text:
Why doesn't the density of water add up?
If the density of hydrogen is 0.09 g/ml and the density of oxygen is 1.429 g/ml, how is it that the density of water is 1 g/ml?

Note that the unit for the density of the gases is not correct. At normal temperature and pressure, the density of hydrogen is 0.09 g/l (not 0.09 g/ml), and the density of oxygen is 1.429 g/l (not 1.429 g/ml).

Any "mixture", with a chemical reaction or without, can only be expected to have the "average" density if the particles in both are of similar form and size and there is no enthalpic interaction. So really never with a chemical reaction.

The densities of hydrogen and oxygen actually do approximately add up to the density of water if you use a state (temperature and pressure) where all three compounds may be considered to be an ideal gas (whereas at normal temperature and pressure, hydrogen and oxygen are gases, but water is liquid).
For example, at a temperature of $T=200\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$, all three compounds are gases with the following densities:
$$\begin{align}
\rho(\ce{H2})&=0.05122\ \mathrm{kg\ m^{-3}}\\
\rho(\ce{O2})&=0.81329\ \mathrm{kg\ m^{-3}}\\
\rho(\ce{H2O})&=0.46031\ \mathrm{kg\ m^{-3}}
\end{align}$$
You also have to take into account that hydrogen molecules ($\ce{H2}$) consist of two hydrogen atoms, oxygen molecules ($\ce{O2}$) consist of two oxygen atoms, and water molecules ($\ce{H2O}$) consist of two hydrogen atoms and one water atom:
$$\ce{H2 + 1/2O2 -> H2O}$$
Thus, the density of water can be estimated as
$$\begin{align}
\rho(\ce{H2O})&\approx\rho(\ce{H2})+\frac12\rho(\ce{O2})\\[6pt]
&=0.05122\ \mathrm{kg\ m^{-3}}+\frac12\times0.81329\ \mathrm{kg\ m^{-3}}\\[6pt]
&=0.45787\ \mathrm{kg\ m^{-3}}
\end{align}$$
which is quite close to the real value of $\rho(\ce{H2O})=0.46031\ \mathrm{kg\ m^{-3}}$.

Explanation
Density $\rho$ is defined as
$$\rho=\frac{m}{V}$$
where
$m$ is mass, and
$V$ is volume.
The mass of a pure sample is given as
$$m=n\cdot M$$
where
$n$ is amount of substance, and
$M$ is molar mass.
The volume of a gas can be estimated using the ideal gas law as follows.
$$\begin{align}
p\cdot V&=n\cdot R\cdot T\\[6pt]
V&=\frac{n\cdot R\cdot T}{p}
\end{align}$$
where
$T$ is temperature,
$p$ is pressure, and
$R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$[source] is the molar gas constant.
Therefore, the density of an ideal gas at a given temperature and pressure only depends on the molar mass of the gas.
$$\begin{align}
\rho&=\frac{m}{V}\\[6pt]
&=\frac{n\cdot M}{\frac{n\cdot R\cdot T}{p}}\\[6pt]
&=\frac{M\cdot p}{R\cdot T}
\end{align}$$
And the molar mass of water can be calculated from the molar mass of hydrogen and oxygen as
$$M(\ce{H2O})=M(\ce{H2})+\frac12M(\ce{O2})$$

This certainly does not help to correct the OPs obvious, grave misunderstanding.

@karl and, worse, it confuses the key issue by making everything about gas-phase density which is a completely different matter than liquid or solid density.

Actually, this is a wonderful method. I am tempted to ask, does this work for all gaseous compounds? or are there some known exceptions?


text:
Negative charged intermediate in acidic conditions - what's the problem?
Why positively charged intermediates are preferred over negative charged intermediates, when formed during a reaction taking place in acidic conditions?
For example:

Image source: Organic Chemistry by David R. Klein
Here, per Organic Chemistry by David R. Klein says,

That is, the mechanism does not exhibit any negative charges, which is
  consistent with acidic conditions.

Why?

What do you think happens to negatively charged species under acidic conditions?

What happens to a positively charged one? I am missing somewhere.

Well, cations containing hydrogen are usually quite acidic and anions basic, aren't they?

Acidic conditions — whether you are using the Brønsted or the Arrhenius definition of acidity — is characterised by readily available protons. Remember the general dissolution ionisation of a Brønsted acid in water  as shown below:
$$\ce{HA + H2O <=> H3O+ + A-}\tag{1}$$
The proton is readily liberated and transferred to species which are more happy to accept it — for strong acids, that means it is transferred to water, a really weak base. A proton typically gladly attaches to anything that has a negative charge, because those species are typically more basic than water in this example. Therefore, postulating an enolate or any other oxygen-centred anion in acidic solution is not considered correct since there will be enough protons to quench it.
However, this does not mean that no anions can exist in aquaeous solution. For example, a typical method to transform alcohols into iodides is refluxing in aquaeous hydrogen iodide. Here, $\ce{HI}$ is a very strong acid that very readily dissociates to give iodide anions, $\ce{I-}$. These, even though negatively charged, do exist in acidic conditions and are able to attack nucleophilicly. That’s all due to the high acidity of $\ce{HI}$.
$$\ce{R-OH + H3O+ + I- <<=> R-OH2+ + H2O + I- -> R-I + 2 H2O}\tag{2}$$

source: Stack Exchange

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sort: votes


text:
Enol or Keto form?
Which will be the major form among the two tautomeric forms?

According to me, it should be enol form due to resonance stabilization as well has five membered ring formes througj hydrogen bonding. On the other hand, there is C=O, with greater bond energy in the keto form. Just want a confirmation.

yeah you are correct!! enol form is more stable. It has resonating structures, a fiver membered ring, hydrogen bonding. Thus making it stable

The enol form has potential to form a giant delocalized pi system granted that everything is sp2 hybridyzed.

Whether a diketo compound is present in solution in its diketo or its enol configuration is not something that can be said a priori — it strongly depends on the solvent and whether it is able to hydrogen bond or not.
If you have a solvent that is able to donate hydrogen bonds, the compound has a much greater tendency to adopt the diketo form since all its hydrogen bond accepting lone pairs are saturated with solvent hydrogen bonds. These can adopt a practically linear $\ce{O\bond{...}H-Solv}$ configuration which typically signifies a strong (or low energy) hydrogen bond.
The intramolecular hydrogen bond is rather strained, rather than having a bond angle $\angle (\ce{O\bond{...}H-X}) \approx 180^\circ$ the angle is approximately $108^\circ$. Therefore, it is a lot weaker. This type of hydrogen bond is more favourable if the solvent is not able to donate hydrogen bonds at all. In this case, the enol structure is also slightly more favourable due to the larger conjugated system.
The question can only definitely be answered by NMR studies; depending on what the structures look like a crystal structure may also help.

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text:
Is there a silicon equivalent to a carboxylic acid?
Can terminal Si on a siloxane chain have double bond O and OH?

Si generally doesn't support double bonds. At least severe steric effects would be needed to get such thing.

Is this good enough? http://www.sigmaaldrich.com/catalog/substance/silicicacid12345134398211?lang=en&region=US

That silicic acid is good.  And has that double bond: probably more acidic than a carboxylic acid.  I wonder if there is a sodium salt.  Thanks Zhe.

@WillK we can make salts, but in an ionic lattice probably $\ce{SiO3^{2-}}$ would polymerize to make tetrahedrally coordinated chains; the double bonds are split into single bonds by this polymerization. Some silicate minerals have this structure. If we take the ortho acid $\ce{H4SiO4}$ as the partent acid, then $\ce{SiO4^{4-}}$ remains monomeric and there are minerals with this too, but again there are no double bonds.

Acids with a functional group $\ce{Si(O)OH}$ can indeed be formed. One such example is given by Xiong et al.1, illustrated in the abstract as compound 4 in the graphical picture below:

As can be seen in the illustration, the silanoic acid is not isostructural with a carboxylic acid. The silicon is tetracoordinate rather than tricoordinate, and although a pi bond is drawn the bond in the silicoxy group is probably more accurately described as $\ce{Si–O-}$. The negative formal charge is passed to the borane electrophile shown while the ligand $L$, a pyridine derivative, is hydrogen-bonded to the dissociable proton. The compensating positive charge is distributed in the nitrogen atoms of the ring. Compounds 2 and 3 show other possibilities involving a silicon-sulfur bond.
Reference
Xiong, Y., Yao, S. and Driess, M. (2010), Silicon Analogues of Carboxylic Acids: Synthesis of Isolable Silanoic Acids by Donor–Acceptor Stabilization. Angewandte Chemie International Edition, 49: 6642-6645. https://doi.org/10.1002/anie.201002970

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text:
What Is A Hydrogen Bond?
I know this might be a bit of a silly question, but what is a hydrogen bond? I always thought that (say in water for example) the hydrogen was slightly more negative than the other ion (oxygen). But hear me out, this is because the shared electron (between the hydrogen and the oxygen due to their covalent bond) is slightly closer to the hydrogen's positive nucleus (than the oxygen's positive nucleus).
And since both are gaining an electron, that additional electron had more of an effect on the overall hydrogen ion (and less so of an effect on the oxygen ion) than a charge of exactly -1. Therefore if the charge of the hydrogen ion is now is slightly less than -1 and the charge of the oxygen is slightly more than -1, then the hydrogen is more negative than the oxygen...
Wait, if both the hydrogen and the oxygen ions are negative, why would the attract other negative ions from other water molecules, surely they'd repel. Obviously there is some flaw in my understanding of this and im just hoping someone can help me out.

https://en.wikipedia.org/wiki/Hydrogen_bond

H in H2O has a partial "positive" charge and not "negative" charge. The Oxygen atom has a partial negative charge.

http://chemistry.stackexchange.com/questions/49961/what-is-a-hydrogen-bond

@MrAP What does the electronegativity mean and why does mean that the oxygen has to 'hog' the shared electrons (giving it its partial negative charge)?

Electronegativity refers the tendency of an atom to attract electrons towards itself. If the atoms forming a molecule of a compound have a high electronegativity difference then an ionic bond will be formed between them; if the difference is moderate then the bond will be a polar covalent bond, with the more electronegative atom attracting the shared electron pair towards itself and hence has a partial negative charge while the other atom has a partial positive charge, i.e., there is a charge separation. If the difference is low then the bond formed will be a non-polar covalent bond.

I always thought that [in water] the hydrogen was slightly more negative than the [oxygen]. […] [T]his is because the shared electron (between the hydrogen and the oxygen due to their covalent bond) is slightly closer to the hydrogen’s positive nucleus (than the oxygen’s positive nucleus).

You do not consider the relative size of atoms to determine their partial charges. It is not relevant by itself whether an atom is large or small; the only thing that matters is how well the nucleus can attract electrons — this is a value tabulated as electronegativity. The size of an atom in itself, and therefore also the distance from middle of a covalent bond to the nucleus of either atom is irrelevant; however, the size of an atom plays an indirect role since additional shells of core electrons reduce electronegativity by shielding the positive charge of a nucleus. It is not required to perform a formal analysis of how much influence the core electrons have, since — as I mentioned — the elements’ electronegativities are tabulated in periodic tables and similar resources.
Going strictly by electronegativity, oxygen has $3.5$ and hydrogen has $2.1$. Thus, the bonding electrons are polarised much more towards oxygen; this means that oxygen has an electron surplus (partial negative charge) while hydrogen has an electron deficit (and therefore a partial positive charge).

