text:
Where to find the ChEMBL21 database in CSV format?
I come from an artificial intelligence (AI) background (and not from chemistry).
However, my Ph.D. thesis is about Deep Reinforcement Learning for drug design.
From one of the most famous papers in this regard [1, p. 6]:

The generative network was trained with ~1.5 million structures from the ChEMBL21 database.

I was looking for ChEMBL21 database to begin experimenting with it on my own, but could not find it.
Can anyone provide me with a link to this same database (in CSV format if possible), and maybe some advice to better benefit from?
Reference

Popova, M.; Isayev, O.; Tropsha, A. Deep Reinforcement Learning for De-Novo Drug Design. Sci. Adv. 2018, 4 (7), eaap7885. DOI: 10.1126/sciadv.aap7885.

Contact the author(s).

Official download links for ChEMBL in various formats are given here: https://chembl.gitbook.io/chembl-interface-documentation/downloads. Be careful with the version number; it appears they are now up to version 29.


text:
Acidic strength comparison of squaric acid and rhodizonic acid
These two are dibasic acids with the following structures (squaric acid and rhodizonic acid in order).

Empirical evidence proves that squaric acid is more acidic than rhodizonic acid.
It is a fact that in most cases the compound whose conjugate base is more stable is more acidic than the other. While comparing the anions of the two, I assumed that since rhodizonic acid would have more equivalent resonating structures, it must be more stable.
So where did I go wrong in my method of comparison?

@Mithoron could you please elaborate your point, as a high-school student who has just started studying organic chemistry, I was taught it is one of the major factors while comparing stability, especially if they are equivalent resonating structures.

@Mithoron I came across the structure of croconic acid and its acidic strength "quite fits" into the expected order, i.e. is more than that of squaric acid. But why do we have that "number of mesomeric structures" criteria fails for rhodizonic acid. Can we consider the more number of oxygen atoms to be responsible for this?

In neutral rhodizonic acid water solution, not all carbonyl groups are available for delocalization. Some are hydrated i.e. there are gem-diol ($\ce{C(OH)2}$) groups instead of carbonyls (1). This may be surprising, but it's a generic behavior of compounds with conjugated carbonyl groups and electron-withdrawing groups bound to $\ce{C=O}$ in general. Because of that cyclohexanehexone wasn't even detected, while it's "hydrate", dodecahydroxycyclohexane was made over one and a half century ago!
Why then lower analogs, like squaric acid, aren't hampered like that? Having less carbonyl groups can be indeed better - endiol group has big stabilising effect - it donates lots of electron density and makes other C-C bonds partially double. One could employ some of Michael Lautman's reasoning - only for neutral molecule, not conjugated bases - while carbonyls next to endiol group are directly conjugated, further ones are "more vulnerable" to being hydrated.
(1) Gelb, R. I.; Schwartz, L. M.; Laufer, D. A. (1978). "The structure of aqueous rhodizonic acid". Journal of Physical Chemistry. 82 (18): 1985–1988. doi:10.1021/j100507a006.

I would suggest that you check your work.  If you look at the image below, only the the two oxygens marked with an asterisk are accessible by resonance.
This means that the resonance hybrid of has the negative charges evenly distributed across all four oxygen atoms.

So how do we get the negative charge in the top oxygen atom?

Deprotonate the other oxygen atom.

I should cut you some slack, as what you drawn was also one of my ideas in here, but dianion is symmetric, fully delocalised and can even be considered aromatic. You didn't consider structures with charge separation on carbonyl groups - they are important here.

Of course... I was trying to suggest the thinking pattern, rather than giving the answer.

@MichaelLautman I came across the structure of croconic acid and its acidic strength "quite fits" into the expected order, i.e. is more than that of squaric acid. But why do we have that "number of mesomeric structures" criteria fails for rhodizonic acid. Can we consider the more number of oxygen atoms to be responsible for this?

source: Stack Exchange

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

sort: votes


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Organic solvent that won’t attack either PET or butyl rubber?
I am looking for a suitable organic solvent for dissolving dried-on polyurethane adhesive on a substrate of nylon/butyl/polyester.  Acetone is ruled out because it attacks polyester (PET) and Toluene is ruled out because it attacks butyl rubber.  From what I can tell, nylon (PA) seems to resist everything!
It’s been a while since I took organic chemistry.  Does anyone know of an organic solvent that fulfills the following mandatory properties at standard temperature & pressure:

It MUST dissolve polyurethane (PU)
It must NOT dissolve nylon (PA)
It must NOT dissolve polyester (PET)
It must NOT dissolve butyl rubber

How about chloroform?

I don't think it's possible in the way you imagine.

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


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Size and maximal power output of a battery of electrochemical reactions
For a battery working and chargeable on electrochemical reaction, how does the size of the battery affect its maximal power output?
Specifically, With any chemicals/materials available and current technology level, is it possible to build a micro-battery which is tiny in size yet still "powerful"? How powerful can a tiny battery be, and what chemical reactions/materials should it be built upon, if this is realistic?

Are you specifically asking about wattage per unit volume?

See also Energy_density_vs_power_density // EDLC = Electric double-layer capacitors.

@ToddMinehardt Yes that could be relevant and actually a very precise description of what I asked. Also I wonder how wattage per volume is affected by the type of materials.

2 major factors affecting power density are the rate of charge transfer and the conductivity of active electrode materials + electrolyte.. // For Li-ion cell family is typical that higher power density means lower energy density.

Size matters in batteries. The bigger the electrode surface area more current can be drawn. A very simple example is that of ten 1.2 V pencil cells connected in series at home vs. a 12 V lead-acid battery. Which one can start a car? The lead-acid battery-although the voltage of both batteries is the same.
Not only does size matter, but the rate of electron transfer also matters in batteries. Poor "lead-acid" cell, despite its all negative environmental image, still has the fastest (I don't know the record) electron rate transfer to the electrodes, hence you can draw more current out of it.
So microbattery is not feasible for large sustainable currents. A cardiac pacemaker is also a very small power source but its out is also very low. You can search about electric power sources for space programs. They must be lightweight and reliable. If I remember correctly, each "kg" adds a million to the cost of launching.

Thanks very much for the answer! I also wonder apart from lead-acid, what other materials have good rate of electron transfer? What are the best known ones in terms of this?

I am not the downvoter, but I can spot a few misspellings/misstypes in your post you might address. Please take a second to review it. Your answers are generally well received due to their content but a little copyediting before posting would be appreciated.

You might also address the OP's original query, which is repeated here in the comments. As it stands, you've not touched on all points, which detracts from your answer.

I agree with Todd. You can find plenty of plots showing the tradeoff between energy and power density, but few explanations. Also, I came across at least one somewhat old ref suggesting that this can be overcome in small devices, and that such a device, scaled up to be as large as a phone battery, could start a car.

There is a very nice 1200 paged handbook called  Handbook Of Batteries by David Linden, Thomas B. Reddy. If the OP is serious about this research, he should get it.

@BuckThorn Thanks very much! I am curious if such small device capable of overcome the tradeoff can be build and have any practical applications. I assume there must be technical difficulties to really apply it? Since even not used to power start a car, such "powerful" while "energetic" battery sounds very convenient and useful, yet it doesn't seem to exist (or does it?). And BTW, do you still, by any chance, remember the old reference you mentioned?

Aside from pursuing suggestions in this thread, you might gain more concrete answers on the electrical engineering SE site.


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Efficacy of Intramolecular Friedel-Crafts reaction on =o, c=o bonds of ring structures?
Updated Q:
Inquiring to understand the efficacy of intramolecular Friedel-Craft like reaction on stability on the ring structures with =o, c=o bonds.
Keen to know if ring structures are challenged

solely based on specific species of Lewis acid used in reaction
acting as electrophiles or
based on the changes in the pH (by any acid) that corresponds to pKa dependent stability of the core-structure (as shown in the rough sketch below).
or as a combination of both? (with the reasoning that in FC, as species like Al3+ may not just remain spectators... but catalysts)

Previous:
AlCl3 is a Lewis acid. In Friedel-Craft reaction, purpose of AlCl3 is to produce electrophile, which later adds to benzene nucleus. This electrophilic aromatic substitution allows the synthesis of monoacylated products from the reaction between arenes and acyl chlorides or anhydrides.
The products are deactivated and do not undergo a second substitution. Normally, a stoichiometric amount of the Lewis acid catalyst is required for both the substrate and the product form complexes. AlCl3 (and other Lewis acids like it) will coordinate to halogens, and facilitate the breaking of these bonds.  In doing so, it increases the electrophilicity of its binding partner, making it much more reactive.
LaCl3 and FeCl3 are mild Lewis Acid. According Lewis Acid, the more deficient of electron to complete octate, the more acidic in strength. AlCl3 have vacant orbital to fulfil octate, where as FeCl3 have already octate form so AlCl3 is more acidic than FeCl3. What about LaCl3?

Can the Lewis dot structure be shown for LaCl3 as an electrophile in this reaction? (it's important in this context)

Can FeCl3, LaCl3 substitute AlCl3 in above Friedel-Craft reaction?

What impact can Friedel-Craft have on the following reversible reaction? If FC doesn't apply to it, then on what account AlCl3(aq) Lewis acid (electrophile) can result in it?

FeCl3 has been used with anisole: https://pubs.acs.org/doi/pdf/10.1021/ed073p272

In its initial form (October 16th), your question was about «Can $\ce{FeCl3}$, $\ce{LaCl3}$ substitute $\ce{AlCl3}$ in above Friedel-Craft reaction?»  After multiple edits and answers already received, at present (October 19th), the question now is about efficiency and selectivity of catalysts and conditions of Friedel-Crafts reactions.  In my perception, this is no edit to clarify a question.  Instead, it is moving the target a question addresses. Please refrain from this. Instead, rise a separate question, including your lines of thought.

@bonCodigo - Please refrain from multiple edits that turn a query into a moving target. If you have other questions please ask them separately.

Your question stays fuzzy, and the addition of the scheme does not add much clarification.  The arrow between mesomers is not the one about a reaction, nor about a chemical equilibrium.  (Typically, R and R' designate substituents at the periphery of a chain, or a cyclic core structure; but not about what different spaces within a chain. This just as a side note.)
Assuming you would like to submit alcohol 1 to conditions of a Friedel-Crafts reaction with $\ce{AlCl3}$

the alcohol would react with the Lewis acid to yield $\ce{HCl}$ (like a hydrolysis) and alkyl chloride 2.  In principle this might lead to an intramolecular Friedel-Crafts alkylation 3. However it is more likely that Lewis acid $\ce{AlCl3}$ yields a complex with the Lewis basic carbonyl oxygen of the ester 4 (and, similarly, on 5, too).
Or, do you intend to use a large excess of $\ce{AlCl3}$ to proceed further?  If you intend to open the ester (similar to the Harworth reaction between maleic anhydride and benzene), the then simultaneous presence of an acyl chloride and now two alkyl chlorides will render the reaction messy.

The R and R' represents extended rings with polar groups. The particular species tend to have few different pKa values. When environment becomes acidic, they expect to have the ring open at c-o-c and becomes coo- (as above). This is reversible relative to the pH of the solution (corresponding pKa of the species). I just wanted to confirm from an organic chemist point of view, that this can actually happen and the closest reaction I could think of was FC (intra-molecular). So I posted the question.

what reaction (even if it's not FC) can result in opening and closing of that particular position of the ring?

