General Chemistry Question Thread

[QofQuimica]

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All users may post questions about MCAT, DAT, OAT, or PCAT general chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what general chemistry topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level gen chem.
-particular MCAT-level gen chem problems, whether your own or from study material
-what you need to know about gen chem for the MCAT
-how best to approach to MCAT gen chem passages
-how best to study MCAT gen chem
-how best to tackle the MCAT physical sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

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If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university general chemistry TA teaching experience. In addition, I teach general chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 14 on PS, 43 overall.

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the PS section of the MCAT, and 36 overall.

-Sparky Man: Sparky Man has his Ph.D. in physical chemistry. He scored 14 on the PS section of the MCAT, and 36 overall.

-GCT: GCT scored in the 99th percentile on the PCAT. He has also taught introductory physics and general chemistry.

 

[QofQuimica]

Can a mod clarify if what the OP said in above is correct? I learned that increased electronegativity of the conjugate base = more acidic (e.g. HCl > HBr > HI), but HF bucks the trend because of its strong bond, not because of its electronegativity?

Actually, HI is the strongest haloacid, followed by HBr, then HCl, and then HF (HF is not even a strong acid). Electronegativity does help the conjugate base stabilize the negative charge, but it is not the only factor that matters. Bond length and atom polarizablility, for example, also matter. Here is another Q from the organic thread that you might also find helpful:

http://forums.studentdoctor.net/showpost.php?p=3989250&postcount=315

 

[QofQuimica]

In an electrolytic cell, does there need to be two seperate cells? or can the anode and cathode be together in one beaker without a salt bridge?

Please do not post in the explanations threads; those are only for didactic posts. You shoud post your questions in the questions thread; I went ahead and moved this one for you.

In answer to your question, no, it is not necessary to have separate cells for electrolytic reactions. This is because they do not occur spontaneously, so if you don't apply energy to the system, the reaction will not occur even if the reagents are all in the same flask together. Galvanic reactions, on the other hand, do need to be separated if you wish to have control over the reaction (which is spontaneous) and get any useful work out of it.

I'm going with no... you need two separate cells (not just one beaker).

Again, this is incorrect for electrolytic cells. See above.

 

[legobikes]

Another quick question: forming bonds is exothermic, therefore breaking them is endothermic. I wanted to know why ATP, as an exception, releases energy when hydrolyzed to ADP/AMP (I mean, what exactly makes its bonds 'high energy') and if there are other major exceptions.

Thanks!

 

[CanadianPre-Med]

Another quick question: forming bonds is exothermic, therefore breaking them is endothermic. I wanted to know why ATP, as an exception, releases energy when hydrolyzed to ADP/AMP (I mean, what exactly makes its bonds 'high energy') and if there are other major exceptions.

Thanks!

The bonds are actually of low energy. Since they have a low energy of activation, they are broken easily and release a lot of energy.

 

[QofQuimica]

Another quick question: forming bonds is exothermic, therefore breaking them is endothermic. I wanted to know why ATP, as an exception, releases energy when hydrolyzed to ADP/AMP (I mean, what exactly makes its bonds 'high energy') and if there are other major exceptions.

Thanks!

That's a good question. You are correct that breaking bonds is endothermic; however, ATP hydrolysis is exothermic overall. To see why, take a look at the structure of ATP that I pasted below. You will notice that there are a lot of negative charges in the molecule (four to be exact), and they are all in very close proximity to one another. This makes the ATP molecule reactive (i.e., at a high potential energy). You've no doubt learned in physics that like charges repel one another, and the relief of that negative-negative charge repulsion upon ATP hydrolysis is the main reason why breaking the phosphate bonds in ATP releases more energy than it requires.

 

[Creightonite]

Question from AAMC test #6, Question #53

You have N2 and 3xH2 --> 2xNH3

The introduction of the catalyst will cause the ammount of ammonia at equlibrium to:

A) to increase
B) to remain the same
C) to decrease
D) change in the manner dependant on the equilibrium constant

The answer is B. I thought that catalyst lowers activation energy. Based on a formula, then increases the rate constant and by default will make Keq larger. How is my logic wrong?
Thank you for taking your time to respond.

 

[87138]

Question from AAMC test #6, Question #53

You have N2 and 3xH2 --> 2xNH3

The introduction of the catalyst will cause the ammount of ammonia at equlibrium to:

A) to increase
B) to remain the same
C) to decrease
D) change in the manner dependant on the equilibrium constant

The answer is B. I thought that catalyst lowers activation energy. Based on a formula, then increases the rate constant and by default will make Keq larger. How is my logic wrong?
Thank you for taking your time to respond.

Catalysts do not change the equilibrium constant. It speeds up BOTH the forward and reverse reactions, without changing the equilibrium constant or the equilibrium concentrations.

