text:
Calculating Delta H for Non Ideal Gasses
We have to calculate delta H for a gas that follows the Van der Waals equation of state:
$$\left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m-b\right)=RT$$
I know that $H = U + pV$ but I don't know how to approach this question.
Since I also know that $\Delta H = \Delta U + \Delta (pV)$, does that mean that I can calculate this simply by calculating $\Delta U$ (which I know how to do) and then $\Delta (pV) = p_2V_2 - p_1V_1$? Is that an appropriate method?
Because previously in other questions I have also seen it done as
$\mathrm{d}H = \mathrm{d}U + V\mathrm{d}p + p\mathrm{d}V + \mathrm{d}V\mathrm{d}p$ so I am wondering when is which equation applicable?

You seem certain that you know how to calculate $\Delta U$.  So please tell us how you would do it for a Van der Waals gas.

I've edited your post with LaTeX to make it look more presentable - you can learn how to do this here. Your vdW equation of state was incorrect; you should have used the molar volume $V_m$ instead of $V$. I corrected it. Also, in the last equation, the product of two infinitesimals should not be in the equation.

By the product rule, $\mathrm{d}(fg)/\mathrm{d}x = f(\mathrm{d}g/\mathrm{d}x) + g(\mathrm{d}f/\mathrm{d}x)$; you can "multiply throughout by $\mathrm{d}x$" (it is mathematically incorrect, but it gives you the right result) to get: $\mathrm{d}(fg) = f \mathrm{d}g + g \mathrm{d}f$. It can be more properly derived by considering the properties of differentials but that is another story.

Are you familiar with the following equation?:  $$dH=C_pdT+\left(V_m-T\left(\frac{\partial V_m}{\partial T}\right)_P\right)dP$$

Ok, but in order to calculate ∂Vm/∂T don't I need to rewrite the equation of state in terms of V? And how can I do this?

Good point.  A better choice of an equation to start with is:$$dU=C_vdT-\left(P-T\left(\frac{\partial P}{\partial T}\right)V{m}\right)dV_m$$

A better choice is:  $$dU=C_vdT-\left(P-T\left(\frac{\partial P}{\partial T}\right)_{V_m}\right)dV_m$$


text:
Why does chlorine oxidise iron to iron(III), not iron(II)?
I was asked to predict the product of the reaction between iron metal and chlorine gas:
$$\ce{Fe +Cl2->} ?$$
The product here is supposed to be $\ce{FeCl3}$. But how would we know if the product is $\ce{FeCl2}$ or $\ce{FeCl3}$? Why should it be $\ce{FeCl3}$ and not $\ce{FeCl2}$?

Because it can. Even iodine can oxidize iron to iron(III), if you have the right co-reactant anc controlled conditions.

The enthalpy of formation ($\Delta H_\mathrm{f}^\circ$) of $\ce{FeCl3}$ is $\pu{-399.49 kJ mol-1}$, while the $\Delta H_\mathrm{f}^\circ$ of $\ce{FeCl2}$ is $\pu{-341.79 kJ mol-1}$. This means that $\ce{FeCl3}$ is $\pu{57.7 kJ mol-1}$ more stable than $\ce{FeCl2}$, a considerable amount. This means that it is more thermodynamically favorable for $\ce{FeCl3}$ to form than $\ce{FeCl2}$, likely due to the larger lattice energy.
Furthermore, in the +2 oxidation state, one electron remains paired in the $\mathrm{3d}$ orbital. When $\ce{Fe}$ is in the +3 oxidation state, however, it has a half filled $\mathrm{3d}$ orbital, a state which is known to be particularly stable, which you can read about further here.

That's true, but I think there is more to the question. See, when you encounter some simple reaction, do you literally reach for the reference book to check the enthalpies? I don't think so. Then how do we really predict things like this? I imagine an ordered row of redox potentials (without numbers, just the order) and think in terms like "OK, chlorine is strong, it will drive iron all the way to +3..." But then again, maybe I am idealizing it, maybe we just remember all simple reactions and use the reference book for the not-so-simple ones?

This does not really answer it - it doesn't really matter which is more thermodynamically stable.

It is thermodynamically stable for diamond to spontaneously turn to graphite at room temperature. This does not mean we observe it to do so.

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text:
Finding equilibrium concentrations in solution from initial reagent concentrations and equilibrium constant

We have the following reaction:
$$\ce{A + B<=>2C +D}$$
The initial concentration of $\ce{A}$ and $\ce{B}$ are $\pu{1M}$ each. $K_c$ is $\pu{1.0E8}$. Find the equilibrium concentration of $\ce{A}$.

This is what I have tried:
The initial concentration of A,B,C,D are 1,1,0,0 respectively
The final concentrations of A,B,C,D have to be ($1-\alpha$), ($1-\alpha$), $2\alpha$, $\alpha$ respectively. Therefore
$$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
$$10^8=\frac{(2\alpha)^2\alpha}{(1-\alpha)^2}$$
$$10^8(1-\alpha)^2=4\alpha^3$$
But even on solving this cubic equation on a calculator, the values are incorrect.

No need for the cubic equation. Just firstly assume that $\ce{A}$ and $\ce{B}$ reacted completely, then $\ce{[C]}=\pu{2M}$, $\ce{[D]}=\pu{1M}$, and from $K_c$ you can calculate $\ce{A}$ which is the same as $\ce{B}$.
That's the approximation usually used with this kind of problem.
The answer should be $\pu{2E-4M}$.

This problem can be solved iteratively, if you are looking to increase precision. A first iteration ("zeroth order solution") was presented in another answer, where it was first assumed that approximately all reagent is converted to product before using the result to back-compute the residual (very small) concentration of reagent at equilibrium.
In fact you can rewrite the cubic equation into the following form:
$$\alpha = 1-\sqrt{\frac{4\alpha^3} {K_c}}$$
where $\alpha$ is the final concentration of $[\ce{D}]$ or the concentration of $[\ce{A}]$ consumed at equilibrium (as defined in the OP). From this definition we can write that the remaining concentration $\beta=[\ce{A}]_\textrm{fin}$ is
$$\beta = 1-\alpha$$
or
$$\beta_1= \sqrt{\frac{4(1-\beta_{0})^3} {K_c}}$$
This expression was used in another solution by using the approximation $\beta_0 \approx 0$. It can more generally be used to solve for $\beta$ iteratively starting from an initial guess (not necessarily $\alpha_0 =1$). For large $K$ convergence is pretty quick. In this case one iteration is enough ie if we set $\beta_0=0$ then
$$\beta_1=[\ce{A}]_\textrm{fin} = \pu{2.0E-4}$$
but even if we make a very bad initial estimate convergence is still pretty fast:
$$ \begin{array} {c|c} i & \beta_i\\ \hline  0 &  0.5 \\1  &  \pu{7.0711e-5}   \\2   & \pu{1.9998e-4} \\3  &  \pu{1.9994e-4} \\4  &  \pu{1.9994e-4}  \end{array}$$

Your approach seems correct.  What is the "correct" numerical value of the answer according to your source?
I plugged in your equation to Wolfram Alpha, but using $x$ instead of an explicit $10^8$ as the factor on the left-hand side.
Since it is a cubic equation, there are (in general) three roots.  In chemistry problems that involve cubic equations, usually two of the roots will violate physical principles of the problem.  In this example, the physically meaningful answer must (i) be real, i.e. have zero imaginary part, and (ii) satisfy $0 \le y \le 1$.
Numerical noise can make exactly satisfying those constraints, especially point (ii), somewhat difficult.  You will note that the approximate answers supplied by Wolfram Alpha are:

$0.999368 - 1.164153 \times 10^{-9} i$
$1.00063 - 1.047738 \times 10^{-9} i$
$2,499,998$

The extent of reaction must be between 0 and 1, so obviously the answer of ~2.499 million is not physical.  The second answer has (a real part) of 1.00063, which is also higher than one and thus not physical.  Thus, only the first answer of 0.999368 is physical.  You may be wondering about the $-\pu{ 1.164153E-9} i $ term, i.e. the imaginary part of the answer.  It arises from numerical noise in the root-finding algorithm.  It can be safely ignored.
Thus, the numerical answer for $y$ is $\approx 0.999368$.  That means that the equilibrium concentration of $A$ is $(1-y)^2$, or $\pu{3.99E-7}$ molar.
This makes chemical sense, as we expect that reactions which have very large equilibrium constants will go very nearly to completion.

my answer section says that the answer has to be $2 \times 10^4$. I'm hoping that's a printing error.

If you start off with one molar each of A and B, then that is two molar total.  The reaction forms three moles from two, so the maximum concentration of  anything formed by the reaction is 3.  So 20,000 is certainly wrong.

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Stereochemistry of product in Diels-Alder reaction

I have a few questions on this problem I solved:
1) Why does carbon "b" on methyl vinyl ketone attack carbon "5" on the lactone? Why not carbon "2"?
2) The answer key says that the end product should have "trans" stereochemistry. The internal ester group should be opposite the carbonyl group as shown in my picture. Why is this the case? Does this have to do with the endo rule?

Your diagram shows "b" connected to "2", yet you're saying the correct result is "b" connects to "5".  Is that correct?

Note that the product is a cyclohexene, which can't adopt a true chair conformation as shown in the last drawing.

1) The regiochemistry of the product can be predicted by drawing resonance structures of the diene and dienophile, allowing us to assign partial charges to the reactive centers. Doing that shows us that 2 and a are partially positive while 5 and b are partially negative. We'd then predict that 5 attacks a and b attacks 2. This answer has a figure showing prediction of regiochemistry this way.
2) We would predict that the product is the endo- product, which would have the bridging ester on the opposite side of the cyclohexene relative to the methyl ketone. This has to do with the endo-rule, which is classically explained by secondary orbital overlap between the the lobe of the HOMO on carbon 4 interacting with the lobe of the LUMO on carbon c. Keep in mind that this trend does exist, but mixtures of endo and exo are common.

Figures will be added when I have time later.

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Are a meso compound and its mirror image isomers?
How are meso isomers even isomers? By definition, meso isomers are compounds that are superposable on their mirror images, but doesn't that mean that the mirror image is just the same compound?
Can someone explain the concept of meso isomers and diastereomers to me?

Have a look at Wikipedia, https://en.wikipedia.org/wiki/Meso_compound and then tell us what part you're not getting.  (Not meant to be sarcastic but we'd only say what was in there anyway).

Go back and check what the word ‘isomer’ actually means.

meso-compound
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.[1, 2]
isomer
One of several species (or molecular entities ) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.[1, 3, 4]

Your notion is correct in so far as the meso-compound and its mirror image are not isomers of each other since they are actually the same compound.
For example, meso-tartaric acid (preferred IUPAC name: (2R,3S)-2,3-dihydroxybutanedioic acid) can be superimposed on its mirror image “(2S,3R)-2,3-dihydroxybutanedioic acid”; i.e. both structures actually describe the same compound. Thus, they are not isomers of each other. (And hence, “(2S,3R)-2,3-dihydroxybutanedioic acid” is not a correct IUPAC name.)