Another thing to consider is that a water molecule is a neutral molecule. Therefore, positive and negative partial charges must be somewhat evenly distributed. There is no overall surplus of electron, i.e. there is no overall charge. For an overall charge, you would have to turn to species such as the hydroxide ion $\ce{OH-}$. However, even therein the hydrogen is partially positively charged due to oxygen’s much larger electronegativity. This is not always given: in a peroxide ion $\ce{O2^2-}$, each oxygen atom will carry one single negative charge to give a symmetric molecule.

Taking this all into account, it makes sense to consider a hydrogen bond as an electrostatic interaction between a positively charged hydrogen bound to a negatively charged atom and another strongly electronegative, negatively charged atom of a different molecule.

I understand most of that, but with the peroxide ion, I would've thought that each oxygen atoms have an additional electron each, making two oxygens with a charge of negative one each, making the peroxide ion have a charge of -2. But if you look at the two oxygens collectively, there are 14 electrons and 16 protons, so the overall charge must be +2.

I understand that in  a molecule the oxygen should be considered an atom rather than an ion, but if it's has more electron (8) than it would natural have (7) then why is it not a negatively charge ion, as opposed to a neutral atom?

@LucasDyson-Diaz an entity is only called an ion if it features a full charge. Also, an atom can only be an ion by itself if it is not part of a molecule. A molecule can only be a molecular ion in its entirety; individual atoms of a molecule are never considered ionic.

@LucasDyson-Diaz Concerning the charges of the atoms in peroxide (and that of the entire ion): you're miscounting. Two oxygens at eight electrons each plus two additional electrons totals at $18 = 16 + 2$. This can also be derived from the ion's Lewis structure.

The assumption you are making that is leading you astray is that hydrogen, covalently bonded to a more electronegative atom (hydrogen bonding is not exclusive to oxygen), is negatively charged. A partial positive charge that interacts intermolecularly with neighboring molecules resides on the hydrogen atom, at which electrons reside less frequently than the Bohr model would suggest.

Sorry @David Shaked, I'm saying that the hydrogen is negatively charge (which you say is wrong) and you're saying that it is in fact "partially positively charge". How do you get to understanding that? I would've thought that if an electron is being gained to a neutral hydrogen atom (since electrons are negatively charged) then the hydrogen ion become negative.


text:
Lewis structure for SO4 (2+)
This was a question on a review sheet for Lewis structures.
Now I could draw $\ce{SO4^2-}$ with no issues, but I can't make sense out of $\ce{SO4^2+}$ because I can't even make bonds to all $\ce{O}$ without exceeding the number of bonding electrons.
This is on a high school AP review so it is possible that there are exceptions to the normal rules, but I can't see how any of them apply. I could just chop electrons off the oxygens and leave it depleted, but I don't imagine that is a stable form of a cation which would exist?
A quick check doesn't turn up anything so is this a real thing or was it just a typo?

Looks like a typo.

How about
$\ce{O=O+-S+=O+-O-}$ ? I don't know if it exists but it is a closed shell species and has a delocalized pi bonded structure that might give it some stability.

A better resonance structure of your suggestion would be $\ce{O=\overset{+}{O}-S-\overset{+}{O}=O}$ which has less formal charges, is more symmetric and is still conjugated.


text:
How to distinguish palmitic acid from lauric acid?
I'm reacting either palmitic acid or lauric acid ($\ce{C16H32O2}$ or $\ce{C12H24O2}$) with $\ce{NaOH}$, sodium hydroxide.
I'm wondering if there's any way to tell which one of those two acids it is after I've already reacted it with NaOH. It has to be a relatively short process, as I won't have much time. I'm doing this for a lab.
In case you feel uncomfortable about what you may call "helping me cheat", please note that my assignment states that I can have online help:

You may look up information to complete your flow-chart, just be sure to cite your sources...

Note: it doesn't have to be NaOH, it could be any base, I just picked that one. If you think another one would work better, recommend that as well, please.

How about removing an aliquot, acidifying it, isolate the crystalline acid and take a melting point.

Consider the reaction which happens, what is produced and then what is different about the products? The key has to be in their properties.
The saponification reaction which takes place will produce a salt, in this case it will make sodium palmitate and sodium laurate. These two salts have very different solubilities and this makes a marked difference in the lathering ability. Sodium laurate is higher and thus lathers more readily.
For more details :
Solubility data
Saponification
Scientific Soapmaking: The Chemistry of the Cold Process
By Kevin M. Dunn


text:
Can ATP be synthesized and consumed?
I do not understand why you could not fill a bowl full of adenosine triphosphate (ATP) and eat it. Is there a way to make ATP other than inside a cell and apply it to your body by injection, ingestion, topical absorption, inhalation, or another method?

Sure, but going into the body via the digestive system isn't the same as the already-established route(s) bestowed upon you by the powers of evolution. Google "ATP capsules" and you'll see numerous products targeting the muscle-building crowd. The claims are somewhat dubious and, besides, digested ATP will likely by hydrolyzed to ADP + P$_{\rm i}$ and not be of much use as ATP.

By the way, if you mean eating only ATP.. that would make not much sense. It can be vaguely thought as an electricity powering you cells' minifactories, that also need the fuel or precursors (carbohydrates, proteins/amino acids, fats, …).

https://chemistry.stackexchange.com/questions/65595/can-atp-be-synthesized-and-consumed/102048#102048

I'm sure you could, but you'd also have to eat a lot for it to affect your general energetics (delivery issues noted in other comments notwithstanding).  Your body uses around 60 kg of ATP per day. https://bionumbers.hms.harvard.edu/bionumber.aspx?&id=105606&ver=9

To answer the question in the title:
Yes, ATP can be synthesised, isolated and you can even eat it. It would be very expensive to do so but considering others put gold on most of their food that’s not a reason in itself not to. While eating too high a dose of ATP is not beneficial as per Paracelsus’ law, small amounts are certainly not harmful.
However, ATP is rarely shuffled into or out of cells. I’m not sure whether the human body provides mechanisms to shuffle ATP across cell membranes. I highly doubt it, since ATP is typically produced cell-internally by mitochondria and what gets shuffled around via the blood stream is primarily glucose. Most transport mechanisms are adopted to that.
Furthermore, even assuming that ATP does get shuffled across cell membranes in significant amounts, eating a crumb of it would first place it into the stomach. There, in the strongly acidic environment there is practically no way that ATP won’t get hydrolysed into first adenotine and diphosphate and then further broken down. Even if some of it magically manages to get through the stomach unharmed, the slightly alkline pH of the intestines are no better and the digestive tract is swarming with enzymes designed to break down practically everything. Thus, you can be sure that ATP has been fully metabolised before the body has a chance of taking it up into the blood stream.
Some of these problems can be overcome by injecting ATP subcutanously or intravenously but the problem with getting it into the cell remains.

Expanding upon the comment above detailing how exogenous ATP isn't that great of an idea given the path of delivery, it's also important to note that our body focuses more on consumption, storage, and mobilization of fuels. The actual ATP production and utilization happens on a cell by cell basis, but fuels are moved around the entire body extracellularly.

What if you inject it?

ATP is not very orally active as it is rapidly metabolized to adenosine and further to uric acid.  There are many clinical trials that have used ATP by IV infusion at a rate of 50 microgram per kilogram per minute for up to 96+ hours.  However both IV and oral administration has shown an increase in RBC concentrations of ATP.  This is likely due to increased uptake of adenosine that is then converted into ATP.
The AIDS Cure Research Collaborative, Inc which is the organization that I am affiliated with is exploring the use of ATP in a comprehensive cure strategy for those infected with HIV.  You have one molecule that can exert many effects.  ATP produces a strong pro-inflammatory response while the generation of adenosine by the system exerts a strong anti-inflammatory effect.  The field of purinergic signalling is fascinating and yet fully understood.

Interesting! Have these clinical trials been published and if so, can you include citations? We strongly encourage citations for scientific data where possible.

Without credible sources, this is an unsubstantiated claim, that is nothing more than terrible pseudo-science.

Here is a source authored by Marc Wagner: DOI: 10.2174/157016211796320289 and the webpage of the Collaborative. Not making claims about credibility, though.

Abstract

In various pathologic circumstances depolarized mitochondria are thought to precipitate cell death by avidly consuming cytosolic ATP. However, for as long as the inner mitochondrial membrane remains intact the reversal potentials of the adenine nucleotide translocase (ANT) and that of F0–F1 ATP synthase are strategically positioned so that they oppose import of cytosolic ATP into the matrix of respiration‐impaired mitochondria. This arrangement also seems to protect against a hysteretic consumption of cytosolic ATP accumulating in the mitochondrial matrix, in view of the depolarization caused by inhibition of F0–F1 ATP synthase by the endogenous protein IF1, yielding fast ANT reversal rates.
Mitochondrial consumption of cytosolic ATP: not so fast.


text:
How does an additive like aluminum oxide make a material like polyurethane harder/more resistant to wear?
I have wood flooring which is coated with a polyurethane finish. According to the marketing, aluminum oxide has been incorporated into the polyurethane to make it more durable, but that's all that is said about it.
I understand that aluminum oxide is a hard substance, but I don't understand how it could be incorporated into a polymer to improve wear resistance. If it occurs as microscopic beads or granules of alumina within a polymer matrix, how does that confer durability? I can understand how a coating of alumina on top of the polymer would provide durability, but how could that even work?
So, my question is: What form does the alumina take in the end product and how, in that form, does it improve durability?

Alumina is extremely hard and very wear resistant, likely harder than anything which would come in contact with the floors. Dirt for example would mainly contain silica based abrasives which are far softer than alumina and hence alumina additives, of sufficient size would provide scratch resistance when present in sufficient quantities.
If you want the exact details, they are going to vary from one manufacturer to another. If you do a search for patents you can see various procedures some of which are pre-mixed, some of which require onsite mixing, for example :
Surface coverings containing aluminum oxide
That kind of post-mixing is very common in concrete, though there the aggregate usually isn't alumina (could be just decorative as well or dual purposed as is common locally) and is far bigger. In general you want the aggregate to be at least as large as, and ideally bigger than the particle which is trying to cut it. If the aggregate is much smaller it won't resist wear strongly as it will be plowed out with the substrate.
Effect of abrasive particle size on abrasive wear
of hardfacing alloys
If you are curious about a specific product then just call the manufacturer and ask to speak to the engineering department. However as they are all likely patented products, you can even search manufacturer specific and turn up results.


text:
How is the process of converting calcium carbonate (CaCO3) into calcium oxide (CaO)?
I need to provide accurate information in relation to the production process of calcium oxide from calcium carbonate.
$$\ce{CaCO3 -> CaO + CO2}$$

From wikipedia article of calcium oxide

Calcium oxide is usually made by the thermal decomposition of
  materials, such as limestone or seashells, that contain calcium
  carbonate ($\ce{CaCO3}$; mineral calcite) in a lime kiln. This is
  accomplished by heating the material to above 825 °C (1,517 °F), a
  process called calcination or lime-burning, to liberate a molecule
  of carbon dioxide ($\ce{CO2}$), leaving quicklime.
$$\ce{CaCO3(s) → CaO(s) + CO2(g)}$$
The quicklime is not stable and, when cooled, will spontaneously react
  with $\ce{CO2}$ from the air until, after enough time, it will be
  completely converted back to calcium carbonate unless slaked with
  water to set as lime plaster or lime mortar.


text:
Why does N-phenylhydroxylamine give Fehling's test?
I have studied that Fehling's test is given by aldehydes and alpha hydroxy ketones. I read in a book that it is also given by phenylhydroxylamine converting it to nitrosobenzene.
Does it gives the test as it has an alpha hydrogen or any other reason? Are there any more exceptions like this?

