text:
Meaning of w in the equation w = -PΔV
The equation for PV work is given by
$$w = -P_\mathrm{ext}\Delta V$$
What work does $w$ imply - the work done by the system or on the system?
Considering that the convention in chemistry is to consider things from the point of view of the system, $w$ should denote the work done by the system. However, by that definition, work done by the system would be positive during compression, rather than the work done on the system.
So does $w$ actually denote work done on the system?

Does this answer your question? Why do chemistry and physics have different sign convention in thermodynamics?

There exist two conventions in thermodynamics, one used by theoreticians, one by engineers.
The theoreticians think that all energies entering the system must be positive, heat or work. So when a work is done on a gas system, the gas is compressed, and the work is positive. But as the volume of the gas decreases in this compression, $\Delta V = V_{final} - V_{initial}$ is negative, and if we want the work to be positive one must write : $\Delta w = - p\Delta V$.
The engineers have a different approach. They think that the gaseous system is part of a machinery with pistons and connecting rods. We heat this gas so that the heat added to the machinery is positive. But the piston must produce some useful work, which is positive. And when it does it repels the piston, so that the volume increase : $\Delta V$ is positive and the work is also positive $\Delta w = + p\Delta V$.
Fortunately both approaches leads to the same final result. For theoreticians, the internal energy $U$is the sum of all that enters the system. $\Delta U = \Delta Q + \Delta w = \Delta Q -p\Delta V$
For the engineers, the internal energy is the fraction of the energy (heat) that is "lost" in the system, and that has not been used for producing work. So $\Delta U$ is : $\Delta U = \Delta Q - \Delta w = \Delta Q - p\Delta V$
So whatever the convention for defining the sign of the work, the change of internal energy is the same.


text:
When does copper leach to water? Conflicting faqs
I have heard conflicting reports:

Copper leaches to water only when the water is high in acid or the Copper is corroded.
Copper leaches to water any time, especially when it is new copper or the water is hot.

Which one would be correct?

It depends on what you consider as leaching quantitatively.

The second answer would be correct in an analytical context as the auto-ionization of water decreases in pH as temperature increases. And new copper does not have the surface of reacted material to shield the metal underneath, as such, scenario 2 always occurs, scenario one occurs faster, however, it's conditional statement that the water be acidic and the copper be corroded are invalid.
Cheers,

Since #2 is true, how do cities that use copper piping and keep the water from getting to much copper?

The amount of copper dissolved is not toxic, just adds the characteristic “city water” flavor

What about a espresso machine with a copper boiler? Would the heat make the water, along with sitting water for hours,  absorb more copper and be toxic?

Don't they sell copper tea kettles? This implies copper does not dissolve appreciably, even in hot water.
But have you seen houses with those copper rain gutters and downspouts that pour out rainwater onto concrete patios? You know, where the concrete turns that nasty-looking green color? Could be due to acid rain. Now acid rain may be acidic, but I would not say highly acidic. Maybe copper just dissolves a little and gets immediately precipitated by the alkaline concrete. So, lots of copper surface and lots of water can give evidence of copper dissolution over some period of time.
So, different results; but that's life.

Copper roofs do not lose copper; they build up a greenish patina which is a complex combination of copper oxides ,carbonates , hydroxides  , etc.

I'm surprised that copper stains are not more prevalent, but the second picture on google showed stains on concrete from runoff that looked like from a downspout. I added it to the answer.

source: Stack Exchange

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

sort: votes


text:
Will a gold coin react with an aqueous magnesium nitrate?
I really have no work to show, I don't even know how to approach answering this... so if someone can push me in the right direction, thanks.
But I did have the idea that it might have something to do with how gold isn't a strong reducing agent.

Who downvoted :( ... Whoever did, was it a bad question?

Why checking a gold coin with magnesium nitrate ? This compound has not too many applications, not too many uses, not too many interesting properties

The most common solution known to react with gold is a 1:3 mixture of nitric acid and hydrochloric acid classically known as "Aqua Regia." As such, it is unlikely that magnesium nitrate alone will react with gold as it is very stable.

OK. That makes sense... Does gold's stability correlate to its oxidizing abilities?

They are synonymous; as you stated earlier, gold is a bad reducing agent, meaning it is not easily oxidized, in a metallurgical sense, this means it is a very stable metal.

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


text:
Is this calculation of the amount of required hydrogen gas correct?
I was a bit hesitant to post this question for it may appear like a "homework help" type of question. But upon rethinking it felt like a genuine question. I was studying stoichiometry with my son (from the OpenStax Chemistry 2e book) and suddenly I had an idea to use these calculations in a real chemistry experiment that I am going to perform at home (of course, I truly believe in a "safety first" approach). I wrote it all up here on GitHub, but reposting here to keep it all at one place:
A practical problem: It was decided to inflate a common latex balloon to 5cm radius by the hydrogen gas produced during a single-displacement reaction:
$$\ce{2HCl(aq) + Zn(s) -> ZnCl2(s) + H2(g)}$$ that uses an N/10 i.e. 0.1M HCl solution (available in many drugstores, typically in $\pu{200ml}$ bottles). How much reactants are needed?
Solution: We need to inflate the balloon to about $\pu{5cm}$ radius. Assuming the balloon to be a sphere gives us the volume of balloon, $V_b$, as:
$V_b = \frac{4}{3}\cdot \pi r^3 = \pu{523.599cm^3}$.
Assuming the density of hydrogen at STP as 0.0000899 $\pu{g/cm^3}$, the mass of hydrogen is $\pu{523.599 \times \space 0.0000899 g} = \pu{0.0471 g}$. Since 1 mol $\ce{H_2(g)}$ has a mass of $\pu{2 \times \space 1.008 g} = \pu{2.016 g}$, we need
$\pu{\frac{0.0471}{2.016} mol} = \pu{0.0233 mol}$ $\ce{H_2(g)}$.
Balancing the reaction, we see that to produce 1mol $\ce{H_2(g)}$ we need 2 mol HCl(aq) and 1 mol Zn(s).
This means that to produce 0.0233 mol $\ce{H_2(g)}$ we need $2 \times 0.0233 = \pu{0.0466 mol}\ \ce{HCl(aq)}$. We have a 0.1M HCl (aqueous) solution which means that $\frac{\text{mol of solute}}{\text{litres of solution}} = 0.1$. Clearly, we need $\frac{0.0466}{0.1} = \pu{0.466 L}$ or $\pu{466 ml}$ HCl.
The amount of zinc needed = $0.0233 \times \text{molar mass of zinc} = 0.0233 \times 65.38 = \pu{1.523 g}$.

Looks good to me. I would have calculated 523.6/22400= 0.023375 mole H2 for simplicity, but you got it right.

editing tip: use \pu{} for physical units and \ce{} for chemical formulae/equations.

@Kedar do not apologize to anyone or feel guilty about posting a genuine question. You are trying to help your son and this should be enough to convince anyone interested in science.

Kedar's calculation may be right. But it must be realized that the reaction will be rather slow. Even with a $1 M$ solution, the reaction is slow. The reaction of $1.5$ $\pu{g}$ metallic zinc with such a diluted $\ce{HCl}$ solution ($0.1 M$) will last more than one hour. Anyway, in order to inflate such a balloon a higher pressure than the atmospheric pressure must be applied. If not, the balloon will keep its original dimension. How to increase the pressure of $\ce{H2}$ in such a balloon for inflating it, if the hydrogen is produced by the reaction of zinc on $\ce{HCl}$ ?

