Physics Question from the "New MCAT Preview Guide for 2015" from AAMC..

[hplssrmantcxox]

https://www.aamc.org/students/download/266006/data/2015previewguide.pdf

So these questions are from here. (These bio passages aren't too bad but it definitely is way more time consuming to read and feels like extended verbal just like everyone's been saying about their exam recently.... ) I figure I might as well gauge what the questions will probably be like on my upcoming exam 2 weeks from now since I've heard dreadful things lately about the horrendous experimental (or not so experimental) passages..


Not based on any passage:
13) The radius of the aorta is about 1.0 cm and blood passes through it at a velocity of 30 cm/s. A typical capillary has a radius of about 4 × 10^–4 cm with blood passing through at a velocity of 5 × 10^–4 m/s. Using this data, what is the approximate number of capillaries in a human body?
A. 1 × 10^4
B. 2 × 10^7
C. 4 × 10^9
D. 7 × 10^12
Answer: C
(Skill Level: 2 - i.e. this is supposed to be easier-ish.. -_-)

So I'm blanking completely... Is there a formula I'm supposed to use.. I've been staring at it trying to fudge with the numbers and still not getting anywhere This is a relavant question for the current exam which I'm taking in 2 weeks and I'm getting really irritated that I'm blanking out on an easy units question..

Anyone wanna help and show their unit conversions? Or formula they used if applicable?

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Page 84 - Passage 2 (Physics) - Hard Water Ions passage..

Hard water contains cations that form precipitates with soap or upon boiling. The principle hardness ions
are Ca2+, Mg2+, and Fe2+. There are two major drawbacks of hard water. First, the M2+ ions reduce the effectiveness of common soaps, which contain the sodium salts of organic acids with long carbon chains. An example is sodium stearate, C17H32CO2Na (MW = 306). The reaction between soaps and hardness ions yields insoluble precipitates through Reaction 1. Removal of stearate
from the solution eliminates the effectiveness of the soap.
M2+(aq) + 2NaC17H32CO2(aq) → 2Na+(aq) + M(C17H32CO2)2(s)
Reaction 1

Second, hard water produces boiler scale, a layer of insoluble carbonates formed by Reaction 2 that lines
the inner walls of pipes and hot-water boilers. Deposits of this type are especially bad in hot water and
are poor conductors of heat.
M2+(aq) + 2HCO3–(aq) → H2O(l) + CO2(g) + MCO3(s)
Reaction 2

It is important that the cations responsible for hard water be removed before the water is heated or used
for washing. Water softening, the removal of hardness ions from water, can be accomplished in several
ways. One method is the ion exchange process in which water is passed through a column containing
solid sodium aluminosilicates. Sodium aluminosilicates are high surface area three-dimensional
extended solids with –ONa groups at the surface.
M2+(aq) + Na2AlxSiyOz(s) → 2Na+(aq) + MAlxSiyOz(s)
Reaction 3

7) What happens to the pH of a soapy solution as a result of the introduction of hardness ions?
A. The pH increases as [H+] increases.
B. The pH is not changed since no acid-base reaction occurs.
C. The pH decreases as [OH–] decreases.
D. The effect on pH depends on the identity M2+
.
Answer: C
...Why is it C? Obviously you can cross out A since it's factually wrong. and D sounds wrong too. But I'm still confused as to why it's considered an acid/base reaction -_-

10) Which experimental approach can be used to analyze the metal content of soapy precipitate
produced by Reaction 1? Dissolve the solid in a known volume of :
A. 0.1 M NaHCO3(aq), then titrate with standardized 0.1 M HCl(aq) using an indicator.
B. 0.1 M NaOH(aq), then titrate with standardized 0.1 M HCl(aq) using an indicator.
C. 0.1 M NaCl(aq), then titrate with standardized 0.1 M NaOH(aq) using an indicator.
D. 0.1 M HCl(aq), then titrate with standardized 0.1 M NaOH(aq) using an indicator.
Answer: D

Okay so M(C17H32CO2)2 (s) + HCl --> M2+ + Cl- + C17H32COOH right ? What am I missing -_- why does titrating that with NaOH help you "analyze the metal content"? Isn't it just neutralizing the acid?.... I'm definitely missing something .. Is it asking how do you measure the metal concentration from titrating with NaOH?...

