text:
What can this chemical be? (Qualitative Analysis)
I have an unknown white powder, which is an ionic solid, and I need to identify what it is by using qualitative analysis. I've run quite a few tests, but I am having a lot of trouble identifying the anion. The data I've collected is below.
General Properties:

It is a white powder, and I know that it is an ionic solid.
It's density is around $\pu{2g/mL}$ (this is not an exact number)
It appears to be a hydrate (as it loses mass when heated over a bunsen burner)
It is fairly soluble in water. It definitely isn't very insoluble, but I have to add a significant amount of water in order to get it to dissolve.
The powder has a very strong smell (although it's hard to describe).
A solution of the solid is colorless.

Cation Test

A flame test produced a deep orange / orange-red flame

Adding $\ce{Na2CO3}$ to a solution of the solid results in the formation of a white precipitate

Adding a base, such as $\ce{NaOH}$ to a solution of the solid results in the formation of a white precipitate

Based on this data, I am fairly confident that the cation is calcium, but please correct me if there are any other metals that fit the pattern described above or if additional tests are necessary.
Anion Test
This is what has really been messing with me, because I can't seem to get a conclusive test for any anion.

Adding $\ce{AgNO3}$, $\ce{BaCl2}$, or $\ce{Pb(NO3)2}$ to a solution of the solid did not result in the formation of any precipitate, regardless of how much I added.
Adding $\ce{H2SO4}$ results in the formation of a white precipitate (and no bubbles or anything). However, I'm fairly sure that this precipitate is not related to the anion, but rather producing $\ce{CaSO4}$ (from the cation).
Adding $\ce{HCl}$ or $\ce{HNO3}$ to a solution of the solid did not result in any bubbles (neither did adding $\ce{H2O}$ directly to the pure solid)
Adding Ferric Nitrate to the solution just retained the orange color of the ferric nitrate. A dark red color or precipitate was not observed.
Adding $\ce{HCl}$ did not result in the smell of vinegar.
Adding $\ce{AgNO3}$ followed immediately by $\ce{HNO3}$ also did not result in a precipitate.
Adding ammonium molybdate and nitric acid to the solution and heating it caused the solution to turn yellow, but did not yield a precipitate (even after centrifuging). This test was done twice with different ratios of ammonium molybdate to nitric acid, and the exact same result was observed both times.
The brown ring test was conducted, and came back with a negative result. It is possible that this test was done wrong, because the acidified ferrous sulphate was added drop by drop on top of the solution, rather than by rolling it down the side of the test tubes, which may have interfered with test results.

Are there any tests for ions that I have missed? I've done almost every test I know trying to identify the anion, but no dice. The results for the ammonium molybdate test are also confusing to me, because I expected either a yellow precipitate (to confirm phosphate) or no reaction, but the test seems to have gone halfway by just turning yellow. Any guidance on what I should do next would be very much appreciated.

If it were the only reasonably soluble phosphate of calcium, monocalcium phosphate, then the phosphate would have dropped out with the addition of Pb(NO3)2. Regarding solubility, do you know about how much water you had to add to how much solid? Did you ever notice a change in smell or the appearance of an oily layer upon adding acids to the solution? Did you observe any boiling, popping, melting when heated with bunsen burner? Maybe most importantly, did you quantitatively determine the weight loss on heating (masses before and after heating)?

For the solubility, I do not have any quantitative measurements, but the volume of water I have to add is easily 10x the volume of the solid present. I can give more details in a few days. When adding acids to solution there is no change in appearance of the oily layer - it just seems to sink. I haven't smelled the solution after adding acid. When heating the solid, there is a white fume/vapor emitted that rises up, but there was no boiling, popping, or melting. The mass of the solid before heating was 0.26g, and after heating it was reduced to 0.21g (so a 0.05g difference).

Is it hydroscopic (i.e. adsorb water from air to form solution) ? 2) Could it loose oxygen on heating and not water? Or maybe some other gas?

@permeakra 1) I'm not sure if it is hydroscopic, how would I test for that? One thing I did notice is that after a few days, the powder in the container started to slightly clump together. Would this indicate anything?

I don't believe it would lose any gas, because it didn't bubble or anything. Most likely, everything that was lost was water.

@L.R.J. 1) leave a small quantity on open air overnight and see what happens 2)incorrect. thermal decomposition of perchlorates produces oxygen, but they won't produce gases when mixed with (diluted) acid.

Possibly a salt with an organic acid other than acetate? They smell pretty strongly and might give an oily-looking layer

OP is not active or else I should've asked for the concentration of sulfuric acid.

Oye @airhuff, it will be $\ce{Pb(NO_3)_2}$ and not as what you have written, just as the other such compound will be $\ce{Pb(NO_3)_4}$ and not Pb(NO3)4,

What about a possible $\ce{F-}$ anion? While the nature of the cation might be guessed based on the flame test, since the $\ce{AgNO3}$ test gave no precipitate, it might have produced $\ce{AgF}$ (soluble in water).
This said, by looking at the densities of $\ce{M-F}$ systems with a cation with orange flame and density around $\pu{2 g ml-1}$, I'd say it's probably $\ce{NaF}$.
You might test for fluoride with a selective electrode, if you have one.
Otherwise, here are listed some other analytical methods for the identification of fluoride anions:PubChem: Sodium Fluoride-Analytical Methods

Maybe, but only if the OP is wrong (as you suggest may be true) about $\ce{Ca^2+}$ being the cation. What bothers me most is that $\ce{PbF2}$ is only soluble to about 600 ppm, which you'd think would have formed a precipitate.

You are right, actually I'm not familiar with the lead test.
Anyway, I excluded calcium simply because the density is quite different from the one provided (3.18 g/ml).

What is the best way to conclusively determine if it is Calcium or Sodium? Would adding Sodium Oxalate do the trick?

My idea is quite nasty since it produces HF (maybe someone has a better idea): by reaction with H2SO4, NaF would yield N2aSO4, and CaF would give CaSO4. Other than having different appearance, Na2SO4 is highly soluble in water 28 g/100 ml, while CaSO4 is poorly soluble: 0,2 g/100ml. You could use this property to differentiate them.

@L.R.J. Atomic Emission Spectroscopy.

@QuantumAMERICCINO That's usually considered cheating in inorganic quant analysis. :-)

It could be calcium chlorate $\ce{Ca(ClO_3)2}$ or perchlorate $\ce{Ca(ClO4)_2}$. The corresponding anions do not react with $\ce{Ag+, Pb^{2+}, Ba^{2+}}$. These salts are soluble in water. The density is not a useful criteria, as a powder has always a much smaller density than a whole crystal.

@Maurice That's true that density is not that of an important factor but did OP measured the density by means of an instrument or calculated it? If he measured it, then it is what it is (Alas, OP is not active). If we are not considering that density, it could be any soluble salt that is matching other criteria which will create more confusion. For pinpoint identification of the salt, we should assume that density is what it is, around 2g/ml.

@The_Vinz "I'd say it's probably NaF". Sodium gives an intense yellow color in flame test which contradicts OP's result which is deep orange to red color.

Wouldn't the low solubility of CaF2 (0.016g/L) exclude it? The low solubility of NaF is also a problem: 42.2g/L, or 4%.

To all supporting $\ce{NaF}$: That fails the last two cation test with $\ce{Na2CO3}$ and $\ce{NaOH}$.

Calcium formate is a white-to-yellow or off-white crystalline powder, has a density of $\pu{2.02 g/ml}$, solubility of $\pu{\sim 17\%}$, decomposes at $\pu{300 ^\circ C}$ . It irritates eyes severely and has a stinging taste, therefore could have an "hard to describe odor" or effect on the nose. One mental conflict I have is that I would expect that measuring the density of a powder with a density of $\pu{2.02 g/ml}$ would actually give a value substantially lower than 2, but maybe the OP's experimental technique was exceptionally good.
When heated, calcium formate decomposes to calcium carbonate (Ref 1). The remainder $\pu{0.21 / 0.26}$ from the original heating is $\pu{80.7 \%}$; whereas the mw of $\ce{CaCO3 (100)}$ divided by the mw of calcium formate ($\pu{138}$) is $\pu{76.9 \%}$. Close enough: perhaps not all of the material decomposed. If not all decomposed, the heating was probably not severe. A test: acidify the burned product with $\ce{HCl}$, if it is $\ce{CaCO3}$, it will bubble. Another test: severely heat the burned product, see if it goes to $\ce{CaO}$ by checking pH ($\ce{-> 13}$) and seeing a further weight loss to $\pu{56\%}$ of the original.
A pH test on the original material could be informative: one SDS reported $\pu{7.5}$ for calcium formate, whereas an SDS for calcium acetate reported $\pu{7.6}$. The difference is small, but suggests that if the pH is closer to neutral than $\pu{7.6}$, the anion would be of a stronger acid than acetic (which formic is).
Ref 1: http://www.geosc.com/Assets/Files/Products-Docs/P-C-Product-Docs/Trimet-Products/CAF-TDS-US-Format.pdf

I had the same thought, but the solubility of Ca formate is higher than what OP observed. Calcium oxalate is a bit more dense and much less soluble, but should have given precipitate with barium, except that the low solubility of original material means very low anion in the solution to which barium was added

@Andrew: The OP says: "For the solubility, I do not have any quantitative measurements,". Dissolution is often slow to get to the published maximums. "Easily 10x the volume" means 1cc of solid (2 g) in 10-15 cc H2O = 20 - 13.3%, av = 16.6%. Quite close to 17%.

I do not expect CaO to give pH>13.  Calcium hydroxide is not sufficiently soluble.  Based on the solubility product for the hydroxide, actual pH with CaO would be between 12 and 13.

Hmm, if it was calcium formate, then it would be release formic acid vapor on reacting with sulfuric acid (from ref.), which would have a distinct smell but OP didn't report any vapor or bubbles or any smell from that reaction. But again, that white ppt could be CaSO4?

@Oscar Lanzi: You are absolutely correct. I rounded up for effect.

@Nilay Ghosh: Formic acid doesn't have a "distinct" odor (it could be "hard to describe"), and being released on such a small scale (OP is talking about 2 grams for density) doesn't really have overpowering consequences.

Ok, forget about the smell. I did a quick book check. Vogel Qual. analysis says: "With dil. H2SO4, formic acid is released having a pungent odor. With conc. H2SO4, carbon monoxide is released". So, there is a reaction happening irrespective of concentration of the acid. If there is a gas evolving, there's bound to be bubbles in reaction pot. But, OP didn't report any bubbles. Only if we knew the conc. of the acids, that would really make a difference (See the comments below Oscar's answer).

@Nilay Ghosh: "Is released" doesn't imply bubbles explode and attack your nose. Wikipedia: "Formic acid is a colorless liquid having a pungent, penetrating odor at room temperature, not unlike the related acetic acid." I suggest that the test produced dilute formic acid, and the odor was not dramatic (vinegar doesn't have a dramatic odor unless there is a large surface, and  bp of formic is ~same as acetic). BUT, I grant you, this is one of those tests I would make sure about an odor or not. If it's not formate, then I'm stumped. (CaS2O6.4H2O has density 2.18. How's that?)

