text:
How to calculate partition coefficients?
Below is a transcript of this doubtful question:

A weak monobasic organic acid $\ce{HA}$ is soluble in both water
  and $\ce{CHCl3}$.  $\pu{500.0 cm3}$ of a solution of $\ce{HA}$ in $\ce{CHCl3}$, with a concentration of
  $\pu{0.057 mol dm-3}$,
  is shaken well with $\pu{500.0 cm3}$ of water and allowed
  to attain equilibrium at $\pu{27°C}$.  An aqueous layer and
  $\ce{CHCl3}$ layer then separate out; the pH of the aqueous layer is
  found to be 3.21 under these conditions.
The dissociation constant of $\ce{HA}$ in water at $\pu{27°C}$
  is $\pu{1E-5 mol dm-3}$.

Calculate the partition coefficient at $\pu{27°C}$
  for the partitioning of $\ce{HA}$ between water and $\ce{CHCl3}$.
In a second experiment, a further $\pu{500.0 cm3}$ portion of the
  same $\ce{HA}$ solution in $\ce{CHCl3}$ in which the concentration
  of $\ce{HA}$ is $\pu{0.057 mol dm-3}$, is shaken well with
  $\pu{500.0 cm3}$ of a $\pu{0.027 mol dm-3}$ aqueous $\ce{NaOH}$
  solution and allowed to reach equilibrium at $\pu{27°C}$.
  Calculate the $\mathrm{pH}$ of the aqueous layer under these conditions.
State the assumptions you make, if any, in the above calculations.

My solution for Question 1.

Since pH is given, $\ce{[H^+](aq)}$ is determined by

\begin{align}
\mathrm{pH} &=-\log[\ce{H^+~(aq)}]\\
[\ce{H^+}~(\ce{aq})] &= 10^{-3.21}\\
    &= 6.16 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}\\
\end{align}

By appropriately applying $K_\mathrm{a}$, the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer is $0.038~\mathrm{mol\,  dm^{-3}}$.
Next, $[\ce{HA}~(\ce{aq})]$ present in the organic layer can be determined by subtracting the $[\ce{HA}~({\ce{aq}})]$ present in aqueous layer
from the initial concentration of $[\ce{HA}~({\ce{aq}})]$, which results in $0.019~\mathrm{mol\,  dm^{-3}}$.
Finally, I got $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.019~\mathrm{mol\,  dm^{-3}}} = 2$.

I hope that this is correct, but  I am not quite sure.

My incomplete solution for Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ is

$$\ce{HA}~(\ce{aq}) + \ce{NaOH}~(\ce{aq})  \longrightarrow \ce{NaA}~(\ce{aq}) + \ce{H2O}~(\ce{l})$$

The number of moles of $\ce{NaOH}$ present is $0.0135~\mathrm{mol}$.

I am confused with the concentration of $\ce{HA}$ that I needed to take for this calculation. Initially I have found the concentration of $\ce{HA}$ present in organic layer as $0.019~\mathrm{mol~dm^{-3}}$. Do I need to use that here or $0.057~\mathrm{mol~dm^{-3}}$?
What would be the assumptions that I need to assume to take out this calculation?

Next, [HA (aq)] present in the organic layer can be determined by subtracting the [HA (aq)] present in aqueous layer from the initial concentration of [HA (aq)], which results in // Typo the last bit should be from the initial concentration of [HA (CHCl3 )], which results in // Also $K_D = 2.0$ when considering significant figures

For part 2 you know that there are 0.019 moles total of (a) HA in CHCl3 (b) HA in H20 (c) A^- in H2O // You also know how HA will partition between CHCL3 and water. // You also know how H+, A- and HA relate in water via Ka // You also know that $\ce{Na+ + H+ = A^- + OH^-}$ so you have enough equations to solve for all the unknowns.

Equilibrium are really dependent on the chemical activities of the species not their concentrations. So an overriding assumption for both parts is that the activities of the chemical species is the same as the concentration. That is a reasonable for these solutions. As a rule of thumb, for solutions with concentrations more than 0.1 molar the assumption is dicey.
I think you're very close to the right solution for part 1, but I'd state it a bit differently.

Since pH is given, $\ce{[H^+]}$ in the aqueous phase is determined by

$[\ce{H^+}] = 10^{-\mathrm{pH}} = 10^{-3.21} = 6.2 \cdot 10^{-4}~\mathrm{mol\,dm^{-3}}$

The mantissa of the pH only has two significant figures, so should the $\ce{[H^+]}$

We know that in the aqueous layer the charges have to balance so $\ce{[H^+] = [A^-] + [OH^-]}$, but $\ce{[A^-] >> [OH^-]}$ so we can assume that $\ce{[H^+] = [A^-]}$
Using $\ce{[H^+] = [A^-]}$ the $K_\mathrm{a}$ equation can be solved for the $\ce{[HA]}$ present in aqueous layer which gives $0.038~\mathrm{mol\,  dm^{-3}}$.

$$\ce{[HA]} = \dfrac{\ce{[H^+][A^-]}}{1\times10^{-5}} = \dfrac{(6.2\times10^{-4})^2}{1\times10^{-5}} = 0.038$$

Next, the $\ce{[A-]}$ in the aqueous phase is insignificant compared to the $\ce{[HA]}$ in the aqueous phase, so the moles of $\ce{HA}$ present in the organic layer can be determined by subtracting the moles of $\ce{HA}$ present in aqueous layer (0.038*0.5=0.019)from the initial moles of $\ce{HA}$ (0.057*0.50 = 0.029), which results in a final concentration of $0.020~\mathrm{mol\,  dm^{-3}}$ of $\ce{HA}$ in the organic layer.
Finally, $K_\mathrm{D} =
   \dfrac{[\ce{HA}_\mathrm{aqueous~layer}]}{[\ce{HA}_\mathrm{organic~layer}]}
   = \dfrac{0.038~\mathrm{mol\,  dm^{-3}}}{0.020~\mathrm{mol\,  dm^{-3}}} = 1.9$

The $\ce{[HA]}$ in both layers is known to two significant figures so the $K_D$ value should have two significant figures too.

For Question 2.

The reaction between $\ce{HA}$ and $\ce{NaOH}$ in aqueous solution is, as you noted,

$$\ce{HA + NaOH -> Na^+ + A^- + H2O}$$

Assume none of $\ce{NaA}$ migrates into the organic phase.
Assume that no $\ce{NaA}$ forms in the aquaeous phase either, which is to say that both the $\ce{Na^+}$ and $\ce{A^-}$ ions are completely solvated.
Since $\ce{[Na^+]}= 0.027$ and the volume is 0.5000 $\mathrm{dm}^{-3}$, the number of moles of $\ce{NaOH}$ present is $0.0135$.
Started with 0.0295 moles of HA total, so the solution will be acidic.
Since the charges must balance $\ce{[Na^+] + [H^+] = [A^-] + [OH^-] }$, but $\ce{[A^-] >> [OH^-] }$ and $\ce{[Na^+] >> [H^+]}$ so $\ce{[Na^+] \approx [A^-] }$ which means that there are $0.0135$ moles of $\ce{[A^-]}$ in the aqueous solution, and 0.016 moles of $\ce{HA}$ split between the aqueous phase and the organic phase.
From above $0.016 = \ce{0.5[HA]_{aq} + 0.5[HA]_{org}}$ and using $K_D$ we can determine $\ce{[HA]_{aq} = 0.021}$ and $\ce{[HA]_{org} = 0.011}$
We can use the $K_a$ to solve for $\ce{[H^+]}$

\begin{align}
1\times10^{-5} &= \dfrac{\ce{[H^+][A-]}}{\ce{[HA]}}\\
\ce{[H^+]} &= \dfrac{1\times 10^{-5}\ce{[HA]}}{\ce{[A^-]}} = \dfrac{(1\times 10^{-5})(0.021)}{0.027} = 7.7\times10^{-6}\\
\mathrm{pH} &= 5.11
\end{align}

Carboxylic acids in organic solvents normally form dimers, this will make the calculations much harder. I think that it may be impossible to get a good answer to this question with what is provided in the question.

I think the question is a bit clever and may needs deeper analysis, a monobasic organic acid may have carboxylic acid group and amine group, both can exist in aqueous form by dissociation or by hydrogen bond. At the aforementioned equilibrium and the dilute initial concentration, there will be ionic forms in water and molecular forms in aqueous form. The sum HA and ions in that equilibrium is in aqueous form, and could be some molecules in organic phase if their sum is less than the initial moles. So here seems no significant quantity is in $\ce{CHCl3}$ phase. On my point of view.

You can't just add an amine group for this type of problem.


text:
Who devised the term elements?
Who devised the term elements? Who was the one who brought the term elements to us? Also please give a bit information on background of the man.

One answer is that the ancient Greeks, specifically Plato, devised the term "element".  Quoting from wikipedia article on Classical elements:

Plato seems to have been the first to use the term "element (στοιχεῖον, stoicheion)" in reference to air, fire, earth, and water. The ancient Greek word for element, stoicheion (from stoicheo, "to line up") meant "smallest division (of a sun-dial), a syllable", as the composing unit of an alphabet it could denote a letter and the smallest unit from which a word is formed. A similar alphabetic metaphor may be the origin of the equivalent Latin word elementum (from which the English word comes), possibly based on the names of the letters 'l', 'm', and 'n', though the validity of this idea is debated.

If you had a more modern, chemical definition of "element" in mind for your question, then another possible answer is Robert Boyle.
Quoting from wikipedia again, this time from its article on chemical elements:

In 1661, Robert Boyle proposed his theory of corpuscularism which favoured the analysis of matter as constituted by irreducible units of matter (atoms) and, choosing to side with neither Aristotle's view of the four elements nor Paracelsus' view of three fundamental elements, left open the question of the number of elements. The first modern list of chemical elements was given in Antoine Lavoisier's 1789 Elements of Chemistry, which contained thirty-three elements, including light and caloric.

I can't resist adding a reference to my favorite English-language modern chemistry text that omits the word "element" entirely.

For most of its being, mankind did not know what things are made
  of, but could only guess. With the growth of worldken, we began
  to learn, and today we have a beholding of stuff and work that
  watching bears out, both in the workstead and in daily life.
The underlying kinds of stuff are the firststuffs, which link
  together in sundry ways to give rise to the rest. Formerly we
  knew of ninety-two firststuffs, from waterstuff, the lightest and
  barest, to ymirstuff, the heaviest. Now we have made more, such
  as aegirstuff and helstuff. [...]

If the Normans had never invaded England in 1066, we might not have "elements" at all, but "firststuffs" instead.  Either way, it's just a debate about the right translation for "stoicheion", a concept which Plato seems to have given us.

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text:
Calculating the electrons an atom wants to gain/lose to reach a noble gas
I'm trying to understand ions.
From what I understand, an ion is when the atom gains or loses electrons. More electrons means it is negatively charged (anion). Less electrons means it is positively charged (cation).
My problem is with the following premise:

Many atoms gain/lose electrons with the hope of having the same number of electrons as the closest noble gas in the periodic table.

