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How to interpret spectroscopic data to obtain the structural formula?
I am currently struggling with an assignment question regarding structural determination using both $\ce{^1H}$ and $\ce{^{13}C}$ NMR data.

The first thing I have done is calculate my double bond equivalences which I believe is 3. I believe the IR data corresponds to a $\ce{C=O}$ bond which is supported by the number of DBEs. As it states this is a cyclic compound I believe that the there is a ring containing a double bond and a carbon double bonded to an oxygen which accounts for the 3 DBEs.
Putting together this information which I believe to be correct with the spectroscopic data above is where I am really struggling.
I am quite confident in saying that the 6.8 triplet is due to an aromatic carbon with two hydrogens on the adjacent carbons so I have a fragment there that I believe is correct but I cannot get anything more than that.

Your calculations of the double bond equivalents seems correct, and there is also another hint towards a $\ce{C=O}$ double bond: the corresponding $\ce{^{13}C}$ chemical shift.
Also, while it says you have $\ce{^1H}$ decoupled $\ce{^{13}C}$ data, this is clearly wrong, because they’re showing you multiplets. The $\ce{^{13}C}$ data is $\ce{^1H}$ coupled data, so you can use the information there to deduce the number of carbons with which number of hydrogens.
A third hint given to you is the amoun of signals: You see signals that add up to six carbons and eight hydrogens, meaning there is no symmetry in your molecule.
The 6.8 hydrogen could, by its chemical shift, be aromatic, but you do not have enough DBE’s for an aromatic ring (you would need at least 4). Do you know other types of protons that give chemical shifts in similar ranges, and are consistent with both your DBE count and the structure element(s) you have already found out?
Further ideas to think of:

You have certain integrals on your hydrogen signals, especially one that stands out (not the $6.8\,\mathrm{ppm}$). What could this integral mean, and how does it affect the rest of the structure?
Make a list of the carbons you have and the amount of hydrogens attached to them.
How can you turn all that into a cyclic compound?
Once you have, you can check if the multiplets of the hydrogen signals can be fitted to that structure. Hint: You seem to have no H–H couplings other than across three bonds. (So no long-range couplings.) And no H–C couplings other than one bond length.

 Right, in this first spoiler tag I’m just gonna state the different signals. Check the second one, if you really want to know the solution.
 The carbon at 209.6 (singlet) is most likely not an aldehyde but a ketone. If it had a proton connected, it would need to be a doublet.
 The singlet at 1.8 is a methyl group, belonging to the carbon at 12.1.
 The triplet at 6.8 needs to be next to a double bond (strong low-field shift). The corresponding carbon needs to be 157.9 (doublet).
 Since it’s a triplet, it couples to two hydrogens. There is no other hydrogen on a double bond, but there is a triplet and a quartet left. One of these two must couple to the proton integral one at 6.8. Luckily they are both integral 2. A little further reasoning tells us, it must be the quartet, because the other triplet (2.4) has to couple to two protons, which have to couple back, therefore no other possibility.
 There is another carbon at a double bond (142.1) and it has no hydrogens attached to it (singlet). It needs to have the methyl group on it.

 The structure I deduced is 2-methylcyclopent-2-en-1-one.

So far I think I have it that the (1.8 ppm singlet) could be in a position where it is at the end of a carbon chain because I believe it looks like a methyl group? Because it is a single does that mean it is likely part of a fragment such as CH3-(C=O)-C-? But the C13 data definitely has a shift that indicates an aldehyde so I don't know how the two different NMR data fits in. Also so as you have suggested I have made a list of my carbons and their attached hydrogen but I think I have something wrong because I cannot see how this can be cyclic. I will keep trying though thank you!

@Michael Are you sure that the $\ce{^{13}C}$-data implies an aldehyde? And yes, 1.8 ppm is definitely a methyl group, but remember that you have a cyclic molecule …

Sorry I had to have a quick look but I'm a bit unsure so now I have had a look at the structure and can better understand the steps working backwards thank you. I think I need to practice lots more of these and probably start off on basic ones. Thank you very much, I will try to solve more problems like this using the method you have posted! :)


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Water reacts or doesn't react?
I know that when you have $\ce{HCl}$, $\ce{H2O}$, and $\ce{NaOH}$ that you produce $\ce{H2O}$ and $\ce{NaCl}$.
However some people say water does react and so would think that this 2 step reaction is correct and the 1 step reaction is wrong.
$\ce{HCl + H2O + NaOH -> H3O+ + Cl- + NaOH -> 2 H2O + NaCl}$
However others say that water does not react and think that the 2 step reaction is wrong and the 1 step reaction is correct.
$\ce{HCl + NaOH -> H2O + NaCl}$
I personally think that both of these happen and so water consumed = water produced and thus there is no change in volume. So does water react with the $\ce{HCl}$ via the 2 step reaction or is it just a spectator molecule waiting around until there is more water and salt?

What happens if you blow a stream of dried hydrogen chloride gas on  dried pellet of sodium hydroxide?

The answer to this depends on how the reagents are mixed. For one, you can have the reaction without water, as pointed out by @Klaus in the comments. If you blow dry HCl on dry NaOH, they will react without involving water in the reactants.
$$\ce{HCl + NaOH -> H2O + NaCl}$$
This is probably not the case though, and if you're dealing with solutions, then yes, water will react. If you start by adding the HCl to pure water, you'll have
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
This representation is prefered over the simple dissociation $\ce{HCl -> H+ + Cl-}$ because it shows that you don't have "free" protons in solution. You could actually even say that this reaction has already happened since, considering you're using hydrochloric acid it's already dissolved, and this step would be just a dilution.
If you start with the NaOH solution, you could argue that water does not react since
$$\ce{NaOH -> Na+ + OH-}$$
but that only happens in the liquid state (usually in solution). Also, you can't ignore the ions solvation (which is one, if no the most important factor which affects the solubility of the sodium hydroxide salt) and equilibrium
$$\ce{H2O + OH- <=> OH- + H2O}$$
And what this means is that you don't have fixed hydroxil ions and water molecules, so you can't simply ignore the solvent and assume it's just a spectator molecule waiting around.
To write the two step reaction, it's not a good idea to do it in one line, because as you can see on the first step you have NaOH on both sides, which is something that doesn't make much sense. It's better to do it like this:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
$$\ce{H3O+ + Cl- + NaOH -> 2H2O + NaCl}$$
It's usually a good idea to represent salts in their ionic form when dissolved, which would be
$$\ce{H3O+ + Cl- + NaOH -> 2H2O + Na+_{(aq)} + Cl^{-}_{(aq})}$$

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Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid?
Why is the melting point of p-toluenesulfonic acid monohydrate higher than that of anhydrous p-toluenesulfonic acid?
Compare for example the melting point of benzenesulfonic acid hydrate, which is lower than for anhydrous benzenesulfonic acid.

The hydrated forms might be an azeotrope which means there's a content of water present that is difficult to remove via distillation techniques which may be how the P-toluene sulfonic acid was prepared.
The anhydrous form would require further processing.
The moisture present in the hydrate forms may affect the melting point if it were acting as an azeotrope with the P-toluene sulfonic acid, also it wouldn't be as high purity as the anhydrous form which would display the true melting point.

Hello, and welcome to Chem.SE! This answer is a bit short, so preferably it should be expanded on or turned into a comment.

Sorry its my first time on this site It said add comment on my phone I wasn't trying to say anyones wrong or anything just some input thanks for the feedback :)

Sorry, what's the relationship between melting point and azeotrope? If it likes what you said, why is  the melting point of benzenesulfonic acid hydrate lower than for anhydrous benzenesulfonic acid?

when a compound is sublimed with water present the compound can melt and sublime with water at a lower temperature that it's true melting point. This is why compounds with melting points  higher than water can be distilled with water. I learnt this in distillation theory. I'm guessing your using a melting point apparatus but I think the same thing could happen subliming a hydrate salt in a melting point apparatus. Correct me if I'm wrong I'm only meaning to comment :)

I think it depends on the the compound whether the hydrate makes the m.p higher or lower. So the azeotrope is the hydrate  salt which is the compound and water which is different to the anhydrous form possibly because it's an azeotrope mixture which won't sublime at its true melting point.

When I learnt about this it was classed as a phenomenon because high m.p compounds could be distilled with water at a much lower temperature whether the missile or i mmiscable. The same phenomenon could be occurring with m.p op of the compounds in question .

The solubility of p-toluenesulfonic acid is $670 \mathrm{g}.\mathrm{L}^{-1}$ so the solution which contains this acid and water is an azeotrop.
Then the melting temperature depend if you keep pressure or temperature constant in the laboratory.
If you take a Handbook I have not here, I think you'll find a lot of interesting things about that.

How do the solubility of chemical influence  whether it is an azeotrop and its melting point?

If there are not miscible you have not an azeotrop but an heteroazeotrop. The solubility is just to justify you have an azeotrop.

This is what I mean by a similar thing happening as what occurs when distilling. A heteroazeotrope is probly it.

This applies to miscible and immiscable substances.

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Acid catalysed hydration of allene
In the acid catalysed hydration of allene, acetone is obtained as the major product. This is because of electrophilic addition of $\ce{H+}$ across a double bond to give a vinyl carbocation, and subsequent addition of water and then tautomerism to yield the ketone. What I'm confused about is that why does not an allyl carbocation form, which would be resonance stabilised?(leading to  allyl alcohol?)

related http://chemistry.stackexchange.com/questions/2774/resonance-in-carbocation-derived-from-allenes

Good question!  As the following figure illustrates, there are 2 ways we can protonate an allene.

(image source)
The top line in the drawing starts by protonating at the central allene carbon.  This generates a very unstable primary carbocation.  Initially, this primary carbocation is not stabilized by the adjacent double bond.  It is not until after this carbocation rotates 90° that its p orbital is lined up with the p orbitals in the double bond and allylic stabilization can occur.  Since there is no immediate allylic stabilization when the carbocation is formed, and since it is a primary carbocation, the pathway leading to it is very high energy.
On the other hand, protonation at the terminal allene carbon (bottom line of figure) produces a relatively stabilized secondary carbocation.  Formation of a secondary carbocation is a much lower energy process.  Therefore, protonation at the terminal allene carbon is lower energy and is what will occur.
Edit: Response to OP's comment

I always thought that a positive charge appearing on sp3 carbon (which
  then becomes sp2) is more stable than a positive charge appearing on
  sp2 carbon

Generally we think of vinylic carbocations as high energy.  That is true, but what is meant by "vinylic" carbocation is different from what we have here.  Here there is an empty p orbital.  In a vinylic carbocation the empty orbital is not a p orbital, but rather something like $\ce{sp^2}$ or $\ce{sp}$ hybridized.

is the secondary carbocation more stable because of inductive effect,
  the rotational barrier in the primary carbocation, or both

The secondary carbocation is more stable for the same reasons that we have the  tertiary > secondary > primary carbocation stability order.

