text:
Very slow chemical clock reaction
I'd like to preface this with a declaration of my understanding that this is not necessarily possible.
I was wondering if anyone knew of a chemical clock reaction where the period before the first change (the induction period?) is very long (between 3 hours and a day) and the change is à la Landolt or Old Nassau reactions (i.e they cross a concentration threshold to create sudden color change).
Bonus Question:
Is there any remote possibility that a reaction like this can happen in pigments or dyes once they are absorbed by a cloth?

I definitely remember about reaction that took about 2 hours and only visible changes started to appear at the end of 2-nd hour, about 10 minutes to end of time period. Don't remember the details, but reaction was autocatalytic, so at the end it runs like crazy and very slow at the beginning. The idea is quite simple, so reactions of this type should be common. Problem: usually reactions used are exotermic, meaning that reaction mixture for autocatalytic process can easily end on your ceiling. Unfortunately, I cannot provide you with specific example at the moment.

I assume you are aware of the common clock-reactions, such as the iodine clock, etc. By varying concentration of reagents and pH of the solution, you could run several repetitions to generate a trend-line, and from that extrapolate the estimated pH and reagent concentration to provide you with the time delay you are looking for. The iodine clock is a two-step reaction as outlined below:
In the first, slow reaction, iodine is produced:
$$\ce{H2O2 + 2I- + 2H+ -> I2 + 2H2O}$$
In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate:
$$\ce{2S2O3^2- + I2 -> S4O6^2- + 2I-}$$
By decreasing pH in reaction 1 and increasing the thiosulfate level in reaction 2, you should be able to lengthen the ‘clock time’. Let me know how things turn out!

I am wondering if really long timed clock reactions would be affected by ambient oxygen dissolving into the solution and oxidizing some species.


text:
Modeling perturbations of a quantum mechanical system
I'm wondering how to properly perturb a quantum mechanical system.
I'm looking for a way to express the return to equilibrium of such a perturbed state, where the perturbation is in the Hamiltonian. The unperturbed Hamiltonian is $H_{\text{u}}$ and the perturbed Hamiltonian is $H_{\text{p}}$.
I can compute the eigensystem for both $H$ and thus find the relevant energy spectra and eigenfunctions, and can add the time-dependence factor $e^{-iEt}$, but I do not know how to generate the final system from its component, isolated systems. I would guess an inner product to be of use here, but...
What theory/mathematical methods do I need to learn to perform such simulations?
Something I want to model is a molecule in a polarized state, that is, with an asymmetric electron distribution.

I don't understand exactly what is your question:
If you start in an eigenstate of $H_p$, say $\varphi_j$ and then your Hamiltonian changes like $H(t) = H_u + f(t) (H_p - H_u)$ where $f(0) = 1$ and $f(T) = 0$ -$T$ being the final time of the perturbation- then in order to follow the evolution of $\psi(t)$ all you need to do is to solve the time dependent Schroedinger equation
$$i\hbar \partial \psi(t) / \partial t = H(t) \psi(t)$$
where $\psi(0) = \varphi_j$. There are many numerical procedures to do so. For instance if you are interested, as it seems, in following the dynamics in the basis of the eigenstates of $H_u$, then you expand $\psi(t) = \sum_k a_k(t)
\phi_k$, where $\phi_k$ are eigenstates of $H_u$, and you solve a set of ordinary differential equations.
If $f(t)$ is very fast ($T << (\Delta E)^{-1}$, where $\Delta E$ is a characteristic energy of the system, say the energy difference between adjacent eigenstates) then you can use the sudden approximation; if $f(t)$ is very slow you can use the adiabatic approximation, if $H_p - H_u$ is small you can use first order perturbation theory, etc.
In any case it is important to realise that in Quantum Mechanics there is no relaxation unless you couple your initial state to a continuum and there is no return to equilibrium unless you do the same (couple to continuum) and work with the proper formalism that allows you to define the (thermal) equilibrium, which are the density matrices, unless by equilibrium you are referring to the initial state. Again, in ordinary Quantum Mechanics $\psi(t)$ will oscillate and will not relax to the initial state $\varphi_j$ (or $\phi_0$); at most, it will return to the initial state as part of an oscillation, called revival.

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Why does cesium have a positive electron affinity?
When an electron is added to cesium, why is energy released?

All elements have a positive (first) electron affinity, except noble gases, for which it has not been measured conclusively, but might be either slightly positive or slightly negative.
You can more clearly see the reason why electron affinity is positive if you think of the opposite process: detaching an electron from the negative ion X– to form neutral atom X. Because the anion X– is stable, it takes energy to take away the electron.

"" Because the anion X– is stable, it takes energy to take away the electron."" This is a chemical version of a syllogism.

I agree with @Georg. Why is the X- ion more stable?

@Ben the anion stable in the sense that the electron fills a bound state… an energy level whose energy is negative

""that the electron fills a bound state"" why is it "bound"? You go on with syllogisms.

I guess
Isaac’s main problem arises from the usual chemical
experience dealing about ions in water. Then of
course Cs- is not a real reaction product.
Electron affinities deal about reactions of atoms
and electrons in vacuum!
The cesium atom "well" to plunge in is not very deep,
but better than nothing and there is no better
alternative for the electron  in a vacuum.
The true reason is the atomic orbital energy for
an additional electron on cesium, not easy to calculate.

Atoms want to become more stable, which is normally characterised by energy release as they achieve a more stable, lower energy. If cesium accepting that electron makes it more stable, then energy is released. As F'x mentioned in his answer though, not all electron affinities are exothermic. For instance, Beryllium's first electron affinity is positive, i.e. it is an endothermic process, because Beryllium's outermost occupied shell is an $s^2$ orbital. Hence, the additional electron would have to be added to the $p^6$ orbital instead, which is further away from the nucleus and therefore does not experience as much electrostatic attraction. Furthermore, the $s^1$ and $s^2$ are also "shielding" the nucleus, such that the outermost electron experiences even less electrostatic attraction.
For understanding why most atoms prefer are at a lower energy when more stable, I picture a person that has consumed a significant amount of caffiene and is hyperactive since he has a lot of energy. He becomes less hyperactive or "more stable" as he loses that energy. Remember, this is only an analogy and analogies are not perfect!
I would also like to point out that I, and I assume your question as well, am referring to to first electron affinities.

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What is the chemical structure / hybridization of sulfuric acid?
I was looking at the chemical structure of $\ce{H2SO4}$.

Intuitively, I would have expected this molecule to be square planar in accordance with $p^2d^2$ or $sp^2d$ hybridization, but instead it is shown to be in a tetrahedral geometry consistent with $sp^3$ hybridization. Why is this?
I guess that an alternative way of asking this question is: What is the hybridization of the sulfur atom in $\ce{H2SO4}$?

From reviewing my p-chem books, I believe that sd3 orbitals are certainly possible. Then again I last looked at molecular orbital theory 6-7 years ago.

Sorry, I should have been clearer. I don't rule out these hybridizations in an absolute sense, but in the precise case of sulfur. I don't know much transition metal chemistry, but I'm pretty sure sulfur doesn't hybridize sd3. So considering its sp3, can all the electrons fit?

You may find the wikipedia article on hypervalency interesting. Suffice to say that orbital hybridisation is not a cogent description of many molecules.

Good read. I'm asking the question primarily because I definitely don't see how the electrons can fit yet, this is how the molecule is displayed. So either my intuition is completely off or the interwebs are wrong.

Orbital hybridizations were invented to make valence bond theory place nice with known geometries. You should use VSEPR to determine geometry, and then extrapolate to hybridization, if you believe such a thing.

I don't see this as complex, The 4 Oxygen atoms are short 2 valence electrons, That's a total of -8 electrons in total, 2 Oxygen satisfy their shell with Sulfur and Hydrogen, and 2 Double bond 2 Sulfur, Since sulfur hypered and formed 6 single shells with 1 electron, But the hydrogen's can get pinched off against (aq)

It is much easier to explain it on a simpler example – the $\ce{O3}$ molecule. It has the structure of resonance hybrid of $\ce{O=O+-O-}$ and its mirror. And of course, the central atom has hybridization state $sp^2$. One bond here is a normal covalent bond, and another bond is dative: an electron pair is donated onto a vacant orbital of an $\ce{O}$ atom with all electrons paired. In the $\ce{H2SO4}$ molecule, two bonds are simple covalent ($\ce{S-OH}$ ones), and two are dative ($\ce{S-O}$ ones). A common concept of electron unpairing to my knowledge is proved broken by quantum chemistry calculations and spectral experiments for hypervalent compounds of $\ce{P}$ and $\ce{S}$.
A little more interesting as an example is the $\ce{XeF2}$ molecule, where three-atom four-electron bonds $\ce{F-Xe-F}$ are formed; that can be thought of as a reasonance hybrid of structure $\ce{F-Xe+\ F-}$ and its mirror $\ce{F- \ Xe+-F}$
Of course, this scheme is still far from perfect, as reality is much more complicated, but if you do not wish to take a course in quantum chemistry, it should be enough. However, I'll recommend to search for "MO LCAO model": it is quite simple and very useful. It is also often used in advanced chemistry books and articles.

Permeakra's explanation is spot on but I didn't really get a good understanding of the result until I visited the Wikipedia page on Sulfate.
The central hallmark of the question and the confusion was why a ten electron configuration would adopt a sp3 configuration more reflective of an octet. This was a source of confusion that goes back to Gilbert Lewis who originally proposed the below structure fitting of a sp3 hybridization (model 2):

Linus Pauling steps in to propose that two of the d orbitals play a role suggesting that there should be a sp3d2 hybridization. The issue then arises with the role of Pi bonds and how they fit in the structure with the proposal that the occupied p orbitals overlap with the empty d orbitals.
Ultimately from computational calculations it turns out that the sulfur is indeed charged and there is very little pi character. As concluded in How relevant are S=O and P=O Double Bonds for the Description of the Acid Molecules H2SO3, H2SO4, and H3PO4, respectively?, model 1 with S=O bonds is incorrect even though this is how we teach highschool and college chemistry.

I guess that an alternative way of asking this question is what is the
  hybridization of H2SO4?

