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Calculating nearest neighbor distances
For cubic materials, there are equations in which to calculate nearest neighbor (NN), second nearest neighbor, etc. Can this be done with tetragonal crystal structures? I want to calculate NN, 2NN, and 3NN of $\ce{TiO2}$ rutile with a tetragonal crystal structure but am unsure how to do it. Any suggestions/help would be appreciated!

There is no equation for cubic materials. What you probably have in mind is the formula for the simple cubic structure having one atom per unit cell. Rutile is very much not like that. True, it is tetragonal, but to say that is the same as to say nothing. You will need atomic coordinates, symmetry elements, and on top of that, some textbook on crystallography. Then (and not before) you'll have your first distances, etc.

@IvanNeretin Could I just use coordination analysis to create the radial distribution function and take the first peak for 1NN, second peak for 2NN, etc?

It depends on what do you mean by coordination analysis.

@IvanNeretin If I create the RDF of my material, it is sufficient to use that to calculate the NN distances right?

Well, then these are just different ways to say the same thing. How are you going to create an RDF?

@IvanNeretin I have software to create the RDF of my system. Forget about how I calculate it. I am just asking: IF I have the RDF, I can take the first peak as 1NN, 2nd peak as 2NN correct?

Yeah, kinda like that. Wait a sec. You do realize that rutile contains two types of atoms and their environments (and hence RDFs around them) are quite different, right?

@IvanNeretin Yes I realize that they are made up of oxygen and Titanium atoms. Essentially I want to perform some sort of cluster analysis on an irradiated model, and most of what I have read defines the spacing between certain types of defects in a crystal given their 2NN distance. With two atom types, how would you go about calculating 2NN?

Like I said before: look up the atomic coordinates, symmetry elements, and then draw the diagram manually, or maybe rely on the software.

@IvanNeretin I just added the RDF into my question. I would interpret this to mean that 3NN = ~2.5 Angstorms, whereas 1NN and 2NN are around 1.9-2

What kind of interatomic distances are included in this RDF? Ti-O, Ti-Ti, O-O, or all of the above?

@IvanNeretin All of the Above

Then probably the distances around 1.9-2 are Ti-O, the next peak is O-O (which is altogether irrelevant if you want the neighbors of Ti!), then goes Ti-Ti, or maybe another O-O... see, it is not that simple, after all.

@IvanNeretin So I took your advance and looked at the crystal. I see that 1.94 and 1.98 peaks correspond to O-Ti, 2.5 peak is O-O, 2.7 peak is also O-O, 2.9 peak is Ti-Ti. I do not have a lot of experience using NN. How would you go about choosing what cutoff distance to use?

First, you can obtain CIF-file from COD, then load it with Olex2 (free, available on Windows, Linux, MacOS) and execute command envi <r>, which will print a list of the atoms about special position within a sphere of radius $r$. For example, envi 3 ($r = \pu{3 Å}$) yields in the following assigned pairs:
    O    SYMM  Ti    Ti    Ti    O     O     O     O     O     O     O     O     O     O     O
Ti  1.95 1_554 -     -     -     -     -     -     -     -     -     -     -     -     -     -
Ti  1.95 I     98.8  -     -     -     -     -     -     -     -     -     -     -     -     -
Ti  1.98 2_544 130.6 130.6 -     -     -     -     -     -     -     -     -     -     -     -
O   2.54 4_454 49.4  49.4  180.0 -     -     -     -     -     -     -     -     -     -     -
O   2.78 1_544 93.4  150.2 44.5  135.5 -     -     -     -     -     -     -     -     -     -
O   2.78 1_545 150.2 93.4  44.5  135.5 64.4  -     -     -     -     -     -     -     -     -
O   2.78 3_655 150.2 93.4  44.5  135.5 89.1  54.3  -     -     -     -     -     -     -     -
O   2.78 3_654 93.4  150.2 44.5  135.5 54.3  89.1  64.4  -     -     -     -     -     -     -
O   2.78 1_554 45.5  96.2  117.2 62.8  111.5 159.1 106.5 73.5  -     -     -     -     -     -
O   2.78 I     96.2  45.5  117.2 62.8  159.1 111.5 73.5  106.5 64.4  -     -     -     -     -
O   2.78 3_554 45.5  96.2  117.2 62.8  73.5  106.5 159.1 111.5 90.9  125.7 -     -     -     -
O   2.78 3_555 96.2  45.5  117.2 62.8  106.5 73.5  111.5 159.1 125.7 90.9  64.4  -     -     -
O   2.96 2_543 40.6  139.4 90.0  90.0  57.8  122.2 122.2 57.8  57.8  122.2 57.8  122.2 -     -
O   2.96 2_545 139.4 40.6  90.0  90.0  122.2 57.8  57.8  122.2 122.2 57.8  122.2 57.8  180.0 -

But this is only based on our guess for allowed $r$ value and is rather speculative. Answering your secondary question from the comments – "How would you go about choosing what cutoff distance to use?" – I would suggest to use Voronoi-Dirichlet (VD) polyhedra set [1] for every crystallographically inequivalent atom (e.g. for Ti and O pair in this case). You can do it with the ToposPro (Free, available on Windows), using Dirichlet subroutine.
Titanium:
Central atom:Ti1 0.500 0.500 0.500 Rsd:1.220
Atom:1.949 < r < 3.487  <r>=2.570   Top: 1.621 < R < 1.785  <R>=1.742
CN=6:0:4 NV=16 V=7.607/20.812 S=23.299 Cpac=0.509 Ccov=3.131
G3=0.083992444
Face distribution: {3/4 6/6 }
Vertex distribution: {3/16 }

  N  Atom   x      y      z     Dist.  SAng.
  1  O  1  0.3048  0.3048  1.0000  1.94857  16.77659
  2  O  1  0.6952  0.6952  1.0000  1.94857  16.77659
  3  O  1  0.3048  0.3048  0.0000  1.94857  16.77659
  4  O  1  0.6952  0.6952  0.0000  1.94857  16.77659
  5  O  1  0.1952  0.8048  0.5000  1.98000  16.44522
  6  O  1  0.8048  0.1952  0.5000  1.98000  16.44522
 *7  O  1 -0.1952  0.1952  0.5000  3.48704   0.00080
 *8  O  1  0.8048  1.1952  0.5000  3.48704   0.00080
 *9  O  1  0.1952 -0.1952  0.5000  3.48704   0.00080
*10  O  1  1.1952  0.8048  0.5000  3.48704   0.00080

VD-polyhedra for Ti atom with NN denoted with the bonds:

Oxygen:
Central atom:O1 0.305 0.305 0.000 Rsd:1.413
 D(CP):0.050  ( 0.2971 0.2971 -0.0000 )
 D(VDP):0.069  ( 0.2941 0.2941 -0.0000 )
Atom:1.949 < r < 3.487  <r>=2.788   Top: 1.621 < R < 1.820  <R>=1.747
CN=6:8:4 NV=32 V=11.805/52.029 S=28.544 Cpac=0.328 Ccov=2.138
G3=0.085774387
Face distribution: {3/4 6/14 }
Vertex distribution: {3/32 }

  N  Atom   x      y      z     Dist.  SAng.
  1  Ti 1  0.5000  0.5000 -0.5000  1.94857  16.77659
  2  Ti 1  0.5000  0.5000  0.5000  1.94857  16.77659
  3  Ti 1  0.0000  0.0000  0.0000  1.98000  16.44522
  4  O  1  0.6952  0.6952  0.0000  2.53648   3.12774
 #5  O  1  0.8048  0.1952 -0.5000  2.77800   4.78494
 #6  O  1  0.1952 -0.1952 -0.5000  2.77800   4.78494
 #7  O  1  0.8048  0.1952  0.5000  2.77800   4.78494
 #8  O  1  0.1952  0.8048 -0.5000  2.77800   4.78494
 #9  O  1 -0.1952  0.1952 -0.5000  2.77800   4.78494
#10  O  1  0.1952  0.8048  0.5000  2.77800   4.78494
#11  O  1  0.1952 -0.1952  0.5000  2.77800   4.78494
#12  O  1 -0.1952  0.1952  0.5000  2.77800   4.78494
*13  O  1  0.3048  0.3048 -1.0000  2.95870   1.71934
*14  O  1  0.3048  0.3048  1.0000  2.95870   1.71934
 15  O  1  0.6952 -0.3048  0.0000  3.32530   2.57704
 16  O  1 -0.3048  0.6952  0.0000  3.32530   2.57704
*17  Ti 1  1.0000  0.0000  0.0000  3.48704   0.00080
*18  Ti 1  0.0000  1.0000  0.0000  3.48704   0.00080

VD-polyhedra for O atom with NN denoted with the bonds:

Briefly, we are looking for interactions that are not labeled with # or *; to sum it up, there are the following sets of NN:

6 NN for Ti: 4×O@1.94857 Å, 2×O@1.98000 Å;
6 NN for O: 2×Ti@1.94857 Å, 1×Ti@1.98000 Å, 1×O@2.53648 Å, 2×O@3.32530 Å.

Reference

Blatov, V. A.; Shevchenko, A. P.; Serenzhkin, V. N. Acta Cryst A, 1995, 51 (6), 909–916 DOI: 10.1107/S0108767395006799.

I really appreciate this detailed answer of calculating the different NN distances. Is there a general consensus on which one of these to use for cluster analysis of a given property? Each value will produce vastly different results.

Also In this case, the 2nd nearest neighbor for Titanium would be 1.98, and for Oxygen would be 2.53 correct?

I always relied on VD-approach for this matter. It provides algorithmic, descriptive and unambiguous result.

@JacksonHart Oh, my bad, I posted wrong number in conclusion for O (already fixed, listing was correct). It is 1.98 both for Ti and O.

I am not sure that VD would work in my case. First, I have to use WS analysis to calculate vacancies, and then I want to determine the size of clustering of vacancies. So I need to load in the MD dump file, perform WS analysis, and then specify the cutoff distance between vacancies, which would form a void.

Also, you use 6NN as 6 nearest neighbors. In this case though, wouldnt the second nearest neigbor for O atoms be 1.98, or do you say 1.94 because there is more than 1 atom with this distance? Just not sure the interpretation of the scientific jargon

@JacksonHart Well, it may be not your cup of tea then, but keep in mind that VD approach is rather universal (e.g. nothing prevents you from doing the same for MOFs and molecular clusters) and system-agnostic; all you need is an established structure.

@JacksonHart Yes, I treat atoms with same $r$ as equivalent, so what I mean by 2NN is the atom with higher $r$ (second subshell, if it can be called so).

Can VD filter out non-defect atoms? Is there a WS subroutine that I can run first?

@JacksonHart What does WS in WS-analysis stand for, for me to be 100% sure we are on the same page?

Sorry, Wigner-Seitz Defect Analysis to determine vacancies and clusters. If I just look at my lattice, one of the Ti-Ti distance is ~4.5 Angstroms. They are next to each other though and just one unit cell over. If they were both missing, they should be a cluster. Maybe I should choose this distance. Could you extend the radius of the analysis you ran so I can see just a bit further out?

@JacksonHart Sorry, I'm unfamiliar with WS analysis and I'm afraid I cannot give you a direct answer whether ToposPro is the right tool for you except for suggesting to read the manual (sorry for RTFMing you:( ) Regarding the radius, in Olex2 you can use any number you want (e.g. envi 5 in your case), but in VD method it's pointless as it relies on solid angle values of VD-polyhedra for a given atom. If the atom is "cut-offed" (as here, as $\Omega(\ce{Ti-Ti})=0$), then it's considered as non-bonded atom

@JacksonHart But sure, if you have a structure with vacancies, then you get different environment and totally different picture in solid angle distribution, possibly resulting in Ti-Ti interaction (vdW, I suspect).