There are ways to quite reliably determine what this solid is. As mentioned above, the counter ion most probably should be $\ce{Ca^2+}$, however there are many organic or inorganic anions that still fit the criteria. The $\ce{AgNO3}$ tests provide insight that its a non-coordinating counter ion.
Ways to determine the anion - however this all takes infrastructure that may or may not be accessible:

Most straightforward way would be to acquire NMR of the sample in $\ce{D2O}$. This will reliably tell you, if there are H atoms, C atoms (and their symmetry), 19-F NMR will reliably tell the presence of a fluoride atom or other fluoride containing counter ions (such as $\ce{BF4-}$ etc.)
There are many ways to create adducts of organic anions for GCMS analysis.
A HPLCMS or UPLCMS analysis will provide You with several pieces of information:
UV-Vis spectra will provide insight whether aromatic structures are part of the counter ion, and the MS will provide the m/z of any anion that is heavier than 50. Fortunately almost every single weakly coordinating anion is fairly heavy and should show up.

These methods should reliably provide the answer for the question asked.
As for the simple answer, $\ce{Ca(BF4)2.xH2O}$ may be the answer:

Tetrafluoroborate salts are soluble and known to form hydrates.
Tetrafluoroborate salts are not the most stable and can actually release $\ce{HF}$ gas in low quantities resulting in strong, irritating odor.
This salts is more consistent with everything examined.

One idea is to retry the brown ring test with the proper procedure, as the question admits that this had not previously been used (acidified ferrous sulfate was added wrongly).  The other evidence is consistent with calcium nitrate, so we need to get that test right.
Calcium nitrate also decomposes at 500°C and above, releasing nitrogen dioxide leaving the solid oxide behind (only alkali metal nitrates, and not all of those, give the nitrite with this reaction).  You would then see the brownish nitrogen dioxide gas.

I was also suspecting a nitrate salt given that the ring test wasn't performed properly but shouldn't that salt react with barium chloride to form calcium chloride + barium nitrate both of them of which are white ppt (?) or shouldn't it react with conc. sulfuric acid/hydrochloric acid(OP didn't mention the concentration though) to form nitric acid which is indicated by evolution of red nitric oxide fumes? OP didn't report any precipitate or bubble or fumes formation in his reaction process.

I do not know this.  The question does not specify adding concentrated acids.

That's where OP missed critically. Did everything except testing it with conc. acids (we don't know, OP is not active). btw, calcium nitrate has a density 0f 2.5 g/ml as opposed to OP's salt which is 2 g/ml. Shouldn't that be a problem?

@nilay the density depends on whether the OP accounted for less than 100% packing in the powder.


text:
How is the the law of chemical equivalence defined and what are its limitations?
I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products. But I'm facing difficulties in understanding double titrations. Moreover, this doesn't work for equations where the n-factor of all the reactants and products don't match.
Let's take an example:
$$\ce{KIO3 + KI + H2SO4 \to I2 + K2SO4 + H2O}$$
Here, the gram equivalence of $\ce{H2O}$ is $2$ which should match with that of $\ce{KIO3}$. But the gram equivalence of $\ce{KIO3}$ is $5$.
Similarly, $\text{g.eq}_{\ce{H2SO4}} \neq \text{g.eq}_{\ce{I2}}$ and so on. When the n-factors don't match, the gram equivalents don't match. Here, we can consider gram equivalent to be $\text{number of moles} \times \text{n-factor}$. Again, this n-factor is what is posing a problem.

Can someone define the Law of Equivalence and state its limitations because of which I seem to find problems as I've mentioned above?

That'll be a better way, you can always ask separate questions, asking too many questions in a single question is generally frowned upon.

Basically the law of equivalence wants you to balance the equivalents that are involved in your reaction. I'd like you to view you reactions in two parts, the oxidation and reduction halves. I see:
$$\begin{align} \ce{I- -> I2} \tag{oxidation} \\
\ce{IO3- -> I2}\tag{reduction} \end{align}$$
In other words, we've just discovered a (reverse disproportionation?) comproportionation reaction. The change in the oxidation state of iodine in the oxidation part is $-1 \to 0 $ ($\Delta_{o.s.} = 1$) and for the reduction half it's $+5 \to 0$ ($\Delta_{o.s.} = 5$). So far so good.
Usually, the questioner would provide you with the number of moles of each species and you're done with number of moles times the n-factor. Nevertheless, let's try to give it a shot. I'm balancing this equation:
$$\ce{1KIO3 + 5KI + 3H2SO4 \to 3I2 + 3K2SO4 + 3H2O}$$
This means for $1$ mole of $\ce{KIO3}$ to react, you'll need $5$ moles of $\ce{KI}$. As this is a proportion, converting it to an absolute value, I'd say you've $1n$ moles of $\ce{KIO3}$ and $5n$ moles of $\ce{KI}$.
So if I try to equate the gram-equivalents, I'll have $1n \times 5 = 5n \times 1$. In other words, $\text{Eq.}_{\ce{KIO3}} = \text{Eq.}_{\ce{KI}}$

I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products.

Law of equivalence is even more fun, for a balanced chemical reaction:
$$ \ce{aA + bB -> cC + dD}$$
Equivalents of every species will be equal. Mathematically,
Equivalents of A = Equivalents of B = Equivalents of C = Equivalents of D
will hold.

Why aren't the gram equivalents of $\ce{H2SO4}$ equal to that of water?

You need to strip your equation of the spectating species. For instance $\ce{H2SO4}$ is primarily here because it acts as a medium-provider, it makes the solution acidic. Neither $\ce{SO4^2-}$ nor $\ce{H+}$ underwent any oxidation/reduction so applying the equivalent concept makes no sense in this case.

Your answer is really helpful. But I have a doubt, aren't the equivalents of each and every reactant and product same irrespective of whether they belong to a balanced chemical equation or not?

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text:
What would this unsaturated cyclic compound be called?
I came across this compound recently, and havent been able to find out what it's name could possibly be:

My guess was maybe it's "cyclohexylene cyclohexane", but not quite sure. Can anyone tell me what this compound could be called?

Chemspider tells me that the name should be

1,1'-Bi(cyclohexylidene)

http://www.chemspider.com/Chemical-Structure.121798.html?rid=e40cd595-d5e5-450e-bdb3-3e558a9da92d
While PubChem mentions:

Cyclohexylidenecyclohexane

https://pubchem.ncbi.nlm.nih.gov/compound/Cyclohexylidenecyclohexane

My guess was maybe it's "cyclohexylene cyclohexane"

There's a flaw to this thinking, to name this, I'd suggest you choose one of the cyclohexane rings to act as the word-root. Now you've to just add the name of the substituent.
Recall that (source: http://www.ebi.ac.uk/chebi/searchId.do?chebiId=65236) also click this link for an example (this website particularly has propylidene group as the example):

An alkylidiene group formed from propane by removal of two hydrogen atoms from the same carbon atom, the free valencies of which are part of a double bond.

So, the substituent in this case would be called "cyclohexyl idene " (without the space). And hence, the name should be cyclohexylidenecyclohexane.

1,1'-Bi(cyclohexylidene) appears to be a preferred name, however it is not very suitable for speech communication, because is may sound like some bicyclo- compound.

Btw, it's better to link via the ID: https://pubchem.ncbi.nlm.nih.gov/compound/138159. They list alternative name Delta1,1'-Bicyclohexane, which is very weird (and likely wrong), maybe incomprehensible misuse of "delta convention".

Assemblies of two identical rings linked by a single bond are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows:

P-28.2 RING ASSEMBLIES OF TWO IDENTICAL CYCLIC SYSTEMS
P-28.2.1 Ring assemblies with a single bond junction
Assemblies of two identical cyclic systems joined by a single bond are named by one of two methods:
(1) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding parent hydride enclosed in parentheses, if necessary. Parentheses are used to avoid confusion with von Baeyer names;
(2) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding substituent group (for names of substituent groups, see P-29), enclosed in parentheses, if necessary.
Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
The name biphenyl is retained as 1,1'-biphenyl.

For assemblies of two identical rings linked by a double bond, the formation of names is restricted to a single method:

P-28.2.2 Ring assemblies with a double bond junction
When two cyclic systems are linked by a double bond, method (2) described in P-28.2.1 is the only recommended method. Method (2) has also been used in which the presence of a double bond is indicated by the Greek letter Δ and the point of attachment of the ring is given by superscript locant numbers. This method is not continued in these recommendations; accordingly, ring assemblies of three or more identical cyclic systems interconnected by double bonds must be named by other methods (see P-31).

Therefore, the preferred IUPAC name of the compound that is given in the question is 1,1'-bi(cyclohexylidene). The name Δ1,1'-bicyclohexylidene is no longer recommended.

It's called cyclohexylidenecyclohexane.

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text:
What is the correct structure for trans-3,4-dichlorocyclopentene?
I have to draw the structure for trans-3,4-dichlorocyclopentene.
What position should I draw the double bond at? Should I draw it as molecule 2? Because there's a trans.
Or should I draw it as shown in molecule 1 because it didn't state?

I feel that molecule 2 in the diagram is correct. What do you think? Where are my mistakes?

Use logic. Why the first atom is the first, how do you think? Why we didn't start numbering from someplace else?

The issue here really is that the name 'trans-3,4-dichlorocyclopentene' is not really correct, as it is unspecific. It lacks stereo-descriptors, and the locant for the double bond.
Assuming that the double bond starts at the first carbon, which is usually the case if it is omitted, there are two enatiomers still possible:

(3R,4R)-3,4-Dichlorocyclopent-1-ene

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=12430197.
(3S,4S)-3,4-Dichlorocyclopent-1-ene

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=131015828.

So the structure you have drawn as 1 is (3R,4R)-3,4-dichlorocyclopent-1-ene.
The structure you have drawn as 2 doesn't really exist, as the double bond is planar, but ignoring that it would be 1,2-dichlorocyclopent-1-ene.

See for details: National Center for Biotechnology Information. PubChem Compound Database; CID=10606793.

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text:
Why is p-toluidine more basic than m-toluidine?
https://en.wikipedia.org/wiki/Toluidine
In m-toluidine the $\ce{CH3}$ group is closer to $\ce{N}$ so shouldn't that increase the electron density on $\ce{N}$ and make it a stronger base?

Related: Ortho-effect in substituted aromatic acids and bases

In p-toluidine you are not actually considering the hyper conjugating structures due to which the compound becomes more basic than m-toluidine which only has it's +I effect and I would like to suggest you that as far as I know resonance effect "usually" overpowers the other effect!


text:
What is the optical nature of the tris(ethylenediamine)cobalt(III) cation?

The molecule I've drawn has neither a plane of symmetry nor a center of symmetry, even though some books refer to this as optically inactive.
Can you please elaborate why is this so, or those books are wrong.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. We prefer to not use MathJax in the title field, see here for details. I actually have no idea what your question is, have you already read the article on Wikipedia?

@Martin - マーチンthanks for information, probably I referred some wrong artical

Ethylenediammine or $en$ is an symmetrical didentate ligand. You can remember a rule of thumb for this if you like:

In the case of $3$ symmetrical didentate ligands, i.e. $[\ce{M(AA)3}]$, the compound exists in only one form, which is optically active

So the complex you've mentioned will be optically active, no doubt.
Also, as a reference, I'm quoting the wikipedia article at https://en.wikipedia.org/wiki/Tris(ethylenediamine)cobalt(III)_chloride#Stereochemistry

The complex can be resolved into enantiomers that are described as $\Delta$ and $\Lambda$.