However, two further stereochemical configurations are possible, namely (2R,3R)-2,3-dihydroxybutanedioic acid and (2S,3S)-2,3-dihydroxybutanedioic acid.

These structures are non-superimposable mirror images of each other, i.e. they are enantiomers, and the above-mentioned meso-tartaric acid is a diastereomer of both of them.
Therefore, there are a total of three isomers of tartaric acid:

(2R,3R)-2,3-dihydroxybutanedioic acid,
(2S,3S)-2,3-dihydroxybutanedioic acid, and
the meso-compound (2R,3S)-2,3-dihydroxybutanedioic acid.

References
[1] Basic terminology of stereochemistry (IUPAC Recommendations 1996). Pure Appl. Chem. 1996, 68, 2193
[2] meso-compound. In IUPAC Compendium of Chemical Terminology, (Gold Book)
[3] Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994). Pure Appl. Chem. 1994, 66, 1077
[4] isomer. In IUPAC Compendium of Chemical Terminology (Gold Book)

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Why does magnesium oxide not react with water?
There is a question in my textbook which asks "which of the reactions are not possible to occur". The right answer is $\ce{MgO + H2O}$ and my chemistry teacher confirmed it. Why is that so? I thought that it would form $\ce{Mg(OH)2}$.
Does the reactivity also depend on the temperature?

MgO has a high lattice energy of about 4000 KJ/mol. Also MgO is insoluble in water because the atoms are tightly held. MgO has a rock salt structure.

That's true, but hardly explains anything. Again, CaO has the same structure.

I'm inactive at this time, and regardless wouldn't be able to answer the question with sufficient rigor it deserves. Perhaps it is of some help to note that an older Russian textbook blames the apparent lack of reaction on the formation of almost insoluble (probably denser as well) passivating layer of $\ce{Mg(OH)2}$. Calcium hydroxide is more soluble (by some 2500 times, I think, near room temperature). Obviously, this simply pushes the comparison from the oxides to their hydroxides (to some extent) which might be of limited use.

Like a lot of things in chemistry, "it depends".
Magnesium oxide is commonly produced by calcination of magnesium hydroxide or carbonate. The magnesium hydroxide or carbonate decomposes at a fairly low temperature whereas magnesium oxide is itself stable to both decomposition and melting  beyond 2000°C. So, we can choose from a wide range of calcining temperature, and as explained here we can use this to dial in the reactivity we want.
At a low calcining temperature the magnesia is quite reactive, even hygroscopic (the "magnesium oxide" film formed on magnesium in ambient air is thus usually magnesium hydroxide); this "light-burned magnesia" is used where easy reactivity is required as in water or soil treatment.  And, yes, the reaction with water gives magnesium hydroxide.
Calcination at higher temperature gives less reactive "hard-burned" magnesia, or in the extreme case the totally unreactive "dead-burned magnesia" used in furnace refractories.

First off, to not overlook the obvious, both reactions ares thermodynamically favored.
$$\ce{MgO (s) + H2O (l) -> Mg(OH)2 (aq)} \\
 \Delta G_\pu{298K}^\circ = \pu{-833.7 kJ mol-1 -( -569.3 kJ mol-1 + -306.7 kJ mol-1) =  - 42.3 kJ mol-1}$$
$$\ce{CaO (s) + H2O (l) -> Ca(OH)2 (aq)} \\
 \Delta G_\pu{298K}^\circ = \pu{-(-754.2 kJ mol-1 + -306.7 kJ mol-1) + -1011.7 kJ mol-1 =  - 49.2 kJ mol-1}$$
Thus thermodynamics is not the reason and thus kinetics must be to blame. For starters combusted magnesium is going to produce partially amorphous $\ce{MgO}$ which will reduce the activation energy to hydrolyze magnesium oxide, but why is it that solid magnesium oxide reacts slowly with water allowing me to use crucibles made of the stuff, while calcium oxide reacts quite vigorously despite both being thermodynamically favored?
I believe the answer is in the Cation-anion radius ration (C/A ratio). The C/A ratio for $\ce{Mg^2+}:\ce{O^2-}$ is $0.683$ ($\pu{86pm: 126pm}$) which correspond to a 6-fold nearest neighbors configuration while the C/A ratio for $\ce{Ca^2+}:\ce{O^2-}$ is $0.905$ ($\pu{114pm: 126pm}$) which corresponds to an 8-fold nearest neighbors configuration. Yet both compounds are in a rock salt 6-fold nearest neighbors configuration. AS a result the oxygens  of magnesium oxide are tightly packed "touching" each other while the oxygens of calcium oxide are space apart farther than two ionic radii. This allows the oxygen atoms of calcium oxide to "make room" for water and increase the coordination of $\ce{Ca^2+}$ to accommodate the additional atoms whereas the oxygens in magnesium cannot.

magnesium forms a layer of $\ce{MgO}$ readily on it's surface when exposed to air, thus preventing further reaction, when you put $\ce{Mg}$ in water, which is actually the $\ce{Mg}$ with the $\ce{MgO}$ crest on it, it is essentially inert to water as the oxide crest won't react, and or allow access to the interior where pure $\ce{Mg}$ lies. But if it were pure $\ce{Mg}$ and Water, the result sure will be $\ce{Mg(OH)2}$
Also for a reaction to occur between a solid and liquid, atoms of the solid need to dissolve on the liquid to interact with the liquid itself or a component of liquid as a mixture. It is a fact that $\ce{MgO}$ is not very soluble in water, hence restricting it's interaction with water
In response to comment by Ivan Neretin
IvanNeretin @IvanNeretin , the solubility of $\ce{MgO}$ is $\pu{0.0086g/100mL}$ while $\ce{CaO}$ reacts directly with water, it is the natural property of $\ce{MgO}$  to not to react with water, while $\ce{CaO}$ naturally reacts with water. I can deduce one thing from this $\ce{MgO}$ is stable in water, while  as $\ce{CaO}$ is instable, and reacts to form a stable a stable $\ce{Ca(OH)2}$

This doesn't explain why MgO doesn't react with water.

@orthocresol, I edited the answer, to a broader scope

That still leaves much of the question open. See, $\ce{CaO}$ is almost the same as $\ce{MgO}$ (would anyone call it soluble?), but it reacts with water quite vigorously.

So it all boils down to "natural property"? Well, this is a good answer to any chemistry question, but still it leaves some people not quite satisfied.

@IvanNeretin  that is quite true

All I would say is to check this link https://youtu.be/qaBUHS5dJhQ   that shows MgO dissolving in water.  It may be due to the temperature (it is freshly formed)

@Beerhunter, yes it is due to temperature MgO does react with water at elevated temperatures

So the reaction is possible as stated in the question. Nothing was mentioned about ambient conditions.

@Beerhunter , as nothing was mentioned in the question I took it as room temperature. Thanks for pointing out :)

I will attempt an answer that may overlap some of the other answers, but it seems pretty clear to me that although the reaction producing magnesium hydroxide does take place ( if you test it with litmus paper, it will turn blue showing the alkalinity), the reaction is unremarkable as MgO is barely soluble in water. There is a strong attraction between the doubly charged magnesium and oxygen ( unlike the singly charged sodium and chloride). A lot more energy is needed to pull apart the ions in MgO, making them much less soluble in water than NaCl. Calcium has one more energy level than magnesium, making it more reactive in general, so its oxide will naturally be more reactive as well.
Another proof that magnesium oxide does produce a base, is that is used for medicinal purposes to neutralize acid in heartburn medication and is also used to treat paper against acid.


text:
Are Organic Solvents used in Addition Reactions catalysts?
Say for example you are performing a halogenation reaction with dichloromethane (DCM) as the solvent:
ethylene + chlorine $\stackrel{DCM}\longrightarrow$ 1,2-dichloroethane
I have 2 questions:

Is the DCM acting as a catalyst to the reaction, or just a solvent? And if just a solvent, why do we have to use an organic solvent, why not water?
Are solvents like DCM necessary if the reactants are liquids already, or only for gaseous or solid reactions?

Thank you

As for not using water, try dissolving some non-polar organics in water, then you'll see why. Also, water+chlorine would spark a whole bunch of side reactions we don't want to have.

In the example you give, the dichloromethane is acting as a solvent. Solvents are usually chosen to be inert under the reaction conditions so that they do not react and produce unwanted products.
There are examples of reactions in which no solvent is used, or in which the solvent is a reagent. An example could be forming an ester from a carboxylic acid, using an excess of the alcohol as both solvent and reagent, in the presence of an acid catalyst.  That said, solvents are usually used as they severely effect the outcome of a reaction. Polar solvents (THF, diethyl ether) promote very different reactions to non polar ones (hexane) due to the stabilisation they provide for intermediates during the reaction, dilution can also play a role, deciding whether inter or intra-molecular reaction occurs.

Thank you very much, I was wondering though, are solvents like DCM necessary if the reactants are liquids already, or only for gaseous or solid reactions?

To answer the second part of your question, solvents are also used as heat sinks. If no solvent is used, the heat given off from a particular reaction can lead to it being uncontrollable, potentially runaway. If engineering controls are used, no solvent is fine (usually in an industrial setting). However,  in a laboratory setting, the heat evolution could cause flashing off (instant vaporisation) ,  in this instance of DCE,  causing a large expansion of material changing from liquid to gas. DCM in itself may not be a great heat sink, but if the reaction is very dilute, the specific heat capacity is supplemented by the volume/mass present.


text:
How to sequester Na+ which prohibits the formation of a micro-emulsion in solution
I am mixing alkaline solutions ($\ce{Na2CO3}$) with crude oil containing carboxylic acids to create micro-emulsions. The carboxylic acids react with the alkaline moieties to create an in-situ surfactant which creates the micro-emulsion.
Up to here everything is going well. I produced the micro-emulsion I wanted. The next step is to have these micro-emulsions taking place when there are also contaminants ($\ce{Na+}$ and $\ce{Ca^2+}$) also present in the alkaline solutions, which prevent the micro-emulsion from forming.
I added EDTA to sequester the $\ce{Ca^2+}$ and it seems to have worked. What can I do about the $\ce{Na+}$?

This doesn't make sense - two parts of your question are self-contradictory. How sth you use to create emulsion would prevent it from forming. Btw. "micro" emulsion?

Not sure I understand entirely your question, but 15-crown-5 may be used to remove Na+ ions from a solution.

