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When an enzyme is diluted with water, what is slowing down the rate of reaction?
I'm specifically talking about the decomposition of hydrogen peroxide by bovine catalase. The number of catalase molecule obviously doesn't change, so there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, and doesn't change the pH or temperature, then shouldn't the rate of reaction remain the same regardless of concentration, since every environmental conditions & the number of catalase molecule in the solution before dilution is the same after it is diluted?

[…] there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, […]

Q: Why is dating between a fixed number of guys and girls more difficult on a large square packed with nuns than in a small club (without the nuns)? ;)
A: It's more difficult to bump into each other since the girls a more far away and there are legions of nuns in between!
You are right in that the absolute amount of catalase molecules does not change upon dilution, but the concentration does!
Reactions are about collisions of the reactants and in the case of dilutions, there are more collisions with innocent bystanders than with an other reactant.

I've only got rudimentary of chemistry knowledge, so I'm paraphrasing your idea into terms I can understand. What's happening is the water molecule is interfering with the active site of the enzyme, preventing it from working at full capacity, thus lowering the rate of reaction. Therefore, more diluted solution means more water molecules means more interference, and an overall lower rate of reaction. Is this a correct understanding?

@LtotheV Exactly! Although the role of water molecules isn't necessarily an active preventing by blocking the active site (there will always be some water molecules there, e.g.  due to hydrogen bonding), It really is more about finding less catalase and $\ce{H2O2}$ molecules in the same volume element. Assuming a constant movement of molecules in solution, it just takes longer for the reactants to form a collision complex while making their way through the water. And since rate is conversion over time...

If you diluted the catalase, while keeping the peroxide concentration fixed (by adding both water and peroxide to the solution), then you're right, the total reaction rate (measured in reactions catalyzed per second, within the whole solution) would not change.
However, by adding just water to the solution, you're diluting both the enzyme and the peroxide.  The lower the peroxide concentration is, the longer it takes for each catalase enzyme unit to bump into another peroxide molecule after reacting with the previous one.


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How to reduce the turbidity of a water sample?
A task requires us to experiment water quality.
How would you reduce turbidity of a clay water sample with an effective and cost-effective method?

What will you test? Will the addition of precipitants interfere with your analysis? Did you have a look at this article?

we just need to purify the water using a cost-effective method

Is your intent to clean the water, or are you also looking to identify anything else, like impurities. What level of purification are we talking about? Should the filtered water be used for consumption or even other analytical experiments? What do you understand as cost effective? Distillation is certainly one possibility, but it is hard to judge how cost effective this can be considered. To get ultrapure water, it might be your only choice (repetitive distillation).

we are looking to purify the water sample. we need to conduct an experimental investigation of a water sample (with clay particles) to determine the most cost effective method for purifying the water sample.

Apart from the article linked in a comment above, possible options are

wait and let settle, then decant
use an established flocculant, such as potassium alum $\ce{KAl(SO4)2*12H2O}$

but however, if you decanted the mixture, wouldn't there still be a brownish colour liquid when you seperate the liquid and the solid?

@BobDylan It depends - and you asked for a cheap solution. It doesn't get any cheaper than that ;)

Another question. How did you come up with that formula? Why is there a 12H2O? And what does the dot mean in front of the H2O

@BobDylan I'm not quite sure what you you mean with "come up"? If it's about how to make it look that way, have a look at this and this. As for the $\ce{\cdot 12H2O}$  notation, please have a look at Water of crystallization on wikipedia. I hope that helps.

Expensive method is to use heat to evaporate the water, unless you have cheap heat (e.g. sunlight).
Cheap way is to push or suck it through a filter. You can buy bottles with filters in places that sell camping supplies.

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short/extensive training on computational chemistry
I want to take a training on computational works in chemistry. It should be of a short duration. Where in Europe I can find it?

I'm voting to close this question as off-topic because it is not about chemistry, but about making a career in chemistry. I believe it is better asked at academia.se.

Did you check the offerings of the national chemical societies and their special interest groups?
In Germany, that would be the CIC - Chemie Information Computer group of the GDCh - Gesellschaft Deutscher Chemiker.
Cosmologic provides TURBOMOLE workshops.
In 2014, the Max Planck Institute for Chemical Energy Conversion organized a Summer School during which a course on QM/MM with Frank Neese's ORCA was given. I don't know if they plan something similar this year.
The NSCCS has some upcoming events.

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Is there a visible change to baking soda when it reacts with odour?
For example, would a box of baking soda in a fridge lose weight as it absorbs odor? How long would it take to drop 100 g?

Let us assume that baking soda works by not simply absorbing, but reacting with acidic/odiferous substances via an acid-base reaction. So, for example if you had rancid butter in the fridge and it gave off the nasty-smelling butyric acid, that acid would react with the baking soda to give sodium butyrate plus carbonic acid (which breaks down to carbon dioxide and water). Let us assume that the carbon dioxide and water would evaporate. That would be a net gain of weight since the butyric acid molecule has a bigger mass that carbon dioxide and water.
Most odiferous substances will likely have a higher molecular weight than carbon dioxide plus water, so I believe over time the baking soda will increase in weight. This will also be true for those cases where there is no reaction - but simply absorption.

My gut tells me that the weight changes due to moisture balance will be more than those due to gas absorption or reactions.  However under controlled conditions the the weight change due to chemical changes should be detectable.


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How can placing a lone pair in axial position differs from that in equatorial in number of electron domains?
I need help understanding the following statement in my book (regarding trigonal bipyramidal geometry):

Placing a lone pair in an axial position would put it at 90 degree to three other electron domains. Placing it in an equatorial put is at 90 degree to only two other domains, thus minimizing the number of strong repulsive interactions

What does it mean by "90 degree" to two/three "domains"?

Have you tried to draw it?

Does this answer your question? Why do the lone pairs of electrons first occupy axial positions before occupying equatorial positions for species with octahedral geometry?

Here's a drawing of the situation.

In the example with the axial lone pair of electrons, when they say, "90 degree to three other electron domains", they just mean that the axial lone pair is situated 90° relative to 3 other substituent bonds - the 3 equatorial substituent bonds in this case.  When the lone pair is located in an equatorial position, then it has a 90° interaction with only 2 other bonds - the 2 axial substituent bonds in this case; the lone pair is 120° away from the 2 other equatorial substituents.


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Identifying the structure of an amino acid from NMR spectra
I have the 1H, 13C, and DEPT spectra of an unknown compound, listed in text form below, but I have also attached original images of the spectra. I know that it is an amino acid that has a cyclic structure and a relative molecular mass of 145. I'm supposed to deduce the structure, but am having some trouble with it:
1H NMR (in D2O): 4.2 ppm (1H, s), 3.85 ppm (1H, dd), 3.2 ppm (2H, m), 2.1 ppm (1H, d), 1.8 ppm (3H, m) (image of spectrum)
13C NMR: 174.27 ppm (quaternary), 61.92 ppm (CH2), 54.09 ppm (CH2), 38.69 ppm (CH/CH3), 32.87 ppm (CH/CH3), 28.04 ppm (CH/CH3) (image of 1D 13C spectrum; image of DEPT spectrum)
From the spectral data, I can tell that the compound doesn't have an aromatic ring, but beyond that I don't know how to proceed. Unfortunately, although it would be really helpful, I'm not able to determine the coupling constants in the 1H spectrum.