Quoting from "REDUCTION OF AROMATIC NITRO COMPOUNDS TO N-PHENYLHYDROXYLAMINES - A Comparative Study of the Yields Obtained when
Aromatic Nitro Compounds are Reduced, under Similar Conditions, to N-Phenylhydroxyiaminee by William Henry James"

N-Phenylhydroxylamine is unstable. It is readily affected by light,
  especially white light. It is also affected by moist air, being
  readily oxidized to nitrosobenzene. It has strong reducing power,
  reducing almost quantitatively Fehling's Reagent to cuprous oxide, and
  causing the formation of a silver mirror when warmed with an
  aramoniacal solution of silver oxide.

So, I guess that its instability and its reducing power are the cause of the effect explained.
Reference

James, William Henry. Reduction of aromatic nitro compounds to N-phenylhydroxylamines. Thesis (M.A.), Boston University, 1932. Boston University Libraries OpenBU


text:
organic hard surf wax?
I made some organic surf wax with 4 parts beeswax and 1 part coconut oil. The wax turned out to be sticky but had no tack.
I'd like to make wax for warm water that forms a layer of hard bumps on the board when wax is applied.
Not sure how I can achieve this w all organic items.
Any suggestions?

Carnauba wax is a high-melting wax that should be suitable. In the Wikipedia article cited,
"Carnauba wax can produce a glossy finish... It was commonly used in its purest form as a coating on speedboat hulls in the early 1960s to enhance speed and aid in handling in salt water environments. It is also the main ingredient in surfboard wax, combined with coconut oil."


text:
What happens to magnesium in water with acid?
I have a pure hunk of magnesium metal submerged in water. After several hours/days, the metal becomes coated with something.
Question 1:
What is the coating?
I've been told it's magnesium oxide. This would make sense to me if it was exposed to air, but it's submerged in water. Is it still MgO? I'll assume there is enough free oxygen in the water to react slowly.
If I add an acid to the mix, the coating gradually disappears. I use some malic acid crystals. Where does the MgO go?
Question 2:
What is in the water when the coating disappears?
As the coating disappears to expose the metal, the magnesium reacts more and more vigorously offgassing hydrogen, which I understand would leave magnesium malate in the water. But what about the MgO? Does the acid react with MgO? And either way, where does it go? Is the MgO just dissolved invisibly in the water or is it reacting to form something else?

Consider that the coating is magnesium hydroxide, $\ce{Mg(OH)2}$. Does the whole picture make more sense now?

What is the coating?

As pointed by Klaus Warzecha, there is no magnesium oxide.  It is  magnesium hydroxide, $\ce{Mg(OH)2}$  and that's the coating.
$$\ce{Mg + H2O -> Mg(OH)2 + H2}$$

What is in the water when the coating disappears?

The magnesium hydroxide reacts with malic acid to form magnesium malate which will be precipitated at the bottom.
$$\ce{Mg(OH)2 + malic~acid -> Mg-malate + H2O}$$
This is a classic acid-base reaction.
Thus, the coating gets dissolved, the magnesium is now exposed to water, it reacts vigorously with it offgasing more and more hydrogen.

There's some problems with the first equation.

@DHMO Yes, i did not wrote the hydrogen gas.  Editing the answer.

You don't even need an "acid" to set off the reaction. Some transition metal salts, like copper(II) salts, hydrolyze to make enough acid to get things going.

In water Mg oxidizes and the half-reactions are:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
$$\ce{2H2O + 2e- → H2 + 2OH-}$$
As the E($\ce{Mg}$/$\ce{Mg^{2+}}$)>E($\ce{H2O}$/$\ce{H2}$), Mg oxidizes and water reduces.
Metal ions can change their speciation depending on the pH. In acidic conditions $\ce{Mg^{2+}}$ dissolves and associates with malate (which in solution is deprotonated) and end up precipitating because the malate is in high concentration and the solution is saturated. And, in neutral o basic conditions they precipitate as metal hydroxides which are quite insoluble.
What you have to take into account is that metal ions can precipitate in basic conditions and dissolve in acidic ones.
Also, water has a reduction potential and has to be taken into account in rusting processes with metals. You can look for more information with Pourbaix diagrams where you can see the changes in speciation of metals in front of pH and the reduction potential of several metals.
Here's the Pourbaix diagram of Mg:

I hope this helps!


text:
Why isn't Cahn Ingold Prelog principle used to name the isomers of complex compound having square planar geometry & different ligands attached to it?
For a complex compound with coordination number 4, say having square planar geometry, and say all the ligands attached to the central metal atom are different, then we know that we will get 3 geometrical isomers.
Now my doubt is like in organic chemistry if we had 4 different groups around a carbon atom, we still named those isomers as Z and E isomers (by Cahn Ingold Prelog principle) . Similarly why can't we name the isomers as Z and E for the case of complex compound with square planar arrangement having four different ligands attached to it?
That is why can't we name those 3 geometrical isomers as Z and E isomers of the complex compound by Cahn Ingold Prelog principle?

You have three isomers, but only two designations, Z and E. Furthermore, what about this sort of complex where $\ce{L^1} > \ce{L^2} \cdots$ in the CIP system? I'm sure there is a IUPAC way to name such complexes, but it's not going to be the Z/E system.

Presumably we do not use $E$ and $Z$ because there are three isomers so we would need three labels.

According to IUPAC, the priorities of the ligands in such nontetrahedral compounds are still based on the CIP rules; however, additional rules may be needed in some cases. A general method for the notation is described in IR-9.3 of the Red Book (2005) and in P-93.3 of the Blue Book (2013).

@BenNorris Can these 3 isomers defined as cis- and trans- isomers?

Part I
The CIP rules in fact are used to determine the priority. However, a different notation system has been developed for inorganic nomenclature of coordination compounds.
For the square planar geometry, SP-4-x descriptor (an instance of ‘polyhedral symbol’) is used, where x is a CIP-based priority number of the ligand in trans (opposite) position relative to the ligand with the highest priority number (1).
In $\ce{[Mabcd]}$ complexes (ligand priority a > b > c > d, priority numbers 1, 2, 3, 4, respectively), for the 3 possible isomers, the descriptors are (forgive me the use of left/right superscript/subscript typography to depict square planar geometry, for the sake of simplicity):

$\ce{^\style{font-weight:bold}{a}_cM^b_\style{font-weight:bold}{d}}$    SP-4-4
$\ce{^\style{font-weight:bold}{a}_dM^b_\style{font-weight:bold}{c}}$    SP-4-3
$\ce{^\style{font-weight:bold}{a}_dM^c_\style{font-weight:bold}{b}}$    SP-4-2

For $\ce{[Ma2b2]}$ (priority numbers 1, 1, 2, 2) and $\ce{[Ma2bc]}$ (priority numbers 1, 1, 2, 3) complexes, where only 2 isomers are possible, the cis/trans notation can be used as well. The E/Z notation from organic chemistry is not used at all for coordination compounds central atom configuration.
Examples:

$\ce{^{\style{color:grey}{(1)}{ }I}_{\style{color:grey}{(3)}{ }Cl}Pt^{Br{ }\style{color:grey}{(2)}}_{F{ }\style{color:grey}{(4)}}}$
(SP-4-4)-bromidochloridofluoridoiodidoplatinate(2−)
(hypothetical)
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }NH_3}Pt^{NH_3{ }\style{color:grey}{(2)}}_{Cl{ }\style{color:grey}{(1)}}}$
(SP-4-1)-diamminedichloroplatinum(II)
trans-diamminedichloroplatinum(II)
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(1)}{ }Cl}Pt^{NH_3{ }\style{color:grey}{(2)}}_{NH_3{ }\style{color:grey}{(2)}}}$
(SP-4-2)-diamminedichloroplatinum(II)
cis-diamminedichloroplatinum(II)
cisplatin
$\ce{^{\style{color:grey}{(1)}{ }Cl}_{\style{color:grey}{(2)}{ }Ph_{3}P}Ir^{PPh_{3}{ }\style{color:grey}{(2)}}_{CO{ }\style{color:grey}{(3)}}}$
(SP-4-3)-carbonylchloridobis(triphenylphosphane)iridium(I)
trans-carbonylchloridobis(triphenylphosphane)iridium(I)
Vaska's complex (a different IUPAC name descriptor claimed there is disputed)

(References: see Loong's comment below the question.)
Part II
Now, your question is self-contradictory. You would like to use E/Z (two) descriptors for three different isomers.
Maybe you wanted to ask something like
Why are there 3 geometrical isomers of square planar [Mabcd] complexes, but only 2 isomers of tetrasubstituted ethene C2abcd?
(Note the latter statement is not true)
It's because for $\style{font-weight:bold}{\ce{[Mabcd]}}$, the four different ligands are at the vertices (corners) of a square. Which can be abstracted as a ‘closed loop’, i.e. a cycle graph. The combinatorics can be abstracted to free circular permutations, $P'_{n}={1 \over 2} (n-1)!$,
for $n=4$: $P'_{4}=3$.
For $\style{font-weight:bold}{\ce{C2abcd}}$, the situation is somewhat different. This is a combination of positional isomers $\ce{xyC=Czw}$, where the set $\{x,y\}$ is $\{\text{a}, \text{combination}_2(\{\text{b},\text{c},\text{d}\})\}$, and $\{z,w\}$ is a complement. There are (concidentally?) 3 of them (${n \choose k} = {{n!} \over {k! (n-k)!}}, {3 \choose 2} = 3$):

$\ce{abC=Ccd}$
$\ce{acC=Cbd}$
$\ce{adC=Cbc}$

(Here the different sortings of substituents at each carbon does not matter.)
However, the isomerism one is talking about when using the E/Z descriptors applies to geometrical isomers of a single positional isomer (take e.g. $\ce{abC=Ccd}$). Now, the sorting of substituents at each carbon does matter. What makes the isomers different is matching vs non-matching sort order at each carbon:

$\ce{abC=Ccd}$    ≡    $\ce{baC=Cdc}$    (ascending-ascending, descending-descending – matching)
$\ce{abC=Cdc}$    ≡    $\ce{baC=Ccd}$    (ascending-descending, descending-ascending – non‑matching)

which are 2 possibilities, and correspond to E and Z configurations, respectively (provided that the lexicographical order corresponds with the CIP priority order).
So, the total number of C2abcd isomers is 6 (3 positional isomers, each has 2 geometrical isomers). One think of the alkene as an object similar to the square planar metal complex, where the C=C can be arranged in the center in two orientations, horizontal and vertical, hence 2× as many possibilities than ${\ce{[Mabcd]}}$.
(I hope exactly zero mathematicians and chemists were harmed during the reading.)