I think he means emulsions with "micro-droplets".


text:
Possibility of R-Si-OCH3 reacting with alkanol group in polysaccharides
For example, the aminosilanizing agent 3-aminopropyltriethoxysilane (APTES)

It has been frequently used for silanization of silica or glasses for various purposes. The surface funcation group of silica contain Si-OH which will react with APTES under certain conditions:

Both Si and C are group 4 elements
R-C-OH where the C-OH is less polar than Si-OH bond
Take glucose as the example, it contains 5 external C-OH bonds. So is it possible for APTES to react with glucose? If yes, what condition should it be?


text:
Does a Na2SO3 solution oxidize to Na2SO4 in an oxygen rich atmosphere?
Let's assume I have a pure solution of water and $\ce{NaOH}$. When this solution comes in contact with an atmoshpere containing inert gases and $\ce{SO2}$, then $\ce{NaHSO3}$ and $\ce{Na2SO3}$ will be formed into the solution.
The question is, if the atmosphere now contains also $\ce{O2}$, would these reactions happen?
$$\ce{2Na2SO3 + O2 <=> 2Na2SO4 + 2e-}\\
  \ce{2NaHSO3 + O2 <=> 2NaHSO4 + 2e-}$$
What would happen to the freed electrons?

That's quite an atmosphere you have there... Well, the redox reaction can and will happen; as to the freed electrons, there would be none.

Well actually I do not have such an atmosphere, it was just to simplify the question compared to the original problem. How can I figure out the equilibrium constants for these reaction? I cannot find the H and S of $\ce{NaHSO3}$ anywhere. Could you suggest a software to simulate these equilibrium constants? That would help a lot with my work.

I don't have these constants or $\Delta G$'s at hand, so keep searching. As for software, it must ultimately rely on the same reference data.

Do you know other reliable databases apart from NIST Chemistry Webbook and WolframAlpha?

No, not really... Well, you may have my educated guess: the constant would be pretty large, so consider the reaction going almost completely. But it would not be fast; you can't ignore kinetics.

Would you not have 1/2 O2 +  1/2 H2O + 2 e- -> 2 OH-? You can look up the relevant reduction potentials to find the equilibrium constant. Granted it's not exactly standard conditions but it's a start.

@orthocresol Could you please elaborate a little? I am not expert in this. Thanks

You can also try to find values in the CRC Handbook of Chemistry and Physics. I think the suggestion by @orthocresol is right, this is certainly a reaction that will happen in an aqueous solution. If you have oxygen and sulfur dioxide in the atmosphere, you will also have to consider $\ce{2SO2 + O2 <=> 2SO3}$.

I'm not sure what is the stable form at room temperature, since we have both sodium sulfite ($\ce{Na2SO3}$) and sodium sulfate ($\ce{Na2SO4}$) in our lab and they're both stable, in air. I do know that the sulfite will oxidise to sulfate with the addition of bleach.
That said, your redox reaction is not quite accurate. In sulfite, sulfur is in the $\ce{S^4+}$ oxidation state. In sulfate, it is $\ce{S^6+}$. What happens in the reaction is that the sulfur gives away two electrons, which then go to an oxygen that changes from $\ce{O^0}$ to $\ce{O^2-}$.
One way of writing it would be $\ce{S^4+ + 1/2O2^0 -> S^6+O2^2-}$. Total charge is preserved and there are no free electrons.


text:
Rust removal with phosphoric acid created more rust
I wanted to remove the rust on numerous steel parts with phosphoric acid. After removing all the grease in trichloroethylene, I put my parts into a 50 % solution of phosphoric acid ad 40 °C for 5 minutes.
When all the rust was removed, I rinsed the parts with clean water to remove all the acid. At that time, I decided to not neutralize the solution in $\ce{NaOH}$.
In about 5 minutes all the cleaned parts exposed to air rusted. The parts into the water became greenish (I guess it is $\ce{FePO4}$ residue).
I would like to understand why I get this ultra fast oxidation and how I can prevent it to occur.

What steel was it? Also why you used such harsh method?

Some regular steel I guess not an stainless one obviously. I have 3-4 hundred parts to treat. Doing all by hand is a tough work. Also I think I like harsh methods :)

There is exactly your problem explained in the second link of this post: https://www.chemicalforums.com/index.php?topic=102183.0

Rust removers, based on phosphoric acid, contain also tributylphosphate and perhaps some corrosion inhibitors. They are not supposed to get rinsed out. The surface gets rather dark violetish and can be painted.


text:
Is there any porous material that can have narrow parallel and "perfectly straight" cavities/holes through it (or bored through it)?
I'm basically looking for a material that can allow photons to pass through its cavities, but only those photons that travel at an angle (almost) exactly perpendicular to the materials surface. Non-perpendicular photons should be absorbed by the material upon impact as much as possible.
I can imagine perhaps a stack of graphene sheets, or a bunch of parallel carbon nanotubes? Or maybe a high precision laser drilling through ...something else? Clearly I'm no chemist or material scientist, so I'm hoping to tap into the knowledge of this community.

This is hardly chemistry. If your holes are small enough to be of interest to chemistry, then the photons don't pass through the holes. They pass through the material, if they can.

@IvanNeretin How large does a hole need to be for a photon to travel through it without being affected by the surrounding material? If this should be moved to physics.se, so be it. I posted this here, because I figured there was a strong material science aspect to this question...

I am not qualified to answer. Yes, I think the question belongs to physics.se.

Some microporous membrane filters might come close to what you are looking for. For example, see www.sterilitech.com.


text:
Are there more positive azeotropes than negative azeotropes?
During a lecture on Flash and Distillation, my professor (which worked as a Chemical Plant Engineer for more than thirty years, specifically on separation processes) pointed out that the most frequent type of azeotropes he encountered in his work was positive-type. At first it seemed just common practice among azeotropic distillation et similia, however page 6-180 of CRC Handbook of Chemistry and Physics says:

Nearly 90% of the binary azeotropic data show a pressure maximum (positive).

So, I'd like to know why it's true or not that nature could be somehow biased toward a specific or more frequent type of azeotrope.

Summary. Positive-deviation azeotropes mean that like interactions are much stronger than unlike interactions, whereas negative-deviation azeotropes mean the opposite. The former case is more common because most substances interact better with themselves than with other substances---like dissolves like.
(So I've only just rephrased the question at a lower level: one could equally well ask why like should dissolve like, but believing this statement on faith is perhaps more palatable than just accepting that there are more positive-deviation azeotropes.)

First, a few definitions.
A binary solution showing positive deviations (negative deviations) to Raoult's law is one in which the dew-point curve in a pressure vs. mole fraction graph lies above (below) the line connecting the pure vapor pressure of the two species.
A binary azeotrope is positive deviation (negative deviation) or minimum boiling (maximum boiling) if it shows positive deviations (negative deviations) to Raoult's law and exhibits a local maximum (minimum) in the dew-point curve in a pressure vs. mole fraction graph. From solution thermodynamics, one can prove that the vapor and liquid mole fraction of either species is equal at these extrema.

Physically speaking, positive deviations (negative deviations) from Raoult's law imply that like interactions are stronger than (weaker than) unlike interactions in the solution, for the solution's dew point pressure is greater than (less than) that of an ideal solution.
Azeotropes are extreme forms of these positive deviations (negative deviations), when the dew-point curve further exhibits a maximum (minimum), such that at some composition the solution's dew point pressure is actually greater (lesser) than the pure vapor pressure of either species.
We expect, therefore, that positive-deviation azeotropes are more common than negative-deviation ones because it is more likely for two species to have stronger like than unlike interactions.

Very detailed and interesting answer. I infer that this behavior can be explained with statistical mechanics and mixing entropy, it would be quite nice to lay out the formal proof one day


text:
Why does shampoo lather much more the second time it is applied?
When I shampoo my dirty hair, the first time it does not lather very much. The second time it lathers a lot. Why is this?

related http://chemistry.stackexchange.com/questions/33797/what-are-the-active-ingredients-of-shampoo-and-how-do-they-work

Quoting a reddit post from chemist nallen:
The short answer is that the dirt and oils from your hair compete for the surfactants making them less available to form lather, which is small bubbles.
To better understand the mode of action, you have to know a bit about the formulation of shampoos and the nature of dirt and oils.
Dirts and oils deposit on hair and fibers because they are at least partially hydrophobic (not soluble in water.) The surface of hair, skin, etc... is more hydrophobic than water (well of course!), so these dirts have greater adhesion for the surface than for water (the water actually pushes them out to minimize surface area.) This is why water is ineffective for removal, it won't pull the dirt off the surface by itself. Surfactants associate with the hydrophobic surfaces to make them more dispersible or soluble in water, allowing them to be rinsed off.
So what happens when there is more surfactant than there is hydrophobic dirt surface?
The answer is that the surfactant orients to the surface of the water, because air is hydrophobic, and the water wants to push the hydrophobic portion of the surfactant out of the water molecule matrix. This orientation on the air-water interface is how skin of bubbles is formed. When the water/surfactant solution undergoes sufficient shear mixing, bubbles are formed, and that is lather.
Lather typically forms when there is surfactant in excess of the hydrophobic surface. More dirt means less free surfactant, and therefore less lather.
Now, it's not purely that simple, because the amount of lather that a surfactant develops varies with the type of surfactant, some are better than others.
Lather does not mean that the shampoo is cleaning better, it's purely cosmetic. Typical shampoos are primarily SLES/Betaine, that is sodium laureth sulfate and cocamidopropyl betaine. The SLES is there for cleaning, and the betaine is there for lather, it's specifically added to produce lather because, well, people like it. There are some other benefits, but that's the primary one. If you used SLES by itself is would clean just as well, but people just don't like poor lather, so it would not sell.

The first time, the dirt, grime, and oil in your hair bonded to the detergent in the shampoo so it doesn't lather as much. The second time, there is not as much oily substance so it lathers better.

This is basically a summary of the accepted answer, without all the interesting detail. It does not add anything else to the question.


text:
How to determine the calorimeter constant from its heat capacity

A bomb calorimeter contains $600\;\mathrm{mL}$ of water. The calorimeter is
   calibrated electrically. The heat capacity of the calorimeter is
   $785\;\mathrm{J\,K^{-1}}$. The calorimeter constant would be closest to:
A. $3.29\;\mathrm{kJ\,ºC^{-1}}$
B. $4.18\;\mathrm{kJ\,ºC^{-1}}$
C. $4.97\;\mathrm{kJ\,ºC^{-1}}$
D. $789\;\mathrm{kJ\,ºC^{-1}}$

My (rather mindless) attempt is as follows:
$$
E=mC_PT\to E/T=mC_P\to C_{\mathrm{cal}}=mC_P=(600)(8.314)(10^{-3})=4.9884\;\mathrm{kJ\,ºC^{-1}}
$$
The closest answer to my result seems to be C ($4.97\;\mathrm{kJ\,ºC^{-1}}$), however I know I'm wrong.

I'd go with (A) - sum the heat capacity of the water (600 $\times$ 4.184) and the heat capacity of the calorimeter.

But I don't understand how we can add $0.785 kj/K$ to $2.51 kj/ºC$ to obtain $3.29 kj/ºC$. Aren't they different units?

See this Wikipedia article - "the magnitude of the degree Celsius is exactly equal to that of the kelvin."

To give a precise answer, the following assumptions are necessary and must be clear:

bomb calorimeter works at constant volume ($V=const$);
both water and calorimeter itself are at thermodynamic equilibrium before the experiment and during the measurement, in particular their temperatures $T_w$ and $T_c$ are equal before the experiment and during the measurement;
the system is compound by calorimeter itself plus water;
the system is an isolated one;
pressure is 1 bar.