The book uses Potassium as an example. Looking at my table, the closest noble gas to Potassium is Argon, which has $18$ electrons. Therefore Potassium wants to lose one electron...
$$\ce{K^+}$$
Nice, I get that. But then I noticed something odd about all the examples in my book: all the elements they picked are elements conveniently one or two steps away from a noble gas. They never picked an element in the middle of the table like, say, iron.
Iron has $26$ electrons. Its closest noble gas is Argon, which has $18$ electrons. So iron wants to lose $8$ electrons (becoming a cation):
$$\ce{Fe^{8+}}$$
Is this correct? My book literally avoids picking elements in the middle of the table, so I feel there's something weird about this.
Also, I noticed that in every example, the book mentions the Group of the element (Potassium is group 1A, etc...). How is that even relevant? All you have to do is count the number of electrons, no?

The explanation is simple: the story is that nice and smooth, when you have to loose one or two electrons. Things in the middle don't just lose 8 electrons or get 5 extra. Extra charge means extra energy: an ion with a large charge is actually getting unstable because of the coulomb energy. So whenever ionization happens with the elements in the middle, 1) they loose less electron / smaller formal charge,l 2) they are not purely ionic substances, but stabilized by other mechanism (e.g complexion with water).

Check out this answer, which goes into some detail about why the octet rule exists, and also why it is more of a "loose guideline" than a rule.

The book's explanation about a noble gas configuration is somewhat accurate, but fairly incomplete.
The elements on the right and on the left of the periodic table (the alkali (earth) metals, the halogens, the chalcogens (the group that starts with Oxygen) and the pnictogens (Nitrogen group)) have electron configurations that make it somewhat easier to lose/gain electrons to "look" like noble gasses (this is why alkali metals and halogens are so reactive: the destabilization from the charge is very minimal compared to the stability of the electron configuration).
However, as you have observed, the book goes to some effort to avoid talking about transition metals. There is a reason for that.
Noble gas configurations are a subset of the stable configurations of electrons. In reality, what's actually being aimed for is an element with no incomplete electron shells.
What is an electron shell? The actual quantum mechanical definition may be a bit more complicated than you need, but for your purposes, it suffices to say that it's how electrons will tend to be distributed around a nucleus. Note that this is extremely distinct from saying that the electron "orbits" the nucleus.
If you look at a periodic table, there are four general regions to look at:

The labels "s", "p", "d", and "f" all refer to the various electron orbitals, and the blocks all indicate what the outermost (or "valence") shell is.
So while elements in the s and p blocks can lose a small number of electrons get complete shells (which will resemble the noble gases), transition metals are a bit more complicated. And by a bit more complicated, I mean that there are very rough rules in place that have almost more exceptions than they have actual examples of following the rule.
But in very general terms, a transition metal will lose electrons. If they are in the first half of the transition metals in their row, they can lose a number of electrons less than or equal to the number of d electrons plus the number of s electrons. This maximum number of electrons lost will make them resemble the noble gasses in configuration, but it's only appreciably stable when the transition metal is in the first half of transition metals in its group ($\ce{Mn^{7+}}$ is well documented in ionic compounds, but $\ce{Fe^{8+}}$ doesn't exist a lot because it has too many electrons to try to lose to comfortably fall into a noble gas configuration). But transition metals can lose electrons up to this point. The issue is, most transition metals can adopt multiple charge configurations for various reasons, so generally, when writing an ionic compound with a transition metal in words, you need to specify the charge of the transition metal (actually, its oxidation state, but in an ionic compound that's mostly splitting hairs). So you can have "Iron (II) Oxide", which is a black powder and "Iron (III) Oxide" which is rust. These compounds will have different chemistries because of the charge on the transition metal.
Suffice to say, your textbook is avoiding transition metals because transition metals are weird.

Also, I avoided talking about the second half of the transition metal elements: this is because the way the electrons are lost in there are a bit wonkier (as some transition metals in there will do some sincerely weird things to their electron configuration.) The explanation requires some quantum mechanics.

$\ce{Mn^{7+}}$ doesn't exist stably as a bare cation so saying that it is well documented as an example of a noble gas configuration is highly misleading.

I thought a counterion to it would have been implied. Let me edit that in.

I'm not talking about counterions, I'm talking about the fact that manganese in the +7 oxidation state is always strongly bonded to something else. Take $\ce{MnO4-}$ as an example. The manganese 'ion' is strongly bonded to four oxygens and does not exist as a bare $\ce{Mn^{7+}}$ cation.

I think we're meaning the same thing, but talking past each other. When I "counterion" I generally mean "ion of the opposite charge that the first ion is bonded ionically to"

So basically $Fe^{8+}$ is the right answer (should they ask me) but, it's just weird/uncommon..?

No. $\ce{Fe^{8+}}$ does not exist. Iron is in the second half of the transition metals of its row. The reason why this is the border has to do with quantum mechanics, Dark Lord.

It's doubtful to the point of impossibility that they'd ask you to do anything with transition metals unless the teacher/professor gives you an explicit list of the charges/oxidation states that the metal in question might take on. (When I was in high school, my teacher basically gave us a list of these charges and said "these metals can have these charges. Memorize them and deal with it.")

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Dispersion of Lithium Ions in Nature (Ratio Similar to Isotopes?)
I'm working on a project that identifies average characteristics of different elements. I'm stuck on lithium, I can't find a way to ratio lithium I and lithium II that symbolizes how it is in nature. I tried looking around on the internet but I haven't found any mention of ratios to ions only ratios to isotopes. Does anyone know of anything that might help?

Doesn't really answer the question. If this means anything I could use the ratio in plasma conditions.

Alright I'm very frustrated, not with this site but with my project, does anyone out there know or understand how lithium ions are dispersed in nature?

I'm not entirely sure what you're after.  Are you trying to find  $\ce{Li+:Li2+}$?  It's going to be nearly infinite (or nearly zero, if you reverse it.)  Aside from in plasmas, lithium is almost always in the +1 oxidation state.

The reference to isotopes in the question is very confusing if you really are asking about the ratio of lithium(I) ions to lithium(II) ions.  Outside of high-energy plasmas, particle accelerators, or the vicinity of stellar cores, there aren't going to be many lithium(II) ions on Earth or in the universe.

I believe I resolved my problem, the attributes I was looking at was spectra and combining the lines of both ions. The problem was I was comparing data given by two authors that used different units, so what I did, which I think solves my problem, was divide the sums of the given intensity by the emission probabilities for which I received the ratio of Li I to Li II as roughly 33 to 6 or 18% Lithium II. More than I or apparently anyone else thought so I may have still made a mistake, either way thanks to everyone who tried and helped.

Just to add, the resultant rgb value did match approximately with the hue of a common lithium flame test, so I am confident in my answer, thanks again. I need to start using stack exchange more often.

authors: نورالدين عبد الرحمن


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Is there an abundance threshold of a 'synthetic' element for it to be considered natural?
I always (possible naively) thought that elements are either natural or synthetic, either one or the other.  But, according to the Wikipedia page about Synthetic Elements, it states that:

All elements with atomic numbers 1 through 98 are naturally occurring at least in trace quantities, but the following elements are usually produced through synthesis. Except for Francium, they were all discovered through synthesis before being found in nature.

These are shown in pink in the image below:

Image attribution: "Periodic table, good SVG" by Incnis Mrsi - Own work. Licensed under CC BY-SA 3.0 via Wikimedia Commons.
I understand that the amounts of these elements exist naturally in trace quantities, not necessarily on Earth, for example, according to the Jefferson Lab page about promethium:

Promethium does not occur naturally on earth, although it has been detected in the spectrum of a star in the constellation Andromeda.

This leads me to my question:
Is there a natural abundance threshold of a 'synthetic' element for it to be considered natural?

Is there a natural abundance threshold of a 'synthetic' element for it
  to be considered natural?

$\mathrm{> 0}$
In other words, any element that can be detected outside of the laboratory (e.g. in the natural environment) is a naturally occurring element.  This would include the 80 stable elements along with the 18 radioactive elements with very long half-lives, and the shorter-lived radioactive elements produced upon their decay.
Therefore there are 98 naturally occurring elements; all of the others are synthetic, artificial, man-made, whatever you choose to call them.
This Wikipedia article provides nice background, especially the following sections:

the introductory section, especially the last paragraph therein
Occurrence and origin on Earth
List of the 118 known chemical elements - Occurrence

Now, if we lived at the center of a supernova, our threshold might change (not for long, though)...

Jon, the threshold would change or the the number of "naturally" occurring elements would change? Please elaborate if the former.

Well, we would be getting lots of every element and isotope conceivable (under, or in, the sun). For one, brief moment of glory... But, seriously, a supernova will make lots of everything. So, it has all happened "naturally", we are just billions of years removed, so what is left to be "natural" isn't a lot.

The radioactive elements which are regarded as occurring "naturally" have at least one of the following properties:

At least one isotope has a half-life sufficiently long that if it is created in a supernova, significant quantities will persist for a long time afterward.  Many isotopes of uranium fit this pattern.
They are decay by-products of other naturally-occurring radioactive isotopes.

Trans-uranic elements are considered artificial because the their short half-lives mean the probability of any remaining from a million or so years ago is essentially nil, and there are no heavier natural elements that could replenish the supply.  Radon is naturally occurring because, despite its short half-life, the supply gets replenished through the decay of heavier elements.  Technetium has a much longer half-life than radon, but since no naturally-occurring radioactive isotopes of anything else produce it when they decay, any supplies that may have existed will have dwindled away to essentially nothing by now.

I have never heard of a formal specification for such a threshold. Qualitatively speaking, though, I believe this threshold could be effectively considered "enough that we can make useful stuff with it." To my mind the key phrase in your blockquote is (emphasis added):

"... the following elements are usually produced through synthesis."

'Production' implies making stuff. 'Measurable in trace quantities by ICP-MS' is not helpful for making stuff.
Worth noting: since different amounts of different materials are needed to make useful things, it's possible that a hard, quantitative abundance number might not even be useful to define.


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How to identify which of two given electrodes with corresponding solutions is reducing more?
For the sake of the question, let’s assume that you are provided with two labelled rods of the pure metals $\ce{A}$ and $\ce{B}$ and two unlabelled bottles containing $\pu{1.0 mol dm-3}$ aqueous solution of $\ce{A^m+}$ and  the other bottle containing $\pu{1.0 mol dm-3}$ aqueous solution of $\ce{B^n+}$.
I want to find out, which metal out of the given two is more reducing, and identify each solution that is given in the unlabelled bottles. What are the procedures that I should follow?
If I immerse $\ce{A}$ in a solution of  $\ce{A^m+}$, I guess  there will be no change. Because if I do so there would be an equilibrium prevail with the ions in the solution and the electrode, and I think this is true for $\ce{B}$ also.
But I don't know how to identify the one which gets reduced more, yet I know that the anode gets reduced. So I am confused about that.
I feel that the only clue I have is immersing $\ce{A}$ in  $\ce{B^n+}$ and vice versa for $\ce{B}$ also. But  really I don't know what  would happen if I do so.

