text:
What is chemical formula of Bauxite?
My book says

Aluminium is extracted from its ore bauxite $\ce{Al2O3.nH2O}$

What does "n" stands for?
Does it mean that there is no fixed amount of water in bauxite?

That's right, it can be different.

So Corundum, Gibbsite and diaspore are also form of bauxite?

Yes an ore can have many minerals.

The Bauxite wikipedia page shows that Bauxite is a mixture of oxides and hydroxides so the formula you have is a general one and so n can be on of several integers 1 to 3.

Background:Chemical formula is short hand representation composition of a substance.
There is difference between an ore and a mineral. An ore is an impure form of mineral, so it can't be given exact chemical formula. Whereas a mineral is pure compound(in general, and can be used to obtain pure elements like metals) so it has an exact chemical formula.Minerals which can used to obtain the metal profitably are called ores.All minerals are not ores but all ores are essentially minerals too.
Bauxite is an ore of the mineral called Alumina( Al2O3) which can be used to extract aluminium. So bauxite is group of hydrated aluminum oxides, Al(OH)3 with possible additional Al and (OH) so it can be written as Al2O3.nH2O where n represents that it can have different number of water molecules per Alumina based on concentration of mineral in the ore on an average.

How can Al(OH)3 be written as Al2O3.nH2O when Aluminium and (OH) vary from compound to compound?

Al(OH)3 means hydroxide of aluminium, an hydroxide is an hydrated oxide so we can write it as Al2O3.nH2O which also basically means the same

$\ce{Al2O3.nH2O}$ is just a general (and slightly incorrect, see below) way of writing the formulae of the aluminium oxy-hydroxides that occur in bauxites. These are solid minerals, so there is no "$\ce{Al2O3}$" molecule in them. Instead, it is an (almost) infinite array of Al and O ions in a three dimensional structure. Therefore, what it basically means is that you have three cations of aluminium for every two anions of oxygen. $\ce{Al2O3}$ means exactly the same as $\ce{AlO_{1.5}}$ or $\ce{Al4O6}$ in this context. We just use $\ce{Al2O3}$ because it's simpler: smallest numbers with no fractions.
In the case of the hydroxides, they are also crystalline minerals. You can see some nice pictures of gibbsite, böhmite and diaspore when they are well crystallised. Their formula are:

Gibbsite: $\ce{Al(OH)3}$ which is the same as $\ce{AlO_{1.5}.1.5H2O}$
or $\ce{Al2O3.3H2O}$.
Böhmite and diaspore: $\ce{AlO(OH)}$ which is the same as $\ce{AlO_{1.5}.0.5H2O}$ or $\ce{Al2O3.H2O}$

Bauxites usually contain mostly gibbsite, which is unfortunate economically because it contains the most water in it. When you dig the stuff out, you want the most aluminium per tonne of material, and dilution by water is undesirable.
It's also important to note that even though we write the formulae of the minerals as $\ce{Al2O3.nH2O}$, the water is not hosted in molecular form in the crystal. You do not have proper $\ce{H2O}$ molecules in the structure of the mineral. Instead, you have hydrogen cations attached to some of the oxygen anions. There are, however, minerals where the water is in molecular form. One example is gypsum: $\ce{CaSO4.2H2O}$.
Usually, as a rule of thumb, the way a formula is written can give you an idea on how the water is hosted. If the formula is written with $\ce{(OH)}$ in it, it's probably in the form of hydrogen bonded to oxygen. If it's $\ce{.H2O}$ then you probably have molecular water. This is also important for the physical properties of the crystals. Minerals with molecular water are usually very soft and will lose their molecular water with gentle heating (up to 150 °C or so). Minerals with hydroxide groups in them require much higher temperatures to de-volatilise, around several hundreds of °C.
So, to sum it up, writing the formula as $\ce{Al2O3.nH2O}$ gives the impression that the aluminium hydroxides have molecular water in them, while they actually do not. However, it's a convenient "catch-all" way to refer to all of them together.
Also related:
What is the meaning of the "dot" notation in chemical formulas?

In "you have three cations of aluminium for every two anions of oxygen", I think, the numbers will be interchanged.


text:
What determines the boiling point of esters of the same number of carbon atoms but different structural formula?
The boiling point of esters increases with molecular weight (and therefore, number of carbon atoms) because of more dispersion forces.
But if you had two different esters with the same number of carbon atoms but different structures (e.g. methyl-ethanoate and ethyl-methanoate), what determines which one will have a higher boiling point?
Does it have something to do with the number of carbon atoms in the alkyl group in comparison to the acyl group?

Well, of course, as that is the only difference allowed in your question! :-/ It will depend on which one is more polar, as usual with polar organic compounds. I'll guess that a shorter alkyl chain on the carbonly side will give the higher bp. Not bets.

Honestly: This is a homework question, right? And you're too lazy to figure it out yourself?

Not at all. I tabulated the different boiling points for groups with the same number of carbons and found that, for some, the longer alkyl groups had higher boiling points and vice versa for other compounds. That's what got me thinking... @Karl

Why didn't you say so? Please put that table into your question!

In this case you have consider the degree of branching of the isomer, this influences the strength of intermolecular forces. Think of how tree logs with lots of branches sticking out cannot stack together with each other as easily as a pile of logs with no branches. Branched chain isomers have less contact and therefore weaker intermolecular forces and lower boiling point and straight chained isomers have better contact and thus stronger molecular forces and higher boiling point.

"In this case .." ?!? This is a very general comment, which is true, applicable to the problem, but does not address the original question.

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Reaction between chlorobenzene and sodium methoxide to produce anisole
$$\ce{C6H5Cl + NaOCH3 -> C6H5OCH3 + NaCl}$$
So this reaction undergoes $\ce{S_N2}$ mechanism, but $\ce{C6H5Cl}$ is not likely to undergo such mechanism, so I doubt if anisole would be formed or not?

Aryl halides are much less reactive towards nucleophilic substitution reactions than alkyl halides . So no reaction will occur

Chlorobenzene is inert towards most forms of nucleophilic substitution. In particular:

A $\mathrm{S_N1}$ mechanism is not likely here because the carbon is $\mathrm{sp^2}$ hybridised, the $\ce{C-Cl}$ bond is strong, and the dissociation of $\ce{Cl-}$ is therefore not very likely.
$\mathrm{S_N2}$ substitution doesn't occur because 1) the $\ce{C-Cl}$ bond is strong 2) the $\sigma^*$ orbital is in the middle of the ring and is inaccessible to a nucleophile.
The $\mathrm{S_NAr}$ mechanism could be followed if there is an electron-withdrawing group ortho or para to the chlorine atom, but in chlorobenzene, there is no such EWG.

I would like to suggest the possibility of a benzyne mechanism, i.e. an elimination-addition mechanism. The elimination step follows an $\mathrm{E1_{cb}}$ mechanism:

The benzyne intermediate is very electrophilic and nucleophilic attack by methoxide ion proceeds rapidly to give anisole.

Although this reaction is well-documented with both $\ce{NaOH}$ and $\ce{NaNH2}$, I can't find any real-life examples with $\ce{NaOMe}$. Most examples of chlorobenzene → anisole reactions involve some kind of catalyst. However, if it were to happen, it would probably occur via this parthway.

Aryl Halides are less reactive than alkyl halides and Nucleophilic aromatic substitution on aryl halides only possible if strong electron drawing group is present at ortho and para position to chloride.
Due to their low reactivity toward nucliophile, the reaction usually occur at higher temperature. For example  1-chloro-4-nitrobenzene the reaction occur at high temperature.,

Nucleophilic substitution reaction probably occur through the slow addition of nucleophile to the ring, and second step fast removal of Cl group.

Does the diagram of 4-nitrochlorobenzene contain a typo?

Without an EWG ortho or para it's probably not going to proceed via SNAr. More likely benzyne in this case.

There is still a typo in the diagram. The middle structure should have a $\ce{C=N}$ double bond and two $\ce{-O^-}$ on the nitrogen.

The C---Cl bond of chlorobenze acquires partial double bond character due to the +M effect of Cl-atom. As a result bond energy of C---Cl bond becomes high which leads to low reactivity of chlorobenze towards nucleophilic substitution. So chlorobenze doesn't react with sodium methoxide to form anisole at ordinary reaction conditions. To prepare anisole by Williamson's synthesis the appropriate choice of reagents is Sodium phenoxide and chloromethane. This reaction follows SN2 mechanism because of least steric hindrance around the partial postive charged Cl-atom of chloromethane.!

The C---Cl bond of chlorobenze acquires partial double bond character due to the +M effect of Cl-atom. As a result bond energy of C---Cl bond becomes high which leads to low reactivity of chlorobenze towards nucleophilic substitution. So chlorobenze doesn't react with sodium methoxide to form anisole at ordinary reaction conditions. To prepare anisole by Williamson's synthesis the appropriate choice of reagents is Sodium phenoxide and chloromethane. This reaction follows SN2 mechanism because of least steric hindrance around the partial postive charged Cl-atom of chloromethane.

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Are all bond-forming processes exothermic?
Are all bond-forming processes exothermic?
If so, why don't helium form bonds with neon?

Related: Is Bond Formation “Strictly” Exothermic?

Downvoter: how could I make this question better?

Yes they are, but
(1) a reaction only occurs spontaneously if it is exergonic at the given temperature, and
(2) it might not be the kind of bond you wanted. He and Ne like it much better with a van-der-Waals bond between them, and
(3) creating the spiecies that can bond can take a large amount of energy. Vaporising a solid, splitting diatomic gases, forming some unheard of He+Ne radicals, etc., and
(4) if the bond is endothermic, it simply won't form, because it is not a bound state.
Point 2-4 are basically redundand, just different ways to look at it.

why don't helium form bonds with neon?

