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Stereocenter or not?
Consider 2,3,4-pentanetriol.Is the middle carbon atom a stereocentre?I know how to count the stereoisomers, but is the middle carbon a stereocentre by definition?Let us say we have a 2R and 4S configuration.Now if we change the orientation of the OH group at the 3rd carbon, we do get a different compound,but both the compounds are optically inactive, so it doesn't seem it should be called a stereocentre.If we consider 2R,4R, or 2S,4S if we change the configuration at C3, we get the same compoound.So I feel the carbon at C3 should not be a stereocentre.But it seems that in the first case, we are changing the C3 configuration and getting a new diastereoisomer, so C3 might be a stereocentre after all.Which one is correct?If C3 is a stereocentre, what would be its R,S nomenclature?

This earlier answer deals with a similar situation and should prove helpful.

The carbon atom in question is called pseudo-asymmetric. Check it out on the IUPAC page.


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thermoplastic polymers
I'm looking for polymer to fit injection molding (thermoplastic):
The final injected part (small part of volume approximately $\pu{300mm3}$) should be transparent, because a stainless steel tube will be inserted to a hole inside the part and adhesive using UV will be accomplished to attach the tube to the part.
Good mechanical properties of the polymer would be a bonus. For example, polycarbonate can be used, but I'm looking for additional material.

Hello and welcome to Chemistry.SE. Feel free to take the short [tour] or visit the [help] if you have any questions about the site. Regarding your question, I'm not quite clear on what you are looking for. For example, you said polycarbonate can be used: what is it that is desirable about polycarbonate for your project, and what is undesirable about it that makes you look for something else? Also, you refer to "good" mechanical properties, which is pretty unclear. I think if you [edit] your question to clarify just what your needs are, you'll have  better chance of getting a useful answer.

Lots of polymers absorb UV, so the UV-set adhesive inside the part may be the hardest issue.

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Why would I not see a reaction between aluminum and tin(IV) chloride?
I was performing an experiment where I was supposed to construct a modified activity series between certain elements. I decided to use chlorides with the other metals for single replacement reactions so I could compare them all in relation to one another for my modified activity series.
When I placed about one gram of aluminum into about five milliliters of tin(IV) chloride, I was not able to see any reaction taking place. However, aluminum is higher in reactivity on the activity series so I know the aluminum had to react to replace the tin, but why would I not be able to see it?

Did you activate the Al before adding it by etching the surface with acid? Al has a tenacious oxide surface layer that prevents it from reacting in many cases.

This is what I was afraid the reason was.  I did not now if it would lose that layer when added to the solution or not.  If you make this an answer I will vote it up for rep.

You need to activate the $\ce{Al}$ to remove the tenacious oxide layer that protects it from reaction. This can be done by a quick wash with dil $\ce{HCl}$, rinse with $\ce{EtOH}$ and finally $\ce{Et2O}$ before adding to the tin chloride.

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Measuring pH using pH paper makes solution change color
I have an unknown substance dissolved in water, and it's milky white. When I put in a pH paper, the paper shows that it is slightly basic at pH ~8. However, after putting in the pH paper, the solution changed color from milky white to a very faint blue. What possible compound could this be?

What about the volume of the analyzed mixture? I doubt that anyone can guess what the substance is, but my guess about the color is that the indicators on the pH paper might have been partially dissolved in the mixture. And, since the color of litmus paper at a pH = 8 is blue, it could be a possible explanation for the color, especially if you are dealing with a small volume of liquid.

It was in a microwell, so the volume was very small. I think the pH paper most likely dissolved in it, so your explanation is probably correct


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In an electrocyclic reaction with two pi bonds that result in a trans product, is a mix of products seen?

When I look at the highest occupied molecular orbital, the two methyl groups will end up being trans to each other. I realized based on which conrotation direction both methyl groups take that two different trans products are formed that do not overlap but the lecture only gave one of these products. Am I missing a concept here or missing something obvious?

The"other" trans-dimethylcyclobutene is the enantiomer of the one you've drawn above. The 2 trans products are formed in equal amounts to yield  a racemic mixture.


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Why does naphthalene have a higher melting point than both water and diphenylamine?
We learn that polar compounds have higher melting and boiling points than nonpolar compounds due to strong intermolecular forces such as hydrogen bonds and dipole-dipole interactions.
Naphthalene is an extremely simple and nonpolar compound, yet it has a significantly higher melting point than diphenylamine (a similar enough structure that is not only heavier, but has a dipole-inducing amine) and water (famously super polar).
What's going on here?

High symmetry, and molecular mass, pi-stacking.

@Mithoron Could you go into a bit of details about each of those in your answer? Like, why does high symmetry and pi-stacking increase melting point (from a molecular perspective)?

related https://chemistry.stackexchange.com/questions/8777/why-does-sif%e2%82%84-have-a-higher-melting-point-than-sf%e2%82%84

https://en.wikipedia.org/wiki/Stacking_(chemistry)

Pi stacking is explained in answer to this question https://chemistry.stackexchange.com/questions/60055/pi-electron-stacking-how-does-it-work/60095#60095

In comparisons such as this water needs to be excluded as it is exceptional in almost every respect due to its highly directional and strong hydrogen bonds.
As a broad generalisation the melting of solids is more dependant on the nature of the repulsive forces between molecules than on their attractive ones. Experimentally the enthalpy of melting ( Latent Heat or just Heat of Fusion) is approximately ten times less than the corresponding heat of vaporisation which indicates that repulsive forces play a greater role in melting than in vaporisation: in a solid molecules tend to be slightly closer together than in a liquid, thus may experience larger repulsive forces. Van-der-Waals forces tend to have weak orientational dependence, but repulsive forces do not as they are short range and  depend on the asymmetric shape of the molecules and can have a large effect.
Thus melting points are rather dependent on the geometry of the molecules, if they can pack well into a lattice then a high melting point is expected. If the molecule has a dipole this can cause both attractive and repulsive interactions depending on the relative positioning of molecules; head to tail can be attractive if molecules are face to face, but repulsive if in line one behind the other and vice versa if their relative orientation is changed.
In the case of naphthalene and the amine, clearly the naphthalene can stack nicely and so increase its attractive interaction relative to repulsive ones, whereas the amine is twisted and cannot so easily pack meaning that repulsion can be stronger and attractive interactions weaker (even with dipole) and melting points lower. As an example consider the  $\ce{C6H14}$ isomers, n-hexane mp -95 C, 2-methyl-pentane mp -154 C but boiling points are 69 & 60 C respectively and $\ce{C8H18}$ isomers, n-octane mp -57, 2-me-heptane mp -120 and bp 126 & 119 C respectively.
( Attractive forces largely determine the enthalpy of vaporisation, and by Trouton's rule their boiling points. The enthalpy of vaporisation is closely related to the (attractive) cohesive energy in a liquid.)


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Does "soluble in alcohol" imply ethanol?
Sources like Wikipedia and SolubilityOfThings say that copper (II) acetate is soluble in alcohol.
Does that mean any alcohol, or does "alcohol" in this context mean ethanol?

Well, unless noted otherwise it should mean that. This compound should be also soluble in methanol or glycol, but not in long chain alcohols.

Sound like they meant ethanol, but the substance will be soluble in other alcohols too.

The term "alcohol", without modification (i.e. "propyl" alcohol, etc.), in the context of something like a solubility table, almost always refers to ethanol (as it does in your sources for copper (II) acetate solubility).
Note, for example, that both pubchem and Wikipedia list alcohol as a synonym for ethanol. This is not the case for other alcohols like isopropanol.


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Selectivity of Hydroboration Oxidation

Will the addition happen across both double bonds? Ordinarily you add H and OH syn and the OH anti-Markovnikov to one double bond
Hypothetical product:

$\ce{BH_3} $ can attack any of the two double bonds. Now it all depends on how many moles are being reacted in the vessel. Let us suppose there is only one mole. In the below process there is formation of chiral carbon and hence there is a pair of enantiomer.
Now we consider the other one.
Here there is no chiral center and hence no optical isomerism.
Hope this was helpful to you....

Is the answer mixture of a single enantiomer? If so, I have a solution.

Theoretically, BH3 can add across either double bond. However, it's more likely to react at the less hindered double bond. I think that it's more likely that it reacts with both double bonds, rather than it reacting only at the more hindered one.

But aren't both possible as they haven't mentioned the number of moles reacting?

I personally feel that the addition should be on the left double bond as it has least number of neighbouring alpha hydrogens to give it a sort of hyperconjugative delocalisation ,and thus is relatively more reactive than the double bond at the right extreme.

That's true but in that case there will be no optical isomerism due to lack chiral centre.

Which in turn implies that option d is meant to be correct.

It is anti addition and see

Small amount of diastereomer is formed but isn't enantiomer a major product


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Why is BDE low in Iodine, but not behaves like Fluorine?
When halogens have a low Bond dissociation energy, BDE, their free radicals are stable.
$\ce{Cl-Cl}$ (BDE: $\pu{58 kcal/mol}$) and $\ce{F-F}$ (BDE: $\pu{38 kcal/mol}$). $\ce{Cl}$ free radical is less stable than $\ce{F}$ free radical. But why that doesn't happen with iodine ($\ce{I-I}$, BDE: $\pu{36 kcal/mol}$). Iodine free radicals are unstable.

Where have you found the information that the iodine free radicals are unstable? It does not sound right for me.
BDEs of chlorine, bromine and iodine decrease down the group as the size of the atom increases: 243, 192, and 151 kJ/mol. The BDE of fluorine, is however, lower (158 kJ/mol) than those of chlorine and bromine. Here you were right! But the free radical stability depends also on other factors. It increases with increasing the atom size and with decreasing the electronegativity. Additionally the stability of the free radical decreases as the orbital is held closer to the nucleus. Thus, as we can expect, the iodide radical will be the most stable, and the fluoride - the least./
Btw correlation does not mean causality! (:

I agree with you, but is not the Iodine free radical more radioactive than Fluorine one? I mean, Iodine can react with any Hydrogen, meanwhile Fluorine is selective. Thanks.

Radioactivity is from the other chapter of the book and has nothing in common with radicals. And about reactivity ... well I would say it will be completely opposite to your statement, which is kind of vague btw. I highly recommend to you go through general and inorganic chemistry to sort the major principles out. Btw may you please answer my previous question? Will appreciate it.


