text:
The shapes of molecules
According to this question/answer, isolated atoms are spherical – and don't have, for example, p-orbital lobes sticking out. The reason is that each complete set of p-, d- and f-orbitals taken together is spherically symmetrical – and the overall wavefunction will be a superposition of all possible states. That superposition will be spherically symmetric, so the atom will be, too. (Sorry for paraphrasing, simplifying etc!).
So my question is: apart from the orbitals involved in making e.g. sigma and pi bonds, is 'the rest' of the atom spherically symmetrical? I have done this rough picture.
You will notice that down the left, the p-orbitals are shown (even though in an isolated atom they would not exist like that – I've kept them so they can show potential 'bonding sites'). But on the right, in the 'finished' molecules, all but the orbitals involved in bonding disappear, into a spherically symmetric shape as before.
I realise there are several to many simplifications here, but this is for a general audience, and I just want to give a sense of molecular orbitals without getting anything horribly wrong. (In every artwork I find online, people only seem to focus on the orbitals involved, as if they are separate from the rest of their respective atoms.)
I'd really appreciate any help and thoughts. Thank you.

Electrons interact, so of course if one orbital is not symmetric (e.g. if it's bonding to another atom), this will to some extent also break the symmetry of all other orbitals.  And I'm not sure if I follow the logic of this "isolated atoms are spherical", firstly because I don't see how you can prove this statement.

Good point. Darn. Do you look at my illustration and balk? Or do you think it's at least useful and acceptable, for a general audience, as I say? Thanks for your answer.

Who exactly is your audience?

Well ... it's a general audience, from teenage up, with little prior experience but hopefully an interest – and I will have built up to this, of course, via a general introduction to quantum mechanics and atomic orbitals etc. Thanks again.

Given an external electromagnetic field, the orbitals do split and even an otherwise isolated atom is no longer spherically symmetrical. For a mostly magnetic field, this is called the Zeeman effect, and for an essentially electrostatic field it is the Stark effect.
Of course, this begs the definition of an "isolated" atom. As @Karl mentions, how do you measure the shape without interacting with the atom? Any incident photon, for example, would be "letting the cat out of the box."

Thanks. Yes, in reality, I suppose there is no such thing as a truly isolated atom! I'm just keen to show the molecular orbitals in context, i.e. with the rest of the atom also present, for a general audience. Do you think the illustration is at least useful?

The illustration is quite useful! If used in a presentation, I would split it into multiple frames, because it presents so many important concepts. Best wishes with what should be an edifying presentation.


text:
How to know the aluminium sulfate solubility as function of the pH?
I'm trying to find a diagram or a reference for to know the solubility of aluminium sulfate $\mathrm{Al_{2}(SO_{4})_{3}}$ in differents pH values. Something like:

If the question only is about the diagram and the reference:

Rubin, A. J., & Hayden, P. L. (1973). Studies on the Hydrolysis and Precipitation of Aluminum (III). page 57.

But the interpretation is not clear for me, so here is a new question

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text:
Degree of hydrolysis
A book which I am reading has this topic on hydrolysis of salts.
Suppose $\ce{NH4Cl}$ is dissolved in water. It is a salt of a strong acid and a weak base. $\ce{NH4+}$ ions being the conjugate acid of a weak base is strong and reacts with $\ce{H2O}$ to give $\ce{H+}$ thus making the solution acidic.
$$\ce{NH4+ + H2O <=> NH4OH + H+}$$
Now for calculating degree of hydrolysis $h$ it takes the initial concentration to be $C$. This means $C(1-h)$ of $\ce{NH4+}$ remains and $Ch$ each of $\ce{NH4OH}$ and  $\ce{H^+}$ is formed. Therefore, $K_h = \frac{Ch^2}{1-h}$. In the next step $1-h$ has been approximated as $1$. While calculating degree of dissociation of weak electrolyte we do a similar thing but that is because degree of dissociation is too small compared to 1 but here, I cannot understand how can degree of hydrolysis be very small compared to 1 when $\ce{NH4+}$ is a strong acid.

I am aware that the question looks like it’s asking something very different, but I believe that the core of your problem is the one answered in the question I am about to link as a duplicate. Or, to put it in one sentence: what in the world makes you think that ammonium be a strong acid?

I cannot understand how can degree of hydrolysis be very small
  compared to 1 when $\ce{NH4+}$ is a strong acid.

That is wrong.
You also wrote:

It is a salt of a strong acid and a weak base.

Well, the salt of a strong acid and a weak base, i.e. the conjugated acid of that weak base is a weak acid.
In the case of the ammonium ion, $pK_a=9.25$, very weak indeed.
Also, the hydolysis reaction equilibrium is (because free $\ce{H+}$ don't exist in water and $\ce{NH4OH}$ is old notation, long obsolete):
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$h$ is negligibly small because this equilibrium leans very strongly to the left, so $[\ce{H3O+}]\approx 0$.

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text:
Has IUPAC been inaccurate in their 1994 definition of kinetic stationary state?
Summary of IUPAC definition
$\def\d{\mathrm{d}}$In IUPAC recommendations from 1994[1, 1166–1167], the authors discuss the process
$$\ce{A <-->[$k_1$][$k_{-1}$] X\\ X + C ->[$k_2$] D}$$
A kinetic stationary state or steady state is defined very reasonably:
$$\frac{\d\ce{[X]}}{\d t}=0.\tag2$$
They go on to emphasise that condition $(2)$ is not equivalent to $\ce{[X]}$ being constant, 'not even approximately'; simply that the change in $\ce{[X]}$ is small compared to changes in concentrations of $\ce{[A]}$ and $\ce{[D]}$ because $\ce{[X]}$ itself is small. If $\ce{[X]}$ were constant in a stationary state, they claim it would lead to the following contradiction.

Say reactant $\ce{C}$ is in excess. Assume that $\ce{[X]}$ is constant. Then the law of mass action $v = k_2\ce{[X][C]}$ would reduce to some $v = k'\ce{[C]} \overset{\mathrm{excess}}{\approx} k\in\Bbb{R}$. In other words, production of $\ce{D}$ could continue at a constant rate even after $\ce{A}$
has run out.

The document concludes that $\ce{[X]}$ cannot possibly be constant in steady-state conditions.
How I would resolve the apparent contradiction
I do not see a way around $\ce{[X]}$ being constant. Indeed, for any reasonable function[a] $\ce{[X]}:t'\to \Bbb{R^{0+}}$ equation $(2)$ implies
$$\int_0^t \frac{\d\ce{[X]}(t')}{\d (t')}\d(t')= \ce{[X]}(t) = r \in \Bbb{R^{0+}}\tag3$$
where $t'$ was simply a dummy variable.

Resolution to contradiction: If $\ce{C}$ is in a large excess, $\ce{[X]}$ is small, and $\ce{A}$ has run out, it is simply not possible for steady-state conditions to hold. In other words, it is our stationary state approximation $(2)$ which fails to hold for $\ce{X}$. So we can easily keep condition (3) for a steady state.
More accurately, if $\ce{A}$ has run out, the only possible stationary state would have been $\ce{[X]} = 0$. This is because, at a stationary state,
$$\ce{[X]} = \frac{k_1\ce{[A]}}{k_{-1} + k_2\ce{[C]}} \overset{\ce{[A]} = 0}{=} 0.$$
If $\ce{[X]} = 0$, it is unfeasible for the reaction to continue at non-zero constant velocity since $v = k'\ce{[C]} = 0$ (where $k' = k_2\ce{[X]}$).

Question

Is / is not the concentration of $\ce{X}$ constant, approximately or otherwise, in a kinetic stationary state (steady state)?

IUPAC states $\ce{[X](steady\ state)}$ is not even approximately constant because it would lead to contradictions.
I claim that $\ce{[X](steady\ state)}$ is constant, and contradictions surface when a steady state itself is impossible given certain constraints (or the constraints themselves are contradictory).

[a] Among other things that it is diffentiable everywhere in its domain (i.e., reaction rate is defined for every time $t'$).
[1] Muller, P. Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). Pure and Applied Chemistry 2009, 66 (5), 1077–1184. DOI: 10.1351/pac199466051077. pages 1166–1167

The trick to understanding this mathematically is through singular perturbation.  At short times the PSSH is invalid; at longer times it is.

Also exactly what is meant by "approximately" is very important. $100 is approximately $0 to Bill Gates, but not to me.

@CurtF. Are you busy at this time? Would you consider expanding your comments into an answer (with mathematical argumentation)? Or perhaps join me in chat?

I deleted my answer for now. I shot from the hip and was talking at you not understanding what you wrote. After really reading  the IUPAC definition I'm having trouble understanding how their #1 definition applies to a steady state. For #1 I think that the use of steady state for [X] is flawed. The #2 definition is ok. // I'm chewing on this for the moment. I'll repost with a new answer and a very different analysis.

@MaxW Thank you for the update. It could also have been that my question was badly worded or a bit misleading. I look forward to your answer. Take your time, there is no rush.

IUPAC should say they are defining the "steady state approximation" rather than the steady state.
There are two approximations in this technique.
First, that final reaction step is not an equilibrium (reverse reaction rate is zero).
Second, that the intermediate is so unstable that its concentration (and therefore the rate of change of the concentration) is approximately zero.
Compare the IUPAC definition to the following 1954 U.S. National Research Council explanation:

STEADY STATE APPROXIMATION
The most generally applicable of these simple methods is the so-called "steady-state approximation." A steady state may be defined as a condition in which the rates of change of the concentrations of the several intermediates are very small compared to the rates of change of the concentrations of the reactants and products. This condition is realizable whenever the ratio of the concentrations of the intermediates to the concentrations of the reactants is very much less than unity. When this condition is not attained, the method is not applicable; however, it should not then be necessary since the (larger) concentrations of the intermediates could be measured by experimental means. ... The steady-state approximation consists in setting the rates of change of each of the intermediates equal to zero and in solving simultaneously the resulting algebraic equations.

Overall, if the situation was truly, not just approximately, steady state, yes the concentration of the intermediate would be constant, but the point is that the "stead state approximation" is useful even when the concentration of the intermediate, though always near zero in absolute terms, decreases to say half of its original concentration over the course of an experiment (as the starting material, A, decreases to half its original concentration).
Additionally, a very small $\ce{[X]}$ does not purely-mathematically imply a small rate of change of $\ce{[X]}$, because $\ce{[X]}$ could rapidly oscillate within a small, near-zero, range, resulting in brief instances of large rate of change.  Instead, $\ce{[X]}$ being small over a given period of time really just places a limit on how long of a time the rate of change can exceed a certain value.  For example, if $\ce{[X]} < 0.001\ \mathrm{M}$,
$|\mathrm{d}[\ce{X}]/\mathrm{d}t|$ must not exceed $1\ \mathrm{M/s}$ for more than $0.001\ \mathrm{s}$.