Inductive effects, $\ce{sp^3}$ and $\ce{sp^2}$ carbons are electron releasing towards an $\ce{sp}$ carbon due to electronegativity differences
Resonance effects from hyperconjugation involving hydrogens on the carbon next to the carbocation center.

OMG!! I had accepted the statement made by OP when I was told about this nature in allenes (long back). This is a wonderful explanation. +1000!

@ron I always thought that a positive charge appearing on sp3 carbon(which then becomes sp2) is more stable than a positive charge appearing on sp2 carbon(which then becomes sp)(even though sp3 carbon may be primary.). In this case, is the secondary carbocation more stable because of inductive effect, the rotational barrier in the primary carbocation, or both?

@Abhishek  I've edited my answer to address these points.

@ron Understood it all perfectly well now, thanks


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Nomenclature of complexes: Cyano vs Cyanido ; Chloro vs Chlorido
What is the correct nomenclature for $\ce{CN^-}$ and $\ce{Cl^-}$ ligands in coordination compounds?
Wikipedia says the coding name(?) as cyanido and chlorido while this suggests cyano and chloro.

Maybe the website that says chloro was created before 2006 and then wasnt updated. Thank you.

It was also cyano over here. but I couldn't add more links.

The IUPAC Red Book is the official reference for recommended inorganic nomenclature. The full change to the additive nomenclature system was made in the 2005 update:

Some new recommendations represent breaks with tradition, in the interest of increased clarity and consistency. For example, the application of the ending ‘ido’ to all anionic ligands with ‘ide’ names in additive nomenclature (e.g. chlorido and cyanido instead of chloro and cyano, and hydrido throughout, i.e. no exception in boron nomenclature) is part of a general move to a more systematic approach.

This is discussed in more detail in section IR-7.1.3 on page 112. Basically, anionic ligand names used as prefixes get changed from '-ide', '-ate', and '-ite' to '-ido', '-ato', and 'ito', respectively. Neutral and cationic ligands are usually used unchanged except for things like aqua, ammine, carbonyl, etc.

Official answer.  The "aMMine" part is one to keep in mind as I get confused between it and aMine. Thank you.


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Identifying phycobiliprotein based on wavelength
I have the absorption spectrum of a specific phycobiliprotein. Its maximum wavelength is 620.6 (nm),this is the only data I have. How can I know what type of phycobiliprotein it is?

Did you try looking it up? Many vendors sell these things and have spectral data available. For instance: http://www.interchim.fr/ft/2/28310A.pdf

Its maximum wavelength is 620.6 (nm)

I'm afraid that this is pretty much a link-only answer, but based on   Spectroscopic Properties of C-Phycocyanin and of Its $\alpha$ and $\beta$ Subunits, published in J. Biol. Chem., 1973, 248, 5679-5685 (PDF) there's a good chance that the phycobiliprotein is C-Phycocyanin.


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Lewis Structure on (CH3)2NO?
I've been taking an chemistry class online (for fun only) and one of the questions in the homework threw me a bit. Here's the wording:
"Draw the Lewis structure for a nitrogen atom attached to. two methyl groups ($\ce{CH3}$) and one oxygen atom."
OK, so I know that $\ce{NO}$ is a free radical due to the odd combined valence, i.e., the octet rule isn't satisfied. Will that carry over when the $\ce{NO}$ bonds with the two methyl groups?
Here's the sketch of what I did:

Lastly, what's the name of this compound? This is not part of the HW question; I'm just curious.

Name of the compund: Dimethyl nitroxide
Lewis structure:

The properly way to determine the Lewis structure, based on this example, is:

Total valence electrons: $4\cdot2 + 5 + 6 + 1\cdot6 = 25$
Total electrons needed for octets/doublets: $8\cdot4+2\cdot6 = 44$
Total shared/bonding electrons: $44 - 25 = 19$ (In other words, there is only nine single bonds and one electron left.)
Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 25 - 19 = 6$ (In other words, the are only 3 pairs of lone electrons.)

Furhter information on how to draw Lewis structures, please follow the link
Further information of related Lewis structures and how this knowledge is applied on real top science, I recommend seeing this reference: Moreira, G.; Charles, L.; Major, M.; Vacandio, F.; Guillaneuf, Y.; Lefay, C. and Gigmes, D. Beilstein J. Org. Chem., 2013, 9, 1589 - 1600.

In your first step, if you were just losing an electron then the middle structure would have a positive charge; but you've lost a hydrogen too - you need to factor that in.

I agree with @ron, but I also see, that the image is directly taken from the Beilstein, who does make the mistake in the first place. || I personally would like it more, if you would write out the reference with names: Beilstein J. Org. Chem., 2013, 9, 1589–1600.


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Possible ionization and fragmentation of chemicals
Is there any generic method where I can know/calculate/generate the possible ionization and fragmentation of a specific chemical, for example, hydrogen chloride and water?
This question is actually related to the mass spectrometer, where I am asked to identify the possible species/ions of a sample. Therefore I would like to know if there's any method to know (calculate) the outcome without memorising everything, for example the outcome of HCl.

Agreeing with Mithoron, we may mean at least two things when we talk about ionization. First, we may mean the conversion of neutral molecules into ions by the acquisition or loss of electrons, for example in a redox process or in a mass spectrometer: $$\ce{ CH4 -> CH4+ + e-}$$ Second, we may mean reaction of a neutral molecule (often with the solvent) to produce ions: $$\ce{ HCl + H2O -> H3O+ + Cl-}$$ The word fragmentation in your title makes me think you want the first type of ionization, but the example of HCl in water makes me think you want the second type. Could you clarify?

Do you want to predict outcome or analyze it? In case of HCl there's not much to predict - you get HCl+ and it fragmentates into H and Cl+ or H+ and Cl atom. MS is rather used to identify unknown more complicated compounds.

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Separate precious metals from catalytic converter
I am looking to separate the small quantities of platinum and palladium in a catalytic converter. However since these are very inert metals I am unsure how to proceed about doing so. Melting the converter is an option I have considered but I am unsure whether I will be able to reach high enough temperatures in order to obtain a decent amount of metal, so I am leaning more towards performing reaction.

As its phrased right now, it's unclear as to what you are asking. What exactly is your question?

@JohnSnow I believe the question is if it is possible to chemically/physically separate the various metals present in a catalytic converter. The OP has raised the issue of chemical inertness and melting points being hard to attain.

For the sake of this being a Q & A site, I would still request a rephrasing so it's clear what is being asked, rather than making assumptions.

By what methods could I separate the precious metals in a catalytic converter?

How are you going to source a catalytic converter? The value is such that manufacturers recycle them so they are not just discarded. And it is illegal to steal them from someone else's car (though often done when scrap prices are high enough). And the recyclers will have the capacity and means unavailable to lone chemists.

Let me try to talk you out of it ;-)
The precious metals in a catalytic converter sit in a honeycomb ceramic that contains a lot of synthetic Cordierite. This material isn't only heat-resistant, but also quite hard.
Industrial recycling of catalytic converters typically starts with

dismantling (decanning) the ceramic monolith using guillotine shears
breaking down the ceramic to smaller pieces in hammer mills connected by a closed tubing system
milling the material to dust in a ball-mill blender

You might want to wear a filter mask during these steps.
Industrial recycling almost exclusively uses pyrometallurgical tretment (melting the metals) then.
Having ruled that out, there are two other options:

hydrometallurgical treatment
In order to separate the metals from the ceramic, you can either try to

dissolve the ceramic in hot alkaline solution in a closed reactor under pressure (to reach temperatures above 100 °C). This leaves the metals intact.
leach the metals out by using strong oxidizing acids, such as mixtures of concentrated nitric and hydrochlorid acid. You have the metals in solution now.

What are your plans for personal protection during these steps and for the waste water management?
biomobilization
There is some ongoing research to replace the unpleasant conditions of hydrothermal treatment with microorganisms that can leach the solid metals and store them. As an example, have a look at an article by Helmut Brandl, Stefan Lehmann, Mohammad A. Faramarzi, and Daniel Martinelli, published in  Hydrometallurgy, 2008, 94, 14-17. The authors examined, whether metals could be converted to cyanido complexes, such as $\ce{[Pt(CN)4]^{2-}}$, by cyanide-producing microorganisms. A PDF of their article, titled Biomobilization of silver, gold, and platinum from solid waste materials by HCN-forming microorganisms is available at no cost from ZORA (Zurich Open Repository and Archive). In the case of platinum, however, the results were rather disappointing.

That sounds like a great deal of opportunities for disaster!

If I understand correctly, the DIY process for Catalytiç converters, hard drive platters and gold from circuit boards and RAM are nearly identical and not very accurate at obtaining a pure product. Muriatic Acid will dissolve aluminum and many other metals, but not Platinum Group metals, Removing as much pollutant material as possible  pregame is extremely important, after the material has broken down in the probably dark green acid there will be shiny flakes of metal floating around, those are undissolved platinum group(Gold, Platinum, Rhodium, Palladium, and tungsten) metals, just filter and rinse... To separate the platinum group metals, Aqua Regia is required to break them down and different chemicals specific to each metal can be added to precipitate and filter out individual compounds that can be further reacted to obtain the metal in the valuable form, for instance, adding ammonium chloride to Aqua Regia containing dissolved PG metals will precipitate Platinum Chloride, filter it out and add another chemical to precipitate a different compound containing another metal. This is all from personal experience so I don't have anything to cite for reference, but will warn that muriatic Acid alone produces painful fumes, while dissolving aluminum and cadmium and any other metals it will produce harmful respiratory vapors so I would do this outdoors unless you have a fume cabinet, adding hydrogen peroxide to muriatic Acid while dissolving will speed up oxidation and reduce total time to dissolve, also evolves more hydrogen which tends to gather in ceilings that have 4 walls around them and explode when concentrated enough at the flip of a light switch or when near a running fan.

I guess the term muriatic acid had to be obsolete even in early 80s when I was studying at high school.

I call it like I see it, that's what it says on the bottle I get at orschelns for $10.81. I know there's other names, it's the most watered down HCL available I'm sure

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What substances form crystal structures?
Just getting interested in this stuff. People are saying I should crystallize salt, for example. What substances besides salts form a crystal structure?

Crystallize a leaf? I'm not entirely sure what you mean. Perhaps your definition of crystallization is different from the normal chemical one, which is the formation of a solid with a regular atomic lattice and long-range order. Are you trying to encase objects in a transparent solid?

I mean how you can make salt crystals, alum crystals, etc.  I'm just stupid.

