text:
Why is water "the universal" solvent?
This is an old question that our textbook tried to answer but worsened the situation.
Many things are soluble in water. So many, that studying solutions will always require studying aqueous ones. It is true that many non-polars like waxes are not very soluble in water, yet I have never run into a solvent as "good" as water.
But how were we answered when we asked that "why is water a good solvent"? They said since water is polar so the attraction between for example $\ce{O}$ and the positive ions is so much blah blah blah!
So either there are "great" solvents like water out there or there are other things about water that make it the master solvent that are beyond me.

Is there a solvent as "versatile" as water?
Can this kind of solvent be non-polar?
If the answer to the questions above is no, what is special about water?

To directly address where the phrase comes from:

Water is called the "universal solvent" because it dissolves more substances than any other liquid. -USGS

What are ideal qualities of solvent?
The strength of a solvent can be attributed to the strength of its intermolecular forces like london forces, dipole-dipole forces, ion-induced dipole, and hydrogen bonding. These are forces of attraction and repulsion. Solvation occurs when a molecule is surrounded by the solvent, so when there are strong intermolecular forces, stronger solvation occurs.

The hydrogen bond
Water makes use of the hydrogen bond, a type of intermolecular force experienced when hydrogen is attracted to the electronegative atoms nitrogen, oxygen, or fluorine. Hydrogen bonding is the strongest intermolecular force.

Hydrophobia
Water is a great solvent for hydrophilic molecules but hydrophobic molecules are, by definition, not easily disturbed by water. This makes a clear exception to the "Universal Solvent" idea. I haven't read any literature claiming it to actually be the best solvent, but it does work well in many chemical situations.

TL;DR: Water is incredibly easy to get and work with.

In full
Water is a good solvent for polar compounds[citation needed], and the reasons for this are laid out pretty well by John Snow, but that's not really what makes it the universal solvent.
Instead, a series of other, incidental properties makes it a popular choice:
Availability
There's a lot of water[citation needed]. It literally falls from the sky. This means that for most applications, it's the absolutely cheapest solvent available.
Reactivity
Water is a reasonably stable molecule. Few things you can dissolve in water are expected to react significantly with it. It won't readily burn or combust, even in vapour form, and its high heat capacity and prevalence (see above) means that it can effectively douse most runaway reactions that might occur, such as fires (there are times when adding water to a fire is a bad idea, especially perhaps in a lab, but even those times the problem can be solved by adding a lot of water).
Toxicity
Unlike some other common solvents, water is not toxic[citation needed], and is not a bio- or eco-hazard. This means that unused solvent can be disposed of without special precautions, and if the solute can be disposed of trivially, the solvent can as well.
Purity and stability
Water is everywhere[citation needed], giving it another unique property: Unlike many other solvents, water won't absorb atmospheric water vapour, decreasing its purity as a matter of course. (It'll still exchange water with the atmospheric vapour, of course, but it doesn't significantly change its make-up.)
Distilled 100 % water will remain more or less pure unless something is actively added to it, which, while not unique, is useful and somewhat uncommon. Water molecules will also not spontaneously decompose under most sane conditions.
Operating temperatures
Water is liquid at a wide range of temperatures, stretching from 0°C to 100°C. Under laboratory conditions of ~1 atm and 20-25°C, water is a well-behaved and practical liquid.
Furthermore, while the liquid phase is available at a wide range of temperatures, the solid and gaseous states are not out of reach when using even extremely basic equipment. Cooling water to well below freezing, say -18°C, or to boiling, are both trivial activities, using equipment that is not only available in every lab, but even most homes.
In short...
Water isn't always the best tool for the job, but it's almost always a good or great one, at least when working with polar compounds. A common alternative to water is ethyl alcohol, which has similar polar properties and operating temperatures, but ethanol does have weaknesses in the other categories mentioned here, i.e. its reactivity, its absorption of water vapour, and its price.

There's a lot of water[citation needed]. It literally falls from the sky. This means that for most applications, it's the absolutely cheapest solvent available.

Well, let's say this to people dying because of the lack of water in 2/3 of the world ^^

@Babounet: Lack of clean water. The problem isn't (usually) getting the water, it's making it potable. General sanitation upgrades and availability of low-cost desalination and purification methods is already helping resolve this problem, besides which, the question of water inequality is beyond the scope of this site. One can of course politicize anything, but sometimes, it's good to take a quick breather to consider if one should.

This ignores the history of the term, and implies a different meaning of the word universal than the meaning meant by that history.

@KRyan: First of all, you might want to look up what an etymological fallacy is, but that's neither here nor there. Water is regarded as a universal solvent because it's universally used, in the sense that all chemists use it extensively, not in the sense that everything that exists is actually in an aqueous solution, because that would be absurd.

But that's not the sense that @KRyan is referring to; they're not talking about the history of how universal was used, but the history of how "universal solvent" was used, which is to say it was used for alkahest. Alkahest doesn't exist, and the closest thing chemistry has to it is water.

@JonHanna It seems to me as if the OP is asking why water is so widely used as a solvent, which is the question I answered.

Yes, I agree you have answered the question. But I think some of your statements might be mildly misleading.

@KRyan Such as? I'm more than happy to receive notes and corrections.

The lede implies that water's ability as a solvent are irrelevant, which is not the case, and that universal solvent as "able to dissolve anything" is no part of the meaning or usage of the phrase, which is also not the case.

@KRyan In modern vernacular, able to dissolve anything doesn't enter into it, because we have the science of chemistry now. As for water's abilities as a solvent, it's a solid B+. Good, not great.

Yes, except that the reason the phrase is used at all rather than one that is more accurate is because of the history.

@KRyan Its pre-scientific, irrelevant history. Chemistry is full of words like that, you just have to deal. (The best example is how oxygen and hydrogen are swapped. Oxygen is the stuff from which water is made, hydrogen is what [typically] makes something an acid.) Besides which, its modern usage is pretty accurate; water is universally used.

Yes, it is pre-scientific history, irrelevant to almost everything. Except, ya know, questions about the usages and meanings of phrases that come directly from that history. As you have it, you imply that the phrase "universal solvent" as applied to water is completely divorced from the phrase's history and has nothing to do with the earlier meaning, when in fact the only reason the phrase exists or is used at all is because of the earlier usage. Knowledge of history and origin is useful. Your answer and argumentation here seems to suggest we should forget that history exists. I disagree.

@Williham Totland could you site the literature from which water is called universal for its "universality, not its solvent skills" please? I have provided in my answer a source that contradicts that statement.

@JohnSnow: Look to DavePhD's answer for an example: For dissolving metals, mercury is a better solvent. Besides which, for dissolving hydrophilic compounds, stuff like hexane is a better solvent. Water is a good, and typically safe, general-purpose solvent, but for many, if not most, applications, better solvents exist.

No, I'm saying that you are making a declarative statement about the phrase that this question is centered around, and you should provide a source for that statement.

Just to provide an alternative answer:
Consider a solvent miscibility table like the one linked.
What is the least miscibile solvent? Water! Water is the worst solvent. Water is immiscible with 17 out of 30 of the other listed solvents.
There are 6 solvents in the table which are miscible with all the other solvents: ethanol, acetone, tetrahydrofuran, n- & iso-propanol and dioxane.
Why is water such a bad solvent? Hydrogen bonds.  Water molecules are strongly hydrogen bonded to other water molucules.  To dissolve something else, these hydrogen bonds need to be broken.
For another prespective, look across the periodic table.  What does water stably dissolve? Does water dissolve gold, silver, copper, tin, lead, aluminum, and many other metals and metaloids like antimony, silicon, and germanium, as mercury does?
There is nothing universal about water, except that there is a lot of it on Earth and we need it to live.

I'd venture to say that there's a lot of water everywhere except Earth, tho', but to each his own.

I think it should be noted that there is a difference between solubility and miscibility, but this is a good contribution.

"miscible" means mutually soluble in all proportions

Since water is polar as you note, and "like dissolves like" remains a good principle, I think the answer is that lots and lots of substances are polar. Most (ionic) salts are, for example. Even substances that are polar covalent are at least slightly polar and so somewhat soluble (and with agitation, more soluble). So you have to be really non-polar (e.g. hydrocarbons, "oils") to not be soluble in water.

It's worth noting some of the history behind the term "universal solvent," and why it is used even though water isn't truly universal or even necessarily the most versatile solvent available.
Before chemistry existed, and for that matter before science existed as the practice we would recognize today, alchemy was an important protoscience. A number of important breakthroughs and refinements came about due to the work of alchemists, and for all that it often got buried in mysticism (and, at times, fraud), alchemy began the research into the chemical properties of matter. Chemistry developed largely in reaction to alchemy, rejecting the mystical underpinnings, and applying the scientific method.
Within alchemy, there was a belief in the existence of a "universal solvent," which would be part of the process for producing the philosopher's stone and/or transmuting lead into gold. Numerous alchemists claimed to know how to make it, while many others considered producing it to be one of their major, overarching goals, the most difficult or secret step in obtaining whatever powers they believed they could achieve through alchemy.
Obviously, no chemical with the properties described for this universal solvent has ever been found. However, because of the critical use of water as the default solvent in chemical work, it has often been referred to with the title. Other answers describe better than I can why water is a particularly useful solvent.


text:
How do I calculate the density of propane assuming an ideal gas?

Assuming an ideal gas, calculate the density in $\mathrm{g/cm^3}$ of propane, $\ce{C3H8}$ at $0.000~^\circ\mathrm{C}$ and $1.000~\mathrm{atm}$. (3 significant figures)

I have figured out that I need this formula:
$$\frac{n}{V} = \frac{p}{RT}$$
I have a total of $44.09562~\mathrm{g/mol}$
I am stuck at this point.

