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Oxalic acid structure
What is the reason why oxalic acid has its second $\ce{-COOH}$ group flipped?
I'm guessing it's something to do with steric repulsions between the $\ce{-COOH}$ groups but I'm not sure.

Draw the charges, then apply Coulomb law.

Like ron already stated, the reason for that arrangement is the possibility to form more intermolecular hydrogen bonds. For oxalic acid there are two modifications which depict this quite clearly.

From charge considerations itself it is not obvious, that the periplanar alignment should be better. However one can see, that there might be a little less repulsion. Given here are natural partial charges from NBO/BP86/cc-pVDZ.

A rotational scan reveals that the rotation barrier is just around $7~\mathrm{kJ/mol}$ and the difference between these are just around $2~\mathrm{kJ/mol}$ (BP86/cc-pVDZ). The dihedral angle of $0^\circ$ refers to the hydroxyl moieties being on the same side.

The barrier results mainly from breaking the partial $\pi$ bond between the two carbon centres.

The $\ce{O...H}$ intramolecular distance is around $230~\mathrm{pm}$ and in gas phase and higher temperatures hydrogen tunneling should be also considered.
In aqueous solution, exchange rates between water and oxalic acid are very fast, so that all oxygen atoms should be equal - there is no difference observable at this time scale.
\begin{aligned}\ce{
(COOH)2 + 2H2O &<=> HOOC-COO- + H3+O + H2O\\
               &<=> (COO^{-})2 + 2H3+O\\
}\end{aligned}

More structure stabilizing hydrogen bonds, both intra- and intermolecular, can be formed in the conformation where the carbonyl groups are anti-periplanar.

The carbonyls' dipole moments minimize energy by pointing in opposite directions.

This is not necessarily a minimization of energy. 1) The OH group also has a significant dipole moment, and 2) in polar solution (like water) you minimize energy by maximizing molecular dipole moment.

No argument.  Vacuum, solution (in what?) and solid state all have their special - and exploitable - footnotes.

In a polar solvent, oxalic acid is like carbon dioxide; all the C-O bond lengths are the same.


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Hardness and softness are terms used to describe which aspect of acidic behaviour?
What do hardness and softness mean?
Is it something to do with the polarisability.

The terms softness and hardness are used in HSAB theory to distinguish Lewis acids and bases by charge density, polarizability, electronegativity, and, in the case of molecular compounds, HOMO and LUMO energies.
Charge density is inversely proportional to atomic/ionic radius and proportional to oxidation state (in case of a neutral atom) or charge (in case of an ion). Hence charge density increases with decreasing radius and increasing oxidation state.
Polarizability is the tendency of the electron shell of an atom/molecule to become deformed in an electric field, which can originate e.g. from the charge of an adjacent ion or the partial charge of a dipole. An electron cloud of an atom or ion is more easily deformed with increasing volume, as it is more diffuse, and its outer electrons tend to be more loosely bound. Therefore, larger atoms and molecules tend to be more polarizable, and higher polarizability is often accompanied by lower charge density or oxidation state.
In the case of molecules, when a Lewis acid reacts with a Lewis base, electrons from the highest occupied molecular orbital (HOMO) of the Lewis base are transferred to the lowest unoccupied molecular orbital (LUMO) of the Lewis acid. How well this transfer of electron density works depends on the relative energies of the acid LUMO and the base HOMO. The stability of the reaction product will increase with the energy difference between its HOMO and LUMO (source).
The HSAB theory predicts that hard Lewis bases react preferentially with hard acids, and that likewise soft bases form stronger bonds with soft acids. For example, $\ce{Li+}$ and $\ce{BF3}$ are hard acids and $\ce{F-}$ and $\ce{NH3}$ are hard bases due to their small size and high charge density/low polarizability of the ions. $\ce{BF3}$ therefore readily forms Lewis acid-base adducts with fluoride and ammonia, and $\ce{LiF}$ is less soluble in water than $\ce{LiCl}$ and $\ce{LiBr}$. $\ce{Ag+}$ is an example for a soft acid (large ion with low charge), and the halide ions become increasingly softer bases with increasing ionic radius. Silver therefore forms stronger bonds with heavier halides like $\ce{I-}$ and $\ce{Br-}$, and this also explains the decreasing solubility of the silver halides in water, which is classified as a hard solvent. In the case of soft acids and bases, the difference of electronegativity tends to be smaller, and the bonds have more covalent character (e.g. silver iodide) than those between hard acids and bases (e.g. lithium fluoride) with larger difference in electronegativity (source).
The shift of reaction equilibria can also be predicted with this concept, like in the following example (reference):
$$\ce{HgO + H2S \rightleftharpoons HgS + H2O}$$
This equilibrium will be shifted to the right side, because $\ce{Hg^2+}$ as a soft acid prefers to bond to the soft base $\ce{S^2-}$. Likewise, oxygen as a hard base forms stronger bonds with hydrogen (hard acid).

$+1$ for actually restricting HSAB to domains where it is actually useful.

I still don't quite understand why hard acids would prefer hard bases and vice versa..

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Modelling planar ring molecules and their evolutions
This is the next step of my previous question.
I am trying to simulate a circular chain (ring) of molecules which are planar (two dimensional). What I do is take a planar unknot and embed it into a two dimensional lattice. Then I calculate energy of the unknot and try to evolve the unknot so that it reaches the canonical form. By definition it then has the minimum energy. To have a realistic definition of the energy I need to study real life planar unknots.
Can anyone please give me an example of the molecular chain which can stay as planar with it's minimum energy state is planar. A published study on the evolution of the geometry of the molecule will be highly helpful.

Could you elaborate a little more what you actually mean and what you try to accomplish. I also do not understand what 'evolution of geometry' should be?

@Martin, I have added some background.

Do you mean something like the heme ligand? And what do you mean by evolution of the geometry?

@Martin, by 'evolution of geometry' I meant changes in shape.

What are you looking at? Molecules in gas, liquid or solid phase? Are we dealing with solutions? Molecules change their shape depending on the environment constantly (rotations, vibrations). One has to consider chemical equilibria. Only to name a few. I do not think you can call that evolution. If you are referring to simulations, where in a given timeframe the orientation of the entities change, that is something else. It is really hard to understand what you are asking for. I guess I am not the only one and I would be glad to help.

@Martin, for the time being I would like to investigate  molecules in a solution. I am trying to keep it as simple as possible. I think polymer rings in a solution is the simplest scenario, right?

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What is the difference between the reactivity of Lead subacetate and Lead II acetate undergoing a hydrothermal process?
I am attempting to adapt a hydrothermal method for synthesizing $\ce{PbS}$ nanocrystals to my school that contains a mixture of cationic and anionic surfactants sodium dodecyl sulfate (.05 mmol, SDS) and cetyltrimethylammonium bromide (.05 mmol, CTAB) and ethylenediaminetetraacetic acid (.2 mmol, EDTA) as a chelating agent that diminishes the reactivity of lead ions in solution. The lead source within the method is Lead II Acetate and the sulfur source is an organosulfur compound with the formula $\ce{SC(NH2)2}$.
The company that we bought our chemicals from sent us Lead subacetate and while I wait for them to send us the right one back I am wondering if using the appropriate ratio of moles of Lead subacetate which has 3 moles of lead per 1 mole of lead subacetate to the documented lead II acetate which contains 1 mole of lead would lead to lead sulfide $\ce{PbS}$ nanocrystals.
The structure of subacetate contains two crystalline basic acetates containing hydroxide ions in addition to acetate ions. Does the subacetate create a conjugate acid that lowers the pH?

Compared to lead(II) acetate, lead subacetate contains additionally $\ce{2 Pb(OH)2}$. The hydroxide can either be neutralized by protons from EDTA or take part in the hydrolysis of thiourea:
$$\ce{SC(NH2)2 + OH- \rightleftharpoons OC(NH2)2 + HS-}$$
$$\ce{HS- + OH- \rightleftharpoons S^{2-} + H2O}$$
The pH of the solution should be higher unless all hydroxide is neutralized or consumed by the hydrolysis reaction.


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How to find trace methanol content in fermented fruit/distillate?
Suppose we have some fermented (optionally also distilled) fruit/fruit juice. The sample therefore contains water, sugars, ethanol etc. Which analysis method is the simplest to find out whether the toxic methanol is also present?

Methanol, formed by degradation of pectines from the skin of fruits, as well as higher alcohols is volatile and does not show significant absorption in the UV.
Consequently, gas chromatography is the method of choice to determine these components in the fermentation mix.

Is this something that can be done on the run? Or at home a least?

@TomášZato No, gas chromatography usually isn't something to be run at home. The prices for the equipment and the operation costs are too high.

Maybe try NMR?
But if the amount of the methanol is too small, you won't see so beautiful signal:

Nice answer! What program generates such a nice NMR-spectrum?

@andselisk I got this image from this site: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm to simulate NMR there are a lot of web-pages: https://web.chemdoodle.com/demos/simulate-nmr-and-ms/ it can predict OH (most of them can't) for example: www.nmrdb.org/new_predictor/ As software I prefer Mnova http://mestrelab.com/ or Chemdraw http://www.cambridgesoft.com/software/overview.aspx - both are not free.

I'm aware of both Mnova and Chemdraw, but to my knowledge  none of them can produce this pixel-perfect plot. But thank you anyway!

@andselisk Ususally, I prefer EXPERIMENTAL data bases, something like: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi for small compound it is much better than prediction.  Here is small comparison http://imgur.com/a/EIjLW


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Boron trihalides acidity
A boron trihalide ($\ce {BX3}$) is a Lewis acid, because it can accept an electron pair, as per Gilbert N. Lewis' definition.
Now my question is-

Arrange $\ce {BF3}$, $\ce {BCl3}$, $\ce {BBr3}$ and $\ce {BI3}$ from the weakest to the strongest acid properties.

Relative Lewis acidities parallel exothermicities of a Lewis base adduct-forming reaction. The adduct is a dative bond tetrahedral boron anion. The worst pi-bonding substituent has the highest energy versus product valley, maximizing exotherm difference. Order radii of the participating pi-orbitals by their worst matches.

hint; most important factor in BX3 is the bonding between X and B which is going to determine the acidic strength , and here we can clearly see that in BF3 there is the partial double bond character between atoms which is due to back bonding (2pi-2pi), and gradually back bonding weakens and lewis acidic character increases so your final order will be. $$\ce{BF3}<\ce{BCl3}<\ce{BBr3}<\ce{BI3}$$

The following order is experimentally found:
$$\ce{BF3}<\ce{BCl3}<\ce{BBr3}<\ce{BI3}$$
This stands in contrast to what is expected when the electronegativity of the halides is considered. A more electronegative halide should be able to stabilize the negative charge at $\ce{B}$ in the Lewis acid-base complex better, and this would suggest that $\ce{BF3}$ would be the strongest Lewis acid.
However, the nature of the $\ce{B-X}$ bond needs to be taken into account. The change of hybridization from $\ce{sp^2}$ in $\ce{BX3}$ to $\ce{sp^3}$ in $\ce{[BX4]-}$ involves the loss of partial $\pi$ bonding which results from the overlap of the $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{X}$. The $\ce{p_z}$ orbitals of $\ce{B}$ and $\ce{F}$ are of similar size, because both elements are members of the same period. Therefore, they have the best overlap and form the strongest $\pi$ bond. Because the reaction with a Lewis base would require some energy to break this bond, $\ce{BF3}$ is a comparably weak Lewis acid. The larger $\ce{p_z}$ orbitals of the heavier halides have increasingly less overlap with the smaller boron $\ce{p_z}$ orbital. Therefore, the $\pi$ bonding is weaker and formation of the tetrahedral Lewis acid-base adduct via rehybridization becomes energetically more favorable.

d if uou can add back bonding concept answer will look more good

order is same as that of backbonding

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text:
Dissociation of ligand
If a ligand K[Ag(CN)2] is dissociated.. Which one is correct dissociation equation?
$$\ce{K[Ag(CN)2] -> K+ + Ag+ + 2CN^-}$$
or
$$\ce{K[Ag(CN)2] -> K+ + [Ag(CN)2]^-}$$

Dissociation $\ne$ ionisation.