Maybe to focus more on the question: what kinds of things can you make crystals from:
Typically a pure element or an inorganic or organic compound can be made in crystalline form. It might be a fine powder or small granular crystals. If grown under controlled conditions, you could get large crystals. If it is a gas or liquid at room temperature, it could perhaps solidify to a crystalline form at lower temperatures.
On the other hand, an impure, indeterminate, or polymeric material, or a mixture of compounds, very seldom will form crystals. Still, there are exceptions. Inorganic salts often crystallize from water solution as "hydrates" - with extra water molecules in the crystal lattice of the salt. Other solvents also sometimes co-crystallize with inorganic or organic substances. Neutral compounds can co-crystallize together as clathrates, such as methane hydrate - a crystalline structure containing both methane and water that forms at high pressures and low temperatures.
Even a complex or high molecular weight substance, such as an enzyme or virus particle can form crystals. In the realm of nanotechnology, new materials are being engineered specifically to aggregate into crystal-type structures.

Crystallization in the chemical sense, as previously noted by Nicolau Saker Neto in comment form, is when a solid forms "with a regular atomic lattice".
Ionic compounds such as salts (e.g. NaCl, KCl) would be my first choice, but there might be others I don't know about.
The easiest method I know to do this is by supersaturating.

Fill a container (such as a cup or a beaker) with water or other suitable solvent.
Add salt (or other solute) while stirring until the solution becomes saturated and there is a noticeable amount of undissolved solute at the bottom of the container.
Heat the solution. This increases the solubility of the solute because the particles move farther apart when at a warmer temperature. This is due to the fact that heat and particle motion are linked. This allows a greater amount of solute to dissolve.
Keep stirring to make sure the solute that was previously collected at the bottom of the container is fully dissolved
Allow the solution to cool. This reduces the solubility and the solute will crystallize.

Salt is an easy example. Add more salt to water until it can't dissolve anymore, heat it, wait for it to cool, and you will have your salt crystal.

Thanks :).  What things can I make crystals with tho?

I don't understand. I listed all the materials you need: cup, solvent, solute. You can make salt crystals by using table salt and water.

My question:  what things could I replace table salt with and still get a Crystal


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Does water vapor in the air oxidize copper?
My friend is installing copper pipes in his house, which is going to be in a humid area, and he says that copper will only be oxidized by the air and the excess water vapor in the humid air won't affect it. However, it is to my understanding that since water vapor is $\ce{H2O}$, which contains oxygen, there will be more oxidation occurring, and thus more corrosion to the pipes because of more oxygen present in the air. Is this correct?

Though copper may have a very thin oxide layer after exposure in air, the green color that will eventually appear is verdigris,  "usually a basic copper carbonate, but near the sea will be a basic copper chloride" [Wikipedia]. Yes, water will speed the corrosion (not simple oxidation) of copper.

Although $\ce{H_2O}$ contains oxygen, it is there in the most reduced state $\ce{O^{2-}}$ and therefore cannot act as oxidizer. It is the oxygen in the air $\ce{O_{2}}$, which is the main and strongest oxidizing agent.
On the other hand, presence of water as a solvent can speed up some reactions. For a short overview of copper corrosion, see e.g. Why does copper oxidize. As the corrosion products form compact (or even protecting) layer, the corrosion itself is not a problem for copper, as compared to e.g. steel corrosion.


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Comparing purity grades for ethanol
I am trying to determine if the following two grades of ethanol are equivalent:

Ethanol absolute, 99.8% Normapur AR (Analytical Reagent)
Ethanol alcohol, pure ACS reagent, ≥99.5%

Information on the technical specifications of these grades would also be helpful for comparison.

I've seen the Normapur AR grade on bottles obtained from VWR/Merck. You might want to search their online catalogues to find a certificate.

I think my big question is "how are you using the ethanol?" Generally ACS reagent grade is designed for reactions, solvent, etc. So it may not be great, say, for spectrophotometric measurements.

The important aspects of reagent purity are the level of purity with respect to compound(s) X, and the purity with respect to compound(s) X required for your particular use case.
For example, if you are conducting a reaction that requires anhydrous reagents but is unaffected by trace metals, then you would want to see a certificate of analysis or equivalent showing an acceptable limit for water in the reagent. Whereas if you require sub-ppm levels of trace metals, then that's the parameter of interest to you and you probably don't care if there is 0.5% water present.
In other words, a 99.5% pure ethanol product and a 99.8% ethanol product could be drastically different in their composition with respect to your needs; i.e. the apparently slightly more pure compound could have orders of magnitude more trace metals present than the slightly lower overall purity product. Furthermore, terminology regarding reagent purity is often similar between different vendors (i.e. pure, ultrapure, milipure, etc.), but frequently has very different implications on the aspect of purity that is important to you.
Summary, TL;DR:
Although you can frequently make generalizations about the meaning of "ACS Reagent" grade or "[pick-your-prefix]pure" grade, you need to inspect either the specifications on the bottle or certificate of analysis for each reagent to know for certainty that it meets your needs. The simplest reliable source for such information is the manufacturer, most of which have online resources providing this information for their products.


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Can compounds found in concrete leach into soil against gravity?
I spent a while trying to figure out which stackexchange site in which to post this question, forgive me if I got it wrong.
A number of scholarly articles indicate that concrete can leach calcium compounds under the right conditions. Could they ever wick upwards against gravity through dry or water-saturated soil?
An example would be a raised bed of soil with wooden sides and a concrete floor.
P.S. I also wonder about plastics leaching upwards, but this is admittedly most likely a different question.

Welcome to Chemistry.SE! Seriously, I don't know. Regardless gravity, I had expected some material flow (dissolved salts) from wet soil into the concrete due to capillary forces. But again, I'm definitely no expert on that field.

Gravity is not much of a thing when you are under water.

Gravity is a small force on ions when compared to water flow and diffusion; lead ions won't sink in a graduated cylinder. But rather than ponder the drift upwards of ions from the concrete, consider the drift downwards of ions from the soil. For instance, the soil may be acidic, and will almost always have more water in it than the concrete has, so the acids in the soil will be much more mobile than the ions in the concrete, and will fairly easily drift into the concrete, neutralizing $\ce{Ca(OH)2}$ or even dissolving $\ce{CaCO3}$. Whether these calcium ions will travel upwards... well, yes, but  calcium compounds may adsorb on soil particles and travel even slower. So the soil may be as much a barrier as the concrete is a poor source of ions.
Concrete may lose some ions (especially $\ce{Ca(OH)2}$) in contact with flowing water, but even more, it tends to pick up $\ce{CO2}$ and the $\ce{Ca(OH)2}$ becomes $\ce{CaCO3}$, which is less soluble. So the soil right next to the concrete may be a little more alkaline than the soil at the top surface.


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Preparation of 6-bromo-1-hexanol
I am refluxing $\ce{HBr}$ with 1,6-hexanediol for 16 hours at $80\ ^\circ\mathrm{C}$ in toluene. How can I be sure that more than $90\,\%$ of the product is brominated only at one terminal? If the amount of double brominated product is in more than $10\,\%$ of the total yield, then the product mixture will be useless for me.

Would having unreacted 1,6-hexanediol be acceptable? If so, use it in excess.

@ringo the ° is not typeset well in math equations in mathjax, it's probably better to use ${}^\circ$.

jerepierre what I am concerned about is the by product 1,6-dibromohexane which I can't allow to be there.

In an article by Katharina C. Kress, Martin Kaller, Kirill V. Axenov, Stefan Tussetschläger, and Sabine Laschat, published in
Beilstein J. Org. Chem., 2012, 8, 371-378 (DOI), this procedure is described. From the Supporting Information, I read that the authors obtained the product in 72% yield after

column chromatography on silica gel (hexanes/EtOAc, 1:1)

I wouldn't be astonished if that's as good as it can get.


text:
How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry
I stumbled-upon this compound:

I'm wondering how to determine the configuration of the double bond. Is it E or Z? How do I decide on the priorities when the substituents can only differ in the configurations on the chiral centers?

In your example, the stereodescriptors R and S are already given. The remaining question is about the double bond which gives rise to cis-trans isomerism.
In the hierarchical Sequence Rules, we finally find

P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.

Source: Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)
Therefore, the configuration would be ‘E’ since ‘R’ precedes ‘S’.
However, the ‘E’ and ‘Z’ stereodescriptors do not distinguish between double bonds whose descriptors are reflection invariant from the geometrically enantiomorphic double bonds whose stereodescriptors are reflection variant.
Following the recommendations for enantiomorphic double bonds, the configuration is ‘seqTrans’.

I remember that when all the basic rules end up in a tie (including the ghost atoms), the following priority is used:

$R > S$
$cis > trans$
$Z > E $

So R is given higher priority and the above mentioned molecule will be considered as $E$ and not $Z$.

The authoritative article for CIP rules is Spezifikation der molekularen Chiralität. According to the article, the double bond has the E configuration. The reason for this is that R has priority over S.


text:
Is it wise to use ECPs on light atoms for electronic structure calculations?
So, I have a big organic molecule consisting of $\ce{H}$, $\ce{C}$, $\ce{O}$, and $\ce{N}$ atoms and the goal is to find its equilibrium geometry and US/Vis spectrum. The molecule is so big that I was thinking about reducing the complexity of the calculations by using effective core potentials (say, Stuttgart RLC) on light atoms ($\ce{C}$, $\ce{O}$, $\ce{N}$). However, I'm not sure how trustworthy the results would be comparing to using full-electron bases, thus, I'm looking for some reviews if there are any.

What do you mean by "big?" Is it big enough that linear scaling DFT methods would help?

@GeoffHutchison, by "big" I simply mean big enough to make my calculations too slow. TD-DFT, for instance, take few weeks. :D I obviously want calculations to proceed faster but without introducing physically horrible approximations. If it possible, of course.

@GeoffHutchison, ~300 light atoms. I tried to turn on the use of the fast multipole method in Gaussian right now and I can see a 20% decrease in computational time. But I want more! :)

you might consider using more nodes. I don't know how parallel TD-DFT is in Gaussian (or other packages) but there is obviously a significant improvement in speed when running parallel calculations on large jobs.

@GeoffHutchison, I see almost now speed-up running on 64 cores vs 32 cores in GAMESS and NWChem, ORCA simply fails to run on 64 cores. And Gaussian setup I have access to is limited to 1 node (16 cores).

I think you will find that the time savings are minimal. Modern integral screening will eliminate virtually all inter-atom core-core integrals anyway and if you only have a 1s core then the cost of evaluating intra-atom core-valence integrals will be comparable to evaluating the ECP-core interaction.

Hmmm... I did some benchmarks, of course. Each SCF cycle with Stuttgart RLC ECPs is about 60% faster that with Dunning/Hay "double zeta valence" basis set (DZV) in GAMESS (US).

Don't know the Stuttgart ECP very well.  How similar is it do DH?  Same contraction, etc.?  Was this with DFT or RHF?

Hybrid DFT, PBE0 to be more exact.

DZV has (10s, 5p)/[3s, 2p] contraction scheme for C, N, O, while Stuttgart ECP has (4s, 5p)/[2s, 3p].