Because any bond you could imagine, except a vdW-Bond, would be endothermic, i.e. not a bond. https://en.wikipedia.org/wiki/Noble_gas

So there are indeed endothermic bonds.

Those are not bonds, because they do not bind at all!

So you're saying that if I used magic to form this molecule: $\ce{He-N}$ then the single bond would not be a bond?

There is no such thing as magic. EoD.

I meant, if this molecule existed, the single bond would not be a bond?

"the single bond would not be a bond?" doesn't make any sense. Once you move away from covalent bonding, there is almost never such a thing as a single bond, double bond, triple bond, etc.

i think that helium and neon are both noble gases and hence do not form bonds(because bonds are meant for those atoms who have more electrons or less electrons).yes bond forming is exothermic(courtesy https://answers.yahoo.com/question/index?qid=20090822024552AAB8qjv).


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Reaction between free radicals and water
Free-radical substitution does not occur in aqueous solution as free radicals react with water (from school textbook). What does the reaction between free radicals, say Br, and water produce?

You might be referring to electrophilic addition like in the video - I am referring to radical substitution. In your first link there is such a reaction but with Cl radicals instead. Was wondering what happens when these radicals react with water.

You're right. I deleted my unhelpful comments.

My best guess would be HBr. Br2 + Water disproportions to HBr and hypobromite. That reaction is sped up by light, so it's probably a radical reaction.

see this:-http://chemistry.stackexchange.com/questions/42696/how-can-bromine-water-be-called-a-solution

Bromine radical reacts with water
$$\ce{Br* + H2O -> HBr + *OH}$$
OH radical reacts with bromine
$$\ce{*OH + Br2 -> BrOH + Br*}$$
to give a new bromine radical, and so on, until the bromine is totally consumed.


text:
Why is benzenediazonium fluoroborate water insoluble and stable at room temperature?

Benzenediazonium fluoroborate is water insoluble and stable at room temperature.

Why is this salt, water insoluble? Also I am told that benzenediazonium salts are stable only at low temperature(<5°C) and decompose at higher temperature, so what is the reason for such exceptional stability of Benzenediazonium fluoroborate?

It's not "exceptionally stable" it is rather unstable.  As you note, it must be stored at low temperature and decomposes rapidly at room temperature. The instability is due to the formation of an extremely stable nitrogen molecule upon decomposition of the diazonium salt.

@ron I am asking why Benzenediazonium fluoroborate is stable at room temperature even though other benzenediazonium salts(like chlorides) are unstable at room temperature.

because the benzenediazonium cation and the fluoroborate anion are approximately of same size that results in its stability.

@NilayGhosh so is benzenediazonium fluoroborate crystalline?

After a bit of research on benzenediazonium fluoroborate, I came up with an (complete?) answer although it is based on some approximations and/or speculations.

Benzenediazonium fluoborate is a versatile organic salt used to create various other organic compounds like by helping to introduce aryl, arylazo, arylhydrazono, or amino groups; forms fluoroarenes upon heating and are building block for heterocycles. They have a much higher stability than the corresponding chlorides and shock-insensitive. Generally decomposes at decomposes at 114-116 °C (Source).
So, what causes it much higher stability? Maybe it is their ion size. The benzenediazonium cation and fluoroborate anion are of same size resulting their stability[citation needed?]. Or maybe it is due to the resonance of benzenediazonium cation.

But when you apply heat to the salt, it is bound to decompose, maybe because of two reasons. One is that exceptionally stable nitrogen is formed on its decomposition (i.e nitrogen is formed from nitrogen cation) and second is the $\ce{N-F}$ bond is long and weak. From the information regarding structure of benzenediazonium fluorobotate, it is evident (Here):

Benzenediazonium tetrafluoroborate, $\ce{C6H5N2+•BF4−}$, crystallizes
  in space group P21/a with unit cell dimensions a = 17.347(2),
  b = 8.396(1), c = 5.685(1) Å, β = 92.14(1)°, Z = 4. The structure was
  solved by direct phasing methods using the program SHELX 76. The
  parameters were refined by full-matrix least-squares to a final
  R = 0.063 for 1346 observed reflections. $\ce{C—N}$ and $\ce{N≡N}$
  bond lengths are 1.415(3) and 1.083(3) Å, respectively, and the bonds
  of the benzene ring vary from 1.371(5) to 1.383(4) Å. There are
  three $\ce{N-F}$ close contacts of ≤ 2.84 Å and the positive charge
  appears to be shared between the nitrogen atoms.

Below is the abstract of the mechanism of benzenediazonium tetrafluoroborate decomposition to give an insight (full pdf behind paywall is here):

The thermolysis of benzenediazonium tetrafluoroborate was studied by
  thermogravimetry in dynamic mode. The decomposition of
  $\ce{[ArNN]+BF4−}$ in the solid state with the formation of
  $\ce{C6H5F, BF3, C6H6}$, and $\ce{N2}$ starts at T>348 K. The speed of
  the thermolysis was estimated gravimetrically and by infrared
  spectroscopy, considering the change of the intensity of the
  absorption band at 1498 cm−1, which corresponds to fluorobenzene. The
  maximal rate of thermolysis observes at the 366.5 K. A kinetic scheme,
  which includes the formation of a neutral complex
  $\ce{[C6H5δ+⋯BF4δ−]}$, is proposed for the thermolysis of
  arenediazonium tetrafluoroborate. The decomposition of the complex
  with the formation of free-radical intermediates explains the chain
  character of the thermolysis.

There is large hydrophobic part in diazonium ion and in $\ce{BF4-}$ ion, size is large enough that hardly any hydration energy would be released to cause the dissolution. Hence the salt is insoluble. (Here)

I'm tempted to downvote. " N−F bond is long and weak" because it's almost not a bond, or rather highly ionic bond; BF4 anion is weakly coordinating and this itself is stabilising factor. It's not prone to losing F- which happens in decomposition.


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Can electrons be present outside orbitals. If yes how does this affect chemical reactions
In my physical chemistry textbook it is written (Or typed) that orbitals are regions where probability of finding electron is high (90%-100%). But since orbitals are regions of probability is it possible for electrons to be present outside orbitals for just a small number of atoms in a reaction if so how does this affect reactions that take place with these small number of atoms

They'd probably return to the orbital in like the next nanosecond.

Misunderstanding in italics: Orbitals are graphically presented as the region where the propability of finding the electron is 90% or 95%, but not 100%.

I think that your textbook is essentially correct. Typically for illustrative purposes orbitals are drawn with a 90% or 95% radius meaning that this is the region where that much of the electron density can be found at any time.
The wavefunction does extend to infinity, but the scaling of the decrease with distance is very severe, typically it is $\approx exp(-Zr/(na_0))$ where $Z$ is the atomic number, $n$ the principal quantum number, $r$ distance from nucleus and $a_0$ the Bohr radius  $\ce{5.29 10^{-11}}$ m. Thus for, say, a 3s H atom orbital at 1 nm the wavefunction has decayed to 0.0018 of its maximum value, the probability of being between $r$ to $r+dr$ is the square of this value, i.e. the chance of being at a larger distance is tiny.
Thus we can assume that molecules can be reasonably described by orbitals with a cut-off value, i.e. atoms are discrete. Recent experiments using atomic force microscopes show that the picture of small discrete atoms is correct, as have numerous x-ray diffraction crystallographic experiments over the last century. Thus although an electron in an atom in its ground state could 'in principle' be anywhere in space the chance that it is any significant distance, i.e. many nanometers from its nucleus is effectively zero. Thus we have well defined molecules, whose shape as I have mentioned can be determined experimentally. This means that molecules exist as identifiable entities!
It also means that most molecules have effectively to collide for a reaction to occur, otherwise electrons in one molecule do not feel the influence of those in the other one sufficiently strongly to react. Thus the number of atoms that you mention might react, when the electron is at a very large separation (compared to average separation), is so small as to have no significant effect on a reaction. (Intermolecular forces do extend for some distance (nm) but their energy is small compared to bond strength and generally do not lead to chemical reaction.)
Two additional point should be mentioned, (a) in metals conduction electrons are not localised on any atom, although core electrons are and (b) in the gas phase highly excited atoms/molecules (called Rydberg atoms/molecules) can be produced by narrow band laser excitation to have electrons in highly excited levels only a few wavenumbers below dissociation. In this case the electron's wavefuntion's mean distance from the nucleus can be microns, ($\ce{10^{-6}}$ m ); as large a bacterium, quite remarkable.

There are more levels of abstraction involved. At first, orbitals are simplification and they are best described as "one electron wavefunction", ie. area of space where one electron can reside (or two, if you think of spatial orbitals, as opposed to spin-orbitals).
Being expressed as a wavefunction, the orbital (and therefore electron residing in it) has nonzero probability in whole space (I mean, whole universe). But the probability decays very fast.
If you plot a probability density, see image below, you can clearly see the behaviour. If, on the other hand, want to plot the 3d surface, which are often shown in textbooks, you have to state some cutoff value. Often it is probability density of 0.9, ie. 90%. By this you obtain the typical pictures of orbitals.

I would deliberately leave your question without answer, as they stem from severe confusion caused by your textbook.