Three different types of dissociation can occur here, the two mentioned by the OP:
$$\ce{K[Ag(CN)2] <=> K+ + Ag+ +2CN-}$$
$$\ce{K[Ag(CN)2] <=> K+ + Ag(CN)2^-}$$
And one where only one cyanide ligand dissociates from the silver:
$$\ce{K[Ag(CN)2] <=> K+ + AgCN + CN-}$$
However, as Klaus points out in his comment dissociation $\ne$ ionization. Ionization is a chemical reaction that generates ions where there were none before. Consider the ionization reaction of ammonia $\ce{NH3}$ in water:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
I'm not sure if this type of reaction is possible, although the cyanide anion is the conjugate base of a weak acid $(\ce{HCN}, \ \text{pK}_a =9.21)$. Thus, any liberated cyanide ion could react with water (assuming this is an aqueous solution!), though I would hardly call this an ionization (as there is no net increase in ions).
$$\ce{CN- + H2O <=> HCN + OH-}$$

Just a note: This wiki page seems to be in contradiction with your ionization reactions. Ionization occurs only when photons are incident or electrons are hit to the atom. Dissociation deals most probably with the splitting of the salts into corresponding ions, particles, or radicals. So, I don't think the last two equations are ionization "reactions" instead they might still represent dissociation. I agree with all the other parts of the answer:)

The last two equations are acid base reactions, i.e. proton transfers. When dealing with dissociation reactions like the mentioned ones, one must consider water (or another solvent) or pretend it is gas phase. But then these would be ionisations as there are ions produced from neutral molecules.


text:
Searching for a solvate that changes its solvation property
I'm looking for a solvate that changes its solvation property when being dried.
One example of such a material is gypsum:

at temperatures up to 250°C it becomes γ-anhydrite which will slowly
hydrate to a hemihydrate.
at temperatures above 250°C it becomes
β-anhydrite which will not hydrate again.

However the temperature range is too high to use it as example in my thesis.

Info on hydrates here:  http://www.lycaeum.org/mv/mu/MgSO4.html

http://nvlpubs.nist.gov/nistpubs/jres/086/jresv86n2p181_A1b.pdf
http://www.phasediagram.dk/binary/magnesium_sulfate.htm
Sodium sulfate hydrates,
http://iopscience.iop.org/0022-3727/41/21/212002/fulltext/
http://www.springerimages.com/Images/Chemistry/1-10.1007_s00269-008-0256-0-0
DOI:10.1209/0295-5075/102/28003
http://ej.iop.org/images/0295-5075/102/2/28003/Full/epl15377f5_online.jpg
http://www.phys.tue.nl/nfcmr/PhD-Saidov-2012.pdf
====
http://patentimages.storage.googleapis.com/US7638109B2/US07638109-20091229-D00001.png
http://www.genchem.com/properties.asp
sodium carbonate


text:
Calculus and physical chemistry
Is calculus and differential equations really that useful in physical chemistry? How does knowing that helps me in physical chemistry? And how much of calculus must I know in order to do physical chemistry topics such as thermodynamics? Because I am currently a chem major and am wondering what maths courses would be relevant. I currently completed single variable calculus and intend to go on to multivariate calculus as well as differential equations. I wonder how useful that'd be.

University library, physical chemistry texts, look; Atkins, etc. While you are there, you must pass theoretical chemistry, too - all those bras and kets, psi. If you do not love what you are doing beyond all rational justification, go into management. 'Tis better to give than to receive, especially when whips are involved.

Indeed, my university requires that Chemical Engineers take math all the way up to partial differential equations.

Calculus is the basis of physical chemistry. Have a look to Maxwell relations,Clapeyron equation. How stated Uncle Al you should take a book of physical chemistry and look into, furthermore some books (see Atkins) have an appendix showing you the mathematical skills required to understand the content of the book. Here what you should know only for thermodynamics in my opinion:

algebra
logarithms
vectors
calculus
Partial derivative
differential equations

If you go deep into the atoms more mathematical skills are required.

My university required calculus (2 courses) and differential equations (1 course) to complete the chemistry degree.


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Easy method of analysing iodine content in supplements?
Most health-food stores carry supplements supposedly containing iodine for people with iodine deficiency. I recently bought some kelp tablets that are supposed to have iodine but as a consumer how can I be sure that there is any iodine in them at all, let alone the claimed minimum daily requirement.
How do I know that I'm not just taking green-colored sawdust? Is there a cheap way to test a kelp tablet for its iodine content? If not are there testing labs that I could send some kelp to for testing that isn't going to cost me thousands of dollars?

Take some potato starch dissolved in water (you can easily get it from the water when you boil potatoes), and put a tablet into the water. If the water becomes bluish black, there is iodine in the tablet.
Yes, there are testing labs that will determine the amount of iodine in the tablets which will cost somewhere around \$100 to \$200.

The deeply coloured iodine-starch complex actually relies on triiodide, which itself comes from the reaction between iodide anions and molecular iodine. I imagine the iodine compounds present in the tablet are in the form of iodide or iodate, with no molecular iodine. Therefore, without the addition of a mildly oxidizing or reducing agent, there will probably be no formation of triiodide ions, and the starch mixture could remain colourless/white even in the presence of iodine compounds.

@NicolauSakerNeto Actually, any halide (e.g. $Cl^-$ from table salt) will complex with $I_2$, it doesn't have to be $I^-$.

Ah indeed, there are several interhalogen species which share some similarity. But still, the starch colour test should require the presence of a trihalogen anion (iso or hetero), which shouldn't be present in the tablet, and their formation requires the presence of a molecular halogen, which would also likely have to be added or produced in situ.

@NicolauSakerNeto Did you noticed that the OP stated kelp tablets. When kelp grows, it accumulates $I_2$ from the seawater and there is about 0.4% $I_2$ present in dried kelp. Since $I_2$ is volatile, the processing of kelp is done in a manner to minimize the lost of $I_2$.

Oh I knew seaweed was rich in iodine, but was unaware it was stored in the form of molecular iodine; I figured it was present as an anion such as those used to iodize table salt. Please excuse my comments then!

Just for reference there are at least other two methods, not so safe and easy to perform, but that can help you:

treat with a little bit of $\ce{HNO3}$ 1 ml of your dissolved sample add 0.5 ml of $\ce{AgNO3}$ 0.5 M. If $\ce{I-}$ is present, you should see a yellow precipitate: $\ce{AgI}$.
more dangerous!: put in your test tube 50 mg of your substance add 1 mL of concentrated sulfuric acid. Heat the test tube. Violet vapors will appear if $\ce{I-}$ is present.

Kelp is exceptionally good at scavenging the Periodic Table, as in Fukushima radwaste now bathing the US West Coast.  Iodized salt takes care of your iodine requirement, and generic multivitamins, and seafood and seaweed (sushi) overall.  You would need work at iodine deficiency unless you live in the Midwest with an idiosyncratic diet.  You can buy iodide pills at a pharmacy.  If a food supplement bottle does not have USP or NF in bold letters on its label, you should not put it into your mouth - no certified assay of contents, amounts, or bioavailability.
Faith-based engineering is rife with tests of faith.

I'm a vegan and I don't eat fish, I have palmar and plantar fibrimatosis and according to some holistic healing guru's, these diseases are caused by lack of iodine, (and in my case, 35 years of alcohol abuse) :-(


text:
Confused with oxidation numbers
If -2 is a oxidation number and 2- is the charge, then what is - by itself, is it a oxidation number or charge? As in $\ce{Ag+}$.
Also do you see $\ce{Cu + 2Ag+ -> Cu^{2+} + 2Ag}$ as a redox reaction?
Because I don't see any increase or increase in the oxidation number of any element in this equation. All I see is charges

According to the IUPAC gold book, oxidation state is:

A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules: (l) the oxidation state of a free element (uncombined element) is zero; (2) for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;...

Let's consider your reaction $Cu + 2Ag^+ \longrightarrow Cu^{2+} + 2Ag$ The superscripts represent charges. Reaction equations typically don't include oxidation states. In the reactants, $Cu$ has no charge and thus an oxidation state of 0 (rule 1). $Ag^+$ has a charge of +1 and thus an oxidation state of +1 (rule 2). In the products, $Cu^{2+}$ has a charge of +2 and thus an oxidation state of +2 (rule 2). $Ag$ has no charge and thus an oxidation state of 0 (rule 1).
In molecules, oxidation state can be thought of the charge of an atom when all bonds are treated as ionic. For example, in $H_2O$, Both $O-H$ bonds are polar and, since $O$ is more electronegative, all bonding electrons can be considered to belong to $O$. Thus, $O$ has 2 bonding pairs and 2 lone pairs for a total of 8. $O$ has 6 protons so it's oxidation state here is $6-8=-2$. Each $H$ has no electrons and 1 proton for an oxidation state of +1. The total oxidation state of the molecule is $2(+1)+-2=0$, which matches its charge.

To add, and to answer OP's ultimate question, "do you see this as a redox reaction?", the answer is yes. The left-hand side (Cu 0 oxidation #, Ag 1 oxidation #) --> right-hand side (Cu 2 oxidation #, Ag 0 oxidation #), Cu has oxidized (lost two electrons, increasing its Oxidation Number) and Ag has reduced (gained one electron, decreasing its Oxidation Number)

In a chemical equation, the oxidation number is not provided, so any numbers are either the number of molecules, or the charge on the ions.
To determine the oxidation number, it is the number of electrons removed or added to the element to get to the stated charge. For example; $Cl^-$ has an oxidation number of -1; $Ca^{2+}$ has a oxidation number of +2; and in $SO_4^{2-}$, sulfur has an oxidation number of +6.


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Chirality of buta-1,2-diene
My teacher says buta-1,2-diene is symmetric and hence optically inactive but I think it's the opposite, because allenes with an even number of double bonds are optically active because they have no plane of symmetry. Which answer is correct?