That's an excellent point @Maurice! I had not thought about actually inflating the balloon as much! Looks like doing such an experiment is going to harder (in a homelab setup) than I initially thought. I have found a couple of resources: https://www.engr.uvic.ca/~struchtr/2002balloons.pdf and https://physics.stackexchange.com/a/362449/143069 which I will now study. Science is never easy :-| ...


text:
Are precipitation and crystallization both analogous between Chemistry and Meteorology?
A comment under the Space SE question Is there really precipitation on Mars? Does frost count? brings up the use of the term "precipitation" in chemistry and we wonder if precipitation in weather and in chemistry are in fact analogous insofar as they both start as small particles suspended in the 3D medium (atmosphere or solution) and only fall to the ground or the bottom of the container if/when sufficiently large to make headway against Brownian or other motion.
In chemistry, on Earth in gravity, if a material from solution conformally, over all surfaces vertical and horizontal (e.g. the sides as well as the bottom of a container) is that still sometimes considered precipitation? Or is that crystallization from solution, and analogous to frost formation but not snow or rain?
Images of frost formation: http://www.its.caltech.edu/~atomic/snowcrystals/frost/frost.htm

It would seem that the accepted opinion is that Mars had surface water in the past. So precipitation could have taken place.

@MaxW thanks but let's keep that conversation on the other post (where it is most welcomed!) and not here. This is strictly a question of terminology.

In chemistry, on Earth in gravity, if a material from solution conformally, over all surfaces vertical and horizontal (e.g. the sides as well as the bottom of a container) is that still sometimes considered precipitation? Or is that crystallization from solution, and analogous to frost formation but not snow or rain?

Precipitation, as typically used an understood in chemistry, is really a very very old terminology (~ several hundred years). It refers to the formation of a solid by adding or mixing chemical reagents or even by cooling. There is no such requirement that the precipitate has to settle to the bottom. For example, colloids may not settle for ages. Precipitation involves crystallization, but not all precipitates are crystals (organic substances can be precipitated out by adding suitable solvents to a solution).
Now there are special ways to generate precipitates in solution. There was an experiment which we taught in classical analysis. The experiment goes like this. Add urea to an iron salt. No precipitate forms and there are no seeds to begin with. Let it boil slowly. With the passage of time, the entire solution becomes basic and it would uniformly produce precipitate iron hydroxide.  Since it was a in situ chemical decomposition of urea to produce iron hydroxides in solution, this process occurs very uniformly in the solution i.e., there is no localization where two reagent were being mixed.  This yielded uniformly sized particles. Such a process is called homogeneous precipitation in solution. Contrast this situation where we would add NaOH solution to iron salts. It will be impossible to avoid localized mixing.
Meterologists use the term precipitation as well, but as per the unabridged Oxford Dictionary (can't provide a link as it is by subscription), the full word is atmospheric precipitation.
In case you are interested, the etymology is

Etymology: < classical Latin praecipitāt-, past participial stem (see
-ate suffix3) of praecipitāre to throw or cause to fall headlong, to ruin, destroy, to fall headlong, to suffer ruin, come to grief, to
hasten, to rush, in post-classical Latin also to cause to be deposited
as a solid from a liquid solution (a1490, 1652 in British sources) <
praecipit- , praeceps (adjective) headlong, sheer (see precipe n.).
Compare Middle French, French precipiter , précipiter precipit v.

Thank you for this informative answer! If my take-home summary of it were the following? "In Chemistry the term "precipitation" is quite broad and inclusive, but when used in meteorology we can't necessarily draw any specific analogy."

Yes, correct. In chemistry, it is a wide and broad term.

This is fun, so I've just asked In chemistry do we speak of precipitation in media other than "normal" liquids?

See precipitation and crystallization in chemistry context, and precipitation in meteorological context.


text:
Can precipitation occur in states of matter other than liquids?
This answer to Are precipitation and crystallization both analogous between chemistry and meteorology? discusses the history and use of the term "precipitation" in chemistry.
From what I remember in school I was introduced to precipitation in the context of reactions in liquid media and usually water.
In meteorology the term refers to a phenomenon in air where droplets or crystals form from some molecules mixed with others. I don't know if we can call the nitrogen/oxygen mix a solvent and the water vapor the solute, but some of the resulting droplet suspensions are indeed called aerosols.
But my question here is constrained to chemistry. Do we speak of precipitation in media other than typical liquids? Are there gaseous precipitations? Are there precipitations in supercritical situations? Do they happen in glasses or in (presumably pretty hot) solids?

I've cleared the comments on the question about whether the two questions should be merged into one. Personally, I think that they are better asked separately, as was done. I do recognise that's subjective and open to debate, but if anybody wants to have a lengthy discussion, please consider doing it in chat.

As stated in my previous answer on precipitation, formation of precipitates in supercritical is a common issue. Ask those who do sub/supercritical fluid chromatography. Sometimes you inject a methanol soluble compound into a chromatography column and the mobile phase is supercritical carbon dioxide (often mixed with a small amount of methanol) and the compound precipitates on the head of the column. So there is no restriction on using the term precipitate in fluids other than liquids.
Wikipedia also talks about precipitation in solids as

In solids, precipitation occurs if the concentration of one solid is
above the solubility limit in the host solid, due to e.g. rapid
quenching or ion implantation, and the temperature is high enough that
diffusion can lead to segregation into precipitates. Precipitation in
solids is routinely used to synthesize nanoclusters

The issue of calling a mixture of gases as a solution or not, it purely a matter of semantics. There is no rule or law which restricts that the solvent and the solute must have (strong or a certain type of) interaction in order to be labelled as a solution. Even gases interact with other gas phase molecules. Microwave spectroscopists know that very well. Many authors use the term gaseous solutions. Even one IUPAC's terminology related document uses this term. Heavens don't asunder apart if a trivial gas mixture is termed as a solution. Keep in mind that if you look up IUPAC definition of solvent, it does not include the usage of solution for gases. Only solids and liquid solutions are defined. So it is better to follow IUPAC's nomenclature.
In short, keep the meteorological precipitation away from chemistry.

In "chemistry", we generally have a good idea of what we mean by precipitation: something falls down out of a liquid and we filter it off. Then we extend the definition to meteorology for rain falling out of the sky. And although precipitation in solids has been mentioned, its importance has not. Well, maybe because much of the solid precipitation occurs in metallurgy - but, really, it all depends on the chemical interactions of individual atoms and largely needs to be explained with chemical insight!
For instance, precipitation hardening steels are a family of corrosion resistant alloys which can have three or four times as great yield and tensile strengths as the common types 304 and 316. They are used in the oil and gas, nuclear and aerospace industries. The alloys contain Cu, Mo, Al and/or Ti in addition to Fe, Cr and Ni. https://www.twi-global.com/technical-knowledge/job-knowledge/precipitation-hardening-stainless-steels-102
The first such alloy, 17-4 PH (17% Cr, 4% Ni), was developed by Armco Steel in the 1960's.
The chemistry involved can be considered to involve a solute in the solid Fe, Cr, Ni solvent, which doesn't flow much at all, but which allows migration of the solute, which is often C. Cool the "solution" slowly, and you get a precipitate of crystalline  cementite $\ce{Fe3C}$; cool the metal rapidly (quench), and you get a frozen, more or less homogeneous alloy. Varying successive heat treatments can cause partial solution or partial precipitation, all of which can have enormous effects on mechanical properties, machinability, weldability and corrosion resistance.

