text:
How to explain the periodic trends in boiling points in groups?

Observing the trend of boiling points of the compounds listed, choose the appropriate terms to fit into the blanks:
\begin{array}{lr}
\hline
\text{Compound} & \text{b.p.}/\pu{°C}\\
\hline
\ce{H2Te} & −2 \\
\ce{H2Se} & −41.5 \\
\ce{H2S} & −60.7 \\
\ce{H2O} & 100 \\
\hline
\end{array}
The ___ the compound, the stronger the ___ present between molecules. The stronger the ___, the higher the boiling point of the compound?
More polar, smaller, London dispersion forces, larger, less polar, dipole-dipole and ion-dipole forces

I noticed that the electronegativity for each compound increases from top to bottom, so does higher electronegativity implies higher boiling point?
I've also tried putting more polar the compound, the stronger the dipole-dipole and ion forces. No luck, I still got it wrong.

You are certainly on the right track. Think about what the higher electronegativity means to the bonds in the molecule; this will give you the first word to insert, that you have correct. I would further think you are right by assigning dipole-dipole and ion-dipole forces to the second gap. The third gap uses in my opinion the more general term for these forces.

I think it is wrong to say there should be a trend as we study the trends of individual atoms, not molecule and compound itself. Boiling points depend on the intermolecular force which depends on the polarity of the whole molecule, which in turn depends on the constituting species, so there should not be specific trend. It is only the matter of comparison between the species in a molecule. Moreover, there is not a single type of intermolecular forces which bonds molecules such as dispersion, metallic or hydrogen: all have different strength.

The LARGER the compound, the stronger the LONDON DISPERSION FORCES present between molecules. The stronger the LONDON DISPERSION FORCES, the higher the boiling point of the compound.


text:
Regarding allylic free radicals and anti-Markovnikov product
I was searching a question "What is the primary product when cyclohexene reacts with bromine in the presence of UV light?" and in a website they said:

3-Bromocyclohexene will be the product because when an alkene reacts
  with $\ce{Br2}$ in presence of UV light, free radical substitution
  reaction takes place. In this case, a stable cyclohexene allylic free
  radical will form which will further react with bromine free radical
  to form 3-bromocyclohexene.

A reaction that I already studied was propene + $\ce{HBr}$ in presence of $\ce{H2O2}$
and the answer is:
The reason is that carbon radical is present on secondary carbon where the neighboring atoms can donate electron density to stabilize the radical.
But based on the first reaction I speculate that the answer should be:
because here an allylic free radical is formed.
So which is correct?

Both allylic substitution and anti-Markovnikov addition are possible reactions under free radical conditions. What exact conditions favor which reaction is something I expect most chemists would have to look up. Your web source may be correct (although the specific conditions are not mentioned), but I would recommend established references or journal articles as more reliable sources of information.

@iad22agp  Thank you very much for clarifying.

Do you notice that in your proposed allylic bromination with HBr, you are consuming $\ce{Br^.}$ and generating $\ce{H^.}$ radicals that don't go anywhere?

@orthocresol Homolytic bond cleavage of peroxide gives 2OH free radical.One of the OH free radical react with HBr and forms H2O and Br free radical so H free radical is not generated right!

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text:
How to derive the total number of molecular orbitals based on the given number of atomic orbitals?

If four orbitals on one atom overlap with four orbitals on another atom, how many molecular orbitals will form?

Why are there eight molecular orbitals and not four? Is not, in molecular orbital theory, each atom's orbital overlaps with another atom's orbital to form a bond?

If you have eight orbitals to start, you will have eight orbitals to finish.

Can you give me an intuitive explanation as to what's going on?

Technically speaking, each individual atom has infinitely many orbitals for its electrons, most of which are unoccupied. Infinity + Infinity = Infinity.

The question itself deals with molecular orbital theory and by extension with the approximation of combining atomic orbitals to form these molecular orbitals; in quantum chemistry this approach is called LCAO (Linear combination of atomic orbitals).
Let us look at one of the simple cases, the dihydrogen molecule. You have one 1s orbital at the first hydrogen, i.e. $\chi_A$, and one 1s orbital at the second hydrogen, i.e. $\chi_B$. You can now combine these orbitals to form a bonding $\eqref{eq:bonding}$ and an antibonding $\eqref{eq:antibonding}$ combination to create the molecular orbitals $\phi_{A+B}$ and $\phi_{A-B}$.
\begin{align}
\phi_{A+B} \equiv 1\sigma_\mathrm{g} &= \chi_A + \chi_B
  \tag1\label{eq:bonding}\\
\phi_{A-B} \equiv 1\sigma_\mathrm{u} &= \chi_A - \chi_B
  \tag2\label{eq:antibonding}
\end{align}
A scheme of this can be found on Wikipedia created by the user CCoil.

This does in principle mean that for every bonding configuration of a molecular orbital you will always have an antibonding configuration. In other words the number of atomic orbitals is the same as the number of molecular orbitals.
In your case, when there are four orbitals on one atom and four orbitals on the other atom, you will end up with eight molecular orbitals. A quite complicated case of bonding that shows this is the $\ce{C2}$ molecule, which has previously been discussed here: Bonding in diatomic C2, a carbon-carbon quadruple bond?

As it usually said, the number of orbitals is conserved in LCAO-MO. Intuitively it should be clear that the number of electrons is conserved when forming MOs from AOs. And since each orbital (atomic or molecular) holds up to 1 or 2 electrons (depending on whether are we talking about spin or spatial orbitals), the number of orbitals should also be (at least) conserved, for otherwise you might have not enough MOs to place all the electrons that populated AOs.
Specifically, for OP's example 4 spatial AOs of one atom can hold up to 8 electrons, so 4 spatial AOs of two atoms can hold up to 16 electrons. Now, if only 4 spatial MOs are built out of these 8 spatial AOs, they could hold only up to 8 electrons and there is no room for any of the possibly remaining 8 electrons.

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What are the disadvantages of using the InChI format?
What are the disadvantages (if any) in producing and using the InChI format of chemical compound representation in chemoinformatics? Are there any particular problem one can face when using InChI?

It really depends on what you are using them for. In an earlier question I tried to demonstrate the known shortcomings of InChI: http://chemistry.stackexchange.com/q/14066/4945

You have to evaluate shortcomings as compared to something. You can make a list and say "but InChI doesn't represent X" where X is ranging around

mixtures
Markush groups
polymers from repeat units
reactions
toasters

The point is: so what? All representations have some limitations. But limitations aren't shortcomings unless you have a different representation that doesn't have those limitations and that is not traded off by other limitations or bloat or other disadvantages.
What do you have to compare InChI with? In my world: SMILES and with MOLFILEs. InChI is most closely related to SMILEs. SMILES is easier to just type, but it isn't canonicalized and it doesn't have the layers of specificity that allow you to find approximate equivalences so easily as with InChI. InChI connection layer is very similar to SMILEs in principle, but it is a lot harder to read because you need to replace the numbers with the atoms from the preceding formula layer. But it is easy to convert InChI to SMILES, except where InChI is better. For example, parenthesized groups, so you represent the carboxyl group -C(=O)OH as -C(O,O)H, which isn't a thing in SMILES.
Where InChI shines is that many isomerisms are abstracted or pushed into more detailed layers. There is no other format that does this as well or at all as InChI, which is why it is good. But InChI is conceived as an "identifier" and some of the details are not at all represented, most specifically the atom positions. That's where MOLFILEs are so easy to work with, if you want to just display the molecule. But in that regard, InChI is no worse than SMILES.
So, you should be more specific in terms of what your goals are that you are asking for "disadvantages" and you need to stay very critical of text that berates InChI or any other standard with a bunch of rationalizations for why not to choose standard A (because the author is committed to standard B). I know this sort of superficial argument is being made a lot in any informatics sector, people like their baby and berate the other's baby. Especially be weary of those who say "don't use InChI, InChI is bad" and then "you really should follow the work on XYZ, because XYZ will be able to represent everything including the toaster."
InChI has it's firm place when the purpose is to represent, refer to, identify basic molecular structures. That's it. If you want mixtures, reactions, polymer repeat units, consider using InChI as the core building block and then build higher order structures around it. For example, obviously there is no reason to push reactions into the InChI representation, just make a simple structure that lists the reaction participants on both sides, their stoichiometry and the enthalpy, activation energies, etc. etc., these can be done in higher order data structures much better than shoving that information into a single InChI++ string. Likewise for polymers, represent each repeat unit as an InChI and then combine them. If your goal is such higher order molecules, you would profitably use InChI as the representation of those core building blocks.
Now that said, I give you my personal pet peeve about InChI, i.e., that is in scope of current InChI where I think it has, let's say, pitfalls, not necessarily that it made the wrong choice. And that is, ionic bonds and coordination bonds. Especially ionic bonds are represented as two separate structures, recognizable with the period . in the formula separating the multiple structures. This is actually a rudimentary representation of mixtures. So in my view, it should not even have introduced this dot notation but simply forced people to represent the parts of these ionic mixes as separate InChIs. I come from the biochemical / physiology / pharmaceutical / medical world where such ionic bonds usually do not even really exist in nature, as the ions are all dissociated in solution. But all that can be disputed. Same for metal coordination or chelate bonds, you want to say where the metal ion associates with, and I think that's not in InChI. So in my world it would be better not to have any of this mixing and coordination inside InChI and we would use higher order structures for it instead, but always use InChI as the basic building blocks.

I object to almost everything being posted here. I find this post also written very unfriendly and not helpful to the project. Also, if you are associated with the InChI project you must disclose this association. I do not want to go into any more detail about this, but I do not think that this answers the question. There are documented limitations of the algorithm, which could have been discussed here objectively. Oh, I almost forgot: I down-voted.

I have no affiliation with InChI. If you want to give a better answer, you are free to do so. If you object but you have no details to say then you simply protest.

I cannot give an answer to the question, that is why I did not answer it. The post I linked clearly states the documented shortcomings of InChI. Something you seem to ridicule within the first paragraph of your answer attempt. Yes, I use attempt here, because I do not think it answers the question. Your attempt basically boils down to: Don't use it for something that it was not designed for. Which is legitimate, but not really an answer to the question. So in summary: Yes, I do protest. For that much text, there is - in my opinion - not enough objective discussion of the problem.

As you can tell from my answer, I'm a no-b/s kind-a guy. So, you get nasty, I get nasty back. The question was "disadvantages of using InChI" and "problems we can face when USING InChIs" Your vehement reaction was that you feel attacked by my Toaster joke. But that is because you are committing the what-about-the-Toaster mistake, and you are apparently emotionally biased, you want to find some negativity about InChI and you use the lack of Toasters. You need to check your list first "as compared to what?". E.g., InChI doesn't NORMALIZE tautomers (yet) but MOLFILE doesn't normalize anything!

I am neither attacked nor offended by the remark about the toaster, even though it is an odd comparison. I see purpose and benefits in using InChI as a system, algorithm, software. I am also quite positive that future version will address the current limitations. They are still limitations and they should be discussed as such. Just stating don't use it for situations it is not designed for without addressing which situations these are is in my opinion not helpful. It is (objectively) a problem that InChI cannot distinguish tautomers and one should be aware of that when working with these.