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Why won't removing the salt bridge cause the battery to recharge?
If I understand correctly, the salt bridge in a battery is used to move positive ions towards the side that is becoming more negative, and negative ions towards the side that's becoming positively charged, to prevent the reaction from quickly reaching equilibrium. The ions in the salt bridge never leave the salt bridge.
But if we removed the salt bridge, wouldn't the forces between the $\ce{SO_4^{2-}}$ ions or $\ce{Zn^{2+}}$ ions cause the reaction to go in reverse and cause them to reattach to the electrodes?

Why is this up for a bounty? Related: https://chemistry.stackexchange.com/questions/12332/is-there-always-a-need-for-a-salt-bridge-to-let-a-galvanic-cell-work-continuousl

The ions in the salt bridge never leave the salt bridge.

That's not true.  The ions literally have to move to balance the charge of the gain / loss of the electrons at the electrodes.  (You even allude to this in your question.)  One of the ways that a battery can run out is that the bridge is used up.  You could views this by using a deeply colored ions (e.g. MnO4-) to see the migration.
While the salt bridge is in place, there is no build up of charge.  Electrons moving through the wires are offset by the ions moving through the salt bridge and into the respective half-cell.  Thus, if you remove the salt bridge, there is no net charge that will cause anything to reverse; there is no 'force' that will cause ions to revert.
If you want to reverse the reaction (and charge the battery), you have to force electricity back in the other direction, using more voltage than what you got out of the battery.  This doesn't work for all batteries.  (I.e. this is why only some batteries are rechargeable).  As you force electricity backwards through a battery, any movement of ions from the salt bridge means that the half-cell might now be contaminated with chemicals other than the ones you want in your half-reaction (unless you designed your salt bridge just right).  Thus, the "right" ions may not go back to being uncharged (or vice versa).

How the bridge used up? I think the solution ions moving across the bridge.

"used up" was a bit inexact.  Salt bridges have to be designed in such a way as the ions don't interfere with the half-cell reactions.  If the ions in salt bridge are depleted (all migrated into the half-cells), ions from the reactions can migrate between the half-cells.  If these ions interfere with the reaction in the new container, voltage will change.  Most intro chemistry texts just barely go over salt bridges and their construction, b/c they are more concerned with the basic idea of batteries.  But, in the construction of long lasting batteries, salt bridges have to be carefully crafted.

I think you are misunderstanding the purpose of the salt bride in the voltaic cell. It acts to prevent a charge imbalance as the electrons are sucked over to the cathode. The negative ions flow through the electrolyte to complete the circuit. Just like cutting a wire to break the circuit, removing the salt bride would stop the flow of electrons.
The battery wouldn't recharge because the ions would stay in solution and the solid metals would stay as they are. Nowhere for the electrons to go without the connection.

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Why bother to standardize a solution?
I have been preparing lessons for some of my students about preparing acidic and basic solutions, and I keep finding pre-made labs in which the students "standardize a basic solution" by titrating it with an acid of "known" concentration.
The problem is that the students create a solution of NaOH by measuring a certain mass and adding it to a volumetric flask, filling with water to the desired volume.  But the students will do the same thing with an acid (potassium hydrogen pthalate, for example).  They will measure out a specific mass, do the calculations to determine the molarity of the KHP solution, and then do the titration to determine the molarity of the NaOH solution.  But if the students took the mass of the NaOH in the beginning, can they not just use that information to determine the approximate molarity of the solution?
As the title says, why bother to standardize?  If our NaOH solution had some room for error in measurement and calculation, surely titrating against another solution of "known" concentration would only introduce more error, no?

Usually, if we're doing these for basic application or experimentation, standardization doesn't really matter. But when it comes to anything analytical where you start to involve calculations, standardization is a must. This is done with NaOH because it's hygroscopic and readily sucks up the moisture in the air. So what is being weighed isn't totally NaOH, but also the moisture that it has absorbed. So almost always the concentration will be lesser than what is sought to be prepared because of this. Tedious as it may be (because you also have to heat up the KHP for around an hour before titrating), standardization brings you closer to the true concentration, but not exactly on the mark. At least in the end, calculations using the standardized concentrations will be analytical. This is why they never introduce standardization in general chemistry courses, I only got to grips with it when I took up Analytical Chemistry, and back then it was really tedious to do.
For example, I actually experienced this before where we had to prepare a 1M stock of NaOH, but after standardization it turned out to be around 0.8 (maybe due to the humidity in the lab at the time).

Fascinating take on it.  My textbook is a high school textbook from about 2008, I think.  Almost every lab the publishers provided with the book involves standardization, which left me awfully curious, as I had only done it once before myself.

I think it'd be something great to teach your students. Then you could follow it up with another experiment that uses the standardized solutions like the titration of Vitamin C for example if they're at that level.

I don't doubt they're at that level, but my budget isn't at that level.  Maybe next year.  Thanks for the great idea!

Nope, it will not introduce, it will decrease the error.
There are a list of compounds which are used for preparation of various standard solutions and appropriate protocols. The chemical structure for such compounds does not depends on external factors (following the protocol) and can be used to determine the concentration precisely. Additionally more sensitive methods can be used to determine the final concentration of standard solutions, e.g., UV-vis spectroscopy.
To understand why you don't want to calculate the concentration of sodium hydroxide solution based on its weight and why this method will introduce more error, just let solid NaOH sit on a bench top for a little bit (please use a weighting paper/dish for it). Let me know what did you find out! (:

I'm guessing it's hydroscopic, and that might introduce some mass error.  I had not thought of that.  Since my NaOH is from the late 90s, it might be wise to do this after all...


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Mechanism of acid-catalyzed addition of water across an alkene
My book tells me that when 1-propene reacts with water and sulfuric acid, I get 2-propanol. The proposed mechanism is that one of the electron pairs from the double bond of 1-propene travels to sulfuric acid, de-protonating it. At the next step, water comes in as a nucleophile and attacks the carbocation intermediate - the terminal carbon gets hydrogen added on it because its the carbon with the most hydrogens. We get CH3CH(H2O)CH3. Now water comes in as a nucleophile again and de-protonate a hydrogen to form 2-propanol.
Now when I asked someone on my ship who used to study organic chemistry at a university, she said that she remembered that it was one of the electron pairs from the double bond of 1-propene that travels to H3O+ instead.
Is either proposed mechanism acceptable?

Do you mean that propene attacks the oxygen or the hydrogen of H3O+?

No, the person on your ship is incorrect.
An electron pair can only be transferred to electrophiles, which are species having a vacant orbital present on them. $\ce{H3O+}$ doesn't have a vacant orbital, so that can't happen. Here is a description as to why:

However,  $\ce{H+}$ ions, which have vacant orbitals can accept the pair and have it shift, while the other carbon sharing the double bond loses its electron, making it electrophilic, and can be approached by a nucleophile, in this case, $\ce{H2O}$:


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Help me proving Dalton's law of partial pressure formula
How do I prove

$${{\displaystyle p_{i}=p_{\text{total}}y_{i}}}$$

where $y_i$ is the mole fraction of the $i^{th}$ component in the total mixture of $n$ components ?
Here is what I know
Mathematically, the pressure of a mixture of non-reactive gases can be defined as the summation:
${\displaystyle p_{\text{total}}=\sum _{i=1}^{n}p_{i}}$
or
${\displaystyle p_{\text{total}}=p_{1}+p_{2}+p_{3}+...p_{n}}$
And mole fraction $y_i= \dfrac{n_i}{n_{\text {total}}}$
I don't see how to relate , please help.

After having some discussion with @JohnRennie
Let $P_i=n_i \dfrac{RT}{V}$ then $P_i=\dfrac{n_i}{n_{\text{total}}}\dfrac{n_{\text{total}}RT}V$
$P_i=x_i P_{\text{total}}$
Where $x_i$ is mole fraction of $i^{\text{th}}$ particle.

Ahh, this is what he must have meant. Good answer. +1

Recall the ideal gas law equation
$$pV=nRT$$
Lets say we have $m$ components in the gas mixture, with number of moles of each being $n_1$,$n_2$,$n_3$,...,$n_m$. Let $n$ denote sum of all moles. Let partial pressure of each be $p_1$,$p_2$,$p_3$,...,$p_m$, and $p$ denote the total pressure.
From the gas law, we get:
$$pV=nRT$$
$$pV=(n_1+n_2+n_3+...+n_m)RT$$
$$pV=n_1RT+n_2RT+n_3RT+...+n_mRT$$
$$pV=p_1V+p_2V+p_3V+...+p_mV$$
$$p=p_1+p_2+p_3+...+p_m$$
I think that's a satisfactorily enough proof.

Sorry but I have to prove that formula highlighted one with using concept of Dalton's law and mole fraction.

That is the definition of partial pressure. John Dalton defined a term called partial pressure in that way. You can't prove definitions.


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What is the cause for Crystal Field Splitting Parameter?
Taking the example of octahedral complexes, the theory says that since the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are closer to the ligands than the $d_{xy}$ ,$d_{yz}$ and $d_{xz}$ orbitals, they reach a higher energy level than the $d_{xy}$ ,$d_{yz}$ and $d_{xz}$ orbitals.(due to greater interaction with the ligands) I can't understand why the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are closer to the ligands. I agree that they are along the bonding axes, but with some rotation in space I can have other d-orbitals along the bonding axes. For example, if I rotated the octahedral arrangement of ligands about the z-axis, then the $d_{xy}$ orbital could be along the bonding axis instead of the $d_{x^2-y^2}$ orbital.So why can't other combination of d-orbitals (like the $d_{xy}$ and $d_{z^2}$ ) reach a higher energy level instead of the $d_{z^2}$ and $d_{x^2-y^2}$ during splitiing?

You choose the x y z however you want. If the x y z is chosen as the direction of the ligands, then you have the regular splitting scheme. If you choose different names, differently orbitals will be the high and low energy ones. This is only a naming problem

According to this theory ligands are considered point charges. These points acquire positions as those you've presented above. This leads to some d orbitals being closer to the ligands. That means that a higher amount of energy is needed for an eletron to be placed there, because it would be closer to the negative charged ligand.
However, which orbital is going to be more affected, is a matter of an "agreement" you may say. In order to be easier to explain situations, structures etc it is inclined to use a conformation such as the one you've shown. It would be much more difficult to discuss matters if everyone considered different naming and configurations.