Let's see if a picture can help.  Newman projections can be handy in analyzing the stereochemistry of allenes.  On the left is a Newman projection of methyl allene (buta-1,2-diene).  There is a sigma plane of symmetry that is perpendicular to the screen and contains the 3 allenic carbons plus the methyl group.  Any molecule with a plane of symmetry is achiral.  On the other hand, look at the Newman projection to the right, 1,3-dimethyallene (penta-2,3-diene).  There is no plane symmetry in this case, only a $\ce{C_2}$ axis.  Draw the mirror image of the pictured allene and you will find that it is not superimposable on its mirror image. 1,3-dimethylallene is a chiral molecule.

2,3-butadiene is not a IUPAC name. It should be 1,2-butadiene. I think you are talking about 1,3-butadiene (natural rubber). In any case, neither of the molecules have a chiral center; therefore, no optical activity.

Buta-1,3-dien is not an Allene, so I think the OP is actually talking about Buta-1,2-dien.

Yes, exactly martin

OK, my mistake. But there is no chiral center since C1 has 2 hydrogens. 2,3-hexadiene would be chiral.

As is penta-2,3-diene (1,3-dimethylallene).

Also, natural rubber is comprised of isoprenes (i.e. isopentadienes), not butadienes.


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Addition of NaCl and CaCl2 to snow
In cold countries, it is a common phenomena to add salt to melt or clear snow. However, only specific salts are being commercially. So my question is that why only sodium chloride and $\ce{CaCl2}$ are used for addition to snow, to melt/clear it?

Salts will lower the melting point of snow or ice.
$\ce{NaCl}$ is chosen because it is cheap.
Edited: Salt with more total ions is more effective. $\ce{CaCl_{2}}$ is the one of the most available salt with 3 total ions. $\ce{MgCl_{2}}$ is another option.
Some crude $\ce{NaCl}$ produced from sea water directly contain $\ce{Ca}$ and $\ce{Mg}$ salt and can be used to melt ice.

But depression in freezing point is a colligative property which should only depend on the number of solute particles and not the nature of the solute. Right?

For ideal solution, there is the Blagden's Law: $\Delta T_{f} = K_{F} * b * i$ where $K_{F}$ is a constant, $b$ is the molarity and i is the van 't Hoff factor, which is 2 for $NaCl$ and 3 for $CaCl_{2}$.

@IanFang, minor correction: $b$ is molality, not molarity.

@GregE. you are right. Thank you for correction.

Sodium chloride is used because it is cheaper than most of the others salts.
Sodium Chloride $\ce{NaCl}$ / Industrial Salt : US $65-95 / Metric Ton
potassium chloride industrial grade: US $90-130 / Metric Ton
Prices taken from http://www.alibaba.com/


text:
Is sodium chloride really odourless? If yes, what do I smell?
I just had to read some general descriptions of sodium chloride and it was always classified as odourless (e.g., by Wikipedia). However, large amounts of table salts (around 1 kg) have a clearly perceptible odour to me. By table salt I mean salt sold for the purpose of cooking – I have not been taking notes or making experiments, but as I had the experience on dozens of different occasions, I would guess that the salt does not need to be iodised.
If I had to describe the smell, I would first say salty. It’s somewhat similar to the smell of the seaside, but the latter is much richer and has additional components. From the ten smell categories proposed here, only sweet and chemical fit somehow. It also has something of petrichor.
Is the classification as odourless just an approximation or do I smell additional ingredients (such as iodide) or impurities that cause table salt not to be pure sodium chloride? In the latter case: What causes the odour?

I see the same thing, too. Especially when I come near to sea.

What kind of smell do you detect? Can you describe it?

@YashasSamaga: See my edit. Now I have to explain, why I have a bowl of salt next to my computer.

The smell of the sea is due to the presence of dimethyl sulfide. I am not quite sure why someone would add dimethyl sulfide to common salt.

@YashasSamaga: I am not saying that it’s precisely that smell. It’s just something my brain associates with it.

Can you approximate the smell to more compounds? Is sea water the only matter you can think of? The issue which this question is that the answer is not well defined. It depends on the source of the common salt. But I like chemical hunting :P

I kept a moist litmus paper above a box filled with $\ce{NaCl}$ for 24 hours. The litmus paper turned slightly red. It could've been chlorine gas but that looks unlikely. There is some acidic component in my table salt.

Your table salt is not just $\ce{NaCl}$; there are countless other compounds which make up the table salt.
Here are a few:

$\ce{NaCl}$ - the main component
Iodine in the form of $\ce{I-}$ and $\ce{IO_3^-}$ salts (salts of potassium and sodium usually).
Iodine Stabilizers: $\ce{I2}$ sublimes at room temperature; therefore, stabilizers are added to prevent the loss of iodine. A commonly used stabilizer is dextrose.
Iron is added in the form of ferrous fumarate.
Folic acid and other vital vitamins are also added.
Fluoride salts are also added in many countries.
Anti-caking agents such as $\ce{Na4[Fe(CN)6]}$, $\ce{CaCO3}$, $\ce{MgCO3}$, $\ce{Ca3(PO4)2}$ and fatty acids are added.

From the big list above, fatty acids usually have an odor. While solids have a negligible vapor pressure, some molecules do leave the surface. Therefore, you can still smell solids.

The main component of the sea smell is dimethyl sulphide ($\ce{CH3-S-CH3)}$. In the ocean, dimethyl sulphide is produced as a byproduct by the bacteria which decompose phytoplankton.
The manufacturer of the common salt you use might be using sea water as the raw material. The sea water might not have been thoroughly cleaned which could have left traces of compounds found in sea water in your salt. You might want to try buying salt manufactured by a different company.

The manufacturer of the common salt you use might be using sea water as the raw material. – I observed this with many salts, some of which are explicitly not made from sea water.

My salt smells too but it smells like some chemical (and I need to get very close to it to recognize the smell). It does not have the distinct strong smell of the sea water, however.

Since odour is generally coming from vapour / gas form of a material, and NaCl has zero vapour pressure, it should be odourless.
However:

Moving around solid materials, dust, small particles can fly around that can be dissolved in the nose and interact with the odour sensors.
You almost always have something else together with salt. And if it is not pure, you cannot be sure what makes the odour, the salt or the impurity. Eg. sea salt often contains remains of algae.

If the dust particles moving in space are the cause of the smell, then shouldn't the OP detect that smell everywhere? Since OP is specifically asking about the odor of common salt, that should imply that random air particles aren't the source of smell in their case.

I am talking about small particles of salt and the impurities of salt in the particular case. Maybe it was not clear, sorry.

Of course NaCl has no vapor pressure (a property of liquids), but it does have some sublimation pressure (a property of solids). These are often conflated, but looking at the NIST property data the point seems moot: http://webbook.nist.gov/cgi/cbook.cgi?ID=C7647145&Mask=4&Type=ANTOINE&Plot=on

@GavinKramar , the link you gave gives the Antoine equation parameters for the vapor pressure of NaCl. The CRC Handbook gives both sublimation pressure and vapor pressure for ice. I've rarely seen the term sublimation pressure used. Is there is an official, say IUPAC, preferred terminology?

@GavinKramar Since it is an ionic material, we can agree that sublimation pressure is effectively zero for room temperature, too. Odor arising from sublimation is much more typical to molecular materials, like iodine or fats, organic acids.


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Finding the equivalent point of a weak polyprotic acid when reacted with a strong base
I have a graph of a titration of a weak polyprotic acid with a strong base.
I graphed $\mathrm {pH}$ versus amount of strong base added.
Now since the acid is polyprotic, how do I determine which equivalence point is the final one?
Also not sure how to find the initial concentration of the weak acid? I know the volume I put in, as well as the volume of strong base at any given time.

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question. For more information, see the site's homework policy for how to ask homework questions. Thanks! It would also be very nice to know which reactants you used and you could maybe post your graph.

There are multiple equivalence points in any titration graph. Polyprotic acids have even more.

A simple way to determine whether an equivalence point in a titration is the last one is to figure out the volume of titrant added for the first titration point, then realize all subsequent equivalence points will be at exact integer multiples of this volume, under the assumption that the analyte contains only one acid (a mixture of acids won't behave the same). When an integer multiple of the first equivalence point volume fails to be an equivalence point itself, then there will be no further equivalence points.
For example, let us imagine a generic polyprotic acid with formula $\ce{H_{k}A}$, being titrated with a strong monoprotic base such as $\ce{NaOH}$. Starting from the fully protonated acid, let us neutralize each ionizable hydrogen in steps. Let the acid  $\ce{H_{k}A}$ being titrated have an initial concentration $c_{acid}$ in a solution of volume $V_{acid}$, and let the titrant have a concentration $c_{base}$ and a total volume added $V_{base}$.
By definition, to reach the first equivalence point ($V_{base}=V_{eq\ 1}$), we consider the reaction which represents the neutralization of only the first proton:
$$\ce{H_{k}A + OH- -> H_{k-1}A- + H2O}$$
Each one mole of acid  $\ce{H_{k}A}$ must surrender exactly one mole of hydrogen ions $\ce{H+}$, each of which will react with exactly one mole of hydroxide ions $\ce{OH-}$. Thus:
$$n_{\mathrm{released}\ \ce{H+}}=n_\mathrm {acid}=c_\mathrm {acid} \times V_\mathrm {acid}$$
$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{base}=c_\mathrm {base} \times V_\mathrm {eq\ 1} $$
$$n_{\mathrm{released}\ \ce{H^+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 1}$$
$$V_\mathrm {eq\ 1} = \frac{c_\mathrm {acid} \times V_\mathrm{acid}}{c_\mathrm {base}}$$
Now, to reach the second equivalence point ($V_{base}=V_{eq\ 2}$), we consider the reaction which represents the neutralization of the first and second protons:
$$\ce{H_{k}A + 2 OH- -> H_{k-2}A^{2-} + 2 H2O}$$
Each one mole of $\ce{H_{k}A}$ must surrender exactly two moles of $\ce{H+}$, each of which will react with exactly one mole of $\ce{OH-}$. The relevant equations are now:
$$n_{\mathrm{released}\ \ce{H+}}=2\times n_\mathrm {acid}=2\times c_\mathrm {acid} \times V_\mathrm {acid}$$
$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{eq\ 2} $$
$$n_{\mathrm{released}\ \ce{H+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  2\times c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 2}$$
$$V_\mathrm {eq\ 2} = 2\times \frac{c_\mathrm {acid} \times V_\mathrm {acid}}{c_\mathrm{base}} = 2\times V_\mathrm {eq\ 1} $$
Notice how the added volume to reach the second equivalence point is exactly twice that added to reach the first equivalence point. This pattern repeats, i.e. the third equivalence point will happen at $V_\mathrm {base}=3\times V_\mathrm {eq\ 1}$, and the $k$-th equivalence point will happen at $V_\mathrm {base}=k\times V_\mathrm {eq\ 1}$. The equivalence points are equally spaced.
In short, if the titration of a pure polyprotic acid shows the first sudden pH swing after adding $\pu{15 mL}$ of the base titrant, then $V_\mathrm {eq\ 1}$ equals 15 mL, and the only possible next equivalence points are at $\pu{30 mL}$, $\pu{45 mL}$, $\pu{60 mL}$, etc. If during the titration another sudden pH swing is observed at $\pu{30 mL}$, but not at $\pu{45 mL}$, then you can deduce that there is no third equivalence point, and hence the acid is diprotic.

