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Determining mole fraction of carbon dioxide knowing Kp only
I've been assigned the following problem at school:

Given the following reaction:
$$\ce{C (s) + CO2 (g) <=> 2 CO (g)}$$
Find the mole fraction of $\ce{CO2}$ provided that, in equilibrium, $K_\mathrm{p} = 14.1$.

Our teacher's solution is $x(\ce{CO2}) = 0.324$ (mole fraction).
I suspect that the problem as is provides insufficient information to arrive at the solution above. Assuming that the total pressure of the system is $\pu{10 atm}$, the resulting mole fraction is indeed $0.324$.
Is the problem truly incomplete or can it be solved without knowing the total pressure?

You seem to be comfortable enough working out the answer when you're given a pressure. Why not try working it out without explicitly choosing a pressure? Use $p$ as a symbol instead of putting in something like 10 atm, and if you find that all the $p$'s cancel out in your answer, then it follows that you don't actually need to know the pressure.

I have attempted this. The final expression is: 14.1x = (1-x)^2*PT. The total pressure does not seem to cancel out. Here, x is the mole fraction. X seems to be a function of PT. The value of X changes with other values of PT.

Took me a while to understand what that equation meant, but yeah, you are correct! So, the question is indeed incomplete. [More common notation would be something like $p_\text{tot}$; PT suggests some pressure multiplied by temperature...]

You can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details.

Ok! Thanks for the quick reply! My bad. I'll keep this in mind for the next question.

Does you teacher at least give you the units of Kp?

@ChesterMiller $K_\mathrm{p}$ is supposed to be dimensionless, isn't it?

@andselisk Actually, it is expressed in terms of the ratio of the partial pressure of each substance to the pressure of each substance in its pure reference state.  If the reference state happens to be 1 bar, and all the partial pressures are in bars, then the units cancel.  If the units of the reference state are other than 1 bar, the pressure of the reference state must be included in the expression for Kp in order for it to be dimensionless.  Otherwise, Kp will have units, and the partial pressures must be expressed in these units (assuming moles products and reactants don't match exactly).

@ChesterMiller I'm not sure I follow, but what prevents the conversion of pressure units into bars with subsequent cancellation as the reference state is included by definition in $K_\mathrm{p}$?Of course, it's not always denoted as such, but it's always there, just as we speak of $K_\mathrm{c}$ in terms of concentrations keeping in mind they are multiplied by activity coefficients. Besides, I've never seen a respectable source showing $K_\mathrm{p}$ with any units (maybe I haven't seen everything in my life yet:) ).

@andselisk That's because all the partial pressures are assumed to be in bars and the reference pressure is in bars.  Try solving the same problem with the partial pressures expressed in mm Hg instead.  You'll get a different answer (unless, fortuitously, the moles of reactants in the formula equal the mole of products).

@ChesterMiller Why would I want to use mmHg scale in this case? It make as much sense as converting a temperature difference in degrees Celsius into degrees Kelvin by adding 273.15 to it:)

@andselisk Well, suppose the total pressure were measured in mm Hg or psi.  In that case, you would calculate partial pressures as mole fraction times total pressure, and you would have to convert back to bars.  I feel like this is leading down a rabbit hole.

@ChesterMiller I think you right both about the rabbit hole and unit conversion, in a sense. The thing is that there are too many textbooks and results published with $K_\mathrm{p}$ taken too liberally and it brings up a lot of confusion. I might also be totally wrong since you have more experience with physical chemistry in general than I do (judging from your answers:) ).

@ChesterMiller My teacher did not provide the units of $Kp$

Then, in that case, the partial pressures must be expressed in bars.

In this case, total pressure "P" must be given to solve for "XA" (equilibrium molar fraction of CO2); however, that's because this reaction has Δn=1≠0.
If a different reaction was given where  Δn=0, the problem could be solved without knowing "P".


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High content of potassium reported by spectrometer in silverware items
We tried to analyze a few silver items (800/1000 fineness) using a hand-held  ED-XRF spectrometer, which is normally used for analyzing industrial alloys etc. The operating element range of the device was Mg-U.
The most surprising result of the experiment was a very high reported content of potassium, often higher than silver. For example, one of results (listing just elements with content >= 1%):
| element                           | content |
|-----------------------------------|---------|
| K                                 |   36.0% |
| Ag                                |   27.5% |
| (sum of elements lighter than Mg) |   24.7% |
| Cu                                |    5.0% |
| Sn                                |    4.5% |

Such high potassium content was pretty consistent across all our silver samples, but not present in others (like stainless steel)
Clearly it's not possible that the alloy could contain that much potassium. Where could this result come from? Can potassium be somehow bound to the molecules at the thin surface layer? Or is there some explanation for this phenomenon?

The spectrometer is reporting the Ag L x-rays as potassium.

To give some numbers to MaxW's comment, you can see on this webpage the x-ray absorption energies of basically all the elements.
For example, the K-edge of potassium is $3.61~\mathrm{keV}$. The L-II edge of silver is $3.52~\mathrm{keV}$ and the L-1 edge of silver is $3.81~\mathrm{keV}$. So, it seems like you are somehow automatically assigning the peaks and scanning over a very large range of energies. The problem is that this will run into problems if you assume that all transitions are out the $1s$ orbitals (K-edge) because there are L-edge transitions for heavier elements which are comparable in energy to those from lighter elements.
So, without more information about how the transitions are being assigned, it seems very likely to me that MaxW was right to say these are L-edge transitions from Ag.

To put the other answer in a graphical context, I took the X-ray emission spectra of K and Ag from some place online*, and plotted them together, crudely resized to fit the same scale. Some of the Ag L-lines (I think it's Lβ 1 and/or 2) are at exactly the same position as the K Kɑ line. So, the software sees a certain amount of counts at K, and identifies it. Good software knows that there's silver there and the K is spurious. Badly written software does not.
The solution is usually simple: just untick potassium in the element identification menu (or whatever it is called on your device).

*The specific source does not matter. Any web search for "silver/potassium EDS spectrum will look similar.

Splendid illustration and good answer, but could you please put a reference to "some place online"?

You also must remember the % transmission vs Energy through the window of the OP's x-ray detector. No mention of what type of detector either, so detector efficiency vs energy needs to be considered also.

@MaxW indeed. But regardless of the unknown configuration, any ED detector will not have the sufficient resolution to distinguish Ag L-lines and K K-lines.

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Why does my synthesized 4-bromoacetanilide smell like almonds?
So, I synthesized 4-bromoacetanilide in the lab by brominating acetanilide. Acetic acid was used as a catalyst. Sodium bisulfide was used to neutralize the reaction product. Then it was washed with water and recrystallized in ethanol. After that I let it dry for a week. Then I added a little water by accident but decided to store it anyways, and a few weeks later I opened the stored product and I smelt almonds. It wasn't even a weak odour, it was much like benzaldehyde, but not that strong. How would this even be possible? I can't figure it out.
2-bromoacetanilide might have been formed. I did search a little bit, and found that the decomposition of 2-nitroacetanilide - a similar product - results in hydrogen cyanide (almonds smell; very toxic). But it cannot be that, can it? Wouldn't I be like dead by now? Sniffed it three times. Or is it not really toxic at very low concentrations, but smellable?
The smell was kindof pleasant, less aromatic-like than benzaldehyde. Not strong, but also not weak when I held it close. People say HCN smells like burned or roasted almonds, but it was more like a sweet but bitter maybe ester-like odour. Something probably most similar to marzipan.

SDS Source

Just because two compounds are different doesn't mean that they don't smell the same...

4-bromoacetanilide isn't supposed to smell like almonds though. It's one step away from being paracetamol, and acetanilide itself doesn't smell either.

I haven't heard anybody using the word 'bromizing' until I checked google. I use the word 'brominating'.

That's because I'm Dutch, not sure how that got in my mind but I guess it's closest to the Dutch word: bromeren.

I remember how "Brombeeren" (German name for blackberries) confused me at first. I was stupid enough to ask whether there are "Chlorbeeren" though, so... (German is not my first language, just in case:) )

It might be hydrolysis product of 4-bromoacetanilide, 4-bromoaniline, which is described to have “sweet odor”.

You may actually be pretty correct, thank you. I had expected hydrolysis to form paracetamol, but apparently not.

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heating methane in a vacuum chamber
Using radio waves to heat, and electromagnetic fields to contain....could methane be heated to much higher temperatures than via simple oxygen/combustion, thus producing a much higher specific impulse for, say, a rocket engine?
Not talking plasma drive temps here (millions of degrees), but several thousand degrees above normal combustion temperature.

There are lots of methods of production of plasma and it's containment, but I don't see a reason to ask about it here...

Thank you. I apologize. I am not a student, and have no educational background in chemistry at all. I found this site, and was curious as to potential methods to get better, more efficient thrust from earth to orbit propulsion systems.

Beside all, when a rocket eject exhaust is ejecting what it has brought up.  In your case the gas would be simply a medium.  How would you power your radio waves of whatever system heats up the methane? That would be the true specific impulse. Either you fall in alternative engine as the plasma drive category and regime of application or is better to burn the fuel. Power regimes differ a lot.

I'm voting to close this question as off-topic because it's rather plasma physics/propulsion stuff.

Methane decomposes to hydrogen and carbon at high temperatures.

Considering methane is $T_\mathrm d$ symmetric, con it even be excited by microwaves?

The problem is where the energy for heating comes from not the gas or the mechanism of heating
Rocket engines often use burning fuels to produce thrust because chemicals are a very good store of energy and burning often produces hot gaseous products.
Yes, you can produce hot gases by other means (there are electrically driven plasma rockets and you could, maybe, heat gases with microwaves or similar radiation). But the problem you have is where to get the energy for that process? A self-contained rocket has to have an energy source and that source has to be compact and a slight as possible (a long electrical cord to an electric socket isn't practical in orbit or in space). Just about the best way to store energy is various combinations of reactive chemicals (eg oxygen and kerosine) and the easiest way to turn that into thrust is burning. The only other approach know to work well is to put a nuclear reactor on the rocket as an energy source but that has obvious disadvantages for other reasons.
Rocket scientists have spent a lot of heroic effort to optimise mixtures of (often dangerous) chemicals to achieve better thrust. For a good summary read John D Clark's Ignition where he traces the history of trying to make use of extreme oxidising agents like chlorine trifluoride to get better performance.


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How do we recognise a Hybridised Sp3 orbital?
As far as i know the sp3 orbital is a combination of orbitals. But how do we know if we have a hybridised orbital or not?