Let us continue this discussion in chat.

I see, just makes it hard determining a cutoff distance for a diatomic system to define what a cluster of defects is. For instance, how far do to vacancies have to be before they are not considered a cluster? These questions are already hard to answer in monotomic systems, and more challenging in diatomic

I already mentioned in the chat that WS-analysis is turned out to be pretty much the same as VD-analysis, I just never seen this name before (shame on me) :D. regarding the cluster stuff, it looks like a separate question to me, I also not quite sure how to answer it shortly – I don't know the uniform criteria either, sorry.

@JacksonHart I just checked the manual, and yes, another subroutine called IsoCryst from ToposPro can do what you want: "Analysis of molecular crystals, including H-bonded ones and supramolecular complexes. Searching for clusters (also of nano-size) in intermetallides and other structures."


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If an electron moves closer to the nucleus in a given value of n, does it lose energy?
If you look at a radial probability chart for a given electron sublevel (excluding n=1), there is a probability of an electron in that sublevel being closer to the nucleus. The electron should lose energy if it is closer to the nucleus because it experiences a greater force of attraction to the nucleus, but how is this possible if all electrons in a given sublevel have the same energy?

Stop thinking of probability as probability. An electron is not here or there; it is everywhere at once.

No, it doesn't. If you believe in electron orbits imagine it to be elliptical like in B-S model.

You are trying to apply macroscopic understanding to the world of quantum mechanics. This thinking is doomed to fail. The energy of an electron occupying a specific orbital is that orbital’s energy. Full stop. Unless it changes its orbital it does not change energy.

The electron does not occupy the orbital, it IS the orbital!

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Quantum Chemistry: Small imaginary frequencies
I am running a freq calculation on a geometry optimized structure using the ORCA quantum chemistry program. The output of the freq. calculation contains small neg. frequencies. Normally, these are considered to be saddle points, and thus would lead to the conclusion that the geometry does not represent a ground state.
freq.             IR Int.
-17.6900          0.000
-16.8900          0.000
-10.6600          0.000
6.4000            0.000
11.7200           0.000
... more small frequencies with IR Int. 0
30.3700           0.000
36.7000           19.3544

However, I am wondering if this is really the case, as the output also assigns an IR intensity of 0.000 to these modes, and states that "The first frequency considered to be a vibration is 14", which is the first positive mode. I am a little bit stuck, as obviously a negative frequency shouldn't appear in the IR spectrum, so does ORCA just removes those frequencies from the IR, even though they are there? Or does the IR-Intensity of 0 imply that the negative frequencies are actually just a precision error of the calculation, and I can still assume the structure to be a ground state.
Normally, the criterion is "no negative frequencies". On the other hand, not much energy should be stored in these modes, so I wonder how this will affect the computed energies.

If my distant memory from quantum chem classes is correct, small imaginary frequencies occur all the time because I forgot. But we should only care about those imaginary frequencies above some arbitrary threshold I also forgot, because only those are able to show us saddle points. I distinctly remember at least one practice calculation we did where we were told to ignore the negative frequency.

What method? If DFT, what functional, what grid, what kinds of RI, ...?

@pentavalentcarbon Is that really important? I tried AM1 and PM3 both give small imaginary frequencies. Calculations should only be semiquantitatively. The aim is to calculate Proton affinities, and a publication by Dewar has shown AM1 to perform quite well for this kind of task. Systems are big (around ~140 atoms), so I only consider semiempirical methods.

Yes, it can be. You just gave me a crucial piece of information; big and/or floppy systems often require tighter optimization convergence criteria (and therefore more cycles) than smaller systems. This probably isn't noise, it's probably an unconverged geometry.

@pentavalentcarbon OK, thanks for the hint. I am currently rerunning the calculation with tight scf and tight optimization settings. The imaginary modes correspond to a weird bending-stretch motion.

@pentavalentcarbon That also explains why I had most problems with the bigger systems, now it makes sense : ) Thanks!

Though I am a big fan of ORCA, I would not recommend using it for semi-empirical methods like AM1 or PM$x$. MOPAC does just fine for those. Further, simply ignoring these frequencies can lead to huge inaccuracies, because if one additional vibrational modes makes it to the positive (say, in a different conformation), you suddenly get more energy. The cleanest thing to do is "get only positive frequencies". Some investigators have also simply flipped the sign (yeah, I know).

just flip the sign? nice. I will try tighter convergence criteria first though

It's probably okay, but an indication you don't have a true minimum in the geometry.
It's important to calculate frequencies to know if you have a true minimum of the potential energy surface. Presence of imaginary frequencies are indeed indications of saddle points.
That's simple theory. Imaginary frequencies indicate problems in the curvature with respect to that degree of freedom - it's simply not a minimum.
If you want a true minimum, you'll need to improve convergence and numerical accuracy.
Now the practical side...
I'm not familiar enough with ORCA to know if the translational and rotational degrees of freedom are removed from the frequency list. Some programs list these, and some do not. Obviously these include 5 or 6 degrees of freedom which are inherently low energy and due to numerical instability or incomplete geometric convergence can sometimes be imaginary frequencies. Your list includes three very small imaginary frequencies, so they could be translations or rotations.
(I also mention this, because ORCA indicates the first vibration is at ~14 $\mathrm{cm}^{-1}$ and the code is likely removing the lowest frequencies as translations and vibrations.)
For even moderate-sized molecules, and particularly for large molecules, it can be very, very hard to converge geometries to true minima.
Consider your list - with multiple vibrational frequencies of 14-30 $\mathrm{cm}^{-1}$. Those are incredibly low energy breathing modes. Gradients and energy changes along those dimensions will be small and a great number of optimization steps will be required.
This is one reason that geometry optimization methods "give up" if the gradients get very small and displacements get very small. It may not be a true minimum, but in all likelihood, it's very, very close and would require a large computational effort to get the true minimum.
Note: Don't read this and think "oh, I don't need to bother." Your needs may vary. If you're doing accurate work, you absolutely need to push hard on geometry convergence.

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How to read these graphs and what are they called?
How does one interpret/read the following graph (found here) as well as on wikipedia or

What are these called?

According to your first link, this is from an x-ray diffraction image of human obesity protein, or leptin. It's called a ribbon diagram (and maybe other things, not my area of expertise).

Well, I'd call them coordinate/arbitrary axes...if you're talking about the arrows at the bottom left ;)

https://en.wikipedia.org/wiki/Ribbon_diagram

This depiction is a ribbon diagram of a protein structure. It shows the protein backbone (alpha carbons and the peptide bonds) smoothed out to a continuous line. In certain areas where the protein adopts alpha-helices or beta-sheets, these are indicated by the elaborate spiral in your picture (alpha-helix) or a wide, flat arrow (beta-sheet; not present in your structure). These depictions are supposed to give you an idea of what the protein looks like.
I would assume the two arrows in the bottom to be two of the coordinate axes of the crystal since ribbon diagrams are typically obtained from crystal structures. However, I cannot know without additional information (e.g. in a footnote of a picture).


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Handling vectors for double-bond graphics
I'm working on a small molecular graphics tool for visualising ball-and-stick style molecules. I'm having trouble handling the linear algebra to position the two bonds in the double bonds. My linear algebra is, unfortunately, very weak. I can follow notation but struggle with working my own way through logical algebra.
If I'm overcomplicating this significantly then I'd appreciate some guidance on how so. Here is my working so far:

I currently have the atomic coordinates of each atom. All of the above atoms are in a plane due to the $sp^2$ nature of atom $i$, though the orientation of the plane is arbitrary.
$\textbf i$, $\textbf j$ and $\textbf k$ are atomic coordinates and $\textbf a$ and $\textbf b$ are vectors between them.
The orientation of the double bond must mean that both of the lines in the double bond are positioned in the plane, and I have quantified the plane as:
$\textbf{a} \times \textbf{b} \cdot (\textbf{r} - \textbf{r}_0) = 0$
where $\textbf r$ is any vector in the plane, and $\textbf{r}_0$ is any other point in the plane.
I can also say that the points which I am looking for on each atom make vectors with the centre of respective atoms at $90 ^\text{o}$ to $\textbf b$, and that the radius at which these points are positioned is constrained to a constant value (and therefore so is the length of said vector). If we call this vector $\textbf l$ we therefore have:
$(\textbf{a} \times \textbf{b}) \cdot (\textbf{l} - \textbf{i}) = 0$
$| \textbf{l} | = \text{const}.$
$\textbf{l} \cdot \textbf{b} = 0$
where $\textbf{a}$, $\textbf{b}$ and $\text{i}$ are all things which are numerically available to me.
I feel as though, based upon geometrical constraints, these algebraic relationships are enough to find only the 4 points which I want. But my amateur self has no idea how to do that.

I guess there should be one more constrain that would rotate a multiple bond plane along $\mathbf b$ so that projection of that plane to a  computer screen is always maximal (so that visibility of multiple bonds is rotation-agnostic). Don't ask me how to implement it, it's just a suggestion:)

Yeah I thought of that as an option but discarded it for now, in favour of always having the double bond in the $sp^2$ plane.

I'm not sure I exactly know what you want, but such code is implemented in Avogadro and in other places.
The first thing to realize is that the unmarked atom in your plane is also your friend. You can calculate the plane of best fit using all 4 atoms.
You're making your work too hard, though. If you had a single bond, you simply draw a line or cylinder between atom i and j.
So the question at hand is how to displace two cylinders slightly in the plane of the bond. There are a few ways to do this. (Avogadro computes the plane of best fit for the molecule. Some people project towards the plane of the screen.)
My recommendation is to realize that atom k in your diagram is in one direction, and the unmarked atom is in the other. Let's call that other atom l. You can compute the vector between them: k - l. In practice, atom i is midway along this vector. So you want to displace the b vector in your diagram by a small amount, say 0.1 in one direction along that k - l vector, and 0.1 in the opposite direction.


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How can I remove a TiO2 coating from my car windows?
I've tried all the common solvents including hydrogen peroxide, chlorine, acetone. I tried a high speed sander with a rare earth polish. I have spoken to the head of the chemistry dept. at Cornell University and she said she did not have a remedy that would not ruin the glass. From the MDS here's what's in it:
1.Polymeric mixture with nano   5%-10%
2.Pexi Polymer   5%-10%
3.Anti static additive    5%-8%
4.Anti rust additive   3%-6%
5.Complex micro emulsion    2%-5%
6.Organofunctional siloxane blend    4%-8%
7.Synthetic nano driver polymeric fusion   10%-20%
8.Hydrogen hydroxide 40%-65%
Any suggestions would be greatly appreciated. The owner/developer of this coating won't take responsibility and help me get this stuff off my windows of my new car. Thank you.

"Hydrogen hydroxide" Buhahaha :D Nice joke...

The "head of the chemistry dept. at Cornell" is my favourite in this post. Troll?

"high speed sander", "rare earth polish" - nice.

@Mithoron . Hi. This is not a joke. The ingredients I listed, including hydrogen hydroxide, were taken word for word from the data sheet the developer of this product gave me when gave me this product to protect the glass from mineral deposits from my city water.