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


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What is the process involved in performing a mass spectrometry experiment?
I have heard that mass spectrometry is very time consuming: why is this?  What is the process involved in performing a mass spectrometry experiment?  What steps are the most time consuming?

Please do not cross-post your questions.  I think it's more appropriate over here and I have recommended that the Biology version be closed.

You submit the sample, it is shot in, the databased answer is e-mailed back, your PI gets billed.

This is hard to answer, because in my experience mass spectrometry isn't that time consuming at all. The experiment itself doesn't take that long. In this experiment the sample is ionized and charged fragments are accelerated in an electrical or magnetic field. Depending on their mass/charge ratio they will end up on specific locations in the detector.
The process around the experiment itself is dependent on your specific lab, but typically the mass spectrometer is operated by only a few people that do all the samples of the entire department. Because of that, there will often be a queue which could be several days. Additionally, not all mass spectra can be digitally identified as coming from a certain chemical, especially for larger molecules there are often various options that require human interference for the identification. In our lab this human interpretation is left to the person owning the sample, but if that part is also done by the mass spectrometer operators (I know in some labs this is the case) then I can imagine that this last step will be time-limiting.
To summarize: the experiment is short, the procedure around it (dependent on your specific lab) can take quite some time.

The problems are mainly in the pre-analytical steps, in most of the cases you can't analyze the sample as it is. You have do perform extraction, digestion and so on. Then if you want to have quantitative result you have to use standards. The calibration can be a quite difficult process.
So the time required depends from your sample and what and how you want achieve from your analysis.


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Why does the equilibrium law work mathematically and what does it really mean
For example consider:
$$aA + bB \rightarrow pP + nN$$ where $a, b, p$ and $n$ are moles.
The equilibrium constant is $$ \frac{[P]^p [N]^n }{[A]^a [B]^b}$$ Where does this come from? I don't understand why anyone would multiply the concentrations of the products together and then divide it by the product of the reactants together. There doesn't seem to be anything in my prior experience to justify that. And why are the exponents the number of moles? What is the physical meaning of an equilibrium constant?

Check other questions on the site to see if something helps. For example, try this.

I've given a full derivation here.


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Configuration of a water molecule?
Water molecules are described as an oxygen atom covalently (ignoring for the monent protonization/deprotonization) bonded to two hydrogen atoms with an angle of ~104° between the hydrogen atoms (like Mickey Mouse). I can understand this with multiple water molecules (the stacking would - for want of better descriptive powers - surround the oxygen's relative negative charge.)
Would a single water molecule retain this configuration? It would seem that in space, it should look like CO2?

It is probably useful to google for "why does water have a bond angle of 104.5". Once you understand why the 104.5 angle occurs you will also be able to answer your own question

The linked Google search mainly finds "Why is the bond angle of water 104.5$^\circ$ and not $109.5^\circ$. If using the VSEPR model, the nonbonding electron domains on water (the lone pairs) produce some "repulsion" contributing to the overall shape. The repulsion is minimize is tetrahedral arrangements. If using quantum mechanics, the shape of water arises from the minimum energy solution to the multielectron wavefunction describing the bonding in water.

The HOH bond angle for an isolated water molecule is accurately known to be 104.5°  (104.52°±0.05° from Rotation‐Vibration Spectra of Deuterated Water Vapor J. Chem. Phys. 24, 1139).
In the liquid phase, values of 105.5° (calculated) and 106° (experimental) are reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.
The gas phase water dimer (two otherwise isolated water molecules hydrogen bonded to each other) has also been considered and an angle of 104.7° is reported in the above reference.
For ice Ih (common form of ice), values ranging from 106° to 108° have been reported.
Attempting to explain the trend, the intermolecular hydrogen bonding in the liquid and solid phases weakens the intramolecular covalent O-H bonds, and from a VSPER approach causes the electrons of the O-H bond to be more like the lone pair electrons.

Omg, I forgot to accept an answer! Please forgive the awful delay. This was very helpful, and though I'm not of the ovine family, I feel very sheepish!

As shown by Uncle Al there is some (old) experimental data on the bond angle of water in this article: Shibata and Bartell J. Chem. Phys. 42, 1147 (1965).
For water in vacuum they find:
For $\ce{H2O}$: r_g (OH) = 0.976±0.0030 Å, mean HOH angle α_g = 107.2°±3°
This means that there is no statistically significant deviation from the value at atmospheric pressure.

(-1) for "liquid water in vacuum".  Clearly, the reference is about gas phase water, and the reference recognizes that the angle is much more accurately known from spectroscopy.  The gas phase water bond angle is 104.52°±0.05° from J. Chem. Phys. 24, 1139 (1956).

@DavePhD You are right. Changed it. Anyway, that was besides the point from my perspective (although that's no excuse for the careless error). I only wanted to point out that $107.2\pm3$ is not a statistically significant deviation from  $104.5$

I agree it's not statisitically significant and commented on Uncle Al's answer too.  But does the reference you cite really say what pressure the experiment was done at?

DOI:10.1063/1.1696094
   You want the electron diffraction structure of water in vacuum.
DOI:10.1038/248405a0
   Liquid water
For $\ce{H2O}$: r_g (OH) = 0.976±0.0030 Å, mean HOH angle α_g = 107.2°±3°
For $\ce{D2O}$: r_g (OD) = 0.970±0.0025 Å, mean DOD angle α_g = 104.2°±3°
It looks like the HOH angle slightly opens in vacuum as opposed to condensed phase.  That is 1965.  See if you can find a more recent study.

$107.2\pm3$ includes $104.5$ so I don't see a reason to assume that the HOH angle opens up in vacuum.

Thank you very much! That answers my question, and paired with the reading, helps me to better understand the significance of it.

@Michiel  If you were conversant with statistics, the angle opens in vacuo.  Find a more recent electron diffraction, ground state microwave, or vibronic spectroscopy structure for tighter angle constraints.  To criticize is to volunteer.

@UncleAl I am certainly aware of statistics and you are wrong on this one - if you give $107.2 \pm 3$ that means that you trust the measurement enough to say that it can be any value within the range $104.2-110.2$. It can very well be the case that a newer experiment with a smaller error margin would result in e.g. $105.5\pm 1.5$ without contradicting the current measurement.

But, instead of arguing, let's have the other users decide. I've written an answer with the same information, but different conclusion

The gas phase HOH angle is 104.52°±0.05° from J. Chem. Phys. 24, 1139 (1956).  Contrary to the above answer, in the gas phase the HOH angle is slightly smaller than in the liquid and solid phases.


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Fraction of concentration of benzene in air water
We have a 120 ml water sample with an unknown concentration of benzene. If the bottle size is 160 ml (with 40 ml of air) and benzene has reached equilibrium, what fraction of benzene is left in the water?
We are only given Hcc (Henry constant - Cair/Cwater) 0.225 and Log Kow (to which I believe is needed for a later question) 2.53.
I'm not sure where to start. We do not know mole fraction so we cannot find partial pressure using Henry's law. Also this rules out Raoult's Law.
Also we can't use the Ideal gas law since we do not have a lot of the variables. I'm stumped. But we can make assumptions such as T for example.
Any suggestions?

You don't need the original concentration or any of the variables you are missing. Henry's Law gives you everything you need. The usual formulation of Henry's Law at equilibrium:
$$c_{i,s}=k_{H,i,s}c_{i,s}$$
Where $c_{i,air}$ is the concentration of solute $i$ in the vapor above the solution, $k_{H,i,s}$ is the Henry's Law constant for solute $i$ in solvent $s$, and $c_{i,sol}$ is the concentration of $i$ remaining in solution.
We can rearrange Henry's Law to give you what you need - a ratio of the amount of benzene in the air to the amount in the water:
$$k_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}$$
The other snippet of information you need is the relationship between moles and concentration. There is a reason you were given volumes.
$$n_{air}=c_{air}*V_{air}$$
$$n_{sol}=c_{sol}*V_{sol}$$
You only need the ratio of $\dfrac{n_{air}}{n_{sol}}$, so you really only have one unknown. It's algebra time!
$$K_{H,i,s}=\frac{c_{i,air}}{c_{i,sol}}=\dfrac{\dfrac{n_{air}}{V_{air}}}{\dfrac{n_{sol}}{V_{sol}}}$$
$$\dfrac{K_{H,i,s}V_{air}}{V_{sol}}=\dfrac{n_{air}}{n_{sol}}$$


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How strict is the "to excite electrons the energy must equal the energy state difference" fact?
We are always told that to excite an electron from one state to a higher energy states, for example from the valance band to the conduction band, the energy must equal the energy difference between the two energy states.
The conduction band consists of many allowed energy states, separated from the valance band by an energy gap. For the element Si, this energy gap is (at room temperature) about 1,1 eV. The range of energies for visible light is 1.8 eV to 3.1 eV. Since Si indeed conducts electricity at room temperature, is it then assumed that there is an allowed energy states with exactly the same energy as the energy difference between the two states? The conduction band, or the valance band for that matter, is not continuous, but discrete. Or is it sufficient that the photon energy "roughly equals", or is "close enough"? If so, how close is close enough?
I hope my question is clearly stated.

There are two things here mixed: 1) there is a conservation of energy, which requires the states to have a given energy (with the error allowed by Heisenberg uncertanity principle) and 2) the fact that semiconductors and such have bands instead of separate electronic states, i.e. there is a broad, continuous  margin whet the allowed transition energies are. Maybe you want to focus some part (1st?) in the question, instead of mixing them up.

In fact there are many factors to consider, so the answer to your question depends on what type of material you are analyzing.

First of all you should know that even if the absorption level are discrete the absorption and the emission of a material are never a single value but a peak. If you consider the absorption or the emission of a gas you can assume them discrete value (even is not completely true), and so you can call them lines (see e.g. Lyman series), but if you condense the matter the interaction between the atoms/molecules broad the absorption/emission range and so you have peaks. If you are talking about electrons you are dealing with UV-VIS spectra in this case if you look at solid or a liquid UV-VIS spectra you can see that you don't have a well define peak but a continuous absorption, in this case you have bands. So it really depends of the state of the material you are analyzing.
There are other factors that determine the broadening of the absorption to multiple wavelengths. Form the previous explanation you can derive that more an atom is isolated, less interaction occurs, and the absorption is more limited to determinate wavelength, because you have similar "oscillators". But if you increase temperature and pressure you increase the probability that some atoms/molecules interact, and have different energy levels. The effect is the same: you are broadening the absorption band, respectively you will have a _temperature and a pressure broadening.
So it seems that if you have a single atom this absorbs at a determinate wavelength. Even in this case is not true, the explanation is not so intuitive as the previous ones because you have to understand Heisenberg uncertainty principle but roughly it states that you can't determine precisely the energy states of an atoms. This last broadening is called Natural broadening.