The solution is to immerse each rod in both of the solutions (separately of course). Let's imagine that $\ce{A}$ is the more reducing metal. We can consider what happens in each case:

If we immerse $\ce{A}$ in a solution of $\ce{A^m+}$ nothing will happen.
Similarly, if we immerse $\ce{B}$ in a solution of $\ce{B^n+}$ nothing will happen.
If we immerse $\ce{B}$ in a solution of $\ce{A^m+}$ nothing will happen because the electrode potential for $\ce{A|A^m+}$ is more negative (more negative means its a stronger reducing agent) than for $\ce{B|B^n+}$ so the reaction potential is negative and the reaction is not feasible.
However, if we immerse $\ce{A}$ in a solution of $\ce{B^n+}$, $\ce{B}$ will precipitate out on the surface of $\ce{A}$. This is because $\ce{A|A^m+}$ has a more negative electrode potential than $\ce{B|B^n+}$ and so the reaction potential will be positive. $\ce{A}$ will reduce $\ce{B^n+}$ ions to form $\ce{B}$ metal and $\ce{A}$ will be oxidised to form $\ce{A^m+}$:
$$\ce{A (s) + $\frac{m}{n}$B^n+ (aq) -> A^m+ (aq) + $\frac{m}{n}$B (s)}$$

In short, the most reducing metal is the one which causes precipitation of the other metal out of its solution.

Thanks a lot. But what  happen if I immerse the two metal rods in same solutions, let's say we immerse them in a solution of $B^{n+}$ ( assuming A is more reducing). Then would there be the same scenario as you explained ?

I have  one more doubt about what you have said.How can you say "so the reaction potential is negative and the reaction is not feasible."

For the first comment, do you mean what happens if you immerse both rods in one solution at the same time?

For the second, the reaction (or cell) potential is related to the Gibbs energy change. If the reaction potential is negative, $\Delta G$ is positive, which means that the equilibrium constant is less than one. Technically this means that some precipitation will still occur, particularly if the cell potential is close to zero but broadly speaking, if the reaction potential is negative, the reaction is unfeasible. This gives the relationship between the three quantities.

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text:
Why is solid phase peptide synthesis limited to 70 amino acids?
I'm reading document about Solid phase protein synthesis (SPPS) from Wikipedia, and according to the document:

SPPS is limited by yields, and typically peptides and proteins in the
  range of 70 amino acids are pushing the limits of synthetic
  accessibility. Synthetic difficulty also is sequence dependent;
  typically amyloid peptides and proteins are difficult to make. Longer
  lengths can be accessed by using native chemical ligation to couple
  two peptides together with quantitative yields.

I'm a bit not understanding why this synthesis method is limited to 70 amino acids in length. Anyone please help me to explain what is the issue?

The issue is one of yield. Each step has a certain yield percentage of yield, that the synthesizers will maximize as much as possible. However, given that it is fairly impossible to generate 100% yield all the time, the amount of desired protein you get over time will deteriorate exponentially with length.
For example, say that each step gives you 95% yield. If you have a protein that is ten amino acids long, your final step will give you .95^10 yield, which is about 60%.
However, with the same yield per step, but a 70 amino acid chain, the amount of protein you want drops to 2%.
Even if you are at 99% yield per step (a difficult feat), your final yield will be under 50% for a 70 amino acid chain.

Thank you for your easy to understand answer. But can we reuse the 5% unreacted peptide to use in the next reaction and thus we won't lost anything?

The yield is not just a function of unreacted peptide. The peptide may have undergone a completely different reaction entirely, or it could be lost in transferring to another beaker. In the lab, the situation is ambiguous enough that we cannot be certain. In solid phase synthesis, we try to give a huge excess the amino acid we're trying to react relative to the peptide, but that may or may not work as much as we would want, depending.

Good answer. If anything, this is on the optimistic side, as for each residue added, there are at least two synthetic steps: deprotection of the terminal end (usually removing Fmoc from the N-terminus) and then the coupling to the next residue.

Very good answer, so to improve this method for create a bigger protein, we will need to reduce the number of reaction (or using other substance to have more percent of efficiency)  ?

No. You wouldn't use this method at all for a bigger protein. Improving the efficiency would be far too expensive relative to its effectiveness. To make a larger protein, you'd just make a bacteria do it.

@BreakingBioinformatics yes, i got it, thank you so much, you're amazing!

You need to perform one reaction per amino acid to couple it to the previous one, and that reaction isn't 100% efficient. You always get a small amount of unreacted peptide, or some other side product. Those small inefficencies add up for longer peptides because you're doing so many reactions.
I created a quick graph showing the yield assuming 98% efficiency of the reation:

As you can see, the yield drops significantly for longer peptides.
There isn't really a hard limit on size, but you get terribly inefficient for long peptides. And at a certain point it is much easier to synthesize the peptide biologically. Proteins are routinely synthesized in large amounts using bacteria like E. coli, unless you need to do certain modifications you can only do with solid-phase synthesis, going the biological route can be much cheaper than using chemical synthesis.

Adding to the already good answers:
Each added amino acid in solid phase peptide synthesis actually consists of three steps:

coupling of the new peptide
capping any unreacted peptide (optional)
deprotecting the newly-added amino group.

The first step is clear. Capping is required because not all free amino groups will be coupled to the new peptide and you want to prevent hard to remove side-products in your mixture. Deprotection then liberates the amino group of the new amino acid so it can form another peptide bond.
This now means that when you couple 70 amino acids, you are in fact looking at 140 or 210 reactions. Even if the deprotection step gives you $99.9\:\%$ yield, you still lose substance every coupling step — and this is where the other answers’ calculations step in.

Does it mean that it's possible to create a bigger protein with this method, but we will need more and more material ? Can i say that : if we have enough material, we can create an unlimited protein length that we want?

@DavidWang technically, yes. In practice, it is a lot easier to create larger peptides either biologically (i.e. letting bacteria do the job) or to synthesise multiple fragments of a length maximum 50 and fuse them together later. (convergent approach)

Yeahh, i like this answer,  i thought about join multiple fragments today, i think that is the best idea to create a protein without using bacteria.


text:
Are aqueous solutions of oxidizing agents safe?
In the "reducing agent" article in Wikipedia it says

Good reducing agents are reagents that deliver $\ce{H2}$.

Wouldn't that make water unsafe to dissolve oxidizing agents?

I don't quite see how the statement about reducing agents is related to the safety of dissolving oxidising agents in water. Perhaps you could clarify this.

@bon I think OP means that water may deliver $\ce{H2}$ and is as such unsafe to dissolve oxidizers in.

Water doesn't deliver $\ce{H_2}$ though...

If the oxidizing agent is sufficiently strong, e.g. fluorine, yes. "[W]ater spontaneously combust[s] under a fluorine jet."
However, there are only a few oxidizers, such as fluorine, chlorine trifluoride, and dioxygen difluoride that can displace oxygen from water. More likely you'd encounter a strong reducing agent, such as Group I metals, that would attach to the oxygen in water, liberating hydrogen.

A 3% solution of hydrogen peroxide bleaches hair pretty well, and it is a neutral solution. In the lab, you can get a 35% $\ce {H_2O_2}$ solution.

The safety arises from the solution's characteristics. Some oxidizers produce highly alkali solutions which pose personnel risks from chemical burns. This nature occurs to a lesser extent when dry chemical contacts the skin or mucus membranes ext. as the agent acquires water from the body and in essence forms saturated solutions at the contact sites.

It doesn't really answer this.

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


text:
calculate the pressure at equilibrium

Given that $K=18$ for the equilibrium below, calculate the total pressure in a sealed vessel initially containing only $\ce{NH4I(s)}$, that reaches equilibrium.
$$\ce{NH4I(s) \leftrightharpoons NH3(g) + HI(g)}$$
a) $4.2\mathrm{~atm}$
b) $8.5~\mathrm{atm}$
c) $2.1~\mathrm{atm}$
d) $\mathrm{18~atm}$
e) $\mathrm{6.3~atm}$

Using $K = \ce{[NH3][HI]} = 18$, I found $\ce{[NH3]=[HI]=}~\frac{4.24mol}{L}$, and therefore $8.48~\mathrm{mol}$ of gas is produced, but I don't know how to find the equilibrium pressure without temperature and volume.

I think what is given here is $K_p= p(\ce{NH3}). p(\ce{HI})$ and not $K_c$. So, What you have here are the partial pressures of the gases. As you have a sealed vessel, so the sum of the partial pressures of the gases gives the total pressure, 8.5 atm


text:
Lemon juice, the ultimate anti-foaming agent?
As an amateur soda maker, I have noticed an odd phenomenon: lemon juice sodas do not foam. Virtually every juice I have tried foams to some degree when mixed with carbonated water. Lemon juice does not. I have read "folk" explanations for this as being due to the "acidity" of the lemon juice, but this is not correct because other acidic juices like lime and orange juice foam. Why is it that lemon juice has an anti-foaming action?
Further information: it should be noted that "pasteurized" (ie, boiled) lemon juice does foam. Only fresh-squozen lemon juice is anti-foaming. Therefore, what is doing the anti-foaming may likely be some volatile compound.
My understanding of anti-foaming is that it is caused primarily by localized depression of surface tension, and that liquids having heterogeneous species promote anti-foaming because of the surface tension differential between those species at the surface. Therefore, one possible theory is that the lemon juice has some volatile oil which populates the surface of the water only in spots and is causing the surface tension breakage. Even if this theory is correct, it remains to determine what this mysterious volatile oil is that is the cause.

Maybe limonene? But then fresh orange juice probably wouldn't foam either, unless it's the stereoisomer specific to lemons...

I think oil is a good candidate here. I have noticed that small amounts of oil in food can spread over a large area on the surface of water and quickly destroy foams in carbonated beverages. However, all fruits you mention have a large amount of essential oils, so I'm not sure why only lemon should work. Try with a drop of vegetable oil and report the results!

Another idea (that I am presenting without evidence because I just made it up, but hey it seems plausible to me) is that its the sugar in fruit juices that is responsible for foaming.  Lemon juice is very low in sugar/starch, so maybe it is less that lemon juice is "anti-foam" and more that there is just no foam in the first place.  A test would be to make a mixture of equal parts lemon juice and another juice, and to test for foaming.  Lots of foam means that lemon juice is not "anti-foam", just non-foaming.  And no foam means that lemon juice really is "anti foam".

A lemon has an oil that is present mostly in the skin. For instance if you squeeze lemon into hot tea in a styrofoam cup and let the tea sit, then you'll see a surface roughness at the liquid level. The oil actually dissolves the styrofoam. The oil acts a defoamer. So with a styrofoam cup you're suppose to use "commercial" lemon juice which has no oil (lest the cup dissolve and spill the liquid).

What is missing from this answer is the identity of the "oil".

@ShakaBoom - After 3 years?!? // The oil isn't going to be a single compound. There are probably a small number of major components and many more minor components. I'd also guess that there are differences between different types of lemons.