As you have already figured out that we have to use the ideal gas law $PV= nRT$, which is you have modified as $$\frac{n}{V}= \frac{P}{RT}$$
To find the density, we can further modify the equation as,
\begin{align}
\frac{m}{MV} &= \frac{P}{RT}\\
\frac{\rho}{M}&= \frac{P}{RT}\\
\rho &= \frac{PM}{RT}\\
\end{align}
Now we just have to replace the parameters with their respective values.
The value of temperature should be converted to Kelvin as $(0 + 273)~\mathrm{K} = 273~\mathrm{K}$.
Likewise, the value of gas constant $R$ should be chosen as  $0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}}$.
\begin{align}
\rho &= \frac{1~\mathrm{atm} \cdot 44.09562~\mathrm{g~mol^{-1}}}{{0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}} \cdot 273~\mathrm{K}}}\\
&=1.9683~\mathrm{g/L}\\
&=0.0019683~\mathrm{g/cm^3}\\
\end{align}

You can use other methods for getting the answer since you are already given 44.09562 g/mol .

R above is out by 10, Use 0.08206 L atm K-1 mol-1

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text:
What are the molecular orbitals of the hypothetical linear H₃⁺ molecule?

What would be the wave function of the lowest energy molecular orbital of a hypothetical linear $\ce{H3+}$ molecule?

According to the LCAO method, I feel the lowest energy MO will be $\mathrm{1s(A) + 1s(B) + 1s(C)}$, where e.g. $\mathrm{1s(A)}$ is the wave function of the $\mathrm{1s}$ orbital of one of the Hydrogen atoms. This has $0$ nodes and has the lowest energy.
Isn't this correct?

Why are you asking about hypothetical linear H₃⁺ molecule, when there is real triangular H₃⁺?

@Mithoron I can't change my homework question, can I ?

The lowest energy MO is : $\psi_1 = (1/2)(\phi_A+ \sqrt2\phi_B+\phi_C)$.
Then comes the MO: $\psi_2 = (1/\sqrt2)(\phi_A- \phi_C)$.
The highest energy MO is  $\psi_3 = (1/2)(-\phi_A+ \sqrt2\phi_B-\phi_C)$.
Where  $\phi$ denotes the atomic orbital $1s$ on each hydrogen atom. $A$, $B$ and  $C$ denote hydrogen atoms, where  $B$ is the central one.

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text:
How does grease become airborne?
When cooking in the kitchen you usually use some oil or butter to bake your meat and potatoes in. After some weeks without cleaning (which happens every now and then) the surroundings of the stove will get a greasy, sticky layer. Since this only happens relatively close to the stove, say a 1-2 meter radius, I'm assuming that it is caused by my cooking.
My question is: how does the grease get so far away from the stove?
I know that sometimes oil will spray a bit when you add something water-containing to it, but the grease is actually also in places that are not in the 'line of fire' (on top of the ventilation hood for example). Therefore, the grease must be getting airborne somehow, but I don't understand how, because evaporation of the vegetable oil or butter seems highly unlikely to me. Does anyone know what is going on?

Let's reword it:

Traces of oil and grease are distilled off at temperatures below their boiling points.
The temperature in the frying pan is above 100 °C.
The food does contain significant amounts of water.

This sounds like the conditions for steam distillation, except that you're collecting the mixture of water and water-insoluble material everywhere in the kitchen and not in a flask.

In addition to steam distillation some fat is sprayed into air by the sizzeling of water droplets falling into hot fat.


text:
Search materials by HOMO, LUMO and work function
I need to search organic semiconductors for organic photovoltaics (OPV) by their HOMO, LUMO, work function at 300K (as I understand, WF cannot be derived from HOMO and LUMO because semiconductor can be differently doped and Fermi level depends on concentration of dopants) and deposition methods (such as CVD, PVD, spin-coating, electrodeposition, etc)..
Is there some free site where I can either download a machine-readable table of needed data or make a search based on that criteria? What is it?

Indeed, Prof. Alán Aspuru-Guzik has placed his "clean energy project" database on the web in a searchable form.
At the moment, it covers ~2.3 million compounds.

Use of the data requires citation of the appropriate literature sources. The data in this database is released under the Creative Commons Attribution ShareAlike license and therefore any academic publications associated with the CEPDB should be appropriately cited.

As you point out, the database only covers HOMO and LUMO orbital eigenvalues. While this may not be the most useful for experimental applications (i.e., you really want ionization potentials and first excited state energies or electron affinities) it should be a helpful "first screen" to find interesting lead compounds.

I have already found this, but there is no work function data in it, so we are not able to align bands properly.

@KOLANICH The work function is a materials property (i.e., it will depend on the crystal structure packing). As of yet, there is no way of doing rapid computational screening for this. Unless there are experimental databases in the field, which I am not aware of, there is nothing which meets your needs.

authors: Макс Сергеев


text:
What defines an element's taste?
A useful post by @Martin indicated that probably the naming of Sweetwater town is because of the sweet tasting lead compounds in it's water.
Then my question arose. I know that the taste of any material is dependent on which tongue buds it provokes. Now, my question is, which chemical property of an element defines or is directly related or is causing the chemical property of taste? (I deliberately avoid asking this for all chemical species to prevent very broad answers, if you think you can give not-too-long answers, my question is about all chemical species.)
Note that by a responsible chemical property I mean things related specifically to an atom, like e.config.

TL;DR: Don't taste pure elements. They either taste of nothing, taste foul, or kill you. Or all of the above.
Edit: Also, for clarification, elements don't have a taste. Taste is a biological reaction to chemical interactions happening with an element, not a property of the element itself. So, like everything else in chemistry, the taste is decided by the electron configuration of the element.

Taste as we know it happens in chemical reactions between our taste buds and the chemical species in question. Most of the taste is actually governed by the nose, where very specific receptors interact with very specific chemicals to send very specific signals to the brain, the chemistry of which is, while not trivial, also not interesting. It's your basic, run of the mill presumptive test.
In the taste buds, however, the receptors look for certain specific characteristics of the species being tasted. As an example, acids are sour, indicating that our sour taste buds are actually pH meters (although there might be more to it than this).
Similarly, most sugars have similar characteristics, that can be detected by a relatively simple chemical detector.
If, however, you go around tasting elements, you'll find two very interesting things.
First, you'll find that you're dying from all kinds of poisonings. At a rough count, all of the periodic table is poisonous in its pure form, with only a few exceptions (carbon, oxygen, hydrogen, nitrogen, silicon, titanium, noble gases, gold and platinum, possibly a few others).
Second, you'll find that the pure elements don't taste much at all, if anything. Taste is a completely biological concept, and these elements simply aren't found in nature, meaning that the tongue and nose have no frame of reference.
Most metals will taste metallic, due to tarnishes forming on their surface, mostly oxides, which all have the familiar, metallic taste. The non-metals are mostly fairly reactive, and taste absolutely foul as they react to the air in your moist mouth.
Gases like chlorine and fluorine may react and form chloride and fluoride, which taste sour, and then dissolve your mouth and nose, and then kill you.
The more radioactive metals will seem to taste of blood, but that's actually just acute radiation poisoning setting in, actually filling your mouth with blood. If they didn't kill you, they'd probably taste metallic like the rest of the metals, but they do, so they don't.
Finally, the alkali metals like lithium and its friends will taste bitter, also kill you.
"But wait!", I hear you scream, "What of, say, lithium in pharmaceuticals?"
When elements like lithium are ingested, they are done so in a rather more controlled fashion. You don't get a brick of pure, elemental lithium and told to go crazy when you go crazy, but rather, delivered in pills, in conjunction with a delivery agent, in this case, lithium citrate.
Most elements have some biological role or other, but if they are in the kind of quantity you could taste, they are most likely poisonous, and in many cases, fatally so.

Strangely, I have licked iron and lived

I don't know how many variables determine the taste but we recently found out that materials can have completely different properties just by the amount. We know that with different structures of atoms materials get completely different like diamond and graphit but, we know at least since the discovery of graphene that the periodic table might be much more complicated than we had imagined.
Every element gets completely different properties with just the quantity of atoms. If you take for example a really small amount of an element with a spoon it glows yellow, then you take with an other spoon a bigger amount and it glow red, then you take a bigger amount it glows blue. Currently it doesn't make any sense why this happens since you don't change the atomic structure of the material.
In the future the periodic table will get a 3rd dimension with every element becoming different with just the amount. This is not about the structure of the atoms or something, just the amount.
For the colors of matrials we have the wave lenght of the light as a reference but for the taste there are to much variables since we taste with our nose and our tongue.
http://periodictable.com/Posters/PlaceMatBackSide3000.JPG
http://periodictable.com/Posters/LetterSizeBackSide1500.JPG

This is simply not correct. The differences between allotropes of carbon are structural, not do with "amount".

Yes this is what I said and additionally with the amount.


text:
Why is methanol toxic?
There are two points of view for the answer of this question:

The biological view, the only one that I faced during my research, states that since it can trigger perilous conditions like metabolic acidosis, cause a neurologic sequelae, etc. methanol is toxic.
The chemical point of view in an answer, which is the one I seek, is the reactions that occur in the body with methanol as a reactant.

Why should it be a reaction? Because I assume something will be toxic for humans when it reacts and disables some of their bodies' vital compounds; e.g.: Hydrogen cyanide reacts with enzymes, renders their active sites useless, and finally causes heavy reduction in bio-chemical reactions that results in quick death.
So, the question is: What reactions use methanol in the body that makes it toxic for us?

There's a third point of view wherein methanol is said to encourage harmful parasite behavior (allowing them to complete their life cycle without an additional host, and migrate to other organs, or some such). Parasites are said to excrete toxins, and the presence of toxins in more organs may cause issues. This isn't a direct conversion of methanol to toxins, however. So, I'm not sure how on-topic this thought is, but if true, it relates to your question.

Related: This BioSE question

Methanol isn't particularly toxic in and of itself, although it's no walk in the park.
If methanol flowed through the body without being broken down, it would cause roughly the same kind of harm as ethanol, i.e. intoxication.
The real culprit is one of its metabolic products, methanoic acid, also known as formic acid.
To understand how formic acid, present as the formate ion, is toxic, we look to Wikipedia:

Formate is toxic because it inhibits mitochondrial cytochrome c oxidase, causing the symptoms of hypoxia at the cellular level, and also causing metabolic acidosis, among a variety of other metabolic disturbances.

Edit: As DavePhD points out, an intermediate product in this process is formaldehyde, or methanal. While formaldehyde is also toxic, it is rapidly metabolized to methanoic acid.
Reedit: The deeper, more historical reason that this happens is that methanol isn't readily available in nature, meaning that few species have developed biochemical tools to deal with it. There simply hasn't existed an evolutionary pressure to deal with methanol.