Can you give any reference to the priority list followed during the numbering of the groups? Is it the same as the CIP rule followed in organic chemistry? Also is the same rule valid for octahedral geometry(take [M(ABCDEF)] for example)


text:
Use of [H] to denote a reducing agent
In the reduction of an aldehyde (e.g. ethanal) to a primary alcohol, a reducing agent is needed:
$$\ce{CH3CHO + 2[H] -> CH3CH2OH}$$
My book states that one hydrogen (the $\ce{H-}$ ion) is from the reducing agent, and the other hydrogen (the $\ce{H+}$ ion) is from the solvent.
However, elsewhere it more generally notes that "$\ce{[H]}$ denotes a hydrogen from a reducing agent".
Does the $\ce{[H]}$ notation refer only to hydrogen atoms from the reducing agent, or, as this example implies, can it also refer to hydrogen atoms from other sources (such as the solvent)?

Any element in square brackets indicates that it is in atomic state I.e it's a free atom.

@JM97 I disagree. If it is a free atom, why not simply write it as a free atom? The brackets instead denote what OP wrote: [H] is any reducing agent. As such it is better to not include it in balancing reactions and instead write $$\ce{CH3CHO ->[\ce{[H]}] CH3CHO}$$ Whether in the end $\ce{H}$ comes from the solvent, or reducing agent directly, or the reducing agent via the solvent is hard to determine.

There's an old concept called a "nascent state" whereby an element produced in situ supposedly provides higher reactivity. For example, hydrogen generated in a reaction mixture was called "nascent hydrogen" and often denoted as $\ce{[H]}$, and was thought to be a better reducing agent than bubbling gaseous $\ce{H2}$ in the reaction mixture.  To my understanding, however, the idea of nascent states is not well supported, and so is falling in disuse.

@NicolauSakerNeto The concept of a lone atom present on a surface in situ (an adatom is well established and frequently used, however, in electrochemistry and materials science at least. Perhaps also in catalysis, at active sites? I have less of a sense for 'nascent atoms' in homogeneous kinetics/catalysis, however.

@hBy2Py I've really only read about this topic lightly, but what I gather is that by establishing the existence of some special nascent state for an element, one neglects a better microscopic description of the reaction system (e.g. surface catalysis by adsorbed single atoms/molecules, for which the surface is also vital, not just the element) and its mechanistic implications, and that's what's frowned upon. These more modern concepts have superseded the nascent state.


text:
How do we know the concentration of hydroxide ions in this derivation?
In the derivation of the pH of a solution of a salt prepared from a weak acid and strong base, I used two equilibrium reactions:

$$\ce{ CH3COOH(aq) + H2O(l) <=> H3O+(aq) + CH3COO-(aq)} \qquad k_{eq} = K_a \\
    \ce{ 2H2O(l) <=> H3O+(aq) + OH-(aq) } \qquad k_{eq} = K_w$$

Reversing the first equation and adding it to the second gave me the "hydrolysis reaction":
$$ \ce{ CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq)} \qquad k_{eq} = \frac{K_w}{K_a} $$
My textbook uses this third equation to derive the pH of the salt solution by asserting that the equilibrium concentrations of the reactants and products of the third reaction are:
$$ \ce{[CH3COO-(aq)]} = C-C\alpha \\
   \ce{[CH3COOH(aq)]} = C\alpha \\
   \ce{[OH-(aq)]} = C\alpha $$
What I am confused about is, why is the equilibrium concentration of hydroxide ions $C\alpha$, and is only governed by the third reaction? Why
does the second equilibrium not contribute any hydroxide ion concentration? That is to say, the derivation assumes the water equilibrium does not exist, and all the hydroxide ion concentration comes only from the third reaction.
Also, even if we suppose some $x$ concentration of hydroxide ions and use it in the 'equilibrium constant expressions' of both the second and third reactions, we have more unknowns then conditions. So that's a problem too.
All in all, I need to know that, why, inspite of hydroxide ion participating in two equilibria, is the hydroxide ion concentration decided by only one of the reactions. And even if approximation is involved, if we try to be accurate, why do we have fewer conditions than variables?
Thank you.

In a solution of acetic ions, say from sodium acetate, some acetate will react with water to produce acetic acid and hydroxide anion (your "hydrolysis" equation"):
$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
Now, let's say you dissolved 1.00 moles of sodium acetate in 1.00 L of water. For the sake of convenience, you would describe this solution as $\mathrm{1.00\ M\ }\ce{CH3COONa}$. However the actual concentration of acetate ion is less than $\mathrm{1.00\ M}$ because of the reaction with water.
Enter $C$, the "total concentration of acetate". This concentration is the sum of all concentrations of species that are related to or derived from acetate. In this case the relevant species are the acetate ion and acetic acid, so $C = [\ce{CH3COO-}]+[\ce{CH3COOH}]$.
What is $\alpha$? $\alpha$ represents that percent dissociation or percent ionization of the acetate ion. Therefore, $C\alpha$ represents the concentration of acetate that has ionization (reacted with water). In this case, since every acetate that reacts with water produces acetic acid, $C\alpha = [\ce{CH3COOH}]$.
How does this derivation lead to $C\alpha = [\ce{OH-}]$? You need to look back at the reaction equation for the process. For each acetate ion that reacts with water, one acetic acid and one hydroxide ion are produced. Based on stoichiometry, the concentrations of acetatic acid and hydroxide ion must be the same.
How do we determine $[\ce{OH-}]$? We plug the three expressions for the three species into the law of mass action and solve for $\alpha$ (we know $C$):
$$K=\dfrac{K_w}{K_a} = \dfrac{[\ce{CH3COOH}][\ce{OH-}]}{[\ce{CH3COO-}]}=\dfrac{(C\alpha)^2}{C-C\alpha}$$
Now, if there was some additional hydroxide $x$ already present, say because the pH was fixed at 12.00, how do we handle that? The expressions for the concentrations of acetate and acetic acid are still valid. $C$ is still the total concentration of acetate + acetic acid. $C\alpha$ is still the concentration of acetic acid. $C-C\alpha$ is still the concentration of acetate ion. However, the concentration of hydroxide ion is different in this case. There is some concentration from the initial conditions $x$, and there is some increase from the ionization of acetate ion $C\alpha$.
If the initial hydorixde concentration is from the autoionization of water, we can account for that:
$$\begin{aligned}
[\ce{CH3COO-}] &= C-C\alpha \\
[\ce{CH3COOH}] &= C\alpha \\
[\ce{OH-}] &= 1\times 10^{-7} + C\alpha \end{aligned}$$
However, except for the most dilute solutions, the contribution from the autoionization of water is usually orders of magtitude smaller than $C\alpha$ and gets lost in the insignificant figures. Let's compare the same calculation for our $1\ \mathrm{M}$ acetate solution. Note that you get the same outcome regardless if you include the hydroxide concentration form $K_w$.
$$\begin{array}{|c|c|c|} \hline
 & \text{without autoionization} & \text{with autoionization}\\ \hline
C & \mathrm{1.00\ M} & \mathrm{1.00\ M} \\
[\ce{CH3COO-}] & C-C\alpha = 1-\alpha & C-C\alpha = 1-\alpha \\
[\ce{CH3COOH}] & C\alpha = \alpha & C\alpha = \alpha \\
[\ce{OH-}] & C\alpha = \alpha & 10^{-7} + C\alpha = 10^{-7} + \alpha \\
K & \dfrac{K_w}{K_a} = \dfrac{10^{-14}}{1.74\times 10^{-5}} & \dfrac{10^{-14}}{1.74\times 10^{-5}}\\
K & \dfrac{\alpha^2}{1-\alpha} & \dfrac{(10^{-7} + \alpha)(\alpha)}{1-\alpha}\\
\alpha & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\
[\ce{CH3COO-}] &  1.00-2.40 \times 10^{-5} \approx 1 & 1.00-2.39 \times 10^{-5} \approx 1 \\
[\ce{CH3COOH}] & 2.40 \times 10^{-5} & 2.39 \times 10^{-5} \\
[\ce{OH-}] & 2.40 \times 10^{-5} & 10^{-7} + 2.39 \times 10^{-5} = 2.40 \times 10^{-5}  \\ \hline
\end{array}$$

Thank you for your answer! I understood it, and though helpful, I actually meant to ask something slightly different. For example: $\ce{A<=>C+D}$ and $\ce{B<=>C+F}$. Here C is formed from both, the 1st and 2nd reaction. So the final conc. of C has contribution from 2 reactions. Similarly, $\ce{OH-}$ should have contribution from both reactions, the 2nd and the 3rd. But we assume contribution from only 3rd reaction. Why?

Well now it seems clear, just one more thing that bugged me: for the autoionization calculation, a contribution of $10^{-7},$M was assumed. But how can we assume that? I mean, why $10^{-7}$? Why not $10^{-6}$? Sure, neutral water has $10^{-7}$ conc. but that isn't the case here... wouldn't the contribution change or something?

Perhaps, can we visualize adding the salt after the water equilibrium is attained? That would make sense. Is this correct?


text:
Solubility comparison of hexan-1-ol and cyclohexanol in water
Is hexan-1-ol or cyclohexanol more soluble in water? The answer is supposedly cyclohexanol. However, I found a statement in a book:

Isomeric alcohols have solubilities in the order: primary alcohol > secondary alcohol > tertiary alcohol, due to decrease in polar character.

By this statement, I guessed the answer to be 1-hexanol, as it is a primary alcohol whereas cyclohexanol is a secondary alcohol. What mistake have I done?

Hexanol H13C6OH and cyclohexanol H11C6OH are not isomeric.

When comparing solubilities, it is helpful to see actual numbers: 1-hexanol is soluble 5.9 g/L; cyclohexanol is soluble 36 g/L in water at 20C. Big difference, but not overwhelming. The solubilities are still relatively small, far from miscibility.

First I'd have to agree with Karl's comment that 1-hexanol ($\ce{C6H13(OH)}$) and cyclohexanol ($\ce{C6H11(OH)}$) are not isomeric. So when you're compare apples to oranges you can come to unfounded conclusions.
The gist in this case is however dependent on the "polar surface area" to the total surface area of the molecule. The molecules will have about the same polar surface area, but the cyclohexanol will have less total surface area. Thus cyclohexanol dissolves better in water.

In general, is very difficult to predict solvation some experimental models are used such as Abraham solvation equation.
I suppose that the mistake you have done is linked to the fact that you haven't considered one of the most important factors: the molecule structure.
In this case, you are comparing a cyclic compound with a linear one, if you could have a look at the potential surface of the two molecules you will see something like this:
What makes the alcohol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The ratio between hydrophilic and lipophilic areas is greater in cyclohexanol compared to 1-hexanol.


text:
Why is the carbon-fluorine bond stronger than the other haloalkanes?
What effect makes  C-F bond  stronger than  other halo-alkanes?