Don’t forget precipitation of gas bubbles in nuclear reactor materials, just to add another twist. Much of OPs problem seems to come from the general vs technical uses of ‘precipitation’.

Precipitation occurs in most metals.  It is very important in : steels, nickel alloys , aluminum alloys . and - ringer-  Be Cu. Precipitation hardening  of low  carbon  ( weldable ) steels has revolutionized the pipelines ; using Cb , = Nb for younger  engineers and Europeans .

This topic and the related one in your other question had 2 aspects: phenomenological and terminological ones.
Meteorological and chemical precipitation are different phenomena.

The former is the water phase change from gas to liquid or solid, that is (almost) independent on air presence, with vapour saturated pressure criteria, usually considered in context of falling on Earth surface.
The latter is based on limited solubility criteria of a compound in a liquid.

Solubility of a compound in liquids depends on this compound, the liquid and temperature.
The maximum compound concentration  in a gaseous phase depends on this compound and temperature, but does not ( almost (*) ) depend on the kind of gas, its pressure or presence at all.
As gases do not dissolve compounds, but are just bystanders, we cannot speak about dissolving and precipitating from them in chemical sense.
Precipitation is applicable to solids with difficulty, as processes of dissolving and precipitating are mostly kinetically frozen.
Diffusion process in hot solids is definitely considerable, OTOH it is extremely hard to observe it. But it does happens and is observed in some circumstances, like steel or alloy processing.

(*) Almost: Saturated vapour pressure slightly depends on the overall pressure by a generic way. Compound phase equilibrium implies equality of compound chemical potentials in both phases. Liquid chemical potential slightly increase with total pressure, what is followed by slightly increased saturated vapour pressure.
This effect may be dramatic near liquid critical conditions, as seen for $\ce{N2O + O2}$ mixture Entonox, where the mixture is at the storage conditions gaseous, but pure $\ce{N2O}$ would condensate.
See also SE article and Wikipedia.
Other then this, gases do not affect the equilibrium vapour pressure, but may cause equilibrium disturbance by side reactions, turning vapour molecules into different compound or a compound adduct.
The extreme example is putting liquid ammonia and hydrogen chloride at $\pu{-40 ^{\circ}C}$ side by side. Both vapours would never reach equilibrium due forming solid ammonium chloride.

Molecules moving by free flight in gases is a well known fact, which you can easily find and would not ask others to find it for you. For flying in vacuum, molecules need not other molecules flying nearby.
E.g. the mean free flight path of molecules in air is typically 70 nm, what is 2-3 orders more than molecule size.

Note that I was serving in the Czechoslovak army as an enlisted air force meteorologist in 1989-1990.

Excellent answer, thank you! fyi as a follow-up I've just asked Why exactly does high pressure (~72 bar) nitrous oxide's critical temperature drop so far when oxygen is add to achieve a 50% v/v mixture? (Entonox) which links back here.

I think you mean the mean free path of molecules in air is 70 nanometres! I was trying to picture what it would look like if I had to wait for oxygen molecules to fly into my nose so I could breathe.

@user253751  Surely nm, it was easily missed typo. :-)


text:
How does the Hammett acidity function work and how to calculate it for [H2SO4] = 1,830? Any references?
I'm modelling the concentrations in the Belousov Zhabotinsky reaction. For this model, I need to convert [BrO3-] to [HBrO3]. I discovered in several papers that the following correlation applies:
$$ [\ce{HBrO3}] = \frac{h0}{h_0+0.2} * [\ce{BrO3^-}]_0 $$
See for example this dissertation on page 28 or this paper on page 2.
I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M).
My $[\ce{BrO3^-}]$ was 0,2411 M.
I researched the Hammett Acidity function, but I still don't completely understand the concept. What I do know is that $h_0$ substitutes for $[H+]$ and that $h_0$ has the unit of M.
Can anyone explain the concept of this function in a bit more detail and explain in what way it exactly relates to pH (and therefore to H+)?
On the internet I found this paper, which says that $h_0$ = 3 when $[\ce{H2SO4}]$ = 1.5M. Are there any other tables or references with $h_0$ related to the concentration of $\ce{H+}$ or $\ce{H2SO4}$? There is one more question on SE about Hammett acidity. This question links to tables with $h_0$ for certain % of H2SO4. Can I perhaps use that? How does $[\ce{H2SO4}]$ relate to a % of H2SO4?
Is there a way to calculate the $h_0$ when $[\ce{H2SO4}]=1.830 M$?

Just plug pH-meter in there and used the values you get - H0 is approx. = pH for diluted sol.

Oke, thanks @Mithoron! So in this case, I assume H3O+ = 1.830 M, my pH = -log(1.830) = -0.262 and I can assume h0 = -0.262? Because when h0 = -.262, the [HBrO3] would be 1,01 and that's about four times the amount of BrO3- I added... But well, you're perhaps the wrong person to ask this, whilst you didn't devise the formula Thanks anyway!

As Karsten mentioned, you assume wrong, more so H2SO4 is not the only thing affecting pH in such mixture.

Hi Mithoron, thanks for pointing that out again. The formule here is about the initial concentrations, so you kind of cancel out the other things affecting pH in such a mixture. I'm just secondary school and there we assume that H2SO4 completely becomes H+. As this thing isn't at a very advanced level, I'll for now just assume H0=[H+].

I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M).

While $\ce{H2SO4}$ is a strong acid, $\ce{HSO4-}$ is not ($\mathrm{p}K_\mathrm{a}$ around 2). So your pH should be about zero, and you might not need to use the Hammett acidity at all.

Hi Karsten, thanks for thinking along. The thing is that I need to calculate the HBrO3 concentration. One of the ways I could do this is by using the aforementioned formula. Because of that I do need to use the Hammett Acidity function. Would that mean that my Hammet acidity function is roughly zero? And is there a way of calculating that?


text:
What is the mechanism of endothermic dissolution?
My real question is in fact "How can dissolution happen if it's endothermic?" However, there
have been many questions regarding this exact topic and the answers about Gibbs free energy
determining spontaneity of the reaction doesn't help my intuition at all. The majority of answers
list the increase of entropy as the cause of endothermic dissolution, but in my opinion, entropy
doesn't make sense as the cause.
Let's take a similar problem, namely the melting of a solid. If we place a solid with lower
temperature into surroundings with higher temperature (enough for the solid to reach its melting
point), it's not that liquification would cause the temperature of the surroundings to drop, but
rather that the transfer of heat from the surroundings to the solid would cause liquification. The
acquired heat allows the solid to increase its entropy, not the other way around.
So, if increased entropy is the result of endothermic dissolution, what is the cause? Or, what is
the mechanism that allows entropy to be increased?
Let's take the dissolution of NaCl in water as an example. This process is endothermic by around
4 kJ/mol. My guess is that, since some ions in the solid have higher kinetic energy than average, this energy
will be sufficient to overcome the 4 kJ/mol barrier and leave the solid, taking some of its thermal
energy away from it. As a result, the temperature of the remaining solid will drop a little. Since
the temperature of the solid is now slightly lower than the temperature of the surrounding water,
heat will be transferred from the water to the solid until the temperatures are equal. This will
result in lowering the temperature of water and in the increase of entropy.
Is my thinking correct? If not, why?

quote: So, if increased entropy is the result of endothermic dissolution, what is the cause?
The external temperature difference, set up using an external mechanism - work done in making this system is equal to the heat content it has, which increases the entropy after dissolution.