Note that formic/methanoic acid is essentially ant poison. Imagine being stung by 100 ants...

The enzyme alcohol dehydroganase converts the methanol to formaldehyde in the body. Formaldehyde is then converted to formic acid.
Formaldehyde can cause blindness before being converted to formic acid, while formic acid causes acidosis as Williham Totland points out.
See Biochemical Aspects of Methanol Poisoning for more information.

Note that the normal purpose of of alcohol dehydrogenase is to convert ethanol into acetaldehyde (which is also toxic, but there's a metabolic pathway to convert it into harmless acetic acid.)

@Mark yes, this is why ethanol has been given as a treatment for methanol poisoning.  Ethanol competes with the methanol for binding to alcohol dehydrogenase, and the methanol can exist the body, for example through breathing, without being metabolized to toxic formaldehyde and formate.

I just wanted to point out that everything is toxic in the right proportion. THe problem is with the dose. I think you could drink a bit of methanol without getting particulary harmed. Do not do it, obviously.
The LD50 is very low at 5g/kg.
Edit: Nothing to do with methanol, but it is a good infographic

Source: http://www.compoundchem.com/2014/07/27/lethaldoses/
As a comparison with ethanol, I believe (not sure) that the oxidation product of these two compounds, ie, acetic acid and formic acid are not equally reactive. The former (metabolite of ethanol) is less reactive than the latter (metabolite of methanol). At least from an organic point of view this makes sense. But maybe there is another more biological reason.

There is a more biological reason: Ethanol presents in serious quantities in nature, available to animals, usually in the form of rotting fruit. Methanol is comparatively rare, so the body doesn't have any reasonable defences against its metabolites.

@WillihamTotland Great apes are more tolerant of ethanol than most animals.

@200_success That's kind of interesting. Tragilarously, elephants are known to go on wicked binges, first getting smashed, then smashing stuff. (That's tragically/hilariously, by the way. Pachyderm alcoholism is tragic, the thought of a drunk elephant is, frankly, hilarious.) Also, birds wont to eat fermenting berries still on the branch, getting completely loopy and, again, hilarious.

@200_success Anyway, answer updated to clarify.

Might be worth noting that even a sub-lethal dose of methanol (such as a small accidental gulp during 'Fire-eating' can result in permanent blindness.

118 * 240 mL = 28 L >> 6 L...using "coffees" is maybe not that great as a unit.

Ethanol is only 2 g/kg safer according to this page, though that goes for rats and doesn't count partial death.

@AlteredState When an LD50 value is given, for example in an MSDS, it is almost always for an non-human animal. For humams "Ingestion of from 70 to 100ml of methanol is usually fatal" Biochemical Aspects of Methanol Poisioning, Biochemical Pharmacology, vol. 11, page 406.  People have died from as little as 30ml, and blindness takes even less.  People have gone totally blind from drinking only 5ml and even just acidentally spilling methanol on their skin.

@AlteredState At my discretion, I have edited out the fatal dosage information because it might attract Googlers looking for such information and I don't want to provide information that could help someone hurt themselves.

Talking about ethanol: if you are drinking very fast, than yes ~ 0.5 liters is almost lethal, but if you are drinking all night, you can drink more and just get hangover and maybe some intoxication.

Old thread but...
Methanol toxicity is attributed to formic acid.
But formic acid itself is a "low-toxicity" food additive E 236 (authorized in animal feed only though, at ~1%, afaik), which is a bit confusing.
I assume formic acid derived from methanol/formaline metabolism has different fate-distribution-bioavailability than ingested formic acid. Or it's simply a matter of dosage.

authors: Gäűţåm Ñïrâj


text:
Grignard reagents react with epoxides but dissolve in THF, what happens in oxetane (four membered ether ring)?
Ring strain causes the epoxides to react and THF does not react because the ring strain is not large enough. Does the four membered ring react or not? Does it favor reaction with the harder or softer nucleophile perhaps - thus only reacting with some Grignard reagents?

In a 1979 paper entitled Copper-catalysed reactions of Grignard reagents with epoxides and oxetane, Huynh, Derguini-Boumechal, and Linstrumelle report that the copper-catalysed ring opening of oxetane by various Grignard reagents gives the alcohols "in acceptable yields":

$$ \begin{array}{cccc}
\ce{R} & \ce{X} & \text{Solvent} & \text{Yield (%)} \\
\hline
n\text{-}\ce{C4H9} & \ce{Cl} & \text{Ether} & 75 \\
\ce{C6H5} & \ce{Br} & \text{THF} & 52 \\
\ce{CH2=CH-CH2} & \ce{Br} & \text{Ether} & 50 \\
\ce{(CH3)2C=CH-CH2} & \ce{Cl} & \text{THF} & 50
\end{array} $$


text:
How will my studies in mathematical statistics be applied to and assist me with problems in chemistry?
I'm currently working towards a major in both chemistry and mathematical statistics as a part of my science degree, and I got to wondering:
How will these two go hand in hand and help each other?
Now, I know there's the obvious applications of statistical inference to any experimentation I do, but I'm talking about less obvious relationships. Could my knowledge of solving SDEs ever help me with my experiments? What about continuous Markov processes? Will I ever even think about a discrete random variable again if I go on to get a PhD in chemistry? Also not looking for the standard "well we use statistical physics to model electrons" example, either (welp, unless you can offer some nice specifics). Surely there's other ways that these skills I've found can interact with each other?

Most definitely yes.
I took multiple statistics classes as an undergraduate and they have helped me considerably. (I can argue fairly that it helped me get tenure.)
A few suggestions from my own work:

Statistical design of multivariate experiments: Many scientists think you should test one variable at a time. No. I taught a grad student (in another group) about optimization theory and proper design and three months later, a great paper came out.
Data mining and exploration: Many chemistry data sets are extremely complicated, with cross-correlated factors. Understanding how to draw empirical inferences, find the most important factors, etc. is enormously helpful across a wide range of chemistry.
Principal component analysis: An emerging area in spectroscopy and imaging is taking complicated spectra and performing PCA to help with quantifying chemical components in mixtures and following everything from cancer growth to materials design.
Multivariate regression and ANOVA: My group has created numerous quick statistical screens using multivariate surrogates for complex simulations or experiments.
Cross-validated analysis: Again, the understanding of training and test sets, bootstrapping, etc. is very useful.

Now, you also mention SDEs and Markov processes. These are crucial in chemical kinetics, including areas like reaction-diffusion systems. Monte Carlo methods are also highly used in studying dynamical processes.
The list would be enormous. Basically, I think it's a great skill to have and you'll use it throughout your career.

Oh, I almost forgot optimization theory. We've been using stochastic evolutionary algorithms heavily. People use genetic algorithm optimization to find parameters for computational models (e.g., DFT), etc. As I said.. the list is enormous.

Investigate the umbrella term 'Chemometrics'. this covers several (if not all) of the areas covered in this answer, but also other areas and techniques. A useful piece of software called MatLab is often used in chemometrics.

Analytical chemistry of complex mixtures and anything -omics (metabolomics, foodomics, petroleomics, etc...) will use chemometrics somewhere, which often involves finding latent variables (interpreting PCA and factor analysis results). Still hot topics with lots of development. Computational chem often uses statistics as does chemoinformatics and QSAR. In addition to MATLAB, the statistical package R is highly useful (what I use).


text:
The difference between peptide bonds and the bonds between polypeptides?
I was doing some tests for the multiple-choice final we've got ahead. And it was on me to count the peptide bonds in an Insulin hormone with 51 aminoacids arranged in two polypeptides with 30 and 21 aminoacids. (these are not true in reality) the number of bonds were 49, not 50, and that means the bond between two polypeptides doesn't count as a peptide bond. Additionally, I know that polypeptide bonds make the proteins' molecular structure, as it is now. (just look at that shape.) PEPTIDE COVALENT BONDS CAN NEVER cause that kind of 3d orientation in space. So there must be some fundamental difference between those bonds. What is it?
 My research couldn't find any results as simple things have jammed the internet.

Aside from covalent bonds (amide and disulfide), the sturcture of a protein is determined by hydrogen bonds, salt bridges, and less specific interactions such as hydrophobic and hydrophilic effects.  Hydrophobic portions of the protein chain tend toward the interior of the folded protein and hydrophilic regions to the exterior, in aqueous solution.

The term peptide bond refers to amide bonds. The 21 amino acids in insulin's A-chain are covalently linked by 20 amide bonds. The 30 amino acids in insulin's B-chain are linked by 29 amide bonds. The two polypeptides are covalently linked by disulfide bonds, which are not amide bonds.
The 3D structure of proteins is a complex topic and a field of intense research. It's worth pointing out though that even polypeptides (with no disulfides or other covalent linkages) have a more complex 3D structure than you might expect. I suggest you look into alpha-helices and beta-sheets as an entry into the topic.
Took some info from the Beta Cell Consortium.

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


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Why is the iodide anion a good nucleophile but a poor base?
The $\mathrm{pK_{aH}}$ of $\ce I^-$ is very low which indicated that it is not favourable for it to bond with a proton. However, why would it be likely to bond with any other atom (mostly carbon when considering it's nucleophilic properties).

You should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent.
Also let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\ce{HI}$) lack of stability; it's relatively easy to ionize its proton due to the weak hydrogen-iodide bond.
On the other hand, the high nucleophilicity of iodide ion in water reflects the high rate of reaction between iodide ion and some electrophilic atom (other than hydrogen).
The relative lack of basicity of the iodide anion in water is precisely what makes it a good nucleophile. Its lack of affinity toward protons - i.e. the protons of water - physically frees $\ce{I^-}$ to attack electrophilic atoms (not limited to carbon). There is little hydrogen bonding between iodide ion and water.
On the other hand, the relative basicity and thus strong hydrogen-bond forming capabilities of $\ce{F^-}$ makes it a poorer nucleophile in water. Rates of reaction between fluoride anion and some electrophilic atom are slowed because the fluoride anion is trapped (strongly solvated) by water molecules. And there will also be a sizable number of fluoride anions that have completely abstracted a proton from water to make $\ce{HF}$, which isn't nucleophilic at all.
In a polar aprotic solvent, however, where there are no positively polarized hydrogens for a nucleophile to hook onto; $\ce{F^-}$ becomes a much stronger nucleophile. Actually,  $\ce{F^-}$ becomes a better nucleophile than $\ce{I^-}$, likely due to $\ce{F^-}$'s greater charge density.