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[ebasappa]

why are these questions so hard.... you think there will be questions like these on our exam? :\

 

[BoxVersionAce]

I'm not entirely sure I could explain any of the answers. For what I can gather together on the spot, they are what I would guess based on my knowledge of equilibrium/acid-base/physiology. Not entirely sure on any of these, but I guess the first one I actually know.

The first question can be seen as more of a physics question I suppose. I would try to apply the volume flow continuity equation, assuming ideal fluid flow in the circulatory system. vA=vA In other words, lets assume that the volume flow rate starting at the aorta is equal to the volume flow rate in the capillaries. Now, from physiology we can vaguely imagine the aorta as a single tube and the capillaries as MANY MANY tubes lol. The volume flows are additive! You can kind of draw a pic in your head and imagine this. Lets call volume flow Q.

Q(aorta)=Q(capillaries)

A1v= A2v + A3v + A4v...

A2, A3, A4, and so on represent each individual capillary tube. Add up all the flow rates of each tube and through the continuity equation it should equal the flow rate from the original aorta. Assume that the velocity in each capillary tube is the same and that the area is the same as well. This can yield the equation:

A1v1=nA2v2

n=number of capillaries/capillary tubes=answer

Use the numbers they gave you. Make sure the all the units are the same, either all in cm or all in meters. Because you're dividing A1 by A2, you don't really have to pay attenition to the pi constant in the pi(r)^2 since they cancel out. Using meters:

(r1)^2v1= n(r2)^2(v2)

(.01)^2(.3)=n(.000004)^2(.0005)

 

[ebasappa]

I'm not entirely sure I could explain any of the answers. For what I can gather together on the spot, they are what I would guess based on my knowledge of equilibrium/acid-base/physiology. Not entirely sure on any of these, but I guess the first one I actually know.

The first question can be seen as more of a physics question I suppose. I would try to apply the volume flow continuity equation, assuming ideal fluid flow in the circulatory system. vA=vA In other words, lets assume that the volume flow rate starting at the aorta is equal to the volume flow rate in the capillaries. Now, from physiology we can vaguely imagine the aorta as a single tube and the capillaries as MANY MANY tubes lol. The volume flows are additive! You can kind of draw a pic in your head and imagine this. Lets call volume flow Q.

Q(aorta)=Q(capillaries)

A1v= A2v + A3v + A4v...

A2, A3, A4, and so on represent each individual capillary tube. Add up all the flow rates of each tube and through the continuity equation it should equal the flow rate from the original aorta. Assume that the velocity in each capillary tube is the same and that the area is the same as well. This can yield the equation:

A1v1=nA2v2

n=number of capillaries/capillary tubes=answer

Use the numbers they gave you. Make sure the all the units are the same, either all in cm or all in meters. Because you're dividing A1 by A2, you don't really have to pay attenition to the pi constant in the pi(r)^2 since they cancel out. Using meters:

(r1)^2v1= n(r2)^2(v2)

(.01)^2(.3)=n(.000004)^2(.0005)

Woops, didn't see that the second velocity was in m/s haha...

 

[BoxVersionAce]

Woops, didn't see that the second velocity was in m/s haha...

Yeah that tripped me up at first. The cool thing about a multiple choice test is that even if you forget some conversion, you can still be somewhat assured of the answer pretty quickly by looking through the choices. Unless they all have 4x10^n -___-

 

[hplssrmantcxox]

I'm not entirely sure I could explain any of the answers. For what I can gather together on the spot, they are what I would guess based on my knowledge of equilibrium/acid-base/physiology. Not entirely sure on any of these, but I guess the first one I actually know.

The first question can be seen as more of a physics question I suppose. I would try to apply the volume flow continuity equation, assuming ideal fluid flow in the circulatory system. vA=vA In other words, lets assume that the volume flow rate starting at the aorta is equal to the volume flow rate in the capillaries. Now, from physiology we can vaguely imagine the aorta as a single tube and the capillaries as MANY MANY tubes lol. The volume flows are additive! You can kind of draw a pic in your head and imagine this. Lets call volume flow Q.