There are ways to determine bond lengths (via X-Ray diffraction or, Neutron diffraction) and if bond lengths are found to be equal in a place where we predict it to be by hybridisation, you can say the orbitals are hybridised.

Not an answe but maybe it highlights a bit about sp-hybridizations https://chemistry.stackexchange.com/questions/89214/draw-a-diagram-to-show-the-shape-of-chloromethane-and-explain-why-it-has-this-sh/89224#89224


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How to accurately determine mass transfer using balances?
I had troubles the other day using a balance that goes to the tenth of a milligram place (i.e. .0005 vs .001 g), when trying to determine how much solid I transferred. I would take some amount of mg, record it, and then set it aside. I'd take the receiving flask, tare it, and add the solid to said flask. The # I would get would either be wildly lower or slightly lower. I am very sure I am not mispouring anything.
So I am wondering, is it possible these sensitive balances cannot handle things with larger masses, or if the receiving flask is not perfectly flat, would that mess up the mass? Similarly, would the location where I put the flask or weigh paper on the balance matter (e.g. dead center, to the left, right, etc.)? Is there another thing I could do? technically I could take the difference in weigh paper with solid vs without, but that doesn't tell me what got into the receiving flask.

Like many manipulations in chemistry, there is technique involved in weighing. Did you touch anything with your bare hands? Was the balance pan swinging? Is the substance being weighed hygroscopic? If you're weighing to the milligram using weighing paper just seems wrong.

Accurate weighing requires you to be aware of enormous amount of factors. It is not possible to cover them all in one answer. For starters, where do you put your balances? Are they levelled? Are there any heat or vibration sources nearby? Are there any sudden air flows? Is your mains power reliable? Were the balances calibrated properly? Is the ambient temperature or humidity too high or too low? Do they change much? Is there much static electricity (depends on the materials you are working with and on the ambient humidity)? Do you use gloves? How much do you weigh? I mean, not you weigh, but isn't it something less than the minimum weight requirement for your balances? Et cetera, et cetera.
Returning to your questions: in general, sensitive balances can handle large masses which are within the specification. However, the measurement uncertainty depends on the mass. Then, if the receiving flask is not perfectly flat, there should not be any problem unless the flask is wobbly or unsteady. The location where you put the flask should be as close to the center of the weighing pan as possible. It matters.
As for the other things you could do, I'd like to recommend you to check out some weighing practice manuals. You can start with the balances manufacturers - Mettler has brief Good Weighing Practice whitepapers, maybe others like Sartorius have something similar. Analytical chemistry handbooks should have something to offer. Then, check out the manual for your balances. Aren't you trying to weigh something more accurately than the accuracy of the balances (remember, accuracy $\ne$ resolution)? Then, if you don't, just ensure that you are following the recommended weighing practices as closely as humanly possible.


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How to get rid of calcium carbonate and other salts formed by calcium hydroxide in air?
My basic question comes from the idea that in olden days people used lime to paint houses, so after it is dried out it becomes white paint which sticks to the walls. So, how to get rid of it? The reaction is
$$\ce{Ca(OH)2 + CO2 -> CaCO3 + H2O}.$$

We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. So I rolled back the formatting to a more appropriate form.

Acetic acid will do.

@IvanNeretin reaction please.


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Compressibility Factor Graph - Which gas attains a deeper minimum?

As in the given graph, how do I predict which curve among that of hydrogen, methane, and carbon dioxide attains a deeper absolute minimum? (greater deviation from Z=1). In fact, how do I make this comparison for any gases for that matter, and on what factors does this depend?
Is there any temperature dependence of the above?
If nothing can be said in general, I'd at least like to know how to make the comparison for ammonia and methane, as that has previously been asked in a contest years ago.
I'm looking for more of qualitative comparison, but all sorts of answers are welcome.

There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.
If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,\; P_r=P/P_c,\;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'

$Z$ vs $P_r$ at different reduced temperatures
For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume  and $T_C=8a/(27Rb), \; P_c=a/(27b^2) ,\; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.

All that you said in the answer can be used to understand Z vs P graph at different temperatures for a particular gas but i fail to understand how can I use this information to get what the OP has asked for?

These plots apply  to many gasses at the same reduced temperature & pressure  as they are in reduced units. Gases/vapours  such as water, nitrogen, ethylene, carbon dioxide, methane, iso-pentane and butane and probably many others fit  well.


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Disiamylborane on alkyl halides
I am having problem predicting the last product in the following reaction:

$$\ce{CH#CH ->[Na/NH3 (l)][(Excess)] (P) ->[DCl] (Q) ->[(i)Sia2BH][(ii)CH3COOH] (R)}$$

I am getting an alkyl chloride as product $\ce{Q}$ and cannot understand how it will react with $\ce{Sia2BH}$. Or, I think I am doing some mistake in the formation of product $\ce{P}$. I do know that alkynes on reaction with $\ce{Na/NH3}$ shall give trans alkene (Birch reduction). Will the "excess" part affect the reaction differently?
$$
\begin{align}
\ce{CH#CH &->[Na/NH3 (l)][(Excess)] CH2=CH2} &(\ce{P}) \tag{1} \\
\ce{CH2=CH2 &->[DCl] CH2D-CH2Cl} &(\ce{Q}) \tag{2} \\
\ce{CH2D-CH2Cl &->[(i)Sia2BH][(ii)CH3COOH] ?} &(\ce{Q}) \tag{3}
\end{align}
$$

I think the DCl step is the work up of the sodium/NH3 reduction step and will give the deuterated alkene

That's quite an unclear question. It's more common for workup after every step to be assumed / implicit, and based on that your interpretation would be correct. But @Waylander's interpretation is the only one that makes step 3 sensible.

so the product P will be NaC≡CNa, and Q will be DC≡CD, and then Q as CHD=CHD?

I did read today that Acetylene reacts with sodium in liq Ammonia or Sodamide to form sodium acetylides. I guess, NaC≡CNa shall be formed and the end product will be what @Waylander said.


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Alcohol or Carbonyl - Which is more reactive towards PCl5?
We're aware that ketones and alcohols both react with phosphorus pentachloride, to produce geminal dichlorides and alkyl chlorides respectively.
What happens in the compound shown below, when treated with $\ce{PCl5}$? Which group reacts first? (which is to say, what happens if I were to add just one equivalent of $\ce{PCl5}$?)

Please help me out here, I'm not able to qualitatively predict an order.

Reaction of alcohols with PCl5 and PCl3 (for reaction with alcohol)
Mechanism for conversion of ketone to dichloride with phosphorus pentachloride
(for reaction with ketone)
Now as per the question as it is not given in excess so it will react with any one of the alcohol or ketone. It's quite hard to tell which one of them will react first because both of them show totally different kinds of reactions. Like ketones prefer neucleophile attack and alcohols prefer substitution and elimination.  But if you look at the mechanism quite closely you would find all the favourability of the reaction depends on the lone pair donation tendency of oxygen.  Now as you know, in sp³ hybridization state the s -character is nearly 25% and in sp² hybridization state s-character is nearly 33% and as s-character increases electronegativity increases
Why a higher s character increases a carbon atom's electronegativity?
and the tendency to hold the electron pair increases so for ketone it is quite difficult to donate it's electron pair to PCl5. So in that perspective I can say alcohol is quite more reactive than ketone in this case.

@arya_stark, I think from a thermodynamic perspective, ROH would be more reactive. In both cases the same product POCl3 is being formed but with the ketone a strong C=O is being broken which is unfavourable.


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pH of very dilute acids
From doing some research on the site I have found that many people have posted about solutions of $\ce{HCl}$ where $[\ce{HCl}] = \pu{1e-8 mol dm-3}$
Here they are able to deduce
$$[\ce{H+}] = \frac{[\ce{HCl}]_0}{2} + \sqrt{\frac{[\ce{HCl}]_0^2}{4} + K_\mathrm{w}}$$
where $[\ce{HCl}]_0$ is the initial concentration of  $\ce{HCl}$ dissolved.
Can we make a generalisation of this and say if $\ce{H_xA}$ is a strong acid that will dissociate fully when dissolved in water? Then if $[\ce{H_xA}]_0 < \sqrt{K_\mathrm{w}}$ then the $\mathrm{pH}$ of solution can be expressed by
$$\mathrm{pH} = -\log_{10} \left(\frac{x[\ce{H_xA}]_0}{2} + \sqrt{\frac{(x[\ce{H_xA}]_0)^2}{4} + K_\mathrm{w}}\right)$$

there is an answer here  https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355

Your formulation us correct for an acid with one dissociable proton ($x=1$) below pH 7, and if you assume ideal solution behavior it even works beyond your upper bound.  Typically you should take $x=1$ because, mostly, even most strong acids dissociate one proton.  However...
One limitation, however, is when you have an acid with two or more dissociable protons.  If the second proton dissociates before you get to pH 7, then you need to add that into your calculation (and then it gets a lot more complicated).  Among common acids sulfuric acid is the one that will do this; in the limit of extremely dilute solutions (0.001 molar or less, roughly), the pH of a sulfuric acid solution will be about 0.3 unit lower than your formula with $x=1$ because of that second dissociable proton.

What would the formula for sulphuric be then?

You'd have to solve a cubic equation.  I said it would be complicated ... .

what is the second proton didn't dissociate before pH=7?

Put $x=1$ then, that is all there is.  The remaining undissociated protons are part of the "A".

Could you elaborate on that?


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Explaining why s and p orbitals do not interact
While going through molecular orbital theory, my chemistry teacher mentioned that the reason why there is no significant interaction between $\ce {s}$ orbitals on one atom and $\ce {p_x}$ and $\ce {p_y}$ orbitals on another atom is that the $\ce {s}$ orbitals interact with both lobes of the $\ce {p}$ orbitals, having both in phase and anti phase interactions with these $\ce {p}$ orbitals, taking the z axis to be the internuclear axis.
However, I read that orbitals which do not have the right symmetry simply cannot interact, i.e. their overlap integral is necessarily $\ce {0}$. Thus, is the reasoning given by my chemistry teacher actually valid?

There is no contradiction behind your "however". It is the same in other words.