 

[Krazykritter]

Question from AAMC test #6, Question #53

You have N2 and 3xH2 --> 2xNH3

The introduction of the catalyst will cause the ammount of ammonia at equlibrium to:

A) to increase
B) to remain the same
C) to decrease
D) change in the manner dependant on the equilibrium constant

The answer is B. I thought that catalyst lowers activation energy. Based on a formula, then increases the rate constant and by default will make Keq larger. How is my logic wrong?
Thank you for taking your time to respond.

For a given rxn w/ given quantities of substrate, the only thing a catalyst does is lower activation energy. This will make the rxn occur more quickly, but equilibrium is equilibrium.

All this question is testing is to see whether or not you know what a catalyst does. The definition of a catalyst is anything that lowers the energy of activation without being consumed itself in the rxn. Therefore, a catalyst changes the rate of rxn but has NO EFFECT on the Keq. Hope this helps.

 

[Cloudcube]

No. Remember, a mole is a count of something; it's not a measure of the mass of stuff that you have. So I can have more moles of chromium than I do of dichromate. Let's use a silly example and see if this helps make the concept clearer: Say I have a dozen eggs. That means I have 12 of them, right? Each egg has one yolk, so I have a dozen yolks, too. But if you ask me how many albumin molecules I have in a dozen eggs (albumin is the major protein in egg whites), each egg contains many, many molecules of albumin. Let's say for the sake of this example that every egg contains 1000 albumin molecules. In that case, if I have a dozen eggs, and each one has 1000 albumins in it, then I have 12,000 albumins, right?

Ok, so analogously, let's say I have 10 molecules of dichromate. Each molecule of dichromate has 2 chromium atoms in it. So for my ten molecules, I would have 20 chromium atoms, right? Ok, so now think bigger. Let's say I have an entire mole of dichromate. (Remember that a mole is 6.022 x 10^23 molecules.) Still, every dichromate molecule contains two chromiums. So I must therefore have 2 x 6.022 x 10^23 atoms of chromium, which is equal to two moles.

I think I get it. Thank you very much QofQuimica!

 

[Creightonite]

I was a bit confused by this as well. From what I gathered in the EK Inorgo Book is that a catalyst changes the rate constant indirectly by changing the energy of activiation. Here's the formula for the rate constant: k=zpe^(-Ea/RT) where Ea is activation energy. So I'm assuming the catalyst does it indirectly by lowering Ea and not directly chaning the rate constant. I find this funny because on page 30 of EK Chem it says "The value of the rate constant depends upon pressure, catalysts, and temperature." Don't worry, I missed a few of EK questions becaues of confusion from the lack of details.

this is how catalyst increases the rate constant. I understand that at any given time there were be more products than in a reaction without catalyst. So equilibrium will be reached faster... but it does not change the relative number of products at the Eq. Am I right?

 

[87138]

this is how catalyst increases the rate constant. I understand that at any given time there were be more products than in a reaction without catalyst. So equilibrium will be reached faster... but it does not change the relative number of products at the Eq. Am I right?

In case it helps clear anything up at all, this is taken verbatim from page 656 of "Chemistry: The Central Science", 10th edition, Brown/LeMay/Bursten:

"The activation energy of the forward reaction is lowered to the same extent as that for the reverse reaction. The catalyst thereby increases the rates of both the forward and reverse reactions. As a result, a catalyst increases the rate at which equilibrium is achieved, but it does not change the composition of the equilibrium mixture. The value of the equilibrium constant for a reaction is not affected by the presence of a catalyst."

 

[harrypotter]

I thought I had finally gotten a handle on this but a question on a practice test proved me wrong.

Take Ti[22], the solutions say the electrons from 3d^2 will participate in a bond. Those are the valence electrons.

Now I looked up valence electrons and it was defined as electrons in the outermost shell. .. Wouldn't the electrons in the outermost shell be 4s because the principle quantum number of 4 is bigger than 3?

I'm pretty sure this is simple but I can't get it. Please help Q~ 🙂

 

[tik-tik-clock]

I thought I had finally gotten a handle on this but a question on a practice test proved me wrong.

Take Ti[22], the solutions say the electrons from 3d^2 will participate in a bond. Those are the valence electrons.

Now I looked up valence electrons and it was defined as electrons in the outermost shell. .. Wouldn't the electrons in the outermost shell be 4s because the principle quantum number of 4 is bigger than 3?

I'm pretty sure this is simple but I can't get it. Please help Q~ 🙂

Ti is a transition state element. And for all transition state elements, the 3d shell comes before the 4s shell,so, [Ar] 3d^2 4s ^ 2 likewise, in the 5th row the configuration would be [Kr] 4d 5s

I just remember this as a property of all the transition elements, there's some pretty wierd logic behind why this happens, but we dont need to know that.. So i d just remember it as it is..Hope I understood your question correctly..