Initially the system is at temperature $T_1$. Let's imagine that an object at $T_o>T_1$ is put inside the chamber of the calorimeter. Temperature of the system increases and, once reached thermodynamic equilibrium, it stops at a precise value $T_2$.
Since $V=const$, heat transferred from object to system is:
\begin{equation}
Q_V=\Delta U=\Delta U_{calorimeter}+\Delta U_{water}=(mc_V\Delta T)_c+(mc_V\Delta T)_w
\end{equation}
where $\Delta T_c=\Delta T_w=T_2-T_1$.
We know that heat capacity at constant volume is defined as:
\begin{equation}
C_V=\left(\frac{\partial U}{\partial T}\right)_V\approx \left(\frac{\Delta U}{\Delta T}\right)_V
\end{equation}
So, reshaping the first equation, we obtain:
\begin{equation}
C_V=\frac{\Delta U}{\Delta T}=(mc_V)_c+(mc_V)_w=(C_V)_c+(\rho Vc_V)_w
\end{equation}
Adding the following data:

$\rho_w=1000\;kg/m^3$;
$(c_V(300\;K,1\;bar))_w\approx 4.134\;J/(kg\;K)$ (source: Perry's Chemical Engineers' Handbook)

and carrying out the conversion: $V=600\;mL=6\times10^{-4}\;m^3$, we obtain finally:
\begin{equation}
C_V=787\;J/K=0.787\;kJ/K
\end{equation}
So right answer is A.


text:
flory ( or theta) temperature
I know the flory temperature is the unique temperature at which the attractions and repulsions of a polymer in a solution cancel each other.
but what does this mean in practice? does it mean that this is the temperature at which the polymer and the solution mix best? and if you move away from this temperature it mixes less well?

I'm going to use Theta-Solvent and Theta-temperature since that is what Wikipedia uses, and Wikipedia usually has the preferred names.
In order to get a polymer to the Theta-Condition two factors must be considered.
(1) Solvent
A Theta-Solvent is a poor solvent for a polymer is one such that the interactions between the polymer segments and the solvent are minimized. This results in polymer segment interactions to be maximized.
A good solvent for a polymer is one such that the interactions between the polymer segments and the solvent are maximized. This results in polymer segment interactions being minimized.
(2) Temperature
The Theta-temperature is the temperature for the particular theta-solvent
which minimizes interactions between the polymer segments and the solvent.
Point
The combination of using a theta-solvent and being at the theta-temperature satisfies the theta-condition.
This makes experimental determination of important quantities such as the root mean square end-to-end distance or the radius of gyration much simpler.
Note
This assumes that polymer will self-coil into its lowest energy configuration. For polymers with same unit repeating this is not unlikely.
For proteins in biological systems this self-arrangement is unlikely. Proteins do not have a "repeating unit." Therefore biological systems have a mechanism to fold a protein in a certain way.

At theta temperature, there are no interactions between the polymeric chains and the solvent. This means that the polymer doesn't mix well with solvent (The solvent is qualified as a poor solvent of the polymer). So, every polymer chain behaves as an ideal chain. The importance of the theta temperature and the theta solvent is to measure the characteristics of the polymer chain, such as the bond length and the bonds angle by cancelling the effect of the surrounding molecules of the solvent.

Incredibly late to the party here, but does every polymer-solvent system have a theta temperature?

@megamence I'm wondering that too. Perhaps it should be a question of its own.


text:
Does Intercalation and Deintercalation of Lithium ions occurs in Half Cell?
The important feature of Lithium ion batteries is the ability to  intercalate and deintercalate lithium ions in  anode and cathode . However when we use  half cell, using lithium metal foil as anode, intercalation and deintercalation process occurs in this case too. How the lithium ion moves and what process cause the current to flow in discharge?

When you're using a lithium metal anode there is no intercalation really, just reduction of the lithium ions to lithium metal:
$$\ce{Li^+ + e^- -> Li^0_{(s)}} $$
The cathode reaction is still the same.


text:
Active transport of Ca2+ ions against chemical and electrostatic gradients, assisted by ATP hydrolisis
I have a homework question that I am not able to solve. No solution is available, so I was hoping someone here could help. The question is this:

Calculate the maximal concentration ratio that can be obtained for
  transport of Ca$^{2+}$ ions if two ions are transported per ATP hydrolyzed
  at 310 K, and they are transported to a potential that is 70 mV higher
  at the receiving side. Assume that the $\Delta G$ for hydrolysis of ATP = -45 > kJ\mol.

Relevant formulas:
\begin{align}
\Delta G_{\text{electrostatic gradient}} &= ZF\Delta \Psi\\
\Delta G_{\text{chemical gradient}} &= RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}}\\
\Delta G_{\text{hydrolysis of ATP}} &= -45000~\mathrm{J/mol}\\
\end{align}
Moving two positive $\ce{Ca^2+}$ ions against both a concentration gradient and an electrostatic gradient implies that $G_{\text{chemical gradient}}$ and $\Delta G_{\text{electrostatic gradient}} > 1$. The negative value for $\Delta G_{\text{hydrolysis of ATP}}$ means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, $\Delta G_{\text{chemical gradient}}$ is multiplied by 2. Also (I presume) the concentration of $\ce{Ca^2+}$ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.
The combined expression for $\Delta G$ for the unassisted transport is
$$\Delta G = RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi$$
Rearranging this yields
$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$
I am not sure what to plug in for $\Delta G$, but it needs to account for the push from ATP. I believe this is done by summing the two free energies affecting the transport, ergo the sum of $\Delta G_{\text{electrostatic gradient}}$ and $\Delta G_{\text{hydrolysis of ATP}}$. However, this gives me a concentration ratio much smaller than 1.
Can anyone see where I am incorrect?

At equilibrium, the entire process (transport of $2~\ce{Ca^2+}$ ions and hydrolysis of $1~\text{ATP}$) involves no change in G.
\begin{align}
0 &=\Delta{G_\mathrm{total}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2\Delta{G_{\ce{Ca^2+},\text{ concentration gradient}}} + 2\Delta{G_{\ce{Ca^2+},\text{ electrostatic}}}\\[2ex]
0 &=\Delta{G_\mathrm{ATP}} + 2RT\ln\frac{C_\mathrm{in}}{C_\mathrm{out}} + 2ZF\Delta \Psi\\
\end{align}
So basically I'm saying you made two mistakes, one a sign error and one a factor of 2 error.

Thank you! So the system is at equilibrium? That seems counter-intuitive because of the concentration gradient and the charge gradient (at equilibrium I would think that these gradients were zero?).

@AndersMB The ATP hydrolysis and Ca2+ transport are coupled, and when the maximum concentration is reached, the system is at equilibrium.  The concentrations stop changing with time at equilibrium.  The equilibrium would be different if the ATP reaction were not be coupled to the transport.

ΔGelectrostatic gradientΔGchemical gradientΔGhydrolysis of ATP=ZFΔΨ=RTlnCinCout=−45000 J/mol
Moving two positive Ca2+ ions against both a concentration gradient and an electrostatic gradient implies that Gchemical gradient and ΔGelectrostatic gradient>1. The negative value for ΔGhydrolysis of ATP means that hydrolysis of ATP is a spontaneous process, or a "work-yielding" process. The energy released from hydrolysing ATP is essentially what drives the transport process. Since we are transporting "4 charges" per ATP molecule, ΔGchemical gradient is multiplied by 2. Also (I presume) the concentration of Ca2+ ions is larger on the inside of the cell than on the outside, implying that the concentration ratio is larger than 1.
The combined expression for ΔG for the unassisted transport is
Rearranging this yields
$$\frac{C_\mathrm{in}}{C_\mathrm{out}} = \exp\left\{\frac{\Delta G - 2ZF\Delta\Psi}{RT}\right\}$$

Welcome to Chemistry! Please use better formatting on your answer. As it stands this is almost unreadable. For information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance.


text:
Is there a consensus how to report coupling patterns greater than quartets?
When analysing NMR spectra, every student at one time during their chemistry studies learns about vicinal coupling and that it produces the characteristic signal shapes of proton spectra. The following multiplets are usually taught and considered consensus:

singlet (s)
doublet (d)
triplet (t)
quartet (q)
multiplet (m)

But that doesn’t even adequately cover all hydrocarbons; quintets and septets need at least be considered. If, furthermore, there are heteroatoms with a half-integer spin of larger than $1$ present, not even these will suffice.
In my lab experience, I often came across a septet of doublets, which ACD termed sptd. However, I just found an internet source proposing the abbreviation sepd.
Is there a consensus on how coupling patterns with more than four of one coupling should be abbreviated? If so, what are the official abbreviations?

ACS guidelines for authors are not very helpful. My personal experience in standard organic synthesis is that anything beyond quintet is not reported as such which shouldn't be happening now that powerful software is widely available.

There is no absolute consensus for reporting abbreviations for splitting patterns. Even Magnetic Resonance in Chemistry, a journal that I generally take as the best guide for reporting NMR data, does not have recommended abbreviations for multiplicities above quartets. Further, there are general inconsistencies in reporting formats within this journal, with abbreviations such as s, sept, spt being used for septet. Additionally, the actual term used to describe the splitting varies. For instance, both pentet and quintet are commonly used to describe a 1:4:6:4:1 splitting.
What is consistent, however, is that the abbreviations are defined within the paper.
Commonly used terms and abbreviations that I have come across include:
singlet (s)
doublet (d)
triplet (t)
quartet (q)
pentet (p, pent, pnt, pentet) or quintet (q, quint, qui, qnt, quintet)
sextet (sext, sxt, sextet)
septet (s, sept, spt, septet) or heptet (hept, hpt, heptet)
octet (oct, octet)
nonet (non, nonet)
While splittings of higher multiplicity are less common within mainstream $\ce{CHN}$ organic molecules (septet for isopropyl substituents are a common exception, and 'apparent' splittings are excluded), they are more common with heteronuclear substitutions and organometallic systems, where spin numbers greater than 1/2 are frequently encountered. Reporting 'apparent' splittings of, say, $\ce{(CH3)2C}H\ce{CH2R}$ as a nonet (where $J_\mathrm{ab} \approx J_\mathrm{ac}$) is not strictly correct according to the $2nI+1$ rule, although in non-NMR journals still occurs, and is certainly taught at undergraduate NMR level.
As for reporting multiple couplings (doublet of septets, for example), again there is no strict rule to adhere to, although the commonly adopted practice is to report the larger coupling first. So, a doublet coupling of $\pu{10Hz}$ and septet coupling of $\pu{2Hz}$ would be a doublet of septets. If it were doublet $(\pu{2Hz})$ and septet $(\pu{10Hz})$, one would report a septet of doublets.

