text:
Reaction of magnesium with the surrounding air
In my lecture notes for inorganic chemistry, it is stated that magnesium has high affinity for oxygen but is rather stable at room temperature in air.
Can anyone explain why this is the case?

In fact we could have the reaction O2 + 2Mg --> 2MgO , so can you have some point missed. Because they can have ionic repulsion, but if I am wrong correct me.

I believe that magnesium exposed to air quickly ends up with a coating of magnesium oxide on its surface.  This oxide coating is quite uniform and does not allow any further oxygen to reach the surface.
Several metals, but NOT iron, have this property of a protective oxide coating.  In iron the coating forms a spongey surface easily penetrated by further oxygen.

Magnesium's oxide layer is not protective- its not tenacious like aluminium oxide on aluminium. Its simply that virtually all metals form an oxide layer almost instantaneous with cold dry air- the activation potential is small because the surface of metals has high energy due to deformed metal bonds to form the surface. After the initial oxide is formed, the reaction always slows down as a greater activation energy is needed for it to continue- both iron and magnesium are still corroding but just its not obvious.  Also don't confuse iron oxide (which would initially FeO) with rust Fe2O3.xH20- with the normal thickness of FeO on iron its just about clear.


text:
Is (S)-3-hydroxypalmitoyl-CoA a hydroxy fatty acyl-CoA?
Is there anything special about (S)-3-hydroxypalmitoyl-CoA comparing to those with different number of carbons, e.g. (S)-3-hydroxyoctanoyl-CoA, (S)-3-hydroxydecanoyl-CoA, (S)-3-hydroxylauroyl-CoA?
I am not a chemist at all; I am looking at the ChEBI ontology and wondering why (S)-3-hydroxyoctanoyl-CoA, (S)-3-hydroxydecanoyl-CoA, (S)-3-hydroxylauroyl-CoA are descendants of hydroxy fatty acyl-CoA, while (S)-3-hydroxypalmitoyl-CoA is not (it is only a descendant of long-chain fatty acyl-CoA).

This just looks like an arbitrary distinction they made in the ontology between 'short chain' acyl-CoAs and 'long chain'. As you can see, the 3-hydroxypalmitoyl chain is somewhat longer (16C) than the others. It stands to reason that the creators of the ontology have decided to subdivide hydroxy fatty acyl-CoAs into those with shorter hydroxyl fatty chains:

hydroxyoctanoyl:   8C
hydroxydecanoyl:  10C
hydroxylauroyl:   12C

and long:

hydroxypalmitoyl: 16C

As to where the dividing line is, the hydroxytetradecanoyl-CoA (14C) homologue is not 'long chain', whilst there is no entry for hydroxypentadecanoyl-CoA. The octadecanoyl homologue belongs to its own class (of 3-hydroxyoctadecanoyl-CoAs, of which there are only two conceivable enantiomeric members), and there does not appear to be icosanoyl homologue in the database. There may be other members of the 'long chain' family if you search around, who knows?
In short, it appears to be arbitrary.

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How is the magnetic property of an element affected when the element is part of a compound?
Data on this site conveys that Neodymium is paramagnetic. Yet the Neodymium magnets out there make extremely strong magnets. I'm guessing these are not pure neodymium magnets but some kind of compound.
Is it possible to predict the change in magnetic property of a compound given the knowledge of it's participating elements?

Magnetism appears in many very different forms that have very different origins.
All elements show some diamagnetism, even though often overshadowed by other, stronger, forms of magnetism, because it is a property of the bound electrons.
The diamagnetic contribution caused by the inner electrons that are not part of a chemical bond will therefore remain unchanged if the compound bonds chemically. The diamagnetic contribution from the electrons participating in the bond will of course change, so even for the most simple form of magnetism, it is not possible to predict the diamagnetic behavior of a compound from that of the elements.
Paramagnetic materials have a permanent magnetic moment from uncompensated magnetic moments, typically electron spins (there are exceptions where a paramagnet is formed from larger units, like magnetic particles).
In a very simplified way, you will find unpaired electrons typically on the outer levels of elements, where they are influenced and participate in chemical bonds. This is even the case for the strongly localized d and f electrons that are characteristic for the strongest paramagnets like $\ce{Gd}$.
So clearly, paramagnetism is influenced by the chemistry.
Both diamagnetism and paramagnetism are extremely weak effects that normally don't play a role in everyday life.
Now for the 'real' forms of magnetism, ferromagnetism (ferri- antiferro and ferri etc.): All of these types of magnetism are collective phenomena, i.e. they are the result of the internal interactions of a large system, through conduction electrons (itinerant), superexchange, double exchange, etc.
Due to the collective nature, it is not possible to predict the behavior of a compound from that of the physical parameters of the elements. (Of course, you can simulate the magnetic behavior of more or less any material, but I think that was not the question)
You could not call $\ce{Fe}$ itself a ferromagnet, the ferromagnetism is a property of a huge number of $\ce{Fe}$ atoms that are acting collectively. This ensemble is very sensitive to changes, often even a small temperature change (e.g. at 20°C for $\ce{Gd}$) will completely change the magnetic properties. A chemical reaction or even simple alloying, will completely change the magnetic properties.
The macroscopic magnetic properties of a material (susceptibilitiy, coercivity, remanence, saturation magnetization, frequency dependence of parameters etc.) do not allow to predict the properties of a compound that contains this material. So you can't predict the coercivity of  $\ce{NdFeB}$ given the coercivities of  $\ce{Nd}$, $\ce{Fe}$ and $\ce B$.
It is possible to simulate the complete electronic structure of a compound with Density Functional Theory. This simulation will tell you the magnetic parameters of the material, including the ordering temperatures in the case of ferromagnetic materials, to a varying degree of accuracy. It is a very computationally intensive approach, to compute the properties of a new material is a research project, requires some ad-hoc assumptions and will need to be experimentally verified as it is not yet 100% reliable.
This computational approach does not make any use of the macroscopic physical properties of the elements in a compound, it directly solves the many body problem of electrons and nuclei.
In summary, it is not possible to predict the change in magnetic property of a compound given the knowledge of the magnetic properties of its participating elements, but you can simulate the electronic structure of a compound, and that will tell you how the compound will behave magnetically.
Note: Neodymium magnets aren't made out of $\ce{Nd}$ metal. They are made from sintered $\ce{Nd2Fe14B}$. $\ce{Nd}$ is only a small fraction of the entire material. This is a very complex material. The exceptionally high coercivity, i.e. the resistance against a change in the magnetization, is the result of high magnetocrystalline anisotropy of $\ce{Nd2Fe14B}$ and the interfaces between the grains that pin the magnetization. This is something you would not think about given just the magnetic parameters of $\ce{Nd}$, $\ce{Fe}$ and $\ce{B}$.

I''m a little confused about the last paragraph stating it is not possible to preduct the property of a compound from knowledge of it's participating elements, and then stating it is possible to simulate the electronic structure. Doesn't it amount to the same thing?

Hi Everyone, edited the answer to make it more clear that you can simulate the magnetic properties, but that simulation does not make use of the magnetic properties of the constituents of a compound.

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What does ingredients being "biologically active" mean?
From Wikipedia

Cosmeceuticals refers to the combination of cosmetics and
  pharmaceuticals. Cosmeceuticals are cosmetic products with
  biologically active ingredients purporting to have medical or
  drug-like benefits.

I was wondering what  ingredients being "biologically active" means?
In cosmetics that are not cosmeceuticals, are their ingredients not "biologically active"?

Chemists usually marks as biologically active compounds, that have clear effect on living being in small amounts. For example, methane is not biologically active, but potassium cyanide clearly is. Of course, the borderline is quite blurry and can move if context changes.

Also, in the cosmetics industry, such phrases may be marketing and have nothing to do with real or perceived benefits.

In pharmacology, assuming that's what you are asking about, biological activity or pharmacological activity describes the beneficial or adverse effects of a drug on living matter. Of course, the substance doesn't have to be a drug, it can be natural, like snake venom.


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What exactly is happening when sodium, potassium, or other alkali metal explodes in water?
There are lots of videos on YouTube showing sodium, potassium, etc. exploding when dropped into water (this, for example).
I understand that when an alkali metal is exposed to water, a violent exothermic reaction occurs where a hydroxide and hydrogen gas are produced, but why and how does the sample of metal end up detonating and fragmenting?  Physically speaking, how can a block of metal seemingly blast apart from the inside when the reaction occurs on the surface of the sample?
The Wikipedia article on alkali metals explains this, but I still don't seem to understand how this would result in an explosion.  Diagrams would be very helpful.
The key of the question is how does the release of hydrogen gas and energy result in an explosion?  Is there thermal runaway, and is the metal physically destabilized in some way during the reaction?

Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the surface of the droplet increases in area rapidly forming a spiny, porcupine like shape as fingers of the molten metal are shot into the liquid at astonishing speed. The larger the surface area gets the faster the reaction occurs, leading to a runaway effect. The study was published in Nature Chem. (Ref.1).
References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, "Coulomb explosion during the early stages of the reaction of alkali metals with water," Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).

Science also published an article on this after the study came out in Nature and provided a link to a video here: http://news.sciencemag.org/chemistry/2015/01/why-popular-chemistry-experiment-blast  Absolutely incredible finding after all these years.

For those stuck behind paywalls or that like videos: High speed camera reveals why sodium explodes!

However, a note of caution. The experiments dealt with a liquid sodium/potassium alloy. The sciencemag.org link reports that the behavior of solid sodium is inconsistent, and often produces no explosion at all.

EDIT/Retraction: This answer was written prior to the publication of the research described in the accepted answer, and is now known to be incorrect. Science!

I assume this is due to mesoscopic defects -- if you think about a piece of metal on a molecular scale, it's not going to be presenting a totally flat sheet of atoms at its surface. Hydrogen gas forming within cracks in the metal can force those cracks to open wider, and this then exposes more surface to react, and produces more cracks.
This process will result in the force of the generated hydrogen ripping the bulk metal apart if it happens quickly enough. If the heat generated from the reactions then also starts boiling water in those same cracks and defects, that'll only add to the effect.

