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Energy in molecules
A textbook states:

Fats also provide an efficient way to store energy over long time
  periods, since they contain over twice as much energy per gram as
  carbohydrates

What's the chemistry beyond this? What topic should I learn in order to understand how fats and carbohydrates store energy?

Fats contain more hydrogen and carbon than carbohydrates. Look at the picture below of the carbohydrate. Notice all the oxygen atoms?
Carbohydrates http://tse1.mm.bing.net/th?&id=JN.kQKbzP5TAGgr7um%2bQI6f4Q&w=300&h=300&c=0&pid=1.9&rs=0&p=0&r=0.
Now look at the picture below of a triglyceride, or basically fat. Notice all the carbon and hydrogen atoms?
Triglyceride http://tse1.mm.bing.net/th?&id=JN.tBMcQZj7LoB48KotpahRzg&w=300&h=300&c=0&pid=1.9&rs=0&p=0&r=0
Carbon and hydrogen can be reduced, or can lose electrons, by oxygen through oxidization.  Hence, fuels like gasoline or natural gas are mostly hydrogen and carbon. The oxygen in carbohydrates cannot be reduced, but instead contribute to the oxygen used to reduce carbon and hydrogen. Thus, the oxygen in carbs and triglycerides are worthless.
This is the same reason why ethanol, a renewable alternative to gasoline, is less powerful than gasoline. Ethanol is an alcohol, so it contains an oxygen atom within its molecule. Gasoline is mostly alkanes, alkenes, and alkynes, all of which do not have oxygen within their formulas.
If you wish to learn more on this subject, I would suggest learning more on oxidation and reduction, and on thermochemistry.


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Is there a certain amount of titrant that is ideal?
For an AP chem lab dealing with determining the amount of citric acid in orange juice by titrating with NaOH, one of the discussion questions goes like this:

Choose an amount of beverage to be titrated that will require at least 10 but less than 20 mL of titrant. Explain why this range of titrant is optimal.

This question makes no sense to me. Why would there be an "optimal range" of titrant? I would think that I should fill the buret to full capacity if enough titrant is available. Would it not be better to have excessive titrant than not enough of it?
We completed the lab, and the titration required 23.05 mL of 0.1 M NaOH solution to titrate 20.00 mL of orange juice. If we had followed the instructions, we would have added less orange juice, and used less NaOH, but what is the benefit in doing so? Budget cuts? If it was about not wasting solution, why would there be a minimal requirement of 10 mL of titrant?

Excessive amount of titrant doesn't make it more precise, it's a waste.

In a perfect world there would be no optimal range. We do however live in a world full of errors and practical considerations.
Why use over a minimum amount?
Let's assume for demonstration, that each drop has a volume of 0.05 mL. Of course this is not an accurate value as various factors would affect this.
The titration has reached its end point when one last drop changes the colour. If your titre value is very small e.g. 1 mL, then each drop is 5% of your titre.
If only part of the drop is needed to reach the end point then all the extra part of the drop is inaccuracy, which in this scenario could be up to 5%! If you have the misfortune of accidentally adding one drop too many you are introducing an even larger error. This will impact concentration calculations based on the experiment.
Why not just use 'a lot' then?
There are a couple of things here:
1) As noted by Mithoron, this would be a waste. The accuracy gained from requiring additional titrant will level off rather rapidly.
2) If you use more than one full burette's worth there is an error penalty.
3) Using extra titrant is a waste of your time. Consider a 50 mL burette, if it needs to be filled up after each titration, that would be inefficient. I'm assuming you're doing repeats so this has real merit.

Thanks! The more of the test substance, the smaller the pH change each drop is worth, but too much of the test substance is a waste of solution and time.  I assume for the same reason it would be more accurate to use a smaller molarity NaOH solution, so one drop is not excessive?

Also, because we were using an pH probe to determine the titration curve, instead of an indicator, the minimum amount of test sample is unnecessary, correct?

@IsaacLiu Provided you aren't diluting your initial solution to make it more dilute and thus gaining an error. The minimum amount still applies, probably more so with a pH probe as you need to put the probe somewhere.

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Why is the amount in a mole equal to the Avogadro's number?
This question just popped in my head as I was working with electric force problems involving atoms. Why is the amount in a mole equal to the Avogadro's number? Do atoms/molecules/etc. tend to group in this exact amount? Why do we use this number as a unit of measurement?

It's simply defined that way.

If this is an arbitrary, though mutually agreed upon, number, how did Avogadro come up with the number? Are there any benefits to using his number as opposed to a randomly large number?

related: http://chemistry.stackexchange.com/questions/35563/what-advantages-does-the-mole-have-over-counting-large-numbers-using-si-prefixes?rq=1, http://chemistry.stackexchange.com/questions/11169/why-is-the-definition-of-the-mole-as-it-is?rq=1, http://chemistry.stackexchange.com/questions/5321/why-is-the-mole-a-unit-of-measurement?rq=1

Avogadro articulated the concept that is Avogadro's number, but didn't pick the number. From Wikipedia:

Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

https://en.wikipedia.org/wiki/Avogadro%27s_law
The Number is a logical extension of this law: So what is that number in relation to the volume and what is the mass of that amount of material?  By agreement, a mole is defined as the mass, in grams, equivalent to the atomic weight of a substance. This was laid out before Avogadro's Number was actually determined.  Then it was simply a matter of measuring the mass of a single atom to determine how many atoms were required to have a Mole of a material.  Over time, Avogadro's Number has been refined to more and more accuracy.

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How to calculate concentrations in a diluted mixture of multiple solutions?
I have a homework problem as follows:

A solution was prepared by mixing $25.0\ \mathrm{mL}$ of $0.208\ \mathrm{M}$ $\ce{Fe(NO3)3}$ and $4.50\ \mathrm{mL}$ of $0.00190\ \mathrm{M}$ $\ce{KSCN}$ and diluting to $50.00\ \mathrm{mL}$ with solvent, $0.10\ \mathrm{M}$ $\ce{HNO3}$. Calculate the initial concentrations of the thiocyanate ion and the iron(III) ion before any reaction takes place.

Now, how do I proceed with the question? At first I thought that it should just be the same concentration as the solutions, $0.208\ \mathrm{M}$ and $0.00190\ \mathrm{M}$, because there is only one thiocyanate and one iron(III) ion in the solutions. But the fact that we have an $\ce{HNO3}$ solution confuses me. Am I somehow supposed to use $M_1 V_1 = M_2 V_2$?

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

Nitric acid is like the question said just the solvent, so you are on the right track with your formula, as it states before any reaction.

Am I somehow supposed to use $M_1V_1=M_2V_2$?

Yep!
The problem has given you three different solutions:

Iron(III) nitrate
Potassium thiocyanate
Nitric acid

You know the concentrations of each, and you know (or can calculate) the volumes of each that are mixed to yield the final solution.  What you'll actually be using in the calculation are three different equations:

$M_{1i}V_1 = M_{1f}V_f$
$M_{2i}V_2 = M_{2f}V_f$
$M_{3i}V_3 = M_{3f}V_f$

$M_{\#i}$ is the initial concentration of each of the three chemicals in their separate starting solutions, and $M_{\#f}$ is the final concentration of each after mixing, that the problem is asking for (except for the $\ce{HNO3}$).


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Paper Towel Soaked with Isopropanol/Water Mixture Burns without Damaging the Paper. Why?
Video: https://www.youtube.com/watch?v=_qFN1hDZwp8
If you choose not to view the video, in the first demonstration a paper towel is soaked in isopropanol. As expected, the paper towel burns. However, in the second demonstration the paper towel is soaked in isopropanol and water. Surprisingly, the towel burns without being damaged.
My chemistry professor tells me that it has something to do with the energy cost of a phase change $\left[\ce{H2O}\ (l) \longrightarrow \ce{H2O}\ (g)\right]$; however, I am still very confused.
Could someone please elaborate? Thank you.

I think your professor is incorrect, and it has to do with the thermodynamics and kinetics of the combined vaporization/combustion process.
The heat of vaporization of water $\left(40.7\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite is considerably less than the heat of combustion of isopropanol $\left(1830\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite.  These numbers indicate that approximately $45$ water molecules must evaporate for each isopropanol molecule combusted in order to hold the temperature constant. This  corresponds to a $\sim2\%$ mol/mol solution, which (a) is not consistent with the $\sim50\%$ v/v solution he actually used, and (b) would probably be too dilute to burn at room temperature the way it does in the video. (Someone more motivated than I could probably confirm this using flash point data.)
I think, instead, the behavior depends on the following:

Vapor-liquid equilibrium stipulates that a liquid phase can never be at a temperature greater than its boiling point$^\dagger$.
It is the vapor of combustible substances that actually burns, not the liquid (see, e.g., this OSHA training document)
Up to a concentration of about $60\%$ alcohol, the vapor phase in equilibrium with a water-isopropanol liquid mixture has a considerably higher concentration of isopropanol as compared to the liquid (source):

For the present purpose, to read the diagram choose a temperature on the $y$-axis and track across the $x$-axis, noting where you cross the red and blue lines.  These correspond to the equilibrium liquid and gas phase compositions, respectively, at that temperature, given a mixture that's not too lean or too rich in isopropanol.  For example, take $82\ ^\circ\mathrm C$: you hit the red line (liquid) at about $20\%$ isopropanol, and the blue line (vapor) at about $53\%$ isopropanol.
Since the boiling point of isopropanol is $82.6\ ^\circ\mathrm C$, #1 means that as long as there's liquid left soaked into the paper (even $100\%$ isopropanol!), the paper can't burn.  You can actually see this in the video: he ignites the paper at around 0:57, but it doesn't actually start to char until about 1:08, ten seconds later.
The reason why the paper does then start to char in spots is that those spots have locally run out of liquid isopropanol to keep them cool.  There remains a cloud of burning gaseous isopropanol all around the paper (clearly!), which is more than enough to burn the now-dry paper.
When there's water mixed with the isopropanol, though, the vapor-liquid equilibrium comes into play.  Per point #3, based on the phase diagram, the isopropanol likes to be in the gas phase a lot more than the water, so once the vapor is ignited and burning (point #2), the heat produced tends to evaporate more of the alcohol than the water.  This feeds the flame for a while, but eventually it sputters out, with plenty of water left on the paper to keep it cooled well below its char point.
$^\dagger$ Mixtures of liquids technically don't have boiling points, as their behavior is more complex, but the concept is useful for making the point.


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How to test for heavy metal in water?
Is there a way to test if water has toxic heavy metals. Samples to both the water and people who drink it are available. Preferably, it should be done with at hold materials. I could also borrow materials from a high school chemistry lab potentially.
I have already tried boiling, but that just showed that it is hard water, which luckily isn't dangerous. Any ideas?