According to my chem textbook:
"Molecular orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the positive lobes above and below the molecular plane."
Your instructor might be pointing out to the bond-symmetry thing in MO formation between pz and s orbitals extrapolated in most chemistry textbooks.
But orbital overlap is a result of constructive and destructive interference. As s orbitals are spherical, they have infinite possibilities of orbital overlap with other orbital shapes and thus even overlap with d orbitals (as explained by VSEPR: Valence Shell Electron Pair Repulsion Theory in a similar manner).
Thus s and px and s and py orbitals do overlap; if you take px to be of lower energy than py and pz, s-px overlapping occurs in boranes (formed by boron, which has one electron in 2px orbital, and hydrogen, which has one electron in 1s orbital), that often dimerize. This is also the case with methane, where px, py-s overlapping occurs, assuming that in sp3-hybridized state px, py orbitals in C bond with H 1s orbitals.


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Why does the surface of an acid appear to have a stronger "dissolving power"?
In this Periodic Videos YouTube video by Brady Haran and the University of Nottingham, we see a piece of gold being dissolved in aqua regia, while Professor Poliakoff explains why aqua regia can dissolve gold even though its constituent acids can't.
The piece of gold was placed partway into the acid. I noticed that the acid appears to attack the gold the strongest at the liquid's surface, and the gold was first "dissolved through" along a line at the surface, leaving a little plate of undissolved gold dangling in the acid by one corner. You can see this best from about 4:10 to 4:30 into the video.
I was wondering if there's some effect at the surface of the liquid that causes this.
(I apologize if this is more physics than chemistry,  and also for my Google-fu not being enough to find ready answers.)

The bubbles formed rise to the surface and grow there until they are released into the air. 2. The piece of metal might be thicker at the bottom.

I do have a possible expl. that possibly I elaborate tomorrow. Rather than accumulation is right at the surface that the contact between metal and aqua regia is easily renewed plus potential.

@aventurin Are you suggesting the bubbles cause an optical illusion, or are you more in line with Alchimista's answer?

Besides mechanical aspects

bubbles might form a sort of layer at the immersed part while at the air interface they evolve and detache easily renewing the metal/acid (oxidizer) contact

a more fundamental reason can be° surface excess. Normally we treat interfaces as very thin and therefore negligible (in mass or volume) as compared to the bulk.
However at an interface the chemical potential (and all related thermodynamic properties) differs from that of the bulk owing to the fact that a layer can be assumed to have the same number of entities per unit of surface.
While such a layer at depth is surrounded by the very same, at the interface it sees two different environments.
This aspect is quite familiar in physical phenomena, such as surfactivity, the shape of droplets, and even the tip effect in electrostatic. Another example  possibly familiar to many chemists is the difference in product of solubility / solubility for solids at very different sizes (digestion of precipitates to attain bigger grains/crystals is based on this, tough intermixed with the merely geometric fact that small particles dissolve faster).
As for in water the interface has higher potential and this results in spherical droplets to minimize it, in your specific case you can see the aqua regia doing the same by being more reactive. It is like the oxidizer trying to escape the air interface (tough a new one is continually formed, of course) by reacting more energically than the layers below, if this pictorial helps you.
If you dig for "potential at acid/air interfaces" or "pH of an acid at the air interface" and so on, you can find some material dealing with example somehow related to your example. For instance I have found study of the dissociation of weak acids (ie pKa at the interface vs bulk), The vast majority is however studies in which the very interface is at focus, like in monolayer LB films, and the surface excess comes naturally at play whatever explicitly considered or not.
(For instance a title such as "On the dissociation constant of carboxylic acids in monolayer thin films" implicitly assumes that some difference exist as compared to the bulk of the solution).
° can be as for I do not know if the evident effect you have spotted is mainly due to the "stirring effect" of evolving bubbles or to the higher activity of the interface, or their mix.


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What do D and L stand for in the Fischer-Rosanoff convention?
In R and S configuration, the R stands for rectus and the S stands for sinister. Similarly, do the D and L descriptors in the Fischer-Rosanoff convention stand for something?

Your question title is inconsistent with the text. See e.g. https://chemistry.stackexchange.com/questions/48840/dl-vs-dl-notation/49015#49015

Dextro (right) and levo (left). (I learned it as "laevo", but I don't see that spelling in a quick Google search.)

In Rosanoff's paper of 1906, the two series were designated as Delta and Lambda from which D/L was derived. For a brief history, see http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/history/Fischer/rosanoff.html


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isomerism-position versus chain
For nitriles carbon of nitrile group is part of the carbon chain. So what is the isomerism shown by pentanenitrile and 2-methylbutanentrile? Chain isomerism or position isomerism?

Pentanenitrile and 2-Methylbutanenitrile should be chain isomers of each other, since the parent carbon chain itself is different for these two compounds. They shouldn't be named as position isomers.


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What is exactly meant by position of equilibrium?
I'm confused with the terms "position of equilibrium" and "equilibrium shifting towards left/right". I do have a vague picture that if a reaction shifts to the right, for example, it basically means that the reaction is now biased towards the products. But I'm not thorough with it. Does it mean that the rate of forward reaction is now faster than the rate of backward reaction, which would mean that the reaction isn't at equilibrium anymore?

You're basically talking about the concept of the reaction quotient.

https://chemistry.stackexchange.com/questions/107116/whats-the-difference-between-extent-of-reaction-and-position-of-equilibrium

The term "equilibrium shifting" is mostly used for a reaction that was already at equilibrium, but we changed something (in the simplest case adding one reactant or product to the equilibrium mixture so the reaction is no longer at equilibrium). The "equilibrium is shifting" is a shortcut for saying there is a net reaction to re-establish equilibrium.
The term is used both in cases where the equilibrium constant does not change (change in one of the concentrations) and in cases where the equilibrium constant does change (change in temperature, for example).

Does it mean that the rate of forward reaction is now faster than the rate of backward reaction, which would mean that the reaction isn't at equilibrium anymore?

Yes, for some reason the reaction is out of equilibrium, otherwise there would be no net reaction (no "shift").

I'm confused with the terms "position of equilibrium" and "equilibrium shifting towards left/right".

"Position of equilibrium" could be roughly equated with the set of all concentrations at equilibrium.
Taking a change in temperature as an example, you start with one set of equilibrium concentrations ("old position of equilibrium" corresponding to some reaction quotient $Q_1$). As you change the temperature, the reaction is no longer at equilibrium because the equilibrium constant changes (lets say from $K_{T_1}$ to $K_{T_2}$), so $Q_1$ is equal to the former equilibrium constant $K_{T_1}$, but not to the current equilibrium constant $K_{T_2}$). There is a net reaction until you reach the new equilibrium ("new position of equilibrium" corresponding to some other reaction quotient $Q_2$). At the first temperature $T_1$ after equilibrium is established, $Q_1 = K_{T_1}$. At the second temperature, $Q_2 = K_{T_2}$ once the equilibrium is re-established.
For adding or removing one of the species, you start with the old position of equilibrium, then the addition or removal of a species changes Q, which no longer equals K. As a consequence, there is a net reaction that changes Q back to K. Now, you have a new set of concentrations, but the reaction quotient is back to what it was at equilibrium. I'm not sure whether to call this a new position of equilibrium, or to say the reaction went back to equilibrium (this might be confusing because the equilibrium corresponds to a third set of concentrations - first set: original equilibrium concentrations; second set: concentrations right after the manipulation; third set: concentrations after the net reaction re-establishes equilibrium).

For a graphic illustration of these concepts, see here: https://chemistry.stackexchange.com/a/177682/72973


text:
Urea + water yields ... urea + water?
I added urea to water and, predictably, observed a temperature drop (10 C).  I wanted to place the beaker under (reasonably) high vacuum (for a classroom) and recover the urea as a dry substance.  Well, I didn't.  It's been a couple of months now, and there's white scale all the way up the inside, and crystaline filaments on the bottom, with an unchanging fraction of the original H2O.  Is the substance still urea?  There's been no odor.  Is it ammonium carbamate? In either case, can I use vacuum to dessicate it?

Let me see if I understood correctly - you put the aqueous solution of urea in a container (e.g. a vacuum desiccator), pulled a vacuum, then closed the container and left it to sit? If so, then it will take ages to dry. You need to continuously remove the moisture either by pumping the solution down constantly or by exposing the solution to a flow of dry air. Also, urea is very soluble in water, so removing the last bits of moisture will be difficult without heating. If you want to recover the urea to perform another endothermic dissolution, it may not work as well due to residual water.

A vacuum desiccator does provide continual extraction. Nowhere does the author of the question state "closed the container and left it to sit". That may be the case, but it's an unfounded assumption.

As water evaporates, the vapor pressure above the more concentrated urea solution drops. Either a better vacuum pump is needed, or perhaps gentle warming, to extract more water.
You might also try chilling to (partially) separate the solution. At ~260 K, achieved in many home refrigerators, much of the urea will have crystallized from the remaining saturated solution... just don't mix this up with food in the freezer.

You each stayed with "urea" and, by implication, stayed with physical change only, not a chemical one.  That resolution was one of the answers i was looking for.  Moishe, correct: open beaker.  Thanks for the reminder about vapor pressure.


text:
How to effectively wash the precipitate or crystals on the filter and why?
My lab manual says that the precipitation on filter should be washed with a few small portions of the washing liquid. Why can't I just completely fill the filter funnel with the solvent instead of steadily add new tiny portions in it?
Also, how effective this washing process is? How many times should I wash the precipitation before it's completely free from impurities?
What filter funnel of what diameter of the paper filter should I use? Isn't the biggest one the is also the most effective?
Finally, why do I have to study limits when I want to become a chemist, not a mathematician?