So it seems that as the reaction proceeds, the hydrogen pushes into the cracks and the increased surface area for the reaction produces even more hydrogen and heat, causing the cracks to expand rapidly in a feedback loop and leading to an explosion.  This means there is thermal runaway, since more heat leads to higher rate of reaction, and this combines with the increased hydrogen production and surface area for reaction to tear apart the metal, possibly at a rate in excess of the speed of sound.  Am I right to say this?

@DragonLord: As I understand it, yes.

Additionally, the amount of heat produced is enough to melt sodium and potassium, and the interior gets super heated.

@BenNorris: Now that's thermal runaway.

I don't think any clever effects need to be invoked to explain this phenomenon. The simple fact that the heat of the reaction is always enough to melt the blobs of metal involved is enough. Once you have a mobile molten drop of metal (which is often on fire), very little energy is required to cause the drop to split up so the violence of the continuing reaction is often quite sufficient to cause the drop to split and disperse.
You can easily verify that drops of liquids don't need much energy to cause them to disperse and, if you were still allowed to play with mercury, you could also prove this applies to liquid metals. Drops of mercury will easily split into smaller drops when small amounts of energy are added (e.g. letting a drop fall a short distance onto a surface). These smaller drops are a lot harder to contain than their parent (which is one reason why mercury contamination is so prevalent and hard to remove in old labs).
So in short, exothermal reaction + liquid metal = readily dispersed drops of burning metal.
PS
While what I say in this answer is broadly correct, the recent research (see accepted answer) add additional explanatory power to the detail of the reaction. The reaction is more violent that my simple mechanism would explain and the high-speed camera footage demonstrates that more factors were at wok than most people expected.

The reaction between alkali metals and water produces hydrogen, which will rise up and mix with the oxygen in the air above. If the reaction generates enough heat, it can ignite this mixture - hydrogen and oxygen will detonate when mixed in the correct ratio. This explosion will either directly spray out the metal (which by then will probably be molten) or force it down into the water where it will produce even more hydrogen gas and steam, propelling it back up and outwards.

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Source for thermodynamic and/or transport properties
I'm working on some combustion problems using Cantera and I can't find references for the thermodynamic or transport properties for many of my chemical species. I'm looking for NASA-7 or NASA-9 curve-fits for the thermodynamics, although if there is another equation format that exists, I will make it work.  And I also need properties for transport that Cantera calls "well depth", "diameter", and "rotational relaxation".
Specifically I am looking for: HClO4, HCl (transport information), ClO3, ClO2 (transport information), ClO (transport information), Cl2 (transport information), Cl (transport information), and ClOH.
So I suppose the real question is where can I find thermodynamic and transport properties for gaseous species with chlorine? Finding data on hydrocarbons and air is simple, and I found data on liquid refrigerants but nothing on gaseous chlorine species.
Note: I checked NIST and JANAF and neither had the information. I can find constant values, but running a very detailed reaction mechanism with calorically perfect equation of state and power or Sutherland law transport isn't worth the computational cost.

Never heard of Cantera, but it looks nice!

It's an open source replacement for CHEMKIN. Can't argue with free!

I have a CHEMKIN files with HCL:
HCL                1   344.7       3.339     0.000     0.000     1.000  !(singh)

344.7 should be $\varepsilon/k$ in K and 3.339 should be $\sigma$ in Angstroms.
! Species: HCL              CAS Number: 7647-01-0
! Name:    Hydrogen Chloride
! Source:  SNLL fit to 1986 JANAF tables, 4/21/89
! H0(298K) =      -22.0600 (Kcal/mole),  S0(298K) =       44.6500 (cal/mole-K)
HCL                42189CL  1H   1          G   300.000  5000.000 1000.00      1
 2.75533500e+00 1.47358100e-03-4.97125400e-07 8.10865800e-11-5.07206300e-15    2
-1.19180600e+04 6.51511600e+00 3.33853400e+00 1.26820700e-03-3.66691700e-06    3
 4.70399200e-09-1.83601100e-12-1.21315100e+04 3.19355500e+00                   4

Also, Poling [1] has some data on HCl, Cl$_2$, ClNO. You could estimate the other properties using group contributions to define critical properties and then corresponding state principle methods (like Chung's [2]) to get your temperature/pressure dependencies. You can then make your own curvefit for your conditions of interest. Not as good as curvefits from experimental data but it should get you in the ballpark...
[1] B. E. Poling, J. M. Prausnitz, and J. P. O’Connell. The Properties of Gases and Liquids, Fifth Edition. McGraw-Hill, 2001.
[2] T. H. Chung, M. Ajlan, L. L. Lee, and K. E. Starling. Generalized multiparameter corresponding state correlation for polyatomic, polar fluid transport properties. 27:671–679, 1988.


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Can silica gel beads cause silicosis if crushed?
Can silica gel beads (which I believe are amorphous) when crushed to dust cause silicosis?

search the web. Always search the web. http://en.wikipedia.org/wiki/Silicosis#Silica . However, despite stated inability of amourphous silica to cause silicosis, any form of mineral dust is not healthy for lungs.

No, it won't. Silica beads dust may cause irritation to the skin and eyes. Unlike silica crystalline,  synthetic amorphous silica gel is  indurated, and so does not cause silicosis.


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Introduction to modeling chemical reactions
I am looking for introduction to modeling of chemical reactions. I think there is the base approach, where concentrations of chemical species are given, plus ratios of each possible reaction / outcome. Is there a paper lightly explaining internal mechanics of such reactions, and their modeling via system of differential equations?

Are you interested in models of reaction kinetics (evolution of concentrations) or reaction thermodynamics (equilibrium composition)?

The kinetics. Basically, I'm looking for some background about collisions, and other possible mechanical explanations of simple reactions like in the following: http://chemistry.stackexchange.com/questions/500/what-is-probability-of-3-possible-products-between-two-chemical-species

The topic you are asking about is called reaction dynamics, and is closely related to chemical kinetics (the former is more focussed on atomistic events, reactive and nonreactive collisions, while the later usually describes the mathematical treatment of the kinetic laws governing species concentrations). It was the topic of the 1986 Nobel Prize in Chemistry.
It is covered by most physical chemistry textbooks, and more in depth by specialized textbooks on that matter. It is sometimes taught together with basics of statistical mechanics or statistical thermodynamics. Check your favorite (or local) university for a list of reading material on this topic!


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Why is my NaCl solution seemingly saturated, when I followed the recipe for an isotonic solution?
I prepared an isotonic solution of table salt in water following this recipe: 1 leveled coffee spoon of salt in 0,5L of water.
9g/L would be an isotonic solution, a leveled coffee spoon is commonly given as 5g - close enough. I use water from the tap, which was freshly boiled in a water cooker. The water is not chlorinated, I don't know the hardness.
However, even given hours of time, a significant amount of salt doesn't go into solution - visible as turbidity/precipitation. How can this be?
I need the solution to flush my child's nose, so I don't want to use too much salt.

Check the composition of your table salt -- some types contain less soluble drying agents, I think.

And your coffee spoon may be very inaccurate.  However since the solubility of sodium chloride is about 359 g/L (per Wikipedia) you'd have to add about 180 g of $\ce{NaCl}$ to a half liter to saturate it at 25 degrees C.  Clearly something is wrong here.

Salt added to food could be sea salt, common table salt or iodized table salt.
Common Table salt typically contains anti-caking agents (desiccants) such as magnesium carbonate, calcium silicate, dicalcium phosphate, tricalcium phosphate, calcium carbonate and sodium alumina silicates.
Iodised table salt may contains added potassium iodide for people who may otherwise have low consumption of this nutrient which is essential for proper thyroid function. Since iodide will slowly oxidize in moist air, stabilizing agents such as sodium thiosulphate or dextrose are also added. Some will also contain a base such as sodium bicarbonate, disodium phosphate or sodium pyrophosphate to also inhibit oxidation of iodide. There may also be added sequestering agents which bond to trace metal ions present as impurities to prevent them oxidizing the iodide.$^{[1]}$
The presence of magnesium and calcium will add "hardness" to the water which could cause precipitates such as calcium carbonate and magnesium carbonate to form.

thanks - my table salt contains magnesium and calcium carbonate as anti caking agents, that should be it.


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Why are solid phase organics often conjugated to HCl?
Many uncharged organic compounds are sold in solid phase with HCl conjugate - e.g. the antihistamine diphenhydramine.HCl or the derivitizing agent pentafluorobenzyl hydroxylamine.HCl, among many others. Assuming the compound will be dissolved at relatively low concentrations in a buffered medium (for instance when a drug enters the bloodstream) the HCl shouldn't have much chemical effect. Why is it there - does it aid crystallization, or perhaps it's leftover from synthesis and is difficult to separate?

I would also suggest looking at http://www.quora.com/Why-are-so-many-drugs-in-hydrochloride-form/answer/Christopher-VanLang

Does it aid crystallization?

Yes.
All of the drugs that you see as "hydrochlorides" are alkaloids - a class of basic natural products containing one or more amine functional groups. The dimethylamino group in the upper right of the structure of diphenhydramine (below) is a prime example.

Many of these compounds are not crystalline (at best amphorous, at worst waxy or liquid). These compounds are expensive to purify, since the inexpensive method of recrystallization is not an option. Chromatography is an expensive and wasteful process on the industrial scale.
Reacting alkaloids with HCl (or other acids - for example dextromethorphan HBr) usually produces a crystalline compound. These crystalline compounds are easier to purify.

There is a second more important benefit associated with the hydrochloride salts - they have higher bioavailability than the freebases. The freebase is nonpolar and poorly soluble in water. The hydrochloride salt has an ionic group, which increases water solubility.

The Wikipedia page on hydrochloride says the following:

Converting insoluble amines into hydrochlorides is a common way to make them water- and acid-soluble. This is particularly desirable for substances used in medications. Many pharmaceutical substances are prepared as hydrochlorides so that they may be quickly released in the gastrointestinal tract; the body usually absorbs a hydrochloride within fifteen or thirty minutes. Very often hydrochlorides of amines have longer shelf-lives than their respective free bases.

The hydrochloride salts of most alkaloids are also more crystalline, and thus easier to purify by crystallization.

And they are weakly acidic (similar as ammonia chloride is) instead of being rather alkaline.

I think another argument for the use of HCl here, as opposed to HF, HI, or HBr, is simply that chloride is physiologically inert in concentrations found in medical drugs, as opposed to the other halogenides. In other words, it won't poison you.