I agree that IUPAC should say they are defining the "steady state approximation" rather than the steady state. // I disagree that Second, that the intermediate is so unstable that its concentration (an therefore the rate of change of the concentration) is approximately zero. It really should be that the rate of change of the intermediate is insignificant to the rates of change of the concentrations of the reactants and products. It is a significant figures argument, not that the intermediate isn't changing concentration.

@MaxW I agree with your phrasing "the rate of change of the intermediate is insignificant to the rates of change of the concentrations of the reactants and products", but I also think my phrasing "is approximately zero" is equivalent to your phrasing "is insignificant".

I up voted your answer, but I'll make a different point. There are two definitions in the IUPAC text. You're talking about the first definition which is what the OP was questioning. The second part of the IUPAC definition for a flow reaction would have a true steady state.

@LinearChristmas yes, I agree, [X] could rapidly oscillate, even though it is very small, resulting in a large rate some of the time.  There needs to be an additional, so far unstated,  assumption to rule that out.

Sorry, I managed to delete my comment before you replied. My own example was a poor (since the constant will not change sign), but yes the main point stands. If you add this point to your answer, I will accept it.

@LinearChristmas ok, I added something more about that

Ok, this isn't an answer, but below is the text in question from Muller's Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). The definition contains two parts.
I agree with DavePHD that the first definition, which is for a batch reaction, would be better called what is known as the STEADY STATE APPROXIMATION
The second definition is for a flow reactor which is at a true STEADY STATE. In this case all the reactants, intermediates, and products would have constant concentrations. That is to say the relative fluctuations (%) in those concentrations would be on the order of the precision with which the flows could be controlled.
Text is below

steady state (or stationary state)
(1) In a  kinetic analysis of a complex reaction involving unstable intermediates  in low concentration, the rate of change of each such intermediate is set equal to zero, so that the rate equation can be expressed as a function of the concentrations of chemical species present in macroscopic amounts. For example, assume that $\ce{X}$ is an unstable intermediate in the reaction sequence:
$$\ce{A <-->[$k_1$][k_{-1}] X}$$
$$\ce{X + C ->[$k_2$] D}$$
Conservation of  mass requires that:
$$\ce{[A] + [X] + [D] = [A]_0}$$
which, since $\ce{[A]_0}$ is constant, implies:
$$\dfrac{-d\ce{[X]}}{dt} = \dfrac{d\ce{[A]}}{dt} + \dfrac{d\ce{[D]}}{dt}$$
Since $\ce{[X]}$ is negligibly small, the rate of formation of $\ce{D}$ is essentially equal to the rate of disappearance of $\ce{A}$, and the rate
  of change of $\ce{[X]}$ can be set equal to  zero. Applying the steady
  state approximation  ($d\ce{[X]}/dt = 0$) allows the elimination of $\ce{[X]}$ from the kinetic equations, whereupon the rate of reaction is expressed
$$ \dfrac{d\ce{[D]}}{dt} = -\dfrac{d\ce{[A]}}{dt} = \dfrac{k_1k_2\ce{[A][C]}}{k_{-1} + k_2\ce{[C]}}$$
Note: The steady-state approximation does not imply that $\ce{[X]}$ is even approximately constant, only that its absolute rate of change is very much smaller than that of $\ce{[A]}$ and $\ce{[D]}$. Since according to the reaction scheme $\ce{d[D]/dt = k_2[X][C]}$, the assumption that $\ce{[X]}$ is constant would lead, for the case in which $\ce{C}$ is in large excess, to the absurd conclusion that formation of the product $\ce{D}$ will continue at a constant rate even after the reactant $\ce{A}$ has been consumed.
(2) In a stirred flow reactor a steady state implies a regime so that all concentrations are independent of time.

I think definitions in

A GLOSSARY OF TERMS USED IN CHEMICAL KINETICS, INCLUDING REACTION DYNAMICS
  (IUPAC Recommendations 1996)
  edited by KEITH J. LAIDLER,
  Pure & Appl. Chem., Vol. 68, No. 1, pp. 149-192, 1996.

are much better.

Pre-Equilibrium or Prior Equilibrium
The mechanism of a  reaction may involve two  or  more  consecutive  reactions: If any step  except  the first is rate-controlling,those  steps  that  precede it  are  essentially at
$\ce{A <=> B <=> C <=> ... <=> X -> Y ->Z}$
equilibrium, and there is said  to  be  a pre-equilibrium,  or a  prior  equilibrium; for example
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\text{rate-controlling step}$
  $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\blacktriangledown$
  $\ce{A <=> B <=> C <=> ... <=> X -> Y ->Z}$
$\text{|}\blacktriangleleft\text{---- pre-equilibrium ------} \blacktriangleright\text{|}$
Steady State (or Stationary State)
If during  the  course of a  chemical reaction  the  concentration of
  an intermediate remains constant,  the intermediate is said  to be in
  a  steady state.
In a  static system a reaction intermediate   reaches  a  steady
  state if  the  processes  leading to its formation and those  removing  it are approximately in balance.  The steady-state  hypothesis leads to  a great  simplification  in reaching an  expression  for  the  overall  rate of a  composite reaction  in terms of the rate constants for the individual elementary  steps.  Care  must  be  taken to apply the  steady-state  hypothesis  only to
  appropriate  reaction  intermediates. An intermediate such as an atom or a  free  radical,  present  at low concentrations, can usually be  taken  to obey the  hypothesis during the main  course of the  reaction.
In a  flow  system a  steady  state may be established even for intermediates   present   at relatively high  concentrations.

If possible, please add quotes around what is from another source, mark what is edited (e.g., the word 'approximation'), and what is your own comment on the matter. (+1)

@LinearChristmas - OK I tried quoting with > character which indents and shades. That what you had in mind?

Yes, that's exactly what I meant. If you still bear with me, wouldn't it be more correct to also have full-headed arrows in the first steady state definition? I believe the harpoons explicitly signify an equilibrium which does not have to be met. (If you disagree, feel free to argue.) Though I admit, mhchem currently doesn't render the proper symbol too well (Win8.1/Firefox Quantum).

@LinearChristmas - I think this is the reaction to which you are referring. $\ce{A <=>[$k_1$][k_{-1}] X}$ // You want this instead $\ce{A <-->[$k_1$][k_{-1}] X}$?

@LinearChristmas -- Actually if the harpoons explicitly signify an equilibrium which does not have to be met that might be closer to the truth for the steady state approximation. Hmm... // Just checked and IUPAC text uses full arrows both ways.

Yes, the latter arrows signify that there need not be (but could be) an equilibrium (at the time of a steady state). If there was an equilibrium, this would be the fast equilibrium approximation. (imo)

@LinearChristmas - Thanks for pointing out problem. I absolutely spaced on that.  It has been 40 years since I took kinetics in grad school.


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Do light and heavy water form an azeotrope?
(Interested in the general principle too, not just for water and its variants, but with particular emphasis on it.)
When you distill a weak ethanol-in-water solution, you get a condensate that's ethanol-enriched, but that only takes you asymptotically up to 95%, because ethanol and water form an azeotrope with that composition. Is there a similar "enrichment limit" one can achieve in enriching water with deuterium, using only distillation at a fixed pressure? (I know one can "break the azeotrope, but I don't want to get into this.) Or could one, given enough stages in a cascade, approach 100% enrichment arbitrarily closely?

What makes you think there might be one?

This is an interesting question, but you'd have to use deuterated ethanol as well. Mixing CH3CH2OH with D2O would yield a mixture of CH3CH2OH, CH3CH2OD, D2O, HDO, and H2O.

Azeotrope forming or not : it is hard to achieve as the boiling points differ by about one degree Celsius.

@MaxW I'm not really interested in ethanol; I just used it as an example of an azeotrope-forming co-solvent with water. But yeah, the general form of my question stands: would ethanol and deuterated ethanol (any/all of them) form an azeotrope with each other??

@Gert only ignorance of any knowledge to the contrary. I'm not aware of any reliable way to know, a priori, whether an arbitrary pair of liquids forms an azeotrope. Are you aware of such an algorithm?

@BerndJendrissek: there exists no algorithm that predicts the existence of an azeotrope.

sigh To get azeotrope you need to have significant difference between compounds and this isn't the case.

@Gert That's what I thought, so I'm slightly puzzled why you seem to think it's odd to wonder if there is an azeotrope between water and its deuterated variants?

This is a good question, but no, there is no azeotrope in any isotopic mixture, as an azeotrope requires non-ideal mixing of two components to form a  minimum or maximum critical point on the plot of total vapor pressure as a function of composition as shown in Figure 1 below. Isotopic Mixtures have a near-zero enthalpy of mixing, making isotopic mixtures the closest thing to an ideal liquid mixture. Because of this ideality of mixing the boiling of protium-deuterium oxide mixtures looks much like the phase diagram in Figure 2 which follows Raoult's law, though this is actually quite exaggerated compared to semi-heavy water mixtures.  Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of $\ce{D2O}$ or $\ce{H2O}$ and be unobservable due to the "doped" species existing as semi-heavy water ($\ce{DHO}$).  This is further evidenced by the fact that other physical properties of isotopic mixtures follow a linear curve when measured as a function of concentration.

Figure 1. Generic Plot of vapor pressure as a function of composition

Figure 2. Generic Boiling phase diagram for ideal solutions
To answer your question about distillation only enrichment, it is possible though impractical.  Commercial enrichment of $\ce{D2O}$ first uses hydrogen sulfide in the Gridler-Sulfide process, once enriched to around 30%, distillation is used to finish the process to the desired isotopic purity.

+1 You might add that a mixture has to deviate quite a bit from ideality before the vapour pressure curve can get a minimum or maximum. No chance for isotopomeres.

+1 Can you expand a bit on "Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of D2O or H2O" (and assume that the pairs are H2O/HDO and HDO/D2O)? While the non-ideality of mixing is certainly very small (and mind you, the H/D chemical difference is much bigger than U-235/U-238), the difference in boiling points is also very small, so you wouldn't need much non-ideality for an azeotrope to form.

@BerndJendrissek I put that in there to preempt that one critic that always asks "what about at 99.99995876%".  In order for there to be an enthalpy of mixing there has to be some sort of interaction of the mixed species. For $\ce{H2O}$/$\ce{D2O}$ this is the formation of $\ce{HDO}$.  If you have only $\ce{HDO}$ and one other species of water then there is no reaction to be had from mixing, thus no enthalpy, thus no azeotrope.