The cyclization 2 to 3 is likely, because it would be an intramolecular reaction and the ring formed includes five carbon atoms.  If 4 opens, I speculate the alcohol moiety of the ester equally is replaced installing a second alkylhalogen potentially engaging in FC type reactions.  Formation of a cycle of now 4 carbon atoms is more costly in energy, but some competition with the other group is possible.  It then becomes more likely that the FC reaction favours an intermolecular pathway, especially with the then present acyl chloride (cf. FC acylation).

Yes, $\ce{FeCl3}$ is a well known Lewis acid that is used in place of $\ce{AlCl3}$ for FC reaction. See the paper linked by user55119 and also this question.
Other $\ce{MX_n}$ types of reagents used in FC reactions are $\ce{BF3}$, $\ce{BeCl2}$, $\ce{TiCl4}$, $\ce{SbCl5}$, $\ce{SnCl4}$, $\ce{SeCl4}$, $\ce{TeCl4}$, $\ce{InCl3}$, $\ce{NbCl5}$, $\ce{IrCl3}$ or $\ce{RhCl3}$.
Regarding lanthanum, they are mostly used in the form of triflates ($\ce{La(OTf)3)}$). Other lanthanoids used as triflates are neodymium, ytterbium, europium and samarium along with yttrium.
Reference

Rueping M, Nachtsheim BJ. A review of new developments in the Friedel-Crafts alkylation - From green chemistry to asymmetric catalysis. Beilstein J Org Chem. 2010 ;6:6.  doi:10.3762/bjoc.6.6 (There is a periodic table which mentions which metal salts can as be used as FC reagents)
Striving towards improved Friedel-Crafts acylation catalysts, International Nuclear Information System (INIS), Scott, N.M.; Deacon, G.B, 1998 (link)

I am updating my question ^^. It is useful to state the following: are these catalysts used in aq form or solid?

@bonCodigo solid anhydrous form

@bonCodigo I agree on the point that Friedel-Crafts reactions require anhydrous conditions.  Still, many proceed in solution for easier transfer of heat and substance, and because some of the Lewis acids mentioned either are a liquid (e.g., $\ce{TiCl4}$, $\ce{SnCl4}$), or are easier to work with/to purchase as a reagent solution (e.g., $\ce{BF3}$ as adduct with diethyl ether, $\ce{BF3\cdot OEt2}$).

@Buttonwood I am particularly asking this question in the basis of aq form of the catalyst acid. Please see the update to the question.

$\ce{AlCl3}$ and $\ce{FeCl3}$ are hydrolyzed in contact with water, and are not usable for Friedel-Crafts synthesis

@Maurice in light with your comment, could you shed some light on part 3 of my question?

Your question 3 has nothing to do with a Friedel-Crafts reaction...

@bonCodigo In same tune like @Maurice, $\ce{AlCl}$ decomposes by action of water (to generate corrosive $\ce{HCl}$).  First intentional encounter in the elementary lab classes org chem, to initiate the work up.  But $\ce{TiCl4}$ and $\ce{SnCl4}$ used for e.g., other FC reactions, these decompose by water, too. 3) What is this for a drawing added?  If you use the arrow for mesomers, which may alter bond order but not create new / delete old bonds (which were a reaction) between atoms, nor alter the atom count.  But the formula on the right has one C and two H more than the on the left.

@Buttonwood is it fair to say that AlCl3 like Lewis acids will not have any impact on species shown in part 3? However what circumstance AlCl3 like Lewis acids may result in mesomers/dimers (reversible based on altered pH) of a species like above?

@Buttonwood would you be keen to chat in the chatroom?

@bonCodigo since your edits had outdated my answer, I might delete this answer.

I wish I could accepted two answers or more. I upvoted yours as it serves previous part of the question. Let's keep yours? I am thinking perhaps, I need to post a new question with the actual title of the question above.


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Why sodium metabisulphite is not an oxidising agent?
We know that the maximum possible oxidation state of sulphur is $+6$. Whereas the oxidation states of sulphur in sodium metabisulphite are $+3$ and $+5$. Can't these oxidation states reduces to a lower oxidation state? If it is possible can you give me a reaction for better clarity. Because my book is saying that it cannot act as an oxidising agent. Is this true?  Thanks in advance

The sulfur could be reduced, actually, as in thermal decomposition yielding some elemental sulfur. It's just that normal laboratory conditions do not favor such reactions.

Okay thats fine! Can't we use sodium metabisulphite as an oxidising agent in labouratory like potassium permanganate or potassium dichromate.

A brick is a poor substitute for paper. I mean, you kinda can write on it, but that's just not very useful. Same thing here. Metabisulphite can act as an oxidizing agent, but it is nowhere near permanganate or dichromate.

Metabisulphite can slowly oxidize the zinc powder, according to the following reaction : $$\ce{2 Na2S2O5 + Zn -> ZnS2O4 + 2 Na2SO3}$$ This produces the zinc dithionite $\ce{ZnS2O4}$, where the two sulfur atoms have the oxidation number +$3$.
Ref.: R. E. Morcom, Inorganic Chemistry, Teach Yourself Books, St. Paul's House, London EC4P 4AH, Second Edition 1973, p. 163.

Says who? Answers like this are more likely to draw upvotes if a reference is given.

@OscarLanzi I agree about references. But OTOH all chemists ( applies to common life as well ) with some life experience collected a lot of knowledge based on studying and empirical experience without remembering or being aware of the source. Such source may still exist, but may be difficult to impossible to track it down.

(+1) Or just replace zinc with cesium: if the cesium gets oxidized, and it will, then we know the oxidizing agent, if not the equation!


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Physical meaning of enthalpy change when internal pressure Is not constant
My chemistry textbook states the following "general enthalpy" formula for the change in enthalpy when the internal pressure is not constant:
$∆H= ∆U + ∆(PV)$
How is the internal pressure involved in this equation? The equation for enthalpy change under constant internal pressure accounts for the heat energy that was used to increase the volume, and in this case the work done by the system only depends on the external force and the difference in height. So what does the ∆P term mean here?

Enthalpy at variable pressure keeps being the state function, just its change is not equal to exchange heat at the same temperature. State functions need not to have particular  physical meaning. You can invent one, that is totally useless.

This has nothing to do with any process or process path.  Enthalpy is a function of state (a property of the material comprising the system), and depends only on the internal energy, pressure, and volume in the initial and final states.  For certain kinds of processes(such as constant pressure),  the change in enthalpy between the initial and final states can be related to the heat involved.  For most other processes, no.


text:
Changing Variables in Thermodynamics
I have a question about changing variables in the context of thermodynamics, but I suppose this would extend to any set of variables that have defined and nonzero partial derivatives on a given set of points. First I should define the variables.
$T$ is temperature, $U$ is internal energy, $H$ is enthalpy, $F$ is Helmholtz free energy, $G$ is Gibbs free energy, $P$ is pressure, $V$ is volume, $S$ is entropy, $C_V$ is heat capacity at constant volume, and $C_P$ is heat capacity at constant pressure.
The four functions $U$,$H$,$F$, and $G$ are called thermodynamic potentials or thermodynamic functions. Each one has a set of variables, called natural variables, from which you can derive other thermodynamic variables through partial differentiation in a much cleaner way than using with variables other than these natural variables as the starting point.
For pure substances, the natural variables for $U$ are $S$ and $V$, the natural variables for $H$ are $S$ and $P$, the natural variables of $F$ are $T$ and $V$, and the natural variables of $G$ are $T$ and $P$.
$U=U(S,V);H=H(S,P);F=F(T,V);G=G(T,P)$
So, for a pure substance, the natural variables of the internal energy $U$ are entropy $S$ and volume $V$. The total differential of $U$ is then
$$dU(S,V)=TdS-PdV=\left (\frac{\partial U }{\partial S}  \right )_VdS+\left ( \frac{\partial U}{\partial V} \right )_SdV=\frac{\partial U}{\partial S}(S,V)dS+\frac{\partial U}{\partial V}(S,V)dV$$
The question that I have came to mind when I was proving the relations
$$C_V=\left ( \frac{\partial U}{\partial T} \right )_V=T\left ( \frac{\partial S}{\partial T} \right )_V$$
$$C_P=\left ( \frac{\partial H}{\partial T} \right )_P=T\left ( \frac{\partial S}{\partial T} \right )_P$$
When I worked through this, I simply started from the thermodynamic function that was most convenient. For the relation for $C_V$, I started with $U(S,V)$ and did the following:
$$dU(S,V)\rightarrow dU(S(T,V),V)=TdS(T,V)-PdV=T\left [ \left ( \frac{\partial S}{\partial T} \right )_VdT+\left ( \frac{\partial S}{\partial V} \right )_TdV \right ]-PdV$$
$$dV=0\Rightarrow dU(S(T,V),V)=T\left ( \frac{\partial S}{\partial T} \right )_VdT=\left ( \frac{\partial U}{\partial T} \right )_VdT\Rightarrow C_V=T\left ( \frac{\partial S}{\partial T} \right )_V$$
For the relation with $C_P$, I did a similar thing. The natural variables of the enthalpy $H$, for a pure substance, is entropy $S$ and pressure $P$. The total differential for enthalpy is
$$dH(S,P)=TdS+VdP=\left ( \frac{\partial H}{\partial S} \right )_PdS+\left ( \frac{\partial H}{\partial P} \right )_SdP$$
To derive the relation for $C_P$, I went from $dH(S,P)$ to $dH(S(T,P),P)$, where
$$dH(S(T,P),P)=TdS(T,P)+VdP$$
After that I did essentially the same procedure as the one I did for $C_V$.
What I'm wondering now is, how would I do a two-variable change of coordinates starting from the natural variables of a thermodynamic function? In each of the above examples, I sort of used entropy $S$ as a dummy variable and made it a function of a variable that I wanted to write the thermodynamic function in terms of, and a variable that the thermodynamic function was already written in terms of, that happened to be one of its natural variables.
But what if I wanted to write the total differential for internal energy, $dU(S,V)$, in terms of another set of variables outside of the natural variables of $U$, like $dU(T,P)$?
Would I have to do something like this?
$$dU(S,V)\rightarrow dU(S(T,P),V(T,P))=TdS(T,P)-PdV(T,P)$$
$$=T\left [ \left ( \frac{\partial S}{\partial T} \right )_PdT+\left ( \frac{\partial S}{\partial P} \right )_TdP \right ]-P\left [ \left ( \frac{\partial V}{\partial T} \right )_PdT+\left ( \frac{\partial V}{\partial P} \right )_TdP \right ]$$
What if I wanted to write the thermodynamic function in terms of another thermodynamic function/functions with their own set of natural variables? An example would be going from $U(S,V)$ to $U(H,F)$, where, in terms of natural variables, $U=U(S,V)$, $H=H(S,P)$, and $F=F(T,V)$? How would that work? I know this part might not make physical sense whatsoever, but I want to know just for the sake of the mathematics.
Would I write something like
$$dU(S,V)\rightarrow dU(S(H(S,P),F(T,V)),V(H(S,P),F(T,V)))$$
$$=TdS(H(S,P),F(T,V))-PdV(H(S,P),F(T,V))$$
$$=T\left [ \left ( \frac{\partial S}{\partial H} \right )_FdH(S,P)+\left ( \frac{\partial S}{\partial F} \right )_HdF(T,V) \right ]-P\left [ \left ( \frac{\partial V}{\partial H} \right )_FdH(S,P)+\left ( \frac{\partial V}{\partial F} \right )_HdF(T,V) \right ]$$
and continue to expand the $dH(S,P)$ and $dF(T,V)$? Sorry for the long post, but this has been bugging me for a while.