A better way to ask this might be: "What is the hybridization of the sulfur atom in Hydrogen Sulfate?"
The sulfur atom has a bond with four other atoms in this molecule. Because that the electrons try to stay as far away from each other as possible according to Valence Shell Electron Pair Repulsion Theory (VSEPR), the other atoms will seperate as far as possible, which results in bonds that are 109.5 degrees apart from one another. If this were a square planar arrangement, the bonds would only be 90 degrees apart, which is not as satisfactory as the tetrahedral arrangement would provide.
According to my book, if the number of effective pairs is 4, the arrangement is tetrahedral, and therefore the hybridization required is $sp^3$
Source:
Zumdahl, Zumdahl, Chemistry: An Atoms First Approach
ISBN-13: 978-0840065322

This certainly answers the question, but why?

It has to do with molecular orbitals, and how they hold only a certain number of electrons, and they prefer to spread themselves out evenly because in the quantum world, electrons tend to prefer the lowest possible energy state. Once the sigma bond is occupied, which is the first orbital referred to as the s orbital, the electrons end up occupying the p orbitals, and when electrons are shared through these orbitals they are called pi bonds, they actually form a whole new orbital, a molecular orbital, which allows the sharing of electrons between two atoms. Let me refer you to this website to help visually show you what is happening.

This certainly answers the question but why?

Going to follow up even further. The SP3 configuration only accounts for 8 electrons. From my count, H2SO4 has 10 electrons. While I agree that there are only 4 effective pairs, where are the extra ones?

I ended up looking it up. Sulfate adopts a resonance structure.

In the end there is no way to predict the structure of a given molecule.  Yes, with experience one can guess correctly in many cases, but not in all of them.
The way all this works is to first find the structure and then second work out the bonding scheme that gives this structure.  In principle one can do an ab initio quantum mechanical calculation, but that's not really practical for molecules with a fair number of electrons such as $\ce{(NH3)2SO4}$.  So most often an approximation procedure is used.
There are two simple approximations.  One is molecular orbitals, the other is the LCAO (linear combination of atomic orbitals) approximation.  They describe the same thing but use different language.  The choice of which to use is up to the person wanting the answer.
As an example of all this, try predicting the bonding and structure of $\ce{H3PO4}$ without knowing the answer.

Sulfur has four $sp^3$ orbitals (four $\sigma$-bonds) with oxygen and two $\pi$-bonds ($d$ orbitals).
In the case of $\ce{SO4^{2-}}$ ion, also sulfur also has four $sp^3$ orbitals and four $\pi$-bonds with $d$ orbitals. The electron from hydrogen is moves to sulfur's d orbitals

A structure with four double bonds in the sulphate ion would give the sulphur atom a formal charge of -2. This is not expected to be more stable than a configuration with two double bonds and two single bonds, because in the latter case the same amount of formal charge is more distributed, and lies mostly on a more electronegative element (oxygen). However, due to mesomerism, the bonding is more accurately described as a combination of six similar canonical structures, creating four bonds of order 1.5, with all oxygen atoms having a formal charge of -0.5 and perfect tetrahedral symmetry.

Your d-orbitals are waa~y to close to the p-orbitals energeticly. Remember that 4s has a lower energy than 3d in the ground state of pretty much every single atom except hydrogen.


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What is 'autogenous pressure'?

Compound 1 was hydrothermally synthesized under autogenous pressure[1]

I've seen this term used a few times, but with no real explanation of what it means forthcoming. I suspect it refers to the synthesis being performed at equilibrium pressure inside a sealed container (the aforementioned synthesis is performed in an autoclave at 160 C). Is my intuition correct?
[1] Sun, et al., CrystEngComm, (2012), 14, 3982-3988

Well, my greek/latin intuition is confirmed by Merriam-Webster: http://www.merriam-webster.com/dictionary/autogenous But that doesn't mean chemists/engineers use it with the exact same meaning... Interesting.

@CHM - based off definition 3, I think we can conclude that the autoclave does not require a meal of blood to operate. I hope.

Could the term mean that pressure is not controlled? The autoclave is sealed and heated and the pressure comes from the heating???

@RichardTerrett HA! I didn't even read those definitions, went straight for the greek!

Autogenous pressure synthesis is a form of solvothermal synthesis (also known as “hydrothermal synthesis” if the solvent is water). In most publications I could find, the two terms (“autogenous pressure” and “—thermal synthesis”) are used together, making it clear what it is: you seal your solution in a stainless steel autoclave, and heat it up. The pressure thus generated is called autogenous pressure because you don't apply it externally, but it simply derives from the closed nature of the container and the $(\rho,P,T)$ relationship for the fluid (its equation of state).


text:
Oxidation number for nitrogen in ammonium sulfate
I'm trying to find the oxidation number of $ \ce N$ in $ \ce{(NH4)2SO4}$.
The answer is supposed to be $-3$ but I can't figure out the logic behind this.

Ammonium sulfate is an ionic compound, composed of two ammonium ions ($\ce{NH_{4}^{+}}$) and one sulfate ion ($\ce{SO_{4}^{2-}}$). As such, you can neglect the sulfate entirely from your consideration of the oxidation state of the nitrogen, as well as treating each ammonium ion as a separate entity. Given the total charge on the ammonium cation, it should be easy to see why the nitrogen has the specified oxidation state.

To easily determine the oxidation number on an individual atom in an ion it is necessary to write an equation.
First determine what the oxidation number of your known atoms are.
In this case, we know the oxidation number for H is +1.
Then set this value equal to the overall net charge of the ion.
In this case, it is +1.
Our equation now looks like this:  1(4) = 1, You use the multiplier of 4 to indicate that the ammonium ion has 4 hydrogen.
Next substitute a variable in the equation for the missing oxidation number:
+1(4) + N = +1
Solve for N.
+1(4) + N = +1
N + 4 = +1
N = +1 - 4
N = -3
Thus, the oxidation number for Nitrogen is -3.


text:
Why would the loss of H₂ to space mean that the Earth's geochemistry would be based on acid/base reactions?
I am reading a paper[1] that contains the line:

Most of the $\ce{H2}$ in the Earth's mantle escaped to space early in the Earth's history; consequently, the overwhelming majority of the abiotic geochemical reactions are based on acid/base chemistry, i.e., transfers of protons without electrons.

The connection between the loss of $\ce{H2}$ and acid/base chemistry is not clear to me.
Why would the loss of $\ce{H2}$ to space make acid/base chemistry the default?

Falkowski, P. G., T. Fenchel, and E. F. Delong. (2008) The microbial engines that drive Earth's biogeochemical cycles. Science 320:1034-1038.

I think what is happening here is this... we know we have abundant H2 gas today, so where did it come from if all the H2 gas disappeared in the early stages of the Earth? I'm going to over-simplify the answer to this question. The author is alluding to acid-base chemistry as the explanation for the existing H2 gas, so it seems.

I edited the LaTeX out of the title for indexing purposes, as the markup would end up showing on search engines, etc., otherwise.

The statement you cite contains a reference to “How Earth’s atmosphere evolved to an oxic state: A status report”, D. C. Catling and M. W. Claire, Earth and Planetary Science Letters 237 (2005) 1–20. This paper in turns give some more details on this assertion, which I'll summarize below.
It all turns around the question of how reducing the prebiotic atmosphere was. The paper I cited above contains this quote:

before life existed the atmosphere contained mainly $\ce{N2}$, with negligible $\ce{O2}$, probably more $\ce{CO2}$ than today, and only minor levels of reducing gases such as $\ce{H2}$.

So, in the absence of widely available gases such as oxidative $\ce{O2}$ or reducing $\ce{H2}$, oxygen was mainly present as $\ce{H2O}$ and $\ce{CO2}$. This means redox reactions are not big players in the overall chemical scheme of the periodic atmosphere, and acide-base reactions dominate (by default).


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Methods: Does failing at intermolecular properties mean the same for intramolecular properties?
Let us consider the following:
1.) Method X fails at describing intermolecular non-covalent interactions.
2.) Method X excels at describing molecules of type Y.
3.) Method X ___?___ at describing molecules of type Y_1 containing intramolecular non-covalent interactions.
Is there an existing general consensus on number 3? I haven't found an example in the literature yet which would highlight a clear answer to this.
EDIT for clarification:
Example 1: An non-covalent intermolecular interaction would be the interaction of two stacked benzene molecules.
Example 2: A molecule of type Y could be like an amino acid. This molecule has well defined, strong covalent bonds.
Example 3: An organic molecule with a non-covalent interaction occurring between two functional groups.
So... regarding #3, would we expect the method to do well at characterizing a molecule in Example 3, would we expect the method to fail (as it did in Example 1) or is there no general consensus on the matter which is essentially the same as saying, "Just have to try it and see!"?

I don't fully understand… what do you call "the same property as an intramolecular property"? would you please give an example?

I think the distinction that you are trying to draw is (as you have hinted) better expressed in terms of covalent and noncovalent interactions, rather than inter- versus intramolecular interactions. A technique that fails to correctly describe noncovalent interactions will fail whether or not these interactions are between two moieties on the same molecule, or on different molecules. The usefulness of a technique that fails noncovalent interactions is contingent on whether or not those interactions are significant to the particular properties of a molecule that you are investigating.

I find it odd that a study would cite one paper which determined a method being particularly good at describing one type of molecule (type could be 'well-defined' with 'strong covalent bonds), and then using that as justification to blindly characterize molecules with intramolecular non-covalent interactions without any sort of benchmarking. I just needed to make sure I wasn't thinking crazy.

@LordStryker - What molecular properties were they interested in?


text:
Reaction between potassium permanganate, alum, and sodium hydrogen sulfite
When I was a teenager, I played with a chemistry set. I tried adding potassium permanganate, alum, and sodium hydrogen sulphite (not sodium hydrogen sulphate) to water.
Upon adding the third ingredient, the water would rapidly turn clear, and a brown precipitate would form. After filtering this out, I allowed the water to evaporate, which left a white crystalline deposit. All three ingredients were definitely needed in order for the reaction to happen, which is why I remember it, since none of the other reactions I knew about needed three reactants. (This was a long time ago, so there's a very slight possibility I've mis-remembered the ingredients, but I'm 99% sure they're right.)
At the time, neither my teachers nor my parents could help me work out what kind of reaction it might be, so out of sheer curiosity, I thought I'd ask here. I'm no chemist, so simple explanations would be appreciated.