For qualitative tests to work properly, you need a concentration sufficiently high to observe the effects, e.i. precipitation of heavy metal salts and/or color changes of solution. If heavy metals are present, they are likely to be in small concentrations (depending on how your water is filtered before it reaches your house).
If you are concerned about your water quality and want to check if heavy elements are present, you should contact a trustworthy laboratory that routinely analyzes water samples. They are likely to use trace-element techniques such as inductively coupled plasma mass spectrometry (ICP-MS) and ICP optical emission spectroscopy (ICP-OES), or maybe atomic absorption (AAS) techniques. This way you get accurate and precise quantitative measurements of almost all elements in the periodic table.


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Just how quantized are energy levels?
One thing which has bothered me for a while about the idea of quantized energy levels is that it's clear just how quantized they are.
Let me give an example. If I was looking the energy some electromagnetic wave would need to an $\ce{O-H}$ in water from the 0th vibrational energy level to the 1st vibrational energy level, it would be around $\mathrm{3750~cm^{-1}}$. Now, for some electromagnetic wave to cruise by and give enough energy to make that transition, how exact must the energy of that wave be?
Meaning, does it need to have exactly $\mathrm{3750~cm^{-1}}$ of energy? Well, what if the transition was actually $\mathrm{3750.3~cm^{-1}}$. Then that wave doesn't have enough energy anymore, and the likelihood of some wave having exactly $\mathrm{3750.3~cm^{-1}}$ seems a lot less likely than just being very close to that value.
So what happens in actual transitions that the molecule only takes the exact amount of energy it needs? And what happens to whatever leftover energy there might be from a wave that had enough energy to excite this water to it's first vibrational energy level? Is there ever leftover energy like I just mentioned? Also, do two waves of say $\mathrm{1875~cm^{-1}}$ each ever meet up and excite this same transition? This to me also seems so unlikely because that would involve constructive interference exactly of the two waves.

First off, let's think of the light as photons, instead of waves. That makes it clear that yes, there are two photon and multi-photon processes. As you surmise, the selection rules for these are much more complicated than a single-photon process.
Your main question is about "how exact" an absorption needs to be. In reality, it depends on the system.
Atomic Absorption Spectra
For atoms, there are no vibrational or rotational transitions, only electronic (or perhaps nuclear). So atomic spectra are indeed very tight lines, e.g., the Rydberg formula.
Diatomic Molecules
Now you have one vibration and some rotational levels. The latter are important, since you can go up or down some rotational levels with a vibrational transition. (And if you have an electronic transition, you can usually go up vibrational and/or rotational levels.)
So the spectra is a bit more complex than single lines. For example you can do rotational-vibrational spectroscopy.

In this case, the individual transitions are still pretty sharp - not completely lines, but $0.3\:\mathrm{cm^{-1}}$ probably won't match any particular transition.
Larger Molecules
Now you have very small spacing in the rotational lines, so if you take a vibrational absorption spectra (e.g., IR spectra) you'll see that most molecules have some width to the absorption peaks, so $\pm 0.3\:\mathrm{cm^{-1}}$ will still fit within the peak, just not as strongly.

So what exactly is happening that accounts for the widening of the peaks? I imagine it could be a larger number of things, but how does that still fit under the definition of quantized energy levels?

@jheindel If you look at the spectrum above it's just a bunch of discrete lines. There's the quantisation. The widening of the peaks is just as Geoff explained: in a large molecule, each vibrational energy level will have a number of rotational levels (the spacings between rotational energy levels are smaller). So you get a bunch of peaks with different intensities

There are various other sources of line broadening, most notably Doppler broadening (molecules or atoms moving parallel or antiparallel to the light source (or observer)). Removing this and other broadening sources, ultimately one is left with the uncertainty principle. Just like momentum and position are mutually exclusive (complementary) observables ($\Delta x \times \Delta p \geq \hbar \div 2$), in the energy-time domain, $\Delta E \Delta t \geq \hbar \div 2$).


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Can Fuel Exhaust Be Recycled
When the spark plug of a car ignites the pressurized air/fuel mixture, could you instead send the exhaust to another system within the car to be Recycled?
I think it's a great idea that could drastically lower pollution, that is if it's possible.

We burn fuel for energy, not for fun. Recycling it would require at least as much energy.  Just where that energy is going to come from?

According to some studies (as in my answer) - solar power, but it is not practical, yet, in the small version needed for a car

If you'd have enough power (solar or whatever) close at hand, why would you need to burn fuel in the first place?

One of the major components of exhaust is carbon dioxide, produced by the combustion of hydrocarbons - if in complete combustion, follows the general reaction:
$$\ce{C_{x}H_{y} + O2 ->CO2 + H2O}$$
Other possible emissions include $\ce{NO_{x}}$, particulate matter, carbon monoxide, VOCs etc. The amounts of these emissions vary from car to car (due to a lot of factors, e.g. age, wear-and-tear etc). According to a self study by Audi, in the document Motor Vehicle Exhaust Emissions, on page 6, they determined that asides from $\ce{N2}$, the main emissions were $\ce{CO2}$ and $\ce{H2O}$.
For this answer, the focus will be on the possibility of recycling the $\ce{CO2}$ emissions (and linking to the $\ce{H2O}$ emissions).
This very question is the subject of the Scientific American article Reverse Combustion: Can $\ce{CO2}$ Be Turned Back into Fuel?, the articles details that there are several research groups looking at this problem.
A couple of examples of ideas and experiments that have been put forward include:

In the 1990s a graduate student named Lin Chao at Princeton University decided to bubble carbon dioxide into an electrochemical cell. Using cathodes made from the element palladium and a catalyst known as pyridinium—a garden variety organic chemical that is a by-product of oil refining—he discovered that applying an electric current would assemble methanol from the $\ce{CO2}$.

Inspired by the natural process that removes $\ce{CO2}$ from the atmosphere - photosynthesis, continued research has sought to emulate this - essentially, the current research is effectively looking at a possible solution by taking

$\ce{CO2}$, water, sunlight and an appropriate catalyst and generate an alcoholic fuel

The research reported in the Scientific American article and the National Geographic article Carbon Recycling: Mining the Air for Fuel go along the lines of extracting the excess $\ce{CO2}$ from the air rather than directly from the car's engine or exhaust. The reason for this is the current practicality, with the National Geographic reporting that

The Sandia prototype’s solar collector has an area of about 20 square meters (215 square feet) for a reactor the size of a beer keg, Stechel says. About 300,000 acres (121,400 hectares) of mirrors would be required to collect enough sunshine to make the equivalent of 1 million barrels of oil per day, she says. (The world currently consumes about 86 million barrels per day of petroleum and other liquid fuels, including biofuels.)

The research shows that, yes it is theoretically possible and practically possible on a larger scale to recycle exhaust gases.


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What are positive and negative voltage?
I read about electrode potential today, the book says that:

Metals more reactive than hydrogen will have negative electrode potential value and metals less reactive than hydrogen will have positive electrode potential.

I'm a bit confused by the sign of the voltage, can voltage have negative value? What does negative voltage mean?
Another question is why metals more reactive than H have negative electrode potential?

This seems like two separate questions and so you should ask them as so.

Negative voltage means the same as positive, except the electric current caused by it would flow in the opposite direction.
As for your another question, that's just what it is: having negative potential means being more reactive than H. See, that electrode potential is not an absolute value, it can be measured only relative to something. And the general agreement is to take the standard hydrogen electrode for that "something".
Suppose you'd make an electrode out of your metal immersed in a solution of some salt of itself, couple it to standard hydrogen electrode and see what happens. If a metal is more reactive, it will dissolve, the released electrons will run to the hydrogen electrode and reduce hydrogen there. With less reactive metals, it is the other way around: hydrogen will dissolve and send the electrons running in the opposite direction, to eventually reduce the metal.


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How do Electrons Cross Nodes in Orbitals?
I was wondering in orbitals that have nodes such as the p orbital, how does the electron move from one lobe to the other? I know that people say it is because of the wave/particle nature of electrons but they don't actually explain how they travel across the node without actually passing it.
My teacher says it is because that electrons are able to act like waves and hence like how EM waves can pass through a wall, electrons are able to pass through the node. However when he says that electrons are like a wave, does he mean that it is literally a wave so does an electron have a wavelength, amplitude, frequency, etc? Because I remember reading somewhere which stated that electrons aren't literally like waves. However they can be mathematically described using concepts that apply to waves. Is this true?
From the comments, they state that electrons can be thought as standing waves. After looking at the internet I came across the following image which explained how electrons act as a standing wave:

However this doesn't make sense to me. Don't electrons exist in orbitals not in orbits around the nucleus? If so, isn't this wrong? Also, since electrons move in orbitals, how can there be any end points as it can move anywhere and its movement isn't restricted to a close loop. Hence how can it act as a standing wave?

I think I've seen a very similar question before. In short, yes, it's a wave. Think of it as a standing wave of peculiar configuration, so the concepts of wavelength and frequency do not apply. (There is a so-called de Broglie wavelength, but that's defined for a free electron).

You say, 'in short, yes, it's a wave'. I am curious to know what it is actually.

There is no "actually". The best model we have is that of psi function, which behaves according to Schrödinger equation, which has such and such known solutions. One of these solutions, namely that for a free electron, is quite literally a wave, with wavelength and everything.

Stop thinking of the electron as a particle - the wave function IS where the electron exists...

In support of @JonCuster's comment; the electron doesn't 'move' between the two lobes since it isn't a discrete object. It exists simultaneously in both lobes.

An orbital is not a physical observable. It is indeed just a mathematical construct to solve Schrödinger equation, and not unique. E.g. the p-orbitals could be used the complex set p(0), p(-), and p(+) or as  p(z), p(x), and p(y), which is more familiar. The properties derived from this wavefunction are, however, real. Integrate it's square over all space and you get the probability distribution. It's 0 at the nodes. The electron IS on both sides of the nodes but never exactly at the node. Hence it does not travel across it, it just exists at both sides of it...

Correct if I am wrong, I thought that when you square the wavefunction you get the probability distribution of the electron being at a particular place. Doesn't this just mean that it is possible for the electron to be at different places (i.e different lobes) but in reality the electron is actually only in one position, we just don't know for certain where it is. Because if we were to observe it, we would find that the electron would only be at one position, not at two places at the same time.

