text:
Excited states and electron addition/removal energies within Hartree-Fock, CI, etc
Coming from a DFT background, I'm used to the concept that the DFT eigenvalues do not correspond to excitation energies (i.e., the band gap, ionization potentials, etc.). To correct for these it's common to go to many-body approaches like the GW approximation to the self-energy, which is said to explicitly deal with quasiparticle addition/removal energies.
I'm curious how excited states are calculated within Hartree Fock or CI. Can one do a ground state calculation in HF or CI and then get excitation energies from the difference in these ground-state eigenvalues? Or is this similarly not possible and one needs to explicitly consider excited states in order to get the excitation spectrum (band gap, band eigenvalues for removal/addition of electron,etc.).

There are actually quite a few methods post-HF that you can do to calculate excitation energies. The simplest one you will want to look at is Configuration Interaction Singles (CIS). To do this, you express your Hamiltonian in the basis of singly-excited HF determinants and diagonalize it. It doesn't involve correlation and is roughly the same cost as a HF calculation. The matrix elements are given in Szabo and Ostlund,  Modern Quantum Chemistry pg 236.
You are already familiar that for DFT, the most common way to get excitation energies is through a time-dependent DFT calculation (TDDFT). You can do a time-dependent Hartree-Fock (TDHF) as well, though you may see this in the literature as the Random Phase Approximation (RPA). They are essentially the same thing. TDHF and TDDFT are pretty much identical save for the fact that TDDFT is based off Kohn-Sham density, and TDHF is based off of Hartree-Fock density. The equations are almost the same, and it isn't that much more expensive to compute.
If you want to include electron correlation in wavefunction-based methods to a greater degree, look for EOM-CCSD like methods. These are based off a HF, followed by a coupled-cluster ground state calculation, followed by an extension to get excited state energies.
My graduate research uses and develops a lot of these methods. In fact, I wrote about these methods not too long ago --- you can read more about it here: http://bit.ly/1anA9dZ

I should also add that EOM-CC methods allow for direct calculation of ionization and electron addition energies directly. For EOM-CCSD, you'll see these as IP-EOM-CCSD (for ionization potential) and EA-EOM-CCSD (for electron attachment).

+1 for mentioning EOM-CC methods.  These methods are so sweet (yes that is my scientific description of them).  (FYI, the equations don't have any motion built into them so "Equations of motion" is a bit misleading.  Has to do with some traditional thing).

Thanks for your answer. I took a look at p. 236 in my copy of Szabo but will need to read some of the earlier material before fully understanding it. While perusing Szabo I also came across the discussion on p. 127 on Koopman's theorem. I think this partially answers my question as well -- that the orbital energies themselves don't give the EA or the IP unless you assume that the system doesn't relax upon addition/removal of an electron. Thus presumably the orbital energies in Hartree Fock, like DFT, don't have a rigorous meaning as quasiparticle energies.


text:
Can F2/HF be separated using acid/base solutions in series?
I am trying to verify the results of EPA method 26A (http://www.epa.gov/ttnemc01/promgate/m-26a.pdf). The Method claims to be able to separate and quantify a mixed halogen/halide gas stream. Although I believe the chemistry holds true for Cl₂/HCl and Br₂/HBr, I'm not so sure about F₂/HF. Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂. I am dubious that the F₂ results are real and am trying to develop a chemistry argument to support that position, if possible.
In summary, the Method takes a mixed gas stream (F₂/HF) and passes it through a 0.1 N H₂SO₄ solution, then a 0.1 NaOH solution; sodium thiosulfate is later added to the alkaline potion. The theory is that the HF is captured in the acid solution, while the F₂ passes through unreacted. In the alkaline solution, the F₂ reacts to form HF and the hypohalous acid (HOF, I assume). The thiosulfate converts the HOF to another HF. When the acid and alkaline solutions are analyzed for fluoride ion, the quantity is assumed to be the HF and F₂ in the original sample, respectively.
Considering that F₂ is so reactive, I would think that any F₂ in the gas stream would react with the water in the acid solution and that theoretically no F₂ would make it to the alkaline solution. Is this a reasonable assumption?
If any F₂ did get to the alkaline solution, I am not sure that the conditions are favorable for the creation of HOF; F₂ reacting with water to produce HF seems more probable. Unless the acid solution is ignored by the F₂, I don't think that a true separation is taking place and any fluoride detected in the alkaline solution would represent blow through or some other error. Does this seem like a reasonable conclusion?
Thanks in advance for any input!

I have no idea whether this is feasible or not, but I imagine you could figure it out with some theoretically simple controls. First pass a mixture of the gasses (preferably in known amounts) through the dilute acid solution and measure fluoride. Then take the same mixture and directly add it to the base solution without thiosulphate and measure fluoride. Finally, take the mixture and pass it though a basic solution with thiosulphate. It should be possible to figure out if the fluorine gas makes it through the acid solution, and also if thiosulphate actually does anything.

I believe the better option to catch $HF$ is to freeze it from the gas stream, as it has relatively high (~19 celsius) boiling point. $CO_2$ cooled trap will do it for sure and ice-cooled trap may do it as well. Fluorine is really active and does react with water. AFAIK, water burns in fluorine.

Reading the standard closely- its says only identify hydrogen halides in Chlorine and Bromine only.

The answer is no, because as you remark $\ce{F2}$ will react with water and produce $\ce{HF}$.  So it is not the case that $\ce{F2}$ passes through unreacted.
I'm puzzled by this sentence:

Despite the Method not mentioning F₂, it commonly produces data for both HF and F₂

Can you clarify what you mean by "it produces data"?

When the Method is performed on a mixed halide/halide acid stream, halide ions can potentially be in both the acid and base solutions; ultimately you get two halide ion concentrations. Test report data I've reviewed show F ion in the base solution and conclude the presence of F2; from this discussion, I do not believe that is the case. However, the Method is often used by EPA and industry as if it can quantify and separate a mixed F2/HF exhaust stream.

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Which ions are accounted for in total and net ionic equations?
There are a number of things I don't understand about ionic equations.
First off, when you do net ionic equations, is it correct that you're only focusing on precipitates? The way I understood it, you put "no reaction" if there are no solid or gaseous products. To clarify, if all of the products of the reaction are aqueous, we look at it as if no reaction had taken place.
Then I went and looked in my book, and that made it seem like all products that weren't covalent were spectator ions, i.e., covalent molecules were the only ones you note.  Now I'm even more confused.

Welcome to Chemistry.SE! You are not the first one who has struggled with spectator ions, complete and net ionic equations. Please feel free to use the search function of this site. In addition, this video on net ionic equations might be helpful.

After reading into the matter (as a non-native English speaker, I've never heard of total and net ionic equations) here goes my small little guide to ionic equations:
Total Ionic Equations
A total ionic equation is where you start with a balanced equation and write all the ionic compounds in their ionic form. For example, in the following reaction
$$\ce{2 Na3PO4 + 3 CaCl2 -> 6 NaCl + Ca3(PO4)2 v},$$
one of the products precipitates (noted by the downwards pointing arrow). So we will not write it in ionic form for the total ionic equation, which is the following:
$$\ce{6 Na+ + 2PO4^3- + 3Ca^2+ + 6 Cl- -> 6 Na+ + 6 Cl- + Ca3(PO4)2 v}.$$
Net Ionic Equations
Net ionic equations can be derived from total ionic equations: We just eliminate the compounds that appear both on the right and on the left side of the reaction arrow. Thus we get for the above total ionic equation:
$$\ce{2 PO4^3- + 3 Ca^2+ -> Ca3(PO4)2 v}$$
The ions that were eliminated are termed spectator ions, because they do not participate in any notable chemical reaction.
If something is still unclear, maybe have a go at Kristy M. Bailey's
website (Oklahoma City Community College), which explains it pretty neatly.

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text:
Electronic configuration of tellurium
Below is a copy of the method I am using to determine electron configurations.

Question:

Write the full and short electron configuration for tellurium.