The fact that the solution turned clear is a pretty good guarantee that this was a redox reaction: the permanganate oxidised some other species, and in turn was reduced, probably to manganese dioxide, the brown precipitate you noticed. Unfortunately, neither of the other reactants you mention are reducing agents!
Do you think it's possible that the third reagent was actually sodium hydrogen sulphite, NaHSO$_3$? If so, this would make a lot more sense: the (hydrogen)sulfite ion would have been oxidised to (hydrogen)sulfate. This reaction requires an acidic medium, and the reason the alum (potassium aluminium sulfate) was necessary was to lower the pH.
These individual steps have the following equations:

Al$^{3+}$ + 4 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 4 H$^+$
MnO$_4^-$ + 4 H$^+$ + 3 $e^-$ $\rightarrow$ MnO$_2\downarrow$ + 2 H$_2$O
HSO$_3^-$ + H$_2$O $\rightarrow$ HSO$_4^-$ + 2 H$^+$ + 2 $e^-$

(note that the last two are half-equations -- they can’t occur on their own, but I’ve written them separately for clarity) giving an overall reaction of

Al$^{3+}$  + 2 MnO$_4^-$ + 3 HSO$_3^-$ + 3 H$_2$O $\rightarrow$ [Al(OH)$_4$]$^-$ + 2 MnO$_2\downarrow$ + 3 HSO$_4^-$

The white crystalline deposit in this case would just be alum again, more or less. You have some sodium in your solution now, from the sodium hydrogensulfite, so you'd get some sodium aluminium sulfate as well, but the only anion in your solution will still be (hydrogen)sulfate, and my guess is that with aluminium this will crystallise out as the sulfate.

Yes, I am actually pretty sure it was sodium hydrogen sulphite, not sulphate. My curiosity has been satisfied - many thanks!


text:
Product/(s) of the following reaction
I was wondering what the product will be in the following reaction:
$$\ce{H3COOCH3 ->[\ce{H+}]} \:?$$
I was thinking about $\ce{CH3COOH}$  being a product but I do not know the mechanism.

This is a specific case of the general reaction of acid-catalysed ester hydrolysis, which should be discussed exhaustively in any good organic textbook. This reaction is closely related to base-catalysed ester hydrolysis, aka saponification. I think any detailed answer to this question is quite redundant, however hopefully giving you the name of the reaction will point you in the right direction.

My school text book unfortunately does not provide the reaction mechanism. :(

here's a good overview: link

Wrong answer/comment. Without water no hydrolysis will happen.

If you're sure that the initial formula is $\ce{H3COOCH3}$ (and that's a big if), then I disagree with Richard's answer in comments. $\ce{H3COOCH3}$ is dimethyl peroxide. While peroxides are not the most stable of molecules, and decompose rapidly (often by oxidizing other compounds), they do not spontaneously transform in acid conditions.
If the formula were $\ce{H3C–COOCH3}$ (note the added C), then yes, it's be acid-catalysed ester hydrolysis.

Whoa, totally missed that. Well spotted. I was mainly looking at the acetic acid product.


text:
Why are alcohols with longer chains less polar?
In my self-study, I recently came across the following question:
"Choose the solute of each pair that would be more soluble in hexane ($\ce{C6H14}$). Explain your answer.
(a) $\ce{CH3(CH2)10OH}$ or $\ce{CH3(CH2)2OH}$ ..."
Undecanol is more soluble in hexane because it is apparently less polar than propanol. Further Internet searches revealed that alcohols decrease in polarity as the chain length increases (assuming a basis in an alkane; don't know if this rule generalizes which I guess is a sub-question) but it is still not clear why. None of the sources I found explained this. The way I imagine things, the carbon-hydrogen bonds should add a bunch of zero vectors to the the O–H vector that both molecules share, giving the same polarity in both cases. But this is apparently wrong.
So why is undecanol less polar than propanol then? Does it have something to do with a more advanced bonding theory?

If you're looking at the overall polarity of the solvent, you're not only consider the value of the individual dipole moment on a single molecule, but also their “density”, i.e. the number of these dipoles per unit of volume. You are right that the dipole moment of an individual linear-chain alcohol will, in first approximation, be independent of the chain length. However, in a given volume, you'll fit fewer of those long molecules, and the polarity of the overall solvent is thus lower.

In the context of a single molecule, the distance between the poles is inversely proportional to the polarity. The farther apart they are, the less polar the molecule is if they have the same electronegative potentials. This Link explains it well.


text:
Isolating salt from sea water using filters
Could a filter be used to isolate salt from sea water? Could we say that the water molecules are smaller than the molecules of salt dissolved in the water. I have no relation to chemistry or physics, so pardon the question if it's too naive.

Yes, this is possible and is a process called reverse osmosis. It is based on a membrane which lets water pass but not the salt ions (and other water contaminants). From Wikipedia:

The membranes used for reverse osmosis have a dense layer in the polymer matrix -- either the skin of an asymmetric membrane or an interfacially polymerized layer within a thin-film-composite membrane -- where the separation occurs. In most cases, the membrane is designed to allow only water to pass through this dense layer, while preventing the passage of solutes (such as salt ions). This process requires that a high pressure be exerted on the high concentration side of the membrane, usually 2–17 bar (30–250 psi) for fresh and brackish water, and 40–82 bar (600–1200 psi) for seawater

Water desalination features prominently among applications of reverse osmosis:

The largest and most important application of reverse osmosis is to the separation of pure water from seawater and brackish waters; seawater or brackish water is pressurized against one surface of the membrane, causing transport of salt-depleted water across the membrane and emergence of potable drinking water from the low-pressure side.

Thank you for your detailed feedback. Just wondering if this technique could be used on a large scale and if it requires any specially costly components or materials? thanks again.

@EmmadKareem, reserve osmosis is essentially how all of the oil-producing countries in desert lands produce a significant portion of their water. Yes, it is done at large scales.

@EmmadKareem Reverse osmosis is rarely used outside of desert countries because the large amount of energy needed makes it significantly more expensive than other sources of fresh water (when available).  Efficiency of the process is increasing; but about a year ago I read an article stating that current designs were approaching engineering/thermodynamic limits and additional large gains were unlikely.  http://arstechnica.com/science/2011/08/desalinization-is-this-as-good-as-it-gets/

@DanNeely It may be currently uncommon to use desalination outside of desert countries, but it isn't unknown: London has a desalination plant. And this article from the register argues the economics are favourable even there compared to the cost of reducing leakage.

@matt_black It's not an either or decision. Leaking pipes wash holes into the underground of your city. That will be very expensive on the long run.


text:
Moment of inertia for a triatomic molecule
What does the moment of inertia for a molecule mean? How is it useful? Most importantly, how would I go about solving it for a triatomic molecule?

blank shot: did you read wikipedia article on the subject?

The moment of inertia is defined for a molecule as it would be for any other solid: you consider the nuclei as points with mass, ignore the mass of the electrons, and you can then calculate either the scalar moment of inertia or the tensor of the same name.
The moment of inertia of a molecule is involved in the quantification of its rotational movements. Of the top of my head, the two areas of chemistry where this plays a big role are:

rotational spectroscopy: the rotation of a molecule is quantized, and the value that the angular momentum can take are related to the moment of inertia of the molecule.
the rotational entropy of a molecule is linked through statistical mechanics to its inertia tensor.


text:
Is tin and tinware toxic?
I tried to find info on tin toxicity but there is not much.
Wikipedia here says that it is not very poisonous; but there could be cases.
My question is: is it dangerous to say have a coffee boiled in a tin container? I know that coffee has some acidity, so it can probably react with tin.

It is known for metals to disassociate and react, for example, the process of rusting. High heats definitely increase reaction rates in alot of chemicals. I wouldn't recommend it for coffee cooking, but for food storage yes, given the proper conditions and lined ofcourse for prevention and reaction reduction. But im only a bio student.

You are right to be suspicious! In fact, the CDC's Public Health Statement for Tin at https://www.atsdr.cdc.gov/phs/phs.asp?id=541&tid=98 issued by Agency for Toxic Substances and Disease Registry appears to be a well-kept secret!

It is lucky pewter is not toxic as it has been used to some degree for eating utensils for over a thousand years. And all those who have used pewter beer mugs would likely be dead. Apparently adding a couple percent copper to tin to make pewter , makes it safe. ( Warning : serious level of sarcasm in this comment).

And copper cookware made before about 1960 is tin coated on the inside . And "Gourmet " /expensive copper cookware is still coated with tin today.

The Wikipedia article you quoted clearly says:

The low toxicity is relevant to the widespread use of tin in dinnerware and canned food.

Moreover, tin cookware, which used to be very common in earlier centuries, is still sold for outdoor cooking purposes. It is thus without a doubt deemed safe to use under current health regulations.

As a side note, the pH of coffee is around 5, which is only weakly acidic (same as unpolluted rain, and less acidic than tomato juice or beer):

There are two serious typos in that chart.  The two places where it says "hydrofluoric acid" should be replaced with "hydrochloric acid".  The stomach secretes HCl, not HF!  And HF is not a strong acid, so it seems unlikely it could achieve a pH of 0.  From what I've read, the pH of concentrated HF is ~1.

The Wikipedia article (https://en.wikipedia.org/wiki/Tin_poisoning ) on Tin poisoning seems to be abbreviated and does wonderfully extoll the safety of inorganic tin.  The widespread health issue of organic tin compounds is not referenced, or also importantly, the CDC's Public Health Statement for Tin at https://www.atsdr.cdc.gov/phs/phs.asp?id=541&tid=98. See my answer below for perhaps a more balanced perspective actually incorporating various sources.

You can not use pewter/tin for outdoor cooking , it melts. Surprising and disappointing to me as a kid; maybe that is part of why I became a metallurgist.