Every time the precipitation on filter is washed, there is a certain constant volume $V_\mathrm{ret}$ of washing fluid retained on the filter. If the initial concentration of impurity is $c_0$, then initial amount of impurity on filter is
$$n_0 = c_0V_\mathrm{ret}$$
and after washing with a new portion of the fluid with the volume $V_\mathrm{add}$ concentration of impurity is reduced to
$$c_1 = \frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}c_0.$$
Analogously, after $n$-th washing act
$$
c_n = \left(\frac{V_\mathrm{ret}}{V_\mathrm{ret} + V_\mathrm{add}}\right)^n c_0
\qquad\text{or}\qquad
c_n = \left(\frac{1}{1 + \frac{V_\mathrm{add}}{V_\mathrm{ret}}}\right)^n c_0
$$
and, assuming we have a total volume of washing liquid of $V_\mathrm{tot} = nV_\mathrm{add}$, the former formula becomes
$$c_n = \left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0$$
Now, lets assume we have a moderate amount of impurities (e.g. $c_0 =\pu{0.1 mol L-1}$) that we are trying to get rid of by washing with different amounts of the same volume of the same solvent (say, $V_\mathrm{tot} = \pu{100 mL}$). However, we use different filters, for instance:

a large one (retains $1/10$ of the washing liquid, or $V_\mathrm{ret} = \pu{10 mL}$);
a middle one (retains $1/20$ of the washing liquid, or $V_\mathrm{ret} = \pu{5 mL}$);
a small one (retains $1/100$ of the washing liquid, or $V_\mathrm{ret} = \pu{1 mL}$),

and, of course, different portions of the liquid (various $n$). Lets plot the concentration of the impurity for each case as a function of washing frequency:

First and foremost, it's obvious that it is always better to wash with small portions many times, rather than two or three times with large quantities of washing liquid. Second, the smaller the filter, the less impurity will be left after the same number $n$ of washing acts.
Now, can we get rid of the impurity once and for all by washing with an extremely large number of small portions? In fact, lets use an infinite number and take the following limit
$$\lim_{n\to\infty}c_n = \lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0},$$
which is, in fact, one of the notable special limits, sometimes called the second remarkable limit:
$$\lim_{x\to\infty} \left(\frac{1}{1+\frac{a}{x}}\right)^x = \frac{1}{e^a}$$
so that in our case we can conclude it's impossible to completely get rid of the impurity only by washing:
$$\lim_{n\to\infty}{\left(\frac{1}{1 + \frac{V_\mathrm{tot}}{V_\mathrm{ret}}\frac{1}{n}}\right)^n c_0} = c_0\exp{\left(-\frac{V_\mathrm{tot}}{V_\mathrm{ret}}\right)} \neq 0$$
Conclusions

It is impossible to completely remove all impurities by washing the crystals/precipitate on the filter even by washing it infinite number of times.
It is more efficient to wash with small amounts of liquid many times than only a couple with large ones.
Use filter of the right size: not too big to retain the traces of impurities, but big enough to maintain efficient liquid transport through its surface. When possible, use pleated filter paper.

Appendix
The figure have been plotted with pgflots. Feel free to play around with it or reuse for your lab manual. Source code:
\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Effectiveness of washing methods ($c_0 = \SI{0.1}{\mole\per\liter}$)},
        xmin=0, xmax=20,
        ymin=0, ymax=0.001,
        xlabel = $n$,
        ylabel = {$c_n$/\si{\mole\per\liter}},
        minor tick num=4,
        grid,
        legend cell align={left},
    ]

    \addplot [
        domain=1:20,
        samples=200,
        solid,
        thick,
    ]
    {0.1*(1/(1 + 10/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.10$}

    \addplot [
        domain=1:20,
        samples=200,
        dashed,
        thick,
    ]
    {0.1*(1/(1 + 20/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.05$}

    \addplot [
        domain=1:20,
        samples=200,
        dotted,
        thick,
    ]
    {0.1*(1/(1 + 100/x))^x};
    \addlegendentry{$V_\mathrm{tot}:V_\mathrm{ret} = 1:0.01$}

    \end{axis}
\end{tikzpicture}

\end{document}

Good for what the explanation covers, but too much washing liquid tends to be bad since the desired compound is typically at least somewhat soluble in the washing liquid.

@MaxW Hey, stop poking my spherical cow in a vacuum with a stick :D On a serious note, you are right, the desired product is going to be dissolved too, but who needs it if it's still contaminated? In fact, if plain old washing is the only available choice, then this math works out pretty good for the reasonable $n$, especially when the washing liquid is not just a solvent, but, say some electrolyte solution, containing either cations or anions of the recrystallized compound.

LOL. Everyone learns quickly in organic chemistry that there is "product" and "pure product". Washings and recrystallizations  take a large bite out of the experimental yield to get a relatively pure product.

@andselisk The way you defined $V_\rm{tot}$ makes it a function of $n$ as well. =) Bad luck!

@LinearChristmas In a similar manner pi can be defined as $π = Ar^{-2},$ but it doesn't mean it changes from circle to circle.

@andselisk Is $V_\rm{add}$ a constant value here?

@LinearChristmas $V_\mathrm{add}$ remains a constant for a series of additions. $V_\mathrm{tot}$ remains a constant across all series. Here, I assume we are adding equal volumes $V_{\mathrm{add},i}$ for each $n_i.$

@andselisk What is a "series of additions", rigorously? What parameters are fixed here beforehand to be able to say that we are looking at one particular (fixed) "series of additions"?

@LinearChristmas An $i$th series of addition is adding a finite equal volume $V_{\mathrm{add},i}$ exactly $n_i$ times. Each new addition takes place once there is no more filtrate leaving the filter and crystals. Both $V_{\mathrm{add},i}$ and $n_i$ are picked up in such a way that the total volume of washing liquid $V_\mathrm{tot} = n_iV_{\mathrm{add},i}$ remains constant for all combinations/series. Probably I should've added this to the answer—I thought this stuff is self-explainatory.

@andsekisk Thank you for the explanation. Now, assumingly when you say "$V_\rm{tot}$ remains constant across all series" you mean that it is independent of the choice of $i$, correct? In other words, for all $i=1, 2, 3, \dots$, the quantity $V_\rm{tot}$ is constant?

@LinearChristmas Exactly that. I tried to model a situation when a person has a given amount of solvent in stock and wonders how to use it as effective as possible. I apologize if my post looks cumbersome.

@andselisk No need at all to apologise, and were you to do so regardless, then apology accepted! I think this answer has a worthy effort. I also apologise because there are a few more comments to come :). (1/2)

@LinearChristmas Yep, I should think about how I can improve the writing. Thank you for pointing out this issue in logic.

@andselisk You say in your answer that "[A]fter the $n$th washing act, $c_n =$ .." Based on our discussion in the comments, do you mean to say "After the $k$th washing act in the $i$th series ($1\leq k\leq n_i$), $c_k =$ .."? (2/2)

@andselisk No problem! I think the post is saveable ;). If you're interested, here's something I might try. First, say that $n_i$ is a strictly increasing sequence of natural numbers, i.e. $n_i < n_j$ for all $i < j$. Then as $i \to +\infty$, $n_i \longrightarrow +\infty$. Now, denote $c_{n_i}^\ce{A}(k)$ as the concentration of impurity $\ce{A}$ after the $k$th addition in the $i$th series. Here $k$ is any natural number such that $1 \leq k\leq n_i$. Now you can do $$\lim\limits_{i\to+\infty}{c_{n_i}^\ce{A}(n_i)}$$ instead of the limit over $n$. Perhaps you have a better idea than mine.

*(By "sequence", I mean an infinite sequence).

Actually, this notation might be more compact: $c_{i}^\ce{A}(k)$, not $c_{n_i}^\ce{A}(k)$, and then later $c_{i}^\ce{A}(n_i)$ for the limit. Sorry. (Of course, $i$ is just a "numbering", not anything physical in contrast to $n_i$, so brevity with a cost).


text:
What is the maximum distance over which van der Waals forces act?
The VDW radius places a lower limit on the distance between atoms mutually attracted to one another. I am looking for the upper limit:  what is the name of the concept or theory that describes the maximum distance over which the various attractive forces (Keesom, Debye, London...) become operable / take effect?
This question is similar to the following Q, which didn't get fully answered:
Range of distance for Van der Waals force

I think editing https://chemistry.stackexchange.com/questions/107870/at-what-distance-does-chemical-bonding-initiate would be preferable to asking new question.

There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.
The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.
The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.
Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.

this clarifies things quite a bit, thanks.  Further:  in looking at protein synthesis in a cell, for example, there must be some "tipping point" distance at which atoms/molecules in proximity to each other become mutually attracted due to VDW forces.  Beyond this distance, they don't "see" each other and thus remain separated?  I understand the exact distance value varies.

VdW forces are infinite in range so there is no such tipping point. However they decay rapidly enough with separation that their contribution becomes negligible at long range and can safely be neglected. Compare this with the coulomb interaction which does not decay rapidly enough, and so should not be truncated

As jheindel has previously pointed the interaction has no limit. What you are probably looking for is the distance at which the interaction energy becomes smaller than thermal energy at a given temperature. When the interaction is smaller then thermal motions govern what happens not the intermolecular interaction. Thermal energy at $300$ K is $k_BT = 1.38\cdot10^{-23}\cdot 300 = 4.1\cdot 10^{-19}$ J.
The interactions all have the general form $E=V/(\epsilon r^n)$ where $V$ is scaling term: for example for charge - charge interaction $n=1$ and $V=q_1q_2e^2/(4\pi\epsilon_0)$ where $q$ are the charge signs, $e$ the charge on the electron and $\epsilon _0$ the permittivity of free space $8.854\cdot10^{-12}$. Finally $\epsilon$ is the solvent dielectric constant, cyclohexane = 2, water = 78.
The interaction energy between two charges, say sodium and chlorine ions, in vacuum separated by 0.3 nm is approximately $190k_BT$, but in water this only $2.4k_BT$ so the distance at which the intermolecular interaction is important depends very much on the conditions. The reason that $\epsilon$ has such an effect is that the high dielectric solvent attenuates the electric field around an ion or dipole.
There are other types of interaction ion-dipole, dipole-dipole, induced dipole-induced dipole and so on that have energies that depend, where appropriate, on the magnitude of the dipole, polarisability, the relative dipole orientation and various powers of separation $1/r^n$. For example for dipole-dipole interaction with fixed dipole directions the relative dipole angle is important and $n=3$ but for freely rotating dipoles the dipole angle is averaged out resulting in $n=6$ (Keesom energy). London dispersion force (induced dipole-induced dipole) depends on each molecule's polarisability and has $n=6$,in comparison charge-dipole interaction has $n=4$.
If you look these up to calculate particular interactions note that many authors quote values in vacuum and so ignore the solvent dielectric constant. You can include this by changing $\epsilon_0\to \epsilon_0\epsilon$.

Indeed there are various decriptions of VdW forces. For some, this class includes dipole-dipole, dipole-instantaneous dipole, and instantaneous dipole-instantaneous dipole. Of course, these forces are not all of the same intensity.
But for all, whatever the exponent of r, there is no limit, and this is why one speaks of energy instead, as the one needed for putting the two chemical species at an infinite distance.
By the way, I would suggest that when one is giving an answer, references would be given as well.
And here, I suggest:
Physical Chemistry, McQuarrie
The chemical bond, Linus Pauling ;-)


text:
Mastering inorganic chemistry for university
I want book recommendations for inorganic chemistry. I want to learn it in a way that every behavior or equation of elements seems logical and can be explained.
Is there any book of that sort? It would be better if it mentions minute details as well.
Note : I have completed by secondary school.