Can you clarify what is the question? Is it about if they can be expressed in other variables? Or is there any meaningful derivative?

@Greg I'm essentially confused about natural variables somewhat. I'm wondering if you can do a change of variables from the natural variables of a thermodynamic potential; for example $dU(S,V)$ to $dU(T,P)$, by writing $dU(S(T,P),V(T,P))$. Is this allowed?


text:
Are there any pathways that can change a solution to all colors of the rainbow?
The challenge's like this. Suppose you have a beaker containing a liquid of transparent or any color. You add some drops of another liquid and the beaker turns into, say, red after some swishing. You add some amount of yet another liquid and the solution now turns into cyan. The process is repeated for at least 5 times, giving us a 'chameleon' beaker that eventually displays all the rainbow's colors.
Let's call that a pathway, for I don't know a better word. So... do pathways like that exist? Provided:

You can do as many steps as needed. Having way more than 7 colors is even more fun!
At each step, the added liquid, the beaker's solution and the result should not share the same color. For example, if we add some blue drops to a transparent solution and it turns blue, then the thing is not really impressive. Adding some purple drops and turning a whole beaker of orange into purple is somewhat OK.
Precipitation is acceptable but not encouraged.
Dramatic effects like smoke or sizzling are great! If they're not dangerous.
The 'added thing' can be liquid or solid or really, any substance.

Did you consider using Reichardt's dye?

That's the 1st time I heard of the thing, thanks! Could you demo a particular pathway?

You might want to check An Easy and Versatile Experiment to Demonstrate Solvent Polarity Using Solvatochromic Dyes. For example, adding water to a solution of the dye in acetone shifts the colour from green to blue to violet to purple.

Red cabbage "soup" is an indicator that covers abut the whole visible spectrum. It has the advantages of being cheap, safe, and can even be prepared at home. Though in genera( it's not a good idea to eat in a lab, the indicator itself is potable.

@Loong unfortunately I don't have access to that article.

@DrMoishePippik it only lacks orange, yellow & cyan (light blue)! How can we proceed from here?


text:
TCP and KCP product
Q:

According to me, the kinetic controlled product(KCP) should be

While the Temperature controlled product(TCP) is:

What should be the answer to this question because no temperature has been specified. I believe if nothing is given we assume the reaction to be happening at room temperature. So in this case if room temperature is considered, what should be the major product? KCP or TCP.

I'm afraid there is no way to reliably answer this question. If no temperature is specified, there's no scientific reason why you should assume it's at room temperature. There may be a pragmatic reason why you do that, which is to solve your ambiguous homework questions / exam questions correctly; but that's not chemistry, that's exam strategy, and is not relevant for Chem.SE.

If you were to only ask: what is the result at room temperature? then that will possibly be answerable. But in that case, you should edit it to focus only on that, and not go through this detour about how the question doesn't specify a temperature. Then ping me and I can reopen it. By the way, please avoid using nonstandard abbreviations like 'KCP' or 'TCP' without explaining what they are, especially in titles. Assuming that "KCP" stands for kinetically controlled product, your title also contains an extra word: it reads "kinetically controlled product product".

@orthocresol agreed but for example we commonly say AC Current even though it litreally means alternating current current, it's a convention I believe. But, nevertheless, will try to avoid next time. Also yes I think this question can be closed, it is ambiguous.


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Titration of CH3COONa with HCl and pKa determination from half equivalence point
I am trying to experimentally determine the $\mathrm{p}K_\mathrm{a}$ of acetic acid by titrating sodium acetate with $\ce{HCl}$. The $\mathrm{pH}$ curve starts in the alkaline region (since sodium acetate is alkaline) and the equivalence point is acidic since at the equivalence point $\ce{CH3COOH}$ is present.
$\ce{CH3COO- + HCl -> CH3COOH + Cl-}$
So, from the half equivalence point : $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$ (of acetic acid) or $\mathrm{pOH} = \mathrm{p}K_\mathrm{b}$ (of sodium ethanoate)
The literature value of the $\mathrm{p}K_\mathrm{a}$ of acetic acid at $\pu{25^{\circ} C}$ is 4.75.
This means that my experimental half equivalence point should be around $\mathrm{pH}$ 4.75 which will NEVER be the case since during the titration till the equiv. point the curve lies on the basic region.
What am I doing wrong? Is there any chance the Henderson/Hasselbach equation won't apply ?

For better site experience, you can find useful How can I format math expressions or chem. equations on Chemistry SE. ( Not to be applied to titles )

The equivalence point of titration of $\ce{CH3COONa}$ by $\ce{HCl}$ definitely is not in basic region, as it will end with  $\ce{CH3COOH}$ and $\ce{NaCl}$.

Hey, welcome to Chemistry SE.  Why would you do it this way?  Why not titrate acetic acid with a strong base directly?

Yes. The half-equivalence point is the point where the solution contains the same amount of $\ce{CH3COOH}$ and of $\ce{CH3COO^-}$ ions. So it should occur theoretically at pH = $\mathrm{p}K_\mathrm{a} = 4.74$. But it will be a little bit  different if the concentration is not $1 M$, because of the activity coefficients being different from $0$.

When you titrate a base like ammonia (pKa of conjugate acid around 9.5) with hydrochloric acid, you can see that it buffers in the basic region and the equivalence point is in the acidic region:

When you try the same with acetate (pKa of conjugate acid around 4.8), if buffers in the acidic region, and the equivalence point is in the acidic region:

(Sorry, bad copy and past job. The start of the curve should be in the basic region, with an initial sharp drop to the buffer region)
Here is a better graph, courtesy of EdV:

As you will find, it is difficult to distinguish the buffering region from the region past the equivalence point. Other than that, it is possible to do this, and there is nothing wrong with that.

I ran the titration curve for 0.1 M acetate ion solution versus 0.1 M HCl, in 0.01 mL acid increments. The titration curve is here: https://chemistry.meta.stackexchange.com/a/4757/79678. If you have any use for it, feel free to use it and annotate, etc., as you please. I did not bother to run the ammonia versus HCl, but it is trivial to do so. I upvoted, by the way. I will leave the figure up for a couple of days. Ignore the M-B stuff: it was for a physics SE comment.

@EdV I added it to my answer, imagining how you went to the lab to do the actual titration - I guess it is a simulation.

Yeah, pretty easy to do the calculations. Nice that the pH at the half-equivalence point, at 12.5 mL added HCl, was the expected 4.74.

Weird that the only upvote is from me. One thing I have noticed is that pH problems, including titration curves, tend not to get much love here. Oh well.


text:
Why Na2CO3 is conjugate base
The defintion of conjugate base is the base formed by removing 1 proton from acid. For example if $\ce{H2SO4}$ is acid then it dissociates as $\ce{H+ + HSO4-}$.
Now we consider $\ce{NaHCO3}$. If we remove one proton we are left with $\ce{NaCO3-}$. But we know $\ce{Na2CO3}$ is the conjugate base of $\ce{NaHCO3}$.But for that to happen we need to break $\ce{2 H+}$.I mean it becomes $\ce{2H+ + Na2CO3}$. But for conjugate base we need to remove $\ce{1 H+}$. That's why i am confused.

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We did remove 1H+ from HCO3-.

First mistake : Removing one proton from $\ce{NaHCO3}$ is not possible. And it will not make $\ce{NaCO3^-}$. Second mistake : $\ce{Na2CO3}$ is not the conjugate base of $\ce{NaHCO3}$, because $\ce{Na2CO3}$ is not a base. It is made of a base, namely the ion $\ce{CO3^{2-}}$ (plus two other $\ce{Na^+}$ ions), but it is not a base.

$\ce{NaHCO3}$ and $\ce{Na2CO3}$ are not molecules, but ionic compounds consisting of $\ce{Na+},$ $\ce{HCO3-},$ respectively $\ce{CO3^2-}$ ions. . There is no ion like $\ce{NaCO3-}$, unless in extreme conditions like kryogenics or interstellar space.
$\ce{CO3^2-}$ is the conjugate base to $\ce{HCO3-}$, because of :
$$\ce{NaHCO3(s) ->[H2O] Na+(aq) + HCO3-(aq)}$$
$$\ce{HCO3-(aq) <=> H+(aq) + CO3^2-(aq)}$$
$\ce{NaHCO3}$ and $\ce{Na2CO3}$ are the respective conjugate acid/base pair only as secondary consideration, transforming the approach of involved ions/molecules into the approach of formal molecules and bystanding ions.

I don't understand how you wrote $NaHCO_3$ reacts to produce $H^{+}$ and $H{CO_{3}}^{-1}$ ions,where did $Na$ go?

You are right :-) I have somehow combined 2 correct reactions into 1 incorrect.


text:
Why are HOMO and LUMO energies invariant under spin-flip?
Atomic theory says singlet and triplet energies for the same orbital differ because of exchange interaction. However, molecular orbital (MO) theory describes the energy between the HOMO (highest occupied MO) and LUMO (lowest unoccupied MO) as the same regardless of whether triplets or singlets are being considered. Why is this the case?

Where did you get that "MO theory describes the energy between HOMO and LUMO as the same regardless of whether triplets or singlets are being considered"?

when reading papers on TADF molecules, where spin state is relevant, only a set of HOMO and LUMO are usually presented. I get how HOMO is the same but not why they only show one LUMO

I think maybe you're confusing MO diagrams with MO theory. The theory predicts different energies that are often not captured in the simplistic diagrams. They have to be simplistic because otherwise you'd have to redraw them every time you changed the electron configuration. That's why we typically use a diagram of system states rather than MOs for something like TADF. For example https://upload.wikimedia.org/wikipedia/commons/thumb/9/9e/Photoluminescence_Pathways.jpg/500px-Photoluminescence_Pathways.jpg Note that the lines here are states of the system (S or T) not individual MOs

That makes sense. In that case, when papers depict the simplistic LUMO and HOMO distributions of particular molecules, what transition is that meant to represent?

It would be helpful if you showed a specific diagram that is confusing you


text:
Chemical product to restore a rubberized coating that became tacky / sticky?
I have several Dell Latitude E6520 laptops whose the palmrests became tacky / sticky.
This problem also concerns other models sharing the same "rubberized" palmrest coating
(E6320, E6420, E6430, E6530, a.s.o.).
On a forum, someone mentioned that this was caused by a degradation of the neopren coating caused by cleaning alcohols (e.g. isopropyl alcohol), at least when not diluted enough. I'm not a chemist and cannot confirm if the exact nature of the surface is neopren, but his answer makes sense to me. At least cleaning does not make the palmrests less tacky.
Some people have tips and tricks to make the palmrest less tacky, like applying wax.
I'm not truly convinced by this solution, and wonder if there is some product or family of compounds which, by cold application, would trigger kind of "soft vulcanization" to harden the "ruberrized" coating again and make it more slippery.
Thank you for your indulgence towards my possible jargon mistakes.

Basically, the question is if polymers that returned to a more elastomer state can be hardened again by vulcanization through simple methods. My question has been rephrased here: https://chemistry.stackexchange.com/questions/158821/can-a-damaged-neoprene-polychloropren-or-rubber-coating-be-cured-hardened-us

4 and 1/2-in grinder with oh buffing wheel that doesn't work switch to 600 grit we can keep going down on grit size so it'll definitely get it sooner or later it won't be sticky

I'v seen the same effect on computer mices, sometimes so badly that a good functioning device had to be trashed...

