the reaction is at equilibrium.  If it’s at equilibrium, then the forward and reverse reactions of Step 1 have the same rate.  Mathematically, that means:

rate_f = rate_r

k₁[A] = k_-1[B][C]

The “-1” on the rate constant is indicating that it is the rate constant for the reverse reaction of Step 1.  This is a common convention in kinetics.  Thus, to find [B], we can rearrange:

[B] = k₁[A]/k_-1[C]

Plugging into our rate law from before, we get:

rate = k₂[B][D] = k₂(k₁[A]/k_-1[C])[D] or k₁k₂[A][D]/k_-1[C].  This matches with answer choice C.  We could get picky and point out that C is in our rate law, and it’s not a reactant of the overall reaction.  That’s true, but you’ll see that we actually cannot get any further with [C] than this step.  Ideally, we’d know more information and be able to remove [C] from our rate law, but we can’t do that in this particular scenario.

Le chatelier's

This makes sense- but I just want to be sure I am understanding this correctly. Adding or removing a substance that is NOT in the equilibrium expression, such as a solid, has no effect on the state of the rxn at equilibrium- so le chatlier's pricinple does not apply?
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Yes.  Raising or lowering the amount of a solid or liquid will not disturb the equilibrium because these terms are not even part of the Keq of the reaction.

Nomeclature and functional groups, #5

Hey Alex,

Just wondering if you could explain to me why the alkene isn't a site
of electrophilic attack.  I know it has a cloud rich in electron
density, but couldn't the extra electrons move to the neighboring bond
to bump off the Br?  So basically why the answer isn't D.

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This question reads:

Which of the functional groups on the following molecule are susceptible to nucleophilic attack?

Looking first at (a) - the Br attached to the secondary carbon is electron-withdrawing, so it renders the carbon at a electrophilic (thus, suscpetible to nucleophilic attack).  For the carbonyl carbon at (c), there is similar logic since the oxygen is electron-withdrawing.  As a side note, carbonyl carbons are the most common electrophile on the MCAT.

For (b) - the main problem with the logic above is that there is no driving force for this sort of reaction to occur.  All reactions in organic chemistry need to have reason to run.  Here, an incoming nucleophile would find it very difficult to be attracted to the reaction site; it has a large electron-rich cloud that will strongly repel any negatively-charged (or even neutral) nucleophile coming in.  Even if the nucleophile were attracted, there is an energy barrier to overcome here.  This reaction would require breaking the double bond, and then undergoing some E2-like reaction mechanism.  Still, any incoming nucleophile would be very strongly repelled from the sp2-hybridized carbons at (b).

Thus, a and c are the reaction sites susceptilbe to nucleophilic attack.

FL 9 - 1, 21, 41

I had a couple questions from the physical sciences section of Full-Length 9. The first one is actually #1. I chose answer c, and I still can't understand why the answer is b. Wouldn't the elastic part be easier to stretch? My next question is number 21, I just can't put together an equation-based relation that includes velocity. Also, could you explain number 41? Thank you!!
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Question 1:

For this question, we need to analyze the graph.  The graph is stretch versus time (NOT force), for a system with a viscous component and an elastic component in parallel.  Well, if we give a certain CONSTANT force to a spring over time, what happens?  It stretches until it reaches a certain deformation from equilibrium, and then just stays there.  Think about hanging a mass from a spring – that’s a constant force on the spring which causes deformation until the mass is at translational equilibrium.

What would happen to a viscous object without an elastic force?  Well, it would stretch and stretch and stretch… and keep stretching until (probably) it eventually broke.  But either way, it doesn’t reach equilibrium because it’s not elastic.  So it just keeps stretching.

Why is it asymptotic, then?  Well the spring will only stretch until it’s in equilibrium with the force.  At that point, it won’t stretch anymore.  Thus, the stretch of the viscous component (which should be infinite) is limited by the stretch of the spring.