The main problem is to think on a part of the system, or taking the system as composed by the subject you are looking at. A piece of ice normally does cool down the normal environment. At least nearby.... That is heat flowing in.

Have you heard about entropic explosion ?

Since the temperature of the solid is now slightly lower than the temperature of the surrounding water, heat will be transferred from the water to the solid until the temperatures are equal

It is fine to think that way as a rough draft in rationalizing this process. The only issue is that temperature is a macroscopic quantity, and mechanism tends to be a particular (molecular) concept. So you could say that the individual solid particles (e.g. ions) will have slightly lower kinetic energy than average, and collisions with individual solvent molecules will balance this out.
Or you could start with solvent molecules near the solid-liquid interface that already have a higher kinetic energy than the average become part of the solid, thus lowering the average kinetic energy of the remaining solute molecules.
Or you could say that the water molecules solvating the ions use parts of their kinetic energy in the desolvation process, directly lowering the temperature of the solvent.
Because all the species have a distribution of kinetic energies around the average, all these scenarios are correct.

As a result, the temperature of the remaining solid will drop a little.

I don't think the bulk of the solid will have a lower average kinetic energy, so I don't think the temperature of the solid will drop faster than that of the solution. I would think that there is a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand. This is maintained as long as there is a net transfer from solution to solid. Once equilibrium is reached, of course, this will include thermal equilibrium.

this energy will be sufficient to overcome the 4 kJ/mol barrier

"Barrier" sounds like an activation energy, something you can step over. Here, I would rather talk of "energetic cost" or "step up", something that you don't get back. The solvation energy is higher than the lattice energy, so there is a cost for taking the ions out of solution and adding them to the ionic solid.

Thank you for the answer! If I may, I have a few points to clarify. Firstly, if an ion with kinetic energy equal to the average kinetic energy got separated from the rest of the solid, the average KE of the bulk solid would stay the same. But if it was an ion with more than the average KE, shouldn't its separation take away energy that was, so to say, meant for other ions, leaving the bulk with less KE on average? Furthermore, could you please clarify what you meant by "a temperature gradient between the solid-liquid interface on the one hand, and the bulk solid and liquid on the other hand"?

@HeatherB With the temperature gradient, I meant that the surface of the solid will be a bit colder than the inside of the solid. Also, the solution near the solid will be a bit colder than the rest of the solution. That's because nothing is happening anywhere except at the surface of the solid.

@HeatherB "KE meant for other ions". Let's take a simpler example: a gas in two containers connected by a tube. By chance, a particle with high kinetic energy goes from the right to the left, and one with low kinetic energy goes from the left to the right. This will result in a small temperature difference. These temperature fluctuations are normal, and tiny given the large number of particles in most systems.


text:
Crystal Field Splitting Energy and Metal Energy Level
Second and third row metals are stated to often have higher crystal field splitting energies than first row metals. The argument is usually made that the more diffuse 4d and 5d orbitals are capable of better overlap and decreased pairing energy. However, two concepts do not make sense to me:

Why do more diffuse orbitals lead to better sigma overlap? In organic chemistry it is often taught that size mismatch between orbitals contributes to poorer overlap, not better.

A 4d orbital will be higher in energy than the 3d orbital in the same group. From an MO theory perspective, I would expect this to mean that the energy difference between the ligand group orbitals and metal d orbitals increases and results in less splitting.


text:
Can I mix pressures and concentrations when finding equilibrium constant?
In order to find the equilibrium constant (or reaction quotient) I know I can find the value as the products over reactants raised to coefficient exponents. But, in all of the problems I have had to date, I have had either concentrations of solutions or pressures of gases. Now I have a problem with both. Can I mix the two?
Here is the chemical reaction:
$\ce{2 Al (s) + 6 H+ (aq) -> 2 Al3+ (aq) + 3 H2 (g)}$
I am told that the concentrations of the aqueous solutions are $\ce{[Al3+] = 9.3701}$ M; $\ce{[H+] = 4.6}$ M; and hydrogen pressure is 0.0271 atm.
If I can "mix" concentrations and pressures, then my $K_{eq}$ or (Q) is:
$Q = \frac{9.3701^2 * 0.0271^3}{4.6^6}$
$Q \approx 1.84 E{-7}$
If I can't mix them then I need to do some kind of conversion between concentration and pressure. I'm not sure how to do that. A conversion might be based on $K_p = K_c(RT)^{\Delta n}$ but I don't have enough information to make a conversion using that formula.
So, the bottom line, can I mix concentrations and pressures and, if not, how do I make a conversion from one to the other?

The values used are indicative of activity and so can be either concentration or pressure

Thank you. If you put it in an answer, I will mark it so. Though this is likely viewed as a "simple problem", I am sure I am not the only person who has ever been a bit confused by it.


text:
Copper vs Stainless Steel - which is more resistant to corrosion?
I keep reading conflicting information on this.
For normal water with a normal ph balance, over time which more resistant to corrosion - copper or stainless steel?

You should quote these conflicting readings.

I upvoted the question because it seems so simple, but allows so much discussion about special cases, all of which can and do happen.
First, copper pipe was used for house plumbing because it is so easy to work. Then copper became too expensive, or plastic became cheaper, I'm not sure which. So copper must pass the test of non-corrodability.
Now stainless steel is used in some fixtures for the sink, shower, etc. (Some is only chromium plated.) But stainless steel seems quite durable in "normal" water with a "normal" pH (~7?).
So, it's a tie.
However, if you scratch the stainless steel with a piece of copper tubing, the stainless may well begin to corrode, because the copper will act as a cathode, and by definition, the stainless steel becomes the anode and corrodes away (but slowly). So even if the two metals are almost equal in corrosion resistance (there are many varieties of stainless steel, some of which are exrtemely resistant to corrosion) treating them as equals is not good for corrosion resistance.
And salt water (it's normal...) is especially tough on stainless steel. It would seem that copper would have an advantage there - but no - even cast iron is better! https://www.ecoplumbers.com/2015/05/27/is-saltwater-corrosion-affecting-your-pipes/
So you have to be more specific, establishing the conditions, and maybe even do an experiment, before you can make a blanket statement. Sometimes I wish chemistry was more black/and/white, but all the intricacies sure do make it more interesting.


text:
Does hot water dissolve Copper more than cold water?
I read from WA state dept health https://www.doh.wa.gov/portals/1/Documents/pubs/331-178.pdf that hot water disolves copper more copper than cold water.

Use cold water for drinking and cooking. Because hot water dissolves more copper than cold
water, limit consumption of water from the hot water tap.

Is this correct that hot water dissolves copper more than cold?

Reactions are faster and compounds generally more soluble in hot water, so probably yes. BTW, who would ever consume water from the hot water tap ??

@Poutnik - why would you not use water from the hot water tap? It is cold water heated up, just as you might do in a pan, teapot, or coffee machine.