An orbital is not a region of space where the probability of finding a bound electron is high. I suspect that your textbook meant that in the context of its shape an orbital is a region where probability is high. The orbital wavefunction is spatially infinite, and (of course) its square - the probability "density" - is therefore also infinite in spatial extent. In other words, it is a function of the quantum numbers of the system (atom) it is bound to as well as x,y, & z points of space (excluding time since this "picture" is a static one). As a function, it is an obvious contradiction to ask "does the function exist outside the locations where the function exists?". This doesn't really answer the question "does the electron always stay in its orbital?". Well, we know the answer to that is "no". If it did, there would never be chemical reactions! So, we know that an electron can be influenced to leave its orbital. And you may have learned enough about the Heisenberg Uncertainty Principle to expect that there are also spontaneous excursions from the energy of a given orbital, allowing dE to be >0 for a finite time dt. This consideration is not useful in dealing with chemistry using orbital concepts. In other words, transient perturbations from the orbital wave function aren't generally useful when thinking about atoms, molecules, and their reactions.


text:
Finding membrane potential equation using the Nernst equation?
I am very new to electrochemistry and ion selective electrodes. The membrane potential of an ion selective electrode is given by $$E=\frac{RT}{nF}\ln{\frac{c_2}{c_1}}$$
Where R is universal gas constant, T temperature in kelvin, n the number of electrons transfered, F Faraday of electricity, $c_2$ the concentration of internal standard solution and $c_1$ is the concentration of the ion to be calculated.
But according to the Nernst equation
$$E= E^\circ -\frac{RT}{nF}\ln k$$
Can someone explain how to come up with the membrane potential equation using the Nernst equation?

The Nernst Equation is $E = E^\circ - \frac{RT}{nF} \ln{Q_c}$. From that you can derive the approximation you gave. $Q_c$ is the product of the activities of the products over those of the reactants. This can be, for dilute solutions where activity approximates concentration, the typical [B]/[A] for the reaction A→B. So, the only thing you need to "explain" is why E° would be zero. Review the definition of E°.  hint:What is ln(1/1)? See the discussion on concentration cells in:  http://www.chem1.com/acad/webtext/elchem/ec4.html

If the concentration of the solutions is different, then their activities are also different (since concentration is a measure of activity). Why E° value is zero? (please correct me if I am wrong)

$E^\circ$ only depends on the standard states, not on actual states.  It thus doesn't really depend on the activities of anything.  For most ion-sensitive electrode applications, there is no electrochemical cell across the ion-selective membrane.

Also there is a misplaced $nF$ in this answer.  I'll edit to move it, but make sure you agree with this change please.


text:
Which d orbitals are used in sp3d3 hybridisation?
What d orbitals are used in the hybridisation of the molecule $\ce{IF7}$. I know the hybrid state of Iodine is $sp^3d^3$ i.e. 3d orbitals are involved in hybridisation.
Some say they are $d_{(xy)}, d_{(yz)}, d_{(zx)}$ and some say $d_{(x^2-y^2)}, d_{(z^2)}$ and $d_{xy}$ are involved? Also what is the theoretical reason for it. Like I know for $sp^3d^2$ hybridisation only $d_{x2-y2}$ and $d_{z^2}$ are involved because they lie along x,y,z axes and so do the hybrid orbitals like in $\ce{SF6}$ each angle is π/2. Is there some theoretical explanation for $\ce{IF7}$ also which has pentagonal bipyramidal shape?
I know a very similar question has already been asked; How can the Bonding in IF7 be explained using LCAO method?
But it says that bonding in $\ce{IF7}$ cannot be predicted as bonds are predominantly ionic plus the answers mentioned there are using very complicated principles and theories which I am unaware of. I only know the basic theories of VSEPR, Valence Bond and the  hypothetical concept of hybridisation plus very little Molecular Orbital Theory. May someone explain the bonding in $\ce{IF7}$ in a fairly simple manner especially the d orbitals used in its hybridisation. I am not specifically asking for $\ce{IF7}$ but any molecule showing $sp^3d^3$ hybridisation?

What kind of egg is used to paint a house?

Indeed, if you somehow didn't noticed, earlier question asked the same and short answer is "They aren't really used to any significant extent".

Your question is based on false premise that hybridisation describes this molecule accurately, but it doesn't.

I know that but that's true if you are studying very advanced chemistry but I am asking according to my level?

@mithoron Like I said in my question I am not asking specifically for IF7. I do not want to know the exact bonding situation in IF7 molecule I am just asking about the specific kind of hybridisation sp3d3 the d orbitals involved irrespective of the fact that hybridisation descrobes the bonding of this molecule or not.

related http://chemistry.stackexchange.com/a/31606/9961


text:
Dehydration of aldoximes
What is the mechanism for dehydration of aldoximes by acetic anhydride?
My attempt:
I doubt if it's correct.

What do you think the mechanism is? This is a homework question, you should show some effort so we know we aren't just doing your homework for you. You can [edit] your post to include your thoughts/attempt.

[tag:homework] is not restricted to questions coming from actual assignments from professors/teachers. "This tag may be applied to questions which are not homework but are similar to a homework problem. "

And you should show your effort nevertheless.

Let us continue this discussion in chat.

You should never just write C and want it to mean $\ce{CH3}$. Do not do this ever. At the very least, write three bonds not connected to anything to symbolise left-out hydrogens. (You may want to define what ‘unconnected bonds’ mean, depending on where you are.) 2) One of the carbons in your second structure has five bonds. That will not happen unless you are discussing non-classical cations. 3) Other than that, you’re pretty close.

@Jan will in second step proton travel to that much far oxygen?

@JM97 Understand that each and every proton transfer is essentially taking and replacing a proton from the large stock of protons available in solution. Hardly ever is the same proton replaced in a mechanism.

@ okay got it, is this kind of reaction really concerted?

My proposed mechanism is a nucleophilic addition/elimination mechanism without a tautomerism step.
It is all concerted.
The first step, addition, features the oxygen in the (E)-propionaldoxime acting as a nucleophile to attack one of the carbonyl carbons in the acetic anhydride, which pushes electrons the carbonyl oxygen, making it negatively charged.
Then, the negatively charged oxygen strikes back, initiating a chain of concerted cyclic electron rearrangement which cleaves quite a few bonds, resulting in three molecules, completing the elimination step.


text:
Properties of steel and aluminum alloys
I have been trying to compare 304 stainless steel with 7075 aluminum for some personal research. This steel alloy has a specific heat capacity of 500 J/kg-C, while for aluminum the value is 960 J/kg-C. Does this mean it is harder to heat up aluminum than steel? Then, steel has a thermal conductivity of 130 W/m-K, while for aluminum the value is 16.2 W/m-K. Does this mean that steel conducts heat away better than aluminum? Thank you.

Did you mix up the values for the thermal conductivity of type 304 stainless steel and the aluminium alloy?

Heat capacity (specific heat) varies inversely with atomic mass, the Dulong–Petit law. Al is about 27 amu, and Fe is about 56 amu, so as you noted, it would be expected that aluminum stores more heat. An extreme example is lead solder, which has such low specific heat that a calloused plumber's hand can wipe a solder joint with little discomfort.
As for thermal conductivity, that property is associated with the rigidity of crystal lattices, where sound travels as phonons, as in diamond, and with electrical conductivity, where "freely moving valence electrons transfer not only electric current but also heat energy." Aluminum and copper are among the best metallic conductors, so are used for cooking-pan bottoms to spread heat evenly. Since stainless steel has many additions to Fe, such as Ni and Cr, these inclusions provide discontinuities at grain boundaries that further impede heat transfer. Dewar flasks are made of stainless steel or glass, rather than Al, because they are poor conductors of heat.


text:
Which of the following would be a strong electrolyte in solution?
Which of the following would be a strong electrolyte in solution?
KCL
Al(OH)3
PbI2


text:
Mechanism of SN1 reaction
Let us say that we have water reacting with t-butyl chloride. When the chlorine atom departs we have a t-butyl cation which becomes an attractive substrate to the nucleophilic oxygen in water. My question is in the reaction that follows, does the hydrogen break off from water first allowing oxygen to attack the carbocation or does water immediately form a bond with the carbocation and then deprotinates the extra hydrogen?

I think that OH which is already dissociated( because water inherently dissociates to give H and OH so that pH is 7) will replace  Cl , since OH concentration is now decreased then according to le chatliers principle more water will dissociative to give more H and OH and this goes on eventually the mixture becomes acidic.

First one molecule of water attacks the carbocation, then it loses a proton. Check SN1 mechanisms for more information.

@JM97 no, the concentration of free OH- in water is so small, it is much more likely that H2O attacks. This can be confirmed by practically any textbook that describes SN1/SN2 mechanisms.

Surely there are both $\ce{H2O}$ and $\ce{OH-}$ which attacks the carbocation, as well as, heaven forbid, $\ce{H3O+}$ (and countless other species formed by the auto-ioniziation of water).
Usually textbooks record $\ce{H2O}$ as the attacker because it is much more abundant in the mixture.
After that, usually another $\ce{H2O}$ comes and take away the extra $\ce{H}$ with itself converting to $\ce{H3O+}$, but as I mentioned above, this step could also be done by $\ce{OH-}$ among many more species, but the concentration of those other species are too small that they are not worth mentioning.
PS: the last step can also be done by the product, i.e. t-butyl alcohol.


text:
Propan-1-ol Heats Objects Faster than Propan-2-ol. Is this an anomaly?
After conducting a practical revolving around burning alcohols, I found that propan-1-ol heated the 50 mL of water to 60 °C about 1 min faster than propan-2-ol.
The experiment was controlled as much as possible, including the temperature escaping the setup.
I know that propan-2-ol is more volatile than propan-1-ol, so I have a reason to believe that this is tied to the slower heating. On the other hand, I have yet to find any sources to verify this.
If I do not find any more about this, I will have to call this out as an anomaly.
If anyone knows anything about the relationship between the position of the $\ce{OH}$ bond in an alcohol and the heat output/speed of heating, please explain this to me.