OK, so it's the speed up is probably a combination of 2 things: 1. a slightly smaller basis set and 2. fewer grid points for the DFT.  My guess is that the latter is the big factor, which I didn't consider when formulating my answer.  As I recall the grid is pretty dense near the nuclei.

The ECP scheme will not bring substantial savings. If the molecule is really large (1000 atoms), you should better think of more efficient approximations, namely RI methods, RI-J, RI-JK or RIJCOSX. You can find their efficient implementation in ORCA package.
In addition, I would suggest using some more recent basis sets, namely the Alrichs def2.

Thanks, I'm aware of this RI/density fit/greed free DFT, whatever way you call it, but. 1) I'm not familiar with the underlying theory and currently have no time to do so. And I try to avoid using methods if I have no idea how they work. 2) In programs I know how to use RI-DFT is not available for functionals of my choice and I (again) currently not in the position to master a new program. 3) I do see substantial savings and the only thing I want to know is how generally good/bad results of such calculations would be.

Hello, I can see your point. The speeds up evaluation of integrals and the errors are well documented and known (much better than ECP). The wavefunction parameters are therefore also good approximation and you can use them as excellent guess to run full canonical DFT. For geometry optimizations the accuracy is perfectly sufficient (especially for large molecules). The use of RI is straigtforward, see https://sites.google.com/site/orcainputlibrary/ri-and-auxiliary-basis-sets

Now at least I know what I gonna do this weekend. :D I'll try to run Orca...

ORCA in fact is slower than Gaussian/GAMESS. Yes, it benefits from RI trick, but it did not take advantage of molecular symmetry. My molecules are highly symmetric (most of them belongs to $\mathrm{D}_{2}$ point group) and as a result ORCA is slower.

Besides, ORCA does not have a simple way of keeping symmetry during geometry optimization, so ORCA is by no means a solution.

Wildcat: In the case of highly symmetric molecules I fully agree with you, ORCA is not the best package for you. It can somehow preserve the symmetry, but does not take advantage of it during integral evaluation. The reason for this is that even highly symmetric molecules often undergo Jahn-Teller distortion, therefore the number of cases where the symmetry would be useful is very limited.

If you want to make use of symmetry and reasonable parallelization for your TDDFT, you may want to try out ADF.
ADF doesn't use ECPs, but you can freeze the core electrons.

I think the answer is fine, particularly the frozen core, but please try to avoid "advertising links."

OK, good to know. It's just that I also saw a link to ORCA.


text:
Behaviour of element 111
Is element 111 considered to act as an eka-aurum?
Being under the same column of group 11, which usually provides enough evidence for correlation of properties for an eka-element, would this element have the similar properties such as conductivity or inertness with gold?
With the understanding of the correlation between group 11 being minimal to me, is anyone able to predict perhaps the colour, ductility, or other physical properties of this element with reason?

I think http://en.wikipedia.org/wiki/Roentgenium answers it nicely.

As far as I understand, there has been very limited experimental investigation of the properties of roentgenium (Z=111), if at all, simply due to the fact that so few atoms have been produced and they decay so quickly. It is possible to do some chemistry with even single atoms in seconds, but as you might expect it's an extremely arduous endeavour and is riddled with massive error bars. For now, just the synthesis of the element seems to dominate experimental roentgenium research.
There has, however, been ample theoretical investigation of roentgenium's chemical properties, both by extrapolation of behaviour for group 11 elements and by ab initio computational methods, at varying levels of approximation. One difficulty is that roentgenium displays strong relativistic effects, which make accurate calculations harder. Of course, atomic properties are easier to predict than molecular properties, which are themselves much easier to predict than condensed phase properties. So while estimates for ionization energy for atoms may be relatively accurate, physical properties of the pure materials such as ductility or melting point are to some extent little more than informed guesses.
Here is a list of some predicted basic properties of the period 7 $d$-block elements, including roentgenium (source, p. 1691):

That source expands upon some of the general properties of roentgenium. Its electronic configuration is predicted to be $\mathrm{[Rn](5f)^{14} (6d)^9 (7s)^2}$, which is dissimilar to the configuration for gold $\mathrm{[Xe](4f)^{14} (5d)^{10} (6s)^1}$. This is because relativistic effects cause a strong stabilization of $s$ orbitals (now more accurately referred to as $s_{1/2}$ due to spin-orbit coupling) by pulling them closer to the nucleus, while simultaneously destabilising the $d$ orbitals (as well as splitting the $d$ subshell into two groups of degenerate orbitals, two $d_{3/2}$ orbitals and three $d_{5/2}$ orbitals) by pushing them away. This means complete population of the $s$ subshell becomes preferable as relativistic effects get stronger. For example, see the effect here (ibid, p. 1667) in the triad Nb/Ta/Db in group 5.
Roentgenium is expected to be a noble metal, like gold, and in fact from its predicted standard reduction potential, is more noble than gold ($\mathrm{Au^{3+}(aq) + 3\ e^{-} \longrightarrow Au^0(s), \ \ \Delta E^0=+1.52\ V}$, for comparison). That said, once the roentgenium atom is ionized to Rg(III), it can reach higher oxidation states more easily due to less stable filled $6d_{5/2}$ orbitals, and so Rg(V) compounds are expected to be more stable. Interestingly, though in the 6th period transition metals gold (group 11) shows the local "relativistic maximum" with respect to stabilization of the $6s$ subshell, in the 7th period the maximum stabilization of the $7s$ subshell shift to copernicium (group 12) rather than roentgenium. This means roentgenium is expected to be less noble than copernicium.
With respect to physical properties, not much is predicted about roentgenium. It will likely be a very dense metal, with good electrical conductivity. The question about its colour is an interesting one. Wikipedia mentions it is expected to be silvery, even though roentgenium displays stronger relativistic effects, which is likely the cause of the colour of gold. This may be related to the switch in electronic ground state between gold and roentgenium. Unfortunately the source for the claim on Wikipedia is no longer accessible.

All versions of articles on Wikipedia are available via their history (unless deleted by admins, in such case only they can see them).

@Mithoron It seems I made a small mistake. The reference for that sentence in Wikipedia is currently the same one from 2015. While the URL appears to have suffered from link rot, its DOI did and still does link to the freely accessible article.


text:
How to calculate the mass of phosgene in the equilibrium mixture of carbon dioxide and chlorine?

Consider the reaction:
$$\ce{CO(g) + Cl2(g) <=> COCl2(g)}$$
A closed $\pu{5.69 L}$ flask containing a reaction mixture at equilibrium has $[\ce{CO}] = \pu{1.8 \times 10^-6 M}$ and $[\ce{Cl2}] = \pu{6.3 \times 10^-7 M}$. Given that the equilibrium constant $K_\mathrm{eq} = 2.9 \times 10^{10}$, calculate the mass of $\ce{COCl2}$ present in the flask.

How can I solve the question above? I thought that equilibrium constants deal with concentrations and not masses, so I'm not sure how to use it to calculate a mass.

We have an equilibrium mixture for the reaction
$$\ce{CO (g) + Cl2 (g) <=> COCl2 (g)}$$
We should recognize that
$$K_\mathrm{eq} = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$
Note that the reactants and products all have stoichiometric coefficients of 1 so we can write our $K_\mathrm{eq}$ equation out in this way.
Because we are given concentrations of reactants in the equilibrium mixture, we can go ahead and plug in the values that we know and solve for $[\ce{COCl2}]$.
$$\begin{align}
2.9 \times 10^{10} &= \frac{[\ce{COCl2}]}{(\pu{1.8 \times 10^-6 M})(\pu{6.3 \times 10^-7 M})} \\
[\ce{COCl2}] &= \pu{0.033 M}
\end{align}$$
We are asked how many grams of $[\ce{COCl2}]$ is present in the reaction mixture.  Concentration and mass are related via the following equation:
$$[\ce{COCl2}] = \frac{n(\ce{COCl2})}{V} = \frac{m(\ce{COCl2})}{M(\ce{COCl2}) \cdot V}$$
or
$$m(\ce{COCl2}) = [\ce{COCl2}] \cdot M(\ce{COCl2}) \cdot V$$
where $n$ is the amount of $\ce{COCl2}$ (in g), $M$ is the molar mass (in g/mol), and $V$ is the volume (in L).
Let us plug in all of our known values and solve.
$$\begin{align}
m(\ce{COCl2}) &= (\pu{0.033 mol L-1})(\pu{98.906 g mol-1})(\pu{5.69 L}) \\
&= \pu{18.6 g}
\end{align}$$


text:
Heat and water absorption by calcium chloride
Can calcium chloride be effectively used as a desiccant to absorb both heat and moisture?

‘absorbing heat’? How does that work?

Anhydrous calcium chloride is an effective desiccant, but the absorption of water by anhydrous calcium chloride is exothermic; i.e. the reaction releases heat.
$$\ce{CaCl2 + 6 H2O -> CaCl2 . 6 H2O}\quad\Delta H = -97\ \mathrm{kJ/mol}$$

CaCl2 is also deliquescent. It pulls in so much water from the air that eventually you wind up with a concentrated solution of the salt.


text:
Explaining the photoelecton spectrum of krypton
The photoelectron spectrum of krypton is shown below.
(where the x-axis is in electron volts (eV))\
Since both the $3p$ and $3d$ shells are more than half full, based on Hund's rules the lowest-energy level is the one with the largest J, i.e. the $2\mathrm{P}_{3/2}$ and  $2\mathrm{D}_{5/2}$ states should be lowest in energy (higher value in eV). However, based on the peak intensities, this is not the ordering observed (see labeling in diagram). Why is this the case here?

I think lowest J is the lowest energy, see here for example.

This is only true for shells that are less than half full. In Krypton's case, the 3p and 3d shells are full so largest J values are lowest in energy.

Maybe the lowest joules


text:
Exocyclic vs. endocyclic double bonds in E1 elimination

What will be the major product of this elimination reaction?
The exocyclic product has more substituent groups on the alkene which should make it more stable but I have heard that exocyclic alkenes are less stable than their endocyclic counterparts. Why is this the case and is it a strong enough effect to outweigh the stability provided by a fourth alkyl substituent?
Additionally, what influence will kinetics have on this reaction? I think that the endocyclic product should be kinetically favoured because there is less steric hindrance for deprotonation of the intermediate cation at the ring position than at the isopropyl position.

Did you consider rearrangement of the intermediate carbocation?

It's already tertiary so why would it rearrange? The tertiary isopropyl cation shouldn't be any more stable should it?

The stability mainly comes from hyperconjugation, which I would expect to be stronger in the isopropyl moiety.

What will be the major product of this elimination reaction?