The only symmetry elements that can be found in $\ce{[Co(en)3]^3+}$ are the three $C_2$ axes of rotation, one of which is shown in this image (colored in red) on wikipedia: https://upload.wikimedia.org/wikipedia/commons/e/e5/Coen3lel3.png

As a side note, here's another rule of thumb:

In the case of $3$ unsymmetrical didentate ligands, i.e. $[\ce{M(AB)3}]$, the compound exists in two forms, namely fac (stands for facial) and mer (stands for meridional), and both of these forms are optically active.

Important note: There's a minute error in this answer; and that's the distinction between optically activity and chirality: I used the word "optically active" while it should've been "chiral". It's bad I missed it, but I'd like you to read Ben Norris' comment on this below. I'm leaving the answer untouched, editing it rip the comment of its value. I encourage you to learn from my mistake.

This answer (and the other one) correctly explain why $\ce{[Co(en)3]^3+}$ is chiral. However, there is a fine distinction between "chiral" and "optically active". A substance or solution is optically active if it rotates plane-polarized light. The solution needs chiral molecules to do this, but not all solutions of chiral compounds are automatically optically active. A racemic mixture is not since it contains equal amounts of both enantiomers. $\ce{[Co(en)3]^3+}$ could be considered optically inactive if either the enantiomers cannot be separated or the enantiomers interconvert in solution.

@Ben Norris naturally  so :)

Some molecules can be chiral even if they do not have an asymmetric carbon. In these molecules, such as the one you draw, it is the overall shape of the molecule that is important. There has to be no centre of inversion or mirror planes (including rotation reflection $S_n$). In point group terms these mirror planes are called vertical, horizontal, dihedral and a mirror plane is part of the rotation reflection operation. They are labelled $\sigma_v,\, \sigma_h,\, \sigma_d,\, \text{and } S_n$ respectively). Your molecule follows this rule in that it has a 3-fold axis and three 2-fold axes at right angles to the  3-fold axes.  In terms of point groups this is called $D_3$. There are two enantiometric forms.
Chiral molecules belong to the point groups $C_n, \, \text{and } D_n ,\;n=1,2,3\cdots$.
The figure shows the symmetry elements. You can see that here is no centre of inversion or mirror planes. Look at www.molecule-viewer.com for examples of point groups and molecules of all different point groups. An explanation of how molecules rotate plane polarised light is summarised in the answer to this question; Dependence of the angle of rotation on the wavelength of plane polarized light


text:
Inactive versus frozen orbitals in CASSCF
Does it make any difference if I use the frozen keyword instead of the inactive keyword in MOLCAS CASSCF calculations?

read here, chapter 8.33.0.1


text:
Does restriction on spin in orbits exist?
For a valence electron in Antimony (Sb) with the electron configuration:
$Kr(5s)^2(4d)^{10}(5p)^3$
The 3 valence electrons have the quantum numbers $n=5,l=1,m=1$, resulting in the following possibilities for $(n,l,m_l,m_s)$:
$(5,1,−1,−\frac{1}{2});(5,1,−1,\frac{1}{2});(5,1,1,−\frac{1}{2});(5,1,1,\frac{1}{2});(5,1,0,−\frac{1}{2});(5,1,0,\frac{1}{2})(5,1,−1,−\frac{1}{2})$
$(5,0,0,−\frac{1}{2});(5,0,0,\frac{1}{2})$
Then looking at the actual quantum numbers of Sb (from a former exam) it is given as:
$(5,1,−1,-\frac{1}{2});(5,1,1,-\frac{1}{2});(5,1,0,-\frac{1}{2})$
$(5,0,0,−\frac{1}{2});(5,0,0,\frac{1}{2})$
Where did the quantum sets with $l=1$ and $m_s=−\frac{1}{2}$ go? When I look at the valence electrons in Kr the set is with $l=1$ and $m_s=−\frac{1}{2}$:
$(4,1,−1,−\frac{1}{2});(4,1,−1,\frac{1}{2});(4,1,1,−\frac{1}{2});(4,1,1,\frac{1}{2});(4,1,0,−\frac{1}{2});(4,1,0,\frac{1}{2})
(4,0,0,−\frac{1}{2});(4,0,0,\frac{1}{2})$
Summing up the question: Why can the valence electrons in Kr have $l=1$ and $m_s=−\frac{1}{2}$ and the ones in Sb not? Perhaps it is a mistake in the exam?

Look at the electronic configuration: you only have 3 electrons in the p orbitals for Sb, whereas you have 6 for Kr. Thus the valence shell is full for the latter and incomplete for the former.

There are three electrons in the $5p$ so the possible total spin quantum numbers are $3/2,, 1/2$ with multiplicities $2S+1$ of  $4 $ and $2$ which produce quartet and doublet states. In Kr all the orbitals are filled so total spin (and orbital angular momentum) is zero and multiplicity one , so this is a singlet state. If you need to include orbital angular momentum then you will need to look up term symbols and Russell-Saunders coupling.

The term symbols for $p^3$ are $^2D_{3/2}; ^2D_{1/2},,^2P_{3/2}; ^2P_{1/2},, ^4S_{3/2}$.  The orbital angular momentum has values $L= 2,,1,,0$  and has to be combined with the spin terms.  To work all of this  out map out a table of microstates of l and s and eliminate those not possible by the Pauli principle.

Why is it that the total spin quantum numbers are positive, could they not be negative, allowing for $(5,1,1,\frac{1}{2}$ or $5,1,1,-\frac{1}{2}$ @porphyrin

Only electron  and orbital angular momentum matter here. If angular momentum  is $A_1 ,, A_2 $ these are added according to the (Clebsh-Gordon) type series, which is  similar to  adding integer length vectors, so cannot be negative.  To find the series find the max value $A_1+A_2$ then the minimum $|A_1-A_2| $ (which is the absolute value) then fill in gaps (if any) in unit steps; thus if $A_1=5/2, ,A_2 =1$ then series is $7/2, , 5/2,,3/2$


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Can ethyl alcohol fumes be suppressed by a household air purifier/filter?
Can ethyl alcohol fumes be suppressed by a household air purifier/filter? In general, how are the fumes suppressed? How is it done?
Can somebody direct me to some information regarding this? I've been searching but do not have the strongest background in chemistry (I'm a mechanical engineering student). I've done some research on filtering media as well as ethyl alcohol itself.

Generally one would use activated charcoal for organic vapors. You would have to check with your local IH folks to be sure your filter was appropriate for the application.

@JonCuster Thank you for your response.  What is IH?

Industrial Hygene

Ethanol fumes can reach explosive limits. Note that the motors that power the fan in home air purifiers are not explosion proof. Explosion proof fans require very special designs.

No home air purifier removes alcohol vapor from the air. There are a number of ways it can be eliminated, though, depending on quantity, purity and other factors.
Bakeries have to comply with clean air standards for VOC (volatile organic compounds), primarily ethanol, which can cause smog. One way the alcohol is removed is by catalytic oxidation, similar to the way an automobile catalytic converter works. Even hot copper is sufficient to oxidize alcohol.
Another way to remove ethanol is by condensation, which allows the alcohol to be saved for resale, or by washing (scrubbing) the oven gases with water.
Sadly, though this regulation of ethanol emission is necessary, it also removes the sublime aroma of baking bread from the vicinity. sigh...

IH = industrial hygiene. Gas molecules can be reacted or absorbed. I assume your education is sufficient so that we don't have to explain that ethanol in the vapor state is a gas. No "standard household" filter will remove ethanol. The volumes of air that say a 3M Organic Vapor mask could process is inadequate for all but very short exposures. Activated carbon, and possibly some zeolites (molecular sieves) might be used in industrial processes. But for removal from room air? Refrigeration/condensation. Of course, the process selected depends on cost - rather than condensation for example, you could always heat and oxidize it. That type of air treatment is very expensive.

Hi @alphonse thank you for your response, very helpful.


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What happens during oxidative and hydrolytic rancidity of fats?
I understand that when lipid molecules - triacylglycerol - are hydrolysed, the fatty acids 'break' off from the glycerol backbone. And when fatty acids oxidise, these in turn can break into smaller end products. Does hydrolysis always happen first? I have made this image - would it be a fair representation of what happens when oils turn rancid?

I am currently finishing my Ph.D. in rancidity, and therefore I know quite a bit about it. The picture is not a fair representation of what happens, since free fatty acids should be present, though you just show the triacylglycerols. Which rancidity happens first will depend on the matrix and on the storage condition of the oil. Usually, but not always, oxidative rancidity happens first.


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Does it matter what electrolyte we use for a Galvanic Cell?
A typical Galvanic Cell, say a Daniell cell, consitsts of two separate beaker, one containing Zn rod dipped inside aq $\ce{ZnSO4}$ and the other beaker containing Cu rod dipped inside $\ce{CuSO4}$. Does it matter what electrolyte we dip the metal in?
If we change the electrolyte of the beaker containing Zn rod to say $\ce{AgNO3}$. I would expect a displacement reaction to take place where a redox reaction occur between the Zinc rod and the Ag+ ions. No electrons will flow through the external wire. What if I replace the electrolyte with a solution of a metal that is less reactive then the metal dipped in? eg. $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{KCl}$(aq). Would my cell still work?
Also, I don't understand the need of a porous pot OR salt bridge in a galvanic cell. Why would it matter if the solutions are mixed together? Why not just have one single beaker, containing an electrolyte $\ce{KCl}$(aq), and have the two different metals dipped into it. In fact, if the solutions are mixed together, there will not even be unbalance charge accumulation, and the porous pot/salt bridge would not be necessary?

Hello and welcome to StackExchange! Take a look at the [help] to learn how to use this site better. Cheerio!

The electrolyte is very important for the functioning of the galvanic cell. The emf that the cell generates depends on the type and the concentration of the electrolyte used. In the Daniell cell, the reaction would be:
$$\ce{Zn + CuSO4 -> ZnSO4 + Cu}$$
As per the Nernst's equation, you get the emf of the cell to be:
$$\text{E}=\text{E}^\text{o}-\frac{2.303\text{RT}}{\text{nF}}\log_{10}{\left(\frac{\ce{[Zn^{2+}]}}{\ce{[Cu^{2+}]}}\right)}$$
Clearly, the emf of the cell depends on the concentrations of the cell. Dipping the rods in $\ce{NaCl}$ would not make the cell work.
And for why the salt bridge is needed, think about why you would build a cell in the first place. To generate power in a usable way right? If you let the solutions mix, they would react directly, and the energy would be released as heat instead, and not as electrical power. You want the ions to move towards the electrodes and get discharged there.

I don't want to add another answer so I'm commenting here. A salt bridge also: (1) completes the internal circuit of the galvanic cell. (2) establishes charge neutralization of the 2-electrode solutions by supplying them with appropriate ions. (3) minimizes liquid-liquid junction potential.

Also see: Ivan Neretin's comment to my question.