If you squeeze a fresh lemon and then use a strainer that retains any pulp and then add the juice to carbonated water it will foam - I think this indicates that it's the Limonene in the pulp that prevents foaming

If you imagine a bubble as being composed of walls of thin films, then anything which weakens those walls can lead to collapse. (note that no pure liquid will foam, foam is proof that there is surface active agents present.) Anyway, there are several mechanisms to cause bubble collapse: addition of a volatile, addition of a viscosity lowering agent, addition of a particulate which disrupts the film, addition of an agent which lowers the surface tension...Those are all I can think of off the top of my head. Foam has an internal structure but in liquids is usually not stable, due to drainage. Obviously, faster drainage (lower viscosity) tends to more rapidly thin the film (hence disrupt it). Increased evaporation both thins the film (weakening it) and increases stresses on it (shrinkage). There's no way for me to determine which if any of these factors are contributing in your case. One interesting experiment is to store the liquid and see if the amount of foaming changes with time (in a well-sealed (ie glass) bottle). The most effective general defoamers I am familiar with are silicone dispersions which are colloidal and partially soluble. Their slow dissolution disrupts the films, while their low density tends to float them (and the silicone oligiomers are surface active). These types of materials once they are at equilibrium concentration, are much,much less effective. Similarly, once any volatiles reach equilibrium (in the head-space) their contribution will be quite limited. So, while it's really difficult to be definitive with natural products (they tend to oxidize or even become food for microorganisms) a liquid which continues to be non-foaming (with the addition of component X) after hours and days is likely due to something floating on its surface.


text:
How many hydride ions are available in NaBH4?
Different sources have different answers. In the reduction of an aldehyde or ketone to an alcohol by Sodium Borohydride, how many moles of aldehyde are reduced by one mole of Sodium Borohydride?
While my teacher claims it is 1, Solomons-Fryhle claims it to be 4.

Practically speaking, a chemist will use one equivalent of sodium borohydride to reduce an aldehyde or ketone. However, theoretically only 0.25 equivalents are necessary. Borohydride donates a hydride to the carbonyl carbon, generating borane, which is a weaker reducing agent than borohydride. However, borane is a Lewis acid that can react with the alkoxide formed in the reduction. This gives a new borohydride, which can reduce a second molecule of the ketone. This process can continue until all of the hydrides are consumed from the boron center.

Keep in mind that sodium borohydride reductions are commonly performed in alcohol solvents, so the exact nature of these steps could be altered. The alkoxide could be protonated by solvent, and then the solvent's conjugate base could reactivate a neutral boron species.
Also keep in mind that while sodium borohydride isn't VERY basic, it is a little basic, so it will deprotonate alcohols slowly. It's that reaction that causes the practical requirement for more than the theoretical borohydride stoichiometry.


text:
Are my protolysis equations right?
We had the task to solve several salts in water and measure the pH value of the solution. After that we should "create" the protolysis equation and to tell about Chemical equilibrium[s].
I get some possible result equations, but I'm really unsure about them. The [] brackets were used to make the ionic loads easier visible. solving equations:

$\ce{KNO3 <=> [NO3]- + [K]+}$
$\ce{FeCl3 <=> [Fe]^{3+} + [Cl]-}$
$\ce{NH4Cl <=> [NH4]+ + [Cl]-}$`
$\ce{NaHSO4 <=> [HSO4]- + [Na]+}$

my protolysis equations:

$\ce{[NO3]- + H2O <=> HNO3 + [OH]-}$
$\ce{[Fe(H2O)6]^{3+} + H2O <=> [Fe(H2O)5OH]^{2+} + [H3O]+}$
$\ce{[NH4]+ + H2O <=> NH3 + [H3O]+}$
$\ce{[HSO4]- + H2O <=> [SO4]^{2-} + [H3O]+}$

I would say that all equations run bi-directional, because the result of solving equations can vary dependent on temperature for example, and the protolysis equations should be bi-directional because we're having a solution, which isn't a pure acid.
Additionally, I got the question if I can draw conclusions on the pH-change through those protolysis reactions. I would say yes, but not very well, depending on the sum of all hydrogen-containing ions like $\ce{H3O+}$

Your equations look balanced and like good equations for representing salt dissolution as well as proton transfer between acid/base species.  What's your question exactly?

Their directions, and if and how you can draw conclusions about the pH value through the proteolysis equation

yes, they are, all of them are bidirectional.
It is not possible to draw conclusions about the pH value, because of the different abilities to solve in water and because those reaction may cause an increased or a decreased temperature.
(Source: my teacher)

This should not be community wiki. Also I believe this is not quite correct. Depending on the involved species, you can predict if a salt will cause a acidic or basic solution based on the involved species.

Please change the answer, this question is for my school project not important anymore, so I wanted to lower its priority here too because it won't help that much anymore and I seriously don't want to waste your time for a question which isn't that important


text:
What is the quickest way to dissolve steel?
What is the quickest way to dissolve a large chunk of steel? I mean a very quick reaction, assuming that the interaction occurs in a sealed or semi-sealed environment? Reaction times of the order of a few minutes versus tens of minutes, hours or days are the objective.
I'm curious as to how quickly acids like $\ce{HNO3}$ and $\ce{H2SO4}$ can react with steel containers.

Like a safe, strongbox or valut?

Time depends strongly on type of steel.

The fastest way is dissolving the stainless steel electrochemically using $\ce{H2SO4}$ at elevated temperature where the steel block is attached as an anode. Higher current densities lead to higher removal rate.
Alternatively a $\ce{FeCl3}$ with minimal content of $\ce{HCl}$ (up to 3%) heated at 45°C or higher.

Two to look up as "homework ".  Aqua Regia which is a mixture that dissolves gold. Not good enough for engineering and some other rare earth metals. Steel? Maybe. The other is Caro's acid. There's a YouTube video of it dissolving a fly. That's organic matter so it may not be good enough for steel. Worth a look though.

Steel can be dissolved with a diluted solution of nitric acid and water. The chemical make up of nitric acid reacts to the iron in steel, producing iron nitrate and hydrogen gas. As this chemical reaction takes place, the steel begins to dissolve. The dissolving process of steel can sometimes take multiple applications, depending on the size and thickness of the metal......
Experience gained is from a pylon collapsing in Ghana...As some corrosive substances were applied to the pylons


text:
If alloys are homogeneous mixtures, why can't we separate their components?
An alloy is a material composed of two or more metals or a metal and a nonmetal. And, they are usually formed by heating the elements to their melting points, and then cooling them, so that the components mix. Now, why doesn't this works backwards i.e. if we heat the alloy again to melting point of their constituents, and they should separate?

It might have something to do with entropy...

If the liquids are immiscible, they will separate. But entropy (as @tennispro1213 points out) and enthalpies of mixing (in the solid and/or liquid phases) have to be dealt with.

Once the alloy has been formed the atoms from the different metals will have shared there electrons with each other and come to an equilibrium. In this state the metal atoms have formed a complex structure which has a different reactivity or properties than each individual metal did in its original form .

Things have surely been compounded together but I'm not sure if an alloy is ever referred to as a compound. It's not an organic compound I know that.

If you melt the alloy and then use electrolysis, or if you use a redox reaction, you should be able to separate the components.
For example, if you used a basic acid, let's say hydrochloric acid, on a zinc-copper alloy, I'd imagine the zinc would be removed from the alloy and taking the hydrogen's place, forming zinc chloride, leaving hydrogen gas and copper metal.
This is based off of simple redox reaction memories and no personal experience. However, if you're curious, try it out!

But, I am sure that they are not resolvable !

I'm not entirely sure what you mean.

I mean that "you should be able to separate the components" is not true.

Now that I think about it, maybe not. Adding chromium to steel allows it to not rust, so I guess the reactions change. Electrolysis definitely sounds like it would work, though.

I think it means how come the original metals can't be recovered from the alloy just by heating the alloy to the melting point temp respective to metals in their original state.

To my knowledge, this alloy is now its own structure with its own properties, including melting point.


text:
When to use -ate and -ite for naming oxyanions?
I'm now learning about nomenclature. In particular, oxyanions.
Basically, when you have an anion that is a combination of a non-metal with oxygen.
According to my book:

It ends with -ate for the most common oxyanions of the element. It
  ends with -ite for the oxyanions that have the same charge but with
  one less atom of oxygen.
Example:
$\ce{NO_3^- \implies}$ Nitrate
$\ce{NO_2^- \implies}$ Nitrite

Ok, if you give me an oxyanion that ends with -ate, I know that I can simply remove one oxygen atom and I will get the -ite.
But, what I don't understand is how does the book know that $\ce{NO_3^-}$ is Nitrate: how does the book know that $\ce{NO_3^-}$ is "the most common oxyanion for the element". How does it know that a charge of $-1$ and $3$ oxygen atoms create "the most common" Nitrogen oxyanion?
In other words, what does it even mean by "most common"?

I'm actually not sure about this, at first I thought it would be something to do with the group oxidation number since nitrate, sulfate and phosphate all feature the central ion in its group oxidation number. But chlorate ($\ce{ClO3^-}$) begs to differ. I guess, as with lots of chemical nomenclature, it may just be due to "historical reasons". Hope someone else can give a better answer.

@orthocresol your right but there's also per- prefix - https://en.wikipedia.org/wiki/Oxyanion

"Most common" means the forms most frequently encountered in the entire scope of chemistry, which I'm pretty sure corresponds loosely to the most thermodynamically stable oxidation state. It is not always the case that the '-ate' ion is the most stable, however—as an example, see the links for chlorine below).
A detailed answer necessitates noting that the nomenclature extends beyond just '-ate' and '-ite'.  Most p-block elements that form oxo-anions form a whole series of them, each with the central atom in an oxidation state two away from each neighbor.  For all elements except the lightest ($\ce{C}$ and $\ce{N}$) and heaviest ($\ce{As}$, $\ce{Se}$, $\ce{Te}$, etc.) , usually the series is considered to contain four members, though it is not guaranteed that all are stable or able to be characterized (e.g., "bromite" and "hyposulfite").
As noted in the comments to another answer, in addition to the '-ite' and '-ate' suffixes, there are also the 'hypo-' and 'per-' prefixes, where 'hypo-' only is used with '-ite', and 'per-' is only used with '-ate'.  These four combinations are used to span the series of oxo-anions for each element:

Phosphorus: perphosphate ($\ce{PO5^{3-}}$), phosphate ($\ce{PO4^{3-}}$), phosphite ($\ce{PO3^{3-}}$), hypophosphite ($\ce{H2PO2^{-}}$)
Sulfur: persulfate ($\ce{SO5^{2-}}$), sulfate ($\ce{SO4^{2-}}$), sulfite ($\ce{SO3^{2-}}$), hyposulfite ("$\ce{SO2^{2-}}$")
Chlorine: perchlorate ($\ce{ClO4^{-}}$), chlorate ($\ce{ClO3^{-}}$), chlorite ($\ce{ClO2^{-}}$), hypochlorite ($\ce{ClO^{-}}$)
Bromine: perbromate ($\ce{BrO4^{-}}$), bromate ($\ce{BrO3^{-}}$), bromite ("$\ce{BrO2^{-}}$"), hypobromite ($\ce{BrO^{-}}$)
Iodine: periodate ($\ce{IO4^{-}}$), iodate ($\ce{IO3^{-}}$), iodite ($\ce{IO2^{-}}$), hypoiodite ($\ce{IO^{-}}$)

For carbon, 'carbonate' was given preference, per the 'most common' rubric, as the only known oxo-anion. For nitrogen, the use of the prefixes was avoided, for what I assume was the sake of simplicity.  (I would argue that nitrate and nitrite are both common enough in the natural world that without such a deciding factor there would have been stiff competition for the '-ate' suffix.)
Possible irregularities in structure within a series include peroxo-anions at high numbers of bound oxygens (e.g., I believe persulfate and "perphosphate" are both peroxo species) and variable numbers of bound oxygens for a given oxidation state in anions of heavier elements (e.g., metaperiodate, $\ce{IO4-}$, versus orthoperiodate, $\ce{IO6^{5-}}$, both of which contain heptavalent iodine). As well, there is the possibility for other, non-oxo-anion compounds of the central atoms with oxygen such as chlorine dioxide, $\ce{ClO2}$; nitrogen dioxide, $\ce{NO2}$; sulfur dioxide, $\ce{SO2}$, and trioxide, $\ce{SO3}$; and (of course) carbon dioxide, $\ce{CO2}$.
You can find references to oxo-anions of arsenic, selenium, antimony, and tellurium also (links are to the '-ates'), but to my mind, these tend to edge more toward the behavior of oxo-anion-forming metals, which either only have a single appreciably stable oxo-anion (chromate, molybdate, tungstate, etc.), or break this (hypo-)-ite/(per-)-ate paradigm pretty badly (see, e.g., permanganate, $\ce{MnO4-}$, versus manganate, $\ce{MnO4^{2-}}$).
If you really want to blow your mind, take a look at the polymeric oxo-anions like polyphosphate; or at the at-present theoretical-only orthocarbonate; or at iron, for which apparently all three of the known oxo-anions are referred to as 'ferrate'.
Update, 13 Oct 2023: Following up on Oscar Lanzi's comment, it appears that alkaline earth orthocarbonates have both been predicted and experimentally observed—and, you can apparently buy the tetramethyl orthocarbonate ester commercially$^*$.
$^*$No endorsement intended.