 

[Foolins]

Just use that highschool diagram trick

1s
2s 2p
3s 3p 3d
4s 4p 4d etc.....and draw diagonal lines like this /

so you get 1s
then 2s
then 2p 3s
then 3p 4s
etc etc

 

[QofQuimica]

I thought I had finally gotten a handle on this but a question on a practice test proved me wrong.

Take Ti[22], the solutions say the electrons from 3d^2 will participate in a bond. Those are the valence electrons.

Now I looked up valence electrons and it was defined as electrons in the outermost shell. .. Wouldn't the electrons in the outermost shell be 4s because the principle quantum number of 4 is bigger than 3?

I'm pretty sure this is simple but I can't get it. Please help Q~ 🙂

Both the 3d and the 4s electrons are valence electrons for titanium. You can't just go by the principle quantum number when you talk about transition metals like this, because the 3d and 4s orbitals are very close together in energy. In fact, they are close enough that 4s actually fills first before 3d does. (4s is slightly lower in energy than 3d.) Confusing the issue is the fact that once they are both filled with electrons, the 4s orbital becomes higher in energy than the 3d orbital is. So, the 4s fills before the 3d does, and the 4s empties before the 3d does also.

 

[Jwax]

Yes, you need to reverse the sign when you flip the equation, but you do not multiply by the number of moles because EMF is an intrinsic property (independent of mass). If you look in the gen chem explanations thread, one of the posts has an explanation about extrinsic versus intrinsic properties.

Just got to this. Thank you for that; if I hadn't read your response I would have gone into the test saying, Screw their explanation I'm doing it my way! 😉

Also, I 😍 your avatar or whatever it's called. It cracks me up everytime!!!

 

[sssddd]

Hello new question. Are catalysts included in a rate equation, will it ever be? If does is there a limit to how much they affect the rate.

 

[xanthomondo]

Hello new question. Are catalysts included in a rate equation, will it ever be? If does is there a limit to how much they affect the rate.

im pretty sure the only thing you need to know about catalysts is that they increase the rate of reaction by lowering the activation energy, i dont think you will need to calculate anything involving them


if you hadd to use it in an equation you would use the k=Ae^(-Ea/RT), but thats beyond the mcat

and no theres no limit to how much they affect the rate, as long as it makes physical sense

 

[QofQuimica]

Hello new question. Are catalysts included in a rate equation, will it ever be? If does is there a limit to how much they affect the rate.

No. Remember that rate laws are the product of the rate constant k and the concentrations of the reagents that are involved in the rate-determining step. Since catalysts do not get consumed, and are present in way less than stoichiometric amounts, they will not appear in the rate law, at least not directly. Indirectly, of course, they will appear because they will increase the value of k.

 

[87138]

No. Remember that rate laws are the product of the rate constant k and the concentrations of the reagents that are involved in the rate-determining step. Since catalysts do not get consumed, and are present in way less than stoichiometric amounts, they will not appear in the rate law, at least not directly. Indirectly, of course, they will appear because they will increase the value of k.

Is it accurate to say, then, that catalysts increase the forward rate K as well as the backward rate K, while keeping the Keq constant?

 

[deleted106503]

Both the 3d and the 4s electrons are valence electrons for titanium. You can't just go by the principle quantum number when you talk about transition metals like this, because the 3d and 4s orbitals are very close together in energy. In fact, they are close enough that 4s actually fills first before 3d does. (4s is slightly lower in energy than 3d.) Confusing the issue is the fact that once they are both filled with electrons, the 4s orbital becomes higher in energy than the 3d orbital is. So, the 4s fills before the 3d does, and the 4s empties before the 3d does also.

Okay, so for Ti, the 4s is slightly lower tha its 3d valence e-. Is it safe to assume it the same for Nb where the 5s is lower than 4d? Now for the part that has the f block, i.e. Ta in the d block, how does its d valence compare to its s valence? Does the f block make the difference very large compared to the "slightly lower" difference between Ti's s and d since there is no f?

 

[QofQuimica]

Is it accurate to say, then, that catalysts increase the forward rate K as well as the backward rate K, while keeping the Keq constant?

Yes. Catalysts get you to equilibrium faster (little k), but the end position at equilibrium (big Keq) is the same.

 

[QofQuimica]

Okay, so for Ti, the 4s is slightly lower tha its 3d valence e-. Is it safe to assume it the same for Nb where the 5s is lower than 4d? Now for the part that has the f block, i.e. Ta in the d block, how does its d valence compare to its s valence? Does the f block make the difference very large compared to the "slightly lower" difference between Ti's s and d since there is no f?