Q(aorta)=Q(capillaries)

A1v= A2v + A3v + A4v...

A2, A3, A4, and so on represent each individual capillary tube. Add up all the flow rates of each tube and through the continuity equation it should equal the flow rate from the original aorta. Assume that the velocity in each capillary tube is the same and that the area is the same as well. This can yield the equation:

A1v1=nA2v2

n=number of capillaries/capillary tubes=answer

Use the numbers they gave you. Make sure the all the units are the same, either all in cm or all in meters. Because you're dividing A1 by A2, you don't really have to pay attenition to the pi constant in the pi(r)^2 since they cancel out. Using meters:

(r1)^2v1= n(r2)^2(v2)

(.01)^2(.3)=n(.000004)^2(.0005)

argh of course I blank out on the most basic formula ever -_-.... I knew this one too. I thought of this while I napped and then saw your reply and got angry for blanking out the first time haha. Thank you yeah I had missed the m/s thing too, haha.

Anyone know the other two questions though?? I'm usually great with chemistry (93% on my self assessment and have been getting 12's on the PS section of my practice tests and I'm literally like ... with those two other chemistry questions I posted -_-

M2+(aq) + 2NaC17H32CO2(aq) → 2Na+(aq) + M(C17H32CO2)2(s)

What does this reaction have anything to do with the [OH-] concentration..?

 

[LoLCareerGoals]

Page 84 - Passage 2 (Physics) - Hard Water Ions passage..

Hard water contains cations that form precipitates with soap or upon boiling. The principle hardness ions
are Ca2+, Mg2+, and Fe2+. There are two major drawbacks of hard water. First, the M2+ ions reduce the effectiveness of common soaps, which contain the sodium salts of organic acids with long carbon chains. An example is sodium stearate, C17H32CO2Na (MW = 306). The reaction between soaps and hardness ions yields insoluble precipitates through Reaction 1. Removal of stearate
from the solution eliminates the effectiveness of the soap.
M2+(aq) + 2NaC17H32CO2(aq) → 2Na+(aq) + M(C17H32CO2)2(s)
Reaction 1

Second, hard water produces boiler scale, a layer of insoluble carbonates formed by Reaction 2 that lines
the inner walls of pipes and hot-water boilers. Deposits of this type are especially bad in hot water and
are poor conductors of heat.
M2+(aq) + 2HCO3–(aq) → H2O(l) + CO2(g) + MCO3(s)
Reaction 2

It is important that the cations responsible for hard water be removed before the water is heated or used
for washing. Water softening, the removal of hardness ions from water, can be accomplished in several
ways. One method is the ion exchange process in which water is passed through a column containing
solid sodium aluminosilicates. Sodium aluminosilicates are high surface area three-dimensional
extended solids with –ONa groups at the surface.
M2+(aq) + Na2AlxSiyOz(s) → 2Na+(aq) + MAlxSiyOz(s)
Reaction 3

7) What happens to the pH of a soapy solution as a result of the introduction of hardness ions?
A. The pH increases as [H+] increases.
B. The pH is not changed since no acid-base reaction occurs.
C. The pH decreases as [OH–] decreases.
D. The effect on pH depends on the identity M2+
.
Answer: C
...Why is it C? Obviously you can cross out A since it's factually wrong. and D sounds wrong too. But I'm still confused as to why it's considered an acid/base reaction -_-

Let's take a look at the reaction in question:
M2+(aq) + 2NaC17H32CO2(aq) → 2Na+(aq) + M(C17H32CO2)2(s)

As you know all Na+ salts generally completely dissociate in water "(aq)" in the equation further confirms that. That means that prior to M2+ addition there were a bunch of Na+ and C17H32CO2- ions floating around. C17H32CO2- is a conj base of an acid and is thus basic. Na+ has no effect on pH. So we had a basic solution!

Now after the reaction all we have is Na+ (no effect on pH) and a solid (no effect on pH) -> so we have a neutral solution! Going from basic to neutral = lower pH. Answer C.