In the acid-catalyzed dehydration of 1-methylcyclohexanol, both 1-methylcyclohexene and methylenecyclohexane are formed in ca. an 85:15 ratio.  In your example we have two more substituents on the exocyclic double bond which will serve to further stabilize the exocyclic double bond and increase the amount of the exocyclic isomer. I don't know for sure how much it will push the equilibrium towards the exocyclic isomer, but I might guess that the product mixture would be at least 50:50, if not more of the exocyclic isomer - but I would expect a mixture.

what influence will kinetics have on this reaction?

Well you won't see the kinetics unless you go out of your way and look for it.  Factors that influence which isomer is kinetically favored include:

statistical factors - there are 4 ring hydrogens that can be eliminated vs. only 1 on the isopropyl group; this favors kinetic formation of the endocyclic isomer
alignment of the hydrogen with the p-orbital on the carbocation carbon - the hydrogen on the isopropyl group can better align with the p-orbital than those hydrogens in the ring, however the cyclohexane ring can easily distort; so while this isn't a major factor it favors the exocyclic isomer
steric factors - I agree with your analysis, but would guess that the difference is small; still sterics would slightly favor the endocyclic isomer

It seems pretty close to call, but if I had to guess I'd say that the statistical factor wins out (and the steric argument further helps) and the endocyclic isomer would likely be the the kinetic product.
But again, in acid this is a fast equilibrium; the products will quickly equilibrate and the thermodynamically controlled product distribution will be observed.  Maybe if you run the reaction at low temperature with a strong dehydrating agent ($\ce{POCl3}$) you could capture some meaningful kinetic information.

Why is the exocyclic isomer usually disfavoured? Is it purely the statistical factor or is it thermodynamically less stable and if so why?

The statistical factor will only matter if the reaction is kinetically controlled and generally, dehydrations with sulfuric acid are thermodynamically controlled.  Take a look a this useful link showing alkene heats of hydrogenation.  It shows methylenecyclohexene to be around 8 kJ/mol less stable than methylcyclohexene (about right for an 85: 15 ratio) as expected for a di- vs. tri-substituted double bond.  Comparing tetramethyl ethylene to trimethylethylene we see that the former is about 1 kJ/mol more stable.

This should push the product mix from 85:15 towards 50:50

Please correct me if I'm mistaken, but in this case we would dehydrate to give the Zaitzev product (second listed product in the OP) as the major product. Consider also that adding heat as a reagent is common place to aid in dehydration.
Give a quick glance here for a refresher on alkene stability: https://en.wikipedia.org/wiki/Zaitsev%27s_rule

Usually it would give the Zaitsev product but I have heard that the ring changes is and that the endocyclic product may be more stable


text:
Is F2 two fluoride ions or just one?
I'm attempting to write the balanced equation of the combination of lithium metal and fluorine gas. I understand that this will produce an ionic solid because it is a reaction of a metal and a nonmetal. However, I'm confused with the fluoride ion having a charge of -1 and fluorine only occurring as a diatomic molecule ($\ce{F2}$). Is $\ce{F2}$ two fluoride ions or just one?

Oxidation state of $F_2$ is 0 but in $F^-$ it has -1 oxidation state hence the  $\ce{F2 + 2e- ->2F-}$

The fluorine atom is far to reactive to exist on its own, so naturally it bonds with other atoms in its vicinity to stabilize itself. It should be noted that an atom is electrically neutral, meaning that it has an equal number of protons as electrons. A fluorine atom can share electrons with another nonmetal to form a covalent bond. When sharing electrons in a covalent bond, the fluorine is still called an atom, because it has not gained complete control over the electron that is completing its $2p$ orbital. In the $\ce{F2}$ molecule, there are to fluorine atoms.
The fluoride ion is a result of a fluorine atom completely gaining control over an electron, which is usually donated by a metal. Electrons can be transferred from a metal to the fluorine atoms to form metal ions and fluoride ions. The reaction you are talking about would look like this:
$$\ce{2Li^{(0)} +F2^{(0)} -> 2Li^{(+1)}F^{(-1)}}$$
I included oxidation numbers to show how each species has gained/lost electrons, though this would normally not be included in the formula.

$\ce{F2}$ will produce  $2\space \ce{F-}$ ions.
You can write the balanced half cell reactions as follows:
$\ce{F2 + 2e- ->2F-}$
$\ce{Li -> Li+ + e-}$
Multiply the second equation by 2 to get $\ce{2Li -> 2Li+ + 2e-}$. This is to ensure that the final equation does not have any electrons on either side.
Now add this equation with the fluorine one.
You'll get $\ce{2Li + F2 -> 2Li+ + 2F-}$ which you can write as $\ce{2Li + F2 -> 2LiF}$

Fluorine gas in the diatomic $\ce{F2}$ is essentially the the covalent pair of fluorine atoms (as fluorine is a non-metal), as seen in the diagram below:

Image source
In the reaction you describe:
$$\ce{2Li(s) + F2(g) ->2LiF(s)}$$
requires the fluoride ion be in an ionic bond with the lithium ion, twice (to balance the equation).

Image source

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

@Martin-マーチン no problems, easily fixed :)

Just pointing that out, I did not know it a month ago, so I want other people to profit from my knowledge :D And I think it's easier to read ;)

It is a very good point too - I learnt something new and I agree, it is clearer

Fluorine in its elemental form is $\ce{F2}$, a neutral molecule. Every fluorine has seven valence electrons and after formation of the molecule, there is one covalent, i.e. electron sharing bond, and the remaining six valence electrons at each fluorine can be regarded as lone pairs.
When elemental fluorine reacts with a metal it will form an ionic compound just as you stated. For this to happen, both elements involved in the reaction have to be ionised, i.e. they have to gain or lose electrons. This is called a redox reaction.
In this particular case, lithium will transfer one of the electrons to fluorine, resulting in the following equations:
\begin{align}
\ce{Li &~<=> Li+ + e-}&&|\cdot 2\\
\ce{F2 + 2e- &~<=> 2F- }\\\hline
\ce{2Li + F2 &~<=> 2LiF}\\
\end{align}


text:
Adding a polyvinyl alcohol chain onto polyethylene
For school, my teacher asked what the proper procedure/reaction would be to add a polyvinyl alcohol chain onto a polyethylene chain. Can anyone help? I'll provide additional information if needed.

Do you want to make a co-polymer? or just a mixture of the 2 polymers? Please give more details.


text:
What is the bonding type in hydrides of group 2 elements?
In one of my MCQ questions, there was a statement saying all group 2 elements form covalent hydroxides. I know it is incorrect. I know that $\ce{BeH2}$ is a covalent hydride. What about the rest? Are all of them metallic hydrides? Is it true that metallic hydrides produce the hydroxide and give off $\ce{H2}$ with water?

There is a quite extensive chapter about hydrogen and its compounds in Holleman and Wibereg's book "Inorganic Chemistry". As I only have the german version, I will briefly summarise what I read.
You are of course right, hydrogen forms salt-like structures with alkaline and earth alkaline elements, except for beryllium. Compounds like $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{RbH}$ and $\ce{CsH}$ crystallise in a sodium chloride structure. Every cation is coordinated by six hydride ions and vice versa. The compounds $\ce{CaH2}$, $\ce{SrH2}$ and $\ce{BaH2}$ crystallise in a fluorite structure. The hydride ions are tetrahedrally coordinated by four earth alkaline cations. The cations are cubically coordinated by eight hydride ions. This β-structure is persistent at higher temperatures. Below $780~^\circ\mathrm{C}$ ($\ce{CaH2}$), $855~^\circ\mathrm{C}$ ($\ce{SrH2}$), and $598~^\circ\mathrm{C}$ ($\ce{BaH2}$) it changes to the α-form, i.e. the lead dichloride structure. The magnesium compound exists in a rutile structure. Hydride ions are trigonally coordinated by magnesium ions. The cations are octahedrally coordinated by hydride ions.
However, according to the difference in the electronegativities, you would expect a predominant covalent character. The maximum is $1.3$ for $\ce{CaH2}$ and the minimum $1.0$ for $\ce{MgH2}$, which translates to a maximum $30\%$ and minimum $18\%$ ionic contribution. Maybe this is where the statement of the question originates from. Experimental findings rather support the more ionic nature of the compounds though, i.e. the found lattice energies were supported by calculations assuming a purely ionic structure.
The conclusion of this according to Holleman and Wiberg is that the elements are fused by ionic bonds with covalent contributions. The latter increases from $\ce{CsH}$ to $\ce{LiH}$ and from $\ce{BaH2}$ to $\ce{MgH2}$. The magnesium hydride is already borderline covalent.
The question about their reactivity towards water is mainly answered in the chapters for the elements itself. Magnesium reacts readily with water and releases hydrogen. Because of this it is sometimes used as a drying agent. $$\ce{MgH2 + 2H2O -> Mg(OH)2 + H2 ^ }$$
The other elements react similar. The reaction is highly exothermic, the reaction of calcium hydride releases $228~\mathrm{kJ/mol}$. It is often used as a drying agent for gasses because of this feature. Strontium and Barium also react exothermic, i.e. they release $177~\mathrm{kJ/mol}$ and $172~\mathrm{kJ/mol}$, respectively.

One explanation for the enhanced ionic bonding versus electronegativity difference: covalent bonding depends on good orbital overlap. The magnesium 3s and 3p orbitals are just a little too diffuse to overlap well with hydrogen (and with heavier alkaline earth metals the overlap is progressively worse). In both alkali and alkaline earth metal hydrides covalent bonding is less driven by electronegativity than by orbital overlap.


text:
Purification of Ethanol
In Vogel's organic book, they have mentioned that in the Lund and Bjerrum method of purification of ethanol, magnesium is added, which forms ethanolate. But they said to activate the magnesium first by iodine.
Why is the activation needed and how exactly does this activation process work in the activation of magnesium?

The refluxing of methanol or ethanol over magnesium is indeed the method of choice to dry these solvents.
As far as the iodine is concerned, this is a classical trick, which is also used to initiate Grignard rections. Is is assumed that magnesium iodide is formed, which is soluble in the alkanols. As a result, a fresh, activated metal surface is left behind.
This might even work, when the magnesium metal is covered with a very thin oxide layer.

Top answer here supports this. It reacts at the defects of the passivated surface and is soluble., thus increasing the surface. On another note, reading papers like this (German) make me very thankful that I was born almost a century later... mercury chloride vapour O.o


text:
Calculating Buffer Capacity
I completed a titration of Ammonium Acetate buffer solution and to it I added $\pu{2M}$ Hydrochloric Acid.
I measured the initial $\ce{pH}$ of the buffer solution before any acid was added and I used methyl orange as an indicator which has a $\mathrm{p}K_\mathrm{a}$ value of 3.
My independent variable was the temperature of the buffer.
I was just wondering how I would calculate the buffer capacity. Is it $\Delta\ce{pH}$/Volume?

