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How do the differences in the number of protons result in such great differences in elemental properties?
I understand(I think) the mass and density aspect, i.e. the more protons you have, the more the element weighs, also the denser the atom is. What about everything else(color, for example)? Elements are not just masses and densities.

Unfortunately, I have to disagree with trb456's answer.  It is both the protons and the electrons that make up chemistry. (Neutrons are considered massive "spectators" though they are what keep protons (all + charged) together)
Bonds are formed due to the attraction of electrons and protons. (cf. Slater, Bader, et al.) (The precise shape and nature of the electron density is additionally determined by constraints such as Pauli Exclusion and electron-electron repulsion.)
So if you think about it, there is a very complex interplay upon mixing electrons up with protons. More positive charge in the nucleus tends to keep electrons drawn tight, but in some ways that is counter-balanced by electron-electron repulsion--and the fact that electrons are attacted to other nuclei as well.
So although we ascribe color ("electronic transitions") and reactivity to electrons, this is only part of the story. The protons are also there, and they provide some governance (via Coulombic interaction) over what electrons are permitted to do.

I disagree with your disagreement, but on a friendly basis! The reason is that proton count defines the element, and is thus a constant. It is the fact that electron count can vary for a fixed proton count that make the interesting chemistry. Now of course, the size of an atom does matter for many reactions, such as the affinity to remove or donate an electron. But even in this case, it is still mostly about the electrons in terms of the reaction itself.

One definition of chemistry is the transformation of matter from one form into another. Presumably this happens through breaking old bonds and forming new bonds.  Since bonds themselves come about due to proton-electron attraction, the protons must play an equally important role.  Electrons-Only Chemistry is a remnant of Lewis theory (dot structures) which predates the quantum mechanical description of the atom.  (Friends of course!  This is dialogue. I agree there is more variety in electron density, but protons are also important.)

It's not the protons, it's the electrons.  Electrons are the main determinant of chemical bonding, and it is the bonding of differing atoms that creates the huge variety of chemical properties.
For example, the color of molecules often is driven by the transition elements, starting at atomic number 21.  These metals have the ability to obtain many different oxidation states; i.e. ways to share elections with other atoms that create a certain implied positive charge, and those differing positive charges produce different colors because of the different wavelengths of light that the molecule can absorb.
A really nice video on this comes from the old Chem Study series.  The specific topic is the element vanadium, but they also discuss transition elements generally, complete with a big period table of actual substances.  Here is a link: Chem Study Vanadium


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Finding Pv in Pv=nRT for molecular hydrogen
I'm attempting to write a small program that will calculate the values for the Ideal Gas Law, specifically for molecular Hydrogen (H2) in space.  Eventually, this will grow to be a sort of "simulator", but right now I seem to be having problems with remembering my college chemistry and physics.
Here is a picture of the form that I created:

So in my program (written in C#), I have the following as initial values for those variables:
        P = ?
        v = ?
        n = 2.0158 (weight in grams of H2)
        T = 2.725 (average temp in space)
        R = 1.67E-27 (1.67 * 10^-27)

My problem is that, how does one go about trying to solve for P or v?  Do you assign an arbitrary number for what you suppose the volume of the gas would be in space, and then calculate P based on that?  Or is it the other way around?
Extra, yet entirely unneeded info:
This was brought about because my 7 year old daughter asked me how stars were made.  I gave her the explanation that anyone who's got even a remote interest in it gives to their kids, but thought it would be pretty cool to make a little program that would give her a visual.  Here was my explanation:

"we think that Hydrogen stuck together to make even larger clouds of
  Hydrogen, which increased the total mass of the gas cloud, increasing
  the gravitational pull on other atoms/molecules of nearby gas, etc.
  This eventually got to the point that the gases started to heat up as
  they were squished together by their own mass, but that increase in
  temperature created an outward push (pressure), which got into a
  tug-of-war with gravity acting on the mass of the gas itself.  And
  later, we got a star.  That's how most of us non-scientists think it
  happens, anyway!"

That's a pretty cool genesis for the question, but I don't think gaseous plasmas inside a star strictly follow the ideal gas law...

Yea, I realize that there are things that we do not know, or things that cannot be modeled by something like this, but the goal was just for a small simulation of gasses gathering.

You can solve for either $P$ or $V$.  Choose one, say $P$, and then divide that into the right hand side $nRT$ to get $V$; i.e. $V=\frac{nRT}{P}$.
More seriously, you need to check your units!  You seem to be using Kelvins for $T$, which is right, and $n$ is just a count in moles.  But I do not recognize your value of $R$.  The SI value for $R$ is $8.3144621(75)$, in units of joules per Kelvin-mole.  Joules, in turn, are Pascal-meters$^3$. Personally, I think most people find Pascals too small a unit for pressure.  Far more practical would be Torr-liters, and then $R=62.36367(11)$.  You can examine lots of different unit combinations for $R$ at the ideal gas constant page on Wiki: Ideal Gas Constant

You need one of p or V to solve for the other. If you have a spherical body, then $V=\frac{4}{3}\pi r^3$ and plug in what you want the radius of your star to be.
On a related note, the ideal gas law can't be applied to stars because IGL assumes the particles are infinitely spread such that gravity between particles is negligible - which is obviously not true for a star!

My apologies for not responding earlier, I was out of town and the mobile app for SE does nothing but crash ;-)
Since the IGL can't be applied to stars, would you be able to recommend another way that I could go about it?  I understand that we don't know everything there is to know about star formation, but perhaps there's a way to compare the temperature of the star to some sort of constant at which fusion is started, thus saying "Ok, here's where it starts"?  Darn, and I thought I had a good idea (for once)!

I'm not sure what you're asking. What sort of "constant" are you looking for? (and you might be interested to know that stars are not uniform in temperature - there can be a very wide difference between the corona and the core).

Good point!  I'm not exactly sure what I was shooting for...the goal was to make a sort of simulation for gas clouds forming into stars, but it seems like that is going to be next to impossible to do without another 8 years of physics and chemistry...  I thought that perhaps the IGL would have been a great starting point, because at least from that I would have been able to calculate temperature and pressure, and then using the mass of the cloud, calculate how strongly it would affect nearby clouds (pulling on each other), etc.  Thanks for the help anyway, though ;-(

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What is a "Strong" Acid?
If a reaction requires a "strong" acid - what does this mean? Does it mean any acid of a specific concentration or pH? Are only certain acids classed as "strong"? I have read on Wikipedia that it needs to ionise completely in an aqueous solution - in this case, is a strong acid (for example, HCl) still classed as a strong acid if it is in a 0.00001M solution?

The strength or weakness of an acid is independent of its concentration. A strong acid dissociates completely in solution, so $\ce{HCl}$ at any molarity would be considered strong, and $\ce{HCOOH}$ (for example) at any molarity would be weak. It is more enlightening to consider $K_a$ values, which are essentially the equilibrium constants for the ionization of an acid $\ce{HA}$.
$$K_a=\frac{[\ce{H}^+][\ce{A}^-]}{[\ce{HA}]}$$
This value will either be very small (<<1) or very large (>>1000). Acids with very small $K_a$ are considered weak, and acids with very large $K_a$ are considered strong (the $K_a$ of strong acids is usually not stated and assumed to be infinite).
Addendum: The $K_a$ value is concentration-independent because you are only interested in how well the acid dissociates from its molecular form. If you had $0.0000001\ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCl}$, practically every molecule of $\ce{H-Cl}$ would dissociate into $\ce{H^+}$ and $\ce{Cl^-}$ no matter how many there are. But if you had $12 \ \text{mol}\cdot\text{dm}^{-3}\ \ce{HCOOH}$ (methanoic acid), only a small proportion of the $\ce{HCOOH}$ would dissociate into  $\ce{H^+}$ and $\ce{HCOO}^-$, irrespective of how many you had to start.

Would it not be simpler to just say that a high concentration of hydronium causes acidity? And, would it matter which definition of acid we are using (Arrhenius and Bronsted-Lowry)?
(I am still learning, so this is a legit question that I have)

Nitric acid Ka is about 24, but it is considered a strong acid.  Instead of "assumed to be infinite", a strong acid is more like Ka > 1

Also, having a $K_\mathrm{a} > 1$ is not equal to ‘fully dissociated in solution’. It is ‘fully dissociated in aquaeous solution.’ Tiny, but important difference.

A Ka>1 does not mean an acid is fully dissociated, even in water. If you start with 1 M of nitric acid, the equilibrium concentrations will become (1-a) for nitric acid, and a for the nitrate-ion and hydronium ion (with a the dissociation between 0-1). Solving Ka (value=24, given in the comments above) for a results in a value for a of about 0.96, meaning 96% dissociation for nitric acid, not 100%! Ka>1 just means the equilibrium is shifted towards the products.

What makes an acid strong is also dependent on the context. E.g. lactic acid is considered strong in body fluids because lactate is completely ionized in body pH 7.4.

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Create my own solar cells?
How would I grow my own and build functional solar cells? I've read and researched that they're mostly a crystal structure of something like silicon. I would assume I'd be able to grow my own crystals and sandwich this in a process of some sort with a p-type and n-type, etc.

Do you want to actually get usable power from them, or would you like to just explore the science? If it's the latter, then you may find this tutorial on making dye-sensitized solar cells interesting and relatively accessible (well, at least more so than silicon solar cells!).

This question should be transferred to physics.se

@NicolauSakerNeto thank you. I'm exploring it as a science and not so much for practical functionality. I'm sure there is a micro-manufacturing-like process that would take special tools in order to make it efficient. As a DIY'er, I'm trying to learn how to do it on my own, though, and would like to make silicon ones. Thanks again.

@Georg You've been warned about your comments before.  I have censored this one, as you may have a point.  Please try to be constructive in your comments at all times

@jonsca I thought about placing this question in physics, but since it is manufacturing and not the logic of solar cells, I thought it best in chemistry, since I would think most of the making is chemical in nature; particularly if I'm growing silicon crystals.

@EhevuTov I would tend to agree, myself.  I just wanted to let the comment about transferring stand (without the insult) in case any active Physics.SE members had further justification for such a move.  Thank you for justifying your position, I will leave the question be.