The bond between I and H is weak so I- is a weak base.  I- is a strong nucleophile because it is polarizable, making it faster for its orbitals to overlap with the electrophile.  Remember that basicity is a thermodynamic concept and nucleophilicity is a kinetic concept.
I have been told that I- can be added to reactions as a catalyst because it is a good nuclephile and a good leaving group.  Leaving group capability seems to be inversely related to basicity.

Why doesn't it overlap fast with protons?

What if it does but the rate of the bond between H and I breaking is even faster than the rate of the H-I bond being formed?

Thank you. What stops the bond breaking in the C-I bond being faster than bond making, as it is in the H-I bond?

It probably is faster.

So why is it a good nucleophile?

Because the bond making is faster.  Nucleophilicity refers to how fast the bond can be made, not how stable it is.

I am sorry to keep asking questions, you have been very helpful but I am struggling to grasp this. You are saying that in the case of an H-I bond, bond breaking is faster than bond making but in the case of C-I, it is the other way around? Is this correct? What causes it to be the other way round?

Or am I missing the point?

In both cases the rate constant for bond breaking should be higher than making for most reactions involving a bond with C and I or H and I.  My point is that nucleophilicity refers to how fast a nucleophile/base can react and basicity refers to how stable the products of the reaction are.  Look at fluoride.  Fluoride makes a stronger bond with C than I does but I reacts faster than F.


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Does the ammonium ion hydrogen bond with water?
What about oxonium ions, such as hydronium ion? Do these hydrogen bond with water?
If we see hydrogen bonding as a purely electrostatic phenomenon, then why not? Ammonium ion still has nitrogen bearing a partial negative charge. Of course it can interact with positively polarized hydrogens in an electrostatic way.
However, we also know that hydrogen bonding is partially covalent. So, it seems unlikely that the ammonium ion would be able to form any additional covalent bonds.

Do you mean through as a hydrogen bond donor or acceptor?

@jerepierre either

There is a detailed consideration of this issue in  Hydrogen Bonding of the Ammonium Ion Journal of Molecular Structure 345 (1995) 77-81.
First, the try to be clear about what they mean by "hydrogen bond":

For the purposes of the discussion, a hydrogen
  bond is taken to be a specific short-range interaction
  in which the N-H bonds within the
  ammonium ion are significantly affected by the
  surrounding ions, and not simply electrostatic
  attraction.

They conclude that ammonium ions rotate too freely in water for significant hydrogen bonding to exist.
In gas phase calculations, they find that one water molecule hydrogen bonds to one ammonium ion (in other words, the dimer is hydrogen bonded), based upon the lengthening of the N-H covalent bond, but as you add 2, 3, and 4 water molecules, the N-H bond length returns to the free ammonium N-H bond length and the O --- H length increases.

"They conclude that ammonium ions rotate too freely in water for significant hydrogen bonding to exist." I wonder what this means for the hydrogen bonding of water with itself. Shouldn't water molecules be rotating too freely for H-bonding to exist as well then?

@jerepierre The activation energy for water to reorient (16 KJ/mol) is more than for ammonium (12 kJ/mol), even though ammonium has twice as many H.  They further point out that it takes 44 kJ/mol to remove a water molecular to the gas phase from liquid water, but 363 kJ/mol for an ammonium ion, and emphasis 12 is a small fraction of 363.


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Do Grignard reagents react with amides?
Clearly the amide will be unreactive given the poor quality of the leaving group and poor electrophilcity due to delocalization. However, does it react with organolithiums for example?

Amides will react with three equivalents n-butyllithium to give nitriles via geminal lithium oxyimide intermediates (J. Org. Chem., 1967, 32 (11)). n-butyllithium is a strong enough base to deprotonate the molecule twice at nitrogen and once alpha to the carbonyl group. Grignard reagents may well behave similarly.
I do not think this is a particularly useful transformation though because n-butyllithium is a rather indiscriminate reagent. There are a wide range of other methods available - see this paper if you're interested.
In the special case of a N,O-dimethylhydroxyamide, you can make ketones using either.

For a N,N-alkyl, a N,N-diaryl or a N-alkyl-N-aryl amide that what will happen is that the grignard will react attacking the carbonyl carbon. This will form R''C(NR2)(R')-OMgX from RMgX and R''-CONR2.
What we normally do is to treat the product with dilute acid, this will then convert the initial additon product into either a ketone or an aldehyde.
One common method of making aldehydes is to react Grignards and other similar reagents with DMF, then treating the intial product with aqueous acid.


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Why is water a poor solvent for organic molecules?
Few organic molecules dissolve in water but why is this the case?

Pure water can form two hydrogen bonds per molecule.  The enthalpy of these hydrogen bonds is about 23 kJ/mol of water.  You have to give water a very good reason to rip apart its hydrogen bond network, and the interactions water experiences with the organic compounds often aren't enough, even considering increased entropy, to make this favorable.

Aren't there up to four hydrogen bonds per water molecule, two donated, and two accepted?

@Martin-マーチン If there is one mole of water, I'm saying there will be about 2 moles of hydrogen bonds.  Each proton participates in about 1 hydrogen bond.  For a sulfur 8-member ring, I will say there is one bond per sulfur, and you will say each sulfur has 2 bonds, and we will both be right.  I don't want to double count the enthalpy contribution.

That of course makes perfect sense. I did not look at it this way. Thanks for clearing that up.

Water is not a bad solvent for all organic molecules. In fact many organic molecules are soluble in water. In general the solubility of a substance in a given solvent is determined by the strength of the intermolecular forces between the solvent and solute.
The predominant intermolecular force in water is hydrogen bonding and so molecules with groups which can hydrogen bond  generally dissolve well in water. These groups can have acidic hydrogens of their own such as alcohol or amine groups which can h-bond to water both through their hydrogens and through the lone pairs on their oxygen and nitrogen atoms respectively. Also other groups which have lone pairs on electronegative atoms such as carbonyl groups can h-bond with water and so will generally dissolve, although not as well as alcohols or amines. All of these groups are generally termed to be hydrophilic.
There are also hydrophobic groups which tend not to bond well with water and so do not tend to dissolve. These are generally groups which are non-polar and interact primarily through Van der Waals forces. Therefore they do not form strong interactions with water (because water primarily interacts through h-bonding) and so do not dissolve. These groups tend to be apolar, the most common example being alkyl groups.
So in answer to your question, many large organic molecules with predominant hydrophobic groups do not dissolve in water whereas smaller molecules with hydrophilic groups do tend to dissolve.
This questions has a more detailed discussion of the properties of water as a solvent: Why is water "the universal" solvent?


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Aspartame poisoning: Can it modify aminoacids?
I just accidentally closed a webpage (and did not find it later) which described that somehow, Aspartame poisoning victims had faults in the vital proteins of their body due to the reaction of Aspartame with some aminoacids. I doubted that this is true. However, considering the high solubility of Aspartame in the liquids in the body and its digestibility it is feasible for it or one of the related compounds to be causing this symptom. The info that will be shared is welcome.

Aspartame is one of the most well studied food additives there is. It seems to be controversial in the same way that evolution is—scientific research is overwhelmingly on the side of the substance being safe to consume, even at very high levels, but a Google search on it brings up things like "Aspartame is, by Far, the Most Dangerous Substance on the Market that is Added To Foods" (from a site trying to sell natural health products). A lot of this comes perhaps from the naturalistic fallacy: aspartame is artificial and thus unsafe. A huge number of regulatory authorities have deemed it safe to consume: FDA, Health Canada, EFSA, FSA, etc.
There is also a good deal of bad biochemistry thrown about: aspartame is metabolized into the two amino acids aspartic acid and phenylalanine, as well as methanol. There have been some hypotheses that aspartic acid intake could induce some amount of toxicity, but experimental evidence is lacking. Methanol, of course, is well known to be toxic. The key is, that aspartame is very sweet and only tiny amounts are added to food. So even drinking enough diet cola to hit the FDA's acceptable daily intake of aspartame (which works out to ~7.5 L), the quantity of methanol consumed is only around 300 mg, and things like fruits are larger sources of methanol in people who drink sane amounts of diet soft drinks.
The one condition in which aspartame is known to do harm at reasonable intake is in people who have phenylketonuria (PKU). This might be what you're thinking of by "faults in the vital proteins of their body". PKU is a genetic disease caused by a defective gene for the enzyme phenylalanine hydroxylase which results in an inability to metabolize phenylalanine correctly; thus, consuming phenylalanine from any source is hazardous to sufferers of the disease.


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Can we define a new unit instead of specific heat that is constant with temperature rise?
Specific heat plays an important role in calculating energy and thus, heat, enthalpy and etc.
But as I sighted the definition for calorie, I was surprised.

A calorie is the amount of heat needed to alter the temperature of 1 gram of 14.5 °C water to 15.5 °C water in the pressure of one atmosphere.

Though the unit was redefined later, the question remained still. Why were there those temperatures? It was just then that I realized specific heat capacity varies with temperature differences and it is due to vibration movements of matter and internal energy ....
Two questions: Can there be a unit which is constant with temperature alterations that can be used instead of S.heat capacity? If so, why isn't it replacing SHC?
The resource for the quote The resource for the reason of the increase that happens to SHC with temperature rise

""specific heat capacity varies with temperature differences "" Wrong. specific heat varies with temperature level. The classical 15 °C were choosen simply for acessability in labs. That was before refridgerators etc. 15°C is between tap water and room temperature.

Please note that specific heat capacity is a quantity, not a unit.