@Martin-マーチン, are you sure you're not misrepresenting the tautomer issue with InChI? InChI can distinguish between different tautomers using the fixed hydrogen layer (/f/h) and it is in fact a very cool feature of InChI to treat mobile hydrogens as shared. This makes InChI a superior representation format in my opinion, better than MOLFILEs and SMILEs (although I am sure there is some add-on to either where you can do it anyway). Just for a simple example, a carboxyl group, in SMILES, I'd like to write it as -C(O,O) instead of -C(=O)O because you can't really nail down that =.

That'll be my last reply to this, because comments are not intended for discussion. We can continue in [chat]. I've gotten distracted by tautomers, it is of course a different structure, so InChI is able to handle these. It can't treat isomers and that was something stuck in my head; or maybe I'm wrong about that, too. To be absolutely honest, for the line of work I was interested in, neither of these formats was accurate enough. It's a stringification of a Lewis structure, works well for global patterns, good initial guess, but fails in the details. Yet these details might become important.

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text:
How can the Brønsted-Lowry theory be valid?

How can an element accept a proton? Isn't this practically impossible? An atom can take an electron and place it in its valence shell to complete its octet or duplet, but, the proton sits back in the nucleus with the neutron. So, it must be impossible for an atom to take a proton and bring it in the nucleus which sits deep in the centre.

Look at your equation and think: if the nitrogen atom would take the proton to its nucleus, would it still be nitrogen? Did the hydrogen atom count change?

Hydrogen is nothing but a proton with an electron about it, so H+ is just a proton.
Therefore it is not the element that accepts the proton, it is the molecule.

The Brønsted-Lowry theory is valid. When indicators are used in titration for example, the color of the solution, depends upon how many H's are bonded to the chemical indicator which is related to the acidity or basicity of the solution. Generally, the Brønsted-Lowry theory would tell you that the molecule with the most H's would donate it's proton (H) to the molecule with least H's when this happens NH4+ becomes NH3 upon losing an H, and NH2- becomes NH3, so you got two equivalents of NH3.
But that is not quite the best model as the Lewis-base model is better where the base is the electron donator (NH2-), and the acid is the electron receiver (NH4+). I might have to shift into my virtual machine to draw you a diagram. But anyway, what is happening is NH2- has two lone pairs of electrons and an negative charge, while NH4+ has no lone pairs of electrons and a positive charge ... so there is electrostatic attraction between the two molecules. A lone pair of electrons on NH2- desires to form a stable molecular orbital (as a lone pair orbital is not as stable as a molecular orbital in a bond between two atoms) so the lone pair of electrons abstracts a free H from NH4+, and as this happens a bond breaks from the H being abstracted to its N on NH4+ as H doesn't like to form two bonds, so NH4+ becomes NH3 and NH2- by gaining an H becomes NH3, as well.
Keep in mind that the bond between N-H, C-H, O-H, etc is polarized (not strong) because there is a large difference in electronegativity of the elements N and H, so it's easy to lose an H from whatever element it's bonded to if it feels a pull of free electrons.
New molecules form by the flow of electrons between atoms on one molecule and atoms on another molecule. It's possible to form monatomic ions like when you dissolve H-Cl in water and get the free chloride ion with a full octet of electrons. However, atoms within molecules usually get their full octet by bonding to other atoms from other molecules and sharing the electrons covalently.
Anyway, here is the electron-flow diagram. Those arrows show the movement of electrons. Note that Any of those H's of NH4+ is up for grabs.


text:
Is there any chemical preservative better than sodium benzoate?
I'm developing a fruit juice for humans, I just want a second opinion about allowed food preservatives. Does anyone happen to know if there are any chemical preservatives that can extend the shelf life of fruit juice better than sodium benzoate?

I'm experimenting with various preservatives   | DO NOT. I'm serious. Most preservatives used are slightly to seriously toxic when used in wrong amounts. Using them wrong way is a guaranteed way to troubles.

I'm a chemical engineer and I think I know what I'm doing. I just want to know if any know a better preservative.

Yes - mix the fruit juice with lots of formaldehyde. Oh, you meant better in the sense of still being consumable by a human without adverse side effects? Well, that is a very different question, and while some chemical engineering is good, a lot of toxicology is required as well...

@user3754535 Not relevant. Toxicology/biology background is required.  In case you are interested in what may be used, consider looking for a 'list of permitted preservatives' (ideally several such lists from several different countries should be referenced). However again, no 'experimentation' here, only allowed methods/proportions are to be used.

Did you think about $\gamma$-iradiation using $\ce{^{60}Co}$?

@permeakra I just a simple question,What is the best preservative (allowed of course) for fruit juice? or physical preservative techniques . I can't believe you guys go deep so fast. I you don't know other chemical for this matter I will answer the question myself, I just wanted a second opinion.

@JonCuster I edited the question.

I just want a second opinion about allowed food preservatives.

Allowed is crucial here and it might strongly depend of where the product is to be sold. You might want to have a look at national legislations.

For Germany, the Zusatzstoff-Zulassungsverordnung lists allowed preservatives and their maximum concentrations in different products.
For Canada, Health Canada provides a List of Permitted Preservatives.


text:
How to reduce Tin Oxide
We have a pilot atomization setup where we atomize molten tin to produce tin powder. Unfortunately a lot of tin is wasted and oxidized.
I have found this video on which looks promising. Is this the best way of reducing tin oxide? Or do you know of anything better / effective?
Thanks!

The video shows the "purification" of zinc metal (probably covered with an oxide layer) by heating/reacting it with powdered carbon.
This is a very common process to reduce metal oxides to the corresponding metals. In turn, carbon is oxidized to carbon monoxide and carbon dioxide. The most prominent example is the production of iron from iron oxide in a blast furnace.
However for some metals, this fails and metal carbides are formed instead.
For tin oxide, the reduction with carbon (or coke) is an established industrial process.
Alternatively, the reduction of tin oxide can can be achieved using hydrogen, see Materials Transactions, 2011, 52, 1814-1817 (PDF).

Thank you Klaus! Is there any acidic medium I can introduce this hydrogen by? Or should I stick to creating an atmosphere of hydrogen gas under high pressures and temperatures?

@DBTKNL In the linked document, the reductions were performed at 500-750 °C in under a hydrogen atmosphere (= hydrogen gas) between 0.3 and 1 bar. However, that was on an analytical scale.


text:
Reaction of carboxylic acids with primary amines

$$\ce{RCOOH + RNH2 -> RCONHR + H2O}$$

$\ce{R}$ stands for any alkyl group.
I have two queries regarding this reaction:

I understand that the reaction is an $\mathrm{S_N{AE}}$ (nucleophilic addition elimination type of reaction). But during the reaction $\ce{R-(CO)-N^+RH2}$ intermediate is formed. After this step the $\ce{OH^-}$ which was eliminated acts as base takes away $\ce{H^+}$ from  $\ce{R-(CO)-N^+RH2}$, converting it into $\ce{R-(CO)-NRH}$. But why did $\ce{OH^-}$ extract $\ce{H^+}$ from $\ce{R-(CO)-N^+RH2}$  and not $\ce{R^+}$ ? Isn't $\ce{R^+}$ always a more stable carbocation compared to $\ce{H^+}$ ?
Why does this reaction require a basic medium?

After this step the $\ce{OH-}$ which was eliminated …

No. $\ce{OH-}$ is never eliminated but formally. To eliminate the oxygen required, it needs to first accept a proton from somewhere leading to $\ce{R-C(O^-)(OH2+)-NHR}$. The hydroxide ion is simply much too bad a leaving group for any $\mathrm{S_N}$ reaction; it must always be transformed into a better one.
This leads to a severe problem in the reaction. Basic conditions hinder the reaction because they decrease the degree of protonation on that oxygen. However, under acidic conditions the amine will be protonated in solution ($\ce{RNH3+}$) and a protonated amine no longer has any lone pairs that can attack nucleophilicly. You’ll want neutral to mildly basic conditions for the reaction to work out.
Another thing to consider: the ‘elimination of a proton’ is a simple acid-base reaction. Proton displacements are typically much more rapid than any carbocation eliminations because protons, being as small as they are, can tunnel. Additionally, you should never compare the stability of a naked proton against that of a naked carbocation in these eliminations. A naked proton will not be generated, it will attach to a base immediately — and there are enough bases around: $\ce{OH-, H2O, RNH2, \dots}$ Furthermore, carbon-heteroatom bonds are pretty stable if the heteroatom side does not equal a good leaving group (e.g. $\ce{NHTs}$). Thus, these are hardly ever cleaved.
In practice, you would not use a particularly acidic or basic medium in this reaction. Rather, you would use acid activation techniques such as the Steglich conditions ($\ce{DCC, DMAP}$).


text:
determination of glucose syrup in sugar product
I work with a sugar product that contains saccharose (approx. 70 %), glucose syrup (approx. 15 %), glycerine (approx. 3 %) and some other ingedrients, e.g. water. Glucose syrup consists of 5 % glucose (Dx), 50 % maltose (Dp2), 25 % higher connected sugars (Dp3), 20 % water. I'd like to determine the amount of glucose syrup in the sample.
Fehling is time-consuming and well-known as an empirical method. Because the concentration of glucose, saccharose and maltose ist unknown (three variables and two equations: mass of cupperoxide before inversion and mass of cupperoxide after inversion) I think it's not possible to calculate accurate enough the amount of glucose syrup. The error should be lower than 5 % of the real value.
To measure the Brix-value in order to determine the glucose syrupe concentration is too inaccurate.
An alternative would be to use enzymatic test-kits which are selective for maltose. Unfortunately this test is too expensive and when opened it must be used within a few days or weeks.
My idea is to label a protecting group with an fluorescent group and to put it on the maltose molecule. The protecting group should only bind to the molecule, if the distance between two hydroxy-groups is given.
(Unfortunately I don't know the distance between the two hydrox groups).
Which protecting group could be the solution of this problem?
Or do you know an even easier procedure to determine the glucose syrup amount?

Do you have to perform the analysis periodically? Which instruments are available? Did you consider fluorescence labelling and HPLC?

We could use a UV/VIS spectrometer (but we have no HPLC). We need to do the analysis one day per month or less, but we really like to do it in our laboratory.


text:
Is the Hollywood portrayal of the effects of ingesting HCN correct?
In the 2012 movie Skyfall, the antagonist Raoul Silva rues that the "hydrogen cyanide" he took ("Do you know what it does to you, hydrogen cyanide?") didn't kill him, leaving him horribly disfigured instead.

Firstly; Could it have actually been HCN that he took?
From what I know about this, HCN is a gas at room temperature, and gases aren't typically the poison of choice for suicide pills. To begin with, suicide pills are preferrably kept as small as possible to make it easy to conceal. Keeping the size constraint in mind, simply filling it with HCN gas wouldn't allow the pill to store enough to (seriously) harm you, much less kill you.
From what I gather, suicide pills (WW2 era) often contain a concentrated solution of potassium cyanide, and not HCN.
Well, I'm bringing this up since this is a James Bond movie, so... what're the chances the MI6 could've compressed a significant amount of HCN in the suicide pill (possibly pressurizing it until it's a liquid), without having the easy-to-crush-between-the-teeth pill exploding on account of pressure? I'm not sure how easy it is to liquify HCN, but if it is possible to do so at room temperatures without an exorbitant amount of pressure, could that have been the 'suicide pill' he had taken?
Are there any known instances ( some half-arsed experiments and such) where a gas at room temperature was liquified and stored within such a small volume.
Secondly: Could hydrogen cyanide, or any other simple cyanide compound for that matter, cause such a disfiguration?
More than having tried to kill himself with some sort of HCN pill, I was taken aback to see what it did to Silva.
Just look at what happened to the guy:

Seriously, just look at him:

This is what I found on Wikipedia, on cyanide poisoning:

If cyanide is inhaled it can cause a coma with seizures, apnea, and cardiac arrest, with death following in a matter of seconds. At lower doses, loss of consciousness may be preceded by general weakness, giddiness, headaches, vertigo, confusion, and perceived difficulty in breathing. At the first stages of unconsciousness, breathing is often sufficient or even rapid, although the state of the person progresses towards a deep coma, sometimes accompanied by pulmonary edema, and finally cardiac arrest. A cherry red skin color that changes to dark may be present as the result of increased venous hemoglobin oxygen saturation. Cyanide does not directly cause cyanosis. A fatal dose for humans can be as low as 1.5 mg/kg body weight.