Thanks for this answer, likely the one I will accept. Personally I am well aware of the rules you highlighted in your last paragraph and the last sentence in the last but one paragraph, but they will very likely help future visitors. The case which prompted this question was a septet of doublets I frequently encounter thanks to an isopropyl group with ~7 and ~3 Hz for the septet and doublet respectively ;)


text:
Kinetics of air evolving from water on heating
Is there a simple relationship describing the kinetics of air evolving from water on heating?
For example, if I let water at room temperature and pressure saturate with air, it should have some concentration. Then, if I heat that water up to 90°C, the solubility of air will be much lower by Henry's law. The evolution of air is, presumably, not instantaneous. Is it fast or slow compared to a reasonable heating rate? Does it depend on just how far the system is from equilibrium or on other conditions like the amount of air-water surface boundary?

Is there a simple relationship describing the kinetics of air evolving from water on heating?

There is no simple relationship. The problem is that you can't heat the water absolutely uniformly. Thus all the convection currents in the water would create a mathematically chaotic relationship. There is also the problem of the water being supersaturated - especially if you heat from the bottom.

Is it fast or slow compared to a reasonable heating rate?

This would seem to be sort of a judgement call, but I'd say that it would be faster to dissolve a gas in a liquid (say via bubbling) than heating to expel it. The bubbles mechanically stir the solution rather than just equilibration by diffusion from the surface. Also think of liquid nitrogen. It boils so fast when spilled on the floor that the drops skitter around. The escaping gas insulates the liquid drop limiting the heat transfer to the liquid. So there is a limit as to how fast you can heat a solution.

Does it depend on just how far the system is from equilibrium

Yes. The further from equilibrium, the less likely a solution could be super-saturated.

Conditions like the amount of air-water surface boundary

The time to reach equilibrium surely depends on the surface area to volume. Think of a cylinder of water. A cylinder which has water one inch deep would certainly come to equilibrium faster than cylinder of the same diameter which has water a mile deep.


text:
Rushed publications based on flawed data
In 2014 BICEP2 made a very controversial press release. It was rushed, given before any papers where published, because they feared they would be scooped by another similar experiment. In the end it turned out that their analysis of their data was flawed. It was particularly dramatic because if they had been correct it would have been a significant development.
Are there any similar cases in chemistry? Not necessary an identical situation, but a high profile result that was released in a hurry that later turned out to be incorrect.

You might want to try Retraction Watch.

It looks off-topic here, flagging for migration to Academia SE

The Schön affair comes to mind (a fraud on the part of J.H.Schön, but an honest mistake on the part of his coauthors).

Since it is only my personal opinion, that your question is not on topic here, I am not going to close it. (@Mithoron Please flag for migration after the question has been closed by the community.) I am absolutely uncertain about how on topic it is at [academia.se]. Please inform yourself about their policies. It could be also on topic on [hsm.se], same advice applies.

Very similar version on HSM.

I'm voting to close this question as off-topic because it is about the publication of chemistry results, and not about the technical aspects of the science of chemistry itself.


text:
Making elemental sulfur from pyrite FeS2
Has anyone done this or know of a good experiment that would allow someone to make elemental sulfur from $\ce{FeS2}$ (iron pyrite)?

Please be aware that there's potential danger to liberate $\ce{H2S}$ which is highly toxic and at higher concetration has surprisingly pleasant smell. Also note that you can very quickly lose olfactory trace of it even though it is still present.  I'm not trying to preach sermons here, but speaking from experience of a friend of mine who underestimate this danger and ended up in hospital severely intoxificated.

You may just need to heat it.
$$\ce{FeS2 ->[\Delta] FeS + S}$$
In the wikipedia article of pyrite, it is written that:-

Thermal decomposition of pyrite into $\ce{FeS}$ (iron(II) sulfide) and elemental sulfur starts at 540 °C.

A good experiment for recovering sulfur from iron pyrite is called ' steam-pyrite interactions'.  The following is an excerpt from the original journal:-

The reactions of natural iron pyrite with steam, hydrogen, and carbon monoxide have been studied in a fixed
  bed tubular reactor at 900-1 100°C. A new process has been developed for the recovery of elemental sulfur
  by the reaction of pyrite with steam. In this process, the yields of elemental sulfur and sulfur evolved as hydrogen
  sulfide and sulfur dioxide were, respectively, 84.5 and 13.7% of total sulfur under the following
  conditions: particle size, below 100 ASTM mesh; temperature, 1000°C; reaction period, 1 5 min; space velocity,
  33.2 X lo4 cc/cc/hr; and partial pressure of steam, 1 atm. Under these conditions, the effluent gas also
  contained an appreciable amount of hydrogen. The calcine contained ferrous oxide, ferrosoferric oxide,
  silica, and 1.3% sulfur. The residual sulfur could be further brought down to a minimum of 0.2% in the calcine
  at 1 100°C using a 30-min reaction period. The incorporation of hydrogen in the steam-iron sulfides reactions
  at 1 100°C increased considerably the yield of hydrogen sulfide (59.5%), but only a small amount of metallic
  iron (1.6%) could be obtained.


text:
Using covalent or ionic radius when calculating dipole moment
When the ionic character of a bond is above 50% should I use ionic radius instead of covalent one when calculating the dipole moment?
The formula I have for dipole moment is $\mu = \text{(% ionic character)} \cdot e \cdot d$, where $e$ is the elementary charge and $d$ the sum of the two radii.

Related: Dipole moment - calculation of percentage ionic character

Disclaimer 4/6/2021: I need to re-verify where the numbers in this answer are taken from.

Personally, I think "50%" as a cut-off is very arbitrary to begin with. It's not like I can suggest anything better, since proper dipole moment calculations rely on quantum mechanics.
But, to answer the question:
The whole premise of my answer is that you want to get the sum of your radii to match the actual bond length in the compound. And to do that, you have to look at the context. For example, for $\ce{BF3}$, you might want to use covalent radii since the atoms are covalently bonded. And for a ionic compound like $\ce{KF}$, you'd better use ionic radii or you might run into some problems.
Now, note that the formula you're using is already extremely inaccurate in itself - if you take the covalent radii for B and F, you get $d = \pu{(82+71) pm} = \pu{153 pm}$. And... the actual bond length is very short - it's $\pu{130 pm}$. Clearly, there is something about that compound that the crude model of adding covalent radii cannot account for. Still, it's better than if you were to use the ionic radii: $\pu{(41 + 119) pm} = \pu{160 pm} > \pu{153 pm}$ (sum of covalent radii) $> \pu{130 pm}$ (actual bond length).

"Hey @orthocresol, using $\ce{BF3}$ isn't fair! That compound has crazy strong $\pi$ bonding between B and F!"

Okay. Let's look at $\ce{BCl3}$, $\ce{BBr3}$ and $\ce{BI3}$, where the $\pi$ bonding is weaker.
$$\begin{array}{|c|c|c|c|}
\hline
\ce{X} & \text{Actual B-X bond length / pm} & \text{Sum of covalent radii / pm} & \text{Sum of ionic radii / pm} \\
\hline
\ce{Cl} & 175 & 82 + 99 = 181 & 41 + 167 = 208\\
\ce{Br} & 187 & 82 + 114 = 196 & 41 + 182 = 223 \\
\ce{I} & 210 & 82 + 133 = 215 & 41 + 206 = 247\\
\hline
\end{array}$$
The covalent radii are closer every time.

"That's still not fair... Can't we just use a compound without $\pi$ bonding?"

Fine. Let's talk about the tetrahedral $\ce{BF4-}$ anion. There's no $\pi$ bonding here and as a result, the B-F bond length is increases to $\pu{145 pm}$. But the covalent radii still win.
You can extend the discussion to something like $\ce{SiX4}$. But, honestly, I think that representing $\ce{SiX4}$ as $\ce{Si^4+}$ and $\ce{X-}$ is not that a great an idea, to say the least.
Also, I'm lazy to dig up bond lengths in $\ce{AlCl3}$, so I'll stop here.

"Okay, now what about ionic compounds?"

You'd better use those ionic radii, because guess how those ionic radii are tabulated? They're calculated by measuring the distance between cations and anions in an ionic lattice and assigning a certain part of that to the cation and another certain part to the anion. So if you add them back up, they'll of course give you the internuclear separation that you want.
Let me use the simple example of $\ce{NaCl}$. You'll have to trust that I'm doing the right things here; if you're interested, you can read up on the rock salt structure. The unit cell parameter of $\ce{NaCl}$, denoted by $a$, is $\pu{564 pm}$. This corresponds to two times the distance between $\ce{Na+}$ and $\ce{Cl-}$ ions, so we could say that the actual "bond length" is $\pu{564 pm} / 2 = \pu{282 pm}$.
Now, the ionic radii of 6-coordinate $\ce{Na+}$ and $\ce{Cl-}$ are $\pu{102 pm}$ and $\pu{181 pm}$ respectively. That sums up to $\pu{283 pm}$ - a beautiful match. And the atomic radii? They're $\pu{191 pm}$ and $\pu{199 pm}$ for $\ce{Na}$ and $\ce{Cl}$ respectively. That sums up to $\pu{390 pm}$ - pretty close but not quite as close as the sum of the ionic radii.

Data is taken from this website, Wikipedia:Ionic radius, Shriver & Atkins, Inorganic Chemistry, 6th ed., and Greenwood & Earnshaw, Chemistry of the Elements, 2nd ed. Apparently, the numbers in that Wikipedia figure are taken from R. D. Shannon, Acta Cryst., 1976, A32, 751, which is in fact one of the most widely-used sources of ionic radii (all inorganic textbooks reference it) - so it shouldn't be that bad.

191+199=390, not 290 (NaCl example).


text:
Principles of Thermodynamics
A gas expands and does 82.7 J of work on the surroundings. At the same time, the gas loses 103 J of energy to the surroundings as heat. What is the change in the internal energy of the gas?
would this be ΔE = q + w  (-103) + (82.7) = -20.3.   I get confused with the positive and negative charges.

Ignore the heat part first. If the gas does work on its surroundings, do you think its energy increases or decreases? If you push a block of metal, does your energy increase or decrease?

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Related: Thermodynamics, pressure volume work

The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is
$$\Delta U = Q + W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur).
Therefore, heat transferred from the surroundings to the system is assigned a positive sign in the equation, whereas heat transferred from the system to the surroundings is assigned a negative sign.
Similarly, work done by the surroundings on the system is assigned a positive sign in the equation, whereas work done by the system on the surroundings is assigned a negative sign.
(Note that some books on physics or engineering still use a non-standard definition when discussing the work done by a thermodynamic heat engine.)


text:
Can I melt Barium Titanate powder into a cylinder and preserve the high dielectric?
I would like to make various solid forms of Barium Titanate e.g. cylinders, discs.
Barium Titanate powder seems widely available and has a melting point of 1,625'C. Can I melt it in an iron forge into a mould, and will this preserve the high dielectric constant?
Is there perhaps a low temp way to do this easier e.g. a solvent or additive to achieve the same? Others have tried embedding within epoxy, but with substandard K values (max 27 instead of 7,000 as per the raw powder)

As you know, barium titanate is a piezoelectric ceramic materials. It has its property as a piezoelectric material due to the specific spatial arrangements of its constituting atoms. Melting this material, means that it will lose its crystalline  structure upon heating (knowing also that its Curie Temperature is $T_C=120^o\mathrm{C}$, i.e. above this temperature it loses its ferroelectric property).
On the other hand, processing and shaping ceramic materials are often performed during the preparation of these materials, not after. Heating at temperatures around $T_C=1600^o\mathrm{C}$ for processing ceramics is not a common procedure and it is expensive.
If you want to prepare cylinders and discs from a powder of barium titanate, you can add it to a polymer matrix (wax, epoxy resin,..) in an appropriate mold using adequate processing techniques like injection molding (heating below Curie Temperature). Of course, the resulting dielectric constant of the composite material is less than the one of pure  barium titanate.