@A.K. I guess I'm that one critic :p I disagree with any "obviousness" of H2O/HDO mixing producing no reaction - while any chemical difference between H and D is necessarily small, it's larger than between any other isotopes. I'm happy to accept an empirical claim like "we know that from 0.000014% to 99.99978% there's no azeotrope", but I'm unfortunately not convinced by any of these "molecular arguments" that none can exist.

Distillation alone has been used on an industrial scale to produce heavy water, but it requires a multi-stage process and uses a lot of energy due to the small difference in boiling points. That would seem to confirm that water and deuterium oxide do not form an azeotrope.

Do you have a reference to back this up? I thought that generally in systems of isotopically substituted molecules an infinite set of azeotropic points is possible. And "distillation" refers to the process of low-temperature rectification of liquid hydrogen, followed by combustion of isolated $\ce{D2}$ with $\ce{O2}$.

@andselisk Where did you get the idea of azeotropic points and isotopes? And by far the cheapest processes to enrich deuterium involve water not hydrogen (one process uses water and hydrogen sulphide followed by distillation of the enriched deuterium water, the other just uses distillation of water).

@matt_black From the back of my head. The difference in b.p. between D2O and H2O is only 1.44 °C, and max. separation factor of about 1.05 (whereas it can be >10 for electrolysis). Separation via distillation only is extremely inefficient; it can lead to enriched heavy water content in combination with the methods you've mentioned, but distillation only is a dead end.

It does seem to indicate that there is no light/heavy water azeotrope, but can we exclude the azeotrope being at, say, 99.9993% heavy water?

Aceotropes form between substances of different polarity. If heptane and hexane don't, I can't see it happen for H2O, HDO and D2O. The vapour pressure of the ideal mixture has a constant slope with composition, you have to deviate quite a bit from ideality to get a curve with a maximum/minimum.


text:
What can we deduce about a (complex, organic) molecule's structure from the order of elements in its formula?
According to IUPAC, a hemiacetal is a compound having the general formula  $\ce{R2C(OH)OR'}$ $\ce{( R' ≠ H )}$.
What does the order in which the elements of this formula appear tell us about the connectivity of this molecule? Does it tell us anything besides that a compound with this formula is composed of a carbon, a hydroxyl group, a oxygen, a $\ce{R2}$ group and a $\ce{R'}$ group? Or does it contain any information about how these elements and groups are connected with each other, about the compound's structure?
Surely the order in which these elements appear isn't arbitrary.

The formula tells you that the two R groups are bonded to carbon as are OH and OR'. The (OH) is written that way so you don't see R2COHOR and possibly misinterpret the formula.

@user55119 Is there any reason why it's $\ce{(OH)OR'}$ and not $\ce{(OR')OH}$?

I don't see why not. Perhaps there is a rule. Clarity is important.

Related: What is the meaning of parentheses in a condensed molecular formula?

These molecular formulas have a very limited amount of information which they can give us. As far as I know, there is no order of bonding nor explanation in how for example this hemiacetal is really structured. They are merely created for you to maybe identify such a group if needed one day (lets say in a bigger molecule). And why its probably (OH)OR`and not the other way around is because of the alphabetical order (H before R). My advice: Don't rely on them, they can be used in a number of ways and have, for my knowledge, no fixed usage. In the golden book of IUPAC it even says "Compounds having the general formula..." meaning that yes, this formula generally applies on hemiacetals, but not always.


text:
Aluminium sulphate solubility as function of the pH
earlier I asked about how to know the aluminium sulfate solubility as function of the ph (see the previous question here). I found this diagram on the page 57 of

Rubin, A. J., & Hayden, P. L. (1973). Studies on the Hydrolysis and
  Precipitation of Aluminum (III).

But I'm a little confused about the interpretation. Is the
$\mathrm{Al_{2}(SO_{4})_{3}}$ insoluble between pH 5 and 7?. What will be the correct interpretation of this diagram?

The lack of continuity for the curve indicates that the concentration of $\ce{Al2(SO4)3}$ was below the detection limit of the analytical technique being used to measure the concentration of $\ce{Al2(SO4)3}$ in the solution.
Theoretically there should be a continuous curve.
+1 for asking this as a new question.

Solutions of aluminum sulfate are highly acidic. The Merck Index states that at a concentration $0.2 M$, the pH is $3.0$.

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text:
Material with refraction index similar to Si
I am looking for a material with a refractive index over the visible spectrum , which is close to the refractive index of Si, but which does not oxydise in air.
Do you know of any such material ?
Thank you.
Edit:
Some more information. A optical reflector should be build from Si. I will not go into the detail, since it is irrelevant to the question, but I will provide all necessary information. The list is ordered according to importance:

The complex refractive index should be close to Si over the entire visible spectrum
The material should not oxydize in air
The working conditions should be between -30°C to 60°C.
Not necessary, but it would be great if epitaxy on single-crystal silicon is possible.

$\ce{Si}$ is not transparent and thus can't have refractive index. Did you mean something about reflection? And $\ce{Si}$ doesn't oxidise in air either.

@Gert Every material has a (complex) refractive index, transparent or not, and Si is very reactive against air and immediately forms an oxide layer on the surface.

Si might passivate?

Silicon is one of very few elements (3) with a self-limiting oxide layer.

Over the range of the visible spectrum, ~1.8 - 3.1 eV, or 700-400 nm, the refractive index (nD) of Si varies from about 3.5 - 5, though it is also absorbent and somewhat reflective over that range. Gallium(III) phosphide, $\ce{GaP}$ is one material that comes close to that. It should be fairly stable in air at STP.
Of course, it all depends on your intended use. At what wavelength should nD match that of Si? At what temperature and in what atmosphere will the material be? Does the crystal structure of the material need to match that of Si, so that a regular layer can be deposited? More information is needed for the answer to be of any use.

thank you very much  for the answer ! This is very informative ! I will add more information to the description of the question.

Do you know if epitaxy on single-crystal silicon is possible ?

According to http://ieeexplore.ieee.org/document/5617118/ it is feasible despite 0.37% lattice mismatch between Si and GaP. Not much of a gap...


text:
Why are elimination reactions in organic chemistry known as beta eliminations?
Why are organic elimination reactions known as beta eliminations? Why don't we call them alpha eliminations?

Take 2-bromobutane as an example. The C2 hydrogen is an alpha carbon while the hydrogens are referred to as beta. C4-H's are gamma. It is a generic terminology that fits all cases regardless of the numbering in the compound. There are alpha and beta eliminations.

Addition:  ...while the C1 and C3 hydrogens are referred to as beta.

In an elimination reaction we typically have a leaving group (denoted by "X" in the figure below),  The carbon that the leaving group is attached to is referred to as the "alpha" carbon.  If a hydrogen is removed from the "beta" carbon, then we have formed an olefin and the reaction is called a beta elimination.

[reference]
Sometimes an alpha hydrogen can be removed.  In this case both the hydrogen and leaving group were attached to the same carbon and a carbene is formed.

It would be helpful if you would add to your answer, As to why base  (K OH +alc) attacks beta hydrogen and not alpha hydrogen


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Do 1s & 2s orbitals overlap of different atoms while forming sigma bond? I
Do 1s & 2s orbitals overlap of different atoms while forming sigma bond?
I think they should overlap as there will not be enough energy difference between them.

Not a bad question but I can't think of any compounds that contain a $\sigma$ bond between $1s$ and $2s$. $\ce{LiH}$ e.g. is ionic, not covalent.

@Gert that's awfully narrow minded. The covalent component obviously needs these combinations. In any case where a hydrogen binds to anything other, there will always be overlap between its 1s and the other elements higher s orbital. (Obviously in the confinement of molecular orbital theory.)

"there will always be overlap between its 1s and the other elements higher s orbital" Really? Do you know the MO structure of say $\ce{NH3}$ or $\ce{CH4}$? Most sigma bonds do not involve two s-orbitals. Sigma bonds form between AO where most of the electron density is on the axis connecting both bonding orbitals. It's not so easy to find examples that involve only s-orbitals.

Lithium hydride in the gas phase maybe?

@OscarLanzi: yes, I can see that as a possibility. I was thinking of the $\ce{HX}$ gases but these are $\sigma_{sp}$ MOs.

@OscarLanzi Lithium Hydrride have ionic not covalent bond in any phase.

Hard to believe that, really.  Does a M.O. calculation really put the bonding orbital 100% on an element with electronegativity about 2.1?  Opposite a cation that is polarizing enough so there is not even a bicarbonate with that ion?  But if you say so, the di-atom is fully ionic.  And that's the truth, pfffffft!

@OscarLanzi Ok I searched it, LiH is only 26% ionic, so now we have a example with 1s & 2s orbitals overlap.

Orbital combinations involving inner shell electrons can indeed interact with outer-shell counterparts having the same symmetry.  But, the inner shell orbitals are so deeply buried inside their respective atoms that the amount of overlap, and therefore the effect on bond/orbital energies, is very small.  We find that in general, molecules are well modeled, without unneeded complexity, by including only outer shell orbitals.  Moreover, any overlap involving inner shells, or even with outer shells, usually does not involve just $s$ orbitals.  There can be $p$ orbital combinations and even $d$ orbital combinations having the same symmetry as the $s$ orbital ones, so the higher angular momentum orbitals also mix in.  Only hydrogen and helium atoms with just $1s$ electrons can avoid that.
It isn't just our description of chemical bonding that exploits this property of inner shell electrons.  We sometimes identify elements in a sample by hitting the samples with X rays that knock out inner shell electrons, and measuring the spectrum of absorbed energies.  The energy required to knock out inner shell electrons from an atom is a characteristic of that element, and lack of strong overlap with other atoms makes that energy essentially invariant.

To  cut to the chase: do you have any examples of MOs that are of the $\sigma_{ss}$ type, other than the $\sigma$ bond in $\ce{H2}$?

Added more to 1st paragraph.  We do not have use $s$ orbital interactions in commonly encountered molecules except hydrogen (helium atoms do not generally combine with anything).

I was thinking. Maybe plumbane would satisfy this criteria. The electronegativities of lead and hydrogen are quite similar and thus, interaction between lead's 6s and 6p orbitals and hydrogen's 1s orbital may be significant

In fact, in methane, we should also see overlap between 1s of H and 2s of C. Since hybrid orbitals can be used to describe the bonding in CH4, of course there should be some overlap between the s orbitals of the two atoms.