Pewter/tin alloy begins to melt above 471 F ,  not a good choice for "out door" cooking.

Anything in widespread is safe? Then, why do scientists criticize the widespread use of pesticides (see https://www.theguardian.com/environment/2017/sep/21/assumed-safety-of-widespread-pesticide-use-is-false-says-top-government-scientist ).

Tin cookware is used for outdoor cooking purposes? That's surprising, not least for reasons of weight. Can you link to something like that?

I don't believe it is the metal itself that should concern you, but when it reacts with something in your food it might lose its metallic bond and therefore allow you to ingest tiny amounts of tin via a molecule that includes tin itself. This is perhaps what can potentially happen when food is cooked inside a tin container.
We had a conversation about this last year in general-chemistry when the OCHEM instructor was talking about using metals as containers for dishes during recitation for our lab on cell potential. She was saying that the acids in food might create undesirable compounds when reacting with the metal containers, but that was probably a guess. She was saying something about how pop cans, for example, are made of aluminum, but the insides are lined with a plastic to prevent any undesired reactions (presumably). But then again she also went on to say that the FDA would not allow plastic containers (2 Litre pop bottles for example) until they finally passed off the plastic as a "food additive." Interesting stuff...
As a rule of thumb, I always say that too much of anything is bad for you, and eating food from tin containers might be one of those ways of getting tin toxicity, but this is all speculation here so take it with a grain of salt, or a grain of tin whichever you prefer.

Per this source, Chemical Reviews: Transition Metal-Tin Chemistry, I surmise that a potential safety issue with Tin stems from its ability to create complexes, which can act in distinctly catalytic roles in organic-based reactions.
In particular, the author discusses the co-ordination chemistry around stannylenes, $\ce{R2Sn}$, to explain a considerable amount of tin transition metal related complex formation (note, technically, Sn is a post-transition metal).
Direct toxicity levels of elemental Sn, itself, is likely just a secondary concern upon ingestion.
[EDIT, CORRECTED MEDICAL REFERENCE]
Relating to long term exposure, here are some comments courtesy of the Ecology Center, to quote:

Depending on the form and level of exposure:
  Exposure to high levels of inorganic tin may cause symptoms including stomach aches, liver and kidney problems, and anemia.
Organic forms, known as organotins, are believed to be toxic at lower levels of exposure. Several tin compounds cause nervous system harm, including tributyl tin, dibutyl tin, trimethyl tin, and dimethyl tin (Cooke 2004, Jenkins 2004, ASTRD).
The developing brain is particularly vulnerable. Dibutyl tin is toxic to nervous system cells at concentrations similar to those found in people today (Jenkins 2004).

My understanding of redox chemistry surrounding corrosion suggests that I should not store opened can foods (as this chemically introduces oxygen) in their container, especially those containing citrus acids (which, together with O2, can cause so-called spot-corrosion of steel). A dented or otherwise damaged container may introduce a galvanic based corrosion as well. Note, cited reactions are not immediate, but require days and could eventually introduce a small amount of potentially problematic Fe(II) (or other anodic alloys) into the food product (but not tin ions as it serves in the role of a cathode, see this).
Interestingly from an electrochemical perspective, boiling acidic coffee in a tin-plated container that has a surface defect (from scratching on cleaning) likely introduces ions not of Sn, but of associated alloyed elements (perhaps copper). However, elemental Sn does react with dilute (especially warmed) HCl albeit slowly. Boiling in air with dilute acid found in coffee for extended periods of time, and left to warm over many hours, suggests a possible path for some tin from organic acids, and much more so, for associated alloyed transition metal ions, to be ingested, in my opinion. Also, why is a fragile glass pot the preferred vessel for preparing coffee? Answer: Perhaps a difference that can be even tasted.
The louder some claim that inorganic Tin is safe, being indeed cathodic, makes me think of what metal is dissolving at the anode, Copper or Iron? And, what about organic bound Sn?
High-temperature chemistry suggests heating a tin cup (without water) to the point that it is red hot, could directly create some SnO2 as a product.
I will leave further comments to our resident contributing bio-chemists.
I just found this Public Health Statement for Tin from the CDC's Agency for Toxic Substances and Disease Registry, providing amazingly confirming comments on my ramblings relating to chemistry suggested above, to quote:

Because tin is naturally found in the environment at low levels, we cannot avoid being exposed to it. The major route of exposure to tin is from eating or drinking canned products. Reducing the amount of canned products you eat or drink may reduce your exposure to tin. Since tin concentrations in food increase if food is stored in opened cans, you can reduce your exposure by storing unused portions of canned foods in a separate container. You may be exposed to organic tin compounds by eating seafood from areas that may be contaminated with organic tin compounds or from contact with household products that contain organotin compounds (polyurethane, plastic polymers, and silicon-coated baking parchment paper). Reducing the amount of seafood that you eat from areas that may be contaminated with organic tin compounds and reducing contact with household products that contain organic tin compounds may reduce your exposure to organic tin compounds. If you are accidentally exposed to large amounts of tin or tin compounds, consult a physician immediately.

And also:

Food from unlacquered tin-lined cans contains up to 100 ppm of tin since the reaction of the food with the can causes some of the tin to dissolve in the contents of the can. Greater than 90% of tin-lined cans used for food today are lacquered. Only light colored fruit and fruit juices are packed in unlacquered tin-lined cans, since tin helps maintain the color of the fruit. Tin concentrations in food also increase if food is stored in opened cans.

Note, a focus on organic tin compounds (which is in accord with my opening paragraph) which appears to be particularly problematic. Also, open fruit cans or fruit juice generally sold in unlacquered tin-lined cans, after opening, can be an increased source of organic transition-metal ions, including those of tin. My thoughts on how this occurs, in the case of electrochemically created copper ions (from a galvanic cell with tin) where cuprous can be further oxidized to cupric by O2 and H+. Instrumentally, in the presence of elemental Tin, the following REDOX reaction can proceed:
$\ce{Sn(s) + 2 Cu(II) (aq) -> Sn(II) (aq) + 2 Cu(I) (aq)}$
Source: Page 29 at this reference.
No word from our resident biochemist, so here is a debate on a cookware site. I would believe the comment posted by PaulJoseph (which is actually taken from a prior version of Wikipedia's commentary on the Toxcity of Copper, since edited, which supportingly states the accelerated formation of copper ions and implies with Cu(II) (a 'reactive salt'), a likely attack on Tin as noted above), to quote:

When acidic foods are cooked in unlined copper cookware, or in lined cookware where the lining has worn through, toxic amounts of copper can leech into the foods being cooked.[11]. This effect is exacerbated if the copper has corroded, creating reactive salts.[12] Many countries and states prohibit or restrict the sale of unlined copper cookware. Cited source

which echoes my comments on the electrochemistry presented above. Once the lining on the copper pan with Tin (versus stainless steel) becomes damaged, this could lead to accelerated corrosion of the copper, and possible toxicity issues. A conceivable dangerous scenario is where one cooks food in an acidic sauce (tomato-based) with the pan having a damaged lining. Then, following the consumption of most of the food, a small amount is left open in the pan for days in a refrigerator. This would allow time, based on the REDOX above, for tin ions to problematically accumulate, adding to the copper ions (which can recycle, in the presence of an acid source, oxygen and elemental tin).

The "dot gov website" article you cite uses TIN as an acronym for a medical condition. It says nothing at all about the element "tin".

Thanks Jeff, I noted the mistake and inserted some comments from the Ecology Center.

Pure Tin may be fine as a food container. However, Tin ore is accociated with the ores of Lead and Antimon. Traces of these may find their way into Tinware. I believe there is an metallurgy motive to alloy Tin with Lead as well. Lead tends to react when in contact with acidic food, and it accumulates in the body. This was not taken seriously for a long time.Tinware once thought to be save might better not be used today.

You can actually get copper pots and cookware retinned at, e.g., East Coast Tinning. URL: www.eastcoasttinning.com . So it is still done.

@EdV they use pure Tin i hope.

That is what they say! I have no copper cookware, so I have no actual experience with this.

Tin lined cans are commonly used in processing light colored fruit as the tin is preferentially oxidized by residual oxygen left in the product and the headspace of the can, headspace gasses can be reduced by steam flushing immediately before seaming. If the oxygen levels are high enough a black line will be found at the juncture of the product and the headspace indicating loss of tin into the product, exposing the steel can to the acidic contents. When sufficient amounts of tin become incorporated into the product as stannous oxide some consumers will react adversely complain of stomach ache and loose bowels.
There is a secondary problem with lead  contamination. All tin sources are naturally contaminated with lead especially those originating in Malaysia, lead levels can be reduced if the tin is subjected to a distillation process but this increases the cost of the tin. Lead levels in most tin canned products will not meet the lead tolerances set by California Prop 65.
One further observation, I do not consider ANY product packed in a tin lined can to be "organic" as it will certainly have measurable levels of tin and lead within days of being processed.


text:
deduce chemical properties of a compound from elements
If we know the properties of two elements e.g Oxygen and Hydrogen.
We know the following properties

sound speed
boiling point
melting point
refractive index
etc.

given that information, is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments? i.e deduce its boiling point

You're not being very clear here... Do you want to deduce physical or chemical properties? You asked for boiling point, that is a physical property. (Aside from asserting that we already know it)

is it possible to deduce the properties of a compound such as water H2O without resorting to chemical experiments?

Not yet, assuming you mean extrapolating such information from the molecule's structure
Physical and chemical properties are hard enough to extrapolate from a compound's structure, let along its composition. The entire field of theoretical chemistry has been working on predicting the structure-property relationship for several decades. Great strides have been made, but accurate physical property prediction has not been achieved (though I would love to be proven wrong). In most cases, very accurate computation of properties of all but the smallest molecules takes more computing time than you would need to do the experiment.
Your question implies an interest in deriving such properties from the formula. As Canageek said, this is not possible, given the phenomenon of isomerism. One example I just used with my class is that there are 22 known molecules sharing the formula $\ce{C6H12}$, which is a relatively simple formula. Two of the more common are1-hexene and cyclohexane, which you can see have very different structures and properties (chemical and physical).