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Lennard-Jones potential and vibrational energy level diagram explanation
I don't understand how to interpret graphs like the following:
Forget the difference between the Harmonic and Morse (I think I get that), It's more what the horizontal lines mean and what the boundary line means. I know that vibrational energy is quantized and I can rationalize the shape of the Lennard-Jones potential but I don't understand how they're combined in this diagram. Please could you walk me through what's going on?

Each horizontal line shows an interval of possible internuclear distances for a particular vibrational level: the higher you get in the ladder of vibrational levels, the bigger is the amplitude of vibration, and consequently, the bigger is the corresponding interval of possible internuclear distances.

And, of course, as user1420303 correctly pointed out in his answer all of what I said in my comment above has to be taken as "speaking classically", i.e. in terms of classical physics. In reality, there is a non-zero probability for nuclei to be at a distance bigger than one allowed classically.

This diagram is conceptually horrible, because it is a mix of quantum and pseudo-classical physics.
Forget for a while the vibrations. If you calculate single point energies (within Born-Oppenheimer approximation) for different internuclear values, you'll get something like blue curve.
Now, for the $r_e$ distance between nucleus (again, within Born-Oppenheimer approximation) you can calculate the "vibrational (normally harmonic) frequency" in a pseudo-classic fashion. For this frequency, you can assign energies (of a quantum harmonic oscillator having this frequency). This energy values correspond to the difference between energy values for the horizontal blue lines, and the well (blue).
The widths of the horizontal lines have not physical meaning.
For the sake of completeness, the $D_e$ would be the energy of equilibrium state if no vibrations are considered. $D_0$ when the fundamental vibrational state energy is add. The former energy is not measurable because this is not a real energy for the system.

I agree that the diagram is indeed horrible. I think one should always plot probability densities for each vibrational level on a diagram of such kind (as in the corresponding Wikipedia article for example) to make it less horrible. :D


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What makes ceramic coated pans fail? Is it reversible?
If one reads reviews on ceramic glazed cookware (esp. frying pans) one pattern becomes apparent: the pan is awesome at first, nothing sticks to it, even with little to no oil. Then, after a few months the non-stick properties more or less plummet and the pan may become worse than stainless steel pans.
As to why that is, I've read multiple claims. Some say (or assume) that the non-stick layer erodes/dissolves/gets damaged. But to my humble knowledge ceramic is harder than steel, so that seems somewhat simplistic of an explanation. Some sources say that the coating reacts with oil when overheated (which will happen on high settings of the stove) and that residue is what dimishes the non-stickiness.
That sounded more plausible to me and I wondered if that proccess can be reversed/undone in some way?
So the first part of my question: how does a ceramic pan deteriorate? If it's not plain "erosion", what can be done to restore its properties? I'm afraid to scrub if harder for the fear of scrathing it.


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Hydrogen Fuel Cell and Battery
For hydrogen fuel cell to produce electricity, hydrogen  will need to be supplied to the anode. To split oxygen and hydrogen from water and obtain hydrogen, we need to use electrolysis. To use electrolysis, we need to use a battery.
My question is, why do we bother splitting up hydrogen and oxygen and then feed it into the fuel cell instead of just using battery to generate electricity? Or is it not practical to obtain hydrogen and oxygen from electrolysis and use it to generate electricity from fuel cell? Thank you

Well, you can have car with big pile of batteries or tank hydrogen.

If you have a fuel cell, only hydrogen needs to be carried with the car; the air provides all the oxygen you need. Hydrogen is an extremely light fuel, but there are some issues with current means to contain it (e.g. heavy high pressure container, or massive hydride tanks).
According to Iceland is Bullish on Hydrogen, they've started to use their plentiful geothermal and hydroelectric power to electrolyze water, using the hydrogen to power cars and the oxygen byproduct for other uses.

It is more likely they use the mains to electrolyze the water, but hydrogen is also a byproduct in the steam reforming of crude oil, which produced 11 million tonnes of hydrogen.
The difference is having to put a large number of heavy batteries into the car, or a smaller, lighter (but still reinforced) tank of hydrogen.


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Why should alkylation of enols/carbonyl compounds be done at low temperature with a strong base?
Alkylation of acetone:

This reaction is done with a strong base (LDA in this case). Is this because the protons aren't particularly acidic? Secondly it should be done at low temperatures; why is this so? -78 degrees Celsius is typical.

Why should alkylation of enols/carbonyl compounds be done at low
  temperature with a strong base?

If you start with an unsymmetrical ketone, then there are two different sides of the molecule you could remove a proton from. Let's use methyl ethyl ketone as an example.  Removing a proton from the methyl group leads to what is termed the kinetically favored anion (e.g. this hydrogen is the most acidic and the corresponding enolate anion is the fastest formed anion).  Removing a proton from the methylene group on the ethyl side leads to the thermodynamically favored anion (e.g the enolate anion with the most highly substituted [most stable] double bond).
Alkylation of carbonyls/enols is not always done using strong base and low temperature.  Strong base and low temperature generally leads to the kinetically favored enolate anion, while weaker bases and higher temperatures generally lead to the thermodynamically favored enolate anion.
In your case, acetone is symmetric so it doesn't really matter which side you alkylate.  But, once you form your desired product, methyl ethyl ketone, it can react further to give polyalkylated products.  These secondary products are more stable and thermodynamically preferred.  To reduce the likelihood of thermodynamic control leading to these undesired products, the reaction is run under kinetic control - strong base and low temperature - to maximize formation of the monoalkylated product.  Here is an earlier answer that further highlights kinetic vs. thermodynamic control in alkylation reactions.


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Benzylic Oxidation of Cyclic Molecules Using KMnO₄
What is the product of the permanganate oxidation of 1-methylindane (1-methyl-2,3-dihydro-1H-indene)?

There are two choices of benzylic protons (one proton on the carbon with the methyl group at the top of the cyclopentane, and two protons on the carbon in the cyclopentane at the bottom of the molecule).
I figure that the ketone is added to the carbon that has two benzylic protons, and that the other carbon with a benzylic proton is untouched. Am I correct?

For the homogenous permanganate oxidation of aromatic side chains, as long as there is at least one benzylic hydrogen, then the side chain will be oxidized to the corresponding benzoic acid.  Here are some examples.

(image source)
Look at the third example, the permanganate oxidation of tetralin to phthalic acid.  It is very analogous to your molecule.  Your molecule would also be oxidized to phthalic acid.

@ron what's the mechanism of oxidative cleavage?

@Dissenter  Some experiments, lots of theories, no one really knows.

@ron, I think we need an example of isopropylbenzene oxidation to the acid...

@Zhe the isopropyl case can be interpolated from the above data.

I'm not sure that's true. You have no example of a secondary alkyl case. Based on the series you have, is the requirement that we have a benzylic hydrogen or do we need 2?

@Zhe See p. 143 in this link .


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How do you list all possible reactions and intermediates of a reaction?
Ins this video https://youtu.be/txk-VO1hzBY?t=2391 the speaker claims that $$\ce{H2 + O2 -> H2O}$$ involves 23 reactions and 11 species. How do you list all of that ?