@EhevuTov growing silicon crystals is not a chemical process, neither can be done at home

You have to study existed architectures. First, it is necessary to distinguish between photo-conductive and photovoltaic elements. In photovoltaic elements one needs a mechanism of separation of positive and negative charges formed under an irradiation of the semiconductor. Even a simple p-n junction can do this. But, this is all about efficiency. The solar cell has to absorb as much light as possible without reflections, and, at the same time, the charge transport inside should be without losses that is impossible in thick structures. Also we have to care about the absorption spectrum. To absorb in all spectrum of the solar radiation, people make cascade structures with different materials absorbing in different ranges of spectrum.

Making pure silicon is somewhat like making aluminum, you heat SiO2 (sand, or more specifically, quartz) to a liquid state and hydrolyze with two carbon electrodes. One electrode combines the oxygen with the carbon in the electrode, releasing CO2, the silicon and other impurities remain behind.  When this cools, you have foundry grade silicon, used mostly as a steel additive.  Since this involves extremely high temperatures it's unlikely you'll be doing this in your garage (although it is, literally, possible).  As an alternative, some 'thin film' panels are printed, usually on a plastic backing, you would have to find the printer and the proprietary ink.  Once you had bought a $100,000 printer and very expensive 'ink', you're ready to roll.


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Nomenclature of alkenoic acids - which group has priority: the alkene or the acid?
Right, how do you systematically name them? Also, do you number them from the double bond end or the Carboxyl end?
Ie would it be something like.  2-methyl Hept-3-enoic acid?
Scuse the bad question and I made up the example, sorry! I'm really confused!
Thanks!!

The carboxylic acid moiety takes precedence over the alkene moiety, so the molecule has to be numbered starting from the carboxylic acid carbon. Hence, methacrylic acid, for example, has the systematic IUPAC name of 2-methylpropenoic acid.


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Open Source tools to draw a set of 2d molecular graphs from sdf file
I have a sdf file for around 50 small-molecules for which I am doing structure-activity study. I would like to draw 2d graphs of all the molecules such that the image would fit into a single journal page. For example, in this paper, page number 3, they have listed 2d molecular structure. I have attached the image file from the same paper. I have software tools like MarvinSketch which give me 2d or 3d graph for a given sdf file. But I want the integrated image like shown below suitable for publication.

An SDF file is just a bunch of MOL entries in a single file; $$$$ serves as the separator for the individual entries.
Splitting into individual files AND simultaneous conversion to graphic file formats (PNG, SVG) is possible with Open Babel, either directly on the command line or through scripts/programmes using the bindings available for various programming languages, such as Pybel for Python.
To my knowledge, the direct generation of PostScript files - probably the nicest option when using $\LaTeX$ - is not possible, so you might want to used SVG as an alternate vector-based format.
obabel -isdf lotsofstructures.sdf -m -osvg -xC -xu -xd -O basename.svg
will do the job.
-isdf specifies the input file format, -m does the splitting, -osvg specifies the output file format, and -O basename.svg specifies the prefix and extension for the individual numbered output files that are generated. -xC prevents that terminal carbon atoms are printed explicitly with attached hydrogen atoms. Surely a matter of taste, but I wouldn't want to have lots of $\ce{-CH3}$ in the drawings. -xu prevents the use of colour for heteroatoms and generates a black and white drawing. -xd suppressed the name of the compound to be written and displayed in the SVG file.
This should work for SDF files with 2D and 3D coordinates, but in the latter case I'd suggest to check it carefully; I've seen 3D -> 2D transformations on some molecules that went wrong.

I don't think we have any interest in adding PDF or PS output, although there are a few svg2pdf utilities out there.

Incidentally, if you find cases where 3D -> 2D does not work, please send an e-mail to the list or file a bug report. I'd like to see these and fix them. Thanks!

This looks like a combination of individual SDF images and some sort of Table editor. A program that can take your PNGs rendered from SDF files and create this table layout is $\LaTeX$, or perhaps plain $\TeX$ or (my favorite) ConTeXt.
You have a number of constraints:  a) a free program that b) makes intricate tables and c) handles SDF images. You probably want it to be easy to use, as well.
I don't know of any single open source program that will simultaneously satisfy these conditions. There probably isn't one.
I know for sure that Mathematica can take care of b) and c).  You can get it for $100 if you are academic.  But then you have to learn it.
The easiest way to make this is probably to render all of your images as graphics, and then lay everything out in Microsoft Excel or Word Table Editor, changing the column widths and grid lines to suit you.  (But this is Microsoft Office, and is not open source, so again, b) and c) )

Perhaps LibreOffice will work, and then you have a) b) and c).

given that you have each separate images, can you give me a brief outline how would you do this in LaTex?

I would look into the makecell package and \includegraphics command.

In addition to those listed in Wikipedia we have a product called
"Chemistry Add-in for Microsoft Word"
As its name suggests it is designed to run inside Microsoft Word.
It is Open Source, I am the lead developer in a team of 4 people who are actively involved with the design and coding.
System requirements are quite modest
Windows 7+, Word 2010+, IE11+

PS: We have tried to get it listed on Wikipedia, but our edits get rejected as we "run" the project
Our project home page is at https://www.chem4word.co.uk/

With V3.2 of Chem4Word we have added the ability to draw reactions.
See here https://www.chem4word.co.uk/2022/03/15/all-new-for-2022-chem4word-does-chemical-reactions/
We expect this version to be out of Beta shortly

Check out this Wikipedia page which lists molecule editors, many of which are licensed under GNU GPL.
An inexpensive (50-60 USD) program for drawing chemical structures is Chemdoodle. Despite not being free, you get some user friendliness and features (like reaction building) for the cost.

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How did Scientists Capture The First Images of Molecules Before and After Reaction?

Using a state-of-the-art atomic force microscope, the scientists have taken the first atom-by-atom pictures, including images of the chemical bonds between atoms, clearly depicting how a molecule's structure changed during a reaction. Until now, scientists have only been able to infer this type of information from spectroscopic analysis.  — Source

Are there any papers how this thing actually works?

There are a ton of papers on how AFM works, but I'll have a go at explaining it anyhow.
Atomic Force Microscopy (AFM) is a venerable technology for imaging nanostructures down to individual molecules by physically contacting atoms on surfaces. The whole process is very similar to a finger reading Braille.
What does it mean to physically contact an atom? It means that the AFM works by detecting some kind of repulsion between an atom and a very, very sharp probe due to their respective electron clouds repelling. The forces between surfaces on the nanoscale are a complicated mixture of attraction due to Casimir-Polder/Van der Waals interactions, dipole-dipole interactions, electrostatic repulsion, Pauli exclusion, etc. etc., but the key point is that if two objects get too close together they will experience a rapidly increasing repulsive interaction, which the AFM picks up.
This repulsion is detected through the deflection of a cantilever (a thin, long, elastic beam of material). The probe is mounted on the end of this cantilever, and the cantilever bends when the probe is pushed up against a substrate. The deflection of this cantilever is measured by bouncing a laser beam off the end, the change in position of the reflected beam indicating the force.
The AFM probe has an atomically sharp tip that can localise single atoms on a surface, and the tip is moved around using exquisitely precise actuators so that the tip can be scanned back and forth over a surface, picking up the bumps and valleys of the sample that indicate the positions of atoms. I've seen ads for AFM probes that actually have flared tips - these can obtain even more 3D information about nanostructures as they can hook into nanostructures from the side and measure the texture and relief parallel to the surface.

"There are a ton of papers on how AFM works" The clue in here is that you can see the chemical bounds.

@user8005 - What you are seeing is the molecular electron density. When two atoms are bonded, there is appreciable electron density between the atoms, and you can accentuate the appearance of distinct bonds through gradient mapping. Regrettably, I didn't read the article you linked which states that the technique used was ncAFM, which is a softer technique that doesn't push up against the sample but instead measures modulation of resonance in the cantilever, however the principle of measuring distance/force relationship is the same.


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chemical equilibrium reaction for burned gases in engine
I do not understand how equation(3) occurs.(taken from the book Internal Combustion Engines by John B.Heywood)
$$ $$
Consider a reactive mixture of ideal gases. The reactant species $M_a,M_b$ etc.and the product species $M_l,M_m$ etc. are related by the general reaction whose stoichiometry is given by:
$$\upsilon _aM_a+\upsilon _bM_b+...=\upsilon _lM_l+\upsilon _mM_m+...$$
$$ $$
This can be otherwise written as:
$$ \sum_{i}\upsilon _iM_i=0$$
where the $\upsilon _i$ are the stoichiometric coefficients and by convention are positive for the product species and negative for the reactant species.
Let $\delta n_a$ of $M_a$ react with $\delta n_b$ of $M_b$,etc. and produce $\delta n_l$ of $M_l$,$\delta n_m$ of $M_m$,etc..
These amounts are in proportion,given by the equation (3):
(3) $$\delta n_i=\upsilon _i\delta n$$
1.Does $\delta n$ here signify 'extent of reaction'?
2.I would eventually like to use the number of moles of each species in expressing the chemical potential,with the gibbs free energy already known. But,the part I don't understant is equation 3. What does the author mean by proportional? Could you give a example?
Gibbs free energy is given as:
$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\delta n_i$$
which by equation(3) can be re-written as(WHY??):
$$(\Delta G)_(pressure,temper_)=\sum_{i}\mu _i\nu _i\delta n$$

It's simpler than it might look. As a random example:
$$\ce{3A + 2B->2C + 4D}$$
If I were to ask you how much of $\ce{D}$ would be produced after $5\  mol$ of $\ce{B}$ were consumed, then you would immediately answer $10\  mol$ "due to the stoichiometry". In more elaborate terms, for a $\Delta n_B=n_B^{after}-n_B^{before}=-5\ mol$, one has $\Delta n_D=n_D^{after}-n_D^{before}=+10\ mol$. Any infinitesimal change in quantity of any species $i$ ($\delta n_i$) is accompanied by a shift in the quantities of all other species precisely determined by the stoichiometry of the reaction (the stoichiometric coefficients, $v_i$), such that $\frac{\delta n_i}{v_i}$ is constant for all species $i$. For the above reaction, $-\frac{\delta n_A}{3}=-\frac{\delta n_B}{2}=\frac{\delta n_C}{2}=\frac{\delta n_D}{4}$. The ratio $\frac{\delta n_i}{v_i}$ is defined as $\delta n$, and you can think of it as a counter for "how many times the reaction happened" ($\delta n>0$ if left to right, $\delta n<0$ if right to left).
$\delta n$ is a general change in amount of substances, but the particular case where changes are calculated relative to the equilibrium condition ($\Delta n=n^{equilibrium}-n^{before}$), then it represents the extent of reaction.