Of course you didn't know it's babble.  That makes it funnier :D

Well this isn't going like I'd hoped. First off I'm an illustrator so I have very little knowledge of what the ingredients in this coating are or do other than the many hours I've spent researching this on line of late. That's why I posted this question here in hopes someone who knows much more about chemistry might have a suggestion on how to get this stuff off my windows. I did speak with Melisa Hines at Cornell yesterday and I did hire a glass cleaning company to get this off the glass. They said it was a rare earth polish.

From my understanding the purpose of $\ce{TiO2}$ layer is to absorb UV (ultraviolet) part of solar spectrum in order to reduce negative (destructive) impact on the car internals (e.g. UV-light degrades plastics), also you and your passengers. So again, why do you want to get rid of it? Does it affect the visibility? If possible, please add the photo of the glass and/or car model (ideally glass type and manufacturer, if there are visible labels).

@andselisk. Actually the Tio2 in these coatings acts as a photo catalyst  neutralizing pollutants and dirt that are easy washed away by rain etc. So it's used more for it'self cleaning properties than UV protection in auto and building applications.

@Jeffrey Oh, I didn't know that, thank you! So why do you want to get rid of this coating? It sounds like it's been put there for a good reason.

One can feel dumber just by reading these "ingredients" but your coating is probably mostly polymeric, so you could try strongly oxidating mixtures like piranha solution or  CrO3 in H2SO4 Another option is to burn it out with flame or for more gentle approach decompose with less extreme heat.

@andselisk The windows look normal until they get wet then they reveal  runs and spots from rain,water, the car windshield cleaner etc, that have sat on the coating for any length of time. These runs etc once "set in" have to be smoothed out by repolishing. So the windows appear normal again until you get water on them again and the whole cycle repeats. So the coating doesn't go away it just becomes more uniform after polishing. I have a video showing this but I'm not sure how to get it fro m my computer to this site. I also have all the Toyota glass specs.

@Mithoron In the data sheet under heading Reactivity and sub heading Incompatibility it says Strong Oxidizing agents. So I researched that  and tried hydrogen peroxide and pool chlorine. The peroxide had some effect in the sense that the area I working became less slippery temporarily. What is piranha solution or CrO3 inH2SO4? I don't want to try to burn it off  that would probably melt the rubber gaskets around the windows. The data sheet claims it has a boiling point of 180-240F. Also states the PH is 7.2-7.6 if that helps.

Related https://chemistry.stackexchange.com/questions/84001/removing-polyimide-coating-without-damaging-carbon-nor-fused-silica-optical-fibe This "data" of yours is about solution (FYI "hydrogen hydroxide" is water...) which was used to make the coating. Like anything strong enough to remove coating, could also damage rubber, so window has to be taken out from the door before operation.

The head of the chemistry department at Cornell University is right.
The only advice I have is to wait. Eventually the ravages of time shall triumph over the titan's fiendishness.

I have seen photos of sections of buildings that had been treated with this stuff and after 6 years it was still on there and working. I was hoping to find a way to get this coating off my windows a little sooner than that.


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Polarity of symmetric Lewis structure
Why is $\ce{CF2Cl2}$ polar? The Lewis dot structure shows the molecule as being symmetric. Wouldn't that make it nonpolar?

The Lewis dot structure doesn't show a thing. It is drawn on a planar sheet of paper, and the real molecule is 3D. Ever heard of a tetrahedral geometry?

I encourage you to make that as an answer @IvanNeretin. It seems to be a precise and complete answer to me.

@PrittBalagopal Maybe, but it is no improvement over the existing answer, hence no.

The problem is that the first Lewis diagrams you learn to draw don't take 3D structure into account. A more realistic structure for dichlorodifluoromethane looks like

where the dashed wedged bond is meant to represent a single bond pointing away from you ("into" the plane of the screen), and the filled bold wedged bond is meant to represent a single bond pointing toward you (coming out of the screen). This is more than a traditional Lewis structure though, which is more about notating single/double/triple bonds and any nonbonding electrons.
You could also deduce the 3D structure from the fact that carbon centers with 4 separate bonds are always tetrahedral, and there is no way in a perfect tetrahedron to place either the 2 chlorines or the 2 fluorines exactly opposite each other. This means there is no way for the individual bond dipoles to cancel.


text:
Why is the specific gravity of 0.9 % saline 1.0046?
Medical Normal Saline (0.9% Saline) contains 9 g of salt dissolved in 1000ml of water.
Hence a litre of NS should weigh 1.009 kg.
Hence its specific gravity should be 1.009, but it’s only 1.0046.
(I've measured it, and verified it on-line. It is apparently, a fact.)
I thought I understood SG, it’s a simple concept but I am missing something.
EDIT: after more reading, I learned that dissolving NaCl in water results in a reduction in the original volume. So my question now becomes:
We're dissolving NaCl in pure water at a temperature where 1.000kg mass of water has a volume of 1.000 litres. By my (erroneous) reasoning:
Say the volume decreases by 1 ml, giving a final volume of 0.999 litres.
So the density of the solution would be 1.009 kg/ 0.999 l = 1.010 kg/l
Giving an SG of 1.010 (because water has a density of 1.000 kg/l, as above) If the volume didn't change, it would be 1.009, but it increases because of the reduction in volume.
BUT: The known, published SG of 0.9% w/v NaCl solution is 1.0046.
Can anyone explain why, please? or at least, why the SG is below 1.009, when it should be greater?

OK, you've got 9 g of salt and dissolved it in 1000ml of water. Who said the resulting volume is 1 litre?

One thing to understand is that significant figures play a significant role in solving chemistry problems. A specific gravity of  1.0046 is only about 0.5% more that 1.000 which is pretty close to the density of pure water at body temperature. So if medical saline is 0.9% saline, is it really 0.90000000% or is it maybe really 0.904%? So in a lot of problems we'd assume that 9 grams of salt and 1000 ml of water give "about" 1000 ml of solution, but it isn't absolutely true.

Thanks for the replies, but neither gives me an explanation. By my reasoning ( which is obviously faulty, and I'm trying to find out how), an SG of 1.0046 is HALF that of 1.009.  That is, it would be 9 grams of salt dissolved in TWO litres of water. I know that, when the salt is dissolved, the volume will change infinitesimally, but it won't double!! The total weight WILL be 1.009 kg and surely, the volume will still be very close to 1000 ml, so the density will be 1.009 kg/l ( or g/ml) which when compared to that of water gives an SG of 1.009. Where is my reasoning incorrect?

@ Ivan, my question has evolved. The volume is indeed not the same, it is reduced. Which moves the SG to be greater than 1.009. Whereas the known SG is 1.0046,  i.e. less than 1.009. which makes my reasoning even more wrong!

Reduced relative to what? To the total volume of (1000 ml of water + 9 g of salt)? That't probably true, but the result is still greater than 1 litre.

After reading all of the comments and answers, it must be EMPHASIZED that normal saline, or 0.9% (w/v) NaCl is 9 grams NaCl per 1000 ml OF SOLUTION, not 9 grams NaCl added to 1000 ml water. So, the final volume will be 1 liter, not more and not less. This was correctly pointed out by @aventurin.

0.9% medical saline is defined to contain $\pu{0.90 g}$ sodium chloride per $\pu{1000 ml}$ solution at $\pu{22 °C}$. Its density is $\pu{1.0046 g/cm^3}$ at $\pu{22 °C}$.
The density of water is $\pu{0.9978 g/cm^3}$ at $\pu{22 °C}$.
If the volume would stay constant when you add $\pu{0.90 g}$ sodium chloride to $\pu{1000 ml}$ water, the resulting density would be
$$\pu{0.9978 g/cm^3} + \pu{0.0090 g/cm^3} = \pu{1.0068 g/cm^3}$$
The density of the 0.9% saline shows us that $\pu{1000 ml}$ of it contain
$$\pu{1004.6 g} - \pu{9.0 g} = \pu{995.6 g}$$
or
$$\frac{\pu{995.6 g}}{\pu{0.9978 g/cm^3}} = \pu{997.8 ml}$$
water.
This means that the volume of the liquid phase increases when we add sodium chloride to water, what explains the difference in density from $\pu{1.0068 g/cm^3}$ to $\pu{1.0046 g/cm^3}$.

That’s not how percentage concentrations work. A $0.9~\%$ solution contains $\pu{9g}$ of salt per $\pu{991g}$ of water so that both add up to $\pu{1000g}$.
Also note that you cannot guess or estimate the density based on the density of water alone. It will change when stuff is dissolved.

Thanks, but NS is a w/v concentration, not w/w. It IS 9 g in 1000 ml. By my reasoning ( which is obviously faulty, and I'm trying to find out how), an SG of 1.0046 is HALF that of 1.009.  That is, it would be 9 grams of salt dissolved in TWO litres of water. I know that, when the salt is dissolved, the volume will change infinitesimally, but it won't double!! The total weight WILL be 1.009 kg and surely, the volume will still be very close to 1000 ml, so the density will be 1.009 kg/l ( or g/ml) which when compared to that of water gives an SG of 1.009. Where is my reasoning incorrect?

@S.Mc If it’s w/v it should always be mentioned being one otherwise normal chemists would not know that they are dealing with something entirely non-standard.

In my original question, I stated that the saline was 9g of salt in a litre of water. That's a w/v statement.

All of which is irrelevant anyway. Why does a litre of solution of SG 1.0046 have 9 g of salt in it, and not 4.6g?

https://chemistry.stackexchange.com/users/7475/jan   Jan, can you cast any light, I still can't understand why 1000ml of water with 9 g of salt, doesn't have an SG of 1.009

@S.Mc My answer’s first part is telling you that $0.9~%$ does not equal $\pu{9g}$ in $\pu{1l}$. My answer’s second part tells you that you cannot trust the addition of densities because you don’t know what the final volume of the solution will be. Also, one ping is enough. People do other things than hog SE 24/7 (especially considering your first ping was 6 am on a Saturday).

I sincerely apologise for the pinging, I didn't realise that the site did that! I left the question hanging for over a week with no progress...but I am sorry, truly. I also apologise that I have offended by re-submitting my question with changed wording to simplify it.( Can I merge the replies?).   But, to paraphrase it again: 1) The bag is 0.9% Saline (known in the industry as 'Normal Saline'. Each 1000 ml contains 9g of Sodium Chloride. It says so on the label.I can't state it any more clearly:  a 0.9% w/v solution of Sodium Chloride in water. 2) What do you mean by addition of densities?

SG is obtained by dividing the density of the solution by the density of pure water when both are at  a given temperature.  No addition.  I realise the volume changes, but at this stage I'm only asking for a concept explanation.  In the case of NaCl in H2O, the volume decreases so the density of the solution for a given mass of solute increases. If the volume didn't change then the density would be 1.009 kg/l if 9g was dissolved in 1000 ml. Thus the SG would be 1.009. Now: the volume decreases. Say by 1 ml, so the density is thus 1.009kg /0.999l = 1.010 kg/l.

Maybe not the exact figure, but it is greater than 1.009 kg/l. However, the known SG of Normal Saline is 1.0046. Not greater than 1.009, but less than. Entirely the wrong direction, which I don't understand and have repeatedly asked if someone can help me. I sense some exasperation, but I am a medically- trained engineer, not an ignorant  dunderhead, so pardon me if answers like 'the density changes when you dissolve stuff" don't quite cut the mustard. I have repeatedly, in great detail, expressed my question in several ways, asking for help and so far been dismissed offhand.