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What will happen if ligand orientation is changed?
How we came to know ligand should approach from t2g orbitals in d5 configuration for tetrahedral complexes? What will be the difference in energy if ligand approach from eg?
As far as I can think, The major driving force for formation complex is the Nucleus-Ligand interaction and the opposing minor force is ligand's e-- Metal's e- but important for determining the coordination no. and orientation.  The field of nucleus should be spherically symmetrical so It can't be the reason.
So we've left with only one reason which is Ligand's e- - Metal's e- interaction, but how ?
Does CFT have any answer to it?
A similar question can be asked for Octahedral, but we know it is associated with CFSE, so it plays an important role for orientation.
But what difference will it make in the octahedral complex if we assume that ligand will approach from x', y' and z' (particularly in the case of d4 configuration)?

Let's assume it is a weak ligand
in this case, the CFSE should be -Δ'2/5, as eg orbitals, will be in lower energy state, but if ligand approaches from x,y and z then CFSE will be -Δ3/5.
Is this the reason, that ligand will always approach from x,y and z.
what if Δ3/5 < 2/5Δ' ?


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How to Prepare a 80mL Carbonic Acid-Bicarbonate buffer solution of ph=7.4 in laboratory?
I have the following materials which I can use:

0.1 M sodium bicarbonate solution
Sodium Bicarbonate powder
0.1 M HCL
Distilled Water
Universal Indicator

It will be a great help if you can provide with step by step instructions on preparing this 80 ml Buffer solution at an exact ph of 7.4.
Kind Regards,


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Why the ionization energy of Au is more than that of Cu?
As we go down the group from 3d to 5
d, size increases.
Vanderwaal radius of Cu and Au is 140pm and 166pm respectively. Since Cu is smaller in size, ionization energy of Cu must be greater.
Then why is the ionization energy of Au greater than that of copper?

First off, the 4d element in the group is not gold but silver.

It doesn't matter although I made the edits. What I want to establish is that even though copper has less size relative to Au, its ionization energy is more.

Smaller size does not necessarily correlate with higher ionisation energy. The trends in 1st IE going across Period 2 offer a simple counterexample.
A better (but not 100% foolproof) guideline is to use the orbital energies (see: Koopmans' theorem). The higher the energy of the orbital from which the electron is removed, the lower the IE.
In the case of gold versus copper, the first ionisation corresponds to the removal of a 6s or 4s electron respectively. In gold, the 6s orbital is relativistically stabilised, leading to a larger than expected ionisation energy. See also Geoff Hutchison's answer here.


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Why doesn't frozen sugar solution taste sweet?
I'll try to make this as brief as possible:
Dissolved two teaspoons of table sugar (sucrose) in about 250ml water. Sipped it, and as expected it tasted sweet. I let the rest of it sit in the freezer overnight. Next day, I took out the frozen sugar solution and, well, licked it.
Surprisingly, I could barely taste any sugar in it. It was almost as though I was licking regular ice.
Why is it that I'm not able to perceive any sweetness here?
I was under the impression that since the solution, being a homogeneous mixture of sugar and water, was sweet, the "popsicle" I made ought to taste sweet too (since the sugar would be evenly distributed over the volume of the ice).

Maybe reduced vibration cause slower diffusion through the mucosa and/or lesser rate of collision with receptors?However, nice observation. Popsicles  taste sweeter when they melts; at least to me.

Where is the sugar?
When you freeze a dilute aqueous sugar solution pure water freezes first, leaving a more concentrated solution until you reach a high concentration of sugar called the eutectic concentration.  Now you have the pure water that's frozen out, called proeutectic water, and the concentrated eutectic sugar solution from which the sugar is finally ready to freeze along with the water.  Upon freezing this eutectic composition forms a two-phase eutectic mixture, in which the sugar may appear as veins or lamellae (like veins of some ores among Earth's rocks, though these typically form form a different process).  If that structure is in the interior of the ice cube, likely since you cooled the solution from the outside, then licking the outside you got only the pure water proeutectic component.
See https://en.wikipedia.org/wiki/Eutectic_system for more about this process.
Addendum:  I tried this with store-bought fruit juice which was red in color.  Poured it into an ice tray and froze it overnight in my household freezer.  It appeared to be a homogeneous red mass and tasted sweet, but was also mushy implying some liquid was still present (after overnight freezing for an ice cube sized sample).  The juice was roughly 10% sugar by weight.

Another possible reason could be that taste-buds get kinda deactivated at low temperatures.

@PrittBalagopal they do get numb with frozen foods, but the amount of heat loss needed to balance the temperature drop from licking an ice cube is not enough to cause such effect. It is the same as frozen ice cream. You sense the taste once the bite melts.

Yes, that's why you need to stir and churn continuously if you want to make sorbet.

So sugar water is a non-eutectic mixture? Interesting. And this non-eutectic mixture cools down until enough pure water freezes, transforming the rest (sugar+ yet unmelted water) into a eutectic mixture. I'm just trying to wrap my head around this.

@CopperKettle that's basically how it works.  A binary mixture is eutectic (where it can be so) only at the right concentration of both components.  At any other composition the component in excess freezes out to get the composition to eutectic, then it's freezing party time for everybody.

So, in summary: "The sugar is on the inside of the ice. You're licking the outside, which is nearly all water." ?

@Mindwin But icecream is still needs a lot more sugar to get the sweet taste than a room-temperature treat. It's not quite as bad as what the OP is observing, but it's a very real effect anyway.

@Luaan that is because nobody just gives one single lick to ice cream.

Can you color the sugar (or use something else) to be able to see these veins in the ice?

Water doesn't freeze from the outside: it freezes from up to bottom because ice floats. If the answer were right, liking the bottom of the ice cube should taste sweet - even sweeter than water tasted. Anyway, because of my child experience of using a freezer to make popsicles, I'm afraid @Luaan is right and what the OP experiences is just that frozen food tastes less sweet, even if it's homogeneous.

@Pere I would expect that in a glass of water, the ice would start to form on the edges (surface) of the glass, which would kind of make it freeze from the outside in, with the center-top being the last to freeze (before the glass cracks). But I wouldn't expect it to be simple - depending on the solution and purity of the water before adding the solute, and the thickness of the glass, and the material (and smoothness) of the glass... it can also freeze from the top to bottom. But note that the OP would likely be licking from top; the sweeter part is either in the center or at the bottom.

I'm wondering the concentration of sugar (g/ml) for the eutectic mixture at -4 ℃.  When making the popsicles, how much sugar to start with? My guess is that a saturated room temperature solution is what most people use?? Found this https://link.springer.com/article/10.1134/S1070427207020061,  but it's a paid for publication.

There may be ingredients other than sugar, as well.  In my juice I discovered later that the juice developed a foam when I poured it (not freezing).  Could surfactants be affecting the (size scale of) phase partitioning?

Re above there's also this phase diagram but no concentrations: https://www.doitpoms.ac.uk/tlplib/biocrystal/water-sucrose.php

This phase diagram that seems to answer the question about sugar concentration https://www.doitpoms.ac.uk/tlplib/biocrystal/water-sucrose.php according to that at -10 ℃ a 60% by weight solution of sugar will not produce that ice (pure water)/ sugar-water mixture.  Wow, that's a lot of sugar! Bon appetit! (Although that article is more about frozen cells, yuck!)


text:
Is the reaction between a carboxylic acid and alcohol endothermic?
I would say that it is endothermic, and then by increasing the temperature, one could get more product, disturbing the equilibrium.
Is this true, though? During an esterification, some bonds are broken, releasing energy. The opposite reaction splits the ester and adds the atoms back, which I suppose releases energy which is then bound in the bonds adding back the water molecule.

Do you really think that breaking bonds releases energy?

@Loong ... Of course not, silly me. Breaking them REQUIRES energy, heh. Is it then exothermic toward the right side of the equation?

When in doubt, go find data. Here is a typical such reaction - the esterification of ethanol and acetic acid to make ethyl acetate.
$$\ce{CH3CH2OH + CH3CO2H -> CH3CO2CH2CH3 + H2O}$$
The following thermodynamic values are from the NIST Chemistry Webbook.
$$\begin{array}{|c|c|}\hline
\text{compound} & \Delta _\pu{f} H^\circ \ (\pu{kJ mol^-1}) \\ \hline
\text{ethanol} & -276 \\
\text{acetic acid} & -484 \\
\text{ethyl acetate} & -480 \\
\text{water} & -286 \\ \hline \end{array}$$
Now for some Hess's Law:
$$\begin{aligned}
\Delta_\pu{r} H^\circ &= \sum_\mathrm{products} \Delta_\pu{f} H^\circ - \sum_\mathrm{reactants} \Delta_\pu{f} H^\circ \\
&= (-480\ \pu{kJ mol^-1} + -286\ \pu{kJ mol^-1}) - (-276\ \pu{kJ mol^-1} + -484\ \pu{kJ mol^-1}) \\
&= -766\ \pu{kJ mol^-1} - (-760\ \pu{kJ mol^-1}) \\
&= -6\ \pu{kJ mol^-1}\end{aligned}$$
Hey! This one is exothermic (slightly).
Here are the enthalpies of reaction for some other esters (by no means a complete or even representative list):
$$\begin{array}{|c|c|}\hline
\mathrm{ester} & \Delta_\pu{r} H^\circ \ (\pu{kJ mol^-1}) \\ \hline
\text{methyl acetate} & -9 \\
\text{ethyl acetate} & -6\\
\text{methyl benzoate} & -5 \\
\text{ethyl crotonate} & +1 \\ \hline \end{array}$$
I would thus guess that formation of most carboxylic esters from carboxylic acids and alcohols would be slightly exothermic or slightly endothermic depending on structure.
The reasons esterifications are often heated are 1) to increase the rate of reaction and 2) to distill off water as it forms.


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Are the coefficients and exponents of contracted gaussian same for cartesian and pure form?
The cartesian form of contracted gaussian is something like
$$\phi^{CGF}=\sum_{n} s_nx^iy^jz^k \exp(-\alpha_nr^2)$$
and these is also a pure or spherical harmonic version (which I don't know the details about).
Are the coefficients and exponents same between the two, different for $\ge d$ orbitals, or different for all orbitals.