I had a friend do his undergraduate thesis on this, and he wasn't able to find any trends within the physical property database he was working on. Best was a R^2 of .7 or something, with a ton of visible jitter in the data.

No. Calculating the physical properties of materials is very, very difficult and can't simply be done from the elemental composition.

it will also depend on Geometry of molecule, polarity and Crystals.

Even knowing that the compound is made out of hydrogen and oxygen is not good enough, since there are two well known binary compounds of hydrogen and oxygen, each with different properties: $\ce{H2O}$ and $\ce{H2O2}$.


text:
What is known about the phase of matter at very high temperature and enough pressure?
Are there theoretical reasons for or against the existence of materials that will not become solid at some fixed, sufficiently large temperature when enough pressure is applied? Or is this even the normal state of affairs?
The usual phase diagrams typically only give a finite portion of the thermodynamical state space, probably as more cannot be experimentally tested.
But theory might tell us what happens, one way or another.

I do not know for sure obviously, but something tells me that nothing is harder to make solid than some good old Jovian hydrogen. I would speculate that if the temperature is high enough, adding too much pressure would only cause the atoms to slip and slide all over, possibly even making it a superviscous substance. But if you subscribe to the big bang theory, then you might believe that matter cannot be classified when that much pressure is applied, and temperature has lost all meaning.

There is a point in which it is too hot for a molecule to remain bonded and will break apart, no matter how much pressure is applied. Now with more pressure the electrons and nuclei will start free flowing in what is known as a plasma. And add a little more heat and fission happens.
So basically the reason they don't show you any more of the curve because your molecule under enough pressure and heat is no longer a molecule.

Could you please give a reference for the unset of fission (at fixed very high temperature) if pressure is large enough? I have seen equations of state for aluminum including gas and plasma phases. How would an aluminum atom decay under high pressure?

Its hard to say when fission does happen but a good reference could be the earths core, were fission happens with heavy elements, with a pressure of a few GPa and about 5000degrees Celsius. To correct myself, yes all material will become a solid with enough pressure, pressure which is only possible in celestial body's. Molecules do not retain their chemical and physical properties at such temperatures and act as a plasma.

How would an aluminum atom decay under high pressure? - How can a material be both solid and plasma, as your comment suggests? - I'd appreciate if you'd add references to your answer, so that I can judge for myself.


text:
Why do collisions in elementary reactions of higher-orders appear to be more likely?
So we are told that a unimolecular elementary reaction has a rate law of $k[\text{A}]$ where a termolecular reaction with three unique reagents, $A$, $B$ and $C$ has a rate law of $k[\text{A}][\text{B}][\text{C}]$. Now, other things being equal, and assuming for the sake of argument that the initial concentrations are all greater than 1M, this means that the termolecular elementary reaction is more likely at the outset. But that doesn't make much sense. There may be some statistical mechanics principle that's lost on me but it seems more likely that A bumps into A than, A, B and C come together all at once. So I don't quite follow this reasoning. Is it the case that $k$ is generally quite smaller in these termolecular elementary reactions?

I second J M’s answer: the rate does not depend only on concentrations, but the rate constant is key. However, I will add two more comments:

First, you cannot compare values of rate constants for reactions of different orders… they don't have the same dimension (unit). So, you cannot say things like “rate constants of third order reactions are typically smaller than those of first-order reactions”.
However, you can say (and it is true) that reactions rates of high-order elementary reactions tend to be small, because of the low probability of the molecules crossing path.

Re: units. Good point.

That is exactly the case.  My professor is famous (well within our group at least) for often saying "A rate is a rate constant times a concentration"  You can flood your reaction with a high concentration of reactants, but if the rate constant is 0, nothing happens.
In the case of a termolecular reaction (which usually unlikely), the rate constant is going to be very small.


text:
Are words on my T-shirt removable?
I wonder if the  words on my T-shirt are removable? If yes, what kind of (chemical or physical) ways are there?
They look like paints on the cloth, as the letter "C" has become blurred in the way shown in the picture over the time when I was wearing it. But I admit that I don't know anything about how words are printed on clothes generally.

Try some solvent to dissolve that paint.

It would depend on the type of dye/ink and the type of fabric.  The are several different types of dye removers available.  Dharma Trading Company makes "Dharma's Dyehouse Color Remover" which is Thiourea dioxide (a.k.a. Thiox, Spectralite).

As you say, it appears to be a kind of paint? So perhaps if you get a razor or something sharp you could try scraping and it will probably come off eventually, that would be one physical way. On the other hand if you want to go crazy try putting some paint thinner on there and see what happens, but don't blame me if you ruin your shirt experimenting on it, I have never used paint thinner before but I am guessing it consists of a non-polar compound and dissolves similar compounds (paint) so that it might be washed off with water. This is all hypothesis of mine, I have no idea what will happen except for what a razor can do.
Edit: I believe that Mr. Bean may have already demonstrated this experiment for us.


text:
Do dishwashing detergent and stain remover powder/stick have the similar ability to remove collar rings?
For yellow rings on shirts caused by skin grease or sweat:

Some recommended to use an "Oxyclean" stain remover or stain remover stick
for cleaning.
Some recommended to use dishwashing detergent to brush and wash the
stains, saying "dish washing liquid is designed to remove grease, and
why buy another product (that doesn't work very well -- I've tried)
when you already have one that does?"

From a chemistry perspective, do dishwashing detergent and stain remover powder/stick have similar chemical composition and a similar ability to remove collar yellow rings?
Original dirty collar

After dishwashing detergent and laundry detergent

That is nice to see that it did something, but we can't tell which (dishwashing or laundry) item you used had the desired effect or to what extent, because it seems you just used both and one of them or both did the trick?

@Leonardo: Not sure which, since I used them one after the other. :-(

Also, my original post is asking about comparing dishwashing detergent (possibly followed with laundry detergent although not mentioned) and stain remover (oxy/clean, or stick, which I don't own), not comparing dishwashing dtergent and laundry detergent.

Detergents seem to consist of both polar and non-polar compounds. Probably because they need the non-polar part of the mixture to dissolve the grease, and the polar compound part of the mixture to get washed off with the water so there is not much left over on your shirt. The ideal solution would probably depend on just how bad the stain is (how much grease needs to be dissolved). If your home solutions did not work the first time, try to look for signs if it did anything at all, and it might not be something that will remove all the grease the first go around.
I have no experience using detergents or removing stains, but I am going off a lot of speculation here as to how the process might work.
"Like dissolves like"
According to this article if there are lots of Mg, Ca, or Iron ions in your water (likely) then it could reduce the effectiveness of soaps. A good question to ask yourself is whether or not your water contains many ions such as these.


text:
Is there any scenario in which the size of a molecule increases due to an increase in temperature?
Take an ice cube for example. Heat is applied in a closed container until it is vaporized completely. Will the molecule's size be larger (on average)? Is there a substance that you know of that has odd behaviour such as changing molecular size due to temperature change? Obviously there would be more pressure in the case of $H_2O$. Have you heard of any substances that are exceptions to this rule? I would love to hear about it if you have.
If this question hasn't quite boggled your mind or you are having trouble thinking of an exception, let's say that an average diamond at about room temperature (something with entropy, ie: not a perfect lattice) and we say that this is a single macro-molecule because every carbon atom present in the diamond shares a bond with another carbon in the diamond.
Pretend we heat said diamond up a bit or let it sit for awhile. Are covalent bonds being broken and reformed as time goes on? If we heated it up enough without breaking the lattice completely, would the diamond itself, the molecule as a whole, increase in size at all? Or would the bonds just be broken more often and reformed, and the size change would be so negligible that it could hardly be considered an increase in volume?

@CHM if you go beyong the harmonic oscillator, and introduce higher-order terms in your interatomic potential (say, a Morse potential), you’ll get a different picture…

@F'x Yes, obviously. I only provided the simplest possible model. Your answer is better.

First, let me remark to “the size of a molecule” is not particularly well-defined. I assume in the following that you are interested about bond lengths (which are averages over time of distances between bonded atoms), as seems to be the case in your example. In the simplest case, you can consider an isolated diatomic molecule (think: gas phase N2), and we suppose we're not going to temperature high enough to ionize it or otherwise break its bond. Because the interatomic potential is asymmetric:

you can see that, as the temperature increases and the system is allowed to explore more of the potential energy curve (climb higher up the well), the average bond length will increase and not stay constant. So, for a single bond in an isolated molecule, it is expected that the bond lenghtens with temperature.

Now, one case where the overall “size expansion” can be clearly studied is that of crystalline materials. In solids, the evolution of volume as a function of temperature is named thermal expansion, and is characterized by the parameter of the same name, defined as: $$\large \alpha = \frac{1}{V} \frac{\partial V}{\partial T}$$ Because bonds grow longer with increasing temperature, most materials display a positive thermal expansion, i.e. they expand with temperature. This is, for example, the case for diamond, with $\alpha \approx +10^{-6} \text K^{-1}$.
However, some materials display the oppositive behavior, and shrink upon temperature increase. This phenomenon is called negative thermal expansion (NTE), and is rather uncommon.

This question has been answered from the point of view of vibration by @F'x; however, rotation of molecules is also quantized.  The rotational energy levels are populated according to a Boltzmann distribution.  The higher the temperature, the more-populated higher energy levels become.
Rotation of a diatomic molecule centrifugally stretches the bond length.
The article Centrifugal distortions in molecules: An ab initio approach quantifies the bond lengths of some simple molecules as a function of rotational energy level (J).
The most accurate calculations can be made for $\ce{H2+}$ because there is only one electron.
In the ground state bond length is 1.058 Angstroms.
The bond length progressively increases with increasing J (see Fig. 1), reaching 2.72 Angstroms for J=43.
Beyond J=43, the molecule is no longer stable because the centrifugal force is enough to pull the atoms apart.


text:
Can molecules diffuse upstream when emptying a cup of tea?
I was drinking some tea, and I randomly decided to pour some into the toilet in an unbroken stream; can particles travel "upstream" quickly enough to reach the mug of tea? How fast can diffusion occur in this system?
What if I pour it into highly concentrated toxic waste?  Given that even trace amounts could make it toxic, should I not drink my tea?