If anybody is still interested: this is all about the partial/apparent specific volume of NaCl in water.
Your first and main mistake is assuming that the $\pu{9 g}$ of NaCl are dissolved in $\pu{1000ml}$ of water. Since the total volume is $\pu{1000ml}$ and the NaCl contributes some volume, it‘s less water (see below). Your second main mistake is assuming that adding $\pu{9g}$ of NaCl doesn’t change the volume of the solution, and thirdly, you confuse SG (relative density) with density.
How much volume the NaCl contributes is determined by its partial specific volume. At infinite dilution, you might expect the partial specific volume to be the reciprocal of the density, but it is often less, because the dissolved solute can take up some of the space between the solvent molecules, or it can be greater if the solute molecules tend to repel the solvent molecules, or the ions take up more space then they do when combined. The partial molar volume is greatly determined by intra-molecular forces. Hence it is for all intents and purposes incalculable and an empirical quantity.
Let‘s do the calculation:
You say the solution has an SG (relative density, i.e. relative to the density of water) of $1.0046$, but a cursory search leading to tables of density of NaCl solutions suggests this value is the actual density of the solution at 22 °C (I found a table of density which is cited as having been taken from "Perry's Chemical Engineers' Handbook" by Robert H. Perry, Don Green, Sixth Edition). If so, one liter has a mass of $\pu{1004.6g}$, and hence contains $\pu{1004.6 g} -\pu{9 g} = \pu{995.6g}$ of water.
The density of water at 22 °C is $\pu{0.99777 g ml-1}$, so this water has a volume of $\pu{995.6 g}/\pu{0.99777 g ml-1} = \pu{997.8ml}$.
Since the total volume is $\pu{1l}$, the $\pu{9g}$ of NaCl in the solution have resulted in a volume increase of $\pu{1000 ml} - \pu{997.8 ml} = \pu{2.2ml}$, hence NaCl at this dilution has an apparent specific volume of
$$\frac{\pu{2.2ml}}{\pu{9g}} = \pu{0.244 ml//g}$$
This agrees well with a simple article I found from a student („Solution Density and Partial Molar Volume as Functions of Concentration“ by Melissa Michaels) in which the author uses standard methods to determine the partial molar volume of NaCl at various concentrations based on experimental measurements of the density. NaCl in the solution with $\pu{10.34g}$ NaCl in $\pu{1000ml}$ of solution was shown to have a partial molar volume of about $\pu{13.68 ml mol-1}$ which would be a partial specific volume of $\pu{0.234 ml g-1}$.
The density of NaCl is $\pu{2.16 g ml-1}$ and the reciprocal of this is $\pu{0.4629 ml g-1}$, so the ions seem to be fitting in between the water molecules well and/or attracting them, which makes sense since NaCl is an ionic compound and water is a polar molecule.
These calculations and this statement are based upon the assumption that the true density of the solution at 22 °C is $\pu{1.0046 g ml-1}$.

Expression formatting reference:
Basics/CH+Math expressions/Upright vs italic/Math tutorial/Advanced

To which aspect of the formatting are you referring?

E.g. literal numerical values with units or mathematical/chemical expressions/equations. Take them as handy links.

E.g.  if you write $\pu{1.123123E5 J K-1 mol-1}$  it is displayed as $\pu{1.123123E5 J K-1 mol-1}$. It also ensures the space between the literal value and the unit does not allow line breaks.

1000 mL of pure water weigh in at 997.047 g at 25 °C, ADD exactly 9 g to that and you have 1.006047 kg.
Remember density is temperature dependent and therefore 1000 ml of pure water may weigh more or less depending on which temperature the volume was measured at.
Unless you are working with super cooler pure water, you cannot reach a density of 1.000 g/mL.

Thank you Jeppe. I thought that might be the answer...but the change in density with temperature is the exact reason why Specific Gravity is used. Expressing the density of the solution, relative to the density of water at the same temperature (giving a dimensionless number) cancels out that factor. To simplify my calculations, I just took it to be at the temperature where water has a density of 1.000 g/ml so that the density of the solution is the same, numerically, as the SG.  Nonetheless, to verify, let's say the solution is at 25C:

If the volume didn't change after adding salt, the density is 1006.047/1000 = 1.006047 g/ml and the SG  is 1.006047 /0.997047 =1.009531                                                              If the solution volume decreases by, say, 1 ml then the density would be 1006.047/999 = 1.00705405 g/ml and the SG would be 1.00705405/0.997047 =1.0100366                                                                        so the SG is greater than if the volume didn't change. The fact is that the SG is 1.0046. which is LESS than if the volume didn't change. Still doesn't explain it.

            Consider this statement from Wikipedia  .
                  “ A solution of  9 grams of sodium chloride (NaCl) dissolved in water, to a total volume of 1000 ml (weight per unit volume). The mass of 1 milliliter of normal saline is 1.0046 grams at 22 °C “(Wikipedia Saline (medical)}
     The 1.0046 g/ml is a density value not a specific gravity value.  Specific gravity is a relative density computation based upon a designated reference.   If the designated reference is  H2O at  4 oC , then the  specific gravity would be the same  1.0046 .  However, at  22 oC  the specific gravity of NS is  1.00684 using  0.99777  g/ml  as the designative reference.

    A statistical error may be present – the value of 1.0046  appears to be a False  Null hypothesis that was taken to be True – the definition of a  Type II error .    The original source of 1.0046 is from a   1973   Handbook of Hydrology(1993 McGraw Hill ) book and then  implemented in an online calculator in 2010  ( ref #13 at Wikipedia Medical Saline )  .  In that reference  Normal  Saline density value of 1.0046 was calculated from fitting 4th degree polynomial for temperature and  salinity.  Analytical  methods for analyzing salt water densities in estuaries involving millions of cubic meters of water is considerably different than infusing one liter of  NS an 8-hour period ( less than 50 L of water) !     Also, polynomial fitting , especially high degree poly's , introduces error factors into the estimation , particularly for the values at the bottom a top of the range.   On average Sea water  is  3.5%.
The Specific Gravity of Normal Saline is dependent upon the choice of the density value of NaCl. There seems to be some slight variation of density depending on the source;  2.16 ,  2.165  and  2.17 .   The density of NaCl would have to be  2.05 , far out of range , in order to generate a Sp Gr  of  1.0046

     At  22  degree Celsius H2O Density 0.99777

NaCl 2.16
((1000 -  9/2.16 )0.99777 +  9 )/(1000.99777)  SpGr 1.00485
NaCl 2.165
((1000 -  9/2.165 )0.99777 +  9 )/(1000.99777)  SpGr 1.00486
NaCl 2.17
((1000 -  9/2.17 )0.99777 +  9 )/(1000.99777)  SpGr  1.00487
NaCl 2.05
((1000 -  9/2.05 )0.99777 +  9 )/(1000.99777)   SpGr  1.0046
Based upon the discussion above. -  this author opines that the specific gravity of NS  ( @ 22 oC )  is hereby deemed to be 1.0049.

There is one error in the question.  When sodium chloride is added to 1000 ml of water, the final volume is slightly less than the sum of the volume of sodium chloride plus 1000 ml but should not be less than 1000 ml.
Total volume = 4.17 ml (volume for 9g of sodium chloride based on density of sodium chloride = 2.16g/cm3) + 1000 ml = 1004.17ml
Total weight = 9 g sodium chloride + 1000 g water = 1009g
Density = 1009g/1004.17ml = 1.0048 g/L
Of course temperature has an effect on the water density.

The solution is a mixture of NaCl and Water. 0.9% Saline solution is the mass ratio between the NaCl and Water in the solution.
Known:
*Mass ratio (NaCl/Water) = 0.009
Density of NaCl = 2.165 g/mL
Density of Water = 1.0 g/mL
Total Volume of the Solution = 1000 mL*
Calculating the Properties of NaCl:
*Mass (NaCl) = 0.009x Mass (Water)
Volume (NaCl) = Mass (NaCl)/Density (NaCl)
Volume (NaCl) = (0.009x Mass (Water))/2.165*
Calculating the properties of Water:
Volume (Water) = Mass (Water)/Density (Water)
Calculate the Mass of Water:
Total Volume = NaCl Volume + Water Volume
1000  = 0.004157 x Mass (Water) + Mass (Water)/Density (Water)
Mass (Water) = 995.86 g
Calculate the Mass of NaCl:
*Mass NaCl = 8.96 g
Total Mass (Solution) = 1004.82 g
Solution Density = 1.00482 g/mL
Solution SG = 1.00482*

Density of NaCl is indicated at 2.16g/cc. Given this 9.0g of NaCl added occupies 4.16667ml of the flask. This leaves 995.833333ml of water in the flask. Therefore:
995.833333ml water * 1g/ml = 995.833333g water + 9.0g NaCl in the flask = 1004.833333g in 1000ml = 1.00483g/ml

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


text:
Using quantiative analysis to find empirical and molecular formula

A compound of Ca, C, N and S was subjected to quantitative analysis and formula mass determination, and the following data were recorded.
A $\pu{0.250 g}$ sample was mixed with $\ce{Na2CO3}$ to convert all of the Ca to $\pu{0.160 g}$ of $\ce{CaCO3}$.
A $\pu{0.115 g}$ of the compound was carried through a series of reactions until all of its S was changed to $\pu{0.344 g}$ of $\ce{BaSO4}$.
A $\pu{0.712 g}$ sample was processed to liberate all of its N as $\ce{NH3}$ and $\pu{0.155 g}$ of $\ce{NH3}$ was obtained.
The formula mass was found to be $\pu{156 g mol-1}$. Determine the empirical formula of this compound.

My issue here is that we don't really know what the starting compound is and particularly how much of each Ca, C, N, and S is there. Without knowing that, how can you balance your reaction equation and find the number of moles needed to get the empirical formula?

Normalize all the samples to say 100 grams and work from there. The empirical formula would contain integer values of all four elements. An educated guess could be made given the four elements involved.

Please type your question accurately if you want to receive an adequate answer quicker. It's a pure luck that this question already appeared online before.

The problem you gave doesn't quite work out, so I searched it up; did you perhaps mean $0.115 \space \text{g}$ of the sample were used in finding the sulfur content (you said $0.15 \space \text{g}$ instead).
If you did mean $0.115 \space \text{g}$, then we can figure out that the molecular formula of the unknown compound is $\text{CaC}_2\text{N}_2\text{S}_2$. Let me explain:
First, let's try figuring out how many moles of $\text{Ca}$ there are for each mole of the unknown compound, using the information that $0.250\space \text{g}$ of the compound was used to form $0.16 \space \text{g}$ of CaCO3. Converting everything to moles, this means we used $(0.250\space \text{g})\space/\space (156 \space \text{g/mol})=0.00160 \space \text{mol}$ of the unknown compound to get $(0.16\space \text{g})\space/\space (100.0869 \space \text{g/mol})=0.00160\space \text{mol CaCO}_3$, or $0.00160\space \text{mol Ca}$. This means that one mole of the unknown compound will contain $(0.00160\space \text{mol Ca})\space/\space (0.00160 \space \text{mol Unknown Compound}) \space\times\space (1 \space\text{mol Unknown Compound})=1\space\text{mol Ca}$
We can do this for each of the other procedures to get that for each mole of the unknown substance, we have $1\space\text{mol Ca}$, $2\space\text{mol S}$, and $2\space\text{mol N}$. Now, we can just subtract the masses of $\text{Ca}$, $\text{S}$, and $\text{N}$ from the molar mass of the unknown compound to get the mass of $\text{C}$ in the compound. For each mole of the unknown compound, we get that we have $\text{156 g}-\text{40.08 g}-2(\text{14.01 g})-2(\text{32.06 g})=23.8 \space\text{g}$ of carbon. This is just $(23.8\space \text{g})\space/\space (12.01 \space \text{g/mol})=1.98\space \text{mol} \approx 2 \space \text{mol}$ carbon per mole of unknown compound.
So, since we have $1\space\text{mol Ca}$, $2\space\text{mol S}$,  $2\space\text{mol N}$, and $2\space\text{mol C}$ for each mole of the unknown compound, the unknown compound must be $\boxed{\text{CaC}_2\text{N}_2\text{S}_2}$.