Usually, you don't ;)

You've discovered a secret. In undergraduate chemistry classes, the field of chemical kinetics is treated with a level of detail very far from reality. The truth is that virtually any reaction you can devise is vastly more complicated than we would like to think. This is especially true for gas phase reactions like the one you write down because it is possible for very unstable intermediates to exist for long enough to collide with lots of different species or undergo an internal conversion (when the molecule is large enough to do so) which introduces minute components that could nonetheless make a big difference in the kinetics.
Some people actually attempt to study these reactions without just ignoring the small sideways paths that lead to weird intermediates or weird products. Generally they do this by using what is called a Master Equation. I would encourage anyone to read about these and a very nice overview can be found here: Miller, J. A., & Klippenstein, S. J. (2006). Master equation methods in gas phase chemical kinetics. The Journal of Physical Chemistry A, 110(36), 10528-10544.
What you'll find when reading this is that they care a lot about what the potential energy surface looks like everywhere. This is because one formulation of the Master Equation is as follows:$$\frac{dn_i}{dt}=\sum_j(p_{ij}\cdot n_j(t)-p_{ji}\cdot n_i(t))$$This basically says that we are looking at the change in probability of finding the system in a state $n_i$ as a function of time and we are doing this by describing the probability per unit time $p_{ij}$ of moving from state $j$ to state $i$ as well as the process which goes the other way. Note that $n_i$ is essentially the same as a number density for a large number of molecules, but strictly this is a probability.
You can imagine then that in order to use this equation (which isn't very usable in its current form), one would need a detailed picture of what possible intermediates can form immediately from the state the system is in right now.
Thus, the answer to how you write all these down is that you look very carefully at the state of any system and ask yourself, what intermediates could I possibly imagine forming? Many of these intermediates are very unstable at low temperatures, but these kinds of models are frequently used when studying combustion chemistry or flames, so the number of intermediates can becomes very large and the number of reactions those intermediates can undergo (the number of states $j$ you can jump to from state $i$) is even larger. To operationalize this, you would basically write out all known intermediates, from experiment or theory, and imagine what reasonable combinations you could make of those that eventually lead to your product.
In the video you link, some unusual intermediates are listed, such as $\ce{HO2}$, and it is said that this is a simplified model. Basically what that means is that they had even more possible combinations of intermediates, but for computational simplicity or good experimental reasons, decided that many combinations of those intermediates contributed negligibly to the reaction of interest.
One final note is that it's not really as simple as combining everything you can imagine because you have to consider where a given intermediate came from. For instance if a single molecule is formed by means of a bimolecular reaction, the angular momentum must be conserved. This can lead to situation where certain states you could imagine hopping to are not actually possible because it would violate conservation of angular momentum. Indeed, one could imagine this leading to some reaction mechanisms which are very different from what intuition would dictate. The paper I linked above notes that when angular momentum becomes an important factor, the intermediate usually undergoes an internal conversion to basically store that angular momentum before reacting with other things... So, when you're writing out all of these intermediates don't forget that $\ce{HO2}$ might very well exist as $\ce{HOO}$ or $\ce{OHO}$ (though that seems unlikely I only meant it as an illustration).
I hope this addressed the spirit of what you were really asking.

You generally only list the intermediates of a net chemical reaction if they are relevant in some context of your study, experiment, discussion, etc. For example, if you have developed a model for the reaction of $\ce{H2}$ and $\ce{O2}$ that invokes 23 intermediate reactions, then you would want to explicitly show each of those reactions.
If you are only interested in the net reaction, the stoichiometry, etc. of a reaction that has many intermediates (and all but the simplest reactions will have many intermediates), then you simply give the net reaction.  In your example, the balanced reaction would usually just be given as
$$\ce{2H2 + O2 -> 2H2O}$$
and it is understood that this is the net reaction that may have a very complex mechanism with multiple intermediates species and reactions, but there is generally no need to list them all.


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What is the equation that demonstrates the relationship between molar conductivity and temperature?
Can someone please help me find the equation that defines the relation between molar conductivity and temperature. I have found some information, but it was not really that much complete.


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What does the pressure in the HPLC(High performance liquid chromatography) do?
Well I have to do research about the HPLC. It says, that because of the pressure the stationary phase can be made out of smaller particles and the pillars can be shorter. Also the seperation of the contents is supoosed to be superior, compared to other kinds of liquid chromatographies. Can somebody explain that to me?

The pressure in HPLC systems is not an intended feature, but a consequence of how the system works.
There are several adjustable parameters/properties of a HPLC/UPLC system that will influence on the internal pressure inside the chromatographic column. Some of those are:

Solvent flow:  A greater flow of solvents mean a faster chromatophy. On the other hand, faster runs also mean lower efficiency. An increase in flow leads to an increase in pressure.
Column length: A bigger column will improve the separation efficiency, but also increase the duration of the run. Bigger columns leads to a greater pressure.
Particle diameter: The lower the particle, the greater the separation efficiency, but this also leads to a greater pressure.
Particle pore size: This has the same effect as above, but is not the same thing.

Optmizing this parameters and/or improving the technology related to them is what generally drives the research in the HPLC/UPLC development field, besides of course find the appropriate type of column, solvent gradient, etc. You can't simply make a system where the pressure is as high as possible, or some connectiong will eventually burst. On the other hand, it's not interesting to have a chromatographic run that lasts several hours, or doesn't really separate the molecules, specially when dealing with hundreds or thousands of different analytes.
There is much more to the above in several books and thousands of papers, including very detailed descriptions, experiments and calculations that show a much more precise relation between those and other parameters.


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How to prevent losing solvent into the vacuum pump in rotovap solvent collection?
This is a practical question about the use of rotary evaporators ("rotovap"). So I'm asking for answers from people who actually use the things.
A rotovap typically uses a cold condenser to condense the solvent into a collection flask. But the distillation is typically done under weak vacuum as well. So the solvent sitting in the collection flask seem susceptible to re-vaporizing and getting sucked into the vacuum pump. Now I have a cold trap in front of my pump that should be cold enough (-50C) to trap most solvents under weak vacuum.
I think the answer is that if the vacuum is weak enough to vaporize the solvent under the hot water bath, but not under the cooler ambient air temperature, then the loss to the pump should be minimal. Indeed, this is what the instructions for the rotovap suggest.

Question: In practice, what are the risks to losing solvent into the pump?

We always kept the collection flask cooled with an ice bath; with volatile solvents we used a dry ice/acetone bath. The ice bath was just so easy to do, so why not.

Honestly, I never bothered to actually measure the volume of the condensed solvent in order to compare it with the original volume.
The concept described in the question, that is using a cooling trap with dry ice + acetone, is good practise and typically good enough to retain the evaporated solvent and protect the membrane pump. In rare cases, I have used a second, flat Dewar filled with dry ice + acetone on a "Laborboy" to cool the condensate-collecting flask.

Thanks--this confirms that having a trap in front of the pump, in addition to the condenser cooling, is probably a good idea, along with cooling of the receiver per ron's comment. Perhaps not vital, but why not, as ron says?

@trb456 ron's suggestion is actually a good one! The only "argument" against it is laziness. It requires an additional movement (adjusting the height of the cooling bath for the receiver) when you evaporate from flasks of different size ;-)

The idea of cooling the collection flask sounds good, but is actually useless. Makes no difference to the partial pressure of the solvent at the cooler, unless you let the rotary evaporator run for an extended time after the flask in the hot water bath is already dry.