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Avogadro - How can bonds be hidden and only show atoms?
I am using Avogadro for visualizing some cluster systems. Is there a way to hide the bonds of the system, ie. only showing atoms?
Thanks!

Not in the current version, as far as I know.
If you need it, you could either make a feature request on the Sourceforge page, or you could do something with extensions, such as making a new version of the ball and stick renderer removing the code that renders bonds, or making an extension that adds a button to delete all the bonds.

Actually, it's not too hard. One of the display types in Avogadro is "van der Waals" which will just show atoms - no bonds. So you'd enable the vdW display type and then turn off the "balls and sticks" one.
Yes, you could write code which removes bonds, but why?

The vdW display settings is a space filling model. I believe the OP just wants the ball&stick model without the sticks.

@Martin Obviously that should have been a comment for the OP. However, the description of "cluster systems" makes me believe vdW might be just fine. (For example, I solely use vdW in rendering inorganic crystals.)

You are right, I did think too much about molecules. For example if you have molecules that make up a cluster you barely see anything anymore, because atoms become too big, but then you could still use ball&stick modus.

This is a minor comment, but I don't have the reps for it. (Any mods prepared to change this into a comment?)
One possibility is to convert the structure file to .xyz using babel and re-loading it in Avogadro, if the bond information is to be removed deeply. Otherwise adjusting the bond and atom radii should be able to make the bonds unobtrusive!

As far as I know, avogadro uses babel conversion internally. You will have to change the bond values, no matter if there is any bond information in the file format. Avogadro will draw bonds based on the default values.

Indeed, this will not work. Saving as .xyz will only remove the bonding information in the file. They will be re-perceived by Open Babel upon import. That's not what the OP wants to do.


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What's the physicochemical difference between a chromophore and a fluorophore?
Let's take Hematoxylin, which is a pH indicator and a stain for histology, and therefore a chromophore, but does not fluoresce:

Now let's take Eosin, which stains tissue pinkish, but is also fluorescent:

I would like to know which features of the molecules decide if it is a chromophore or a fluorophore. As far as I can see it (I'm a biologist with some training in organic chemistry), both have conjugated systems, although the system in Eosin spans a lot of more bonds. Is it really just the size of the conjugated system that decides this, e.g., if Hematoxylin had a larger conjugated system, would it also be fluorescent?

A molecule "is" not a chromophore, it may contain a chromophore! Second one cannot say some dye "does not" chromophore in general. It may or may not fluorecse in a certain environment (solvent etc). Whether or not fluorescence orcurres, is a question of electronic and vibrational levels of dyes. Look for some literature on dye lasers.

The physical chemistry of fluorescence is a little bit complicated, and so you should consult a text book for "the reason" that things fluoresce.
One of the things that I have noticed is that fluorescent compounds often have several "heavy halogens" Br and I attached to some poly-aromatic hydrocarbon.
Disclaimer: this is not to say that the halogens are the fluorescing portion of the molecule (it's probably almost the whole molecule, conjugation + halogen), and it does not mean that all compounds that contain Br and I will be fluorescent. It also does not mean that Br and I are required for fluorescence.
I think that a lot of fluorescent molecules were discovered by chance, or even brute-force changing substituents until the molecule exhibits the desired fluorescence.  (It's really too hard to predict with qualitative theory, and electronic structure theory descriptions of fluorescence are also very difficult.)

The key to fluorescence is an extended Pi system. In hematoxylin, the two aromatic rings are far apart, while in Eosin, the conjuated Pi system is uninterrupted and continuous. You need at least four conjugated groups to absorb in the visible light region. Aromatic systems absorb strongly in the UV region. In addition, the attachment of halogens (i.e. Br) reduces fluorescence.
Other things that reduce fluorescence besides halogens are $NO_2^-$ (nitro groups) and flexibility of the molecules. Rigidity increases fluorescence.

Other things that reduce fluorescence besides halogens are NO2 (nitro groups) and flexibility of the molecules. Rigidity increases fluorescence.

Why do these groups reduce fluorescence? Is it due to their high electronegativity drawing electrons and thus somehow disrupting the conjugation?

I incorporated your comment into the answer.  Feel free to edit if you don't think it belongs.  Also, can do you have a reference for the 'need at least four conjugated groups..' statement.

@bobthechemist if you do a rough partice-in-a-box calculation, you find that the box needs to be about 10 angstroms long in order to absorb at the electron-volt order of magnitude (where visible light is). With a C-C bond at roughly 150 pm and a C-atom at roughly 50pm, that's 6 carbons to get to 1000pm. I'll buy four conjugated groups to absorb light in the visible, although the particle-in-a-ring calculations may yield something quite different.

While I agree that an extended pi system helps fluorescence, I do not agree that the reason is that it needs to absorb in the visible portion of the spectrum.


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What is the accuracy of scf=qc when convergence criteria is lowered?
I am trying to run single point calculations with Gaussian 09 for a Mulliken population analysis on large graphene/graphite (940-1411 atoms) sheets/clusters.
I was having difficulty with memory issues (single CPU license) running DFTBA for single point calculations. I have stepped my convergence (beginning with conver=1, and increasing by reading in my previous checkpoint files).  I found my results oscillating and failing.
I decided to try scf=(qc,conver=5,maxcycle=500).  This worked, which concerned me.
It not only worked, but it worked much faster than my failed sp calculations.  I know Gaussian's old default convergence for single point calculations was N=4.  As my calculation was done in less than two hours compared to the days it was taking for DIIS, I have become concerned at the accuracy of the QC using conver=5.  I cannot find anything indicating my approach is bad, but also I cannot find anything syaing my approach should yield reasonable results.
Can anyone please tell me if this approach is reasonably accurate?

Something that I have discovered is that when Gaussian (or other good ab initio software) fails to converge, there is something sketchy about what one is trying to calculate. Sometimes DFT calculations just aren't appropriate, especially in cases that require multiconfigurational SCF for even a zero-th order description.
In this case, given the sheet-like character, I suspect the problem is that the molecular orbitals are so close together, and so similar, that there are serious linear dependency problems that lead to all sorts of havoc in the linear algebra routines and convergence accelerators.
If you are trying to simulate an infinite sheet, then I suggest that you try plane wave codes like those found in solid state chemistry (Gaussian's PBC code, NWChem, abinit, etc.)  These sorts of calculations have gotten easier in recent years, but sometimes they require serious horsepower.
There really does seem to be a problem with SCF convergence, and this could be because you are approaching some kind of band structure limit. I have always used QC as an absolute last resort when I must do something wacky.
(One other thing, you have "capped" your peripheral benzenoid rings with hydrogens?)

Thank you for your reply.
First I must ask, why do use QC as a last resort?  I tried as such expecting an error due to memory.  Everything I have read suggests it is more memory intensive, but at the same time more reliable.  It was because of that I became very concerned about the speed of my single point runs (roughly 1/4 of the time my SP-DIIS calculations).
If you avoid QC for other reasons, please let me know.

Next, yes I have tried different configurations of the substrate.  I started with a simple single layer graphene; capped my edges of that graphene with hydrogen; am currently running a double layer capped with hydrogen (so far DIIS only); and have graphane and a pure double graphene to try.
My geometry is set up according to accepted theory (although it is completely pristine).

I think that QC calculates the orbital Hessian w.r.t. the coefficients.  That's awfully expensive, when there are other methods which could get the job done faster. My point is that there seems to be something pathological with your molecular system, which is why DIIS doesn't work, and why you have to tweak convergence criteria even on QC -- the most robust convergerer known to humankind.

another question: is Gaussian complaining about linear dependencies?

My challenge is to accomplish my calculations using the given software on a single cpu and with the memory limitations.  I had read publications where they were quite successful doing something similar, using DFTB and SWNT (albeit much smaller surfaces).  Our sheets need to be at least 32x32 Angstroms.  When my convergene was oscillating I too suspected an issue with molecular orbital overlap, which is why I began "capping" my ends and set up a fully saturated graphane.  I will certainly give the PBC code a try.  Thanks again for your reply!

No, it is not mentioning anything about linear dependence. In doing the full print, I can see the steps and they are most definitely stuck in an oscillation.

Eric, your first comment on what (or rather how) it does its calculations is what had me the most concerned.  For a system of my size and a robust calculation, it shouldn't have converged so quickly.

Eric...You have given me another idea of how to double check the calculations.  Let me give that a go.  I know folks don't normally complain when calculations are fast, but it was a huge red flag for me when it took less than four hours to complete on a 940 atom system (using QC that is).  Unless being stuck in the oscillation caused the problem (extreme long run time prior).
The only reason I have had to tweek my convergence criteria for QC has been for memory limitations.  I have found I can overcome quite a lot by stepping my convergence to where I want.

I will take another look at my geometries.

You need to be careful.  If you are running into linear dependency problems Gaussian may not always tell you.  It likes to drop the suspect basis functions when it can instead.  Be sure to check your output file at the beginning and end to make sure that the number of basis functions used did not change.  This is very important to take note of.

@Linda You may have a more symmetric structure, even if it is a larger molecule.  This might be much faster to compute and converge.

@EricBrown, I am not sure I follow your last statement.  I went back through my basic structure today (no hydrogen termination) and cleaned it up (set all of my bond lengths to exactly 1.418).  Is that what you were suggesting, or are you saying I may have too much symmetry?

@Linda I think you definitely want H terminations (should be like benzene around the periphery) Otherwise there will be trouble.  Symmetry:  it is possible that a large structure with high symmetry (e.g. D2H) will compute faster than a smaller structure with lower symmetry (e.g. Cs).   You may want to make sure you are using the highest Abelian symmetry group as possible.