In addition to Martin's answer, there is at least an old recommendation
by E. B. Sandell and T. S. West in Pure Appl. Chem., 1969, 18, 427-436 (DOI), which states:

Buffer capacity or buffer index. The capacity of a solution to resist changes in pH on addition of acid or base, which may be expressed numerically as the number of moles of strong acid or strong base required to change the pH by one unit when added to one litre of the specified buffer solution.

oh okay thanks!  Any suggestions how I should calculate that?

As far as I can remember from my tutorials, the formula was
$$\beta =\frac{n(\ce{H+})}{\Delta\ce{pH}},$$
meaning: How many protons have to be added to produce a change in the pH of one unit. But there might be different definitions. The IUPAC does not provide an official one.

oh okay thanks!  Any suggestions how I should calculate that?

@Sushanth Take a sample of your buffer and add a known amount of acid to it and see how the pH changes. Ideally this should be done with an pH electrode.

Perhaps a derivation of the definition I was taught. Consider the Henderson-Hasselbalch equation: $$pH=pK_a+\log\bigg(\frac{[A^-]}{[HA]}\bigg). $$ Given the definition provided by Klaus above and the fact that a weak acid's $pK_a$ is unique, the only term that could possible alter the $pH$ by an amount of $\pm1$ is $\log\big(\frac{[A^-]}{[HA]}\big).$ We decide to choose the case of the $pH$ increasing by $+1$ (it does not matter which direction---either more acidic or more basic---the reaction proceeds, we simply want the magnitude of the numerical change to be 1). If the $pH$ increases by 1, then the solution becomes more basic. Consequently, the term $\log\big(\frac{[A^-]}{[HA]}\big)$ changes by $\log\big(\frac{[A^-]+\beta}{[HA]-\beta}\big)$, where $\beta$ denotes the buffer capacity. We know this term must equal $+1$ because a perfect buffer has equimolar solutions of $[HA]$ and $[A^-]$. We obtain the following equation: $$1=\log\bigg(\frac{[A^-]+\beta}{[HA]-\beta}\bigg). $$ Solving for $\beta$ yields the buffer capacity equation $$\beta=\frac{10[HA]-[A^-]}{11}. $$


text:
Why does my professor use both the upper, and the lower, case literal of a letter to symbolize the same concept?
In one of our slides, the professor has written

$ \Delta E_{sys} = -\Delta e_{sur}$

Where $E_{sys}$ symbolizes the energy of a given system, and $e_{sur}$ symbolizes the energy of the surroundings of that system.
On a few occasions, she's used both the upper, and the lower, case literal of a single letter as symbols of a single concept.
Is there a notational convention that requires the use of an upper case literal of a letter in some circumstances and the use of its counterpart in other circumstances?

She is simply inconsistent. I would never you lower case $e$ for energy, especially in an expression where usual capital $E$ is already used for it.

Of course, there is a convention. In physics $E$ usually stands for energy, $e$ for elementary charge. You professor is inconsistent, to say the least. In my opinion using lower case $e$ for anything rather than for elementary charge is wrong by itself, and using it for energy is simply ridiculous. Besides, I don't see any reason for doing so.
P.S. During my university years I had a professor who randomly picked letters for physical quantities every lecture. :D Today I think of it with a smile, but at that time I hated both his lectures and him personally.

“lower case e for anything rather than for elementary charge is wrong” – that's a bit of a strong statement, seeing as loads of people write $e^x = \exp(x)$. The lowercase e is also used for unit vectors, and I don't see anything bad about using it in other contexts as long as it's properly disambiguated.  — In the context of the OP's question, the use $e$ seems indeed questionable though.

@leftaroundabout, well, mathematical constants has to be printed in Roman (upright) type, so the base of natural logarithm has to be  $\mathrm{e}$ and not $e$. The only issues is that it is difficult to do on a paper/board, but it is another story. Vectors has to be printed in bold or with an arrow on top (especially useful on a paper/board) so unit vectors has to be $\boldsymbol{e}{i}$ or $\vec{e}{i}$ but not $e_{i}$.

In France we are free to use upper are lower, I saw $E$ or $e$ a lot of times and also $\Delta_r g$ sometimes...

@Shadock, well, it's too bad of you, French. :D You'd better follow the International Standards.

Me I use the upper letter but my teachers... bad teachers :P

@Wildcat One may be advised, in certain style guides, to set mathematical constants in upright letters and so on but "have to" is clearly false.

@DavidRicherby, I'm not talking about "certain style guides", I'm referring to an International Standard (ISO 80000). Everyone is strongly advised to follow it, so "have to" might be exaggeration but only a slight one.

I believe you are right, that the teacher is sloppy. Used in the correct way, there can be a difference, see intensive and extensive properties.

The quantities and equations used with the International System of Units (SI) are called the International System of Quantities (ISQ). Many of the quantities as well as their recommended names and symbols are listed in the International Standard ISO/IEC 80000.
They are considered almost universally accepted for use throughout the physical sciences.
According to this standard, the recommended symbol for the quantity ‘energy’ is an upper case $E$. (The quantity symbol is always written in italic type.)
Furthermore,

When, in a given context, different quantities have the same letter symbol or when, for one quantity, different applications or different values are of interest, a distinction can be made by use of subscripts.

This means, if you need more than one energy, you may for example write $E_1$ and $E_2$ or as in your example $E_\text{sys}$ and $E_\text{sur}$.
(Note that only subscripts that represent a physical quantity or mathematical variable are printed in italic type. All other subscripts are printed in roman type. Thus, writing $E_{sys}$ or $E_{sur}$ would be wrong.)

+1 for mentioning the ISO Standard, though, I would not pay a lot attention to descriptive subscripts not being in Roman type since the equation was probably written by hands on a board.

@Wildcat Sure but the typesetting advice still applies to the text of the question.

I tend to just fix these when I see them - I've seen sinx before, some people just don't know how to put things in roman.


text:
Solute mole fraction for hydrated magnesium sulfate dissolved in water

Problem: 70 grams of pure, hydrated $\ce{MgSO_4}$ is dissolved in 90
  grams of $\ce{H_2O}$. If the mole fraction of the solute is 0.039,
  find the molecular formula of the solute.

My flawed attempt at a solution:
Let the solute be $\ce{MgSO_4.xH_2O}$. In $70~\mathrm{g}$ of solute, the amount of solute is $$\frac{70}{120+18x}~\mathrm{mol}.$$
Likewise, in $\pu{90 g}$ of solvent, we have $$\frac{90}{18} = 5~\mathrm{mol}.$$
Plugging these values into the formula for mole fraction and equating it to 0.039, I thought I would get the value of $x$ as 7. Instead, I seem to be getting approximately 12.
Could somebody please help me solve this question? In general, when we have a hydrated solute in an aqueous solution, do we include the water of crystallisation in the total mass of available water while calculating the number of moles of water?

I think the idea here is that once the solid  $\ce{MgSO_4.xH_2O}$ dissolves, its water content becomes part of the solvent. Does that help?

@NicolauSakerNeto I couldn't understand Sir. Could you please show me through calculations?

You've done a good job.  You just need to take into account the following equation: $\ce{MgSO_4.xH_2O(s) \xrightarrow{water} MgSO_4(aq) + xH_2O(l)}$. That is, the total amount of solvent is not only the water initially added, but also has to take into account the water that was freed from the solid when dissolving. Thus, $n_{solvent}=5+\mathrm{something}$. If you can figure out what that something is, you'll be able to solve your problem. Try it out a bit more, if you don't manage I can provide some additional guidance.

@NicolauSakerNeto Sir, but then shouldn't the mass of the solute reduce from 70 g? Originally, $MgSO_4.xH_2O$ had a mass of 70 g and as part of it (the water of crystallistion) is lost to the solvent hence the mass of the solute should reduce. Therefore, the number of moles of the solute should also reduce.

The "solute", really, is only the $\ce{MgSO4}$, as the crystallized water in the hydrate becomes indistinguishable from the solvent once dissolved. The mass of  $\ce{MgSO4}$ is indeed smaller than the initial amount of $\ce{MgSO_4.xH_2O}$, but by the stoichiometry of the dissolution in my previous comment, the number of moles of  $\ce{MgSO4}$ is the same as $\ce{MgSO_4.xH_2O}$.

Alright, let’s take this one step at a time.
First of all, you want to determine a molecular formula, and among your known data you have a mole fraction. This suggests it’s a good idea to work with moles to solve the problem. As such, the first step is to convert the amount of water added and magnesium sulfate hydrate used. Calculating the number of moles of water added is easy:
$$90\ \mathrm{g}\ \text{water added} \times \frac{1\ \mathrm{mol}\ \ce{H2O}}{18\ \mathrm{g} \ \ce{H2O}} = 5\ \mathrm{mol}\ \ce{H2O}$$
The number of moles of $\ce{MgSO4.$x\,$H2O}$ is unknown because even though you know how much mass was dissolved, you don’t know its formula completely. However, $\ce{MgSO4}$ contributes $120\ \mathrm{g\ mol^{-1}}$ to the molar mass, and each additional molecule of water in the hydrate stoichiometry adds a further $18\ \mathrm{g\ mol^{-1}}$, so that the molar mass of $\ce{MgSO4.$x\,$H2O}$ is $120\ \mathrm{g\ mol^{-1}} + x \times 18\ \mathrm{g\ mol^{-1}} = (120+18x)\ \mathrm{g\ mol^{-1}}$. With this molar mass, the number of moles of $\ce{MgSO4.$x\,$H2O}$ is:
$$70\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O} \times \frac{1\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}}{(120+18x)\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O}} = \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}$$
From the definition of mole fraction:
$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{total}}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}$$
Now, one is tempted to write the following and solve for $x$:
$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}\ }{\frac{70}{120+18x}+5 \ }=0.039$$
However, this equation is incorrect. The problem is that the following process was not taken into account:
$$\ce{MgSO4.$x\,$H2O(s) \xrightarrow{water} MgSO4(aq) + $x$\ H2O(l)}$$
In other words, when the magnesium sulfate hydrate dissolves in water, the water molecules contained in the solid are released and become an indistinguishable part of the solvent; the final mixture is a simply a solution of the solute $\ce{MgSO4}$ in water. Therefore, the number of moles of solvent is not simply $5\ \mathrm{mol}$ as calculated at the start, but rather a larger value, which should include the number of moles of water released when the solid dissolves.
From the dissolution equation, we see that $1\ \mathrm{mol}$ of $\ce{MgSO4.$x\,$H2O}$ generates $1\ \mathrm{mol}$ of $\ce{MgSO4}$ and $x\ \mathrm{mol}$ of $\ce{H2O}$ from crystallized water. However, we know the amount of moles of $\ce{MgSO4.$x\,$H2O}$. This means the following:
$$n_\text{solute}=n_{\ce{MgSO4}}=1 \times n_{\ce{MgSO4.$x\,$H2O}}= \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4}$$
$$n_\text{solvent} = n_\text{added water} + n_\text{crystallized water} = n_\text{added water} + x \times n_{\ce{MgSO4.$x\,$H2O}} = \bigg(5 + \frac{70x}{120+18x}\bigg)\ \mathrm{mol}\ \ce{H2O}$$
Now we can finally apply these values to the mole fraction equation:
$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}}{\frac{70}{120+18x} + \bigg(5 + \frac{70x}{120+18x}\bigg)}=0.039$$
Solving for $x$ provides the answer $x=7.03$, which means there are seven moles of water per mole of $\ce{MgSO4}$, and thus the formula for the solid is $\ce{MgSO4.7H2O}$, representing magnesium sulfate heptahydrate.


text:
Why is the gram no longer a function of the atomic mass number?
According to Wikipedia,

... the definition of the gram is not (as of 2011) mathematically tied to that of the atomic mass unit

Why did we change that?