My final answer was:
Full Configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4
Short: $[\ce{Kr}]~\ce{5s^2 4d^10 5p^4}$

However, checking with google, it seems the answer is actually: $[\ce{Kr}]~\ce{4d^10 5s^2 5p^4}$
Is the silly "follow the yellow brick road" thing faulty? Or is it that order doesn't matter in this situation?

related, but not a duplicate http://chemistry.stackexchange.com/questions/7942/what-is-an-intuitive-explanation-for-electron-configurations

The order does not matter. Preferably, it should be arranged by the principal quantum number.

I haven't done that in decades (or so), but I'd bet that you did right!
The result from google only lists the final outcome, not the order in which the orbitals are filled. Except for some examples $\ce{La}$, $\ce{Ac}$, etc., the Aufbau principle is reliable. So: Follow the yellow brick road :)

Thank you. I've found this image and it has cleared some misunderstandings that I originally had. I thought that all the level three orbitals (spd) had to be filled before a level four could be filled. However, with your explanation, I can see now why 4s must be filled before 3d can be filled.

The Aufbau principal is flawed because there are many instances in which the electrons within a given sublevel (say 5s) are not all at close enough energy for that sublevel to fill completely without SOME electrons being more stable at higher "average energy" sublevels first.  So what ends up happening is that the 5s starts filling before the 4d if there are not enough electrons to put some in both, but if there are enough to put some in both, some go into the 4d before some go into the 5s.
Certain numbers in the d sublevels are more stable than others, particularly having one electron in each orbital (5 total).
In the case of Tellurium, the 5s sublevel may have a lower average energy than the 4d, but the 4d starts and finishes filling before the 5s as long as there are enough electrons to put some in both sublevels.
http://www.rsc.org/eic/2013/11/aufbau-electron-configuration
Current electron configuration tables show the 4d lower than the 5s when there are electrons in both sublevels, but when there is one electron in the 5s and none in the 4d the 4d may not be shown.
http://periodictable.com/Properties/A/ElectronConfigurationString.al.html
Overall, there are 2 issues.  One is that 20 out of 70 natural elements don't follow the Aufbau pattern.  Some deviate a lot.
The other is that the Aufbau pattern itself is wrong because xs sublevels should really be placed higher than (x-1)d sublevels, ex. 4d comes before 5s in terms of the order that they tend to fill when there are enough electrons to put some in both.


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How is titanium metal obtained from natural resources?
How does titanium occur in nature?
Is it available as a native metal or only in the form of ores?
Which precesses are typically used to convert the natural resources to the solid metal?
Any hints on where to find further information are welcome. Thank you very much in advance.

http://www.chemguide.co.uk/inorganic/extraction/titanium.html#top for purification. I have no idea where you'd find it.

Ok thank you for your answer. Hope it is useful

It is very useful and a great geeky website which I like thank you very much

This question appears to be off-topic because it is about obtaining a material, and not about the properties of said material.

@ManishEarth To be honest, all he would have had to ask was "where does Titanium typically reside" which would convey 'properties' rather than 'obtaining said material' yet I don't see how this would affect the right answer in any way.

@user123126 This site isn't about being "geeky", whatever that means! I took the liberty to rewrite the question completely. I hope that's ok with you.

@LordStryker true. We may want to revisit that policy on meta :)

I think the edits made it answerable (as I was able to)

Titanium is found as an ore, according to the Titanium Occurrences Wikipedia page, specifically, the 2 most economical ores are:
Ores of titanium
Rutile $\ce{TiO2}$ and Ilmenite $\ce{Fe^{2+}TiO3}$
Extraction of titanium metal from ore
According to the chemguide page Titanium,

Titanium can't be extracted by reducing the ore using carbon as a cheap reducing agent. The problem is that titanium forms a carbide, TiC, if it is heated with carbon, so you don't get the pure metal that you need. The presence of the carbide makes the metal very brittle.

Rather, the ore is chlorinated, then reduced as shown in the schematic from the webpage Titanium Exposed (the chemical processes are included):

Chemguide states that either $Mg$ or $Na$ can be used in the reduction stage, which adds to the cost of production as they need to be extracted themselves.
Asides from the links used in this answer, another good resource is Chemistry of Titanium (Professor John Lancashire - University of the West Indies)

Sponge is only partway to metal ;  It usually goes through a vacuum -  arc remelt to make the ingots, blooms or billets.  ( I am not familiar with Ti industry terminology.)


text:
What chemical properties make LSD so psychoactive?
Lysergic acid diethylamide (LSD) is a well known hallucinogen and (usually) illegal psychoactive drug with the structure below:

The drug is active is doses measured in micrograms in contrast to many other active compounds where the dose is milligrams or even grams. The effective dose of Mescaline, a related psychoactive, is something like 200mg (more than 1,000 times more than for LSD).
Why is LSD so active in small doses? Do we know which features of the structure that contribute to its activity?

The diethylamide "unit" dramatically potentiates lysergic acid. LS-Dimethylamide is significantly less potent, as are other/different amide or amine units. The differences in the amide or amine units also result in different experiences (spiritual, visual, auditory, etc.). Of course, it's different for everyone, but a loose analogy could be to the differences between beer, wine, bourbon, and moonshine.
Now, as the DrAzulene noted, there is an indole associated with many hallucinogens. Indoles are pretty important, but mescaline and ecstacy (for example) don't have an indole group, whereas serotonin, tryptophan, wellbutrin, and many other indole-containing molecules are not hallucinogenic. But mescaline and ecstacy (again, for example) have amine and methylamine groups, respectively. Other derivatives of these (and other) psychedelic drugs (designer drugs) may use diethylamide, ethylmethylamine, or other amine groups.
So, in my humble estimation, the xyz-amine group seems to act like a key to interaction with receptors, to allow communication or attachment, and by default, delivery of the affecting "body" of the molecule.
The combination of interactions with specific and a diverse array of neural receptors is what imparts the profile of hallucinatory characteristics. Another loose analogy would be like how sugar, salt, and MSG all interact with different taste receptors. Aspartame, for example, interacts strongly with the sweet-receptors. If one molecule could interact strongly with these three flavor receptors, it would have a very pronounced impact on your tongue.
LSD interacts strongly with at least 21 different neural receptors, and chief among them, all (five known) dopamine receptors as well as many (at least 14) serotonin receptors, including the classical (3 known) 5-HT2 receptors: 5-HT2a, 5-HT2b, and 5-HT2c. The 5-HT2 receptor familiy has the most pronouced impact on psychedelia - as well as depression and schizophrenia.
Phenylalkylamines (that don't contain indoles), like ecstacy and mescaline, primarily interact with only the 5-HT2b receptor and none of the dopamine receptors (interesting because phenylalkylamines tend to resemble dopamine). They interact with serotonin receptors and other miscellaneous neural receptors, instead. But generally, the list of receptors is shorter (less than 10), and the primary way that they work (by blocking serotonin reuptake so that the brain releases more serotonin) are mechanisms that cause release of the natural chemicals from the brain to produce euphoria, so they are (pound for pound) less euphoric and less psychedellic. The way that brain produces feeling of euphoria in response to phenylalkyamines is due to the flood of serotonin (produced by your own brain), resulting from phenylalkyamines' blocking the reuptake of serotonin. The increased levels of serotonin cause a release of dopamine (from your own brain). Dopamine is generally the source of euphoria and the surge of dopamine resulting from the surge of serotonin is essentiall the reward stimulation.
Again, LSD intacts directly with all five kinds of dopamine receptors, all three psychedellic receptors, and many more... and there's a very good probability that there are receptors still undiscovered.

It's the indole portion of the molecule. the lowest two rings...if you remove those, the compound has no such properties. Indoles, the side chains of tryptophan , an amino acid, are suitable for binding to receptors that affect emotion, perception, cognition, sleep/wake cycles etc. You can see how the indole looks similar to both DMT (dimethyltryptamine--another hallucinogen) and the benzene ring bears similarity to psilocin, the major constituent of psilocybin (the active principle in magic mushrooms). Tryptamines are common hallucinogens.

Very interesting question. I always thought that most of the hallucinogens would somehow act as agonists for serotonine receptors in the frontal cortex and thus trigger visual enhancements, but i've never been so much into bio- and neurochemistry.
Anyway, I just found Psychedelics and the Human Receptorome at PLOS ONE, which might be an interesting survey, but I'm too tired to read and comment on it.