Orthocarbonate salts are now known. The Wikipedia article currently mentions calcium and strontium salts, and it turns out that magnesium oethocarbonate is experimentally known too. All are synthesized under conditions matching Esrth's deep mantle and thus may exist there.

Historical naming conventions dictate that the "-ate" suffix refers to the group containing the cation with a higher oxidation state than that in the corresponding group designated with the "-ite" suffix.
In the case of nitrate, nitrogen carries a +5 charge; in nitrite, nitrogen is a +3. For chlorate, chlorine is +5; in chlorite, it's +3. And so on.

However, there is also the prefix per- which denotes an oxidation state higher than just an -ate.

And also 'hypo-', which denotes an oxidation state lower than '-ite'. E.g., hypochlorite, $\ce{ClO-}$.

@bon I was taught that "per-" is an abbreviated form of "hyper-", which pairs nicely with "hypo-".

@JasonPatterson This seems quite likely.

The "per-" prefix is due to Thomas Thompson (and is not an abbreviation for "hyper-" but rather denotes the highest oxide). See M.P. Crosland's excellent text Historical Studies in the Language of Chemistry.

Gets fuzzy when you start messing with the peroxo derivatives...


text:
What is the difference between threo/erythro ephedrine?
I want to know how to easily make the difference or just understand the importance of stereocenters of ephedrine ?

Are you asking about how the chemical is processed by the human body?

I was asking about ephedrines  stereochemistry part. thank you Martin

1.Ephedrine exhibits optical isomerism and has two chiral centres, giving rise to four stereo isomers. By convention, the pair of enantiomers with the stereochemistry (1R,2S and 1S,2R) is designated ephedrine, while the pair of enantiomers with the stereochemistry (1R,2R and 1S,2S) is called pseudoephedrine.
Source :https://en.m.wikipedia.org/wiki/Ephedrine

Again in order to distinguish threo from erythro.. Just memorize the word :TOES =Threo - Opposite Erythro - Same
e. g.,

hi!( 1R 2S and 1S2R),( 1R2R 1S2S)  are enantiomers,right? @user16976

(1R,2S and 1S,2R)is a pair of enantiomers and (1R,2R and 1S,2S)is another pair of enantiomers.

Please blockquote any text that you take directly from the source.  Thanks!


text:
How to dissolve a strong resin in difficult situation?
The resin is made up from two components:
Component A: silicic acid, sodium salt (less than 10% sodium silicate)
Component B: Diphenylmethane-4,4'-Diisocyanate, isomers and homologs (MDI,Polymer-MDI)
It has been cured in place, however it became stuck to a piece of equipment made out of rubber.
It is stuck several meters down a line so access is very limited.
Is there anything that would be strong enough in small quantities to dissolve or even loosen this adhesive?
Any thoughts on a mixture of water 90%, ammonia 8% and detergent 2%?
Would I need a large quantity of this mixture or how effective would it be?
Or a mixture of the same except using sodium carbonate instead of ammonia?
Any help or suggestions would be greatly appreciated.
Thanks

Amides may be, theoretically, hydrolyzed by concentrated sodium hydroxide. Silicic acid esters usually are readily destroyed by exscess of fluorides, like ammonium hidrofluoride... though it is not a health-friendly compound.

It might help to add a highly polar organic solvent like DMSO or HMPT to help wet and swell the polymer so it will react more efficiently with the base.

cyclohexanone dissolves things like cumar, sylvalite, and NI100 resins


text:
Through bond interaction in biradical dioxo compound and its correlation diagram
I'd like to investigate into the ring opening, the migratory reverse reaction of the electrocyclic ring-closure of Dimethyldioxirane:

As a first question I'd like to ask if I have drawn the through bond interaction correctly?

Following scheme is wrong, because the p orbitals are perpendicular to the O-C-O as @ron clearly pointed out in his answer in Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?

Next I tried to draw the orbital correlation diagram.
I guess for this correlation diagram I don't have to draw the through bond interaction, yes?

In the dioxo biradical product I've drawn a smaller energy gap between $\pi$ and $\pi^*$ orbitals as for $\sigma$ and $\sigma^*$. I guess this should be correct.
Because there is an additional through bond interaction for the product I've drawn a small exothermicity. I see no activation energy for the disrotatory ring opening. So the disrotatory ring opening must happens very fast (without a lot of thermal energy, may be even at room temperature).

Finally, I tried to draw the state correlation diagram:

For the photochemical reaction there is a transition from $S_0$ to $S_2$. Because of the avoided crossings the conrotatory correlation is steeper falling down, therefore I expect most of the electrons direcetly going in the pericyclic minimum and do there Interal Conversion to the $S_1$ surface and afterwards the excited product gives off a photon going back to the $S_0$ state. I expect products in excited states are not very long living and therefore undergoes emission of a photon as Internal Conversion under fluorescence here.
Another way out from the pericyclic minimum can be the directly falling to the transition state of the $S_0$ state and giving the product to a good quantum yield.
As third option some part of the $S_2$ states can go through the disrotatory pathway and directly emitting light from $S_2$ to $S_0$. As written in the first point I think this way is less probable because of the less steeper falling down the energy surface.

Is this all correct?

I would like to help here, but I am not sure where to begin. First of all, I want to point out that the reaction you have shown is not an electrocyclic reaction, because it does not involve the central saturated carbon atom. It is rather a simple homolysis reaction. The terms "conrotatory" and "disrotatory" are misapplied, since the two oxygen atoms do not have functional groups on them, and the two modes of ring opening are indistinguishable.
Dioxirane does have an electrocyclic ring opening reaction available to it - namely cleavage of a carbon-oxygen bond to form "acetone-O-oxide". Perhaps you should consider that as an alternative pathway. Does that help?

Yes. It's a homolysis reaction. I guess the first picture with the through bond interaction ist correct. For this old exam exercise only the homolysis reaction has to be considered. I know O-O bonds are weaker than C-O bonds. In the exam it was required to draw the correlation diagram. I therefore thought, everytime I draw an orbital correlation diagram, I can also draw a state correlation diagram. The question now is, to which symmetry the reaction coordinate belongs to?


text:
Conformations of Cyclohexane
Is there any such thing as a twist chair conformation?
Also, the boat and the chair conformations are achiral, while the twist boat conformation of cyclohexane is chiral and dissymetric.
What about the half chair conformation? Am I right in thinking that it is achiral because it has a plane of symmetry?

What plane of symmetry does the half-chair have? It looks like it has an axis of symmetry, but no plane.

Short answer, no. There is not a "twist-chair."
To convince yourself, it probably helps to make a physical model with a chemistry model kit. If you try to twist the chair, you can't do it without significantly moving at least one atom, in which case, the conformation is basically a "skew" or "twist boat." Try it.
The boat and chair conformations are indeed symmetric and achiral. The twist-boat or skew conformation has point group $D_{2}$ and is indeed chiral.
A normal half-chair will have an axis of symmetry and is chiral.
An envelope conformation will have a plane of symmetry and thus be achiral.
I've found that some of the best depictions of six-member-ring conformations come from pyranose sugars. Because of the hydroxyl groups and oxygen atoms, there are 38 distinct conformations (2 chairs, 6 boats, 6 skew-boats, 12 half-chairs, and 12 envelopes).
From Wikipedia:

Obviously, in less-substituted cyclohexane, there are fewer unique shapes, but it helps to see the different ring-puckering or pseudo-rotation shapes.

My book gave the "envelop" as "half-chair" - so I thought the half chair was achiral. The book is normally reliable - I wonder why the mistake. Are four atoms arranged linearly in the half chair? Does that create immense strain - 180 degrees in place of 109.5?

@Charles No, the 4 atoms are not linear, but they are planar. The difference between the half-chair and the envelope is that the half-chair has one atom above and one atom below the plane of the ring, while the envelope has 5 planar atoms and one atom above the ring.

How are those 4 atoms arranged in the plane of the ring - how does the half chair look if you tilt it a bit from the angle you've shown it at?

Here's an image showing 1-2-5-6 plane. I can send a PDB file if you like.


text:
Can I make Ferrofluid without evaporating Ammonium?
I want to make ferrofluid, which is ferromagnetic particles suspended in a carrier fluid. Three methods I have seen involve using magnetic ink (for MICR codes), cassette tapes, or ferric chloride as the iron source. The last method is the only one that makes ferrofluid with nano-scale ferromagnetic particles. AFAIK, the process boils down to using ammonia to precipitate magnetite from a ferric/ferrous chloride mixture, coating the precipitated particles with oleic acid (acting as a surfactant), and suspending the coated particles in a carrier oil such as kerosene.
The steps (based on this):

Oxidize iron with ferric chloride to form ferrous chloride: $\ce{2FeCl3 + Fe -> 3FeCl2}$
Precipitate magnetite: $\ce{2FeCl3 + FeCl2 + 8NH4OH -> Fe3O4 + 8NH4Cl + 4H2O}$
Add oleic acid, evaporate the ammonia, and end up with surfactant coated magnetite nanoparticles?
Add kerosene to magnetite suspension => ferrofluid

There is another example here that has slightly different steps:

Dissolve $\ce{Fe^2+}$ and $\ce{Fe^3+}$ in $\ce{HCl}$ and mix
Precipitate magnetite
Repeatedly:

Separate precipitate from liquid with a magnet
Decant the liquid, leaving the precipitate
Add distilled water

Add $\ce{N(CH3)4OH}$

It seems to me that the first process generates a 'smoother' ferrofluid. I mean to say that it looks more like a homogenous fluid, whereas the ferrofluid in the videos from the second link looks more like a sludge.
I have two questions. First, what is actually happening in the first process when the oleic acid is added? From the first article, it sounds like the oleic acid reacts to form ammonium oleate, and then adding heat decomposes the ammonium oleate, releasing ammonia gas and dissolving the oleate ions, which then coat the magnetite particles. But that sounds a bit odd. Second, would the following work (create a suspension of magnetite nanoparticles suspended in a carrier):

Obtain a mixture of $\ce{FeCl2}$ and $\ce{FeCl3}$ via either method.
Precipitate magnetite
Separate the magnetite precipitate from the ammonium chloride via decanting (as in the second process)
Add oleic acid (and heat?) to coat the magnetite particles
Add kerosene to magnetite suspension => ferrofluid

It seems like removing the ammonium chloride via washing would be easier than via evaporation. Less heat, less time, less nasty fumes. But I don't really know what's going on when the oleic acid is added, so I'm not sure if my proposed method would actually work.