Cleaning a sticky surface is tricky, unless you know what you are removing, or whether you will be putting more plasticizer into the surface,
The palmrest gets dirty, right? That's why it gets cleaned off. Well, cleaning may not get all the skin oils off, and they might be plasticizing the neoprene to a higher degree of stickiness. It is probably good to get most of the skin oils off.
Then try dusting the surface with talcum powder, in such a way that it does not get into cracks and crevices in/on the keyboard. Other fine powders could work, like chalk. If you need some reasoning to add more "dirt" to the palmrests, consider it to be filler for the rubber which is still a little too gooey.

I don't want to clean the sticky surface, nor to remove the coating. It is not sticky because of dirt, but because likely the cleaning agent (e.g. isopropanol) for these second hand laptops altered the polymer, bringing it closer to its initial elastomeric state. So, my question is basically if vulcanization is possible again by applying a compound and using a hot air gun (something like when repairing a bicycle inner tube). I rephrased my question (see link in the comment of my original post). Thank you for the talcum powder idea. I may try it if re-vulcanization is not possible.

It is unlikely that rubber, whether hydrocarbon, neoprene or silicone, could be devulcanized by isopropanol, under any conditions. The palmrest may be coated with a paint, which may have, as binder, a resin, or even low molecular weight rubbers, and plasticizer. Solvent softening by iPA can't be reversed by rebonding because an active crosslinker would also likely overbond and embrittle the coating. In fact, the likely cause of tackiness is that the original resin/rubber has continued to crosslink, shrink and extrude plasticiser to the surface, which becomes oversoftened (and tacky).


text:
What's the difference between a 'centered' and 'side' double bond on a structural formula?
It's hard to explain since I don't know what you'd call this but here's what I mean:

You can see that the oxygen is drawn with two bonds centred on the diagram while the nitrogen has a single centred line with a line to the side.
So, what's the difference?

There are many recommendations about double bonds in the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). In particular, the general rule GR-1.10.1

If a double bond has more substituents on one side than on the other, the double bond should be offset to that side.

explains your second example (N-methylbutan-2-imine). And GR-1.10.2

Double bonds with two or more substituents on one end and no substituents on the other should be drawn with the two segments of the double bond centered relative to its atoms. Double bonds of this type are necessarily acyclic, and are most commonly found in carbonyl and acid functional groups.

explains your first example (butan-2-one).
Reference: Brecher, J. Pure Appl. Chem. 2008, 80, 227-410; doi 10.1351/pac200880020277 (open access).

Anyone know what these recommendations are good doing for? Plus 1 to both question and answer as for I wasn't aware.


text:
Why is the formation of the nitro isomer more favorable under acidic conditions
In the reaction between chloropenataamminecobalt and nitrite either the NO2 or the ONO linkage isomer is formed. This preference seems to be mainly due to the pH of the solution with acidic conditions favoring the N-bound nitro over the O-bound nitrito. does anyone know what causes this I'm thinking it has to do with the formation of nitrous acid in the solution.

Provide some illustration. Difficult to understand what you're taking about.

I figured it out for those wondering. Under acidic conditions, the nitrite molecule forms nitrous acid via an oxygen-hydrogen bond. This makes it so the only atom available for bonding is the nitrogen and thus the nitro (N bonded) isomer is favored.

Do you have any reference to back your answer?


text:
Which ion has a greater paramagnetic property?
I just recently began studying the Molecular Orbital Theory in Chemistry and I just learnt about the paramagnetic and diamagnetic nature of molecules based on their unfilled and filled orbitals. Now if I talk about the $H_{2}^{+}$ ion and the $H_{2}^{-}$ ion, both of them have an unfilled orbital.
In case of $H_{2}^{+}$, the electronic configuration is: $\sigma 1s^1$ and case of $H_{2}^{-}$, the electronic configuration is: $\sigma 1s^2 \sigma^{*}1s^{1}$. This shows that both the ions are paramagnetic. Now my question is, how to tell which is more paramagnetic than the other, is there a way to quantify the paramagnetic properties, and if yes, how?

The paramagnetic properties of atoms and molecules are mostly determined by the properties of the electron (in principle, a rotating molecule also is susceptible to a magnetic field, but the interaction associated with this effect is negligible in most cases). In addition to electron spin angular momentum, an electron can also have orbital angular momentum. To understand which energy level will have the strongest interaction with a magnetic field, you need to know a little bit about angular momentum coupling. The example you give is actually rather simple, because s states have no orbital angular momentum and - ignoring interactions between different states - the strength of the interaction is only determined by the electron spin and roughly the same for both ions. Note also that the Lorentz force on these ions is about four orders of magnitude larger than the magnetic interaction of the electron spin with a magnetic field.
If you want to know more about this, you could search for "Zeeman effect".


text:
Recycling flow if the chemical process of acetylene and hydrocyanic acid
I'm having some issues trying to solve this problem, because the concentration of inert in the recycling flow that I calculated is bigger that the concentration of the inert that is enterying to the system, so I'd like to know if you can help me with it (the section b).
Acrylonitrile is produced industrially by the reaction of acetylene and hydrocyanic acid in the gas phase:
$\ce{C2H2(g) + HCN(g) -> CH2=CHCN(g)}$
Acetylene and hydrocyanic acid, both with a purity of 95% by weight, enter the system in a molar proportion of 10: 1 and 85% of the entering hydrocyanic acid is converted into the reactor, giving acrylonitrile. All the acrylonitrile of the reaction product and the unreacted gases are recirculated to the reactor.
Since the raw materials contain a gaseous inert, there is a purge so that the concentration of inert in the recirculation does not exceed 6% by weight. For a feed of 200 kg / h of fresh pure acetylene entering the system, calculate:
a) The acrylonitrile production in kg / h.
b) The recycling flow (kg / h).
c) The composition of the stream leaving the reactor.

Your process is not clear. You say that all the acetylene, HCN and acrylonitrile crossing the condenser are recycled in the reactor. How and how much acrylonitrile is separated ? And what is P ? And how is P separated form R after the condenser ?

Let's see your calculations so far.


text:
Understanding weak acid-strong base titration
Today I stumbled across a website which defined the titration of weak acid and strong base as the following:
"The titration of a weak acid with a strong base involves the direct transfer of protons from the weak acid to the hydoxide ion. The reaction of the weak acid, acetic acid, with a strong base, NaOH, can be seen below. In the reaction the acid and base react in a one to one ratio." (https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Acid_with_a_Strong_Base)
CH₃COOH + OH⁻→ CH₃COO⁻ + H₂O
But, I clearly remember my professor for analytical chemistry saying that the neutralization of a weak acid and a strong base occurs due to the shift of equilibrium:
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Where the strong base (NaOH) would continue to remove hydronium ions from the equilibrium, so that more and more of conjugated base, acetate, would form.
So the first part of my question is; what definition is correct?
Furthermore, indeed we usually formulate following equation to approximately quantify the change in concentrations of acid and base:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O  +  Na+
This way, we find the moles of acid/base and divide by the total volume to get the pH using Hendersson-Hasselbalch equation. But is it correct this way? Are there more accurate methods? The weak acid should dissociate first before being consumed, is that correct?
Anyway, this is not a question about how to calculate the titration curve of this example, rather it is about understanding the nature of the conversion. Thanks in advance!

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The neutralisation happens by both reaction paths. Hydroxide ions react with both hydronium and acetic acid.

Your second equation is an equilibrium. Both acetic acid and acetate ions are always present in a solution, although their proportion may vary. When $\ce{CH3COOH}$ is dissolved into water, a fraction of the molecules are transformed by reaction with water, producing some $\ce{H3O+}$ But if these ions are destroyed by adding $\ce{OH-}$ ions, more molecules $\ce{CH3COOH}$ have to react with water. Acetic acid is transformed into acetate ions to compensate for the consumption of the first $\ce{H3O+}$ ions. So more and more molecules of acetic acid are transformed into ions. The final result is similar to the first equation : more and more $\ce{CH3COOH}$ is consumed by adding $\ce{NaOH}$ and more and more acetate ion and water is produced from this process.

I would 100% agree. But that would be hard to quantify using Hendersson-Hasselbalch wouldn't it? The same websites just uses ICE and quantitative neutralisation like equation 2.

Hendersson-Hasselbach's equation is valid for a solution that does not change in time. It does not apply in the present problem.

How so? Why doesn't it. I learned it this way in university and don't see why

H.-H. equation is just rewritten equation for the acid dissociation constant $K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$, which is obviously true if and only if the reaction is in equilibrium.

@Maurice Sir when we reach the point where there is very very less acetic acid in the container , then how will the equilibrium be there ? In next few moments of dropping more naoh

I am sorry. I do not understand your question. The equilibrium constant is a constant. it will not change with the amount of acetic acid, or of NaOH.


text:
Why is sodium carbonate a diacidic base?
We know,the acidity of bases is defined by the number of replaceable $\ce{OH-}$ ions,or in case of oxides,it is the number of moles of $\ce{HCl}$ it reacts completely. But we know $\ce{Na2CO3}$ is diacidic. How is it determined?

Related: Why Na2CO3 is conjugate base

Carbonates, $\ce{CO^{2-}_3}$, formally derive from $\ce{H2CO3}$. You may postulate two deprotonations:
$$\ce{H2CO3 + H2O <=> HCO^-_3 + H3O+}$$
and
$$\ce{HCO^-_3 + H2O <=> H3O+ + CO^{2-}_3}$$
Thus, hydrogencarbonates are monobasic; and carbonates are dibasic.

However, note that $\ce{CO2}$ gas may solute physically in water (like, for example, oxygen may dissolve in water, too) without a chemical reaction which would create a chemical bond, or delete a chemical bond in the $\ce{CO2}$ molecule.  Instead, there is a chemical equilibrium between $\ce{CO2}$ and water and the hydrogencarbonate:
$$\ce{HCO^-_3 + H3O+ <=> H2O + CO2}$$
In contrast to sulfuric acid consisting of $\ce{H2SO4}$ molecules, those of carbonic acid ($\ce{H2CO3}$) decompose in the presence of water to yield $\ce{CO2}$.

Does that mean we will have to make it react with water to ensure acidity?Like $\ce{Na2CO3}$ with $\ce{H2O}$?

«Monobasic», «dibasic», «tribasic» (e.g., $\ce{H3PO4}$); these are assignments on paper, the mentioned postulated about deprotonations possible at all.  If e.g., $\ce{HCO^-_3}$ acts like a base, or like an acid, however depends on the presence of a reaction partner which either is able to donate a proton, or is able to accept one proton. This then is a question of pKa of the reagents present. Even more so since water may be either acid, or base (amphoteric character), and that water undergoes some autoprotolysis ($\ce{2H2O <=> H3O+ + OH^-}$), too.


text:
How much heat is lost to the calorimeter when hot water is mixed with cold?

If a student adds $\pu{100.00 g}$ of hot water at $\pu{50.00 °C}$ to $\pu{100.00 g}$ of cold water at $\pu{25.00 °C}$ and measures a heat change of $\pu{10707.37 J}$ for the hot water and $\pu{4778.86 J}$ for the cold water, how much heat was lost to the calorimeter?

My thinking is assuming no energy is lost to the calorimeter. The amount of energy lost by the hot water should be equal to that gained by the cold water. It seems to me that the difference between $\pu{10707.37 J}$ and $\pu{4778.86 J}$ is the amount lost to the calorimeter.

Given the numbers, what are your lines of thought to determine the temperature of the two waters mixed together if there were no heat dissipation to the calorimeter?  Add these in first place to document your attempts to solve this question on your own to show where you struggle.