From the Psych. and the H. Receptorome article that you mentioned... "This diversity of receptor interaction may underlie the qualitative diversity of these drugs." and "LSD has the strongest interaction collectively with the five dopamine receptors (D1, D2, D3, D4, D5), the ten assayed 5-HT receptors, and the four assayed 5-HT1 (serotonin-1) receptors."


text:
Are there natural or artificial enzymes that break down metals?
Do natural or human engineered enzymes exist that can speed up the break down metals (element, compounds & alloys)?
If no, then what biological or chemical construct can break down metals in the same way that enzymes break down food in a person's stomach?

No chemistry will ever be capable of literally breaking down the elements further than the single atom; chemistry can't do anything more than adding/removing electrons to it. Transmutation is a no-no!

I was actually just wondering if there were enzymes of some sort that could breakdown metal objects (e.g. kitchen utensils, aluminum cans or bottle caps) into very small particles like a fine powder.

Silly answer
There is no enzyme that grinds down an Airstream, but an alligator might do the job!
Serious (but disappointing) answer
In nature, most metals seem to exist as oxides, sulfides, carbonates, halides, etc. and not as bare metals. Consequently, there was/is probably little evolutionary benefit in tackling the bare metals.
Breaking down something to into very small particles, as you described it, is a mechanical process that leaves the oxidation state of the metal unchanged. It is not a chemical transformation.
Nothing cool with enzymes an metals?
There's a lot! Two topics come to my mind:
Chemosynthesis
There's a bunch of bacteria that can oxidize $\ce{Fe(II)}$ to $\ce{Fe(III)}$

at different pH levels
in the presence and absence of oxygen
in the presence and absence of light

Acidithiobacillus ferrooxidans and Leptospirillum ferrooxidans - look at their family names ;-) - operate at $pH\ < 4$, use the oxygen from the air and munch away pyrite.
Metal hyperaccumulation
Some plants and bacteria harvest enormous amounts of metal ions even from highly contaminates soils and store them. The metals are held by by the action of proteins. If you're interested, here is a freely available article on PLOS One Pathogens with lots of further references.

This is getting silly!


text:
What prevents the anode from corroding in electrolytic organic reactions?
An electrode can be plated by having a solution full of electrolyte, using the the metal that will receive the plating as the cathode, and the metal doing the plating as the anode. Current is put through the cell and the anode dissolves oxidized metal ions into the solution, which gravitate over to the cathode, and then "plate" onto the cathode.
My question is, what prevents this process from happening when doing an electrolytic organic reduction or oxidation? Imagine you plate an electrode with a specific metal to be used as an anode in a different reaction, what prevents that anode from immediately corroding and having the metal ions fall off into the solution when current is applied? Does the presence of an organic species to be reduced/oxidized take priority over the anode corrosion, thus preventing it?

One would control the composition of the electrodes (mercury, platinum, Ebonex, boron-doped diamond coating) and the potential of the cell so that the organic reaction went forward while the electrodes remained inert.
Metal oxidation is facilitated by Lewis base ligands that drive reaction by stabilizing the ionic products.  Don't have them in there.


text:
Finding the number of cyclic isomers algorithmically
Is there an algorithm for finding the number of cyclic isomers a given compound might have?
I've found a general algorithm for all isomers: link, but I'm not sure how to apply it for cyclic isomers only.

Do you care only for structural isomers, or do you need spatial isomers too? I'm sure including the latter makes it a much harder problem.

Only structural isomers.

I wouldn't be astonished if A. T. Balaban has published something on this. Apart from his great book on pyrylium salts, he has spent decades on the application of graph theory in chemistry.

Wow, that code listed is awful. It really hurts my eyes!

Depending on the level of complexity, you may want to consider statistical approaches (treating isomers as populations) too.

You can form a cyclic isomer many times, though unstable. For ex. the cyclo-propane is highly unstable because of ring strain. How would an algorithm take care of this?

If I remember correctly, several years ago I stumbled over the SMOG program addressing this question (description here).  The software requestested sum formula and a selection of allowed structural elements, cyclic substructures were among them.  Maybe the algorithms implemented are not the first ones to answer such a question, yet publications like this showcase continuing interest in this field, too.


text:
Does the position of the functional group on the terminal carbon change the nomenclature?
Consider two alcohols:
$$\rm{CH_3 - CH_2 - OH }$$
and
$$\rm{CH_3 - CHOH - H }$$
In the second alcohol, the hydroxyl radical instead of sticking out from the molecule, is bonded below the terminal carbon, like this
$$\rm{H}$$
$$|$$
$$\rm{R- C- H}$$
$$|$$
$$\rm{OH}$$
Now, this doesn't affect the nomenclature (as far as I know), both will be called $\rm{1-ethanol}$. Clearly, these are isomers. But IUPAC nomenclature is supposed to be completely unambiguous. Or did I go wrong somewhere and there actually is a difference in their naming?
PS: Is there any way to produce structural formulas more efficiently in latex on chemistry stackexchange?

I never tried to draw structures in LaTeX here on Chemistry.SE and to my knowledge, chemfig isn't supported. For me, BKChem with export to png does the job.

Actually, both of these molecules are exactly the same. The way you write it on paper is just an approximate representation and does not represent the actual structure of the molecule.
By writing the hydroxyl group below the carbon or jutting out from it with a single bond makes absolutely no difference to the actual structure of ethanol.( Ethanol has no isomeric alcohol.) Therefore, it does not matter how you represent it on paper, it will be the same molecule.
But in certain chiral compounds, by using some predetermined rules of representation, writing different groups on different sides does change the compound. But this is true only in cases where stereoisomeric forms of the compound exist. Then, writing the group in a different way changes the compound to one of its stereoisomer. For example, Z form is written with higher priority groups on one side and the opposite is followed while representing the E form. Fischer projection, Sawhorse projection and Newman projections to represent molecules are a few examples where the representation of molecules on paper is sensitive to the way in which different groups are attached to the compound, But not here.

They are not isomers, they are the same molecule, ethanol (the "1" is superfluous, there is no 2-ethanol), CH3CH2OH


text:
What is chemistry considered the study of?
I am asking this because I have a review question in my physics class which gives me a multiple-choice question.  I have heard and seen elsewhere on the internet (such as wikipedia) that chemistry is the study of reactions and the properties of matter.  That is usually a satisfactory answer, but for whatever reason this question makes me choose between reactions, properties, and matter.  Now I find that the general definition I accepted contains basically all of those, as far as I can tell.  While I don't necessarily need the answer, my curiosity has been aroused and I want to know why the question makes me choose between those answers.
The question is, what is the very basic idea behind chemistry?  Would it be classified as the study of reactions, properties, or matter (and you can only select one)?

Is it possible that several answers are correct in that multiple-choice exam?

Unfortunately, no (it's part of an online review exam and the questions use radio buttons, not checkboxes)

Chemistry is what chemists are doing.

Personally, I think this question is rather ill-defined. I have yet to see a good answer to this question in general that is a single sentence long, much less one that can be answered by "It is the study of ______(single word)"

Chemistry is the science of matter, especially its chemical reactions, but also its composition, structure, and properties. Chemistry is sometimes called "the central science" because it bridges physics with other natural sciences such as geology and biology.

Could you choose one?  I'm starting to get the feeling that question I was asked really was nit-picking now.

chemistry is the science that seeks to understand the behavior of matter by studying what atoms and molecules do.

I answered the question with "properties" and it was marked wrong, so I know that it's not the answer.  I'm just going to rule the question as unfair and ask my teacher about it because at this point it has become a guessing game and I surely can't be helping anybody on SE with this question.  Thanks for helping.

Please be courteous and cite your sources.  At a minimum please link to them.

I didn't notice anything discourteous about my behavior, so if I was rude then I apologize.  In my statement above I said that my question wouldn't help anybody else, not that people were being unhelpful to me.  I don't have sources to cite except for the review test my teacher gave me on elearning.

@Stopforgettingmyaccounts...The correct answer to the question is matter because this encompasses the other answers. But the reason is simply that in english things that happen to matter and properties are subsets of the broader category matter. And chemistry is a broad subject.