Nicely done! OP should've indeed typed the question more accurately as with the wrong mass the formula would've been $\ce{CaS_{1.5}N2C_{3.3}}$, which doesn't make sense.

I edited original question to denote actual mass of the sample for S content, it was clearly an OP's typo.

Summarizing the approach (nicely illustrated in an answer by Brandon Lee) towards solving similar quantitative analysis problems, it is always a good idea to present any reaction or set of reactions as the following conversion scheme where stoichiometric coefficient about analyzed compound is normalized to 1:
$$\ce{E^i_{$x_i$}R -> $\frac{x_i}{a_i}$ E^i_{$a_i$}R'} \tag{1}$$
where $\ce{E^i}$ is the $i$-th element being quantitatively determined; $x_i$ and $a_i$ are stoichiometric coefficients of the given element in unknown compound and obtained product, respectively; $\ce{R}$ and $\ce{R'}$ are the rests.
For simplicity, lets abbreviate analyzed compound $\ce{E^i_{$x_i$}}R$ as $\ce{X}$ and $i$-th product $\ce{E^i_{$a_i$}R'}$ as $\ce{P_$i$}$:
$$\ce{X -> $\frac{x_i}{a_i}$ P_$i$} \tag{1a}$$
According to the principle of mass conservation amounts of the $i$-th element must be preserved:
\begin{align}
n_i (\ce{X}) &= \frac{x_i}{a_i} n(P_i) \tag{2} \\
\frac{m_i(\ce{X})}{M(\ce{X})} &= \frac{x_i}{a_i} \cdot \frac{m(\ce{P_$i$})}{M(\ce{P_$i$})} \tag{3} \\
x_i &= a_i \frac{m(\ce{P_$i$}) \cdot M(\ce{X})}{m_i(\ce{X}) \cdot M(\ce{P_$i$})} \tag{4}
\end{align}
where $M$ is molecular weight and for the analyzed compound $\ce{X}$ is determined as
$$M(\ce{X}) = \sum_{i = 1}^n{x_i \cdot M(\ce{E^i})} \tag{5}$$
which allows to find the latter coefficient $x_n$ indirectly after all analytical routines performed for $n-1$ elements:
\begin{align}
M(\ce{X}) = \sum_{i = 1}^{n-1}{x_i \cdot M(\ce{E^i})} + x_n \cdot M(\ce{E^n}) \tag{6} \\
x_n = \frac{M(\ce{X}) - \sum_{i = 1}^{n-1}{x_i \cdot M(\ce{E^i})}}{M(\ce{E^n})} \tag{7}
\end{align}
For example, using the first step ($i = 1$) $\ce{E^i} = \ce{Ca}$; $\ce{R} \equiv \ce{S_{$x_2$}N_{$x_3$}C_{$x_4$}}$; $\ce{E^i_{$x_i$}R} \equiv \ce{Ca_{$x_1$}S_{$x_2$}N_{$x_3$}C_{$x_4$}}$; $\ce{R'} \equiv \ce{CO3}$; $a_1 = 1$ equation (1) can be written as
$$\ce{Ca_{$x_1$}S_{$x_2$}N_{$x_3$}C_{$x_4$} -> $x_1$ CaCO3}$$
and coefficient $x_1$ is therefore determined according to (4) as follows:
$$x_1 = 1 \cdot \frac{\pu{0.160 g} \cdot \pu{156 g mol-1}}{\pu{0.250 g} \cdot \pu{100 g mol-1}} = 1.0$$
Same applies to $x_2$ and $x_3$ for sulfur and nitrogen; in order to find $x_4$ (carbon content) equation (7) is used.


text:
How do you tell which molecule has a higher melting point?
In the instance of comparing $\ce{NaCl}$, and $\ce{Al2O3}$; which one has a higher melting point and why?
I have learned the following inter molecular forces, however the question the chapter is in does not explain about inter molecular forces, and is actually in the next chapter.

London Dispersion Forces
Dipole Dipole
Hydrogen Bonds
Ionic Crystal Lattice

I can imagine that London Dispersion Forces will prove that $\ce{Al2O3}$ have a higher melting point due to higher complexity; however, the chapter the question is in is about Properties of Chemical Compounds which talks about the trends on the periodic table ect.
Also, how does Ionic Crystal Lattice affect the melting point in this instance?

Neither NaCl nor Al2O3 consist of molecules, so the first three points are irrelevant.

Generally, for ionic compounds, we can calculate the lattice energy for it using a simple formula. The higher the lattice energy, the higher the melting point. This is because, for an ionic salt, melting means the breaking of ionic bonds. Therefore, this relationship holds.

@IvanNeretin I thought London dispersion affects Ionic and Covalent molecules?

There is no such thing as ionic molecules.

@IvanNeretin London dispersion forces, under the category of van der Waal forces: These are the weakest of the intermolecular forces and exist between all types of molecules, whether ionic or covalent—polar or nonpolar.

@IvanNeretin https://www.khanacademy.org/test-prep/mcat/chemical-processes/covalent-bonds/a/intramolecular-and-intermolecular-forces

https://chemistry.stackexchange.com/questions/47628/crystal-structure-vs-melting-point

Frankly a lot of these "comparisons" depend on knowing the answer then forming a rational argument to "justify" the assumed "common knowledge."
I just know that NaCl will form a molten salt fairly readily. Alumina is a refractory so it will melt at very high temperatures.
One thing that points to the difference is the solubility in water. NaCl readily dissolves, alumina does not dissolve to any appreciable extent. NaCl has ionic bonds, whereas alumina is more covalently bound. But you have to know the solubilities, nothing in just the compounds' formulas gives that information.
You also have to appreciate that both of these compounds will form 3D lattices not discrete molecules. But again this is a "fact" that we would all readily assume to be true, but the problem statement gives us no justification for such a belief. You just have to know that there are not discrete molecules in either compound.
Thus the explanation is really more of an educated guess based on the nature of chemistry rather than being absolutely logically rigorous.

I’m pretty sure the deduction OP should have based their reasoning on should come from the charges of the ions involved and thereby the strength of the crystal lattice.

@Jan yea I guess. I do not think they think we know by common knowledge melting points of compounds at this level, so I'm assuming it must be something else

There is no simple, general rule for judging the melting points of compounds
The trouble with judging melting points is that solids have a wide range of different chemical structures. Some are covalent solids made from infinite networks of covalent bonds (alumina, graphite) some are essentially ionic salts (sodium chloride, copper sulphate) some are made from neutral molecules (freebase cocaine, candle wax) and so on. It is the structure that really determines the melting point.
Within a given class (say ionic solids) there may be some generalisable rules (alkali metal halides might show some explicable trends based on ion size, for example). But given a formula and no knowledge about the structure there will not be any rules that are helpful.
Alumina is a solid best thought of as an infinite network of covalent bonds; common salt is an ionic solid consisting of sodium and chloride ions. Alumina is a refractory and strong solid because of this structure, with a very high melting point. Common salt has a high melting point but is much lower than alumina. Some ionic salts with large floppy ions are liquid at room temperature.
The forces holding the solid together determine the melting point but you need to know the structure before you you know what the key forces are. Covalent bonds are strong in covalent network solids. Ionic bonds are not as strong but are still far stronger than the forces that hold together discrete molecules in molecular solids like candle wax or menthol. Knowing the structure tells you a lot; knowing the formula doesn't (always) tell you what you need to know.


text:
What is occuring on the quantum level when a molecule rotates plane polarized light?
What is occuring on the quantum level when a molecule rotates plane polarized light?
Also, why do enantiomers then rotate light in opposite directions? I would think that the electromagnetic waves would interact with the transient electric and magnetic fields within the molecule itself as it propagates and therefore the light changes direction as it passes through the molecule. However with this idea there seems to be a problem with the fact that all the molecules wouldn't be oriented in the same directions and thus scatter the light in all directions rather than a net rotation in one direction of the light. I am looking for an explanation using electromagnetism if it can be explained at that level or if it cannot, the quantum mechanical answer would be fine as well.

Possible duplicate of Molecular chirality and optical rotation

This definitely isn't a duplicate.

I agree with @pentavalentcarbon. The post Mithoron linked is related to part of your question (why does the random orientation of a solution not cancel out the net rotation), but it doesn't address the physical mechanism by which optical rotation occurs.

I'm not sure if it counts as a duplicate if it's on a different SE, but see how do optically active compounds rotate plane polarized light.

@a-cyclohexane-molecule A version of that answer in terms a chemist could understand would make a decent answer here (especially if you can avoid any quantum mechanics or math).

As mentioned in the comments, the answer to your $2^{nd}$ question regarding why optical rotation isn't simply cancelled out in solution due to random orientations is given in ManishEarth's answer to Molecular chirality and optical rotation
For your first question, I will try to explain what actually causes rotation when light impinges on an optically active molecule. For chiral molecules, it simplifies the picture to think of each enantiomer as a screw of negative charge that is either right or left handed. Plane or linearly polarized light can always be decomposed into circularly polarized clockwise and counterclockwise components. When the light collides with the molecule, the clockwise component and counterclockwise components interact differently with the electric field of the molecule, leading one of these components to travel faster than the other, which alters the angle of the plane polarized light. A good visualization of this effect (for a whole solution rather than an individual molecule) can be found at http://cddemo.szialab.org/.
As an interesting aside, you might imagine that because we have a reasonable understanding of why optical rotation occurs, we should be able to predict, either heuristically or from a first principles calculation, the magnitude and direction of optical rotation for a molecule just based on its structure. This, thus far, has not been the case. Not only are there very few ways of heuristically guessing the magnitude/direction, but quantum mechanical calculations at high levels of theory have proven unable to even consistently obtain the correct direction of optical rotation. Given a particular chiral molecule, we still can't predict which direction, let alone magnitude, that a particular enantiomer will rotate light.
My main point in this little tangent is that even something as seemingly simple and in such widespread use as optical rotation is still not very well understood and is a difficult area of active research.


text:
How nickel plated copper strip can be welded? What is the optimum thickness of nickel layer?
I have a trouble with  (resistance) spot welding of nickel plated copper strip to 18650 battery. I tried to weld 0.1mm, 0.2mm and 0.3mm copper strips with 5 micrometers and 10 micrometers nickel plate thickness. Actually i have tried all of the combinations, that is i have tried to weld 0.1mm with 5 microns to 0.3mm with 10 micrometer or 0.2mm with 10 micrometers to 0.1 with 10 micrometers… Additionally, i adjusted power of spot welding machine, i tried with full power and half of the maximum power… Unfortunately all of the trials have failed.
According to you what is the main problem in my case? What is the optimum nickel plate thickness for spot welding application and how it is determined?