1 min faster than what? 1 min vs 2 min is quite different from 100 min vs 101 min.

You are aware of the difference in combustion enthalpy? -2021 kJ/mol compared to -2006 kJ/mol?

The enthalpy of combustion $\Delta_\mathrm cH$ of a substance can be calculated from the enthalpy of formation $\Delta_\mathrm fH$. For a compound containing only carbon, hydrogen, and oxygen, the general combustion reaction is
$$\ce{C_$a$H_$b$O_$c$ + ($a$ + 1/4 $b$ – 1/2 $c$) O2 -> $a$ CO2(g) + 1/2 $b$ H2O(l)}$$
The corresponding standard enthalpy of combustion is
$$\Delta_\mathrm cH^\circ=-a\Delta_\mathrm fH^\circ(\ce{CO2,g})-\frac12b\Delta_\mathrm fH^\circ(\ce{H2O,l})+\Delta_\mathrm fH^\circ(\ce{C_$a$H_$b$O_$c$})$$
This equation applies if the reactants start in their standard states and the products return to the same conditions. The standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$). The definition of standard state makes no reference to fixed temperature; thus, the standard enthalpy of formation and the standard enthalpy of combustion remain functions of temperature. The most widely used reference temperature is $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$.
The following values for the standard molar enthalpy of formation at $p=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ are taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
$$\begin{align}
\Delta_\mathrm fH^\circ(\text{propan-1-ol},\ \mathrm l)&=-302.6\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\text{propan-2-ol},\ \mathrm l)&=-318.1\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{H2O, l})&=-285.8\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{CO2, g})&=-393.5\ \mathrm{kJ\ mol^{-1}}
\end{align}$$
During your experiment, the products certainly did not completely return to the initial temperature. Anyway, for the difference in enthalpy of combustion of propan-1-ol and propan-2-ol, the exact state of the products is not relevant (provided that the conditions are reproduced) since the produced amounts of $\ce{CO2}$ and $\ce{H2O}$ are identical. The remaining difference is caused by the difference in enthalpy of formation of propan-1-ol and propan-2-ol.


text:
Are there common diluents for oils that will not cause an emulsion when added to water
If I combine a viscous oil into solution in a carrier in a ratio of between say 1:10 or 1:100 to make it low viscosity but not separate into layers.
I then place a few drops onto a larger quantity of water (think indoor water feature) to spread out as a thin film. Boiled water with a gas barrier may be less likely to start going green.
My thoughts are 100% isopropyl alcohol, 100% acetone or 95% denatured ethyl alcohol but not sure if they will cause my oil to go in to emulsion rather than float on top.

Is there a carrier that will dissolve into the water but leave
the very thin oil on the surface?
Do I need to use a non-polar hydrocarbon that can evaporate instead of
a polar solvent that would move into the water?

The carrier should be safe enough to handle and any residue non toxic to your lawn or kitchen drain when disposing of the water later and any vapours not highly flammable or toxic to inhale. It is not intended for consumption.
EDIT:
Addition to clarify for comments.
I apologize that am not able to give the specific application at this point.
The water not going green was an attempt at illustrating a possible application (I did not realise it might not even work), another similar application would be the layer of paraffin oil (iirc) that was sprayed on stagnant ponds to control mosquito larva (though mechanical sealing of their breathing tubes iirc), the sprayers were used as an aid to dispersal but I don't want a sprayer, the right carrier for me might even work for this application though no my goal.
The aim is to have some way to dispense a thin layer of oil on to the surface of still water.  The final layer will be very thin and the amount of remaining oil would be less than the dispensing amount.
The oils will be of organic origin, possibly semi viscous.
The carrier is intended to get out of the way in a few minutes.
The carrier is to ease dosing and assist with spreading otherwise I could just mix it with lots of oil have thick film and large doses, not ideal.
I am indifferent of whether the carrier goes into the water or evaporates as long as it is not horrible stuff.
I will avoid noxious, toxic, explosive and greenhouse type chemicals, we are talking about drops and no likely ignition sources so flammability is not a big concern.
Part of this problem is to avoid having to require shaking, mixing or other special handling or storage of the dispensing mixture.
I am worried that if I use water as a carrier with an emulsifier to prevent separation, on application the oil will not float on the surface but will disperse into the water pool.
Additional problem is that the vapour pressure cannot be so high that it needs to be held under pressure, this is the other extreme where I could just use an aerosol can with propane but that is really far from ideal.
Pentane may be in the right direction, I will look further, are there consumer applications that use pentanes?
I wonder what perfumes would do if dropped onto water, will the carrier drag oily scents into the water or let them float (note to self must run a few tests)? What are the carriers, alcohol (and read some labels)?

Oil is still soluble in water, so I'd guess that acetone or isopropyl alcohol could drag the oil into water and leave it there. So I'd prefer pentane or other alkane. Almost all organic solvents are flammable. The only non-toxic and non-flammable are CFC's, which should be avoided.

What type of oil are you referring to?  Mineral oil?  Or vegetable oil?  Also, what do you mean by "boiled water with a gas barrier may be less likely to start going green"?  Air is more soluble in most any oil than it is in water.It's also unclear what your final goal is.  Do you want a thin film of oil on top of water?  If so, why do you need a diluent at all?  Or do you want to dissolve the oil evenly and homogeneously in a water phase?

I have the same question as @CurtF., if the goal is to create a thin film of the oil on top of the water, why must you dilute it first in a carrier? Can't you just add it to the top of the water?

@JM For ease of dispensing, I wish to have only a little of the oil remain on the water, it is slightly viscous and will not be easy to dose in small amount from a practical point or be sure to cover if one were to spray it on neat.

@ssavec If you can think of a oil in pentane product that I could get off the shelf to test (or a convenient consumer source of pentane or other likely alkane in Finland) and write it up in an answer I would accept it.

The blowing agents for polystyrene foam are frequently mixtures of pentanes and are readily available in the US, I don't know about Finland.


text:
determine at which v and J the rotational and vibrational energy differences are equal
I was presented with this question and I started working on it but I ran into a problem.
We've got $\Delta E_{rot} = 2B_e(J+1) $  and $\Delta E_{vib} = \hbar \omega$
We get to the point that we've got  $J= \hbar\omega/(2B_e)-1$ but we know that J should be a lot larger than 1 so we don't quite know what we've done wrong. Can anyone help us?
We are given that $\omega _e =4400$ and $B_e = 60.9$

What are the units of $\omega_e$ and $B_e$? I assume they are given in cm$^{-1}$.  If so, your formulas (in J) should read $\Delta E_\text{rot}=2hcB_e(J+1)$ and $\Delta E_\text{vib}=hc\omega_e$.

Aah, We missed the speed of light in the vibrational energy. We had hc in the rotational energy formula but removed it. Thanks.


text:
Predicting product of amine elimination

From section 'Nitrogen Containing Compounds'

I spent quite a lot of time here but could not figure out the answer. It's not mentioned in my school text book either. The solution given in my book is:

I'd like to know:

Is the answer provided (c) correct?
How do I know that heating that compound in the presence of a base results in the formation of that product? We've got plenty of cases in our textbook (Tollen's Test, Fehling's Test and the aldol reaction, to name a few) where we heat organic compounds in the presence of a base, but none of them go by what happened in this question.

Which book is it by the way?

In chemistry, you don't solve things like that; you just recall the right class of reactions (Hofmann elimination in this case) and apply that knowledge. You can hardly expect to rediscover that from the first principles. Knowledge about other classes of organic compounds and their behavior in the presence of a base is not particularly helpful either. It is not until much later that those seemingly random pieces of data start to fit into a bigger picture.
As for the mechanism, it is all there in your second picture. Indeed, that $\ce{OH-}$ would readily snatch a proton from an ammonium cation $\ce{NH4+}$, only there is no ammonium, it is a substituted ammonium, and we don't have any H linked directly to N. So we go for the next best thing and grab a proton from some carbon in $\beta$ position, simultaneously with the formation of a double bond.
Note the reversed regiochemical preferences as compared to Zaitsev-type elimination, which has a different mechanism, usually involving the appearance of a carbocation. (Not that it matters much in your case, since we only have one option anyway.)

The presence of larger leaving groups like $\ce{–NR3+}$ result in Hoffmann product than smaller leaving group like halogens. This is because the positive charge on leaving group increase the basicity of β-protons, and in the presence of bulky base like potassium tert-butoxide the Hoffmann product is the major compared to Zaitsav product. Also, if the basicity of base was increased without increasing the bulkiness of base, in that case Hoffmann product is also a major product.
This is complex situation which theory tells us, but the actual situation will be known when someone conducts a particular experiment.

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


text:
Atomic Orbitals
I read in a textbook that after an electron jumps to a higher energy level when supplied with high temperatures or current, they again fall back to their respective energy level by losing energy and in doing so, release photons.
First question, how does this happen with details?
I had this thought:
Consider the lithium atom whose outermost subshell is the 2s. Thus atom is supplied with hundreds of degrees of temperature. The valence electron is bound to get excited. It's suppoed to jump to a higher energy level. How can it when lithium only has 2 energy levels? Furthermore, do all possible orbitals preexist in every atom in the universe? Like does lithium also have the 4p subshell but it's simply empty?