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Finding rate constants for parallel reactions
A reactant $\ce{X}$ is converted into products $\ce{Y}$ and $\ce{Z}$
according to the following first order parallel reactions:
$\ce{X->[k_1]Y}$
$\ce{X->[k_2]Z}$
$\ce{[X]}$ has decreased from $\pu{1.00 M}$ at $t = 0$ to $\pu{0.549 M}$ after $\pu{10 min}$.
Calculate the rate constants $k_1$ and $k_2$, knowing that the yields of $\ce{Y}$ and $\ce{Z}$ obtained are $25.0$ and $\pu{75.0 mol\%}$, respectively.
So far, I have the ratio between $k_1$ and $k_2$ $=0.33$, which I found from the yield ratio but I'm unsure what to do with this ratio.
I also have derived the equation $[A] = [A]_0 \cdot e^{-(k_1+ k_2)t}$

By some simple mathematics, you can reach the following equations (the first of which you have reached):
$[X]=[X]_0\cdot e^{-(k_1+k_2)t}$
$[Y]=\frac{k_1}{k_1+k_2}[X]_0\cdot e^{-k_1 t}$
$[Z]=\frac{k_2}{k_1+k_2}[X]_0\cdot e^{-k_2 t}$
The $1:3$ ratio that you have arrived at is correct,  i.e. $\dfrac{k_1}{k_2}=\dfrac{1}{3}$.
Substituting this in the equation for $\ce{X}$, we get
$[X]=[X]_0\cdot e^{-4k_1 t}$
All that is left to do now is to substitute and solve, which leads to the answer:
$(k_1,k_2)\equiv(0.015\ (min)^{-1},0.045\ (min)^{-1})$
There you go!
PS: I've used $[X]_0$ for representing the initial concentration of $\ce{X}$.


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Can hydrogen on platinum reduce carboxylic acids and esters?
Can $\ce{H2/Pt}$ reduce carboxylic acids and esters?
I am confused about the reducing nature of hydrogen in presence of platinum. In some books, it reduces carboxylic acid and ester while in others, not. I have checked Finar and Boyd books for it.

This is based off my memory, as far as I can remember what my teacher taught me.

Note, instead of $\ce{Pt}$, you can go for $\ce{Ni}$ (or Raney Nickel) or even $\ce{Pd}$ as a catalyst.
Functional groups such as acid halide, cyanide, nitro, aldehyde, ketone, alkene, and alkyne are easily reduced by $\ce{H2/Ni}$.  Some other groups can be reduced with $\ce{H2/Ni}$ as well, but if someone's asking you a question in an exam, they want you to refer to the list of groups which are easily reduced (more in the reactivity order below).  So the answer in your case would be a no.
Next, here's a reactivity order:
First, the groups which are easily reduced:
$$\begin{align} \text{acid halide} > \ce{R-C#C-R} > \ce{R-C=C-R} > \ce{R-CHO} > \\ &\text{ketone}>  \ce{R-CN} > \ce{R-NO2} > \ldots \end{align}$$
Then the groups that are reduced only with difficulty:
$$ \ldots > \ce{R-X} > \text{benzene} > \ce{R-COOH} > \text{amide} > \text{ester} > \text{acid anhydride}$$
$\ce{R-X}$ would require temperatures around $\pu{50^\circ C}$, benzene would require a temperature of about $\pu{100^\circ C}$ and groups from carboxylic acid to acid anhydride would require temperatures around $200-\pu{250^\circ C}$.


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Why doesn't tetraiodine exist?
I've run into a question that says after doing the Lewis structure of $\ce{IF3}$, which I did, explain why $\ce{I4}$ doesn't exist.
I am confused as to how it can be explained. I've thought that maybe it has to do with iodine's valence orbitals, both hybridised or not being too large and as a result the different orbitals would have to overlap and superimpose. But again I'm not sure.

There is an $\ce{I4^2-}$ anion.

I think due to the large atomic size of Iodine atom, the central Iodine can't accommodate other 3 Iodine atoms (as well as 2 lone pairs of electrons) . There is strong probability for the ligand Iodine atoms to suffer steric repulsion and hence the molecule $I_4$ doesn't exist!

@chail10 https://en.wikipedia.org/wiki/Polyiodide iodine can coordinate 3 other iodine atoms. Also elements aren't capitalised.

@Mithoron Yupp, I wasn't sure about that and that's why I didn't post it as answer! Thanks for the reference.

While it's true that there are polyiodides, the $\ce{I4^2-}$ ion has a completely different geometry from $\ce{IF3}$ molecule. The ion is most easily regarded as two iodine molecules stuck end on and the ensemble is bound because two additional electrons are popped into the in-phase combination of each side's $\ce{I(p_z)-I(p_z)}$ antibonding orbital. This is bonding between the fragments and antibonding within the fragments. Since the individual molecules can't do anything with the extra two electrons, it's best just to stay together and stabilize them together.
As to why there isn't a $\ce{I4}$ that is isostructural to $\ce{IF3}$, this is because the stability of interhalogens is heavily influenced by the ionic character of bonds. Recent work[1,2] has shown that the bonding in halogens is dominated by "charge-shift" interactions in which the ground covalent state is stabilized by ionic states in which both electrons of the bond have been transfered to either atom (e.g. $\ce{F-F}$ is stabilized by $\ce{F+-F-}$ and $\ce{F^--F+}$ states). This stabilization is due to highly electronegative atoms. The stability of higher interhalogens relies on this "charge-shift" stabilization. Since I is a lot less electronegative than F, this charge-shift stabilization is much weaker (and this is also the reason why nearly all of the higher interhalogens have fluorine as a bonding partner). Since this charge-shift stabilization is absent, the 4 iodines are much happier to break apart into 2 $\ce{I2}$ molecules.

Sason Shaik, David Danovich, Wei Wu, and Philippe C. Hiberty, Nature Chemistry 2009, 1, 443 - 449. DOI: 10.1038/nchem.327
Daniel S. Levine, Paul R. Horn, Yuezhi Mao, and Martin Head-Gordon, J. Chem. Theory Comput. 2016, 12 (10), 4812–4820. DOI: 10.1021/acs.jctc.6b00571

While it's charge-shift for F2, fluorides are described with 3-center 2-electron bonds. Still your argumentation stays valid, I guess.

This is a (modern) valence bond theory vs MO theory discrepancy in nomenclature. Both are equivalent and equally valid representations, as you note.

3c 4-electron, sorry, and yeah, it looks like.

@levineds Agreed. I'm reading Weinhold's Valency and Bonding currently, and 3c4e bonding looks an awful lot like a subset of charge-shift bonding to me.

$\ce{I4}$ certainly is a minimum on the potential energy surface, albeit not isostructural to $\ce{IF3}$, as the former has $D_\mathrm{3h}$ symmetry, not just  $C_\mathrm{2v}$ like the latter. On the (admittedly ridiculously low DF-BP86/def2-SVP) level of theory it is even lower in the electronic energy than the dissociated $\ce{I2}$ molecules. Clearly an explanation solely based on electronic effects, like that "charge-shift explanation", but not limited to it, cannot be a sufficient reason. And just because it has not been observed, it does not mean it cannot exist.

While that is a local minimum, it has no lifetime. At a much better level of theory ($\omega$B97M-V/cc-pvdz with the Stuttgart Dresden small core relativistic pseudopotential), the $D_{3h}$ $\ce{I4}$ structure is 30 kcal/mol higher in energy than two $\ce{I2}$ molecules. Coupled with the large entropy gain from separating into two molecules, the Gibbs free energy of separation is quite downhill. Also, the lowest vibrational mode of $\ce{I4}$ (13 cm-1) is the reaction path to separation, so it is kinetically facile. The explanation given can fully account for the non-existence of this molecule.

To clarify, it has no lifetime at any reasonable temperature because the barrier to fragmentation is very small, and it is many 10s of kcal/mol above separated 2$\ce{I2}$ and so not a populated state (per the thermodynamic Boltzmann distribution that $\ce{I4}$ and 2$\ce{I2}$ would obey).


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Is it possible for a compound to be soluble in a mixture of solvents but not the individual solvents?
Some weeks ago one of my friend needed to use turpentine. As a good but ignorant chemist on that point I did some research about it. I found (at the beginning) in the French wikipedia page of turpentine :

"L'essence de térébenthine est un mélange de molécules variées,
  comprenant en particulier des terpéniques, des acides et des alcools.
  Ce liquide est insoluble dans l'eau, dans l'alcool absolu et dans
  l'éther, cependant il est légèrement soluble dans un mélange
  eau-alcool."

Which translated in English gives,

"Turpentine is a mixture of different molecules, particularly terpenes, acids, and alcohols. This liquid is insoluble in water,
  absolute alcohol, and ether, but it is slightly soluble in a mixture of
  water-alcohol."

I found no reference about that. Is it really possible? It sounds particularly strange to me. And obviously if that is possible, how?

An analogous situation might be the dissolution of metals in aqua regia.

Remember that solubility and insolubility are nebulous terms. Even if a compound is  insoluble a "tiny" amount will actually dissolve. So it would better to think of the compound being more soluble in a mixture of solvents than either solvent alone.

@MaxW I know that. But I was not asking a question as a nitpick ^^ I mean at least measurable :)

@a-cyclohexane-molecule Aqua regia isn't really an example of this. In that case, two different chemical reactions are taking place: nitric acid oxidizes a small amount of gold while HCl solubilizes the oxidized gold atoms. Chloroauric acid (the result of these two acting together) is soluble in water, which is the only solvent per se in the aqua regia. No chemical changes are occurring in the asker's question.

@Hexacoordinate-C - I'm an analytical chemist. You can "measure" extremely small amounts in solution. Much too small to directly weigh on a balance.

@MaxW measurable means measurable :-) But you may think the question would not have a lot of sense if I was speaking about only one molecule (not two nor three, only one!) in a mixture of solvents ;)

The short answer is: yes, this is possible. Unfortunately, solubility is a fairly complex phenomenon to explain simply. Let's start with some examples where solubility is higher in a binary mixture than either solvent alone.
For a solid-liquid-liquid example: phenanthrene-cyclohexane-diiodomethane.[1]
For a liquid-liquid-liquid example like the one you were asking about: trifluoromethane-tetrafluoromethane-ethane at low temperatures and high pressures.[2]
In the Hildebrand Theory of solutions, there is a quantity called the "internal pressure", loosely how much these compounds want to break apart. In this model, this is usually something like the molar energy of vaporization divided by the molar volume. If the internal pressure of the solid is between the internal pressures of the liquids (as in this case), the solid is more soluble in the mixture than in either one individually. Another way to asking your question is about phase separation in ternary mixtures. You can find many examples of curved lines in ternary phase diagrams which are due to the ternary mixture not being the sum of its parts.
Qualitatively, you could imagine that some solvent solvates some parts of the molecule well and other poorly and vice versa for the other solvent. Hence, together they solvate better than each individually.

Gordon, L. J.; Scott, R. L. J. Am. Chem. Soc. 1952, 74 (16), 4138–4140. DOI: 10.1021/ja01136a054
Paas, R.; Peter, K. H.; Schneider, G. M. J. Chem. Thermodyn. 1976, 8 (8), 741-747. DOI: 10.1016/0021-9614(76)90053-7

Looking up Hansen parameters may be more relevant than Hildebrand for this question


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Is Jell-O flavored plastic?
Per this article, "Gelatin is a thermoreversible, cold-setting polymer." Further, Jell-O is flavored gelatin.
Is the phrase "thermoreversible, cold-setting polymer" equivalent to the term "plastic?"
Is Jell-O, therefore and despite the unappealing sound of it, flavored plastic?