I don't seem to find anything on 'severe-tissue-damage-and-jawbone-dissolution', so I'm still skeptical.
I mean, sure, HCN is a weak acid, but that is not what a weak acid does to you.
So did Raoul Silva bite down on a cyanide pill (of sorts...) or did he gargle a bottle of drain-cleaner and just forget about it?
Is there anything realistic about that scene, or is it just another movie gaffe?

Pressurised hydrogen cyanide capsule in your tooth? I hope MI6 agents don't have to fly that often. Oh wait, they do...

Guess they just confused HCN with KCN.  KCN will form HCN in slightly acidic media, so its not that far from the truth...

@paracresol hydrofluoric acid is a weak acid too ($\mathrm{p}K_\mathrm{a}=3.2$) and you can dissolve a body, glass and a lot of things. So tell me what a weak "must" do, and I'll answer you ;)

HCN is a high vapor pressure liquid at room temperature, bp  25.6 °C (78.1 °F).  It makes it way more dangerous, because it lingers.

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

(+1) It's the same question that came to my mind when I watched the movie, just after the "eek" moment. It sorta spoiled the action flow: my brain entered the "scientific BS, meh!" mode.

Is there anything realistic about that scene, or is it just another movie gaffe?

I have never heard about anything similar caused by $\ce{HCN}$ or $\ce{KCN}$.
However,

bleeding of the gums,
loss of teeth, and
osteonecrosis of the jaws (= the bone tissue literally rots away)

are known symptoms of certain intoxications:

Bisphosphonate-associated osteonecrosis of the jaw was observed in the treatment of patients with cancer
Radium jaw was observed in painters (Radium girls) that licked their paintbrushes while applying radioluminescent paint to dials
Phossy jaw was observed in (mostly female) workers in the matchstick industry of the late nineteenth century that worked with white phosphorus.
Back in the days, the lives of the patients often could only be saved by removing the jawbones surgically. (If you feel for it, look for some photographs online.) Since one of the tags is chemistry-in-fiction, you might be interested that this was partly covered in the episode Become Man of the BBC series Ripper Street.

Yes, the movie is “accurate” in a sense. Since it is stored in a capsule, it is a liquid as long as it is held below 78°F. As for what happened to his face, chemical burns mixed with hypoxia can account for the dead tissue and lack of healing.

The problem is that there are multiple films which depict cyanide poisoning, I am sure that they will not all show the same things.


text:
What is a mole?
Is my understanding of the concept of a mole correct?
The mole is a simply a number ($6.023 \times 10^{23}$). It's just like what a dozen would be like for eggs, except that the mole applies to atoms and molecules. If we were to count the number of atoms in 12 grams of carbon-12, it would be a mole.
Molar mass is the mass of all the particles in a mole of substance. Molar volume is the volume occupied by all the particles in a mole of substance.

Mole applies for everything you want even eggs. This is just so big that it is better to use it in chemistry than for count the eggs. Some people dicided to say one thousand for 1000 others chose one mole for $6.022 \times 10^{23}$ so yes this is only a number.

The molar volume $V_\mathrm{m}$ is the volume of a substance per mole of substance.
This is different from the "volume of one mole of substance".
For example, the volume of one mole of water is approximately $18~\mathrm{cm^3}$. But the molar volume of water should be $18~\mathrm{cm^3~mol^{-1}}$.
Saying that the molar volume is the "volume of one mole of substance" is incorrect, just like saying that the speed is the "distance travelled in one second" is incorrect. The speed is the distance travelled per unit time and therefore has units of $\mathrm{m~s^{-1}}$. I have had this "argument" with others before and I will stand by what I say.
Furthermore, there are no restrictions on the temperature, pressure or other conditions whatsoever. There is no restriction on what the compound is so the other answer is absolutely wrong on this.

The mole is not a number but the name of an unit: the unit of amount out substance. Amount out substance is the number of elementary entities divided by the Avogadro constant.
IUPAC's related definitions are somewhat recurrent in my opinion.
We can say that the amount of substance of 12 grams of carbon-12 is $1\,mol$.
Molar mass of something is the value of the variable that represents its mass divided by the corresponding to its amount of substance.
Molar volume of something is the value of the variable that represents its volume divided by the corresponding to its amount of substance.
Example: if you have 24 $g$ of carbon-12, its mass is represented by the variable $m = 24 g$. Its amount of substance by $n= 2\,mol$, so the molar mass of your solid is:  $m/n = 12\,g\cdot mol^{-1}$.
The mass of an amount of your solid which correspond to 1 $mol$ is 12 $g$, which is different to $12\,g\cdot mol^{-1}$.
You should check IUPAC Gold Book  whenever you are in doubt about chemical related definitions.

A mole is defined as: 6.023 x 10^23 elementary units of "something." That something can be atoms, molecules etc. So yes, I think you got that right.
Molar volume is defined as the volume occupied by one mole of an ideal gas at STP (standard temperature and pressure) which is approximately 22.4 liters.

Your definition of the molar volume is wrong.


text:
Intrinsic Reaction Coordinate - Converting To Reaction Coordinate with Units of Distance
I've located a transition state and ran two IRC calculations (one to get to the product and one to get to the reactant). Is there some type of way of approximating the IRC to something that has units of distance (Angstroms, Bohr, etc.)?

Intrinsic reaction coordinate (IRC) is not some internal coordinate of a molecular system (such as, bond length, bond angle, etc.), rather it is a curvilinear coordinate that describes the intrinsic reaction path (IRP), which is the reaction path along the direction of the gradient. Only for some simple reactions one can visually (and approximately) identify IRC as one of the internal coordinates, but in general that is not possible.


text:
What is special about the molecule luciferin that it can emit light?
A classic example of bioluminescence; the protein luciferin is oxidized by the enzyme luciferase, releasing energy as light.
What is it that gives luciferin this ability? Why can't just about any protein display bioluminescence?
Are there more chemical reactions where electromagnetic waves are produced? Also, are there any chemical reactions which produce high energy electromagnetic waves, such as X-Rays?

I don't get what you mean by the "higher energy waves" bit ._.  But as far as any sort of luminescence is concerned, the emitted light is of a higher wavelength (and therefore lower energy) than the incident beam of light.

I didn't know it was a kind of luminescence because it takes place in the dark. But with higher energy waves I mean xrays etc.

X-Rays? If it's a nuclear reaction you want, then there must be plenty of examples out there. But X-Rays from a chemical reaction? That's highly unlikely.

Well, if all highly unlikely events didn't took place than we perhaps wouldn't exist....;-)

Have you studied any quantum chemistry/spectroscopy? Do you know about electronic energy levels, and transitions between them?

[...] the protein luciferin is oxidized by the enzyme luciferase

Luciferin is not a protein, but a benzothiazole with a thiazole attached to the carbon atom between nitrogen an sulfur.
Upon oxidation at the thiazole ring, an oxetanone is formed. This four-membered ring breaks, releases carbon dioxide and turns the thiazole into a thiazolone. The resulting molecule is called oxy-luciferine.
Amazingly, the oxy-luciferine generated through this reaction is in an excited state! While typically excited states are the result of the absorption of a photon, here excited state is generated chemically!
Anyway, the fate of this excited state is as to be expected: it relaxes to the ground state by emitting a photon.
In general, these reactions are known as chemoluminescence.
Another known example is the peroxyoxalate reaction of glow sticks.
Phosphorescence at ~450 nm has been observed from the comproportionation of cyclohexanoxy radicals in polymer matrices, when hydrogen atom atom transfer between two radicals produce cyclohexanol and cyclohexanone (in an excited triplet state).

But is for this reaction any light needed as in luminescence?

@Marijn No. The excited state is reached in the prior chemical reaction in the dark. No photon is absorbed before.

Is oxy-luciferine the only known molecules which is chemically got in an excited state?

@Marijn No. 1,2-dioxetane dione formed in the peroxyoxalate reaction (see my answer) is a similar case.

Is it known what precisely causes the excited state? Or with other words why is it so rare; what is needed for it?

@Marijn You need the right match in the electron structure; remember, you can't just add x energy to an electron - you need a rather precise amount of energy to shift it to some excited state. So you need a chemical reaction with the right amount of leftover energy (that is, energy not immediately absorbed by some other ingredient), and a molecule/atom with the right electron structure to absorb that amount of energy to form a "singly" excited state, and the energy must correspond to a visible photon (mostly strongish single bonds, like C-C).

@Marijn Note that photoluminiscence is already quite rare, and it only needs one of those conditions - an atom (or molecule) with an electron that will drop from an excited state with energy in the visible band. Chemoluminiscence is much more complicated, and thus even rarer.

Klaus gave a nice summary on luciferin. The second part of your question was:

Also, are there any chemical reactions which produce high energy electromagnetic waves, such as X-Rays?

Any energy released with an emitted photon must previously have been added into the molecule (or atom) in question. For example, if singlet oxygen is created chemically, it will relax (slowly; the mechanism is, naturally, spin-forbidden) back to triplet oxygen; the energy difference corresponds to red light which is why a faint red glow can be observed with darkness-adapted eyes. For most compounds, the emitted photon corresponds well to the energy difference between a one-electron excited state and the ground state. These can often be calculated quantum-chemically.
For the vast majority of molecules, the energy difference between HOMO and LUMO — the one most likely to get excited by a random proton, and thus also the most likely one to get emitted — corresponds to middle-ultraviolet, near-ultraviolet or visible wavelengths. For example, most phenyl rings have an absorption at a wavelength around $250~\mathrm{nm}$. In some very simple molecules, this difference may extend into the slightly far ultraviolet region.
Typical X-rays have a much smaller wavelength. It helps considering how they are generated in a bench-top manner: a surface (typically a metal) is bombarded with accelerated electrons which knock out core electrons from their atomic orbitals. These holes are filled by the relaxation of valence electrons. For example, if an electron is displaced from the 1s orbital, an electron from a 2p orbital may replace it. If that occurs, energy is released as a photon and the corresponding radiation is termed $\mathrm{K_\alpha}$: $\mathrm{K}$ signifies the shell into which the electron relaxes and $\alpha$ denotes the first relaxation that can fill this void.
The $\mathrm{K_\alpha}$ transition is typically the most important and one of the most intense while all K-series transitions are those with the highest energy. This energy is often close to a wavelength of $0.1~\mathrm{nm}$ — note how far away that is from your average near-UV HOMO-LUMO transition.
Thus, the energy difference that corresponds to X-ray emission is typically that of core orbitals which is much larger than anything present out in valence orbitals even if bonding/antibonding interaction differences are concerned. No chemoluminescent-type emission will be able to emit X-rays unless you’re doing something really bizarre (e.g. bombarding with electrons).
Camara et al. published what can classify as ‘something bizarre’ from the former paragraph: rolling of sticky tape, the researchers found triboluminescence with photon wavelengths right up to ultraviolet ranges — in vacuum.[1] They published a nice graph that reveals how the required peeling force rises and then suddenly drops with concomitant emission of light. Under ambient pressure, the force differences are much smaller and the photon energy released equally. They attribute this behaviour to static electricity discharge, the static electricity building up during the peeling and then suddenly discharging. They were even able to use the X-rays to record an X-ray of one of their fingers.[1]
Remember though that this is only possible in a vacuum ($10^{-3}~\mathrm{torr}$); under atmosphere, the charge separation is probably neutralised much faster and at much lower energies probably due to the interaction of air molecules.