They are the same.
Your equation can be rewritten as
$$ \phi_\mathbf l^\text{CGF}(r,\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l}\sum_ns_nr^l\exp(-\alpha_nr^2)\equiv\Phi_\mathbf l(\Omega)R_l(r) $$
where $l=l_x+l_y+l_z$ is the total angular momentum, $\Phi_\mathbf l(\Omega)$ is the angular part, and $R_l(r)$ is the radial part, and they are separable not only in mathematical form, but also for practical purposes. (This discussion omits the norm, which is not separable to radial and angular contributions.) When specifying a Gaussian basis set in quantum chemistry, one talks only about the radial part, independent of the angular part. The angular part is really considered an implementation detail, because its form usually doesn't have any significant effect on the results.
For completeness, two most common angular forms are Cartesian:
$$ \Phi_\mathbf l(\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l} $$
and spherical harmonic:
$$ \Phi_\mathbf l(\Omega)=Y_l^m(\theta,\varphi) $$

I presume normalization constant for >p orbitals will be different?

@packet0 Yes, the norm is not separable to the radial and angular part. I'll make this clear in the answer.


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I can't choose the set of 3 chemicals with each of which Si will react (out of 5 possible sets)
This is one of the questions used in the Russian State Chemistry Exam. One needs to select the set of chemicals in the second column such that each of them would react with the element or chemical in the the first column.
I mean, it should react with each element in the set in order for that set to be suitable.

It seems to me that $\ce{Si}$ has no suitable set among those five.
It does not react with $\ce{Na2CO3}$ (that excludes set 1), with $\ce{HCl}$ (set 2), with$\ce{MgO}$ (set 3), with $\ce{H2SO4}$ (set 4) and with$\ce{Al(OH)3}$ (set 5).
I used Chemguide and a Russian school curriculum-based website to refresh my memory of $\ce{Si}$ activity.
Am I wrong in regard of any of the five chemicals? Maybe it will react with $\ce{MgO}$, for instance?

$\ce{Si}$ does in fact react with $\ce{HF}$ and $\ce{NaOH}$. The reactions would be:
$$\ce{Si + 4HF -> SiF4 + 2H2}$$
$$\ce{Si + 4NaOH -> Na4SiO4 + 2H2}$$
$$\ce{Si + 2MgO -> SiO2 + 2Mg}$$
You can conclude these, since $\ce{SiO2}$ is quite stable, and also reacts with $\ce{HF}$ and $\ce{NaOH}$ (You may have heard that $\ce{HF}$ dissolves glass), so definitely $\ce{Si}$ will react way better than $\ce{SiO2}$.

Yes, but it does not react with MgO, and that excludes set 3.

My bad, I forgot to mention its reaction with MgO, I'll add it now.

I have added the reaction, the one with $\ce{MgO}$ is also called the Pidgeon Process. More about it: https://en.m.wikipedia.org/wiki/Pidgeon_process

Thank you! I knew that it reacts with Mg to produce MgO so I assumed the reverse reaction was unlikely.

I have edited your post with MathJax. You can learn how to use MathJax to write chemical equations here:https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

Silicon also react with the chemicals from set 2.
$$\ce{Si + 3HCl ->[300°C] SiHCl3 + H2}$$
$$\ce{Si + O2 ->[>900°C] SiO2}$$
$$\ce{Si + 2Cl2 ->[300°C] SiCl4}$$
Maybe you doubted the reaction between silicon and hydrogen chloride forming trichlorosilane.
Source of reactions: webelements.com


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What is the basis for naming the p-orbitals with reference to the Cartesian coordinate system as x, y, and z?
There are 3 p-orbitals per subshell, which are $p_x$, $p_y$ and $p_z$. As well as there are 5 d-orbitals, which are $d_{xy}$, $d_{yz}$, $d_{zx}$, $d_{x^2-y^2}$, and $d_{z^2}$.
My Question: What is the basis for choosing the axes as $x$, $y$, and $z$? Can we rearrange the axes, suppose lets replace $x$ with $y$, $y$ with $z$, and $z$ with $x$. Then we will have the d-orbitals to be $d_{yz}$, $d_{zx}$, $d_{xy}$, also you would get $d_{y^2-z^2}$, and $d_{x^2}$, orbitals. Why do we choose that particularly?

We prefer to not use MathJax in the title field, see here for details.

The choice of axes is arbitrary in this case (isolated atom with no external influence).
You could choose about any set of co-ordinate axes, and that wouldn't really affect the behavior of any of the electrons there ;)
However, in the presence of external influences, such as a magnetic field, it's customary to take the positive axis in the direction of the field as the positive z axis. Though ultimately, the choice of notation is left to you.

EDIT: Since you seem to have an issue here, I'll mention this as well:
Quoting from chemguide.co.uk
A)  $\mathrm{d_{xy}}$,  $\mathrm{d_{yz}}$,  $\mathrm{d_{zx}}$

The names tell you that these orbitals lie in the x-y plane, the x-z plane and the y-z plane respectively.

Each orbital has four lobes. Notice that each of the lobes is pointing between two of the axes - not along them.

B) $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2}}$

Although these two orbitals look totally different, what they have in common is that their lobes point along the various axes. That's different from the first three where the lobes pointed in between the axes
The $\mathrm{d_{x^2 - y^2}}$ orbital looks exactly like the first group - apart, of course, from the fact that the lobes are pointing along the x and y axes, not between them.

The $\mathrm{d_{z^2}}$ orbital looks like a p orbital wearing a collar! The main lobes point along the z axis.

So the reason why they call it $d_{z^2}$ is because of the magnetic field applied?

I'm sorry but I'm still not quite satisfied. I am asking why they call the particular orientation of the ${d_z^2}$ as ${z}$? Why not call it as ${x}$, with its name being ${d_x^2}$ instead?

yes it could be $d_{x^2}$ but its useful to use the same convention to distinguish the orbitals and so we use $d_{z^2}$ .


text:
Why is OTs a good leaving group and how does it work?
I've been told that $\ce{OTs}$ is a good leaving group for substitution reactions, but I'm not sure how it works or why it is a good leaving group. Can someone please explain?
$\ce{Ts}$ comes from Toluenesulfonyl chloride, and when the $\ce{Ts-}$ ion replaces and H in an alcohol, it becomes an $\ce{OTs}$ group.

Tosylate groups ($\ce{CH3C6H4SO3-}$) are good leaving groups because their conjugate base forms are quite stabilized by resonance:

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


text:
Does firing a proton beam on water make it acidic?
Lets say I fire a proton beam upon the surface of water. Lets assume I do the experiment in an air-free environment and that the proton beam doesn't reflect off the surface of water.
Now, protons, which are simply $\ce{H+}$, should enter the bulk of water and probably combine with a $\ce{H2O}$ molecule and form $\ce{H3O+}$. As per the definition of acidity and pH:
$$\ce{pH}=-\log [H^+]$$
If the pH reduces, the solution would become acidic as seen here. But at the same time, the water becomes charged, so I am not sure whether this is applicable.
So, in short, does water become acidic due to a proton beam?
Edit: I do not have access to proton guns and it's certainly not on sale on eBay. I only attempt to theorize on what may happen.

sigh What would happen would depend on energy of protons. Also sth separating water from vacuum would be needed.

Put water in a stainless container. Place Mylar foil for a beam entrance (easy and the standard for in-air ion beams. Irradiate with MeV protons. Ground the chamber for charge neutralization. Now it is a balance of solvated protons vs formation of H2. Not hard to do, if you have an ion accelerator handy.

I think you've answered your own question. The water would become acidic but the positive charge would cause reactions with the surroundings.

Now, I have two conflicting answers, and no way to choose which is correct. Quite a dilemma indeed!

No, probably not.
pH represents the concentration of hydrogen (/hydronium) ions in water, requiring a negative ion complement. pH could be thought of roughly as a response to things other than water or it's components that are also dissolved in the water. If you try to artificially add protons to the water, the water will reestablish the original equilibrium concentration quickly.
Protons stopping would likely acquire electrons most of the time, find each other, and combine to molecular hydrogen. As the amount of "inserted" hydrogen increased, bubbles of hydrogen gas would tend to form in order to establish a partial pressure of hydrogen 'above' the liquid in order to maintain a finite dissolved fraction.
I found a table of numbers and a form of Henry's law here that can be used to estimate the partial pressure of H2 gas 'above' the water that would be in equilibrium with the amount of dissolved H2.

If you wanted to perform the experiment however, this could be done in a fairly straight-forward way with a low energy proton accelerator and its associated machine shop, instrument builders, and graduate students.
Proton beams from a Tandem or other MeV range accelerator can be focused to much smaller than 1 millimeter in diameter, and slowed by absorbers so that they stop within a spread in range of less than 1 millimeter as well.
A cubic millimeter of water has $\frac{1}{18}10^{-3}N_A$ molecules, or about $3.3 \times 10^{19}$.
A beam current of 10 pna (particle nanoamperes) for an hour (just for an example) would represent $10 \times 10^{-9} \times 3600 sec \times 1 coul$ or about $2.2 \times 10^{14}$.
Numerically that suggests a resulting pH of around 5.2 if the protons remained as H+ ions. However, this is not going to happen because of the reasons stated above.
This might be measured in situ with micro titration for example by introducing a small amount of indicator at the end of the experiment and collecting transmitted light with a microscope objective. There may be other ways as well.
Assuming the last 5 MeV of energy of the beam were dissipated in the water, the heating is $5 \times 10^6 volts \ \times \ 1 \times 10^{-8} amperes$ or about 50 milliwatts, which should be possible to dissipate across six square millimeters of surface area without catastrophic results.
Practically, you'd have to use a clever window to isolate the water from the beam line vacuum without chemical contamination by reaction or sputtering. Ice might be an excellent choice, except that you'd have to use a lower current for a longer time to handle the heating. Diamond-like carbon (DLC) might perhaps be another option, both for chemical as well as heat and charge conductivity reasons.

While you have answered my question of whether proton beam can penetrate water or not, you haven't answered whether as a result water becomes acidic or not?

@PrittBalagopal OK  I've completely rewritten the first paragraph of my answer... yikes this is an interesting question!

@PrittBalagopal let me know if you'd like further clarification.