Sorry, but I don't know. And I promise that you don't need to know that for the MCAT, either. 😉

 

[grapeflavorsoda]

When energy is transferred between a system and its surroundings it can be transferred in the form of work or heat. To understand how work is involved a derivation may help. Consider a gas which is confined to a cylinder with a movable piston. Furthermore, consider a lead shot of some weight on top of the piston so that at rest the weight is balanced by the pressure of the gas. If we remove some lead shot, the gas will push upward through some differential displacement. We can relate this to the work by:

dW = Fds
= (pA)(ds)
= p(Ads)
=pdV
​

dV represents the differential change in volume. Integrating from an initial to final state to solve the integral yields the following equation which relates pressure and volume changes to work:

W = P [V final - V initial]​

Don't worry about having to derive this equation, just know the equation. This derivation tells us that when a gas expands the work done is positive and when the gas is compressed the work done is negative. Some books actually say the opposite. However, the important thing to remember is to use one sign convention and to assign positive and negative quantities appropriately--the answer will come out this way.

So how do we use this equation in thermodynamics?​

There are many different processes but for MCAT purposes know the difference between a closed process and a open process. In a closed process, in the example above the piston would be locked to prevent the gas from expanding, no work can be done because the volume is held constant. Because the volume is held constant energy can be transferred in the form of heat. This means that the change in internal energy, from the first law of thermodynamics, is equal to the heat changes that occur. Furthermore, since no work is taking place, enthalpy changes are also equal to zero. However, in a open system, work does take place. A open system is one at constant pressure. Now a material, such as a gas, can expand or compress leading to changes in work. Furthermore, at constant pressure enthalpy changes can take place--this is because a gas can expand or compress (not only a gas but any material).

Ultimately, the difference between an open and a closed system is the work done. Energy is transferred between the system and surroundings in both

the differential of work function used in thermodynamic has one significant implication that gen. chem or undergrad. p.chem most likely to skip. it is only valid for so called "infinitesimal process."

why? because differential function is a linear map that approximates a continous function(it is defined this way actually) and for a physical process to fit into continuous function it must occur in so called "infinitesimal process."

thermodynamic does not define nor mention something called closed and open work.

only difference between an open and a closed system is that in open system material exchange with surrounding can occur but not in closed system.

in isochoric system no work can be done. but even in that process enthalpy change corresponding to a process does not have to be zero;

H = E + PV, so unless the process involved is both isochoric and isothermal, enthalpy change is not necessarily zero(you just could heat up the system and you will end up with positive value of enthalpy change).

on a side note, enthalpy was defined such that heat at constant pressure (thus isobaric condition)can be expressed in terms of state variables.

 

[Creightonite]

What's the deal with Faraday law and redox reactions in general? I do not exactly understand. For example, how many faradays is needed to do this:

1) Me--> Me2+

vs.

2) 2Me--> 2Me2+

i gues, it is needed 2F for the first and 4 Faraday for second reaction?

 

[christian15213]

All pre-health test students should memorize this list of strong acids and strong bases. Any acid not on this list should be considered to be a weak acid. Most other bases are also weak bases, but realize that there are also some strong organic bases. I have listed some examples.

Strong Acids:
-H2SO4
-HNO3
-HCl
-HBr
-HI
-HClO3
-HClO4
*note that HF is NOT a strong acid, and its omission from the list is not an accident!!!
**only the first proton on H2SO4 is strong (i.e., completely dissociates); the second one is weak.

Strong Bases:
-Group I elements with hydroxide (ex. NaOH, KOH)
-Group II elements with hydroxide [ex. Ca(OH)2]
-some organic bases, including alkoxide ions, sodium hydride, Grignard reagents, and LDA

See this is kinda what I mean... just broading this a bit and giving a sort of definiation of each acid and why it is strong medium or week... and what it is typically used for i.e. oxidizer, reagent, and or catylist... etc...

 

[tracy34]

How do you determine how many microFardays is needed to oxidize one mole of metal?

 

[yenisha]

How do you deal with "percent by mass" problems, as in this one:


A 10.0 % aqueous solution of fructose by mass has a density of 1.53 g/ml. What are the molarity and the molality of sucrose in this solution?

 

[Dave_D]

How do you deal with "percent by mass" problems, as in this one:


A 10.0 % aqueous solution of fructose by mass has a density of 1.53 g/ml. What are the molarity and the molality of sucrose in this solution?

Oh it's not that bad, just use a volume that makes things easy. So do this, they tell you the solution has a density of 1.53 g/ml. The easiest volume to use is the liter so convert to that. So the density is 1530 g/L. (I just multiplied by 1000 since there's 1000ml per liter.) Ok, so if we have 1 L of solution it has a mass of 1530 g, 10% of that mass is the fructose or 153 g of fructose in 1 L of solution. From here it's pretty straight forward.