Yep, that's a perfectly reasonable description. I would expand to "flavored, edible plastic" but YMMV.

More of a flavored hydrogel with physical properties that include plasticity. But that's splitting hairs.

I figured "edible" was inherent to being a foodstuff. I was less certain about the equivalencies of "thermoreversible, cold-setting polymer" and "plastic;" I feel like there might be differences in the terms I wasn't familiar with; I'm not a chemist. I only knew that plastics were polymers, but I didn't know if polymers were plastics. And if it's plastic, why can't I bend it into a position and have it remain that way?

You kind of do bend it into a position and have it remain that way when you add the plasticizer (water) to the gelatin and cool it in whatever-shaped container you've put it in.

So the plasticity is more of a temperature-controlled reflow?

Technically, "plastic" is anything that is moldable, such as clay, as well as organic polymers. Ceramics is also known as a plastic art https://en.wikipedia.org/wiki/Plastic_arts . If you can make a mold of gelatin, it's plastic.

If someone wants to write this up as an explanatory answer, I'll accept it...


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Does a red hot metal reflect light in the same way as a cold metal?
I know that a metal at room temperature looks shiny because of the free electrons on the surface, but what about a hot red metal? Would it also reflect light in the same way as it does at a colder temperature? Even if it emits some red light due to it being hot?

Many metals would oxidize from the surface when heated on air, and hence lose their luster. Short of that, they would reflect all right.

The light that you see from hot metals is incandescence.
Incandescence is caused by high temperature. The high temperature promotes electrons to higher energy levels. These electrons then relax and emit light.
On the other hand, the "light" (shiny) that you see from room temperature metals  is caused by re-emission/reflection of other light. Light from another source is absorbed by the metal and re-emitted, or the light from another source is reflected from the surface of a metal.
The reason as to why metals look shiny have to with plasmons. If I recall correctly, most materials will look "shiny" at a certain wavelength range. It just happens that metals are shiny in the visible wavelength range.
What determines what type of light a material will reflect is determined by its plasma/plasmon frequency. Temperature does seem affect the plasmon/plasma frequency. Several studies confirm this.
In general, it appears that the plasmon/plasma frequency decreases with increasing temperature. In other words, the material is able to reflect a narrower range of light as the temperature is increased. This would mean that if you had a material that is shiny in the visible light range, you could make it become dull by heating it. Of course, this is a simplification that doesn't take into account other factors.
However, I wasn't able to find an equation that directly relates temperature and plasma frequency. How the plasma frequency changes with temperature seems to depend on the material and the size of the material (nanoscale, bulk, etc). Furthermore, there are ab initio studies that suggest that the opposite must happen: the plasmon frequency increases with temperature.

While this is a nice answer, I don't think if this is what the OP wanted. He asked, would the metal still retain it's lustre even while heated to red hot temperatures.


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How to measure chemical potential
I wonder if there is a way to measure the chemical potential of a substance in a two-component mixture and also to account for its dependence on the number of moles. Explaining it further, suppose there is a mixture of a light component (like methane, C1) and a heavier one (like C20). Given temperature and pressure, one could add some amount of the light component to the mixture and that would rise the chemical potential of that first component. Thermodynamics tell us that the chemical potential can be expressed by
$$d \mu_i = - \bar S_i dT + \bar V_i dP + \sum_{m=1}^{n} \left(\frac{\partial \mu_m}{\partial N_i}\right)_{T,P,N_j[i]} d N_m $$
Being pressure and temperature constant, the increase in the chemical potential by adding some amount of light component depends entirely on the summation of $({\partial \mu_m}/{\partial N_i})_{T,P,N_j[i]}$. That is precisely the point: how could I measure or calculate that derivative or, alternatively, variations in the chemical potential by adding mass to the mixture?
Of course, one could use a cubic equation of state or the like and calculate fugacity and chemical potential etc. But I am interested in the more fundamental aspects of thermodynamics. One of them is that, given P-V-T data and specific heats, in addition to the composition, one is supposedly able to calculate all the thermodynamic properties of a mixture or of a component.
Now, the practical side of this question is this: if there is a two-phase mixture of C1 and C20, under ordinary T and P, the vapour phase is almost pure C1, but the liquid phase is a mixture of both. For given P and T, composition can be measured somehow. Given an increase in pressure, say $\Delta P$, with coexisting phases, there must be an increase in the concentration of C1 in the mixture, $\Delta x_1$ (or $\Delta N_1$, if the number of moles is to be used instead of concentration). The change in composition can obviously be calculated by equating the chemical potential of C1 in the vapour and in the liquid phases. While calculating the change in chemical potential in the vapour phase is quite easy because it is pure C1 and
$d \mu_1 = \bar V_1 dP$,
the chemical potential in the liquid phase depends on the above mentioned derivative:
$$d \mu_1 = \bar V_1 dP + \sum_{m=1}^2  \left(\frac{\partial \mu_m}{\partial N_1}\right)_{T,P,N_2} d N_m $$
Finally, by "measuring" chemical potential I mean any way of obtaining it from other variables that can be measured by ordinary instruments in a lab. And, just to mention, I have used the notation of Modell and Reid, "Thermodynamics and its applications".

Dear fellows, I've got no answers so far. Is this question too complex or too badly written? Or perhaps it is not suited for this forum?


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Need clarification on pH titrimetry .
I found somewhere in this site that question The reason behind the steep rise in pH in the acid base titration curve
when I tried to follow accepted answer, at some point I got stuck.

Answer says, that steep rise in PH must happen when d/dV([H+])( V is volume of added base in this case) is maximum. I disagree. It also depends on current [H+] concentration; maybe at some point d[H+]=1, and at another d[H+]=2, but if [H+]=0.1 in first case and [H+]=3 in second, then d/dV(pH) will be maximum in first case.
He differentiates this :

and gets this :

I tried to solve, but got different results.So, if I'm correct, can you give me mathematical proof why d/dV(pH) is maximum at equivalence point ?

Since $pH = -log(a_{[H^+]})$, $[H^+]_{initial}$ is irreverent when discussing the equivalence point. You can always show a full titration curve over all pH for a given acid-base titration regardless of initial proton concentrations. Therefore, equivalence point is independent of initial concentration. That being said, the fact that the equivalence point occurs when $\frac{pH}{dV}$ is at a maximum  has to do with the shape of the pH vs Volume graph. Since titration curves look similar to $y = tan^{-1}(x)$, the maximum slope will occur at the equivalence point (that would be (0,0) for inverse tangent). This can be further confirmed by taking the second derivative and comparing where the graph crosses 0 and comparing to the maximum of $\frac{pH}{dV}$.
As for the derivative, using a derivative calculator to ensure the correct answer, I got:
$2[H^+]-(\frac{C_{A}V_{A}}{V_{A}+V_{B}})+(\frac{C_{B}V_{B}}{V_{A}+V_{B}})$ which simplifies to $2[H^+]+\frac{C_{B}V_{B}-C_{A}V_{A}}{V_{A}+V_{B}}$. A simple way to determine if these are the same is to use a CAS (computer algebra system) and divide the answer provided by one of these answers. If the result is 1, the difference is only in simplification.
I hope this answers the question adequately.


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Planar transition-state of proline racemization facilitated by proline racemase
The racemization of proline can be facilitated by the enzyme proline racemase.

Since the active sites of enzymes complement the conformation of the transition-states of the molecules they act upon, the most effective enzyme-inhibitors often resemble these transition states.
This enzyme can be inhibited by the transition-state (TS) analog pyrrole 2-carboxylic acid.

The strength of the bond between the TS-analog and the enzyme, is approximately 160 times stronger than the proline-enzyme bond.
When I was first reading about this, I did not believe it because of the negative charge on the alpha-carbon. I decided to draw the resonance structure of the transition state, to find a double bond between the alpha and carboxyl carbons.
My guess is that this resonance structure would make the transition state relatively planar (as is pyrrole 2-carboxylic acid).
But my question is, how "planar" is the transition state, and how might one determine the exact bond angles at the alpha-carbon?
TLDR: How "planar" is the transition state of proline isomerization, and how might one determine the exact bond angles at the alpha-carbon?
(P.S. Sorry for the bad scaling of my ChemDraw pictures, I don't know how to fix it.)

Looks like the picture was transparent. I made it opaque and slightly shrinked it. To shrink it, add either m (for medium) or s (for small) at the end of the file name. For example, if it's https://i.stack.imgur.com/zCuY3.png just make it https://i.stack.imgur.com/zCuY3m.png instead.

@PrittBalagopal Use t (as in tiny) instead of s. Because using s will crop the image to a square (even if your original image was a rectangle) while t would leave the proportions untouched.

Unfortunately the only way to be confident that you know what the bond angles are is to have an x-ray structure. As general rule s(a) The effect of the negative charge will almost certainly be negated to a greater or lesser extent by a nearby amino acid and (b) the protein way well also distort the shape of the substrate while it is in the active site. Sorry to be vague but anything more precise  would be pure guess work.

@porphyrin Since the transition-state does not exist for a long period of time, crystallising an ES-complex would be impossible, correct? But, I guess, given evidence of the fact that pyrrole 2-carboxylic acid is able to permanently bind the enzyme would suggest that the transition state of proline racemization probably is (almost) planar.

I agree, in fact I believe that we do not know, from direct experiment  the structures of any transition state, even in solution.  So you have a real problem! I meant of the inhibitor in protein as a starting point as this may be measurable. Presumably the fact that this is more stable reflects the fact that the protein does not have to distort the inhibitor as much as it does the substrate so as to fit it into the active site.

I have not looked this enzyme up, but is there any covalent modification of the amino acid during the course of the reaction, e.g. by thiamine pyrophosphate (a known cofactor in amino acid racemisation)?

(Note also that the anion you have drawn is probably an intermediate, not a transition state.)

As others have written in the comments, there are large parts of your question that can't be answered by experiment. However, it is possible to come up with a mechanistic hypothesis (and even better, also an alternative hypothesis) and gather evidence that rejects one hypothesis and is consistent with another.
Structural data supports the mechanism you sketched for the PLP-independent racemases. The structure of proline racemase in complex with the transition-state analog (PDB ID 1W61) shows fairly symmetric binding interactions between the enzyme active site and the ligand:

In the figure (a screen grab from https://www.rcsb.org/3d-view/1W61?preset=ligandInteraction&sele=PYC), the inhibitor is labeled [PYC]700, and it has similar interactions on the left and the right (for example two main chain hydrogen bond interactions with the carboxylic acid group each, and one cysteine residue each in the vicinity of the inhibitor.
The binding pocket is fairly symmetric (view from the "backside" with respect to previous image):

In the paper reporting the structure, the authors point out "two Cys residues optimally located to perform acid/base catalysis through a carbanion stabilization mechanism". They also suggest that the sulfur atoms sandwich proline when it binds, "squeezing" it into a conformation closer to planar than it would be in solution.
However, as the proline-bound structure of a different, putative racemase shows, the active site does accommate proline with a non-planar alpha carbon:

http://www.rcsb.org/3d-view/6J7C?preset=ligandInteraction&sele=PRO

When I was first reading about this, I did not believe it because of the negative charge on the alpha-carbon. I decided to draw the resonance structure of the transition state, to find a double bond between the alpha and carboxyl carbons.