Yes, the concentration of $\ce{H3O+}$ will increase which will lower the pH value of the water. This is theorie based.
In reality you won't be able to measure the change. The mount of protons you need to significantly change the pH will also effect the charge of the water as you mentioned. The charged water will affect your method of measuring AND the protons will leave the water because of it's positive charge.
To neutralize the water you could add elections, which will cause the protons to form hydrogen gas. Or you could add a negative ion like chloride, which would have been the same as adding $\ce{HCl}$ to the water.


text:
Recrystallization Process Questions
Should a crude compound should be dissolved completely in cold solvent and then heated up?
I think non-polar solvents are best choice for recrystallization of polar compounds, no?
Recrystallized product is then isolated using vacuum filtration
Can scratching the inside of the flask can help initiate crystallization?

If you chose to recrystallise by variation of temperature, you have to find a liquid that dissolves your product much better when heated (or even heated to reflux), than at room temperature.  A qualitative test may be performed with a small quantity of substance, for example a tip of a spatula, that you fill in a test tube, followed by a small amount of liquid.
Does your compound dissolve well in the liquid already at room temperature? This is then not a good choice of sole solvent for a recrystallisation, as much of the product will be retained by the liquid.  Intentionally I use liquid, because while you likely find a commonly used solvent suitable for recrystallisation, some other chemicals not typical perceived as "solvents" may be useful, too.
Depending on your equipment available, you may extend the accessible temperature range towards lower temperatures by use of an ice bath, or a freezer.  Be aware, however, some of the solvents that are liquid at room temperature may solidify then, too (glacial acetic acid, for example).
Often, you face a situation like "ethanol (as one solvent) has a too high solvent capacity for X, and water (as an other solvent) has a too low solvent capacity -- regardless if warmed, or heated to reflux".  Then it is time to use a mixture of mutually miscible liquids as solvent of recrystallisation, to tune the solvent properties.  This is why, for example, aqueous ethanol (ethanol with an intentionally added proportion of water) is used as solvent of recrystallisation, too.
There are some seemingly obvious choices of solvents of recrystallisation, like methanol and ethanol for not so polar organic compounds, and other somewhat surprisingly well working examples (acetic acid for less polar compounds); hence, recrystallisation offers a spectrum of science, craftsmanship, art, and obviously some luck, too.  Note that recrystallisation aiming for purification may be well a different one as aiming for nice crystals (for the purpose of single crystal diffraction analysis, for example).
If you made your choice, retain a small quantity of non-recrystallised compound which may be helpful later.  Mix your solid (or tar-like) product with the solvent / solvent mixture at room temperature and heat it up; if you want to heat up to reflux, consider a reflux condenser.  At high temperature, all visible solid should be dissolved.  If this is not the case, there is to little of solvent present and carefully add more.  After each addition of some new solvent, wait a little bit until the temperature of your mixture is the same as prior to this addition. (Caveat: If your product to recrystallise is a mixture, one component may dissolve better, than the other.  If so, consider a fractional recrystallisation or other means of purification first.)  Then allow the solution a slow cooling to room temperature, for example switch out the heater (and stirrer, too) of the oil bath yet keep the round flask in the oil bath.
Scratching the inner walls of the flask with a glass rod may be dangerous as you accidentally may pierce the wall (and consequently loose product to the oil bath).  A long Pasteur pipette is a better choice, too much pressure applied will break it, but not the flask.  But in general, it is one good method to generate crystal seeds for nice crystals.  (The optimal temperature for crystal seeding need not be the same as the optimal temperature for growing larger crystals.)
If possible (vide supra), complete the crystallisation by cooling below room temperature (ice bath) as this often lowers even more the solvent capacity.  If accessible, collect the crystals at this / near this low temperature, too.  If you want to rinse your crystals with a "bad solvent" (i.e. with low solvent capacity), equally cool it in advance by an ice bath.  If you crop your crystals by vacuum suction, prior cooling the equipment (Buchner filter, suction flask) by an ice bath / in a freezer may be helpful, too.
As a final note (here), do not throw away the mother liquor / filtrate of your recrystallisation too early.  It may contain a considerable amount of product.  Concentrate it (for example, at the rotary evaporator), and repeat the recrystallisation for a second crop.  Do not merge crops of subsequent crystallisation until an analysis reveals them as of sufficient similarity, like comparable purity.


text:
Color of complexes d-d transition or electron exchange
I always have the problem of finding whether a complex has colour due d-d transition of electrons or if it is due to electron exchange. Is there a general way to tell whether a complex has d-d transition or electron exchange?
Is it the right method to check if a metal has d electrons and if it does. Its color must come from d-d transition and if not the color must come from charge transfer?

By electron exchange do you mean electron transfer of the form Fe(II) to Fe(III) for example?

No.. From metal to ligand or from ligand to metal.

That's usually called "charge transfer" not "electron exchange" https://en.wikipedia.org/wiki/Charge-transfer_complex#Inorganic_charge-transfer_complexes

in which case d-d transitions are usually weak and at long wavelength, metal-ligand transitions are usually more intense and may swamp d-d in some cases. But there is no hard and fast rule, you have to study each case carefully.

So if I'm given a complex compound. I would have to check if it has d-electrons or not and find out if d-d transition is taking place?


text:
Why is carboxylic acid hard to enolize?
Why do we need PBr to help enolize a carboxylic acid. My textbook doesn't give a reason for this at all. Does it have to do something with the acidity of carboxylic acid. Thanks.

Think about the pKa of the carboxylic acid protons compared to the pKa of the enolisable proton (hint: the acid is more acidic).
Once the carboxylic acid is turned into the carboxylate, enolisation is not possible, as the carbonyl cannot simultaneously conjugate with the carbon and the oxygen.
The use of phosphorus tribromide is known as the Hell-Vollhard-Zelinski reaction, it allows carboxylic acids to be directly brominated at the alpha position without initial protection of the carboxylic acid via in situ formation of the acyl bromide which hydrolyses upon workup.

Second paragraph isn't ok, there's no problem with conjugation. You'd need to get dianion first, though.

They don't form a linearly conjugated system. At any one time the carbonyl can only conjugate with the oxygen or the carbon, not both. I'm not saying its impossible, the carbon just becomes significantly less acidic and hence significantly less enolisable (hence why nobody ever really uses a carboxylic acid to enolise).

-OH group in carboxylic acids have a much better resonance than an acid chloride due to difference in the size of halogen atoms vs carbon; this difference even increases with Br.
This lack of resonance from LP of Br onto carbonyl group in acid chloride makes Enolization possible, which is very difficult to be in -COOH.
Using PBr3 to convert an RCOOH in RCOBr brings in Enolization, which can give Substitution with Br-Br.


text:
Concentrating acetic acid from vinegar
Vinegar seems to kill poison ivy or at least severely stunt their leaves. I am trying to increase the concentration of acetic acid and I put a bottle of vinegar in the freezer. I have looked at it and it has turned cloudy. I'm thinking if I poured this through a coffee filter I should be filtering out solid acetic acid. I might do this in the freezer because it might take some time. Does anyone think this will work? I might be getting water ice instead. Either way I would be separating the two. I could repeat this step and concentrate it further. Would I not really be filtering out anything? On another blog someone suggested buying aluminum acetate which has more acetic in it instead, but this might not be as effective.

Yes, you are freezing out acetic acid, which has a melting point of 17C and your freezer is probably on the order of -10C. Your plan to filter it should give you very concentrated acetic acid, which could be a safety hazard. I would read this and this before planning how to proceed. You will probably want to dilute the acetic acid by about 3 parts water to one part AA. Note that this is not going to be an efficient solution and some AA will be left unfrozen in solution.

Thank you for your answer. I had bought a small  bottle of lab grade acetic acid once which was pretty high %. I have experience working with this. Besides my throat has gotten more sensitive to things than it once was. I know to keep stuff like this at a distance. I used to keep the bottle in the bottom cabinet of a camper trailer and it would get down to the 20s sometimes. With that concentration it would look like it turned to gel or a chunk of ice in liquid at much higher temperature than that-much like the name glacial acetic acid.

If I understand correctly, what you described can be seen as a form of fractional freezing.
This is done in order to make alcoholic beverages more alcoholic (Eisbock is an example): the beverage is left to partially freeze and the ice is removed.
Since ethanol has a lower melting point (-114 °C) than water (0 °C), this effectively removes more water than ethanol.
But in your case, since acetic acid has a higher melting point (16 °C) than water, the solids will have a higher concentration of acetic acid and you would want to keep that instead of the liquid.

Yes, you are freezing out acetic acid, which has a melting point of $\pu{17^oC}$ and your freezer is probably on the order of $\pu{-10^oC}$.
Your plan to filter the solid acetic acid should give you very concentrated acetic acid, which could be a safety hazard. I would read this and this before planning how to proceed. You will probably want to dilute the acetic acid by roughly 3 parts water to one part acetic acid while it is still solid, or wile it is melting to a slush. Note that this may not be a very efficient solution and some acetic acid will be left unfrozen (dissolved) in the original solution.
Regarding your overall goal to kill poison ivy:
You can greatly increase the contact area of the solution with the plant by adding some detergent, say dish-washing soap.  Additionally, you can increase the toxicity of your solution to the target area in an environmentally friendly manner by dissolving a couple cups (sorry for the non-SI units, it's just how I do it) of table salt per gallon of the acetic acid solution. This will result in "poisoning" the area of the soil at the base of the plant, inhibiting it's ability to survive your chemical attack. You may find that a commercial 10% acetic acid solution is effective when used in this manner.

Your hint of adding detergent to increase the permeability of acetic acid across the leaves brings back to my mind this recent publication: "Demonstrating the Effect of Surfactant on Water Retention of Waxy Leaf Surfaces" by Chiu et al. (J. Chem. Educ., 2017, 94, 230-234, DOI: 10.1021/acs.jchemed.6b00546).

I thought of adding detergent, but thought it might have some constituents that are basic and would neutralize the acid somewhat. I've heard of salt. I kinda didn't want to create a buildup of that in the soil. I know its probably negligible though. I've added linseed oil and turpentine. These will form a film on the leaves.

@discountbrainsurgery , the effect of the small amount of detergent that you would need to get the effect is negligible compared to the amount of acetic acid present, but it sounds like you have that effect covered with your creative use of those oils. We just started adding salt to the solution this spring, so I can't speak directly to problems of long-term buildup. Partly it depends on the rainfall amount where you live, but sodium chloride is so water soluble it shouldn't accumulate much from season to season. I will say it seems to work much better on our weeds with the salt added.