 

[Tanner82]

Oh it's not that bad, just use a volume that makes things easy. So do this, they tell you the solution has a density of 1.53 g/ml. The easiest volume to use is the liter so convert to that. So the density is 1530 g/L. (I just multiplied by 1000 since there's 1000ml per liter.) Ok, so if we have 1 L of solution it has a mass of 1530 g, 10% of that mass is the fructose or 153 g of fructose in 1 L of solution. From here it's pretty straight forward.

not so fast, just had a question exactly like this in p chem... the volume can change due to IM interactions when two solutes are mixed, so that reasoning cannot be taken as accurate

 

[Tanner82]

oh ok the density of the solution, thought it was solvent... anyways a question similar to this I had in pchem lab.....

what is the mass% of H202 to H20?

I know....
moles H202
volume H202
Molarity H202
density pure H202
density pure H20

 

[yenisha]

I've spent a lot of time trying to balance this......and I'm still working on it! Can anyone help?


NaClO4 + C12H22O11 --> NaCl + CO2 + H2O

 

[Nutmeg]

I've spent a lot of time trying to balance this......and I'm still working on it! Can anyone help?


NaClO4 + C12H22O11 --> NaCl + CO2 + H2O

This is pretty straight forward. The Na:Cl ratio is the same on both sides, so you don't need to look at them independently. You need to see next that all of the water in C12H22O11 goes to water, and that in that molecule, the H:O ratio is 1:2, just as in water. So C12H22O11 makes 11H2O. What's left is 12 Carbon atoms. Each carbon atom needs another 2 oxygen atoms to make CO2, because no oxygen is left from the C12H22O11 (it all went into water). So to get the needed 24 atoms of Oxygen for your 12 molecules of CO2, you need 24/4 = 6 molecules of NaClO4. What you have left over is 6NaCl.

Altogether:

6NaClO4 + C12H22O11 ---> 6NaCl + 12CO2 + 11H2O

Each side has 6 Na, 6 Cl, 35 O, and 22 H.

 

[yenisha]

Thanks DaveD, Tanner82, and Nutmeg=) 😍

Now another *silly* question I am having trouble with:


1000mg of HC9H7O4 is dissolved in 500 ml of water. What is the pH of this solution? (Ka of 3.3 x 10–4?)

I know that I'm supposed to apply Henderson Hasselbach to this one, but how do I figure out the concentrations of the conjugate base and the acid?

 

[spicedmanna]

Thanks DaveD, Tanner82, and Nutmeg=) 😍

Now another *silly* question I am having trouble with:


1000mg of HC9H7O4 is dissolved in 500 ml of water. What is the pH of this solution? (Ka of 3.3 x 10–4?)

I know that I'm supposed to apply Henderson Hasselbach to this one, but how do I figure out the concentrations of the conjugate base and the acid?

I'm having trouble visualizing this organic compound. Is it an acid, base, or is it both? Given that only one Ka value is given, I'm going to assume that it's a monoprotic organic acid.

In that case, this problem is a typical ICE chart problem. Set it up this way:

HA ---> H+ + A- (some acid dissociates in soln by X amt)

I) [HA] | 0 | 0
C) -X | +X | +X
E) [HA]-X | +X | +X

So, writing the equilibrium equation, it looks like this, by substitution:

Ka = [H+][A-]/[HA] = (X)(X)/([HA]-X) = (X^2)/([HA]-X) = 3.3E-4

The concentration of HA is given (let's assume that the total solution volume is 0.500L):

1.000g of HA (108g/mol) in 0.500L ---> [HA] = (1.000/108)/0.500 = 1.85E-2

Since [HA] >> Ka, we can drop the "X" term in the denominator of the equation (it won't significally effect the answer):

3.3E-4 = (X^2)/[HA] = (X^2)/(1.85E-2)

Rearranging the equation to solve for X:

sqrt[(1.85E-2)(3.3E-4)] = X = [H+] = 2.5E-3

Oops! X might be significant after all and might need to be taken into account. IF this was for a chemistry class and not for the MCAT, we would need the quadratic formula, otherwise I'd stick with the estimated value.

First, let's rearrange the equation:

(3.3E-4)(1.85E-2 - X) = X^2
(6.1E-6)-(3.3E-4)X-(X^2) = 0

X = [(3.3E-4)-{sqrt[((3.3E-4)^2)+4(6.1E-6)]}]/(-2)
X = 2.31E-3

Our estimate was off by 2E-4, so if the pH answer you are looking for is 2 significant figures and not three or greater, then you could have stuck with estimate.

Now since we finally have [H+], we can find pH thusly:

pH = -log [H+] = 2.64

I'm not sure about the math, but the method is sound.

Good Luck!