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Alloy-component weight based on desired molecule
I'm a gold smith and I want to start experimenting with non-standard metal-alloys. Now high school is a long time ago for me so while I think I've got this right, I'm just not as sure as I should be when working with precious metals costing 40 dollar a gram. So could anyone confirm that what I'm doing is right:
I have two metals
Metal X, with an atomic-weight of 20
Metal Y, with an atomic-weight of 3
If I want to create an intermetallic alloy of $\ce{XY2}$, where one atom of X is combined with two atoms of Y and I'd want 10 grams of that alloy. I would alloy (20*10)/26=7.692g of metal X and (6*10)/26=2.308g of metal Y, correct?

I think this is perfectly fine here, great question!

Yes, you are correct.
7.692 g / (20 g/mol) = 0.3846 mol X (or 2.316 * 10^23 atoms X)
2.308 g / (3 g/mol) = 0.7693 mol Y (or 4.633 * 10^23 atoms Y)
Moles are just a count of atoms, so if you have two moles of Y per mole of X, you have two Y atoms per X atom.


text:
In practice, how dangerous is liquid oxygen as an oxidizing agent?
The Wikipedia page on liquid oxygen tells us

Liquid oxygen is also a very powerful oxidizing agent [...], if soaked in liquid oxygen, some materials such as coal briquettes, carbon black, etc., can detonate unpredictably from sources of ignition such as flames, sparks or impact from light blows.

I have heard similar warnings about LOX in texts about amateur and semi-professional rocketry, but I know nobody who has ever handled stuff like this.
In practice, say I spill some LOX and soak paper lying around (that maybe has been warmed by a desk lamp), will that burst into a fireball?
Maybe more to the point, how do I assess such risks short of conducting foolhardy experiments? Is there a way to estimate the maximum effect a given amount of spillage on a given material may have?

Please note that I'm not foolish enough to toy around with such things based on advice I read on the Internet, so post away, you're not getting me in trouble.

Even a high concentration of oxygen in the atmosphere is dangerous, as the early Apollo program found out the hard way. The very least LOX can do in create high atmospheric concentrations.

This is a very vague question to which you already seem to know the answer. The answer to "In practice, how dangerous is liquid oxygen as an oxidizing agent?" is
very. I know of no way to quantify this. Evidence for it is contained both in your question and the comments.

read this http://en.wikipedia.org/wiki/Oxyliquit

“I know nobody who has ever handled stuff like this” should be an good indication as to its dangers :)

A mixture of LOX and common lab solvents very likely explodes spontaneously, within minutes. While cooled with LN2 (that's typically how this deadly mixture is foolishly, accidentially fabricated in the lab) it's probably slightly more stable. Do not bet your digits on it!

I'll have to acknowledge my question might be too vague for a straightforward answer, as can be seen by the various comments. The correct answer, as given by Paul J. Gans, probably is

[V]ery. I know of no way to quantify this.

However, using the comments as a starting point, I read some more around the topic, and I would like to expand on the comments to the question and to summarize what I found out.
Since there's no unit to measure "dangerous" with, we'll first have to get more specific. In the context above, we can offer the following dimensions:

ease with which LOX + X is set on fire or detonated, the easier, the more dangerous
predictability of ignition, the less predictable, the more dangerous
energy output of ignition, the more energy the more dangerous
speed of reaction, the faster the worse

The core problem with all of the above is that the LOX is not a fuel and will, on its own, do nothing at all. So before we can even speculate on possible reactions, we will have to narrow down the fuel options our spilled LOX might encounter. When we also want to account for oxygen-enriched atmosphere by boiled off LOX, a danger in itself, even without spilled, as matt_black correctly points out, the fuel can be almost anything burnable in the room.
That pretty much takes care of 2), predictability, which is low, so keeping the whole setup safe is difficult. -> Danger: high
Ease of ignition depends on the fuel, and the number we're after is the activation energy (Wikipedia). I found various Internet pages mentioning LOX spill accidents that turned tarmac into an explosive that can be set off by walking or driving over it, although I have not found an actual report of such an accident. So apparently, there are common materials that combine with LOX to a mixture requiring a very low activation energy. -> Danger: high.
Energy content and speed of reaction also depend on the fuel, but the fact that it's being used as rocket propellant and that LOX was used to build explosive charges in mining (so called Oxyliquits (Wikipedia), via permeakra's comment) means that we can expect plenty of both for a number of materials. The article also emphasizes the unpredictability, even of things that are meant to explode. -> Danger: high.
Finally, various safety guidelines inspire vivid imaginations of some scenarios:

The whole floor of an office is known to have caught fire when oxygen vapors contacted a lighted cigarette butt. -- Saftey Precautions for Oxygen Plants
If liquid oxygen spills on asphalt or other surfaces contaminated with combustibles, do not walk on or roll equipment over the area of the spill. Liquid Oxygen, Environmental Health and Safety


text:
Are there notable organic compunds which look very similar but have very different properties?
I've seen people arguing for some questionable "miracle cure-all" alternative medicines not scientifically, but by their own logic. A common claim is, that if a compound looks very similar to an important compound found in our body, it must be very beneficial. By looks very similar I mean the drawings of their structures look similar, as in, similar as images.
I found this reasoning alone to be unscientific. (correct me if I'm wrong). I would like to explain it to people, but it's hard to do if they lack even basic understandings in chemistry, and have claims like "I think it might happen to be like this, so it's a proof that it's like this".
So, I'm looking for a counter-example. Are there famous and well-known (organic) compounds which look similar on a structure drawing, but have vastly different properties, for example one is either beneficial or a necessary component of our body, and the other is highly toxic?
I'm sorry for the possible unclear terminology, English is not my first language and I didn't find any better term to say "structure drawing" for something like this

trb456 and Mad Scientist have raised the point that enantiomeric forms of a bioactive molecule may have markedly different activities, however there's more to it than that. A molecule that has a similar shape and surface charge distribution to another biologically important molecule is liable to bind to the same protein receptor sites, but in different ways. This may be a good or a bad thing and is a crucial observation of modern drug design.
As an example, the potent hallucinogen DMT (all image credits go to wikipedia):

and the fairly pedestrian neurotransmitter serotonin:

Are strikingly similar in the presence of both an indole and ethylamine group in both molecules.
Perhaps the ur-example of two extremely similar molecules with tremendously different effects in the body are:
Estradiol (female sex hormone):

Testosterone (male sex hormone):

The reasoning that a molecule that looks similar to a biologically important species must be beneficial is flawed. A molecule that looks similar to a biologically important species may just as easily strongly bind to and jam up the active site of something really important. This is how carbon monoxide, sarin, disulfiram and glyphosate work.

this is a great answer with terrific diagrams! Yours was the type of answer I was expecting to this question, so I'm glad it's here. My answer was just to point the even more extreme conclusion that sometimes shape is the only thing that matters. For people without much knowledge of chemistry, that's likely a very surprising result.

One example of disastrously different effects of very similar molecules is thalidomide. There are two isomers, only differing in their stereochemistry. The (R)-isomer is responsible for the sedating effect it was initially used for. The (S)-isomer lead to horrifying birth defects in the children of woman that took that drug.

It is somewhat more complicated, as the isomers are racemized quickly in vivo, so even if you administer only the (S)-isomer, it will still be converted to the (R)-isomer in the body.

@vsz: I'd say the examples by Mad Scientist and F'x are even better than you asked for. These compounds are not just similar, they have the same atomic structure. Only the spacial geometry differs, to disastrous results. Hopefully this gives you good educational material for the people you have in mind!

I think this is mainly an urban legend @Mad Scientist. See for example: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5778007/
"This idea was challenged by the findings that the (R)-enantiomer is also teratogenic in a rabbit model11,12 and that interconversion of the enantiomers could occur under physiological conditions."
The difference between those compounds seems to be not as stark as you say.

Recall that carbon in an organic molecule has 4 bonds in a tetrahedral shape.  If each of those bonds connects to a different group, then there are two configurations of the bonds which have the same formula but a mirror image geometric relationship in space.  Such a carbon (or molecule) is called chiral.  Chiral molecules have the same formula but different spacial structures and thus are different molecules.
For example, carvone has two enantiomers (mirror images):  R-(-)-carvone smells like spearmint, while S-(+)-carvone, smells like caraway.  This is because different olfactory receptors in the nose fit the molecules differently.
Here's a video about carvone and chirality from the good folks at Periodic Videos.

Good example. Maybe even more remarkable is Naproxen: one enantiomer is used to treat arthritis pain, but the other causes liver poisoning with no analgesic effect. (source)

In biochemistry branch searching for new drugs, it was common to search for new active substances using chemical similarity. It does have reasoning behind it.
Most drugs acts, binding themselves to some protein site, in the place, where normal ligand of the protein will bind. However, chemical similarity is not enough to be a drug. Sure, a substance must bind to given protein strongly, and imitating typical protein ligand is wonderful for it. But to be a drug, substance must differ from natural substance to modify protein behavior, usually blocking its activity or activating in heavily.
Because of it, drugs are often searched in modified natural protein ligands or their structural analogs, but usually modified. For example, there is well known group of beta-blockers, that mimics noradrenaline. However, they have to indestructible in vivo and bind strongly, preferably to some, but not all types of Adrenergic receptor, so all of them has structural twist.
In addition, theoretically it is possible to find a drug in chemically unrelated structures, if they exhibit required groups in right positions or can bind to target protein in some other way.
As for your question, the most infamous example is previously reported thalidomide. However, it is not the only one. For example, most aminoacides has to enantiomers, but only one of them is usable for our bodies
To get basic understanding of the subject I recommend to dwell the wiki starting from this article

1-propanol and 2-propanol are isomers but have different boiling points: 370 K and 356K. As I remember my undergraduate studies correctly, my distillation was to separate a mixture of these two alcohols. The metabolism of the two alcohols are different.
Further readings:
http://en.wikipedia.org/wiki/Propan-1-ol
http://en.wikipedia.org/wiki/2-propanol


text:
What is the specific structure of Poly(allylamine hydrochloride) (C3H7N)n HCl?
I'm working with Poly(allylamine hydrochloride), and I found that the chemical formula is
$$\ce{(C_3 H_7 N)_{$n$}. HCl}$$
But I need a diagram like below, but more explicit. I mean with the carbons, where they are exactly?
 (source)
I found a web page, with a lot information of my molecule. What I was looking for is that kind of diagram (the one with 3D aspect). That is the structure of the allylamine hydrochloride, now I want to know what is the structure of the poly(allylamine hydrochloride).