Are there any matrices you would recommend particularly for a high Bt:matrix ratio? What is the purpose of the heating step in the mold?

I recommend epoxy resin, polyester resin or wax. Heating is for cross-linking the thermosets.


text:
Can calcium carbonate scorch in a pan?
Today a friend showed me something we could not explain:
She had a pan which she used for years to heat water. On the bottom it had a thick layer of calcium carbonate. She forgot the pan on the hot plate. So all the water evaporated.
Then the calcium carbonate layer at the bottom got really black. Is this possible if it was pure $\ce{CaCO3}$ or must there have been other organic compounds to explain the black colour?
May I have an explanation what happened chemically as well?

Black is due to organic, no doubt.

So it is impossible that is was the CaCO3?

Why, it's totally possible that it was mostly $\ce{CaCO3}$. But black is due to something organic (probably just a tiny amount of it).

Like the comments have said, black residue is usually from organic material that has scorched.
If the pan somehow got to $840 ^{\circ}$C, then the calcium carbonate would decompose into calcium oxide and carbon dioxide, neither of which would lead to a black residue.
$$\ce{CaCO3 -> CaO + CO2}$$
Organic materials are mostly made of hydrocarbons. When you burn the organic matter, you're left with pretty much all carbon. This occurs when there is not enough oxygen to completely burn all of the carbon - when there is enough oxygen, some of the carbon combines with it to form $\ce{CO2}$ and $\ce{CO}$ and the hydrogen combines with oxygen to form $\ce{H2O}$. When there is a lack of oxygen, some carbon is left over, and this is what we see as the black substance.


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Hemp and Cotton
Hemp fibre has a higher tensile strength and toughness than cotton. Could anyone explain how their different chemical compositions and molecular structure (E.g. beta sheets if any present, bonding,etc.) cause this.
I know cotton is made up of a larger percentage of cellulose, which confuses me as more cellulose is supposed to make fibres strong. I can't find any information using the internet, so it would be extremely helpful if anyone could point me to any that would be relevant.

Based on what I can find out there from the vast world wide library on the Internet, hemp fiber contains higher amounts of hemicellulose and lignin compounds than cotton. Based on that and the polymer properties thereof, to me these are the obvious reasons why. . .

composition wise they differ the most, and below are a few resource links I found as you requested

I'd research these areas more to see if it adds up for you to make better sense since you specifically asked for pointers.  I hope you find this helps you somewhat based on the wording of your inquiry/question per my interpretation.
Sources

http://www.hindawi.com/journals/amse/2013/325085/
http://www.researchgate.net/publication/239773617_Hemp_fiber_and_its_composites_-_A_review
http://www.ipme.ru/e-journals/MPM/no_11111/suardana.pdf
http://www.ncbi.nlm.nih.gov/pubmed/16808870
http://www.hemptraders.com/v/vspfiles/templates/Jade2/images/textile3.gif
http://www.hemptraders.com/v/vspfiles/templates/Jade2/images/textile2.gif

EDIT:
I think the big difference molecular composition wise for hemp compared to cotton is as mentioned above "hemp fiber contains higher amounts of hemicellulose and lignin". So to be more specific in terms of bonding and structure with this regard, you should compare those chemical structures and bonding but perhaps their polymer properties is the most obvious reason.
Also per your comment regarding silk as an example, I wanted to point a few things out as well that may be worth you reading up on for a better understanding of general material classifications.
Please check the link at the bottom too for more detail on cellulose, hemicellulose, and lignin.
GENERAL MATERIAL CLASSIFICATIONS

natural cellulose
regenerated cellulose
regenerated plant protein

http://stl.bee.oregonstate.edu/courses/BFP/Class_Slides_W2011/BFP_Lecture6.pdf

This is perhaps better suited in the comments section. Could you please expand on your answer so that it completely answer's the OP's question or otherwise write this up as a comment.

Hey, sorry to bother you again but could you actually be a bit more specific in terms of the bonding and structure of both cotton and hemp?  I cannot quite get my head around it. e.g. silk is made primarily from fibroin, which has hydrogen bonded beta sheets of polyalanine chains

@chemlove29310 I added an EDIT to my answer, you may want to consider starting a new question for more specifics in case additional questions start falling out of the scope of your original question.  I hope this helps further clarify but for you to get a better understanding, keep breaking everything down where there may be confusion and get granular with your research.  You understanding and comprehending this yourself is better than the quick answer or someone else's answer in the long-run.

Cotton fibers are composed principally of cellulose, while hemp fibers contain, besides cellulose, hemicellulose.
Cellulose is a linear polymer with glucose repeating units. Its chains are long, packedwith inter-chain hydrogen bonds.So, it's a highly crystalline polymer.
On the other hand, hemicellulose is a highly branched polymer. Its presence increases the strength of the hemp fibers by allowing the internal redistribution of stress that occurs in the fibers. In fact, when replacing the flexible hemicellulose-cellulose-hemicellulose bond by a more rigid cellulose-cellulose bond, the stress is concentrated at the individual bonds and lowers the fiber strength.


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Vaporizing NOx (NO and NO2) with hot metal?
So essentially I'm looking at a product that my intuition tells me is kind of similar to a perpetual motion machine (Impossible).
Catalytic Converters use platinum, rhodium, and palladium to convert Hydrocarbons, NOx, and CO.
NOx turns into Nitrogen and oxygen.
Hydrocarbons turn into Water and carbon dioxide
Carbon monoxide into carbon dioxide.
THIS PRODUCT (see link) claims to "vaporize" "toxic fumes". It doesn't seem to explain what they turn into. just that they go away. (in the report it appears that some of it is turning into oxygen as that number goes up?)

Isn't NOx and CO already considered a vapor? how does one further vaporize it?
Simply using hot steel to convert gases seems strange but using hot platinum to convert gasses is also pretty strange when you think about it.
Is this actually a legitimately effective product?

http://www.impulsengine.com/newproducts/how-it-works_ev_muffler.shtml
http://www.impulsengine.com/newproducts/pdf/Impulse_EV16_Muffler_Emission_Results.pdf

Thin wall channels (this would be the "steel" plates perhaps) in the device are coated with a wash coat of aluminum oxide which is porous and increases the surface area. This allows more reactions to take place since the converters also contain precious metals such as platinum, rhodium, and palladium e.g. converting NOX into nitrogen and oxygen, hydrocarbons into water and carbon dioxide, and so on.
These catalytic converters use oxidation and reduction reactions to convert the "toxic" fumes—I'm not certain they are stating to work any differently with this regard than other catalytic converters do though. Oxidation is the loss of electrons and reduction is the gaining of electrons at the molecular level. This in addition to the precious metals help promote the transfer of electrons into the conversion of "toxic" fumes to "non-toxic" fumes.
Furthermore, these can also contain oxygen sensors on them measuring and reporting back to the Engine Control Module the oxygen levels in the exhaust. This will help tell other sensors to adjust air-to-fuel ratio helping to keep the catalytic converter running at optimum efficiency.
I would say if this device has a patent and has held up to the tests thereof, then it's likely legitimate and accurately effective as advertised by the manufacturer.


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H NMR of 4-chloroacetophenone
For the H NMR of the compound I am wondering why peak (A) is the farthest out of all of them? I think that it is due to the electro negativity of oxygen. Am I correct? If so, why would the presence of oxygen push the peak to the left?

The further to the left a peak in an NMR spectrum is, the more deshielded it is. Deshielding or shielding are terms to describe how much a certain nucleus is shielded from the outer magnetic field that is applied during NMR measurements. Shielded nuclei will experience less of a magnetic field and thus resonate at a lower frequency meaning lower ppm-values; and vice-versa. In general, shielding can be explained by more electrons being in the vicinity of an atom because the electrons — in layman’s terms — can be thought to create an electric current induced by the outer magnetic field, which in turn generates an opposing magnetic field and lowers the effective magnetic field at a nucleus.
If we have an electronegative atom close by, that atom will draw electrons towards it, reducing the electron density around the nucleus we are observing and thus deshielding it. That not only works for electronegative atoms alone, but also for electron-withdrawing groups like the carbonyl group.
In your specific example we have two electron-withdrawing groups that will deshield neighbouring protons. However, the deshielding that the acetyl group is responsible for is stronger than that of the chlorine atom. So both pairs of protons are moved downfield (to higher ppm values) when compared to toluene or para-xylol, but the protons closer to the acetyl group are deshielded more.
It ultimately boils down to oxygen’s electronegativity in combination with an $\ce{sp^2}$ carbon that is not attached to any hydrogen, but it may be easier to think of carbonyl groups being electron-withdrawing per se.

Building on Jan's very useful answer, it is often easy to consider these substituted benzenes by understanding the chemistry associated with the substituents; both $\ce{-Cl}$ and $\ce{-C(O)CH3}$ are electron withdrawing groups, and it is this property rather than straight electronegativity that controls the chemical shifts in this case. Halides are weakly electron withdrawing, and acetyl groups are reasonably strongly electron withdrawing. If we consider benzene at $\delta = 7.34$ as our starting point, we can see that both sets of aromatic signals have a deshielding response, but one more so than the other.
EW subsituents on a benzene ring will have the greatest influence on the protons on the ortho position and para position, with the least influence on the meta position. This can be justified by drawing out resonance structures:

I'll leave you to work out the effects of an electron donating group. Often, a good sign for working out whether you have an electron withdrawing or electron donating group attached to your ring is to note the relative shifts for the ortho and meta protons; ortho are upfield of the meta for EDG and downfield for EWG.
Now, you have two subtituents which are para-substituted. Knowing that the meta position is the least affected by substituents (in most cases — there are exceptions), it can be easy to rationalise the relative effects of the deshielding influence by considering the relative strengths of the electron withdrawing groups. An acetyl group has much greater electron withdrawing potential than chlorine, and hence the protons ortho to this group experience greater deshielding, and have a larger chemical shift.


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Relationship between Productivity and Yield
I am trying to formulate an optimization problem. Can you help me understand the definition of productivity and yield? Why is there a trade-off between them during the reaction?
Thanks in advance.

Depending on your application and the specifics, to keep it simple. . .