For your specific case you should look at the the UV-VIS spectrum, you will not see a sudden step next to the absorption band but a gently curve. This means that there are less and less atoms absorbing in this region because statistically is less probable that atoms can be perturbed so much to absorb that wavelengths but there are still chances that some atoms absorbs even at wavelength far from the mean.
Have a look to wikipedia spectral lines page for more infos.


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Addition of salt to ice
Salt is very commonly added to ice even at room temperature to prevent it from melting. Can anyone explain how does it really work?

Uncle Al is correct.  Colligative properties are properties of solution in which only the ratio between number of particles of solute and solvent are considered (often expressed in different forms, such as moles of solute per kilograms of solvent for molality); the identities of either are irrelevant.  So we could dissolve sodium chloride in water or magnesium bromide, but the general properties will be the same (though note that the magnitude of freezing point depression and similar properties will be slightly different depending on the substance).
In particular for salt water, the point at which the solution will freeze is depressed.  To understand this, we must understand what occurs when pure water freezes.  In general terms, when a liquid freezes, its normally hectic and randomly moving particles form a stable, packed structure that we call a solid; a hexagonal structure in particular for water at standard conditions.  The specifics are more complicated, but this is the main idea.
However, if you introduce a solute into the pure water, such as salt, the solute particles are getting in the way of the formation of the packed structure.  As a result, the temperature must be lowered slightly further in order for the solvent (i.e. water) molecules to properly overcome the solute molecules acting as obstacles for the formation of the solid structure.
In fact, when the solvent freezes, more often than not, the solute particles are often excluded, either pushed outside or trapped inside small pockets.  In the case of salt water, a eutectic system is created, if you're interested in learning of them.  Salt water itself cannot freeze into salty-ice or otherwise.

Colligative properties, then freezing point depression.  The melting point of ice is lowered -1.86 C/molal of solute particles.  One molal is one mole of particles/kg of solvent (NOT solution).  Addition of salt lowers the melting point of water, causing the ice to melt, absorbing the latent enthalpy of fusion, until the system temp lowers to its equilibrium value.
If you want your ice to stick around, thermally insulate it from the surrounding environment.  You must interdict conduction, radiation, and convection.


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What is the appropriate procedure for determining the quantum numbers of an electron?
How do I find the set of quantum numbers for a specific electron in an element? For example,

Calculate the set of quantum numbers for the 19th electron in chromium.

The electronic configuration of the 19th electron is 1s2 2s2 2p6 3s2 3p6 4s1. From here on, how do I decide what would the principle quantum number be?

This sounds like a homework problem.  Could you please write out your attempt at solving this question?  For example, what are the quantum numbers and how do they relate to atomic 'structure'?  One example of this would be that the principal quantum number, n, is the electron shell of an atom.

the electronic configuration of the 19th electron is 1s2 2s2 2p6 3s2 3p6 4s1.
from here on, how do i decide what would the principle quantum number be?
@LordStryker it isn't a homework problem.. i'm just trying to clear up my concept which is pretty much messed up :/ i put in the example to make it more clear as to what direction would i like the answer to proceed..

Okay so, from your example (1s2 2s2 2p6 3s2 3p6 4s1), the principle quantum number for the 19th electron is what?  You have it written right there in your provided electron configuration.  N refers to something.  How about the angular momentum number (L)?  This number refers to the shape of the orbital.  S orbitals have different shapes from p orbitals (sphere vs. dumbell).  You have the answer written out in your provided configuration.  The only other number that remains (if neglecting spin), is the magnetic quantum number(m_l).

Note: We're getting closer. Try using this site as a reference. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

the website was really helpful! :)

1s2 2s2 2p6 3s2 3p6 4s1 is the electronic configuration of the 19th electron of chromium.
since the last shell is 4, therefore n=4,
l=0(as there is only one orbital in the s sub shell) ,
m=0 (since we're still in the s sub shell only)
and s=+1/2 or s=-1/2(arbitrarily assigned values)
note:+1/2 or -1/2 are used to distinguish between electrons as no two electrons can have the same electronic configuration, but here the question isn't that specific, so we need not bother.

Electron spin is not necessarily arbitrary in the absence of another electron.  Interestingly, the nucleus is a point of reference for spin to be defined and there is a very real, physical consequence with respect to which spin the electron has.  This comment is included just for fun.


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Why is a singlet state called singlet and a triplet state called triplet?
I kind of get the idea of singlet and triplet states. But why are they called singlet and triplet (what is the single and what is the triple in these cases)?
I feel that I am missing something obvious!

The terms arose back in the early days of quantum physics when spectral lines that were expected to be singlets were actually observed to be more complex (doublets, triplets, etc.).
An electron can have a spin quantum number of $+\frac12$ or $-\frac12$.  For a system that exists as a singlet, all spins are paired and the total spin for the system is $S=0$.  If we have a single electron, $S=\frac12$.  If we have a triplet system with 2 unpaired electrons, $S=1$.  Also associated with these electrons/ systems is a spin-angular momentum vector, $L$.  Quantum mechanics tells us that $L$ is allowed to have $2S+1$ distinct values.  So
for a singlet, $L = 2(0) + 1 = 1$     only one state exists (singlet)
for a doublet, $L = 2\left(\frac12\right) + 1 = 2$   two states exist (doublet)
for a triplet, $L = 2(1) + 1 = 3$     three states exist (triplet)

An exceptionally well given answer! I hope you do not mind, me putting some more formatting into it (If it will be accepted by peer review).

It might be worth adding that spectral lines were only observed to be split in the presence of magnetic (Zeeman effect)  or electric (Stark effect) fields. Initially some of the observed lines could not be explained at all  hence were called 'anomalous'. After  spin was understood it all did make sense and no lines are any more called  'anomalous'. In molecules it was not until the mid 1950's that triplet states were identified with the then new techniques of time-resolved spectroscopy developed by Porter and called 'flash-photolysis' but is now called 'pump-probe' spectroscopy.

A pair of electrons, being fermions, must have antisymmetric wave function, i.e. if $\psi(\xi_1,\xi_2)$ is a wavefunction describing the system, where $\xi_1$ are position and spin of electron 1 and $\xi_2$ is position and spin of electron 2, then $\psi(\xi_2,\xi_1)=-\psi(\xi_1,\xi_2)$.
In the first approximation, spin degree of freedom can be separated from orbital degrees of freedom, so that the wavefunction becomes $\chi(s_1,s_2)\phi(x_1,x_2)$, where $s_i$ is spin of $i$th electron, and $x_i$ is position of $i$th electron. Here $\chi$ is spin part of wavefunction, and $\phi$ is orbital part. To preserve total antisymmetry of the wavefunction, $\chi$ and $\phi$ can be either symmetric, or antisymmetric. If one is symmetric, the other must be antisymmetric.
The spin of a single electron can be up $\uparrow$ or down $\downarrow$. I.e. simplest options for a two-electron system could be $\uparrow\uparrow$, $\downarrow\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$. But the latter two don't honour indistinguishability of electrons. To correctly include indistinguishability of electrons, we should take symmetric and antisymmetric linear combinations of these spin states.
Now we have four options, split in two variants:
a) Antisymmetric orbital and symmetric spin part of wavefunction

$\uparrow\uparrow$
$\downarrow\downarrow$
$\downarrow\uparrow+\uparrow\downarrow$

b) Symmetric orbital and antisymmetric spin part

$\downarrow\uparrow-\uparrow\downarrow$

From here we can see that symmetric spin part of wavefunction gives rise to three different states — these are triplet states. If spin part of wavefunction is antisymmetric, there's only one such state — it's the singlet state.
When one makes spectroscopic measurements with not very high resolution, states with different spins but same orbitals will appear to have the same energies, so the spectral lines will appear the same. But if you put your system in magnetic field, you'll see that the spectral lines split according to spin multiplicities: spin-singlet states will remain single lines, while spin-triplets will split into three different spectral lines. This is the origin of such naming.


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Why is dialysis tubing impermeable to sucrose?
This question was posed a couple years ago but not answered adequately. You can view the older thread here. The cellulose membrane commonly used in high school and university labs is widely available from suppliers such as Wards, Carolina, Flinn, PASCO, etc. The specifications are either available online or were obtained by contacting the companies and in each case the molecular weight cut off (MWCO) was between 6,000 and 16,000 Da. Regardless of the specifications these products are used to conduct osmosis demonstrations by placing 0.2-1.0M sucrose on one side of the membrane and to create water potential. In addition to having conducted this exercise myself the results are validated in countless examples online. The mass/pressure results from osmosis and a benedicts test indicate sucrose is not moving across the membrane.
Considering Sucrose has a MW of ~342 Da the previous explanations seem insufficient;
1) The pore size of the membrane is variable. True, but according to specs at least 10% of the pores are larger than the specification and even with variation of 0-6,000 Da (worst case) it would still be like throwing a tennis ball through a garage door 10% of the time.
2) Interactions between solute molecules, shape, and solute-solvent causing a larger solvodynamic radius. Sucrose doesn't have any charges I'm aware of, it has a simple structure, and even covered in water molecules it would still be much smaller than the pores.
Any suggestions or alternative explanations?

Benedict's reagent detects reducing sugars (free aldehyde group) by reducing soluble blue Cu(II) to insoluble red Cu(I) oxide.  Sucrose aquoous solution is inert unless it is pre-hydrolyzed by heating with strong acid catalyst.
That certainly is a poser.  One possibility (SWAG, not fact) is regenerated crosslinked cellulose is overall not permeable to neutral solutes.  Its size selection is active toward dissociated solutes and proteins that are charged.  One can test this with anti-freeze.  The sodium fluorescene color should freely diffuse.  Tiny molecule ethylene glycol (EG) should be retained.
https://www.chemetrics.com/Glycol
  Test kit and how it works:  Iodate cleavage  of 1,2-diol to aldehyde, then aldehyde detection with Purpald.

One could iodate cleave permeated EG (or propylene glycol, glycerin, etc.) then use Benedict's solution.  Serial dilution of standard EG solution for control and calibration.  One expects any dialysis membrane to leak somewhat.  Major passage of uncharged 1,2-dihydroxylic solutes smaller and larger than than sucrose is the SWAG's falsifying observation.
http://chemsee-foods.com/antifreeze-detection
http://www.chemworld.com/Ethylene-Glycol-Test-Kit-p/k-4815.htm
http://www.hach.com/glycol-visual-test-kit-model-eg-1/product?id=7640218461
  EG in food and fluids tests.
http://www.astm.org/Standards/E202.htm
  Testing standards


text:
How to separate lead condensate from aluminum
A common problem in shooting sports is precipitation of lead onto aluminum suppressor baffles.  (The lead presumably comes from both lead styphnate priming compound and from the base and bearing surface of lead bullets that vaporize during propulsion.)
If the lead precipitate is allowed to build up a contiguous layer it requires mechanical removal, which tends to damage and degrade the aluminum alloys used.
A common solution for ferrous baffles is to strip the lead with H2O2-boosted acetic acid, but that goes after aluminum almost as fast as lead.
What practical electrochemical methods could be used to strip the solid lead from the aluminum?
(Bonus if you can propose a method that would also leave titanium baffles intact!)