As the enzyme is stepping through catalysis, the enzyme active site binds to the molecule in different charge states, and itself changes charges as the cysteine residues accept and donate protons. In the cited paper, it is left open whether the reaction proceed in a concerted manner (one proton abstracted on one side while one is donated on the other) or via a carbanion intermediate. They do point out the the pKa of the alpha carbon is  "in the range 21–32", arguing against a long-lived intermediate.

But my question is, how "planar" is the transition state, and how might one determine the exact bond angles at the alpha-carbon?

As you can read from the excellent comments, there is no good experiment to get at the exact structure of the transition state, and it is not a given that the transition state is planar. However, if the carboxylate group is on one side of the ring before the reaction and on the other side of the ring after the reaction (and stays attached to the alpha carbon throughout), at one point on the reaction coordinate the structure around the alpha carbon has to be planar (or some version of trigonal bipyramidal for a concerted mechanism). Therefore, the active site has to be able to accommodate that structure.
It would be possible to study this system using computational methods, to at least get a sense of the energy landscape in the absence of enzyme. The calculation with enzyme is complicated by the fact that the enzyme undergoes fairly large conformational change upon ligand binding.

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text:
What substances in ash may create a green colour and are also soluble in water?
I have an ash that is high in $\ce{K2O}$ ($ \gt 30 \text{%}$), $\ce{SiO2}$ ($  \gt 10 \text{%}$) and $\ce{CaO}$ ($ \gt 10 \text{%}$), with $\ce{P2O5}$ ($5 \text{%}$) and $\ce{MgO}$ ($3 \text{%}$) also present. When the ash is heated to above $700^\circ \pu{C}$ for $16$ hours for decarburizaton, a color change is observed, from beige to a shade of turquoise, which at higher temperatures ($1300^\circ \pu{C}$ for $10$ mins) becomes a deep shade of green during cooling (the green substance appears to be condensing through the sample).
Whatever substance that is causing this green color appears to be inhibiting melting within the sample (melting temperature should be around $1200^\circ \pu{C}$, according to ash fusion testing), and is soluble in water. I've found through XRD that the ash has significant peaks that correspond to $\ce{KCl}$ in the form of sylvine, but I'm not sure that there is enough chlorine to cause such a deep color change, or that it could cause the color change at $700^\circ \pu{C}$.
So, my question is, does it seem possible that $\ce{KCl}$ is what is causing this color change, and are there other possible candidates that I should test for? (For instance, I've read that potassium manganate will dissolve in water, resulting in a green color too, but I'm not sure how feasible that is).

A little impurity can go a long way in the color world.  Replace 1% of aluminum atoms in $\ce {Al2O3} $ with chromium and you get ... https://en.m.wikipedia.org/wiki/Ruby.

Yeah, that's part of my problem: the ash I'm looking at has so many components! At first I thought it might have been a phase change to Amazonite due to some lead impurities, but after seeing the colour change appear to be from condensing I'm not so sure any more. Any ideas that might narrow  my options a bit are appreciated!

Is it $\ce{FeSO4}?$ Wikipedia says: All the iron(II) sulfates dissolve in water to give the same aquo complex $\ce{[Fe(H2O)6]^2+}$.

I haven't got a sulphate analysis of the ash yet, but oxide analysis only accounts for some 65% of the ash, and carbonates only 3%, so that's definitely something I'd like to look into. Thanks!

Most sulfates decompose at the temperature where you see the deep green color.  Chlorides are more thermally stable and have already been found by X-r a y diffraction.  Iron or even nickel chloride, maybe?

We've just finished some hot-stage XRD scans, and found that we created a pretty thick blanket of KCl over the top of the sample during cooling. I'm starting to think that this KCl coming into contact with moist air while the sample is still hot is what's causing the colour change. There are other peaks to look at though, so I'll have to see how iron and nickel fit if at all. Thanks for the help!


text:
How can I convert weight percentage of calcium chloride hexahydrate to calcium chloride dihydrate?
I have a chemical formula with $27\%$ $\ce{CaCl2.6H2O}$ and $43\%$ water with the balance of $30\%$ being other chemicals.
I want to convert the $\ce{CaCl2.6H2O}$ and water percentage to $\ce{CaCl2.2H2O}$. I understand molecular weight percentage of $\ce{CaCl2}$ is higher in $\ce{CaCl2.2H2O}$ so I expect the percentage of calcium chloride will increase but I'm unable to calculate by how much.
Can you please show me how to calculate?

Do you understand that you're trying to do $\ce{CaCl2.6H2O -> CaCl2.2H2O + 4H2O}$? Just assume that in the beginning you had $\pu{100 gm}$ of the total mixture. So that makes the weight of $\ce{CaCl2.6H2O}$ present in the intial mixture to be:

 $27\%$ of $\pu{100 gm = 27gm}$.

Next, calculate the number of moles of $\ce{CaCl2.6H2O}$ you had in the beginning by:

 dividing the mass you've just obtained by the molecular mass of this compound.

Now according to the stoichiometry of the reaction I've given you, when $1$ mole of $\ce{CaCl2.6H2O}$ converts, you get $1$ mole of $\ce{CaCl2.2H2O}$. Can you calculate the mass of $\ce{CaCl2.2H2O}$ thus obtained? So now you've the mass of $\ce{CaCl2.6H2O}$ and $\ce{CaCl2.2H2O}$.
To calculate the percentage of $\ce{CaCl2}$ in $\ce{CaCl2.2H2O}$ all you've got to do is:

 $\displaystyle \frac{\text{Mass of } \ce{CaCl2}}{\text{Mass of } \ce{CaCl2.2H2O}} \times 100\%$

You can carry out this step for $\ce{CaCl2.6H2O}$ on your own and then finally calculate the percentage change.

Hi Berry,Thank you so much for your really well laid out answer.

Hi @Berry Holmes,Thank you so much for your really well laid out answer. OK bear with me as i'm new at this. Is the number of moles of Cacl2.6H20 27/219 =0.123 ? To calculate mass of CaCl2.2H20 147 g .? therefore is the % of cacl2 in cacl2.2h20 111/147? = 75% ? And so is the % of cacl2 in cacl2.6h20 108/219 = 49.3%. So therefore75%/49.3% = 1.52. Does this mean Berry to convert 27% ofCacl2.6H20 to a % of Cacl2.2H20 = 27% * 1.52 = 41.07 % Cacl2.2h20 = 27% Cacl2.6H20 ?????. Thanks for your efforts Berry I really appreciate it  :)

@jane If your calculations are correct I see no flaw in your reasoning, and that's a good method; I hope you're getting a correct answer! Also, I'm pretty sure you're running low on time, but the next time you ask a question here please add whatever you've done (that's called showing your effort) and we'll all be glad to help you!

thank you for your reply and feedback. I still feel like I am fumbling my way through. Was my calculation figure above of 75/49,3 = 1.52 the correct figure to  convert CaCl2. 6H20 to CaCl2.2H20 ?

Thank you so much. You see even if my reasoning and hopefully  calculations are correct i dont know if they are actually correct. Hope to hear from you and sorry to waste your time Berry

@ Berry Holmes - workings : CaCl2. 2H20 =Ca - 1x40.1gm, Cl 2x35.5 = 71g, H20 =2X18 =36gm. So % Cacl2 = 71 + 40.1 / 40.1 + 71 + 36 =111.1/147.1 =75.56%

CaCl2. 6H20 = Ca - 40.1gm, Cl 2x35.5 = 71 gm, H20 = 6*18 = 108gm. So % CaCl2 = 111.1 / 40.1 + 71 + 108 = 111.1/219.1 = 50.7% (NOT 49.3 ABOVE)

Therefore 75.56/50.7 = 1.49, so to convert 27% of CACL2.6H20 to a % of CACL2.2H20 = 27% * 1.49 = 40.24%  @Berry Holmes how does this look to you ?

@jane I'm sorry for not being able to check your work right now, but please bear with me, give me a day I'll revert back once I get home.

no rush @Berry Holmes. I am looking forward to your words of wisdom


text:
How is rotenone a polyphenol?
How is it a polyphenol if it doesn't have a hydroxyl group?
If it isn't which subclass would you put it in, and which functional groups does it contain?

It is not a polyphenol as it does not contain hydroxyl group. It has isoflavonoid like structure(contains ketone as functional group).

How is rotenone a polyphenol?

It is not. Wikipedia says: Polyphenols are $\ldots$ characterized by the presence of large multiples of phenol structural units.
And if you're to pull up an image which gives you the structural formula of rotenone you get (Image source: https://upload.wikimedia.org/wikipedi...)

Do you see any phenols in this structure?

If it isn't which subclass would you put it in

According to this article I've found:

Article source

In the late $1920$’s, the chemical structure of rotenone was determined. Rotenone is a member of a class of compounds of related molecular structure referred to as isoflavones.

And finally,

which functional groups does it contain?

With a keen eye, if you look at the structure carefully, you'll at least be able to spot a ketone.

Thanks, thats what I too thought, but an article I read mentioned it as a polyphenol. Im still horrible at chemistry though, so can you tell me if the two benzen rings count as a functional group (like an aromatic hydrocarbon), or does that also get ruled out because it isn`t a hydrocarbon?

@user45681 I'd call that double bond (top-right) a functional group. As for benzene, this Purdue Univ's handout says benzene rings are functional groups but one of the answers here says otherwise: https://chemistry.stackexchange.com/questions/41953/what-is-the-functional-group-derived-from-benzene-called :(

OK, would you just call the functional group top right a methyl group, or would you have to point out the double bond? Is that what makes it special? It`s still a methyl right?

@user45681 Oh wait, I'm sorry, yes since this has benzene rings in a bonded form, then they can be called functional groups.

@user45681 That methyl group is a separate functional group (called an alkyl) and that double bond is a different functional group.

OK, is there no other name for alkyl that doesn`t have three hydrogen? To summarize: retonone is an isoflavone, with a methyl group top right, two benzene rings and a ketone (in the middle).  Is this right?

@user45681 I didn't get that, methyl itself means $\ce{-CH2}$ , $\ce{=CH2}$ is called a methylene group. Yes that's right, also all the $\ce{R-O-R'}$ groups you see there are ethers.

Ok, got it (I keep mixing up names form Norwegian to English). Thank you so much for your help Berry!

@user45681 The pleasure's all mine, also you can accept one of the answers here to mark this question has been solved. To do that please click on the green tick mark.

It looks like ChEBI classifies it as an isoflavone which are ketone-containing hetro ring systems with an oxygen.


text:
What is the name of this 3 open tube and where can I purchase one?
I would like to do some plasma activated water tests.
Does anyone know the name of this three open tube along with where I can purchase one?
The screenshot came from this video if you need a better angle to look at it.  Here's the video, the screenshot came from
YouTube

http://www.sigmaaldrich.com/labware/labware-products.html?TablePage=17213889

The flask pictured in your question is simply a three necked round bottom flask. They're available commercially in a variety of sizes with different configurations of ground glass joints (personally, I've always found it easier to buy single neck round bottom flasks and then get our glass blower to append the joints where I want/need them).


text:
Where can I find (citable) FTIR spectra of CuCl and CuCl2?
I can't seem to find any published/citable infrared (FTIR) spectra in the 4000-400 cm-1 range of CuCl and CuCl2. Does anyone know of a source available, either free or paid?