If question is not up to the mark you make close or down vote it but please mention the book which you think might help me.

"in a way that every behavior or equation of elements seems logical and can be explained." Welcome to chemistry - a somewhat related series of special cases.

Concise inorganic chemistry by J D Lee is a great book for this. It explains everything easily but in great detail.

I do not understand why this answer was downvoted?

Can you please hilight it's features that makes it;s different ?


text:
Acidic tendency of copper(II) ion vs. calcium(II) ion
In my Chemistry textbook, a table is shown wherein ions originating in various salts are categorised according to their acidic or basic tendencies. Among others, $\ce{Cu^2+}$ and $\ce{Ca^2+}$ are present in the table; however, the copper ion is classified as being acidic, whilst the calcium ion is classified as being neutral.
I don't really see how they differ, apart from the amount of electrons they have in non-valence shells and the subshell that holds their valence electrons. Either way, since this is in the context of Brønsted–Lowry and not Lewis, I have no clue how the textbook got to its conclusion; does it have to do with copper parttaking in any redox reactions, or perhaps a certain stability calcium has?

It's still Lewis needed to analyse this - it's acidity of metal aquo complexes.

Cu2+ has an incomplete 3d orbital (4s0 3d9) and can therefore be included in the chemistry of transition metals. As it can be coordinated (form dative covalent bonds) with ligands to create complex ions, for example H20 in hexaaquacopper(ii), it acts as a Lewis acid.
But in terms of Brønsted–Lowry theory, perhaps the textbook is referring to the fact that when transition metals are dissolved in water they can donate an H+ ion to another H2O molecule to form H3O+. This can be observed with Fe3+:

[Fe(H20)6]3+ + H2O -> [Fe(H20)5(OH)]2+ + H3O+
The complex ion is acting as an acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a base by accepting the hydrogen ion.

Well-explained! So, I'm assuming calcium doesn't form an aqueous complex, even though it is positively charged?

@Mew Ca does form aquo complexes, they are much weaker acids though.


text:
Are enzymes with superior catalytic properties more sensitive to temperature changes?
I am wondering whether enzymes with superior catalytic properties are more or less temperature sensitive than average enzymes.
I know that enzymes lower the activation energy, Ea. So the more catalytic superior an enzyme is the lower the value of Ea. Then the Arrhenius equation can be used to describe how the rate constant of an enzymatic reaction changes with temperature. But how can I use this information to decide whether the more catalytic enzyme is more or less temperature dependent?

There is no obvious direct connection between catalytic properties and stability of an enzyme.
You are trying to make a connection, but let's consider your statement:

Then the Arrhenius equation can be used to describe how the rate constant of an enzymatic reaction changes with temperature.

According to the Arrhenius equation, all reactions (catalyzed by an enzyme or not) become faster as temperature increases. The higher the activation energy, the more pronounced the increase. However, if the enzyme unfolds, it will no longer catalyze the reaction.

But how can I use this information to decide whether the more catalytic enzyme is more or less temperature dependent?

The biggest temperature dependence occurs at the temperature where the enzyme unfolds. There is the idea that enzymes have to be flexible to be effective catalysts, so there might be a correlation between optimal temperature for the activity and the unfolding temperature. The best way to figure this out would be to compare enzymes from mesophiles and thermophiles in terms of optimal temperature for catalysis vs melting temperature.


text:
Why doesn't lithium form dilithium crystals? (Calling Mr.Spock)
Hydrogen is in the first column of the periodic table. At STP it is a diatomic gas. When cooled and/or under pressure it forms a crystalline solid of di-hydrogen molecules. Calculations indicate that under high enough pressure it may form a monatomic basis metallic crystal, and scientists are actively looking to find evidence of this. Results so far seem inconclusive.
However, lithium is also a first column element. But its solid is a monoatomic basis metal and NOT a dilithium crystal as we would expect based on hydrogen. Why not?

https://chemistry.stackexchange.com/questions/70678/do-li4-or-li8-molecules-exist https://chemistry.stackexchange.com/questions/91279/do-alkali-metals-not-form-diatomic-molecules

Happy to!  Any advice on what is not specific?  I expect solid lithium to look like solid hydrogen.  Solid hydrogen is a lattice of diatomic molecules, but lithium is a metal with a single atom basis.  Does this wording work better?

@PaulYoung No because all the links suggested do answer the question.

As far as molecular structure goes H2 is better compared with halides - single bond completes el. shell.

I do think Mithoron's second link answer the Q, as well as his comment


text:
How many stereoisomers are possible for 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane?

How many stereoisomers (not counting different conformers) does 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane have?

I think the answer should be 3. There are two geometrical isomers (cis and trans) considering the two methyl groups on the ring, and the trans isomer will also show optical isomerism as it doesn't have a plane of symmetry. But in my book, the answer is given as 2. Is this a misprint?

Comments are not for extended discussion; this conversation has been moved to chat.

In this structure the left hand methyl groups can be on the same side of the ring. This is one stereoisomer. This molecule is not chiral and there is only one isomer with the methyl groups on the same side of the ring.
Initially I assumed that there was only one other isomer: the two methyl groups on the opposite side of the ring. But building models sometimes wrecks intuition. The picture below shows the two possible version with trans (relative to the plane of the ring) methyl groups:

So my initial view was wrong: the molecule is chiral and that gives three total stereoisomers.

Yes, not threo and erythro, but meso and d/l, slide

If you know your tartaric acids, you know this one.


text:
Can hexavalent chromium be created by exposure of chromium alloys to air at room temperature?
I believe that Cr on its own (as in chrome coating) oxidizes as Cr(III) when exposed to air, and similarly I think that it is the same for Cr as a component of stainless steel.
Is it always the case that an alloy containing chromium will oxidize as Cr(III) when exposed to air at room temperature, or are there alloys which might form Cr(VI) instead?
The situation which brings this question to mind is the case of a computer hard disk platter. In the case of a "head crash" of the hard disk, layers of the platter are ground to dust. The layers as I understand are composed of various things: Some kind of carbon polymer as an outer covering, magnetic layers of things like Co-Pt-Cr or Co-Ni-Cr (I found a source which suggests "traces of boron or tantalum" are also added to this layer), and also layers composed of things like Cr-Ti or Cr-Ni. I'm wondering whether any of these layers would cause hexavalent chromium to form.

Can Cr (VI) be formed? Certainly! But a few Cr(VI) ions probably would not be significant, or even measurable.

Ah, so even though generally Cr(III) is formed in these cases, sometimes Cr(VI) ions still formed at room temperature in Cr and its alloys (this is the case even with solid Cr or stainless steel)?
Does this mean that other oxidation states of Cr could be formed as well, and it is just that Cr(III) is only formed almost exclusively?

There are ~6*10^23 Cr atoms in a mole, ~52 g. Thermodynamics is based on statistics, which gives the most likely outcome. However, individual Cr atoms and O2 molecules may have kinetic energy far beyond the average and might form unusual oxidation states. Your question asks "Is it always the case that an alloy containing chromium will oxidize as Cr(III)...", so of course there might be a few exceptional cases.

Thanks for the clarification! I suppose that since the presence of Cr in an alloy does not change its kinetic energy, Cr will tend to oxidize in air as Cr(III) regardless of the alloy? (I hope "tend to" is the correct terminology in light of the information you've given me).
I guess I am hampered here in my question because I'm not sure though why Cr tends to oxidize as Cr(III) rather than as Cr(VI) in air to begin with, I suppose because for Cr to reach the +6 oxidation state using only air requires that the atoms be unusually charged?


text:
Why does a lowered dielectric constant cause RNA/DNA precipitation in ethanol precipitation?
For ethanol precipitations, as mentioned in https://en.wikipedia.org/wiki/Ethanol_precipitation, given that the attraction of water molecules to charged ions like phosphates is based on Coulombic attraction, shouldn't the water-phosphate attraction be increased in magnitude by the same ratio as that by which the phosphate-cation in increased in magnitude? I.e., should there not be a lack of increased "preference" for phosphate-salt cation ionic bonds (and thus precipitation) over water-phosphate hydrogen bonds (and thus polynucleic acid solvation) as a result of the increased permittivity to charge-charge interactions that ethanol brings over water?
An argument based on the relative higher amounts of cations and ethanol, and lower in water, would make sense in increasing the likelihood of a cation-phosphate bond over a water-phosphate hydrogen bond, but then any mention of the dielectric constant in explaining ethanol precipitation seems entirely superfluous.

For ethanol precipitations [...] shouldn't the water-phosphate attraction be increased in magnitude by the same ratio as that by which the phosphate-cation in increased in magnitude?

The interaction between negatively charged nucleic acid and positively charged counter ions is an ion:ion interaction. The interaction between ions and water is an ion:dipole interaction.

An argument based on the relative higher amounts of cations and ethanol, and lower in water, would make sense in increasing the likelihood of a cation-phosphate bond over a water-phosphate hydrogen bond, but then any mention of the dielectric constant in explaining ethanol precipitation seems entirely superfluous.

The presence of water between ions of opposite charge diminishes their long-range attraction (the dielectric constant is related to how well solvated ions are "shielded" by the oriented dipoles of the solvent).
Ion:dipole interactions are short range and rely on the dipole to be properly oriented. Adding ethanol to the aqueous DNA solution does not make the ion:dipole interactions of water with ions stronger. In contrast, it does make the ionic interactions stronger because ethanol is less effective in shielding charges than water.
It is important to compare dielectric constants of water and ethanol because both molecules are uncharged and polar (and capable of ion:dipole interactions), but water is the better solvent for ionic compounds for the reasons discussed above.

[from comments] I don't understand why [we add] ethanol. Ion-dipole interactions and ion-ion interactions are both interactions based on charge and should thus both follow Coulomb's law. Ion-dipole interactions are certainly weaker, and hence shorter range, but that doesn't seem like it should exempt them from Coulomb's law. Dipole orientation is something I hadn't considered.

A dipole has a positive and a negative partial charge that add up to zero. Imagine you are the cation that interacts with a dipole. When the dipole is near you, the negative partial charge will be right next to you, and there will be a large attraction. The positive partial charge will be a bit further, so there is a smaller repulsion. Overall, it adds up to an attractive force. Now move the dipole away from you, maybe by a factor of 10. Now, the positive charge will attract much less, and be at approximately the same distance as the negative charge, so the repulsion and attraction will almost cancel out.
Take a look at pentavalentcarbon's answer to a question on intermolecular interactions. There is a nice table with the dependence of interaction strength on distance.