So, I've been trying to make Ferrofluid with powdered Iron (III) Oxide (Synthetically created, 1 micron) using Citric Acid, Heat, and Water.  It didn't work.  I'm trying the same thing with Ammonia today, if it works I'll let you know!

To answer your first question, I believe that the ammonium oleate already being a relatively soluble salt allows its deprotonation such that the resulting oleate ions are able to bond to the magnetite in suspension (both by electrostatic attraction and hydrophobic interactions). Subsequent heating is done simply to remove dissociated $\ce{NH4Cl}$ from the water in the form of $\ce{NH3}$ and $\ce{HCl}$ gas. Might be worth noting that your initial addition of aqueous $\ce{NH3}$ (aka $\ce{NH4OH}$) for the co-precipitation is in excess. A video by NileRed suggests neutralising this excess ammonium (hydroxide) instead with the use of hydrochloric acid, instead of using high heat which would expel a large amount of nasty fumes. Cleaning of the magnetite can then be done with deionised water to remove the dissolved salts.
To answer your second question, no, adding oleic acid post-decanting would not work. Using the logic I've stated, adding pure oleic acid would not coat the magnetite particles, as oleic acid by itself will not deprotonate in water until it has reacted with ammonia to form the ammonium oleate soap.
Even if you were to add aqueous ammonium oleate, doing so post-decanting doesn't seem logical; normally, surfactants are added as soon as the magnetite has fully precipitated, without stopping the stirring, because you'd want the surfactants to coat the magnetite nano/microparticles (MNPs) while they are still dispersed (before they begin aggregating). In your case, having the intermediate decanting step will definitely introduce excessive, irreversible  magnetite aggregation and the resulting MNPs will not be of high quality. Anyways, as previously mentioned, the decanting step is not necessary because the ammonium chloride salt will not contaminate your final magnetite sample, considering that it can be easily washed with deionised water, and even short-chain alcohols to remove excess oleic acid.
To summarise, your last paragraph is correct in saying that ammonium chloride can be rinsed more easily from the magnetite sample than removed through evaporation. Using an acid might be a good way to do it as proposed by NileRed.

Yeah, NileRed's video definitely clarified a lot of things for me, but it's good to see it written out in detail. His process seems more accessible than the boiling ammonia process, but it's probably still not something I can reasonably do in my garage given that I have no experience beyond high school and college classes.


text:
Where exactly do the orbital angular momentum get their numbers e.g. +1/2
I was thinking about orbital angular momentum today and  wondering how exactly do the numbers come about that quantize them. I was doing a little research on how they come about and I  came across this image many times.
How exactly does a chemist/physicist measure this to get the +1/2,-1/2 number. Are there other values that are possible?

And besides, you're talking about orbital angular momentum in your post, but use pictures for spin angular momentum. They have different eigenvalues since for orbital angular momentum there is an additional restriction: the Born interpretation requires cyclic boundary conditions to be satisfied, thus, only integer values of $l$ are possible.

Well I've only gone over them in general chemistry and my chemistry book doesn't expand on the matter so I didn't know where to start other than google. The page I found mentioned orbital angular momentum and showed these pictures. I wouldn't call it laziness as I'm trying to learn over the summer, but thanks :)

Well, I suggest then that next time you include your background in the question. And it is likely that your chemistry book doesn't discuss this matter more closely simply because it is not yet time to do so. You indeed must have a solid background in quantum mechanics to understand this whole business about the angular momentum.

$$
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\ramuno}{\mathrm{i}}
\newcommand{\ket}[1]{\,|{#1}\rangle}
$$
Classical Mechanics
In classical mechanics, the angular momentum, $\vec{L}$, of a particle with respect to a chosen origin is given by the vector product of its position vector relative to the origin, $\vec{r}$, and its linear momentum, $\vec{p}$,
\begin{equation*}
    \vec{L} := \vec{r} \times \vec{p} \, .
\end{equation*}
By expanding $\vec{r}$ and $\vec{p}$ over a standard basis $\{ \vec{i}, \vec{j}, \vec{k} \}$ of a Cartesian coordinate system as $\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$ and $\vec{p} = p_x \vec{i} + p_y \vec{j} + p_z \vec{k}$ and then writing the cross product as the formal determinant [1] and expanding it one could obtain
\begin{equation}
    \vec{r} \times \vec{p}
    =
    \begin{vmatrix}
     \vec{i} & \vec{j} & \vec{k} \\
           x &       y &       z \\
         p_x &     p_y &     p_z \\
    \end{vmatrix}
    =
    (y p_z - z p_y) \vec{i} - (x p_z - z p_x) \vec{j} + (x p_y - y p_x) \vec{k} \, ,
\end{equation}
which implies that components $(L_x, L_y, L_z)$ of the angular momentum in terms of which it can be expressed over a standard basis as $\vec{L} = L_x \vec{i} + L_y \vec{j} + L_z \vec{k}$ are given as follows
\begin{equation}
 L_x = (y p_z - z p_y) \, ,
 \quad
 L_y = (z p_x - x p_z) \, ,
 \quad
 L_z = (x p_y - y p_x) \, .
\end{equation}
The magnitude, $L$, of the angular momentum is given as usual as the square root of the scalar product of the vector with itself, $L := \sqrt{\vec{L}\vec{L}}$, or, in terms of components,
\begin{equation*}
    L := \sqrt{L_x^2 + L_y^2 + L_z^2} \, .
\end{equation*}
Quantum Mechanics
Operators
The quantum mechanical analogue of classical angular momentum is known as the orbital angular momentum and the operators for the components of orbital angular momentum can be obtained by simply replacing the components of position and momentum in classical mechanical expressions by the corresponding operators
\begin{equation}
    \linop{L}_{x} = (\linop{y} \linop{p}_z - \linop{z} \linop{p}_y) \, ,
    \quad
    \linop{L}_y = (\linop{z} \linop{p}_x - \linop{x} \linop{p}_z) \, ,
    \quad
    \linop{L}_{z} = (\linop{x} \linop{p}_y - \linop{y} \linop{p}_x) \, .
\end{equation}
and the square of the magnitude of the angular momentum operator can be expressed in terms of these operators as follows
\begin{equation}
    \linop{L}^2 = \linop{L}_{x}^2 + \linop{L}_y^2 + \linop{L}_{z}^2 \, .
\end{equation}
Then the following commutation relations of the angular momentum operators can be easily established:
\begin{equation}
    [ \linop{L}_{x}, \linop{L}_{y} ] = \ramuno \hbar \linop{L}_{z} \, ,
    \quad
    [ \linop{L}_{y}, \linop{L}_{z} ] = \ramuno \hbar \linop{L}_{x} \, ,
    \quad
    [ \linop{L}_{z}, \linop{L}_{x} ] = \ramuno \hbar \linop{L}_{y} \, ,
\end{equation}
\begin{equation}
    [\linop{L}^2, \linop{L}_{x}] = [\linop{L}^2, \linop{L}_y] = [\linop{L}^2, \linop{L}_{z}] = 0 \, .
\end{equation}
Interestingly, in quantum mechanics there exist a different kind of angular momentum, the intrinsic angular momentum, or spin, which does not have a counterpart in classical mechanics.
While the intrinsic angular momentum components operators does have a different form from that of orbital angular momentum, the commutation relations between them and the square of the intrinsic angular momentum magnitude are exactly the same.
More interestingly, the angular momentum can be treated in such a way that the treatment relies entirely on the commutation relations just introduced, and as such, valid for any kind of angular momentum.
Thus, for generality we can use the letter $J$ for any kind of angular momentum reserving the letter $L$ specifically for orbital one.
\begin{equation}
    [ \linop{J}_{x}, \linop{J}_{y} ] = \ramuno \hbar \linop{J}_{z} \, ,
    \quad
    [ \linop{J}_{y}, \linop{J}_{z} ] = \ramuno \hbar \linop{J}_{x} \, ,
    \quad
    [ \linop{J}_{z}, \linop{J}_{x} ] = \ramuno \hbar \linop{J}_{y} \, ,
\end{equation}
\begin{equation}
    [\linop{J}^2, \linop{J}_{x}] = [\linop{J}^2, \linop{J}_y] = [\linop{J}^2, \linop{J}_{z}] = 0 \, .
\end{equation}
Eigenvectors and eigenvalues
According to the commutation relations above, $\linop{J}^2$ commutes with all three components $\linop{J}_{x}, \linop{J}_y, \linop{J}_{z}$ while the components do not commute with each other.
Thus, $\linop{J}^2$ and one (and only one) of the components operators, conventionally chosen to be $\linop{J}_{z}$, admit the set of simultaneous eigenvectors, and using the so-called ladder operators technique it can be shown that,
\begin{align}
    \linop{J}^2 \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\
    \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, ,
\end{align}
where $\ket{j(j+1),m}$ is a simultaneous eigenvector of $\linop{J}^2$ and $\linop{J}_{z}$, $j$ could take non-negative integer and half-integer values,
\begin{equation*}
    j = 0, 1/2, 1, 3/2, \cdots \, ,
\end{equation*}
and $m$ ranges from $-j$ to $j$ in integer steps
\begin{equation}
    m = -j, -j+1, \cdots, j-1, j \cdots \, .
\end{equation}
Then, as I already said in my comment with respect to half-integer values of $j$, for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible, but where cyclic boundary conditions are not relevant, as for the intrinsic angular momentum known as spin, the half-integer values might be appropriate.
Epilogue
I'm still not sure what this question is about, but if it specifically about the spin of an electron, i.e. "How do we now that its component in any direction is $\pm 1/2$ in the units of $\hbar$?", then we know it both from the experiment (the Stern–Gerlach one) and the theory (the quantum field theory). And yes, for electron only these values are possible, but for other particles there might be different possibilities. Thus, for instance, a photon have component of the spin in any direction  $\pm 1$ in the units of $\hbar$.

[1] Here, "formal" means that this notation has the form of a determinant, but does not strictly adhere to the definition; it is a mnemonic used to remember the expansion of the cross product.


text:
About combining a cation and an anion
I’m trying to learn about ions. There is a slide with examples about combining cations with anions. It goes like this:

$$\text{For} \ \ \ce{Al^3+} \ \ \text{ and } \ \ \ce{O^2-} \ \ \text{you get} \ \  \ce{Al2O3}$$

Wait what? So you have an atom of aluminum (in this case a cation because it loses three electrons) and an atom of oxygen (an anion since it receives two electrons). You combine them. Why do you now have TWO atoms of aluminum and THREE atoms of oxygen? Why does combining a cation and an anion suddenly spawn new atoms? And what happened to their charges?
The next example is this:

$$\text{For} \ \ \ce{Ca^2+} \ \ \text{ and } \ \ \ce{Br-} \ \ \text{you get} \ \  \ce{CaBr2}$$

Alright. I see a pattern. For some reason, if the charge of an atom is positive or negative, such number will define the number of atoms for the other element. So since calcium has $2{+}$, there would be two bromine. And since bromine has $-1$ there will be one calcium. I don’t get why, though.
And a third example,

$$\text{For} \ \ \ce{Na+} \ \ \text{ and } \ \ \ce{CO3^2-} \ \ \text{you get} \ \  \ce{Na2CO3}$$

Ok so this is a bit weird since there are now three elements involved.
But according to the pattern, since sodium has charge $+1$, there should be just one $\ce{CO3}$ molecule. And since $\ce{CO3^2-}$ has a charge of $-2$, there should be two of sodium.
Great, I suppose it’s an useful pattern, but my questions are:

Why? Why does losing/gaining an electron cause the other element to gain/lose atoms?
Does this pattern apply only when you combine a cation with an anion? What if it is the other way around? (An anion with a cation, or does it make no sense at all?)
After combining a cation with an anion, do both atoms lose their charge? Observe how the results no longer have $+$ or $-$ superscripts.