@jonsca: To me it looks more like, "If I pour some of liquid X into liquid Y, can any of liquid Y diffuse back up the stream into my vessel of liquid X?"

@Aesin That makes sense.  If you can edit and formulate this into more of a "chemistry" question, be my guest.

Aesin seems to have the correct idea formulated in a more mathematical point of view.  Sorry.. I was describing a layman's point of view.

(Note that diffusing concentrations of discrete classical particles cannot be continuous in a fixed volume - I removed that bit for that reason.)

did you come from here? also you might want to look at posting this question here But to be honest, I would say yes your coffee is toxic, the minute the supposedly sneaky airborne feces particles entered your drink as soon as you walked in to take a leak!

A bit oddly worded, but it seems clear to me. Also, I'd say diffusion is as much chemistry as physics, since we use it all the time. I mean, that is why we stir things, right?

I'm rather late to this one, but ... technically diffusion/convection is the purview of chemical engineering. Now that it exists, Engineering.SE would probably be the proper home for this question.

Particles/solutes are always diffusing in all directions. The question to ask is, "are they diffusing fast enough to overcome being swept along by the flow?"
Short Answer: No. Don't try it, though.

Long Answer:
One governing equation that can be used to answer this question is the following:
$$
\overline v\!\left(\vec r\right)\cdot \overline\nabla C_i\!\left(\vec r\right) = D_i\nabla^2C_i\!\left(\vec r\right)\tag{1}
$$
This is the (simplifed) steady-state mass transport balance for species $i$ at a given point $\vec r$ in space. The left-hand side represents transport due to convection; the right-hand side encapsulates transport due to diffusion. $\overline v\!\left(\vec r\right)$ is the fluid flow velocity, $C_i\!\left(\vec r\right)$ is the concentration, and $D_i$ is the diffusivity. For concision, the explicit dependence on $\vec r$ will be omitted below.
Given that we're most interested in the transport in the direction of the flow of the tea, if we approximate the stream as a cylinder of uniformly flowing liquid with velocity $v_z$ and assume that concentration variations perpendicular to the flow are unimportant, Eq. $\left(1\right)$ becomes:
$$
v_z{dC_i\over dz} = D_i{d^2C_i\over dz^2} \tag{2}
$$
if we choose the $z$-axis as lying along the stream of tea.
The most useful way to use Eq. $\left(2\right)$ is to non-dimensionalize it. That is, to re-write it so that all of the variables and coefficients have no units. For this problem, $z$ and $C_i$ are the variables to be non-dimensionalized, whereas $v_z$ and $D_i$ are parameters to be grouped into non-dimensional groups.  So, let's define:
$$
\eta = {z\over L}; \Theta = {C_i\over C_i^*} \tag{3}
$$
$L$ is some length scale representative of the system of interest. Here, the best choice is probably the height you're holding the cup over the surface of the toxic waste. $C_i^*$ is a representative concentration of a chemical species of interest, such as the nastiest player in that concentrated toxic waste. To substitute these definitions into Eq. $\left(2\right)$, I'll rewrite them as $z=\eta L$ and $C_i = \Theta C_i^*$:
$$
{v_zC_i^*\over L}{d\Theta\over d\eta} =
  {D_iC_i^*\over L^2}{d^2\Theta\over d\eta^2} \tag{4}
$$
The terms with derivatives in Eq. $\left(4\right)$ are now dimensionless. To make the coefficients non-dimensionless, divide both sides by $D_iC_i^*/L^2$:
$$
{v_zL\over D_i}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{5}
$$
The new grouping of parameters in Eq. $\left(5\right)$ has a special name: the Péclet number, abbreviated $\mathrm{Pe}$.
$$
\mathrm{Pe}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{6}
$$
The magnitude of $\mathrm{Pe}$ indicates the relative importance of convection relative to diffusion in a problem like this one. If the Péclet number is large $\left(\mathrm{Pe}\gg 1\right)$, it means that convection dominates the mass transfer behavior; if it's small $\left(\mathrm{Pe}\ll 1\right)$, then diffusion is the key phenomenon. If it's in the vicinity of unity $\left(\mathrm{Pe} \sim 1\right.$, or approximately $\left.0.3\leq \mathrm{Pe}\leq 3\right)$, then both diffusion and convection must be considered together.
So, can pouring tea into toxic waste poison your tea?
Well, setting aside the questionable wisdom of bringing a drink into a toxic waste area in the first place, what are some representative numbers for $D_i$, $v_z$, and $L$?  The upper limit for the diffusivity of a solute in water is about $2\times 10^{-6}\,\mathrm{cm^2\over s}$ (see, e.g., here). Since we want to know if diffusion can outrace convection, we should give it the best chance possible and selecting this maximum value will do that. (That is to say, maximizing $D_i$, which is in the denominator of the definition of $\mathrm{Pe}$, will tend to minimize $\mathrm{Pe}$.)
Similarly, selecting the minimum practical values of $L$ and $v_z$ will also tend to minimize $\mathrm{Pe}$, giving us a worst-case scenario for whether or not you face a gruesome death from toxic waste poisoning.  Let's say, optimistically(???), that one inch is the minimum height you can hope to pour your tea from without it splashing back into the cup. If we (rather loosely) approximate the  velocity of the tea stream at the toxic waste surface as that of a freely falling particle, the velocity will be $v_z=\sqrt{2gL}$.
So, our estimated Péclet number is (Google Calculator result):
$$
\mathrm{Pe} = {v_zL\over D_i} = {\sqrt{2gL^3}\over D_i} = 8.96\times 10^7
$$
This is a huge Péclet number. There's no way any of that toxic waste will diffuse up into your tea. (Please don't test this, though. Splashing, breathing toxic fumes, falling in, etc. are very real possibilities.)
As a further note, observe that in combining terms to form Eq. $\left(5\right)$, the reference concentration $C_i^*$ cancelled out completely. So, in this case, the answer is the same regardless of how concentrated that toxic waste is.

was it those crazy tea fanatic Cuban scientists again?
The particles can supposedly travel "upstream" if you were to say, pour it into your toilet as closely as your hand can get without a hazmat team busting down your door. But I would be more concerned about the sneaky particles that instead diffuse through the air itself, and not the stream you are considering...
Also, note that the more highly concentrated solution Y is, and the more pure solution X is, the more likely it is for solute in solution Y to end up in solution X. Higher concentrations flow into lower concentrations. And the closer the two solutions are during the decanting, the easier it is for this counter-current to form. But it is assumed that at more than 1 centimeter height, a counter-current should not form according to that article.

The article referenced is about floating particles. https://arxiv.org/pdf/1105.2585.pdf


text:
Is there an alternative to the half-reaction method to solving red-ox reactions?
The equation for the reaction between a permanganate and iron(II) ions in acidic solution is:
$\text{MnO}_4^- + \text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + \text{Mn}^{2+}$
My book, as well as a certain equation balancer, shows that the answer is:
$5\text{Fe}^{2+} + \text{MnO}_4^-  8\text H^+ \longrightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text H_2\text O$
They check the charges to ensure that it is indeed balanced. My book says that:
$5(2+) + (1-) + 8(1+) = 17+$
$5(3+) + (2+) + 0 = 17+$
And voila, they must be therefore balanced. However, there must be more than one way to balance such a reaction, like many problems. Is there a different, perhaps a more interesting, method than the half-reaction method for balancing this reaction and others?

See this video for a soup-to-nuts description of a matrix-based method that can handle your example. No previous matrix algebra experience necessary. As Leonardo notes in his answer, you'll need to supply the missing species - in this case water as a product and proton as a reactant. Fo course, this is also part of the process of balancing by half-reactions.

There is one other way that I have used, using a linear system to solve this reaction. Although it can be used for other more casual reactions too, such as the combustion of ethanol for example.
As you can see, in this special case (because the reaction is relatively easy to solve already) we see that there must be $H_2O$ molecules on the product side of the equation, because there is no oxygen on the products already, but there is oxygen in the reactants. So start by assuming the products include $H_2O$ with a coefficient of at least 4 (because there is 4 oxygen in permanganate ion).
Also, there is no hydrogen anywhere in the equation, so we can assume that to balance the water (which has at least 4 moles) there must be some hydrogen ions added as reactants to balance the reactants side (it is acidic conditions in this case). As it turns out, since there must be at least 4 moles of water, then there must be at least 8 moles of hydrogen ions to satisfy the convention of having all mole coefficients in the lowest terms.
Now we can start building the linear system:
Since we obviously do not know the coefficients yet, we will assign variables to them. Let a b c d e and f be the molar coefficients of each respective molecule in the following reaction we have established so far:
a$H^+$ + b$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + e$Mn^{2+}$ + f$H_2O$
Now how do we build the linear system for this reaction? We might do it as follows:
Starting with the hydrogen atom, simply because it comes first when reading it from left to right. We write this equation to represent the hydrogen atoms in our reaction:
a = 2f
It is as simple as it looks, there is one mole of hydrogen atoms per mole of hydrogen ions, and there is two moles of hydrogen atoms per mole of water molecules.
We can keep doing this for the rest of the elements in our reaction, until we have three more equations along with the one for hydrogen, because there are three more different elements to account for.
b = e This one is pretty straightforward... This accounts for Manganese atoms.
4b = f Oxygen atoms are now accounted for.
c = d Also straightforward, now iron atoms will be accounted for as well. Notice that c and d are lone variables, they do not show up with the other equations, but at least they must be equivalent
Remember that we established some mathematical axioms when we said that there must be at least four moles of water molecules to satisfy the one mole of permanganate ion. An axiom is just something that we assume to be true, such as f = 4 in this case.
Now look at the equation we established for hydrogen atoms, the first equation a = 2f. Since we are assuming that f = 4 we can therefore safely assume that a = 8 without going against our original axiom. Also if we assume that f = 4 we can also go ahead and solve the equation to account for oxygen atoms 4b = f and solving it gives us b = 1 and the rest is pretty straight forward. If b = 1 and b = e then e = 1.