I don't understand why adding ethanol. Ion-dipole interactions and ion-ion interactions are both interactions based on charge and should thus both follow Coulomb's law. Ion-dipole interactions are certainly weaker, and hence shorter range, but that doesn't seem like it should exempt them from Coulomb's law.
Dipole orientation is something I hadn't considered.

A dipole has a positive and a negative partial charge that add up to zero. Imaging you are the cation it interacts with. When the dipole is near you, the negative partial charge will be right next to you, and there will be a large attraction. The positive partial charge will be a bit further, so there is a smaller repulsion. Overall, it adds up to an attractive force. Now move the dipole away from you, maybe by a factor of 10. Now, the positive charge will attract much less, and be at approximately the same distance as the negative charge, so the repulsion and attraction will almost cancel out

Also, take a look at pentavalentcarbon's answer to this question. There is a nice table with the dependence of interaction strength on distance.


text:
Broken iPhone battery no fire
I work on electrical and Electronics quite often and working on a iPhone the other day with a dead battery,I ruptured it open trying to remove the glue I smelled a funny smell like nail polish remover but no fire or sparks it was completely dead I heard they can release hydrofluric acid but when only when they ignite I been venting the room last few days and removed the phone imidiatly after I ruptured it, is there any danger?

How is this "medical"? I find it a bit stupid to deny an answer to people who ask "should I suspect that I'm in danger?".

The danger from broken lithium ion batteries is primarily their fire risk not their chemical contents. Broken batteries where short circuits occur can generate enough heat to cause the battery to catch fire and this fire will be vigorous as the batteries contain metallic lithium which is fairly flammable in air and will also react vigorously with water.
Some of the electrolytes in the batteries contain non-coordinating lithium salts like $\ce {LiPF6}$ this is presumably where the rumour that they can leak HF came from but this is very unlikely as they are very unreactive compounds. If the battery burns vigorously the smoke might, possibly, contain some HF but this will be the least of your worries if the device is on fire.
Damaged batteries should be disposed of carefully according to standard safety procedures, but, if it isn't already on fire, there is nothing to worry about.

And the nail polish remover smell comes from the electrolyte solvent, typically carbonates or ethers. (Ethylencarbonat, Propylencarbonat, Dimethylcarbonat, Diethylcarbonat oder 1,2-Dimethoxyethan, just cp from de:wp)

Thank you for all the help, I am know how electrical and electronics works but was unaware of what batteries contained chemically, When we were in school they always told us never to rupture one because the fumes are highly toxic so I have always been very careful. Finally I slipped with a plastic spuger and ruptured one and took the phone outside imidiatly thank god it didn’t go off but was worried about the fumes and even more worried about the hf released but apprently that’s only if they ignite, I figured I would ask a chemist who would know the composition of something a lot better then i


text:
Inverse problems in Computational Chemistry: The case of Inverse eigenvalue Problems for molecule structures
Mathematics Background
An inverse eigenvalue problem (IEP) is the problem of reconstructing a matrix
with a special structure from prescribed spectral data. By structure we mean
the pattern of entries that are either zero or nonzero. There are many different types of IEPs with different level of difficulty which depends on the structure of the matrices which are to be reconstructed and the available eigen information. For example consider the following real Jacobi matrix:
$$
A_n = \begin{bmatrix}
a_1&b_1&0&0&0&0& \cdots & 0\\
b_1&a_2&b_2&0&0&0&\cdots&0\\
0&b_2&a_3&b_3&0&0&\cdots&0\\
0&0&b_3&a_4&b_4&0&\cdots&0\\
0&0&0&b_4&a_5&b_5&\cdots&0\\
0&0&0&0&\ddots&\ddots&\ddots&0\\
0&0&0&0&\cdots&b_{n-2}&a_{n-1}&b_{n-1}\\
0&0&0&0&\cdots&0&b_{n-1}&a_n
\end{bmatrix}
$$
Such that $b_i$s are nonzero and $a_i$s may or may not be zero. An IEP about this matrix is this:

Let $A_i$ be $i$th leading principle submatrix of $A_n$ and let $n$ be an even number, for a given Jacobi matrix $J$ of size $n/2 \times n/2$ and list of numbers $\Lambda = \{\lambda_1, \cdots, \lambda_n\}$, find an $n \times n$ Jacobi matrix $A_n$ such that $J$ is its minor principle submatrix and $\Lambda$ is its eigenvalues.

or

For a given list of real numbers $\Lambda = \{\lambda_1, \cdots, \lambda_n \}$ and $\mu =\{ \mu_1, \cdots, \mu_{n-1} \}$, find a Jacobi matrix $A_n$ such that $\Lambda$ is its eigenvalues and $\mu$ is eigenvalues of $A_{n-1}$.

Many different problems are considered by researchers in many papers.
Because graphs are appropriate tools for presenting a matrix structure, many papers concern IEP of graphs. For a given symmetric matrix $A_{n \times n}$, graph of $A$ is a graph $G=(V,E)$ such that $V = \{v_1, \cdots,v_n\}$ and edge $\{v_i,v_j\} \in E$ if entry $[A]_{ij} \neq 0$. In many papers IEP of matrix of graphs (path and broom graph (path graph is actually a symmetric Jacobi matrix), star graph, four different IEPs for paw graph and etc.) are investigated by researchers. Please note that it is also possible to consider IEP of a directed graph, in this case its matrix will not be symmetric anymore.
What I am looking for ...
After a few years of studying IEPs in pure mathematics, I am looking for inverse eigenvalue problems with real practical application. many times, problems are modeled as a matrix and the limits on the problem are moved to the matrix entries or its spectral information and then researchers study the matrix. For example IEP of mass spring system is investigated which is a topic in physics.
IEP of problems in Computational Chemistry are rarely (or even never) investigated and because many entities and molecules in chemistry are look like some graphs (path or cycle), and graphs can also be modeled as a matrix, hence different types of IEPs may happen there. I ask for any help to introduce some field of studies or references or even cooperation to help me in doing this.
Thanks in advance.

Comments are not for extended discussion; this conversation has been moved to chat. In addition, most of them are obsolete.


text:
Total number of stereoisomers of 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane

I don't think there is any change of showing geometrical isomerism across the double bond because the two groups on the right side (two $\ce{-CH_3}$) are the same.
Also I don't think there should be optical isomerism as there is plane of symmetry. Can someone please explain the solution?

There might or might not be a plane of symmetry. This whole thing is not planar.

are there any atoms which are chiral centres?

@H.Linkhorn I don't think so

Would the two carbon atoms joining the two butyl branches together be chiral. It is a different route to get back to the double bond depending which way you go.

The two methyls on the left can be on the same side of the 5-membered ring or on opposite sides. This generates two distinct compounds.

Maybe it would help to add the hydrogen atoms to the structure like in this related structure. For the related structure on chemspider, you can click on the 3D link to rotate it to examine the geometry at each carbon.

Question identical to this one

@matt_black Can you please explain a bit more?

@matt_black what about the enantiomers of the trans configuration?

@IvanNeretin Am I correct in saying that a plane of symmetry will be present in the same side configuration as pointed out by matt_black?

@YUSUFHASAN What trans configuration? It isn't geometrically possible.

@matt_black Can you please show with perhaps figures what you are talking about? It would be more helpful ;-)

@tatan Why don't you think there is a plane of symmetry here? There could or could not be depending on the stereochemistry at the two chiral centers.

Here is a similar molecule to illustrate:
Two of the ring carbons are asymmetric (chiral). This molecule itself (as shown) is chiral. If you change one of the chiral carbons, you get a molecule that is not chiral (mirror plane along the double bond and perpendicular to the ring).

Thanks for your effort. Can you please mark the carbons with a paint tool etc. and make a reference to which particular carbon you are refering to...

Thanks for commending my effort. Could you check the rules for chiral carbons, and do it yourself?


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Why do people add H(core)+F in Hartree-Fock?
I'm trying to implement my own HF code, and it seem usual that, after building the Fock matrix $F$, they add that to the pre-existing $H_{core}$, from the two-point interactions, when computing final energy.
As one example, https://github.com/ipudu/SCFpy/blob/master/SCFpy/scf.py . In lines 135-138, they include Hcore in F. Then in line 156, when they actually return the energy, they add Hcore[i,j]+F[i,j], apparently double-counting Hcore.
As another example, in http://sirius.chem.vt.edu/wiki/doku.php?id=crawdad:programming:project3 , when they compute the energy (see Step #9), they use $$E_{elec}^i = \sum_{\mu\nu}^{AO} D_{\mu\nu}(H^{core}_{\mu\nu} + F_{\mu\nu})$$
again apparently double-counting $H_{core}$. Why? What gives?
My best guess is that this is somehow related to the fact that these are double-filled orbitals, containing both spins. But then, shouldn't $F$ need a factor as well, for all it particles occurring twice as well?
In particular, I have an HF code that agrees with the 'crawdad' link above by copying their steps pretty much verbatim. I'm now trying to extend it into a generalized Hartree-Fock code, but I'm having difficulties, and in trying to understand the problem I became confused by this expression.

Actually, $F$ is double-counting the electron-electron interaction when evaluating the total energey. By adding $H^{\text{core}}$ and using a factor of $\frac{1}{2}$, which I suspect is implicit in $D$, one can obtain the correct energy. This is basically an occurence of the HF energy not being the sum of orbital energies (which are a result of $F$).


text:
EtG formation outside the liver
Ethyl glucuronide (EtG) is formed by conjugation with glucuronic acid catalyzed by the enzyme UDP-glucuronosyltransferase.
My question: Is it possible for EtG to form spontaneously in a urine sample that contains a small amount of ethanol?

No, because the equilibrium for the simple reaction lies on the side of the reactants:
$$\ce{Ethanol + glucoronic acid <<=> Ethyl glucuronide + H2O}\tag{1}$$
The reaction works enzymatically inside the cell because glucuronic acid first reacts with UTP, turning it into a mixed anhydride (phosphate of UDP bound to carboxylic acid of glucuronic acid) plus inorganic phosphate.
$$\ce{UTP + glucoronic acid <=>> UDP-glucuronide + phosphate} \tag{2}$$
Then, the enzyme swaps UDP for ethanol or other molecules.
$$\ce{Ethanol +  UDP-glucuronide <=>> Ethyl glucuronide + UDP}\tag{3}$$
The net difference between hypothetical reaction (1) and the two step process (2) and (3) is the hydrolysis of UTP, a reaction with a negative Gibbs energy of reaction (i.e. equilibrium constant larger than one):
$$\ce{UTP + H2O -> UDP + phosphate} \tag{4}$$
So the formation of UPD and phosphate from UTP drives the two step reaction (it is a coupled reaction).
Attaching glucuronic acid makes molecules more water-soluble, helping to excrete them in the urine 1.