Ah, I see what you did there. Don't get mixed with arbitrary ways of finding the empirical formula for the ionic compounds. Try to understand them better instead. For example, note that ionic compounds are billions of cations and anions together and we only tend to demonstrate the 'smallest' relation between them. You could say $\ce{Na50(CO3)25}$. It would be correct, but non-standard.

@M.A.Ramezani hm... ok I got an idea which might be completely wrong: let's see... when people say $Al_2O_3$ they don't mean that there are two Aluminum atoms and three Oxygen atoms, but they actually mean that there is a 2:3 relation of Aluminum:Oxygen if you were to combine $Al^{3+}$ with $O^{2-}$ because one $Al^{3+}$ will attract $3$ Oxygens whereas one $O^{2-}$ will attract 2 Aluminums therefore the relation is $2:3$ and hence the end result is written as $Al_2O_3$?

Yes, kinda like that. For each three $\ce{O^2-}$ ions you get two $\ce{Al^3+}$. Don't get the impression that this is literally happening in the unit cells though. Ionic compounds tend to form special crystal structures. (e.g. fcc)

@M.A.Ramezani But in that case, if $Al^{3+}$ "attracts" $3$ Oxygens and $O^{2-}$ "attracts" $2$ Aluminums, shouldn't the result be like $Al_2O_3Al^{3+}O^{2-}$? I mean, why do $Al^{3+}O^{2-}$ vanish? Why don't they just stay?

Why do compounds form bonds? I've got three words for you: Less potential energy. I don't remember a case of positive lattice enthalpies$\ldots$

@M.A.Ramezani well I don't know what a positive lattice enthalpy is lol, but I think my doubts are cleared now. Thanks.

@voldemort That's your problem with valence, but for other compounds. Do you understand it all now?

I really like your process of trying to work it out! Trying to find patterns is a great way to look at this (and all science really). I think all you're really missing is some nomenclature, as this appears to be primarily a question of understanding of what's actually going on.
The 3+ alumnium ion ($\ce{Al^{3+}}$) and 2- oxygen ion ($\ce{O^{2-}}$) form a crystal lattice in solid form.
(A crystal lattice is a $3D$ arrangement of ions in a repeating pattern, in which each ion is interacting with more than one other ion)
The total charges need to balance out per "unit cell" (an intermediate-level chemistry concept that means the "smallest repeating subunit of a crystal lattice.")
So this means that you need to balance the total charges by changing the number of ($\ce{Al^{3+}}$) and ($\ce{O^{2-}}$) ions to get the correct chemical formula.
In this case: $(3+) *(2) = (6+)$ balances $(2-) *(3) = (6-)$, giving $\ce{2 Al^{3+}}$ and $\ce{3 O^{2-}}$.
Ionic compounds that have an overall neutral charge are written without the charge superscripts,  thus, $\ce{Al^{3+}_2O^{2-}_3}$ is written as $\ce{Al_2O_3}$
If you're actually looking at precipitation reactions (where the solid salt forms from a solution), you have to balance the equations.
Thus to form $\ce{Al_2O_3}$ the correct ratio would be: $\ce{2 Al^{3+} + 3 O^{2-}} -> Al_2O_3$
Similarly for $\ce{CaBr_2}$, the correct ratio would be: $\ce{Ca^{2+} + 2 Br^-} -> CaBr_2$.
For $\ce{Na_2CO_3}$ you're applying the exact same concept, except you're dealing with an anionic (negative charged) molecule instead of an atom.
Thus, to form $\ce{Na_2CO_3}$ the correct ratio would be: $\ce{2 Na^+ + CO^{2-}_3} -> Na_2CO_3$.


text:
Can negative (abnormally) lattice enthalpies during the formation of ionic compounds exist?

Why do compounds form bonds? I've got three words for you: Less potential energy. I don't remember a case of positive lattice enthalpies $\ldots$ - My comment

This makes me wonder. I remember seeing $\Delta H_{\mathrm{lattice}}$ of "common" ionic compounds like $\mathrm{NaCl}$ and considered them fairly positivea:
\begin{array}{|c|c|} \hline
\mathbf{Ionic~compound} & \mathbf{\Delta H^\circ (kJ~mol^{-1})} \\\hline
\ce{LiF} & 1022^\mathrm{\,b} \\\hline
\ce{NaCl} & 771 \\\hline
\ce{KBr} & 670 \\\hline
\ce{CsI} & 585 \\\hline
\end{array}
Yeah$\ldots$ so judging by the usual trends $\ce{CsI}$ should have the least positive lattice enthalpy in ionic compounds derived only from monoatomic ions. But that's still some lattice enthalpy!
So, could there be negative lattice enthalpies?

Notes:
a: Note that there are two major definitions of lattice enthalpy. Here, we're using the definition relevant to $\Delta H_r$ of the following endothermic reaction: $$\ce{MX(s) → M+(g) + X^- (g)}$$
While the other definition uses the enthalpy of the reverse process as the reference, only the sign might differ. Note that I had the other defintion in my mind when posting that comment.
b: All data obtained at $298~\mathrm{K}$. source

I really want to say no because you have to break bonds in order to ionize the lattice and you aren't forming any (as far as I know) but I don't have any hard theory to back this up.

Yes @Bon $\ldots$ The "no" coming at first to my mind fascinated me even more towards asking this question.

Short answer: Not for "normal" compounds.
First off, let's consider what the lattice enthalpy means. As you've written it, these lattice enthalpy is part of the Born-Haber cycle of formation of an ionic solid from a metal and halogen.
As you see from your table, the energies decrease with:

Small charges (i.e., these are higher for $\ce{Al2O3}$ or $\ce{CaO}$ because of higher ionic formal charges)
Large inter-atomic distances (i.e., increased ion distance between $\ce{Cs}$ and $\ce{I}$ yield lower electrostatic attraction).

The Born-Lande and Kaputinskii equations give some numeric estimates, e.g. (Born-Landé)

$E = \frac{N_AMz^+z^- e^2 }{4 \pi \epsilon_0 r_0}\left(1-\frac{1}{n}\right)$
where:

$N_A$ = Avogadro constant;
M  = Madelung constant
$z^+$ = charge number of cation
$z^−$ = charge number of anion
e = charge on an electron (elementary charge)
$\epsilon_0$ = permittivity of free space
$r_0$ = distance to closest ion
n  = Born exponent, typically a number between 5 and 12, determined experimentally or derived theoretically.

(Note that I've inverted the formula from Wikipedia's version to match your reaction.)
None of these can give you a negative lattice enthalpy.
How could it happen? Well, you'd need something where the solid is less stable than the gas-phase ions. Consider that in an ionic solid, the electrostatic attraction between anion and cation is always favorable. Since gas-phase ions aren't stable by themselves, I don't see how this is possible.

I've been trying to think if an unusual compound (e.g., $\ce{Xe+ F-}$ might create this, but it doesn't line up.

Even without assuming lattice energy formulas, just going by regular thermodynamics, the only way for lattice formation from an ion gas to be exergonic ($\Delta G<0$) while being endothermic ($\Delta H>0$) is if entropy increased going from the gas to the solid ($\Delta S>0$). That is probably impossible, so either thermodynamics under constant pressure isn't enough, or an endothermic lattice can only exist in non-equilibrium conditions or with some form of work being continuously performed on the system.

@NicolauSakerNeto Agreed. I wanted to go from the lattice description, but your comments are easily worth a separate answer. I'd vote for it. :-)

Geoff's answer is enlightening because it shows several of the factors influencing lattice formation energetics and how they cannot be easily made to output endothermic ionic lattices. However, an even stronger statement can be made by forgoing any reference to lattice energy formulas.
Considering thermodynamics as we chemists are most used to, we have a system which is free to evolve under constant pressure and in the absence of external potentials such as gravitational, electric or magnetic fields, and which is given infinite time to reach an equilibrium condition. As such, we can use our good old friend, the Gibbs free energy equation, in its simplest form:
$$\Delta G = \Delta H -T\Delta S$$
For a process to be spontaneous, we say it is exergonic ($\Delta G<0$). This requires one of three conditions to be met:

$\Delta H<0$, $\Delta S>0$ : The process is exothermic and increases system entropy; spontaneous process at any temperature.
$\Delta H<0$, $\Delta S<0$ : The process is exothermic and decreases system entropy; spontaneous process only at sufficiently low temperatures.
$\Delta H>0$, $\Delta S>0$ : The process is endothermic and increases disorder; spontaneous process only at sufficiently high temperatures

Now consider the process where gaseous ions combine to form an ionic solid:
$$\ce{y\ A^{+x}(g) + x\ B^{-y}(g) -> A_{y}B_{x}(s)}$$
The question essentially asks whether this process can be exergonic while being endothermic. For this to be the case, the only condition which has any chance of being met is the third. Therein lies the problem. It is hard to imagine a situation where a crystalline solid with long-range ordering, as is the case with most common ionic compounds, could possibly have a higher entropy than a gas of ions. This is a strong, general argument against the stability and existence of ionic solids with endothermic lattice energies.
This thermodynamic argument, however, does not constitute ultimate proof that endothermic lattices are impossible, just unlikely. For a stable endothermic lattice to be made, one of the constraints mentioned at the start must be relaxed. For example, perhaps thermodynamics under constant pressure is not adequate, but under constant volume (with the associated Helmholtz free energy considerations) it could work.
Another possibility is that additional potentials can be present in the system, in which case other terms must be added to the Gibbs free energy expression. As a loosely related example, water can liquefy/freeze at unusual temperatures under the action of a very strong electric field, even at atmospheric pressure.
Yet another way is if, even though formation of an endothermic lattice is not spontaneous, once formed it could somehow be kinetically stabilized against dissociation. This delicate interplay of kinetics and thermodynamics in Chemistry is certainly thoroughly exploited, but whether it can applied to this particular situation, I don't know.
One more opportunity for the existence of a negative energy lattice can be based on non-equilibrium thermodynamics, either in systems close to (but not quite at) equilibrium, or in systems very far from equilibrium. Thermodynamics in these systems is much harder to describe mathematically, however.


text:
Nomenclature of MnO4^-: is it an oxyanion?
Having a real pain with nomenclature. Look at this substance:
$$\ce{MnO4-}$$
So, do we have an anion of manganese and oxygen, or is that just an anion of oxygen? Anyway, my book says that $\ce{MnO4-}$ is "permanganate".
Since it ends with "-ate", it seems my book thinks $\ce{MnO4-}$ is an oxyanion (because those are the ones that end with things like "-ate" and "-ite").
However, it was my understanding that an oxyanion is when you have oxygen and a non-metal. But manganese is a metal! What is going on here then?
And not just that. Let's assume that $\ce{MnO4-}$ is indeed an oxyanion called permanganate. From what I've learned, if you remove one oxygen atom (getting  $\ce{MnO3-}$), it should become just manganate, right? But, according to my book, manganate is $\ce{MnO2-}$. It had to lose two oxygen atoms and not one to become manganate. Why?