Related: http://chemistry.stackexchange.com/questions/21581

Possible duplicate of Is a carbon-fluorine bond stronger than a carbon-chlorine bond?

From wikipedia article of carbon-fluorine bond,

The high electronegativity of fluorine (4.0 for F vs. 2.5 for carbon)
  gives the carbon–fluorine bond a significant polarity/dipole moment.
  The electron density is concentrated around the fluorine, leaving the
  carbon relatively electron poor. This introduces ionic character to
  the bond through partial charges ($\ce{C^{δ+}—F^{δ−}}$). The partial
  charges on the fluorine and carbon are attractive, contributing to the
  unusual bond strength of the carbon–fluorine bond. The bond is
  labeled as "the strongest in organic chemistry," because fluorine
  forms the strongest single bond to carbon. Carbon–fluorine bonds can
  have a bond dissociation energy (BDE) of up to 544 kJ/mol. (This is why) The BDE
  (strength of the bond) is higher than other carbon–halogen and
  carbon–hydrogen bonds. For example, the molecule represented by
  $\ce{CH3X}$ has a BDE of 115 kcal/mol for carbon–fluorine while values
  of 104.9, 83.7, 72.1, and 57.6 kcal/mol represent $\ce{C–X}$ bonds
  to hydrogen, chlorine, bromine, and iodine, respectively.


text:
Hydrogen Bond Without Hydrogen
Surely, by the same logic, any atom with a moderate electronegativity (carbon, phosphorus, silicon) can form a polarise covalent bond with an atom with a high electronegativity (fluorine, oxygen). Giving that carbon, phosphorus or silicon a partially positive charge (just like hydrogen), allowing it then to form a 'hydrogen bond' with another atom with a high electronegativity (fluorine, oxygen). Is it called a hydrogen bond because it most commonly occurs with hydrogen? Or is it that a hydrogen bond with carbon, phosphorus or silicon is instead called a carbon bond, phosphorus bond and silicon bond? Since you don't need a hydrogen to make an atom high electronegativity partially negatively charge, surely those atoms around atom would be paritally postive to balance the charge?

I'm sure there's far more to it, but I think one of the major contributors to hydrogen bonding is the unusually short donor-hydrogen-acceptor distances, allowed by the anomalously small size of the hydrogen atom and the absence of multiple substituents on the bridging hydrogen atom. Given that the attractive electrostatic potentials in a molecule are very sensitive to the distances involved ($ 1/r^n$ for high n, e.g. n=6), then the greater proximity allows for much deeper potential wells and thus more robust bonding.

Seems like you want https://en.wikipedia.org/wiki/Halogen_bond - a non-covalent interaction involving halogens

See http://chemistry.stackexchange.com/questions/35488/is-hydrogen-bonding-a-type-of-dipole-dipole-interaction - H-bond is only partially a dipole-dipole interaction.

Lithium bonding, too. Lithium acts as a compact electrophilic center much like protic hydrogen.

Just looking at the electronegativities, there seems to be no reason why atoms such as boron, carbon or silicium — which all have similar electronegativities to hydrogen — should not undergo hydrogen-bond–like intermolecular interactions.
There is one key difference that separates hydrogen from all the other nonmetals that readily form 2-electron-2-centre (i.e. standard) bonds at ambient temperature and pressure: the lack of any core electrons.
Atoms like boron and carbon (or worse: silicon) have a low-lying 1s atomic orbital that is generally not considered to take part in bonding. However, its spherical structure surrounds the nucleus and effectively shields its positive charge from the outside world. Hydrogen, on the other hand, has a basically unshielded nucleus consisting only of a positively charged proton.
The positively charged hydrogen nucleus can now interact well with other orbitals on other atoms (typically part of other molecules). Any in-between core electrons will repulse the electrons that the electropositive element is trying to interact with. Hence, the stabilisation is greatest in hydrogen’s case.

In fact hydrogen bond is not a bond but a kind of dipole-dipole interaction so it is an inter-molecular force. Indeed these kind of interactions occur also with other atoms but not with the same intensity.
The hydrogen bond is due to the fact that there is a covalent bond between an hydrogen atom and an high electronegativity atom such as O, F or N. What happen is that because of the small size of the hydrogen atom the dipole is highly concentrated in a small volume, and this results in the high directional strength of the hydrogen bond. With bigger atoms the bond would not be so strong.

To be honest, I am unable to explain why hydrogen is in a special category when it comes to potential bonding, it just is. (The quantum mechanics would tell you that it does hydrogen-bond, and a thorough analysis of the QM might shed some light on it - if you want to go there. Personally, I've always been comfortable with just saying that nothing else comes close to the non-covalent (intermolecular) bonds it can participate in. Two other notes:1. Chemistry is "all about" electron density. Sure, it's quantum mechanical wave function probability distributions, but its all about electron density changes. Electronegativity is a convenient term to "invoke" to "explain" the effects of electron density on changes in bonding. This is very true. But, electronegativity is the "wrong" term. Technically, electronegativity is the property of the element in its neutral ground state. That is Co° and Co(+3) have the same electronegativity...even though we KNOW that the electron density is vastly different around them (both can participate in bonding with organics (organometallics)). So, the term is misused probably as often as it is not (if not more). Especially by beginning students (and their teachers!). I see no real harm in it - except when kids (or adults) go to the electronegativity tables and claim that say C in methane has a different electronegativity than C in ethane or formic acid. It's a slippery slope. 2. Not too long ago (10 years or so, couldn't (I hope!) have been 20), a paper was published showing that halogen bonding was a "thing". It remains to this day to be controversial. The Wikipedia article on h-bonding says it typically has bond strengths of 1-5 kcal/mole. The Wikipedia article on halogen-bonding says the bonds range between 5 and 180 kJ/mole. What's wrong with this picture?!! Well, 1 kcal ~ 4.2 kJ. So, 180 kJ is ~43 kcal. There's no f**king way halogen bonding is 8 times stronger than h-bonding. It's nonsense. X-bonding is so weak, that there's real debate about the usefulness of the concept (if it isn't significant then there would be no use in invoking it). My point is, separating out hydrogen-bonding for special consideration has proved itself to be a USEFUL thing. This is just not so with C, P, Si, O, or F (although the latter two often participate in h-bonding). It's just not very useful to consider say the Si in Me2C=Si interacting strongly with the O in Me2C=O  While if you take a look at the difference in Electronegativity between them (using the tables) you'd not come to that conclusion, I think. Hence: not useful.


text:
How to determine accurate mass of beta-carotene from beta-carotene extract using spectrophotometer?
In this video, a beta carotene extract(the first yellow fraction) is obtained from $0.5$ g of presumably raw spinach. The extract contains beta-carotene dissolved in organic solvent (9 to 1 hexane to ethyl acetate solution). My question is how to determine mass of beta-carotene contained in this extract using spectrophotometer. I am doing a research that requires quantification of beta-carotene in specific mass of spinach but my school does not offer advanced equipment such as HPLC.

The video Beta-Carotene from Spinach from the Eastern Kentucky University Chemistry Lab describes the extraction and purification of $\beta$-carotene by flash chromatography, a method developed by W. Clark Still in J. Org. Chem., 1978, 43, 2923-2925.
Note that this is not HPLC, but preparative column chromatography (on silica) under pressure. It's an awesome method every organic chemist should be familiar with.
Typically, you would run the (flash) chromatography, collect your material in test tubes, check the composition of every test tube by thin-layer chromatography (TLC), combine identical fractions and evaporate the solvent (on a rotary evaporator). This will give you the total mass of each fraction.
In order to have a closer look at these fractions, prepare solutions of defined concentrations in n-hexane and examine them in a UV spectrometer. In n-hexane, the molar extinction coefficient $\epsilon$ of $\beta$-carotene is $139500~\mathrm{L\,cm^{-1}\,mol^{-1}}$ at $\lambda = 451~\mathrm{nm}$.
Supposed that your solutions are not too concentrated, their absorbance will obey to the Lambert-Beer law.
$$E_\lambda =  \epsilon\cdot c \cdot d $$
This means that the absorbance is proportional to the concentration of the absorbing species and that the concentration can be calculated from the absorbance $\mathrm{E_\lambda}$, the thickness $d$ of the cuvette and the molar extinction coefficient $\epsilon$.


text:
Why does thermodynamics only apply to macroscopic systems at equilibrium?
A text says:

The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Thermodynamics is not concerned about how and at what rates these energy transformations are carried out but is based on initial and final states of system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or it moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature fo not change with time for a system in equilibrium state.

Why do the laws of thermodynamics not apply when the system is microscopic or not at equilibrium but in intermediate stage? How can we say that thermodynamics cannot tell anything about rate and method of energy transformations?

https://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics

You have a reaction $\ce{A <=> B}$. You have one molecule of $\ce{A}$ in a closed container. Does the reaction quotient have any meaning here?

Where is that text from? Especially the sentence starting "Laws of thermodynamics apply ..." is utter nonsense.

The mathematics of equilibrium systems are far simpler than others
It isn't strictly true that thermodynamics doesn't apply to systems that are not at equilibrium only that the simple theorems and formulae of equlibrium thermodynamics only apply to systems at equilibrium.
The reason why most discussion and most teaching is about those equilibrium systems is because the mathematical results are far simpler and far easier to teach. They are also widely useful even with systems that are not exactly at equilibrium. Chemists can understand the key features of many reactions, for example, by understanding the equilibrium thermodynamics of the reaction (like the heat of formation of the products). Exceptions can be explained by understanding kinetic barriers in the reaction pathway.
But when details of non-equilibrium behaviours are required, the behaviour of many systems is so complicated there are no easy rules to teach or to talk about.
Non-equilibrium thermodynamics is actually a very large field and key results have won Nobel prizes (Prigogine won the 1977 Nobel in Chemistry for work on key aspects of it). Arguably a great deal of the complexity of the universe and life itself is the result of the complexity and structures that can arise when systems are not at equilibrium. So you could even argue that our ability to understand any thermodynamics is a result of the complexity the can arise from non-equilibrium systems. If the subject is that complex, there are not going to be many simple formulae or generalisations to teach or talk about.
And that is probably the real reason why some people think thermodynamics only applies to systems at equilibrium: anything else is too complicated to talk about so educators and textbook writers tend to avoid the subject.

I'd agree with Li Zhi that this is a very broad question. Let me make a few points and see if they clear up the confusion.
Macroscopic systems - The gist here is that thermodynamics is based on statistics so you need a large enough system (i.e. enough molecules or atoms) for statistics to apply. Thus you can't get the Maxwell-Boltzmann distribution from 3 molecules of gas sealed in some container. You need millions of molecules - a macroscopic system.
initial and final states of system - The foundations of thermodynamics works with the energy difference between some starting point and some end point.
Think of it sort of like rolling a rock downhill. At the top of the hill the rock has some potential energy. Part way down the hill the rock stops and has a lower potential energy. So the difference in potential energy can be calculated by the height difference.
Naively using the rock and hill analogy, thermodynamics says that rocks roll downhill not uphill. Also in order to move the rock uphill you have to input energy into the rock-hill system.
Dynamics - Since thermodynamics is about the energy difference between the initial state and the final state, (basic) thermodynamics doesn't deal with chemical kinetics. But (basic) thermodynamics and chemical kinetics can be combined to become a specialized field of study such as non-equilibrium thermodynamics.