@EricBrown, Well I can't thank you enough!  We followed your advice and for the first time I understand what a symmetrical calculation actually looks like.  Going back through my old files Gauss had switched to nosymm on my old graphene files.  I have been able to leave the system the size we wanted (1476 atoms) and we finally got convergence for our hydrogen terminated sheet (10^-4, running for 5 now), and I was able to do so in one step with no memory issues.  For our final sheet we used geometry for xy instead of defining bond lengths and brought our SD for the same down.  Thanks again!

@EricBrown, I think I have also answered my original question.  I tried the qc on DFTBA with a much smaller system.  The rigors of the qc calculations don't appear to start until N=6 in Gauss.  When I finally have my susbtrate optimized I am going to go back and compare my results to the qc calculation at conver=5, see how well it compares.


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What tools are used to predict the products of a chemical reaction?
I was wondering if there is a decent and non-complex way to predict the outcome of mixing two chemicals together.
I know that molecular dynamics is used in computational chemistry to understand some of these types of problems, but in many cases I have noticed that they seem to be restricted to classical physics and therefore cannot accurately predict any real reactions.
My point in asking is such that I can have a broader understanding of this; However, at the moment, I am interested in the difference between this reaction:

and the same reaction while replacing the triflic acid ($\ce{CF3SO3H}$) with chlorosulfonic acid.
Would there be any differences if these two superacids were switched?
The paper from which I obtained this reaction from is here:
Klumpp, Douglas A., Rendy Rendy, Yun Zhang, Alma Gomez, and Aaron McElrea. "Dicationic Intermediates Involving Protonated Amides: Dual Modes of Reactivity Including the Acylation of Arenes." Organic Letters 6, no. 11 (2004): 1789-1792.

I don't know for sure, but I suspect these two reagents would lead to the same product.  They are both superacids, and extremely harsh.
Whichever is a better Bronsted acid (proton donor) in benzene probably leads to a faster reaction, or better conversion.  But who knows?  Maybe the reactions are so acidic (two plus charges in a small organic molecule!) that they are just about the same.
There may be some superacid expert lurking on chemistry.se who knows this answer, but I think the only way to know is to do the reaction.
You may also want to write the corresponding author of the paper. They may already know the answer. Or, they might be flattered and may find the question interesting enough to go to the lab and do the experiment.
(I don't think computational chemistry will help here.)

Thanks Eric!  That's a great answer, but why do you believe that computational chemistry will not help in this instance?  Is it just too complex?

@chase I think it's too complicated.  The conjugate base may complex with the dication in some peculiar, solvent-separated ion pair.  These just aren't easy to model.

Reactivity of these acids is very different, triflic acid shows mainly acidic properties, while chlorosulfonic is also strong oxidant and electrophile, and also chlorinating and dehydrating agent, which makes it unsuitable for using instead of triflic in your case.
It's not a good idea to even use benzene as solvent for chlorosulfonic acid, because it can get chlorosulfonated (see this question) and substrates in these reactions are more reactive than benzene. In condensation reaction, water is created and $\ce{HSO3Cl}$ would gradually hydrolyse, even if it worked properly.


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Geometrical Isomerism of Metal Complex
Does $\ce{Ni((en)2(Cl)2)}$ show cis-trans form ?
And if No
Why not ?

Sounds like a homework question. Can you post what you think are the cis- and trans- forms of this compound?

The question is asked whether the compound will show geometric isomerism . I think it should show cis-trans  but the answer says no. I can't figure out why , are bidentate ligands always bonded to adjacent sites of the geometry ?

If you draw en, then you can see that there is no way for it to reach all the way around the molecule to bond to a trans position.  (I think it's the smallest bidentate ligand.)

I can draw a cis (w/r/t the Cl ligands) form, so if it does not exist (i.e. has never been observed), then why would that be (which is what the question is really asking)?

Theoretically you can draw both situation; when monodentate ligands replaced with bidentate ligands cis- trans- isomers can be drawed (in ML2B2 type). How did you have an answer that says "no" ?

$\ce{en}$ is bidenate so it has to take two neighbour corners from octahedron (let say equatorial position). The second $\ce{en}$ can bind with 1st nitrogen also on equator but then the 2nd has a choice: it can bind in equator or on axial position. So in theory it would be possible to see two isomers. The reality has however more to do with energy. $\ce{en}$ is very good ligand and $\ce{Ni++}$ makes very stable quadratic planar complexes. For that reason both $\ce{en}$ goes into equatorial position and the weaker ligands $\ce{Cl-}$ occupy axial positions.


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The influence of lemon juice & sugar on the ph-value

If I add equal amounts of lemon juice and sugar into water, does the ph-value remain unchanged, as both cancel each other out?

This question arose in my head as I learned that pouring milk into coffee increases the ph-level of the beverage making it taste less sour (You might want to check out the following video at 5:43: https://www.youtube.com/watch?v=eTYsQUqodeE; it's in German but you will get the overall point, I'm sure).
Now, if I add lemon into my tea, it tastes more acidic, so the ph-value would be more acidic right? If I add additionally some sugar, the tea tastes not sour anymore. So, I wonder if sugar is able to make a acidic drink more alkaline as it tastes less acidic...Or am I comparing apple with pears?
(Sorry, if my question sounds a bit unscientific...)

Welcome rena, your question is fine.  Some of your answers are going to be found in acid/base theory, which you can probably read about a bit in Wikipedia or an introductory chemistry book in order to get more background.  Feel free to leave your question as it stands or add more details/refine the questions if you find some of the information (see for example this article.

These are interesting questions. You can buy over 100 strips of pH paper for about $3 on Amazon. This should answer your questions without a doubt.

You're right about the lemon juice and the milk but wrong about the sugar. Lemon juice is weakly acidic (hence its sour taste) and milk is slightly basic (or alkaline, the terms are used interchangeably). A sugar solution is completely neutral since sugar can't take hydrogen ions out of the water or donate them in.
If you don't have background in chemistry, then let's not talk about pH - I thought pH was confusing when I learned it because "lowering the pH" makes a solution more acidic, and "raising the pH" makes it more basic! (Ironically, the pH scale was originally devised to make the concepts of acidity and alkalinity more accessible to the general public)
Your observations about tea probably have nothing to do with acidity or alkalinity since the foods you've described are not significantly acidic or alkaline (except lemon juice). Tea is slightly bitter because of tannins in the leaves. You've evolved to think of things that taste bitter as unappetizing because they often do not have nutritional value and may even be poisonous.
Sugar (and sweet things in general), on the other hand, is very tasty because you've evolved to know that consuming sugars (carbohydrates) will give you energy. Milk also contains sugars, as well as lipids (fats) which give it its creamy taste, which you've also evolved to find appealing because they give you energy - more than simple sugars.
The creamy taste of milk or sweet taste of sugar are usually enough to cover up the bitter taste of tannins (if the tea is brewed right, that is, but that's an issue for another discussion) and make your tea tasty.

Thank you for your elaborate response. I agree that adding sugar and milk to food always tend to make food more tasty for its sweetness and richness. I heard from people putting a bit of salt (or baking soda) into their coffe powder when brewing depending on the hardness of the water... a little drop of something here and there can sometimes have a tremendous effect on the taste... So I see what you mean; it does not solely depend on the ph-value of a food/beverage for it to taste 'pleasant', there are a lot more factors to consider...

Correction on an old post: Lemon Juice has a pH of 2 making it a strong acid - not weakly acidic, unless you are diluting your lemon juice with water.


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How do chemists decide on which detergent to use?
I found betaine in my face wash. I think it is there because it has a hydrophobic (non-polar) end that can pick up grease, and a hydrophilic (polar) end that means it is attracted to polar solvents, such as water.
My question is why are betaines used over other detergents? What are the factors that contribute to the which detergents are used in face wash (or indeed shampoo/shower gels)?

Good question. To get you started on some background you might want to read the answers to this question: http://chemistry.stackexchange.com/questions/4188/what-is-the-difference-between-hand-wash-and-body-wash-soap

I am guessing that you or the manufacturer mean Cocamidopropyl betaine, which is a surfactant. some points are mentioned here http://en.wikipedia.org/wiki/Cocamidopropyl_betaine such as producing more foam and being less irritating to the eyes and skin

If it's polar, then it won't pick up grease because lipids (fats and oils) are generally non-polar. I don't know specifically how a detergent is chosen, but soaps are chemically the salts of a fatty acid. When choosing a detergent, you want a molecule with a polar head with a long, nonpolar tail.
Recall that water is polar and polar substances dissolve polar substances, and nonpolar dissolve nonpolar. The detergent in a sense forms a "bridge" between the water and the nonpolar substances (dirt, grease, oil, etc.) which would not dissolve in pure water. The detergent molecules form little "cells" (my choice of this word is not a coincidence - this is also how your cell membranes work) with the polar heads facing out and the nonpolar tails mingling with nonpolar dirt molecules inside. This allows the dirt be washed away by water.
As to why certain detergents are chosen, it is probably a balance struck between safety, effectiveness, cost to mass-produce, possibly environmental considerations associated with production, etc.

Geeze. A detergent is always some sort of "long" chain with a hydrophobic (non-polar) group on one end and a hydrophilic (polar) group on the other. So the molecule has one side of the molecule in water and the other side in oil.


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Why is a theoretical titration curve higher than a practical one?
I'm currently in high school and I did two titrations. The first one is a titration of $10.0~\mathrm{mL}$ $0.1~\mathrm{M}$ $\ce{HCl}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:

The blue curve is what I measured and the red curve is what the curve would look like in theory, with these data.
The second titration is a titration of $10.0~\mathrm{mL}$ $0.2~\mathrm{M}$ $\ce{CH3COOH}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:

Again, my measurements are in blue and the red curve is the theoretical model. I noticed that (looking at the second curve in specific) the measured pH value is most of the time lower than in the theoretical model, and close to the equivalence point, the graph is not as steep. Why is this? Not a valid reason would be that the pH sensor does not measure the right value instantly, because near the equivalence point, only one drop of $\ce{NaOH}$ was added about every two seconds. Do you have a more reasonable explanation?