Reference:
C. G. Camara, J. V. Escobar, J. R. Hird, S. J. Putterman, Nature 2008, 455, 1089. DOI: 10.1038/nature07378.


text:
How do you determine steric hindrance?
Steric hindrance is important for understanding regio-selectivity and stable conformation of molecules. I know the general definition, but can somebody explain to me in detail how does phenomenon of steric hindrance come about?
Like, to give a simple, hypothetical example, $\ce{CH4}$ is said to have no steric hindrance. But why does attacking molecule have no trouble squeezing past $\ce{H}$ atoms and attack center carbon atom. I know hydrogen atom is said to be small, but seriously how small is it?
I am asking this because when examples go beyond $\ce{H}$ atom and alkyl atoms I can't judge whether steric hindrance is present or not and thus what products are formed. Like in $\ce{(CH3)3CHOH^+}$, is there steric hindrance. It is a primary carbon cation but instead of two hydrogen then is $\ce{OH^-}$ group?

what do you mean by $\ce{CH3CH3CH3CHOH^+}$ ?

Oh... I was giving a example similar like you know how u have tertiary carbon cartion, nucleophile can't attack it simultaneously with leaving group. So now instead of tertiary carbon cation what if u had a carbon cartion attached to propyl, hydrogen and OH- group. If the Oh- group was  H carbon cartion will be primary and SN2 can occur but with Oh- doesn't that add steric hindrance like as if carbon cation was tertiary?

It not answers my question. I just said what do you mean by this $\ce{(CH3)3CHOH^+}$ because written like this it doesn't exist. Also this is not carbon cartion but carbocation.

Okay. this chemical formula is a hypothetical example. It may or may not exist. But the bigger point I wanted to make is which groups provide more steric hindrance OH-, CH3 and so on....

$\ce{(CH3)3C^+}$ is a carbocation (which exists) it can be from $\ce{(CH3)3CCl}$ which will do an E1 in a polar solvant for example $\ce{EtOH/H2O}$. If you want to know the size of a hydrogen atom you can look on google, even myself, I don't know exactly the size but it's small. The more you add atoms the more it will be big and create steric hindrance. So $\ce{H^+ < CH3^+ < (CH3)3C^+ }$ and so on ...

https://fr.khanacademy.org/science/organic-chemistry/substitution-elimination-reactions/sn1-sn2-sal/v/steric-hindrance-1

Normally we don't discuss quantum mechanics in this context, but an answer to your question requires it.
Taken from:
http://chemistry.umeche.maine.edu/CHY252/HOMO-LUMO.html

What we see here is the LUMO for CH3Cl.  This is where an incoming nucleophile will 'stick' its electrons to form the new bond.  Since the orbital is antibonding, electrons in the orbital will destabilize a bond, in this case the C-Cl bond.  The actual meaning of the size of this orbital, rigorously, is a little unclear.  However you can roughly infer that if a nuclephile gets its electrons here, a reaction is 'likely'.  The reddish region to the left encompasses the hydrogens.  That is also where the nucleophile has to go.
Compare the next from:
http://chemistry.umeche.maine.edu/CHY252/EtCl.jpg

This is for ethyl chloride.  There is an extra minor node on one of the hydrogens.  Because it is small, it is less important than the other regions.  Notice the methyl group is partly obscuring the backside of the orbital lobe near the electrophilic carbon.  There are less ways a nucleophile can approach this to cause a reaction.
I'm no longer practicing so I do not have software to produce analagous images, so you will have to imagine the extrapolation.  1-chloroopropane would not add much more to the steric hindrance, because the 'extra' methyl group would be dangling on the same of the molecule as the methylene group.  However, if we had two methyl groups, such as 2-chloropropane, the hinderance would be much more.  Three groups, as in t-butyl chloride?  Give it up, to much hinderance for a backside attack reaction.


text:
Could diatomic helium exist?
I know that $\ce{He2}$ (homonuclear diatomic helium) does not exist because its bond order is zero. Since the bond order of $\ce{He2+} = 1/2$, that means that the positive $\ce{He2+}$ ion exists, but how does the positive ion exist if the neutral molecule doesn't?

It does not last long when formed.

See Helium chemistry: a survey of the role of the ionic species. Int. J. Mass. Spectrom. 2004, 237, 243

Related: How can an excited helium dimer be formed?, Why do noble gases bond with themselves but not other elements?

Although we know that the helium dimer has a tiny bond dissociation energy, we don't consider it a molecule, as the "bond" does not survive a molecular vibration, even rotation of the molecule is enough to cleave it.
From an answer by Lighthart on "Molecular orbital theory & predicting the stability of a molecule?" I took the liberty to borrow the MO schemes:

In the scheme you can clearly see how we arrive at the bond order of zero for $\ce{He2}$. Both, the bonding $\sigma$ and the anti-bonding combination $\sigma^{*}$ are doubly occupied. Dihelium is only held together by weak van der Waals forces.
In the case of the dihelium cation $\ce{He2+}$, one electron is removed from the anti-bonding orbital. There are now more electrons in bonding than anti-bonding orbitals. This effectively results in a net energy loss or bond energy gain, and the overall bond order increases to 0.5. In a more phenomenological context, three negative charges are now in a field of four positive charges, this gives rise to Coulomb attraction.
Reference: Wikipedia article on dihelium and references therein.


text:
Rate and order of a reaction
The rate of reaction can be measured from a concentration vs time graph of a particular reactant in a reaction.
But, if in your reaction you have different reactants of different orders surely they will each give a different concentration vs time graph and so different rate at any time t. This implies that the rate of reaction can have multiple values depending on which reactant you follow.

The rate of reaction is defined in IUPAC's Gold Book as:
$$r = - \frac{1}{a} \frac{\mathrm d[\ce{A}]}{\mathrm dt} = - \frac{1}{b} \frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{1}{p} \frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{1}{q} \frac{\mathrm d[\ce{Q}]}{\mathrm dt}$$
where $[\ce{X}]$ denotes the concentration of the substance $\ce{X}$, and the lower case letters represent the coefficients of those substances in the balanced equation. The rates for $\ce{P}$ and $\ce{Q}$ are positive, as these are products in the reaction.
You are right in that you can determine the reaction rate by studying the change in concentration of a particular reactant (or product) over time. But, to determine the rate r for the reaction, not just for a particular species, you must take into consideration the coefficients in the balanced equation. The stoichiometric numbers are included in the definition of the rate, $r$, so that it is independent of which reaction species you study. For example if $a = 1$ and $b = 3$ then $\ce{B}$ is consumed three times more rapidly than $\ce{A}$, but $r = -\mathrm d[\ce{A}]/\mathrm dt = -(1/3)\mathrm d[\ce{B}]/\mathrm dt$ is uniquely defined.
Source: https://en.wikipedia.org/wiki/Reaction_rate

If I understand your question correctly, you are asking about a reaction that requires multiple steps to reach completion?
Eg. In a reaction such as Compound A --> Compound B; there are intermediates formed first?
In reactions like these, the overall reaction-rate is determined by which ever step takes the most amount of time to complete - the rate determining step. This is because, no matter how quickly the reaction takes place in steps that occur before the rate determining step, there will only be a build up of "intermediate reactants". Likewise, steps that occur following the rate determining step will not speed up the overall reaction rate, because the reactants will only be made available to later steps once they have been formed by the rate determining step.


text:
Identifying unknown solution with indicators. why is one pKa value ignored and how to treat negatives values?

Three test tubes contain an identical solution of unknown pH.

The first one is tested with thymolphthalein and turns colourless
The second is tested with a-naphtholphthalein and turns blue

What colour would thmyol blue appear in the third tube?
Note: If the ratio of one form to the other form (e.g. acid to base form or vice versa) is greater than $10:1$, the colour observed will be that of the predominant form. However, if between $10$ and $0.1$, the colour will be a mixture.
\begin{array}{lccc}
\text{Indicator} &
\mathrm{p}K_\mathrm{a} &
\text{Colour of acid form} &
\text{Colour of base form} \\
\hline
\text{$\alpha$-naphtholphthalein} & 8   & \text{yellow}     & \text{blue}   \\
\text{thymol blue}                & 2   & \text{red}        & \text{yellow} \\
                                  & 8.8 & \text{yellow}     & \text{blue}   \\
\text{thymolphthalein}            & 10  & \text{colourless} & \text{blue}   \\
\hline
\end{array}

Steps to solve this question:

Find pH of solution via Henderson-Hasselbalch equation:
\begin{align}
  \mathrm{p}K_\mathrm{a}
  &= \ce{pH} - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}
\end{align}
For α-naphtholphthalein, as it appears blue, its basic form dominates and so
\begin{align}
\frac{[\ce{In-}]}{[\ce{HIn}]} &> 10:1 \\
  \mathrm{p}K_\mathrm{a}
  &= 8 =  \ce{pH} - \log_{10} (10)
   =  \ce{pH} - 1\\
\ce{pH} &= 9
\end{align}
For thymol blue
\begin{align}
  \mathrm{p}K_\mathrm{a}
  &= \ce{pH} - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}\\
  8.8
  &= 9 - \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}\\
  \log_{10} \frac{[\ce{In-}]}{[\ce{HIn}]}
  &= -0.2
\end{align}

Where I get stuck:

How do I handle the negative sign in negative 0.2?
If it was positive 0.2, this would tell me that the ratio of $\log_{10} ([\ce{In-}]/[\ce{HIn}])$ is between $10$ and $0.1$, hence a mix between the acid colour and basic colour would result, giving a green solution. But I don't know how to treat the negative.
In my solution manual they ignore the $\mathrm{p}K_\mathrm{a}=2.0$ value of the thymol blue and only use the $\mathrm{p}K_\mathrm{a}=8.8$ value. Why is this the case?

There is nothing special about the negative sign. Also, $\log_{10}$ of the ratio is not the ratio itself; to find the latter, you need to do some more arithmetic. As for the other pKa, why, since our pH is far from it, the corresponding equilibrium is shifted all the way to one side, and is as good as dead for most purposes.

As you noted, the $\mathrm{pH}$ is fairly close to 9.
It's probably quite different from the $\mathrm{p}K_{\mathrm{a}}$ value of $\alpha$-naphtholphthanlein because if you have any of the yellow acidic form around, the solution will take on a greenish tinge. One order of magnitude seems sufficient so, $\mathrm{pH} \geq 9$
The sample with thymolphtalein follows the same logic in that since we don't see any blue, there must be little of the base form, so the $\mathrm{pH} \leq 9$.
9 is very close to 8.8, the $\mathrm{pH}$ where we would expect equal forms of thymol blue. Indeed, $10^{0.2}\approx 1.05$, so both forms are present in solution in roughly equal amounts.
Further assuming that intensity of the two colors is similar, you would have a mix of yellow and blue, also known as green.