Question 21:

We’re supposed to relate the force created by the spring with this scenario of a pilot landing a jet.  Well, anytime you see force, think acceleration!  F_net = ma, but F also represents the force of the spring here, and thus F_spring = kx.  So, we can say F_spring = ma.  We’re given one more key piece of information:  the final velocity is zero.  Acceleration is simply change in velocity per time or Δv/t.  Since v_f = 0, we can say that a = (v_f – v_i)/t = (0 - v_i)/t = -v_i/t.  Plugging into our equation, we get F_spring = m(-v_i/t) = -mv_i/t.  Thus, the force is directly proportional to the starting velocity.

Question 41:

The first thing we see when looking at the answers is the fact that it says that one has higher or lower energy than the other, and differences in principle quantum number (5sto 5d, etc.).  So, let’s start with energy.  E = hf, so the higher the frequency, the higher the energy.  Red light has the lowest frequency while violet has the highest, so that means red light has lower energy than violet.  Here, we’re dealing with red (Sr) and green (Ba).  So that means that strontium has a lower energy jump than barium does.

Well, for principle quantum number, where is strontium?  It’s in the 5s grouping.  Barium, however, is 6s.  Where are these electrons jumping to?  Well, it should jump to the next-highest-energy orbital.  If we start at an s orbital, we’ll just jump to p or d.  So we’ll want to see that electrons jumping from 5s to 5p or d have less energy than those jumping from 6s to 6p or d.  And that matches with answer choice (B).  

AAMC #3, BS 114

I'm having a hard time understanding why developing a leak in the apparatus increases the surface pressure and thus increases the BP of both substances. How do we know the pressure in the appartus is different  from atmospheric pressure to begin with?

When the leak occurs does the vapor pressure of the liquids also decrease because temperature is going to be decreasing? This wouldnt necessarily mean the BP will be different, just that it will take longer to get there right?
Thanks!
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This question reads:  "If a leak develops in the vacuum distillation apparatus, the boiling points of the two components of caraway seed oil will:"

The key, then, is the fact that it says "vacuum distillation."  Remember that vacuum distillation will lower the pressure above the two liquids being distilled, so that it lowers their boiling point.  This is because liquids will boil when their vapor pressure equals the ambient pressure.  By lowering the ambient pressure, liquids will boil at a lower temperature because they don't need as high of a vapor pressure.  Vacuum distillation is used for liquids that have boiling points >125 degrees Celsius.

Thus, if a leak develops, we will expect the pressure inside the apparatus to go up.  It starts lower than atmospheric pressure, because that's the whole goal of vacuum distillation.  When a leak develops, the pressure will go up because air comes in from the surrounding atmosphere.  By increasing the ambient pressure, the boiling point starts going up again.

The vapor pressure of the liquids shouldn't change here just because of the leak.  Indirectly, it will cause them to go up in the end.  Remember that boiling is an isothermal process.  By increasing the boiling point, we increase the vapor pressure needed to cause that boiling.  Thus, assuming that we're supplying heat to the apparatus, the temperature will go up because the boiling point has gone up.  Before, it would hit the boiling point and stop at that temperature since the boiling process is isothermal.  Now, it adds heat until the boiling point -- at a higher temperature and higher vapor pressure -- and then boils at this temp.

Scaled scores

Does scaling the scores according to the percentage correct vary from exam to exam? I've just been noticing differences with the percent I got correct and my scaled score- like on FL 5, my scaled score for physical sciences was 2 points higher than what the percentage I got correct would have said.
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I've said this a few times before -- please be careful with the rubric I gave you in the MSCT III packet.  It gives you an idea of where your score falls, but it is not wholly accurate for the MCAT.  The MCAT varies from test administration to test administration; likewise, our scaling varies from Full-Length to Full-Length.  The scaled scores you are given by the Kaplan tests are quite accurate; it wouldn't be fair for us to either give you a false sense of confidence or to make you feel like you're not doing as well as you actually are.

The scaled score conversion chart should be used from section tests, and remember that it predicts usually +/- 1.  In your case, it was off by 2, but usually it should be accurate -- give or take 1 scaled score point.