What do you call a failure ? The nickel is unlikely to have much affect.  I found a reference that said the electodes  may need to be cleaned ( smoothed) because copper from the work can transfer to the spot weld tool.


text:
What is the reaction mechanism of a selective catalytic reaction with urea?
What is the reaction mechanism of the following reaction:
$$\ce{4NO + 2(NH2)2CO + O2 -> 4N2 + 4H2O + 2CO2}$$
I expect that the catalyst is vanadium oxide and titanium oxide.
Where can I find a detailed description of the reaction mechanism?

Have you consulted any sources before posting here, like SciFinder, Web of Knowledge, Google Scholar? I'd assume that there is a special interest in these systems, but if wouldn't be totally surprised if extensive mechanistic studies are not yet done due to the complexity of the reaction.

https://doi.org/10.1016/S0920-5861(00)00299-6; https://doi.org/10.1016/j.cattod.2006.08.018.

After researching this topic, I've found the mechanism to be:


text:
What is nanoporous structure?
I would be very grateful if you help me to understand what is nanoporous structure. Does this term relate only to the nanoporous materials (as I understand, materials with regular porous structure) or to the such structures as nanotubes, fullerenes, layered silicates and so on? My doubt arises from the fact that the last do not seem for me to have regular structure.

https://en.wikipedia.org/wiki/Nanoporous_materials

Yes, but is nanoporous structure = nanoporous material? Than You.

I guess it's context-dependent and "structure" term can resemble 1) a repetitive motif as a whole, or 2) composition of a subunit (e.g. molecular geometry).  If 1), then I'd say nanoporous structure = nanoporous material, otherwise not.

Ok, thank you very much. I always hardly understand terms.

No prob, but please do keep in mind it's just my vision. I actually thought that "еру" stood for "epytaxy" before I realized it was typed in Russian layout :D

Yes, that was my mistake, sorry :D. Anyway, thank You for the opinion, it is important for me, I am new in chemistry and completely bewildered by some terms, You helped me.

I believe this question is both simple and difficult -- especially for those who are not native English speakers. I'm going to suggest a straightforward answer, but first let me explain the difficulty.
Unfortunately, terminology related to the nanoscale is often misused. But if we can agree on what the nanoscale is, the answer to the question becomes much simpler. So I'd like to respond to this question in two parts: 1) What is nanoporous? and 2) What is a nanoporous structure?
1) What is nanoporous?
This question relates to a more general question, "What is the nanoscale?" In my experience authoring articles in nanotechnology and serving as a referee for ACS Nano, I've struggled with this question.  It seems no accepted authority has defined the term. The closest I've found is this website which says clearly, "...the nanoscale, which is about 1 to 100 nanometers". I realize that good arguments can be made for extending this range to larger values… but surely it must stop short of 1000 nm = 1 micron. If someone knows of a more authoratitive body making a definition, I’d love to hear of it.
Unfortunately, I’ve found a significant number of scientific experts publishing statements suggesting that the nanoscale extends up to the 1000 nm range. This does not make sense to me: that would more properly be called the "microscale".  As a referee I've been surprised at the number of articles submitted for publication which claim nanoscale work when actually the objects of study are at the microscale. The motivation for this is obvious: it’s a lot more impressive to say you’re working at the “nano” than the "micro" scale these days.
So, my suggestion is that if we accept nanoscale means 1-100 nm, then nanoporous means the presence of pores in the 1-100 nm range.
2) What is a nanoporous structure?
Strictly speaking then, a nanoporous structure would be any structure having pores in the 1-100 nm range. The term itself implies nothing about the uniformity of the pores, and that is why we refer to sponges as porous despite very different sized pores. It is true that uniformity of pore size is of great interest to scientists for important reasons, but the term itself does not include that meaning. It is a case of denotation verses connotation. Though scientists have a reputation for precise terminology, they can be just as guilty as others in applying terms imprecisely – and for the same reasons: convenience and self-interest.


text:
Removing polyimide coating without damaging carbon nor fused silica optical fiber
While being manufactured, an optical fiber (fused silica, 125 microns diameter) has been coated with a carbon coating (~50 microns thick) and an external, ~60 microns thick, polyimide coating.
The problem is removing the external coating but not damaging the carbon layer or the glass core. Several procedures are recommended to remove polyimide coatings, being the more usual:

thermal (flame),
chemical (hot sulfuric acid -> within seconds),
mechanical (razor).

After a little research on flame temperatures and other chemical methods (which is a completely unknown field for me), I have found that sodium hydroxide would also attack polyimide while being much easier for me to get than hot sulfuric acid. The questions are: how long could it take to remove the coating along a few cm? Would it damage the carbon layer?
Is there any method I could try?


text:
IUPAC rules on omission of locants in carboxylic acid
I have been lately in debate with my teacher as to why did the IUPAC name the compound with the current official IUPAC name of 2-methylpropanoic acid. (CN: isobutyric acid)
My argument is that:

You can't attach a methyl in carbon 1 (because the bonds in carbon 1 will exceed 4)
You can't attach a methyl in carbon 3 (because it will become butanoic acid)
Placing the locant in a substituent with no other constitutional isomers will be useless when to be interpreted, will be a few more characters longer, and will be less pleasant for the eyes end ears (I mean, methylpropanoic acid a bit more pleasant to see and hear than 2-methylpropanoic acid, right?)

To verify my initial assumption, is the name of such compound 2-methylpropanoic acid? If it is, then we may proceed to the real question in the next paragraph.
So, where in the 2013 Blue Book or any updated IUPAC nomenclature guidelines with less complicated wordings in organic chemistry books (this has seniority over the former) can I see that you must not omit the locant in these cases (and why).
Additionally, how can I request to the Union to alter this notion to make it methylpropanoic acid because of the stated reasons? (It's a bit like ethan-1-ol and ethanol.)

There can always wound up someone not knowing it's redundant and ask  a question here why it wasn't used, I think there was a precedence :D

Where in the Blue Book?

Huh? I mean there are noobs who wouldn't get it's redundant, and ask opposite question for example on Chemistry.SE...

It means that they did not listen/get the idea to their octet rule lesson and the part where the the teacher explain why IUPAC prefers ethanol over ethan-1-ol. Like physics, chemistry also needs cumulative learning.

IUPAC has always been about making names as unambiguous as possible (and you do not need the blue book for proving this). While ethanol is unambiguous, 2-methylpropanoic acid is not. It may be stupid, but you could actually be asked to draw 3-methylpropanoic acid - even though you are right, this would be butanoic acid. Have you ever used ChemDraw? If you "Convert Name to Structure" and just put in methylpropanoic acid, it will tell you that the name is ambiguous, because you could mean 3-propanoic acid as well.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name is 2-methylpropanoic acid. The trivial name ‘isobutyric acid’ is no longer recommended (see Subsection P-65.1.1.2.4).

Concerning your objection

You can't attach a methyl in carbon 3 (because it will become butanoic acid)

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.
The given parent structure is propanoic acid. It includes two kinds of substitutable hydrogen. Therefore, the locant cannot be omitted in substituted propanoic acid, e.g. 2-chloropropanoic acid or 3-chloropropanoic acid.

This principle applies to any substituent. The special situation that replacing the chloro substituent in 3-chloropropanoic acid with a methyl substituent would actually lead to the name butanoic acid instead of 3-methylpropanoic acid does not change the preferred name for 2-methylpropanoic acid.

Nevertheless, there are a few exceptional cases where locants are omitted, but only when there is no ambiguity. For preferred IUPAC names, the nomenclature rules explicitly stipulate when locants are omitted. Most of these rules apply to the omission of the locant ‘1’. One case is the example given in the question:

(It's a bit like ethan-1-ol and ethanol.)

Homogeneous chains consisting of only two identical atoms such as ethane only have one kind of substitutable hydrogen. Therefore, the current IUPAC recommendations include a rule (in Subsection P-14.3.4.2) that the locant ‘1’ is omitted in monosubstituted ethane; i.e. ethanol (not ethan-1-ol) is the preferred IUPAC name. However, if any locants are essential for defining the structure, then all locants must be cited for the structure, e.g. 2-chloroethan-1-ol (not 2-chloroethanol; see Subsection P-14.3.3).

Coming back to carboxylic acids, acetic acid also has only one kind of substitutable hydrogen, i.e. all substitutable hydrogen atoms have the same locant. Therefore, explicitly according to Subsection P-14.3.4.6, this locant is omitted, e.g. chloroacetic acid (not 2-chloroacetic acid).


text:
Should the mean energy be written in angular brackets or with an overline?
How would you write mean energy? I came up with two different ways:

$\displaystyle\langle G_{\textrm{f}}^\circ(j) \rangle$ with the MathJax code \langle G_{\textrm{f}}^\circ(j) \rangle
$\displaystyle\overline{G}_{\textrm{f}}^\circ(j)$ with the MathJax code \overline{G}_{\textrm{f}}^\circ(j)
Or maybe another way?

I am looking for the proper use of this notation in the scientific literature.

When in doubt, simply indicate what you mean in text.

To the question itself: I believe that the bar notation is most common for point values, while the bracket notation is mainly for functions. In any case I would write $\displaystyle \overline{G_{\textrm{f}}^\circ(j)}$.

Depends on the convention used. Different people use different conventions but both indicate a mean.

I would use \overline for representing arithmetic mean, and \langle \rangle for  expected value (notice that their difference is commonly overlooked in chemistry literature). However, the proper use only requires  stating the meaning in text.

Both the IUPAC 2008 Green Book (section 4.2) as well as the ACS Style Guide (specifically Chapter 11) agree that both methods of representing the mean value are acceptable. There might be some style guides / international standards that state otherwise, but for the average chemist these two are generally sufficient.
If ever in doubt, just state the meaning in text.

The international standard ISO 80000-2 Quantities and units – Part 2: Mathematical signs and symbols to be used in the natural sciences and technology agrees.

Good to know, too. I was lazy to check, hence the disclaimer.


text:
What determines the maximum size of salt crystals as grown in flux growth?
Like many other children, I grew single crystal salt crystals in a salt solution. This is a common and fun experiment for children.
However, with both this experiment, and with flux growth methods for single crystals in general, is there a limit on the size of the grown crystal?
That is, if I was to keep dumping in more salt so as to replenish the solute, would the crystal grow forever? And if not, what limits the size of these crystals?

In theory, there are no limitations. In reality as the single crystal gets bigger, it will a) accumulate more and more defects and b) new nucleation centers will occur producing polycrystalline assembly of several phases. This happens statistically as the volume of mother liquor needs to get larger, therefore there will be more concentration fluctuations, not to mention impurities and temperature gradient from the contact with the external system. Crystal growth in gas phase is somewhat different, but similar principles are applied there too.

What are the sources of the defects in this system, just impurities? additionally, why do the new nucleation centers necessitate a separate phase?

Yes, I understand crystal defects. But how does that limit the size of the crystal? I can guess at how that would cause the formation of a separate phase.

Crystals can get quite large, as seen at http://ngm.nationalgeographic.com/2008/11/crystal-giants/shea-text . It just takes a steady source of material and patience... a few thousand years, in this case.


text:
Questions about air-quality testing in high school
The Pinelands Regional High School in Ocean County New Jersey was closed for four days last week for air quality concerns after rooftop construction. Yesterday was the students‘ first day back. There have been conflicting reports of health issues, but it generally seems that the first day back is going OK.
There were two rounds of air-quality testing conducted last week. The first found unacceptable levels of multiple chemical compounds, the second, conducted two days later (after being cleaned), found no dangerous levels of any chemical.
Full results document of each test

Test 1: October 3
Test 2: October 5.