Note that the question makes no sense, if you mix two volumes of water, how would you measure the exact heat dissipated by one volume and that gained by the other? That aside, it is clearly a conceptual problem. To clarify the question, perhaps you might post the source.

I believe there is something wrong with the numbers provided in the question. Your thinking is right, assuming no energy is lost to the calorimeter, the heat lost by the hot water is the same as the gained by the cold water. So, by using the following formula:
$$Q=mc\Delta T$$
We know the following:
$$m(H_2O_{hot})\times c(H_2O)\times \Delta T(H_2O_{hot})=m(H_2O_{cold})\times c(H_2O)\times \Delta T(H_2O_{cold})$$
Let the final temperature be $T_f$,
$$\implies mc(50-T_f)=mc(T_f-25)$$
Over here the unit of temperature (degree Celcius) does not affect the reading since we are subtracting the values (and therefore the difference would be the same irrespective of whether we are using degree Celcius or Kelvin as our unit). Now solving this equation for $T_f$, we get:
$$T_f=\frac{75}{2} ^\circ C=37.5^\circ C$$
Now if we use $Q=mc\Delta T$ to find the value of heat loss for the hot water, we will see that it is not consistent with the values you have provided. This is shown below:
$$Q=mc\Delta T$$
$$\implies Q=100g\times 4.184J\cdot g^{-1}\times (50-37.5)^\circ C$$
$$\implies Q=5230J$$
As seen, the heat change for hot water is $5230J$ worth of energy whereas in the question you have mentioned that the heat change for hot water is $10707.37J$.
I would request you to recheck the question. But yes, as mentioned earlier, your concept is right, in an ideal case the heat change for hot water is the same as heat change for cold water (the signs will be opposite since hot water losses heat whereas cold water gains heat) and if it is not an ideal case, i.e. the calorimeter takes some of the heat energy then the heat lost to the calorimeter plus the heat change for cold water is equal to the heat change of the hot water.

Please post comments to the OP under the original question. I agree that there is something "wrong" with the question that the OP needs to address (then again, the question makes no sense, if you mix two volumes of water, how would you measure the exact heat dissipated by one volume and that gained by the other?)

@BuckThorn I mean to measure the heat (energy) you can use $Q=mc\Delta T$... What I was saying was that the numbers given in the question are wrong. Will keep in mind about the commenting feature.


text:
Why does mercury(II) nitrate form only when using concentrated nitric acid?
In the reaction between mercury and nitric acid, mercury(II) nitrate only forms when concentrated nitric acid is heated up and added to it, if the acid is dilute, mercury(I) nitrate forms.
I understand heat must be added to make the nitric acid properly oxidise with the mercury, but how does the concentration affect the reaction? Is it due to extra water content in dilute nitric acid? How does that affect it?

My guess is that with concentration nitric acid, the reaction first proceed with the formation of the +1 salt but then oxidises to the +2 salt owing to the excess of acid present as compared to mercury.

Oxidation Hg(I)->Hg(II) has higher standard redox potential(data page) than for Hg(0)->Hg(I), so HNO3 may need high enough concentration ( and therefore oxidative power/redox potential) for the former to be preferred.

Why does mercury in dilute HNO3 give mercurous nitrate while hot concentrated HNO3 produces mercuric nitrate?

“Mercury dissolves in oxidizing acids, producing either Hg${^{2+}}$ or Hg$_2$$^{2+}$, depending on which reagent (mercury or e.g., nitric acid) is in excess.” Ref 1

Wikipedia says

“Mercuric nitrate can be reacted with elemental mercury to form mercurous nitrate…(and)…If the solution is boiled or exposed to light, mercury(I) nitrate undergoes a disproportionation reaction yielding elemental mercury and mercury(II) nitrate:”  Ref 2

So, dilute acid is simply acid that is insufficient to oxidize all the mercury to mercuric; excess mercury reduces mercuric to mercurous nitrate by disproportionation. The equilibrium seems labile enough. In my simplified imagination, a mercuric ion just picks up a mercury atom and brings mercury nuclei even closer together than in the metal. The $\ce{Hg-Hg}$ bond length in mercurous nitrate is 254 pm (Ref 3) and 253 pm in mercurous chloride vs 300 pm in the metal. Ref 4

Figure 1. Structure of mercurous nitrate

Figure 2. Structure of mercurous chloride
Mercury will dissolve in nitric acid because the oxidation potential of nitric acid is higher than necessary to dissolve metallic mercury. Electrochemical potentials (Ref 5) give a more quantitative picture of the possibilities (see eq 4):
eq 1: $\ce{Hg^{2+} + 2e-}$ =  Hgº   ..................................0.851V
eq 2: NO$^-_3$  +  3 H$^+$ +  2e-  = HNO$_2$ + H$_2$O   ..........0.94V
eq 3: 2 Hg${^{2+}}$  + 2e-  = Hg$_2$$^{2+}$..............................0.905V
Combining equations 1 and 2 gives equation 4:
eq 4: Hgº + NO$^-_3$ + 3 H$^+$ +2e-  = HNO2 + H2O + Hg${^{2+}}$ +2e-    ......0.089V
The first product of dissolution of mercury will be mercuric ion. Hg${^{2+}}$ is a powerful oxidant, capable of oxidizing metallic mercury. Combine equations 1 and 3; cancel the electrons on both sides:
eq 5: 2 Hg${^{2+}}$  + Hgº  + 2e- = Hg$_2$$^{2+}$ + Hg${^{2+}}$ + 2e-       ............0.054V
So mercuric ion in the presence of excess mercury (or insufficient HNO3) is reduced to mercurous, but the driving forces are small. The labile reaction is well known. Excerpts from Ref 6 give equations for “excess acid” and “excess mercury”:
"eq 6: 3 Hgº + 2 NO$^-_3$ + 8 H$^+$ = 3 Hg${^{2+}}$ + 2 NO + 4 H2O  (excess acid)   and
eq 7: 6 Hgº + 2 NO$^-_3$ + 8 H$^+$ = 3 Hg$_2$$^{2+}$ + 2 NO + 4 H2O    (excess mercury).

…the value for the potential of the reaction (Hg${^{2+}}$ + Hgº  = Hg$_2$$^{2+}$) corresponds to an equilibrium concentration of Hg$_2$$^{2+}$ at 25ºC only 166 times that of Hg${^{2+}}$; hence the equilibrium is easily reversed…”(by solubility variations) “…reducing agents first reduce mercuric ion to mercurous…and…most reducing agents capable of reducing mercuric ion will, in excess, reduce the mercurous ion to mercury as a second step…”

The binuclear structure of mercurous ion is readily accepted, but it was proved meticulously, in five steps, in Ref 7. Interesting observations include that mercurous compounds are diamagnetic, ruling out Hg$^+$ with an unpaired electron. Electrical conductance of solutions of mercurous salts is consistent with a divalent mercury species. Equilibrium experiments between mercuric ion and mercury metal yield a constant only when the mercurous product is dimeric. The reference also mentions the different oxidations of mercury metal due to excess or deficiency of either reagent. But of course, X-ray diffraction is a clincher.
Interestingly, the ground state electronic configuration of neutral mercury is [Xe].4f14.5d10.6s2 Ref 8. The f electrons are supposed to be poor shielders, so the ionization potential of mercury is a little over a volt higher than might be expected. Whatever is going on (like resonance) appears to make metal-metal bonding very attractive.
Ref1:https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Qualitative_Analysis/Characteristic_Reactions_of_Select_Metal_Ions/Characteristic_Reactions_of_Mercury_Ions_(Hg²⁺_and_Hg₂²⁺)
Ref 2.https://en.wikipedia.org/wiki/Mercury(I)_nitrate
Ref 3.http://nopr.niscair.res.in/bitstream/123456789/11011/1/IJCA%2050A(2)%20137-140.pdf
Ref 4. https://en.wikipedia.org/wiki/Mercury(I)_chloride
Ref 5: CRC Handbook, p D134, 62 ed.
Ref 6: Reference Book of Inorganic Chemistry, W.M. Latimer and J.H. Hildebrand, 3rd ed, Macmillan Co., 1951, pp 140-141
Ref 7. Advanced Inorganic Chemistry, 2nd ed, F. A. Cotton and G. Wilkinson, Interscience /John Wiley, 1962, p611ff
Ref 8. https://www.webelements.com/mercury/

Is there anywhere explicitly stated an excess of metallic mercury is required for forming mercurous  instead of mercuric nitrate with dilute acid ? (unless I have missed it in the answer) I ask as notes I have read focus on acid dilution, not excess of mercury.

@Poutnik: The answer is way too long; it really consists of about 7 answers. The one you are looking for is illustrated in equation 7 from Ref 6. Lines 3 and 4 in my answer use the word "excess" rather than "dilute" but is not clear on whether this is REQUIRED.


text:
Where are orbitals located in an atom?
I am having trouble understanding what are the orbitals. Till now what I have understood, I have depicted in this diagram.

Is the diagram correct?? If not, what should be the corrections?

An orbital is sometimes thought of a possible region where you can find an electron inside an atom. We know for sure that an electron lies inside this region, but we cannot pinpoint the location of an electron inside of this orbital. This is because of Heisenberg's uncertainty principle, which states that you simply cannot know both the position and the momentum of a particle with complete certainty.
Now in the model that you have drawn, you have shown the electron as a particle which means you are sticking with Bohr's model of an atom and therefore, it does not make much sense to talk about 'orbitals' per se. Now, where do orbitals come in? Orbitals come in when you accept an electron's dual nature. That is, you accept that an electron can act as both wave and particle depending on the given circumstances. The most significant contribution to this new model of an atom was given by Erwin Schrödinger, who came up with the celebrated Schrödinger equation:
$$-\frac{\hbar^2}{2m_\mathrm{e}}\left(
\frac{\partial^2 \psi}{\partial x^2} +
\frac{\partial^2 \psi}{\partial y^2} +
\frac{\partial^2 \psi}{\partial z^2}
\right) + V\psi = E\psi.$$
The atomic orbitals are technically the wavefunctions $\psi$, obtained by solving this equation. In the course of solving it, you can derive three out of four quantum numbers for the electron. These are $n$ or the principal quantum number, $\ell$ or the angular momentum quantum number, and $m$ or the magnetic quantum number. These sort of describe the orbital of an electron. $n$ gives what we know as the shell number of the electron, most often labelled as numbers starting from $1$. $l$ gives what we know as the subshells. An electron can be in the following subshells: s, p, d, f (for all the elements we have till now). $m$ is used to label the individual orbitals within a subshell. For example, the s subshell has only 1 orbital, the p subshell has 3 orbitals, the d subshell has 5 orbitals, and the f subshell has 7 orbitals. Each solution $\psi$, and hence each atomic orbital, corresponds to one combination of $(n, \ell, m)$.
The fourth quantum number, which is not derived from Schrodinger's wave equation is the spin quantum number, $m_s$. This quantum number basically defines the spin of an electron in order to follow the Pauli exclusion principle. If you are more interested in how these electrons decide their quantum numbers, check that out and also the Aufbau principle and also Hund's rules. So to give you an idea of how the actual picture of an atom might look like:

This image only shows the 1s and 2s subshells, now if we take a look at one 2p subshell, it would look as follows:

This is a $2p_z$ subshell, that is, this p subshell aligns itself along the $z$-axis. Follow this link to use the simulation for yourself.
What you can see in these images, the coloured portions, are the subshells, that is, where the electrons can be. For example, in the first image, you see 2 subshells, the 1s and the 2s subshell and each subshell can contain two electrons, so in the first image, the total number of electrons you see are 4 electrons which can exist anywhere inside the coloured region (you simply cannot pinpoint a particular location as it would be against Heisenberg's uncertainty principle), the same is the case with the second image where the $2p_z$ subshell is shown. This again contains two electrons and they can be located anywhere. Now a real atom is a mixture of all these shells, subshells, and orbitals where the electrons can be located each with a spin to satisfy the Pauli exclusion principle. I hope that clarifies your question and hopefully, I have been able to provide with some more knowledge/resources to understand the topic further.
One interesting note: If I have not already confused you enough, listen to this: In actuality, there is nothing known as 'shells' and 'subshells' all we have are the orbitals. The shells are a simple way to represent the size of these orbitals, the subshells are a way to represent the shape of the orbitals, and the magnetic quantum number (m) is a way to represent the orientation of these orbitals.