If your institution holds a subscription to Elsevier's database Reaxys, than you may find, based on chemical name, molecular formula, or CAS number (the later for example from a catalogue of a supplier of chemicals) links to primary literature in the section of spectroscopic properties. Reaxys tries to cover both organic materials (formerly deposited in the Beilstein database) and inorganic / metalorganic materials (formerly deposited in the Gmelin database).  Otherwise, Scifinder from the ACS equally points towards primary literatures, too.
Beside the databases behind a paywall, freely accessible databases like the NIST webbook may be worth to consult; the entry about $\ce{CuCl2}$ here
mentions in the references already one with IR data of gaseous $\ce{CuCl2}$ (here).

Thank you. But unfortunately I can't access those databases, only paid articles. :( The NIST webbook has entries for both, but CuCl (or ClCu as they put) has no references in it at all. I checked the article you mention (thanks), but it has no data in the 4000-400 cm-1 range (which is what I need, and I will edit the question now to state it, sorry).


text:
Number of cyclic isomers of C4H6
I faced a question

How many cyclic isomers are possible for $\ce{C4H6}$?

I and my friend found the following four.

But the answer key says there are 5. So what's the other one?

Your formula and the figures you've drawn don't match, the formula contains $3$ carbons and each of your figures has $4$ carbons.

We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details.

There's also bicyclobutane.

I find it easier to deal with such questions using the Degrees of Unsaturation...

The main difficulty with this question is deciding whether or not to include trans-cyclobutene.

@EashaanGodbole Link (where I can learn this stuff extensively) please?

@Mockingbird You'll find it easily enough on the internet. Check YouTube too.

@DavePhD Why not trans-cyclopropenes if you wanna talk about hypothetical stuff.

@Mithoron ron lists them too here https://chemistry.stackexchange.com/questions/19475/hydrocarbons-with-only-4-carbon-atoms.  I'm interested if there is a corresponding theoretical potential energy minimum for any of  them (trans-cyclopropenes or cyclobutenes).  Maybe I should ask a question about that.

@DavePhD Well, he listed everything

@Mithoron This question can be useful to the forum if I just change the title with " How to determine molecular/geometric structure from a chemical formula. Mersentery's answer is the one a student will look for.

jee 2008 right .

You can answer such questions using D.U. (Degree of unsaturation)
The formula is $\mathrm{C + 1 - }\frac{H + X - N}{2}$
C = Carbon.
H = Hydrogen.
X = Halogen.
N = Nitrogen.

If you get D.U. to be one, then in the structure there could be:

1 double bond.
1 ring.

Let's take an example of $\ce{C4H8}$ which has D.U. equal to one.

It can have three structures of 1 double bond:

It can have two structures of 1 ring:

Now for D.U. equal to two, the possibilities are:

2 double bond.
2 ring.
1 double bond and 1 ring.
1 triple bond.

Let's take your example of $\ce{C4H6}$ which has D.U. equal to two:

It can have two structures of 2 double bond:

It can have one structure of 2 ring:

It can have four structures of 1 double bond + 1 ring:

It can have two structures of 1 triple bond:

So, the answer is 5 cyclic isomers as you can see above.

How do you decide whether or not the cyclobutene has both cis and trans isomers?

@DavePhD It cannot show geometrical isomerism as far as I can see.

"Now for D.U. equal to two, the possibilities are:
2 double bond.

2 ring.

1 double bond and 1 ring.

1 triple bond." how do u know that?

Drawing angled $sp$ carbons is a sin bordering on blasphemy.

@Mesentery Hypothetically, with respect to the double bond, the two hydrogens  could be on one side and the two carbons on the other (cis) or there could be one hydrogen on each side with the ring stretching diagonally across the double bond.  But I don't think it is known to occur experimentally for a ring this small http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/stereochemie/cyclische_stereoisomere.vlu/Page/vsc/en/ch/2/oc/stoffklassen/systematik_struktur/cyclische_verbindungen/carbocyclisch/cycloalkene/physikalische_eigenschaften.vscml.html

@Mockingbird That I have studied. But you can also make out like D.U. Is 2 means there should be 2 double bonds that are removing the hydrogen from the formula. And it can have 2 rings that all that. You will have two check what are the possibilities that can remove hydrogen from an alkane

@DavePhD Thanks for the information. I have studied that cyclic alkenes does not show geometrical isomerism having carbon less than 8.

@IvanNeretin Apologies for committing the sin :)

@Mesentery well perhaps you will find this article interesting, concerning the limit supposedly being 8 http://pubs.acs.org/doi/abs/10.1021/ja055388i and also http://pubs.acs.org/doi/abs/10.1021/jo00389a067

Now that's much better. But look, a carbon with two double bonds is also an $sp$ carbon.

@IvanNeretin You are referring to the Allene. Ok Thanks.

@Mersentery Is the formula empirical?

@Mockingbird You can have an idea how the formula is made from here. BTW I don't understand what you mean by empirical.


text:
What's the reaction when bleaching hair with lemon juice (citric acid)?
Is the acid working as an oxidizer or reducer with the melanin of the hair breaking down the bonds?

Hair is mainly keratin, the same protein found in skin and
  fingernails. The natural color of hair depends on the ratio and
  quantities of two other proteins, eumelanin, and phaeomelanin.
  Eumelanin is responsible for brown to black hair shades while
  phaeomelanin is responsible for golden blond, ginger, and red colors.
  The absence of either type of melanin produces white/gray hair.

I think the citric acid (in lemon) acts as a bleaching agent. The "bleach" reacts with the melanin in hair, removing the colour in an irreversible chemical reaction. The bleach oxidises the melanin molecule.
The illustration of the molecules perhaps might give a better picture on which groups are oxidised in the process.
The melanin is still present, but the oxidised molecule is colourless. However, bleached hair tends to have a pale yellow tint. It behaves as an oxidising agent at least according to my findings.
Acknowledgements

https://www.thoughtco.com/salon-hair-color-chemistry-602183
http://www.keratin.com/as/as002.shtml

Lemon juice is rather a reducing than an oxidizing agent.

@ave Is it? Because for the particular reaction to work it behaves as as weakly oxidising agent. I'm quite aware citric acid is a better reducing agent in most of reactions, but if that the case in this "bleaching reaction" I stand corrected, in research though I couldn't get substantive evidence to support that theory.

Lemon juice contains citric acid and ascorbic acid. Both are used as antioxidants in the food industry. I doubt that citric acid could oxidize anything but strong reducing agents. I also have serious doubts about the hair lighting power of lemon juice (pure, no light, etc) in a scientific experiment. ;-)

Yes I agree with with, actually this works with powerful oxidising agents like hydrogen peroxide,  however this reaction some argue that it works but the hair has to be exposed long enough in sunlight, but haven't tried I just did a bit of research and put my findings in the answer above.


text:
How many cis-trans isomers does 1-O-stearoyl-2-O-arachidonyl-phosphatidyl-inositol-4',5'-bisphosphate have?
I saw an exam that had a question asking the number of cis-trans-isomers of 1-O-stearoyl-2-O-arakidonyl-phosphatidyl-inositol-4',5'-bisphosphate:

The "correct" answer in the exam is 1023, but this is clearly not correct, because the stereoisomerism of the inositol ring is not called cis-trans-isomerism.
So, obviously only the arachidonyl chain has cis-trans isomerism. But if an isomer has several of the double bonds in trans orientation (not just one), is is still correct to call it a cis-trans isomer of the original molecule. If it is, I would say that this molecule has 15 cis-trans isomers. If it isn't I would say that it has only 4 cis-trans isomers.
Which one is correct?

There can be cis and trans in the cyclic rings. The substituents can be axial or equatorial. Also, 15 is too low a number for the number of stereo isomers. Just consider the double bonds, you have four of them, each has two choices, cis or trans. That gives $2^4$ isomers on its own. Just 16 isomers from the double bond chain, there's way more possible from the ring. Keep trying, you'll get it soon.

The isomerism in the ring is not called cis-trans-isomerism in this case. (See my other question at https://chemistry.stackexchange.com/questions/74853/are-glucose-and-galactose-cis-trans-isomers-of-each-other/74858) And I would not say that the molecule is a cis-trans isomer of itself. Therefore it can't have more than 15 cis-trans isomers.

Perhaps the answer is correct. If by cis/trans isomers the question includes the four stereogenic double bonds of the arachadonic acid chain and the six stereogenic carbons of the inositol ring, then there is a total of ten or 210=1024 isomers. But the answer is 1023 stereoisomers. That is because the question says "cis-trans-isomers OF...", which means NOT including the one shown. [210-1=1023.] Had the question asked "How many stereoisomers OF ...", then the lone stereogenic center of the glycerol moiety would have to be included. Then the answer would be 211-1=2047.

The stereoisomerism of the inositol ring is not called cis-trans-isomerism.

Au contraire! Read the first sentence here in the IUPAC Gold Book. https://goldbook.iupac.org/html/C/C01092.html

Ok. I see your point. But what would you say about this: https://chemistry.stackexchange.com/questions/74853/are-glucose-and-galactose-cis-trans-isomers-of-each-other/74858
There everybody seems to agree that glucose and galactose are not cis-trans-isomers of each other.

I read your link. The problem with using cis/trans with a pyranose is that it is too cumbersome to express all of the relationships. As @NotEvans states cis/trans is usually reserved for rings with two substituents (and "simple" 1,2-disubstituted alkenes).  It is more convenient to use glucose as a frame of reference and refer to D-galactose as the C4-epimer of D-glucose or alpha-D-mannose as the C1-epimer (anomer at C1) of beta-D-mannose. The point of the original exam question was to see if one recognizes elements of stereogenicity. Four double bonds would have been too trivial.

Yes, I understand that that's what the authors of the exam were thinking, but I'm interested in figuring out if they used the terms correctly. However, I have to admit that the IUPAC Gold Book definition seems like it would allow the interpretation of having 1023 cis-trans-isomers for this molecule.

They knew what they were doing.  Time to put the issue to bed.


text:
Why are dipoles "permanent/induced dipole permanent/induced dipole" and not just "permanent/induced dipole" once?
My teacher would always say "induced dipole induced dipole" and while it annoyed my slightly (as you were saying the same word twice) it didn't really bother me, but now looking through some exam papers it's called the same thing "induced dipole induced dipole", so why is it repeated twice? Would I be "wrong" for just saying "induced dipole" once? If so, what's the difference between the two?
(Sorry if this seems like a simple or stupid question, just I'm always curious and the internet doesn't really work for searches like this.)

The correct typographical syntax would be "induced dipole–induced dipole", which should help to clarify why it is being repeated twice. Dropping the adjectives, you get "dipole–dipole", which is easily understood: two dipoles are interacting. Unfortunately, in verbal speech, this kind of nuance isn't there to provide the additional clarification, and many people are sloppy in written grammar, too.