Name and formula in the title do not fit! The monomer is allylamine hydrochloride, a more suitable representation thus would be $\ce{[C3H7N*HCl]}_n$

My intuition is telling me that, based on your empirical formula and the image you provided, all three carbons are at the bends in the line-notation as is conventional in bond-line notation. The structure inside the brackets is repeated virtually indefinitely, as shown with the $n$ at the bottom right of the image. I have circled the alleged carbons in the image.
This adds up if you count the number of Hydrogen atoms bonded to these carbon I believe it will add up to 7 as your formula suggests.
The structure in the link that you provided in your comment does not show us a structure of the hybridization that is, well very good to be honest. For one thing it is not a picture of the polymer, instead it is just a small molecule with a double bond, and the Hydrogen on the Carbon connected to the Nitrogen appear to be in a funky alignment, they should face away from the flat shape the Jmol applet is showing.
As for the structure in the polymer, we should be able to determine from the bond-line notation that if each carbon is $sp^3$ hybridized, then as the bends suggest we have a molecule which bends roughly 109.5° between all of the Carbon-R bonds. This can be visualized as a tetrahedral shape where the Carbon atom sits in the center of the geometric shape and the atoms that it bonds to are at the corners.
visualize hybridizations
Now given that the Nitrogen atom is $sp^3$ hybridized, since it has 3 covalent bonds (and one lone pair that is left off for clarity), it should have bond angles of ~108° between itself and either of the two Hydrogen or the one Carbon atom allowing a 'trigonal pyramidal' shape. Not very easy to imagine after staring at this picture, which is why we use model kits to help us visualize the molecules.


text:
Where did the rest of the elements come from?
There were ninety original elements on the Periodic Table. Over time, new ones were founded and put on. However, they were all either the original elements experimented on. For example, Plutonium was discovered by deuteron bombardment of uranium. Therefore, I would like to know: How were the rest of the elements discovered? It would also be nice if an example could be given. Thanks!

By the "original periodic table" I assume you're talking about the "elements ocurring in measurable quantities in nature".--everything till Uranium sans Technetium and Promethium.
The method of preparation of these elements is pretty much the same -- bombardment. Just the projectile and substrate differ.For example, the (accidental) neutron bombardment of Molybdenum in old cyclotron parts led to the discovery of Technetium.
There are some other sources as well. Analyzing fission products can get you new elements--that's how Promethium was discovered. The process of fission creates smaller nuclei--but there is a shower of neutrons created as well (which can bombard everything in sight). Most of the discoveries have been due to the "shower of neutrinos", because generally all "smaller nuclei" are of already-known elements. Promethium was synthesized by the "formation of smaller nuclei", though.
Since fission occurs in nature (Usually it's just a little bit, though we sometimes have natural fission reactors), we actually have all the elements up till californium existing in minute quantities in nature (source).


text:
Why should I add lithium bromide to my GPC eluent?
I'm running gel permeation chromatography (GPC) of a polar, uncharged polymer (acrylamide-based) in a polar solvent (dimethylformamide). The literature recommends adding a salt to the eluent; for my particular experiment, I'm using 0.05 M LiBr (example). Other salts are reported (Sodium Sulfate, for example) in the literature, both for this eluent/polymer mixture, and for similar polymers and eluents. Why is this done? Most papers simply report the eluent and the additives they used, with no mention of their reasoning. One paper just identifies the eluent, with no mention of any added salts; given the prevalence of the use of salts, I'm not sure if this is an oversight on their part, or if they just didn't use a salt.
If I were working with a polyelectrolyte, I would know that adding a salt would screen the charges along the polymer, reduce the $A_2$ effect, and make my data much easier to interpret. One GPC column manufacturer claims (under "What additives are important...") that dipole-dipole interactions can have a similar effect, but I'm not not sure how much to trust that claim. In addition to confirming or denying that this is the reason, I would also appreciate pointers to any relevant literature.

Could you provide a link to one of the papers describing the addition of salts to GPC eluents? I have never added salts to my GPC eluents, so I am curious about this technique. Maybe it is specific to polyacrylamides?

@BenNorris, that's really interesting that you have never done this. I added some links to hopefully clarify.

Also, I think I may have found a paper that answers this definitively. When I get the time, I will write that up as an answer (but certainly do appreciate any other answers in the meantime!).

well most of my GPC has been done in THF and chloroform, in which most most salts are insoluble, so my experiences might not be typical.

This is my first experience with GPC, so I wouldn't jump to any conclusions. Your experience may be more typical than mine.

The answer to my question appears to be in a 1977 paper by Nancy D Hann (DOI: 10.1002/pol.1977.170150604; full citation below). In it, she compares GPC of polyurethane in DMF with and without added LiBr. The chromatograms with no added LiBr show multiple peaks. The chromatagrams with added LiBr show just a single peak. The addition of LiBr also shifts the peak to later elution volume.
Hann explains these results by the polymer having a strong polar interaction with the pure solvent. This causes a higher effective hydrodynamic radius, resulting in the earlier elution time. The polar interactions also case aggregation of the polymer, giving multiple peaks. Adding a salt breaks up the polar interactions, giving just a single, later, peak.

Hann, N. D. (1977). "Effects of lithium bromide on the gel-permeation chromatography of polyester-based polyurethanes in dimethylformamide." Journal of Polymer Science: Polymer Chemistry Edition 15(6): 1331-1339.


text:
Why don't the electrons move through the electrolyte (instead of the circuit) in a galvanic cell?
I was learning about galvanic cells and I had a problem understanding why electrons do not travel through the electrolyte solutions themselves, instead preferring to travel through metals. Can electrons travel through an electrolytic solution?
Imagine a galvanic cell, without the wiring between the two electrodes and instead we only have the salt bridge. Won't the electrons (although it is very unfavorable to happen) travel through the salt bridge and the solutions? In other terms will the transformation of the electrodes occur?

Not in water. Free electron in water is really unfavorable, so no significant concentration of them can be generated chemically, and it almost immediately reduces water itself to hydrogen (but I heard rumors about generation of solvated electrons in water in very special experiment with short half-life)
In liquid $\ce{NH_3}$, however, solvated electrons can occur, so self-discharge of galvanic cells with $\ce{NH_3}$-based electrolyte may occur through travel of electrons via electrolyte.

But we have ions in water.  So can't the water act as a conductor in this case?

You're questioning the intuitive disconnect caused by most galvanic cell drawings which seem to assume the electrolyte solution in the salt bridge does not conduct electricity, so let's investigate.
Imagine a Zn/Cu$^{2+}$ cell with electrodes 5 cm apart in a 3.5% NaCl solution with a tube (1 cm$^2$ cross-section) of solution as the salt bridge for balancing charge.
The electrical resistance (R = $\rho$l/A) of our NaCl 0.05 m x 1 cm$^2$ salt bridge solution is:
$$\frac{0.2 \ ohms*m}{} * \frac{0.05 \ * m \ (length)}{10^{-4}m^2 \ (cross-section \ area)}= 100 \ ohms$$
Considering the predicted EMF of 1.1 Volts for this cell, the expected current ($I = V/R$) through the salt bridge is: $1.1V/100 \ ohms \ = 0.011 \ amps$
This current may be negligible in a galvanic cell drawing compared to the current through some wire or low-resistance load.  However, this would make a terrible battery for most common purposes as a typical AA battery (3000 mAh) would go completely dead in less than 2 weeks if it actually leaked at this rate!
It seems then your intuition is basically right... until you understand what the models leave out. In real alkaline battery designs, the cathode, electrolyte, and anode are sandwiched together very closely with a very large surface area, yielding excellent conductivity through electrolyte (and therefore very low resistance).  However, these layers are separated by a membrane which allows ions through but has a very high resistance to electric current.
Sources:
https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499  (seawater resistance)
https://en.wikipedia.org/wiki/Alkaline_battery (alkaline battery design)

This is a much more satisfying answer than the accepted one. Thanks!

Electron's can travel through the electrolyte solution, however electrons take the pathway with the lowest resistance, the solution has a relatively higher resistance compared to the outer circuit. Hence the electron takes the path of the outer circuit.

Electrons cannot survive in aqueous state. Being a charged subatomic particle, the electron
has to stay close to protons which are located at the center of the atom. Hence, the electron
can move from one atom to another which are closely-packed, what we have in a solid.

Maybe just think of it as the electrons hitching a ride through the electrolyte on the relatively giant and mobile atoms that make up ions in the solution.  If you really wanted to force those electrons to travel on their own through the electrolyte you'd have to crank up the voltage big time.  Thousands of volts would be needed to provide the electrons with enough energy to rip through the electrolyte (i.e. dielectric breakdown).
Another way to get electricity to flow through an electrolyte, although still not electrons flowing, is to impose an AC current across a cell.  At high enough frequency (usually in kHz range) current can short through the capacitance at each electrode.  This is typically how conductivity measurements are made on liquids.


text:
Can we create a galvanic cell with only half a cell connected to the ground?
I have a question about electrical potential in galvanic cells ,suppose that we have only half a cell wich means that we have only one Ox/Red couple (one metal and the other as ionic solution ...)clearly nothing will happen ,suppose that we put a wire between the metal and the ground .The ground has no voltage while the Red/OX couple has a potential does It mean that the oxidation will happen ? and the electrons will go through the ground ?If the oxidation happens with any kind of OX/RED couple ,and if we take a couple wich the deltaG of the oxidation reaction is very high (so It's very unlikely to happen all by Itself ) won't we be creating a free source of energy ? Thank you !

Can you specify what you mean by a "free source of energy"?  It's vague as it stands.