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


text:
What is the difference between a phosphoryl group and a phosphate group?
I am looking at covalent modification of enzymes and I found out that only one of these types is phosphorylation. But I do not understand what the difference between the a phosphate and a phosphoryl group is in terms of nomenclature.

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

@Mithoron It simply says "phosphoryl group should not be confused with a phosphate group." , but does not explain why.

The difference is basically one oxygen atom. If you consider monoethyl phosphate as an example, the ethyl alpha-carbon bears a phosphate group, whereas the ethoxy oxygen bears a phosphoryl group.

I added a diagram for you, feel free to elaborate on it if you want.

your diagrams they both look pretty much the same just some different colours. But looking https://en.wikipedia.org/wiki/Phosphoryl_group  I see they look quite different. e.g. different number of oxygens.

Phosphates are esters of phosophoric acid, $\ce{HO-P(O)(OH)2}$, when one or more its $\ce{OH}$ groups are replaced by (substituded) alkyloxy (alkoxy) groups. E.g. Ethyl phosphate, $\ce{CH3CH2-O-P(O)(OH)2}$. In biochemistry, where the ‘alkyls’ are very complicated, the names are based on the alcohols: ethanol phosphate.
A compendium of various IUPAC books info, IUPAC Gold Book, lacks ‘phosphoryl’ term, however, for ‘phosphorylation’, states:

phosphorylation
A process (catalysed by enzymes) involving the transfer of a phosphate group from a donor to a suitable acceptor; in general, an ester linkage is formed (…)

See, not ‘phosphoryl’, but ‘phosphate’ group.
Phosphoryl is very different group. Cite from IUPAC Nomenclature of organic chemistry, preferred names (2013):

P-67.1.4.1.1.2 Fundamental acyl groups for substituent derived from mononuclear noncarbon oxoacids
(…)
$\ce{-P(O)<}$
phosphoryl

It's the same in inorganic chemistry. E.g. phosphoryl chloride, $\ce{POCl3}$. Cite from IUPAC Nomenclature of Inorganic chemistry, 2005:

Table III, Suffixes and endings
(…)
oryl
Ending of prefixes for substituent groups formed by removing all hydroxy groups from ‘oric’ acids, e.g. phosphoryl, P(O), from phosphoric acid.

I honestly don't know why the process of introducing the phosphate group is called phosphorylation. I guess is sounded much better than e.g. ‘phosphatization’ to the term designer, and analogous to e.g. ‘methylation’, but he/she did not realize, that ‘phosphoryl’ already has a (different) meaning. Btw, the $\ce{-P(O)(OH)2}$ group is ‘phosphono’ group, see e.g. this answer or this question.
UPDATE
… and the (di)anionic form of the phosphono group is named phosphonato group.

P-72.6.1 Prefixes for anionic centers derived from acid characteristic groups
(…)
$\ce{-CO-O^-}$
carboxylato (preselected prefix)
$\ce{-SO2-O^-}$
sulfonato (preselected prefix)
$\ce{-P(O)(O^-)2}$
phosphonato (preselected prefix)
(…)

There's also biochemistry related document[1], effectively discouraging the use of “phosphoryl” for phosphate group names of biochemicals:

Phosphoric esters, $\ce{RO-PO(OH)2}$, are named
as O-substituted phosphoric acids or as substituted alcohols (Table I).
Thus, choline O-(dihydrogen phosphate) and O-phosphonocholine are
both appropriate names. The latter may be contracted to phosphocholine,
but not changed to phosphorylcholine; “phosphoryl” is defined
as OP and requires, if used,
the naming of all three groups attached to the
phosphorus atom. However, “phosphoryl” is retained
in derived terms such as the names of enzymes
(e.g., phosphorylase) or of processes (e.g.,
phosphorylation).
(…)

Table I (I won't reproduce it for technical reasons), lists many examples of (1) phosphate names, (2) O-phospho/phospho names, (3) systematic names (the first two types marked as recommended for biochemical usage), and (4) abbreviations. No name uses “phosphoryl”; some examples:

D-Fructose 1,2-bis(phosphate)
1,6-Di-O-phosphono-D-fructose;
1,6-Bis(phospho)-D-fructose
D-Fructofuranose 1,6-bis(dihydrogen phosphate)
Fructose-1,6-P2

D-Glyceraldehyde 3-phosphate
3-O-Phosphono-D-glyceraldehyde;
3-Phospho-D-glyceraldehyde
2-Hydroxy-3-oxopropyl dihydrogen phosphate;
2-Formyl-2-hydroxyethyl dihydrogen phosphate;
D-Glyceraldehyde 3-(dihydrogen phosphate)
D-Glyceraldehyde-3-P

(oxidated - carboxylate anionic form of the above aldehyde)

D-Glycerate 3-phosphate
3-O-Phosphono-D-glycerate;
3-Phospho-D-glycerate
(R)-[2-Hydroxy-3-(phosphonooxy)propanoate];
(R)-[2,3-Dihydroxypropanoate 3-(dihydrogenphosphate)]
D-Glycerate-3-P

On the other hand, despite of this recommendation, somewhat related recent document operates with “phosphoryl” name for phosphate unit frivolously.[2]
If I borrow the image from the neighbor answer

these would be properly names phosphonatooxy and O-phosphonato, respectively. And, for protonated non-anionic forms, phosphonooxy and O-phosphono. No “phosphoryl”.
However it's for names where one needs to name these groups as prefixes. The correct name of the structure in the picture still is ethyl phosphate.
References:

Nomenclature of Phosphorus-Containing Compounds of Biochemical Importance. (Recommendations 1976). IUPAC-IUB Commission on Biochemical Nomenclature. Biochem J 1978, 171 (1), 1–19.
Blackburn, G. M.; Cherfils, J.; Moss, G. P.; Richards, N. G. J.; Waltho, J. P.; Williams, N. H.; Wittinghofer, A. How to Name Atoms in Phosphates, Polyphosphates, Their Derivatives and Mimics, and Transition State Analogues for Enzyme-Catalysed Phosphoryl Transfer Reactions (IUPAC Recommendations 2016). Pure and Applied Chemistry 2017, 89 (5), 653–675.

Thanks. From what you've said I understand that phosphorylation involves the addition, to a molecule,  of a phosphate group not a phosphoryl group.  Any idea why https://en.wikipedia.org/wiki/Phosphorylation   "In chemistry, phosphorylation of a molecule is the attachment of a phosphoryl group.....A phosphoryl group should not be confused with a phosphate group....The phosphoryl group plays the central role in phosphorylation...."      Whereas this other link agrees with you   https://en.wikipedia.org/wiki/Protein_phosphorylation  "...phosphate group..."


text:
What is the common denominator of all the lipids?
I've read the lipids article on Wikipedia, but I couldn't understand the answer for my question and I'll explain why.
In biochemistry we have monomers and polymers while the monomer is the basic unit molecule for the polymere. In protein for example, we have amino acid that this is the basic unit of the proteins. Amino acid is characterized as structure of 4 groups: Alpha carbon, amine group, carboxylic group, hydrogen and R chain. this is the basic of all the amino acids.
But when we are talking about lipid as a main family of molecules (like what we have we have with protein) then I can not understand what is the "amino acids" (parallel) of them, i.e. what is the basic unit of the lipids and what is the structure.
Well, I could understand that the fatty acids are the analogous for the amino acids, but the problem is that fatty acids according the article (and other schemes that I saw on google pictures), they are considered as subgroup of the lipids, and that says that not all of the lipids are made of them. So I don't have clue for the answer to my question.

First sentence in the second paragraph of the wikipedia article: "Scientists may broadly define lipids as hydrophobic or amphiphilic small molecules [...]. This category is even broader than fatty acids ;)

Have a look at the definition of lipids in the IUPAC Gold Book. Lipids is as exact as calling a cow a four-legged animal that gives milk ;)

There are several types of lipids. Usually lipid molecules consist of long chains of carbon atoms with few polar atoms, thus making them hydrophobic (water-fearing).  Triacylglycerols (triglycerides), a glycerol molecule attached to 3 fatty acids of  variable length are for example mostly used as an energyy source in the body. However they are not the only 'type' of lipids that exist.  Another type of lipids are for example sterols, such as cholesterol, also water-fearing but have a completely different structure than triglycerides. They're both used differently in the body.

As you note, most biomolecules — or more precisely primary metabolites — exist as polymers that can theoretically be infinitely long and consist of well-defined monomers. This is true for nucleic acids and the building block nucleotides, for proteins made up of amino acids and for carbohydrates made up of monosaccharides.
Lipids, on the other hand, do not have that kind of polymeric possibility. In essence, they are trimers or monomers by definition, depending on how you look at them. Furthermore, all lipids contain the same (with same I mean identical) connective molecule namely glycerin or propan-1,2,3-triol $\ce{CH2OH-CHOH-CH2OH}$. Glycerin has three ‘sockets’, if you wish, namely its three hydroxy groups.
The ‘plugs’ that connect to glycerin always contain an acid functionality: typically this is either a carboxy group (fatty acids) or a phosphate (e.g. phophatidyl choline). The bond between ‘plug’ and ‘socket’ is an ester bond.
Both the fatty acids and the substituted phosphates can be structurally diverse. If you wanted, you could label the different fatty acids ‘monomers’ and the overall lipid molecule a ‘trimer’. However, the most typical difference between fatty acid residues is simply their length ($\ce{C16}$, $\ce{C18}$ or $\ce{C20}$ — typically multiples of 2) and their degree of unsaturation (e.g. stearate, oleate, linoleate and linolenate have increasing degrees of unsaturation). Attempting to teach students more than a few exemplary fatty acid residues or phophate residues is rather pointless since the choice of fatty acid hardly has any effect on the properties of the overall molecule, and aside from generally knowing what constitutes cellular membranes the same can in principle be said for the phosphates. Hence, students are hardly ever bothered.


text:
What is the name of the chemical compound: Cr(CO3)3?
Cr(CO${_3}$)${}_3$
This is an ionic compound. Chromium is a transition metal cation and CO${_3}$ is an anion with a charge of 2-, as said in my chemistry textbook.
And since CO${_3}$ has a charge of 2-, the over all charge of CO${_3}$ would be 6-, since it has a subscript of 3 in the bolded formula above. So my guess is that since Chromium has no such subscript, its charge would automatically be 6-, which should be right because it would cancel out both charges.
Therefore it would be Chromium (VI) Carbonate? Please correct me if I am wrong.

Fairly sure this thing doesn't exist to begin with - it would immediately decompose to $\ce{CrO3 + 3CO2}$.

Scifinder agrees with @ortho. The only compound with the molecular formula $\ce{C3CrO9}$ to be found is tricarbonatochromate(III).

I'm quite sure you have a typo there and if it's not supposed to be 'am I right" question, you should take this to mind - it may be CrCO3 but not this. even if compound with your formula existed it wouldn't be ionic, but covalent.

Honestly I think even the Cr(III) compound $\ce{Cr2(CO3)3}$ doesn't exist. The aluminium and iron(III) carbonates certainly don't. With that said I'd presume this is just a nomenclature exercise and the question setter just didn't bother thinking twice about whether the compound was realistic or not.