In my experience, no further cooling other than the condensor above the collection flask is necessary. Remember that you are selecting a pressure which is low enough to boil the entire liquid at $40~\mathrm{^\circ C}$ (your flask’s temperature) but high enough so that it will recondense on cold water (I’m going to guess $15~\mathrm{^\circ C}$). Dripping down into the collection flask, the solvent will retain that temperature or at maximum reach room temperature again which should be just below a temperature for sufficient evaporation. If it does evaporate, it will be caught by the condensor.
Any solvent ‘pulled through the pump’ should be caught in a collection flask below a condensor on the high-pressure side of the pump. In $99~\%$ of my evaporations, nothing is there and the remaining percent are special cases.

Your cooling water should have around 10°C, and as long as you're not impatient and use a lower vaccum as recommended for your specific solvent, you should not loose much of it to the pump. Using an additional cold trap in front of it is not usually necessary. The manuals don't recommend or require it either.
Unless of course you (mis)use the roti to dry a product from solvent residuals. ;-)
Btw. The common membrane pumps are quite resistant to small amounts of solvent. What breaks first (if) are usually the seals, which are easy to replace. Don't soak it in chloroform, of course, and let it run dry if it got some. Put on empty flasks, leave the manual ventilation valve slightly open, and let it pump for ten minutes or so.


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What happens to a molecule when it is unstable?
I'm just starting OChem and I see that for a SN1 reaction, the rate of the reaction changes with the number of alkyl groups attached to the carbocation that is formed. The textbook (Klein) explains that this is due to the electron-donating nature of alkyl groups which helps to stabilize the carbocation. I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?
Thanks,
Reference: Klein, David. Organic Chemistry as a Second Language, First Semester.

Think of a reaction-energy diagram.  Specifically, think of a two-step reaction energy diagram linked like this one:
.
The carbocation is the intermediate $\mathbf{I}$.  The relative energy for alkyl-substituted carbocations will be lower.  This also practically means that $\mathrm{TS1}$ will have a lower relative energy as well.  The fact that $\mathrm{TS1}$ and $\mathbf{I}$ are of lower energy for a substituted carbocation (vs. a primary or methyl carbocations) means that it will be formed more often under the quasi-equilibrium assumption of transition state theory.  Basically, if the barrier is lower, then then the intermediate can form more often.  And thus the reaction can happen more often.

I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?

I think you are correct as far as your thought goes.  If you could somehow instantaneously create a beaker full of primary carbocations, they might well react more quickly than other carbocations.  (And that's not only via the SN1 reaction we are considering but also via other pathways such as elimination as well.)  But instantaneously creating a beaker full of primary carbocations is way more difficult than creating a corresponding beaker full of secondary or tertiary carbocations.  So (referring to the diagram again) yes, iff you have lots of $\mathbf{I}$, for $\mathbf{I}=$ primary carbocations, reactions would probably be quicker than the case where $\mathbf{I}=$ higher carbocations.  But in reality in the primary carbocation case essentially no $\mathbf{I}$ is ever generated.  So the overall reaction rate is slower.


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How to find pKa given absorbances and pH

If the absorbance of one of the buffer solutions (pH = 4.623) was 0.319 and the absorbance of the basic solution was 0.625, what is the $\mathrm{p}K_\mathrm{a}$ of the indicator? Include activity coefficients in the calculations.

The correct answer is 4.927, but I don’t know how to approach the problem to get the right answer. Thanks in advance!

There is information missing in the question since it does not state what the buffer concentration is. Also, the site is not suppose to provide answers to your homework questions, but we we try to help you understand how to solve the problem. Please edit your question and state what was done in class about this type of problem.

I don't understand the question at all. What is this absorbance? What is the basic solution? Do you know which kind of buffer it is?

Molar absorptivities for the protonated and deprotonated form of the indicator are needed.

If you don't know the what absorbance is, than you should first study absorption spectroscopy.

The following only gives some directions!
The molar extinction coefficient (molar absorptivity) is not necessary to correlate the absorbance and the concentration of the observed species, supposed that the same species is measured at one particular wavelength at different pH values!
Using the Lambert-Beer Law, we can derive that
\[\frac{E_1}{E_2} = \frac{\epsilon\cdot c_1\cdot d}{\epsilon\cdot c_2\cdot d} = \frac{c_1}{c_2}\]
The ratio of absorbances is linear proportional to the ratio of concentrations.

For a pH of 4.623, it seems fair to assume that an acetate puffer is used here. Taking this into account, you could even estimate the ratio of $\ce{HOAc}$ and $\ce{NaOAc}$ used.
The species monitored by UV is either the indicator $\ce{HInd}$ or its corresponding base $\ce{Ind-}$.
Since the absorbance is higher in basic solution, we might conclude that $\ce{Ind-}$ is observed.
\[\ce{HInd + B <=> Ind- +BH+}\]
$\ce{HInd}$ is typically a rather weak acid.
Is it safe to assume that the basic solution only contains $\ce{Ind-}$, while in buffered solution $\ce{HInd}$ is present as well?
Is it furthermore conceivable that $\ce{HInd}$ and $\ce{Ind-}$ constitute a buffer system for which the pH can be calculated using the Hasselbalch-Henderson equation?


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Why do charges result on atoms when coordinate bonds are formed?
I have not been able to get my head around it yet.
For instance in the reaction between ammonia and borontrifloride why is it that a positive and a negative charge results? Is it that ammonia donates is lone pair completely? If not so, if sharing of lone pair donated by ammonia occurs why then should charges result?

http://chemistry.stackexchange.com/questions/4828/formal-charges-how-to-work-them-out

The first thing to note about the charges that form during the formation of a coordinate bond: They are a formality.
The second thing to note about the charges that form during the formation of a coordinate bond: They are a mere formality.
There is little correlation, if any at all, between the formal charges assigned to two atoms that have formed a coordinative bond and the actual charges observed in the compound. Take $\ce{H3N-BH3}$: Nitrogen is so much more electronegative than boron, that the electrons are still closer to the nitrogen. Thus, nitrogen has $\delta ^-$ while boron has $\delta ^+$.
So where does the formalism come from? When drawing Lewis structures, after having filled in all bonds and lone pairs, check every single atom and mentally cleave all its bonds homolytically. Count the electrons, subtract the amount it should have as a neutral atom and the difference is its formal charge. Boron would have four electrons if all four bonds be cleaved homolytically, but it should only have three, so it gets a $1-$ charge.
This is only a formalism so that you can basically derive all structures the same way. It has no physical significance.

There are enough people who were unhappy about the ‘wrong’ charge distributions that they proposed different methods to draw these compounds: some used arrows from donor to acceptor, others just drew a simple bond, knowing that it would be different from a covalent bond. Each method has its advantages and disadvantages and all have survived in at least one niche.


text:
History of acid-base reactions
Basic and acidic materials exist in nature, so it makes sense that our forefathers could know what happens when they react. I would like to know when and how did they learn about this. In particular:
a. What is the earliest time in which people could witness acid-base reaction that causes fire or explosion (as in this video)?  What are the earliest written accounts of such reactions?
b. What are the earliest written accounts of neutralization reactions?
c. When were acid-base reactions intentionally used, e.g. for neutralizations or to create explosions?