The answer to your question is difficult to explain and understand without a correct understanding of charge and the redox reaction. Another way of asking this question is: Why does current not flow if I connect two batteries using only one wire from the positive terminal of one battery to the negative terminal of the other. The common answer you will get is that you need a complete circuit, but that avoids the deeper question of why you need a circuit. To understand it a little better, you could think of a cell as two parts, one tries to pull electrons from its metal terminal and another tries to push electrons into its terminal. However, neither can do so unless the oppositely charged ions that are left behind can be exchanged between the two metals inside the battery. This means that while the cathode does have a negative charge, it will not let the electrons go unless there is a negative ion to pair up with the positive ion that gets left behind inside the cell. This negative ion comes from positive anode ions leaving the solution and attaching to the anode, but more negative ions only get created if the anode receives the electrons needed to attract the positive anode ions.
In your question, the one Ox/Red half might be the cathode. In that case it will not have a potential at all since there is nothing but other electrons to balance the charge of the positive ions created at the cathode as the electrons leave it for the wire. The ground connection acts only as a conductor of the electrons. There is no reason for an electron to leave behind a positive ion just to push other electrons through a wire to ground and recombine with a positive ion. That would require energy since the wire has some resistance. The reason the battery works is because a chemical reaction takes place that releases energy. In other words, the positive and negative ions from each half cell would rather be combined together to reach a lower energy state. This can only occur if the positive metal ions that detach from the cathode as it releases electrons can combine with negative ions in the electrolyte. However, the negative ions in the electrolyte must be replenished and that can only occur if other positive metal ions in the electrolyte are pulled out of solution by the anode, which can occur only if the anode receives electrons.
As you may know, the cells' anode and cathode are typically made of two different materials such as zinc and copper, and the electrolyte is a solution made of a salt containing positive ions of the anode metal such as copper sulfate. As the cell discharges, negative zinc ions are leaving the cathode and dissolving into the solution, while positive copper ions are leaving the solution and attaching to anode. One terminal dissolves while the other grows. As negatively charged electrons move from the zinc cathode to the copper anode via the circuit outside the battery, positively charged zinc ions now in the solution are soaking up or balancing the negative charge created in the electrolyte as the positively charged copper ions leave the solution and attach to the anode that received the extra electrons via the wire from the cathode.
The reason all of this takes place is because the process results in a lower overall energy state after it occurs. In other words, the zinc atoms would rather be combined with $\ce{SO4^2-}$ ions than with each other, and the copper ions would rather be combined with each other at the anode than with the $\ce{SO4^2-}$ in the solution. The only way this can happen is if electrons are moved from the cathode to the anode and the easiest way is via the wire and circuit outside the battery. Without the electron flow, the chemical reaction between the two half cells is inhibited. Conversely, without the chemical energy potential of the reaction between the two half cells, there is no reason for electrons to flow.


text:
Is phenolphthalein used in laminar wood construction panels?
Recently someone is claiming that the laminar wood panels used for building our new primary school contains a "carcinogenic and dangerous compound", namely phenolphthalein.
I couldn't find any evidence of it being used as that on the net.
Could you confirm my findings?
Thank you

I'm not sure this is a question about chemistry. If the only source you have for this claim is that "someone said", it probably is not true. That being said, a question about testing for phenolphthalein in wood laminate would be interesting and on topic. So would a question about the harmful health effects of phenolphthalein (probably not many).

You're right, but my question was the one in the title, namely whether a substance is used in an industry. It's maybe more an industrial chemistry question. Your proposals are more interesting from the chemical pov, but they wouldn't help me much. Asking just about testing for presence is like asking for any other bad substance that comes to mind that could be in the panels. Harmful effects are debated, but e.g. the wikipedia entries are very detailed.
Thank you

Could it be that he meant phenol-formaldehyde resin?
These resins are often used to bond wood chips. They contain unreacted formaldehyde and the resin decomposes slowly, giving off more formaldehyde, which is volatile and very toxic. This is a known health concern. Many companies offer indoor air testing for formaldehyde, you can even buy dubious tests to do it yourself.
My bet would be on such a resin.
There are indeed some resins that use phenolphtalein:
Phenolphtalein might be cancerogenic, even though it was used as a laxative

Phenolphthalein was commonly used as a laxative for most of the 20th century. The use of phenolphthalein in laxatives has decreased since 1997 when the United States Food and Drug Administration (FDA) proposed to withdraw its classification as an over-the-counter drug (21 CFR, Part 310). Phenolphthalein has been previously evaluated in 2-year carcinogenicity studies by the National Toxicology Program (1996). The major route of human exposure to phenolphthalein is via ingestion, dermal contact, and inhalation of contaminated air originating from process units manufacturing the compound.

In any case, phenolphtalein is a powder that is not very volatile and would not pose a big risk.
Commonly the formaldehyde resin is used for particleboard, I have never heard about the phenolphthalein resin being used for this purpose (but this is far from my area of work)
Of course, there is a simple test. Phenolphtalein changes color in a basic solution and looks like this

Dissolve the material in ethanol and add NaOH (drain cleaner), phenolphtalein will turn fuchsia.


text:
SN1 like reaction acid-cataylzed hydrolysis of epoxide
What would be the reaction mechanism for the reaction between epoxide and $\ce{H2SO4}$?

Note that your question could do with some more data. Here is my answer based on what I think you mean:

You ought to get a glycol.

First, a proton from the acidic medium attacks the neucleophile, $\ce{O}$. This creates a positively charged trivalent oxygen. The electrons in one of the two $\ce{C-O}$ bonds transfer to the $\ce{O+}$, breaking the bond and effectively transferring the positive charge to the carbon.
At this point, note that the carbocation thus formed may rearrange.
Now, we have the nucleophilic attack. An oxygen atom (specifically the lone pair of an oxygen) from a nearby water molecule attacks the carbocation, forming another trivalent positively charged oxygen. This is deprotonated, giving the final glycol product.
Here's the same mechanism,  showing movement of electrons (pairs):

This answer is incorrect. Wrt this 2010 journal, note that " In the SN1 mechanism, the existence of the typical carbocation was not observed" This is in direct contrast to your statement: "At this point, note that the carbocation thus formed may rearrange." Please can you fix this. Thank you!


text:
Is this mystery chemical reducing carbon emissions?
My chemistry is not good enough to solve this problem, and it's driving me crazy.
At my job, a sales agent approached me to be their representative in my area for their product.  It's a product that when mixed with diesel or gasoline, it reduces carbon emissions in the vehicle to almost 0.
Now this seems impossible to me.  The burning of hydrocarbons necessarily creates $\ce{CO}$ or $\ce{CO2}$, and the $\ce{CO}$ can be converted to $\ce{CO2}$ in the catalytic converter by oxidizing it.
In the test results, $\ce{CO}$ dropped to almost 0, and $\ce{CO2}$ stayed the same or went down (also my own calculations show the change in $\ce{CO2}$ was not statistically significant with the small sample size - 3 before and 3 after tests on 3 cars).
I've seen the test results (yes, they could have been faked, but they filmed the testing process).  My buddy also personally tested it on his car and got the same results.
Anecdotal evidence shows that the car's mileage is the same.
Can anyone point me to somewhere to try and figure out what is going on here? It sounds too good to be true, and as such, there is no way I'm going to recommend my company represent this product until I know what's going on.
Here are the mean test results (in ppm for CO and NOx, in % for CO2 and CxHy).  Some of them have high variance.  The only statistically significant changes are $\ce{CO}$ reduction in all cars, and all changes in the first car.  The $\ce{NOx}$, $\ce{CO2}$, and $\ce{CxHy}$ changes in car2 and car3 are not significant.
        Before                          After
        CO    CO2    CxHy    NOx        CO    CO2    CxHy    NOx
Car1    3491  9.4    0.39    46.7       33.3  7.8    0.07    9.7
Car2    164   11.5   0.13    2.7        12.9  11.4   0.37    0.7
Car3    2868  11.1   0.18    14         123   10.9   0.06    39.3

I'm suspicious this may just be ethanol, but the manufacturer swears it isn't.
The advertised ratio is 1 oz to 12.5 gallons of fuel which is less than 0.1% (I believe they used double that in the tests).  Too low to be ethanol anyway, right?
Follow-up:
I have confirmed that I was originally mistaken in that all measurements are ppm.  I received a new copy of the original results, and $\ce{CxHy}$ and $\ce{CO2}$ are indeed measured in percent, rather than ppm.  It was incorrectly rewritten in the report.  The numbers are the same, though.

Hang on - you state that the product is supposed to make carbon emissions drop to almost zero, and then state that the CO2 emissions did not drop statistically significantly (ignoring CO for the time being). Can you please clarify what the observations were in terms of total carbon?

I think the CO2 sensor might have been faulty. Those CO2 concentrations are several orders of magnitude too low, and lower than ambient atmosphere (currently around 391 ppm).

The sensor was certified calibrated (although still could be faulty).  It was inserted into the exhaust pipe of the car. "For accurate readings, the O2
readings must be lower than 20.5% and gas temperature must be more than 5 degrees
hotter than ambient temperature. The analyzer was purged with ambient air prior to
each set of readings." Would that make a difference?

Regarding your original comment - catalytic converters produce more CO2 than normal exhaust because one of their functions is to convert CO and unburnt fuel to CO2. The total amount of carbon going in and coming out must be the same. Regarding CO2 measurements - if wikipedia is to be trusted, the concentration of CO2 in petrol engine exhaust is about ten thousand times higher than the 'before' figures you have listed, which leads me to believe that the CO2 sensor is reporting % CO2 rather than ppm CO2.

You are right, I edited to reflect this.  The report I have was incorrect.  The units for CxHy and CO2 are %, not ppm.  I understand about catalytic converters.  My issue is determining how this product can possible achieve these results.  It would be horribly unprofessional of me to recommend this product to my bosses (possible scam?), but at the same time it's hard to let it go.

theoretically speaking, the chemical seems to be able to achieve more deep fuel combustion, giving slightly more energy per volume and so, one can burn little less of it to get same amount of energy. The reduction should be veeeeeeery low, though.

The catalytic converter in your car has to heat up to work effectively. I'd guess several minutes at least. So one way to fake this would be to take the "without additive"  measurements before the catalytic converter heated up and take the "with additive" measurements after the catalytic converter heated up.

The baseline CO emissions in car 1 and car 3 are too high for a car with working catalytic converter. This means you can't trust the test.