Magnetic quantum number can be represented by $m_l$ and spin quantum number as $m_s$

(1) An orbital is not a region; it is a wavefunction. (2) The magnetic quantum number does not take the values 1, 3, 5, 7... it is permitted to take integer values between $-\ell$ and $+\ell$ inclusive. The total number of permitted $m$-values is equal to 1, 3, 5, 7... but the values of $m$ itself aren't that. (3) Nobody really uses KLMN for shells, except for spectroscopists sometimes. You yourself wrote about the 1s orbital, not Ks orbital. Anyway, I corrected these and other (fairly small) technical inaccuracies in my edit.

@orthocresol I have said in some old answers/comments the orbital has several meanings: orbital(1)= wave function, orbital(2)=el. quantum state related to (1), orbital(3)=3D region, related to (1,2) covering probable occurance of el. according to the integral probability and boundary pr. density. // Is it wrong ?

@Poutnik Hmm, I wouldn't say it's entirely wrong. Going by technical correctness, the orbital is the wavefunction. The 3D region, or the isosurface is just a representation of the 3D function (this depiction is necessarily limited by the fact that we can't actually draw 3D functions). It's a bit like a function $f(x) = x^2$ vs a graph of the parabola $y = x^2$; the former is a function, the latter is a depiction. In this 1D case the depiction is entirely faithful so one might possibly argue they're the same thing. In the 3D case not so much.

@Poutnik I have to say, I remember that answer of yours and I'm pretty sure I upvoted it. Reason is because it explains the link going from wavefunction to isosurface. That's the real underlying science and if that's explained correctly, then I probably don't really care so much about what an orbital "is" or "isn't", because the facts are there and the reader may use critical thinking. However, here, the link is not drawn (presumably for simplicity) and so I think it becomes more dangerous to say that an orbital is an area -- that may be indirectly true but it's not explained why this is true.

@orthocresol I see and agree with your late reasoning.

First, electrons do not actually "orbit" the nucleus, because if they were whirling around in circles, they would radiate electromagnetic energy and "fall" into the nucleus.
While you might draw a circle on paper as an illustration, the probability of finding an electron is better viewed as a fuzzy cotton ball, stretched into different shapes for different orbitals.
So there's no harm in drawing illustrations of orbits as circles, don't be confused into thinking the electrons follow circles, or circles with planetary epicycles, in an atom

Orbitals are centred around the nucleus. If an atom has many orbitals, either one orbital is bigger than the others or they overlap.
Moreover, the orbitals are an area where an electron is most likely to be in. You can, like your prototype, draw a dashed line along the orbital.

Orbital is not an area.


text:
Why is C-S-C bond angle in thiirane so small at 48.5 degrees?
Why is the C-S-C bond angle in thiirane (48.5 degrees) so much smaller than C-O-C bond angle in oxirane/epoxide (60 degrees).

Source: https://coek.info/pdf-three-membered-ring-heterocycles-.html

Because the sulfur atom is big, maybe?

How does that cause the bond angle to be smaller?

Because the non-bonding doublets are more cumbersome than the bonding doublets.  Sulfur is a large atom. It has two bonding electrons that are stretched between carbon and sulfur, and so are not really cumbersome. But it also has two non-bonding doublets that repell one another and are attracted by the sulfur nucleus. They also repell the stretched bonding electrons. As a consequence the angle between them is smaller than it is in a ordinary triangle

Similar question, possibly a dupe: Why does bond angle decrease in the order H2O, H2S, H2Se?

This is pure planar geometry, no chemistry required. Try drawing a triangle with one side much smaller than the other two - what happens to the angle between the large sides? I guess the chemistry part comes in knowing that a C-C bond is much shorter than a C-S bond, as your table shows.


text:
Name for long chain of double-bonded carbon atoms
Long ago (over 10 years), I read an article in my father's copy of Chemistry World on a substance consisting of many carbon atoms in a chain double bonded to each other:
$$\ldots\ce{=C=C=C=C=C=}\ldots$$
So, no other elements; a very exotic allotrope of carbon.  As I remember, the article was sceptical of the claims of synthesizing this compound but its existence is not my question.  My question is much more basic: what is the name of this hypothetical compound?  I cannot remember and I have failed to find it in searches (possibly due to my searching skills).

Cumulenes https://en.wikipedia.org/wiki/Cumulene

That's close though this (hypothetical) substance had much longer chains.  A sort of 1D analogue of graphene.

@orthocresol can it be called an allene system?

@Vega No. Allenes have $n = 2$ and other things on the end.

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

https://doi.org/10.1016/j.cplett.2004.01.007

@Mithoron Thanks, that's it.  If you convert your comment to an answer, I would accept it.


text:
What does valence electron mean (in the context of spdf orbitals)?
For a little bit of context, my background is in physics and my understanding of chemistry doesn't go past, say, middle school level. Recently, I decided to self-study chemistry and picked up "Inorganic Chemistry (Housecroft & Sharpe)" but I couldn't get past the first chapter because of the following question:
In the advent of quantum mechanics, our understanding of electron orbital underwent a paradigm shift; they are no longer always in spherical orbit, but their boundary surface take on awkward shapes like "dumbbell" shape or have "rings" around the nucleus.
With this in mind, how then does the idea of valence electron still fit into this picture? How do we define "outer electrons" if the "shells" are no longer nested like Russian dolls? Why do we say that Co ([Ar]$4s^2 3d^7$) has 9 valence electrons but Br ([Ar]$4s^2 3d^{10} 4p^5$) only has 7? Is there a single, consistent definition for valence electron that does not differentiate between s-block, p-block, d-block and f-block of the periodic table?

@Poutnik That was what I am perplexed about. If we go by "top total energy", then we should have said that Co has 7 valence electrons and Br has 5. Why do we consider electrons in both $4s^2 3d^7$ as valence for Co but totally ignore the $d$-shell for Br and only say that electrons in $4s^2 4p^5$ is counted? Using spatial extend as an "indicator" is also problematic since the irregular shape of the orbitals makes it hard to properly quantify this.

@Poutnik Hmm.... So why is $3d^{10}$ not counted towards to valance electron of Br since $3d$-orbital has a higher energy than $4s$-orbital?

@Poutnik Ohhhh... That makes much more sense now! Do you think you can point me to any textbooks/monographs/resources that discusses how energy level changes as the electron count increases?

Moved my comments in the answer.

Valence / outer electrons in context of quantum atomic models are those orbital groups(1) with the top total energies, involved or being able to be involved in interatomic bonds. It usually involves orbitals with the most spatially outstretched electron probability density. Usually ns, np, (n-1)d, (n-2)f, mostly just 2 of these 4. See also the (*) orbital footnote in this answer
See also wikipedia: Valence_electron
The chart overview of orbital energies from Sciencedirect

and the 4th period details from QM class
 from

(1) I meant it as the group, not necessarily the very top only.. Spacial extent is not the indicator, there is just the correlation. The energy order rules becomes complicated, starting with the 4th period of elements.
Electrons in folly occupied 3d orbitals have lower energy than 4s electrons. Orbital energy depends on electron configuration and the order frequently switches.


text:
Attempting to make Rochelle salt but obtained paste
I was attempting to make Rochelle salt by adding sodium carbonate to a mixture of cream of tartar and water, but ended up with a solid paste. I kept the cream of tartar/water mix boiling the whole time, which I guess was a mistake, and used a large pan so that there was only a thin depth of liquid. Things seemed to be going more or less as they should, until it suddenly became a paste.
I am letting it cool, but am wondering what I have obtained. Is it potassium sodium tartrate (without the tetrahydrate that would make it Rochelle salt)? If I re-dissolved what I have obtained in water could Rochelle crystals form upon cooling, or is it useless in that respect?

Free evaporation of solution  at rel. humidity >30% and <84% should be better.// The sudden paste formation can be fast crystallization of oversaturated solution, like Na2SO4 or Na2S2O3 can do.

Why do you think this paste is not Rochelle salt, to begin with?

Thank you for the replies! I've not done any chemistry for 20 years, and am really lacking in the fundamentals. I guess my worry is that the water evaporated before all the carbon was removed in the cream of tartar (which I assume is the point where no further bubbling occurs in when one adds sodium carbonate). I also don't know what happens if one keeps adding sodium carbonate beyond what is needed.

Rochelle salt (sodium potassium tartrate tetrahydrate) is relatively soluble in water (260 g/L at 0ºC, 660 g/L at 26ºC, and melts at 75ºC) (Ref 1), but cream of tartar (potassium bitartrate) is much less soluble (5.7 g/L at 20ºC, 61 g/L at 100ºC) (Ref 2).
Driving the relatively insoluble salt toward the more soluble product should be easy because the sodium carbonate will neutralize the acidity of the tartrate and conversion will be complete. Carbon dioxide will be eliminated even at room temperature, especially with stirring. The boiling and large pan are probably not optimal.
Weighing to very precise proportions would be helpful, but if that is not possible, I suggest you do the best you can to get about 400 g/L of product, starting at room temperature or a little higher - way below 75º. Dissolve some sodium carbonate (53 grams would be half a mole, if you use anhydrous; or use 84g (a mole) of $NaHCO_3$) in about 0.72 L water and add 188g cream of tartar, slowly, stirring to insure reaction and dissolution of the potassium bitartrate. You'll wind up with a mole of Rochelle Salt (282g) in  about 650 mL $H_2O$. This is well below the solubility limit. You can filter thru a coffee filter if you like. Then cool in the refrigerator; if you get no crystals, you can cool more in an ice bath and should get some, or, as a last resort, use the freezer.
However, I suspect you can resuscitate your original experiment more easily: add the minimum amount of warm water (with vigorous stirring) to dissolve your paste completely and either allow it to evaporate slowly or cool it in the refrigerator. I'm sure you will get crystals.
Ref 1. https://en.wikipedia.org/wiki/Potassium_sodium_tartrate
Ref 2. https://en.wikipedia.org/wiki/Potassium_bitartrate


text:
Purpose of Sodium Fluoride (NaF) in Sodium Hydrogen Sulphate (NaHSO4) acid pickling solution?
I am a proud home workshop electroplating fanatic, and the kit I sourced (from another country, with five weeks shipping) contained this bag of "Dry acide salts". I'm trying to source the chemicals locally to make up more of the solution and pickle bigger parts than the 5 litres (50g/litre) of solution I purchased provisions for.
Researching online, plenty of jewlers and people doing home metal preparation for plating are using Sodium Bisulphate but for some reason this kit contains Sodium Fluoride.
What purpose might it serve, can I ignore it? And in case it is useful, what is the ratio of NaF to NaHSO4?