I suspect the (in)homogeneity of the mixture is playing an important role. Though acid-base reactions are often very fast, very small concentration gradients can be difficult to eliminate entirely due to the low speed of diffusion of substances in liquids, precluding the formation of true equibliria. How vigorously was the titrated solution being mixed? Another factor is that the familiar pH scale with $pK_w=14$ only works at 25°C. How well was the temperature controlled during the experiment?

@NicolauSakerNeto Thanks for your comment. I mixed it by hand: that is, I kept shaking the beaker the titration was done in continuously throughout the experiment. I could almost go around the beaker with my thumb and my middle finger, so it wasn't very large. The temperature was approximately $21^\circ$C, so that could also make a difference.

Usually titrations are performed inside erlenmeyers to allow better mixing by hand and to avoid spillage from droplets formed on impact. Another option is to use an automatic stirrer, either mechanical or magnetic. I believe mixing is then a large part of the answer. The deviation seems to be in general more severe for the second half of the titration, which is presumably due to the larger volume of liquid, making it even harder to homogenize. Also you get tired after a while!

The volume has indeed almost tripled at the equivalence point, so that makes sense. One last thing bothers me: in the first titration, the measured value was higher than the expected value, while in the second titration, it was the other way around. Could this have anything to do with the fact I already started with $10.0$mL of $\ce{NaOH}$ added to the solution, to skip a boring part of the curve? That doesn't make sense to me though.

$\ce{HCl}$ and $\ce{NaOH}$ solutions are a bit tough to prepare at an exact concentration, because $\ce{HCl}$ is volatile and $\ce{NaOH}$ is hygroscopic, making both usually slightly more dilute than expected. For proper analytical precision, the concentration is usually tested by a previous titration with a more reliably prepared acid/base solution (a primary standard), such as potassium hydrogen phthalate. It may be simply that you didn't prepare the solutions to quite the concentration you expected.

@NicolauSakerNeto While KHP standardization is likely part of the issue, there appears to be an inflection point both of the experimental titration curves which cannot be explained by titrant standardization.  There's another phenomenon occurring here.

One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.

The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on the pH of your aqeous system. As long as it is sufficiently acidic no carbonic acid can/will react to form hydrogen carbonate or even further to carbonate. But from $\ce{pH}\approx4$ the possibility that hydrogen carbonate will be formed rises. This means for your titration, that you are not only titrating your acid anymore but you are also titrating the dissolved $\ce{CO2}$. The longer you wait for equlibration near the the equlibrium point the more dissolved $\ce{CO2}$ reacts to hydrogen carbonate and distorts your ideal $\ce{CO2}$-free titration curve.
To show you the impact more clearly I created the following figures showing both of your systems with varying carbonic acid concentrations $\left(0.00\ldots0.05~\ce{mol~L^{-1}}\right)$.
At first your system with hydrochloric acid:

And secondly the system containing acetic acid:

There are two remarking differences between both titration curves.

You can determine the equlibrium point of the $\ce{HCl}$-titration (though it's not around pH 7 but at pH 4-5) no matter how many hydrogen carbonate will be formed during your titration.
You have to pay attention during the titration of acetic acid because the inflection points gets distorted above any detectable other inflection point. However it is not that important as it seems because titrations of acetic acid are fast and there is usually not much time to dissolve enough $\ce{CO2}$.

The other points are already mentioned before. Your pH electrode could not be calibrated correctly and your concentrations might not be as precise as you think they are. The titration curve of hydrochloric acid looks like your acid was less concentrated which results in a higher starting pH and in a lower ending pH. The titration of acetic acid indeed looks like the electrode was not properly calibrated because it looks quite good if you just raise the pH values per measuring point and it suffers from dissolved $\ce{CO2}$ which is why the inflection point is shifted to a higher volume.
The mentioned increase of volume that is most times neglected during the calculation of titration curves has no big impact unless the concentration of the base is much lower than that of the acid that has to be titrated.

On the one hand the first four images show that different base concentrations around your ideal concentration don't effect the curve itself very much but only the needed volume to reach the point of inflection.
On the other hand you can see that there is a visible influence on the titration curve if the base is 10 times as diluted as the acid.

Nicolau Saker Neto's last comment is probably what's going on here.  All of your titrations have too low a pH compared to theoretical, and so are less basic than expected.  This suggests that your concentration of $\ce{NaOH}$ is lower than you think, and the fact that $\ce{NaOH}$ is hydroscopic is a good explanation for this (i.e. you have a more dilute solution than you think).  Neto continues to be correct in that the usual procedure to correct this is to standardize your titration solution against a standard that is known to be more reliable with respect to producing accurate concentrations.  $\ce{KHPh}$ is indeed one of these standards, and is used after drying in a drying oven.  So the full procedure is: prepare $\ce{NaOH}$ solution as accurately as possible, do the same with $\ce{KHPh}$, titrate $\ce{NaOH}$ against $\ce{KHPh}$, and record the resulting concentration of $\ce{NaOH}$ accurately for use in future titrations.  Also, $\ce{NaOH}$ loses strength over time with exposure to the air, so you have to re-standardize over time.

That "looses strength with exposure to air" is carbonate forming from $\ce{CO_2}$ captured from the ambient air. Which means that you have a small amount of buffer in the solution. pKA $\ce{HCO_3^2-}$ is 10.3, so this may be the explanation for the bump in the second titration curve.


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Transition Metals' Charges
I came across a compound $\ce{[FeO4]^{2-}}$ but $\ce{Fe}$ has +6 charge according to my calculations. How this is possible ? Or it is possible but my periodic table is kind of non-detailed one. (On the table I got $d$-metals possible charges).

If you write out the electron configuration of Fe, you will find it has $3\text{d}^6 4\text{s}^2$ in its valence level, meaning it can theoretically take on any oxidation state from +1 to +8 (and even some negative states, for that matter).
In practice it is extremely uncommon to find any oxidation states other than +2 or +3 for iron - which explains why a simple periodic table containing only the most common naturally occurring states will omit them - but they are possible, just not very stable or long-lived. The polyatomic ion you have observed is called ferrate.

I've always enjoyed citing catalase as an example of a ubiquitous compound where iron takes on uncommon oxidation values.

Actually iron(VI) is not all that exotic, being in use commercially in water treatment. If we regard each oxide ligand as donating a pi electron pair as well as a sigma pair (a reasonable model for oxide ligands to electron-deficient metal-ion centers), then ferrate(VI) meets the 18-electron rule.


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Valence electrons in helium?
I am having trouble understanding how valence electrons are counted in Helium.
My references are:

Here where it appears that Helium has 0 valence electrons because the shells are finished
Here where it appears that Helium has two valence electrons (which is what I thought)

How many valence electrons does helium have?

You're comparing Wikipedia on the same grounds as a comment in a funny picture? I'm sure you can answer your own question...

@NicolauSakerNeto Appears to be from a test question.

Helium has two electrons in total, and according to the aufbau principle, it adopts the electronic configuration $\mathrm{1s^2}$. This means it has two electrons in s orbitals with a principal quantum number of 1.
The last (and only) level of helium's electronic configuration is $\mathrm{1s^2}$, and therefore $\ce{He}$ has 2 valence electrons.

I would like to add that the "valency" is 0 although it has two "valence" electrons.

Valence electrons

Valence electrons are electrons in the outermost shell of an atom

Valency

Valency is the number of electrons involved in the formation of a chemical bond

Coming to your question
The number of valence electrons of helium is 2 because of its configuration which is $1s^2$
Hovewer, the valency is 0 because it is already stable and none of the outermost electrons are involved in the formation of chemical bonds.

The outermost shell of an atom can accomodate a max of 8 electrons. It was observed that the atoms of elements,having a completely filled outermost shell show little chemical activity. In other words their combining capacity is 0. Of these inert elements, the helium atom has 2 electrons in its outermost shell and all other elements have atoms with 8 electrons in the outermost shell. Their combining capacity of the atoms of other elements,that is their tendency to react & form molecules with same or different elements,was thus explained as an attempt to attain a fully-filled outermost shell.


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Helium empirical wavefunctions
I know that analytic forms of the Helium wavefunctions are not known.  However, are there empirical expressions for the low-lying states of Helium?  I'd like to use them to calculate some transition matrix elements for fun.

Maybe this will help: "Wave Functions for the Excited States of Neutral Helium" Phys. Rev. 139, A373–A378 (1965)

If you want a quick and dirty expressions in terms of hydrogen atom wave functions, you can try this:

For ground state you can get a quick approximation by applying effective nuclear charge.  Then you get a hydrogen-like wave function with a modified nuclear charge between 1 and 2;  solve that and get orbital.  It will be doubly occupied, spin singlet.
The lowest excited states of He are all Rydberg states.  The roughest approximation would be to consider it He+ plus a Rydberg electron, and you can even use hydrogen-like atom formula to get the Rydberg orbital.  The total wave function would be antisymmetrized product of these two orbitals, coupled to be either singlet or triplet.

I am not sure you want transition density matrix or transition dipole matrix.  Either way, its just quite straightforward integration from there.  I think there might be ways to do this analytically, but frankly I do not remember and you can probably try that on Mathematica.

Yes, wavefunctions for helium can be calculated numerically with various levels of approximation with quantum chemistry methods. Be aware, though, that wavefunctions depend on a specific representation of the system (the use of molecular orbitals) but are not physical observables of the system (only the electronic density is an observable).

I think the OP is well aware of this. He is asking for empirical expressions, not methods of calculation. Also, he wants to calculate transition matrix elements and the electronic density is of no use for this.


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Why is Lanthanum a "fluoride ion sensitive electrode" and hence used for measuring "flouride concentration"?
My chemistry text book says that "Atomic Absorption Spectrometry (AAS) can be used to determine the concentration of metal ions, but other methods must be used for anions. One convenient method is to use an ion-selective electrode (ISE)." What is the reason? Why can't ASP be used for determining the concentration of anions?
The author then argues that "Lanthanum fluoride crystal doped with europium" is the element used for measuring the concentration of fluorine in water because "fluoride ions bind to the crystal". What is the reason behind choosing "Lanthanum doped with europium"?

What is the reason behind choosing "Lanthanum doped with europium"?