Related fun fact (sort of): Seneca the Younger (c. 4 BC – AD 65) is quoted saying: 'Anger: an acid that can do more harm to the vessel in which it is stored than to anything on which it is poured'. I am not sure to what extent the ancient Greeks understood acids/bases, but they had the terminology

There's a bit of a misconception on what actually happens in the video linked in your question:
The reactions between alkali metals, namely lithium ($\ce{Li}$) and sodium  ($\ce{Na}$) and water, diluted acid and concentrated sulfuric acid are not neutralisations, but redox reactions, that is transfer of an electron from one reaction partner to the other.
\begin{align}
\ce{2H2O &<=> 2H+ + 2OH-\\
2Na &-> 2Na+ + 2e-\\
2H+ + 2e- &-> H2 ^}
\end{align}
If you take this three reactions as a (mathematical) equation system, with which you're probably more familiar with, and sum it up, the total reaction is
\[\ce{2Na + 2H2O -> 2NaOH + 2 H2 ^} \]
To put this into words:

The metal is dissolved and forms metal hydroxide.
The resilting solution is alkaline (basic), hence the name alkali metals.
The reaction releases a lot of heat!
Hydrogen gas is formed.

Since the reaction releases a lot of heat, the hydrogen gas is burnt with the oxygen from the air, yielding water, This is the explosion that you noticed.
Taken these vigorous reactions into account, it is not astonishing that in nature, sodium and and lithium do not exist as metals, but in the form of various salts. It wasn't until 1807 that sodium metal was actually made in the laboratory, followed by lithium in 1818.
Turning back to your question on neutralisations, let's define what it actually means.
(I apologize if I'm getting to colloquial here!)
The term neutralisation is typically used to describe that two effects don't sum up to result in something stronger but rather incapacitate each other. In the case of acid-base reactions, this translates to:
\[\ce{effective\ acid + effective\ base -> something\ ineffective + water}\]
In water, a typical example would be the reaction of of a strong acid (such as hyrochloric acid) with equimolar amounts of strong base (such as sodium hydroxide):
\[\ce{HCl + NaOH -> NaCl + H2O}\]
Adding the right amount of a strong and corrosive acid to a solution of also corrosive and harmful caustic soda a harmless solution of sodium chloride (table salt) in water.
The only side effect of the reaction is that the solution can get pretty hot, but an explosion is not to be expected here.
Typically, usable explosions do not result from neutralisation reactions but from fast combustion reactions in with large volumes of gases are formed, possibly (but not necessarily) accompanied by the release of heat.
Taking into account that strong acids had to be made by men, I'd be surprised to find that these materials were wasted in simple neutralisation reactions.
You will however find historical accounts on the use of the acids or the bases. Nitric acid was used to separate silver from gold (Scheidewasser). Calcium oxide (burnt lime), the precursor of calcium hydroxide e.g. is used to make cement.

Thanks for the detailed answser. So I understand that in ancient times people would not want to waste acids in neutralization reactions. But, did they know about accidental neutralizations? E.g. is there an account of someone complaining "My precious acid were wasted because an alkalic material was dropped into it", or something like this (not necessarily with the same terms)?

You are very much welcome! I'm afraid I can't answer that part. An expert in the history of chemistry is needed here, which I'm definitely not. Anyway I rather doubt that there are lots of recordings on these findings from ancient times. Mercury was known back then, but did Aristotle write about adding vinegar to cinnabar? No idea. I wouldn't go back beyond the 17th when looking for understandable sources.

Klaus-Dieter Warzec, Why do you, in the above neutralization reaction, designate “salt” to be “something ineffective”? Shouldn’t rather water, which contains the parts (acid $\ce{H^+}$  and base $\ce{OH^-}$ ) that have been made ineffective, have this label? — But the salt ions are basically the same as before the reaction since they are not more or less effective than before.


text:
Chemical equilibrium for simultaneous dissociation reactions
While calculating the adiabatic flame temperature for the following combustion reaction:
$$\ce{\Phi\, C2H6 + $3.5$\,(O2 + $3.76$\,N2) -> $a$\,CO2 + $b$\,CO + $d$\,H2O + $e$\,H2 + $f$\,O2 + $g$\,N2}$$
Where $\Phi$ is the equivalence ratio.
To determine the product composition ratios $a$, $b$, $d$, $e$, $f$ I need to make mass balance on $\ce{C}$, $\ce{O}$ and $\ce{H}$ and then use two $K_{p}$ (Chemical equilibrium constant) relations. Then solve this system of equation simultaneously.
My question is about the elementary reactions of dissociation I should use to apply $K_{p}$ equations, As I could have the following scenarios happening (regarding dissociation):
\begin{align}
\ce{CO + H2O &<=> CO2 + H2}\\
\ce{CO2      &<=> CO + $0.5$\,O2}\\
\ce{H2O      &<=> H2 + $0.5$\,O2}\\
\end{align}
I only need two equations for $K_{p}$ which two should I use and why?

Short answer (After some trials): It does not matter which dissociation reaction you use, The system of equation will be solved anyway since the reactions are not independent.
However, using the first reaction as a dissociation mechanism would generate a Kp equation that's a lot easier to solve than the others.

This question was flagged very low quality. I think it is okay, since it answers the premise. However, could you explain that a little more, I cannot really grasp what you are trying to tell us here. You could at the same time also accept your answer and make this question solved this way.

We should use only those equations which are linearly independent. For example, of the three equations specified eqn.(1) can be written by linearly combining eqn(2) and eqn(3).
 So use eqn (2) and (3) only.


text:
What's the relationship between charge and acidity/basicity?
I understand the acidity of a substance increases with electronegativity and size (valence shell space) of an atom (periodic trend) that contains a hydrogen atom (Brønsted-Lowry). However when looking at following pairs, I am not able to determine which one is a stronger acid.
$$\ce{H3O+ ,\ H2O}$$
Further, I would want to reason, the relationship between the charge and acidity/ basicity.

http://chemistry.stackexchange.com/questions/1341/is-this-the-correct-relative-bronsted-acidity-in-these-four-acids

@Mithoron, you may mark my question as a duplicate in that case. Spot on. I guess I panic and question. Instead I should do more research.

Hmm, it seems there's answer to your q. so maybe... Also http://chemistry.stackexchange.com/questions/24342/what-is-the-ka-of-oh-and-kb-of-h3o is related


text:
Organic reaction of dry ice and Grignard reagent

Which of the following product is formed in the reaction $\ce{CH3MgBr}$ in DryIce/acid?
A) $\ce{CH3COOH}$
B) $\ce{CH4}$
C) $\ce{CH3OH}$
D) $\ce{CH3CH2CHO}$

My answer is A, since dry ice is $\ce{CO2}$ and it has one carbon atom and reactant has one carbon, so in all the product should have two carbon atoms, which is option A.
Am I correct?

You are correct. Grignard reagents $\ce{RMgX}$ (where $\ce{R}$ stands for some hydrocarbon group and $\ce{X}$ is a halogen, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$) react with carbon dioxide to produce carboxylic acids after acidic workup:
$$\ce{RMgX ->[1)\ \ce{ CO2}][2)\ \ce{ H3O+}] RCO2H}$$
$$\ce{RMgX + CO2 -> RCO2- MgX+}$$
$$\ce{RCO2- + H3O+ -> RCO2H + H2O}$$

In general your reasoning when approaching this problem is a good one. If you are reacting two 1-carbon reagents, then you probably have a 2-carbon product.

Just to confirm, acidic workup or adding water, would be fine right?

Acidic workup would be better, since we want our acidic product and not its conjugate base.


text:
How to convert benzamide to benzylamine?