Yield: The amount of product obtained in a chemical reaction
Productivity: The rate of speed a product can be obtained in a chemical reaction

For the compromise, as vague example, perhaps the more of the product you want to yield, the more time it may take thus a longer productivity you'll have.


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I forgot the basics of writing equations, or maybe it is not me?
Sulfuric acid is commonly used for the pickling of iron to remove surface rust before galvanisation.
$$\ce{Fe2O3 (s) + 3H2SO4 (aq) -> Fe2(SO4)3 (aq) + 3 H2O (l)}$$
However, I am not sure if given:
$$\ce{Fe2O3(s) + 3H2SO4(aq) -> ~~?}$$
how I would work this out?

http://chemistry.stackexchange.com/questions/4547/how-to-get-the-valencies-of-elements

You have to have an idea of the chemistry of $\ce{Fe2O3}$ and $\ce{H2SO4}$ - there is no real way around it.
The former is Fe(III) oxide - basically, a typical metal oxide. Fe(III) is a little bit oxidising, but it's not going to oxidise $\ce{H2SO4}$ (it's incredibly hard to do so).
The latter is a strong acid. It might be a little oxidising if it is concentrated, but Fe(III) isn't going to get oxidised. It can act as a dehydrating agent if it is concentrated, but there's no water to be lost.
Chances are, there's not going to be any redox reaction. It's just going to be a simple acid-base reaction to form water and a salt ($\ce{Fe2(SO4)3}$). Notice how the oxidation state of iron does not change - that lets you work out the ratio of $\ce{Fe^3+}$ to $\ce{SO4^2-}$ in the salt.


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Can SLES get too old?
I've been using SLES to break surface tension in some water solutions. However after a year or so it seems like the ability to break surface tension seems to wear off in the solutions. So my question is if this is supposed to happen? Does Sodium Lauryl Ether Sulphate lose its ability to break surface tension over time?
The SLES im using has also begun to smell bad from the tank, which made me curious about it wearing off or getting too old.


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Naming the spin system present in ethyl butanoate
I am to identify the spin system of ethyl butanoate, but I am confused as to how I should start.

I know that chemically equivalent but magnetically inequivalent nuclei should be denoted as $\ce{AA'A''}$ or $\ce{BB'B''B'''}$, etc., whereas magnetically equivalent nuclei are denoted as $\ce{A3}$, etc.
But there seem to be many chemically nonequivalent protons in this molecule.
Should I just put down this: $\ce{A3B2C2D2E3}$? How can I determine if the $\ce{CH2}$ protons are not magnetically equivalent, i.e. couple differently to neighboring protons?

By Pople Nomenclature convention, coupled nuclei for which the chemical shift difference is much greater than the coupling between them (say, greater than 3-5 times) are labelled with letters alphabetically well separated (AX, AM etc). Labelling nuclei with adjacent letters of the alphabet (AB, MN, XY etc) indicates that the chemical shifts are close compared to their coupling, and this means they are typically strongly coupled ie cannot easily be analysed using first order techniques.
As for so many NMR related questions on this site, I refer you to Hans Reich's valuable online resource for further reading.
So, what to call your system. Clearly, this is an example for teaching purposes only. You have correctly identified that you have 5 non-equivalent sites with relative populations 3:2:2:2:3. If you cannot work out, or do not know what the expected approximate chemical shifts would be for each of these guys I can tell you they are well enough separated not to worry about second order coupling.
So now you need to do is determine whether you have any magnetic inequivalence in your system. Again, have a read of this useful page to understand how to test for this, but I think you are fairly safe with this case that there are no nasty surprises.
All that is left is to assign your spin labels to your system. So for the propyl group, none of the chemical shifts are close (compared to coupling) so your labels should be well separated; so not ABC, but something like AJM. Likewise for the ethyl chain; something like RX. I'll leave you to work out how to put the two together.

The geminal protons in the methylene groups are magnetically inequivalent in $\ce{H3C-\color{red}{CH2}-\color{blue}{CH2}-CO2Et}$, I think? (Merry Christmas!)

@orthocresol Yes, but the chemical shift differences sufficient not to cause complicating appearances, so will look like a t and q, and the CH3 looking like a simple t. Ho Ho Ho right back at ya.

authors: یوسف محمد یوسف


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Hybridization, MOT and Paramagnetism
In what way can hybridization or molecular orbital theory be used to explain paramagnetism?
For instance, when something is hybridized to make enough bonding electrons, do all the electrons end up paired? Or are some left unpaired, explaining paramagnetism?

Paramagnetism is a form of magnetism whereby certain materials are attracted by an externally applied magnetic field. Valence bond theory (VBT) and hybridisation doesn't really do a good job at predicting whether a molecule is paramagnetic or diamagnetic (isn't attracted by an external magnetic field). This is why molecular orbital theory (MOT) is so useful as it is successful at predicting whether a molecule is paramagnetic.
For a molecule to be paramagnetic, it needs to have an overall magnetic moment meaning that it needs an unpaired electron. If all the electrons are paired, then the molecule is diamagnetic. So, by seeing whether a molecule has an unpaired electron, we can predict if it is paramagnetic or not.
Now let's consider the example of $\ce{O2}$. Experimentally, $\ce{O2}$ is known to paramagnetic. According to VBT, $\ce{O2}$ should look like this:
]1
As you can see, all the electrons are paired. Therefore, VBT predicts that $\ce{O2}$ should be diamagnetic.
Now let's examine how the electrons are arranged according to MOT. In MOT, unlike VBT, it involves the creation of bonding and anti-bonding MOs. MOs are basically the superpositions of the wavefunctions of atomic orbitals. In $\ce{O2}$ the 2s AOs of each oxygen atom constructively and destructively overlap with each other while their 2p AOs also constructively and destructively overlap with each other The resulting MOs for $\ce{O2}$ looks like this:

Now that we got the MOs, all we have to do is fill them with electrons using the same method that we use for AOs. By doing that we get:

Note that we have 2 unpaired electrons. Therefore MOT correctly predicts that $\ce{O2}$ should be paramagnetic, unlike VBT which predicts that $\ce{O2}$ is diamagnetic.

for an ionic compound (for instance AgCl), could hybridization or MOT be used to predict paramagnetism? from the readings I've done, all the examples are of covalent compounds

I am not that sure as I have only briefly studied MOT. I don't think you are able to apply MOT or hybridisation to ionic compounds as they exist as large crystal lattices and aren't simple molecules consisting of 2 or 3 atoms. But I think you should wait until someone else who knows more about this topic answers.

@user264985 crystal lattices have CFT and LFT theories, which add a lot to it, but ultimately MOT does explain them. However it is not as simple as atomic orbitals. You have to use band theory of solids to explain. Go have a look http://hyperphysics.phy-astr.gsu.edu/hbase/Solids/band.html


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Why do noble gases like Xe and Kr form compounds, despite offending the octet rule?
According to the octet rule noble gases should not form compounds but xenon and krypton form compounds like $\ce{XeF2}$, $\ce{XeF4}$, $\ce{XeOF2}$ and $\ce{KrF2}$ respectively. Why?

This question and its answers should apply to your question. It's a slightly different context but the idea is the same: the octet rule is rather useless beyond Period 2.

Basically, the octet rule only applies to a very specific subset of possible compounds. In general, there is no octet rule.

Nobody told Xe or Kr that they had to follow the octet rule (which is more of a rough guide than a rule). Chemistry has many 'rules' that only apply in limited circumstances. Don't blindly apply rules.

See my answer to the question ortho linked to learn more about the octet rule. Also, all compounds (e.g. $\ce{XeF2, KrF2}$) with noble gases are conforming with the octet rule if applied correctly.

When we say that noble gases are stable it means that they are more stable than the rest. Yet it is to be considered that even these have ionisation potential ;though high. if we make a noble gases react with such ions that help in their bonding then we can have a compound of noble gas. Search this on Google.

The first published report, in June 1962, of a noble gas compound was by Neil Bartlett, who noticed that the highly oxidising compound platinum hexafluoride ionised O2 to O+
  2. As the ionisation energy of O2 to O+
  2 (1165 kJ mol−1) is nearly equal to the ionisation energy of Xe to Xe+ (1170 kJ mol−1), he tried the reaction of Xe with PtF6. This yielded a crystalline product, xenon hexafluoroplatinate, whose formula was proposed to be Xe+
  [PtF
  6]−
  . It was later shown that the compound is actually more complex, containing both XeFPtF5 and XeFPt2F11. - Wikipedia, noble gas compounds

This was the first real compound of any noble gas.


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How to find out the energy gained from hydrolysis of Dimethylsulfoniopropionate?
I am trying to find out what the energy gain from hydrolysis of DMSP (Dimethylsulfoniopropionate) is, versus the energy gain from oxidation of a sugar or carb (e.g. glucose) in a terrestrial atmosphere.
To be clear, I am talking about the 'cleavage of chemical bonds by the addition of water' to DMSP.
This question is motivated by the issue of whether heterotrophic animals living in a reducing atmosphere of an exoplanet which has no free atmospheric oxygen might use DMSP or some similar compound for energy storage in preference to carbohydrates.
DMSP was mentioned by the paper linked below, which suggests that ocean algae get energy from this process and so could alien metabolisms which don't have access to O2. The author states that the energy available from hydrolysis of DMSP is the same in a reducing or oxidising atmosphere but doesn't state what that energy gain actually is in j/mol.
I realise the actual energy storage molecular species used by an actual organism on such a planet is probably going to be something else, but this seems like a good proxy.
"Photosynthesis in Hydrogen-Dominated Atmospheres", William Bains, Sara Seager and Andras Zsom

Some places your question refers to "hydrolysis" and other places "reducing DMSP with free hydrogen".  You need to decide what reaction you are really interested in.  I answered based upon the "cleavage" reaction, which is occurs on Earth.  If you are really interested in a different reaction the answer would be different.

Yes I see your point and have identified my source of confusion. I will edit the question to clarify.

Would you accept theoretical data from a QC program such as Gaussian?

@gsurfer04 - yes I would as long as you can clarify the results in some kind of laymans terms and provide a link to recreate the results (assuming of course the program is open access).

@rumguff I edited my answer to make it more clear.  If it doesn't answer the question clearly, please let me know.

When the literature refers to "hydrolysis" of DMSP, it is referring to the following cleavage reaction, even though no water is involved in the overall reaction:
DMSP --> DMS + acrylate + H+
See for example at page 156 of Biological and Environmental Chemistry of DMSP and Related Sulfonium Compounds: "hydrolysis of DMSP to DMS and acrylic acid".
For the above reaction:
$\Delta G = -11.8$ kcal/mol
Source: MetaCyc Reaction: 4.4.1.3
This means 11.8 kilocalories (49.4 kilojoules) of energy are released for each mole of DMSP that reacts.


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Difference between starch surfactant and CTAB, SDS, Tx-100
I am doing a research on the effect of surfactant on ZnO nanoparticles properties. There is a confusion here between starch surfactant and CTAB, Tx-100 and SDS. I need point on what is the difference between these surfactants in terms of their costs and effects on environment.