You could try hard anodizing Al and Ti (e.g., Calphalon cookware) to give them a non-delicate surface, and go on from there.  Black is a good gun component color.  CVD deposition of titanium boride, ion plating of titanium nitride or carbonitride gives exceptionally survivable hard surfaces.
http://www.brycoat.com/coating-services/pvd-coatings/titanium-nitride-tin.html
http://www.brycoat.com/coating-services/pvd-coatings/ticn.html
Among mostly cheap and harmless diddles, I'd try 1:2 mole ratio of choline chloride  with urea or ethylene glycol ("all natural," etc.).  Mix, the deep eutectic liquefies, soak.  Both eutectics conduct electricity, so you could make lead contamination the anode and plate it out on a closely spaced but not touching copper foil cathode. Test on contaminated waste Al or Ti.  Careful with the voltage - too high and you mobilize Al and Ti.  Avoid contact with skin pro forma, do not ingest, keep away from eyes.
Try vacuum thermal deposition of a micron or so thickness of the Parylene family on the newly fabricated and cleaned metal - intrinsic, especially C, or F.  Parylene polymer surface coats survive the insides of excimer lasers, protecting the excitation electrodes.  If the lead still sticks, chemical cleaning should stop at the polyparylene film.  Or not.  Gotta test it.  You are worried about vapor deposition not abrasion, so an inert barrier film is reasonable.
Google, parylene coating services
For aluminum or titanium vs. lead in chemistry, I'd pull one out of left field to try on test coupons:  Rauchfuss' elixir: elemental sulfur dissolved in N-methylimdiazole (pyridine less so, very smelly, toxic, and don't get it on your skin or breathe the awful vapors).  It goes through polysulfide-forming metals like a buzz saw, giving soluble products.  I would expect aluminum to be inert; titanium is a toss-up.  It should hit lead hard, probably.  Test it small scale.
Inorg. Chem. 30 (11): 2514 (1991)
DOI:10.1021/ic00011a011
http://dx.doi.org/ (resolve a DOI)
Last gasp:  Weapon Shield to clean before and after firing.
Concentrated nitric acid (already a bad idea) passivates aluminum and eats lead.  Toxic and corrosive NOx fumes come off.  However...aluminum plus lead in galvanic contact make aluminum the sacrificial anode.  Homeland Severity does not like people possessing concentrated nitric acid.
Wear gloves and goggles, have ventilation.  None of this stuff is good for you.

Thanks.  Prophylactic coating (as with Weapon Shield) is the current best practice -- along with cleaning frequently enough that no solid deposition layers form.  Hard anodizing is often done but doesn't seem to help with this problem substantially.  Parylenes look interesting, but I have read that PTFE has been tried and failed because propellant temperatures can exceed 650F.  Assuming I can source it I'll certainly like to test the sulfur/methylimidazole.  First search shows methylimidazole to be super expensive.  Any ideas for cheaper sources?  I don't need lab grade.

http://www.chemistry.illinois.edu/faculty/thomas_rauchfuss.htm Talk with Rauchfuss. Pyridine is workable but terrible for the stink and some toxicity. If you stop smelling it you are OD'd.  340 C is beyond Parylenes dying about 200 C, event too high for phenyl silicones. Give deep eutectics a whack, especially with voltaic stripping.  Cheap, non-hazardous, available in bulk. It could be a kit for general civilian and military armory use.

Choline chloride is a bulk chicken feed additive.  Urea is a bulk nitrogen fertilizer. You certainly don't need reagent grade. Choline chloride is 139.62 g/mol, urea is 60.06 g/mol, ethylene glycol is 62.07 g/mol.  Thus one part by weight choline chloride to 0.86 parts urea or 0.89 parts ethylene glycol, both by weight. It works or it doesn't work. As you said, the problem is keeping Al and Ti inert while moving the lead - and disposing of it. You want Pb out net, not dissolved in a gallon of goo.

Looks like two excellent options.  I've contacted Prof. Rauchfuss about his elixir.  I'm drawing a blank on "eutectics" here.  Can you point me to some background on that chemistry?  Also basic parameters to try for the voltaic reaction -- e.g., what range of voltage likely to move the lead but not the light metals, and what that's a function of: e.g., distance to cathode, purities, masses of anything?  And when do the various components need to be refreshed -- e.g., will the cathode be impaired by some thickness of lead?  Thanks!

http://en.wikipedia.org/wiki/Deep_eutectic_solvent http://en.wikipedia.org/wiki/Ionic_liquid http://www2.le.ac.uk/departments/chemistry/research/lil http://www.chemspecevents.com/europe/images/uploads/chemsource/4_-_2_McKenzie.pdf (deep eutectics start about 55% down, including lead) Cell voltage depends on cell resistance in part, and the chemistry. This is not an aqueous cell. Try Pb-fouled metal around a volt oxidizing, see what happens. Closer electrode spacing is a more efficient cell. The cathode can keep accumulating lead. Cu is a better conductor.


text:
Questions concerning redox reactions
What does it mean for an element to be reduced?
What does it mean for an element to be oxidized?
What is an oxidizing agent?
What is a reducing agent?
How do you identify oxidizing and reducing agents?

Consider this a warning.  If you continue to post homework questions with no effort put forth, I will suspend your account.  You have been asked repeatedly to include this information in your questions and you have ignored it thus far.

@jonsca None of these are homework questions (I have said this repeatedly in response to your aggressive accusations).

It doesn't matter if they are formally assigned or not, they are still considered to be "homework" regardless.  These are the guidelines we have laid out for asking homework and homework-like problems on the site.  You have, to my recollection, never replied to one of my comments.

@jonsca I thought I did but unfortunately I did not. Can you please define what is categorized as a homework-like question and what I need to do differently when asking one? As none of these questions are "homework" in terms of my understanding of what homework is, I don't know what I need to do in order to rectify the situation.

I have stated what you need to do differently with the comments that I have left.  Simply add what you have tried already into the body of the question.  "I think reducing agents are X, Y, Z, because W" would allow someone to correct your understanding rather than having to start from scratch in their explanation.  It's also to your benefit to explain what you already know because that will help you in your own understanding.  A "homework question" is anything that would come out of a textbook. We're not trying to be jerks here, but rather attempting to make things fair for everyone.

@jonsca Alright. I am sorry if I came across as being rude, I was just a bit disgruntled that you were threatening to terminate my account...

Suspension is a "timeout" and is different from terminating your account, but I understand that I didn't make that clear.  Apologies for that.  Give the Meta post a read, it really does have a lot of helpful information for asking questions.

Here's a great example of a homework question: http://chemistry.stackexchange.com/questions/10210/relation-between-van-der-waals-forces-and-hydrophobic-interactions

My instructor told me to remember LEO and GER.
LEO - loss of electron is oxidation
GER - gain of electron is reduction.
Sodium metal wants to give up an electron. It looses an electron, so it is oxidized.
Chlorine likes electrons. It gains electrons, so it is reduced.
Sodium is a reducing agent since it donates an electron.
Chlorine is an oxidizing agent since it will grab an electron.
A reducing agent is oxidized and an oxidizing agent is reduced.

Thanks. I was having a hard time remembering the difference.

My instructor used OILRIG, oxidation is loss, reduction is gain. @okarin


text:
How to calculate the concentration of a K₂Cr₂O₇ solution?

How many ml of acidified $\ce{K2Cr2O7}$  solution, which includes 21.6 g of dichromate ions per Litre, are required for the oxidation of 44.4 g of  $\ce{CH3CH2CH(OH)CH3}$?

Obviously 44.4 g of $\ce{CH3CH2CH(OH)CH3}$ means 0.6 moles. So from the stoichiometry of the the oxidation we find out that 0.2 moles of $\ce{K2Cr2O7}$ are needed. I have trouble finding the concentration of $\ce{K2Cr2O7}$ though and thats why I need help.

To calculate the dichromate concentration (in moles/l) of the solution, simply divide the given mass of dichromate per litre (21,6 g) by the molar mass of $\ce{Cr2O7^{2-}}$.

First, you need a balanced equation for the oxidation.  The gives you the mole ratio of dichromate to substrate.  You know how many moles of substrate you have, you know how many moles of dichromate you need.  If you have 21.6 g of dichromate ions (as such)/liter, then you know how many moles/liter you have.  You know the number of moles you need, you know the number of moles/liter you have, you know the number of liters you need.
You will be doing your Jones oxidation in acid, of course.  Collins reagent is a sweeter deal for workup, since MEK is 27.5 g/100 ml water-soluble.


text:
How does glue work?
Simple question.
How does household glue work? How does it attach to almost any other compound? Thanks.

Different glues act with different mechanism, are you interest in any particular glue?

Not really, I'm just interested in adhesion in general, I guess. I'm even interested in how those blue things that teachers use to stick papers on the wall work..
Is it the same kind of bonding that happen at a molecular level? Covalent etc.?

I like this question :) +1

At a microscopic level most of the materials have many microscopic pores with different diameters and shapes, so in most of the cases any non solid material which can harden and which has the ability to go inside the tiny pores can be a glue of a determinate strength simply hardening inside the pores. This is called mechanical adhesion.
Furthermore the adhesion can be favored if some bonds between the molecules are formed so some glues are more effective for some sort of material. Generally van der Waals' force occurs spontaneously between compound with dipoles, this is called dispersive adhesion.
Hydrogen bonds can increase the strength of the glue. Covalent bond between glue and materials don't occurs often, I think that cyanoacrylate (super glue) probably form a covalent bond with the hydroxilated surface of the material (but nobody seems talk about this). In fact a real surface is always cover by other compounds so is very difficult that something interact directly with the surface, and this is why you should always clean (and in most of the cases dry) the surface before apply the glue.

There is glue (sets) and there is adhesive (stays sticky).  Glue is a fluid that wets both surfaces, then sets to a solid by solvent evaporation, reaction, or melt solidifying.  Solvent and melt glues are primarily mechanical bonds, especially for porous surfaces.  Reactive glues can be chemical bonds as well (resorcinol carpenters glue by oxidation, isocyanate Gorilla glue by nucleophilic attack, Super Glue by free radical grafting).  Solvent alone can weld plastic surfaces, such as cyclohexanone for vinyl (plasticized or filled PVC) or organic solvents in general for acrylic, polycarbonate, polystyrene, ABS, etc.
Failure of bonded surfaces can be through adhesion (interface separates) or cohesion (material separates).


text:
Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium
Is there a complete list of all the half equations for $\ce{H2O2}$ - both oxidation and reduction, in acidic and alkaline conditions? I've looked on the internet but can't seem to find a list with all of them.
These are my first attempts:

$\ce{H2O2 + 2e^- -> 2OH^-}$ (in alkaline conditions)
$\ce{H2O2 + 2H^+ + 2e^- -> 2H2O}$ (in acidic conditions)
$\ce{H2O2 -> 2H^+ + O2 + 2e^-}$ (in acidic conditions)

Are these correct and are there any more?