So without chemical kinetics just thermodynamics, is it always valid at equilibrium only ?

Going back to my rock-hill analogy, the difference in potential energy depends only only the height of the rock. The rock is at "equilibrium" only when it is still.

Why is it valud at just equilibrium what is problem with non equilibrium state ? Has it got to do something with entropy ,S

I'm oversimplifying this, but consider the rock. If I want to measure the "total energy" of the rock while moving I need to know how fast  the rock is moving the slope of the hill, the direction of the rock, the friction between the rock and the hill and so on. So it is a much more complicated problem. But is easy to measure the difference in potential energy of the still rock at two different heights on the hill.

So thermodynamics works by starting with some "standard state" and then calculating the energy differences from that standard state to some other state. Chemists could do that 100 years ago. It isn't practical to measure thermodynamic properties in an absolute sense since it isn't practical  to start with all substances at absolute zero.

Even three (or thirty) particles would perfectly follow thermodynamics, IF you managed to keep them together and isolate them from the rest of the world, including their macroscopic container. ;-)

Your questions are really too broad to answer in this forum. One of the strengths of Classical Thermodynamics is that it is "path independent". As your textbook stated, this means it is primarily focused on the starting and end points and not any intermediate states. This could be looked at as a weakness, since it doesn't help much distinguish between alternative paths, or it could be (and is) looked at as a strength since your answer/prediction isn't dependent on (doesn't rely on) the intermediate states (path). In other words, it gains predictive power by limiting itself to path independent states.
There's a lot you can know about a process, given only its initial and final conditions. One of the things this allowed was the development of Classical Thermodynamics BEFORE the discovery of the atomic nature of matter! Just think about that! Before we even knew about atoms, we could predict what the energy balance (not to mention entropy) was of a particular reaction!
Another part of your question states Thermodynamics "doesn't apply" to non-equilibrium systems. First, there's a whole sub-category of Thermodynamics which does deal with systems far from equilibrium, if you're interested. Second, think about what you're asking. You should know (if you think about it) that nothing, nothing is "at" equilibrium...yet. Give it another 10E15 years. So if you actually think that Thermodynamics can't be applied until the Universe comes to thermal (and energetic) equilibrium, you're being foolish. It does apply to systems "near" equilibrium – which is enormously, stupendously useful (if it didn't it would be virtually useless, imho).
The way I look at it you give up the details (the intermediate steps) because otherwise the results would depend on them and if they did then we'd have to know what those steps (states or mechanisms) are in order to apply Thermodynamics – and every time we were wrong about the intermediate steps (which is quite often, they're much harder to pin down because they are occurring on much shorter time scales and are (by definition) much less stable) we'd be wrong about the Thermodynamics – we exchange the intermediate details with more powerful predictions about results.
It's like you asking why the Ideal Gas Law only applies to point particles without volume or with no interactions between them. Short answer: it applies, but not perfectly, to real world gasses. Likewise, thermodynamics applies, but not perfectly, to real world systems near equilibrium.


text:
Reaction enthalpy
In an experiment, we measured the temperature difference in a reaction of $2.5\ \mathrm g$ of $\ce{Na2SO3}$ with $25\ \mathrm{ml}$ of $10{-}13\ \%$ $\ce{NaClO}$ in a calorimeter with $300\ \mathrm{ml}$ of water.
Now I am to calculate the enthalpy of the reaction, via the thermodynamic data provided as well as via through the experimental results.
Using the provided data:
$$\Delta H_\mathrm f[\ce{SO4^2-(aq)}] = -909\ \mathrm{kJ/mol}$$
$$\Delta H_\mathrm f[\ce{SO3^2-(aq)}] = -636\ \mathrm{kJ/mol}$$
$$\Delta H_\mathrm f[\ce{ClO-(aq)}] = -107\ \mathrm{kJ/mol}$$
$$\Delta H_\mathrm f[\ce{Cl-(aq)}] = -167\ \mathrm{kJ/mol}$$
$$\Delta H/\mathrm{mol} = \Delta H(\text{products}) - \Delta H(\text{reactants}) = 333\ \mathrm{kJ/mol}$$
Number of moles of the products $0.0397150\ \mathrm{mol}$
$$\Delta H(\text{reaction}) = \Delta H/\mathrm{mol} \times 0.0397150\ \mathrm{mol}=
13.22510096\ \mathrm{kJ}$$
Using the measured temperature difference:
$$q(\text{reaction}) = C(\mathrm{cal})\cdot\Delta T = 1251.4344\ \mathrm{J/K}\times3.5\ \mathrm K = -3754.3032\ \mathrm{J}$$
$$\Delta H(\text{reaction}) = q(\text{reaction})/\ \mathrm{mol}\ (\ce{Na2SO3})= -3754.3032\ \mathrm{J}/0.01986\ \mathrm{mol} = -189062.1432\ \mathrm{J/mol}$$
Which value is the correct one?

What's the question? And what about the ~25ml water from the hypochlorite solution, and the calorimeter heat capacity?

The results are completely different. The calorimeter heat capacity is C(cal) = 1251,4344

1256 kJ/K is the heat capacity of 300ml water. This does neither include the ~25ml water from the  hypochlorite solution nor the heat capacity of the calorimeter itself. So it's not surprising that the result differs significantly from the expected value.

So it seems that I have problems with understanding the term "heat capacity" - I thought the amount of water in the calorimeter was all I needed and I wasn't aware I had to account for the Hypochlorite as well. Thanks for the hint, I will have another look.

Units added - I have experimentes with different options, but would think that this could only account for a comma shift, but not entirely different values.

@AnnaSarah It still seems rather an odd coincidence that the heat capacity of the calorimeter approximately corresponds to the heat capacity of 300 ml water. Was the value of 1251.4344 J/K (with this unusual number of significant digits) given to you, or did you determine it experimentally (how?)?

The provided values for the enthalpy of formation look good. In the CRC Handbook, I found the following values for standard conditions of $T=25\ \mathrm{^\circ C}$ and $p=100\ \mathrm{kPa}$: $\Delta H_\mathrm f[\ce{SO4^2-(aq)}]=-909.3\ \mathrm{kJ/mol}$, $\Delta H_\mathrm f[\ce{SO3^2-(aq)}]=-635.5\ \mathrm{kJ/mol}$, $\Delta H_\mathrm f[\ce{ClO-(aq)}]=-107.1\ \mathrm{kJ/mol}$, and $\Delta H_\mathrm f[\ce{Cl-(aq)}]=-167.2\ \mathrm{kJ/mol}$

Note that the your starting material is not in the aqueous phase. As such, your thermodynamic data for the reactants does not apply.

I get 460 kJ/mol using your data, after making an assumption. Not too far off from 333 kJ/mol. Here's how we get there.

First law of thermodynamics

If we pretend that your calorimeter is an isolated system (and the calorimeter constant helps us handle the flaw in this assumption) and that no work is done, then the sum of all heat transfers must be zero. You started here, but it looks like you forgot the heat absorbed by the water.
$$q_\mathrm{reaction} + q_\mathrm{calorimeter} + q_\mathrm{water} = 0$$

$q_\mathrm{calorimeter}$

You calculate the heat absorbed by the calorimeter correctly.
$$q_\mathrm{calorimeter} = C_\mathrm{calorimeter} \Delta T = \mathrm{ \left( 1251\ \dfrac{J}{K} \right) \left(3.5\ K\right) = 4379\ J}$$

$q_\mathrm{water}$

Water requires $\mathrm{4.184\ J/g}$ to increase its temperature by $\mathrm{1\ K}$. You have approximately $\mathrm{325\ g}$ of water from the two solutions.
$$\mathrm{q_{water} = ms\Delta T = \left( 325\ g \right)\left(4.184\ \dfrac{J}{gK}\right)\left(3.5\ K\right)= 4759\ J}$$

$q_\mathrm{reaction}$

The reaction must have produced $\mathrm{-9138\ J}$. When you divide by the number of moles you give is $\mathrm{-460.3\ kJ}$. Huh. Wrong sign. Are you certain about your $\Delta T$? Are you sure you calculated it by $\Delta T = T_\mathrm{final}-T_\mathrm{initial}$ and not as the unsigned difference between the larger and smaller values? If I assume your temperature change was negative, I get the correct sign.

I am certain about the temperature, which was rising, giving a positive ΔT. I believe that the calorimeter is not to be considered after all, just the water, but I am not sure.


text:
Can a first degree carbocation form if a methyl shift is possible?
I was going through a reference book when I stumbled upon this question
Is that even possible?

In ethanol with heating? Seems highly unlikely.

@Zhe So it's fine for me to consider this as a false answer?

The question is: ‘can the product be observed experimentally?’ If the answer to that is ‘yes’, then the mechanism the book proposes is practically the only plausible mechanism, unless you call a concerted rearrangement and dehalogenation mechanism a different one. (Meaning the primary carbocation is quenched before it is formed.)

FYI, there is an error in the direction of the curved arrow for the last step. The electrons go toward oxygen, not hydrogen.

In this case, I think it's quite slow, but there are no competing mechanisms, so you're just playing the waiting game. It's not a practical reaction though.

@Zhe I personally don't find it possible even theoretically, I mean there's no evidence of a 1 degree carbocation

Personally, I would never, ever propose a primary cation for these kinds of mild conditions, but I don't have access to the literature any more, so I can't do a search to confirm.

The cation is produced when a methyl group migrate from C-2 to C-1 as the C-Br bond is broken. The simultaneous migration of alkyl group and departure of leaving group to form a tertiary cation  is therefore faster and more favourable than formation of primary cation due to simple loss of leaving group.
The products observed are those that result from further reaction of the so formed tertiary cation. The mechanism mentioned in the pic you've attached is incorrect, 1° carbocation intermediate is not formed at all.
Several comments correctly mention that formation of primary carbocation is not favourable. So, it is important to note that methyl shift and departure of leaving group in this case are simultaneous processes - primary carbocation is not formed at all!
P.S.
Also, I'd like to quote some evidence I found for a very similar reaction of 1-bromo 2,2-dimethyl propane:

Above is the solvolysis reaction of 1-bromo 2,2-dimethyl propane.
Source: Reactive Intermediates in Organic Chemistry: Structure, Mechanism, and Reactions - By Maya Shankar Singh (Wiley Publishers)


text:
Comparing SN2 reaction rate for phenacyl chloride and methyl chloride

KI in acetone undergoes $\mathrm{S_N2}$ reaction with (1) and (2). Compare the rates of the reactions.

Like many kinds of these questions there are many factors involved. One is the steric effect, and, since steric effects plays a central role in inversion of reactant and there is a very bulky group attached to (2), I think (1) should be more reactive. Also, (2) will serve as a better $\mathrm{S_N1}$ substrate as the carbocation generated will be resonance stabilised.
And so using these two reasons I thought (1) is more reactive than (2), but, according to the answer, (2) is almost 500 times more reactive than (1), and I am not able to see the reason for this mammoth difference.