There isn't a simple redefinition that would work in all cases, and this is probably why nobody does this.  Of course, you COULD define such a unit for a given system by defining it as a function of temperature, but it would be applicable only to the specific system.
As you noted, the physics behind the temperature dependence of a material's heat capacity lies in the quantum states which become available for excitation at different energies.  Consider an insulator which stores heat primarily in the lattice vibrations.  As you increase the temperature, more vibrational modes are accessible.  This is often described using the Debye model which gives the heat capacity as such:
$C_V = k(\frac{T}{\theta})^3$
where $C_V$ is the heat capacity, $k$ is a constant, $T$ is the temperature in Kelvin, and $\theta$ is a constant pertinent to a specific solid.[1]
However, a metal can also store a significant amount of heat in the motions of electrons since conduction electrons are not bound to specific sites.  Heat capacity for the electron 'gas' is often described like this:
$C_V' = k' \frac{T}{\theta'}$
where $C_V' \neq C_V$, $k' \neq k$ and $\theta' \neq \theta$.[2]  Depending on $\theta$ and $\theta'$ for a given material, at low temperatures (often cryogenic) the electron contribution can be the dominant mode of storing heat in the solid.  Thus, if you chose a unit which was sensible for an insulator, it wouldn't apply to a conductor (and vice versa).
Even within a specific material, the heat capacity on either side of a phase transition is often different since the phase transition has changed the quantum states available in the material.  Therefore, your new unit would become discontinuous at phase transitions and would need a different mathematical expression for each segment of the heat capacity function, not to mention a different parameterization for each material.  Hence, no real gain.
[1] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 106, eq. 47b
[2] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 193, eq. 37.  Note that folks don't actually use the symbol $\theta'$ in this case, but choosing so makes the explanation above clearer.


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Synthesis of glycerol from methane?
Any ideas on how to get $\ce{C3H8O3}$ (Glycerol) from $\ce{CH4}$?
My theory is that by reacting $\ce{CH4}$ with $\ce{Cl2}$ (or $\ce{Br2}$), getting $\ce{CH3Cl}$, then Wurtz's method to $\ce{CH3CH3}$, adding $\ce{Cl2}$ (or $\ce{Br2}$) again, getting $\ce{CH3CH2Cl}$, then reacting with $\ce{NaOH}$ to get $\ce{CH3CH2OH}$, and finally $\ce{CH3CH2OH + CO2 + H2O}$ and getting glycerol ($\ce{C3H8O3}$), but my way seems rather off to me so I need some help.

From methane?!?!

Your synthesis is unlikely to become a huge commercial success ;)
Methylchloride is a technical product, but usually made by reaction of methanol with hydrogen chloride:
$$\ce{CH3OH + HCl -> CH3Cl + H2O}$$
This works nicely without side products, because

$\ce{CH3Cl}$ has a low solubility in water (around $\mathrm{5\,g\cdot L^{-1}}$)
under standard conditions, $\ce{CH3Cl}$ is a gas (bp around $\mathrm{-28\,°C}$)

The radical halogenation of methane will usually lead to various halogenation products and thus is not a good idea. The same is true for the similar reaction of ethane.
In addition, I fail to see how the reaction of ethanol and carbon dioxide will lead to glycerol.
You should further take into account that there's plenty of glycerol available as a side product in the production of biodiesel.
From a commercial and ecological point of view, methane therefore is better converted to cabon monoxide and hydrogen by steam reformation. Further hydrogen may be obtained from a subsequent shift reaction and the resulting carbon dioxide may be reacted with gycerol to yield 1,2-glycerol carbonate.

Radical chlorination of hydrocarbons was done till the 70ties. As long as there was demand for methylene chloride, chloroform and "tetra", all products of the chlorination  of methane were welcome. Today synthesis of glycerol would start from propene, via epichlorhydrine. But I agree that there is no demand for synthetic glycerol now. There is too much peace and biodiesel around :=)


text:
Same monomers, different water content?
We're currently studying "the chemistry of contact lenses", which is mostly polymers and gels. I just looked at the package of two pure hydrogel lenses made out of the same monomers, but they have different water content? Why?

There are a few things that influence the properties of polymers:

Molecular weight of the chains, i.e. how many monomers are joined in each chain, can drastically alter the polymer's properties. For example, for polyethylene, UHMWPE has extremely long chains and produces a very strong material suitable for bullet-resistant vests, whereas LLDPE has much shorter chains that produce a weaker material that is much more ductile (think about how easy it is to stretch a plastic shopping bag or the like).
Another factor that affects a polymer's properties is branching. Branching is when a polymer is not made up of nice straight chains, but chains that occasionally fork off into several different chains. The difference between LLDPE and LDPE is that LDPE has a lot more branching. This keeps the chains from packing as nicely so LDPE isn't as strong.
The other big factor (other than adding another monomer to make a copolymer) is the introduction of crosslinking. Crosslinking is where different chains are connected to one another, often by addition of a molecule that can bond to a chain at both ends, e.g. rubber vulcanization crosslinks the different chains with linear sulfur chains. Natural rubber is very different than what most people think; it's a soft sticky material that doesn't hold its shape. The polyisoprene chains it's composed of aren't crosslinked and can easily slide past one another, but once crosslinked, the chains are connected and aren't free to move as much. Depending on the degree of crosslinking, you can go from something somewhat soft, like a bouncy ball, to something very rigid, like a car tire.

Hydrogels are not simple linear polymers. While they're mostly made from a single monomer, small additions of related molecules are used to control how they polymerize. N.B. These apply generally to silicones, other polymers may have other branching/termination processes. If the monomer only has two sites at which it can connect to other monomers, naturally the only structure that can be formed is a linear chain. One can control the length of the chains by adding a small amount of a terminator—a monomer that can only make one connection. If a terminator reacts with a growing chain, it stops the growth as it can't connect to anything else. The more terminator added, the shorter chains are, on average. Similarly, if you add a monomer that has more than two reactive groups, branching and crosslinking will occur. Often, a non-crosslinked polymer is supplied as a liquid, then a crosslinking agent is added to harden the polymer. Dow Corning has a great primer on the actual chemistry involved.
What makes a hydrogel special is that it's a polymer network that encapsulates water, i.e. it's an extensively crosslinked polymer that is hydrophilic. If the polymer was just a series of individual chains, it would just dissolve in water since nothing is holding the different chains together. When crosslinked, water can get in, but the mesh of polymer is linked together enough to hold its shape, though it may swell to accommodate the water. Depending on the degree of crosslinking, more or less water can fit within the structure. A less linked polymer can fit more water since there are fewer crosslinks resisting, but is less rigid when hydrated for the same reason (think Jell-O). A more crosslinked polymer won't absorb as much water, but will be much more rigid (like a contact lens). Contact lenses have to balance holding their shape, oxygen permeability, surface hydrophilicity, etc. so there are many reasons why, despite being based on the same primary monomer, two different brands may have a different blend of the other stuff that lets one control the hydrogel's properties.

When a polymer is produced, the final result does not only rely on what the monomers were. For example, there is "extended" polystyrene and "extruded" polystyrene. Or there are polystyrene sulfonate chemical polymers and etc.
Since nowadays, we need polymers to do many things for us (sheathing, sheltering, and sometimes even heat conductivity), we usually tend to modify their structure to make them more temperature-resistant, harder, stronger and ...; and that is because researches on new polymers are not always cheap or successful.
Let's explain with an example. Imagine you have $10000$ ethylene monomers that you will need to make a polymer from. Each of the PE polymers require $100$ ethylene molecules to form. Since you need this thing to be a little bit stronger, you would pressurize it, heat it, liquify it etc. So the result is a set of well-arranged polymer molecules [You can imagine lines of pupils standing in a school's yard, waiting to go to their classes] They are resistant to minor temperature rises and can be used in places where hard polymers are needed. (actually, not yet, but you are currently imagining :))
On the other hand, Imagine the raw "plastic" that is industrially produced with $10000$ monomers and the polymers are made of 100 ethylene molecules. This time it seems as if the pupils are dispersed through the school yard, though in tact with each other. You can easily conclude that this polymer is potentially weaker and is more vulnerable to temperature differences.
The first example is called "extruded polyethylene" and the 2nd one is called "expanded polyethylene". Notice that the monomers are the same, but the result is different in as big of a range that includes usage to specific heat capacity.
This is not the only case. You remember that I mentioned PS sulfonate. Sulfur, in here, is added to polystyrene to make chain-like "links" to polymers. For example, imagine the lines of pupils again. These sulfonates act as some older students that hold the pupils still and do not let them escape. These chains make polystyrene sufonate ideal for being an agent for dying clothes (since it becomes resistant to dissolution).
In your case, Hydrogel monomers are the same, but as I described above, either the how of their production or additional materials that modify the behavior of the Hydrogel or both are different.
Sorry, I didn't find any appropriate applicable images without license and I hope the description is not that vague. Hope I've helped.

I don't konw much about contact lenses. But for hydrogel, the reason may be as followed:
Conventional synthesized hydrogel is chemical crosslinked polymer network. Except the monomer, another reagent named crosslinker with multi-functionality such as the MBAA(N,N'-Methylenebisacrylamide)is needed. Then the swelling ratio or the water content of a hydrogel is strongly affected by the relative amount of crosslinker. Higher the crosslinking density, lower the water content.

It's better if you use capital letter after a full stop.


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Neutralizing sulfur-like smells in car
A full bottle of jewelry cleaning (links product and material safety data sheet below) broke in my car and went unnoticed for a few days.  Now the car has a sulfur-like smell whenever I go into the car.  It dissipates within a few minutes, but then returns very quickly once I leave the car.
The primary chemical components are alcohol ethoxylates, and can decompose into carbon monoxide, carbon dioxide nitrogen Oxides, sulfur oxides and other toxic gases.
I have cleaned the area and had the carpet (rear driver) shampooed from a car detailer, but still the smell remains.
What chemical compound can I use to neutralize the smell permanently (enter equilibrium)?

Product Link (Amazon)
Product Link (Manufacturer)
Material data safety sheet

Did you break a bottle with a cleaner for gold or one for silver?

Klaus, you're pretty amazing. I thought it was for gold, but you are right, it is for silver (my wife just woke up and confirmed).

Brian, don't tell her that it would not have happened with a cleaner for gold ;)

The reason the smell "dissipates" after a few minutes is actually that your nose starts to ignore it.

""The primary chemical components are alcohol ethoxylates, and can decompose into carbon monoxide, carbon dioxide nitrogen Oxides, sulfur oxides and other toxic gases."" Absolute Nonsense

@Georg, I pulled that from the master data safety sheet.