From Proceedings of the Symposium on Oxygen Electrochemistry and Water Research 2002, 36 (1), 85-94:
$$
\begin{array}{lcc}
\text{Formation}&:&~\ce{2H+ + O2 + 2e- &->& H2O2} &\quad E^0=\pu{0.440V} \text{ vs. SCE}\\
\text{Reduction}&:& ~\ce{H2O2 + 2H+ + 2e- &->& 2H2O} &\quad E^0=\pu{1.534V} \text{ vs. SCE}\\
\text{Direct water}&:& ~\ce{2H+ + 1/2O2 + 2e- &->& H2O}\\
\text{Decomposition}&:&  ~\ce{H2O2 &->& H2O + 1/2 O2}
\end{array}$$
($\text{SCE}$ stands for Saturated Calomel Electrode)
Another strange suggested reaction involves the anion $\ce{HO2-}$ (Abel E.: Über die Selbstzersetzung von Wasserstoffsuperoxyd. Monatshefte für Chemie 1952 83(2) 422–39.):
$$\text{Decomposition 2}:\ce{H2O2 + H2O- -> H2O + O2 + OH-}$$

At the cathode occur at the same time the hydrogen gas evolution:
$$\ce{2H+ + 2e- -> H2~} \qquad\qquad\phantom{00} E^0=\pu{0.242 V} \text{ vs. SCE}$$
Meanwhile, at the anode, oxygen gas is evolved:
$$~\ce{2H2O -> 4H+ + O2 + 4e-} \quad E^0=\pu{0.987V} \text{ vs. SCE}$$


text:
Is N-methylimidazole an unusual solvent, or are there more common substitutes for it?
I want to test a process that was documented in the literature using N-methylimidazole.
I am not doing academic research, but rather testing a documented reaction for application to an industrial process, so I don't believe I need lab-grade chemicals.
When I look for this solvent online I can only find lab suppliers (e.g., Matrix Scientific, Acros Organics, Alfa Aeser) selling tiny quantities at very high prices.
Is there something unique to this chemical as a "donor solvent" that is not reproduced by any more common or inexpensive solvent?  I have seen it mentioned frequently in organic chemistry papers, but since I can't find industrial suppliers I assume it's peculiar to research.
The application I'm interested in is chemical stripping of lead using Rauchfuss's Elixir, which consists of sulfur as an oxidant and a donor solvent.  I haven't been able to learn why they used methylimidazole as the solvent in the study.  Obviously I would prefer one more widely and/or cheaply available.

What kind of price per unit are you looking for?

Something significantly less than $100/L.

http://www.chemicalbook.com/ChemicalProductProperty_EN_CB1316726.htm
http://www.lookchem.com/cas-616/616-47-7.html
http://www.guidechem.com/cas-616/616-47-7.html  2000 RMB = $(USD)320
http://ahjinao.en.alibaba.com/product/595605676-209523668/Time_honored_Supplier_of_N_Methylimidazole_CAS_No_616_47_7_.html
http://www.alibaba.com/countrysearch/CN/n--methylimidazole-supplier.html
http://www.buyersguidechem.com/chemical_supplier/N-Methylimidazole.php
Info and suppliers,  Note that a Chinese contract is often not the legal or philosophical equivalent of a Western contract.

That's funny!  But I feel like I'm missing something if I have to go overseas to get industrial quantities of a common organic compound.

http://en.wikipedia.org/wiki/1-Methylimidazole  http://www.basf.com/group/corporate/us/en/brand/1_METHYLIMIDAZOLE America is regulated, legislated, and benefitted out of business. War on Drugs, War on Terror, the coming War on 3-D Printing. Miraculous Plasti-Dip is CA-banned: "solvent-based."  Aluthane, a very real exposure hazard with lower solids content, is OK - "solvent-based metallic paint."

From what I see in the paper they likely wanted a basic solvet, and pyridine may have had a high enough dipole moment.  For quantities this small you are going to pay a bit more than you would in bulk, though as far as organic reagents go the prices are fairly cheap.  Acros and Alfa-Aesar are both owned by Fisher-Thermo Scientific by the way. Try Sigma-Aldrich or TCI America if you haven't already.


text:
Why is mass proportional to the strength of a dipole-dipole attraction? meaning more mass more strength
I don't understand why the mass of a molecule make it attract more to other polar molecules.

Hi, take a look to the question posted by Philipp if this doesn't clear your doubts make one more specific or clarify what you mean! :-)

Increasing mass increases the availability for molecules to attract through London dispersion forces (induced dipole-dipole). A longer chain has more atoms and hence more electrons and therefore experiences stronger bonding.
When you say 'polar' I assume you mean permenant dipole-dipole and hydrogen bonding which is actually not affected by the 'mass' or length of a carbon chain in a molecule. However if you take ethanol and pentanol as an example they both have hydrogen intermolecular bonding on account of the -OH group but pentanol also experiences greater London dispersion forces than ethanol which results in its significantly higher melting and boiling points.
So to recap increase mass is not a direct causal link to charge. It is in fact more mass meaning more atoms meaning more electrons meaning induced dipoles leading to greater intermolecular forces. I hope that's clear (:


text:
Why is the groundstate electronic degeneracy of monoatomic O(g) 5?
A problem in my Pchem textbook states that the electronic degeneracy of an oxygen atom is 5. My reasoning leads me to believe that the degeneracy is actually 6 (S = 1, L = 1, so (S+1)(L+1) = 6. This would indicate there are two unpaired electrons and a pair in the P orbitals...
The only way I can reason a 5 degeneracy is if the ground state has a pair in the d-orbitals, which I do not believe is the case.

S = 1, L = 1, so (S+1)(L+1) = 6

You are using here not just alternative physics, but also alternative math. For, even if the formula were right, you should get $2 \cdot 2 = 4$ and not $6$.
Anyway, the right formula for the degeneracy is $(2S + 1)(2L + 1)$ which being used for the ground $^1D$ state of oxygen atom (i.e. for $S = 0, L=2$) indeed gives $1 \cdot 5 = 5$.


text:
Condition for formation of azeotropes
Consider a binary azeotropic solution of two volatile components $A, B$. I wanted to try and find out the required concentrations of $A, B$ in the solution in terms of their vapour pressures. I tried using Raoult's Law to do this, but the I got an unexpected answer:
Let the vapour pressure of $A, B$ at the current temperature be $P_A, P_B$, respectively. Let their concentrations be $\chi_A, \chi_B$ and their partial vapour pressures over the solution be $p_A, p_B$.
Then, by Raoult's Law,
$$p_A = P_A\chi_A$$
$$p_B = P_B\chi_B$$
By Dalton's Law of partial pressures, the mole fractions of $A, B$ in the vapor phase are given by $\frac{p_A}{p_A + p_B}$ and $\frac{p_B}{p_A + p_B}$ respectively. Since the solution is azeotropic, these two are equal to their mole fractions in solution.
That is:
$$\frac{P_A\chi_A}{P_A\chi_A + P_B\chi_B} = \chi_A \implies \frac{P_A}{P_B}(1-\chi_A) = \chi_B\tag1$$
Similarly,
$$\frac{P_B}{P_A}(1-\chi_B) = \chi_A \tag2$$
Substituting $(1)$ in $(2)$, yields
$$ \frac{P_B}{P_A}\left(1-\frac{P_A}{P_B}(1-\chi_A)\right) = \chi_A$$
$$ = \frac{P_B}{P_A}-1+\chi_A = \chi_A$$
$$\implies P_B = P_A$$
But there are numerous azeotropes where this condition is not satisfied. How come?

It is not that surprising that you got this answer: the existence of azeotropes is due to non-ideal behaviour of mixtures (see for instance the Wikipedia page and the associated diagram). It is not surprising that when you impose an ideal behaviour of the solvent that follows Raoults law, you get some unexpected results. Basically, what your calculation leads to it that for ideal solvent mixture, the only situation in which the gas phase molar ratio can be equal to the solution molar ratio is when both solvent have equal vapour pressures.


text:
Why weren't the electrons attracted to doubly charged Helium particles in the Rutherford experiment?
I have a chapter and in it there was a topic on the Rutherford experiment.
It is written that doubly charged helium particles or alpha particles were thrown towards a very thin gold foil and some of the alpha particles where bounced back because there were positive particles (protons) and positive and positive always repel each other.
If this was the reason for the bouncing back of the alpha particles, then why didn't the electrons, which are much smaller than protons, get attracted from alpha particles?
Especially when the alpha particles are doubly charged.

The gold foil experiment conducted by Rutherford (or more appropriately by Geiger and Marsden under Rutherford's direction) involves a stream of $\alpha$ particles (which are helium nuclei) bombarding a thin gold foil.
Why didn't some electrons go with the alpha particles because of electrostatic attraction? Some probably did. The experiment was equipped with a fluorescent zinc sulfide screen to detect alpha particles ($\ce{He^{2+}}$ ions). This screen might have also detected $\ce{He+}$ ions (but not necessarily in a way that was different than the detection of alpha particles), but probably not free electrons or helium atoms. The experiment was only set up to detect alpha particles, and so that's all that Rutherford, Geiger, and Marsden observed. We'll never know how many (if any) alpha particles picked up electrons in their experiment. We could, however, repeat the experiment, replacing the fluorescent screen with a mass analyze to detect at least the difference between $\ce{He^{2+}}$ and $\ce{He+}$.
More importantly, we known that many alpha particles passed through (and some were deflected) unchanged. Why didn't these particles pick up electrons? The answer to this question is another question: What is holding those electrons in the gold foil?
Gold atoms have 79 protons holding down those electrons. The first ionization energy of gold is 890.1 kJ/mol and the second is 1980 kJ/mol. However, these values are for isolated gold atoms. Gold (like all metals) is held together by metallic bonds with delocalized electrons occupying a valence band. Each electron is not being held by just one gold nucleus. Once the first few electrons are ejected or snatched by the alpha particles, the gold foil will have a net positive charge. It will be much harder to remove negatively charged particles from the positively charged foil.

Is a mass difference between $\ce{He+}$ and $\ce{He^{2+}}$ really measurable?

now can i count the weight of each and every helium particle!!!

@Martin - the mass difference is negligible, but most mass analyzers actually detect mass-to-charge ratio

@Ben Thanks! You are right of course, I just did not think about the huge difference in the charge of the particles.

Sorry, but this explanation is just downright wrong. Positive charge on the foil, if any, will surely try to repel the alpha particles (though that is pretty negligible, as they come with energies of a few MeV), but once they reach the foil, they'll feel as if there were no potential at all. Otherwise we'd have to assume that the potential acts differently on gold atoms within the foil and alpha particle which is already within the foil too. Also, how then alpha particles avoid being neutralized in the source they come from? It is also solid and has no reason to be positively charged.

Consider radium decay alpha-particle energies: Ra-226 4.86 MeV, Ra-222 5.59 MeV. The first atomic ionization energy of gold is 9.22 eV.  We do not expect charge neutralization of the alpha-particle until it slows by collision. It will leave a short path of dense ionization as it transfers energy, continuously being stripped to a bare nucleus until it slows.
The vacuum work function of bulk gold tops at 5.47 eV, giving the same expectation.