With reference to the discussion in the question comments, this product does not appear to appreciably reduce carbon emissions in total. It does however appear to reduce carbon monoxide emissions. As carbon monoxide is a product of incomplete combustion, it is plausible that the additive is helping the fuel burn more completely. Broadly speaking, you would hope that the majority of the carbon in your exhaust gas would come out as carbon dioxide, which is 'merely' a greenhouse gas and not truly nasty like CO or PAHs. Prior to the advent of catalytic converters, carbon monoxide poisoning by vehicle exhaust was a much greater risk.
It should be noted that conservation of mass tells us some powerful things about what to expect from an engine - namely that the carbon that goes in to the engine as a component of fuel must come out as exhaust gasses, particulates, polycyclic aromatic hydrocarbons etc., or else be sequestered inside the engine (though the latter is achievable by putting a banana into the exhaust pipe, it is generally done as a prank).
Reduction of carbon emissions in general is achieved through obtaining more work per mass of fuel burnt, which, when coupled with the mass and other design features of the car confers the fuel efficiency. Unfortunately, the data provided don't tell us anything about that. The more complete burning of the fuel, evinced by the reduction of CO concentration in the exhaust gas does imply some increase in efficiency, however CO is a very small component of exhaust gas, indicating alongside the low concentration of unburnt hydrocarbons that the fuel is already burning virtually to completion.
Without knowing the specific claims made by the manufacturer of the additive, I will not make any judgment of good or bad faith, however my distilled conclusions are:

The product does not dramatically impact total carbon emissions
The product appears to appreciably reduce carbon monoxide emissions

@Bryan - sorry, my answer does sort of walk through some points that I see the edit to your question evinces a good understanding of, however I'll leave my answer unmodified for the potential edification of other future readers.

Thanks for the reply.  Going with what I already know, and your's and permeakra's comments, I'm going to have to decide to either leave this product (somehow it's just trapping the carbon in the engine), or do some some tests to see if it is meaningfully increasing the efficiency and work output of the reaction. I didn't consider PAH emission...

@Bryan - I was sort of joking about carbon sequestration inside the engine. It is not practical to just store away the carbon without a radical re-imagining of the internal combustion engine (for instance attaching something like a rebreather to it, which would require frequent replacement). My conclusion is that it's difficult to say whether the additive is doing anything at all to total carbon emissions, and the chances are good that it is not.

I think the entire claim by the sales representative, the question itself, and the ensuing discussions in comments and answers can be distilled down to two items:

The claim that "It's a product that when mixed with diesel or
gasoline, it reduces carbon emissions in the vehicle to almost 0"
is nonsense, as the OP and further discussions stated. The desirable
net chemical reaction taking place in an internal combustion engine
is simply:

$$\ce{hydrocarbons + O2 -> CO2 + H2O}$$

The only useful measure of a fuel additive's ability to decrease carbon emissions is a comparison of distance traveled per volume of
fuel, with and without the additive. This was only addressed in
the statement "Anecdotal evidence shows that the car's mileage is
the same." Thus, the total carbon emissions were essentially
unchanged by the additive, as $\ce{CO}$ and unburned hydrocarbons
represent a negligible fraction of the total carbon emissions.

Clearly, the "Reduces carbon emissions" catch-phrase is unfortunately an untruth distracting from what the additive appears to possibly do well: reduce $\ce{CO}$ and possibly unburned hydrocarbon and $\ce{NO_x}$ emissions.

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


text:
Anomalous electronic configuration of platinum
Why is the electronic configuration of platinum
$$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$
and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or
$\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$?

The Madelung energy ordering rule gives the energy of the orbitals approximately:
$ \mathrm{1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d <
5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p}$
That would speak for $\mathrm{[Xe] 4f^{14} 5d^{8} 6s^{2}}$ if you follow the Aufbau principle (from German Aufbau = setup). But $\ce{Pt}$ is an exception (as there are some). A rule of thumb is that half-filled shells are stabilized. So that means in the case of $\ce{Pt}$, $\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$.
The real answer is much more complicated. It comes from relativistic effects, electron correlation and shielding effects. There is an interplay between the attraction of nucleus and the electrons, and the electron repulsion between all electrons. The heavier atoms become, the more important relativistic effects become, since the inner electrons are moving much faster as they are in a stronger electric field from the higher charge of the nucleus. Often it is said that the outer orbitals are less compact than the inner ones, which is true if one calculates and analyses them, but a stronger effect is that outer electrons are shielded and therefore the nucleus attraction is weaker. So there is no simple way, learn the exceptions or solve the Schrödinger/Dirac equations.
Also be careful when speaking about orbitals. An orbital is a one-electron wavefunction and they are more a chemical concept than reality in multi-electron atoms.

If half-filled or full shells are more stable, why stop at [Xe] 6s1 4f14 5d9? Why not go all the way and have [Xe] 4f14 5d10?

@user3932000  Because palladium already did that and platinum wants to be unique /s

At this point in the periodic table, s (outer) and d almost have same energy. So, it is better having 4 fully filled and 2 half filled orbitals, than having 5 fully filled and 1 empty ( this empty 6s creates a problem as it is not energetically favored ), since in the first case all of them are stable whereas, in the second the empty 6s is not.
The initial configuration is : [Xe] 4f14 5d8 6s2, then one electron is transfered from 6s to 5d, so that all orbitals become stable, either through full filling or half filling, which is better then having one empty and unstable.
This makes it : [Xe] 4f14 5d9 6s1.
It cannot be the other two because, in both of, one orbital is empty and unstable.

yes but empty orbitals are not unstable. Orbitals don't exist unless electrons are in them.

A quicker, less lengthy answer to your question:
Because it is in Group 10, Period 6 and all elements in that group are required to end with the 5d and 6s orbitals. And since it MUST use 6s, it is natural to put 1 electron in it (simply to show it exists, based on the element's placement on the periodic table). Also, because of the Aufbau principle, the other arrangements you mention simply cannot be.

Why must is use the 6s? Why is is natural to put only one electron in it rather than 2 (or 0)? This doesn't really answer the question.

@bon I think it must use 6s because that is the outermost electron orbital.

@user3932000 There are an infinite number of orbitals it's just that most of them are not occupied. There is no reason why the 6s has to be occupied like the answer is suggesting.

@bon I misunderstood what the OP was saying; I thought they were saying the 5d orbital must be filled by using the 6s electrons. I agree with you that there's no reason why the atom must have the 6s orbital occupied, since it could just as easily be [Xe] 6s0 4f14 5d10.


text:
Simultaneous equilibria
Rephrasing:
The following reactions are a small example of my big system
$$(1)\  \ce{A + B <=> AB}$$
$$(2)\  \ce{AB -> CB}$$
$$(3)\ \ce{A -> C}$$
$$(4)\ \ce{C + B <=> CB}$$
with
$$\ce{\frac{[AB]}{[A][B]}} = 100 = K_e$$
$$\ce{\frac{[CB]}{[C][B]}} = 100 = K_e$$
I know the rate constants for reactions (2) and (3), and I know that
$$\ce{[A] + [AB] + [C] + [CB] = T}$$
with T =  some constant.
Now, although there is a fast equilibrium between $\ce{A + B}$ and $\ce{AB}$, and between $\ce{C + B}$ and $\ce{CB}$, I want to know how the concentration of each form changes with time.
But I dont know how to write down the differential equations of the system and include the equilibrium constant. Could someone please help me out? My only problem is how to include $K_e$ in the system of ordinary differential equations.

Ok.  You have five equations, two are differential equations and three are algebraic.  And you have the explicit assumption that the equilibria are fast enough so that the two equilibrium equations are always satisfied.
You also have five unknowns, A, B, C, AB, and BC.  One method of solution is to use the algebraic equations to eliminate three of the variables.  This should leave you with two differential equations with two unknowns.  If you are very lucky (I've not worked out the algebra) you will get two equations that can be integrated independently.  More likely you will have two simultaneous differential equations.
In either case the result is apt to be a mess.  So I'd cheat.  If this is a textbook problem or homework problem, there is probably a way to combine the equations to get simple enough expressions.  Otherwise, you may have serious troubles solving the set.


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What's that tangy smell on cheap plastic?
Most cheap items (from China mostly) have a strong tangy odor to the plastics (or some resin, but I can mostly smell it on plastics). The smell is always consistent over a broad range or products and I get it for at least a decade. So I do not believe this is a one-off thing and I bet most people have experienced it. The only variation is intensity.
On some really bad products that smell impregnates other things and I failed to remove it with soap, baking soda, acid, scrubbing...
I'm curious to know what exactly is that smell.
Apparently it got some attention already, but there's no consensus, and a lot of crywolf(?) e.g. Campaign to Halt the Import of
Chemical-Emitting
Smelly Plastic from China.

I notice a strong plasticky smell, which I can immediately recognize in many plastic grocery bags, which I've checked and are usually number 2 recycle bags. Also, I noticed that Walmart and Fresh and Easy grocery stores both had higher quality number 2 recycle code plastic grocery bags, which didn't smell...but Fresh and Easy just recently replaced theirs with slightly lighter colored brown, cheaper, smelly plastic bags.

Code 2 is HDPE (High Density Polyethylene). Its monomer, ethylene is not odorless.

I know exactly what you mean! I have several products that came from china that have that exact same smell. It's very distinctive. I find myself smelling many things i buy (cases, pads, etc) and not buying or returning some of them. Recently i bought an WD ELEMENTS 4TB external hard drive and... surprise!!!! i immediately noticed the the smell on the plastic CASE!!!! Really? Even "quality" products now smell like TOXIC plastic? I believe this is due to inferior manufacturing, probably to save costs. There is no other explanation... Now, i'm i going to return an external drive because of smell?

https://dynamics.org/CHINA_PLASTIC/

Without a sample and a gaschromatograph it is hard to say what exactly it can be but... there are at least three well-known sources of odour in plastics:

Some residual of the monomer that makes up the plastic (that is a polymer).
Some residual of some other substance used during the manufatcturing process (catalyzer, co-polymers, modifiers and so on). For example, PETcan sometime contain small traces of terephthalic acid.
Some odourizing substance that is added to the plastic just to cover others smells.