All of the unacceptable results found in round one (Indicated by red flags):
Sample: Room 164

Chemical | Actual ug/m3 | Times greater | NIOSH Res uh/m3

Ethelbenzene | 2.7 | 2.45 | 1.1
1,2,4-Trimethylbenzene | 15 | 2.05 | 7.3

Sample: Upper level - Media

Chemical | Actual ug/m3 | Times greater | NIOSH Res uh/m3

Benzene | 8.9 | 24.72 | 0.360
Ethelbenzene | 37 | 33.64 | 1.1
Xylene (p,m) | 110 | 1.1 | 100
1,2,4-Trimethylbenzene | 180 | 24.86 | 7.3
Naphthalene | 11 | 132.52 | 0.083

The clean results of round two

Chemical | Actual ug/m3 | NIOSH Res ppm

Benzene | 8.9 | 0.360
Ethelbenzene | 37 | 1.1
Xylene (p,m) | 110 | 100
1,2,4-Trimethylbenzene | 180| 7.3
Naphthalene | 11 | 0.083

This report concludes: "TTI has evaluated the results and can confirm that none of the target compounds exceeded limits for either exposure or notification at both the state and federal levels."
Confusion
There are a number of things I find confusing about this.

First, the  first round of tests analyze specific areas in the school, but the second does not identify any area.
Both rounds seem to have measured almost exactly the same levels.
The recommended NIOSH residential levels are measured in ug/m3 in round one testing, but ppm in round two. In round one, for example, the recommended maximum for benzene is 0.360, which according to this tool, translates to 0.00036 ppm. However, on the second test, has a recommended maximum of 0.1 ppm. Also, the same calculator translates benzene’s ug/m3 of 8.9 to 0.0089 ppm, but in the Test two results, it is translated to 0.0028.
Finally, parents are simply unfamiliar with these chemicals and what the levels imply, and how they affect teenagers. It is hard to fight what you can’t see or smell.

Help understanding and interpreting these results, and especially in giving parents some reassurance if it is warranted, would be appreciated. Thanks.

Your links seem to suggest that the company that did these tests is open to questions/concerns. Have you tried contacting them?

First off the acceptance levels the first time are way lower then the second time, you should ask why there is a large difference. (I have no clue, it's quite important since the second time those limits are NOT met!)
I found NO acceptance level for the chemicals (long term exposure.)
THE most accepted guidelines are usually set by the world health organisation and even they do not dare to set limits since it is difficult see http://www.euro.who.int/en/health-topics/environment-and-health/Housing-and-health/publications/2010/who-guidelines-for-indoor-air-quality-selected-pollutants
For common benzene they wrote a small book.
But to put it in perspective, I think that you inhale way more of those chemicals during filling your car. Then staying in the classroom for a week ;) [Citation needed!]
Edit: the concentrations are actually all more elevated then the averages mentioned by literature. Can we please have a copy of the raport? Next time it might be better to measure air quality according to the WHO guidelines since we simply do not know what we are looking at.
"First, the first round of tests analyze specific areas in the school, but the second does not identify any area.
Both rounds seem to have measured almost exactly the same levels."
I agree, this is weird. You should ask where and how this is sampled. But there are regulations for the sampling, let's assume they are professionals and based on previous results this is a worst case measurement.
"The recommended NIOSH residential levels are measured in ug/m3 in round one testing, but ppm in round two. In round one, for example, the recommended maximum for benzene is 0.360, which according to this tool, translates to 0.00036 ppm. However, on the second test, has a recommended maximum of 0.1 ppm. Also, the same calculator translates benzene’s if/m3 of 8.9 to 0.0089 ppm, but in the Test two results, it is translated to 0.0028."
PPM is a horrible unit without actual dimensions just like percentages. Are we talking about volumes, weights , molars or a mix? Anyway, just use the ug/m3 since that's what really Is measured. (I you like you can see the right conversions inside the WHO document http://www.euro.who.int/en/health-topics/environment-and-health/Housing-and-health/publications/2010/who-guidelines-for-indoor-air-quality-selected-pollutants but that's only if you like math ;)
"Finally, parents are simply unfamiliar with these chemicals and what the levels imply, and how they affect teenagers. It is hard to fight with you can’t see or smell."
I totally agree! But don't underestimate the nose ;)
I GUESSE, looking at the numbers there is no extreme dangerous exposure. The second report concludes the the same. The world health organisation does not exactly know what levels are really exceptional dangerous, therefore it's by "law save" as long as there are no observed health problems. But just ask the professionals ;)
This is where my field ends, it's up to you to tell the right people the right information and ask the right questions, (you are probably way better at that then I am ;) good luck!!

Still reading your answer. There is a link to the full results document for both tests right after the second paragraph. I edited the question to make it clearer.

I read the document, it contains some errors. There is no error margin mentioned. To be honest it looks like my work when I was in school. Anyway, I did compare the results found with the averages mentioned by the world health organisation, it's seems only slightly elevated from other indoor situations. Keep it well ventilated and there should be no significant difference with other comparable indoor environments. Thanks for the edits. The question good I was just a bit lazy to be honest ;)

Thank you for the information and insights. Can you please give me a brief description of your qualifications and background?

Actually, if you would be willing to contact me at citizensmediatv@gmail.com I’d appreciate verifying who you are and your qualifications. I’m a journalist and I want to quote you. It would be nice to be able to say more about you than just “he has 11 reputation points on chemistry.stackexchange.


text:
Determining the rate of stratospheric ozone destruction
I am a sophomore. A friend and I ran across a kinetic chemistry exercise. We solved it in a slightly different way and we can't come to an agreement on who is right. We would appreciate some help:

Destruction of stratospheric ozone as determined by using the steady-state approximation.
The balance of ozone in the stratosphere is of critical concern
  because this molecule absorbs ultraviolet light that would be harmful
  to life at Earth's surface. The principal production mechanism for
  ozone is recombination of $\ce O$ atoms with $\ce{O_2}$. The principal
  destruction mechanism is that given below. There is increasing concern
  over alternative destruction mechanisms involving molecules introduced
  into the stratosphere by human activity.
Determine the destruction rate of ozone in the following mechanism.
  $$\begin{align}\ce{O3 + M &<=>[k_1][k_{-1}] O2 + O  + M}\\
\\
\ce{O3 + O &->[k_2] 2 O2}\end{align}$$

Our answers:
We agree on the beginning, so we have :
$$\begin{align}
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} &= k_1[\ce{O_3}][\ce{M}] - k_{-1}[\ce{O_2}][\ce O][\ce M] + k_2[\ce{O_3}][\ce O]\\
-\frac{\mathrm d[\ce{M}]}{\mathrm dt} &= k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M]
\\
-\frac{\mathrm d[\ce O]}{\mathrm dt} &= -k_1[\ce{O_3}][\ce M] + k_{-1}[\ce{O_2}][\ce O][\ce M] +k_2[\ce{O_3}][\ce O]
\end{align}$$
Now here comes the disagreement, we need to use the steady-state approximation to say that one of the above speed is null:
My friend argues that $-\frac{\mathrm d[\ce O]}{\mathrm dt}$ should be approximated to null because it is a transient compound.
Even if in most other exercise, it would be correct, I disagree with him because in the proposed mechanism, $\ce{O_3}$ and $\ce O$ play a symmetrical part. So if we approximate $-\frac{\mathrm d[\ce O]}{\mathrm dt}$ to $0$ we should also approximate $-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt}$ to $0$. Since we want to calculate this specific value, we don't want to approximate it.
In my demonstration, I approximate $-\frac{\mathrm d[\ce{M}]}{\mathrm dt}$ to $0$, arguing that the compound $\ce M$ is on both side of the equilibrium equation and therefore it's concentration will remain constant.
We both manage to find a result:
My friend’s calculation:
$$
- \frac{\mathrm d[\ce O]}{\mathrm dt} = -k_1[\ce{O_3}][\ce M] + k_{-1}[\ce{O_2}][\ce O][\ce M] + k_2[\ce{O_3}][\ce O] = 0
$$
$$
k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] = k_2[\ce{O_3}][\ce O]
$$
We substitute the left member in the speed of ozone:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = 2\times k_2[\ce{O_3}][\ce O]
$$
Because $[\ce O]$ is transient, it is difficult to mesure experimentally, so we expresse it in function of $[\ce{O_3}], [\ce{O_2}]$ and $[\ce M]$ :
$$
k_1[\ce{O_3}][\ce M] = [\ce O] \times ( k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M] )
$$
$$
[\ce O] = \frac{k_1[\ce{O_3}][\ce M]}{k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M]}
$$
And finally we have my friend’s result:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = \frac{2k_1k_2[\ce{O_3}]^2[\ce M]}{k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M]}
$$
Now my calculations:
$$
-\frac{\mathrm d[\ce M]}{\mathrm dt} = k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] = 0
$$
Which allow us to simplify the destruction rate of ozone:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = k_2[\ce{O_3}][\ce O]
$$
We already disagree by a factor 2 on the rate. But if we replace $[\ce O]$ like before, we obtain a much simpler result:
$$\begin{align}
k_1[\ce{O_3}][\ce M] &= k_{-1}[\ce{O_2}][\ce{O}][\ce M]
\\
[\ce{O}] &= \frac{k_1[\ce{O_3}][\ce M]}{k_{-1}[\ce{O_2}][\ce M]}
\\
[\ce O] &= \frac{k_1[\ce{O_3}]}{k_{-1}[\ce{O_2}]}
\end{align}$$
Finally I have:
$$
-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = \frac{k_1k_2[\ce{O_3}]^2}{k_{-1}[\ce{O_2}]}
$$
Which one of us is wrong and why?

There is no reason to set $\frac{d\ce{[O3]}}{dt}$ to zero. If that were true, the reaction would have zero rate. The steady state approximation allows you to say that the intermediate concentration is low and therefore the change in the concentration of intermediate is very close to zero. There's no reason why it can't be produced very quickly and consumed equally quickly. Therefore, your claim is incorrect.

Thank you for your comment. What I don't understand is how $\frac{d[O]}{dt}$ can be considered null without considering $\frac{d[O_3]}{dt}$ to be null at the same time. Unless $k_1 >> k_{-1}$ I don't see how it's possible... The wording doesn't specify values or order of magnitude on $k_1$, $k_{-1}$ or $k_2$.

Ignore the reverse reaction for a minute. Consider buckets one on top of the other each with a hole in the bottom. The top bucket is full of water. The water drops into bucket 2 and immediately drops out. The rate of water leaving the top bucket is significant. The rate of water accumulating in the second lower bucket is effectively zero.

Or perhaps your confusion is that the $\ce{O}$ is consumed by the forward and reverse reactions as quickly as it is produced by the decomposition if ozone?

Well yes, for me $O_3$ and $O$ play a similar role in the described mecanism. My intuition tells me that if we make the graph of $[O_3] = f(t)$ and $[O] = g(t)$, they should look the same. Put in other words $\frac{[O]}{[O_3]}$ should remain constant over time.

Also, when I look back at my result, I find strange that $\frac{d[O_3]}{dt}$ doesn't depend of $[M]$, so my reasoning is probably flawed somewhere. Is my expression of $-\frac{d[M]}{dt}$ correct ?

I don't understand why you want to make the ozone concentration constant. That would imply no reaction. $\ce{O3}$ and $\ce{O}$ don't have the same role at all. One is a reactant and the other is an intermediate. Steady state approximation only applies to one of them. The change in concentration of $\ce{M}$ seems fine, but your intuition about the rate of change of ozone concentration is wrong.