I had a similar, but slightly different question over on the physics exchange.
Relevant to this question: Grabbing electrons would also seem to change the trajectory of the He+ (as would bumping into one of the 79 high-speed electrons per gold atom).  While my question was more about the impact this has on the scatter (as this seems important to the measurement of the size of the nucleus of the atoms), seems like deflections should give clues to pick-up of electrons as well.

Wait...are you asking a question?

@MARamezani:  wait, are YOU asking a question?


text:
Why does water volume decrease when salt is added?
Why does water volume decrease when salt is added? Our teacher asks us in the class but I don't find any strong reason.

What you are experiencing is the phenomenon of partial molal properties - components of a mixture contribute unequally to the properties of the mixture.
Sometimes the components of a mixture do not contribute linearly to the properties of a mixture. A well-known example is the water-ethanol system, which has negative excess volume. When 5 mL of water and 95 mL of ethanol are mixed, the final volume is less than 100 mL. See the following graph constructed by Wikipedia user Wilfried Cordes:

Similar behavior is occurring in your sodium chloride and water system. Now, as for why these phenomena occur, a lot of it has to do with intermolecular forces. For example, you could consider that the hydrogen-bonding and dipole-dipole interactions between water and ethanol are perhaps stronger than between water and water or ethanol and ethanol. Additionally, water molecules are smaller than ethanol molecules. There is also a thermodynamic consideration. This state of affairs must represent minimum free energy (maximum entropy).
How can you relate the interactions of sodium and chloride ions with water to the observation of negative excess volume?

ions almost always dissolve in any solvent, polar, or nonpolar, but not ionic(because then you would form for example sodium potassium chloride(not sure if that is right for NaKCl(following molecular formula of ionic compound rule of cations before anions)). Because of this sodium chloride dissolves in water. You have similar hydrogen bonding to the ions as you do to the ethanol and you get similar characteristics like lower volume of 1 substance due to higher amount of the other and so you would get a similar mole fraction graph for NaCl(aq).


text:
Is there a chemical test for gluten?
I am looking for a chemical test for the presence of gluten: however, having looked online for such a thing, all the results seem to be about 'home testing kits' (presumably for people with a gluten intolerance). Is there a simple chemical test which can be used for a school science project (meaning that the chemicals involved should be easily available), something similar to how iodine can be used to detect starch?

Gluten describes a subset of proteins (glutenins and gliadins) found in wheat grain and making up a large fraction of the protein found in common wheat flour (>70%). Since the protein fraction of wheat flour can represent >10%/wt, the gluten fraction commonly represents >5%/wt of the flour.
Given the requirement for an easily implemented chemical test this eliminates more advanced and regularly used methods such as immunoassays (eg ELISA), and maybe even dye-binding assays such as described by the AOAC Official Method 2011.04.
One frequent chemical method to measure the amount of protein in a food sample is known as the Kjeldahl method, which detects nitrogen (which is absent in the carbohydrate). Other assays for nitrogen would also do the trick.
Büchi manufatures equipment for performing the required assay (Ref. 1), and provides an application note based on standard method AOAC 920.87. Although it is unlikely that you will have a Büchi SpeedDigester K-436, K-439, or Kjeldahl Sampler System K-370/K-371, available in your school lab, you'll still require the chemicals and equipment for performing a high temperature digestion and distillation, which should be similar to that listed in the application note. Note this is not a simple experiment.
Unfortunately the Kjeldahl method is likely to be difficult to implement in some educational laboratories given the required equipment, chemicals and experience. An alternative family of less challenging methods to determine protein are biuret methods (BCA and Lowry assays, which exploit protein-copper chelation and secondary detection of reduced copper), as described for instance on the ThermoFisher Scientific website. These however employ absorbance measurements to determine the amount of chelated or reduced copper.
If you are looking to distinguish gluten against a background of other proteins, you may have to isolate the gluten fraction (see eg Ref 2.). In general your approach would depend on the expected concentration and amount of gluten in the sample. Since this is a question about an educational application, it is assumed that you are not concerned with a non-gluten protein background or with detection of low concentrations.
References

Büchi Application Note 027/2010. Nitrogen and Protein Determination in Flour according to the Kjeldahl Method.

Ravinder Kaushik, Naveen Kumar,Manvesh Kumar Sihag, and Aradhita Ray. Isolation, characterization of wheat gluten and its regeneration properties. J Food Sci Technol. 2015 Sep; 52(9): 5930–5937.
doi: 10.1007/s13197-014-1690-2

Typically when one tests for gluten, the point is to distinguish gluten from other proteins. I assume that's what the OP had in mind for their classroom demo--e.g. show that wheat flour tests postive for gluten, while oat flour tests negative. Given this, it wouldn't make sense to use the Kjeldahl assay as a gluten test, since it tests for total protein of all types (by testing, as you said, for total nitrogen).  Plus it definitely doesn't meet the OP's requirement of a "simple" chemical test.

@theorist Yes, you are largely right, although I think the point is to detect gluten period. There is no explanation as to what background to expect. However for a HS lab I reduced the problem to identifying protein (reductio ad absurdum perhaps). Kjeldahl is too hard so I mention other protein assay choices. In general not easy to beat immunoassays for specificity.


text:
How do I clean a silk tie at the knot?
I would like to know how I can clean a silk tie at the knot. I presume it got filthy, over time, due to "fatty" fingers. I am looking for answers either based on 1) good theory or 2) personal experience. I am not looking for grandma's advice.

I think this is fine for the site, it's an application question about materials and reagents.  You will probably get a better answer eventually.

I'm confused, what do you mean by "at the knot"?

@qwersjc That's the location of the staining. The staining probably happens when I pull up the knot to my neck (over months or even years). Maybe it is sweat.


text:
What happens when C17H35COO-, a soap, reacts with an acidic solution?
My book says that-

In acidic water, the negative fatty acid ions become fatty acids.

However, they did not include a chemical equation to demonstrate this, so I made my own:
$$\ce{C17H35COO- (aq) + HCl (aq) -> C17H35COOH (aq) + Cl- (aq)}$$
But according to this equation, the fatty acid ion is a base, since it gets a proton. So this must be wrong. Can anyone point out my mistakes (if any), and explain what happens when a negative fatty acid ion comes in contact with acidic water, and becomes a fatty acid?

Here's a thought: Try adding vinegar to a solution of soap and water. Will this be enough to break down all soap, and cancel out the effect of the soap molecules totally? Why / Why not?

Your equation is correct. Hydrochloric acid ($\ce{HCl}$) is a stronger acid than stearic acid ($\ce{C17H35COOH}$),i.e. it is dissociated into $\ce{H+}$ and $\ce{X-}$, the anionic rest, to a greater degree. Therefore, it protonates the stearate ion. One can also say that the conjugated base of stearic acid (the stearate ion) is a stronger base than chloride, the conjugated base of hydrochloric acid.

Actually, soap in acidic water just means reactions with the hydrogen ions in acid and whatever other element in the acids is not a matter to be concerned about. For example, sodium stearate $\ce{C17H35COONa}$ decomposes in in acidic water soap:
\begin{align}
\ce{C17H35COONa &-> \underset{\text{(stearate ion)}}{C17H35COO-} + \underset{\text{(sodium ion)}}{Na+}} \\
\ce{C17H35COO- +  \underset{\text{(from any acid)}}{H+} &-> \underset{\text{(stearic acid)}}{C17H35COOH}}
\end{align}
Due to the formation of stearic acid, soap cannot act efficiently in acidic water.


text:
UV radiation on IBr (Radical Substitution)
I came across a radical substitution which Methane reacts with $\ce{IBr}$ to form iodomethane and hydrogen bromide and many more products. I understand the whole process but I don't quite see how the $\ce{IBr}$ molecule will break via homolytic fission instead of heterolytic. Since Bromine is more electronegative than Iodine, wouldn't $\ce{Br}$ takes both electrons and form $\ce{Br-}$ ions?

The presumed stability of $\ce{Br-}$ is overcompensated by the instability of $\ce{I+}$. While some polyatomic iodine cations like $\ce{I2+}$ exist, $\ce{I+}$ is found only to be stable in the gas phase and in the absence of a counterion. Due to the low difference of electronegativities (0,5) of Br and I, the I-Br bond is predominantly covalent, which also makes heterolysis unlikely. Homolytic cleavage of the I-Br bond is therefore energetically more favorable.

Thanks! Considering the instability of I+ the concept makes much more sense now


text:
How can water be electrolysed if it's not ionic compound?

The electrodes have different charges. One is positive ( the cathode ) while one is negative (anode). They will attract the particles in the solution which are charged. Hydrogen to the cathode while oxygen will go to the anode.

I have been taught that water is a covalent bond. Also I have been taught that covalent bonds do no have a charge so how is it possible that this compound is charged if it's covalent?

Did you consider $\ce{2 H2O <=> H3O+ + OH-}$?

Pure water contains only small amounts of $\ce{H3O+}$ and $\ce{OH-}$ which are generated by autoprotolysis (the reaction given by Klaus Warzecha in his comment to your question). Reduction of $\ce{H3O+}$ at the cathode produces molecular hydrogen. At the anode, water is in turn oxidized, yielding molecular oxygen and $\ce{H3O+}$ (reference).
$$\mathrm{Cathode:}\ \ce{2H3O+ + 2e- \rightarrow H2 +2H2O}$$
$$\mathrm{Anode:}\ \ce{6H2O \rightarrow O2 + 4H3O+ + 4e-}$$
$$\mathrm{Overall\ reaction:}\ \ce{2H2O \rightarrow 2H2 + O2}$$

So water contains some ions?

Yes, but at low concentration. At 25°C, the product of the concentrations (ionic product of water) is $[\ce{H3O+}][\ce{OH-}]=1 \times 10^{-14}$

Hm. It might seem like an easy question but do any other compounds contain ionic bonds at low concentrations?

Pure liquid ammonia can also undergo autoprotolysis: $\ce{2NH3 \rightleftharpoons NH2- + NH4+}$

If you use NaCl as the aqueous electrolyte, you get sodium metal and chlorine. Sodium reacts with water to give hydrogen and hydroxide.  Chlorine eats the anode, dissolves in water, and disproportionates with hydroxide to give HOCl and chloride.  NaOH aqueous electrolyte gives hydrogen and oxygen.  Sodium carbonate aqueous electrolyte can be interesting at the cathode versus voltage.
Chemistry tops off around 3 eV bonds.  A mere transistor battery gives you nine volts.  Products depend on electrolyte, concentration, voltage, current density, and electrode material (e.g., overvoltage).
http://nopr.niscair.res.in/bitstream/123456789/21833/1/IJCB%2041B(2)%20423-426.pdf
Electrochemistry can push things hard.

This is all correct, but I fail to see how it relates to the question.

Water electrolysis does not fundamentally electrolyze water.  It creates other species than can net obtain oxygen and hydrogen from reaction with water.  Understanding the process accesses a broad range of answers, not merely one answer.

You have never even hinted that something like water electrolysis might happen at all (and given pure theory this has a probability far away from zero). I very much understand that other ions in solution will catalyse the whole process, but the fundamental hypothesys is that water can be hydrosysed.

So he does water get electrolysed inspire of being a covalent compound?