I've uploaded 3 pages from Griffiths, Introduction to Quantum Mechanics 2nd ed here. You might find it relevant (it's a very good book but much more oriented towards physics). http://imgur.com/MOWuMA6,p37pwpQ,6Aopv68

I am familiar with the basics of quantum mechanics. Thanks for uploading the pages, it really helped me understand now what people mean by that the electron simultaneously existed in two places. So basically what the book was saying that electrons exist in orthodox position before it is measured. The reason why I was so confused was that I was thinking that the electron existed in realist position before it was measured.

related http://chemistry.stackexchange.com/questions/62009/do-electrons-in-an-atom-revolve-around-the-the-nucleus-clockwise-or-counterclock/62054#62054

http://chemistry.stackexchange.com/questions/16128/how-do-electrons-travel-through-nodes

Electrons (indeed all matter and energy) behave both as wave and as particle, depending how the observation is made. The most obvious example is the single-electron interference pattern.
"In 1974 the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi repeated the [interference] experiment using single electrons, showing that each electron interferes with itself as predicted by quantum theory. In 2002, the single-electron version of the experiment was voted "the most beautiful experiment" by readers of Physics World." Wikipedia.

could you also explain the image that I have uploaded on my question? Don't electrons exist in orbitals not in orbits around the nucleus? Therefore since there is no endpoints to its movements, how can an electron act as a standing wave?

If electrons actually moved in circles or orbitals (or any closed path), they would radiate energy until they fall into the nucleus. The diagram is not a physical path but a graph of the probability of finding the electron in a particular experiment.

@DrMoishe Pippik if electrons do not move in circular path what is the significance of angular momentum given by azimuthal quantum number?

It is an analogy. If electrons actually moved, that would create Bremsstrahlung radiation, and, as the electrons lost energy to that radiation, their orbits would decay and fall into the nucleus. https://courses.lumenlearning.com/introchem/chapter/the-bohr-model/

Electron properties are well described by treating them as particles or waves, and the de Broglie equation is useful because it relates wavelength and momentum which are wave and particle properties, respectively.
Remember that there are limits to any analogy!
Electron as standing wave:
You can make standing waves with a slinky toy spring, and it takes more energy to get more nodes. Your diagram with n=8 is quite high in energy, but you can probably get n=2. You can say where the spring is going to move for each standing wave and make the analogy that the spring is the electric field and the movement is the electron.
Your diagram tries to show how a standing wave can be in an orbit by bending the 1D slinky toy analogy into a circle.  This might be useful in understanding that only standing waves make orbits through a constructive interference analogy.
Electron as particle:
If you want to look for the electron particle that has the energy of a standing wave then the probability curve of finding it in 1D is going to look like the slinky toy.
Electron orbitals:
These are representations of the probability curve in a 3D for electrons in orbit.


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What is the conjugate acid of NaOH using the Brønsted-Lowry definition of acids?
To the best of my knowledge, a conjugate acid of a base is the base after it has accepted a proton, or a $\ce{H+}$ ion. In this case:
$$\ce{NaOH + H+ -> Na+ + H2O}$$
Is the conjugate acid of $\ce{NaOH}$ the sodium ion, or the water? Common sense tells me it can't be the $\ce{Na+}$ ion, because it has no protons to donate, so how could it ever be an acid? So I am thinking that the conjugate acid is $\ce{H2O}$. However, wouldn't that mean that the conjugate acid of any base of the form
$$\ce{(something)OH + H+ -> (something)+ + H2O}$$
would be water, and that seems unsettling to me. I also believe that since $\ce{NaOH}$ undergoes the following reaction:
$$\ce{NaOH -> Na+ + OH-}$$
the $\ce{Na+}$ is something of a 'spectator ion' (not sure if that's the correct term), this seems to imply that $\ce{H2O}$ should be the conjugate acid. Basically, I'm really confused, and could use a little help sorting all this out.

In Bronsted theory OH- is a base not NaOH like in Arrhenius theory.

Your first equation is more properly written as
$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$
in aqueous media. However, we can do better if we explicitly show the dissociation of $\ce{NaOH}$ as
$$\ce{NaOH -> Na+ + OH-}$$
and substitute that into the first expression (note that I write $\ce{2H2O}$ as $\ce{H2O + H2O}$) to get
$$\ce{Na+ + \underbrace{OH^{-}}_{base} + \underbrace{H3O^{+}}_{acid} -> Na+ + \underbrace{H2O}_{conjugate\;acid} + \underbrace{H2O}_{conjugate\;base}}$$
where we see that $\ce{H2O}$ is the conjugate acid of $\ce{OH-}$ as well as the conjugate base of $\ce{H3O+}$. The last bit - where water plays 2 roles - is due to water being amphoteric, or able to act as an acid or a base.


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Why do iron(II) ions and chromium have different electronic configurations?
What I know:

Empty 3d orbitals are higher in energy than empty 4s orbitals
Aufbau Principle (electrons always go into an empty orbital with the lowest energy)
Partially/half/fully filled 3d orbital are lower in energy than 4s orbitals
$\ce{Cr}$ is an exception: $\ce{[Ar] 3d^5 4s^1}$. It breaks the "rule" because the extra stability from half-filled 3d orbitals is favoured.
From (1) and (2) => After the $\ce{[Ar]}$ core, electrons will fill 4s before occupying 3d
When forming ions, electrons of the highest energy are removed first.
Iron(II) ions exist as $\ce{[Ar] 3d^6}$ in reality.

What I understand:
I understand why Fe is $\ce{[Ar] 3d^6 4s^2}$. [from (5)]
I understand why Cr is $\ce{[Ar] 3d^5 4s^1}$.
From (4), (5) and (3), I believe for $\ce{Cr}$ the 4s is filled before 3d too, i.e. it is $\ce{[Ar] 3d^3 4s^2}$ before the last electron is added. After the last electron is added to form $\ce{[Ar] 3d^4 4s^2}$, a 4s electron immediately moves to the 3d orbital which is of lower energy.
From (6) and (3), for transition metals, 4s electrons are removed first. i.e., I understand why iron(II) ions may be $\ce{[Ar] 3d^6}$

What I don't understand:
I know $\ce{[Ar] 3d^6}$ is correct for iron(II), but I don't understand why?
Why doesn't iron ions adjust itself as Cr does to achieve the extra stability? Is it because, in this case, it needs to excite an electron to an orbital of higher energy, whereas in Cr, the 4s electron just moves down to an orbital of lower energy?

P.S.

I've read What is the electronic configuration of Fe(II) ion? and I am not satisfied with the answers in that thread.
I've read Do ions also form special electron configurations? (just as Copper and Chromium do), and it only tells me "it is possible for ions to have special electronic configurations too!"

https://chemistry.stackexchange.com/questions/18988/how-does-the-radial-distribution-function-of-vanadium-differ-from-that-of-calciu

For starters, it isn’t nearly as simple as it is usually put in chemistry classes to explain chromium’s ground state electron configuration.
$\ce{Cr(0)}$ and $\ce{Fe(II)}$ have the same number of valence electrons. You might have heard of the concept of isolobality, which, briefly explained, suggests that compounds with similar electronic configurations (isolobal configurations) will react similarly; so you could replace $\ce{F}$ with $\ce{OMe}$ or even $\ce{[Co(CO)4]}$. The concept has its values and its strong simplifications which I will not go into here.
However, even assuming the isolobality concept’s validity, it would not count $\ce{Cr(0)}$ and $\ce{Fe(II)}$ as isolobal to each other. The reasoning is rather simple: One is a neutral metal, the other is a cation with $+2$ charge. For every electron an atom looses, the atom’s orbitals will contract and the atomic (or ionic) radius will decrease. This also changes the energy levels between orbitals and will affect lower d-orbitals differently than higher s-orbitals.
The result is that the $\ce{4s}$ and $\ce{3d}$ orbitals are now so dissimilar in energy, that electrons hopping back or forth is now no longer an option.

Another way to look at the transition in electronic configuration is to consider species with the same number of electrons but different numbers of protons.
Start with $\ce{Cr^0}$ with its electron configuration $\ce{[Ar] 3d^5 4s^1}$.  We add a proton to make $\ce{Mn^+}$, throwing in any necessary neutrons to keep the nucleus stable.  We repeat this process to make $\ce{Fe^{2+}},$ $\ce{Co^{3+}}$, etc., and eventually we get to let us say, $\ce{Xe^{30+}}$.  In all these species the electromagnetic interactions between the 24 electrons are intrinsically the same, but the excess positive nuclear charge is building up and is bound to overwhelm the electron-electron interactions at some point.  By the time we get to the huge excess positive charge in the xenon ion above*, the orbitals become roughly what they would be with isolated electrons, like the familiar hydrogen spectrum multiplied by the effect of a greater positive charge.  The 3d orbital energy level sinks down to meet its shell-mates at 3s and 3p, leaving the higher-shell 4s orbital at a higher level. (* -- actually XPS binding energy data reveal that $\ce{Br^{11+}}$ already shows this effect.)
So at some point the outer electrons all fall into the 3d orbitals and the electronic structure shifts from whatever it was in the neutral chromium atom to $\ce{[Ar] 3d^6}$. In my example I predicted that we might have to go all the way to $\ce{Xe^{30+}}$ (or maybe $\ce{Br^{11+}}$ given the above-cited XPS binding energy information) for this transition, but in reality the 3d and 4s orbitals are so closely balanced in the first row of transition metals that we need not go that far. Adding two protons, and a few neutrons for nuclear stability, to chromium is enough.

" like the familiar hydrogen spectrum multiplied by the effect of a greater positive charge" do you mean it would become similar to a single electron H atom and follow the Bohr's law of orbits?

The energy levels do that, except multiplied by the effect of nuclear charge.

I've had the same question and this is how my chemistry teacher answered:
  3d is a sub energy level of the 3rd energy level and 4s is a sub energy level of the 4th energy level. When electrons fill into the Fe atom they fill according to the Auffbau principle, in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d. But when 2 electrons are to be removed to form the Fe2+ ion, the electrons are not removed from the 3d orbitals. 3d belongs to the 3rd energy level which becomes an inner filled energy level when electrons are present in the 4th energy level. So electrons are removed from the 4s sub energy level, without disturbing the fully filled 3rd energy level.

Thank you for the answer! My question is mainly that, after two 4s electrons are removed, why doesn't one 3d electron go up. As in, Cr is special, why can's Fe(II) be special?

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Good measures ot atomic size
I was looking up the Lennard-Jones parameter $\sigma$ (which, to my understanding, is a measure of molecular size) when I was surprised to discover that the value for the $\ce{N2}$ molecule (3.667 Å) was slightly lower than the value for the $\ce{CH4}$ molecule (3.78 Å). It's my understanding that $\ce{N2}$ molecules are significantly larger than $\ce{CH4}$ molecules, and this is exploited in, for instance, molecular sieves.
My question: is $\sigma$ a reasonable measure of mean atomic radius? And if it isn't, what is a good measure of molecular size, with which I can compare molecules?