I think you misunderstood what M is.  M is the concentration of all molecules in the atmosphere, including O2 and N2.  It is so large that it doesn't change significantly with time.

@Amon Zhe is correct that you shouldn't treat \ce{O3} and \ce{O} as being the same. The steady state approximation applies to intermediates. This also suggests that you shouldn't use it for \ce{M} because M is not an intermediate, but rather a catalyst. Even if you had 2 intermediates, it shouldn't be surprising that using the steady approximation on a different intermediate results in a different answer. Remember, it is an approximation and different approximations will lead to different results.

Why $\ce{O3}$ isn't an intermediate too ? It is produced by the reaction $\ce{O2 + O + M -> O3}$ and the wording says " The principal production mechanism for ozone is recombination of $\ce{O}$ atoms with $\ce{O_2}$"

@Tyberius Two different approximation may lead to two differents results, but in that case at least one of the approximation won't represent the reality. I see that the consensus is to approximate $\frac{\mathrm d[\ce O]}{\mathrm dt} = 0$ but I still have trouble understanding why. $\ce M$ is a catalyst, but why can't I say that $\frac{\mathrm d[\ce M]}{\mathrm dt} = 0$ ? $\ce{[M]}$ seems to remain constant...

@Amon as Chester Miller mentions, you likely can say that $\frac{d\ce{[M]}}{dt}=0$, as this is meant to represent a generic other molecule in the atmosphere and this concentration shouldn't change significantly. But I would argue it isn't the steady state approximation as it is commonly understood, since that refers specifically to intermediates(for a real reaction scheme, there would likely be some pathway by which the catalyst broke down). Different approximations can both be correct so long as they are applied in the limit that those approximations are valid.

Dioxygen is split in the stratosphere by UV light into oxygen atoms. This is the main source of free oxygen atoms. Their production rate depends on the intensity of the UV light.
In steady-state we can assume that the concentration of oxygen atoms is constant.
$$\frac{d[\ce{O}]}{dt} = 0  \tag{1}$$
For the above destruction mechanism we can formulate the rate laws
$$\begin{align}
\frac{d[\ce{O3}]}{dt} &= -k_1 [\ce{O3}] [\ce{M}] + k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}] \tag{2}\\
\\
\frac{d[\ce{O}]}{dt} &= k_1 [\ce{O3}] [\ce{M}] - k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}]  \tag{3}
\end{align}$$
Adding equation $(2)$ and $(3)$ which by $(1)$ is $0$ we get
$$\frac{d[\ce{O3}]}{dt} = -2~k_2 [\ce{O}] [\ce{O3}] \tag{4}$$
With equations $(1)$ and $(3)$ we can express the concentration of oxygen atoms as
$$[\ce{O}] = \frac{k_1 [\ce{O3}] [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]} \tag{5}$$
Substituting $(5)$ into equation $(4)$ we get
$$\frac{d[\ce{O3}]}{dt} = \frac{-2~k_1~k_2 [\ce{O3}]^2 [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]}  \tag{6}$$
So your friend seems to be right.

Hi, thank you for you answer, however it doesn't really answer my question. What allow you to make the approximation $\frac{\mathrm d[\ce{O}]}{\mathrm dt} = 0$ ? Just saying "steady-state" feels like you're waving a magic wand and poof the problem is gone... That's not really satisfactory.

If we can assume that the primary source of oxygen atoms is the flow of UV light, the we have a constant rate of production and, according to the above reactions a variable rate of destruction. When these are equal we effectively have a transformation ov UV to heat (hot Ms) and a constant concentration of oxygen atoms.


text:
What causes the yellow colour in the Briggs-Rauscher oscillating reaction?
The color changes are due to different concentrations of supposedly $\ce{I2}$ (which causes the yellow/amber color), $\ce{I-}$ (colorless) and $\ce{I3-}$ (dark blue).
Since iodine ($\ce{I2}$) is barely soluble in water, I do not understand why in this case it would be.
If one was to put just iodine in water and mix it thoroughly, the color change of said water would not be as visible to the extent of the Briggs-Rauscher reaction.

$\ce{I3-}$ is by no means dark blue. In fact, it is colored the same as $\rm I_2$.

I'm not quite sure I understand the question, but keep in mind there is an equilibrium between $\ce{I-}$, $\ce{I2}$ and $\ce{[I3]-}$ all the time, which means not all I is present at a given form ($\ce{I2}$) at a given moment, so I see no contradiction in observed color intensity and solubility of iodine in water. Also, it's not $\ce{[I3]−}$ that is dark blue, it's triiodide-starch complex.

Pardon, when interacting with starch indeed the dark blue color will be visible.

My question is that how it would be possible for the mixture to have a yellow color. Many sources claim this is due to the I2(aq). My point is that I don't understand how I2 can be dissolved.

Starch test is normally done with neutral iodine and neutral starch; neither ionizes in water. Pray tell us how $\ce{I3-}$ comes about in these circumstances, what is the cation that balances it, and where does the latter come from?

The color is in fact caused by the elemental iodine. At the 'yellow' stage of  reaction the concentration of iodide is low, so the triodide anion (yellow at low concentration and red-brown at high) does not form, hence there is no formation of the blue triodide-amylose complex.
The OP wonders how the iodine concetration can be high enough to give the solution its yellow coloration, given the low solubility of elemental iodine in water?
First let me make a remark that crystalline iodine is very sluggish to dissolve in water to reach its maximal concentration. Even when making potassium iodide - iodine solutions - it takes a lot of stirring, anyone who worked in an analytical lab knows that.
But this is not the answer to the question. You have to consider that the iodine in solution does not come from a solid phase going to the liquid phase - it is generated in a homogenous chemical reaction. Therefore, it can reach concentrations far beyond the maximal solubility, forming a metastable supersaturated solution. This stage of the oscillation does not last long enough to allow nucleation and precipitation of the elemental iodine. The subsequent increase of iodide concentration causes formation of triodide, which along with starch gives the blue color and so on.

The colour changes are due to the indicator. In the case of the Briggs-Rauscher reaction, starch is included which causes a blue colour when complexed with iodine and iodide (high concentrations of $\text{I}_3^-$. When the concentration of $\text{I}_2$ is high, the reaction mixture is yellow due to the colour of iodine.


text:
How do I calculate lattice points per unit volume?
My professor was explaining simple cubic, body centered cubic, and face centered cubic lattices and I didn't quite understand the calculations below:

Simple Cubic: $8 \cdot 1/8 = 1$ lattice point per unit volume
Body Centered Cubic: $(8 \cdot 1/8) + 1 = 2$ lattice points per unit volume
Face Centered Cubic: $(8 \cdot 1/8) + (6 \cdot 1/2) = 4$ lattice points per unit volume

Can someone explain the reasoning behind these values in the equations?

You have to imagine a cube with spheres (the atoms) on the corresponding locations. Now you have to look at how much of each of these spheres is inside the cube (please see Patrick Moloneys answer for very descriptive images).
For Simple Cubic: You have spheres on all 8 corners. Because the spheres are on the corners of the cube, only $1/8$ of each sphere is inside the cube, while $7/8$ of the sphere are on the outside. Now, you have 8 spheres on the corners, giving you 8 times $1/8$ of a sphere. So there is $8 \cdot 1/8 = 1$ sphere inside the cube.
For Body Centered Cubic you have a sphere completely inside the cube additionally to the 8 on the corners ($8 \cdot 1/8$). Now we add the $1$ on the inside, giving you a total of ($8 \cdot 1/8) + 1 = 2$ spheres inside the cube.
Face Centered Cubic has the same 8 spheres on the corners ($8 \cdot 1/8$) and an additional sphere on all the faces of the cube. For each of these spheres, only half is inside the cube, while the other half is not. There are 6 faces, giving you $6 \cdot 1/2$ spheres for the faces. Add these to the corner spheres and you get $(8 \cdot 1/8) + (6 \cdot 1/2) = 4$ spheres inside the cube.

I felt as though this answer was easier to understand, so I marked it as correct. Thanks!!

Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell.
An atom on a face is shared by two unit cells $\implies$ half an atom belongs to each unit cell.

An atom on the edge is shared by four unit cells

and an atom on the corner is shared by eight unit cells

What does this have to do with the equations?
Well lets say we have an sodium atom that is crystallized in a simple cubic unit cell, then there would be an sodium atom on each of the eight corners of the cell. However, only $\frac{1}{8}$ of these atoms can be assigned to a given cell. Hence the simple cubic structure is
$$8 \, \text{(corners)} \times \frac{1}{8}=1 \, \text{atom}$$
if sodium formed a body centered cubic structure, there would be two atoms per unit cell (as seen in the top diagram). It isn't shared with an other cells. Therefore the body-centered cubic structure is given by
$$8 \, \text{(corners)}\, \times \frac{1}{8} + 1 \, \text{(body)}= 2 \, \text{atoms}$$
now finally if sodium was a face-centered cubic structure the six atoms on the faces of the unit cell would contribute three net sodium atoms, for a total of four atoms per unit cell. Hence the face-centered cubic structure is
$$8 \, \text{(corners)} \times \frac{1}{8} + \Big{(}6 \, \text{(faces)} \times \frac{1}{2}\Big{)}= 4 \, \text{atoms} $$
Hope this helps


text:
How to explain the selectivity between methanol and methanethiol in an SN1 reaction with an halogenated hydrocarbon?

Consider the SN1 reaction shown below:

Would different products be formed if $\ce{CH3SH}$ was an additional reagent?

Superimpose both energy diagrams on top of each other.

For the first question, yes, two products would be formed. I'm confused about the second question though.  I know The rate determining step would remain unchanged, as the rate depends on the concentration of the substrate not the nucleophile (SN1 is unimolecular). However, methanethiol is a better nucleophile than methanol. Would the activation energy for the second transition state be less for methanethiol?

Obviously saying "the rate is same but the activation barrier is different" is a contradiction, as both are directly related via the arrhenius equation. The rate is predominantly determined by the rate determining step. If there are several rate determining steps (aka several steps with similar activation energies), the overall rate will be a product of all steps.
So what does that mean for your question?
You stated that methanethiol is a better nucleophile than methanol. This would suggest that the activation barrier in the last step is slightly lower in the case of methanethiol. However, the rate determining first step is unaffected. Hence, the overall rate will be only slighter smaller for methanol.
The ratio of both products will be determined by the last step.


text:
Is there a way to determine the orientation of a benzene-ring ethyl group in synthesis reactions?
On my study guide for synthesis reactions (specifically for the synthesis of disubstituted benzene derivatives), two of the problems look extremely similar.
The question on the guide asks us to synthesize each of these by using benzene as a starting material.

They appear to be the same molecule, just with a different orientation of the ethyl group. And in the solutions section of the guide, both have the same exact synthesis reaction listed.

Is there actually a way to determine the orientation of the substituent that isn't being explained by my study guide? Perhaps something done during the Friedel-Crafts acylation step? Or is this simply an effort to try and trip me up by asking the same question twice?

There is nothing to determine. They are the same molecule.

Unless the temperature is really low, you will not be able to distinguish the two conformation isomers (conformers).
Rotation about single C-C bonds is rarely hindered. It can be, by using bulky groups or a tether, but neither is the case here. I expect the benzyl*-methyl bond and the phenyl*-ethyl bond to rotate freely at reasonable temperatures (say, above 250K).

Arbitrary very high temperature you chose. Why $\pu{250K}$ and not $\pu{100K}$?

I have no data to back up the temperature claim. I was thinking of the typical conditions of the synthesis given in the question.