That carbocation is not resonance stabilised.

Yeah, the carbocation is not resonance stabilised. Try drawing resonating structures (if you can). If you face it anyways, you will end up with a positive charge on Oxygen. Moreover you already know that Halogen exchange reaction goes with SN2, so the carbocation formation is pointless (no SN1).

If you are given this question in an exam it's really easy to identify which is more reactive by hybridization. As in (1) hydrogen is already stable that refers to its not much reactive. we already know something which is stable is less reactive and vice versa then looking at the second compound is not stable hence would be more reactive than one.

First, let me point out that a rate difference of 500 is really not that large.  There are solvolysis reactions with relative rate differences on the order of $\mathrm{10^{10}}$ or greater (1).
The transition state for an $\mathrm{S_{N}2}$ reaction involves hypercoordinate (or hypervalent) bonding.  The transition state is, more or less, a trigonal bipyramid structure and the bonding at the central carbon atom ($\ce{Nu-C-X}$) involves a 3 center-4 electron bond (see the above link and links therein for more detail on this concept). With 5 electron pairs being shared by this central carbon, anything that removes electron density from the central carbon atom will prove beneficial and accelerate the reaction.

$\mathrm{S_{N}2}$ Reaction Pathway
In your molecule #2, a benzoyl group is attached to the central carbon involved in the $\mathrm{S_{N}2}$ process.  You are correct that such a group is larger than a hydrogen and should decrease the reaction rate due to steric effects. However the size of the benzoyl group is not exceptionally large and the steric effect will likely be small. More importantly, the benzoyl group will also exert an electronic effect. The central carbon in an $\mathrm{S_{N}2}$ reaction is $\mathrm{sp^2}$ hybridized [e.g. leading to a trigonal bipyramid structure as mentioned above] and has a $\mathrm{p}$ orbital.  This $\mathrm{p}$ orbital can interact with the adjacent carbonyl group through resonance.  This resonance effect will tend to remove (delocalize) electron density from the central carbon atom.  Also, the carbonyl group will inductively remove electron density from the central carbon atom.  These electronic effects will remove electron density from the central carbon atom thereby accelerating the reaction as observed.
(1) For example, the relative rates of solvolysis of 7-tosyloxynorbornane and anti-7-tosyloxynorbornene in acetic acid are $\mathrm{1:10^{11}}$

I Think I have got you (almost) ...I have just one problem with your statement "However the size of the benzoyl group is not exceptionally large and the steric effect will likely be small" .

As I think Steric effect will not be small after all there is the benzene ring and a carbonyl group both hugely obstructing the attack of  Nucleophile...

But then whenever I come across these questions I always feel like we are using the creationist method(https://i.stack.imgur.com/bS6S0.gif)

Funny and it's easy to see how folks might view things the way the cartoon suggests, but to actually publish a paper a lot of experiments are done to rule out \ rule-in various hypotheses.  As to the size of the $\ce{-CO-Ph}$ group, we can look at something called A-values which are used as a measure of the size of various substituents.  The A-value for a methyl group is 1.7, it is 3 for a larger phenyl group.  I can't find a value for $\ce{-COPh}$, but $\ce{-COMe}$

is only 1.17, $\ce{-COCl}$ is 1.25.  All of these values are smaller than the A-value for a methyl group and I doubt that $\ce{-COPh}$ would be much different from $\ce{-COMe}$ or $\ce{-COCl}$.  So I would expect $\ce{COPh}$ to be sterically rather "small".

Thanks for the great answer ... Interestingly I found the same (but broader) Question (answered by yourself) so should we mark my question as duplicate http://chemistry.stackexchange.com/questions/27635/comparing-sn2-reaction-rates

Can we, in fact, generalise that the more electron-deficient the carbon atom is in the halogenoalkane, the more reactive it is towards SN2?

TanYongBoon Probably, but I'm sure there are many exceptions due to sterics, etc.

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


text:
What is this device? (Constant level device?)
Found this in my lab on a random shelf (water quality lab). It is old enough that nobody knows where it came from. I think it's a constant level device. It has in- and out-flow connections, as well as electrical plugs for some kind of monitoring equipment. The only writing is on the glass, which says "FSTREEM [ASTREEM?] MADE IN ENGLAND". The inside float looks like it is meant to float up and down, while the thin metal wire keeps track of its level via a small electric recorder. What is this thing called, and where can I find more info?

Thanks!

FiStream  https://www.google.com/search?q=old+seismograph&rlz=1C1CHFX_enUS511US512&espv=2&biw=897&bih=873&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiF_eO6z6bRAhVB5GMKHTmTBhkQ_AUIBigB&dpr=1#tbm=isch&q=FSTREEM+glass+&imgrc=_

I sent FiStream an Email with a picture.

Thanks Joseph. Knowing that it is a Fi-Streem product, I called/emailed them and got an answer from their friendly support tech. From his email response: [This] is a "Switch Level Device, front" that is used in conjunction with an older series Fi-Streem (Fisions) Still, which is no longer manufactured. The part number for this item is, 06756."


text:
Does salt on the snow and ice covered roads and sidewalks make the nearby air temperature cooler?
I  noticed when it snows and salt is melting the snow on the roads in the city, that the air temperature feels a lot cooler than the temperature indicated on the thermometer.
A 40 degree fahrenheit temperature feels like 30 degrees. Is it the moisture from melting snow or rain that makes the air temp feel ten degrees cooler or is it the salt brine (melted snow or ice  and salt?  Is it similar to adding salt  to the ice in an Hand powered ice cream churn to make the cream container colder to make ice cream?

As water melts due to adding a salt (AKA melting point depression), it absorbs energy from the environment and the temperature of the salt/ice/water mixture does, indeed, drop. Whether that is sufficient to cool the air above the ice perceptibly, though, is a matter of opinion.
Another effect that might make the air feel colder is increased humidity, particularly as the salt slush is splashed into the air by vehicles.
You could perform an experiment yourself to determine the air temperature above two containers, one with fresh snow and the other with snow and salt. If you do, please post your results as an answer!

It depends on the exactness of your thermometer and your definition of ‘nearby’.
Since adding salt to ice causes the overall mixture to assume a liquid phase rather than the previous solid phase, melting enthalpy is required to liquefy the ice/salt mixture. This melting enthalpy is typically supplied by drawing heat from the surroundings, i.e. due to the melting everything gets colder. This, naturally, also draws heat from the surrounding (close) air. However, there are no direct effects on the surrounding air.
You can read more about the effect of salt added to ice in this question.

No. It merely reduces the melting temperature of the ice. If you've ever noticed, salt water doesn't freeze until a much lower temperature than regular water. It is because of the same reason.


text:
Why do the left- and right-handed enantiomers of a chiral molecule react differently?
So I was researching about thalidomide when a thought struck to me. Why does the left enantiomer react differently than the right? They have the same elements and structure.

Try shaking somebody’s left hand with your right hand. And then try shaking their right hand with your right hand.

Presumably, you mean they interact differently in the body leading to different biological activity. Please update your question if this is so.

With other chiral molecules, the transition states are necessarily diastereomeric. Those transition states thus have different energies, which implies that rates will be different. Also, the products with other chiral reactants are diastereomeric, so equilibrium mixtures will be different.


text:
Could XY crystal be a special case of X2YZ crystal?
There are some types of crystals we call them XY type (i.e. NaCl) or X2YZ type?
Could XY type be a special case of X2YZ type by assigning Y=Z?

Sometimes it can, sometimes it can't. That depends on how different the crystal positions of Y and Z are.

Could you give me some example about why cannot?


text:
Combustion of non-hydrocarbon organic compounds
When we learned about combustion, we learned about the combustion of hydrocarbons and compounds with only hydrogen, oxygen, and carbon. However, how would combustion work with compounds that contained nitrogen or other compounds? For instance, what would the combustion of an amino acid look like in a formula?

Usually, you get nitrogen or nitrogen oxides.

Simultaneous elemental analysis of organic carbon, hydrogen, and nitrogen is classically accomplished by combusting the material under a stream of $\ce{O2}$, usually in the presence of a catalyst to ensure complete combustion, at a temperature of 950-1050$^o$C.  Carbon, hydrogen, and nitrogen are quantitatively converted to $\ce{CO2}$, $\ce{H2O}$ and $\ce{N2}$, respectively.  This is a common method in particular for determining nitrogen concentration in amino acids.


text:
Is the effectiveness of chemical catalysts affected by temperature the same way enzymes are?

Enzymes do not need high temperatures to function while metal catalysts tend to need a bit of an energy kick to get going.

Enzymes will denature, or break, if heated up too far beyond their optimum temperature (for most around 40 degrees, although some bacterial enzymes can work at 100 degrees).

Do chemical catalysts also have an ideal operating temperature?
I am not asking if the rate of the reaction itself speeds up when a chemical catalyst is used. Rather, if the chemical catalyst becomes more effective at certain temperatures.

One small puntualization: we are not talking about "optimal temperatures" in terms of efficiency for the enzymes, but of "limit-temperatures": since enzymes are complex (and fragile) molecular structures, high temperatures tend to break them.
By the way, there are, indeed, optimal temperatures for chemical catalysts.
Just think about the simple reaction of (water), sulphuric acid and ethanol.
If you want diethyl ether, you should keep the temperature at around 130°C. If you want ethene, you should increase the temperature until 160°C. Below 100°C, the catalyst has small or no effect.

Ah right.  The way the Rate vs Temp curve (http://i.imgur.com/9BiTJk7.png)  decreased after a peak made me think it was just an optimal point. So is the gradual decrease of the curve actually showing that more and more enzymes breaking/denaturing after a certain limit? And thanks for the info about chemical catalysts.

Also, is there a reason why chemical catalysts would decrease efficiency at high temperatures? It kind of makes sense that biological ones would. But I am not too sure why a metal catalysts would decrease after a certain temperature. Or are you just saying that metal catalysts have a minimum, not maximum temperature?

This is a somewhat complex question to answer because even your ‘simple’ enzyme example isn’t as simple as it looks. Let me discuss enzyme optimum temperatures first.
For any reaction that is exergonic — and that explicitly includes enzyme catalysis — an increase in temperature raises the reaction rate. This is due to at least two microscopic features:

higher temperature means more and faster movement thus a higher probability of collisions
each collision has a greater chance of being energetic enough to supply the required activation energy

This is condensed in van ’t Hoff’s postulate that an increase of the reaction temperature by $10~\mathrm{^\circ C}$ leads to an increase of the reaction rate by a factor of $2$ to $4$. Therefore, we would initially expect all enzymatic processes to be faster with increased temperatures.
However, the enzyme itself is a highly fragile system that is held together a good deal by noncovalent interactions. These can break apart under elevated temperatures causing the well-balanced enzymatic structure to break down (‘denature’)  into a seemingly random protein chain. This also breaks the catalytic centre meaning that the enzyme can no longer catalyse efficiently. Taking these two effects together, we arrive at the optimum curve of enzyme catalysis.