Are you sure you don't mean molecular radius (if such a thing exists) rather than atomic radius. Similarly the title seems out of place.

Please be aware that there are actually two different definitions for the Lennard-Jones radius. One is the radius at which the vdW force goes from attractive to repulsive, whereas the other is radius at which the attractive energy is greatest. Both are in common use. While Wikipedia uses different symbols for the two radii, not all sources are as consistent/forthcoming, so be careful when looking up the values.

Atoms and molecules, as you surely know, are not tiny hard pebbles that you may measure using a slide gauge, unambiguously and with great precision. Instead, they are clouds of electron density with rather fuzzy edges, like things on an out-of-focus photograph. Their apparent size depends on how you define what "edge" is, or how hard you squeeze them.
This is one of the reasons why we don't have a single definitive list of atomic radii, but a whole variety thereof.
As to the point of your question, yes, $\sigma$ is a good measure of molecular size. But there are also others, equally good and all subtly different. Which one to use depends on your purpose, and may be a hard problem by itself.
Also, remember that (1) the Lennard-Jones potential is defined for spherically symmetrical objects, which your molecules are not, and (2) even for these objects (atoms) it is but an approximation.

Thanks for your answer. Could you suggest a few other measures of molecular size? I'm curious to see how $N_2$ and $CH_4$ compare in other measures.

Sure, be my guest. One: parameter b from the van der Waals equation of state. Two: effective area from the Langmuir adsorption isotherm (this one usually works the other way around, but anyway). Three: volume calculated from van der Waals radii and experimental bond lengths, as if the molecules were made of overlapping solid spheres.


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What is the oxidation mechanism of gunpowder?
I've been recently looking into fireworks and the chemistry of explosives, and I found out that, amongst others, one of the most important reactions in gunpowder ignition is the oxidation of charcoal and sulfur by $\mathrm{KNO_3(s)}$. I am curious as to whether it is known if this reaction occurs between charcoal/sulfur and solid state potassium nitrate or if the potassium nitrate decomposes first and the reaction occurs via gaseous oxygen as an intermediate.
Is there a general mechanism of oxidation between solid oxidizing salts and organic matter/sulfur? Does the same thing happen with $\mathrm{KClO_3(s)}$ or $\mathrm{KMnO_4(s)}$ for example?

$\mathrm{KNO_3(s)}$ can oxidize charcoal without sulfur. It needs a "hot" primer through to get the reaction started.  Sulfur's role is to lessen the ignition temperature.
See section in Wikipedia article.
Since the reaction isn't complete and has a number of possible products bthere isn't an exact chemical formula for black gunpowder burning. The Wikipedia article lists several approximations to black gunpowder burning. What is supposedly the best follows.

Although charcoal's chemical formula varies, it can be best summed up by its empirical formula: $\ce{C7H4O}$. Therefore, an even more accurate equation of the decomposition of regular black powder with the use of sulfur can be described as:
    $\ce{KNO3 + C7H4O + 2S → K2CO3 + K2SO4 + K2S + 4CO2 + 2CO + 2H2O + 3N2}$

I know it can, I meant Charcoal/Sulfur as in Charcoal and/or sulfur, I´m interested of any of the two reactions, I guess they are kind of analogous.

I checked wikipedia before posting, but thanks for the advice. Wikipedia (L

The potassium nitrate does decompose with heat to yield oxygen. The oxygen causes the sulfur to burn which creates more heat. Finally the charcoal gets hot enough that it too burns.

Wouldn't mind a better summary of the reactions occurring and how the ingredients interact.


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How to calculate the composition of a borate buffer with a defined pH using the Henderson-Hasselbalch equation?
I am struggling with what appears to be an extremely easy pH problem that uses the Henderson-Hasselbalch equation. The problem and answer provided by the book is given below. I understand where the book is getting the answer but it appears that they are overlooking something.
From what I understand, the Henderson-Hasselbalch equation allows you to calculate the pH of a solution based on the concentrations of acid and conjugate base at equilibrium. In the problem, you start off with a solution of just the conjugate base. Then you add pure acid to the solution to get a desired pH. However, the book seems to assume that the acid you add completely contributes to the concentration of undissociated acid at equilibrium [HA]. How can this be the case? Some of the acid you add will surely dissociate. Anyways, here's what the book did...

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || Please visit this page, this page and this one. It would be appreciable to use the info provided in the links to include the book's answer; images are not searchable.

Boric acid and sodium borate forms an acidic buffer (A mixture of weak acid and salt of weak acid and strong base). The acid is weak hence the concentration of basic ion from acid is much less compared to the concentration of basic ion obtained from the salt which can ionize completely. Yes. Some of the acid dissociates to form hydronium ion and a basic ion. That's how a buffer works.
Consider a weak acid HA and the salt BA.
Its easy to show that the concentration of $H^+$ due to ionisation of HA is $$[H^+]=k_a\frac{[HA]}{[A^-]}$$
It is reasonable to assume that the concentration of $A^-$ from BA is much larger than that from HA due to complete ionisation of BA. Hence $[A^-]=[BA]$=[salt]
So, $$[H^+]=k_a\frac{[acid]}{[salt]}$$
Taking logarithm, you get the desired equation.

Never mind, I found the answer! I was correct to assume that the initial concentration will not be equal to the equilibrium concentration. However, they will be extremely close because we are working with weak acids. In fact, I found an article published in a chemistry education journal that discusses this so-called "Henderson-Hasselbalch Approximation" and how it is often overlooked in chemistry textbooks. Here is the link: Henderson-Hasselbalch Approximation. I believe textbooks should make sure students are aware that the equation is only an approximate method for determining the pH of a buffer.


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Reactivity of propellanes
Why is [1.1.1]propellane considerably less reactive than either [2.2.1]propellane or [2.1.1]propellane as illustrated in the hydrogenation reactions pictured below?

The heat of hydrogenation reflects the thermodynamic difference in energy between starting material and product. It is a thermodynamic measure of the difference in stability of the products and reactants.  Reactivity, on the other, hand is related to kinetics, a reflection of the activation energy  involved in the rate determining step.
Cyclopropane and cyclobutane have roughly the same strain energy (SE=27.5 and 26.3 kcal/mol respectively). To a first approximation, the strain energy of a polycyclic system is the sum of the strain energies of the individual rings in the system.
Each of the 3 starting materials contains a total 3 (cyclopropane plus cyclobutane) rings.  Since the SE of cyclopropane ~ cyclobutane, we would therefore expect the SE's of the 3 starting materials to be similar and larger than the SE's of the less strained products. In agreement with this expectation the calculated SE's of the 3 starting materials are 109, 106 and 103 kcal/mol respectively (see fig. 2.2 on p. 37 here), indeed very similar.
The SE's of the final products are 17, 41 and 62 kcal/mol respectively (ref, see Table 3.2 on page 72).
If we calculate the strain released during hydrogenation it is in the same order and very close to the actual heats of hydrogenation reported in your figure.

$\mathrm{109-17=92}$ kcal/mol
$\mathrm{106-41=65}$ kcal/mol
$\mathrm{103-62=41}$ kcal/mol

Clearly, in this series of propellanes the heat of hydrogenation is a measure of the strain released in the process.
Because bicyclo[1.1.1] pentane (the right-most product) has a much higher SE than either of the two other hydrogenated products, the heat of hydrogenation given off when it is formed will consequently be much less.
[1.1.1]propellane is an extremely reactive compound (see here for example).  Comparing the thermodynamic heats of hydrogenation in your series of compounds cannot be used to meaningfully predict the kinetic reactivity of these compounds.

The concept of ring strain needs a reference system how it is calculated. This can be chosen completely arbitrary and the numbers have no meaning whatsoever. Strain cannot be measured, too.

Strain energies have been experimentally measured and reported in countless publications. The same techniques used to measure resonance energies can be used here as well; for example heats of combustion and  heats of hydrogenation. Typically, a strain-free reference is chosen.

I took the last statement from the reference you provided, when I tried to find out what theses numbers actually mean. I did not have enough time to dig in deeper into the matter, i.e. finding out what the most common reference is. Since you have used this concept a couple of times before, I trust your expertise: What would be the or a strain-free reference for [2.1.1]propellane?

Probably something like 2-methylhexane adjusted for gauche interactions, etc.


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Hydrogen visible transitions
When dealing with electrons and the electromagnetic spectrum, why are there only four visible transitions in the hydrogen atom?

This all falls back to the hydrogen absorption spectra. There are multiple "series" discovered by different scientists. The series in the visible region is the Balmer series.
Since the electronic energy levels of the hydrogen atom are quantized, there can only be certain discrete transitions.
As it turns out, if you solve the Balmer series (n=2), you get:
$$\frac{1}{\lambda} = \frac{4}{B}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) = R_\mathrm{H}\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \quad \mathrm{for~} n=3,4,5,...$$
You will then get lines at $\lambda$ = 656.3 nm, 486.1 nm, 434.1 nm, 410.2 nm...
Only a few of those fall in the visible. The Lyman series (n=1) occurs in the UV spectra, and the Paschen series (n=3) occurs in the IR.


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Are there any chemicals that combine readily with oxygen?
I am designing an industrial process which can remove oxygen from air.
This chemical should combine readily with oxygen in the presence of heat.
In particular, I am looking for a chemical with the following properties:

Combines readily with oxygen, thus removing the oxygen in air
Inexpensive
Environmentally friendly (nontoxic)

For point #1, heating is available if it is an endothermic reaction.
I am looking at at an operating temperature of ~$\pu{800 ^\circ C}$.

https://en.wikipedia.org/wiki/Pyrogallol for example

Pyrogallol is way too toxic

At what temperature? Carbon or magnesium would remove oxygen at 1,000 K or higher. Alkaline glucose might be used at room temperature, though it reacts slowly (see http://faculty.mansfield.edu/bganong/biochemistry/bluebott.htm). More information would be helpful.

If $\ce{CO2}$ is acceptable in the feed gas, along with $\ce{CO}$ and a bit of left-over oxygen, then carbon (coal, charcoal etc.) might do. It's ignition temperature is ~1,000 K.
Magnesium is more expensive, but leaves less oxygen and has no gaseous byproduct. It's self-ignition temperature is ~800 K.
Many other materials could be used; it really depends on how the gas is to be used.

What about just releasing the gas into the atmosphere?

It depends on the purpose of your process.