@Gert The asker isn't specifying that it must be a pure s-s overlap. He/She only asked if they is overlap (even a very slight overlap).

@TanYongBoon Sorry for that, but I am searching for a pure s-s overlap not between hybrid orbitals, as we can get it in many examples. Pure S-s overlap may be found in LiH as its only 26% ionic as OscarLanzi said in the comments to the question. Will post this as answer when Get confirmed.


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Removing Tungsten Trioxide from Tungsten Electrode
I need a very sharp tungsten electrode, and I think radial electrochemical etching with sodium hydroxide is the way to go. The problem therein (I think) is that the etching will cause an oxide layer to form deeper than the native oxide layer, ruining the electrode.
Are there any reactions close to STP that will get rid of a few layers of $\ce{WO3}$? I found one recommendation that involves using ammonia as a reducing agent at 80 °C, but I can't determine its validity.
$$\ce{WO3 + 2 NH3 -> 3 H2O + W + N2}$$

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site.
Could you link back to that recommendation to see the context?


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Continuous series of inorganic hydrates (from mono- to deca-)
I'm doing some research for university and it has to do with crystals. I'm currently experimenting with $\ce{CuSO4 * 5 H2O}$. I know we refer to the formula as copper sulfate pentahydrate. And also, I'm experimenting with $\ce{ZnSO4 * H2O}$ which is a monohydrate.
What chemical compound can be bought with all the 'versions' of the hydrates? I would like to purchase copper sulfate monohydrate, dihydrate, trihydrate, tetrahydrate, pentahydrate, hexahydrate, heptahydrate, octahydrate, nonahydrate, and decahydrate, but I can't find them all (only pentahydrate which I have bought).
Same with $\ce{ZnSO4}$, I have the monohydrate, but I would still like to purchase all the hydrates for my project. I'm a newbie at chemistry. Please, can anyone explain whether there a chemical compound such as the ones above, which can be bought with all the version of the hydrates?

MgSO4 comes pretty close (though not quite there).

Iron(II) sulfate is a good candidate too.

I think that a colossally strong acid such as $\ce{HSbF6}$ will form a series of well-defined hydrates as the naked proton desperately tries to solvate itself with what it can, but you probably don't want to play around with it...


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How much energy water "captures" at it evaporates in room temperature?
As explained in Why does water evaporate spontaneously at room temperature despite ΔG > 0? water evaporates in room temperature because partial pressure of water is less than "standard" conditions of air in normal atmosphere.
If I need to evaporate or boil 100 g of water to artificially increase humidity of air over time, does it make any difference in total energy comsuption if I slowly evaporate the water or if I boil it rapidly? Common sense seems to indicate that rapidly boiling the water takes much more energy but is it really so? I would not be surprised if slowly evaporating the water would take exactly the same amount of energy (which would cause slight temperature drop in room temperature) but I don't know how to compute or demonstrate this. Does it make a difference if the room is considered big but closed or the room is open to atmosphere? If it makes difference for this discussion let's say that humidity of air in room is 25% RH and the room temperature is 25°C.

If the humidity of air is less than 100%, then water will evaporate by itself with no effort on your part, given enough time. The energy needed for that will be taken from the surroundings. You may add some heat if you want the water to evaporate faster.

I consider the "energy needed for that will be taken from the surroundings" as part of the total energy comsuption. Consider trying to increase humidity of air inside a house when air outside the house is much colder than the air inside the house. If energy is taken away from the room to allow evaporation then more energy is needed to make the room warm again.

Then the total energy consumption for water evaporation is the same no matter what.

Is it safe to say that the claim that warm mist humidifiers use more energy than cold mist humidifiers is false? Pretty much all cold mist humidifier manufacturers are claiming that their product is better because it saves energy.

No. Besides water evaporation, there is some energy spent on heating the air in the room. So the claim can pretty well be true.


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How to calculate resulting pH of mixing 2 fruit juices
I want to make jelly from a mixture of 50% fresh apple juice and 50% juice of another fruit. The resulting pH must be less than 4.6, otherwise the jelly will not be safe from C. botulinum.
Based on my research, apple juice generally has a pH between 3.3 and 4.0 (Let's say 4.0 to be conservative), so it's safe for home canning. The other juice is less acidic (a pH of ~ 6.0).
I plan on mixing equal parts (by volume) of the two solutions (pH 4.0 and 6.0); I need to know if the resulting pH will be less than 4.6.
I found a claim that if you mix equal parts of 2 acids, the resulting pH will always be the lower pH plus 0.3. So if that's true, the answer to my question is 4.3 (safe from botulism, but I need to know if this claim (and my subsequent result of pH 4.3) is correct!

You can't reliably calculate pH this way. There is too much going on at once. The claim you found might be a good approximation when working with pure individual compounds, which the juices are not.

This. You have most of the periodic table in a glass of juice, literally thousands of reactions. Just prepare a sample and measure the pH.

I agree with @SF , there is nothing like measuring the pH; and different batches of juice can have different pH's to start with, so there is no universally valid formula. However, the concept of adding 0.3 sounded intriguing, so I did some calculations, and I don't think the rule is correct. For instance, if you assume that the solution with the lower pH contains the strongest acid, and ignore all weaker acids, then doing a 1:1 mix of the two solutions is like halving the concentration of the strongest acid. But that would result in a pH increase of about 0.15, not 0.3... Bizarre...

Here's an 'exact' calculation. Juice A is a 0.3 M solution of an acid with pKa = 3. Juice B is a 0.3 M solution of an acid with pKa = 5. pH of A = 1.774. pH of B = 2.763. pH of the 1:1 mixture of A and B = 1.927. pH of the 1:1 mixture of A and water = 1.930. 1.927 - 1.774 = 0.153. It would be interesting to see the effect of different initial concentrations of the two acids, of their difference in strength, of the presence of salts of the acids (in cases buffers should form...). Not obvious at all.


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Difference between Enthalpy and Heat transferred in a reaction?
We know at const pressure change in enthalpy is change in internal energy plus work done.
Here by taking enthalpy as heat i can easily think the situation in mind that heat absorbed is due to the two factors
But when enthalpy and heat is used simultaneously i get confused.
I want the logical answer i know the mathematical equations.
I want to think it the way it practically occurs.

What is your precise definition of the standard enthalpy change for a reaction?

@ChesterMiller  net heat absorbed or released after a chemical reaction takes place..

Not during the time the reaction is taking place?  And, what if the final temperature is not equal to the initial temperature?  Does that change the heat of reaction?  And, what if the pressure changes?  How does that affect the heat of reaction?

@ChesterMiller i dnt know all these answers dats y i amasking

For gases, the heat of reaction at a specified temperature is defined as the amount of heat that has to be added to the system to bring about the change from an initial state of pure reactants in stoichiometric proportions at the specified temperature and 1 bar pressure to the pure products at the same temperature and 1 bar pressure.  This is equal to the standard change in enthalpy of the reaction at the specified temperature. The process for bringing about this change may involve doing work on the surroundings equal to the volume change for products minus reactants times the applied pressure of 1 bar.

Then what is the diff btwn enthalpy and heat

For this particular change between initial and final thermodynamic equilibrium states, the change in enthalpy is equal to the heat added.  More generally, heat is energy in transit, while enthalpy is a physical property defined by the thermodynamic equilibrium state of a system.

I got it but when enthalpy not equals to heat can u give me a case ...

and then if both were not equal at any case the differeence of those 2 parameters heat and enthalpy where that energy goes

Examples of when change in enthalpy is not equal to heat:  isothermal expansion of an ideal gas; adiabatic expansion of an ideal gas.  In isothermal expansion, the heat goes to work.  In adiabatic expansion, the internal energy change is the result of work, and, since the temperature changes, the enthalpy change is not zero (even though the heat is zero).

Enthalpy and heat are entirely different things.
Enthalpy is a function of state. If you know the state of a system, you know its enthalpy. If you know the starting and ending states of a process, you can find the enthalpy change.
Heat, on the other hand, is an inexact differential. Knowing the initial and final states of a process is not enough information to tell you the heat transfer. Instead, the heat transfer depends on the particular path taken between the states.
For a simple example of why this is important, consider a heat engine. The process is a cycle, meaning the initial and final states are the same, so a cycle of a heat engine has zero enthalpy change. However, the entire point is that the heat engine converts heat into work, so the heat exchange during a cycle is not zero.
That is why the heat exchanged is equal to the enthalpy change only under constant pressure. We need to give extra information (such as that the pressure is constant) to know the path taken during the process in order to say anything meaningful about heat exchange.


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Is it correct to talk about an empty orbital?
Professor A. J. Kirby mentions:

The properties of an orbital are those of an electron contained in it. It is normal practice, illogical though it may sound, to talk of 'vacant orbitals'.The properties of vacant orbitals are those calculated for electrons occupying them.

Since an orbital isn't defined until it is occupied by an electron, would it still be correct to say that an empty orbital (such as a LUMO) can interact with other filled orbitals?

Orbitals are only approximate functions, efforts to describe things not known precisely.

@Mithoron The orbital functions of a hydrogen or Rydberg atom are very precise descriptions, and so are, with some effort, the more complex approaches for multielectron systems, aren't they?

@Karl There are so many levels of theory that I'd sooner flip out then remembered all of them :D , and these are always approximations, it's only a matter of specific case if they're precise enough.

Simply take "vacant" or "empty" as "potentially occupied" (or overlapping with an occupied orbital, to form a new pair of un/occupied orbitals).

@Mithoron What I meant to say is that for many practical purposes, they are. Although not many people have practical uses for hydrogen or Rydberg atoms. ;-)

Is it correct to talk about an orbital? No.

"The properties of an orbital are those of an electron contained in it. It is normal practice, illogical though it may sound, to talk of 'vacant orbitals'. The properties of vacant orbitals are those calculated for electrons occupying them."

I consider this a bit of a truism, especially the last sentence. IUPAC defines as an orbital:

Wavefunction depending explicitly on the spatial coordinates of only one electron.

which as noted in this answer falls short of considering spin. So let's use that definition for spatial orbitals. (For a spin-orbital, the analogous definition would be: depending explicitly on the spatial coordinates and spin coordinate ...) At this point, let us point out that a spin-orbital is a one-electron wavefunction. It is not an observable. One can observe the density or even the spin-density by means of X-ray diffraction, indirectly by NMR, ESR etc. (Another truism: orbitals of a single-electron systems, such as the hydrogen atom, are an important, but also somewhat trivial exception.) The density can be calculated from a many-electron wavefunction (WF) and a popular way of obtaining such a WF is combining several orbitals. This involves a lot more theory and mathematics than introductory chemistry courses can show.
So one can decide that if no electron is present, there is no orbital and stand on sound mathematical ground. Then again, I can just put an electron there by means of mathematics. Depending on the "computation-chemistry-method", for instance when employing a basis set (which is by far the most common approach), one can calculate properties even for empty orbitals such as the orbital energy (not an observable), which can be used to (approximately) describe electronic excitations (an observable). Computational chemists have resolved to call the empty orbitals "virtual" to bridge the gap between the two opposing views and use them as mathematical tools to describe excited states or to improve the quality of the description of the ground state.