Chemical catalysts come in a number of shapes and varieties. Consider for example:

the catalyst (mainly platinum as I heard at school) in your car’s exhaust
the complex catalytic oxide system in the Haber-Bosch process
transition metal catalysts such as $\ce{[Pd(PPh3)4]}$
acid catalysis, i.e. dissolved $\ce{H+}$ is the catalyst
organic catalysis
…

Most of these will have some equivalent of a denaturation temperature; e.g. once platinum metal (or palladium, if on charcoal for typical organic hydrogenations) melts, it can no longer act as a catalyst for the reaction in the way it could in solid state. Thus, there should be a ‘denaturation equivalent temperature’ for most chemical catalyst systems at some point. Some compounds — the Haber-Bosch catalysts jump to mind — will, however, probably melt at such high temperatures that you would need to live on the sun’s surface for that to happen. Therefore, we could say that no limit exists for all intents and purposes.
Catalysts such as tetrakis(triphenylphosphane)palladium(0) need to dissociate at least one ligand to be catalytically active. An increase in temperature should enhance that dissociation. Therefore, many homogenously palladium-catalysed reactions are indeed sped up by a moderate increase in temperature. However, side reactions also speed up.
Organocatalysts and simple proton catalysis should be among the most temperature-insensitive — the former because all bonds should be covalent, the latter because it is just a reactive, activating species that cannot be poisoned. However, these two require a medium in which they act; the boiling point of the solvent is the highest possible temperature for their reactions.

Thus basically: yes, all chemical catalysts have some maximum useable temperature somewhere, but for most it does not matter for practical intents and purposes. Also, there are no simple models to explain this like in the enzyme case, hence it isn’t typically taught.


text:
If propane with specific parameters were to combust in open air, what is actual volume of CO2 produced
Lets say we have a sample of propane gas having a volume of 2.70 L at standard room temperature and 1.75 atm. This gas is exposed to our air. The equation that governs this reaction is:
$mC_3H_8 + xO_2 + 3.76xN_2 \rightarrow yCO_2 + zH_2O + dN_2 + eO_2$
Here $m,x,y,z,d$ and $e$ are coefficients. I want to find out what the volume of carbon dioxide given this information. The trouble I'm having is finding out what pressure the air is and I strongly suspect that I would use the sea level or 1 atm and the volume of nitrogen and oxygen with which propane reacts with.

Rephrased homework question? Anyway, you are going about this the wrong way, thinking much too complicated.
If you combust m moles of propane, how many moles y of CO2 do you get? Now remember that moles and liters of ideal gas are totally proportional, and you have the answer.


text:
Refractive Index (RI) variation with Temperature & disagreement between two standards
I am comparing two different international standards for the same organic compound (clear and colorless liquid) and for the refractive index one standard says

1.531 to 1.534 (@ 20 °C)

The other one says

1.526 to 1.527 (@ 27 °C)

Now the ranges don't even overlap.
Couple of  questions:

Would the change in testing temperature (20 °C vs 27 °C) be enough to explain this difference in standards? Would an increase in Temperature typically reduce the RI or increase it? By what percent per degree change in Temperature?
Any other testing intricacies for RI that I ought to be aware of? Is there a methodological / protocol difference that may explain this?
Is this a difference large enough to worry about? The difference in ranges is only about 0.3 %. What is the typical least count / accuracy of RI measurements in the lab?

Are the refractive indices specified at the same wavelength? That alone could explain the difference.

@JonCuster Good point. Let me check the text of the standards. So far as I saw there wasn't explicit mention of a wavelength. Is there an assumed default? One standard was the US Pharmacoepea (USP-39) so quite an authoritative document.

Usually at 589 nm, the sodium D2-line

The refractive index of a pure liquid varies with the wavelength of the light (this dispersion causes the splitting of white light into rainbow colours) and with the density and thus temperature of the liquid (this causes the schlieren patterns when heating a liquid on a hot plate).
The section “Index of Refraction of Liquids for Calibration Purposes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. lists values for several organic liquids at various wavelengths and temperatures.
Reference values close to the values given in the question can be found for toluene:
$$\textbf{Index of refraction of toluene}\\
\begin{array}{lll}
\hline
&\lambda/\mathrm{nm} &T=20\ \mathrm{^\circ C} &T=25\ \mathrm{^\circ C} &T=30\ \mathrm{^\circ C} \\
\hline
&667.81 &1.49180 &1.48903 &1.48619 \\
&656.28 &1.49243 &1.48966 &1.48682 \\
&589.26 &1.49693 &1.49413 &1.49126 \\
&546.07 &1.50086 &1.49803 &1.49514 \\
&501.57 &1.50620 &1.50334 &1.50041 \\
&486.13 &1.50847 &1.50559 &1.50265 \\
&435.83 &1.51800 &1.51506 &1.51206 \\
\hline
\end{array}$$
The uncertainty of the refractive index is estimated as $\pm0.00003$.
For example, the refractive index measured for toluene using the characteristic wavelength of a sodium-vapour lamp $(\lambda\approx589\ \mathrm{nm})$ at $T=20\ \mathrm{^\circ C}$ is $1.49693$. When increasing the temperature, the refractive index is decreased to $1.49413$ $(-0.19\ \%)$ at $T=25\ \mathrm{^\circ C}$ and to $1.49126$ $(-0.38\ \%)$ at $T=30\ \mathrm{^\circ C}$.
The range of values given in the question corresponds to a decrease between $-0.26\ \%$ and $-0.52\ \%$ when the temperature is increased from $T=20\ \mathrm{^\circ C}$ to $T=27\ \mathrm{^\circ C}$.

Electromagnetic waves are scattered at electrons, and refraction is interference of scattered waves.
The RI usually goes down with temperature, because it is proportional to the volume electron density in the substance.
So, yes, that looks OK with your two substances, if they are indeed identical. The absolute difference should be proportional to the thermal expansion of the material, 0.3% in seven centigrades sounds reasonable for an organic liquid (thermal expansion coefficient of e.g. ethanol $1.4\cdot 10^{-3}/$K, glycerol $0.52\cdot10^{-3}$/K).

For solids, such a large volume change in only seven degrees would be pretty extreme (normal thermal expansion coefficients are given in units of $10^{-6}$/C).

@JonCuster Right, tnx. However the expansion coefficient of a liquid rises by typically two decades at $T_g$ .

And the OP said 0.3%, not 1.3%, my typo.

good points on liquids, particularly near transition points.

@JonCuster I think the expansion coefficient is typically rather constant between phase transitions. Water is afaik very unusual in that respect.


text:
Primary carbocation in addition of HBr to ethylene?
All mechanisms of the addition of a halogen acid to ethylene (ethene) I can find involves a primary carbocation as the intermediate ion.

However, primary carbocations are generally unstable and unlikely to form. This can indeed explain why the first step is the rate-determining step, but does primary carbocations really form in the process? Is there no other plausible explanations?

What conditions do you consider? For example, bubbling ethene through a polar solvent containing HBr, or a gas phase reaction between HBr and ethene? If you have HBr in a polar solvent, you get H+ that could start the reaction with ethene; however, if you have HBr in the gas phase, you don't get any H+ ions.

You can always consider the cation to be a bridged hydro-ene ion.

The ethyl carbocation is a non-classical carbocation.  See this earlier answer for a description of the ethyl carbocation.

There is quite an evidence that carbocations are involved in this process, and that no concerted pathways exist for this transformation.
While the case of ethyne may be difficult to analyse, lets consider 3-methylbut-1-ene:

If you consider the regular Markovnikov's rule to predict the product, you would probably get this:

While in actuality, the major product is actually this:

You can't really explain why that happens without accepting the existence of carbocations and the following mechanism:

This shows that carbocations are involved in the addition of $\ce{HBr}$ to an alkene.
Note: Since this question is about carbocations, I'd like to mention George A. Olah, a reputed chemist who studied carbocations by preparing them in superacid environments, where they were stabilized and studied using spectroscopid techniques. It was his research that led to the conclusion that carbocations have a planar structure. Have a look at his wikipedia page for more information.

P.S.: I hope you don't mind me resizing your images a bit.

@BerryHolmes I actually wanted someone to do that! Thank you! Do you mind teaching me how to resize the images?

No worries, and here you go: https://meta.stackexchange.com/questions/25051/resizing-an-image-in-a-post


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Is Delta H = Delta U for bomb calorimeter or Delta H = Delta U + Delta n(g)RT?
As we know volume change in bomb calorimeter is zero. So if we use $\Delta H = \Delta U + p\Delta V$ then we get $\Delta H = \Delta U$. But in some reaction there is change in number of moles like $$\ce{NH2CN(g) + 3/2O2(g) -> N2(g) + CO2(g)+ H2O(l)}$$ so here $\Delta H = \Delta U + (-0.5)RT$.

First and foremost your $-0.5$ should have units of moles (i.e. should be $-0.5~\mathrm{mol}$). Don't leave out units. They're important.

$$\Delta H = \Delta U + p\Delta V$$

If you were not taught what conditions this equation holds true, shame on your teacher. If you were taught it, you should probably try to remember that this holds for constant pressure.
In general
$$\begin{align}
H &= U + pV \\
H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
&= (U_2 - U_1) + (p_2V_2 - p_1V_1) \\
\Delta H &= \Delta U + \Delta (pV)
\end{align}$$
Only under constant pressure $p_2 = p_1 = p$ does this simplify to
$$\begin{align}
\Delta H &= (U_2 - U_1) + (pV_2 - pV_1) \\
&= \Delta U + p(V_2 - V_1) \\
&= \Delta U + p\Delta V \\
\end{align}$$
If this is not true (and think about it, if your volume is constant, then what do you think is happening to the pressure inside the calorimeter?), then you may use the ideal gas approximation
$$p_2V_2 = n_2RT_2;\quad  p_1V_1 = n_1RT_1$$
and if you further assume constant temperature $T_2 = T_1 = T$ (which is probably not true, but it is close enough to the truth since the heat capacity of a calorimeter is usually large leading to a very small difference between $T_2$ and $T_1$), then
$$\begin{align}
\Delta H &= (U_2 - U_1) + (n_2RT - n_1RT) \\
&= \Delta U + (n_2 - n_1)RT \\
&= \Delta U + RT\Delta n
\end{align}$$
The further approximation is that the non-gaseous species do not contribute significantly to the $\Delta(pV)$ term, so $\Delta n$ here can be identified with $\Delta n_g$, the difference in the number of moles of gas.
TL;DR
$$\color{red}{\textbf{Learn when the equations you are using are appropriate to use.}} \\
\large\color{red}{\textbf{Don't just blindly use them!}}$$

I wish I could frame the lines in red and put them up on a banner on the top of this site...

While I agree with your 'red' comment we do not help ourselves when teaching and often use a variety of notation, e.g. such as G, $\Delta G$ and dG or create equations out of 'thin air' as $dU=(\partial U/\partial T)_VdT+(\partial U/\partial V)_TdV$. As an example, consider a change at constant volume then $\Delta V =0$ and so by the first law $q=\Delta U$. But $\Delta U$ is a function of state and q a path variable so how can this be? The answer  is that the path is fixed so q is now a path function and should be changed, say, to $q_V$ . So the possibilities for confusion are many.