@orthocresol Well, at least chromium(III) carbonate seems to exist.

https://chemistry.stackexchange.com/questions/63442/what-is-the-name-of-the-chemical-compound-crco33

You are pretty close. The chromium will have a 6+ charge (not 6-). I suspect this may have been a typo, because you are right in understanding the need to cancel out charges to generate an overall neutral molecule.
The only other correction is that there is no need to capitalise. So, the correct name is chromium(VI) carbonate.

The correct name, if it existed.

Actually, the correct name, regardless of whether it exists or not. Even hypothetical chemical compounds need to have correct names. Nothing in this question asked about validity of the compound. Nitrogen pentafluoride needs a name whether it exists or not.


text:
Melting Point Trend For Non Metals?
I noticed that the melting point for non metals decreases across a period (i.e. melting point of nitrogen is higher than oxygen, oxygen's is higher than fluorine, so on). It is significantly more noticeable for chlorine and argon. Why is this? Furthermore, I a question came up in my homework asking sulfur (S8) had a higher melting point than phosphorus (P4). Their explanation was that sulfur is a larger molecule, but I don't see how this affects melting point. Is there a better explanation?
Note - this is the graph I was referring to for melting point: http://antoine.frostburg.edu/chem/senese/101/periodic/images/meltingpoint.gif

as one moves along the period, the atomic radii decrease. You see, this decrease leads to an increased net Z-effective and hence, the electrons come very close. In the case of Nitrogen, which exists as a diatomic molecule - this shrinking in size is less than that for oxygen and fluorine.
    the melting point is determined by the intermolecular interactions as well as the geometry of molecules. greater the number of bonds to break, higher is the melting point. For oxygen and especially Fluorine, there can be a forced compression, but yet again - these small atoms repel each other because of their high electron density. This repulsion lowers the melting and boiling points.
As for noble gases like Argon, they do not need to form any sorts of bond. The only kind of forces acing between two atoms are very weak London forces. These London forces are responsible for the freezing of Helium, at temperature as low as -273K, very close to absolute zero!!
I hope that helps!!

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.

Even absolute zero is not low enough to freeze helium unless you mechanically apply pressure to help things along.  Blame it on quantum mechanics.


text:
Why does halohydrin formation occur?
I am so confused with addition reactions.
My book just gives random reagents and say this that happens without giving reason.
For example, in halohydrin formation with $\ce{Br2}$ and $\ce{H2O}$, why does alkene attack $\ce{Br2}$ first instead of $\ce{H+}$ of $\ce{H2O}$ in an acid base attack. Isn’t an acid base attack generally quicker and requires less energy than attacking a $\ce{Br}$ atom?

Proton transfer would certainly be quicker if there was any incentive for it to happen. Sadly in this case there's zero incentive (arguably negative incentive), since alkenes are such poor bases and water is such a poor acid. Think about the position of equilibrium for such an acid-base proton transfer; do you think that any protonated alkene (i.e. a carbocation) is going to hang around for long?

oh okay. I was suspecting that? So is there like a way we can know how to rank double bond as bases among other non alkene molecules. Like is it stronger base or weaker than molecules with lone pairs or with certain atoms? Just curious.

You've been taught to compare the stability of the conjugate acid, perhaps? The conjugate acid of an alkene is a carbocation - octet deficient, positively charged, carbon - so don't expect an alkene to be significantly acidic, compared to something like $\ce{OH-}$, which has a happy, neutral, octet-satisfied $\ce{H2O}$ as its conjugate acid.

According FMO approach, $\pi$ MO of alkene attacks $\sigma^*$(anti bonding) of $\ce{Br-Br}$. As a result three member cyclic halonium ion gets formed which is very strained. If alkene is unsymmetrical, water( acts as a nucleophile) will attack at that carbon which is lesser in hydrogen and ring gets opened up. Product will be $\ce{R-CH(OH)-CH2-Br}$.


text:
Fe/Ga cell has a negative potential?
In the following homework problem, the answer says:
"$E ^{o}_{cell}$ is negative, which makes the $\ce{Fe}$ and $\ce{Ga}$ half-reactions non-spontaneous."
But I found $E ^{o}_{cell}$ to be positive - NOT negative:
$$E ^{o}_{cell} = E ^{o}_{cathode} -E ^{o}_{anode} =
E ^{o}_{Fe} - E ^{o}_{Ga} = -0.409\,\mathrm{V} - (-0.56\,\mathrm{V}) = +0.151\,\mathrm{V}$$
which means that these half-reactions would actually be spontaneous.
Am I solving this problem wrong or did the writer of this question make a mistake?
After all, $\ce{Fe}$ should be the cathode since it has a more positive voltage. Right?

It says nonspontaneous, so aren't you looking for the combination with the negative voltage?

Yes. But they're saying that Fe and Ga are the combination with the negative voltage. However, after subtracting them, I ended up with a positive voltage. So is their answer wrong?

I'm not seeing that. They're saying that the gallium is reduced and the iron is oxidized, so you should reversing the iron equation and its potential. The sum looks negative to me.

Even if you reverse the iron's equation, you would still end up with a positive cell potential:
0.409−(−0.56)=+0.969Volts.
So I don't think you're right...

I'm still confused by what you're trying to do. The gallium reduction potential is $-0.509, \mathrm{V}$. The iron oxidation potential is $+0.409,\mathrm{V}$. The sum is negative.

Why would you take the sum? The formula is $E ^{o}{cell} = E ^{o}{cathode} -E ^{o}_{anode}$.

Good lord, I just realized what they've tried to teach you... I'll explain in an actual answer. I see what you're confused about.

You need two half reactions for the total redox reaction: one reduction and one oxidation. The convention is that we provide all potentials as standard reduction potentials. If you need an oxidation potential, you're going to need to reverse the reduction half reaction and negate the sign standard potential.
This means that the total potential of the cell is the raw sum of the oxidation and reduction potentials. In your specific example, you want:
$$\ce{Ga^{3+} + 3e- -> Ga}, E_{red}^{\varnothing} = -0.560\,\mathrm{V}$$
$$\ce{Fe -> Fe^{2+} + 2e-}, E_{ox}^{\varnothing} = +0.409\,\mathrm{V}$$
The EMF of the cell is the sum:
$$E_{cell}^{\varnothing} = E_{red}^{\varnothing} + E_{ox}^{\varnothing}$$
And this value is negative.
Now, as a shortcut, you can identify which of the two reduction half reactions needs to be reversed to make an oxidation, and you can add that value--negated--to the potential for the reduction half reaction, as above.
What your teachers or book meant was the following. I've changed the names of the variables so that they don't conflict with the standard notation (which is why we're talking past each other).
$$E_{cell}^{\varnothing} = E_{red,1}^{\varnothing} - E_{red,2}^{\varnothing}$$
where $E_{red,1}^{\varnothing}$ is the standard reduction potential for the reduction half reaction and $E_{red,2}^{\varnothing}$ is the standard reduction potential for the reduction that corresponds to the oxidation half-reaction. This is hellishly confusing, which why I think you are confused. The key here is that
$$E_{\mathrm{cathode}} = E_{red,1}^{\varnothing} = E_{red}^{\varnothing}$$
but
$$E_{\mathrm{anode}} = E_{red,2}^{\varnothing} = -E_{ox}^{\varnothing}$$
Note also that ordinarily, for a spontaneous reaction, iron would constitute the cathode of the cell because it would be the reduction. However, in this case, because the reaction is non-spontaneous, the iron half reaction is actually going to be an oxidation, so that means it's actually going to be the anode. This is the other factor you're confused about. You're trying to apply shortcuts for a spontaneous cell to a non-spontaneous one. Try to understand what the underlying principle is.

Thank you for the detailed answer. This makes much more sense now. I was taught the shortcut before the underlying principle which was what led to this confusion.


text:
How do you find the enthalpy change for a given reaction?

$0.500\,\mathrm g$ of magnesium chips are placed in a coffee-cup calorimeter and $100.0\,\mathrm{mL}$ of $1.00\,\mathrm M$ $\ce{HCl}$ is added to it. The reaction is:
$$\ce{Mg(s) + 2HCl(aq) -> H2(g) + MgCl2(aq)}$$
The temperature of the solution increases from $22.2\,^{\circ}\mathrm{C}$ to $44.8\,^{\circ}\mathrm{C}$. What’s the enthalpy change for the reaction, per mole of $\ce{Mg}$? Assume specific heat capacity of solution is $4.20\,\mathrm {J\,g^{-1}K^{-1}}$ and density of the $\ce{HCl}$ solution is $1.00\,\mathrm {g/mL}$.

I don't understand how to solve this; does this have to do with heat capacity equation? (Or is that just a completely different thing).
$$q = mc\Delta T$$
(Also note this is not a homework problem. The answer should be $\Delta H = -4.64 \cdot 10^{5}\,\mathrm{J\,mol^{-1}}$ $\ce{Mg}$, but I don't know how to get there and would like to understand.)

You're on the right track. The enthalpy change is reflected in the heat released the reaction (you immediately know it should be negative). This heat is absorbed by the water, raising its temperature. You need to back compute the amount of heat using the temperature change and then determine how much heat per mole of magnesium.

Step 1 Calculate $q_\text{solution}$. To do so, you have to have in mind that the mass of the solution is aproximatly equals to $100\ \mathrm{ml}$ of $\ce{HCl}$ plus the mass of $\ce{Mg}$.
\begin{align*}
   q_\text{solution}&=100.5\ \mathrm g\cdot4.20\ \frac{\mathrm J}{\mathrm{mol\cdot K}}\cdot\left(318.0\ \mathrm K-295.4\ \mathrm K\right)\\
   &=9.54\cdot 10^3\ \mathrm J
  \end{align*}
Step 2 Calculate $q_\mathrm{rxn}$
\begin{align*}
 q_\mathrm{rxn} + q_\text{solution} &=0\\
 q_\mathrm{rxn} + 9.54\cdot 10^3\ \mathrm J &= 0
\end{align*}
Step 3 Calculate the value of $\Delta H$ per mole.
\begin{align*}
 \Delta H&= \left(\frac{-9.54\cdot 10^3\ \mathrm J}{0.500\ \mathrm g\ \text{of}\ \ce{Mg}}\right)\left(\frac{24.31\ \mathrm g\ \text{of}\ \ce{Mg}}{1\ \mathrm{mol}\ \text{of}\ \ce{Mg}}\right)\\
 &= -4.64 \cdot 10^5\ \frac{\mathrm J}{\mathrm{mol}}\ \text{of}\ \ce{Mg}
\end{align*}


text:
What are the reactions occurring during the titration of a carbon dioxide contaminated water sample with diluted sodium hydroxide solution?
During estimation of dissolved $\ce{CO2}$ in water, we took a water sample, added few drops of phenolphthalein and titrated it against dil $\ce{NaOH}$.
Questions
What are the reactions that happen:

initially  in the water sample?
after addition of phenolphthalein (if any)?
after addition of NaOH?