Since an orbital isn't defined until it is occupied by an electron, would it still be correct to say that an empty orbital (such as a LUMO) can interact with other filled orbitals?

Let's consider the formation of a bond between two hydrogen atoms in a special way: reversing a heterolytic splitting, which we will briefly compare with one reversing a homolytic splitting.
In order to decide this question, one needs to split the non-existent hair dividing the following two views: a) Since there are no empty orbitals, in the heterolytic case, the proton will distort the orbitals on the anion until the bond is formed. Does the orbital on the former proton now magically appear? b) There is an empty orbital on the proton. However, at large distance, its effect on the anion can be simulated by an electric field, which should not carry orbitals (one can assume that the metal plates creating the field are far away and crank up the charge). If a situation involving an empty orbital cannot be distinguished from one where there isn't one, is it really there?
Of course, the end result of either way of looking at things is the same hydrogen molecule that we also get from combining two hydrogen atoms the standard way. I thus suggest to abandon the view of orbitals carried around by atoms (except in the computational chemist way, as will be outlined below). Rather, I suggest to think of the effective potential felt by a newly added electron - where would it go? Regardless of how the nuclei got to where they are now, where do the electrons go? "Unoccupied orbital" is then a useful shorthand for the relevant regions.

(At this point, I consider the original question answered. I will elaborate a bit on how I got here.)
The last question of the previous paragraph is one way of looking at the algorithm of the Hartree-Fock procedure (the first step of wavefunction-based quantum chemistry, which is comparable to density functional theory, DFT, in this regard). We have some "basis set" (and we do not care what is looks like right now) containing candidates for orbitals, nuclei (i.e. charges and positions) and a number of electrons. The first step is to evaluate the potential/the forces acting on the candidate orbitals, which is nuclei only at this point$^1$. One then linearly combines the candidate orbitals (that's LCAO right there) to form the best choice one can make. Obeying the aufbau principle, one fills the candidate orbitals until all electrons have got an orbital. One then updates the potential (which now also considers the electrons), combines again, fills the new candidates, updates the potential again and so forth until the changes between iterations are small. The result is then evaluated in terms of energy, and possibly electron density or other observables. By doing it this way, one arrives at the familiar MO picture without assuming an electron distribution or specific orbitals filled in a certain way on certain atoms.
Let's get back to the basis set. In the previous paragraph, no assumption was made on its nature. The initial candidate orbitals could be any shape (as long as they are reasonable from a mathematical point of view, for instance, they must be twice differentiable, square-integrable etc.) and many different choices exist. Experience shows that using functions resembling the exact orbitals of hydrogen are very useful for molecular chemistry$^2$. This is because they combine results of decent accuracy with computational efficiency and chemist-friendly starting points of interpretation - such as HOMOs and LUMOs that can be related to individual atoms, flawed though as that line of reasoning may be (as I have hoped to show with this wall of text).
Footnotes:
$^1$ A bit of a white lie: this is only the special situation of the so-called core guess. But done here for illustration.
$^2$ Solid state chemistry is a different matter, where standing waves are the norm.

It depends what you mean by correct.
If we are being technical, it isn't valid to speak of orbitals at all (besides the trivial case of a one electron system, like a hydrogen atom), as they only emerge as an approximation to the real description of the atom/molecule, which is the wavefunction that satisfies the Schrodinger equation. Specifically, orbitals have meaning if we make the assumption that a many electron wavefunction can be formed as the antisymmetrized product of many one electron wavefunctions. This is useful because we can solve the one electron Schrodinger equation exactly, but have yet to develop a way to analytically solve for even a two electron system.
Now the use of orbitals is convenient, not only because it allows us to obtain approximate solutions to the Schrodinger equation, but also because they often facilitate qualitative understanding of certain molecular properties. For example, if we use atoms and their atomic orbitals as building blocks, we can use valence bond, MO theory, or even wavefunction based methods to construct qualitatively correct geometries, absorption/emission spectra, etc for molecules composed from those atoms.
In terms of your original question, it can be useful to consider the LUMO to describe the behavior of a molecule. The only caveat is that the features of the LUMO are meant to describe an electron in it, not an empty orbital. So, hypothetically you could obtain the electron affinity for a molecule by determining the energy of the LUMO (this is a version of Koopmans' Theorem). However, these results would only be valid under the assumption that orbitals that describe the molecule don't change when another electron is added, which is typically a poor assumption.
Similarly, it might be qualitatively convenient to say that the excited state of some molecule is obtained by moving an electron from the the HOMO to the LUMO, but this again ignores the likelihood that the orbitals would relax to better suit the new electronic configuration.
In general, unoccupied orbitals don't have as much meaning as their occupied counterparts. When I, and I assume others, initially learned about orbitals, I pictured them as bubbles that contained electrons, but this is not the case. A slightly better, but still incorrect conception I had was that orbitals were like preexisting orbits or paths in which an electron would travel if it encountered them. But these orbitals are descriptors of the electron itself; they don't exist on their own. The reason an unoccupied orbital is something of a misnomer is that an orbital is meant to describe the probability distribution of an electron, but since an unoccupied orbital isn't describing any of the electrons in the molecule, its unclear what meaning can be drawn from it.

Just to be even more technically correct, is it not valid to speak of orbitals in the one electron case?

@orthocresol you are correct, I should make that distinction more clear. Using orbitals is making the assumption that the many electron wavefunction can be made as an antisymmetrized product of many one-electron wavefunctions (i.e. orbitals). This assumption is exact when we only have one electron, as in this case the exact wavefunction is already a one electron wavefunction. The orbital picture would also be exact if the Hamiltonian only depended on single particle interactions.


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Reversed size dependence in ion solubility
My book says that solubility increases with interionic distance, because the attractive forces between ions gets smaller. However, if one of the ions in a binary ionic system is very large compared to the other, the trend is reversed. For example, CsF has greater solubility than CsCl. Why is this?

Lattice enthalpy inversely depends on size difference.also lattice enthalpy inversely related to solubility.
Thus as size difference increases solubility increases

This can be explained with some crude thermodynamic models. The lattice enthalpy $\Delta H_\mathrm L$ of an alkali metal halide $\ce{MX}$ is given by
$$\Delta H_\mathrm L = \frac{k_1}{r_\ce{M} + r_\ce{X}}$$
and the hydration enthalpies by
$$\Delta H_\mathrm S(\ce{M}) = -\frac{k_2}{r_\ce{M}}\qquad  \qquad \Delta H_\mathrm S(\ce{X}) = -\frac{k_2}{r_\ce{X}}$$
where $k_1$ and $k_2$ are some positive constants and $r_\ce{M}$ and $r_\ce{X}$ are the ionic radii of $\ce{M+}$ and $\ce{X-}$ respectively.
The solubility is therefore a balance between these two quantities: larger (magnitudes of) solvation enthalpies increase solubility, but larger lattice enthalpies reduce solubility. The solubility $S$ is thus related to the difference of these terms. We ignore the entropic contribution, or to be precise, we assume that $\Delta S$ does not vary too much between different salts $\ce{MX}$.
$$S \sim \frac{k_2}{r_\ce{M}} + \frac{k_2}{r_\ce{X}} - \frac{k_1}{r_\ce{M} + r_\ce{X}}$$
Let us now assume that
$$r_\ce{M} = a+b \qquad \qquad r_\ce{X} = a - b$$
where $a,b$ are more positive constants. Here we have implicitly assumed that $r_\ce{M} \geq r_\ce{X}$, but it works the other way round too, so there is no loss of generality. Thus
$$\begin{align}
S &\sim \frac{k_2}{a+b} + \frac{k_2}{a-b} - \frac{k_1}{2a} \\
&= \frac{2ak_2}{a^2-b^2} - \frac{k_1}{2a}
\end{align}$$
This is clearly a minimum when $b = 0$, i.e. $r_\ce{M} = r_\ce{X}$. So, salts with similar cationic and anionic radii are less soluble, and vice versa.

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


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Can someone help me predict the products for this problem?

will it be cyclic? I just don’t understand the methods to determine how to start this.

For this type of problem, you should look at the modes of reactivity available.
The reactant has an electrophilic component (the alkyl chloride) and a mildly nucleophilic component (the alcohol). It has one slightly acidic hydrogen on the alcohol ($\mathrm{p}K_{\mathrm{a}} \approx 15$).
Sodium hydride is a non-nucleophilic base. This means you should immediately look for acid base reactions.
As Waylander pointed out, the best reaction is to deprotonate the alcohol with the hydride. You now have a strongly nucleophilic species in the form of an alkoxide, which can displace the chloride via an $S_{\mathrm{N}}2$ reaction. Stereochemical concerns are important here because it's a bimolecular reaction so pay careful attention to the product stereochemistry.

Can you explain how you got the pKa value for the alcohol?

You memorize it and/or look it up. Probably, here I rounded too aggressively. It should be closer to 16. @SourabhYelluru

Sodium hydride is a strong base, strong enough to completely deprotonate the alocohol. An internal nucleophilic attack will displace chloride giving the epoxide

Will be any kind of stereochemistry inversion?

As your starting material is single isomer your product should also be


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Is the equation dU= TdS - PdV always valid (even for irreversible processes?)
I really don't think it should be.
$dU = dQ + dW$ is the statement of the first law of Thermodynamics.
$dW$ can be put as $-pdV$ and $dQ $can be put as $TdS$ only if the process is reversible. (From the definition of entropy, the $dQ$ in $dS$ = $\frac{dQ}{T}$ is for reversible paths between the inital and final states.)
But when I checked on the net, it says $dU= TdS- pdV$ is always true for any type of process. How? Many say because it contains state variables?
We assumed the process to be reversible while deriving it, otherwise the most general equation should be $dU  ≤ TdS - pdV$, right?
Where ever I see on the net, they used that equation $dU  = TdS - PdV$ as something always valid and as a base to derive fancy new equations with partial differentials and stuff.
For example, in page 2 in this pdf: https://ocw.mit.edu/courses/chemistry/5-60-thermodynamics-kinetics-spring-2008/lecture-notes/5_60_lecture11.pdf

The differential statement of the first law, $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ is valid for all states---but not all processes---for the reasons you have described. When we apply it, we are always assuming that a reversible process connects the two states of interest; irreversible processes are outside the range of applicability of this equation. States and processes get muddled up a lot, because we can always find a reversible process that connects any two states.
The generalization of this equation for all processes is given by $$\mathrm{d}U = \delta q + \delta w,$$ which is just the statement of energy conservation. We cannot, however, write $$\mathrm{d}U \leq T\mathrm{d}S - p\mathrm{d}V,$$ for this would imply that $T$ and $p$ are defined for a system at all points during an irreversible process, which is not true.