If you want to absorb only a small amount of O$_2$ to purify an airstream, you should look up gettering. Titanium metal is one of the most popular if you are willing to heat.
If you are looking to oxidize lots of material, I would look up common reducing agents that are relevant to you in terms of cost, abundance, and toxicity.

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Why is the ideal gas law so ubiquitous?
Why is the ideal gas law usable in most gas problems?
For example, in a problem like this: "Suppose you have $1.00\: \mathrm{mol}$ of a gas at $0\:^\circ \mathrm{C}$, occupying a container which is $500\: \mathrm{mL}$ in size. What is the pressure of this gas in $\mathrm{atm}$?"
The ideal gas law is was applied to solve that problem and the answer was $44.8\: \mathrm{atm}$. But isn't the ideal gas law only applicable to gases which are in high temperatures and low pressure? Because the kinetic energy overcomes the intermolecular bonds and the molar volume is negligible, etc etc.
But the conditions for this "gas" are low temperature and consecutively high pressure which is basically a stamp for real gases so why isn't the real gas law applied to this problem instead?

There are two key points that make the ideal gas law so universal. One is that it is universal: you don't need to know the actual gas to work with the equation. The second is that there is no single real gas law and the various laws that attempt to improve on the ideal situation require you to know the specific gas involved. Mostly the ideal law is a good approximation, unless you are a chemical engineer working with extreme conditions.

You have to go to rather arcane substances and/or extreme conditions until the ideal gas law is not correct to within 10%.

You are right in saying that in your question, the gas is an non-ideal (real) gas. Basically, the reason why we use the ideal gas equation despite knowing that it is a real gas is simply because that the equation for real gases is too complicated and contain more variables which are hard to find given limited data.
The reason for this complicated formula is that several assumptions that we make for ideals gases don't hold for real gases. Recall that an ideal gas is considered to be a point mass - a particle so small that the volume of that particle is negligible. A real gas particle does have real volume. For an ideal gas, the collisions between gas particles was said to be "elastic" - no attractive or repulsive forces exist, and thus, no energy is exchanged during collisions. For a real gas, collisions are non-elastic. So, the ideal gas law must be corrected for this extra forces. There are several real gas laws, one is the van der Waals equation:
$$\left[P + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$
Notice how "corrections" are being made to the pressure term and the volume term. Since collisions of real gases are non-elastic, the term $a(n/V)^2$ is correcting for the interactions of these particles. Since real gas particles have real volume, the $nb$ term is correcting for the excluded volume. The values of $a$ and $b$ are constants, and must be determined experimentally for each gas.
As you can see from this equation that first the values of $a$ and $b$ must be known then you must solve this equation which can be complicated. Even then, when you do finally solve it, you will probably get a result which is similar to the result obtained if you used the ideal gas equation. This is why the ideal gas equation is continued to be used as in most cases it provides quite accurate results while bypassing complicated formulas.

But I assume that there is another reason for the usage of the ideal gas equation rather than just complicated. Otherwise the answer would be not possible.
So let's say that the problem actually gives us the name of the gas. Then I would be able to get a value for the real pressure of the gas using the van der Waal's equation and this value might be close to the value we got using the ideal gas equation.
In summary, are you maybe implying that the deviations caused by real gases properties are so small that are not relevant? Which is maybe why the ideal gas law could still be used?

Well, if they did give you the name of the gas you will be able to use the van der Waal's equation. However you are right to say that it will probably give a result that will only deviate slightly form the result obtained from using the ideal gas law. Which is the reason why the ideal has law is used as it still produces relatively accurate results in most cases while bypassing complicated formulas such as the van der Waal's equation.

The other factor here is that assuming the ideal gas law typically allows a solution using a liner or quadratic equation. It isn't really a problem for a computer to solve the van der Waal's equation, but it is awful pain in the neck to do by hand - especially so if you solving for two or three gasses. So using the ideal gas equation let's you apply the chemistry without grinding through a lot of tedious mathematical equations.


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How to calculate the Gibbs energy for the vaporisation of solid ammonia?
Given that $P_{\ce{NH3}}=\mathrm{10~atm}$ at $298\ \mathrm K$, calculate the $\Delta G^\circ$ of the reaction $$\ce{NH3(s) <=> NH3(g)}$$
I'll begin solving the problem...
I'm using the formula $$\Delta G=\Delta G^\circ+RT\ln Q.$$ The vaporisation of $\ce{NH3}$ is an equilibrium so  $\Delta G=0$ and $Q=K_c$ . Also $K_c=\left[\ce{NH3}\right]$. So $\Delta G^\circ=-RT\ln K_c$.
How do I continue? Is it $K_c=K_p=10$ or $K_c=K_p/\left(RT\right)$?

Actually for this reaction $\Delta G=\Delta G^\circ+RT\ln Q$ holds only when $Q$ is used in terms of partial pressures. This is because the activity of an ideal gas is measured by its pressure, not its concentration. This equation can be derived from the definition of the Gibbs energy.

It says nowhere in the definition of amount (of substance) concentration that the concept must not be used for gasses. It might not lead to the desired result in this case, but it is not necessarily wrong. Also the activity is derived from the chemical potential of a substance and for an ideal gas not measured by the pressure. However, it can be related to the pressure.

Yes,the term concentration can be used.What I meant was that the activity or fugacity of an ideal gas itself happens to be its pressure per unit atmosphere or bar,as per the definition of the standard state.The definition of activity is of course only from the chemical potential,independent of the substance

Your first statement is still wrong. For any substance $\Delta G=\Delta G^\circ+RT\ln Q$ holds for any $Q$. They can be interconverted.

What exactly do you mean by 'they can be interconverted'?I understand that the "Q's" can be written in terms of concentration or pressure,but I don't see how both might give the same result in this case(assuming we are referring to an ideal gas).

Have a look at this question

I know how they can be inter converted,what I mean is that in this case q must be used in terms of partial pressures to get the desired result.


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What is the correctly diagrammed electron probability distribution for p orbital?
This is what my book shows (the darker the more probable):

The obvious places where the electron probability is zero is the midsection of the dumbbell.
My question is how the probability distributed inside the dumbbell.

The probability distribution of an electron in the p sub-shell is determined from its wavefunction. The wavefunction of a system contains all the information about the quantum states of the system. Once of the most important properties of the wavefunction is that the squared modulus of the wave function, $|ψ|^2$, is a real number, interpreted as the probability density of the particle. Here is the wavefunction for the 2p orbital:

So if you square this function you will get the probabily distribution which looks like this:

As you can see, this corresponds with the p orbital drawn in your book, as there is a node in the middle, on either side is high probability density and at minus and positive infinity probability density is actually very, very small despite being shown as 0 on the wavefunction. The reason for this is that you need to remember that the orbital shows where an electron will be 90% of the time, not where it will be at all times. This means that the electron can be found outside the orbital.

I'm afraid your last paragraph is wrong. Radial density without angular won't give you such shape

Indeed, the angular distribution is missing.

At infinity the probability density is not zero. It is infinitely small, but significantly different from zero.

Thanks for pointing that out. I wasn't sure to say if the  probability density is non-zero at infinity as the graphs that I have uploaded go to zero at infinity rather form an asymptote.

@Martin-マーチン That expression means it converges to zero (asymptotic behavior)...

FWIW, I would also say that it simply refers to the limit which is exactly zero. Strictly speaking, saying that a function has a value "at infinity" is not the best practice (usually). The infinities are sometimes appended to the real number line making something like $f(\infty)$ bearable but even then in this case the value would again be exactly zero. Happy to be proven wrong though.


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Why doesn't every pure enantiomer racemise?
After having been acquainted with enantiomerism for nearly six months, I have one standing doubt.
Consider the structure of alanine. Suppose I have a pure mixture of exclusively 'S-alanine'(left in the picture) in a flask.

There are hundreds, if not, thousands of molecular collisions in the flask per second. Why can not a bump to one of the S-alanine change it to R-alanine? This way, the mixture should racemise itself if left standing for some time. But this does not happen. Why? Can someone please explain?

If I somehow had a tank full of left hands, and I agitated the tank, you might get lots of different conformations such as a tight fist, an open palm, etc... but there's no way I'm going to turn a left hand into a right hand (a different configuration) without chopping off the fingers and gluing them back the wrong way. Similarly, racemisation (at a chiral centre) requires the breakage of a single bond, which is not very likely to happen. When it does occur, there is often some mechanism to explain it, e.g. keto-enol tautomerism for alpha carbons. Sorry for the morbid analogy.

@orthocresol You should write this up into an answer.

Because the activation energy for the enantiomeric H-Atom to jump to the other side of the carbon atom's sp3 tetraeder (by whatever mechanism) is so high that it is just not available.
You're right, statistically, every energy is available (with very bad odds). But very likely other processes, and more destructive ones, have a lower activation energy. Your compound will likely degrade and rot away long before it racemises. But even that, for alanine at room temperature, does not happen on any timescale that you can observe.


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Better functionals to work with transition metals
My Masters in inorganic chemistry is focused in magnetic properties of transition metals; great part of the work is done with DFT methods using softwares like ORCA and NWChem, but since my University has a lack of resources, we are performing computations with not much computational power. That is why we are focusing in results and using functionals like PBE and B3LYP (since many papers are showing good results).
Nevertheless, as said by Martin in this thread, today we have "better and more efficient functionals" available, because PBE and B3LYP are very popular, but kind of old. That is why I’d like to ask: what are good, modern, functionals to work with transition metals (mostly Ti…Zn)?
Aditional information:
We are working with systems  that have no more than 200 atoms, most of the time, usually dimers.

You can pass some time reading this question. I hope I can come back to this question soon, but essentially the answer will most likely be: It's not easy.

Thank you, Martin, and great text here. I always like your answers.

Depends what you mean by 'better'.  Getting a 'better answer' for the wrong reasons may be serendipitous and inconsistent.  Getting a 'reasonable answer' for a 'better reason' is somewhat preferred by others.  For instance, my group likes to use PBE0 for a lot of metals because we simply don't like over-parameterized  functionals.  In the words of von Neumann, "Give me four parameters and I can fit an elephant."

Better for what properties? Also, you realize that many times often the problem is not the functional, right?

I can't speak for the OP, but I would say the question is about finding reliable functionals that can give accurate ground-state geometries and predict the right spin state. Getting relative thermodynamics is important, and maybe a few comments about common pitfalls (e.g., convergence).

I am also a ChemE Masters student working on organometallics (second TM row mostly)! I found that B3LYP works perfectly fine for my purposes. Though if you want other suggestions ive heard mention that PBE0 or Mo6 also work well for transition metals.