"Sorry if this feels like a simple or stupid question"  In the words of a certain someone from the Physics.SE "There are useful questions, and  there are questions that aren't so useful...but there's no such thing as a stupid question". This is a pretty good question, and I for one, appreciate that you took the time to post it here. All questions are welcomed at Chem.SE (except poorly formatted HW questions), so feel free to post here. Looking forward to your future contributions to Chem.SE. Cheers! ;)

Because it takes two to tango.
Dipoles interact with each other. A Lone dipole has nothing to interact with (other than an electric field, but if we ignore some externally applied macro field, there is nothing for a lone dipole to interact with).
So molecules with an inherent dipole (like water or chloroform) interact with each other. One molecule's dipole interacts with the other molecule's dipole. So you would never say "dipole interaction" only "dipole-dipole interaction". The repeated word is because it takes two dipoles to interact.
The same pattern applies to non-polar molecules with little or no inherent dipole. For example, benzene. Benzene has no built-in dipole, but the electrons in its bonds are fairly polarisable (which basically means it is easy to induce a dipole in them). So benzene molecules do interact but via London or van der Waals forces which are much weaker than the reactions of molecules with inherent dipoles. But one way to describe those weaker interactions is (simplifying a lot) to see them as forces created by fluctuation and temporary quantum-mechanical induced dipoles. Imagine the electrons in one benzene molecule stray from perfect symmetry in their distribution and create a temporary dipole; this may interact with the electron clouds in other benzene molecules an induce other temporary dipoles (or interact with other existing temporary dipoles). Either way the interaction is an "induced dipole-induced dipole" interaction.
So however you understand the interactions, to describe them properly you have to repeat the phrase twice.

You can also have "permanent dipole-induced dipole interactions", where, for example, the permanent dipole on the water molecule interacts with the electrons on the benzene and causes an induced dipole, which then has a dipole-dipole interaction with the water dipole. (The fact that the interaction can be mixed means that it's not redundant to specify "permanent" or "induced" twice - it's providing valuable information that the type of dipole is the same.)

Because it takes two items to have an interaction.
Two permanent dipoles interact.
Two induced dipoles interact.


text:
Is there intramolecular hydrogen bonding in salicylaldehyde?
Is there intramolecular hydrogen bonding between the hydroxyl and carbonyl groups?

I think personally is that it does but what is concerning me is that I have got no adequate answer or response to my suggestion, so if anyone could help on this one, that would be great.

It can indeed hydrogen bond, this can (to a certain extent) be observed by NMR. Due to the internal hydrogen bonding, the phenolic proton is surprisingly consistent, and shows no dependence on concentration.

Image reproduced from Basic one- and two-dimensional NMR spectroscopy, Wiley, 2010.

A simpler way to check for the presence of hydrogen bonding is to compare the acidity of salicylide with its para isomer.

Case 1: No hydrogen bonding
Let us assume that there is no hydrogen bonding for the following analysis.
The $\ce{-CHO}$ group is an electron withdrawing group ($-M$ effect). The effect of acidity due to mesomeric effect at both ortho and para positions are identical. Therefore, we can ignore the effect of mesomeric effect for our analysis.
However, the $\ce{-CHO}$ group is slightly electron withdrawing group due to inductive effect. The strength of inductive effect varies with distance: farther away you are, the lesser is the effect. As the aldehyde in ortho position is closer to the carbinol carbon, effect of inductive effect on the carbinol carbon is more pronounced when compared to the case where the aldehyde is in the para position. As a result, the positive charge developed on the carbinol carbon is higher when the aldehyde is at ortho position. Therefore, the acidity of salicylaldehyde must be higher than that of its para isomer.
If there was no hydrogen bonding, then we should expect the acidity of salicylaldehyde to be higher than its para isomer.
Case 2: Hydrogen bonding
Salicylaldehyde forms intramolecular hydrogen bonding and its para isomer forms intermolecular hydrogen bonding. The effect of acidity due to intermolecular hydrogen bonding is lesser than the effect of hydrogen bonding due to intramolecular hydrogen bonding. This is because intermolecular hydrogen bonds can be broken easily as they are between two molecules and the molecules are in thermal motion.
If the effect of difference in hydrogen bonding in the two isomers is stronger than the effect of difference in inductive effect shown by the aldehydic group in the two isomers, we should expect a reversal of acidity order of that obtained in the previous case, i.e: salicylaldehyde is a weaker acid than its para isomer.
Experimentally measured values:
Molecule                   pKa
-------------------------------------
Salicylaldehyde            8.37
4-hydroxybenzaldehyde      7.61

As the measured values suggest, intramolecular hydrogen bonding exists in salicylaldehyde.


text:
Methods to extract sucrose from tea after it's been stirred in?
Sugar is often labelled  "white poison" now-a-days and a lot of negative health consequences happen due to increased sugar consumption.
Keeping this in mind, if I were handed a cup of tea (with sugar already stirred in), what methods/processes can I employ to separate/remove the sugar from it?

Why would you add "white poison" to your tea or other foods in the first place if you want to reduce sugar in your diet? Seriously, you're not going to find an after-the-fact means of reducing sugar content that is remotely as cheap, simple and safe as using low-sugar foods to begin with.

The best way is to buy some green or black tea as you prefer and drink it alone. Then you don't need to drink other chemicals you used to removed the sugar... also if you like lemon tea or vanilla tea but a lemon or vanilla and use it as flavour for you tea. Now you can still ask GIVAUDAN to get the perfect tea.

@Hexacoordinate-C I always prefer to have green tea or coffee without suger. Question is about some chemical methods to solve a problem. Please look into my improved question with clear use case.

Have to wait for an hour? All taste of hot served tea will be gone. :( But anyway, thanks for your practical solution suggestions. :)

https://chemistry.stackexchange.com/questions/40905/is-it-possible-to-separate-remove-sugar-from-common-drinks-such-as-juice

Add yeast and lt it ferment the sugar


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Is this enough information to solve this problem?

Calculate the volume of a $\pu{0.100M}$ solution of $\ce{HNO2}$ and the volume of a $\pu{0.200M}$ solution of $\ce{NaNO2}$, that must be mixed to prepare $\pu{300ml}$ of a solution with a $\mathrm{pH=3.35}$.

Since this is a titration problem shouldn't I be given one of the volumes ?

If you're given the volume of one of the species as $x$ (say) then all you've to do to find the other species' volume is $300-x$. Then question, then, would be meaningless.

You need to use the Henderson-Hasselbach equation:
$$\pu{pH = pK_a + \log\frac{[Base]}{[Acid]}}$$
The $\pu{pK_a}$ of $\ce{HNO2}= 3.15$
Therefore,
$$10^{\pu{pH-pK_a}} = \pu{\frac{[Base]}{[Acid]}}$$
$$\pu{10^{3.35-3.15} = \frac{[Base]}{[Acid]}}$$
$$\pu{1.58 = \frac{[Base]}{[Acid]}}$$
Implies that the numerator must be $1.58$ times that of the denominator.
The base is $\ce{NaNO2}$; the acid is $\ce{HNO2}$. The base is two times as concentrated as the acid.
This means, for every $\pu{mL}$ of acid you use, you need $\pu{1.58/ 2 = 0.79 mL}$ of base.
If,
$$V_T = V_A + V_B$$
And,
$$V_B = 0.79V_A$$
Then,
$$\pu{300 mL} = V_A + 0.79V_A  \Rightarrow V_A = \pu{167.60 mL}$$

You can use $\ce{chemical formula goes here}$ for writing compounds $\pu{units here} for physical units and \Rightarrow for $\Rightarrow$.


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Will a film of oil on water retard evaporation (of water) and prevent air and atmospheric impurities from diffusing into it?
I vaguely recall hearing that a layer of oil on water will slow down the evaporation of water, as well as serving as a barrier that would prevent (or retard) the diffusion of air and atmospheric impurities into it. How accurate is this claim?
Some Background:
I want to make zinc/air batteries. $\ce{KOH}$ is commonly used in the electrolyte, but $\ce{CO2}$ in the air can react with the $\ce{KOH}$ and form carbonates which hinders the battery's performance. It seems like oil on top will shield the electrolyte from atmospheric $\ce{CO2}$. The $\ce{H2O}$ also must be kept from evaporating, which would otherwise result in changes to  concentration. $\ce{KOH}$ is also very hygroscopic and tends to attract water from air which would also cause variations in concentration.
When the batteries get discharged, $\ce{ZnO}$ is a final product. This should almost certainly precipitate and collect on the bottom of the tank since its considered insoluble in water.
Solar panels harness a vast amount of energy from the sun during the Summer. Storing surplus enrgy for use in Winter is a daunting task. A method that's been doing rounds these days: One could collect and reduce the $\ce{ZnO}$ back to $\ce{Zn}$, deposit it on plates/electrodes, and store these plates for Winter use. The plates could be put back in the batteries and the electrolyte (a fresh solution of $\ce{KOH}$) would have to be replenished.

Water "dislikes" oil and oil "dislikes" water". So if a molecule of water would like to go through the oil it will be hard for it. Then water doens't try to escape and stay under the oil. It might have a very tiny amount of it which find a way to escape but this is very very slow. For the diffusion of gas in oil I don't know how much is it possible for the air to go through it. I need to find some publications.

Long story short, it is "yes" to both questions. Oil would hinder the exchange between water and air, both ways.

Who edited my question? Now it doesn't mean the same as I typed it. I'll have to do some balancing of the equations. The O in the ZnO may come entirely from the air and it may be the electrolyte solution returns back to a solution of KOH once the ZnO is removed-no replenishing necessary. What happens is the Zn turns into Zn(OH)2 which disassociates into ZnO and OH  ̅  . There are rechargeable zinc/air batteries on the market which are not suitable and apparently not available to me anyway. Their configurations won't work well in this solar application-self discharge too easily.

Regarding the edits: Click on the "edited X hours ago" link to see the edit history and who edited what. Keep in mind they have nothing to gain except trying to help you, but they may well have made a mistake. All you can really do is to [edit] it back to how you like it.

It wasn't clear what was meant when the editor said "making the rounds". Now I assume this same idea has been discussed before. It would be silly of me to think only I had thought of this. The cathode side of the battery may be the difficult one; methods have been developed to keep the zinc from depositing there. Could someone point me to previous discussions on this?

I don't know about oil (fun fact, putting oil on water was a traditional approach to calming waves, thus the phrase "pour oil on troubled waters").  But cetyl alcohol has been used to decrease evaporation from reservoirs, which might be what you heard about.  For instance, here is a report on this, with lots of references.  They used a mix of steryl and cetyl alcohols (long-chain alcohols that will form a monolayer on the water).

Interestingly, I know of another chemical called aspartic acid which is made by Flexible Solutions Inc which is biodegradable for the same purpose u mentioned. I own some stock in this company. They are now claiming it slows evaporation by 45% to 47%. I think it lasts only for a few days though. They might get some contracts from San Diego for their reservoirs and are expecting more from Brazil. Anyway, I'm going in a different direction. I'm amazed at how many amp hours u can get out of 1 mol of oxygen in this reaction. I will show my figures.

A oil film definitely slows down evaporation. For example, when the meibomian glands don't work properly, there isn't enough "oil" in tears, and people who have a problem with these glands are subject to dry eyes.