So the answer is no, that dU=TdS−pdV equation is only valid for reversible processes, not irreversible ones, right? I don't get why all these links say it's valid for all processes.. -- https://www.google.com/search?&dcr=0&source=hp&ei=PkUcWu_9BMfTwQKF4KCgCA&q=dU%3DTdS%E2%88%92pdV&btnK=Google+Search

@Rick, yes, that's correct: the differential statement of the first law is only valid for reversible processes. I presume the reason for the confusion is because states and processes are frequently (incorrectly) conflated.

Have a look at the second page here: https://ocw.mit.edu/courses/chemistry/5-60-thermodynamics-kinetics-spring-2008/lecture-notes/5_60_lecture11.pdf

@Rick, yes? That page states that the first law is always correct and valid for (states of) a closed system, regardless of the nature of the process, and that agrees with what my answer has stated.

Ohhh.....so it's always valid as long as we know we can connect the 2 states with a reversible path. Kinda like how we can always write dU as nCvdT even if our process is not isochoric. Did I understand that right?

@Rick, yes, that's correct. But $\mathrm{d}U = nC_v\mathrm{d}T$ (without any pressure dependence) only for an isothermal process or an ideal gas (or some other system where the energy is independent of pressure).

Actually, I was trying to answer this question https://physics.stackexchange.com/questions/370926/du-dq-and-du-tds-but-dq-not-always-equal-to-tds-why But I got really confused, can you answer what's going on in that question please?

@Rick, this is better suited as a new question, instead of as a comment on my answer to this question. But you seem to have answered it fine.

I don't know, I guess it'll be marked as a duplicate if I ask it again.. What I'm asking is: We just discussed that equation dU=TdS−pdV is always valid. If we have an irreversible isochoric process, dV should be 0, so dU= TdS. -----(1) also, dU = q + w always, here w is zero, so dU = q ----------(2) If you combine (1) and (2), we should get q = TdS, This is what's confusing, it clearly isn't, right? Since it's an irreversible process.

@Rick, this will be my last reply to your comments. I think that my answer is complete and will resolve any difficulties you encounter if you read it closely. I have stated that $\mathrm{d}U = T\mathrm{d}S-p\mathrm{d}V$ is valid for all states (but only reversible processes), and you are trying to apply it to an irreversible process. The premise is wrong, so your conclusions are wrong.

Ohhhh....ok. I understood. I was in a really confused state, just calmed down and re-read your answer carefully. I understood now. Thanks alot. I just realized how stupid my many follow up questions look now :P Sorry for bothering you with all that. Thanks alot!!

Does the conditions used to derive the equations apply too? Like in this case we use a constant Pressure reversible process to derive the equation dU=TdS-pdV. Do we have to find a constant pressure reversible process between two states to be able to use the equation dU=TdS-pdV?

Sorry, but I don't see where you rely on a constant-pressure reversible process to derive the first law.

In an isothermal process why do we need an expression? dT is going to be zero anyways and dU will also be zero since dT is zero.
replying to this: "But dU=nCvdT (without any pressure dependence) only for an isothermal process or an ideal gas (or some other system where the energy is independent of pressure)."

@optimusprime I suspect I meant to say isobaric. Even in an isothermal process, there may be changes in the energy due to changes in pressure.
Edit: Actually, it may be isovolumetric, since U can't be a function of purely intensive variables...

@a-cyclohexane-molecule I understand that $dU=TdS-pdV$ can be then only applied to calculate an internal energy difference between 2 equilibrium state. What about the relation for irreversible processes $dU=\delta Q_\mathrm{irr}+\delta W_\mathrm{irr}=T(dS-d_i S)-p_\mathrm{ex}dV$ (where $d_i S$ is the entropy production)?

@Guiste I'm not familiar enough with that equation to discuss it at length, sorry. If you want to calculate the change in energy for an irreversible process, you can still use dU = TdS - pdV. If you want to calculate the entropy production for an irreversible process, you would probably use the framework of stochastic thermodynamics and compute path probabilities.

As I understood your answer, you can only use $dU=TdS-pdV$ if initial and final state are in equilibrium. Otherwise I think it makes no sense, because the basics of this equation is that it is enough to know 2 thermodynamic variables to express the change of another one ($U$), and such a relation between thermodynamic variables only holds in the equilibrium state.

For the equation I mentioned above, we know that $\delta Q_\mathrm{irr}<TdS$ and $\delta W_\mathrm{irr}=-p_\mathrm{ex}dV$, do you agree with that? We can then write $\delta Q_\mathrm{irr}=TdS-Td_i  S$ to indicate the irreversible entropy production. As I understand it this equation should be applicable to calculate the internal energy change for an irreversible process, independent if the final state is in equilibrium or not.

@Guiste You're correct to state that dU = TdS - pdV implies that both the initial and final states are at equilibrium and that it would not work for a system with an initial/final state out of equilibrium. In your scenario, the final state may not be at equilibrium. In that case, I'm not sure $U_\text{final}$ makes any sense, because it's quite possible that the energy is not well-defined for such a non-equilibrium state. I don't feel qualified to speak about your irreversible process equation, because there may be subtleties that I'm not aware of.

For any two processes (reversible or irreversible) connecting two equilibrium states that are described by the state variables $U,T,S,V,p$ it is always true that $$dU=TdS-pdV.$$ If during the processes the transferred heat and work performed are $\delta Q$ and $\delta W$ then it is always true that $$dU=\delta Q + \delta W.$$
Both equations are just the 1st law of thermodynamics applied to any processes connecting two equilibrium states described by the given thermodynamic variables. The 2nd law comes into play by postulating an equality $TdS = \delta Q_{rev}$ for any reversible processes and an inequality $TdS > \delta Q_{irrev}$ for any irreversible process. Note that this does not say anything about quasi-static processes. The only requirement is that the beginning and end of the processes have well defined thermodynamic parameters and during the process one should be able tell the amount of work and heat transferred.
Since for all processes $dU=TdS-pdV = \delta Q + \delta W$ and $$TdS \ge \delta Q$$ we also have the inequality $$pdV \ge \delta W $$ between any two equilibrium states.
If now we demand that during any instant of the process between the end equilibrium states we are able to define the thermodynamic variables $U,T,S,p,V$ then during the whole process we will have these (in)equalities.
Occasionally, you do see the inequality $dU < TdS-pdV$ but in this form $p$ is not the state variable we call system pressure, it is instead the external pressure needed to overcome both the system and some friction or similar cause to affect the compression $dV$.

$dU = TdS-pdV$ is not always valid.  Furthermore, the reason it's not generally valid isn't because of the distinction between reversible and irreversible processes.  There's an easy workaround for that*. Rather, it's because it only applies to closed systems.  If the system is open, it is necessary to use the more general expression $dU = TdS-pdV+\sum_i \mu_i dn_i$.
*Here's the workaround for irreversible processes:  $dU = TdS-pdV+\sum_i \mu_i dn_i$ can be used to calculate $dU$ for any process, reversible or irreversible, since if there is an irreversible path that connects two states, it is always possible to find a reversible path that connects them.
Likewise, $dU = \text{đ}q + \text{đ}w$ is not always valid for the same reason $dU = TdS-pdV$ isn't: it only applies to closed systems.
The following will illustrate why the idea that we must use $\text{đ}q \text{ and } \text{đ}w$ for irreversible processes is an artificial one:  To calculate $\Delta U$ for one of the most common types of irreversible processes we study, namely chemical reactions in open systems, we typically don't use an expression for $dU$ that contains $\text{đ}q \text{ and } \text{đ}w$. Instead, we rely on $dU = TdS-pdV+\sum_i \mu_i dn_i$.  That's because $dU = \text{đ}q + \text{đ}w$  can't be straightforwardly extended to reacting open systems.  I.e., the following equation is not valid, period:
$dU = \text{đ}q + \text{đ}w + \sum_i \mu_i dn_i$
For more on this, see:
Jan T. Knuiman, Peter A. Barneveld, and Nicolaas A. M. Besseling.  On the Relation between the Fundamental Equation of Thermodynamics and the Energy Balance Equation in the Context of Closed and Open Systems.  Journal of Chemical Education 2012 89 (8), 968-972.  DOI: 10.1021/ed200405k


text:
Assigning the IUPAC name of this compound

Going by the rules that were given to me by my professor:

I first decide the seniormost group and its parent chain. In this case, cyclohexene carbonitrile.
Then, I decide direction to traverse the ring to assign locants by the first point of difference. Accordingly, I move towards bromine.

The final name I obtain is thus "2-bromo 3-hydroxy cyclohex-4-enecarbonitrile". The answer given in my workbook however is "6-bromo 5-hydroxy cyclohex-3-enecarbonitrile". Apparently, they decided to traverse the ring in reverse direction.
Obviously, this must be some special case. Please explain the exact rule involved here. Thanks!

Actually, the preferred name is 6-bromo-5-hydroxycyclohex-3-ene-1-carbonitrile.

For the numbering:

P-14.4 When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority: [...]
(c) principal characteristic groups and free valences (suffixes); [...]
(e) saturation/unsaturation: [...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order [...]

As correctly stated, carbonitrile therefore gets priority for the low locant, in this case '1'. The next group to be considered is the alkene. Since the locant '3' is lower than the locant '4', the numbering is therefore chosen as follows:

Numbering aside, Loong's comment points out two other issues with your name.

There are hyphens, not spaces, between substituents.

The locant '1' in carbonitrile is also required, because of P-14.3.3 (see Loong's comment). There is no rule saying that it can be assumed to have the locant '1'.