 

[jace's mom]

*

 

[jace's mom]

What I've gathered so far:

1) Rutherford demonstrated that an atom has a dense, positively charged nucleus that makes up only a small part of the volume of the atom.

2) Planck developed the quantum theory, which says that energy is emitted from matter in the form of electromagnetic radiation. This energy is emitted in bundles called quanta. The energy value of a quantum is determined by E=hf, where h is Planck's constant and f is the frequency of the radiation. This can also be written (I think) as E=hc/lambda, where c is the velocity of light and lambda is the wavelength of the radiation.

3) Bohr used Rutherford's work to theorize that the hydrogen atom consisted of a central protein with an electron that passes around it in a circular orbit. The circular part isn't really valid, as we now know, because Bohr didn't look at atoms with more than one electron. Classical mechanics would tell us that the velocity and radius (and thus the angular momentum and kinetic energy) of the electron could take on infinitely many values. However, Bohr used Planck's ideas and basically said that angular momentum and kinetic energy depend on the quantum number.

4) Bohr's model can be used to explain atomic emission and absorption spectra. At room temperature, atoms tend to be in ground state, where the electrons are at their lowest energy levels. However, they can be excited to higher energy levels. When electrons go from higher to lower levels, they emit energy, and when they go from lower to higher, they absorb energy. Each element has a unique atomic emission spectrum (and I'm assuming a unique absorption spectrum as well) that acts like a fingerprint. For each element, wavelengths of emission and absorption spectra directly correspond. Bohr's model of the hydrogen atom looked at the atomic emission spectrum for hydrogen, defining the Balmer series (n>2 to n=2, visible) and the Lyman series (n>1 to n=1, UV). When the spectrum of hydrogen is calculated using Planck's quantum theory, it matches closely with the transitions predicted by Bohr's model (the energy of the emitted photon is the difference between the higher-energy initial state and the lower-energy final state).

Do I have the general idea down well enough? It's been 8 years since I took gen chem in college. Also, how might I be expected to apply this? Is just understanding the concept enough, or are there formulas/math problems I will need to calculate? I apologize if this question seems too simple, but I feel like I'm just beginning here. Thanks!

 

[QofQuimica]

What I've gathered so far:

1) Rutherford demonstrated that an atom has a dense, positively charged nucleus that makes up only a small part of the volume of the atom.

2) Planck developed the quantum theory, which says that energy is emitted from matter in the form of electromagnetic radiation. This energy is emitted in bundles called quanta. The energy value of a quantum is determined by E=hf, where h is Planck's constant and f is the frequency of the radiation. This can also be written (I think) as E=hc/lambda, where c is the velocity of light and lambda is the wavelength of the radiation.

3) Bohr used Rutherford's work to theorize that the hydrogen atom consisted of a central protein with an electron that passes around it in a circular orbit. The circular part isn't really valid, as we now know, because Bohr didn't look at atoms with more than one electron. Classical mechanics would tell us that the velocity and radius (and thus the angular momentum and kinetic energy) of the electron could take on infinitely many values. However, Bohr used Planck's ideas and basically said that angular momentum and kinetic energy depend on the quantum number.

4) Bohr's model can be used to explain atomic emission and absorption spectra. At room temperature, atoms tend to be in ground state, where the electrons are at their lowest energy levels. However, they can be excited to higher energy levels. When electrons go from higher to lower levels, they emit energy, and when they go from lower to higher, they absorb energy. Each element has a unique atomic emission spectrum (and I'm assuming a unique absorption spectrum as well) that acts like a fingerprint. For each element, wavelengths of emission and absorption spectra directly correspond. Bohr's model of the hydrogen atom looked at the atomic emission spectrum for hydrogen, defining the Balmer series (n>2 to n=2, visible) and the Lyman series (n>1 to n=1, UV). When the spectrum of hydrogen is calculated using Planck's quantum theory, it matches closely with the transitions predicted by Bohr's model (the energy of the emitted photon is the difference between the higher-energy initial state and the lower-energy final state).

Do I have the general idea down well enough? It's been 8 years since I took gen chem in college. Also, how might I be expected to apply this? Is just understanding the concept enough, or are there formulas/math problems I will need to calculate? I apologize if this question seems too simple, but I feel like I'm just beginning here. Thanks!

Looks like you've got a good handle on it to me. 🙂 Understanding concepts is MUCH more important than memorizing formulas or doing calculations as far as performing well on the MCAT is concerned. This is because on the MCAT, most problems, including PS problems, do not require much if any calculation. So if you find yourself spending a long time trying to work out a math problem, you aren't doing it right. The testmakers know that you don't have a calculator, and they know that you don't have enough time to solve complicated mathematical problems. The MCAT really doesn't reward people who plug and chug, and in fact, relying heavily on that strategy is bound to lead to a poor PS score. That being said, you probably should memorize that E = hf and c = (lambda)f, because those are useful equations to know and they might not give them to you.