Sodium fluoride may react with small amounts of insoluble compounds like silica $\ce{SiO2}$ or silicates present in glass. It will produce some fluorosilicates which are soluble in water, according to the global equation : $$\ce{SiO2 + 6 NaF + 4 NaHSO4 -> Na2SiF6  +  4 Na2SO4 + 2 H2O}$$ Matter of fact, this reaction is the sum of two intermediate reactions, namely :
$$\ce{1)~~~~ HSO4^- + F- -> HF + SO4^{2-}}$$
$$\ce{2) ~~ 6 HF + SiO2 -> H2SiF6 + 2 H2O}$$

The weak hydrofluoric acid produced helps with rust removal. Where I live, a dilute hydrofluoric acid is sold, in some grocery stores, specifically for rust stain removal. The product is Whink Rust Stain Remover. Does a great job dissolving aluminum metal, as I discovered as a kid with a basement lab in the old days.

Does Maurice's answer suggest that the correct ratio is 1.5:1? (6NaF+4NaHSO4?)

If it was me, I would low ball the NaF: it is toxic and you do not need hazardous waste issues anyway. The sodium bisulfate is commonly available as the chemical used to lower pH in swimming pools, so it safer than using sulfuric acid. Stay safe!

@LeeHambley No, he doesn't and I don't see a way to know it for sure without checking specific product data (perhaps someone might know some generic proportion, though).


text:
Tin(II) chloride electrolysis problems: (1) Why is the tin dendritic? (2) Why does the unconnected metal in the middle also act like an electrode?
I'm watching this tin(II) chloride electrolysis video:
https://www.youtube.com/watch?v=coPGazAaVBE&ab_channel=FlinnScientific
Two phenomenon puzzled me, and I couldn't find a good explanation on google-
(1) Why is the tin metal dendritic when it's formed on the cathode?
(Left hand side)
I searched for tin, and I only found that its crystal structure is "body-centered tetragonal" or " face-centered diamond-cubic", which I'm unable to link with the needle-like appearance. Is there a reason?
(2) Why does the unconnected paperclip act like an electrode?
(In the middle, it has dendritic metal on its right, and little amount of white precipitate on its left)
I thought Sn(II) ion would have to receive electrons to be reduced to tin metal. How can it be reduced in the middle of the petri dish, with no wire connected to it?

Regarding your second question, note that the tin chloride solution is a solution of an electrolyte, so it conducts electricity. It has resistance and conductance. So any point  in the solution is at a potential intermediate between that of the cathode and anode. The little wire near the center has a potential that is negative with respect to the anode, but positive with respect to the cathode. So tin dendrites grow from the right (anode-facing) side of the little metal wire toward the anode. And the precipitate forms on the left (cathode-facing) side, etc.

By the way, my hypothesis can be tested. Replace the 9 V battery by a DC power supply, adjustable from 0 V to 9 V or so. Start over with 0.5 V applied and observe that no tin dendrites occur. Increase the applied voltage in small increments, e.g., 0.2 V. Stop increasing the applied voltage when tin dendrites just start to appear slowly. Then put the little wire in the center. Nothing should happen: the potential is too low half-way between the cathode and anode. Now double the applied voltage or go a little higher. Dendrites should slowly grow on the little metal wire.

The demo of electric field on a small wire is amazing (Q. no. 2). I never thought of this experiment.

@EdV I thought of the potential difference, but I thought "electrolyte is conducting electricity" means they go to electrodes and receive or provide electrons. Do they really carry the current?
Doesn't the Sn(II) ion need to receive electrons somehow, to be reduced to tin metal? What brings electrons to the paperclip ( could it be other Sn(II) ions, which turned into Sn(IV) ions?)

Sure the ions carry current! I have never done this demo before, but I have a jar of granulated tin, hydrochloric acid (“muriatic acid”) is sold in hardware stores where I live, and I have adjustable DC power supplies, etc., so I will try the experiment next week (this week will be busy for me). As for your question about the paperclip, perhaps it is comproportionation? This will be a fun little experiment, however it turns out! Thanks for asking the question, which I upvoted first!

Sorry, but I would still like to know the detail: where does the Sn(II) ion get its electrons? Is it other Sn(II) ions? If this is the case, why does it need the paperclip? Wouldn't they just meet in the middle of the solution (without the paperclip), and engage in redox reaction?

@EdV The more pertinant point is that Tin(s) is a good conductor. Solution conductivity needs to be higher by an order of magnitude or so, in order to get uniform growth.

@StianYttervik Could you please elaborate the point? Is there some paper review or even books that are on this( the growth of metal from electrolysis) ? I would like to know the detail.

@Wang Just something I remember from electrochemistry. The idea is that the electrochemical reaction happens where the potential is greatest. If your solution conductivity is less than the solid, the point of preference is the one closest to the anode: You'll get dendrites. If, however, your solution has greater conductance than the solid material you'll get growth in a shape that mimics the anode surface. Flat anode - flat deposition. ish.

@StianYttervik Thanks, but I said nothing about dendrite growth versus uniform growth. But speaking of this, the video presenter make a point of tin dendrite growing at right angles from existing dendrites. In a silver tree demo, the silver ‘dendrites’ are more fractal-like. So what drives the right angle tin dendrite growth habit? I would have expected growth from the sharp points rather than the non-sharp sides, as per the electric field issue.

@EdV A-ha. I see now. I misunderstood you. My apologies. Though, since the comment garnered some followup replies I'll let it stand. Regarding the fractal shape that is possibly a preferred orientation of crystal growth, in it's somewhat rare tetragonal body centered arrangement.

I have done this experiment plenty of times with my students. It is one of my favorites. The results are surprising, and often difficult to reproduce. It depends on the current. With a low current $(I < \pu{0.05 A}),$ long, bright and thin metallic needles are obtained starting from the cathode. They sometimes touch one another in the middle of the box, like in your experiment.
Anyway, after a couple of minutes, the needles grow so much that they join the electrodes. This must be avoided, because it stops the electrolysis process. In this case, the “bridge” should be broken by hand with a glass stick.
If the current increases $(I > \pu{0.1 A}),$ the needles are not so thin. On the contrary, they tend to be tightly packed. If the current is rather high $(I > \pu{0.2 A}),$ they even look like gray and dull moss… Deceiving!
On the anode, the $\ce{Sn^{2+}}$ ion is oxidized into $\ce{Sn^{4+}}$ according to
$$\ce{Sn^{2+} -> Sn^{4+} + 2 e-},\tag{R1}$$
but this ion is quickly hydrolyzed into weekly acidic solution, producing a white deposit of $\ce{SnO2·nH2O}$ according to
$$\ce{Sn^{4+} + $(2 + n)$ H2O -> SnO2·nH2O + 4 H+}.\tag{R2}$$
It is possible to prevent this white precipitate by dissolving $\ce{SnCl2}$ in a $\pu{2 mol/L}$ $\ce{HCl}$ solution, which reverses the last equation.
Something interesting can be observed at the end, when you think the experiment is finished. At this moment, be careful. Don't touch the solution, but disconnect the wires connecting the electrodes to the power supply. And quickly insert these plugs into a $\pu{50 mA}$ ammeter. You'll see that some current is produced for a couple of seconds. Your small tray is working like a galvanic cell!
Last but not least. At the end, when the tin has invaded the whole solution (except a tiny zone around the anode), try to save the tin from the tray with your hand (gloves!). Press it on a blotting paper or a filter paper. When dry put it in a tiny test tube, and heat it with a Bunsen burner. It quickly melts, and the liquid can be thrown on the lab bench: it makes a bright and shiny metallic stain or platelet. Looks like silver! Now try to bend this thin platelet with your hands near the ear. It makes a surprising noise, showing that bending the tin platelet breaks tiny tin crystals.

(+1) Nice! How did you learn about the short-lived galvanic operation?

I did it so many times, for many years, that one of my students once asked me to try !

@Maurice, Since you have been teaching this experiment for ages, do you recall  the second part of the experiment where the metal rod is placed in the center and then electrolysis occurs. It is a beautiful experiment. I am wondering if Ostwald ever noted it in his book on Electrochemistry: Theory and its history.

Nice observation on the different tin appearance at different current, and the galvanic cell! Do you know the theory on the appearance difference at different current? I would like to read that.

@Wang. I don't know the theory on the appearance difference at different current. It may be due to a competition between the Brownian movement of the ions, their migration rate, their ionic mobility, and the attraction due to the electric field, which increased when the inter electrode distance decreases.

Is the sound at the end related to "tin cry"?

@WaterMolecule Yes. And other metals, such as zinc and indium, “cry” as well.


text:
Appearance of drying hemoglobin-and-salt droplets
I initially posted these or similar images at the bio site because they looked to me like organic/mold contamination in a slide (low power) containing hemoglobin and salt. As the slide dried these fibers emerged and joined into arches on the drying edge, eventually forming thorn-like nests draped with heme-containing debris and sometimes NaCl crystals. A seemingly astute commenter there suggested it was "cracks," and the speed at which these things grew made me reconsider. Now I think they are inorganic and caused by dessication--but not cracks exactly. More like a condensation of the protein and salt into forms that resemble hyphae. I saw basically the same with lysozyme and salt. Does anyone have experience with these structures and an idea of what they are?
The top image is shortly before disintegration due to drying out. The bottom is of the initial fiber-like arches that appear near the periphery of the slide. Power is 150x, the colors are as seen, the red material draped over the thorns in the top image are I think just heme-containing debris.

Strictly a guess--I think they are tunnels that form in the liquid as entropy forces a rearrangement due to loss of water; cylindrical cracks, as it were.


text:
What do sequence numbers in PDB files actually mean and why don't they match the sequence?
I am studying the 3D structure of the LDH from x-ray crystallographic imaging I was pointed to from Is there any stereospecific enzyme in PDB that catalyzes an anabolic reaction and has an entry showing both reactant ligands?. It's 4ND4 and 4ND3 in the PDB and the paper about it is Cook, Senkovich, et al. Biochemical and structural characterization of Cryptosporidium parvum Lactate dehydrogenase. And I am trying to exactly link the detailed molfile model that I have extracted (and touched up) with the help of How do I obtain a 3D MOLFILE for the lactose synthase structure, especially the catalytic center?.
The key picture in the article is this:

When trying to exactly relate this to the actual sequence of the LDH enzyme I am finding puzzling issues. 4ND3 and 4ND4 refer themselves to UniProt Q5CYZ2. But the sequence is different. This raises a bunch of questions that perhaps I do not understand about PDB files in general, hoping someone could shed a light on this. From the PDB file https://files.rcsb.org/view/4ND3.pdb:
REMARK 999 SEQUENCE
REMARK 999 PEPTIDE SEQUENCE FOR THIS ENTRY WAS TRANSLATED FROM DNA SEQUENCING
REMARK 999 OF ACTUAL CLONE USED FOR PROTEIN EXPRESSION. AUTHORS DO NOT KNOW IF
REMARK 999 THE SEQUENCE DIFFERENCES ARE PCR ERRORS OR MUTATIONS CORRESPONDING
REMARK 999 DNA POLYMORPHISM OR THERE WERE ERRORS IN PROTEIN DATABASE.
DBREF  4ND3 A   17   337  UNP    Q5CYZ2   Q5CYZ2_CRYPI    17    337
DBREF  4ND3 B   17   337  UNP    Q5CYZ2   Q5CYZ2_CRYPI    17    337
SEQADV 4ND3 ALA A  202  UNP  Q5CYZ2    VAL   200 SEE REMARK 999
SEQADV 4ND3 ALA B  202  UNP  Q5CYZ2    VAL   200 SEE REMARK 999
SEQRES   1 A  321  MET ILE GLU ARG ARG LYS ILE ALA VAL ILE GLY SER GLY
SEQRES   2 A  321  GLN ILE GLY GLY ASN ILE ALA TYR ILE VAL GLY LYS ASP
SEQRES   3 A  321  ASN LEU ALA ASP VAL VAL LEU PHE ASP ILE ALA GLU GLY
SEQRES   4 A  321  ILE PRO GLN GLY LYS ALA LEU ASP ILE THR HIS SER MET