Compare this with "butanoic acid", for example, where the locant '1' for the -oic acid suffix can be dropped. This is because

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

This rule is very prescriptive on when the locant '1' can be dropped. Since "butanoic acid" is a monocarboxylic acid derived from an acyclic hydrocarbon (butane), the locant '1' can be dropped, and the PIN is butanoic acid instead of butan-1-oic acid. My personal interpretation of this rule - or what I think is the rationale behind it - is that this is because "butan-2-oic acid" simply does not make chemical sense.
However, in this case, "6-bromo-5-hydroxycyclohex-3-ene-2-carbonitrile" is still a name which can generate a valid structure which is different from yours.

This numbering is obviously not in line with IUPAC rules, because it does not assign low locants to the nitrile, but that's beside the point.

I think even for functional groups for nitrile,which doesn't that necessarily have to be terminal position needn't be specified in case of cyclic compounds as we can always start counting from it unlike in the case of a straight chain compound where we cant always start from functional group

No, that's not how it works.

The locant 1 in 6-bromo-5-hydroxycyclohex-3-ene-1-carbonitrile is a result of rule P-14.3.3 “In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant '1' in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. …”

This is because alkene has higher preference than alkyl halides. So when counting clockwise position of ene is less than that when counting anticlockwise. Hence the answer is as stated.
Check out Rule 6.2 of basic principles on Wikipedia.


text:
How is "dietary cholesterol" processed in a human body?
Person eats an egg that is high in animal cholesterol ("dietary cholesterol").
What do chemical reactions play a role in breaking down it in stomach?

It isn't broken down.

Why? Any references?

I'm not getting your question, it's simply absorbed in intestines. Maybe you ask how it's separated from proteins? Or metabolised?

OK. It's absorbed in intenstines. Does it enter blood stream after absorbtion or metabolism?  I have basic knowledge in this topic, and want to learn more. I googled it and the search results are not explaining the process.

Most lipids, such as cholesterol (and triglycerides), are nonpolar and therefore very hydrophobic molecules. In order to be transported in watery blood, such molecules first must be made more water-soluble by combining them with proteins produced by the liver and intestine.
Lipids and proteins associate noncovalently to form lipoproteins (globular micelle like particles that consist of a nonpolar core of triacylglycerols and cholesteryl esters surrounded by an amphiphilic coating of protein, phospholipid, and cholesterol):

Lipoproteins have been classified into five broad categories on the basis of their functional and physical properties:

1. Chylomicrons, which transport exogenous (dietary) triacylglycerols and cholesterol from the intestines to the tissues.

Chylomicrons therefore function to deliver dietary cholesterol and triacylglycerols to muscle and adipose tissue and dietary cholesterol to the liver.

2–4. Very low density lipoproteins (VLDL), intermediate density lipoproteins (IDL), and low density lipoproteins (LDL), a group of related particles that transport endogenous triacylglycerols and cholesterol from the liver to the tissues (the liver synthesizes triacylglycerols from excess carbohydrates).
5. High density lipoproteins (HDL), which transport endogenous cholesterol from the tissues to the liver.

Circulating HDL acquires its cholesterol by extracting it from cell-surface membranes and converts it to cholesteryl esters through the action of lecithin-cholesterol acyltransferase (LCAT), an enzyme that is activated by apoA-I. HDL therefore functions as a cholesterol scavenger.

This enzyme transfers a fatty acid residue from atom C2 of lecithin to cholesterol with the concomitant formation of lysolecithin
The cholesteryl esters are hydrolyzed by a lysosomal lipase to yield cholesterol, which is subsequently incorporated into the cell membranes. Any excess intracellular cholesterol is reesterified for storage within the cell through the action of acyl- CoA:cholesterol acyltransferase (ACAT).
References

Principles of Anatomy and Physiology 13e Tortora, Derrickson
Biochemistry Voet and Voet: Lipoproteins

Based on the answer, I understand now that dietary and liver-generated cholosterol both are carried by blood to build the cell membranes around human body.


text:
Is it possible to make a disaccharide from two fructose monomers? If so, what is its name?
Maltose is a disaccharide made from two molecules of glucose; and sucrose (table sugar) is made from glucose and fructose. Is it possible to combine two fructose molecules to make a disaccharide?
If so, what is that called? Or if not, what prevents the glycosidic bond from forming?

I don't really see the need to close this question.

@orthocresol it is not a homework question either. I am just a curious layman.  I looked up disaccharide charts, and not a single one had two fructoses listed.

I was curious how two fructoses would compare in sweetness to other disaccharides

This is not at all as trivial as it seems, and I certainly disagree with the close verdict of the community. I would recommend detaching the request for a comprehensive list of disaccharides from the question. There is a fairly comprehensive table on Wikipedia. I couldn't find anything more comprehensive. In general I do not doubt the existence of a fructose dimer, as there are fructooligosaccharides of the form $\ce{GF_{n}}$, and decomposition will probably yield some kind of di-fructofuranose... it's all hard to find though.

See for example Peter Forgo et. al.,  Microchemical Journal 2013, 107, 37-46. DOI: 10.1016/j.microc.2012.06.017

@Martin-マーチン yes, unfortunately, the wikipedia article had the most comprehensive list, and it seems to me, to be quite far from unabridged. Anyway, the question has been edited to remove that request.

@Martin-マーチン also, from more googling, I have learned that polysaccarides composed of fructose are called fructon; however, this is not precisely what I was looking for as it refers to a wide range of chemical structures.

I also disagree with closing this question. In my opinion, the question was even fine in its original scope. We could have had a community wiki answer, with columns, e.g., common trivial name, monosaccharides, linkage, PIN, and reference to IUPAC (or similar) document, or published article (+ perhaps picture). Of course, the incentive for answerers would have been lower, but over time the table would probably be better-referenced and more comprehensive than on Wikipedia. Anyhow, I am voting to reopen this question as is.

Google fructofuranosyl fructofuranoside. Also fructofuranosyl fructofuranose. You will find some.

in agave plants the typical sugars are fructans (fructose-polymers) like inulin, levan, agavin - with sometimes really complex structures comparable to starch. As starch breaks down into glucose, maltose, etc. you have also a need for break-down process e.g. in the tequila/mezcal industry which is usually a thermal hydrolisis (no enzymes like in beer). This hydrolisis is never complete so you have dimers and trimers (or higher) fructose-sugars, which are even metabolizable by yeast-strains. They are referred to as nistose or kestose (https://pubchem.ncbi.nlm.nih.gov/compound/1-Kestose#section=Top).
I hope this answer helps you to finally resolve your question - check out also user55119 coment.

A molecule of two fructoses is called inulobiose or difructan (PubChem: 439552).
It occurs naturally in the fungus Aspergillus.
https://pubchem.ncbi.nlm.nih.gov/compound/Inulobiose
see also
Matsuyama (1982) J Biochem 92(4):1325-8. doi:10.1093/oxfordjournals.jbchem.a134051.PMID:675724


text:
How to construct a QST2 input file for Gaussian16 for multiple reactants and products?
I am attempting to determine the transition state structure for the following reaction using the QST2 method on Gaussian 16:
$$\ce{CH3 + N -> HCN + H2 }$$
However, I am struggling to put together my input file. How do I format my z-matrices for multiple reactants and products? Here is what I have so far (though it is not running successfully):
%chk=sample2.chk
%mem=1500MB

#T UB3LYP/6-31G** opt=(qst2, redundant)

CH3 + N -> HCN + H2 reactants

0 3
C
X 1 1.0
H 1 r 2 90.0
H 1 r 2 90.0 3 120.0
H 1 r 2 90.0 3 -120.0

r 1.0828

CH3 + N -> HCN + H2 reactants

0 3
N

CH3 + N -> HCN + H2  products

0 1
C
X 1 1.0
H 1 rh 2 90.0
N 1 rn 2 90.0 3 180.0

rh 1.0587
rn 1.1326

0 1
H
H 1 r

r 0.741

I'm pretty new to this type of calculation on Gaussian so any help would be much appreciated!

The reverse reaction, which has the same transition state, breaks a triple bond. Even CC will likely have trouble with that. Consider computing the relaxed PES for extending the CN bond in the presence of H$_2$, since the reaction is a single atom coming from infinity.

TL;DR: QST2 will not work.
For QST2 calculations you need one reactant structure and one product structure, the program will then attempt a transition state guess. The essential point is, that the supplied structures have to have identical ordering (in Cartesian coordinates), or identical z-matrices, although I am not sure that this works.
For more on how to run these types of calculations, see Finding transition states in Gaussian – focus: Electrophilic Addition Reaction.
QST2 (and QST3 by extension) only make sense, if you know that you will have a single transition state - they are especially difficult to use for dissociation or association reactions because of the implied supermolecular approach. The minima $(\ce{CH3 + N})$, $(\ce{HCN + H2})$ will be very shallow and not well converged. From these structures it would be very difficult to guess the transition state, so you would at least have to supply a guess for it (QST3).
Additionally you are going from a "quintet" state (doublet + quartet) or "triplet" state (doublet + doublet, or doublet - quartet) to a singlet state (singlet + singlet). You could probably force your reactants into a broken singlet state (doublet - doublet), then it's a problem of how much sense this makes. In the other cases it is doubtful that the program will get a good guess for the electronic state of the TS.
DFT is maybe not the best methodology to investigate this kind of reaction in the first place.
In any case, here is an input file (tested with Gaussian16 rev. A.03) that will at least start running. SCF will not converge. Also reactants and products have not been optimised beforehand (which you must do). This is only for illustrative purposes.

%chk=bp86svp.chk
#P BP86/def2SVP
EmpiricalDispersion=GD3BJ
DenFit
opt(QST2,CalcFC,maxcycle=5)
gfinput gfoldprint iop(6/7=3)
nosym
pop=full

N + CH3

0 1
C        0.000000000      0.000000000      1.891780658
N        0.000000000      0.000000000     -2.108219342
H        1.029433990      0.000000000      2.255740536
H       -0.514716995     -0.891515987      2.255740536
H       -0.514716995      0.891515987      2.255740536

HCN + H2

0 1
C        0.649013613      0.189100642      0.000000000
N       -0.817143840      0.083131316      0.000000000
H        0.964811208     -1.929636661     -0.499461795
H        0.964811208     -1.929636661      0.499461795
H        1.696363276      0.452628980      0.000000000

Thank you very much for your response! However, I am still confused as to how I am supposed to format the Cartesian coordinates in this way? I am relying on databases which would provide the coordinates for each species separately so how do I combine them into a single coordinate?

@pennypeat Use the molecular viewer and combine them in your best guess orientation. Optimise the supermolecular structure (might not converge because it's a very shallow pot. hyp. surf.), check for imaginary modes. Do same for product super molecule. Remember that the ordering must be identical.


text:
Why does hot oil fry things, but hot water does not?
When a slice of potato is dropped into hot oil, it gets fried but the same in hot water, gets ‘cooked’. What is it about oil that gives the slice of potato the crispy feature while water only makes it soft?