Question
  Benzamide can be converted to benzylamine using
A) $\ce{Br2,\ KOH}$
   B) $\ce{PCl5}$
   C) $\ce{LiAlH4}$
   D) $\ce{NaBH4}$

My answer:
Reducing agent is used so C and D options both must be correct but since its single choice answer, I do not know which is more appropriate. Any help?
Also, what does option A and B reagents do to amides?

You're on the right track. $\ce{R-CONH2}$ is an electrophile. Is it more or less electrophilic than a ketone? Your reductions are nucleophilic additions to the amide, aren't they? Which of the hydrides is the stronger reagent here? Would the comparison of the electronegativities of boron and aluminium be helpful here?

Less electrophilic! So weaker hydride donor is used and so is the answer $\ce{NaBH4}$ ? @KlausWarzecha

+1 You have correctly assigned the electrophilicity and nucleophilicity but somehow came to the wrong conclusion :) For a ketone or aldehyde, I would indeed use $\ce{NaBH4}$: Offer a reagent that is just strong enough. The less electrophilic carbon in the amide requires the more reactive $\ce{LiAlH4}$ to add a $\ce{H-}$ there: In your face, sluggish amide ;)

@KlausWarzecha Wouldn't the HSAB principle say otherways?

Thank you @KlausWarzecha What is HSAB, Chemexchange?

@KlausWarzecha Any help?

@ChemExchange It wouldn't BH3 is softer acid than AlH3, and H- is soft.

@Mithoron That wikipedia link says that hard refers to small species. Isn't H small?

@Mixcels It is - this theory is rather tricky - H has low electronegativity for an anion and H- is easily reducible.

@Mithoron So soft means which lose their charge easily? Table1 says H- is a soft base. According to the EN, you are correct but I don't understand why the theory says "hard refers to small".

It is basic rule - F- is small, hard; I- big, soft. Here's another link http://faculty.uml.edu/ndeluca/84.334/topics/Hard-Soft%20Acid-Base%20Theory.ppt

@Mithoron Looks like H- is an exception to it as it looks small but also soft. Thank you for your time.

$\ce{H-}$ is not small at all. Hydrogen atom is as big as oxygen. Its anion is even bigger. Imagine one positive charge vs two negative charge. Thus, $\ce{H-}$ is a very typical soft base: big and easy to be distorted. In this question, hard soft effect is really overwhelmed by the electronic effect as $H$ in $LAH$ has more partial negative charge than $H$ in $\ce{NaBH4}$.

I didn't know about the size. I thought since H was small so would be H-. Thank you for your update @IanFang

Thank you everyone. The above comment discussion helped me out. Thanks @Mithoron

LiAlH4 is corect answer.
It is stronger reducing agent, whereas NaBH4 is mild reducing agent

The more reactive $\ce{LiAlH4}$ is required for the reduction of the amide to the corresponding amine (source). The carbonyl carbon of an amide is less electrophilic compared to a ketone, because electron density is donated by the amide nitrogen via resonance.
Bromine and hydroxide react with an carboxylic acid amide to yield an amine with one fewer carbon atom (Hofmann degradation). $\ce{Br2}$ and $\ce{KOH}$ react to form $\ce{KOBr}$ in situ, which transforms the amide into an intermediate isocyanate ($\ce{R-N=C=O}$). Hydrolysis of the isocyanate yields a carbamic acid $\ce{R-NH-COOH}$, which decomposes to the amine $\ce{R-NH2}$ and $\ce{CO2}$. As described in this paper, the reaction of $\ce{PCl5}$ with carboxylic acid amides produces amidochlorides $\ce{RCCl2NH2}$ as intermediates, which break down to the corresponding nitrile $\ce{RC#N}$ and $\ce{HCl}$.

Is that carbamic acid decomposition instantaneous?

@Mixcels Yes, it is. Carbamic acid and many of its derivatives are generally unstable (http://en.wikipedia.org/wiki/Carbamic_acid).

authors: Inñöçèñt Wåñ


text:
How many electrons does each shell hold?
why some times 2nd shell onward it can hold 8 electrons and other times it is like they can hold K2 L8 M18 N32? when do i have to use first method(K2 L8 M8 N8) and when to use second method (k2 L8 M18)? I'm really confused.

Hmm, have you learned about orbitals?

yes I have learn it already. I remember my teacher telling me that each shell can hold certain amount of electron (K2 L8 M18) But when i was looking through text book, in an example it says that potassium have electronic configuration 2,8,8,1 why it cant have 2,8,9?

Have you learned about the subshells as well? The first shell only has $s$, the second one has $p$ and $s$, the third one has $d$ as well, and so forth. Also, you might take a look at here, if it helps.

No i haven't not yet.

@student11 See my answer here, I have explained in detail the exact concern that you have.

This is a great question that allows us to learn from the periodic table. The previous answer asked if you have learned about orbitals, and if you haven't this is probably why you are having difficulty.
If you look at the periodic table, you will see that the elements are arranged in two ways, vertically (group) and horizontally (period), there is good reason for this. Electrons can only occupy space based on mathematical probability; each of these probability functions is what we refer to as orbitals. Each orbital can only hold a maximum number of electrons ( the equation for total electrons is 4l+2 electrons where l is a quantum number starting at l=0 for s, l=1 for p etc.) in the valence shell (for 1, it can hold 2 Starting on the left side of the periodic table with Hydrogen we know that it is in group 1, which means that it has 1 electron in its valence orbital. Element number 2 (He) is a bit of a special case because although it only has 2 electrons in its valence orbital, it is full because the orbital number is 1. What I mean by this is that if you look at period 2 elements, starting with Li, you see that it falls under Hydrogen, and has 1 valence electron. This is because the 1s orbital is full with 2 electrons, and the additional electron goes into the next lowest energy orbital which is called the 2s orbital. Be has 4 electrons, 2 of them in the valence orbital, and 2 of them in the 1s orbital. When we move to B(oron) the 1s and 2s orbitals are full, and we start to fill the 2p orbital. The p orbital can hold a total of 6 (4l+2 where l = 1) electrons and so the boron atom has the following configuration (1s)2(2s)2(2p)1. What you see is that the highest energy orbital (in this case 2 (s + p) is the valence orbital and so B(oron) has 3 valence electrons, but also has 2 electrons in the much lower energy 1s orbital. Think of these orbitals like layers of an onion, where the further from the center, the greater the energy of the orbital. There are rules to how the orbitals are filled and the rules are dictated by quantum mechanics but generally speaking the orbitals are filled as follows 1s 2s 2p 3s 3p 4s 3d 4p 4d using what is know as the Madelung rule. s orbitals can hold 2 electrons, p orbitals can hold 6 electrons, d orbitals can hold 10 electrons. As you work your way from low atomic number on the period table to high you will see why it is organized the way it is. The break between H and He is because of electrons, when you move to period 2 you start filling valence shell 2, when you move to period 3 you start filling valence shell 3 and so on.  I hope this helps, and let me know if you need clarification on anything!


text:
Is Polymorphism accompanied with an energy change?
Differential Scanning Colorimetry (DSC) is a technique used to identify Polymorphism by revealing two melting points of a pure crystalline structure.
DSC shows a peak associated with the formation of the crystals. Hot Stage Microscopy (HSM) shows that only one of the polymorphs forms at the crystallisation temperature, the second polymorph forms between the crystallisation temperature and the melting temperatures.
Is the formation of the second polymorph not accompanied with a change in energy? If not, then why would the two structures have a different melting point seeing as no energy has entered or exited the system?


text:
What is the electrical conductivity of pool water?
What is the electrical conductivity of pool water? I tried to find some information, unfortunately without success. I need to know it to make an underwater electronic project.