The thing about chemistry being the science of "matter" is that "matter" needs to be properly defined. If matter means particles with rest mass greater than zero, then approximately 85% of the matter in the Universe by weight is dark matter, for which chemistry has nothing to say. Excluding dark matter, by amount of particles ~99.999% of those with rest mass are neutrinos. I think calling chemistry the study of matter is a short but not very well self-contained explanation.

@NicolauSakerNeto Of course it isn't very self-contained.  However one could debate for hours what exact concepts chemistry extends to.  However nobody here really wants that.

While the answers so far are not wrong, I'd like to add a refinement. Chemistry is the study of matter at the level of electron interaction. The basic question in chemistry is about chemical bonding, and as far as we know this is a function of electron interaction. In contrast, radioactivity is also about matter, but is not at all about electron interaction and is not generally considered chemistry (it's considered a part of physics). Also, the chemistry of isotopes (i.e. differing neutron count) is not generally different across different isotopes. This is good, because we need solvents with are deuterium based ($\ce{D=^2H}$) as opposed to hydrogen based for NMR spectroscopy. If we go lower than electrons, we are into quantum mechanics, which is again physics. So electron interaction is a good level at which to characterize chemistry.
EDIT: I realize the original question was for a one-word answer. Frankly, that's not the greatest question, but it's generating answers that make it more interesting, and it's why I answered.
Let me give a counterexample to my own answer to show the difficulty: gases. We learn a lot about gases in chemistry which has nothing to do with chemical bonding; e.g. all the gas equations. I'd argue that most of this is not really chemistry, it's physics. Understanding how temperature, pressure, and volume interact dynamically is a very physical sort of study, not all that different from studying light, or magnetism, or force.  So why do we study gas physics in chemistry? I suspect it's because we use gas physics a lot in chemistry! You need to understand the gas laws to do any real experimentation involving actual gas chemistry; i.e. how would you measure gas as a reactant or product without knowing the gas laws?
So to be clear, if you want one word, I agree with "matter". But I think if you want to get at what chemistry is about, it's not so simple.

Indeed I felt that the other answer lacked a mention of electron interactions. However, I think a refinement is needed; chemistry is the study of interactions concerning electrons bound to nuclei. This still isn't a perfect definition, but it narrows it down more.

"In contrast, radioactivity is also about matter, but is not at all about electron interaction and is not generally considered chemistry"  -- Nuclear chemistry anyone?

@LordStryker: Well, the radioactivity observation (i.e. more physics than chemistry) came straight out of a good college level chemistry textbook. You can disagree, but as the answers are showing, this turns out to not be an easy question. I'm well aware of the old chestnut that chemistry is the study of anything you can drop on your foot (matter), but quantum mechanics has blurred some boundaries. In particular, I don't see a reason why we are likely to find a good one word answer. Matter is probably as close as we can get, and it's incomplete, in my view.

@trb456 "this turns out to not be an easy question" - You're overthinking it and the OP asked for one of the three options to be chosen.  Besides, science overlaps in so many ways that you CAN'T draw clear boundaries between what is strictly say 'chemistry' or 'physics' or 'math' etc.  Why would anyone want to anyway?  It just seems silly.

This is a little review of the online definition I've found:
Oxford Dictionaries definition:

The branch of science concerned with the substances of which matter is
  composed, the investigation of their properties and reactions, and the
  use of such reactions to form new substances.

Cembridge Dictionaries definition:

(the part of science which studies) the basic characteristics of
  substances and the different ways in which they react or combine with
  other substances.

I like the Treccani, that here I've badly translate:

The science concerned with the properties, composition, identification, preparation and the different ways in which the natural and
  artificial substances react.

More mystic the Wikitionary definition:

The branch of natural science that deals with the composition and
  constitution of substances and the changes that they undergo as a
  consequence of alterations in the constitution of their molecules.

The point I find fundamental are that Chemistry concerned with matter and compounds and historically studied the methods to:

investigate the forces between them and their properties
separate and isolate them
analyze, identify and quantify them (and this is why I like Treccani definition because include analytical chemistry)
understand and predict the ways in which they combine and react
synthesize and produce new materials
find the  way to produce the new material industrially

Chemistry is the study of chemical systems (i.e. matter).

The meaning of the word chemistry fluctuates with time.
According to Thomson, chemistry is derived from Egyptian word which meant "natural philosophy."  This definition over time changed into the art of working with metals.  About 300 AD it came to mean the art of making gold and silver.  This art was passed from the Greeks to the Arabians.  Supposedly around this time the Arabic 'al' prefix was attached to chemist to form the word alchemist.
Most alchemist who lived between about 900 AD and 1500 AD were in search of a substance, called the philosopher's stone, which converted metal into gold.  Around this time alchemy also came to mean the "art of preparing a universal medicine."
In the 1600's alchemy fell into discredit and was the butt of many satires. With the passing of alchemy, chemistry was organized as a science. In the mid 1700's chemistry and mathematics became infused.  At this point in time chemistry was a division of natural philosophy which was defined as "changes in natural bodies not accompanied by sensible motions."
The reality is the definition of the word chemistry has been changing as we better understand our world for thousands of years.  Any definition is valid only for this point in history.
T Thomson -  A system of chemistry (1810)


text:
In cleaving ethers with both SN1 and SN2 mechanisms possible, which occurs first?
What are the products when tert-butyl ethyl ether is cleaved with concentrated $\ce{HI}$?
My answer is iodoethane and tert-butanol, but the correct answer is ethanol and 2-iodo-2-methylpropane.
I know that the ethyl side goes by an SN2 mechanism because it is a primary R group, and the t-butyl group would be SN1 because it is tertiary. However, why would the SN2 reaction not happen first?

One "side" of a molecule doesn't react by one mechanism and the other "side" by another mechanism. Generally speaking, a molecule reacts by a single mechanism.  Now, the mechanism can change depending upon the other reactants that are present as well as the reaction conditions.  In the case of t-butyl ethyl ether, if the reaction were carried out with $\ce{KI}$ under non-acidic conditions an $\mathrm{S_N2}$ reaction would occur and your answer (iodoethane and tert-butanol) would be correct.  On the other hand, if the reaction were carried out under acidic conditions then an $\mathrm{S_N1}$ reaction would occur.  As you pointed out a tertiary carbocation is more stable than a primary carbocation so the expected products in this case would be ethanol and 2-iodo-2-methylpropane.  Apparently, the reaction conditions in your reaction must have been such that the reaction occurred by the $\mathrm{S_N1}$ mechanism.

Please see page 21, second slide in this pdf. One side is cleaved via SN2, and the other is cleaved via SN1.
Furthermore, no conditions were stated in the question beyond having KI. I don't understand why I should presume the mechanism is SN1 and SN1 only.

I looked at p.21.  While it may be literally true that"One side is cleaved via SN2, and the other is cleaved via SN1", it is not true in a chemical reaction sense.  T-Butyl methyl ether is only cleaved via an SN1 process on p.21.  A resultant product, methanol, can then go on to react further via an SN2 reaction in a second, independent reaction.  If it happens like this in the lab, then you get a mixture of products from the separate SN1 and SN2 reactions.  Usually reagent concentrations and reaction conditions are chosen to favor one pathway over the other.  But the key is that the original

ether reacts only by the SN1 mechanism

Why does the acidic condition favour the SN1 mechanism?

@TanYongBoon It is not so much that acidic conditions "favor" the SN1 pathway as that protons are necessary for the pathway to occur.  Without          a proton source there is no easy way to protonate the ether oxygen and then generate SN1 carbocation intermediate.

@ron Also, ethers do not necessarily have a preference for SN1. You were saying that it only reacts by SN1. Did I understand you wrongly?

@TanYongBoon  Ethers can react by SN1 and/or SN2 pathways depending on the reaction conditions.  The two pathways are in competition, but if there is no proton source, then the SN1 pathway is dramatically slowed down.

I wouldn't say that a molecule reacts via one mechanism. A substrate will always react via all possible pathways, it's the energetics that determine which one is most often used. Otherwise there never would be any side products and all chemical reactions would be specific in stereo chemistry.