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Can anyone give me detailed logical solution to my questions?
What is 1 amu? From where it is derived? Why it is compared with carbon-12? Why it was once compared with hydrogen-1? and now its not? Can anyone please explain me logically? And what would have happened if other elements were used?
May be my question is wrong. If it is, can anyone point my mistake out?

What do you mean by "detailed logical solution"? And did even see https://en.wikipedia.org/wiki/Atomic_mass_unit ?

I think I read a bit but I am not getting

What is 1 amu? I think it was originated from hydrogen-1? I think, it is like hydrogen-1 has a value of 1 amu? So, why oxygen 16 was later used?

how 1 amu is derived?

Related: Units of mass on the atomic scale and Why was atomic mass scale changed from Oxygen - 16 to Carbon - 12?

Please tell us exactly what you didn't understand. Repeating what Wikipedia said is useful to none.

Thanks but why did not they consider a single isotope of hydrogen?

@Ahmad The atomic mass of H-1 is about 1.00778 amu (as used in chemistry until 1961) or 1.00783 u (since 1961). Thus, changing the reference for the relative scale to H-1 would have changed all numerical values by about 0.78 %, which probably would not have been acceptable.

Originally, atomic mass had been just  relative and without any units. In 1803, John Dalton decided to use hydrogen-1 as a basis for atomic mass, but Wilhelm Ostwald disagreed later on, and said that relative atomic mass should be measured relative to 1/16 of an oxygen-16 atom. This was, however, before elemental isotopes were discovered.
When isotopes of oxygen were discovered in 1923, relative atomic masses had different representations based on the isotopes used. In physics, pure oxygen-16 was given an RAM of 16 amu, whilst chemists gave this value to naturally occurring, isotopically-weighted oxygen.
These different values led to errors in computations, and so in 1961, the reference standard was changed to carbon-12. Carbon-12 was used to stop any more divergence in values for RAM. The new (and current) unit became u, the "unified atomic mass unit".
Therefore, the atomic mass unit is 1/12 of carbon-12, and so the mass of carbon-12 is 12.

Thanks but why they did not consider a single isotope of hydrogen?

It's low mass leads to inaccuracy, and so a heavier element was used so that error decreases.

why would it? or why they did not consider a single isotope of oxygen? What is the point of amu, since we can determine the mass in amu just by getting the nucleon number?

A low mass increases percentage error, making it less accurate. As I said in my answer, they did consider oxygen, but before isotopes were discovered. We use amu as a unit for atomic masses. The problem with just using nucleons number is the existence of isotopes.

So. we can consider one specific isotope instead of a specific element? Like oxygen-16, what is to do with isotopes?


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Rendering and visualization of orbitals of an atom?
I am looking for program that would display and render the orbitals of an atom.
Preferably display all orbitals. Specifically Argon.
Its not duplicate. I am looking for visual appealing rendered orbitals (hydrogen like) rather than exactness of the levels. And I would like to know which program will do it ?

There are many programs for 3D graphics, but that's not chemical stuff.

Yeah, I like how the accepted answer for the question they pointed you to was basically "No, you cannot visualize an atom" - which is BS.  You can visualize it.  Can you visualize the many-electron wave function? No, but you can plot the orbitals, and chemists do so all the time, understanding that they are just approximations to reality.  Assuming you don't have expensive software like Gaussian, you can use this freeware - which gives you 3D versions of these 2D plots.

Thank you. The orbital viwer program is ok it has 3D but i dont see a way to put several orbitals on one atom and display them.  I can generate hydrogen like orbitals in matlab but the rendering is not so good.


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PV=nRT approximation on other planets?
In Chemistry class yesterday, I learned that real gases with low atomic masses behave like an ideal gas at high temperatures and low pressure. Since on earth at sea level the pressure is close to $1\ \mathrm{atm}$ (which is a low pressure) and temperature is close to $288.15\ \mathrm K$ (high temperature), I can use the equation $PV=nRT$ to approximate properties of the gas. (I got these values from Wikipedia, under Standard Sea Level).
$$\left(\text{pressure}\right)\cdot \left(\text{volume}\right)=\left(\text{number of moles}\right)\left(R\right)\left(\text{temperature}\right)$$
In physics we learned that pressure is the force per unit area. On earth the atmospheric pressure is earths gravitational pull on the air. The gravitational pull of any object on earth is $G_\text{pull}=\left(\text{mass}\right)\cdot \left(9.8\ \frac{\mathrm m}{\mathrm s^2}\right)$. Because the gravitational pull is $G_\text {pull}=\left(\text{mass}\right)\left(\text{acceleration due to gravity}\right)$.
Since different planets have different masses the acceleration due to gravity on these planets are different. Below I listed the values:
$G_\text{Mercury}=3.59\:\frac{\mathrm m}{\mathrm s^2}$
$G_\text{Venus}=8.87\:\frac{\mathrm m}{\mathrm s^2}$
$G_\text{Mars}=3.37\:\frac{\mathrm m}{\mathrm s^2}$
For planets like Mercury, Venus, Mars the approximation of $PV=nRT$ still can be used since they would have even lower pressures than Earth. But what about planets like Jupiter and Neptune
$G_\text{Jupiter}=25.95\:\frac{\mathrm m}{\mathrm s^2}$
$G_\text{Neptune}=14.07\:\frac{\mathrm m}{\mathrm s^2}$
Jupiter and Neptune have more pressure at ground level than that of Earth. So can $PV=nRT$ still be used? If yes, how well? If not, is there a different formula that can be used to find the volume and number of moles on these other planets?
Any input is appreciated. I am just being curious and didn't find anything on Google.

You can use it, it just becomes less accurate. There are many other equations of state that try to account for deviation from ideality. https://en.wikipedia.org/wiki/Equation_of_state By the way, there is no difference whether you are on Mars or Jupiter or Venus. A gas at high pressure on Earth is the same as a gas at high pressure in Jupiter.

@orthocresol Thanks, I didn't know about the other equations of state. Is their a correction factor to make it more accurate that can be added to the equation? Or is $PV=nRT$ still the best approximation formula I can use to calculate properties of gasses on other planets?

@orthocresol Oh and Boyles and Charles Law seem to be a simpler version of $PV=nRT$ If $nRT=k$, you get $PV=k$ which is Boyles Law. And if $nR/P$ is a constant you get $nR/P=k$, $V/T=k$. Which is Charles Law

If you're interested about the parameterisation (i.e. correction factors), you might want to read up particularly about the van der Waals equation of state http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/waal.html And you are right about Boyle's and Charles's laws. They are both encompassed by the ideal gas law.

Venus might actually have a little higher surface pressure than Earth.

The differences in acceleration due to gravity is not the main factor in comparing how accurate the approximation is for each planet. The main factor is the mass of gas each planet's atmosphere contains.
Mercury has almost no atmosphere. According to the NSSDCA — Mercury Fact Sheet  (accessed 2023-05-22) the total mass of all gas in Mercury’s atmosphere is less than $\pu{E4 kg}$ and the pressure is less than $\pu{E-14 bar}.$ The approximation will be very accurate.
According to the NSSDCA — Venus Fact Sheet (accessed 2023-05-22) the mass of Venus’s atmosphere is $\pu{4.8E20 kg}$ and the pressure is $\pu{92 bar},$ so $pV = nRT$ will be much worse of an approximation than on Earth.
Mars has a small atmosphere, less than Earth and more than Mercury. Jupiter, Saturn, Uranus and Neptune are gas giants with no distinct surface.

This only addresses the question for "open air" scenarios, right?  If we are looking at closed systems then the answer is that the planet's atmosphere has no relevance, and gravity is only relevant to the degree that it can create a measurable pressure gradient within the system in question.

@feetwet Right, it only applies to the conditions at the surfaces of the planets, "outdoors".

This answer is so wrong. PV=nRT on earth, but gas density at high altitudes isn't the same as low altitudes because of gravity. The simple equation ignores differences in altitude. If you are using a sphere with a volume of one cubic meter sphere that is fine. But if you container is one $1 mm^2$ then it would be a million meters high and gravity would have an effect if the tube was vertical. If the tube was horizontal then gravity could be ignored.

Another interesting factoid along this line. Helium boils off the earth because at high altitude collision frequency is reduced.  So a helium atom gets knocked with enough energy to reach escape velocity and if it doesn't hit a nitrogen or oxygen molecule which are trapped by the earth's gravity, then it escapes into outer space.

Obviously when there is a great range of altitudes there is also a change in temperature due to height. Imagine the atmosphere with zillions of layers. How much sunlight reaches each layer varies. Since the density of the gas varies due to gravity. Thus temperature of layer depends on how much sunlight gets to that layer and how much gas is in the layer to absorb that sunlight.

Just adding some reasoning:
The equation is better applied when the intermolecular forces can be neglected, this is when (i) the concentration of gas is small, (ii) pressure low and (iii) chemical potential who considers gas proclivity to react with same/other species (inert $\ce{N2}$ vs. oxidant $\ce{O2}$). For intermolecular interactions there are virial corrections and so on (Compressibility Factor, Van der Waals forces).
Planets with no atmosphere and relatively small in size (i.e. rocky planets like Mercury, Mars) don't have high concentration of gases over their surface, besides pressure and attraction exerted by gravity are low, which make the gases behave as ideal. Bigger gas planets (Jupiter, Neptune) are often more turbulent and the gas behavior depends more on the distance of these to center of the planet, and corrections should be talking into account.

Apropos:

Comparing this equation with the similar equation of state for the Earth's atmosphere, you will notice that the gas constants are different, 1149 for Mars and 1718 for Earth. These numbers are different because the Martian atmosphere is almost entirely composed of carbon dioxide, while the Earth's atmosphere is a mixture of 78% nitrogen and 21% oxygen.


text:
Is it incorrect to use Fischer Projections for carbons that don't have stereocenters?
For example:

Is a cross specifically used for stereocenter representation?

According to IUPAC it is not incorrect for using it to depict carbohydrates being stereogenic or not. However here is an excerpt from the link listed below:

The use of
  Fischer projections in non-carbohydrates is problematic. Any drawings
  of this type are necessarily ambiguous since it is unclear whether the
  author of the drawing intended to make a Fischer projection or simply
  happened to draw a diagram in that orientation. Molecules depicted in
  the style of Fischer projections are likely to be recognized as such
  when they contain at least two acyclic stereo- genic centers oriented
  vertically, although such interpretation is not guaranteed. Molecules
  containing a single stereogenic center pose even greater problems
  since it is often convenient to separate the four ligands by equal
  angles whether or not a single stereoisomer is intended. Accordingly,
  although Fischer projections of carbohydrates are acceptable, similar
  depictions of non-carbohydrates should be used only when the viewer of
  the diagram is certain to understand that the diagram was intended to
  represent a Fischer projection. Fischer projections should be avoided
  where ambiguity is possible, including when used in computer
  applications.

See more here (page 37-38):
http://www.iupac.org/publications/pac/2006/pdf/7810x1897.pdf