I turned the nob up in my refrigerator to more than half way to max and found most of the contents of the bottle frozen. There was a little liquid inside. I poured this into a container. I have a wet bulb pH meter and I measured it. I got a pH of nearly 1.5. I don't think vinegar is anywhere near this. This is pretty acidic. Someone on stack exchange presented a phase graph showing the several regions depending on temp and % acetic acid where there is liquid water and solid AA and another where there is ice and liquid AA.

I looked up the pH of regular vinegar and found it is typically 2.5. The pH I measured from what I poured out was 1.5 and still dropping. That's a whole factor of 10 or this solution is 10 times stronger. So, from 5% vinegar this is 50% acetic acid. If I repeat and separate this from the remaining liquid I doubt I'll get as strong a concentration which  is already established knowledge.

@discountbrainsurgery , unfortunately the pH and concentration calculations are much more complex, mainly because AA is a weak acid.  Being a weak acid just means that not all of the compound dissociates (as it will in a strong acid like HCl) into a proton and acetate. In fact, for AA only a fraction of a percent of it will dissociate, and theoretically concentrated AA should have a pH of ~2.4. How are you measuring pH? Do you have a low pH standard? See here and here.

My pH meter is a hand held one I had gotten off the internet. It says ATC and 'electcity' on it. I assume it has the full range from 0 to 14. I only took 2 semesters of general chemistry. I know the book mentioned Lewis acid and another definition and the instructor mentioned that too, but what we did problems on was the pH or the -log of the number of H ions that have disassociated in an aqueous solution.  I thought this is exactly what u said. Weak acids don't disassociate as much. Nitric acid I believe is 14 molar.

Yes, there are 2 types of acid base theory, and Lewis is one of them. The one we're dealing with here is  Brønsted–Lowry, in which pH is -log[H+], as you said. I have a feeling I may have confused you with 2 very different terms: strong vs concentrated, or quantity. A weak acid only gives up a small fraction of it's protons (H+), regardless of the concentration, and a strong acid, like nitric, readily gives up all of it's protons. Nitric acid is stable up to a concentration of about 14 molar, but it is an oxidizing acid with some more complex chemistry. More about pH meter next...

Oh btw, I looked up the pH of the 5% vinegar that's sold and they said it is 2.5 roughly. So, you're saying concentrated AA has the same pH as vinegar? I never measured my vinegar.

Have you ever tested a calibration solution? I know you can readily get pH standards as low as pH=2. I suspect such a standard would read low on your system based on your 1.5 reading for AA, although it might be correct within a few tenths of a pH units anyway. I do have to correct myself from earlier: conc AA should have a pH around 1.7-1.8. The pH of 1M (M=molar) AA is around 2.4, 0.1M AA is about pH=2.9, 0.01M AA is about pH=3.4. That means you are getting little additional acidity by concentrating, though it still could be much more toxic to plant. Again, plz careful with the conc stuff!

I have two standards that came with the meter to be mixed in a specified amount of water. Actually, I haven't checked the meter in a while. In any case regular vinegar seemed to be pretty distasteful and detrimental to poison ivy last year or 2 years ago. Seems like last year I didn't have much to contend with. So, if I dilute what I have to what I think is 10% or 15% it will be even more effective.


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influence of neucleophiles in substituition reactions

in the image the substitution  reaction of 1-bromobutane and 2-bromobutane is shown. In presence of NaCN 1-bromobutane reacts fast and with the other it becomes slow, which is understandable since CN-  is a poor leaving group compared to Br-.
But OH- is a poorer leaving group compared to CN-(pKa of HCN and water is 9.1 and 16 respectively) , so shouldn't the substitution reaction of 1-bromobutane with water be even faster than it's reaction with NaCN instead of slowing down?
the image was taken from: http://www.masterorganicchemistry.com/2012/06/27/two-types-of-substitution-reactions/ and a detailed discussion is given here


text:
On what factors does the true reaction Stoichiometry depend when there are multiple valid options?
Consider the following unbalanced reaction:
$$\ce{XeF4 + H2O -> Xe + XeO2 + HF + O2}$$
There are six variables to be solved, but only four equations linking them. I solved the set of equations using Gauss elimination and arbitrarily set the stoichiometric coefficient of $\ce{XeF4}$ as $3$:
$$\ce{3XeF4 + 6H2O -> $(4k)$Xe + $(3-4k)$XeO2 + 12HF + $(4k)$O2}$$
where $k$ is the free variable, the parameter, with $0<k<\frac 34$ (since coefficients must be positive).
Clearly, for different values of $k$, it gives radically different yet balanced equations. The very yields are different for fixed amount of $\ce{XeF4}$ and $\ce{H2O}$. This issue has been noted in a previous question.
My textbook, however, gives the balanced equation:
$$\ce{6XeF4 + 12H2O -> 4Xe + 2XeO2 + 24HF + 4O2}$$
clearly setting $k=\frac 12$. The linked question above does not offer any explanation for why a particular value of $k$ should be favoured. So, my exact question is: Why set $k=\frac 12$? Why was $\frac 12$ preferred over any other value for $k$:

Is it determined by experimentally measuring the stoichiometry?
Is there a theoretical explanation for it?

related https://chemistry.stackexchange.com/questions/37684/write-the-chemical-equation-of-this-reaction

chemical equations are not unlike math equations, multiplying each side by same number will give a valid equation. Similarly, addition of two (or more) equations side-wise with subsequent simplification gives a valid equation. However, if the equation depicts an actual chemical process is another matter, a well-balanced equation may be irrelevant for this reason.

@Mithoron In the answer to that question, they just say that the balanced equations are different. I want to know which is correct and why? Which factors determine that? And it is not a duplicate because he mentions just 2 possibilities and I found all the solutions.

I would change the title of your question to better reflect the change you made. Something along the lines of "How to determine true reaction stoichiometry among multiple valid options".

There is a meta discussion on the closure of this question: Possible overreach on marked duplicate

While I never quite figured this paper out and it doesn't give a concrete example of carrying this process out, I think this paper addresses your  question. For just finding these solutions, check out this paper or summaries by Fred Senese or myself

Recently found another great resource on this topic that I haven't gotten a chance to read all the way through. Section 7 seems to address your concerns: http://www.chemical-stoichiometry.net/CRS_tut.pdf


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H1 NMR integrations
I know this may be a bit of a dull question considering the integration on hydrogen NMR are usually quite precise but I was wondering If integrations in a linear depsipeptide compound could be thrown off by anything such as electron withdrawing inter or intramolecularly. The compound in question is a has a chemical formula of C77,H88,N4,O11,S4.curently the integration of all other protons are correct, however, the protons on two methyls on a valine, and the protons on a boc and tertbutyl group are off by a factor of three.

As deviation by a factor of three for some of the protons seems large (hence +1 for this question), does this observation persist even while 1) changing the pulse program during the NMR data acquisition (easier to do) as, for example, in Weizman, J. Educ. Chem. 2008, 85, 294 (DOI: 10.1021/ed085p294)?  Or/and 2) while changing the solvent used for NMR (may be trickier, may coil / decoil the peptide more or less than by now)?  3) Is there alternative analytical evidence one BOC / one tBu group  is present, and not three; like by LC-MS with a soft ionisation (CI, rather than EI 90 eV)?

There are lots of reasons why integrals may be out, even to a factor of 3. It would help if you posted the conditions of your acquisition; pulse program, pulse angle, number of scans, repetition delay, solvent, concentration. And by a factor of three - do you mean 3 times more or less than they should be?


text:
Why does hydrogen burn with a pop sound?
Our teacher made us to do an experiment with zinc granules and hydrochloric acid.
$$\ce{2HCl + Zn -> ZnCl2 + H2}$$
We collected the hydrogen gas that was evolved in a test tube and on ignition, it burned with a "pop" sound. Why does hydrogen show this behavior? Why doesn't it burn more quietly like other fuels (gasoline, coal, wood, etc.)?

We may agree that during the combustion, regardless if the one of wood, gasoline, or in your example, hydrogen gas, is an exothermic reaction.  In a simplified view, products of these reaction are gaseous water; in the processes burning organic matter varying relative amounts of carbon monoxide and carbon dioxide are to be considered, too.
The gaseous products formed take up a volume, much more than their starting materials in the condensed state (liquid, or solid). Because of the local heat, these gases aim to expand rapidly.  In contrast to burning wood for a bonfire, the volume offered by a test tube is confined except for the mouth of the test tube, where a decompression to reach ambient pressure may take place.  This small scale explosion initially occurs at the interface of air's oxygen at the mouth of your test tube filled with hydrogen, yet this frontier actually propagates up to the closed end of the test tube, too.  Sometimes, especially with lengthier test tubes, this may be seen from the outside, too.
The combustion of hydrogen is worth a special note, because -- as correctly spot by airhuff -- formally there is volume contraction along
$$ \ce{2H2 + O2 -> 2H2O}$$
which is more than compensated by the thermally caused expansion of the gaseous water.
Gasoline driven combustion engines work by the same principle; spark plugs locally ignite a mixture of finely dispersed fuel and air, triggering a rapid combustion (explosion) where the reaction products again aim to expand. As they operate at a larger scale as the test tube experiment, the generated sound is much more intense, and these vehicles need a silencer, too.
While your bonfire equally yields $\ce{CO2}$ and water, these products may expand easily; there is no pressure generation, hence there is no sudden expansion (at least, not at this scale), and hence we may perceive the bonfire as (in comparison to the other examples) as peacefully calm.
I suggest to take a view on videos about explosions of hydrogen gas, and an appealing demonstration how it looks in the inner of such an Otto engine here.  In the later example (the display of the combustion starts around 2:51 min and again at 6:00 min), gasoline was replaced and a high speed camera looks through a special lid, yet the example still shows nicely the propagation of this frontier of explosion.

Big +1 for the first video illustrating the rapid expansion of the explosion (plus it's just really cool). The only nitpick I have is the first sentence of the second paragraph about the expansion due to the phase change. In the case of burning $\ce{H2}$ and $\ce{O2}$, we're converting gas to gas, and contracting from 3 moles down to 2 at that (2 hydrogens plus one oxygen going to just 2 waters) so I don't think that's relevant. But the rest of that paragraph hits it; the rapid thermal expansion in a confined area. Anyway, nice answer.

@airhuff Your comment is completely justified.  Had I written the chemical equation first...