 

[Daharsh]

Hello. I'm having problems with a chemistry problem:

Write the net ionic reaction for the neutralization of HF(aq) by NaOH(aq).

And the apparent answer:

HF(aq)+NaOH(aq) - NaF(aq) + H2O(l)

Is incorrect.


Please help me!

 

[spicedmanna]

Hello. I'm having problems with a chemistry problem:

Write the net ionic reaction for the neutralization of HF(aq) by NaOH(aq).

And the apparent answer:

HF(aq)+NaOH(aq) - NaF(aq) + H2O(l)

Is incorrect.


Please help me!

Remember, in a net ionic equation you cancel ionic species from the overall equation when they are present on both sides of the chemical equation (e.g., an ion that appears on both the reactant and product side):

1) HF is a weak acid (electrolyte), so it stays intact.
2) NaOH is a strong base (electrolyte), so it separates out in solution as Na+ and OH-
3) NaF, I believe is soluble, so it separates out into (Na+) and (F-)

Write it all out based on the above:

HF (aq) + (Na+) + (OH-) ---> (Na+) + (F-) + H2O (l)

Cancel out the ionic species that appear on both sides of the equation:

HF (aq) + (OH-) (aq) ----> (F-) (aq) + H2O (l)

Voila!

 

[jace's mom]

but the fact that I can't see what I'm doing wrong with this one is driving me nuts! I suspect I'm not understanding something about the use of the formula, but it's the only one I have in my book (and it's given without much explanation and no examples). Please help!

"Calculate the fraction of atoms in a sample of argon gas at 400K that have an energy of 10.0 kJ or greater."

So 10 kJ = 10000 J

And -Ea/RT = (-10000J-mol-K)/(8.314*400J-mol-K) = -3.1

So f = e^(-Ea/RT) = e^(-3.1) = .045

However, the book says the answer should be .023

Where am I going wrong? I felt like I was gaining momentum...really getting a grip on the different formulas involved in kinetics, and then I hit this. Thanks for your help!

 

[QofQuimica]

but the fact that I can't see what I'm doing wrong with this one is driving me nuts! I suspect I'm not understanding something about the use of the formula, but it's the only one I have in my book (and it's given without much explanation and no examples). Please help!

"Calculate the fraction of atoms in a sample of argon gas at 400K that have an energy of 10.0 kJ or greater."

So 10 kJ = 10000 J

And -Ea/RT = (-10000J-mol-K)/(8.314*400J-mol-K) = -3.1

So f = e^(-Ea/RT) = e^(-3.1) = .045

However, the book says the answer should be .023

Where am I going wrong? I felt like I was gaining momentum...really getting a grip on the different formulas involved in kinetics, and then I hit this. Thanks for your help!

Hmm. This formula is used to calculate the fraction of collisions that have a high enough activation energy to overcome the barrier to reaction. You don't have a reaction here, and your energy isn't an activation energy. Your answer is about double what the book has. I wonder if it's as simple as them dividing the answer by two because each collision is between two molecules? BME, if you are reading this, what do you think?

 

[jace's mom]

Yes, I thought it was strange that it wasn't activation energy. However, that's exactly how the problem reads, and there's no picture/diagram/reaction/example, etc to accompany it. I'm probably thinking about this far too much. On a more general note, what (if anything) should I know about this formula for the MCAT? How might I see it applied?

jace's mom (who will be quite happy to get through the Chemistry section and move on to something she likes...Biology!)

 

[gridiron]

Hmm. This formula is used to calculate the fraction of collisions that have a high enough activation energy to overcome the barrier to reaction. You don't have a reaction here, and your energy isn't an activation energy. Your answer is about double what the book has. I wonder if it's as simple as them dividing the answer by two because each collision is between two molecules? BME, if you are reading this, what do you think?

Yeah, that would be my guess as well because collision theory is based on the fact two or more particles collide. What is weird is that if you use the maxwell boltzmann distribution to calculate the fraction collisions that have more energy than the activation energy you get zero (the equation is e raised to (-Ea/KbT) where Kb is the boltzmann constant.). For MCAT purposes, I don't believe it is necessary to memorize this equation. It would probably appear in passage form and the passage would probably hint as to how to manipulate the equation---in this case divide by two if we consider two particles.

 

[jace's mom]

Got it. Thanks for your help!

By the way, you have no idea how comforting it is that it took a consultation between the two of you to answer the question. Each time I post in here, I'm afraid to come back and find every Tom, Dick, and Harry on SDN has replied saying "Duh! Divide by two, you *****!" At which point I will accept that I just might be the first person in the history of the MCAT to score a zero, and I will strongly consider quitting this pursuit and working for Taco Bell instead. 😀

 

[QofQuimica]

Got it. Thanks for your help!