Karsten, these reactions are spontaneous only because of the energetic contribution from $\mathrm{UTP \rightarrow UDP + P_i}$, right?

@hBy2Py yes, right (Gibbs energy contribution)

See edit - I tried to clarify the differences between the urine and liver contexts a bit. Please revise or revert as you desire.


text:
Changes in pH when adding substances other than acids or bases
If I start out with 100 mM acetic acid, what substance (to a final concentration of 1 M) could I add to change the pH considerably? I was thinking of NaCl or ethanol or acetone. Substances added should not be acids or bases, i.e. a 1 M solution of them should have a pH close to neutral. I'm asking because I am trying to create a demonstration on how pH calculations are just approximations under non-ideal conditions.
The acetic acid should not undergo an chemical reaction (other than acid/base). I would use a pH meter to monitor the pH. I would have a control where I increase the volume to match the experiment where I add a substance that influences the pH (or dissolve the substance in 100 mM acetic acid to keep the acetic acid concentration constant).
I am looking for changes in pH based on non-ideal behavior. The concentration of acetic acid should remain the same, and there should not be any neutralization reaction. In other words, the calculate pH should stay at 2.87 (assuming ideal conditions) while the measured pH would change.

This is an interesting question. The easiest way is to the change the properties of the solvent. Add something like DMSO.

If your initial solvent is water, the easiest to increase the concentration of aceitic acid and, therefore, increase the pH would be to somehow decrease the volume of water. Possibly look for a way to evaporate water. Distilling the solution might be tricky since the boiling point of acetic acid and water are close (118 celsius). Another way might be to use a dehydrating agent to eliminate water. The best dehydrating agent I can think of are all strong acids (sulfuric and phosphoric). However, hot aluminum oxide might also work (if it doesn't have any side reactions with aceitic acid).

I'm not entirely sure, but probably some neutral buffer or a zwitterion with $\mathrm{pI}$ close to neutral would satisfy the requirements as either of those separately a) would have a pH close to neutral; b) effectively "capture" protons and keep $\mathrm{pH}$ close to 7.

It becomes harder and harder to make suggestions as you edit the question:) Since  changing the concentration of acetic acid is not allowed, should the system remain homogeneous, e.g. is it allowed to add a secondary phase (such as non-mixing organic solvent or solid) while measuring the $\mathrm{pH}$ of the aqueous one?

@andselisk I appreciate the thoughtful comments, you are helping me to write a better question. Adding a second phase would be fine (there is a second phase already - air - and it might get me into trouble if CO2 starts dissolving in my solution).

A bit different from what you're asking, but how about a demo of NaOH absorbing water from the atmosphere?  A simple one would be to make up a 1.0M solution in a 1.000L volumetric the night before the demo, then have the demo being about estimating the volume difference (based on the neck diameter and height above the volumetric mark) and hence the pH difference.  In this case, the demo is less about added solutes, and more about the property of one solute and how that affects the solution requiring standardization.

Here's another idea: what about adding or removing heat (i.e. showing the temperature dependency of kw)?  It's not the exact scenario you ask for, but, in terms of showing that pH is based on ideal conditions, it should make the students question the validity of the kw = 1E-14 assumption.

You question sounds like you really misunderstood a lot of the science about pH. The pH is DEFINED according to the concentration of H+/OH-, so if there is a discrepancy between your prediction and reality, then your prediction is wrong. Also "substances added should not be acids or bases" does not make sense because all substances are acids or bases when compared to another substance.

You could use an alcohol or acetone but they are slightly acidic (though much less so than acetic acid; also alcohols react to form esters). Now effects from acidic nature can't completely avoided as even a molecule as non-acidic as methane still has a pka of 50. I think an aliphatic ether such as tetrahydrofuran (THF) or diethylether would be the best substance to add as the ethers are only very slightly basic and are miscible with water.
More simply:

what substance could I add to change the pH considerably?

Why does it have to be a substance? The $K_\mathrm w = 10^{14}$ at $\pu{25^\circ C}$. If you add heat the $K_\mathrm w$ will increase and thus the pH will increase.

Another option is addition of the neutral concentrated solution of some salt.
It will affect activity coefficients of the conjugated acids/bases.
E.g., if a phosphate buffer is used, $\ce{HPO4^2-}$ would be affected more then $\ce{H2PO4-}$.

From Wikipedia, https://en.wikipedia.org/wiki/Thermodynamic_activity:

When a 0.1 M hydrochloric acid solution containing methyl green indicator is added to a 5 M solution of magnesium chloride, the color of the indicator changes from green to yellow—indicating increasing acidity—when in fact the acid has been diluted. Although at low ionic strength (< 0.1 M) the activity coefficient approaches unity, this coefficient can actually increase with ionic strength in a high ionic strength regime. For hydrochloric acid solutions, the minimum is around 0.4 M.

So you mix a solution with a pH of about 1 with a neutral salt solution, and the solution gets more acidic.


text:
Is an alcohol-based spray cleaner ok for cleaning an aluminum heat exchanger?
Simple alcohols such as ethanol can, apparently, corrode aluminum. However, this corrosion reaction seems to be very slow at room temperature and below. Is an ethanol based, non-foaming spray cleaner, propelled by LPG and with an ethanol concentration of, say, about 20%-30%, some (about 2%-3%) alcohol ethoxylate (this acts as a surfactant) and a slightly basic pH (between 7 and 8), really ok to spray on a cold parallel-flow type heat exchanger without fear of any damage? Or should an alcohol-free cleaner be preferred in any case for that job?
My belief is that the corrosion can only happen when the aluminum object is actually submerged in pure alcohol. That little alcohol conveyed by the spray probably evaporates before having the chance to do any damage, but i'd like to understand the surfactant role better, i.e. if any added surfactant agent can act as a corrosion inhibitor.

"My belief is that the corrosion can only happen when the aluminum object is actually submerged in pure alcohol."  And you're right. Just don't keep booze in an aluminium bottle.

I doubt that ethanol is more harmful to aluminium than water.

I'll say it depends. This guy has tested it, and found nothing: https://www.youtube.com/watch?v=xkob6fBNrOg  ;-)

And here http://www.diva-portal.org/smash/get/diva2:720728/FULLTEXT01.pdf (Master thesis of Jenny Linder at KTH Stockholm) is the reason: As long as there is water present, the infamous ethanol corrosion is inhibited. I take back what I said about the booze above. ;-)

(disclaimer: I just read the abstract. Whoever want's to check out the whole thing is welcome to write it into an answer here. No time here. ;-))

When aluminum metal is exposed to the oxygen of the air at room temperture, it forms a protective air-tight layer of aluminum oxide. So as long as nothing is present that will disrupt the oxide layer on aluminium it is OK for aluminium at 20 oC to be in contact with air.
If an aluminium alkoxide is exposed to water then the aluminium alkoxide will react with the water to form aluminium oxide and the alcohol. So in alcohol which contains some water it is not possible for the aluminium to dissolve in the alcohol.
But if the alcohol is anhydrous then it will be able to enter a scratch in the aluminium and start to react with the metal thus causing corrosion. For titanium that anhydrous methanol is very corrosive.
So as long as some water is present and air is present then at room temperture

I'm afraid it's not that easy. Alcohol is probably not reacting vigorously with aluminium metal, but it is definitely altering the (re)formation of protective oxide layer making it less dense at least. OP mentioned there is elevated pH, so that even a weak base can actually react with aluminium. Several government standards (e.g. Russian, I know for sure) forbid using tanks made of aluminum and aluminium alloys for transporting and storing pure alcohols (and booze, too, of course).

"so that even a weak base can actually react with aluminium" – This was absolutely not what the OP said.

Read again: "a slightly basic pH (between 7 and 8)"

Read again: "Is...slightly basic...OK?". Nothing in common with your comment.

No, not OK. I take the plain text info provided by OP and make a rather obvious statement that aluminium is prone to corrosion in basic medium. If you are so sure in your answer, feel free to support it with proper literature references.

Everything is done in contact with the atmosphere so that slight basicity (the one possessed by the cleaner when inside the spray can) i mentioned is likely to end up being neutralized by the air's CO2 content anyway.

@andselisk I have never heard that a pH of 8 could be appreciably corrosive to Al. At pH 10 I would certainly start getting careful, but 7-8? Have you got a source for that?

@Karl Many general textbooks would state that Al is prone to corrosion in both acidic and basic (especially basic) media. Of course, corrosion rates wouldn't be high for the slightly basic solutions, however the corrosion current is never 0 when pH deviates from neutral (see e.g. https://doi.org/10.1016/0010-938X(96)00037-6).


text:
Is there a definitive model of the Elements as a Periodic Tetrahedral
i.e. Is there a specific reason why these diagrams exist? Is the "neatness" of the design simply a graphical representation of the mathematical constants, or is there some underlying reason why a person would envision this?
To be clear: I am asking if there is a physical phenomenon being portrayed or is this just a mathematical trick that is pleasing to the eye?
I have seen several depictions and they vary considerably in their complexity.
Example: https://www.youtube.com/watch?v=f7o0kah6soA

Here is a site that is less of an incoherent jumble of science jargon.
(although it still says nothing about the underlying reason for applying it)
http://weise.symmetry-us.com/pythagorean-approach-in-chemistry-tetrahedron/

https://chemistry.stackexchange.com/questions/901/what-are-the-alternatives-to-the-periodic-table-of-the-elements?rq=1

I'm quite skeptical of how useful the linked diagram is. My skepticism greatly increased with the incoherent jumble of science jargon in the video.