Presumably the reason the explosion is not bigger is that the explosive chain reaction depends on exact pressure/temp and composition. If you want to try a demo on this. Find a thin tin used for  say granulated coffee, make a 2mm hole in lid and one in base for a rubber tube. Place on retort stand, fill with hydrogen from cylinder until you are sure its full,  a minute or so,  then remove tube close cylinder and get cylinder well away. Light hydrogen at hole in lid. Will burn then go out as it explodes. Lid is light , flies off, no damage. Have done this is lectures several times. Great fun.


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What gases would be produced from the electrolysis of distilled water and sodium bicarbonate?
I have two questions: one practical question and one conceptual question.
1) I'm trying to set up an electrolysis experiment and I am trying to make sure I am going to end up with the gases that I think I will. I am planning on using stainless steel electrodes, and sodium bicarbonate (baking soda) as the salt, dissolved in distilled water. I am not using NaCl, as I believe it would produce some chlorine gas as well.My thought is that using sodium bicarbonate, I would get hydrogen and oxygen gas, but I am unsure if it would produce some CO2 as well...
Basically I am not sure what the intermediate reactions are, and I am not sure how the baking soda breaks up and what you are left with. I am trying to produce hydrogen and oxygen gas, and I want the 2 to 1 ratio, so I want to make sure I am not getting other gases as well (I don't care so much about any solids produced, so long as they are not dangerous.
2) Also a side question that I have been wondering about: I have been looking at different electrolysis experiments online, and I have found two different types. From what I understand, one type uses two electrodes that are separated from each other, and the current is allowed to pass through the solution, which conducts electricity due to the ions and breaks up the water molecules into hydrogen and oxygen. The second type I have found is composed of multiple plates to increase surface area.
 (https://drive.google.com/file/d/0BzXcmrdDUUGNOFc2cUJldnBJRnc/view)
From what I understand, they are spaced so that they alternate between positive and negative, though I saw some that use neutral plates (not sure what those are for). I believe hydrogen and oxygen bubbles are then formed on the surface of the metal. What is really happening in each setup (where are the bubbles being formed...) and are they really different at all, or is it just about the size of the gap? And what is the purpose of the neutral plates?
Thanks so much for your help!
-Abe

This seems rather far from broad.  It asks what the reaction product of electrolysing a sodium bicarbonate solution would be and what the purpose of neutral electrodes would be in a cell.


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Why doesn't the reaction require a dehydration step after aldol formation?

During aldol formation, I've learned that there is a second addition of water in many cases to cause a dehydration to occur that results in a double bond at the alpha and beta position to the carbonyl. Why does this specific reaction automatically dehydrate? Is it due to the benzene ring?

That certainly helps because you now have an extended $\pi$ system.

@Biomed Boy
What do you mean by addition of 2nd molecule of water? Actually, that was not a clear statement. If you could show it in some of the mechanisms, it coul have been better to understand your doubt/question completely.
There is dehydration usually always associated with Aldol, if substrate has 2 or more (alpha) H atoms, even in absence of a Ph ring: in acid catalyzed as well as base catalyzed Aldols, though their mechanisms differ.
The reason is always a conjugated (with carbonyl) bond, which may be facilitated by a Ph ring, as in the given case.
The dehydration steps do involve:
I. In acid catalyzed Aldol, it is usual E2

II. In base catalyzed Aldol, it is E1cB

Why do they vote down, for no apparent reason?

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy or which need further clarification, like you imply with "Actually, that was not a clear statement."


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Platinum engagement ring as oxidation catalyst
I know platinum is used in the catalytic converter to oxidize pollutants.

For a pollutant ($\ce{N_xO}$, $\ce{CO}$, etc., your choice), I'd like to get an order-of-magnitude rate of catalysis, assuming partial pressures found in a major city (NYC?)
At room temperature (or body temperature)
Let's assume the ring has 1 $\mathrm{cm^{2}}$  of surface area (I'm not rich, I assume  1 $\mathrm{cm^{2}}$ is reasonable)

Obviously using bulk platinum metal is an cost-ineffective method of catalysis, so it doesn't seem to be well studied. I know reaction is more efficient at high temperature, and while my girlfriend may figuratively be from Venus, I'd like to know the catalysis near STP. And perhaps a support is needed to prevent catalyst poisoning.

Hello and welcome to Chemistry.SE. Feel free to visit the [help] if you have any questions about the site and how it works. Regarding your question, the conditions inside a catalytic converter are drastically different from that around a ring finger ;) Seriously, the air quality surrounding a person wearing a platinum ring will be indistinguishable from that around anyone else in the area.

I would assume the rate of catalysis is zero, pretty much exactly zero.


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How to determine relative polarity (basic procedure)?
I am taking AP Chemistry, and have noticed that I can identify a polyatomic polar molecule, but struggle to determine which molecule is the most polar given a set of several polar molecules.
What would be a basic procedure/set of questions to ask myself in determining which molecule is the most polar? I hope my question is not too general...
Thank you!

Difference in electronegativity values between the two atoms sharing the bond should give some idea about the bond's polarity.

Determine which molecules are polar out of the given options.
Determine the polar bond (This will be the bond causing the molecule to be asymmetrical)
The molecule with the polar bond that has the greatest difference in electronegativity is the most polar. For example a carbon-oxygen bond is more polar than an oxygen-fluorine bond because the difference in electronegativity for oxygen and carbon is greater than the difference between fluorine and oxygen.

You can find electronegativity values in the following table or a table like it, I am not sure if you are provided with one. If not you can normally judge electronegativity by the how fare an atom is from fluorine, as seen in the table below. The size of the molecule also affects polarity, for example polarity decreases with the size of alcohols, but large complex molecules should not come out at your level.


text:
Chemical kinetics with the reaction of tert-butyl bromide with azide ion

The reaction of tert-butyl bromide with azide ion in aqueous solution
  is proposed to proceed by the following mechanism:
$\ce{(CH3)3CBr(aq) <=>(CH3)3C+(aq) + Br–(aq)}$
$\ce{(CH3)3C+(aq) + N3–(aq) \to (CH3)3CN3(aq)}$
Assuming that $\ce{(CH3)3C+ (aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants,
  what is the rate law for this reaction, where $k_1$ is the forward
  reaction rate of the first equation, $k_{-1}$ is the reverse reaction
  rate of the first equation, and $k_2$ is the forward reaction rate of
  the second equation?
(A) $\ce{k_1[(CH3)3CBr]}$
(B) $\ce{k_2[(CH3)3CBr][N3-]}$
(C) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-]}}$
(D) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-] + k2[N3-]}}$

My thoughts: First of all, I read that $\ce{(CH3)3C+ (aq)}$ is a steady state concentration, which meant that $\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = 0,$ which prompted me to write out differential equations for each reactant. Thus, I got a system of differential equations:
$$\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} =-\frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$0=\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} - \frac{d\ce{[(CH3)3CN3 (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}] -k_2 \frac{d\ce{[(CH3)3CN3 (aq)]}}{dt}$$
$$\frac{d\ce{[Br- (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$\frac{d\ce{[N3- (aq)}]}{dt} = -k_2\ce{[N3-(aq)]}$$
But I was unsure of how to use these differential equations to actually find the rate law. Could anyone provide any suggestions?

You really don't have to write all that differential equations.
Note: I'll write all my steps on a paper, since MathJax formatting is a little time-taking at the moment.
Consider the rate laws for the two reactions:

Since the $\ce{[(CH3)3C+]}$ isn't changing with time, we get

From which, you get

Plugging these in the Rate2 which I mentioned, you will get,

Which is the same as what's given in option (D).
I hope I have satisfied your queries.

Sorry, but how did you get your first rate law?

Those rate laws are directly derived from the reaction. For example, in an elementary reaction: $$\ce{mA + nB -> C}$$ The rate law would be $$Rate=[A]^m+[B]^n$$

According to my book rate would be $k[A]^m [B]^n$ is that just a typo on your part? Additionally, how did you derive the rate law for reactions in equilibrium?

Sorry, my bad, that plus was not meant to be. (I cant edit that comment now, but I think you get it)

You should really take a look at this so you can correctly format your future posts.

@Display Name, for reactions at equilibrium, consider it has two reactions, one forward and one backward. Suppose for the reaction: $$\ce{mA + nB <=>[K_1][K_2] pC + qD}$$ We consider the rate of reaction as difference between the forward and backward rates. In that way, you will get the rate law to be: $$Rate=K_2[C]^p[D]^q-K_1[A]^m[B]^n$$

@PrittBalagopal Wouldn't the rate be the negative of what you wrote, because we are measuring the rate of formation of product, not reactant?

No, rate of reaction is defined as $$\text{Rate}=\frac{d\text{[Reactant]}}{dx}$$


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Why is sodium potassium tartarate used in Fehling's solution?
Why is this salt used specifically in Fehling's solution, instead of a simple disodium or dipotassium salt of tartaric acid?

It's simply common.

Common in the sense, easily available or easily synthesised?

https://en.wikipedia.org/wiki/Potassium_sodium_tartrate

Fehlings soluition is alkaline. Mixing cupric ions with an alkaline solution would result in formation of cupric hydroxide of very low water solubility. By adding tartrate ions, the cupric ions will form a complex with the cupric ions which will keep them dissolved in the alkaline solution. The type of cation of the tartaric acid is of no importance other than for solubility reasons.
The same principle is used in Benedicts´s solution, where tartaric acid is replaced by citric acid or a salt thereoff.

Does replacing one sodium or potassium ion from a disodium or dipotassium salt cause that much change in solubility?


text:
How does cross-linking between polymer chains increase the melting point of the polymer?
https://i.stack.imgur.com/uUcQr.jpg
Suppose we have this polymer chain. Now, we cross link some chains ( not all ) , so some chains get linked, and others remain as they are. What we now have is a 'mixture' of polymer chains which are linked, and those which aren't. My question is how does this increase the melting point? I've read that intermolecular forces between chains is replaced by the covalent bonds of cross-linking, which are much stronger, and thus melting requires much more energy. But melting doesn't involve the breaking of covalent bonds unless the structure we have is 'one giant molecule', like Diamond, in which there are no 'intermolecular forces' because it's just one molecule.
In our case, we most certainly will have more than one molecule, and so the melting point should depend on the intermolecular forces between the chains. As these forces are largely unchanged, the melting point should be unaffected. When we heat this molecule, the weakest of the forces ( intermolecular ) should be broken first, and thus the melting point should remain unchanged.
I've been confused about this for about a week now, and I'm surprised google cannot find a similar question.