@JonCuster In a shortcut, if 2 are doing the same, the result is not the same. The result very much depends on the source, distance, pipe state and the way of usage the system. In one extreme, in is like from the kettle. In other extreme, it is something between being a biological ticking bomb of Legionella or various patogenic Protea (like the one wandering along the nasal neuron path to the brain and kills you in 24 hours), or sensoric disaster.

Actually, the topic of this question involves an important health care issue surrounding exposure to copper sources in varying conditions.
For example, in case of acidic drinking water with copper plumbing, to quote a 1980 article in Archives of Disease in Childhood:

The boy's copper toxicosis was attributed to the ingestion of water which had a high copper concentration. Although the family's well water contained little copper, it had a pH of 3.8-4.8, and dissolved copper from the domestic plumbing. Three explanations may be offered... Thirdly, he had been fed on cows' milk diluted with water, and it is known that the availability for absorption of copper in milk is greater than in other foods.[4]

Generally, applying heat is known to increase the speed of chemical reaction and relatedly, increased copper poisoning is found in cultures employing copperware, especially in combination with common milk consumption, to quote a source:

Indian childhood cirrhosis:
One manifestation of copper toxicity, cirrhosis of the liver in children (Indian childhood cirrhosis), has been linked to boiling milk in copper cookware.

So, caution and education is especially required in environments with potential copper exposure involving acidic conditions, warming and apparently, foods (like milk), noted to promote copper accumulation.


text:
What are the products of alkene oxymercuration with diols?
I was wondering what would be the product of the reaction of alkene with a diol (such as glycol) in alkoxymerucuration/demercuration, i.e. $\ce{(CF3CO2)2Hg/NaBH4}$?
I tried to look online but I couldn't land on a paper or website that goes through such reaction. I also tried Scifinder, I marked both compounds as reactants, but zero results stemmed from the reactants!
Any input?

I'm thinking if these possibilities (I have more possibilities in my mind, but for example I selected three) are structurally accurate and energetically stable to be produced in such reaction and which is favored?

To classify them more ... to single and di addition, here is what I'm thinking:

Oxymercuration is often run in the presence of excess water or alcohol, but only one RO/OH adds. Therefore I presume in the presence of a diol you get mono-addition by one of the 2 available OHs, probably the least hindered. https://en.wikipedia.org/wiki/Oxymercuration_reaction

@Mithoron so Alkoxymercuration/demercuration (CF3CO2)2Hg/NaBH4) won't force this reaction? if it doesn't what can force this reaction? I want to see whether such reaction will add seperate glycol on each carbon of the double bond, or the glycol will form a cyclic with the two carbon of the double bonds?

I think the top left option is most likely. I see no reason why you would get di-addition to the alkene

@Waylander I updated the question with single and di-addition possibilities, I'm not sure if excess of glycol or other factors will move the reaction toward single or di-addition or toward open or cyclic formation in either case.

In your original scheme, the product in the upper left would likely have the hydroxyl attach to the benzylic position. Two of the remaining three products involve C-C bond formation. How is that accomplished with oxymercuration?

@user55119 that's my question: could glycol move the reaction toward single or di-addition or and could it proceed toward open or cyclic formation in either case. For instance if it is a single addition, could the glycol, after forming -OR, form a cyclic orientation?

Alkyl mercurials react via radical processes but not to form the products you show.

There are so many structures here that it is at times difficult to keep things straight. The best way to approach your question is to take advantage of what is known about oxymercuration. If you enter "oxymercuration" at Search on Chemistry at the top of the page you will find a wealth of information at your disposal right here on ChemSE.
To start, the alkene is the (E)-propenylbenzene 1 and the diol is butan-2,3-diol 3, devoid of stereochemistry. Formation of mercurinium ion 4 with mercuric trifluoroacetate (2) is drawn as unsymmetrical because in the transition state for addition the carbon bearing the greater positive charge is the site of SN2, C-O bond formation. Mercurial 5 is a racemate, ignoring any stereochemistry of 3, which is what is formed after the first reaction. The second reaction, reduction with NaBH4, proceeds via radical 6 to form alcohol 7. A minor product may be bis-ether 9 arising from alkene 1 and mercurial 5. For mercurial 5 to form 1,4-dioxane 8 requires an oxidation of the carbon bound to mercury and a reduction of Hg+2 to Hgo as shown by the red arrows in structure 5. How do we know that this process does not occur? Because if it did, NaBH4 would not be required! Consider the oxymercuration of alkene 1 in the presence of water. If what you are suggesting were true, then 1 would form a diol rather than the observed mono-ol prior to the participation of NaBH4.


text:
Why is RO water more corrosive?
According to several articles from the American Water Works Association, permeate from RO should be stabilized in terms of pH and hardness to avoid corrosion of the distribution piping.
How does corrosion differ from that of normal water? Would RO water corrode stainless steel faster than normal water if both had the same chloride concentration?

It is a good practise to explain abbreviations used; even if they are common in your part of the world, there will probably be users, who are not familiar with them.

What is RO water ? Rarely Overheated ? Really Outdated ?

Water passing RO does not have the same chloride concentration.
The raw RO water contains almost no ions, aside of those that passed by leakages of the osmotic membrane.
This disbalances the water natural $\ce{CO2/HCO3-}$ pH buffer, leaving the ionic alkaline buffer part before membrane, so water becomes more acidic with corrosion aggressive $\ce{CO2(aq)}$.


text:
Determining the number of triphenylphosphine ligands in nickel(II) complex

A complex is synthesised by combining hydrated solutions of $\ce{NiCl2}$ $(\pu{0.6 g})$ with $\pu{1.35 g}$ triphenylphosphine $(\ce{PPh3}).$ After cooling the product precipitates as a solid with formula $\ce{NiCl2(PPh3)_x}.$ Determine $x.$

I first found the molar ratios
$$n(\ce{NiCl2}) = \frac{\pu{0.6 g}}{\pu{129.59 g mol^-1}} \approx \pu{4.6 mmol}\tag{1}$$
$$n(\ce{PPh3}) = \frac{\pu{1.35 g}}{\pu{262.29 g mol^-1}} \approx \pu{5.1 mmol}\tag{2}$$
The smallest ratio these go into gives $x \approx 1,$ but I don’t think this is right.

Have you taken into consideration the possibility of having nickel chloride as a hydrate? x=2 for the hexahydrate.

Oh gosh what a silly mistake to make,  I didn’t take that in thank you!

Nothing looks silly here. Only one issue: the question contained contradictory information: trimethylphosphine $(\ce{PPh3}).$ I assume it was supposed to be triphenylphosphine judging by the molar mass you are using. $x = 2$ looks the most correct answer to me as it both justifies a typical C.N. = 4 for nickel, but still leaves enough room for two bulky triphenylphosphine ligands.


text:
How to calculate the refractive index of mixtures?
In "How to Calculate the Refractive Index of a Formulation", the article outlines a method to find the refractive index of a mixture but the sources don't lead to any evidence regarding its validity. I would like to know the name of the formula used and if it has any limitations because I haven't been able to find any techniques to calculate the refractive index of non-binary mixtures other than this article.

Well, obviously, this is just a crude approximation.

The link in this earlier answer may be helpful.