Lithium (just like any other atom) has infinitely many orbitals and energy levels, most of them empty. Whether they exist when empty is a philosophical question. On the other hand, in what sense do they exist even when filled? An orbital is but a mathematical abstraction, after all.

Who said lithium only has 2 energy levels?

Have a look at the wikipedia  or Hyperphysics web pages on the hydrogen atom spectrum. These give the basic ideas about energy levels which can then be extended to hydrogenic atoms (with one outer electron) such as lithium.

Be gentle with me, I am going to try to explain orbitals in a very simple way using many similes. I do this with primary school kids so many of the words I will use aren't literally accurate but I hope the overall picture is as accurate as possible (without knowing quantum math). Please fell free to make suggestions where the overall picture deviates from "practical" reality. Please, don't bother to comment about the semantics of individually problematic words.
So:
The first thing is, don't think of electrons as particles. That may work well for certain models or math with specific purposes, but in general it actually makes them harder to understand.  Second, electrons don't orbit in orbitals. That's just an unfortunate name leftover from old theories. Finally, electrons will basically be as "lazy" as they are allowed to be. They will endeavor to come to rest and "spread" their "charge" in a position and geometry around the nucleus that is as close to the nucleus as possible, but as far from any other electrons as possible. With one exception:
If given the chance, an electron will team up with another electron in their own orbital to hold the "space" with half the individual "effort". They can pull this one-time trick off by adopting complimentary wave functions thereby staying out of each other's way. This lets each of the two electrons shed some energy individually, but only two can team up this way in any given orbital space.
If there is just one electron in an atom, it is easy for it to come to rest: it "spreads" into an equidistant spherical layer (or crust) around the nucleus. It won't get any closer because its "charge" can't be "compressed" into any smaller of a space without more of a "push", which requires more energy, not less. As more electrons are added (or rather, attracted by a bigger nucleus with more "pull"), they all jockey to find that sweet-spot geometry to spread themselves into where they are as close to the nucleus as possible while being as far from each other as possible. This creates more and more complex geometries as more electrons are added.
These orbitals don't really "exist" when they aren't filled, they are simply the predictable places electrons can possibly come to rest in this rule-based system, depending on the circumstances. Sort of like Lego bricks: You can only attach them to each other in places where the studs and holes line up, there is no in-between position.
So to get to your specific question, if you add energy to the lone electron in the 2s orbital in lithium, it will no longer be able to "fit" itself in that space, and it will jump up to another, "bigger" orbital that can contain its full "charge". It isn't that this new orbital was already there, it was simply the predictable next-best resting place.
But it won't stay there for long. Electrons constantly seek to find the lowest energy place to hang. As soon as you stop feeding it energy it will choose to shed its excess energy and get as close to the nucleus as possible again, falling back down to its original resting place. Interestingly, that energy will come out as light. This is how neon lights and the aurora borealis work.
Further:
This is also why certain atoms create certain molecules so well: if one electron is trying to fill up its orbital all by itself, it will gladly team up with another electron that will help it do the job with less energy. By overlapping two orbitals from different atoms, both containing electrons going it alone, each electron can actually drain off a little energy and be more lazy. It's like if two balloons found a way to overlap and share a little of their air: by overlapping they could drain off the volume of air equivalent to the volume of the overlap because it would be redundant. BUT, pretend those balloons have to be full, no matter what. So if you want to pull them apart again you must add that air back so they can both be full on their own again. That is why you need to add energy to break molecules apart. Those shared electrons need to step up and start holding their orbitals with no outside help again.
One of the coolest things is, bringing in another atom with its own electrons completely disturbs the geometries of resting places of all the outermost electrons in the first atom, so they have to re-jockey to find a new "equilibrium" to accommodate the "disturbance". This creates totally new geometries. These are called hybridized orbitals.

there are an infinite number of atomic orbitals for each atom. Because most of them are unoccupied, they are usually not considered. When electrons gain energy, they are able to move to an orbital with a higher energy by absorbing energy. Scientists use results called emission spectra and absorption spectra in order to understand from which energy shell to which shell the electron jumping has occurred.
You also asked about preexistence. Well yes they technically do preexist. Because the results of these experiments involving the spectra mentioned above point to the fact that there are only specific energy promotions. And this can be used to calculate the energy of the orbitals. This is only for an isolated atom.
If you put many atoms together, they form what is known as an energy band. Here, you have many many more energy jumps, so they are just grouped together into a band.
Usually heating isn't the method used for such. Heating is used for ionisation energy. This is energy required to remove electrons completely from the atom.


text:
Mechanism of action for Citric Acid as a cleaning agent for limescale?
This may be a silly question, but I am not that certain I understand this process because of the low solubility of Calcium Citrate.
How does citric acid actually clean limescale? How much limescale can be cleaned (dissolved in water) in a container with 1 liter of water and limescale under room temperature?
Thanks!

Limescale is essentially calcium carbonate. This question might be helpful.

Citric acid has a low first pKa (3.13) to protonate limescale (mainly $CaCO_3$) and it can chelate anions like $Ca^{2+}$. Both will help removing limescale.
You describe calcium citrate as poorly soluble in water, but the figures show it is much more soluble than calcium carbonate: 0.95 g/L vs 0.013 g/L at 25°C (so 73 times more soluble).

@Martin-マーチン The question you describe deals with HCl which will produce Calcium Chloride which is highly soluble in water. My question is different because Calcium Citrate is not very soluble in water.

I would say it's soluble enough to the extend that colloidal particles will be washed away.


text:
Most useful reagent for gravimetric analysis of silver
My question reads as follows:
"What's the most useful reagent for the gravimetric analysis of silver from the options below:
A. Sodium nitrate
B. Barium carbonate
C. Potassium sulphate
D. Ammonium chloride"
I can rule out A and C as they do not form a silver precipitate with a solution containing silver. However, both B and D can form a silver precipitate so I am not sure which of them would be the most useful.

You are missing something: before you even put the silver in, NH4Cl is soluble, while BaCO3 has already precipitated (insoluble)!
This means that Cl- will precipitate more completely with NH4Cl.  Therefore, D.

Don't know how I didn't see that. Thanks anyway.


text:
MO diagram of Zundel cation
Here is the structure of the Zundel cation (courtesy Wikipedia):

The outermost shell electron count is $16$.
May I have the MO diagram of this cation?
This seems very complicated so I don't know where to start.
Thanks.

Strictly speaking, drawing a MO diagram is only possible for two entities.
That being said, I will attempt to show you the MOs and order them according to their energy.
At PBE0-D3/def2-TZVPD, using NWChem 6.6, I found the following minimum geometry (vacuum):

Figure 1: Minimum geometry found for Zundel cation ($\ce{[H2O...H...OH2]^{+1}}$) with natural atomic charges, angles (in degrees) and distances (in angstrom)
MOs 1 and 2 correspond to $\ce{1s^2}$ core electrons on oxygen and orbitals 3-10 give us the states for the 8 remaining valence electron pairs.
MO 11 is the lowest unoccupied molecular orbital.

Figure 2: Molecular orbital diagram for structure from Figure 1 at PBE0-D3/def2-TZVPD
Since the structure has no symmetry, no symmetry was assigned to any MO and the diagram above has simple labels only.
Energies (in hartree) can be found further below.

Below are images of each MO:

Figure 3: MO #3 (cutoff 0.04)

Figure 4: MO #4 (cutoff 0.04)

Figure 5: MO #5 (cutoff 0.04)

Figure 6: MO #6 (cutoff 0.04)

Figure 7: MO #7 (cutoff 0.04)

Figure 8: MO #8 (cutoff 0.04)

Figure 9: MO #9 (cutoff 0.04)

Figure 10: MO #10 (cutoff 0.04)

Figure 11: MO #11 (cutoff 0.4)

As a side note, we may obtain a finer view of the MOs by representing them as linear combinations of natural bonding orbitals (NBOs):
CMO: NBO Analysis of Canonical Molecular Orbitals