"Decomposition" in MSDSes is something to happen at temperatures above 200 °C, not relevant here. The sulfurous smell comes from the thiourea, which reacts with moist air.

I guess that it actually was a silver cleaner, not one for gold or precious stones.
The latter mostly contain nonionic surfactants, which are made by the reaction of water-insoluble long chain alkanols (C12 - C15, from natural fatty acids) with oxirane. These washing agents dissolve well in water and help to remove grease and fingerprints from the jewelery. Their action isn't much different from that of a dishwasher for manual cleaning.
These cleaners don't smell the way you described and the car retailer probably used similar cleaners for the car.
Silver cleaners are a different league. The do contain the same surfactants (washing agents), but in order to actually remove the tarnish (silver sulfide) from the silverware, they do also contain a bit of acid (typically sulfuric acid) and thiourea, which probably is responsible for the smell.
Repeated shampoeing probably is the best method to get rid of the thiourea without ruining the carpet.

Why wouldn't you recommend a basic detergent or shampoo to neutralize the acid?

@Nick Acidity did not seem to be the problem, thiourea is. Since this is everyday-chemistry, I'd always suggest the most skin- and material-friendly method, which seems to be solubilisation of the thiourea plus dilution of the acid rather than neutralisation.

Ammonia might break up the thiourea, but then ammonium sulfide would form and last! Some hydrogen peroxide or hypo bleach might solve the problem, but beware any textiles.

Homemade Carpet Fresh
I would recommend a home made dry cleaning using diatomaceous soil compound. This will eliminate the problem of toxins and erosion of the metal interior under the carpet. Use some DAP mixed with your favorite dry detergent and vacuum it up. If plaster of Paris is not available basically any spackle will work. Mix 1/1


text:
Why is sand, sand; and why is glass, glass; While both mainly contain Silicon Dioxide?
I know that there is the famous process of heating material and converting them to glass; but what I don't know is, what is the chemical process of the creation of glass? Is it crystallization?
How could it result in so much visual differences?
What is the description of such a radical process?

Is it crystallization?

You are correct. The main difference is that sand is crystalline and glass is not—it is amorphous.
The main component (> 95%) of common yellow sand is quartz (the mineral whose composition is SiO2). Note that not all sand is quartz. There are white sands containing calcite (CaCO3) and black sand (containing various heavy minerals). But the most common sand is indeed quartz sand: SiO2.
Glass, the type you see in your everyday life, on the other hand, is not composed of pure SiO2. It has a bunch of other additives such as Na, K, B, and others. This is done to modify the properties of the glass and make it more suitable for human use. It doesn't matter much though for our discussion.
So if they are made of the same thing, why the difference? The answer is cooling rate. If you cool molten SiO2 slow enough, the atoms have enough time to organize themselves into a crystalline structure. In the case of pure SiO2, this is a network of SiO4 tetrahedra: One silicon atom surrounded by four oxygens. If it cools too fast, then the crystalline structure does not form. It may be completely amorphous, or form into a sub-microscopic array of SiO2 crystals in various structures (CT-opal for example).
What determines the cooling rate? Well, in the case of glass it is a matter of minutes. You've seen glass making: The glass is molten and very quickly it solidifies to a solid. In contrast, most of the quartz sand you're seeing is actually broken fragments of rocks called granite. This type of rock has abundant quartz in it, and it forms deep underground (as in 10s of kilometers) at very slow cooling rates. While a glass maker can take his glass and let it cool in the atmosphere or in water, molten silicate magma ("glass") deep in the Earth is surrounded by rocks that are in the hundreds of degrees. This slow cooling facilitates crystallization of the SiO2 into quartz rather than glass. How slow is this? At least tens of years, more commonly hundreds or even thousands of years. This is much slower than the seconds and minutes in glass making.
Slow crystallization of SiO2 in the Earth is not the only process. There are actually processes that can take molten silicate magma and cool it very rapidly. This commonly happens in volcanic settings: A volcano throws molten magma (again, made mostly out of SiO2) and throws it into the atmosphere where it can cool very rapidly. It's even faster if the magma erupts into water. The result is volcanic glass: sometimes called obsidian.
The difference in the visual appearance is actually not due to the presence or absence of a crystalline structure. It's mostly related to grain size. If you take window glass and shatter it so it's all small grains, it will look just like sand. On the other hand, if you take quartz, and make it big and smooth (just like window glass), it will look like this:

(By Parent Géry (Own work) [Public domain], via Wikimedia Commons)

Take sand, melt it, let it cool and you have (sort of) glass.
Glass is an amorphous solid. It appears to be solid, but is, in reality, a viscous liquid.
Sand is essentially the same, although broken up into very small pieces. Although sand could contain quartz (crystallized) fragments, it doesn't change what we observe about the sand.

How could it result in so much visual differences?

Glass is clear since the sand it is made from has chemicals added to lower the melting point and to make it transparent (by removing some impurities). Sand is composed of everything in the earth, so it is not as transparent as glass. Also the small particles deflect light so it is difficult to see a pattern through sand.

Also, sand is not amorphous. Sand is crystalline.

@Michael Not every particle of sand is crystalline. Pick up any metamorphic rock. How much of the rock is crystalline?

Proper metamorphic? 99.99%.


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what is the white fuzz left behind on basement floor after puddle evaporated
My basement is quite dry (knock on wood!).  Recently I brought into the basement for the winter a table from outside.  The table had some ice stuck to it which I couldn't remove, so I simply let it melt on the floor of the basement.  After the puddle that formed evaporated, an interesting artifact was left behind along the the perimeter of the puddle.  It's a raised white line of fuzzy material.
Here's the puddle's perimeter

And here's a close-up of the line that's left behind with a clothespin for scale

Is this efflorescence?  I did some googling, and this is what seems most likely, but I don't know much about chemistry and thought I would ask those who would actually know.  Is it something else?
Thanks.

@theo well, that may be a bit extreme. Anyway, it's something that is probably crystalline and highly soluble in water. NaCl is my best guess. Another way would be to put it on a blue flame and see if it burns yellow-orange.

I will happily go put some on my stove's burner and report its burning color.  Sounds like more fun than tasting it.  :)  I'll let you know about vinegar too.

@Michael - It was a little hard to tell with the burning.  I scraped some onto a knife and stuck it in the blue flame.  The flame did turn orangish-yellow.  However, it wasn't clear it was because of the substance: when I stuck a plain knife in the flame it also changed color to the same orange-yellow.

@Michael When I poured a little vinegar on it, it fizzed and disappeared.  Calcite, perhaps?  Pouring vinegar on an area of the basement floor without the substance did not cause fizzing.  Nor did pouring vinegar on a small platter of table salt.

@Michael I need to revise my earlier comment: I tested it again in a flame.  This time with a knife that originally did not change the color of the flame, and it clearly did when I added the white substance to it.  I think perhaps the first time I tried with a "plain knife" I had not adequately washed off the substance.

also, being a bit reckless, I tasted it.  It's salty, but not quite like table salt.  Almost a stingingly-sharp saltiness.

Three pieces of information from the comments:

It fizzed and disappeared in vinegar.
It had a yellow-orange color in blue flame.
It's salty (note that it's usually a bad idea to taste random stuff you find on your floor).

The only white crystalline material that is a plausible candidate for being there and fizzes in acetic acid (ie vinegar) is calcite - $\ce{CaCO3}$. It's also a mineral that has 'retrograde solubility' meaning it dissolves more readily in cold water. This is probably how it got to the water in the first place. This is also why it precipitates on your electric kettle. It is the same stuff.
The salty stuff that burns yellow-orange is obviously $\ce{NaCl}$, aka table salt. It also probably dissolved with the vinegar as well. Why is the salty taste not quite like table salt? Well, first of all it is mixed with calcite. But, calcite has no taste. It's possible that there is some $\ce{KCl}$ or $\ce{MgCl2}$ in there as well. They are slightly more bitter than $\ce{NaCl}$. This is the stuff they use in low sodium salt.
There's another possibility - it could be any of the epsom salt family, $\ce{MgSO4.$n$\,H2O}$.
As to would this be dissolved in the first place, you can only speculate. You say that it came from a table that was outside. Did people eat there? Could people had spilled salt on it? Is the airborne dust in your area contain a lot of calcite? What about your basement? Could someone have spilled some salts on the floor, maybe even years ago?

so am i correct in understanding your answer to say that the substance is a mix of both CaCO3 and NaCl?   would make sense to me...

Most likely, yes.

Thanks - it's all just out of curiosity, not of any real need to know.  is it the evaporation that pulls the minerals from the floor?  or are they more likely already present in the water?


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How are catalysts discovered for a particular reaction?
I want to understand how are catalysts discovered for a particular chemical reactions say for example $\ce{Fe,~Mo}$ for Haber's Process reaction of $\ce{N2}$ & $\ce{H2}$.
Reactions as examples with some background (story of discovery,discoverer or method) would better serve the purpose.
For this question keep it limited to Catalysts involved in Haber's Process.

Specifically for Haber process finding of iron catalyst was a result of wide screening of many metals in attempt to find cheaper alternative for well-known at time ruthenium catalyst.


text:
Phase rule and degrees of freedom
In the formation of 3-bromo-1-butene and 1-bromo-1-butene, from $\ce{HBr}$ and $\ce{C_4H_6}$, in a gas phase:
$\ce{HBr + C_4H_6 \to 1BrC_4H_7}$
$\ce{HBr + C_4H_6 \to 3BrC_4H_7}$
the number of degrees of freedom is given by: $F=C-P+2$
I'm trying to rationalise this but finding it quite confusing. For example, can we state that there is only 1 phase, i.e. a gas phase?
And the components, is the isomer of one independent of another? I'm taking there to be 3 individual components.
My answer is then 4 degrees of freedom, which seems okay, given $p$, $t$, $X$, $V$ could vary?
I'd appreciate your help in this, thanks.

Let's do it step by step.