My speculations

Inside the sample water
I think dissolved $\ce{CO2}$ reacts with $\ce{H2O}$ to form $\ce{H2CO3}$. But then does all $\ce{CO2}$ get converted into $\ce{H2CO3}$?
After addition of phenolphthalein
In an acidic medium the equilibrium shifts to the left and the weak acid phenolphthalein ($\ce{HIn}$) mostly remains undissociated and hence colourless.

$$\ce{HIn <- H+ (excess) + In-}$$

After addition of $\ce{NaOH}$
According to my book after $\ce{NaOH}$ is added to the sample the following occurs:
\begin{align}
\ce{2NaOH + CO2 &= Na2CO3 + H2O}\\
\ce{Na2CO3 + H2O + CO2 &= 2NaHCO3}\\
\end{align}
From my understanding
\begin{align}
\ce{NaHCO3 &= Na+ + HCO3-}\\
\ce{HCO3- + H2O &= H2CO3 + OH-}\\
\ce{OH- + H+ + In- &= H2O + In-}\\
\end{align}
As the $\ce{H+}$ concentration decreases equilibrium shifts to the right
$$\ce{HIn -> H+ + In-}$$
This $\ce{In-}$ ion imparts the pink colour which at pH 8.3 becomes susceptible to our eyes.

At the beginning some of the dissolved $\ce{CO2}$ got converted into $\ce{H2CO3}$. They remain beyond the scope of estimation, isn't it?
Are my speculations correct?

Your thinking is in principle correct, let me just add a few more things, that may help you further.
What you are effectively titrating is carbonic acid, or more precise the hydrogen carbonate ion. Carbonic acid is not very stable in water, an most of the carbon dioxide stays the way it is, i.e. dissolved and interaction via van der Waals forces. The main equilibria are as follows (and derivations thereof):
$$\begin{align}
\ce{CO2 + 2H2O &<=> HCO3- + H3O+}\tag1\label{major}\\
\ce{HCO3- + H2O &<=> H2CO3 + OH-}.\tag2\label{minor}\\
\end{align}$$
In a contaminated water sample $\eqref{major}$ is by far the more predominant reaction.
When you now add sodium hydroxide solution you shift $\eqref{major}$ to the right side, as you are removing hydrogen carbonate ions via
$$\ce{HCO3- + OH- <=> CO3^2- + H2O}\tag3\label{conversion}$$
As you correctly identified, phenolphtalein is a weak acid by itself and the colour change happens via
$$\ce{H2In + 2OH- <=> In^2- + 2H2O}.\tag4\label{indicator}$$
Once you have added enough sodium hydroxide solution to completely shift the equilibrium $\eqref{conversion}$ to the right, you will start deprotonating the indicator molecule via $\eqref{indicator}$.
It is important to understand, that all these reactions happen at the same time and they keep on happening. This is the basic characteristic of a dynamic equilibrium. Since you are changing the composition of these equilibria the overall composition changes and the system adjusts, see Le Chatelier's principle.


text:
Oxidation of malonic acid

Question
  How many grams $\ce{CO2}$ will result when burning $\pu{100 g}$ malonic acid?
  The formula for malonic acid is $\ce{C3H4O4}$ so 3 $\ce{CO2}$ molecules should result from oxidating one malonic acid molecule.
  $$\ce{C3H4O4 + 2O2 -> 3CO2 + 2H2O}$$

My attempt
$$44.01\cdot\frac{100}{104.0615}\cdot3 = \pu{126.88 g}$$
with $\pu{44.01 g}$ being the mass of a mol $\ce{CO2}$ and $\pu{104.0615 g}$ for a mole of malonic acid.
However, this doesn't seem to be right. Can anyone hint me for the correct solution?

Welcome! Could you comment why it doesn't seem right, please. Apart from a little inconsistent use of significant figures, I can't see anything wrong either.

@Martin It is kind of an online riddle and my answer isn't accepted

Why can't it be right? You're converting $\ce{C3H4O4}$ to something that has the same equivalent weight as $\ce{C3O6}$, which is has a higher weight than malonic acid.

You actually are correct, the value is greater than 100g because when you burn a hydrocarbon, it is actually an exothermic redox reaction with oxygen.  Meaning that the carbon in the compound is combining with atmospheric oxygen.  So even if you use 100g of malonic acid, you could end up with more than 100g of CO2.  Your answer is correct.

Using Bunsen burner for this reaction will most likely result in the reaction you've shown. But please keep in mind that even aqueous solution of malonic acid upon heating above $\pu{70 ^\circ C}$ easily undergoes decarboxylation:
$$\ce{HOOC-CH2-COOH ->[\Delta] CH3-COOH +CO2}$$
If your "burning conditions" are rather mild, then you get three times less carbon dioxide: $m(\ce{CO2}) \approx \pu{42.3 g}$.
Apart from that, your result look fine to me as well.


text:
What phase of matter is shaving cream?
Our teacher asked us if shaving cream is a solid, liquid, or gas. She didn't quite know herself.

Possible duplicate of Is toothpaste solid or liquid?

Shaving cream is a special type of mixture called a colloid, consisting of tiny air bubbles dispersed in liquid. This particular type of colloid can further classified as a foam. Since the mixture itself consists of two distinct phases, it doesn't really make sense to classify the entire mixture as one phase. If you had to pick one though, it is most nearly a liquid since that is the dispersion medium.


text:
Why Does an Area in Phase Diagram Considered as Having Degree Of Freedom=1?

Why did the area in the center are Considered as Having 1 Degree Of Freedom. Aren't we able to change the concentration and the temperature at the sametime without changing the Phase?
Here is the link to the document https://www.google.co.id/url?sa=t&source=web&rct=j&url=http://www.colby.edu/chemistry/PChem/notes/GibbsPhaseRule.pdf&ved=0ahUKEwjw7Laq7M3QAhXGQI8KHXwKB3sQFggcMAE&usg=AFQjCNHWRoz8CuB0BSOoLJm-K0iPayNvgQ&sig2=vYhsTSV_woeOd2_GPPLdEw

The Gibbs phase rule is $f = c - p + 2$, where $f$ is the degrees of freedom, $c$ is the number of components, and $p$ is the number of phases.  Your diagram says $P = \mathrm{cst}$, which I interpret to mean that pressure is constant.  That specification lowers the number of degrees of freedom by one, so $f = c - p + 1$.
Inside the "center" region, liquid and vapor phases coexist.  Thus there are two phases, and (as with every point on this particular diagram) two components, thus $f = 2 - 2 + 1 = 1$.
Your question is why.

Aren't we able to change the concentration and the temperature at the same[ ]time without changing the [p]hase?

Essentially, no, you can't do that.  Inside the center area, liquid and vapor coexist at equilibrium.  And -- this is the key point -- that equilibrium is another constraint that lowers the degree of freedom.  The compositions of the both the liquid and the phase vapor are not independent of temperature, given the constraint that the liquid is in equilibrium with the vapor.
The composition of the liquid phase will be given by the lower curve.  And the composition of the vapor phase is given by the upper curve.
Temperature as the single degree of freedom
So if you decide to use temperature as your (first and only) degree of freedom, then the composition of both the liquid and the vapor is fixed.  Say you pick an arbitrary temperature $T_{arb}$, which must obviously be between $T_{bB}$ and $T_{bA}$.  This temperature could be represented by a horizontal line on your chart.
The composition of the liquid phase is given by where that horizontal line for $T_{arb}$ intersects the lower curve on your chart.  And the composition of the vapor phase is given by where that same line intersects the upper curve.
Thus now that we've specified temperature, the compositions of both phases is also specified, and (given that pressure is constant) thus the system is fully specified.  No more degrees of freedom.
(Liquid) composition as the single degree of freedom
Instead of picking a temperature then let's say you pick an arbitrary liquid composition $x_{Aarb}$.  Then, given that vapor and liquid are in equilibrium, then the temperature is specified.  Your specified $x_{Aarb}$ is a vertical line on your chart.  The temperature of the system is given by the value of the ordinate where your $x_{Aarb}$ line intersects the lower curve on the graph.
Now that you know temperature, you can find the corresponding $y_{Aarb}$ by drawing a horizontal line at the newly identified temperature and finding where this horizontal line intersects the upper curve.
Extensive vs. intensive properties
The goal of the Gibbs phase rule is to describe all the intensive degrees of freedom of a system.  It doesn't tell you about extensive degrees of freedom.  The extensive degrees of freedom tell you "how much" of the system you have.  In this system, the extensive degrees of freedom are the absolute amounts of each phase.  (The compositions of each phase are intensive properties.)  In general for $p$ phases there are $p$ extensive degrees of freedom.
As a simple example, consider a single component system (e.g. pure water) at a constant pressure and in equilibrium with its vapor.  The Gibbs phase rule says $f = c - p + 1 = 1 - 2 + 1 = 0$, i.e. that there are no degrees of freedom.  But there can be any amount of water, from one picomole to one examole, and the equation is still true.  Since there are two phases, there can be one picomole of steam and one examole of liquid water, or vice versa.  The Gibbs phase rule isn't going to tell us how much of each phase we have.
The same idea holds in the two-component case that you are asking about.  The total amounts of liquid and vapor can vary in the center region.  The relative abundance of each phase would be given by the "lever rule" as described in this nice electronic lecture note from the University of Omaha.


text:
Are triglycerides soluble in polar solvents?
In my textbook it is said that they are soluble in non-polar solvents (which makes sense), but also soluble in polar solvents (halogenoalkanes and some ethers like ethoxyethane) It is said that the polar solvents can induce a dipole in the fatty acid chain. If this is the case, why does water not induce a dipole in the chain?

"halogenoalkanes and some ethers like ethoxyethane" aren't particularly polar.

Are you sure about that? You don't think carbons bonded to some of the most electronegative atoms produces a dipole?

Dipole of molecule and polarity of solvent are different things. 2) By not particularly, I mean more polar than alkanes but much less then DMSO or HMPA, or water.

Polar is not a black-and-white distinction. There are weakly polar compounds (like ether), medium polar compounds (dichloromethane) and strongly polar compounds (dimethyl sulfoxide). Fats (i.e. triglycerides) can dissolve in weakly and maybe moderately polar solvents but not in strongly polar ones.
While a dipole can be induced in triglycerides to some extend, most of the long nonpolar side chain can only give weak induced dipoles. Thus, weakly polar solvents may still dissolve these by dipole–induced dipole interactions but the triglycerides are just too nonpolar for strongly polar solvents.
Water is one of the most polar solvents and additionally able to hydrogen bond. While triglycerides can accept hydrogen bonds around the glycerin moiety, they are just not polarisable enough for water to act. Hence, water prefers to form its own phase.

I think it depends on the number od carbon atoms in fatty acid chain. For example butyric acid is a carboxylic acid, like fatty acids, and it is partially soluble in water. All fats and oils are partially soluble in cold methanol, ethanol, acetone, but soluble when they are heated, so there is also this effect.
Also, it depends what do you think is polar or non-polar enough, for example you mentioned ether. Ether is considered non-polar rather than polar, but yet it is more polar than hexane or pentane for example.
And yes, polar solvents can induce dipoles in non-polar compounds, that is dipole-induced dipole interaction, type of van der Waals forces. But those interactions are not favorable in the case of water for example, since water is "trapped", molecules of water would rather build hydrogen bonds instead of dipole-induced dipole interactions. That is why water escapes from nonpolar molecules and nonpolar molecules build micelles. Some compounds and solvents are emulsifiers so they distribute micelles, and prevent them to form emulsions, so it seems like fats remained soluble.


text:
Use of Na2S2O3 before starch in Winkler's method
In Winkler's method why is some amount of $\ce{Na2S2O3}$ added before addition of starch solution turning the deep straw coloured solution pale straw?
Is it to generate $\ce{I-}$ ions needed for formation of $\ce{I3-}$ in the $\ce{I2}$ containing solution? This $\ce{I3-}$ inturn reacts with the starch to form blue-black complex.
However once we didn't add $\ce{Na2S2O3}$ and the addition of starch solution did turn the solution it blue-black.