This is just a weighted average, also known as the rule of mixtures.

Does this answer your question? Is the refractive index of a solution of four substances is a median of the four subtances' refractive index?


text:
Lewis dot structure of H3NF2
Consider a hypothetical compound $\ce{H3NF2}$. What will be its Lewis dot structure? I'm stuck, since Nitrogen cannot exhibit pentavalence, or at least its not stable (unlike phosphorous in $\ce{H3PF2}$). Will the compound be ionic?

You seem to be acting on the premise that any random text string represents a compound. This is not true.

Why discuss the structure of non-existent compounds ?

In principle, such a compound is possible by allowing an ionic structure, to wit $\ce{[NH3F+][F-]}$ where the cation has the usual tetrahedral structure for an ammonium ion, with nitrogen at the center and three hydrogen and one fluorine atom attached to it.
Whether this compound exists is another matter.  We may expect such a B-L acidic cation to protonate the fairly strongly basic fluoride ion so that our proposed $\ce{[NH3F+][F-]}$ would break up into $\ce{NH2F + HF}$.
However, the cation $\ce{NH3F+}$ does exist with more inert counterions such as trifluoromethylsulfonate (triflate), $\ce{CF3SO3^-}$.  Spectroscopic data are reported with the triflate salt as well as a couple salts involving more highly fluorinated ions in the table below from Olah et al. [1].

Reference

G. A. Olah, K. K. Laali, Q. Wang, G. K. Surya Prakash, Onium Ions (New York:  John Wiley and Sons, 1968), p. 15.


text:
What is the significance of using cold conc H2S04 followed by boiling water for hydration of alkenes?
I tried searching the internet but I failed to find a answer  satisfactory to my understanding.
I know if I react the alkene with cold conc. H2SO4 I get an alkyl hydrogensulfate which upon reaction with boiling water gives alcohol,but I can't understand the need for these conditions like what if we reverse them.
I think that since alkene is nonpolar we react it with sulphuric acid to make the carbocation instead of direct reaction with water,but I dont understand the formation of alkyl hydrogen sulphate.If a carbocation has been formed, water can attack it easily forming an alcohol (assuming that HSO4- is weaker nucleophile than water).So why don't we just add both sulphuric acid and water combined?

Sulfuric acid is an oxidising agent as well as a strong acid. Using hot sulfuric acid will result in oxidation of the alkene giving a complex mixture.

then why is there the need for boiling water,H2SO4 is non-nucloephillic acid so H20 is free to act as a nucloephile

Hydrogensulfate is not that great a leaving group - remember it is still acidic and will protonate the water

Oh..so here is what I have concluded (correct me if I am wrong)   In hydration of alkenes we react alkenes(non polar) with sulphuric acid first(as direct reaction of water is not possible)~the cold and concentrated conditions favours maximum production of alkyl hydrogen sulphate without oxidizing alkenes.Then we react alkyl hydrogensulfate with boiling water (perhaps boiling water makes hydrogen sulphate a better LG?) and thus water reacts with it forming h30+ and OH- which then attacks on alkyl carbocation forming alcohol.

wait, shouldn't Hydrogen sulphate be a better leaving group as it is weak conjugate base of a strong acid?

Hydrogen sulfate will protonate the water meaning that you are trying to make a species leave that is already negatively charged so it has to leave as a dianion. Compare with methanesulfonate which is an excellent leaving group

so heat favours removal of dianion?

Heat makes it possible.


text:
Stability of molecule based on ring size
During a discussion on chemically valid molecules based on atom valency, I was told that in molecules like
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
although the sum of the bonds that are incident to each atom do not surpass their valence capacity, it would have to be very far from the optimal geometry as predicted by VSEPR to have the represented atomic bonds. In addition, it was suggested that molecules with 3-member rings in general would be very energetically unstable. I was wondering if this is true and if you could expand more on this. Pointing me to the appropriate literature would also be appreciated.

EDITS:
Here is another set of molecules from an article that pass validity based on valency. Are these invalid just because of having 3-member rings as well?

@Mithoron 1,1,1-propellane is a thing, though.

Cyclopropane is very strained, but it does exist (and is used as an anaesthetic). Many derivatives exist but whether many fused cyclopropanes exist is probably stretching things. I'm guessing, though.

:/ Somehow permeakra got my point. It doesn't matter if there's cyclopropane ring. Even fused isn't necessarily a problem, but there's a pyramidal carbon there! And hydroxyl next to it... Good luck trying to make it!

Cyclopropane ring itself isn't all that bad, it has enthalpy of formation of 53.30 kJ/mol, same as ethylene.  It's not excellent, but it is't much. Fused cyclopropanes are not great, but byciclobutane exists. The problems start when one considers strongly strained double bonds and pyramidal carbons (when all four bonds are directed into same half-sphere).  Some compounds with pyramidal carbons are isolable (like 1,1,1-propellane), but usually because of some specific gimmick of this particular system slowing down its decay.

@permeakra So as a rule of thumb, I was told to forbid all 3-member cycles, 4-member cycles containing a double bond and all cycles containing a triple bond. Do these constraints make sense?

@Blade Not really except for triple bond. Triple bond in small (<8) cycles is really unstable. But for 8+ membered cycles it's perfectly OK. There is not, regrettably, wide-scale rules of thumb.


text:
What are the reaction products in my dishwasher from a steel colander?
In my dishwasher I noticed that a steel colander and a coffee maker (the Moka Pot) both became discolored after being in the dishwasher. The metal is clearly reacting with something. They aren't made of aluminum (I don't think) and I've never seen this reaction before when I do dishes in the sink.
To clarify: the metal on my coffee pot and the colander was relatively smooth and shiny, and then after they were in the dishwasher they became grayer, and the texture was clearly rougher. That said, i was able to remove the discoloration with a Scotch-brite plastic pot scrubber sponge or Brillo pad, with a little elbow grease.
So, the chemicals I saw on the back of my dishwashing liquid were:
NaCl, Na$_{2}$SO$_{4}$, Citric Acid, Na$_{2}$CO$_{3}$, Sodium percarbonate, sodium silicate and a couple of organics (it says enzymes, but I don't know how much of that is marketing hype). I looked up the rinse aid (it's Finish brand) and I came up with alcohol ethoxylate, sodium polycarboxylate, sodium cumene sulfonate and tetrasodium EDTA, Methylisothiazolinone and methylchloroisothiazolinone, and CI Acid Blue dye.
I do not know the xact metal composition of my colander or coffee pot, though I suspect it's is stainless steel I don't know which particular type and what the exact mix of carbon, chromium, etc. is. I can say that the colander is one that is quite old; it's age is measured in decades. The coffeepot not so much; I got that maybe a decade ago? At least that. Both are made of a gray steel that has a slight "rough" feel; it's not the super-shiny modern variety. The Colander in particular seems a type of metal I recall being quite common with cheaper kitchen stuff in the 70s and 80s.
In any case, I wanted to know what the following might be:
-- The reaction products of the dishwashing chemicals dissolved in the water; clearly there is something going on
-- If I need be concerned about toxicity, since this is something that touches food. I would assume there is less of an issue if I scrub the discoloration off. And it's not likely to be in concentrations that are harmful one-off, but why ingest weird stuff if you don't have to and all that.
Any insights are appreciated; this is something I was rather curious about (and may make a good demo for a chemistry class).
ETA: In the images you can see the darker gray on the metal -- that was the discoloration.
ETA II: apropos of blacksmith37 's comment/ answer I checked if there was any magnetism and the fridge magnet does not stick so it's either aluminum or a steel that is non-magnetic.

enter image description here

I am confident both are aluminum ; Steel would be magnetic, austenitic stainless is non-magnetic ,but normal relatively shiny or bright in consumer products. All steels are about 3 X more dense than aluminum if you could judge the density. Aluminum is subject to corrosion in solutions of salts, acids ,and bases. But I couldn't pick one of your components as the most likely cause. That is, I expect it is corrosion of aluminum you are seeing. I have had a very similar colander  but washed it by hand because it was fairly large to put in a dish washer.