"Normal" table is pretty much as good is it gets - it's based on actual periodicity. The ones you show aren't so it's rather "pleasing to the eye" thing then any serious stuff.


text:
Order for acidic strength of hydrides across a period
I have learnt that in a group , say group $15$, the acidic strength increases on going down the group as
$NH_3 < PH_3<AsH_3<SbH_3<BiH_3$
This is due to decrease in the thermal stability on going down the group. But we can also reason it as follows : on going down the group , the electronegativity of the central atom decreases , and the bond between the central atom and hydrogen becomes less polar , so the bond strength also decreases hence it becomes easier to lose a hydrogen atom thus increasing the acidity.
Now if we compare the hydrides of a period, that is
$PH_3 , HCl , H_2S$ , according to the above logic , the order should’ve been $PH_3>H_2S>HCl$ because the electronegativity increases across a period. However , the correct order seems to be the reverse. Could you please explain why ? Is there a flaw in my reasoning?

The reasoning you gave for the increasing acidity of hydrides of group $15$ down the group is not the best one. You claim that less polar bonds are easy to break which is absolutely wrong. Non-polar covalent bonds are very difficult to break (think of beaking  $\ce{C-H , N-H , O-H }$ bond, which is easier to break ?)The logic can be modified as follows.

Actually the basicity of those hydrides decrease down the group, which in turn results in the increase of acidity of them. The reason for decrease in their basicity down the group is the increase in inability to donate their lone pair to accept the proton. In $\ce{NH3}$, the lone pairs recide in the $sp^3$ hybrid orbitals, whereas for other hydrides like  $\ce{PH3 , AsH3 , BiH3}$ the lone pairs recide almost in pure $s$ oritals (i.e. the hydrogen atoms are bonded to almost pure $p$ orbitals). So, as the $s$ character of the orbital where the lone pair is residing increases, it becomes closer and tightly bound to the nucleus, and thus difficult to protonate. Thus the basicity of group  $15$ hydrides decreases down the group, alternatively, the acidity increases.

Coming to the case of  $\ce{PH3 , H2S}$ and  $\ce{HCl}$, two factors govern the acidity of them. One is the  ease of bond-dissociation, and other is the solvation of the medium. When you consider the gas-phase acidity, i.e. no solvation is present, the bond-energy will only be the determining factor and bond energy is given as,  $\ce{P-H < S-H < Cl-H}$ (easily justifiable by  decreasing size of central atom), which tells us in gas phase the acidity of  $\ce{PH3}$ should be the highest. But normally when we talk about acidity, we talk about $pK_a$ in water. Whenever there is a medium, it stabilises the dissociated species via solvation (in water , this is hydration). As. the size of the conjugate base decreases across the period, the extent of hydration energy released will dominate over the bond-dissociation energy. Chloride is the smallest ion among them having higher electronic charge density, and $\Delta H_{hyd} \propto \frac{1}{r} $, It is highest for hydrolysis of  $\ce{HCl}$, making it a stronger acid among three.

The paragraph on s character increasing down the group,is it based on Drago's empirical rule?

Also, for basicity increasing down the group,   I generally use the rationale that as more orbitals(d and f) become available, the atom will be able to diffuse it's electrons in those empty orbitals, and hence they will be unavailable for functioning as Lewis base lone pairs. But since you have presented a different argument,is my reasoning justified?

@Yusuf The heavier elements because of their vacant orbitals form the coordination compounds. Nitrogen does not form coordination compounds. I’m not sure if the lone pairs of electrons will be promoted to the empty orbitals .

@Aditi Yep, this promotion of electrons will be absent in nitrogen,hence it's lone pairs will be more susceptible(available) to Lewis basicity. Also, I'm talking of the elements per se, not in their complexed forms. After complexation,it will exhibit entirely different properties.

@aditi I would like to suggest another dimension to look at it...acidity means the ability to evolve H+...now it's well know that EVERYTHING in this universe wants stability...now if ph3 evolves H+ it leaves Ph2-...now observe that phosphorus being more electronegative draws the electron cloud with hydrogen towards itself...causing high charge density on it in addition to the already present negative charge...but not being VERY electronegative it's unable to hold it...this thing is known as inductive effect you shall study in organic...but on the other hand look at cl-...no additional charge due to inductive effect as no hydrogen is present...plus it's far more electronegative than phosphorus...so it will accommodate negative charge far better...now you might be thinking how electronegative atoms accomodate negative charge better than comparitively electropositive ones...well it's cause greater effective nuclear charge causes them to attract electrons towards themselves...They LOVE charge...but remember...this is the case in a solvent...IN CASE WE ARE DISCUSSING WHAT IS HAPPENING IN VAPOUR STATE...acidity of PH3 is very much due to bond length...AS H-O bond is very strong hence in protic solvents (in this case water) will remove h from compounds...in that case cl- is more stable and hence HCl is most acidic...but otherwise h-cl bond is weak due to smaller size of cl and hence cl doesn't kick away h while PH3 and H2S do comparitively more.


text:
Why are the protons in the nucleus not repelled by each other?
Since we know that like charges repel each other and the protons in the nucleus have equal and like charges, but they are held intogether instead of being repelled. Why?

They indeed are repelled, and that with great force. But there is another force, even greater...

There’s the strong nuclear force that holds them together . It’s even stronger than gravity.

Not stronger than gravity when you accumulate three solar masses

Protons in nucleus no doubt are repelled by each other. But we know that an atom is stable. The reason for this is that the protons and the neutrons(together called nucleons) are attracted to each other by a strong for called nuclear force. This force acts only in the distances of orders of angstrom or picometre. As mentioned earlier, not only protons are attracted by this force but the neutrons too are. If the distance increases this force decreases drastically. Also this force depends on the no of particles. This is the reason why heavier elements have higher n/p ratio.

What's particularly interesting is that the electrical repulsion between two protons only barely exceeds the attractive (residual) strong nuclear force between them. If the strong nuclear force were a few percent (~2 to 6%) stronger relative to the electromagnetic force, two protons would spontaneously bind to form the diproton, a stable isotope of helium ($\ce{^2_2 He}$). This would have massively altered the course of the evolution of the Universe (completely different big bang and stellar nucleosynthesis). Chemistry could be virtually non-existent!

And so would we.


text:
Is my chlorine sample real?
I bought a pressurized liquid chlorine ampoule online and I was surprised how thick it was.  It was more like oil than water.  Here's a YouTube video showing someone else with a very similar sample.
I trust my seller but I’m wondering why is it so thick? It doesn’t flow easily.

The only thing thick is the wall of that ampoule.

I have no idea how big the ampoule is but be very careful, chlorine gas is very bad for your lungs

At $\pu{20 °C}$ dynamic viscosity of liquid chlorine is about $\pu{0.35 mPa·s}$ (see Figure 4 from [1, p. 7]), whereas dynamic viscosity of water is about $\pu{1 mPa·s}$, e.g. water is nearly 3 times more viscous at room temperature than liquid chlorine, which should not appear that thick.

Figure 4. Viscosities of chlorine gas and liquid.

Note that your perception of viscosity and "thickness" may be distorted due to capillary effect if your chlorine sample is sealed in a tiny ampule.
References

Chlorine: Principles and Industrial Practice, 1st ed.; Schmittinger, P., Ed.; Wiley-VCH: Weinheim; New York, 2000. ISBN 978-3-527-29851-8.

Capillary effect can make a fluid appear to be more viscous?


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Oxidation of sodium thiosulfate by iodine
When $\ce{Na2S2O3}$ reacts with $\ce{I2}$, it undergoes a redox reaction as
$$\ce{2 Na2S2O3 + I2 -> Na2S4O6 + 2 NaI}$$
Sulfur is being oxidized from a $+2$ state to a $+2.5$ state. Why isn't it getting oxidized any further, say to $3+$, to form a compound like $\ce{Na2HSO3}$?
I do notice that there's no hydrogen involved in the equation, but maybe the reaction occurs in an acidic medium?
Could someone please tell me whether the above scenario is possible or not and why?

Whether compound $\ce{X}$ is going to be reduced/oxidized and to what degree is mostly dictated by the corresponding redox potential for the given medium. The thiosulfate anion $\ce{S2O3^2-}$ is used in quantitative analysis and in iodometric titration in particular since it is a reasonably strong reducing agent [1 pp. 714-717]:

$$\ce{S4O6^2- + 2 e- <=> 2 S2O3^2-} \qquad E^\circ = \pu{0.169 V}$$
$$\ce{2 S2O3^2- + I2 -> S4O6^2- + 2 I-}$$

However, stronger oxidants such as chlorine are capable of oxidizing thiosulfate further down to sulfur(VI), e.g. forming a sulfate instead:

$$\ce{S2O3^2- + 4 Cl2 + 5 H2O -> 2 HSO4^- + 8 H+ + 8 Cl-}$$
Bromine, being intermediate between iodine and chlorine, can cause $\ce{S2O3^2-}$ to act either as a 1-electron or an 8-electron reducer
according to conditions. For example, in an amusing and instructive experiment, if concentrated aqueous solutions of $\ce{S2O3^2-}$ and $\ce{Br2}$ are titrated, and the titration is then repeated after having diluted both the $\ce{S2O3^2-}$ and $\ce{Br2}$ solutions 100-fold, then the titre will be found to have increased by a factor of exactly 8.

Note, however, that it is not entirely correct to state that sulfur in thiosulfate has oxidation state $+2$. In fact, there are two nonequivalent $\ce{S}$ atoms with oxidation states -1 and +5. Same goes for tetrathionate anion $\ce{S4O6^2−}$ where two central $\ce{S}$ atoms have oxidation state $0$ and the two terminal ones $+5$.
References

Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.


text:
A mixture of 4 compounds is extracted with HCl. Which one would go to the aqueous layer?

So this has to do something with the H+ ion bonding with the compound. I dont know how to approach it. The answer given is 2

Only one of those is basic enough to protonate

This is a classic example of an acid-base extraction. If a mixture of organic compounds in some organic solvent is treated with an aqueous acid, any bases in the original mixture react to form their conjugate acids. These conjugate acids tend to have ionic functional groups that increase their solubility in water.
The key skills required to solve this problem are 1) identifying the principal functional group in each structure and 2) recognizing the characteristic acid-base properties of each function group.
These compounds are (in order), a sulfide, an amine, an ether, and a ketone. While you could envision perhaps three of them being polar and suitable hydrogen bond acceptors (sulfuer is approximately as electronegative as carbon, and sulfur does not participate in hydrogen bonds), only one is a base - the amine. The characteristic acid-base behavior of functional groups could be constructed each time you need it, but may best be memorized. There are only a few common functional groups that typically behave as acids or bases in aqueous solutions.

Acids: carboxylic acids, phenols, thiols, sulfonic acids, 1,3-dicarbonyl compounds, imides
Bases: amines, imines, amidines, guanidines