Given what we have found out so far, lets take a look at our equations again:
a = 8
b = 1
c = d
e = 1
f = 4
So,
8$H^+$ + 1$MnO_4^-$ + c$Fe^{2+}$ ==> d$Fe^{3+}$ + 1$Mn^{2+}$ + 4$H_2O$

So now we can look at the charges to figure out the correct coefficients for iron(II) and iron(III) ions, which should be the same of course (c = d). As it stands currently (ignoring the iron ions), there is a total charge of 7+ on the reactants side, and 2+ on the products side. These should be the same for the equation to be balanced, so we can start by creating another axiom and saying that c > 1. If there are 2 moles iron then our charges would be 11 on the reactants and 8 on the products. Incrementing it to 3 would increment the charge on the reactants side by 2+ while the product's total charge would increment 3 due to the charge of the ions themselves. Eventually things are balanced when the moles for iron reach 5 making both sides have a total charge of 17 and everything is nicely balanced.

You can add one more equation to your system that relates c and d to the others - the charge balance equation: $a(+1)+b(-1)+c(+2)=d(+3)+e(+2)+f(0)$. This still only gives five equations for six variables, which will not provide an exact answer, but will provide the system of answers that include all multiples of the one above.

@Ben Norris You are absolutely right, I had not tried this yet, but it does make the last variables readily solvable when you simplify the equation c=d to a single variable such as c, and use your equation substituing the previous values to give: 7 + 2c = 3c + 2. Giving us c = 5. Thanks for your contribution!

One can always choose one coefficient.  This is usually done to put things in "lowest integer terms".  Sometimes in doing the linear equation method you will still have insufficient equations to solve the problem.  In such a case you have an overall equation made up of two (or more) independent equations.  To see this take the balanced equation for the production of water from its elements and the same for the production of carbon dioxide from its elements and add them after multiplying each by an integer.

To simplify, if you do not have enough information to solve the system then assume your best guess by assigning a variable a value such as 1. This is the synonymous method of guess and check that often happens in classrooms. In the case of water, 1 molecule of hydrogen means half a mole of oxygen, but we want integers so 1 mole of oxygen is enough information to fill in the rest.

There is oxidation numbers method. Look it up.
Shortly, you give any hydrogen atom (in a compound which also has one different element, aside from oxygen) +1 charge, any oxygen atom (in a compound which also has one different element, aside from hydrogen) -- -2 charge. Using these and molecule charge you determine unknown atom's charge. Difference between charges of the same element on the left and on the right should equal difference between charges of another element on the left and on the right, multiplied by -1. Then you balance a reaction so that charge is conserved.
Of course, before that you need to use the information given to complete your reaction with compounds, that are initially absent. For example, in this case you see, that there should be oxygen on the right. Also it's said that reaction proceeds in the acidic solution, which means protons are used (add them on the left side), which means they will react with oxygen, which means on the right should be not oxygen, but water.
Walkthrough:
So we have
$MnO^−_4+Fe^{2+}⟶Fe^{3+}+Mn^{2+}$.
Obviously, we need to add $4O^{-2}$ on the right side and $H^+$ on the left side (and, consequently, on the right side too):
$MnO^−_4+Fe^{2+}+H^+⟶Fe^{3+}+Mn^{2+}+4O^{-2}+H^+$.
As we establish by the oxidation numbers,
$Mn$'s charge (in $MnO_4^-$) equals $-1 - -2(4) = 7$. While on the right side it's charge is $+2$. $2 - 7 = -5$.
Iron also changes it's charge, so we see, that we need to find a coefficient to make a difference between charges of Fe on the right and Fe on the left equal $+5$. That would be $5$:
$MnO^−_4+5Fe^{2+}+H^+⟶5Fe^{3+}+Mn^{2+}+4O^{-2}+H^+$
Now only thing, that's left, is to add a coefficient to the hydrogen, so water can be formed:
$MnO^−_4+5Fe^{2+}+8H^+⟶5Fe^{3+}+Mn^{2+}+4H_2O$.

Could you add to each step of your method the respective example for the reaction in the question, so that you provide a complete walkthrough for the OP and show how one can arrive at the correct result?


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multielectron spin-orbit coupling and energy level
For multi-electron atoms for which the Hamiltonian (including the spin-orbit coupling) reads
$$H=\sum_i T_i-Z\sum_i V_i+\sum_i V_i^{\text{s.o.}}+\sum_{i>j}V_{ij}$$
The $T_i$ are the kinetic energy of the individual electrons, $V_i$ are the nuclear Coulomb attraction, $V_i^{\text{s.o.}}$ is the spin-orbit coupling, and $V_{ij}$ are interelectron Coulomb repulsion.
In http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_atomic.pdf (page 103), using Wigner-Eckert, they are able to diagonalize the spin-orbit in terms of the total orbital and spin angular momenta of all the electrons.
Although it says it is beyond the scope of the course, I would still like to know how to do it.  I am pretty familiar with the Wigner-Eckert theorem in its basic form (that the matrix elements of a tensor-operator can be written in terms of a C.G. coefficient times a reduced matrix element).  Thanks.


text:
What is the 'ancestral term' in the LS coupling scheme?
On the physics.nist.gov site: http://physics.nist.gov/Pubs/AtSpec/node06.html
The table titled "Atomic structural hierarchy in LS coupling and names for the groups of all transitions between structural entities" contains the hierarchy of atomic LS coupling.  But there is a rather interesting footnote in the table: The letter $\gamma$ represents any additional quantum numbers, such as ancestral terms, necessary to specify a particular term.
What's ancestral term?  Is there an example of this in the literature?

There is some more information here: http://en.wikipedia.org/wiki/Term_symbol#Notes , but I'm also not an expert on this.

"Ancestral Terms" are terms such as parent and grandparent terms that couple to form the final term.
For example, if you see $3d^7(^4F)4s4p(^3P°) ^6F°_{9/2}$
this means 7 3d electons couple to form a $^4F$ grandparent term
the 4s and 4p electrons couple to form a $^3P°$ parent term and
the grandparent and parent terms couple to form the final $^6F°$ term.
The above is one of several example given at the NIST page Notations for Different Coupling Schemes


text:
Will alcohol or soap damage plastic or rubber?
I have read conflicting advice on various sites about the appropriate methods for cleaning plastic and rubber items and would like to ask people who actually know what's going on what they would recommend.
I realize some of my assumptions may be false:

I am (perhaps wrongly) assuming that most plastics would have similar properties as it relates to cleaning with alcohol or soaps/detergents and that you don't need me to give you the actual chemical structure of a specific polymer.
I am (perhaps wrongly) assuming that most soaps and detergents would act similarly on either a plastic or a rubber item.  If this is false please give your understanding of what an "average" soap or detergent would do.

I am not assuming that plastics and rubber will respond the same to either alcohol or soaps and detergents.  If they are different, please note which will have what effect.
I am not assuming that synthetic and real rubber will act the same.  If they are different and you know the answer for both, please mention the difference.
I am hoping that it's okay to ask "applied" questions here.  This seems like the best of all stackexchange.com sites, not just because none of the less theoretical sites are deal with this sort of question, but also because it really boils down to a matter of chemical reactions.
Update: by "alcohol" I mean rubbing alcohol, or isopropanol.

Can you perhaps reference the conflicting advice you have already found? It may help your research efforts to start with a specific case. What item do you have in mind that needs cleaning, are you trying to use soap to clean hand grease off a plastic fridge door handle for example?

Asking about the mechanisms through which this might occur would make this a stronger question, I think.  We're happy to have the "applied" questions, but they do should be grounded in some type of theory or else we would be unfaithful to the scope.  See if you can make some minor changes to it in that regard.

Will soap damage plastic or rubber?

Probably not. Most plastics and rubbers are resistant to aqueous solutions and ionic compounds (including most surfactants and soaps).

Will alcohol damage plastic or rubber?

Maybe. There are two problems, in that neither alcohol nor plastic refer to specific compounds, but are rather generic terms encompassing hundreds of compounds each. The number of possible combinations is endless. As Leonardo suggests, the best approach is either to look in the manual or experiment in a small inconspicuous part of the product.
However, assuming that by alcohol you either mean ethanol, which is the drinking kind of alcohol, or isopropanol, which is the alcohol in many cleaning products, then I can give you some advice based on common plastic types. Both are similar enough in their solvent properties that we can generalize.
Alcohol will damage some plastics, but not all. I will list my answers by the resin code, or "recycling symbol", found on most plastic items:

Poly(ethylene terephthalate), PET or PETE - PET is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening due to the dissolution of plasticizers.
High-Density polyethylene, HDPE - HDPE is resistant to most things.
Poly(vinyl chloride), PVC - PVC is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.
Low-Density polyethylene, LDPE - LDPE is resistant to most things.
Polypropylene, PP - PP is resistant to most things.
Polystyrene PS - PS is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.
This stands for "other", but the most common is polycarbonate, which is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.

Rubber - most rubbers are probably resistant to alcohols, but prolonged exposure will cause loss of elasticity

Wow.  That's an awesome answer.  My main concern is a whole load of cables, mostly USB cables.  Some of the cables and cords (notably on any Apple products) have a sort of rubbery feel, but I have no idea how to tell just what they are made out of.

Rubber is made constituted of Buta-1,3-diene polymer

@BigGenius Rubber is also a generic term for partially crosslinked thermoplastic elastomers with certain properties in common "natural" or latex rubber is polyisoprene and some rubber is poly(1-3-butadiene). Most dienes can be polymerized into a "rubber" material, with the properties deriving from the mixture of linear, branched, and crosslinked polymers in the material.

What about TPU?