@ron I found this old question and posted an answer of my own. Could you please take a look and see if my answer makes sense? Thank you :)

@ron My teacher told that if the HX used was anhydrous, then SN2 dominates. Is it true? No matter is it tertiary or else, it always go through SN2 if the HX is anhydrous . Thanks. Please reply.

@An_Elephant  Very few things are "always" in chemistry.  If you use methanol or ethanol you'll still get some SN1.  Using a non-polar, aprotic solvent would push things towards SN2.

@ron  Sorry. I should have said it better. I wanted to know that it's SN1 or SN2 only for major product. So, restating, the major product will be SN2 product? Thanks very much

I would like to elaborate more on Ron's answer to make it easier to understand. In particular, I would like to elaborate on the significance of acidity on the reaction mechanism.
Ron mentions that:

Acidic conditions favour the $\mathrm{S_N}1$ mechanism while non-acidic conditions favour the $\mathrm{S_N}2$ mechanism.

Acidic conditions
Initially, I had a lot of problems understanding this too. Upon discussion with my friend Ian, I was able to understand this fully. I was puzzled as to why there was a preference for one mechanism over the other in the different conditions. Let me explain why this preference exists in detail.
When using the reagent $\ce{HI}$, the conditions are acidic, meaning that there is an abundance of proton-donating species in the solution. Under such conditions, the oxygen atom in the ether is protonated. This gives the oxygen atom a formal positive charge, causing it to be more electron-withdrawing, resulting in larger partial positive charges on the two carbon atoms bonded to it. This also provides impetus for the breaking of the $\ce{C-O}$ bond to break. Now, we have two options, the $\ce{C-O}$ bond to the t-butyl group or that to the ethyl group. Since this bond breaks to form a carbocation, it necessarily means that a more stable carbocation would be preferred. Thus, this $\ce{C-O}$ bond would break to form the t-butyl carbocation. Eventually, the products ethanol and 2-iodo-2-methylpropane are formed.
This reaction follows the $\mathrm{S_N}1$ mechanism due to the protonation event which allowed the $\ce{C-O}$ bond to break more easily.
Non-acidic conditions
When  using the reagent $\ce{KI}$ in non-acidic condiitions, there is no abundance of proton-donating species in solution. Under such conditions, the oxygen atom in the ether is not protonated significantly. Thus, the impetus to break the $\ce{C-O}$ bond to form the carbocation is not present. The mechanism of the nucleophilic substitution reaction would thus be a concerted one (i.e. $\mathrm{S_N}2$) as the $\ce{C-O}$ bond cannot break on its own but only breaks as a nucleophile forms a bond to the carbon. It thus follows that the nucleophile attacks from the opposite side. Now, note that the t-butyl group is rather bulky and thus, it would sterically hinder the approach of the nucleophile. However, the ethyl group is less sterically-hindering. Thus, the nucleophile would attack the carbon of the ethyl group, forming iodoethane and tert-butanol.
Without the protonation, the reaction can only proceed in a concerted manner.

I hope I got what you meant right? @ron

Sn1 and sn2 are extreme cases. In reality there will always be a mixture of these mechanisms. And if you have competing mechanisms, they will still both happen, one to a lesser degree, because of energetics of the potential energy surface.

@Martin-マーチン Sorry I did not intend to mean that they react only via a particular pathway. Notice that I used the phrasing "favour the [mechanism]".

This answer is based on what I have learnt in class $12$ of Indian curriculum (which has not been updated for over several decades, so might be slightly erroneous).
There are a few (non-exhaustive) "rules" for cleavage of ethers by $\ce{HI}$, in general reaction conditions -

If both sides have primary or secondary alkyl carbons (dimethyl ether, ethyl isopropyl ether), favoured pathway is $\mathrm{S_N2}$.
If any one side has a tertiary alkyl group (di tert-butyl ether, ethyl tert-butyl ether), favoured pathway is $\mathrm{S_N1}$ in aqueous medium. This is due to high stability of carbocation + solvation by water molecules.
If any one side has a benzyl group, favoured pathway is $\mathrm{S_N1}$ in aqueous medium. (Same reason as in 2)
With anhydrous $\ce{HI}$,  favoured pathway is $\mathrm{S_N2}$ (I think it is the only pathway) regardless of substrate, as it is much less polar than $\ce{H2O}$.

Source - NCERT for class 12, part 2 and Organic Chemistry for JEE Advanced, part 2 by Dr. KS Verma (Cengage publications)

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


text:
What is the differences between partial pressure and vapour pressure?
Was looking at Henry's law and Raoult's law
constants and there seemto be lots of
equations involved.
Henry's law involves partial pressure and the latter involves the vapor pressure.
Wondering what the difference is?

In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature.

What does this mean? For example, if we have a mixture of gases $A$, $B$ and $C$ in an isolated room, then, according to Dalton's law, the pressure exerted by the gases will be the sum of their partial pressures :
$$P = p_A + p_B + p_C$$
where $p_A$, $p_B$ and $p_C$ are the partial pressures of each gas. Also, if we have a moles of $A$, b moles of $B$ and c moles of $C$, we can express the partial pressure of each gas as below:
$$p_A = \frac{a}{a+b+c} P$$
$$p_B = \frac{b}{a+b+c} P$$
$$p_B = \frac{c}{a+b+c} P$$

Vapor pressure or equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

As the definition states, if we took an equilibrium as $\ce{A(l) <=> A(g)}$, where $A(l)$ and $A(g)$ are the liquid and the vapor state of the compound $A$ . The vapor pressure of $A$ is the pressure exerted by the gaseous state ($A(g)$) over the liquid state ($A(l)$).
Now, if we have a mixture of $A$, $B$ and $C$, we will have the equilibrium between the liquid and gaseous state : $\ce{(A, B, C)(l) <=> (A, B, C)(g)}$ and the vapor pressure $P_v$ (I noted $P_v$ so there won't be any confusion between it and the total pressure $P$ I mentioned before).
According to Raoult's law, the vapor pressure $P_v$ is the sum of the products between the vapor pressures of each compound taken separately and the molar fraction of the compound in the mixture (i.e. in the solution) :
$$P_v = x_A p_{v^A} + x_B p_{v^B} + x_C p_{v^C}$$
This may look just like Dalton's law, but there are different pressures involved.
The partial pressure of a gas is the pressure exerted by a gas in the volume occupied by a mixture of gases, while the vapor pressure of a gas is the pressure exerted by a gas over its condensed phase.
Although :

The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial pressure.

The vapor pressure still refers to the pressure of a gas over its liquid (or solid) phase in a phase transition (condensed phase$\ce{<=>}$non-condensed phase), while the notion of partial pressure can be extended to a mixture of gases only.
All the definitions are linked to Wikipedia, and they can really cause confusion throughout chemistry learners..

Partial pressure is the pressure exerted by the individual gas in the mixture of different gases at the same temperature.
Vapour pressure is the pressure exerted by the condensed(cooling) gas by lowering the temperature.

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I don't understand your definition of vapor pressure.  It can either mean the total pressure of a mixed gas above a mixture, or more often, the equilibrium pressure of the vapor of a pure compound in equilibrium in with a pure liquid of the same compound.


text:
Sigma bond strengths S-S, S-P, P-P
I want to know how the following sigma bond strength values are calculated.
Like S-S = 1.0, S-P=1.71 And P-P=3.0.
How these 1.0, 1.71, 3.0 came?

Are there supposed to be units on any of these numbers?

Just to clarify, these values are relative bond strengths, not absolute bond strength values, which would have units of energy, i.e. Joules, kilocalories, or electron volts. Another detail missed in this question is the fact that these values refer to the bond strength of these types of orbitals directly overlapping along the same axis, as shown in the diagram below.

You'll notice that these relative bond strengths corresponds the amount of overlap occurring in the orbitals. This gives you a qualitative description, but to actually calculate the relative overlaps we can use quantum mechanics. Remember that these orbitals are really just the probability distributions electrons and can be solved analytically by squaring the wave function. We have a mathematical description of both S and P orbitals. By integrating over the volume in which they overlap, we can calculate the relative overlap between S-S, S-P, P-P, and any other orbital overlaps we might be interested in.

Does Sigma bond not imply that the overlapping is taking place along the same axis?