So it references Q5CYZ2, and the sequence there is:
        10         20         30         40         50
SEFLFSKHLF IFFILKMIER RKIAVIGSGQ IGGNIAYIVG KDNLADVVLF
        60         70         80         90        100
DIAEGIPQGK ALDITHSMVM FGSTSKVIGT NDYADISGSD VVIITASIPG
       110        120        130        140        150
RPKDDRSELL FGNARILDSV AEGVKKYCPN AFVICITNPL DVMVSHFQKV
       160        170        180        190        200
SGLPHNKVCG MAGVLDSSRF RTFIAQHFGV NASDVSANVI GGHGDGMVPV
       210        220        230        240        250
TSSVSVGGVP LSSFIKQGLI TQEQIDEIVC HTRIAWKEVA DNLKTGTAYF
       260        270        280        290        300
APAAAAVKMA EAYLKDKKAV VPCSAFCSNH YGVKGIYMGV PTIIGKNGVE
       310        320        330
DILELDLTPL EQKLLGESIN EVNTISKVLD NAPAAGA

and what I'm finding is none of these sequence numbers match exactly starting from I32. You can see in the UniProt sequence there is I at 31, not 32. But it's not a constant offset either, D53 is at 51. And what about the prefix SEFLFSKHLFIFFILK before the MIERRKIAVI... in the PDB file and associated FASTA sequence?
The answer seems to lie in the ATOM list, which has in the 6th column a sequence number, that starts with 17, and that means it skips right over the SEFLFSKHLFIFFILK to start with M. But why? UniProt isn't identifying SEFLFSKHLFIFFILK as a signal peptide or anything. But then Q5CYZ2 is also not really vetted, but it also seems long to me to be a signal peptide.
SCALE3      0.000000  0.000000  0.005376        0.00000
ATOM      1  N   MET A  17      32.275  60.760  98.001  1.00 65.51           N
ATOM      2  CA  MET A  17      30.819  60.627  98.290  1.00 65.33           C
ATOM      3  C   MET A  17      30.216  59.589  97.341  1.00 62.08           C
ATOM      4  O   MET A  17      30.381  59.683  96.125  1.00 64.10           O
ATOM      5  CB  MET A  17      30.122  61.990  98.139  1.00 68.91           C
ATOM      6  CG  MET A  17      28.988  62.249  99.129  1.00 72.14           C
ATOM      7  SD  MET A  17      27.333  61.845  98.525  1.00 76.22           S
ATOM      8  CE  MET A  17      26.946  63.315  97.574  1.00 75.10           C
ATOM      9  N   ILE A  18      29.536  58.595  97.907  1.00 56.99           N

now, it gets even weirder, in this list there are gaps, like here it skips over 21 right from 20 to 22. And the same phenomenon happens at the 47, 82, 104, and who knows later too.
ATOM     35  NH1 ARG A  20      18.415  54.478  92.556  1.00 41.01           N
ATOM     36  NH2 ARG A  20      18.134  54.811  94.808  1.00 41.94           N
ATOM     37  N   ARG A  22      25.730  54.160  90.284  1.00 34.87           N
ATOM     38  CA  ARG A  22      26.279  53.881  88.963  1.00 33.99           C
...
ATOM    212  O   ALA A  46      20.706  50.735  87.589  1.00 25.85           O
ATOM    213  CB  ALA A  46      23.964  51.086  86.776  1.00 28.99           C
ATOM    214  N   ASP A  48      21.675  52.759  87.757  1.00 29.47           N
ATOM    215  CA  ASP A  48      20.536  53.481  87.203  1.00 30.89           C

But why? Why would it say that A is at 46 and D at 48, when UniProt and their own FASTA sequence say that D follows A directly at that point? What is the meaning of this gap? Is it merely an arbitrary numbering in the PDB file? So that if I want to associate to a UniProt sequence, I would have to re-number and have to say I31 and D51 instead of I32 and D53? Is that it? Or are they claiming there are actually a missing amino-acid here which they haven't identified?
I would guess I'd have to re-number and there is no actual missing AA.
But now it gets even stranger:
ATOM    400  CE  MET A  72       4.872  44.082  88.838  1.00 33.49           C
ATOM    401  N   PHE A  73A     11.326  42.924  91.678  1.00 31.09           N
ATOM    402  CA  PHE A  73A     12.685  42.370  91.742  1.00 31.48           C
ATOM    403  C   PHE A  73A     13.463  42.779  92.989  1.00 31.81           C
ATOM    404  O   PHE A  73A     14.583  42.317  93.195  1.00 32.21           O
ATOM    405  CB  PHE A  73A     13.489  42.810  90.520  1.00 30.78           C
ATOM    406  CG  PHE A  73A     12.939  42.321  89.212  1.00 30.96           C
ATOM    407  CD1 PHE A  73A     12.580  40.990  89.041  1.00 30.80           C
ATOM    408  CD2 PHE A  73A     12.831  43.184  88.130  1.00 31.13           C
ATOM    409  CE1 PHE A  73A     12.090  40.542  87.827  1.00 31.23           C
ATOM    410  CE2 PHE A  73A     12.347  42.740  86.913  1.00 30.95           C
ATOM    411  CZ  PHE A  73A     11.978  41.416  86.761  1.00 31.09           C
ATOM    412  N   GLY A  73B     12.888  43.658  93.806  1.00 32.52           N
ATOM    413  CA  GLY A  73B     13.569  44.166  94.988  1.00 31.46           C
ATOM    414  C   GLY A  73B     14.728  45.089  94.656  1.00 31.53           C
ATOM    415  O   GLY A  73B     15.667  45.211  95.440  1.00 31.87           O
ATOM    416  N   SER A  74      14.668  45.740  93.496  1.00 30.43           N

suddenly there are two at position 73, F and G. What does that mean?
I am thinking that the truth is these numbers are all meaningless and I can just re-number them and in reality we have M, F and G, S in that sequence, no variance or alternative at position 73, the FASTA sequence has -VMFGST-.
So what is going on here? Numbers are arbitrary and if I want actual meaningful sequence numbers I just re-number. Or is there a meaning to these gaps and double assignments 73A, 73B? And even if there is no actual meaning, I'd sure like to know why they did it in this complicated way? I guess it is an artifact of the technology somehow?

It's not uncommon to cleave off unstructured regions in order to improve expression or crystallization or stability. Presumably someone found that leaving off 1-16 gave higher yield of protein or better crystals.

For the other numbering issues, they note that their sequence is derived directly from the DNA construct used to produce it, which was made by PCR amplification of something that is supposed to match Q5CYZ2. The difference could be errors in the original Q5CYZ2 sequence deposition or arose during their manipulation of it. In situations like that, it is common to adjust the numbering based on alignment so that residue numbers correspond between two structures as well as is possible.  Because LDH has been studied for so long, they may also attempt to conform to historical residue #s.

As @Andrew mentions in the comments, there is a weird tradition for depositing enzyme structures where the authors adjust the residue numbers so that the active site residues have the same residue numbers as famous related enzymes. For instance for serine proteases, the catalytic triad would always have the same numbering. If you do that, you have to introduce gaps in the numbering or residue numbers like your 73A and 73B (for deletions and insertions with respect to the aligned reference sequence).
Here is the active site of a related thermophilus structure. Notice that the active site residues are R171 and H195 as well. If you look at the underlying sequences, the numbering might be different.

Source: https://hal.archives-ouvertes.fr/hal-03011827/document (googled "R171 N140 H195 LDH")
The reference sequence is that of Squalus acanthias, shifted by one (see Abad-Zapatero, C., Griffith, J.P., Sussman, J.L., Rossmann, M.G. (1987) J Mol Biol 198: 445-467). The historical PDB structures are 1LDM, 3LDH, and 6LHD (and obsolete 2LDH and 4LDH). There is also a historic structure of protein from pig (5LDH) from the same lab. These are the structures that led to the discovery of the Rossmann fold.
As for the mismatch of the Uniprot sequence and the PDB sequence at the N-terminus, the source of the Uniprot sequence is a shotgun DNA sequencing project, and they did not know where the translation start is. Usually, these protein sequences should start with a methionine.
Here is the sequence alignment with a different related sequence obtained with a different method which supports methionine as the first amino acid in the sequence. Notice the Iowa strain has two mutations, R->K and K->E near the end of the sequence.

Wow, this is super interesting! It seems to be saying that different enzymes use similar motifs so similar that a few gaps and additions turn one into the other. Or maybe not different enzymes, rather, because your other one is also LDH. But I guess that is what is necessarily conserved even as mutations may occur. And the fact we can use the same conventional site numbering really makes this point very clearly.


text:
Does pressure affect rate constant?
I would like to know if pressure affects rate constant.
Suppose pressure affects the rate constant. Does it depend on partial pressure (like: if I add an inert gas to the system), or only increasing the pressure of at least one of the reagents?

In solution there is a pressure dependence via the transition state as $d\ln(k)/dP=-\Delta V^/RT$, where $\Delta V^$ is the difference in molar volume between transition state and reactants. Typically thousands of bar are needed and the volume change is $\approx 5 \to 15$ mL/mole so small. In peroxide radical decomposition , pressure is observed to affect reaction via packing in solvent cage.

Ideally, pressure affects just the reaction rate, not the rate constant, like for $\ce{A + B -> C}$:
$$\frac{\mathrm{d}C}{d\mathrm{t}} = k_1 \cdot p_\mathrm{A} \cdot p_\mathrm{B}$$
In case the are some non-linear effects and the reaction rate does follow the simple equation like above, with integer/half-integer exponents, there are possible several formal aproaches:
Formulating reaction rate function like:
$$\frac{\mathrm{d}C}{d\mathrm{t}} = k_2 \cdot f(p_\mathrm{A}, p_\mathrm{B})$$
or, if deviation is just slight, it could be written as:
$$\frac{\mathrm{d}C}{d\mathrm{t}} = k_3(p_\mathrm{A}, p_\mathrm{B}) \cdot p_\mathrm{A} \cdot p_\mathrm{B},$$
or there can be derived different $k_1$ values optimal for particular partial pressure ranges.

Thank for your replay
So to confirm, the rate constant may alter slightly in some reactions with having different partial pressures.
Aslo I would like to be 100% sure: concentrations cant affect the rate constant

I would rather say the rate constant at given temperature does not change, unless the used reaction rate equation is just an approximation of the true reaction rate equation .

I once had a question that asked which the rate constant was NOT dependent on: Pressure, Temperature, catalyst, concentrations.
I know rate constant was dependent on temperature, catalyst. I believe that rate constant did not depend on concentration, so I thought that concentration should be the answer but then that will mean rate constant pressure is dependent

Well, you can express gas composition in both equivalent ways, in partial pressures or in molar concentrations. Rate constants cannot be dependent on one set of the quantities and independent on the other.