My advice: Look up manuals where applicable, or experiment with methods if you cannot find good answers, hopefully, you can figure out what exact material it is you are cleaning.
What I can tell you for sure is that I have a small 'plastic' refrigerator which, by quickly looking up the model No. and finding this order website, and the manual, it is made of Polyurethane. And the manual indicates under "cleaning and maintenance" that I should "wash the inside (polyurethane) material with a solution of warm water and two tablespoons baking soda per quart water." And, specifically, do not use "soaps, detergents, scouring powder, spray cleaners, or the like because it may cause odors."
Perhaps about.com has answers you wish, but I prefer the good old owners manual for the cold hard truth. By the way, I went ahead and used dish soap and cold water from my hose (I am a very lazy person). It smells slightly.. goofy, and that was a couple days ago. So I think they know what they are talking about. Note that the manual says nothing about Polyurethane!
I hope that helps clarify something or another.


text:
Inherent error in mass and volume?
In one of my chem classes I had to do a lab that involved finding the mass and volume of a substance.  I did three trials and for my mass I got 9.3g, 9.3g, and 9.4 g.  For my volume, I got 9.7ml, 9.9ml, and 9.9 ml.
I am being asked to find the inherent instrument error calculation.
For volume, I used a graduate cylinder which has 1 % vol/reading as an inherent error.
For mass I used a one place electronic balance which has error of 0.1g/reading
How would I find the error in mass and the percent error in mass?

Is the percent error formula you are looking for something like the formula this website service offers? Your lab instructor should have given you the formula, perhaps you missed that. Do you have a reliable lab partner you can ask? The error in mass would probably be something to do with how far each measurement deviates from the average.
If for example I used a 250 ml Erlenmeyer flask which had marks at only 25ml increments. I would then say that I could make a 'guess' as to what the volume is at if I were trying to measure something less than 25ml or between two marks. But the error for that instrument would then be somewhere around +/- 12.5ml error. My general-chem teacher was fine with that, maybe you should ask yours?
I was told that the convention is to expect at least 2% error in the last significant digit if not explicitly defined. When taking a measurement with a device like a graduated cylinder, you always measure down to the most precise "known" value, and then the last significant digit is your own judgement.


text:
What determines the electronegativity of an element?
Is it possible to work out without post-Bohrian mathematics?

There are various slightly different electronegativity scales.  The two most common are the Pauling scale and the Mulliken scale.  Of the two the Mulliken is the simplest.
The Mulliken scale is based on two quantities, both measurable.  One is the ionization energy $(E_i)$, the energy required to remove an electron from a neutral atom.  The other term is the electron affinity $(E_a)$.  That is the energy released when the neutral atom gains an electron.
The Mulliken electronegativity $\chi$ is then given by
$$
\chi = (E_a + E_i)/2
$$
It is that simple!
For more details and explanation see
http://en.wikipedia.org/wiki/Electronegativity

Is it possible to work out the two variables with the Bohrian model, or are experimental results much better?


text:
Calculating the heat capacity of a calorimeter
I've been struggling with this problem that I can't seem to figure out. I think I know how to solve it but I think there's missing information.

The combustion of 1 mole of glucose $\ce{C6H12O6}$ releases $\pu{2.82\times10^3 kJ}$ of heat. If $\pu{1.25 g}$ of glucose are burnt in a calorimeter containing $\pu{0.95 kg}$ of water and the temperature of the entire system raises from $\pu{20.10 ^\circ C}$ to $\pu{23.25 ^\circ C}$. What is the heat capacity of the calorimeter?

I think I need the specific heat of glucose (which I haven't found yet), but also I don't know why they give me the heat released by 1 mole of glucose.  I need to know the method of how to solve such problems.

please give your comment on the accuracy of the measurement of the heat capacity of the calorimeter. explain any assumptions used calorimeter heat capacity in the experiment? and why a cooling curve (T versus t) was used to determine the mixing temperature? and do you expect your value for the specific heat to be too high or low?why?
and what is your unknown metal?

$12.5\ \mathrm{kJ}$ of heat was absorbed by the surroundings.
I found this by using the mcat formula and the specific heat capacity of water (4.18 J/(g °C)):
$Q=m\cdot c\cdot\Delta T$
$Q = 950\ \mathrm{g} \times (4.18\ \mathrm{J\cdot g^{-1}\cdot {^\circ C^{-1}}}) \times (23.25\ \mathrm{^\circ C} - 20.10\ \mathrm{^\circ C}) = 12508.7\ \mathrm{J}$
If you wanted to use this whole formula for solving the calorimeter's specific heat capacity, you would need to know the mass of the calorimeter as well, which is not given.
What your book is probably asking is for what is called the "calorimeter constant". This is given in units of $\pu{J/^\circ C}$ notice that it does not include mass.

Note: Sometimes "the calorimeter's specific heat capcity" is used instead of referring to the calorimeter constant, but in this case we cannot find a value which will include mass in the units, so I think it is more clear to use the term "calorimeter constant."

You can determine the constant by this formula: $$Q_\text{cal} = C_\text{cal} \times \Delta T_\text{cal}$$
Where $Q_\text{cal}$ is the energy absorbed, $C$ is the constant and $\Delta T$ is the same as the change in temperature of the water.
You may calculate $Q_\text{cal}$ by using this formula: $$Q_\text{cal} = -(Q_\text{water} + Q_\text{glucose})$$
It may also help to think of $Q_\text{water}$ = $Q_\text{surroundings}$ and $Q_\text{glucose}$ = $Q_\text{system}$
To find $Q_\text{glucose}$ I did: (glucose has lost energy, it is negative value)
$-2820\ \mathrm{kJ} \times 0.007\ \mathrm{mol}$ and $Q_\text{water}$ is simply the $12508.7\ \mathrm{J}$ positive because $\Delta T$ is positive for the surroundings (the system/glucose lost energy)
$Q_\text{cal} = -(12508.7\ \mathrm{J} + (-19740\ \mathrm{J}))$
So my final answer is then: $2.3\times10^3\ \mathrm{J/{^\circ C}}$
It is important that heat capacities are positive, think about what it would mean if this were a negative value.

In the laboratory, it is necessary to do a calculation such as this one before using a calorimeter for anything. Normally it can be done by heating a piece of nickel or something, recording the temperature of the metal and the water, and then dropping the metal into the calorimeter to find the final temperatures, and then calculate the calorimeter constant. You can then proceed with further experimentation using that calorimeter, but only after this constant has been found can you find the specific heat capacity of other materials.

First of all, I have to say a massive thank you, because I wouldn't have thought this that way, (specially because I'm quite bit confused about the specific heat-specific heat capacity thing). Second, I'm guessing, if heat capacities were negative, wouldn't it be violating laws of thermodynamics?

I just found out there are actually systems where the heat capacities are negative, and even though it goes beyond my knowledge, I can't think of how would it be possible, increasing the temperature by losing energy, seems irrational to me.

I think that is some in-depth physics topic about systems. A system might be able to present the idea of a negative heat capacity, but try to stay focused on actual materials in this case. If a material had a negative heat capacity it would do the opposite of what boiling water does. If water was -4.18 then to raise water by 1 degree celsius, you would need to extract energy stored in the molecules, so putting water in the fridge would cause it to boil in a sense. This is impossible. Like putting water in the stove and waiting for it to freeze...

You missed a very critical assumption. The problem assumes that the heat capacity of the calorimeter is only due to the water which it contains. For a good calorimeter that is mostly true, but not entirely true.


text:
Why must an oxidizer not touch rubber?
Why is it that oxidizers (specifically $\ce{KClO3}$) are at risk of causing a fire when they come into contact with the rubber stopper used to contain them in their container, but can be slowly heated if they do not contact the rubber stopper?
Essentially, why is it that oxidizers are incompatible with rubber?

Note that an oxidiser is not combustible, it is a supporter of combustion. You need both a combustible material and a supporter if you want fire, heat is not a magical creator of fire. I'm not sure as to what the exact reaction is (oxidation of rubber)

A lot of rubbers possess C=C double bonds. Alkenes are very easily oxidized by KMnO4 or KClO3, therefore in the case where you're not aware of the chemical nature of the rubber, it's best not to expose it to strong oxidizers.

$\ce{KClO3}$ and $\ce{KMnO4}$ are special kind of oxidizers, that donate oxygen very, very easy. Mixtures of this oxidizers with some reducers can self-combust on friction. Rubber stopper on regular friction produce fine powder of rubber and oxidize.
The reason for it is a special property of this oxidizers, that give out oxygen of surprisingly low heating.


text:
How to extract pure caffeine from tablets?
I have some caffeine tablets and I want to extract the pure caffeine from the rest of the ingredients (caffeine 200 mg. Inactive ingredients in order are corn starch, dextrose, dicalcium phosphate, D&C yellow #10 aluminum lake, magnesium stearate, microcrystalline cellulose, silica gel).
I assume the result would be aqueous solution but any method which works would be great.

these? If so, they should be yellow, judging by that artificial coloring ingredient.

If pure caffeine is all you want, and you're just mentioning these tablets because you have some and they seem like a good source, you might as well just buy some food-grade, from somewhere like here. It's pretty cheap and pretty widely available (at least from online stores) in the US, Canada, and the UK, at least. Careful with quantities, though, it's dangerous stuff.

@Leonardo Yes very similar to those but not quite identical http://i.imgur.com/LVvIR.jpg (This is the front of the tablets.)

@AbstractionOfMe - The caffeine could probably be obtained by one or more separatory funnel extractions with an organic/aqueous solvent system, assuming you can perform one safely.

@Aesin - Reading that page: "Do not consume [...] more than 1200 milligrams caffeine anhydrous per day." That's equivalent to over 3.5 litres of Monster...

I made this question a little more generic.  I think this is a common experiment in some intro level organic labs, so it might have some good information for someone.  Otherwise, I think we'd like to discourage people from experimenting with something they might consume.

The best option would be an organic extraction with dichloromethane. All the other ingredients (except possibly D&C yellow #10 aluminum lake) will be insoluble and therefore unaffected. Filter, let the dichloromethane evaporate and you should have your caffeine.
EDIT: I agree with J M. This caffeine IS NOT EDIBLE. You could easily become intoxicated by the small amounts of dichloromethane that will remain. Also be careful when handling the dichloromethane. Vapors are also toxic and if the liquid is heated or shaken in a closed recipient it might explode. Use caution.

If you do this, please DO NOT CONSUME the caffeine afterwards.  That cannot be stressed enough.

re: "You could easily become intoxicated by the small amounts of dichloromethane that will remain." NO! it should be "You could easily become poisoned by the small amounts of dichloromethane that will remain."