Oh, you are correct. I totally missed that in the question.


text:
Why do some dentists claim mercury "leaks out" of amalgam fillings?
From a chemistry point of view, if dental amalgam is a compound of mercury and other metals, and compounds are elements which are chemically bonded, how would it be possible for mercury to "leak out" of fillings if no further chemical reaction would be taking place?

But the condition "no further chemical reaction" is not fulfilled in mouth.

You are labouring under a misconception about the exact meaning of the word "compound".
Normally when we describe, for example, ethanol as a compound of hydrogen, oxygen and carbon we don't mean that the component elements can "leak" out of it without a chemical reaction taking place.
Mercury amalgams are casually described as "compounds" but not in the same sense as ethanol. Amalgams are mercury alloys with varying compositions but they don't, usually, consist of strongly bonded molecules like ethanol with strong chemical bonds preventing the molecules falling apart. They are closer to mixtures than what we normally think of as compounds. Having said that they are not just mixtures as they can have very different bulk properties than their components.
The issue with mercury amalgams used in fillings is that they are mostly, but not completely, inert. Some mercury will evaporate from them and some chemical and physical processes can increase the amount that evaporates. But good estimates suggest only micrograms/day which is less than typical workplace exposure limits so probably not a source of legitimate worry. The dentists who still use mercury fillings are likely exposed to more than this and this is a better reason to limit their use. The subject is controversial with a lot of disagreement among studies. This is well summarised by the Wikipedia article on the controversy.

'Leak' is an imprecise term here. Amalgam itself could (and probably does) wear away with attrition, and the 'powder' would be ingested. Certainly, fillings have worked loose and been swallowed. The effect of stomach acid on amalgam is probably not negligible.

Mercury vapors are released. It has long since been proven that people with mercury in their teeth constantly have small portions of mercury vapor evaporating from the tooth. There's many videos of this taking place. It's not a mystery.
https://www.youtube.com/watch?v=hqIdGwAMxxs

Why do some dentists claim mercury “leaks out” of amalgam fillings?

Because it does.
http://www.fasebj.org/content/4/14/3256.abstract

While this answers the title question, it does nothing to address the questions regarding the chemical principles in the body of the question.


text:
What does "EM grade" mean
I'm looking at purchase orders and I see purchases for chemicals like the following.
Glutaraldehyde, EM Grade, 25%
What does the "EM Grade" mean in this text? It says that it's recommended for histological or immunological techniques, but how?

It is highly purified, typically distilled into glass ampoules and sealed under an inert atmosphere.  Suitable for use in *E*lectron *M*icroscopy.

EM grade is a label for chemicals used in electron microscopy.


text:
How well (or poorly) does natural leather insulate?
What is the general thermal conductivity of animal skin leather?
I understand there are many different animal's skins (cow, sheep, buffalo, deer, pig), but in general, does natural leather insulate well (compared to, for scale, an esky for food insulation), or not?

Interestingley enough, there are papers entirely devoted to the subject of measuring the properties of living skin. The paper linked gives a value of $7 \times 10^{-4}~\mathrm{\frac{cal}{cm\, s\, °C}}$ for human skin, or in everyday units used nowadays: $0.29 ~\mathrm{\frac{W}{m\, K}}$. This is only for the dermis, the part of skin that then gets transformed into leather.
Now, this website has a comprehensive list of many different thermal conductivity values, and dry leather clocks in at $0.14 ~\mathrm{\frac{W}{m\, K}}$, which is about 2 times less the conductivity than that of living skin. What a difference the tanning process can make!
Given the similarity  observed in the review paper between man, beef and porcine tissue, I think it is fair to say that the value for leather will not change greatly across species.

I feel that my answer is somehow insufficient, since it's gotten so many upvotes but hasn't been accepted. Is there something that's missing, @foregon?

sorry about that, your answer was fabulous I think it was just me not finishing looking into the matter and I tend to wait until I definitively find the answer until I can ark something answered (if it is not clear) but upon  cursory look of yours again it is indeed sufficient, thank you for the superb research and links!!

I never worked in that field and thus had to google it myself ;)
Apart from the general observation that leather is indeed a good heat insulator, there seems to be an article in the Journal of the AQEIC (Asociación Química Española de la Industria del Cuero) addressing the question quantitatively.


text:
Anthocyanins as pH-indicators: stop the titration at red, purple or blue?
If you titrate some hydrogen bromide solution (0.1 M) to some sodium hydroxide in order to determine the exact concentration of hydrogen bromide (0.1 M), and you use anthocyanins as a pH indicator. This anthocyanin will be red if you have a low pH value and purple if it is slightly acidic until it is slightly basic. It will be blue if the solution is basic.
Do you have to titrate until the color of the solution is blue or purple?
Equilibrium from red to purple: $$\ce{H2X+ <=> HX}$$
$K_a = 1.047*10^{-3}$
Equilibrium from purple to blue: $$\ce{HX <=>X^{−}}$$
$K_a = 3.16 * 10^{-8}$

If you titrate some hydrogen bromide solution to some sodium hydroxide [...]

It's probably a bad idea to start with the anthocyane in strongly alkaline solution. Like other pyrylium compounds, flavylium cations do not tolerate (strongly) akaline conditions very well.
The nucleophilic opening of the pyrylium ring to a chalcone is irreversible under the conditions of the titration.

First of all, I don't understand why you would titrate the acid of unknown concentration to your base of known concentration, I've learned it the other way around.
Now, following my proposal, we have our solution of $\ce{HBr}$ waiting to be titrated with $\ce{NaOH}$. Since sodium hydroxide is a fairly strong base, we can assume total neutralization:
$$ \ce{HBr + NaOH -> NaBr + H2O}$$
Operating under this assumption, equilibrium is reached at $\text{pH}=7$, so we have to know when that happens. How do we find out where the indicator changes colour?
Well, we calculate the $\mathrm{pK_a}$ values of the indicator:
$$ \mathrm{pK_a} = -\log \mathrm{K_a}$$
The second equilibrium (from purple to blue) switches at $\text{pH} = 7.5$, which is about where we need it to. So you need to titrate until you get the blue colour.
Note: It is not always easy to see these colour transitions, I myself have overtitrated quite a few solutions because the colour shift was too subtle (and my eyes too untrained).

How did you calculate pH at which the second equilibrium swithes did you use Henderson–Hasselbalch equation?

The pKa is 7.5, so at pH=7.5 the switch happens.

I'm sorry I didn't catch it, I thought pKa indicates how much an acid is dissociate not at which pH a specie dissociate...

It's the pKa of the indicator. The "a" merely refers to its property as an acid.

mmm I know that is the pKa of the indicator for this I'm not sure how you can correlate it with a system composed from this with additional $H^{+}$ ions from another acid... I will meditate about it! Thanks however!


text:
How to demonstrate the safety of wearing leather shoes in the chemistry lab?
Quite often (from my own experience anyway), one of the safety rules in a school (high school or university) laboratory is that the students must wear leather shoes.
I like to demonstrate some of the safety rules, so my question is, what is an effective and safe experiment to demonstrate the safety of wearing leather shoes while experimenting in the chemical lab?
*This is in comparison to say sneakers or canvas shoes

Tangential I know, but a rule like this may not be up-holdable in the case of, e.g. vegan lab users.

@Aesin interesting point, I must admit, I have never encountered that

Most institutions I know of (in the US) would have trouble enforcing the leather part, because of 1) price and 2) moral/ethical objections by certain groups of the population.

@BenNorris in some places, such as here in Australia, it is enforced rules.  I am just interested in the scientific experiment above.

Interestingly, the European standard EN 13832-2:2006 for rating footwear based on chemical resistance specifically excludes footwear with leather outsoles -- presumably because leather's actual resistance to materials varies significantly.

The simplest thing to do is to sacrifice two pairs of shoes, one with leather uppers and the other with canvas/cloth/whatever. Put something inside the shoes to represent "feet". Do something nasty to each pair, like pour concentrated acid over them (using secondary containment to avoid spills and splashes!). Then, extract your indicator and compare (assuming the different in damage to the shoes was not enough). You can also try lighting them on fire (though canvas shoes probably have fire-retardant compounds added), or seeing how long it takes water to soak through (leather will win here).
And, whatever you do, record it to video so that next time you only need to play the video and not buy more shoes. If this is for your students, I would make sure you appear in the video to increase "authenticity".

love it! A very simple but extremely effective method!  Now to choose what kind of nastiness I will do to them...