Usually, tangy smells are a symptom of some kind of  ester

I can't believe i'm the only one who knows that smell by heart. it's pretty common on cheap products. you could go to any hardware store and get any cheap screwdriver set and sniff, or you can give your address and i can send a sample of my worst offenders (some prototyping board jumper wires that i'm not brave enough to open the bag a second time). edit: changed my search terms, there are more plastic-smelling crazy ppl out there http://dynamics.org/Altenberg/CURRENT_AFFAIRS/CHINA_PLASTIC/ :)

Well, in the case of violin case, they suspect tetrachloroethylene, that is a chlorinated compound. In other cases (Samsonite), they suspect polycyclic aromatic hydrocarbons (PAHs) (no chlorine...) that are totally different compounds. It is not easy to tell for sure what the source of a smell can be without a professional analysis. Please note that both these compounds are banned (as residuals in commercial products) in EU and in Italy (where I live) since the mid '80s. BTW: the current fashion here seems to be to add some cheap orange flavour to plastic to cover other smells.

if anyone have a gaschromatograph or other relevant equipment i'm willing to send a big sample. The hobby electronic jumper cables i got from amazon are still going strong with the smell! years later! also, all the item reviews mention the overpowering smell, so i think i don't even have to send the sample myself, i can just buy a new one and tell amazon to deliver it to you/your lab.

https://dynamics.org/CHINA_PLASTIC/

This question is still valid. I bought a bagpack to try this Temu this Temu https://www.temu.com/se-en and it impregnated clothes and even the room with closed doors. I fear it is toxic. Difficult to describe. It smells a bit acidic, a slightly bit of "rotten eggs", even if that tone is not that strong, but there is an aromatic note. It is hard to stand. It is inside the bag, not outside. So i wonder what the volatile substance is, and if carcinogenic. Bad me that I indulged on try the cheap and I contributed to pollution, but first time ever I smell something like this.


text:
Physical state of hydronium
I am doing my chemistry homework and I am supposed to write the products when strong acids react with water, but I am unsure of the physical state for hydronium ions. Should it be liquid (l) or aqueous (aq)?
Example:
$\ce{HI(aq) + H2O(l) -> ?}$
Should it be
$\ce{H3O+(aq) + I^{-}(aq) }$ ?
Or should it be
$\ce{H3O+(l) + I^{-}(aq) }$?

This is a bit touchy.  The problem is that a bare proton, $\ce{H^+}$, won't exist in water.  It attracts electrons too strongly.  Once this was recognized, chemists started to write the bare proton as being bound to a water molecule, thus $\ce{H3O^+}$.  However, studies also showed that water molecules are typically linked together in clusters containing a fair number of water molecules held together by hydrogen bonds (Google for this if you are not familiar with it).
Thus today's version really is $\ce{H+_aq}$ but that looks awful, so it is as you note $\ce{H3O+_aq}$ is more often used.
The important point is that how you write it isn't really important.  It is the realization that the bare proton attaches itself to any nearby source of electrons.
How you should handle your homework depends on what your teacher does in class.  I'd follow that example.

The lecture slides didn't have any chemical equations on them. I will go with (aq) on your advice. Thanks!


text:
Is electrochemically activated water a real technology?
In essence what I'm asking is if the applications of Electrochemically activated water (ECA water) as a bactericidal-disinfectant is real and profitable (especially this last part). The principle at first seems very dull just take salt water through an electrolysis process and you get anolytes and catholytes, with disinfectant proprieties. Why is that only now is it viable?

I've never heard of this, but it sounds fishy.

Far from 'only now' becoming viable, the production of disinfectant from electrolysis of salt water (brine) is not only 'real' it is how bleach sold on supermarket shelves is produced! Industrially, this is called the  'chor-alkali process' and is well established for over 100 years. Both chlorine and sodium hypochlorite are manufactured industrially by electrolysis of brine. Where gaseous chlorine presents a significant hazard however, liquid sodium hypochlorite is offered as an alternative, however it has a limited shelf-life and degrades rapidly with temperature. Recent innovations in terms of design and manufacture of ion-exchange membranes has opened up new possibilities for small-scale on-site production of socium hypochlorite which is especially useful for remote communities.
On the question of economic 'viability' or profitability, like any question of economics, it depends upon the available technology and specific circumstances. For example, large power stations typically have the need to disinfect huge volumes of water which passes through their cooling towers. Since they literally 'generate electricity at cost', it is often more economic for such power stations to produce disinfectant on-site using eletrolysis than to truck in volumes of bleach manufactured at a chemical plant. Of course, this does not necessarily mean its 'economically viable' for you to make your own bleach at home right now, although this may change in the future.
The on-site generation of hypochlorite for disinfection may not only be useful for cooling towers on power stations but also for treating wastewater on board ships and in in remote locations. Since the chemical is produced 'on demand', there is no need for the extra waste created by packaging, shipping and storage of large volumes of hazardous chemicals.

Thank you for your answer. What I intended initially was not to question the usefulness and availability of standard products like bleach. The links are pretty clear. Well, in fact I didn't even know that bleach was a product of such process so my question was kinda vague. What I am interested about is on the fringe applications of ECA like for example: http://www.vbinstitute.org/technologies/
First that institute sounds fishy and I've seen that some "scientific" papers were not published in real journals but magazines. https://bitcointalk.org/index.php?topic=110884.0

The wikipedia page says all:
"The electrolysis of such salt solutions produces a solution of sodium hypochlorite, which is the most common ingredient in store-bought household bleach. The resulting water is a known surfactant (soap) and sanitizer."
That is: the effect of electrolysis is equivalent to adding a (very) small amount of bleach. There is not real reason to prefer this energy-consuming process to the tradionat bleach.

As described by others before, sodium hypochlorite ($\ce{NaOCl}$) is a strong disinfectant.
The technology for the production is established:

electrolysis of brine ($\ce{NaCl}$) to yield chlorine ($\ce{Cl2}$)
disproportionation of $\ce{Cl2}$ under alkaline conditions:  $\ce{Cl2 +2 OH- -> OCl- + Cl- + H2O}$

If you want halogen-free disinfection of water without bleach, there are two options:

ozone treatment
uv irradiation

Both technologies are widely used too and have a solid scientific background as well.
And you can even go "low-tech" to save lifes: A successful initiative of the Swiss Federal Institute of Aquatic Sciences and Technology, SODIS (Solar Water Disinfection), uses sunlight and PET bottles to kill microbes and produce drinkable water in developing regions of the world.

And then, you can rely on snake oil technology.


text:
What are the products in this SN1 reaction?
I am asked to identify the organic/inorganic products of an SN1 reaction between 3-iodopentane and ethanol. I have identified correctly that the substrate is the 3-iodopentane, the leaving group is the iodide, and that the nucleophile is the ethanol. My problem is that I do not understand what the products will be, and I am unsure why.
I decided that the products should be 3-pentanol and ethyl iodide (or iodoethane) because iodide is a good leaving group, and the oxygen takes two electrons and leaves ethane to bond with the pentane group where the iodide left off. Now it makes sense that the ethane and iodide react to create ethyl iodide.
Unfortunately, I am wrong in this case, and I just can not put my finger on why. What is the error in my thinking and what is the concept that I am missing?
I realize that my major organic product will be the ether compound which I am having trouble naming, and the other compounds are a protonated ethanol plus the iodide. The most confusing part now is the protonated ethanol and the iodide, why not just create the ethyl iodide?

OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$
So it's SN1, so the $\ce{I}$ is going to leave spontaneously and give you the carbocation: $\ce{(CH3CH2)2CH+}$
The oxygen is in the ethanol so that will attack the carbocation giving the oxonium ion: $\ce{(CH3CH2)2CH-O+} ~ \ce{H-CH2CH3}$
Which will then be deprotanated by something acting as a base. There's probably some water around, but I suppose saying protonated ethanol and iodide is fine. This gives the final product of: $\ce{(CH3CH2)2CH-O-CH2CH3}$
For ethyl iodide to form, one of two things would need to happen:

SN1 mechanism: $\ce{OH-}$ would have to spontaneously leave forming a primary carbocation. It is SUPER DISFAVORED, so much so that it straight up won't happen.
SN2 mechanism: $\ce{I-}$ would have to act as a nucleophile AND $\ce{OH-}$ would have to leave. Iodide is an OK nucleophile, but OH- is a pretty bad leaving group. Both things considered this pathway is unlikely.

This was a pretty weak question by your (I am assuming) professor. This reaction would be horribly slow, and probably would barely go. Now, if they used sodium ethoxide instead of ethanol you get a nice Williamson ether synthesis and form the same ether and NaI, much more favored thermodynamically.

I just realized you probably meant why doesn't $\ce{I-}$ perform an SN2 on the protonated ethanol and have water leave. There are a couple of halfway decent chemical explanations for this:

Assuming there is any water around (there is pretty much always water around in the real world (especially with ethanol), unless remarkably heroic measures are taken to avoid it), the protonated ethanol will quickly react with it and form hydronium, ethanol and $\ce{I-}$. (Protonated ethanol has a $\mathrm pK_\mathrm a$ of about -2.4 and hydronium has a $\mathrm pK_\mathrm a$ of about -1.7.)
Assuming there is not water around, you are right it will probably react and form small amounts of ethyl iodide, however this will generate water which will stop this from continuing for the above reason.
You could also write the mechanism where the iodide deprotonates to form the ether (and avoid this whole fuss). BUT $\ce{HI}$ has a $\mathrm pK_\mathrm a\approx10$ where as oxonium proton $\mathrm pK_\mathrm a$s are usually between -7 and -2, so this would NOT be an accurate way to write the mechanism (in my opinion), but some professors get a little hand wavy here and allow it.

I know those are kind of sketchy workaround reasons, but honestly it won't be formed in any real quantity. There is probably another (probably better) argument from a thermodynamic stand point, but I don't know the bond energies off hand.


text:
How should I identify my leaving group in SN1 reactions if there might be more than one option?
Before I can even begin to contemplate the products of a given SN1 reaction, I need to be able to determine what my leaving group will be, which should be a species with a strong conjugate acid, and as I understand it the species should be relatively stable once it has left the substrate. Please correct me if I am making any mistakes in my understanding.
A hydroxide ion will be a terrible leaving group, since its conjugate acid is water, and hydroxide is not very stable by itself.
So when I am asked this question "Draw the major organic product(s) for this SN1 reaction:

How do I know what will be the best leaving group? Hydroxide is not very good, so perhaps $\ce{CH_3OH}$ will work better? How do I know it is not $\ce{CH_3CH_2OH}$ instead?
Is this only one decent option, and I am over thinking this?

this example, to be honest, is bad since OH is bad leaving group. also, H-Br is used for hydrogenation of alkenes. on the other hand, this Sn1 reaction can be done by first toscilating it with TsCl to make a very good leaving group!