10) Which experimental approach can be used to analyze the metal content of soapy precipitate
produced by Reaction 1? Dissolve the solid in a known volume of :
A. 0.1 M NaHCO3(aq), then titrate with standardized 0.1 M HCl(aq) using an indicator.
B. 0.1 M NaOH(aq), then titrate with standardized 0.1 M HCl(aq) using an indicator.
C. 0.1 M NaCl(aq), then titrate with standardized 0.1 M NaOH(aq) using an indicator.
D. 0.1 M HCl(aq), then titrate with standardized 0.1 M NaOH(aq) using an indicator.
Answer: D

Okay so M(C17H32CO2)2 (s) + HCl --> M2+ + Cl- + C17H32COOH right ? What am I missing -_- why does titrating that with NaOH help you "analyze the metal content"? Isn't it just neutralizing the acid?.... I'm definitely missing something .. Is it asking how do you measure the metal concentration from titrating with NaOH?...

I am not as sure about this one, but I'll try:

Option A:
I would guess it'd go like this (if it dissolves at all):
M(C17H32CO2)2 (s) + NaHCO3(aq) -> M2+ + HCO3- + Na+ + 2C17H32CO2-
Now if we start titrating we don't know where to stop. There are 2 basic anions in the solution with different pKbs. What pH is the equivalence point at? We don't know their relative concentrations. (all this is assuming solid even dissolves, which I am not sure about).

Option B:
M(C17H32CO2)2 (s) + NaOH(aq) -> M2+ + OH- + Na+ + 2C17H32CO2-
Same thing again 2 basic anions with no knowledge of relative concentrations.

Option C:
M(C17H32CO2)2 (s) + NaCl(aq) -> M2+ + Cl- + Na+ + 2C17H32CO2-
I am not as certain about eliminating this one. Cl- is essentially a non-base and should not affect the pH. Here we just need to titrate away C17H32CO2- and if we know its pKb, it should be doable. Maybe solid does not dissolve well here or Cl- is still very very slightly basic? Also there might be some miniscule common ion effect due to Na+ presence, but I doubt it is much. Just makes this a little more iffy answer.

Option D:
M(C17H32CO2)2 (s) + HCl --> M2+ + Cl- + C17H32COOH
Pretty sure HCl will dissolve just about anything. Now we have pure C17H32COOH acid in water. M2+ and Cl- do not affect the pH. Every drop of basic NaOH will go toward neutralizing the acid. Once we see pH that we expect to see for an acid of this pKb, we know we hit it with enough moles to reach equivalence point.
Then we compare moles of base we poured in and that is how many moles of acid there was in the solution. Divide by volume and you got yourself a molarity. Best answer imo. Though C is closish too. Now just use coefs to get concentration of the M2+ ions.

 

[Incontrol]

So here is how I thought of answer 7:

7) What happens to the pH of a soapy solution as a result of the introduction of hardness ions?

C17H32CO2- is basic (or more correctly conj. base), when Hard Metals are added we have less free C17H32CO2- (because the it is now either attached to a hard metal or because of the common ion effect that reduces Na+ solubility). Because we have less base, the pH of the solution will be acidic.

for number 10.. I have no idea, but I have a question on it... will M2+ metals have low solubility with hydroxide?

 

[LoLCareerGoals]

So here is how I thought of answer 7:

7) What happens to the pH of a soapy solution as a result of the introduction of hardness ions?

C17H32CO2- is basic (or more correctly conj. base), when Hard Metals are added we have less free C17H32CO2- (because the it is now either attached to a hard metal or because of the common ion effect that reduces Na+ solubility). Because we have less base, the pH of the solution will be acidic.

for number 10.. I have no idea, but I have a question on it... will M2+ metals have low solubility with hydroxide?

Ca2+, Ba2+, Sr2+ will be soluble, others not.

 

[Sial8tr]

Let's take a look at the reaction in question:
M2+(aq) + 2NaC17H32CO2(aq) → 2Na+(aq) + M(C17H32CO2)2(s)

I was just looking at this. And, due to the M2 removing the Na, it causes the new complex to be a SOLID rather than AQUEOUS. That reduces the [OH].

FML