Leading (> 5%) NBO Contributions to Molecular Orbitals
======================================================
 MO   3 (occ): orbital energy =  -1.356442 a.u.
               0.426*[  5]: BD ( 1) O 4- H 7
               0.400*[  1]: BD ( 1) O 1- H 2
              -0.399*[  2]: BD ( 1) O 1- H 3
               0.396*[  4]: BD ( 1) O 4- H 6
               0.396*[  3]: BD ( 1) O 4- H 5
               0.336*[  9]: LP ( 2) O 1(lp)
 MO   4 (occ): orbital energy =  -1.327547 a.u.
               0.444*[  1]: BD ( 1) O 1- H 2
              -0.440*[  2]: BD ( 1) O 1- H 3
              -0.437*[  3]: BD ( 1) O 4- H 5
              -0.434*[  4]: BD ( 1) O 4- H 6
               0.280*[  9]: LP ( 2) O 1(lp)
              -0.272*[  5]: BD ( 1) O 4- H 7
 MO   5 (occ): orbital energy =  -0.859519 a.u.
              -0.606*[  5]: BD ( 1) O 4- H 7
              -0.502*[  9]: LP ( 2) O 1(lp)
               0.369*[  3]: BD ( 1) O 4- H 5
               0.359*[  1]: BD ( 1) O 1- H 2
 MO   6 (occ): orbital energy =  -0.839700 a.u.
              -0.562*[  4]: BD ( 1) O 4- H 6
               0.511*[  2]: BD ( 1) O 1- H 3
               0.471*[  3]: BD ( 1) O 4- H 5
               0.425*[  1]: BD ( 1) O 1- H 2
 MO   7 (occ): orbital energy =  -0.838909 a.u.
               0.533*[  2]: BD ( 1) O 1- H 3
               0.514*[  1]: BD ( 1) O 1- H 2
               0.481*[  4]: BD ( 1) O 4- H 6
              -0.469*[  3]: BD ( 1) O 4- H 5
 MO   8 (occ): orbital energy =  -0.696009 a.u.
              -0.549*[  9]: LP ( 2) O 1(lp)
               0.502*[  5]: BD ( 1) O 4- H 7
               0.302*[ 10]: LP ( 1) O 4(lp)
              -0.301*[  8]: LP ( 1) O 1(lp)
              -0.253*[  3]: BD ( 1) O 4- H 5
               0.244*[  1]: BD ( 1) O 1- H 2
              -0.227*[  4]: BD ( 1) O 4- H 6
 MO   9 (occ): orbital energy =  -0.636884 a.u.
               0.692*[ 10]: LP ( 1) O 4(lp)
               0.687*[  8]: LP ( 1) O 1(lp)
 MO  10 (occ): orbital energy =  -0.623778 a.u.
              -0.620*[  8]: LP ( 1) O 1(lp)
               0.614*[ 10]: LP ( 1) O 4(lp)
               0.344*[  9]: LP ( 2) O 1(lp)
              -0.319*[  5]: BD ( 1) O 4- H 7
 MO  11 (vir): orbital energy =  -0.201773 a.u.
               0.263*[121]: BD*( 1) O 1- H 2*
               0.262*[123]: BD*( 1) O 4- H 5*
              -0.257*[122]: BD*( 1) O 1- H 3*
               0.257*[124]: BD*( 1) O 4- H 6*
              -0.232*[113]: RY*( 1) H 7(ry*)

Figure 12: Atomic numbering for the NBO Analysis of Canonical Molecular Orbitals above


text:
M/P nomenclature for axial-chirality

Synthesis and Chiral Resolution of C3-Symmetric Tribenzotriquinacenes, Eur. J. Org. Chem. 2016, Early View article
DOI: 10.1002/ejoc.201600890

I came across this graphical abstract with M and P overlayed. After reading the article it became clear that M and P were being used as stereochemical descriptors, presumably of some kind of axial chirality/screw sense.
When I've previously looked, I've often seen either (+)/(-) (where + is the right-handed screw and - is the left) since this is actually describing the physical property of the molecule as measured, in the same way that (+)-proline describes the way in which the molecule interacts with plane polarised light. In many cases (R) and (S) are also used to describe axial chirality (BINAP, for example).
A quick google of the issue seems to suggest that (-) --> M and (+) --> P (cf. Wikipedia, confirmed by the IUPAC goldbook), but I was wondering if there is something more systematic going on here (i.e. a (+)-molecule doesn't always have an (R)-configuration).
In my head, the fact that there are descriptors like M and P (cf. E/Z and R/S) suggests that one should be able to look at the molecule and assign these by some kind of priority rule rather than by physical measurement (and by extension, does M/P actually predict +/- in a way that R/S doesn't).

See also helicity in Basic terminology of stereochemistry (IUPAC Recommendations 1996)

A general relationship between the sign of optical rotation and any specific designation of priority-based stereochemical assignment is tenuous at best. There's no reason nature should think about symmetry breaking as naively as just ranking atomic numbers.

I realise this: "A quick google of the issue seems to suggest that (-) --> M and (+) --> P (cf. Wikipedia, confirmed by the IUPAC goldbook), but I was wondering if there is something more systematic going on here (i.e. a (+)-molecule doesn't always have an (R)-configuration).
"

The (+)/(-) notation refers only to the sign of the experimentally measured optical rotation. This notation alone does not tell you anything about which enantiomer you have.
The M/P notation refers to the spatial arrangement of atoms (see https://en.wikipedia.org/wiki/Atropisomer for an explanation).
What is the relationship between (+)/(-) and M/P? NONE.
You cannot say a priori if a certain compound for which you know the spatial arrangement of atoms will give you a positive or negative optical rotation.
Similarly, you cannot deduce the spatial arrangement of atoms from an optical rotation measurement.
There are methods to establish the relationship between the two descriptors, which means assigning the absolute configuration of a chiral molecule.
People have come up with rules to relate the two (e.g. binaphthyls, biphenyls), and these rules often work. More specifically, I am talking about the so-called Exciton Chirality method by Harada and Nakanishi (see DOI: 10.1021/ar50056a001).
The reasons why it works (often) or fails (sometimes) go probably beyond the scope of the question, but if you are interested I can maybe look for some paper explaining the problems.


text:
How to write half equations based on oxidizing/reducing agents pairs?
In most videos, people use oxidation numbers to find the oxidizing agent and the reducing agent, It's a great way to use, but our teacher never used it before he would give us a reaction and ask us directly to write the half reactions and what are the oxidizing and reducing agent on each. Using oxidization numbers I can easily find the oxidizing/reducing agents and write half reaction for each.
However, sometimes we don't have the final reaction, we only have oxidizing/reducing agents and are asked to write half equations. For example:  Given the pair ($\ce{CO2/H2C2O4}$) as an example, should I write the half equation for the $\ce{CO2}$ to turn into $\ce{H2C2O4}$, or the inverse?
Another question is how can I find oxidation numbers on compound elements such as: $\ce{C6H12O6}$?

Firstly, you need to understand that there is a maximum and minimum oxidation number.
For metals, the minimum is usually $0$ while the maximum is usually the number of their outermost electrons. For example, $\ce{Na}$ which has $1$ outermost shell electron would have a minimum of $0$ and a maximum of $1$.
For non-metals, the minimum is usually the number of their outermost shell electrons minus $8$, while the maximum is usually the number of their outermost shell electrons. For example, $\ce{S}$ which has 6 outermost shell electrons would have a minimum of $-2$ and a maximum of $6$.

For the pair $\ce{CO2}$ and $\ce{H2C2O4}$:
Firstly, note that the most electronegative element in each compound is $\ce{O}$. You can learn more about electronegativity here.
Therefore, it takes the minimum, which is $-2$.
Secondly, usually hydrogen is $1$ (in most covalent compounds) or $-1$ (in most ionic compounds). In this case, hydrogen is $1$.
It follows from calculation (that the sum of the electronegativity of each element in a compound must be zero) that the $\ce{C}$ in the first compound has oxidation number $4$ while that in the second compound has oxidation number $3$.
Since the oxidation number has dropped, this is a reduction reaction.

We write the unbalanced equation as a starter:
$$\ce{2CO2 -> H2C2O4}$$
Balancing the number of $\ce{C}$.
Now, we are in an acidic environment, so we add $\ce{H2O}$ to balance the oxygen and $\ce{H+} to balance the hydrogen. The oxygen is already balanced, so:
$$\ce{2H+ + 2CO2 -> H2C2O4}$$
Note that the charges are still not balanced while the elements are balanced, so we add electrons:
$$\ce{2H+ + 2CO2 + 2e- -> H2C2O4}$$
Thus our half equation is complete.

The method to find the oxidation number of each element in $\ce{C6H12O6}$ is detailed above, and I leave this for you as an exercise.

PS: in an alkaline environment, we add $\ce{OH-}$ to both sides of the equation to eliminate the $\ce{H+}$, i.e. (the following is wrong because we are in an acidic environment thanks to the acid $\ce{H2C2O4}$ (acetic acid):
$$\ce{2H+ + 2OH- + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$
$$\ce{2H2O + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$

Great answer, but what I can't get into my mind is that why we wrote a half reaction for CO2 to turn to C2H2O4 instead of writing a half reaction  to turn the C2H2O4 to CO2 ? My basic question is that how do we know which element turn to which other element given the agent pair (ox/red) ?

@AnisSouames You can also write a half reaction to turn $\ce{C2H2O4}$ into $\ce{CO2}$:
$$\ce{C2H2O4 -> 2H+ + 2CO2 + 2e-}$$
So when to use which equation? Well, when you are converting $\ce{CO2}$ to $\ce{C2H2O4}$, use the first equation. If vice versa, use this equation.

@DHMO. I believe when you wrote "It follows from calculation (that the sum of the electronegativity of each element in a compound must be zero)", you meant that the sum of the oxidation numbers ...must be zero. Yes?


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Salt + Sodium Bicarbonate releasing water in carpet
I had some fleas issues so I read about putting salt and sodium bicarbonate in the carpet and in the edges with the purpose to reduce the humidity and dehydrate eggs and larvae. After few days, the weather changed, to humid, rainy and foggy. UK weather... Suddenly, out of nowhere I found all the floors and carpets humid and specially in areas where the mixture was more abundant, there were quite a lot of drops of water.
I wonder why. Probably the hygroscopic property of salt and sodium bicarbonate, in a change in humidity or atmospheric pressure released the water accumulated during several days. I have no idea.

After the salt and the sodium bicarbonate absorb the water, where else do you think should the water go to?

I've never heard of sodium chloride/sodium bicarbonate being an effective dessicant. Pure sodium chloride is not hygroscopic at all.
However, in air sodium bicarbonate slowly decomposes to sodium carbonate which in fact is hygroscopic. If sodium carbonate, and hygroscopic impurities in table salt, such as magnesium chloride or calcium chloride, deliquesce in air of high humidity to give small droplets then these droplets start to grow due to a high concentration of dissolved salts (including sodium chloride and bicarbonate). The effect is reversible at low humidity or higher temperature.


