# AAMC C/P Qpack spoiler Passage 8

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#### treeclimbingmonkey

##### Full Member
2+ Year Member

Hi I know this answer has been asked in the past but I still don't understand why C can't be an option. What does the AAMC answer mean by "motion of photons is certainly not random". Doesn't the particle nature of light consider photon movement random as all other particles are?

Was also confused with this question:

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For question (1), I wouldn't overcomplicate it—they're just getting at the idea that light emitted from a given source is moving in a certain direction. Although light can be deflected in a somewhat random way (this what happens in light scattering, for example), it would be going too far to say that the motion of light is overall random. Brownian motion doesn't apply here, for instance. Another way to think about it is this: if (C) were correct, a classic illustration of a prism (or lens), like this one, would have the light waves bouncing all around within the prism before emerging at random places.

For question (2), the first key point is understanding that this detector works by reducing the gas. This enables choices A and B to be eliminated, because they don't involve anything about reduction or redox reactions. The question then boils down to "can CH4 be reduced"? The confusion here might stem from how the AAMC approaches this.

They break down this question into (a) can H be reduced in CH4? and (b) can C be reduced in CH4? The answer to (a) is "yes, maybe, theoretically" and the answer to (b) is "no." The reason why the answer to (b) is "no" is because the oxidation number of C in CH4 is -4 (because each H has an oxidation number of +1, and the oxidation numbers have to sum to 0 in this uncharged compound) -- and you can't get lower than -4 as an oxidation number for C. Since the answer to (b) is "no", D is correct. Choice C is wrong because H is not at its lowest oxidation state in CH4. In CH4, the oxidation state is +1, and H can have a lower oxidation state of 0 (as in H2). Since choice C contains a factually incorrect statement, it cannot be correct.

A couple things about this question, though...first, I'd focus on the C atom first, because when we talk about oxidation/reduction in organic compounds, the focus is almost always on carbon. If you can direct your attention accordingly, you might be able to pick out choice D as correct more quickly and not waste your time agonizing about choice C. Second, it's a good reminder to get familiar with how to calculate the oxidation state of carbon in organic compounds, because doing so helps you cement your understanding of the reduction-oxidation hierarchy for organic compounds—and that's key for the MCAT, because redox reactions with organic substrates are at the core of the metabolic pathways that the MCAT tests.

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For question (1), I wouldn't overcomplicate it—they're just getting at the idea that light emitted from a given source is moving in a certain direction. Although light can be deflected in a somewhat random way (this what happens in light scattering, for example), it would be going too far to say that the motion of light is overall random. Brownian motion doesn't apply here, for instance. Another way to think about it is this: if (C) were correct, a classic illustration of a prism (or lens), like this one, would have the light waves bouncing all around within the prism before emerging at random places.

For question (2), the first key point is understanding that this detector works by reducing the gas. This enables choices A and B to be eliminated, because they don't involve anything about reduction or redox reactions. The question then boils down to "can CH4 be reduced"? The confusion here might stem from how the AAMC approaches this.

They break down this question into (a) can H be reduced in CH4? and (b) can C be reduced in CH4? The answer to (a) is "yes, maybe, theoretically" and the answer to (b) is "no." The reason why the answer to (b) is "no" is because the oxidation number of C in CH4 is -4 (because each H has an oxidation number of +1, and the oxidation numbers have to sum to 0 in this uncharged compound) -- and you can't get lower than -4 as an oxidation number for C. Since the answer to (b) is "no", D is correct. Choice C is wrong because H is not at its lowest oxidation state in CH4. In CH4, the oxidation state is +1, and H can have a lower oxidation state of 0 (as in H2). Since choice C contains a factually incorrect statement, it cannot be correct.

A couple things about this question, though...first, I'd focus on the C atom first, because when we talk about oxidation/reduction in organic compounds, the focus is almost always on carbon. If you can direct your attention accordingly, you might be able to pick out choice D as correct more quickly and not waste your time agonizing about choice C. Second, it's a good reminder to get familiar with how to calculate the oxidation state of carbon in organic compounds, because doing so helps you cement your understanding of the reduction-oxidation hierarchy for organic compounds—and that's key for the MCAT, because redox reactions with organic substrates are at the core of the metabolic pathways that the MCAT tests.
Thank you very much for the thorough response!

Thank you very much for the thorough response!
Of course, happy to help!!

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