C/P Section Bank Part 2

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C/P Section Bank Part 2

Post by ragtaguzoo » Wed May 13, 2020 2:20 pm

17. I didn't understand the explanation in relation to the equation or R3 = zero?

18. How are we supposed to know if it's Ans C or D? I'm not familiar with Phosphatides or Phosphonic acids?

19. Why are the other answers wrong?

20. I didn't understand the thermodynamics and kinetics of the answer?

23. How does the answer correlate to the last paragraph? How did they derive that from the graph in the passage for compound 1? Why is 30-40 mL constant through every graph?

24. I got stuck on the e and ln part of this question. I got to the 12ln part. Put couldn't understand how to get to e^12. Maybe you can tell me where I went wrong with my calculation or reasoning. I attached a pic of my work for this problem below.

25. Shouldn't it want to be 3d5, because it is more stable due to it being half-filled? Doesn't an element always want to have a half-filled or completely filled subshell?
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Re: C/P Section Bank Part 2

Post by NS_Tutor_Mathias » Mon May 18, 2020 7:25 pm

Hey, sorry this has been sitting here for a week. This forum has been eating more and more of my replies (and I've definitely been logged in!), but here goes again:

17. Be careful here. The voltmeter reads zero when R3 is adjusted. This doesn't mean that the change in voltage for the circuit is zero (that would mean no current flows!), but that there is no potential difference between R1 and R or R2 and R3.

This is why the AAMC explanation gives:

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R1*I = R*I0
R2*I = R3*I0
Which, when those two equations are divided by each other yields:

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R1/R2 = R/R3

R1/R2 * R3 = R

18. You don't necessarily have to be familiar with phosphonic acids as a name, just to be able to deduce from the name that such a compound would have readily dissociable protons (the pictured phosphatide does not) and be familiar with the structure of membrane lipids (in this case, phosphatides). There won't be made-up answer choices on the MCAT, but there may very well be things that are not explicitly in scope but that are fair to reason out.

a) The macromolecule refers to compound 1 and 2. It is not conjugated at all, let alone extensively conjugated. This cannot be the cause of fluorescence.
c) I'm not even sure the excited state of which electron or molecule this may be referring to. This is largely a nonsense answer.
d) This would be reflection, not fluorescence.

Imagine a large activation barrier to changing the shape of the liposome (because doing so may require breaking it apart and/or fusing it). In such a case, whatever size of liposome we start with will, after mixing, be much the same. Some mixing shouldn't disturb this balance. This is analogous to a reaction being under kinetic control, in that it doesn't proceed even though it would be thermodynamically favorable, simply because the activation barrier can't be easily overcome.

Imagine a much smaller activation barrier now. If I now mix liposomes of different sizes, they'll inter-mingle, fuse and break apart and make average-size liposomes. This is analogous to a reaction under thermodynamic control.

This one is, underneath the fancy words, an easy logic game:
1. All size-exclusion chromatography seems to first elute at different volumes depending on the concentration used.
2. The final wash fluoresces heavily, because you are literally just rinsing things out.

Now, only a single elution is performed, and only at a single concentration (0.15 mM most likely), since the two solutions given are mixed. So we expect a spike in fluorescence at around 15-20 mL, and a final one on washing. Only C and D even fit the profile at all, and of the two, D has it's spike better positioned and more in proportion with the intensities seen in figure 1.

Laws of logs and simply putting both sides to the base of e.

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12 = ln([ADP]/[ATP])
e^12 = e^ln([ADP]/[ATP])
e^12 = [ADP]/[ATP]  
Remember that a natural log is "the power I must put e to in order to get the number expressed within the parentheses". So if A = ln(x), then e^A = e^ln(x) = x.

This is easier to visualize if we say x = e^3.
A = ln(x) = 3 (I have to put e to the third power to get e^3)
e^A = e^ln(x)
e^A = x
e^A = e^3 (because x = e^3)

See how these give equivalent expressions.

25. Absolutely not. That is a sloppy heuristic given when chemistry is taught and not a real physical law or property. Apply it to chromium and copper, and in a pinch any other cases you find where it is explicitly stated to apply, but do >NOT< under any circumstances generalize this. As a rule, consider it such that the 4s shell will lose electrons before the 3d.
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