This question is like the snake that eats its own tail. There is no functional good enough for metals. To my limited knowledge, fitting too many parameters -which happens with metals- feels like DFT is not first principle anymore and could be called "empirical" (which is also not correct). You allow a number of parameters to to be fit(1,2,more!!) to a specific system. Those dozens of parameters -at the end- may give you desirable results but you "trained" the set of energies according to your will which is not that realistic. You train your dragon not in the jungle but in a cage. The result is never the same. This "fixing" not only contradicts the spirit of DFT but is also a weak form of empirism. It is like adding more "dark force" to the community. Imagine in what extend many of those "good", modern functionals affect research daily. To end, maybe other comments will come up with accurate functionals about Ti or Zn with papers in good journals etc but without being against DFT, I would just keep a small basket.

Up-voting for "You train your dragon not in the jungle but in a cage."

Having looked through several review articles over the last few days, I disagree with your comment about "no functional good enough for metals." Yes, there are some functionals that have a lot of parameters, but there are clearly modern functionals without "dozens" of parameters, e.g. $\omega$B97-xD

Your best bet with this sort of research is to see what other functionals people have used on similar systems. Try and find validation for the functionals against real world data for similar systems.
If you use a functional & basis set which hasn't been benchmarked, for lack of a better term, against the type of system you're modelling and real world data, it becomes hard to trust any results, no matter how computationally expensive.
Are you looking at crystals & macromolecules, or individual molecules/dimers?
When you talk about a functional, there is also the basis set to be considered.
A paper from Cramer and Truhlar* appears to be well worth a look, its pretty exhaustive in examining DFT and TMs. You are probably better off looking at stuff like this than asking on here, given the limited expertise on here for your specific systems.

* Christopher J. Cramer, Donald G. Truhlar, Phys. Chem. Chem. Phys., 2009, 11, 10757–10816

Since many people can't access that paper, it would certainly help if you're willing to give a quick summary on Cramer & Truhlar's suggestions.

I think the OP asked the question because they were looking for some suggestions on what functionals people have used (beyond PBE and B3LYP). Obviously if benchmarking is possible, it's preferable.


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Why there is an exception in melting and boiling point in p block?
Why is the boiling point and melting point of 15th group and 16th group has an exception?
We know that as molecular mass increases boiling point and melting point also increase. So, down the group 15 and 16, as molecular mass increases, melting and boiling point should also increase but melting and boiling point of antimony is more than that of bismuth. (even though molecular mass of bismuth is more)
Similarly, melting and boiling point of tellurium is higher than polonium. Why is it so?

Your "rule" holds for the non-metals in groups 18,17,16,15. In group 14,13 it's total nonsense, and also beneath the diagonal. Look at the structure of those elements. (Di-)Atomic gases and one liquid, more complex molecules, huge macromolecules, and metals. Why should they all follow this one oversimplicistic rule?

@Karl I'm retagging this question; and I believe that's a fine response in itself. Just add why that doesn't work for metals; and huzzah!

http://chemistry.stackexchange.com/questions/33815/trend-in-the-melting-point-down-the-group


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NO2 gas was detected in FTIR
We have a laser chamber full of $\ce{F2}$ gas (no air should be in the chamber) and several parts including steel, wires, ceramics, zinc.... This is for laser generation so the temperature is pretty high in the chamber. We have detected $\ce{NO2}$ gas in our FTIR specturm which caused our laser to fail and there’s a good chance that the $\ce{NO2}$ may be associated with the failure. The questions are

How might $\ce{NO2}$ be formed in a chamber? If it’s from air, why would $\ce{NO2}$ be formed and not $\ce{NO}$ or $\ce{N2O}$? Why would $\ce{NO2}$ persist after $\ce{N2}$ and $\ce{O2}$ are removed by chamber refills with fresh $\ce{F2}$?
How does $\ce{NO2}$ interact with fluorine? What other species might be formed when they react?
Can $\ce{NO2}$ be dissociated by UV light (i.e. 193 nm)? What is its UV absorption spectrum?

I'd guess that you have an air leak. "Flushing" only helps if the source of the $\ce{NO_2}$ is fixed. $\ce{O_2}$ would definitely be dissociated by a 193nm laser to produce two oxygen free radicals. Normally with extra oxygen that would make ozone. I'm guessing that the oxygen must be low in concentration and nitrogen relatively high. So perhaps most of the oxygen is $\ce{OF_2}$? So $[F_2] >> [N_2] >> [O_2]$

If you somehow manage to get nitrogen to react with oxygen, as long as there is oxygen present, $\ce{NO}$ will combine to form $\ce{NO2}$, especially at high temperatures. $\ce{NO}$ is the most likely primary reaction product, because it is even harder to form $\ce{N2O}$ from $\ce{N2}$ with oxygen iirc. I can’t help you on the other questions but I would be greatly surprised if no-one could find a UV spectrum.

Fluorine reacts with practically everything, especially at high temperatures, including zinc, wires, some ceramics, polymers, etc., so the system you describe may be prone to contamination. I don't see any obvious source of NO2, but it could easily form from nitrates, if present, in an acidic environment, which could result from fluorine attacking metals present.


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Interaction between Ozone and Carbon Dioxide
We are trying to predict what will happen when carbon dioxide gas is introduced into a contact chamber that serves as an ozone contact chamber. These contact chambers presently oxidize compounds found in municipal pool water along with chlorine at below 5 ppm. There are 3 contact chambers (one for each pool) and the temperature range is from 104 to 80 F. We are switching from hydrochloric acid to carbon dioxide to keep the total alkalinity from dropping when the acid is introduced.

"Predict what will happen" is exceptionally broad. If you can narrow your question to specifics, that will help a great deal. For instance, we don't know what the compounds are in the municipal pool water. See this link for a guide to asking questions here on chemistry.SE.

Ozone and carbon dioxide will be pretty much indifferent to each other. Indeed, ozone is known for its ability to oxidize almost everything, but $\ce{CO_2}$ is already a final product of oxidation. "What is dead may never die", they say. Likewise, $\ce{CO_2}$ may never be oxidized.

Thank you for taking the time to respond Todd and Ivan. the other compounds in the municipal pool water would include body oils, salt from perspiration and what ever a particular bather might introduce.

Again, I would say any reaction between those two is highly unlikely. Both may be short-lived, but for independent reasons.


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Why is there a sudden drop in the concentration of ammonia in this graph?

So initially they are mixed with the mole ratio of $\ce{H2}$ and $\ce{N2}$ being 3:1 respectively. It seems as the reaction proceeds, both these quantities decrease and the quantity of ammonia increases until a point at which $\ce{NH3}$ stagnates. What happens at this point?

https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle

In the graph the following reaction will occur: $$\ce{N2 +3H2 <=> 2NH3}$$
This reaction is an equilibrium reaction and you can see that there is exponential decay for the reactants while the product is formed. After a while the concentrations do not change any more, so the system is in equilibrium. This is at the end of the time span "a".
At the beginning of time span "b" the system is disturbed, as the ammonia is completely removed. The system is changed and the equilibrium has to form again. Hence the concentration of the reactants is further reduced while new product is formed. Like santiago already pointed out in the comments, this is known as Le Châtelier's principle. At the end of time span "b", the system is in equilibrium again.

The first step to this question is writing the chemical equation for this reaction: $$\ce{N2(g) + 3H2(g) \rightleftharpoons2NH3(g)}$$ As you see, this reaction is an equilibrium reaction, meaning that not all the products will form reactants, but rather react until they reach a certain ratio. This ratio is known as the equilibrium constant, K. The equilibrium constant is given by: $$K = \frac{[A]^a[B]^b}{[C]^c[D]^d}$$ [A] and [B] are the concentrations of the products at equilibrium while [C] and [D] are the concentrations of the reactants at equilibrium. a, b, c, and d are the stoichiometric co-coefficients of their respective letter. The equilibrium constant is the same value for a particular reaction at a constant temperature and doesn't depend on the initial concentration of the products or reactants. In this case, the equilibrium constant is given by: $$K = \frac{[NH_3]^2}{[N_2][H_2]^3}$$ One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by: $$Q = \frac{[A]^a[B]^b}{[C]^c[D]^d}$$ Note how this is similar to the equilibrium constant, but here, [A], [B], [C] and [D] are the concentration of the substance at a specific time during the reaction, not at equilibrium point. This means that if Q is smaller than K, the reactants will react to form more products so that Q will increase so that it will eventually equal K and when Q is larger than K, vice versa. When Q equals K, the system is at equilibrium.
So now lets apply this concept to this graph. Initially, there is no ammonia but there is hydrogen and nitrogen gas present. This means that Q = 0 which is smaller than K as K is non-zero. This means that nitrogen and hydrogen gas will react to form ammonia. At the time corresponding to dashed line, the system is at equilibrium as the concentrations of the substances aren't reacting. This means that Q = K. Now, the sudden drop in concentration of ammonia indicates that some of the ammonia is taken out from the system. This means that Q will become smaller and hence, be smaller than K. Therefore the reactants, hydrogen and nitrogen gas, must react further to form more ammonia until reaches equilibrium again.


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What is the state of the oxygen free radical in the following reaction?
$$\ce{O2(g) ->[h\nu][\nu\,<\,240~nm]2O^{.}(?)}$$
Oxygen is split by UV light below 240 nm to form oxygen radicals, which then go on to react with oxygen molecules to form ozone.
Firstly, is the oxygen a free radical?
Oxygen has six outer shell electrons, but my teacher mentioned that by the way orbitals work, they don't actually all pair off, rather you end up with two pairs of electrons and two unpaired electrons, hence making oxygen a free radical, though I'm not too sure. But because the free radical is so highly reactive, I'm not sure whether I'm even allowed to assign it a state, because straight after being formed it pretty much immediately reacts.
Going by process of elimination, it can't be aqueous, because it isn't dissolved in water, and I don't think it is solid, because solids are structurally rigid. I am therefore inclined to say that it is a gas, but my main problem is that each reaction only creates two oxygen free radicals, which immediately disappear, so it shouldn't actually show any properties of being a gas.
So what is the state I should assign the oxygen, or does it simply not need one?

According to the reaction $$\ce{O2\,(g) ->[h\nu][\nu\,<\,240~\mathrm{nm}] 2O}$$
photolytic cleavage of oxygen yields two radicals. These will most likely be in the high spin triplet state. And yes these are free radicals according to the definition of the IUPAC Goldbook, as they have two unpaired electrons.
I assume you are talking about state of matter and then of course, you are right. Oxygen radicals can exist neither in aqueous solution, nor liquid, nor solid. Even in gas phase the concentration of oxygen radicals will be very low as they are an incredibly short lived species. They might exist in plasma, too. Keeping that in mind it is probably unnecessary to assign a state to the oxygen radical.