Related: http://chemistry.stackexchange.com/questions/27450/finding-out-the-concentration-of-oxygen-in-water/27469#27469

@KlausWarzecha thanks but it doesn't answer the question.

I know :D Therefore it is just a comment, not an answer ;)


text:
Use of pipette during experiments for determination of oxygen by Winkler's method
We were performing estimation of $\ce{O2}$ by Winkler's method and we were told by the instructor that while adding $\ce{MnSO4}$ and $\ce{NaOH}$ / $\ce{KI}$ the pipette tip should be well below the surface of the solution.
When asked as to why this should be done, he said that if the reaction took place at the surface then there's a possibilty that the atmospheric $\ce{O2}$ may get dissolved.
Is it true?

[...] the pipette tip should be well below the surface of the solution [...]

Usually, one would indeed keep the pipette above the solution to avoid that a (minor) part of the solution forms a drop (or film) on the tip of the pipette and is carried away.
Here, when the amount of oxygen dissilved in the solution is to be determined, another effect prevails.
If you add your reagents from greater height, you might create a splash, just like throwing a rock into a lake. As a result, you will mix some oxygen from the gas phase into the solution, which might taint the measurement.

I was always taught that, no matter what the solutions are, never immerse the pipette.  Always touch it to the inside wall of the target container as close as possible to the receiving solution's surface.


text:
Deriving the equation for oxygen estimation by Winkler’s method
We were doing $\ce{O2}$ estimation by Winkler's method
The procedure and reactions involved in this experiment are:

Carefully fill a 300-mL glass stoppered bottle brim-full with sample water.
Immediately add 2mL of manganese sulfate to the collection bottle.
Add 2 mL of alkali-iodide reagent (NaOH + KI) in the same manner.
Stopper the bottle with care to be sure no air is introduced. Mix the sample by inverting several times.
Add 2 mL of concentrated sulfuric acid.Then mix well.
In a glass flask, titrate 100 mL of the sample with sodium thiosulfate to a pale straw color. Titrate by slowly dropping titrant solution from a calibrated pipette into the flask and continually stirring or swirling the sample water.
Add 2 mL of starch solution so a blue color forms.
Continue slowly titrating until the sample turns clear.

$$\begin{align}\ce{MnSO4 + 2NaOH &-> \underset{{white ppt.}}{Mn(OH)2} + Na2SO4}\\[0.6em]
%
\ce{2Mn(OH)2 + O2 &-> \underset{{orange-brown cloudy ppt.}}{2MnO(OH)2}}\\[0.6em]
%
\ce{2Na2SO4 + 2H2SO4 &-> 4 NaHSO4}\\[0.6em]
%
\ce{MnO(OH)2 + NaHSO4 + 2KI &-> \underset{{golden-brown colour}}{I2} + MnSO4 + K2SO4 \\&\quad\quad + 2Na2SO4 + 3H2O}\\[0.6em]
%
\ce{I2 + 2Na2S2O3 &-> 2NaI + Na2S4O6}\end{align}$$
And at the end of the experiment we needed to calculate the $\ce{O2}$ concentration in mg/l by the following equation:
$$c(\ce{O2})\ \mathrm{(mg/L)} = \frac{(V_1 \times N \times E \times 1000)
}{[V_4 {(V_2-V_3)/V_2}]}$$
Here,
$V_1=$ Volume of titrant $\ce{Na2S2O3}$ (ml)
$V_2=$ Total volume of water sample (ml)
$V_3=$ Volume of ($\ce{MnSO4 + KI}$ solution) added (ml)
$V_4=$ Volume of analyte taken in titration
$E=$ Equivalent weight of $\ce{O2}$
$N=$ Normality of $\ce{Na2S2O3}$ solution
So how can I derive this equation?

Reference followed: Ghosh, K. C. & Manna, B. (2005) Practical Zoology

A different book though had a different formula :
Conc. of O2 (mg/l)= 0.2 x (Volume of sample water in ml) x (Strength of $\ce{Na2S2O3}$)

Are you sure is $V_2 - V_3$?

@Another.Chemist Yes I copied it from my book.

[Concepts]

Normality, N: The normality is the relation between the sustance equivalence and the liters of a solution. The sustance equivalents are referred to the charges per mole of a sustance. The number of equivalents (NOE) of a sustance is obtained as follows:

Acid equal to number of hydrogens.
Hidroxide equal to number of hydroxils.
Salt equal to number of cations.

To determine the normality of a solution, these are the steps you must follow:

Weight Equivalent Gram, WEG:
\begin{equation}
    WEG=\frac{Molecular\ weight\ of\ solute}{NOE}
   \end{equation}
Weight Equivalent Number, WEN:
\begin{equation}
    WEN=\frac{Mass\ of\ solute}{WEG}
   \end{equation}
Normality, N:
\begin{equation}
    N = \frac{WEN}{L}
   \end{equation}
Doing some basic algebra, you find that:
\begin{equation}
   N=M\cdot NOE
  \end{equation}
Where M stands for Molarity.

Dilution factors, DF: It is employed when the solution concetration of your sample is too concentrate. In order to determine the concentration of the substance you want to find, you have to dilute the initial solution. This is done through the dilution factors. I.e. you have 25 ml of a solution 10 M and you are asked to dilute it 100 times. To do so, you have to take an aliquot (A portion of a large whole) of 0.250 ml, pour into a volumetric flask of 25 ml and then fill it with destilled water to capacity. Then, you have a diluted solution of 0.1 M. In general, a dilution factor is expressed as:
\begin{align}
   Dilution\ factor=\frac{Total\ volumen}{Sample\ volume}
  \end{align}
Dilution: Dilution is the process of decreasing the concentration of a solute in solution, usually simply by mixing with more solvent. To dilute a solution means to add more solvent without the addition of more solute. This can be represented as:
\begin{align}
   V_1\cdot M_1 &= V_2\cdot M_2\\
   M_2&=\frac{V_1}{V_2}\cdot M_1
  \end{align}

[Answer]
Now, if you put all together:

If you look at the first parenthesis at the right of the equa, having in mind that E is WEG and doing some basic algebra:
\begin{align*}
   V_1\cdot N\cdot E\cdot 1000&=V_1(L)\cdot\frac{WEN}{L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{\frac{Mass\ of\ solute}{WEG}}{L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{Mass\ of\ solute}{WEG\cdot L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{Mass\ of\ solute}{L}\cdot 1000\\
   &=V_1(mL)\cdot\frac{Mass\ of\ solute}{mL}\\
   &=Mass\ of\ solute\cdot\frac{V_1(mL)}{mL}\\
   &=Mass\ of\ solute
  \end{align*}
The square bracket is just a dilution factor.
If you put together the last two steps with the dilution section, you find a similarity.

[History]
If you are interested in the origins of this method, I recommend you the following paper:

Winkler, L. "Die Bestimmung des in Wasser Gelösten Sauerstoffes". Berichte der Deutschen Chemischen Gesellschaft. (1888) 21, 2, 2843–2855.

I cannot give you a brief handout of it, my german is basic. But, I want to point out that in page 2845, you will find an equation similar than the indicate in dilution section.
[How to understand the Winkler method]

The Winkler method for the determination of dissolved $\ce{O2}$ involves adding an excess of ‘white’ $\ce{Mn(OH)2}$ which contains $\ce{Mn^{2+}}$ ions.

For every molecule of $\ce{O2}$ two molecules of $\ce{I2}$ are formed. The quantity of iodine present can be found though titration of the sample with sodium thiosulphate.

[Bibliography]

Shashi,  B.P. A complete review on: effluent testing and treatment in pharmaceutical industry.

How do I find out the equivalent weight of O2 here? How do I derive the number of equivalents (n) from the reactions?

Your 2nd reaction is a redox one. Therefore, the manganese is oxided by loosing two electrons and the oxigen is reduced by acepting 2 electrons.

Then, look at weg formula.

I get that.  But as Na2S2O3 is reacting with I2 and not O2. I wonder how's this formula V1 S1 =V2 S2 going to calculate the conc of O2 present.

On your presented reactions, just go stoichiometrically backwards, having in mind that all reactions are redox.

I don't get you.


text:
Difference between neutral and basic activated alumina
Sigma Aldrich sells two different types of activated alumina for column chromatography: neutral and basic.
Apparently basic alumina is more active and better at adsorbing anions.
Why is this the case?


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Which potentials are suitable for molecular dynamics simulation of simple explicit solvents?
Which potentials (force fields) are suitable for adequately simulating (mixtures of) explicit solvents? Interesting rheological properties include viscosity, surface tension, heat capacity.

I would look at polarizable force fields like AMBER/GAFF or CHARMM.

If you are still interested in this, it would probably be a good fit on the new Matter Modeling SE.


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Why doesn't Epsom salt dissolved in water act like sulfuric acid?
People take baths in it so it clearly doesn't. But if epsom salt is the salt of sulfuric acid when it dissolves the chemical ions present in sulfuric acid should also be present in the bath water and in abundance enough to melt you.. right?
I also remember that water autodissociates into hydronium and hydroxide, so my other question is why wouldn't plain water simultaneously be a acid and a base and also just melt you?
If there's a trillion (or whatever) water molecules next to your skin and even a tiny fraction of them turn into an acid and some of them turn into a base, there'd cumulatively be a lot of acid on your skin. I can't imagine adding salt which would also ionize could help things.

Well, as you might have guessed, it's the $\ce{H+}$ in sulfuric that is dangerous, not the $\ce{SO4^2-}$. So, Epsom salts, being $\ce{MgSO4}$, aren't very dangerous.
In the same vein, table salt ($\ce{NaCl}$) is perfectly OK despite hydrochloric acid ($\ce{HCl}$) being rather nasty.
And $\ce{H+}$ is only corrosive in high concentrations, mind you - otherwise we'd be burning ourselves drinking orange juice, which itself has a small concentration of $\ce{H+}$. That also answers why water doesn't kill you: only a tiny fraction of it (roughly 1 in ten million molecules) actually undergoes autodissociation at any one time. It's not about the actual number of $\ce{H+}$ ions, but rather the concentration. If you dilute $\ce{HCl}$ enough you can pretty much just drink it. (Please don't be stupid and take this as a licence to do a home experiment and drink $\ce{HCl}$.)

Yeah but wouldn't the magnesium 2+ ion suck some hydroxides off from some otherwise stable dihydrogen monoxides and thus increase the hydrogen ion concentration (if only temporarily)? Also how did you do super scripts and bold your chemicals?

That can happen but again it is a question of to what extent it happens. The answer is that it happens to such a small extent that you might as well imagine it not happening. People have tabulated data that quantitatively describe that reaction, and someone else might do the number crunching (I'm not free now), but they'll tell you exactly the same thing: that happens to only a tiny extent. TLDR: It's not a black and white thing where "has some $\ce{H+}$ = death". For formatting see here.

Okay I think that answers it, also thanks man!


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How does Lindlar's catalyst work?
I can not find a mechanism online. I understand that some poisoning effect causes the catalyst to be unable to reduce an alkene to alkane; however, if the catalyst is poisoned, how does it continue to reduce alkynes to alkenes?