What does "pool water" mean? Water in a swimming pool? From where does that water come?

@DrMoishe Pippik - Thank you for your answer. Yes, water in a swimming pool, for instance in a water park. I know this kind of water is chlorinated, but I also read that other substances are used instead of chlorine. Because of that I ask about "pool water” not "chlorinated water". Could you help me to figure it out?

Depends on the pool.

The pH should be around 8.4 and the conductivity in the range of 2.5 - 3 mS/cm. The lower value is appropriate for freshly prepared water. The conductivity is used for the determination of TDS (total dissolved particles).

This is a fine answer in principle, and cheers to it, but again we'll recommend you ask on Engineering.SE so that you might get some grasp on the ways in which your system is going to change over time under these conditions.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why [MnCl6]^2- has got 4 unpaired electrons instead of 3?
In $\ce{[MnCl_6]}^{2-}$, $\ce{Mn}$ has +4 oxidation state which gives it a $\ce{3d^3}$  $\ce{4s^0}$ configuration. But how is it having 4 unpaired electrons?

@DhrubaBanerjee You can also visit this page and this one  for a better tutorial.

are you sure it was 2- , I mean the charge can be 3- on the complex.

The complex you are asking about has an odd number of electrons, there is no way it can have four unpaired electrons. With three unpaired electrons it is in a quartet state though. Can you refer us to the source of your doubts, maybe it is a typo in a textbook.


text:
How to calculate an equilibrium constant from concentrations and a reaction equation
2.50 mL of 0.0150 M A, 3.50 mL of 0.0250 M B, and 4.00 mL of distilled water were added to a 50-mL beaker. At equilibrium the concentration of C is 2.05*10^-3M at 25degrees celsius. Calculate Kc for the following reaction.
\[\ce{2A + B <=> 3C}\]

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Sorry im new to this website. How would i go about answering this though im confused.

Do you know the answer, or you have no clues about what to do?

Welcome to Chemistry.SE! If you want to get familiar with the site, please take the tour which explains a lot.

I took the liberty to remove the confusion from the title of your question and maybe my answer gives some directions too. In the end, the work is yours - but so is the success ;)

Start with the equation for the equilibrium constant

For a reaction $\ce{a\cdot A + b\cdot B <=> c\cdot C}$, the equilibrium constant $K_c$ is given as
\[K_c = \mathrm{\frac{[C]^c}{[A]^a\cdot[B]^b}}\]

Get familiar with the concept of ICE charts.

Determine the concentrations of A and B for the moment when you have poured together the solutions of A and B and diluted the mixture with water. (Don't forget about the water!)
Imagine that you've worked incredibly fast and no reaction has taken place! This is the Initial situation.
Now have a look at the reaction equation to figure out what happens when A and B react to yield C. Denote the change of the concentration of C as $x$ (or $3x$) if you think that is easier to handle. How does the formation of $x$ moles of C change the concentration of A and B? This the Change part.
Can you use the information from the Initial and the Change part to express the concentration of the reactants in the Equilibrium, knowing that $[\ce{C}]$ is $2.05\cdot 10^{-3} \mathrm{mol\cdot L^{-1}}$?
(Yes, that's the E part of the ICE).

Feed these concentrations, actually "some expressions with $x$" in the equation for $K_c$ and solve.
Success ;)


text:
Why leveling effect is reduced when strong acid is dissolved in a weaker acid than water?
I am reading a book that basically states what is the title.
This is something I observed when I was doing ICE charts: when acid and base neutralize, we always assume complete neutralization.
Interestingly, when we put acid in water, which can also be considered a Brønsted-Lowry base, we called it ionization of acid, and thereof acid ionization constant.
Leveling effect happens when multiple strong acids dissolve completely and we don't know which is stronger.
Here is the question: does complete neutralization always take place with a base and an acid, why?

your talking about strong acids - strength of acid depends on solvent

Can you make it an answer?

http://chemistry.stackexchange.com/questions/6751/how-does-one-tell-if-a-specific-molecule-is-acidic-or-basic, http://chemistry.stackexchange.com/questions/28234/what-is-the-reason-for-strong-acids-bases-dissociating-in-water?rq=1

http://chemistry.stackexchange.com/questions/13855/hydrochloric-vs-sulfuric-acids-which-are-stronger-acids


text:
Explain the formation of replicators to a layman
I am reading The Selfish Gene by Richard Dawkins.
I am on chapter two.

He speaks of the observation of the formation of amino-acids when you simulate environmental conditions of primordial earth.
UV light + water + carbon dioxide + methane + ammonia + a couple of weeks time = amino-acids
I understand that amino acids are an organic compound that serve a lot of important functions in our bodies.
Presumably, they are important to all life, because from the context I can derive, they seem to be somewhat of a precursor to life itself.

Dawkins then says:

Processes analogous to these must have given rise to the 'primeval
soup'  which biologists and chemists believe constituted the seas some
three to  four thousand million years ago.

He goes on and eventually states:

At some point a particularly remarkable molecule was formed by
accident. We will call it the Replicator.   ....
Actually a molecule that makes copies of itself is not as difficult to
imagine as it seems at first, and it only had to arise once. Think of
the  replicator as a mould or template. Imagine it as a large molecule
consisting of a complex chain of various sorts of building block
molecules.  The small building blocks were abundantly available in the
soup  surrounding the replicator. Now suppose that each building block
has an  affinity for its own kind. Then whenever a building block from
out in the  soup lands up next to a part of the replicator for which
it has an affinity,  it will tend to stick there. The building blocks
that attach themselves in  this way will automatically be arranged in
a sequence that mimics that of the replicator itself. It is easy then to think of them joining up to
form a  stable chain just as in the formation of the original
replicator. This  process could continue as a progressive stacking up,
layer upon layer.  This is how crystals are formed. On the other hand,
the two chains might  split apart, in which case we have two
replicators, each of which can go  on to make further copies.
A more complex possibility is that each building block has affinity
not for  its own kind, but reciprocally for one particular other kind.
Then the replicator would act as a template not for an identical copy,
but  for a kind of 'negative', which would in its turn re-make an
exact copy of  the original positive. For our purposes it does not
matter whether the  original replication process was positive-negative
or positive-positive,  though it is worth remarking that the modem
equivalents of the first  replicator, the DNA molecules, use positive-
negative replication. What  does matter is that suddenly a new kind of
'stability' came into the world.  Previously it is probable that no
particular kind of complex molecule was  very abundant in the soup,
because each was dependent on building  blocks happening to fall by
luck into a particular stable configuration. As  soon as the
replicator was born it must have spread its copies rapidly  throughout
the seas, until the smaller building block molecules became  a scarce
resource, and other larger molecules were formed more and  more
rarely.

As someone with a limited understanding of Biology and Chemistry, I have a couple of questions.

We have observed that the chemical conditions outlined at the top of this question seem to yield more complex organic compounds with time. Do we know (or have theories) as to why this is, or is it merely something we have observed and re-created?