It does not need one. Short-lived reactive intermediates never form a phase of their own. There are just too few of them at any given moment.
On second thought, however, that lone oxygen atom does exist inside some phase, and that is the gas phase. So it would not be entirely inappropriate to assign it the gas state. With some other reactions, it could have been different. There are examples of reactions between gases that really occur on the solid surface of certain catalyst. In such a case, the intermediates would not fly in the air among the gas molecules, but sit on that surface. (Not that I would then call them "solid", though.)
Now, the radicals with two unpaired electrons use to be called diradicals, and that's what your atomic oxygen is.
BTW, the molecule $\mathrm O_2$, despite being quite stable by itself, is also a diradical, for unrelated reasons.


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Nucleophilic Addition of Hydrohalic Acids
Hydrogen cyanide shows nucleophilic addition at carbonyl compounds' carbon to form a cyanohydrin. Alcohols, similarly, form acetals or ketals. Will hydrogen chloride - or hydrohalic acids in general - show a similar reaction to form alpha halohydrins?

No, they don't.

Is there a reason for that? Could you explain it in the answer if there is, please?

As you noted, hydrogen cyanide will react with carbonyl compounds to form cyanohydrin and alcohols to form ketals/acetals. The list can be expanded to include aminals after attack of a nitrogen nucleophile, or thioacetals after sulphur attacks.
So why does it not happen with hydrogen halides? Well, in the case of the above mentioned, the reverse reaction is usually not favoured. In fact, ketals are used as protective groups for ketones (usually the reaction with ethan-1,2-diol or propan-1,3-diol); and propan-1,3-dithiol is similarly used for umpolung reactions. Likewise, the cyanohydrin has a new carbon-carbon bond formed, and these are relatively stable. To remove the ketal protective group, a rather strong Brønsted acid is often required along with an excess of water.
But in the case of a halide, that can just easily diffuse away again as a happy little halide; carbonyl double bond will be reformed. In fact, this reaction can be (and is!) used in exactly the opposite direction: liberate a hydroxide on the same carbon as a halide, and the halide will be eliminated to create the double bond.


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Does fumaric acid form maleic anhydride upon heating?
Maleic acid (the cis isomer of the dicarboxylic acid $\ce{C4H4O4}$) readily forms its anhydride on heating. Clayden's organic chemistry textbook mentions this as a point of distinction between it and fumaric acid (the trans isomer), which fails to dehydrate to its anhydride upon heating. However, I have discovered that at temperatures above 200 $^{\circ}\mathrm{C}$, fumaric acid forms maleic anhydride as well.
Which is the correct answer? I don't think it is likely that fumaric acid can form the anhydride - even at higher temperatures - given the trans configuration and restricted rotation.

Some  calculations (values from NIST) reveal that the formation of maleic anhydride and water from fumaric acid is an endothermic reaction ($\Delta H\approx +65.28 \frac{kJ}{mol}$) but the change in entropy is positive ($\Delta S\approx 134.24 \frac{J}{mol\cdot K}$). This means that the reaction becomes spontaneous for temperatures above $213^\circ C$, but this is only "mildly" spontaneous ($\Delta G\approx -5 \frac{kJ}{mol}$ at $523\ K$). So, the use of a dehydrating agent to remove the water from the reaction mixture and drive the equilibrium forward (Le Chatelier's principle) is necessary.

Fumaric acid does form maleic anhydride due to partial combustion. From wikipedia article of fumaric acid:

Fumaric acid converts to the irritant maleic anhydride, upon partial
  combustion.

Also, dehydrating fumaric acid with a  dehydrating agent also forms maleic anhydride. This paper describe how to convert $\ce{C^{14}}$ labelled fumaric acid to maleic anhydride.

Maleic anhydride-2-$\ce{C^{14}}$ is readily obtained from fumaric
  acid-2-$\ce{C^{14}}$ through dehydration using phosphorus pentoxide
  getting 96% yield under controlled condition.

I agree with your reasoning but it really seems that fumaric acid forms the anhydride.
According to the Wikipedia page on maleic acid
the isomerization process (from maleic to fumaric) is rather easy.


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What property allows ozone to absorb uv light
Why can ozone act as a barrier of UV light, so that we don't get irradiated by the sun's energy, and not any old molecule? In a recent exam I sat, one of the questions was "Name one physical property of ozone and contrast it with oxygen" and under exam pressure I spurted this onto the page: "Because the coordinate covalent bond in ozone is weaker than a traditional covalent bond, the bond is free to 'wiggle' about a bit. This means that the ozone can absorb a range of frequencies of photons, as a completely rigid molecule like oxygen would only have very specific frequencies which it could absorb." How much of this 'guess' is grounded in fact, or in other words, what is the actual mechanism that allows ozone to protect us from UV rays and not any old molecule?

Ozone is less stable than oxygen. When UV radiation reaches ozone layer, the ozone molecules decompose to form oxygen molecules which are more stable. So even before UV radiation can reach earth's surface, its energy is used up in ozone layer for forming more stable oxygen molecules.

Now you may think that due to UV radiation ozone layer depletes quickly. But in reality, reverse reaction also takes place. A free oxygen atom combines with a normal oxygen molecule to form ozone. The free oxygen atom comes from decomposition of ozone.

@Aditya Dev With links and some extra stuff you could make answer with it.

Shortwave UV light (100-240nm) has enough energy to knock an electron out of an $\ce{O_2}$ bond creating an excited molecule $\ce{O^{excited}_2}$ which decomposes to two free radicals of oxygen. Each of the free radicals then combines with different oxygen molecules to create a net of two ozone molecules.
The crux here is that the electron is not "knocked out" of the oxygen molecule, the electron is just moved to a higher energy orbital of the same oxygen molecule. Since orbitals are quantized it takes a very specific energy to make the transition happen.
Now there is a second reaction. Longwave UV light (240nm - 350nm) has enough energy to knock an electron out of an $\ce{O_3}$ bond creating an excited molecule $\ce{O^{excited}_3}$ which decomposes to a free radical of oxygen and an oxygen molecule.
Thus oxygen protects the ground from shortwave UV, and ozone from longwave UV. Neither alone would protects us from all UV as effectively.
Now the rest of the story is that all molecules bouncing around in the air are not in the same energy state due to atom-atom vibrations. So the UV absorption occurs as a "band" rather than as a "fine line" like in the absorption spectra of an atom in free space.

Firstly, ozone is an unstable compound and decomposes at about $573K$ to form oxygen. When UV radiation reaches ozone layer, ozone molecules present at the layer absorb its energy and decompose to form oxygen which is more stable than ozone.
One ozone molecule decomposes to form one oxygen atom and a normal oxygen molecule. Hence the energy of UV radiation is absorbed in ozone layer even before reaching earth's surface. That's how ozone layer protects us from UV radiation.
Now, this does not deplete ozone layer since reverse reaction can also take place. That is, high energy UV radiation hits oxygen molecules and each decompose into two free oxygen atoms. Each oxygen atom combines with one oxygen molecule each and forms ozone molecules.
You should also note that the bond order of ozone is $1.5$ and that for oxygen is $2$.
You may also want to read about depletion of ozone layer. Here is a link: Ozone hole

This is not some intrinsic property of ozone. In fact, any molecule would absorb a range of frequencies of photons. It just so happens that O3 covers the very range that is not covered by other components of our atmosphere, and thus saves us from the detrimental effects of UV rays of that range.

The reason ozone "filters" UV light at certain wavelengths is due to creation and destruction of ozone via UV light.  This is the reason we have an ozone layer in the stratosphere.
UV light creates ozone at wavelengths less than ~240 nm.  While UV light from ~240 - 315 destroys ozone.  In the process of creating and destroying ozone the UV light is absorbed in those wavelengths.
http://www.oxidationtech.com/ozone/ozone-production.html#answer2


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Flavor components of high end vodka?
I always thought vodka was just alcohol and water, and that the ideal vodka was one lacking all flavor components characteristic of the grain or potato mash that it was distilled from. I once saw a vodka process diagram showing multiple distillation and charcoal filtration steps--suggesting an effort to remove all flavor components. But looking at the prices being charged for high end vodkas, I am wondering what the draw is. Are there flavor components in these high end vodkas that make them that much more desirable than pure alcohol that people pay more than $50 a bottle for them?

a good (not hi-end) vodka is made using specially threated water, sometimes mineral or artesian water. This is worth some extra money. No idea about hi-end vodka, though.

The New York Times ran a blind taste test in 2005 which declared Smirnoff to be hands-down the best vodka. Based on that I'd be skeptical about there being something special about a "high-end" vodka's flavor that makes it worth paying for. However, the article did note that different vodkas do have distinct flavor profiles, so there is some taste difference between different vodkas.

High end vodka is primarily good vodka+branding. You pay extra to get that designed bottle. Perhaps the manufacturer also does something extra that they claim makes the vodka special. Whether it actually does depends on what they have done.
Good vodka has most flavours removed. Some flavours - specifically those that both have a boiling point close to that of alcohol and the ability to pass through an active carbon filter will remain.


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Axial or planar chirality in spiro[3.3]hepta-1,5-diene
The two spirocyclic compounds drawn below are enantiomers. My question is, do they exhibit axial or planar chirality?

3D representations:

Note that, according to IUPAC recommendations concerning stereogenic centers at spiro fusion atoms, the two bonds associated with one of the rings should be drawn as plain bonds, while the two bonds associated with the other ring should be drawn with one solid wedged bond and one hashed wedged bond.

Look here a detailed explanation of 2,6-dichlorospiro[3.3]heptane. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RSspiro45/spiro3-1-1.html#dichloro33 See Figs. 5 & 6

Useful Definitions (from the IUPAC Gold Book)

chirality plane - "A planar unit connected to an adjacent part of the structure by a bond which results in restricted torsion so that the plane cannot lie in a symmetry plane. For example with (E)-cyclooctene the chiral plane includes the double bond carbon atoms and all four atoms attached to the double bond; with a monosubstituted paracyclophane the chiral plane includes the monosubstituted benzene ring with its three hydrogen atoms and the three other atoms linked to the ring (i.e. from the substituent and the two chains linking the two benzene rings)."
planar chirality - chirality resulting from a chirality plane.
chirality axis - "An axis about which a set of ligands is held so that it results in a spatial arrangement which is not superposable on its mirror image. For example with an allene abC=C=Ccd the chiral axis is defined by the C=C=C bonds; and with an ortho-substituted biphenyl the atoms C-1, C-1', C-4 and C-4' lie on the chiral axis."
axial chirality - chirality resulting from a chirality axis

Interpretation

According to the above definition, a chirality plane must involve restricted torsion (restricted rotation) about a bond.
A molecule can contain both a chirality plane and a chirality axis; an example would be an ortho-substituted biphenyl where the restricted rotation is responsible for chirality (atropisomers).