I just tested them with a magnet; you are correct there is no magnetism and both might well be aluminum. I have not been able to gauge the density yet.


text:
Acidic KMnO4 oxidizes H2S to S. Why doesn't the oxidation proceed further to sulfur oxide, sulfur dioxide or sulfur trioxide?
I have some idea about the question. I think it has to do with oxidation/reduction potentials but I am not able to find any resource for the exact values or how to make sense of them in this context. Any reference would be appreciated.

The Chemistry SE  site policy recommends to use just plain text for the question titles, for indexing/searching reasons.

Instead of writing e.g. H_3PO_3 in MathJax, write e.g.  \ce{H3PO3 + 2 KOH -> 2 H2O + K2HPO3}. \ce{} invokes mhchem   MathJax extension. It additionally uses expected upright font. See $$\ce{H3PO3 + 2 KOH -> 2 H2O + K2HPO3}$$

Yes, the reaction can further proceed to sulfur dioxide and sulfur trioxide, but only appears for certain, per my research, to occur in alkaline conditions.
Here is a supporting opinion to this effect, which cites the corresponding reaction in alkaline conditions as:

$\ce{3 S(2-) + 8 KMnO4 → 3 SO4(2-) + 8 MnO2 }$

that implies the further conversion of $\ce{S}$ to $\ce{SO2}$ to $\ce{SO3}$, which in alkaline conditions is represented by sulfate formation.
Note, the cited source notes only in acidic conditions, to quote:

Neutral-Acid pH:
$\ce{3 H2S + 2 KMnO4 → 3S + 2 MnO2 }$

Wikipedia on KMnO4 supports this pH effect, to quote:

The Mn-containing products from redox reactions depend on the pH. Acidic solutions of permanganate are reduced to the faintly pink manganese(II) ion (Mn2+) and water. In neutral solution, permanganate is only reduced by three electrons to give manganese dioxide (MnO2), wherein manganese is in a +4 oxidation state.

Interestingly, the corresponding reaction replacing KMnO4 with either H2O2 or HOCl, can results in $\ce{SO2}$, ..., in acidic conditions.

Thank you Ajkoer for your answer. It is very informative and an interesting read. Unfortunately (or fortunately) it challenges my knowledge. You have said the following things that seem confusing:

Although $MnO_4$ is a weaker oxidizing agent in alkaline conditions, it is still able to turn $S$ to $SO_2$ and $SO_3$.

Adding peroxide or hypochloric acid works the other way (by making the solution even more oxidizing) to obtain $SO_2$ or $SO_3$.

I checked out the references and indeed your argument is in agreement with them. I think something more interesting is happening.

I wonder if it is the enthalpy of salt formation/solvation that pushes the S to Sulphate form in alkaline conditions.


text:
What nomenclature do names like fumarate and malate derive from?
When reviewing the citric acid cycle, I find names like glutarate and succinate to be helpful in thinking about the structure of the molecules because I was taught about the sequential dicarboxylic acid naming scheme: oxalic (2 carbon), malonic (3C), succinic (4C), glutaric (5C) etc.
But I often struggle with fumarate and malate because I have not learned the numbering or nomenclature system that these names are derived from. Can someone point me in the right direction so that I can learn the background behind these names?
Knowing that will be helpful to my recall of the molecular structure. Then I can stop memorizing and start understanding.

As per the OED, the etymology of  maleic is related to that of malate. Loss of water from the latter gives the cis-form of the former.

malic, maleic, and malonic acids have all the same origin : malus = apple in latin

Unfortunately, you will have to memorize the structures and the names because there is no chemical structure hidden in those names.
From Elsevier's Dictionary of Chemoetymology

fumaric acid $\ce{C4H4O4}$, derived from the genus name Fumaria (fumitory), from fumus (Latin:
smoke)

Fumaria plant

malic acid $\ce{C4H6O5}$, derived from malum (Latin: apple)

The same goes for all other common names like oxalic acid, acetic acid etc.

"Dictionary of Chemoetymology" Price is a bit steep, but would you otherwise recommend it?

My favorite source of historical word origins is the unabridged version of the Oxford English Dictionary. It is also subscription based. OED will most likely have 99% of the words contained in Chemoetymology.

How is Chemoetymology stuctured? Just A-Z?

Yes, A-Z just like a dictionary. If you want more "stories" and the names which are not present in OED is Alex Nikon's "Organic Chemistry- The Name Game" is worth it.

Also *Chemoetymology" has no pictures. I just added Fumaria from the web.

Yea, I'd want a good enlightening read, not a dictionary. Unfortunately, that one is also from Elsevier and therefore equally overpriced, for a Christmas or graduation present. ;-)

Formic comes from the dry distillation of ants, this makes me wonder :))


text:
How can I cluster infrared spectra?
This is sorta a cross-post from my post in StackOverflow. Although, rather than seeking help with the code I'm here asking for help with a chemical/statistical problem associated with the code.
I'm trying to cluster infrared (IR) spectra so that spectra most similar to each other are grouped and processed differently. To do this I am implementing a Density-Based Spatial Clustering of Applications with Noise (DBSCAN) algorithm on a 2D set of data representing the spectra. Since the IR spectra clusters need to have similar shapes and vertical height (the substance is absorbing the same amount of light for spectra in the cluster) I have adapted a 2D system with the following parameters
The mean of the spectra (the average %Transmission, this should correlate to the vertical height) and the mean of the first derivative of the spectra (this should correlate to the shape). These two parameters then give a 2D system that I can use DBSCAN to cluster the IR spectra, it works but it does not cluster them perfectly. Note that the DBSCAN algorithm typically clusters scatter plot points through their Euclidean distance from each other. This is why I have the two parameters which you can imagine being plotted on an X(mean) and Y(mean of derivative) plane, for the clusters algorithm to then cluster them.
I can already make logical sense of the fact that the average of the derivative will not be an accurate measure of the shape of the spectral line, but I can not figure out better method. How can I get a better parameter for representing the shape of a spectral line?

Can you give an example of such spectra, and how you expect them to differ?

Hey @Karl, this is tough since I am a little nervous of posting an image of the spectra as they are unpublished research. Would it help if I give an example using random made up spectra for the purpose of illustrating what I'm doing?

Isn't there an established way to really compare the spectra for all their features? In my opinion you should ask also in Maths. Basically you want to group the plot of similar xy functions, they should know. The fact that the plot is actually a spectrum does not change anything.

Hey @Alchimista, I haven't seen any established ways of doing it yet but they probably exist. I will post this in Maths StackExchange as well, thankyou!