Well, you said it yourself, because "fluoride ions bind to the crystal".
Here's a great explanation from Wikipedia:
In the lanthanum fluoride electrode, the sensing element is a crystal of lanthanum fluoride LaF3, doped with europium fluoride EuF2 to create lattice vacancies. Such a crystal is an ionic conductor by virtue of the mobility of fluoride ions which jump between lattice vacancies.
Here's a rough sketch I made to help you visualizing this. This is a simplistic drawing and the actual crystalline structure of the compound is 3-dimensional.

I must stress this: Both the structure and the proportions are probably nowhere near the ones in my drawing, it is only for didactic purposes.
An electrochemical cell may be constructed using such a crystal as a membrane separating two fluoride solutions. This cell acts as a concentration cell with transference where the fluoride transport number is 1. As transference of charge through the crystal is almost exclusively due to fluoride, the electrode is highly specific to fluoride. The only ion which significantly interferes is hydroxide (OH-). Generally such "alkaline error" can be avoided by buffering the sample to a pH below 7.
We had the pH around 5.5. We also used TISAB for the preparation of water samples. The Wikipedia article on TISAB gives an excellent explanation on it's usage in fluoride ion analysis.

Why ASP cannot be used for determining the concentration of anions?

Well, I cannot give you an authoritative answer on that. I'll ask a teacher of mine this week and edit the answer with an update - if someone reading this knows it feel free to improve this answer or add a new one. In the case of fluoride it's probably because the Lanthanum electrode has pretty decent detection limits (it does). It's a lot cheaper to use it instead of AAS.


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Flame temperature for cooking food
Which has hotter temperature, blue flame or yellow flame?

Blue flame has a hotter temperature. In physics there is a relationship called Wien's displacement law, which states
$$\lambda_{max}T=2.898$$
where $\lambda_{max}$ is the peak wavelength of light emitted (which is the wavelength you see) and T is temperature in kelvins.
This relationship can be rewritten $$T=\frac{2.898}{\lambda_{max}}$$, or just $T\propto \frac 1 \lambda$.
From a table we find that the wavelength of blue light is about 450-500 nm and yellow light is 570-590 nm. Since temperature is inversely proportional to wavelength, the light with the smaller wavelength has the higher temperature.
(Now you might quibble that Wien's law is for idealized blackbodies but it can be used to model most things which emit light because of their temperature regardless.)

It's worth noting that specific emission spectra are separate to this and may swamp the black-body radiation; this is, for example, the origin of elemental flame tests.

That's true, but I neglected that point since I didn't think it would be relevant to the OP's question about cooking.

@Zen I would like for you to consider the possibility that this may not be blackbody radiation, but rather emission from the reactions in the combustion process. (Aesin, is this close to what you were saying?)

If it were emission from the reactions in the combustion process, then the question is unanswerable since the color of the emission products is not related to the temperature of the flame.

I think this is what I'm getting at.  I think the blue part of a flame may be hotter than the yellow part, but I don't think that it is transferable between different fuels.

@Zen: Sorry, yes, it was mostly a general addendum to your answer, but I guess in the context of cooking in particular, the sodium emission flame (from, say, a dirty gas hob) can make your flame look less hot than it is. (Correspondingly, I guess sulphur could make your flame look hotter than it is.)

In fact black body radiation has a small role in the color of the flame:
putting in 700nm (for blue) in the equation yields over 3500 degrees celsius while the temperature of a gas stove flame is about 1700. The colors are a result of, as was mentioned in the comments, ionization of gas molecules . If i remember correctly from analytical chemistry class, the hottest part of the flame is the middle which has a pale color, this results from a certain fuel(gas)/oxidiser(air) ratio that gives complete combustion.
also see Wikipedia


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Liquid with low specific heat and high thermal conductivity (relative to water)?
As the title suggests – does there exist a liquid with a lower specific heat then water but with a higher thermal conductivity?
I thought of this in the context of sous vide cooking: Ideally, the liquid involved would transfer heat to the food quickly (hence high thermal conductivity), but would also respond to quick changes in temperature (hence the low specific heat), reducing the risk / time of heater element overshoot.
So, are there such liquids? (preferable ones that stay liquid between 20 and 85 degrees Celsius).

...How toxic is the liquid allowed to be?

Let's start with anything the won't melt plastic or aluminum, and that doesn't burst into flames when in contact with water or oxygen.

*crosses mercury, gallium, galinstan, caesium off the list, for one reason or another.

Do you actually need it to respond quickly to changes in temperature? Normally the water bath is kept at a constant temperature and the food is inserted/removed when needed. With a higher heat capacity there is less of a risk of overshooting the temperature as it will heat up/cool down slowly.

On reflection, that cooking method is fairly medium independent given its long timescales. What matters is a control system for holding a constant temperature over a long period of time. Startup effects on warming at the start will not matter at all.

Most heat transfer through fluids is actually convection rather than conduction, so you need something that expands a lot with heat and has a low viscosity. Mineral oils or silicone are used as heat transfer fluids when temperatures above 100 °C are needed, but they are much more viscous than water.
However, for your system I guess that heat capacity is actually the most important number, as convection will be sufficient to ensure an equal temperature throughout the system (unless you actively impede it) even with a viscosity slightly higher than that of water.
Eyeballing some of the tables at http://www.engineeringtoolbox.com I’d suggest trying ethanol – its heat capacity is around half that of water, its viscosity is not much higher and it stays liquid over the desired range. Plus of course, unlike some of the more exotic solvents, it’s not a disaster if some of it does leak in to your food. If the food isn’t actually going to be eaten (just for demonstration purposes) then you could perhaps try toluene or petrol (gasoline).

Of course ethanol is highly flammable and this would matter if it was nearly boiling and the flammable vapour was flowing out of your vessel onto a gas flame or a glowing heating element.


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How many protons are required to satisfy hydroxyamine's (NH₂OH) lust for protons?
I ask because I'm trying to write an equilibrium equation for this reaction and I'm wondering whether an extra 2 protons will be required after one is taken to attack the $\ce{OH}$ group and turn it into $\ce{H_2O}$ in order to convert the freed ammonia (with one proton missing) into a conjugate acid.

You make hydroxylamine sound a lot more salacious than a molecule has any right to be.

@RichardTerrett Some people consider atoms sexy, who's to say that molecules don't feel the same way?

subatomic particles^

Well, hydroxylamine is technically amphoteric, meaning it can act as either a Bronsted acid or base. The acidic proton is presumably the one on the oxygen (although the resulting salts are apparently explosively unstable), while the nitrogen is the proton acceptor. I wouldn't expect the oxygen to be basic (which, unless I've misunderstood, seems to be what your question implies), at least not in any measurable way or for any appreciable length of time, given hydroxylamine's instability and tendency to spontaneously decompose in myriad ways. When treated with acid, it forms hydroxylammonium salts, which disproportionate upon heating.
Note: this book was my primary reference; it contains a concise but informative summary of the reactions hydroxylamine undergoes.

Not that hydroxylamine has all that much lust for protons. It is a weaker base than ammonia.


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Acids that donate just one proton

What is the general name for acids that can only lose one hydrogen ion? (i.e. $\ce{HCl}$)
What is the general name for acids that can lose $n$ hydrogen ions? (i.e. $\ce{H_3PO_4}$, which can lose 3)

The answers to these questions are probably mentioned a million times on the internet, but I could not find them.

An acid that can donate one hydrogen ion is called monoprotic (i.e., $\ce{HCl}$). An acid that can donate two is called diprotic (i.e., $\ce{H_2SO_4}$). An acid that can donate 3 is called triprotic ($\ce{H_3PO_4}$). I don't know of any which can donate four.

An acid that can donate $n$ protons is called polyprotic, assuming $n > 1$.

@EricBrown And assuming $n \in \mathbb{N}$. ;-)

@timjver tough crowd!  :-)

Just a side note: most polyprotic acids with more than three acidic hydrogens are organic, I believe, the first example that comes to mind being EDTA. There's only one inorganic tetraprotic acid I can think of off the top of my head, and that's pyrophosphoric acid.

Oh yes, I forgot about organic acids! If you make the chain long and branched enough you can squeeze just about as many carboxylic acids in as you want.


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Air recycling system(CO₂-->O₂) or (CO₂-->fuel)?
I'm not a chemist but I'm working on an eco-friendly project that is supposed to recycle $\ce{CO2}$ from the air and convert it to $\ce{O2}$ or at least a usable fuel source ,the system can use the solar system as an energy source (I know that such system isn't efficient) but for the country in which the project is applicable it would be a great economical benefit (over direct sun power).
-What is the most appropriate chemical reaction to achieve such goal ?
the reaction of course shouldn't emit toxic material(the whole purpose of the project is to benefit the environment, plus the cost of any filter or disposal mechanism isn't affordable)
is there any way to construct a circular system (Regenerative carbon dioxide removal system )
Wikipedia's article doesn't elaborate further.
Please help tag this question(this is my first time on chemistry.stachexchange)

If there is a problem in the question please don't down vote ,instead if you don't mind guiding me to correct what ever is wrong.

Right after I wrote this question I happened to stumble across Wikipedia's article about "Sabatier reaction" ,is this the most efficient method ?

Welcome to Chemistry.SE. We have the MathJax plugin built into our site to hand chemical formulas, equations, and maths. Check out the [faq] for more info.

@mahos - This kind of thing (more specifically, Fischer-Tropsch catalysis using CO2 as a feedstock) is a current active area of research. I feel that your question is a thematic duplicate of this one so I'm going to vote to close on those grounds (not that there's anything wrong with your question, per se), however you should definitely check out the answer, links and discussion over there.

I don't think it's quite a duplicate: the other one asked about one very specific fuel synthesis - Tony Marmont's Air Fuel Synthesis. Whereas this one is a general question.

@EnergyNumbers - fair enough, I withdraw my close vote. Nice answer btw.