You're right, melting point does depend on the intermolecular forces. But the term intermolecular isn't quite applicable in the case of branched polymers like Bakelite, since the whole plastic just a very large molecule. In that case, you have to consider the branching bond energies to melt the plastic.

I see what you mean, but consider the following case :
There are 5 chains of a polymer in a sample, and 3 are crosslinked to each other. Now, if we start heating it up, then the two which were not cross-linked will soon move freely, and the 3 which were cross-linked, though they cannot move independently of each other, can also move freely as a whole. The only bonds we've broken are the intermolecular forces between these chains, so the melting point should be unchanged. Or is it only 'melting' if the 3 cross-linked chains also move freely of each other?

The image is about hydrogels. These do not melt, and also their precursors are usually in aqueous solution before cross-linking, so melting of the polymers is not a relevant question. Is your question about hydrogels or non-swollen (partly) cross-linked polymers?

Cross-linking of polymers reduces the chain length between the links and higher the cross-linking lower the length of these segments this consequently reduces the freedom of chains to rotate, fluctuate and translate converting it to be more rigid.The concept of melting for such structures may not be applicable and it is true that cross-linked polymers do not show melting but just softening at low cross-link density. At high cross-linking the material may not soften but degrade and burn.In the example mentioned above two uncross-linked chains may by entangled, which may show melting but as other three can not melt one may not get clear MP as in case of crystalline polymers.
Dr. B R Gupta,
Retd. Prof. I I T Kharagpur, India


text:
How many crystal forms of HCl have been identified - where to find the unit cells/lattice constants?
This blog about stomach acid shows this image of the "zig-zag" form of crystalline HCl (see also Wikipedia where it looks (to me) like five four HCl's per unit cell), but links to this cubic crystal structure (2nd image) data card instead with only one HCl/cell.
I've read crystallographic "data cards" in a past life, but I have no idea how to search for all the forms of HCl on-line (and non-paywalled) now, in order to get the lattice constants, angles, unit cells etc. to build a 3D model of each form of crystalline HCl in Blender for example.
note: I am really looking for just the dimensions and shape of the unit cell, and coordinates of the atoms within the unit cell as a name and numbers, as one would find on a crystallographic data card. I can do all the plotting - no need for plotting software.

below: Here's the Wikipedia image mentioned above, from here. It certainly looks like the Chlorines are at 0.25 and 0.75 along the a-axis (vertically) and 0.00 and 0.50 along the b- and c-axes, so there seems to be enough information here to work out or infer the complete 3D structure, assuming it's cubic, and further infer that a=b=c=521pm, which suggests a density of this solid of about 1.7 g/cm^3. But remember, when you infer, you make an inf out of e and r.

From the first picture depicted, I would refrain from stating there are five HCl molecules per unit cell.  One thing, while there is some choice in the definition of a unitcell (there are several definitions possible, even under the constrain "it should be the simplest and still complete representation possible"), lower multiplicities occur more often; and more frequently seen values are $Z = 1, 2, 4$.
Second remark, again to the first picture: no, it does not look so as there were five molecules per unit cell.  Here, the projection of the $(a,b)$-plane is depicted (the little green and red axis emerging the origin O bear a label).  To graphically determine the actual multiplicity or the number of molecules present in a unit cell, it is better to "walk around" the unit cell; for example by accessing either *.cif (typical for small molecule structures) or *.pdb (more typical for protein structures) in a appropriate program, like CCDC's Mercury (likely used in the first drawing), or Jmol (used for the second drawing), or Avogadro.  Each of the three programs may be deployed under Windows, Linux, or MacIntosh; and in terms of displaying structures and to perform measurements of distances and angles available for free (as free beer).
It is possible, yet not in every case necessary to go from *.pdb or *.cif via Blender to prepare a dataset for 3D print though.  The "just do it" software that comes to my mind for this purpose is Werner Kaminsky's cif2vrml (windows only).  If you have Mercury (which actually may read both *.pdb, *.cif, beside *.mol, *.mol2, *.xyz, too), there is a built-in file -> print 3D menu, too; allowing you to chose between *.vrml or (then colourless) *.stl file.  Doing so found entry in publications, where examples in the Journal of Chemical Education like DOI: 10.1021/acs.jchemed.5b00597 and DOI: 10.1021/acs.jchemed.6b00293 come pretty close to what you intend to do. (As you are not afraid of larger molecules, consider the dedicated NIH 3D exchange as a valuable source of inspiration, too.)  Both jmol as well as Avogadro (referring to 1.2.0; not to the branch of Avogadro2) equally offer VRML export, the former including a dedicated how-to here.
For Jmol (the second figure presented), there are (at least) two options to read *.cif data.  The one by the GUI typically opens the *.cif, representing both the crystallographic motif and the unit cell (the later may be switched off, if necessary / useful).  Note that the GUI in the top row differs slightly from the one accessible by right-mouse-button click inside the window.  Via the toprow gui, you may access the CLI console, too -- allowing you to access many other functions, too.  For example, if you decompressed the *.zip archive, placed your *.cif into the same directory where jmol.jar lives, typing

load cell test.cif {1 1 1}

will load the data set, test.cif, and will fill the complete unitcell till indices {1 1 1}. For one of the data found at COD, Jmol provided me this representation:

Where to get the *.cif? Literature databases like ACS' Scifinder and Elsevier's Reaxys indicate / link to such primary literature.  Depending on the journals policy, these *.cif data may be freely available in the SI (ACS journals, for example), or not (Tetrahedron family ...).  Clearly, a dedicated database ones like CSD (organic, metalorganic; preferentially small molecules than proteins) is an option.  For the service and potential insight offered, an annual academic subscription is very well invested money (was once around US$ 300-400 per year and full time equivalent).  If you encounter their structure code in a literature reference, but not the *.cif, you may request the data free of charge here; regardless if you / your institution holds a subscription, or not.
For mineral compounds, for example ISCD or the one by AMS.  The baltic COD is much younger than the others (CSD started more than 50 years ago), so the data coverage is not yet this complete, but searchable by keyword and structure, too; yielding by the former (allowing only H and Cl as the only elements present) 5 entries for HCl.
Where to find all the *.cif of HCl? It is not really organic / metallorganic (CSD), nor really a mineral (ISCD).  Perhaps via Scifinder and Reaxys one may retrieve a more round answer.
Addition
One reference found doi:10.1038/213171a0 is based on powder diffraction studies of HCl and DCl, listing these structures known to the authors back in 1967:

It continues with further powder diffraction studies, presenting an orthorhombic model, with four molecules per unit cell:

and a projection very similar to your first figure:

From these two it may be possible to construct (manually) a *.cif file which is then readable to Mercury others.  For this nature paper, Thomson Reuter's Web of Science alone knows about 139 citations, the most recent one in 2015 10.1021/acs.jpcc.5b01653 about predicted supraconducting polymorphs of HCl and HBr in the monoclinic $C2/m$.
Furthermore, a cif file based the Italian 1933 structure report the English wikipedia refers to is accessible in the COD if searching by keyword (H and Cl elements present, excluding O and C, minimal = maximal number of elements = 2).

Please check what I've written: "see also Wikipedia where it looks like five HCl's per unit cell" I can change it to "...looks to me like..." but I have a hunch that's... NOPE!  Now it looks like four! :)  $ 4\times \frac{1}{4} \ + \ 4\times\frac{1}{2} \ + 1 $ (counting the chlorine sites in the unit cell shown there).

@uhoh Initially five, now four ... I'll take a look at it; thanks for your input.

...and I'll need a bit of time to take in this very extensive answer as well.

@uhoh The paper mentioned by you has some limitations, data set based on powder diffraction, not on a single crystal diffraction), which are understandable; Z = 4.  It is cited more than 139 times, as a look into one of the literature databases shows.  Briefly glancing over some abstracts, there are cubic, orthorhombic, monoclinic, triclinic or even disordered structure exp./theo. models, too (http://dx.doi.org/10.1063/1.1743967) around. Further ramifications imaginable, it may dangerously turn into a question put on hold "because to broad" (well, along the parameters used for chemSE).

Well, I didn't ask how many papers there are, or how many entries there are in data bases. I've just asked How many crystal forms of HCl have been identified? If it is not clear to you which are duplicates or which are in error, then your answer should be "I do not know personally", but let's wait and give some others a chance to look at the question before putting it on hold so quickly. It is possible that someone actually knows the answer, or finds a reference where this has been cleared up. I'm guessing the answer is three or two.

@uhoh "Currently I am not able to state how many truely different crystal structures of HCl are known, either experimentally, or theory predicts." I sincerly agree with your suggest.  Good luck.

Thank you for the extensive answer, and also for the additional info from the 1967 paper - for me this is really helpful to go back to the original work - where one can read more about how things were actually being worked out.


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How much additional product forms when small amounts of reactants are added to a reaction at equilibrium?
I am not able to understand how to determine the exact amount of additional product formed when additional reactant is added to an equilibrium reaction.
For example consider the reaction:
$$\ce{CO2 + H2 <=> CO + H2O}$$
and suppose the number of moles of $\ce{CO2}$, $\ce{H2}$, $\ce{CO}$, and $\ce{H2O}$ at equilibrium are $a$, $b$, $c$, and $d$ respectively. $K=1$ for simplicity.
Now how do I determine the amount $\ce{CO2}$ to be added to double the amount of $\ce{CO}$ produced.
I have tried by writing the equation for the equilibrium constant and keeping the amounts of $\ce{H2}$ and $\ce{H2O}$ constant but it didn't work out.
What I had done is (where $x$ is the new number of moles of equilibrium)
$$K=\frac{c\times d}{a\times b}= \frac{2c\times d}{b\times x}$$
This gives me $x=2a$ but after this I'm not able to proceed any further.

Hi, welcome to Chem.SE! This looks like a homework question. Could you please edit the question to show how you tried to solve it? Also please read out homework policy here.

Note that you can use MathJAx to typeset maths and chemistry so that it looks appropriate. I have edited your post to do so. Please visits https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here to learn how to do it for yourself.

You could try with a simpler reaction first, say A=B with initial A as $a_0$ and at equilibrium let amount $x_e$ have reacted. Write down the equilibrium constant and solve for $x_e$. When you have method try with your reaction, its just algebraically messier.

I have been able to do it for simpler reactions involving only one reactant and product but I'm not able to apply the same strategy for more complex ones. @porphyrin