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What chemicals would need to be added to human urea to make it a nutritional substance?
I've been thinking about food supplies, waste management and sustainable living, particularly in highly isolated areas like a Mars colony or a space station, or also in a non-technically advanced situation like urban survival during a major emergency.
One of the major things preventing human urine from being drinkable is the prescence of urea.
An adult supposedly excretes 25 grams of urea per day.
What chemicals (and in what quantities) would be needed to turn this amount of human urea into:

a non-toxic, but non-nutritional ingestible substance?
a nutritional substance?

I am thinking specifically about how to simply add a tablet/capsule to urine and having a nutritionally viable drink within 60 seconds or so.

In a space station you should just distill away the urea.

Ok, what about in a non-technically advanced situation like urban survival during a large scale emergency?

Urea could well be used as a fertilizer to enrich soil with nitrogen, and as such would technically count as a low-tech but long route for turning urea into nutritional substances.

True, but my question is about how to simply add a tablet/capsule to urine and having a nutritionally viable drink within 60 seconds or so.

It won't work as a water supplement: urine is salty and contains other dissolved solids (https://en.wikipedia.org/wiki/Urine). The osmotic pressure would actually remove water from the body. However, urine is not intrinsically poisonous, though it may contain bacteria and viruses that should be killed (e.g. by heating).

@DrMoishePippik Ok, thanks! I've made a new question in response to you comment: http://chemistry.stackexchange.com/questions/58681/what-chemicals-could-be-added-to-human-urine-to-cause-harmful-chemicals-to-preci


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Calculating time needed to coat an object using the process of electrolysis
Suppose we have an object with a surface area of $S=\pu{1.5 dm2}$, and we need to coat it with hard chrome. The thickness of the coat should be $d=\pu{80 \mu m}$. The mixture used for coating is of the following mixture: $\ce{CrO3}$ $\pu{290 g/L}$, $\ce{H2SO4}$ $\pu{2.5 g/L}$. The current density is $J = \pu{50 A dm-2}$. The current utilization factor is $\mu = 15\%$. Using Faraday's law of electrolysis:
$$\frac{m}{M}=\frac{q}{F}$$
and the fact that
$$J_{\mathrm{actual}}=0.15\cdot\frac{q}{S\Delta t}$$ where $S$ represents the surface through which the constant current passes through, I got $\Delta t = \pu{1425 s}$, however the answer is $\Delta t = \pu{2 h 22 min}$.

The trick you're missing here is that you don't account for how many electrons are needed to reduce $\ce{CrO3}$ to $\ce{Cr}$.
Cr has an oxidation state of 6 in $\ce{CrO3}$, so it takes 6 electrons to reduce every chromium atom.
The full form of Faraday's law is $$\frac{m}{M} = \frac{q}{zF}$$  where $z$ is the number of electrons transferred.
Therefore, you need 6 times as much charge as you initially calculated, which works out to 8550 seconds or 2:22:30.


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Finding x and y in Pt(NH3)xCly

Questions 8 and 9 refer to the analysis of a salt whose formula is $\ce{Pt(NH3)_xCl_y}$.

$0.02$ moles [sic!] of this salt required $40.0~\mathrm{cm^3}$ of $2.0~\mathrm{mol\,l^{-1}}$ nitric acid for exact neutralisation.
The number of moles of $\ce{NH3}$ per mole of salt is
A 2
B 4
C 6
D 8
$0.02$ moles [sic!] of the salt were dissolved in nitric acid and excess silver(I) nitrate solution was added. The precipitate formed was filtered, washed and dried. It weighed $5.74~\mathrm{g}$.
The number of moles of chloride per mole of the salt is
A 1
B 2
C 3
D 4

For question 8, I first found the no. of moles nitric acid to be $0.08\ \mathrm{mol}$, which is four times the number of moles of the salt. However, I just can't see how I would get $x$ from this. (The answer to this question is somehow B which makes even less sense since that's the number of moles of nitric acid for each mole of salt.)
For question 9, I tried to write a balanced equation for the formation of the precipitate but I don't think that's possible seeing that we don't know $x$ and $y$. Now, I know that the number of moles of the precipitate (silver chloride) is $0.04\ \mathrm{mol}$ which is twice than that of the salt. Yet again, I don't know how I would get $y$ from this. (The answer to this is also B, which is also the number of moles of precipitate for each mole of salt.)
 It would be appreciated if someone could explain how one can obtain the values lf $x$ and $y$ here.

You're almost there!
For 8:  For every mole of acid added, it reacts with a mole of the base included in the Pt(NH3)xCly, which is the ammonia.  So you can figure out how many moles of ammonia have reacted, and compare this with the moles of salt you started with.
For 9: For every mole of AgCl you have formed, you have used up one mole of Cl from the Pt(NH3)xCly.  So you can figure out how many moles of Cl you have captured, and compare this with the moles of salt you started with.


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Are there metals that react with sulfuric acid but not hydrochloric (or vice versa)?
In school practicals introducing the reactions of metals with acids, it seems to be standard (as in this one from the RSC) to use both hydrochloric and sulfuric acid, but with metals which react to both or neither (e.g. magnesium, zinc and copper). It would be nice to be able to make the point that different things react with different acids, but I'm at a loss for what might fit the bill. Is there anything?

One example is lead. Lead reacts with HCl but not H2SO4.See this:-http://chemistry.stackexchange.com/questions/27954/what-causes-passivity-of-specific-metals-in-conc-nitric-acid/50935#50935

OK, let's clean things up.
Reaction of an acid with a metal is generally a redox reaction. The oxidizing agent are protons generated in the process of dissociation of the acid (let's forget about anions for a sec). The property that decides whether a metallic element can be oxidized by protons is its standard reduction potential. In general, any metal with a negative reduction potential will react with any Brønsted acid (one that produces protons), no matter whether it is a weak (e.g., acetic) or strong acid (e.g., hydrochloric). And if a metal has a positive reduction potential, it won't react with acids. Therefore, if we do not consider other factors, there are no examples to make the point that different acids react differently with a given metal, in fact they all should behave in the same way.
But what makes Chemistry fun are those 'other factors'. The anions produced in the dissociation process of an acid can also act as an oxidizer. For example, nitrate ions produced from nitric acid will oxidize copper, so the metal will dissolve in nitric acid. Sulfate ions are also oxidizing agents, but much weaker than nitrate ions, so copper won't dissolve in diluted sulfuric acid, but a hot and concentrated sulfuric acid will do the trick. And hydrochloric acid in the absence of some other oxidants will not attack copper at all. Another factor to consider is the passivation of a metal. A thin, invisible layer of an oxide or salt can protect a metal from an acid attack. Aluminum, which is a very reactive metal, will not dissolve in oxidizing acids (e.g., nitric), because it will develop a protective layer of aluminum oxide on its surface. Non-oxidizing acids, on the other hand, will easily do the job. Metal salts that are insoluble in water can also create a passivation layer.
Regarding the original question, there are some examples for sulfuric and hydrochloric acid, where they will behave differently due to the formation of insoluble salts. One example is thallium - it dissolves in sulfuric acid but not in hydrochloric acid. An opposite is barium - it dissolves in hydrochloric acid but not in sulfuric acid.

Thanks, good answer! I do actually have a little barium, stored under paraffin, so I could use that, but I'm not sure if it's worth it. Doesn't it react with water anyway? I also found that zinc doesn't react noticeably with acetic acid, unlike magnesium, but I'm not sure I want to deal with the vinegar smell. I guess this is why teachers usually just go with hydrochloric and sulphuric acid with metals where it doesn't make much difference which you use!

@Oolong Unfortunately barium reacts with water quite vigorously. Plus it is toxic (its soluble salts). Thallium is also dangerously toxic. Zinc does react with acetic acid, it all depends on concentration. In general weak acids and diluted acids react slowly. I think you can do a nice experiment where you show how the concentration of an acid changes its reactivity. Or you can go for the classic: show why some metals are called 'noble' because they do not react with 'normal' acids, but react with nitric acid or aqua regia.

I think you will be astounded to learn about aqua regia. Also, have a look at the reactivity series to learn more about the reactivity of various metals. To make your point that different metals react differently, you could reference this reactivity series and show the trend of certain metals and their ions to lose and/or gain electrons.
Following on what @Hanry said, in aqua regia, nitric acid is used to partially dissolve the gold, whereas hydrochloric acid is used to "take advantage" of chloride ions by using them to form a complex with chloride anions to form a complex of tetrachloroaurate(III) anions, thereby allowing nitric acid to continuously dissolve a small amount of gold as the HCl continuously "grabs" that ever-so-small amount from solution, causing it to be captured in the chloride complex. This cycle continues until the gold or either acid has been used up.

Usually when we use these acids, we use them for the H+ (protons) that they will donate to react.  $$\ce{HAcid \ce{->} \ce{H^+} + Acid^- 2x\ce{H^+} + Metal^0 \ce{->} x\ce{H_2} + Metal^{+x}}$$
This is why it seems that different acids react the same way.  However, if you wanted them to react differently, you might try to take advantage of the different non-acid parts ($\ce{SO4^2-}$ and $\ce{Cl-}$).
So for sulfuric acid and hydrochloric acid, $\ce{SrSO4}$ will form, but $\ce{SrCl2}$ will stay in solution. Therefore, sulfuric acid will react with strontium ion, while hydrochloric acid will not. (Note: strontium metal will react with both.)
Edit: I may have misread your question. If you are looking for a metal element, then one metal element that resists reaction would be Au; it is commonly known that it will react with aqua regia (Nitric and Hydrochloric acid), but resists other lone acids. (www.cracksat.net/sat2/chemistry/question-644-answer-and-explanation.html)

$SO_4^{2-}$ not $2+$.