This case (and similar cases) you need to think about the following:

When a leaving group leaves, only one bond is broken. By this logic, $\ce{OH}$ is a leaving group - it can leaving by the breaking of the $\ce{C} -\ce{O}$ bond. The alcohols you mention, $\ce{CH3OH}$ and $\ce{CH3CH2OH}$ require breaking two bonds to leave: the the $\ce{C} -\ce{O}$ bond and a $\ce{C} -\ce{C}$ bond.
What is that acid $\ce{H-Br}$ doing? When you are given a strong acid as a reagent, very commonly the first step in the reaction's mechanism is a proton transfer (an acid-base reaction). Proton transfers are very rapid; they often happen before anything else has a chance to. In this case, the proton transfer is also exothermic (or exergonic if you prefer) - the equilibrium constant is on the order of $10^8$ in favor of the products. What does the product of the initial proton transfer look like? Is there a better looking leaving group on that structure?

I guess I was thinking the hydroxide would be near impossible to leave on its own, but I suppose it reacts with the hydrogen to make water and the leaving group could be replaced by left over bromide. I started giving myself the idea that the 2-3 carbon bond would break and separate the entire thing but that is bad thinking considering the obvious affects of the hydrogen bromide.

Its just the opposite. The only feasible leaving group on this molecule is -OH group. In general, a carbon group will never be a leaving group, because it would leave as a carbanion, and that is like the worst leaving group ever. C-C bond breaks are rare in organic chemistry, and they will definitely never take place as a consequence of a R group leaving the molecule.

To answer this question, you must first inspect the species that are reacting first. What are these species? What is acid? what is base?...not necessarily focusing on the leaving group.
In general, acid-base reactions have extremely high Keq (Ka) and in the system will occur first. It is so rapid that the exchange of a proton is close to the diffusion-controlled rate limit.
That happens first. The alcohol is protonated and oxygen becomes formally +. It is now the leaving group, water. As it leaves a stable tertiary carbenium ion is formed. Bromide, the conjugate base of the acid used then attacks to form the Sn1-type product.
Someone mentioned TsCl...tosylation will not be a good choice. True it will activate the alcohol to be a weaker base. But since this is a tertiary alcohol, aa back-sided displacement is not going to happen to give you an alkyl bromide, you will get the elimination instead of substitution.


text:
Preparation of alkyl halide from alcohols
$\ce{R-OH + HBr -> R-Br + H2O}$
An $\ce{R+}$ carbocation is formed during this reaction.
If $\ce{R}$ is any alkyl group instead of methyl or ethyl, can the beta hydrogen to the hydroxy group be removed to form a double bond, similar to a dehydration??

It can happen.
It practice, you avoid it happening by keeping the temperature low. $\ce{Br-}$ is a better nucleophile than $\ce{H2O}$ is a base, so at lower temperatures, the substitution is faster.
If the elimination happens, alkenes react with $\ce{HBr}$ to form alkyl bromides. As long as acid is present, the carbocation can reform. For example, with isopropyl alcohol
Direct SN1 Substitution
$$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$
Substitution by elimination-addition
$$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$
$$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$
$$\ce{(CH3)2CH+ + H2O <=> CH3-CH=CH2 + H3O+}$$
$$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$
In some cases, two processes may appear to generate different alkyl halides, but this would be no different than if a carbocation rearrangement occurred.


text:
Thiosulphate acidic decomposition
I just came across this reaction
$\ce{Na2S2O3 + 2HCl -> 2NaCl + SO2 + S + H2O}$
I was wondering if it was possible to predict the products of this reaction if I didn't know it (or any inorganic reaction for that matter).
Can we use arrow-pushing (as we do in organic chemistry) to get the products of an inorganic reaction? Can we devise a mechanism to find the products? It seems unrealistic to me how anyone could learn the many reaction of inorganic chemistry.

The arrow pushing formulism is expandable to any reaction of the main group elements involving the forming or breaking of covalent bonds. Organic chemistry is just the largest subset of these reactions. The remaining "inorganic only" reactions are a much smaller set.

If that is so, then can you please explain how I can obtain the product from the above reaction using arrow pushing. Can you give me a link to the image or something, I would like to know.

See http://www.amazon.com/Arrow-Pushing-Inorganic-Chemistry-Approach/dp/1118173988.

The general answer to this question is no, one cannot simply predict the products of any reaction.  How do we know that one of the products is not $\ce{S2O6^{2-}}$ or perhaps even $\ce{S2O5Cl2^{2-}}$?
The best that can be done without any guidance from experiment is to compute the thermodynamically favored products.  One does this (at standard temperature and pressure) by writing down all plausible products and then computing the Gibbs free energy ($\Delta G$) for the reactions.  The one with the smallest $\Delta G$ is the thermodynamically preferred.
But be warned:  there are mechanistic and kinetic factors to take into account.  The thermodynamically favored reaction may, as a practical matter, not take place.  The standard example is diamond --> graphite, which is thermodynamically allowed, but don't hold your breath waiting for it to happen.

So how do chemists possibly make sense of all the reactions ?


text:
Why does chocolate ice cream melt faster than others?
The other day I was dipping out Neapolitan(strawberry, vanilla, & chocolate) ice cream and noticed that the chocolate ice cream tended to melt faster than that of the vanilla and strawberry.  I couldn't remember in which order I dipped the ice cream so I decided to do a controlled experiment with it the next time I dipped it out.
I took three bowls of the same type/shape and dipped out the strawberry first, then vanilla, and chocolate last.  Even dipping the chocolate ice cream last showed that the chocolate ice cream would melt faster than the strawberry or vanilla.
Neither the strawberry nor chocolate use artificial coloring, though the vanilla states that it does.  Is there a reason as to why chocolate ice cream melts faster?  Is it because of the darker coloring that attracts heat or perhaps the low melting temperature of chocolate in general?

The darker color hypothesis sounds reasonable, how can you change your experiment to test this hypothesis?

what about different melting points?

@mdpc Could you explain, in your answer, why you think they have different melting points? Please include some chemistry concepts in your answer.. (I have converted to comment for now)

It's a complicated problem and I'm going to add a hand-waving physics answer.
For most materials, at a given wavelength the albedo and and emissivity sum to roughly unity. So if something is close to white with an albedo of 0.9, it's emissivity in the visible range is about 0.1, and if something is dark, those are reversed with albedo ~0.1 and emissivity ~0.9.
However, in the thermal infrared many non-metallic materials have an emissivity of very roughly 0.9 no matter what color they are in the visible. Even glass and water which are transparent in the visible have a high infrared emissivity.
In direct sunshine a surface perpendicular to the light will receive about 1100 W/m^2 in the visible, and so chocolate ice cream will be heated much faster than vanilla. Strawberry is pink because it absorbs shorter wavelengths but there's not a lot of power there from the Sun's black-body spectrum, so strawberry is like vanilla in the Sun.
However on an overcast day, or indoors, the the light is roughly 50 to 100 time dimmer, or about 11 to 22 W/m^2
(You can prove that by shooting a photo at f/16 with a shutter speed of 1/ISO or 1/ASA. I'm dating myself, but if had ASA 64 film you'd shoot f/16 at 1/60th second outdoors. You'd get nothing if you left it that way and shot indoors (Sunny 16 rule)
So now we're only down to say 11 or 22 W/m^2 in the visible, but that's not all the light! The Stefan-Boltzmann law tells us
$$P = \sigma \epsilon T^4.$$
If you plug in 273K and 298K (0° and 25° C) and $\epsilon=0.9$, you get about 280 and 400 W/m^2. That means that you, the walls ceiling or clouds are heating your ice cream at 400 W/m^2 while it is only radiating 280 W/m^2 back, it's a difference of 100 W/m^2 or way more than the heating at visible wavelengths.
So I would propose that if the phenomenon happens indoors that it's not related to the color, and could be related to the recipe.
Maybe chocolate ice cream has more salt, depressing the melting point, and so it starts melting sooner because it melts at a lower temperature.
So I've just asked Does chocolate ice cream tend to have more salt than strawberry or vanilla?

This appears to be physics more than chemistry. A dark color means it emits less visible light but rather absorbes it with temperature rise. Have your next ice cream in a darker place as to visible and ir emisions to test this.

This is almost certainly a factor, but anecdotally I have also observed that chocolate ice cream is softer, even straight from the freezer. This would likely be due to some change in the mixture of substances in chocolate ice cream versus other flavors. I wouldn't go so far as to say it's any one thing, but it's a different recipe so it will have different properties.


text:
List in order of increasing radius
List in order of increasing radius:
a) Rb, K, C5, Kr
For this one I got C, Kr, K, Rb
b) Ar, Cs, Si, Al
This I got: Ar, Si, Al, Cs
Does anyone know if this is correct? I don't have the solution for these in my book :S

What do you know about the periodic trends in atomic radii? There are two useful generalizations and it looks like you have followed them.

A) Rb:265 pm
   K: 243 pm
   C: 67 pm
   Kr: 88 pm
B) Ar: 71 pm
   Cs:298 pm
   Si: 111 pm
   Al: 118 pm
Where Pm is picometer =1.0 × 10^-12 meter

Could you provide some conceptual reasoning in your answer as to why the order of radii is what it is?

There is a trend that follows an arrow from the lower left of the periodic table to the upper right.  As the atomic number increases along a period, the radius for each element decreases.  Here is a useful image of this:

Notice that, with the exception of the noble gases, you can draw an arrow from the lower left to the upper right.  The justification for what you see here is that as the principal quantum number increases, the probability that the electrons are more distance from the center of the atom also increases.  The higher the principal quantum number, the higher the energy of the electron and therefore its probable distance from the center of the atom.  Therefore, as you move down a group, the radius increases.
However, another factor is that as the number of protons increase in an atom, so does the positive charge which happens to pull the electrons toward the center of the atom.  So, there is the impact of the protons causing a decrease in radius (probability of the electron being closer to the center) and the probability that the electrons are in an orbital that is farther from the center.
The transition metals, however, have d orbital electrons.  These electrons enter into a shell that is closer to the nucleus and the change is less noticeable.  That is part of the reason why they do not seem to change as much as the other atoms, according to my understanding.