# AAMC 8R Q58

Discussion in 'MCAT Study Question Q&A' started by pacer, Sep 5, 2014.

1. ### pacer 2+ Year Member

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Can someone go over Question 58 from the PS section of AAMC 8R?

Correct answer is C - colour of the solution is due to nickel II because Nickel II has unfilled d orbitals.

I know that Nickel II has lost 2 electrons, and thus has a charge of +2 but aren't d orbitals anomolous meaning that electrons move from 4s into 3d to fill them up because a 3d5 and 3d10 is more stable?

I also thought that valence electrons react and since they are in the s and p orbitals, I thought that answer choice A would be correct but it is not.

I am confused. Please explain.

Thank you.

PS I don't have the complete solutions (only the solutions that mention which choices are correct, if someone else has the complete solutions that explain each choice and why it is correct/incorrect, can you please share via PM). Thanks!

2. OP

### pacer 2+ Year Member

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I am also having trouble with Q62.

Correct answer is that fringes on T3 would disappear if S1 is covered. - But there still S2 which is receiving light waves from the source. So would it not be sufficient to produce bright and dark fringes on T3?

Why is choice D wrong? Wouldn't moving T3 farther away prevent the fringes?

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3. OP

### pacer 2+ Year Member

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Another Question from Biology section

Equation given: P = CO x VR
P doubles, VR increases by 50%

I get that CO = P/VR = 2/1.5 = 4/3. So the above mentioned changes cause CO to be 4/3 times of what it was before.

How did they get the correct answer choice that CO increased by 1/3??? I can't see the relation between 4/3 (as calculated and 1/3).

4. ### Marleywh

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I can answer both those questions:

Electrons are moved between empty D orbitals by photons. They jump up when they absorb a photon, and then release photons when they fall back. The energy gaps between D orbitals correspond to wavelengths of visible light. This is why transition metal element complexes have such vibrant colors. The D orbitals are not degenerate because when complexed, the ligands push some of the orbitals (due to their geometry) into higher energy states.

The blood pressure question had me confused for a bit too, but if you see that 4/3 is what the cardiac output value has become in relation to what it was before the changes. You see that it went from 3/3 (1) to 4/3 (the addition of 1/3)

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5. OP

### pacer 2+ Year Member

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I am still a little confused with Q58.

Would electrons move from 4s into 3d to fill it up if we have a 3d8? Or does that only work for 3d9 when 1 electron moves from 4s into 3d to fill it up because it is more stable.

I get that the color is a result of photons being emitted (of certain wavelength), giving off energy. But I don't see how nickel II has unfilled d orbitals and also, aren't valence electrons always in s and p orbitals, so they are the ones being excited or falling back into ground state.

Can someone go over Q201?

6. ### Marleywh

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Yah you only have electron promotion from the s orbital for d4 and d9. That's because making a full filled d orbital is more stable than having a full s in those cases. Also for nickel (II) it means you have removed two electrons. When you ionize transition elements (d block elements) you first remove the s orbital electrons. So nickle (II) has an empty 4s orbital and 3d8 configuration. It has at least two empty d orbitals. I don't think you should worry about the specifics, just know that transitions between d orbitals release visible light. If nickle had p electrons then yes they would be the ones getting excited I believe, but nickle doesn't have any so it doesn't apply.

7. OP

### pacer 2+ Year Member

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I found that there were some random Q's on this tests

32 - do we need to know the equation for angular frequency? It was not in the passage and I don't remember seeing it in my content review.

40 - using steel spheres is not a good idea - I thought it would be because they would be too heavy and would not change in depth easily - which was required for this experiment to work well.

42 - isn't max speed measure at the bottom of the tank - vmax = (2gd)^1/2? I know that in comparison to the wide end, the speed of flow at the narrow end is greater. But why is it better than choice B?

68 - I had never come across the term metathesis reaction????

For VR section - does someone have any tips on dealing with comparative passage (where the author compares multiple theories etc) - I seem to be having a lot of trouble with this particular passage type. I find that since I am not reading for detail, it is hard for me to separate all the details about the 2 or more topic discussed and I end up getting confused on the questions.

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### pacer 2+ Year Member

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Got it, thanks!

9. OP

### pacer 2+ Year Member

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After going through the mistakes I made, I find that most mistakes are due to problems in interpreting the passage information or figure/data presented. Any tips on how I can fix this in a short period of time remaining till test date (2 weeks!).

10. ### Marleywh

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40 - Steel spheres could be very large and very thin, meaning they have a large buoyant force but small force of gravity, so they could still float (think of aircraft carriers).

42 - v = (2gd)^1/2 tells you the speed of the water coming out of the hole. So the speed will be the large at the hole, and will be even larger at the narrow end. Av = Av

I had never heard of a metathesis reaction either, but I eliminated the other options so I had to be that one!

11. OP

### pacer 2+ Year Member

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What is the correct answer to Q111 (VR section)? It is missing from the solutions I have.

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