Use directly shoes could be not so practical and reproducible. I will try to find first to figure out which experiment could be more convincing.  So why don't take many little piece of leather, cotton and synthetic fiber and make some general test (maybe useful even for clothes). These are what is passing through my mind.
Wettability and permeability
One of the  most important thing is that substance drop over the shoes don't get in touch with the skin. So you could measure the contact angle of polar and non polar liquids over different "shoes textile" and find which has the lower one.
For permeability there are also some simple instrument you can buy.
Thermal conduction and inflammability
Even if this is more difficult to perform you could try to make your own system with a little hot plate and some thermocouples to see which materials do not conduct heat or is less flammable. One simple test could be the Clothes iron test (this in fact doesn't exist) you can palace a piece of different shoes textile and place it beneath the clothes iron measure the time at which it burns.
Acid and bases resistance
Why don't you put a little piece of textile over a piece of paper with an acid-base indicator and then put a known amount of acid or base over it if you do it in a glass table you can monitor when the acid reach the other side watching when the paper change color.
Penetrability
Pieces of broken glass could cut the textile so you could perform some test for determinate which textile is more hard to penetrate.


text:
Which mechanism do ethers follow when cleaved by hydrogen halides?
Ethers can be cleaved under acidic conditions by treatment with hydrogen halides (HX, where X = Cl/Br/I). Some of these reactions proceed by SN1 mechanisms and some by SN2 mechanisms. How can I tell which mechanism is being followed?

The first step of this reaction is always a proton transfer:

Then, the substitution follows either the SN2 or SN1 pathway depending on whether R is methyl, primary, secondary, or tertiary.
Methyl and primary ethers undergo acidic cleavage by the SN2 mechanism:

Tertiary ethers undergo cleavage by the SN1 mechanism:

Secondary ethers probably undergo cleavage by both mechanisms competing with each other.
Note that unless R' is an aryl group, the alcohol R'OH that forms will also be converted into a halide by a similar mechanism.
Read more at any of the sites in this Google search.

I suppose you were to say: If R is primary... or secondary... or tertiary... . I don't think you can extend this notation from alcohols to ethers, it is ambiguous which carbon you are referring to.

Hey, my teacher told that no matter what ether is it, it will always go through SN2 if the HX used is anhydrous. Is it true? Does anhydrous HX guarantee SN2? Thanks

@An_Elephant Did you get an answer somewhere?

@Silica19 Yes, the SN2 will be major.

@An_Elephant Will it be SN2 even for a tertiary alkyl group? If yes, can you please share a source? Thanks.

@Silica19 According to my teacher the major will be SN2. Unfortunately, I do not know any online source. Sorry :(


text:
Distilled water to clean greasy stains: Why does that work?
I noticed that distilled water works well to clean away fingerprints on touchscreens or oil stains on a ceramic field after frying an egg.
But water and oil/fat are not supposed to mix, so why does that even work?

As far as I understand, fingerprints are mostly made from sweat and substances dissolved in sweat, so I can see them being removed with water. Not sure about the oil stains though. Are you sure it's not just a sort of confirmation bias?

There may be two different factors at play here.
First, it isn't necessarily the water that is doing most of the cleaning. I presume the water is used to wet a cloth and the cloth then wipes the surface. So the cloth may then physically remove surface contaminants. The role of the water may be just to improve the cloth's surface contact hence improving its ability to physically remove the contaminants.
Secondly, distilled water might do a better job than tap water as it has fewer dissolved solids which would leave other stains on the surface when the water evaporates.
However, the only way to be sure is to do a comparative experiment using different cleaning fluids. You might try tap water, distilled water and distilled water with a surfactant (e.g. a liquid soap). You might find that the best result uses the surfactant but that pure distilled water does a less good, but good enough job. Science isn't about single isolated experiments: we usually need to compare alternatives to know what is best.

In fact distilled water can be acid when expose to atmosphere because $CO_2$ dissolve into forming  carbonic acid:
$$\ce{H_2O(l) + CO2(g) <=> H2CO3}$$
 Although the $pH$ usually in not very low (normally it is around 5) this can help to removing the  greasy stains.
I think also the diffusion of the salts present in greasy stain in water can improve the water cleaning capabilities.
Finally one important thing is that distilled water don't leave residue in the surface. Normal water or other detergents usually contain salt or other substance that can leave halos in the surface. So the final result could appear better even if is not the best substance for solubilize fats.

The glass is almost certainly hydrophilic. (Does a drop of water on flat, i.e horizontal, screen bead up or spread out?). Thus the oil in the finger print is "lifted" from the surface which is wetted with water, and absorbed onto the cloth used to wipe the liquid away.
A dilute solution of white vinegar would probably work better since the vinegar would help "cut" the grease, yet leave no residue.
If you use a soap solution on glass then you have to rinse to remove the soap so that there isn't any residue.

The increased surface tension in tap water affects the ability to penetrate and lift stains, dirt, and grime easily.
The lack of mineral content in distilled water allows it to be more efficient in a chemical and mechanical fashion....
Minus detergent, I would agree with GM's carbonic acid theory - you are bound to be breathing on what you clean, and I suspect that distilled water would bond with carbon dioxide faster and more effectively than tap water - which would likely contain minerals and organic matter that would normally neutralize the acid as is forms.


text:
Shouldn't the pH at the equivalence point always be 7?
I learned in class that the equivalence point in an acid-base titration is reached when the solution contains an equal amount of substance of $\ce{OH-}$ and $\ce{H+}$ ions. However, in a weak acid and strong alkali titration, the pH at the equivalence point (when I added the right amount of acid and alkali to make it a neutral solution) is 9. Also, in the titration of a strong acid and weak alkali, the pH of the solution when I have added the correct amounts is 5.
My question:
If I add 1 mole of ethanoic acid to 1 mole of sodium hydroxide, I should get a neutral solution and the pH should be 7. However, why is the pH 9? This does not make sense to me as a pH 9 suggest that the concentration of $\ce{OH-}$ is higher than the concentration of $\ce{H+}$. But how is that, I reacted the correct amounts?

when you add 1 mole of sodium hydroxyde to 1 mole of ethanoic (=acetic) acid, you will get 1 mole of sodium acetate (try to write down the neutralization reaction). Acetate is the conjugated base of a weak acid and is therefore strong enough to significantly alter the pH: that's why the final pH is 9.

@mannaia: could you please elaborate "strong enough to significantly alter pH"

Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms".

Two kinds of reactions have to be considered:

Neutralisation of the weak acid with the strong base
Secondary reactions from the products

The neutralisation is fairly straightforward:
$$\ce{HA + NaOH -> NaA + H2O} $$
However, and here comes number two from the list: It doesn't just stop there because we still have unreacted acid in the solution and lots of water. So what will happen? The conjugated base of the acid can also react (as a base):
$$\ce{A- + H2O <=> OH- + HA}\; ,$$
thus raising the pH of the solution. This depends on the equilibrium constant $\mathrm{K_b}$ of the conjugated base.
So although you have reacted exactly equal amounts of a (strong) base with the (weak) acid, you also have to consider the (weak) conjugate base of the acid, which also swims around in your titration soup.

What do you mean with "we still have unreacted acid in the solution" ?

It's a phrase used to make you aware of all the other variables that might have something to do with how the solution reacts.

I think it's rather confusing. I would simply say that, at the equivalent point, the acid has been fully neutralised by the base (no unreacted acid present anymore)

@tschoppi: in case of strong acid and strong alkali titration, aren't there conjugate bases and conjugate acids... won't there be a secondary reaction?Also, in the case of strong acid and weak base, will a weak conjugated acid form?

@mannaia The term "neutralized" is slightly ambiguous in this context, as you don't necessarily have pH=7.

@Eliza For most of the cases, absolutely. But always be aware of the kinds of reactions that can take place in your medium, it might even be a (unexpected) precipitation of a salt...

@tschoppi you first used the term neutralisation in your answer ... even by giving the neutralisation reaction.