Unless I misunderstand the question, the gram was never defined that way.
For historical reasons the SI unit for mass is an oddity:

The base unit is not the gram, but the kilogram. This is the only case where the base unit has a prefix. The gram is defined as $\frac{1}{1000}$ of a kilogram.
The kilogram is the only remaining SI unit that is based on a man-made object, the International Prototype of the Kilogram.

At its 25th meeting in 2014, the General Conference on Weights and Measures (CGPM) has (again) declared its intention to abandon this atificial standard and link

the definitions of the kilogram, ampere, kelvin, and mole to exact numerical values of the
  Planck constant $h$, elementary charge $e$, Boltzmann constant k, and Avogadro constant $N_A$ […]

(Source)
One of the possible approaches for the redefinition of the kilogram would be a side-effect of the redefinition of the Avogadro constant with in the Avogadro project.

The decision on the redefinition was made in 2018 and will be in effect May 20th 2019.


text:
Is there a term that denotes 'molecules and molecular constituents'?
In my notes, I've used phrases such as 'atom, particle, or molecule' more often than I'd like. For example, Wikipedia defines atomic mass as

the mass of an atomic particle, sub-atomic particle, or molecule.

I remember concise definitions more easily, so a single word for 'molecules and molecular constituents' would be useful.
Is there such a term?

As a general rule for these definitions, have a the Coloured Books of the IUPAC (International Union of Pure and Applied Chemistry)! In this case, you can find the definitions for molecule and molecular entity in the IUPAC Compendium of Chemical Terminology, known as the Gold Book.

@KlausWarzecha That's a good answer. Care to put it as an answer? I'd accept it.

This is pretty much link-only, but anyway...
When searching for official definitions, nomenclature, or recommended procedures, the organisation providing all these is the IUPAC (International Union of Pure and Applied Chemistry)!
An IUPAC book series, known as the Colour Books, provides the normative definitions and guidelines on various topics. These are the results of what chemists in the different committees in the IUPAC divisions have eventually agreed upon.
To my knowledge, all these Colour Books are available online.

As kindly mentioned by Loong in a comment, the online versions are not necessary the latest ones! The online version of the IUPAC Nomenclature of Organic Chemistry is from 1993. In addition, a 2004 draft copy is legally available.

Organic chemists will often refer to the Blue Book (Nomenclature of Organic Chemistry), inorganic chemists to the Red Book (Nomenclature of Inorganic Chemistry), etc.
As far as your question is concerned, the Gold Book (IUPAC Compendium of Chemical Terminology) provides the definitions for molecule and molecular entity.

This is far more than a 'link only' answer - great explanations of the IUPAC online resources

Unfortunately, the current version of the Blue Book (2013) is not legally available online yet. You can only find the old recommendations (1979 and 1993).

@Loong Yeah they still want you to buy the hardcover book or the online access. Not a very nice thing, since it is very expensive and most organic chemists/ research centers should have access to it.


text:
Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered
I am a second year undergraduate and have come across this when studying stereochemistry and conformations and have been unable to find a satisfactory answer.
I know that $\ce{sp^3}$ carbons prefer anti-periplanar and gauche conformations due to the overlap of filled orbitals with empty orbitals but cannot seem to find an answer for why $\ce{sp^2}$ carbons prefer an eclipsed conformation. I was possibly thinking about the need to avoid $\pi$ electrons however I am not sure.

I think you are asking about preferred conformations for $\ce{sp^3-sp^3}$ single bonds and $\ce{sp^3-sp^2}$ single bonds.
You are correct that the staggered (or gauche) conformation is preferred in simple alkanes.  This is generally explained by noting that

adverse steric interactions are minimized in the staggered conformation, and
stabilizing hyperconjugative interactions are maximized

(From: “Isomers of Organic Compounds”, section 24.2 from the book Principles of General Chemistry (v. 1.0).)
In the case of the $\ce{sp^3-sp^2}$ single bond as found in propene, toluene, acetaldehyde, etc., the eclipsed conformation is now the more stable.

(Michael Di Maso, Jared Shaw, the UC Davis ChemWiki by University of California)
Again both steric and electronic arguments can be advanced to explain this preference.  The steric argument says let's use the bent model to describe the double bond.

Now replace one of those ethylene hydrogens with a methyl group and note that in the eclipsed conformation (build a model if you can) none of the methyl hydrogens eclipse the hydrogen on the adjacent olefinic carbon or any of the bent bonds.  In the staggered conformation all 3 of the methyl hydrogens are eclipsing something, the 2 bent bonds and the $\ce{C-H}$ bond on the adjacent olefinic carbon.  The basis of the electronic argument can be found here.  Personally, for me, the steric argument is easier to understand.


text:
How to determine if a compound is likely to become a solid at room temperature?
Reasoning: My guess would have to be that it has to do with the strength of the intermolecular forces between each molecule. For example, molecular compounds that have a very low molar mass like $\ce{CO}$ and $\ce{CH4}$. Both of theses molecules have a very small molar mass, a very low London dispersion force, and thus a very low boiling point. This in turns makes it easier to break the bonds that hold each individual atom and more than likely cause them to turn into a gas in normal conditions.
Then when you look at ionic compounds like $\ce{MgCl2}$ and $\ce{NaCl}$, it is more likely for these compounds to become solid because of the strong intermolecular force that they have (ionic bonds).
Is my reasoning correct and is there something else that I should consider?

Are you specifically interested in crystalline solids or just solids in general? The title implies the former but there doesn't seem to be any mention of it in the question.

@bon Reading my entry again, it would seem that you're right. I will change it to solids because is what I initially had in mind.

You are pretty much on the mark with looking at the strength of intermolecular forces. If you wanted to make a general comparison, the tendency to form a solid can be arranged by the strength of the intermolecular forces.$$\text{London Forces}<\text{Dipole-Dipole}<\text{Hydrogen bonding}<\text{Metallic}<\text{Ionic}<\text{Network Covalent}$$
Then, to compare within a group, you need some way of measuring the strength of that type of intermolecular force.
A more mathematical way to look at this if you have empirical values for a compound is to look at the $\Delta G_\text{fusion}$ at room temperature. $\Delta G_\text{fusion}=\Delta H_\text{fusion}-T\Delta S_\text{fusion}$, where $\Delta H_\text{fusion}$ is negative (since it will free up energy in forming the solid) and $\Delta S_\text{fusion}$ is negative (since, whether it started as a gas or a liquid, forming a solid will reduce entropy). Letting $\Delta G_\text{fusion}=0$ (reaction at equilibrium), we find that $T_{\text{fusion}}=\frac{\Delta H_\text{fusion}}{\Delta S_\text{fusion}}$. This allows us to see that when a compound becomes solid is a trade-off between the enthalpy released from forming bonds and the energy needed to reduce entropy at a given temperature.


text:
Why are there more fluoride compounds formed with Xenon?
Based on WebElements, of the Noble Gases, $\ce{He}$ and $\ce{Ne}$ do not react with any of the halogens; however:

$\ce{HArF}$ has been detected at low temperatures (thank you to @bon and @Martin in the comments)
$\ce{KrF2}$ can be synthesised at low temperatures.
Xenon can react with fluorine to produce $\ce{XeF2}$, $\ce{XeF4}$ and $\ce{XeF6}$
Radon can react with fluorine to produce $\ce{RnF2}$

I understand that fluorine is the only halogen that seems to react with the Noble Gases (partly due to the related question Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?.  My question is
Why are there more fluoride compounds formed with Xenon compared to the other Noble Gases?

Do you know the three-center four-elcetron bound ?

Also $\ce{HArF}$ has been detected at very low temperatures  and confirmed to have proper bonds between the atoms.

Calculations for HArF show that there is a covalent HAr and an ionic ArF interaction: http://www.nature.com/nature/journal/v406/n6798/full/406836a0.html

Thank you @bon and Martin, I have added this compound into the question

The reason that we discovered more fluorides of xeon than helium, argon, and krypton is quite obvious and you might already know it. As lighter noble gases have more stable shell configuration, it is more difficult to make any kind of compound, including fluorides.
Also, the higher amount of fluorine atoms in a molecule of the compounds means drawing more electron from the noble gas, which is thermodynamically not preferred. Thus, $\ce{KrF4}$ and $\ce{KrF6}$ are far less stable than $\ce{KrF2}$. The quantum chemical calculation confirmed this (Dixon 2007). The same reference said that they are stable, but have not been synthesized yet, and the synthesis would be very difficult.

For radon it is a different problem. Briefly, there might be more fluoride compounds of radon than xenon, but the synthesis is difficult to be confirmed.
Even though radon is less inert than xenon and its compounds were predicted and calculated to be more chemically stable than xenon compounds (Chernick 1962, Thayer 2005), it is difficult to study. Radon is is rare and radioactive (half-life of 3.8 days). Studying chemical compounds in trace amount is a difficult task. There are claims for discoveries of $\ce{RnF4}$ and $\ce{RnF6}$ (also in the above references), but confirming that you have synthesized new compounds in trace amount is extremely difficult because the evidence is not obvious. $\ce{RnF8}$ was calculated to be unstable, but should be more possible than $\ce{XeF8}$.

References:
1 David A. Dixon, Tsang-Hsiu Wang, Daniel J. Grant, Kirk A. Peterson, Karl O. Christe, Gary J. Schrobilgen, "Heats of formation of krypton fluorides and stability predictions for $\ce{KrF4}$ and KrF6$\ce{KrF46}$ from high level electronic structure calculations," Inorganic Chemistry 2007, 46(23), 10016-10021 (https://doi.org/10.1021/ic701313h).
2 C. L. Chernick, H. H. Claassen, P. R. Fields, H. H. Hyman, J. G. Malm, W. M. Manning, M. S. Matheson, L. A. Quarterman, F. Schreiner, H. H. Selig, I. Sheft, S. Siegel, E. N. Sloth, L. Stein, M. H. Studier, J. L. Weeks, M. H. Zirin, "Fluorine Compounds of Xenon and Radon," Science 1962, 138(3537), 136-138 (DOI: 10.1126/science.138.3537.136).
3 John S. Thayer, "Relativistic Effects and the Chemistry of the Heaviest Main-Group Elements," Journal of Chemical Education 2005, 82(11), 1721-1727 (https://doi.org/10.1021/ed082p1721).