Semi-relevant: https://www.youtube.com/watch?v=a4gYv2BK-HQ

Thanks for the link @CurtF. It is interesting to know about fluorinert. But I really do not see the point. In fact, it has me a little more intrigued. The chip out of F.C.-40 was as crispy as the one out of canola oil. But then the former had a lot less fat. I understand why it has less fat but then, how did the slice get fried at all, does it all come down to dehydration? Because even a microwave oven fries things by removing water by getting it to excite..

Related: Why does water boil vegetables, but oil does not?

What temperature is a fryer? What temperature is boiling water?

Oil can get hotter than the b.p. of water, 373 K.
The browning (decomposition) of starches proceeds very slowly (perhaps years for a noticeable change) at 373 K. In addition, the water in the potato will not evaporate away, since the potato is in water (not the water vapor above the liquid, which can be hotter than 373 K).
Since the cooking oil is hotter (e.g. 440 K), starch decomposes (caramelizes) in a few minutes, and dehydration makes the chip crunchy. Bon appétit!


text:
How do I combine z-matrices of multiple molecules into one input stack?
I'd like to run a calculation in Gaussian which requires me to provide cartesian coordinates for reactants and products respectively. My question is how do I combine the cartesian coordinates for two reactants (that I would obtain from a database)? For instance, CH3 + N?
Is there a specific program/procedure that I need to use?
To be more specific, say I would like to combine the z-matrix of $\ce{CH3}$
0 3
C
X 1 1.0
H 1 r 2 90.0
H 1 r 2 90.0 3 120.0
H 1 r 2 90.0 3 -120.0

r 1.0828

with a z-matrix of $\ce{N}$
0 3
N

And I would like to combine the z-matrix of $\ce{HCN}$
0 1
C
X 1 1.0
H 1 rh 2 90.0
N 1 rn 2 90.0 3 180.0

rh 1.0587
rn 1.1326

with that of $\ce{H2}$
0 1
H
H 1 r

r 0.741

In other words, I want to combine these z-matrices (or geometries) of the two reactant/ product species into a single coordinate, so that I can use them for my QST2 calculation. What do I do?

"Combine" is an awfully broad term. For example, copying and pasting the cartesian coordinates of two reactants into one file will surely combine them, though probably not in the way you want.

Thank you for responding! I edited it to be more specific. Could you help me please?

I clarified your request, however, if you don't know how to construct valid input geometries for Gaussian, you should not be playing around with transition state searches, but rather learn the fundamental basics of how to optimise a molecule. What software do you use to visualise your molecules? I am 99% certain, there is a molecular editor included.

The Setup
The first thing you need to do is to get the difference between a Z-matrix (sometimes also called internal coordinates) and cartesian coordinates straight. They are truly different things.
Secondly, take the Z-matrices of a sufficiently large molecule. The basic structure is:
C1 .  .  .
C2 r2 .  .
C3 r3 a3 .
C4 r4 a4 d4
C5 r5 a5 d5
(and so on from this point)

where in this simplified notation, I have omitted the anchor centers for the coordinates. One notices that there are 6 coordinates left out, marked by dots. They can be omitted because three degrees of freedom for translation and three degrees of freedom for rotation of the entire molecule are irrelevant for the calculation of properties.
Now imagine a second molecule tacked onto the other one:
C1 .  .  .
C2 r2 .  .
C3 r3 a3 .
C4 r4 a4 d4
C5 r5 a5 d5
C6 *  *  *
C7 r7 *  *
C8 r8 a8 *
C9 r9 a9 d9

This Z-matrix is invalid. One needs to fill the missing coordinates marked by the asterisks. In a sense, the second molecule does not have the 6 degrees of freedom mentioned above, because any change in them would change the orientation relative to the first one, impacting all properties.
So, what can you do?
Filling in the missing coordinates is hard, at the very least by hand or some other automated fashion. Doing it in an editor is basically the same amount of work as the solution I actually suggest: Create the system of both molecules in one Z-matrix and split that Z-matrix. Remove the superfluous coordinates in the resulting matrices and run calculations as necessary.

Splitting a z-matrix also seems quite tedious though. One notices that there are 6 coordinates left out, marked by dots. They can be omitted because [...] I wouldn't say they are omitted. You simply cannot define them, because you do not have a reference. With one point you cannot define a line (or an angle, or a dihedral) etc..

OK thank you for clarifying that and I apologize for my ignorance but I still don't understand what "splitting w Z-matrix" entails. Would be so kind as to show me an example of splitting? Thank you!

@pennypeat The split would be the opposite of what I did with tacking on a second molecule. I'm afraid you will have to get a molecular editor and play with Z-matrices for a while.


text:
about Markovnikov's rule, is it wrong to think of it with MO theory?
Markovnikov's rule stated 'The hydrogen ends up attached to the carbon of the double bond that had more hydrogens to start with.' And many books give it a reason applying stability of carbocation. Carbocation stability is also important for Hammond postulate and rate of a reaction.
But the way I'd like to interpret it is using molecular orbital theory and hyperconjugation. If say C-H sigma bond in assymetric alkene(ie propene) hyperconjugated with pi* orbital, then the other carbon which have no hyperconjugation effect has more orbital coefficient (more pi orbital electron density). This means that the carbon without alkyl substituent is naturally more reactive to take proton. After that, we have tertiary carbocation, proceeding rest of reaction.
I used this concept a lot when I solve problems and found it very useful. However, I just figured that no books, no questions, no google infos suggested this idea at all. so, is this way of thinking wrong? if it's wrong, how so? if it's right, why nobody use it and why all curriculum stress Markovnikov's rule?
(I tried to find a image of asymmetric alkene MO but i couldn't. There is no image of it. But in a book, Oxford Organic Chemistry by Clayden, page 890 has similar image I was looking for. It's HOMO of 1,1-dimethylbutadien. 4th carbon has bigger orbital than the others as pi2 MO)


text:
What is the product of the titration of potassium permanganate and iron(II) sulfate?
I recently did an experiment at school, where I had to titrate KMnO4 with FeSO4. The solution was colourless, then turned yellowish until the end point was reached and it turned pink in one drop. After letting it sit for a while, the pink tinge disappeared. Can someone please explain what happened and show the products and this mysterious observation? Thanks

What was the pH, or at least what was the medium: acidic/neutral/basic? I suspect you should've added $\ce{H2SO4}$, haven't you?

From your description I'd say you were titrating ferrous sulphate, $\ce{FeSO4}$ solution (the analyte), with potassium permanganate, $\ce{KMnO4}$ solution (the titrant), in acid conditions (dilute $\ce{H2SO4}$ present).
The permanganate ion $\ce{MnO4^{-}}$ is a strong oxidiser and oxidises the ferrous ion $\ce{Fe^{2+}}$ to the ferric ion $\ce{Fe^{3+}}$ very easily. In dilute solution both ferrous and ferric ions are more or less colourless.
In this process the very strongly coloured permanganate is reduced to the manganous $\ce{Mn^{2+}}$ ion, which is also colourless in dilute solution.
When all the ferrous ions have been oxidised, one more drop of the strongly coloured permanganate solution then turns the analyte solution pink and this is your end point.

I recently did an experiment at school, where I had to titrate $\ce{KMnO4}$
  with $\ce{FeSO4}$.

In that case you'd have started from a purple solution (analyte) which would gradually, during the titration, have turned colourless. This method is also used.

As this is a redox reaction, I suggest you write out (or look up) the reduction (permanganate to manganous) half reaction and the oxidation (ferrous to ferric) half reaction, to complete your knowledge of this titration.

+1 for not giving whole answer. Since this is a homework type question the asker should do some of the work... You gave the asker plenty of hints to finish looking up the bit remaining unanswered (chemical equation).


text:
Boiling points of HF, HCl, HI
I'm really confused over how to weigh the molecular mass with intermolecular forces such as hydrogen bonding in HF in terms of figuring out the melting points in relation to each other.
I know that HF has the highest melting point out of HCl and HI, but then HCl and HI both have significantly larger molecular mass.
So how does one properly explain why HF has the higher melting point out of HCl and HI? (Not just saying that generally hydrogen bonds out weigh the molecular mass when considering boiling point)
Is there any mathematical basis?

You seem to be asking about one/both/either/neither of melting points and boiling points. Please clarify by editing the post.

In HF there is a hydrogen bond which always outweighs all other factors when deciding properties related to stability. H-Bonded molecules are very stable

Is there no mathematical basis for this? Surely there's some explanation using mathematics other than actual data.

I don't quite understand what you mean by a mathematical basis. Could you give an example of what you mean by explaining sthg in chem on a mathematical basis?

Is there a quantifiable reason why hydrogen bonds outweigh the other factors?


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Why would two isomers have different solubilities in water?
I was wondering as to why, e.g. $\ce{CH3CH2CH2CH2OH}$ and $\ce{(CH3)3COH}$, do not have the same solubility in water, considering they both have the same molecular formula.

Of course it's different, like every property they have, only molecular mass is same.

What is attached to the C-O-H "part" has different surface areas.

@Mithoron Yeah I do see why they're different. But I wanted to know why are they different? Like an actual reason

While they have the same formula, the molecules have different shapes, which means that they are different in almost any important way with respect to how they interact with other entities. The electron cloud is different. The rotation and vibration modes are different. The reactivities are different. The 'actual reason' that they are different is because they are different.

Your last question in comments is bordering on philosophical. They're different because they're not the same. We look different because we're not the same person. Not sure how useful it is to ask why...

highly related https://chemistry.stackexchange.com/questions/50451/why-do-alcohols-and-ethers-have-approximately-the-same-solubility-in-water-but-d

Isn't this due to extent of hydrogen bonding in both the molecules. 3 degree alcohol due to hinderance is less likely to form  h-bond.

In the most general terms: Molecules consist of nuclei and a surrounding electron cloud. That cloud is dependent on molecular charge and geometry - so it will be different for two different isomers. It follows that its interaction with the nuclei and electron cloud of a third molecule, say, water, will be different.
More specifically, one will find that properties such as

polarizability (related to van-der-Waals forces)
bond dipoles (related to H-bonds and static dipole interactions)
effective electronegativity (related to H-bonds and static dipole interactions)

On a very basic level, any solvation involves the formation of solvent shells. When varying the shape and surface of the solute, those shells will change as well. One will then find shells that work well and others that work less well, influencing the solubility.