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What are things that impact a mineral's color, clarity, hardness?
I don't know what else to write here because my question is the title. Please help.

You really need to read a book on mineralogy. To answer your question more directly, hardness is a characteristic of the mineral. Color can be indicative of the mineral (i.e azurite or malachite) or it can be due to impurities as in the many colors for quartz minerals. Some minerals never occur in transparent crystals, and those that due may lose clarity due to impurities.

Several things influence the color, clarity, and hardness of materials. Some of them are:

Structure: Diamond vs Graphite. The structure, or how atoms are bonded in a mineral can influence both the color and the hardness of a mineral. Diamond and graphite illustrates this. Although both diamond and graphite are made of carbon, their different structures (how they are bonded) makes them different. Graphite is black and brittle whereas diamond is hard and transparent.
Impurities: Ruby vs Purple Sapphire and Steel. The vast majority of crystals have impurities in them. The impurities inside a material can alter the color of a crystal. Ruby and purple sapphire illustrates this. Ruby and Sapphire are mainly composed of aluminum oxide. However, they contain impurities in them. This is what gives them their different color. Ruby contains Chromium impurities whereas purple sapphire contains vanadium. Impurities can also affect the strength of a crystal. Steel illustrates this. Steel is an alloy composed of iron and carbon. The property of steel changes depending on how much carbon is present in it. Steel with high carbon content tends to make it brittle. However, steel with just the right amount of carbon (around 1%) is stronger.


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Liquid that does not spoil for a mineral aquarium
I want to create a decoration - a mason jar "aquarium". I want to take a large mason jar, put some nice mineral specimens in it, fill the jar with liquid and seal it with lid.
I believe I can't use pure water because there is a chance that it will go bad after some time.
So I want to know if there is some liquid that is not too aggressive (in terms of mineral decay) or dangerous that has indefinite shelf life. I want it to last for years.
I know that vodka basically never goes bad, but it's pretty expensive and the alcohol would probably be too harsh for the minerals in the jar.
Additional information: I am considering water with salt.

You'll want as inert a solution as possible, and vodka contains water. Consider mineral oil: it's cheap, transparent to visible light, is generally non-toxic in the setting you suggest (sealed container), and won't react with many minerals . What minerals are you considering putting in our display?

Vodka would be probably less harsh for the minerals than water with salt. Depends on the minerals, though. Most minerals would withstand both just fine. After all, salt itself is a mineral. Thinking of salt crystals in a saturated brine... well, that may looks nice, and may last forever, unless the lid is made of metal and the brine eats through it.

DO NOT USE WATER WITH SALT. This is just about the worst thing you can do. Some minerals will dissolve and corrode with time in salt solutions. Vodka is not too harsh, but the ethanol is volatile and it has water in it. Mineral oil is better. Air is the best: I wouldn't use any liquids at all.

There are three things to take into consideration here: volatility of the liquid you will be using (yes, the jar will be tightly closed, but some candidates might still evaporate slowly and/or have toxic or foul-smelling vapors), stability, and refraction index. Mineral oil is a great candidate, as Todd Minehardt commented. It's really stable, releases no vapors whatsoever, but its refractive index is a little higher than that of water, which might make your aquarium look funny. There are substances with similar properties that might be useful and also readily available, but with a refractive index closer to water's. Amyl alcohol (fusel oil) or ethylene glycol (car antifreeze) are good candidates and shouldn't be too hard to find. Glycerol and propylene glycol are also pretty good options, but their refractive indexes are closer to mineral oil than the other two.


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How to distinguish first excited state and second excited state?
I am learning excited state calculation. In this paper, the author use "$D_0$","$D_1$" represent excited state of nitrate radical.
I know the symbol such as "$D_1$","$T_2$" means first excited doublet state, second excited triplet state. Can I get specific electron configuration of a molecule with these symbol,like ${}^2\!A_2$ symbol in this answer? Or first and second excited state are different in a range of energy? In other word, Can I write a electron configuration use symbols like "${}^1\!A_2$" for $D_1$ state of nitrate molecule?

Unfortunately, no. If you don't want to work it out by hand, the best you can hope for is that the authors explicitly specify orbital occupations with something that looks like an electron configuration. This paper doesn't, the previous paper sort of does.
Here is a paper that doesn't have symmetry labels, but does have configurations. For example, on page 2977, the SCF reference for $\ce{CuO}$ is given as
$$
\ce{ (O 2s)^2 (O 2p\sigma)^1 (O 2p\pi)^4 (Cu 3d\sigma)^2 (Cu 3d\pi)^4 (Cu 3d\delta)^4 },
$$
where the wavefunction symmetry is determined by the singly-occupied orbital (everything else is doubly occupied). $\ce{2p\sigma}$ probably lies along the bond, which would be taken as the $z$-axis when looking at a character table. Go to the $C_{\infty v}$ table, check for $z$, and discover that it's $\Sigma^+$, which is even mentioned in the paper. So there are symmetry labels given here, but the linear groups also have Greek characters for their irrep labels...
One interesting point is that non-Abelian groups are not used in practical quantum chemistry. $C_{\infty v}$ would be reduced to $C_{2v}$, and the $^{2}\Sigma^+$ state would become $^{2}\!A_{1}$.

The ground state radical belongs to the $\ce{D_{3h}}$ point group and has symmetry species $\ce{^2A^'_2}$, there is an excited state also (at $\ce{\approx 15100 cm^{-1}}$) with the same point group with species $\ce{^2E^'}$. However, to work out species for other excited states you will need to know their point group and which electrons have been excited. If the excited state has the same point group as the ground state you only need to look at the highest orbital and one other orbital (the one the electron has been excited from) as all others are doubly occupied and so have A (totally symmetric) symmetry species. Multiply the characters in the 'irreps' (characters in rows of point group table) together to get the symmetry species of the excited state.

Could you "translate" $D_2$ state to wavefunction symmetry irreps as an example?

No not without more information. For example to get to the ground state symmetry species, you must know which orbitals the odd electron is in , then work out its symmetry species, (left hand column in point group table) by performing each of the symmetry operations (top row  of table) in turn;  character is 1 only if the orbital is indistinguishable after any operation compared to that before operation, (else -1) hence work out symmetry species.


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How can one determine the charge of a polyatomic ion?
I'm stuck on determining the charge on various polyatomic ions according to the rule of charge balance. I keep getting 0 for each of these, but wonder if that's not the case.

$\ce{KMnO4}$
$\ce{K2C2O4}$

Can you please tell me the charge for these, and how you determined this?

Well, for starters, $\ce{KMnO4}$ is not an ion, just like $\ce{NaCl}$ is not an ion. The ions in $\ce{NaCl}$ are $\ce{Na+}$ and $\ce{Cl-}$, so the ions in $\ce{KMnO4}$ are... what?

I understand it's a compound. But what I'm being asked is "given the following formulas for these salts and using the rule of charge balance, determine the charge on the following polyatomic ions": KMnO4

@orthocresol: you seem to be implying that it's just a simple matter of looking at the compound, looking at the ions within the compound, and reporting those back?

@MonicaHeddneck Well, perhaps you don't know what the charge on the $\ce{MnO4}$ ion is. That's fine. However you probably should know the charge on a $\ce{K}$ ion, so maybe that lets you calculate the charge on the $\ce{MnO4}$ ion.

Yes, I've determined that the K has a +1 charge, and the O has a  -8 charge. The K and O must add to equal 0, so the charge on Mn must equal +7. However, I'm being asked for the "charge on the polyatomic ions". Possible answers are A) 0; B) 1- C) 2- D)3-  or E) 4- . I assume 0 can't be the answer for all of these ions (really, ionic compounds):but that's what it would be given the rule of charge balance. I guess I'm not sure what I'm being asked for--

Apparantly, your question is phrased:

Given the following formulae for these salts and using the rule of charge balance, determine the charge on the following polyatomic ions: (examples)

That is a bad wording. Consider this better wording:

Given the following formulae of salts, determine which cations and anions exist and which are polyatomic. For the polyatomic ions only, determine their total charge.

Taking the potassium permanganate example $\ce{KMnO4}$:

You have potassium, manganese and oxygens which could (at first assumption) all exist as part of polyatomic ions.
However, deduce potassium from that above list. It does not form polyatomic ions.
Hence, whichever polyatomic ion may exist, it must counteract potassium’s charge.
Hence, the polyatomic ion is permanganate and displays a $-1$ charge to balance $\ce{K+}$. ($\ce{MnO4-}$)


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How to understand activity with different standard state
There is a description of activity in Wikipedia as below. How shall I understand it? Fox example, I have a solution with pH = 7 ({H+} = 10^-7). The molarity based activity of H+ will be a = 10^-7 (standard state is 1mol/L). However the molar fraction of H+ is 10^-7/55.5 (the 55.5 is the molar of water per liter), which is far away from 10^-7. What is the problem of my understanding? Thanks!
The activity is the same regardless of the ways to express composition and the standard state chosen so the above expressions are equal.

I think that you probably need the mole fraction of $\ce{H^+}$ not water if you standard state is defined via a concentration of $\ce{H^+}$.

@porphyrin Thanks! How shall I correct it? My understanding is the molar fraction of H+ in the solution (H+ and water together) which should be  10^-7/(10^-7+55.5).

If you define the standard state to be pH 7 then the mole fraction of protons is 1, then it changes in a sensible way at pH 6 and 8 etc.