The theory about replicators suggests that they are like a chain of building-block molecules each with an affinity for either its own kind, or some other kind. (positive-positive vs positive-negative). -- Which ever was the case, you could end up with a big chain of molecules that had a fractal nature, because these building blocks were freely available in the soup surrounding the replicator. -- If the chain split apart, you would get two identical copies and each part could go off and replicate further.
Now, I suppose this question mirrors the first, but why is it that it is so easy to imagine the affinity relationship? What makes molecules want to stick together with other, specific molecules?

Further more, what makes chains of molecules split apart (replicate)?

Can somebody elaborate further on these points, and on what Dawkins is saying here in general, so as to pander to the cravings of someone who is not particularly versed in the parlance?

Since you're reading about the concept of molecular replication, you might be interested in very relevant and more general chemical term: autocatalysis. A substance that once formed favours the production of more of itself is not unique to life or biochemical compounds.

We have observed that the chemical conditions outlined at the top of this question seem to yield more complex organic compounds with time.
  Do we know (or have theories) as to why this is, or is it merely
  something we have observed and re-created.

A recurring theme in chemistry,and all of your questions, is "lower energy usually equates with more stable structures."  Oversimplifying just a bit (see note on time to reach equilibrium below), we can say that all chemical transformations are also equilibria.  That means that our starting materials are converted to products and our products are converted back to starting materials - reactions are continually happening in both directions.  Once at equilibrium the rate of the forward reaction matches that of the back reaction and the concentrations of starting materials and products will no longer change. At equilibrium the equilibrium constant ($\ce{K_{eq}}$) describing the relative concentrations of our starting materials and products will have the following form
$$\mathrm{K_{eq}={\frac{(concentration ~products)}{(concentration~starting~ materials)}}}$$
and the equilibrium constant is related to the free energy difference between our starting materials and products by
$$\mathrm{\Delta G = -RT\ln K_{eq}}$$
These equations tell us that the more stable a structure, the more it will predominate at equilibrium.
If we take carbon, oxygen, hydrogen and nitrogen an equilibrium will be set up between these elements and the simple amino acid glycine

according to the following equation
$$\ce{4C + 5H2 + N2 + 2O2 -> 2C2H5NO2~(glycine) + heat}$$
Over 500kJ of heat is produced per mole of glycine formed.  This means that glycine is over 500kJ/mol more stable than the elements used to create it!
Bottom line: If a complex molecule is more stable than a simpler molecule, then over time the higher energy "simple" molecule will transform into the more stable (lower energy) "complex" molecule.
Just a word about time.  It may take a long or a short time for an equilibrium to be attained on its own.  Suffice it to say that with catalysts and\or energy, equilibria can be attained quickly.

...why is it that it is so easy to imagine the affinity relationship? What makes molecules want to stick together with other, specific molecules?

Again the answer has to do with lowering the energy of the system and making things more stable.
Some molecules have what is called an "active site".  It may be an odd-shaped kink or pocket in the molecule.  Let's think of this odd-shaped pocket as a lock, only molecules shaped like the key will fit.  Further, there are often other functional groups (perhaps hydroxyl groups) around the active site that will bind (hydrogen bonds perhaps) with the key and stabilize it once it is inserted into the lock. Stabilization means lowering the energy of the system.  Only the key fits into the lock and once inserted it is stabilized - energy is lowered.

Further more, what makes chains of molecules split apart (replicate)?

By now you've probably guessed that the splitting apart step occurs because it lowers the energy of the system.  Perhaps once the molecule reaches a certain size it folds differently - a new lower energy shape is now possible.  A  protein that was bound to the molecule and stabilizing it can no longer bind to it due to this new folding pattern.  The protein is displaced, it is replaced by a different protein that stabilizes the molecule's new folded structure but predisposes it to splitting apart to a smaller size.

At first, I was disappointed to see an answer like "lower energy usually equates with more stable structures." because I immediately want to know WHY this is, but the answer actually made something click in my head. That is, If I want to know that answer, I need to go much deeper, as I believe its more related to questions about the fundamental forces that cause the atoms themselves to form. And like Dawkins says, No matter what level you're on, It all comes down to "survival of the stable"

This answer is just plain wrong. Biological life happens far from thermodynamical equillibrium, where entropy is maximised. Glycine is more stable than the elements, but far less stable than CO2, water and nitrogen.

The situation leading to the formation of amino acids was simulated by Stanley Miller and Harold Urey in their famous experiment and published as A Production of Amino Acids Under Possible Primitive Earth Conditions in Science, 1953, 117, 528-529 (DOI).
Imagine that this formation of amino acids might have happened not just in the sea, but also in shallow pits at the shore that once in a while fell dry and were heated by sunlight. Under these conditions, particularly in the presence of insoluble salts serving as catalysts, amino acids may have undergone condensation reactions, i.e., splitting off water and forming peptide bonds. The resulting molecules are oligopeptides, small brothers of proteines.
Imagine that these pools were filled with water again and that the process repeats. Between oligopeptides of fitting amino acid sequences, hydrogen bonds can form, resulting in $\beta$-sheets or similar aggregates. Upon repetition of the condensation step, free amino acids bound to one chain by hydrogen bonds might preferably undergo condensation in the other. This may be the molecular background of the replicator concept described by Richard Dawkins.

'UV light + water + carbon dioxide + methane + ammonia + a couple of weeks time = amino-acids'

The answers by ron and Klaus are very good, but I'd like to add one additional point which I think is very important.
The key "reactant" in the above equation is, to my mind "UV light".  UV light comes from the sun.  The sun is very hot.  The surface of the sun is at ~5800 Kelvins, or about ~5500 °C, and the light it emits is effectively the equilibrium mixture of photons for that temperature (this equilibrium for photons is called black body radiation).  The high temperature for the sun means that the equilibrium mixture of light frequencies it emits has much more UV light than "equilibrium" light at Earth's temperatures.
You can view the formation of complex molecules from simple molecules via UV light as a partial, frustrated attempt for the simple molecules to equilibrate with the blackbody radiation of the sun, i.e., for the simple molecules to equilibrate at temperatures of 5800 K.  Those high temperatures favor high entropies or "disorders" among the molecules.  You can view a "soup" that contains small amounts of many different amino acids and other molecules (in addition to large amounts of unreacted simple molecules) as having more disorder or entropy than a mix where the simple molecules are present in 100% purity.
An analogy is the frying of an egg.  An egg in the refrigerator is ordered and "simple":.  When you try to equilibrate the egg with blackbody radiation at a higher temperature (i.e. put it on the stove), the simple mix changes considerably and the result is a complex, solid-like gel of egg white and a pasty solid goo of yolk.
Note that your question mainly seems to be about formation of the primordial soup, and not actually formation of the replicators themselves.


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Are the following compounds aromatic?

I know Huckel's rule states that an aromatic species must have $4n+2$ π-electrons. Is the last molecule also aromatic?

Molecule 1, the cyclopentadienyl cation, is not aromatic. It only has 4 π electrons and does not fit the $4n+2$ rule.
Molecule 2, the cycloheptatrienyl cation (or tropylium cation), is aromatic; it has 6 π electrons and fits the $4n+2$ rule with $n=1$.
Molecule 3, azulene, is generally considered to be aromatic. It fits the $4n+2$ rule with $n=2$, and is often referred to as a "nonbenzenoid" aromatic compound. It can also be viewed as a fusion of two other aromatic units, the tropylium cation and the cyclopentadienyl anion.