Let's consider your example where you are titrating acetic acid with sodium hydroxide. You are correct in saying that when equal amount of substance of acid and base is added, the acid will be completely neutralized. The reaction is:
$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$
Hence, at the equivalence point, the conical flask will only contain $\ce{CH3COONa}$ (which is a salt) and $\ce{H2O}$. However, acids and bases are not the only substances which have a pH different from 7. Some salts are also able to be acidic or basic. In the case of $\ce{CH3COONa}$, it is a basic salt. The reason why is because $\ce{CH3COONa}$ actually consists of $\ce{CH3COO-}$ (a base) and $\ce{Na+}$ (an acid). Since $\ce{CH3COO-}$ is the conjugate base of a weak acid, it is strong enough to be able to react with water to produce $\ce{OH-}$ ions:
$$\ce{CH3COO- + H2O <=> CH3COOH  + OH-}$$
Meanwhile, since $\ce{Na+}$ is the conjugate acid of a strong base, it won't be strong enough to react with water.
Therefore, at the equivalence point, even though there is no acid or base present, $\ce{[OH-] > [H3O+]}$, hence the pH will be $>7$.

Why is the conjugate base/acid of a strong acid/base weak?

Lets consider HCl which is a strong acid. When HCl ionises in water, their is the following equilibrium: $\ce{HCl + H2O <=> Cl- + H3O+}$. Since HCl is a strong acid, the equilibrium lies heavily to the right. Therefore, it follows that the equilibrium for the reverse reaction (where Cl- is acting as a base) must lie heavily to the left. Hence Cl- is a weak base. This can be applied to any conjugate acid/base pair, so we can get the general statement the conjugate base of a strong acid is weak and vice versa.

My answer to your question in the headline is: Not always.

Neutralisation and neutral $\mathrm{pH}$ are often different things.
Neutral $\mathrm{pH}$ is when $\ce{[H3O+]} = \ce{[OH-]} = 7$
Neutralisation of a weak acid by a strong base means that you completely have converted the acid to its corresponding base. The $\mathrm{pH}$ at the equivalence point will depend on the acid used.
Neutralisation of a weak base by a strong acid means that you completely have converted the base to its corresponding acid. The pH at the equivalence point will depend on the base used.

I will give you three examples:

Suppose you titrate a strong base with a strong acid:
$$\ce{NaOH + HCl = NaCl + H2O}$$
The equivalence point is reached when all $\ce{NaOH}$ has been converted to sodium chloride. The result, at the equivalence point, will be the same as dissolving sodium chloride in water at the same concentration as you have at the equivalence point. We expect the pH of the solution at the equivalence point to be neutral, i.e. $\mathrm{pH} = 7$.
Why do we expect that? Well, neither $\ce{Na+}$  nor $\ce{Cl-}$ will react with water in any protolysis reaction. [We do not consider any ionic strength effects on the $\mathrm{pH}$ in this reasoning.]
Suppose you titrate a weak base with a strong acid:
$$\ce{NH3 + HCl = NH4+ + Cl-}$$
The equivalence point is reached when all ammonia has been converted to ammonium ions.
The result, at the equivalence point, will be the same as dissolving ammonium chloride in water at the same concentration as you have at the equivalence point.
Now, suppose the concentration of the ammonium and chloride ions were $\pu{0.01 M}$ at the equivalence point. The $\mathrm{pH}$ would then be about $5.6$.
Suppose you titrate a weak acid with a strong base:
$$\ce{NH4+  + OH- = NH3 + H2O}$$
The equivalence point is reached when all ammonium ions have been converted to ammonia. The result, at the equivalence point, will be the same as dissolving ammonia in water at the same concentration as you have at the equivalence point.
Now, suppose the concentration of the ammonia was $\pu{0.01 M}$ at the equivalence point. The $\mathrm{pH}$ would then be about $10.5$.

From this we learn, neutralisation can give $\mathrm{pH} = 7,$ but not always.

You have confused neutrality point, inflection point and equivalence point. The former is where there are equal numbers of protonated and deprotonated solvent species ($\mathrm{pH}~7$ in water at $\pu{25 ^\circ C}$). The inflection point is where the second derivative of the titration curve (amount of base added plotted against $\mathrm{pH}$) is a maximum (maximum buffering) and the third the point where the amount of analyte and titrant are equivalent. If you take acetic acid and start adding base to it even a small increment of added base will produce a large change in $\mathrm{pH}$. The base isn't reacting with the $\ce{HOAc}$, it is reacting with the solvent. The happens until the inflection point $\mathrm{pH}~4.76$ for acetic acid, is approached in which region the base is reacting with the $\ce{HOAc}$:
$$\ce{HOAc + OH- -> OAc- + H2O}$$
and a small change in $\mathrm{pH}$ requires much more base. Eventually, (about $\mathrm{pH}~4.76,$ that is two $\mathrm{pH}$ units above the inflection point) nearly all the $\ce{HOAc}$ is reacted and additional base just goes to deprotonate water, increasing the $\ce{OH-}$ ion concentration and $\mathrm{pH}$ at a rate much faster than near the inflection point. Thus the fact that you wind up at a $\mathrm{pH}$ near $9$ is simply a consequence of the fact that you are well past the inflection point of your acid which is, because the inflection point is at $4.76$, considered a weak acid. Once all the acid has been deprotonated, additional additions of base go to deprotonate water and the pH rises.
The pH at the equivalence point is the pH at which the solution is neutral. Finding this pH requires that we compute the charge of each species in the solution and totalling them up as a function of pH. We then seek the pH value that zeroes the sum. The species are $\ce{Na+}$, $\ce{OH-}$, $\ce{AcOH}$ and $\ce{AcO-}$. The charge on a liter of water is $Q_w(pH) = 10^{-pH} - 10^{(pH - pK_w)}$ which is the sum of the hydronium and hydroxyl ion charges. For the acetate its a bit trickier. We start with the Henderson Hasselbach equation writing it in the form $r_j = 10^{(pH - pK_j)}$ This is the ratio of the concentration (activity, really) of species which have lost j protons to the that of the species which has lost $j-1$. In the case of acetic acid there is only one proton to lose and so there is only one $r$ but in the case of polyprotic acids there is, of course, 1 $r$ for each proton. Now it shouldn't take you long to figure out that if the ratios are $r_1, r_2, r_3...$ the fraction of undissociated acid molecules is $f_0 = 1/(1 + r_1 + r_1r_2 + r_1r_2r_3)$ As r_1 is the ratio of singly deprotonated to undissociated its clear the the fraction of singly deprotonated is $f1 = r_1f_0$. At this point we have all we need to calculated the charge. Assuming 1 mole $\ce{NaOH}$ and 1 mole $\ce{AcOH}$ in 1 L of water the total charge would be $+1 + 1*Qw(pH) -1*f_1(pH)$ So the question now turns to how we find the value of pH which brings this expression to 0. Clearly the fastest way to a solution is to have and Excel spreadsheet compute it and either grope for the answer (pH 9 gives + $4.75·10^-5$ and pH 9.5 gives $4.75·10^-5$ so clearly the answer lies between those 2 pH values. This suggests a bisecting root finder as a path to a solution and that works very well. Excel has the Solver which does the iterations for you and takes you directly to the solution $pH = 9.38$. Note that the solution for 1 mole each in 10 liters of solution is $pH = 9.38$ which makes it clear that the water is playing a significant role here. Not all the $\ce{OH-}$ are going to deprotonate acid. Many are going to shift the waters charge balance.

"are equal numbers of solvent and protonated solvent species" That cannot be right, i.e. it would mean that that in pure water half of all water molecules would be dissociated, which doesn't really fit into $K_\mathrm{w}(\pu{25 ^\circ C}) = 10^{-14} = c[\ce{OH- (aq)}]\cdot c[\ce{H3O+ (aq)}].$ Also, I feel like that the part you are omitting ("but I won't go into that here."), would actually answer the question of why the pH is different from 7 at the equivalence point.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

A more productive comment would be "you left out a key word". I don't need to be schooled in the fact that this leads to an absurd situation, which, of course, it does.  I can easily add in the how - to - determine - pH part if anyone wants me to.