This is quite a hot topic at the moment - the current buzz-phrase is power to gas.
There are quite a few conference papers and technical papers, and at least one dedicated conference, for it.
And it's too soon for there to be a known "best" process.
Audi and others are using a modified Sabatier process.
$$\ce{CO2 + 4 H2 -> CH4 + 2 H2O }$$
Others are looking at a modified Fischer-Tropsch process.
$$\ce{(2n +1) H2  +  n CO -> C_{n} H_{2n +2}  +  n H2O}$$

(image courtesy of the RWE power to Gas Project)
Essentially, you've got to prise the Oxygen off the Carbon, which is an energy-intensive process; and then attach the Hydrogen, which is not. But the Hydrogen typically comes from electrolysis of water, which is energy-intensive.
Not only are the processes being prototyped high-energy, but they're high-exergy too, requiring very high heat and / or electricity.
So it's unlikely that this is a better option than using solar power for solar-thermal or photovoltaics, unless the application is long-distance travel, which would require an energy source with a very high energy density. In which case, there are several renewable-fuel synthesisers being prototyped at the moment, but it's too early to tell which is the best.


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How does pressure shift equlibrium
Here is a question from my practice problems:

I know that the answer is A, but I don't know why.  Is it because A has the most moles of gas on the left? thanks

Shifts in equilibrium position of chemical reactions can be addressed by Le Chatelier's Principle. In short, the response of a system will be such that it counteracts the change that you are applying.
For a closed system, an increase in pressure at equal temperature will result in a decrease in volume according to the ideal gas law:
$$\frac{P}{RT}=\frac{n}{V} $$
This change will cause an upward shift in partial pressures which the system will counteract. In this case the equilibrium will shift to the right for reaction A, because this will lower the partial pressures. Indeed this means that for an increase in pressure the reaction will shift to the side with the lower number of moles.

I think the most intuitive way to think about this is rates of reaction. A reaction involving 2 molecules in the gas phase only happens when they collide. At higher pressures there will be more collisions between molecules.
This means that the reaction with the most molecules will be sped up more, so if a reaction has 2 molecules reacting to give 1 molecule of product, the forward reaction will be sped up more than the reverse reaction by increasing the pressure.
To do the kinetic calculations properly will require all the mechanistic steps involved. The general principle is that that at equilibrium, both reactions are happening at the same rate, but when you increase the pressure you will speed up the reaction which has more moles of gas, resulting in that reaction happening faster than the other reaction, shifting the equilibrium until the rates of the two reactions are balanced again.


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Can dust or fingerprints interfere in ultra-violet spectroscopy
My question pertains generally to any form of spectroscopy, but I'll focus on UV spectroscopy. I am trying to get information on carrying out spectroscopy, but most of it is theoretical. Wikipedia is completely useless. Textbooks generally go into more detail about the underlying sub atomic processes being carried out. However, they do not provide too much details about how spectroscopy works in practice. So my question is this
Will something like a fingerprint, or a speck of dust, or a label on the beaker (not even sure what material the beaker would be made of), or even air for that matter interfere with the absorption spectrum?

You can deal with this through background correction of your spectra, which is a very common practice in spectrometry and implemented in any good spectrometry software.
In the most general sense, background correction consists of recording the spectrum of the spectrometer path without the analyte, and then subtracting that background spectrum from the spectrum of the analyte. In that sense, background correction is a form of difference spectroscopy. The spectrum you obtain from this ideally contains only the peaks attributable to the analyte.
A concrete example is subtracting the absorption spectrum of a solvent in a cuvette from an analyte. You would run a cuvette with solvent but without the analyte and use that as your background, followed by a cuvette with the analyte. Subtracting the first spectrum from the second would yield the signal attributable only to the analyte, assuming that the cuvettes, solvents and atmosphere between the two runs are negligibly different.
Obviously, there are a lot of potential confounds here that need to be accounted for:

Cuvettes have to be made of the same material, have the same dimensions (path length is important!) Cuvettes and such are often designed to be maximally transparent to radiation in the region of the spectrum of interest to maximise sensitivity. This is why they sell uber-expensive quartz cuvettes for UV/vis.
Solvents/substrates have to be of identical composition between runs. You also don't want a solvent or substrate that absorbs strongly in the spectral region of interest for similar reasons to the above.
Cuvettes have to be homogeneously clean - if one is dirty, greasy or scuffed and scatters light more than the other, it may ruin the results.
The atmosphere has to be the same between runs - I've heard stories about quantitative IR spectroscopy of $\ce{C=O}$ stretches being thrown out by an increase in $\ce{CO2}$ due to people breathing (how rude!). You can understand that researchers who look at atmospheric $\ce{CO2}$ concentration need very careful sampling and analysis protocols for reasons similar to this.

Thanks! This is what I was looking for, practical information.

What about things like temperature, mixing/not mixing analyte, different lighting conditions, vibration (on a table for example), general sound levels in the room...Do these all through off the results?

@puk - Temperature, vibration and other factors could conceivably affect results for any number of reasons, however I've never encountered any real-world examples of these factors being meaningful problems. Assuming your spectrometer has a tightly-closing lid, it should be insensitive to ambient light. Unless you are doing some kind of diffusion study, an inhomogeneous solution will likely yield unacceptable results.


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How do I chill water without a refrigerator?
How can we chill water in a plastic bottle without a freezer or refrigerator?
Scenario:
I'm in the office and I have two plastic bottles of water (brought from home). I'm in the office for 3 to 4 hours and this means the bottles I brought from home will get hot (and won't be drinkable in a hot season). Now I want to make it cool so that I can drink chilled water, without access to ice with salt or a refrigerator.
Is it possible?

Have a look at these techniques: http://www.wikihow.com/Cool-Beers-Without-a-Fridge (obviously you can just substitute the beer with water). They are all based on either touching a cooler object or something very similar to sweating.

The simplest solution is to put the water in a Duwer flask/thermos flask in the fridge before you go to work. It will then stay cold all day.

You can use some good type of bottle, which can keep water drinkable for more then 5 hours(normally student use them.)

I'm really not sure this is a chemistry question.
Having said that, evaporative cooling will do the job for you. Here on the Indian sub-continent a clay/earthenware water container works wonders using the technique of evaporative cooling.
In the absence of earthenware, wrap two pieces of thick cloth of the appropriate size around your containers. Keep both cloth wet/moist, and let nature do the rest.
You may also consider taking a bit of pot luck

It is a very good trick forma the beach, with wind blowing

The old system of keeping things cool utilized the evaporation of water, which takes its latent heat from the item you want to keep cool (milk, butter etc. normally). Either using a cloth soaked in water (and dipping into water at its base) or a coarse earthenware vessel, it worked.

It depends on what you can use:

Leave the bottle in the cold air from the air conditioner,

Freeze it at night and wrap in bubble plastic,

Use a peltier cooler.
This last one is expensive and not too good, but I've used the others with good results.

I think this is Physics more than chemistry.

Freezing overnight and putting it in a flask can be very effective, I tend to find the water is still reasonably cool at the end of the day

It's pretty easy. Put one of your bottles in the freezer when at home and let the water turn into ice. Then wrap an old cotton t-shirt around the bottle so that it makes a thick insulating layer around it. Then fasten the cloth with a safety pin. This would prevent the ice from being melted quickly and provide you with cool water for almost a day. With a little enthusiasm you can make it so that you just have to change bottle through pulling old ones and pushing new ones into the cloth without necessarily opening the cloth.


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Can you heat water with additives?
I have been curious about this question for a while. If you want to warm up a large amount of water, is it feasible to do this by adding a substance that has an exothermic reaction with the water? What is the relevant chemistry one has to know to work out how much the water would be heated up by?

This is an interesting question. What do you want to do with the hot water? If you want it to be potable (fit for drinking), that would limit some answers.

Well... Swim in it :)

Theodore Gray, in his book ("Theo Gray's Mad Science", Black Dog & Leventhal Publishers, Inc, NY, ©2009, pp. 234-237) uses 500 pounds of CaO to heat 700 gallons of water in a hot tub, predicting that the temperature would rise from 60 to 100 °F. He cheated a bit: the CaO was incrementally added to water in an inner tub, heating the water (already hot at the start because the outside temperature was 10 °F) in the surrounding tub, where he and his 2 kids were. Details in the reference plus two nice photographs.

From the comments:

What do you want to do with the hot water? Swim in it. I was thinking
  of thousands of liters

That's an interesting idea, but unfortunately, I don't think adding chemicals to a pool in order to heat it is a good idea (especially yellowish chemicals). The water temperature would drop in a few hours, tops, and you'd have to be constantly adding more chemicals to it. At some point, you'd have a pool of chemicals instead of a pool of water. Not to mention that the chemicals would probably end up being more expensive than the electricity or the gas bill to heat up the pool to the same temperature the conventional way.
A nice way of heating up pools is using solar water heating systems, and coupling them with the conventional heating systems, if necessary.
So... Is it feasible? I'd say it isn't. But since you said this was more of a thought experiment... The only scenario I can imagine where the best solution to heat up a pool is to add chemicals to it is: You have a small pool, very little time to heat it up, you want to heat it up to sauna-like temperatures, and although you'll only use it for a couple of hours you are willing to spend a lot of money to do so. In this case, I'd add some water to the pool, and then add the following reagents to it, making the following reaction:
$$\ce{NaOH + HCl -> NaCl + H2O} \\ \Delta H = -56\,200\ \mathrm{\frac{J}{mol}}$$
You can do the math, but I assure you it heats up pretty quickly. We used to neutralize concentrated acid in ~10 L plastic buckets, and if we "accidentally" neutralized too much at a time the bucket would start to heat up to a point you couldn't even touch it.
NaCl is kitchen salt, so no issues with toxicity if you do the math correctly. Salt water also makes you float better, which I think is a nice quality in a pool. There would be salt on the bottom, but I think that's nice too, makes it feel like the sea.

(For comparison with the calcium chloride, this works out at ~830 J/g.)

@Aesin it is intriguing to wonder how much higher this number can go.