$C = 2$. The reason is that you have two components: $\ce{HBr}$ and $\ce{C4H6}$. The third compound, $\ce{BrC4H7}$, is simply a combination of both of them. Remember what is $C$: It's the minimal number of components needed to describe all phases in the system. Since $\ce{BrC4H7}$ can de described as $1\times\ce{HBr} + 1\times\ce{C4H6}$, it is not an independent component.
$P=1$ or $P=2$. It depends on how much the reactions has progressed. Basically, you have one phase on each side of the reaction. On the left, you have one gas phase that has a mixture of both $\ce{HBr}$ and $\ce{C4H6}$. On the right, you one one liquid (?) phase: $\ce{BrC4H7}$. So here are the possibilities:

The reaction has not started yet. All you have in the one gas phase. In that case, $P=1$ and $F=2-1+2=3$.
The reaction is going as written, but not completed yet. In this case not all gas is consumed yet, but liquid exists. Then, $P=2$ and $F=2-2+2=2$.
The reaction is finished. All gas is now liquid. Then we're back to $P=1$ and $F=2-1+2=3$.

So what happened in stage 2? Why did you lose a degree of freedom? Well, it depends on what what the driving force for the reaction. For example: Why did gaseous $\ce{HBr}$ and $\ce{C4H6}$ combine to form liquid $\ce{BrC4H7}$? Was it because of cooling or heating? In this case, the lost degree of freedom is $t$. Was it pressure? Then the lost degree of freedom is $P$.
$X$ is a bit problematic. There is not enough information given in the question. In my opinion, $X$ was fixed to begin with: I assume that once the reaction is complete, none of the reactants remain. Thus their ratio is fixed to 1:1 and $X$ is constrained. Had it not been the case, you would also get differing $P$ in stage three because you would have two phases, not one.

Thank you Michael. So, if I was to keep the ratio of reactants fixed at the beginning, would cause a loss of a degree of freedom? Or would that make any difference?

@Maximilion I edited my question to address $X$


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What is the usage of orbitals more complex than f orbitals?
Every high school learner, in each corner of the world, faces the lesson History of Atom during his courses, just as I did. We learned about s, p, d and f orbitals, though there were no signs of orbitals in molecules.
Then I wondered, are there any other orbitals, simpler or more complex, than the four mentioned? Surprisingly, I learned that there are also usages for orbitals g, h, i and even k and l.
Yes, I use the word "usages". Because I believe, unless something is useful, it will never enter the domain of science. Anyway, I read in Wikipedia (though not much of it I did understand) that these orbitals are used when describing and doing the measurements of molecular orbitals.
Since no element in the periodic table has enough electrons to fill even orbital g (in its base state), in cases of molecular orbitals that have a g defined in themselves, atoms must have been excited. Excitation needs energy, doesn't it? Where does this energy come from? Isn't the formation of new bonds usually exothermic?

I strongly disagree with the notion of "unless something is useful, it will never enter the domain of science". There is a branch called Applied Science, where this is true, but for most other science "usefulness" is not a criterion.

Here, by usage I don't mean sth that humans will make a use of it!:) I mean the notion should be able to enter a specific field of theory, no matter how much "applicable" it is.

I am not sure I follow your train of though entirely... Anyway, a part answer to your question: you can excite your electron quite easily by eg electromagnetic radiation of the right wavelength. I would also assume that when the higher orbitals will be briefly occupied when eg a cation accepts a new electron - but I am no theoretical chemist...

Surprisingly, I learned that there are also usages for orbitals g,h,i
  and even j.

Actually, the letter "j" is not used, so it is s, p, d, f, g, h, i, k, l, etc.
The higher angular momentum orbitals do enter the domain of science, due to excited states of atoms. Transitions to and from excited states are observable through atomic spectroscopy.
For example there is the article Microwave spectroscopy of Al I atoms in Rydberg states: D and G terms

can you please give more description of your answer? I repeat the question: Where does the energy of excitation come from? Or is this just my misunderstanding?

The energy of excitation comes from electromagnetic radiation (light).  It could be visible, infrared, ultraviolet, or microwave depending upon the transition.

One use of orbitals beyond f is in computational chemistry to construct basis sets. It's important to remember that orbitals are entirely a mathmatical construct that chemists and physicists have found useful in conceptualizing chemical properties and so we can use the idea of orbitals outside of their normal context of framing the periodic table. To develop a nearly complete basis set for certain elements, one can use a large collection of orbitals of different angular momentum. As an older example, Nesbet, Barr, and Davidson attempted to compute the energy of a neon atom using a basis set that included up to i orbitals.
As an interesting side note, I found a nice visualization of the orbitals (really the spherical harmonics, but its essentially synonymous in this case) up to n on Sylvain Paris's MIT page.

The letters beyond $f$ can also appear in higher orbital angular momentum states, as, $^1G$ in a low spin $d^2$ complex.


text:
What is the reason behind choosing the specific elements used for the synthesis of heavier elements?
The following has been derived from the wiki page "Extended periodic table" and is one of the many examples d or f block elements being used in synthesizing new heavier elements (or at least this is the pattern I figured out by searching many other examples of the nuclear reactants in the way to synthesize heavy elements):

Attempts to synthesize still undiscovered elements:
\begin{align}
\ce{^254_99Es + ^48_20Ca &-> ^302_119Uue^*} \\
\ce{^249_97Bk + ^50_22Ti &-> ^295_119Uue + 4 ^1_0n} \\
\ce{^{nat}_68Er + ^136_54Xe &-> ^{298,300,302,303,304,306}Ubb^* -> no atoms} \\
\ce{^238_92U + ^66_30Zn &-> ^304_122Ubb^* -> no atoms}
\end{align}

The question is, why is it like this? How did the scientists reached an agreement on which elements to collide? Was it anything especial about those atoms?
P. S. If you are sure that there is an exception (exceptions omit rules; at least in this case), your answer will be accepted.

This is for Physics!

Something special was asked about the atoms. If the answer is a property of an atom, it must be chemistry.

There are different strategies for synthesizing superheavy elements.
Two of the main strategies are referred to as "hot fusion" and "cold fusion".
The first two reactions are good examples of the "hot fusion" strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking.
Experimentally, a very heavy element target (the Es or Bk in the first two reactions) is bombarded by an ion beam of the second element.  The target element is choosen on the basis of being very heavy, availible and not rapidly decaying.
$^{48}_{20}\ce{Ca}$ is choosen for a very special reason.  This is not ordinary calcium.  The most common isotope of calcium has only 20 neutrons, but this isotope has 28!.  It is extremely neutron rich, espeically compare to other relatively stable isotopes in this mass-range of the periodic table.  $^{50}_{22}\ce{Ti}$ is also very neutron rich.
Why are neutron rich nuclei chosen? There are two reasons for this:
First, confining many protons in a small space causes electrostatic repulsion.  As atomic number increases, the stable nuclei deviate further and further from a 1:1 neutron:proton ratio, because having more neutrons keeps the protons further apart and decreases electrostatic repulsion energy.
Secondly, you will notice the asterisk (*) in the 1st, 3rd and 4th equations.  This denotes the compound nucleus which initially forms in the fusion reaction.  It is an extremely ($<10^{-21}$s) short lived intermediate state, not the ground state of a nucleus, but more like a highly excited state.  It's formation is not considered creating a new element.  It is critical to the strategy of forming an element that the intermediate state loss energy through neutron "evaporation".   Notice that in the second equation, the step of the neutrons evaporating is shown, while the intermediate state is not shown.  Generally speaking, in the hot fusion strategy, 3-5 neutrons will evaporate to cool the intermediate state.
For further reading see:
Future of superheavy element research: Which nuclei could be synthesized within the next few years? and
The discovery of the heaviest elements Rev. Mod. Phys. 72, 733

The stability of heavy nuclei is a function of both atomic number and atomic mass.  Therefore, an appropriate number both of protons and of neutrons in the target nucleus must be attained, or else vanishingly short lifetimes can be expected.  (See the Wikipedia article on, e.g., the island of stability, and links therein.)  The 'source' nuclei are chosen to provide proton and neutron counts that are anticipated to yield heavy nuclei with sufficiently long lifetime so as to be measurable.
Of course, it's not as simple as just choosing the right total number of protons and neutrons in your starting nuclei, as you're not guaranteed that all of the neutrons and protons will stay put in the synthesized nucleus, nor that you won't have rapid radioactive emissions that knock you away from your desired composition.  For example, in the various syntheses of dubnium reported at Wikipedia, various numbers of neutrons were ejected as part of the nuclear synthesis 'reaction'.

This answer is informative, but unfortunately is not the main answer for the question. Yet, thanks for sharing your knowledge.

Can you provide further clarification as to the type of answer you're seeking? It's unclear what other sorts of considerations to which you may be referring.  (I assume you're not looking for 'frivolous' reasons, such as "they like their names" or somesuch.)

:D Of course not. I just asked for an expansion of your 2nd paragraph, since it's what's been mainly asked. More references, more examples, more description. You agree that for this type of question, a paragraph of answer is not really enough.


text:
What do atoms look like?
A professor of mine noted that when he was in school, microscopes weren't powerful enough to resolve certain things (I forgot what it was). But current microscopes are powerful enough.
Extrapolating from this trend, he hypothesized that by the time we were his age, we would be able to see atoms.
Wait. Seeing atoms? Do we expect to see discrete little dots for electrons (I don't think so)? What exactly can we expect to "see," if anything at all?

I wonder that's the question that every curious student tackle in chemistry and physics class.

Here are some images from a scanning electron microscope Link

Yes, we can definitely see atoms in modern microscopes.  I'll try not to get deep into the subject of "seeing", but I will note that most people say they are seeing when they use their eyes to observe objects.  That is, they are using an optical system (eye) consisting of a lens and a detector (retina) and some image processing carried out by a computer (brain).  Thus, any instrument which uses some sort of detector and a computer to turn the signals of that detector into an image we feel comfortable looking at qualifies to me as "seeing in a microscope."  If you're comfortable with that then read on.
My favorite microscope which is capable of seeing atoms is the transmission electron microscope (TEM).  Using a high quality TEM, we can easily resolve objects around an angstrom in size, which is the distance between atoms.  To image a single atom, you also need to have a sample which is only a single atom thick at some point.  This is tough to accomplish but can be done.  Today, images of graphene provide arguably the best "single atom" TEM images.  A seminal paper is here (paywall).  You can also search the internet for graphene images such as this.
In the links above, the images show the shapes of atoms as little blobs, and to a large extent the blobs are more a description of the microscope's ability to focus rather than the shape of the atom itself.  However, the atomic force microscope (AFM) excels at seeing the electrons at the surface of an atom.  With AFM, researchers are even able to image the chemical bonds of molecules sitting on a substrate.  In this case, the image shows the bonds as rods and the atom centers are not really visible.  Chemistry textbooks often show the "shapes" of atoms by drawing the electron orbitals, so in a sense these rods are the shapes of these atoms too.
This just points up that the "shape" of atoms is a non-trivial subject, which is very dependent on how you look at them.