Conclusion

Axial or planar chirality in spiro[3.3]hepta-1,5-diene

There is no restricted rotation that brings about chirality in this molecule.  Unlike trans-cyclooctene or a biphenyl, this molecule cannot be made to lie in a plane by torsional motion (at least not as long as we maintain the tetrahedral nature about the spiro carbon). Therefore, this molecule does not possess a chirality plane.
Rather, this molecule is analogous to the allenes. It does have a chirality axis passing through the $\ce{CH}$ group in each ring furthest from the spiro carbon along with the spiro carbon. The molecule can be said to possess axial chirality.

With regard to nomenclature of chiral spiro compounds, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) distinguishes three cases:

stereogenic spiro atoms of the type ‘Xabcd’, where ‘a’ > ‘b’ > ‘c’ > ‘d’
stereogenic spiro atoms of the type ‘Xabab’, where ‘a’ > ‘b’
axial chirality of spiro compounds

The given example spiro[3.3]hepta-1,5-diene contains a stereogenic spiro atom of the type ‘Xabab’. The stereodescriptors ‘R’ and ‘S’ are used when the spiro atom ‘X’ is surrounded by four atoms arranged as equivalent pairs ‘a’/‘a'’ and ‘b’/‘b'’, where ‘a’ > ‘b’. Thus, the two stereosiomers are (R)-spiro[3.3]hepta-1,5-diene and (S)-spiro[3.3]hepta-1,5-diene.

(Note that no stereodescriptor ‘E’ or ‘Z’ is needed to describe a double bond when the stereogenic unit is located in a ring having less than eight members.)
By way of comparison, (2R,4S,6R)-2,6-dichlorospiro[3.3]heptane is a spiro compound with axial chirality.

(Note that the sterodescriptors ‘R’ and ‘S’ are used to describe the chirality centers. The stereodescriptors ‘M’ and ‘P’ can also be used to describe axial chirality of spiro compounds; however, sterodescriptors ‘R’ and ‘S’ are used in preferred IUPAC
names.)
The compound (1R)-5'H-spiro[indene-1,2'-(1,3)oxazole] is an example for a stereogenic spiro atom of the type ‘Xabcd’, where ‘a’ > ‘b’ > ‘c’ > ‘d’.

I don't understand the nomenclature of 2,6-dichlorospiro[3.3]heptane. Only two possible sereoisomers should exist, but in the name three stereodescriptors are represented. Shouldn't the compound be given just an R or S descriptor by the means of assigning configuration to an axially chiral compound: https://en.wikipedia.org/wiki/Axial_chirality#/media/File:Allenes_chirality_depiction.png?

@Marko That is correct. 2,6-dichlorospiro[3.3]heptane has two stereoisomers: (2R,4S,6R)-2,6-dichlorospiro[3.3]heptane = (2P)-2,6-dichlorospiro[3.3]heptane = (2Sa)-2,6-dichlorospiro[3.3]heptane and (2S,4R,6S)-2,6-dichlorospiro[3.3]heptane = (2M)-2,6-dichlorospiro[3.3]heptane = (2Ra)-2,6-dichlorospiro[3.3]heptane. ‘R’ and ‘S’ describe the individual chirality centers. ‘M’ and ‘P’ or ‘Ra’ and ‘Sa’ describe the axial chirality. (‘Ra’ and ‘Sa’ are not used in preferred IUPAC names.)

Then the spiro carbon atom is a pseudoasymmetric center and should be written with a small letter, r or s, right?

@Marko I suppose, your question is justified. By way of comparison, ChemDraw actually calls it ‘(2R,4s,6R)-…’, whereas ChemSketch fails to generate any stereodescriptors at all. However, the example in the current IUPAC recommendations (Blue Book) explicitly reads ‘(2R,4S,6R)-…’. I think, the IUPAC recommendations are correct and that this is not a pseudoasymmetric center since reflection of (2R,4S,6R)-… yields (2S,4R,6S)-…, whereas the stereodescriptors ‘r’ and ‘s’ describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (‘r’ remains ‘r’ and ‘s’ remains ‘s’).

Now that I have had some experience on this site, I would like to get into the good graces of Glorfindel by expanding on how the assignments to (2R,4S,6R)-2,6-dichlorospiro[3.3]heptane presented by Loong are made. I will use the (2S,4R,6R)-enantiomer. The method requires the use of digraphs as shown in illustrations A and C below.  To create digraph A, each bond to the spiro atom (C4) is “cut” and methylene carbon is terminated with a phantom group (red dot, fake C4). What is left is a carbon atom with two pairs of identical groups of the type C(a,a’,b,b’) where a>a’>b>b’. The chirality of all “four” chlorine-bearing carbon atoms (C2 and C6 ) are assigned temporary descriptors (R0>S0) where Cl>CH2-C4>CH2-phantom>H. The R0 groups are assigned the top priorities with the proviso that the first and third priority groups are derived from the same ring. Accordingly, the CIP  method assigns the spiro carbon as having the R-configuration.

Another approach is represented in illustration B that doesn’t require dissection of the rings. The methylene groups attached to both chlorine-bearing carbon are assigned temporary descriptors knowing that chlorine is the top priority and hydrogen the lowest. Both digraphs A and B give the same R-configuration for the spiro carbon independent of which ring is assigned the first and third priorities.
The assignments to C2  and C6 utilizes digraph C.  To determine the configuration of C6, one dissects C2 and C4. The temporary configurations at C2 are assigned as seen in digraph A. Assignment at C4 has the priorities C6-chain>R0>S0>phantom. The chirality at C6 follows the priorities Cl>R0>S0>H leading to an S-configuration at C6. Owing to symmetry, the same analysis applies to C2.


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Aldehyde and ketone nomenclature when aldehyde/ketone group is bonded to an alkyl group

The longest alkane chain is decane and, attached to the fifth carbon, is an ethyl group the end of which is double bonded to an oxygen atom. My best guess at naming it would be:
5-(ethyl-2-al)decane.

Here, the longest chain is again decane, but there is a propyl group and an oxygen atom double bonded not to the ends of the alkyl, but to carbon #2, which leads me to:
5-(propyl-2-one)decane
Now I have no clue as to whether I am right or not because I have never come across such molecules and I have just done what I found to be logical. Have I come to the correct conclusions?

No, chain with functional group is more important

@Mithoron So 3-butyl octane and 4-butyl nona-2-one? Why doesn't the longer chain take priority?

Because there is a hierarchy of priority that starts with functional groups and ends with longest chains.

The characteristic group $\ce{=O}$ corresponds to the substitutive prefix ‘oxo’.
Thus, the substitutive names that you are looking for are ‘5-(2-oxoethyl)decane’ and ‘5-(2-oxopropyl)decane’.
However, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred names for aldehydes and ketones derived from alkanes are constructed using the suffixes ‘al’ and ‘one’, respectively.
Furthermore, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes and then to parent hydrides when different classes are present.
Therefore, the preferred IUPAC names for the given compounds are ‘3-butyloctanal’ and ‘4-butylnonan-2-one’.


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Urea-Formaldehyde cleanup
I work a lot with Urea-Formaldehyde resin/plastic, both in carpentry and textiles. Cleanup while still uncured is simple enough. Cleanup before even mixed is easy enough. What I don't know, is how to cleanup plastic that has set and cured. If it was reasonably safe to guess and try, I would, but in this environment/conditions acetaldehyde formation is very possible, which makes release formaldehyde a problem. Grinding or sanding the plastic is not an option. I haven't seen a MSDS say anything about cleaning up cured plastic. From accidental spills it doesn't seem to disolve in xylol, acetol, or toluol. Anyone have any ideas? As long as there is a preditable outcome I have the resources to test just about anything.

It appears it can be decomposed in the presence of Hydrogen Peroxide, with decomposition being noticeable even at 3% (extremely slowly), but working well around 12%. I didn't do anything beyond basic tests, but a white salt (likely Urea?) will precipitate out, embedding in the pores, if any, of the plastic; this can interfere with this clean up method. It also produces Aldehydes, seemingly mostly Formaldehyde from the odor, but with reactions based on the environment. Some Acetaldehyde was produced in my case, but the rate was slow enough that only small short lived flames were produced.
This was able to be sped up with heating or HCl. I would shy away from adding HCl, as it produced a strong Chlorine smell, without any obvious yellow gas.


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Why does Stokes raman not relax to the ground state while anti-Stokes does?
I'm just wondering why stoke-Raman doesn't go to the ground state. Where does the rest of energy from exciting laser leave?
On the other hand, why does anti-stoke go back to ground state? I know that the overall energy of anti-stoke is higher but why it has to go back to the ground state?
My question is based on this diagram.

As I understand, the anti-stoke has higher energy as it is excited from v1 due to an enhancement. I'm just confused with the last process of anti-stoke when it comes down to the ground state. Why doesn't it come down to the v2 or higher vibrational state like stoke?

Hello, and welcome to Chemistry SE. Feel free to look around, and take the Tour available under 'help' up on the top menu bar. As it stands, your question does not demonstrate much research (even Wikipedia) on Raman scattering. What is unclear?

I'm sorry for this stupid question.
My question is based on this diagram. https://upload.wikimedia.org/wikipedia/commons/8/87/Raman_energy_levels.jpg
As I understand, the anti-stoke has higher energy as it is excited from v1 due to an enhancement. I'm just confused with the last process of anti-stoke when it comes down to the ground state. Why doesn't it come down to the v2 or higher vibrational state like stoke?

You can [edit] your question to make it more precise. || @JonCuster You can directly link to the tour with [tour].

There is no law that a transition must go to the ground state. Now, each molecule has a multitude of electronic states, and within each of them a multitude of vibrational states. Thus each transition between two pure electronic states is accompanied by a variety of related transitions between higher-to-lower and lower-to-higher vibrational states. Which of these are the Stokes lines and which are the anti-Stokes ones, you'd better figure out for yourself.

It is really a matter of definition. Rayleigh scattering does not involve any change to the internal energy of the molecule whereas Raman transitions do. The Raman transitions always involve a change in the vibrational /rotational energy levels.
If the molecule has sufficient vibrational energy in its ground state then anti-stokes scattering can be observed since then the $v=1$ level has some population.The Stokes line can always occur as it starts from $v=0$. The total energy remains the same: initial photon energy = scattered photon energy $\pm$ vibrational energy. In general 'selection rules' have to be considered when analysing which spectroscopic transitions are allowed and which are forbidden.