You’re looking for materials that either react with water with a large reaction enthalpy $\Delta H_\mathrm{r}$, or have a large enthalpy of solution $\Delta H_\mathrm{sol}$ (which really amounts to the same thing but with a slightly different scope).
The problem with this as a general method, though, is that water has a really high heat capacity: take a look at this table of heat capacities, which gives it as approximately 4.2 joules per gram per kelvin.
A common method of producing self-heating coffee cans is to combine anhydrous calcium chloride with water in a closed vessel surrounding the can. According to a vendor of calcium chloride, 100 % anhydrous $\ce{CaCl2}$ has a solution enthalpy of 737 joules per gram when dissolved in water.
Assuming an approximate density of water of 1 gram per millilitre, that means to heat a litre of water by 10 K, you’d need about:
$$ m_{\ce{CaCl2}} = \frac{m_\text{water} c_p \Delta T}{\Delta H_\text{sol}} = \frac{1000\ \mathrm{g} \times 4.2\ \mathrm{J\ g^{-1} K^{-1}} \times 10\ \mathrm{K}}{737\ \mathrm{J\ g^{-1}}} = 57\ \mathrm{g}$$
So, that’s feasible, but you may have a different idea of a large amount of water. And, let’s not forget, you're still adding stuff to the water, so it may not be suitable for what you wanted it for anymore.
Other notable reactions that can heat water include reacting it with lithium metal (~220 J/g), or dissolving sulfuric acid (~90 J/g). You’re looking for materials that don’t hold together particularly strongly (e.g. if they’re a crystal, they have a relatively weak lattice), because you have to break those bonds to dissolve/react it, and also bind to water very strongly, because that’s what gives you the exothermicity. Because of that, ionic materials and materials that form at least doubly charged ions when reacting with water are likely to be better candidates.

Thank you! I wonder if 737 is as high as you can get. You are right that I was thinking of thousands of liters although I only need to heat it by a few degrees. In fact it's mostly a thought experiment but an interesting one I hope.

@marshall: Well, calcium chloride is available in large quantities -- while more rare than NaCl or MgCl, it is occasionally used for gritting roads in winter in some places, for example. Getting it anhydrous in large quantities will likely be trickier, though.

How about mixing calcium carbonate and sulfuric acid with the water? There will be a lot of heat released (and carbon dioxide, of course), and the resulting calcium sulfate will, for the most part, just precipitate out.


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Is there a word for a compound that has both ionic and covalent bonds?
For example, calcium carbide (CaC$_2$) has covalent C‒C bonds and ionic Ca$^{2+}$‒ C$_2^{2-}$ bonds.

I looked up ionic compound in Wiki and no compound is 100% ionic or covalent...

Technically, that is true. However in CaC$_2$ the C‒C bond is mostly covalent and the C‒Ca bond is mostly ionic.

How about ambi-valent or ambivalent.

atomic salt (NaCl) vs molecular salt (NH3-SO4-) vs an atom and a molecule forming a salt (CaC2). You have the third case. Sorry it isn't jargon, but the description exists.

There may be a strong case to call this a "salt", like other ionic compounds.  Very often, the anion (negative ion) of a salt is an ion comprised of different atoms with covalent bonds.  For example, $\ce{ SO_4^{2-} }$ is the sulfate anion, and the bonds between $\ce O$ and $\ce S$ are covalent to some degree.  So $\ce{ Na_2SO_4 }$ is an ionic compound under this definition.  Thus, any compound with some component that is strongly ionic can be called an ionic compound; i.e. a salt, regardless of the bonding structure of the components.

Thanks for your answer. I am wondering if there is a word that will distinguish ionic compounds that do have covalent bonds from ionic compounds that don't.

@Max Radin: I do not know for certain, but it would seem unlikely. There are just too many anions (negatively charged ions) that have lots of covalent bonds ($\ce{SO_4^{2-}},\ce{NO_3^-}$, and so on).  There are even some cations (positively charged ions) with this property ($\ce{NH_4^+}$).  Molecules with these ions are generally considered ionic.

The problem with your case is that it covers a very wide range of possibilities. The simplest cases are probably calcium carbide or any cation’s peroxide/superoxide ($\ce{KO2}$, $\ce{BaO2}$) which are salts for all intents and purposes.
On the other end of the scale consider a protein consisting of hundreds of amino acids with the occasional lysine-glutamate, lysine-aspartate, arginine-glutamate or arginine-aspartate salt-bridge type bond. For all intents and purposes, that is a covalent compound although it includes intramolecular ionic interactions. There are probably simpler cases, too, where a Brønsted base and a Brønsted acid are present in the same molecular backbone and form an ionic interaction between themselves. I am just too lazy to open SciFinder to look for an example.
We can and probably should treat the two extreme ends of the scale separately. We can distinguish between them a priori by asking ourselves whether there is a contiguous chain of mostly covalent bonds between each side of the ionic interaction. If there is no link between the two ionic fragments, we should call the compound an ionic one (usually ‘salt’). If there is one, we are dealing with a molecule that we may choose to further specify as zwitterionic.
There is no real reason to classify these compounds in a separate group altogether. Ammonium chloride differs so little from potassium chloride much like sodium iodide is almost like sodium perchlorate. At least, the difference in behaviour inside these pairs is a lot less than e.g. between sodium sulphide and zinc sulphide (although the latter is already well on the way towards covalency).
On the molecular side, classifying small zwitterions as such is likely enough to distinguish them from uncharged molecules. And glycine, for example, is only zwitterionic ($\ce{^{+}H3N-CH2-COO-}$) in very polar solvents such as water. In less polar and non-protic solvents, the neutral from ($\ce{H2N-CH2-COOH}$) is encountered.


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Assay over 100% (L-Cysteine Hydrochloride Monohydrate)
In reading a COA (Certificate of Analysis), I came across an Assay line:
The Assay is 100.5% where "Lower Limit" is 98.5 and "Upper Limit" is 101.0

This product is L-Cysteine Hydrochloride Monohydrate.  How can the Assay be over 100%?
I have found this statement: "If the substance under test is purer than the standard, the assay comes more than 100 pct."  If that is the case, then the question becomes, where can I find out about this standard?

Can you post an excerpt from the COA? It probably won't help me figure out what's going on, but it might help someone else.

I created a free login at ajiaminoacids.com.  The include the COA under Additional Info.  I don't really have (and maybe should not) post the whole thing.  Is there something specific I can paraphrase?  I did include the entire Assay line so you can see how they anticipate a value higher than 100%.

another comment from a supplier: "we measure the % of compound based on the ions present in the compound reference to a standard. In some cases there may be other ions present in compound also detected in that test. When this result is compared to a standard it shows a higher %."

There is a simple reason to have the purity greater than 100% for this compound.
If the substance was exposed to a dry environment for several hours, a small amount of the  water of hydration could be lost, causing the calculation to have a higher purity.
Also, I believe that the purity is calculated from the perchlorate titration of the substance in acetic acid (I will need to check the USP monograph). There are many sources for error in the calculations: the weight of the L-Cysteine Hydrochloride Monohydrate, the volume of the perchlorate solution, the error from the standardization of the perchlorate solution, etc.
Under the right conditions, you can easily get 101% for the purity.


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Do chemical compounds have spectral lines, or only elements?
I know that elements (ex. Ag) have spectral/emission/absorption lines, but do chemical compounds (ex. NaCl) have spectral lines as well? Ones that when seen you can tell that that compound (NaCl) is made up of certain elements (Na and Cl).

Yes, but for "spectral lines" the gaseous state would generally be the implication. In solids and liquids photon interactions are generally over "bands" of energy not a very discrete wavelength as in a "spectral line."

In icp analysis the sample is heated to a temperature in which you don't have compounds and you detect only elements by their emissions.
For compounds there is absortion or fluorescence in which the sample is at near room temperature mostly.

Of course this too general, and the technique to use depends on what you are looking for and where because each technique has its strengths and weakness.


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Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?
Why do molecules like alkanes with higher relative molecular
mass ($M_\mathrm r$) have stronger intermolecular forces? For example, methane ($\ce{CH4}$) has a weaker intermolecular force than pentane ($\ce{C5H12}$). Pentane has a higher $M_\mathrm r$ than methane and therefore pentane possesses stronger intermolecular force than methane. These are members of the same family in the homologous series, known as alkanes.
My issue is why does pentane have stronger intermolecular force than methane? Why is there a relationship between relative molecular mass ($M_\mathrm r$) and inter-molecular forces. The stronger the intermolecular force the higher the boiling point and the melting point?

There are a lot of reasons for why some molecules have stronger IMF's than others, but the trend of increasing IMF for increasing relative molecular mass ($M_\mathrm r$) is due to an increase in London Dispersion Forces, part of the greater set of van der Waals forces. As you may or may not know, these alkanes are non-polar molecules (that is, they don't have a real internal dipole--all $\ce{C-C}$ bonds are non-polar and the electronegativity of the $\ce{C-H}$ bond is close enough that those bonds are also considered non-polar). This matters because polar interactions are a relatively strong IMF.
Let's take a detour into e- distributions. Typically, electrons have the highest probability of being evenly distributed about the atoms in a molecule, proportional to the electronegativity of those atoms. That being said, that is only the most probable state, and so, of course, there are times when the e- are NOT evenly distributed. Remember how alkanes normally aren't polar? Well, when the electrons get unbalanced in the molecule, surprise! An Instantaneous Dipole is temporarily introduced.
What does this mean? Well, when an alkane is next to another alkane that is subject to an instantaneous dipole, the electrostatic forces from the quasi-polarized alkane cause the other alkane to also attain a lopsided distribution of electrons, causing it to gain a temporary polarity, and this effect both attracts these molecules while also propagating throughout the surrounding molecules.
So now that you have a little understanding of what LDF's or instantaneous dipoles are, why the trend with molecular weight? The short answer is that larger molecules have more e-, and are affected by LDF's more than smaller molecules with less e-. With more electrons and more atoms for those electrons to be distributed over, there is a greater chance for frequent or large instantaneous dipoles, increasing the overall Van der Waals forces.

In the case of alkanes, with increasing molecular mass, the number of atoms in a molecule also increase, which means an increase in the chain length of molecules. A molecule with a large chain length experiences stronger London Dispersion Forces. The reason is that longer molecules have more places where they can be attracted to other molecules. This is the reason why pentane (longer chain molecule) experiences stronger intermolecular forces of attraction than methane.
As alkanes are non-polar, therefore, they will only exhibit London Dispersion Forces.

Larger molecules have larger electron clouds so therefore stronger instantaneous-induced dipole bonds.