Afaik, manganate is actually $\ce{MnO4^2-}$

As far as I know, an oxoanion just refers to an anion which has oxygen covalently bonded to the central atom; the central atom can be pretty much anything apart from the Group 1 and 2 metals, since the only oxygen compounds those form are ionic oxides/peroxides/superoxides. For the 3d transition metals, you have $\ce{VO3^-}$ (vanadate(V)), $\ce{CrO4^{2-}}$ (chromate(VI)), $\ce{MnO4-}$ (manganate(VII)), and $\ce{FeO4^{2-}}$ (ferrate(VI)), amongst others.
I would not recommend reading too much into the nomenclature, because a lot of it is just what chemists are comfortable with after using for a long time - it's like a language that you just get more familiar with over time. But there are some general rules. The 'per...ate' name, such as perchlorate $\ce{ClO4-}$, implies a higher oxidation state than the normal '...ate' compound, which is chlorate $\ce{ClO3-}$ in this case. For chlorine, only certain oxidation states are available, namely the -1, 0, +1, +3, +5 and +7 oxidation states. Apart from 0, which occurs in diatomic chlorine $\ce{Cl2}$, you will notice that these are in increments of 2 - which means that you can generate the whole list of chlorine anions by adding one more oxygen, but keeping the charge of the anion constant: $\ce{Cl-}$, $\ce{ClO-}$, $\ce{ClO2-}$, $\ce{ClO3-}$, $\ce{ClO4-}$. This tendency to form compounds 2 oxidation states apart is very pronounced for the p-block elements, i.e. Group 15, 16 and 17.
However, for d-block elements such as $\ce{Mn}$, this isn't the case because single d electrons are easily lost and gained. So after permanganate, $\ce{MnO4-}$, which features $\ce{Mn}$ in the +7 oxidation state, the oxoanion with the next highest oxidation state is the manganate ion $\ce{MnO4^2-}$ which has $\ce{Mn}$ in the +6 oxidation state. (As far as I know the ion $\ce{MnO2-}$ is not called "manganate".)

An oxyanion is not necessarily a non-metal and oxygen, rather, it is more simply a chemical element and an oxygen. There are a large number of elements that form oxyanions, many of which are listed here.
To quote this Chem.SE post regarding a the difference between manganate and manganite:

Manganese oxidation state nomenclature is a real world mess. Drawing nomenclature parallels to $\ce{Cl(VII)}$ and downward is not valid. Drawing nomenclature parallels to titanium is poor, for $\ce{Ti(IV)}$ is the highest common oxidation state. $\ce{SrO +TiO2}$ gives $\ce{SrTiO3}$, properly called titanate, for it is $\ce{Ti(IV)}$ on both sides.

For someone who is new to chemistry, the only metal oxyanions I would recommend that you learn are chromate ($\ce{CrO4^2-}$), dichromate ($\ce{Cr2O7^2-}$), and permanganate ($\ce{MnO4^-}$).

What a huge mess! In other words, I ought to be learning most compositions out of memory D:

So an oxyanion is basically oxygen with whatever as long as it is an anion? (The oxygen an anion, or the other thing an anion?)

That is correct. Luckily the non-metal oxyanions behave nearly all the same. The metal oxyanions are the exception.


text:
According to this equation for glycolysis, what is reduced and where do the electrons go?
I know that the glucose is oxidized into pyruvate but I don't understand what is reduced or where the electrons go....
Equation:
$\ce{Glucose + 2 ADP + 2 NAD^+ -> 2 Pyruvate + 2 ATP + 2 NADH}$

Look up the 2 reactions ADP -> ATP and NAD+ -> NADH.

This appears to be a homework question, please share your thoughts and attempts towards the solution. It would be nice if you could share what the abbreviations stand for to make this question self-sustained.

NAD+ accepts a hydride ion (proton + 2 electrons) and is reduced to NADH.

Yomen, would you mind elaborating a little bit? This seems really really comment-like to me and it might to other people.

I think its okay, it answers exactly the question how it was asked. The question itself is none of the brilliant ones though.

I think the question "where do the electrons go?" is excellent, and if answered correctly can excite otherwise disinterested students into chemistry. This is because the only way to properly answer the question is to get into the mechanistic details which are at the heart of chemistry. And it's not that hard, given a quality image to point out the specific "electron pair". I've included one of many images that can be found on the web for the mechanism of Glyceraldehyde-3-phosphate (G3P) Dehydrogenase below. The precise answer to the question is that the "electrons come from the bond attaching the hydrogen directly to the carbonyl carbon of G3P.  This corresponds to the same hydrogen in the open chain form of glucose, and that, in turn can spark further questions about that. Unfortunately, despite the central importance of this step, many (I would say most) educators would have trouble answering with this precision. No wonder chemistry gets a bad reputation.


text:
Calculating valence of oxides
Learning about oxides. Basically when oxygen is combined with a metal.
$$\ce{FeO}$$
This is called "iron oxide (II)" according to my book.
Apparently, the II represents the valence. But how come? Look:
$\ce{Fe}$ has the configuration $1s^22s^22p^63s^23p^6\color{red}{4s^2}3d^6$. Since the outermost shell has $2$ electrons, the valence is $2$.
Now $\ce{O}$ has the configuration $1s^2\color{red}{2s^22p^4}$. Its outermost shell has $6$ electrons so oxygen has a valence of $6$ right?
Then how come $\ce{FeO}$ has valence $2$? Shouldn't the valence of $\ce{FeO}$ be like, the combination of the two valences? So $6 - 2 = 4$? (I read somewhere you had to subtract it).
Not only that. I am starting to think that my definition of valence is completely messed up. I just checked in here and apparently oxygen has $-2$ valence? Why? And iron has +2, +3, (+4), (+6)? What does all that even mean?

Ok, finding valencies of elements in simple compounds is very easy, but you have to know some thing to start:
You are treating compound as if it would be composed of discrete molecules
If you have compound/neutral molecule, sums of valencies of both types of atoms should be equal, as valency is number of bonds between atoms.
In case of FeO you need to know that O make usually two bonds (valence = 2) and as both valencies have to be equal Fe also has val = 2
In case of Fe2O3, 3*2=6, 6/2=3 so Fe is trivalent.

$3$ oxygen atoms multiplied by the valence of Oxygen ($2$) results in $6$ and then it is divided by the number of Iron atoms resulting in $3$ - is that how it's done for simple compounds? Well great, I finally get it then. Thanks.

For binary oxides it's that simple, but there are peroxides and superoxides that could be surprising

Valency is the combining power of an atom; so, when we talk of iron having a valency of 2, it can combine either with two atoms of valency of one; or with one atom of valency of two.  Oxygen has a valency of two and so the formula for iron (II) oxide is FeO.
In the history of chemistry valency pre-dates the idea of electron configuration which looks like the source of your confusion.
Hope that is clearer.
Edit: you mentioned that oxygen has a valency of minus 2; that's not quite right.  It has a valency of 2 but it can form an oxidation state of minus 2.  You need to separate out the ideas of valency and oxidation state.  Although the ideas are related, they are distinct.

Why does Oxygen have valency $2$? It has $6$ electrons in its outer shell I think? Or am I calculating this all wrong?

@voldemort one can imagine that oxygen gets 2 electr. to have 8 total - octet

@Mithoron Then why doesn't Iron have a valency of $6$? Since its outer shell has only $2$ electrons, it could gain $6$ to reach $8$?

@voldemort As you wrote iron can have +6 oxidation number and than also valence = 6 I'm afraid that you're trying to learn what they are on one of more difficult cases.

@Mithoron ok so, it CAN have valence 6, but it normally has 2? How do you know which one is the most common / the lne I should use?


text:
Calculating yield percentage of magnetite synthesis
I’ve made magnetite $\ce{Fe3O4}$ in the lab and now I need to calculate yield percentage. How can I do it? I have these data.
\begin{align}
\ce{FeSO4.7H2O(aq) &-> Fe^2+(aq) + SO4^2- (aq) + 7H2O(l)}\\
\ce{KOH(aq) &-> K+(aq) + OH- (aq)}\\
\ce{KNO3(aq) &-> K+(aq) + NO3- (aq)}\\
\ce{Fe^2+(aq) + 2OH- (aq) &-> Fe(OH)2(s)}\\
\ce{3Fe(OH)2(s) + NO3- (aq) &-> Fe3O4(s) + NO2- (aq) + 3H2O(l)}\\
\end{align}
\begin{align}
\ce{FeSO4.7H2O &->} 4.20~\mathrm{g}\\
\ce{KOH &->} 1.90~\mathrm{g}\\
\ce{KNO3 &->} 0.12~\mathrm{g}\\
\text{Magnetite}~\ce{&->} 0.36~\mathrm{g}\\
\end{align}
I know how to calculate the yield for simple reactions but I don’t know how to do it when there are many reactions involved.

Ok thanks, as I said I know how to calculate yield for simple reactions but not for multistep synthesis. (ac) state symbol is the same as (aq) except that in spanish it is common to write (ac) ("aqueous" = "acuosa"), english is not my native language sorry.

then share the result with us -post an answer

Add up the reactions (double the second one so there is sufficient hydroxide) to get a net reaction with your starting materials as reactants and your desired product as one of the products:
$$\ce{3FeSO4.7H2O(aq) + 6KOH(aq) + KNO3(aq) -> Fe3O4(s) + NO2-(aq) + ...}$$
Then, find the limiting reactant, calculate the theoretical yield from it, and compare it to the amount of product to figure out the percent yield.

I think it should be of 3 mol of FeSO4.7H2O and 6 mol KOH since based on the equations, 3 mol of Fe(OH)3 react with 1 KNO3

@Simon-Nail-It Yup, you nailed it.


text:
Pb-based pigments and rhodizonate salts
In a previous question, I learned that we can detect lead with sodium / potassium rhodizonate, giving a nice purple color when reacting with pure (elemental?) lead.
But in the paint, lead is primarily used in the form of salts such as lead oxide, lead carbonate, lead chromate...
What if the lead in my paint is not pure, but a salt? Would sodium rhodizonate react with such salts? (I insist on sodium vs potassium, since it's cheaper to get). And if not, how can I detect these salts?

According to Ashley et al, "a rhodizonate spot test kit [for] in-situ testing of lead in paint... can be used for screening (i.e., qualitative) purposes." The test is sensitive not only to metallic (elemental) lead, but also to lead compounds used as pigments. However, the reference cited also states that the sulfide and rhodizonate tests have a high incidence of false positives.
Sodium, potassium or even cesium salts of rhodizonic acid should be equally effective, since the active species is the rhodizonate ion, but be aware that the sodium salt is less stable in water solution and should be used within a few days.