By the way, you have no idea how comforting it is that it took a consultation between the two of you to answer the question. Each time I post in here, I'm afraid to come back and find every Tom, Dick, and Harry on SDN has replied saying "Duh! Divide by two, you *****!" At which point I will accept that I just might be the first person in the history of the MCAT to score a zero, and I will strongly consider quitting this pursuit and working for Taco Bell instead. 😀

You have no idea how many times I look up answers before I post about them here. 😉 The good thing about the MCAT is that it's an open-book test, JM. So don't feel bad about having to look up answers, because you can. 🙂

 

[jbmcely]

There is a question in my study guide that ask:
Which of the following is a possible set of quantum numbers for the last electron added to an atom of aluminum?

.....n.....l.....m1.....ms
A. 3.....2.....-1.....+1/2
B. 3.....1......2.....-1/2
C. 3.....3.....-1.....+1/2
D. 3.....2......3.....+1/2

Answer given is A

Maybe I am missing something but I thought that the numbers would be 3,1,-1,+1/2 respectively. Since the electron configuration for Al is Ne,3s2,3p1. Any help would be appericated been along time since I have seen gen chem.

 

[jace's mom]

There is a question in my study guide that ask:
Which of the following is a possible set of quantum numbers for the last electron added to an atom of aluminum?

.....n.....l.....m1.....ms
A. 3.....2.....-1.....+1/2
B. 3.....1......2.....-1/2
C. 3.....3.....-1.....+1/2
D. 3.....2......3.....+1/2

Answer given is A

Maybe I am missing something but I thought that the numbers would be 3,1,-1,+1/2 respectively. Since the electron configuration for Al is Ne,3s2,3p1. Any help would be appericated been along time since I have seen gen chem.

I am by no means an expert, but maybe this can help. My Kaplan book says "IIIA-VIIA elements beyond period II might, under some circumstances, accept electrons into their empty d subshell, which gives them more than 8 valence electrons." The Octet Rule doesn't always hold for Al. So choice A could work, because l=2 would be the d subshell.

Of course, it would be great if someone else who actually knows Chemistry could confirm this... 😉

 

[heymanooh1]

There is a question in my study guide that ask:
Which of the following is a possible set of quantum numbers for the last electron added to an atom of aluminum?

.....n.....l.....m1.....ms
A. 3.....2.....-1.....+1/2
B. 3.....1......2.....-1/2
C. 3.....3.....-1.....+1/2
D. 3.....2......3.....+1/2

Answer given is A

Maybe I am missing something but I thought that the numbers would be 3,1,-1,+1/2 respectively. Since the electron configuration for Al is Ne,3s2,3p1. Any help would be appericated been along time since I have seen gen chem.

I am by no means an expert, but maybe this can help. My Kaplan book says "IIIA-VIIA elements beyond period II might, under some circumstances, accept electrons into their empty d subshell, which gives them more than 8 valence electrons." The Octet Rule doesn't always hold for Al. So choice A could work, because l=2 would be the d subshell.

Of course, it would be great if someone else who actually knows Chemistry could confirm this... 😉

Answers B and D are out because the possible ranges of m1 don't correlate with the respective value of l given. That leaves choices A and C. Answer C doesn't work because from the given value of l =3 this designates occupation of an f-orbital. The problem with answer C is that l = n - one which is not possible for the value of n given and filling in f-orbitals is possible only when n = 4 or greater. That leaves answer A.

The last electron should have quantum numbers n=3, 1 = one, m1 can either be -1, 0, or 1 and ms = +/- 1/2 since we don't know which p-orbital is filled and whether in a spin up/down fashion.

For simplicity's sake let's say that the quantum #s for last electron to be filled in aluminum is: n = 3, l = one, m1 = -1, ms = +1/2. This designates the ground state electron configuration in aluminum. Now answer A is possible because it is a set of quantum #s for an electronically excited state of aluminum i.e. the lone p-orbital electron (l = one) goes in to the next highest available orbital which is 3d (l = 2) in this case. Choices B, C, D are not possible even if they were excited states for the reasons given in the first paragraph.

Also, rows IIIA-VIIA on the periodic table refers to transition metal/d-block elements which aluminum is not.

 

[heymanooh1]

Got it. Thanks for your help!

By the way, you have no idea how comforting it is that it took a consultation between the two of you to answer the question. Each time I post in here, I'm afraid to come back and find every Tom, Dick, and Harry on SDN has replied saying "Duh! Divide by two, you *****!" At which point I will accept that I just might be the first person in the history of the MCAT to score a zero, and I will strongly consider quitting this pursuit and working for Taco Bell instead. 😀

Working at In 'N Out in Southern California is better. If you become manager in one of the franchises, you earn 90K and upwards.