Another reason why the fluorides of the lighter noble gasses are less stable is, that the elements are smaller, which leads to a smaller fluorine-fluorine distances and higher repulsion between the non-bonding electrons (Pauli repulsion).

Most of the noble gases compounds involve Xenon, Fluorine and Oxygen. The only known binary compounds of noble gases are fluorides and oxides of Xenon and to a lesse extent of Kryton. For example, $\ce{XeF2, KrF2,  XeF4, XeF6,  XeO3,  XeOF2,}$ etc. but not $\ce{NeF4}$.
Noble gases form compounds with Fluorine and Oxygen only because Fluorine and Oxygen are the most electronegative elements. Heavier noble gases (Xenon and Krypton) are able to form compounds with other elements because  Xenon and Krypton, both have low ionisation enthalpies as compared to Helium and Neon. The noble gas elements do not react with other elements as they are less electronegative and are not able to attract the lone pair of electrons from noble elements. Fluorine and Oxygen are more electronegative elements. They are strong oxidizing agents. They can easily attract the lone pair of noble gas elements.
See also Why are inert gas (especially Xenon) compounds powerful oxidizing agents? and Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?


text:
Is there a chemical reaction which could give off enough heat underwater to cook something?
The mer-people demand cooked food. They don't want to go on land to make it.
Is there a chemical reaction they could use which would create enough heat to cook their food underwater in a safe manner?

Reference for the "silly first paragraph" - say hello "world building": http://worldbuilding.stackexchange.com/q/17735/2466

Welcome to Chemistry.SE! You're not the first writer that comes to Chemistry.SE for some scientific background :-)  The first thing that came to my mind were undersea vulcanoes, just as suggest on Wordbuilding.SE. So, do the mer-people have gills or lungs? In the latter case, they would probably reside in habitats like the Aquarius and would simply use electrical ovens.

To satisfy the original question, it would have to be a reaction that could occur naturally, as fire does in the terrestrial biosphere, so its usefulness could be discovered by primitive merfolk, thereby increasing their available nutrition and giving rise to civilization.  But at the same time, it would have to be not so common or violent in that it would prevent the evolution of intelligent life in this imaginary sea.

Interesting prompt!
There are plenty of reactions which could be performed underwater to release a large amount of energy. In fact, many substances react vigorously with water itself. You may have heard about how alkali metals such as sodium are not compatible with water, and there's a good reason for it!
Explosive reactions aren't particularly suited to cooking food, though, and a sustained flame would be much more useful. Interestingly enough, there was legitimate research performed on making underwater flames, for a very specific purpose: the Olympic torch!
For the 2000 Summer Olympics in Australia, during the classic Olympic torch relay before the start of the games, the torch was actually carried ablaze underwater for three minutes at the Great Barrier Reef. According to Wikipedia, the usual fuel for the burning torch is a mixture of liquefied propane/butane, which burns using oxygen gas in the air as an oxidizer (more details of a torch's construction can be found here). There is very little dissolved oxygen in ocean water, however, so the fuel/oxidizer mixture had to be changed. From the BBC article:

The special flare was developed after nine months by a team of chemists and engineers at Melbourne pyrotechnics company Pains Wessex.
The company's managing director Charles Tegner said designing the torch was a challenge - not only to produce a flare to burn underwater at such a depth, but to burn like the Olympic torch flame as well.
"It had to be clearly visible," he said. "Such flares don't normally exist."
The flare's chemical composition, pressed into a steel tube, produces sufficient oxygen and nitrogen to maintain a very hot flame.
The flare burns so fiercely at more than 2,000 degrees Celsius that this creates enough pressure to keep the water from entering the tube.
Its intensity is produced by a mix of oxygen-generating chemicals as well as the combustible element magnesium in a finely powdered form

I haven't found the detailed composition of the fuel/oxidizer mixture, but this is already a very good hint. Magnesium is an alkali earth metal which burns vigorously with a bright white flame in many conditions, be it in pure oxygen, air, pure nitrogen, or even pure carbon dioxide, and it will also react with water when hot producing hydrogen gas, much like sodium (this is why reactive metal fires are so hazardous; don't even think about using a $\ce{CO2}$ or water extinguisher on them, that just feeds the fire!). In this case, the pyrotechnicians decided to mix finely pulverized magnesium metal with a solid oxidizer, very likely something such as potassium chlorate, $\ce{KClO3}$, which decomposes under heat releasing oxygen gas. The flame was lit in air until the reaction between the magnesium and solid oxidizer started going, and then it was safe to submerge the torch while the reactants lasted. The stunt has in fact been repeated since.
To add a few more options to this answer, there is a general class of reactions with water called hydrolysis reactions, some of which are quite energetic. Here, the starting material tears apart water molecules directly and uses the fragments to modify its structure to a more stable form. For example, this is what happens when phosphorous pentoxide is allowed to come into contact with water. This avidity for water makes phosphorous pentoxide a useful material in a laboratory, as it is capable of scavenging trace amounts of water from a liquid or gas. However, these types of reactions already tend to release much less energy per gram of reactant, compared to combustion reactions.
Lastly, I'll cite one more thing. Even without any chemical reaction, the dissolution of some solids in water is capable of releasing some heat. This happens when starting material has a weaker attraction to itself than to water molecules. This property can be exploited in disposable heat packs. These types of processes usually release even less heat per gram of starting material than combustion or hydrolysis, but at least the process is conceivably easily reversed, provided energy is given to dry the material again.
Therefore, it is possible to get a lot of heat from reactions underwater, though it is quite a more elaborate process than starting a fire in air.
Edit: Now that I saw the original question at Worldbuilding.SE, I realize part of it wondered about how cooking would be done in earlier stages of technological development. In this case, I'm hard-pressed to think of a chemical reaction which happens using materials found easily underwater and releases a lot of energy. The difficulty is great enough that it has actually been suggested previously as a general argument for why technologically advanced or even sentient life would be unlikely to develop underwater, as any hot thermal processing of food or materials would be severely hindered. Except for incredible strokes of luck, hydrothermal vents on the sea floor sound like the most reliable heat source which could be accessed by a primitive civilization.

Your comment about hydrothermal vents being the best option would be a welcome addition to the question on Worldbuilding.

@ArtOfCode I dunno, seems to have been covered pretty well over there, as several answers touched on it.

Hydrothermal vents have been postulated to have contributed to the oligomerization of amino acids into peptide chains.  Great stuff.

This is a good excuse to mention FOOF, which would burn the water itself.

@JDługosz I don't think FOOF would be very good for cooking, all things considered.


text:
Methyl orange coordination compound chemistry
Methyl orange changes colors depending on the acidity or basicity of the solution. But I have read that it forms complexes with certain ions. Can anybody suggest any resource regarding the coordination complex-formation chemistry of methyl orange?

Methyl orange is a zwitterion at low pH and an anion at high pH; it's a model species for coordination chemistry with cyclodextrin, Google should reveal numerous papers about it. It's also a diazo compound; the Wikipedia entry on diazo compounds has a section on metal complexes.


text:
Grignards and anhydrous ethanol

I remember using both 95% ethanol and absolute ethanol quite a few times. You'd need absolute ethanol for a reaction that would fail if water was present, such as a Grignard reaction. I can't remember whether the absolute ethanol was in an ordinary bottle or the type with a rubber septum (where you keep the liquid under nitrogen to avoid contact with air). If the ethanol was to be used as a reagent, a septum bottle would be more appropriate.

http://boards.straightdope.com/sdmb/showpost.php?p=5314683&postcount=10
First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol?
I'm guessing that the poster just forgot all his chemistry. Ironic for someone posting on a message board that proclaims to be "fighting ignorance since 1973."

Second (assuming that one could somehow run a Grignard reaction in ethanol), do you really need anhydrous ethanol for Grignard reactions to work? Why not just use a slight excess of Grignard to account for the water present in 95% ethanol? Seems to me that anhydrous ethanol would be pretty expensive given that the usual distillation methods wouldn't get you from 95% ethanol to 100% ethanol.
Finally, is there a reaction in which using excess Grignard to account for acidic protons wouldn't chemically work in some way?

Absolute ethanol (~99.8%) is not particularly expensive. There are several techniques to skirt around the water-ethanol azeotrope which can be done in very large scale. It's just not quite as common to see for sale, because for most home purposes 95% ethanol is just as good as absolute.

First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol?

As well as being excellent nucleophiles, Grignard reagents are also very strong bases because they essentially function as alkyl anions. Therefore they will react with ethanol to form magnesium ethoxide, the alkane, and magnesium bromide.
Grignard reactions are usually carried out in ether solvents such as diethylether and tetrahydrofuran because they are unreactive towards the reagent and they actually stabilize it by coordinating to the magnesium ion to form a tetrahedral complex.

(image source)

so the poster is just looney?

I think so. I don't see any way you could get around the deprotonation.

This is so true. We recently had a question about iodine's role in the activation of magnesium: http://chemistry.stackexchange.com/q/31824/4945. Therefore I read up on the described method by Lund and Bjerrum. Adding magnesium to the alcohol is used to obtain absolute alcohol, via $\ce{MgEt2}$. Any Grignard would be destroyed by ethanol, just like water would do it.

@bon - after critically reviewing the quoted text, I note that the poster did not specifically come out and say that ethanol was used as a solvent for Grignard reactions. Is there any case in which ethanol would be useful in the general context of a Grignard reaction?

@Dissenter Not that I can think  of but I don't have any experience doing Grignard reactions so there might be something less obvious.

@Dissenter you might use ethanol to quench the excess Grignard in a reaction vessel at the end of a reaction.


text:
Determining the boiling point of a substance given the triple point and critical point
If I have the pressure and temperature of a substance in its triple point, as well as its pressure and temperature in the critical point, how can I determine the boiling point of the substance at 1 atm of pressure?

You can try using various functions of state - it's quite broad topic and one needs much more data about the substance to to do it with good precision.

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

What you need is the Clausius–Clapeyron equation, which is used to relate vapor pressure and temperature (or external pressure and boiling point).
$$\ln{\left(\dfrac{P_1}{P_2}\right)}=\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$$
Step 1: Find the enthalpy of vaporizations $\Delta H_{\text{vap}}$ by plugging in the two sets of temperatures and pressures that you have.
Step 2: Use your $\Delta H_{\text{vap}}$ and the temperature and pressure of the triple point to determine the boiling temperature at 1 atm.

In principle, using the Clausius-Clapeyron equation is correct. However, the implied approximations of this approach ($\Delta_\text{vap} H = \text{const}$ and $\Delta_\text{vap} V\ = V(\mathrm{g})$) are not valid near the critical point. In the given question, one of the reference ($p,T$) points is the critical point; thus, the results might show a significant error.