We can't "see" individual atoms with light. The wavelength of visible light is too large to resolve individual atoms, but that doesn't mean we can't detect individual atoms using various methods.
X-rays and beams of electrons both have about the right wavelengths to detect atoms and some microscopes based on either can produce images that resolve to show atomic positions. Transmission electron microscopes can "see" individual atoms in some circumstance. More importantly, the structure of crystals can be resolved by interpreting the diffraction of x-rays from the crystal lattice, providing detailed analysis of the 3-dimensional structure of crystals. This is almost a routine way of determining the structures of molecules as complex as proteins. In a sense this is "seeing" atoms (more strictly it is detecting the electron density). But it is a very indirect way of "seeing" and, strictly, you are detecting using a large assemblage of atoms not just one at a time.
There are modern techniques that allow individual atoms to be detected and even manipulated. These are based on the use of very small needles with atom-sized tips. Atomic force microscopy or scanning tunnelling microscopy and other variants can use the controlled movement of these fine needles to measure the force of interaction between the atoms on the tip of the needle and the molecules or atoms on a surface. So what you are "seeing" is the atomic forces generated by the electron clouds in individual atoms the they interact with something else that is also atom-sized.
For example this image shows a sheet of graphene:

And this image shows a single molecule of pentacene (with a model of the structure shown below the actual image):

There is even an amusing animated video by IBM (who invented the atomic force microscopy technique) showing a story told by manipulating the positions of carbon monoxide molecules on a surface.
These techniques have even been used to understand reaction mechanisms by observing what happens to individual molecules.
So we can "see" molecules at atoms by using devices that can measure atomic forces very precisely. In that sense we can see what they "look" like at least from the point of view of other atoms interacting with them.

Since the nucleus is concealed within the much larger electron cloud, the shape of an atom is the shape of its orbitals; the 3D pattern of bonds in crystals, etc.  So, if we could somehow see one without using light, an atom would "look like" its electron wavefunction, its surrounding standing-wave of probability.

H orbitals wave function:  a 2D probability plot
Gallery of orbitals: three-D, the hard and shiny version!

If I understand it right, the wavefunction axes of an individual atom in space have uncertain direction.  Therefore unless we somehow probe an atom and "collapse" its orientation, the above wave functions should be seen as smeared-out spherically.
On the other hand, everyone in the 1950s knew that atoms look like this.
The common "three-planet solar system" image has ended up deeply embedded in American pop-culture.  It's wrong, and should be replaced by various plots of 2D wavefunctions, but just try to convince the public!  Heh, also, radio waves don't come out of the tips of broadcasting towers.  And, falling raindrops don't have sharp points.  Instead they're spheres!
:)


text:
Electrolysis of salt water into sodium hydroxide with YSP
I want to electrolize salt ($\ce{NaCl}$) water into a solution of sodium hydroxide for soap making. The salt I have has yellow prussiate of soda (sodium ferrocyanide) added as the anticaking agent. What will happen to the ferrocyanide ($\ce{Fe(CN)_6}$) molecules during the electrolysis process?
I will be using an old battery charger to perform the electrolysis, $20\, \mathrm{V}$, $0.9\, \mathrm{A}$ or $1.8\, \mathrm{A}$ settings.

""yellow prussiate of soda (sodium ferrocyanide) "" I doubt that. Look up "ferro" and "ferri"

@George Sodium ferrocyanide is frequntly used as an anticaking agent added to food grade salt.

If you want sodium hydroxide for soap making you should just buy some: it is a cheap and common chemical. This is because there are effective industrial processes to make it in vast quantities.
Trying to make it yourself is not easy and will cost you far more effort than just buying it. The least of your problems will be what happens to contaminants like the hexacyanoferrate (which is present in such low concentrations it isn't really significant). Have a look at the way industry makes bulk hydroxide and chlorine (e.g. here); it isn't as simple as just electrolysing brine. All the processes involve sophisticated ways to separate the products produced at the different electrodes. You don't get decent quantities of either product if you don't maintain this separation (one method uses flowing mercury; another expensive ion-exchange membranes). In addition to this, uncontrolled electrolysis is dangerous as it produces toxic and explosive gases. You really don't want to try it at home.
Since the most basic components for an effective reaction will cost you far more than the couple of dollars a large bottle of pure sodium hydroxide will cost, buy the chemical and ignore the electrolysis as a route to what you want.

I think your answer is stating the obvious.  OP was probably hoping to find out how to circumvent some or all of the obvious.  I am also curious.

Electrolize salt (NaCl) water without a proton exchange membrane is more likely to produce NaOCl. You need the membrane to keep chlorine and sodium hydroxide from combining to form bleach. The membrane may be made from a polymer which only allows positive sodium ions to pass through it.
Saftey note. The whole process also creates dangerous potentially deadly chlorine gas if not vented properly.
Its probably safer and cheaper to buy your sodium hydroxide from a commercial source as the producer is likely able to capture revenue from selling the sodium hydroxide as well as chlorine, and hydrogen that are made during the process, while buying their electricity at lower than residential rates.
Edit,  I am not sure what will happens to the sodium ferrocyanide. The worse case is it is also broken down, and  that at above room tempurature, cyanide gas will evolve at  the same electrode as the chlorine gas,  underscoring the need for proper ventilation.

Same problem as the above, you are clarifying why it is hard, not how to make it happen.  There are cases where buying is not an option.  What if your underwear on Mars is dirty and you have to go to see the nurse.  You have salt (oil too I hope) and electricity but no nearby shop to buy soap or sodium hydroxide but the Space-X people thoughtfully included a lot of salt as a raw chemical  feedstock. (Pity about the anti caking agent though.)

KalleMP, The OP asked how to make NaOH from salt with electrolysis, not how to make soap to wash underware on Mars. He has all he needs except a proton exchange membrane. The first thing I did was tell him he needs one and why. Sure, I did not write a how to make a proton exchange membrane because that was not the question. I supose if one were to write a "how a dig a hole", one could start with cutting down a tree for the handle, and how to forge steel for the shovel blade, but where does it end, plant a tree and make an axe before he cuts down said tree for the shovel's handle?

You are right.  It looks like there are suitable films available from specialist suppliers.  I was hoping that a salt bridge or some other lateral thought might be an option for someone trying to do this at home.  It is good that you added something about the OPs question to your answer as well.  -  http://fuelcellsetc.com/2012/07/nafion-membrane-thickness-for-electrolyze/

There are many options for a proton exchange membrane, including ones that can be fabricated at home with common materials. The breath of possibilities is largely why i thought it inappropriate to choose one specific type to use given the  op's specific question.


text:
Why can alkali metal hydrides usually not act as nucleophiles towards carbonyl groups?
Think for example of $\ce{KH}$ or $\ce{NaH}$ in enolization reactions. This remark was made in one of my textbook, but I could not remember why this was actually the case. Why can't $\ce{NaH}$ or $\ce{KH}$, being hard nucleophiles, add to carbonyl groups?

What do You mean with "enolisation"?

Jori that's an excellent question. The enolization process that you refer to (e.g. removal of an alpha proton) is very different from carbonyl addition.  Enolization is dependent on the basicity of the reagent, while carbonyl addition is dependent upon nucleophilicity, and NaH is a very strong base.  However, the hydride in NaH has more negative charge on it than with LAH, since the electronegativity difference between Na and H is greater than Al and H, therefore I would have expected the hydride from NaH to also be a very strong nucleophile. I think @Dissenter is on the right track when he

suggests that metal complexation with the carbonyl oxygen may be important.  Indeed, Mg and Al are good at complexing with oxygen, sodium much less so.  Perhaps favorable complexation reduces the activation energy for carbonyl addition through entropic considerations.  Also note that as @Georg has noted, NaH is not really soluble and who knows what goes on in a surface reaction.  Perhaps if the NaH were somehow solubilized, maybe carbonyl addition would be observed.

@Georg See https://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism .

The problem is that neither of these nucleophile are hard. Vanilla hydrides such as potassium hydride or sodium hydride release very soft hydride ions. Hydrogen only has one proton. Giving it two electrons to stabilize as in the case of the hydride anion is a tall order. Note the proton to electron ratio: 1 to 2. Even carbanions which are unstable species don't have such a ratio.
Therefore, the hydride ion has a massive ionic radius and is easily polarized, making it an extremely soft base. Electrophilic carbon atoms in carbonyl groups are usually hard(er).
Plus the 1s orbital of hydrogen is too small to effectively overlap with anything other than another hydrogen. Is your textbook written by Clayden? I distinctly remember Clayden's remark about hydrogen's 1s orbital with regard to its anion's nucleophilicity.

Usual reagents used for hydride delivery/reduction of carbonyl compounds (and related) include LAH, NaBH4, and NaBH3CN, in order of decreasing strength. As far as I know, the lithium ion in LAH plays an integral role in its mechanism of reduction by bonding with the carbonyl oxygen of carboxylic acids.
Also, these aforementioned reducers are in order of decreasing strength because they are listed in order of decreasing partial negative charge on their hydride ions. For example, the nitrile group in sodium cyanoborohydride withdraws electron density from the hydrogens. Hence the reason it is the weakest of the three aforementioned reducers. NaBH3CN will only act on strongly polarized iminium ions rather than plain imines or carbonyl compounds.

Could you add a few words to explain why "Jori's hydrides" are different from the hydride in say, lithium aluminum hydride?

@ron I'll try my best to expand; my knowledge of them is limited.

The most important and primary problem is (un-)solubility!

@Georg can you expand on what you mean by that?

The more polar Hydrides like NaH are not soluble in aprotic solvents! How should they react?

I can find different mechanisms, in some sources, the Li is bonded with oxygen of carbonyl compound and in others $\ce{AlH3-}$ attaches to oxygen atom.