@Karl I have added a minimum working example in the StackOverflow post that might better illustrate what I'm asking for help with.

DBSCAN works well on only a small number of dimensions, so I understand your need to do dimensionality reduction.  But I don't think computing the mean and the mean derivative are the only dimensionality reductions I'd try.  Why not (do preprocessing like aligning and calibrating the spectra and then) do PCA on the full spectra?


text:
Sodamide vs alcoholic KOH
My textbook says

In the presence of strong base such as KOH and at high temperatures, triple bond at terminal C-atom of higher alkynes migrates to give disubstituted alkyne.Therefore alcoholic KOH is used when rearrangement is not a problem,on the other hand terminal alkynes can be produced using NaNH2 in liquid ammonia as those alkynes essentially donot
rearrange.

Why is that so?I know KOH is not as strong a base as sodamide but why alkynes formed by using it are capable of rearrangement while those formed through sodamide are not?
The product is same in both cases,what its got to do with the reagent?And what is the significance of high temperature?

I guess the last sentence in your book should read "as in those, alkynes do essentially not rearrange".

Oh,that makes sense.But  how does the reagant effect it?

Well, what is the mechanism of this rearrangement? (Btw. please give the name of the textbook, and what is the audience?)

That book is indeed bad, from the short look I had. Stuffed to the brim with facts, little in the way of adequate didactics, and a lot of bad grammar to mislead the student. I honestly don't know a good alternative, my highschool chemistry books were in German, and first semester university books would not span the whole field, you'd have to buy two or three. You might try Atkins "General Chemistry". It's regarded as OK to help not too ambitions biology or medicine students over their chemistry, and Atkins is a fine writer.

Anyway, a really strong base like NaNH2 deprotonates the terminal alkyne, and that's that. Actually amides tend to shift triple bonds to the end of an alkyl chain. That's called the "alkyne zipper". I'm not sure how the reaction with KOH should proceed towards a disubstituted alkyne. Subsequent addition/elimination steps in the heat?

I have come up with this possible reasoning(not sure if it's correct) Reaction of alcoholic KOH with vicinal dihallide takes place at high temperatures(100~200°C) due to the fact that KOH is less strong a base and doing second elimination would require added heat,this heat in turn favours rearrangement of the bond.On the other hand reaction of sodamide happens at -33°C (it's a strong base so second elimination will occur at low temp)Without the heat the rearrangement will not occur.(relation of heat and rearrangement is not clear to me )

That's about what would happen, yes. https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/09%3A_Alkynes_-An_Introduction_to_Organic_Synthesis/9.03%3A_Preparation_of_Alkynes-_Elimination_Reactions_of_Dihalides , scroll down to "Terminal Alyknes"  With NaNH2, the reaction is locked as R-CC-Na, with NaOH not, and if you heat the latter, it will rearrange into the more stable disubstituted alkyne.


text:
Does Hybridization change the shape of molecules after bonding to other molecules?
If you have two molecules bonding to each other such as NH and CH2 you have a change in hybridization for both N and C.
NH with its two lone pairs is sp2 hybridized and CH2 is also sp2 hybridized.  As the molecules begin to bond, the Nitrogen changes to sp hybridized and the CH2 changes to sp3 hybridized.  First of all, is this correct?
Secondly, does this shift in hybridization result in a change in the orientations of the atoms?  For example CH2 is bent, but in the combined molecule it should be trigonal planer? I.e. 120 degrees between the H atoms?
Thanks for any help!

You, got it all backwards...

Hybridization is a "concept" that stops making any sense as soon as you investigate it. I've written about it in this answer. Short version: Don't even bother with decoding how hybridization works, you're spending a lot of brain power on something that is ultimately physically nonsensical.

@Antimon it seems to be a hard statement. The same can be told for orbitals. Perhaps is the way the concept is thought and presented that let novices think of some clockwork machinery...


text:
1,5-Pentanediol MS
I'm having troubles assigning the signals m/z = 41 and 56 of the compound 1,5 pentanediol. I tried the basic fragmentations (McLafferty, loss of OH radical, etc.) but I can't figure the structure of these signals.

Thank you.

Worth noting that $\ce{C3H5+}$ has an $m/z$ of 41.039.  Not sure where 56 is coming from.  I'm not sure that $\ce{C3H5+}$ really is the ion in this spectrum, but if it is, one way to get there would be by double dehydration of the alcohols, forming 1,4-pentadiene or an ion derived therefrom, followed by C-C bond breaking.  The MS spectrum of 1,4-pentadiene does have a peak at 41, but admittedly it is not he main peak.

I found this paper https://doi.org/10.1002/oms.1210120113 that helped me a lot. Basically, there is an initial rearrangement and then the fragmentation occurs.


text:
Ion Mass Spectrometry of Polymorphs
Can ion mass spectrometry differentiate between 2 distinct polymorphs? Why?
In my opinion, the spectrum would be the same as only the crystal lattice differs.

Here’s an example. In short, surface analysis with SIMS seems to allow them to calculate the ratio of oxygens bound to calcium atoms from the surface which can be used to distinguish polymorphs

That articles mentions a time-of-flight secondary ion mass spectroscopy (TOF-SIMS). Would it be the same with a simple IMS?

Because I believe that IMS will provide similar Ko values as only the lattice changes in polymorphs.

simple IMS? Not really sure what your referring too could you provide a reference/example of a simple IMS


text:
The best way to designate unsaturation (double bond)
Although it has already been explained here that the use of the "Δ" (Greek capital delta) character is not recommended to designate unsaturation in individual names, IUPAC 1989 Nomenclature of steroids states that it may be used, however, in generic terms, like "Δ5-steroids"; see note 3) on page 10. What are the current trends on whether should we continue or cease using the "Δ" symbol to describe a double bond per se (rather than in the name of a steroid) or just the position of such bond between consecutively numbered junctions, in various relevant compounds, not just steroids? Isn't it a concise and elegant way to denote such a bond, or is there a better, modern, suggested, clear way? Are there recommendations or best practices documents published on that, or research papers? For, example, instead of writing "steroids with a double bond between carbon 5 and 6 junctions" we could simply write "Δ5-steroids", an example given in the note 3) on page 10 of the UIPAC Nomenclature of steroids of 1989? Here is such an example in practice, in a 2020 publication (quote): "couple of Δ5-steroids with hydroxyl group at C3 position". Can we just use the "Δ" symbol to describe the position of the double bond e.g. "3βHSD isomerizes the double bond from the Δ5 to the Δ4 position" or just (quote): "Δ5 → Δ4 isomerization by 3βHSD"? Or should we cease the use of the double-bond position as a common denominator in favor of other similarities in a group of compounds, e.g. "3-keto-steroids" instead of "Δ4-steroids" or "3-hydroxy-steroids" instead of "Δ5-steroids"?

As you care of these things, "HSD3B2 isomerizes the double bond from the Δ5 to the Δ4 position"? does not tell anything clear. Moreover 5,6 to 4,3 position is also a common possibility (and doesn't require special character).


text:
Why exactly does high pressure (72 bar) nitrous oxide's pseudocritical¹ temperature drop so far when oxygen is added to make 50%v/v mixture? (Entonox)
This answer to Can precipitation occur in states of matter other than liquids? mentions the unusual case of Entonox.
The effects of temperature on nitrous oxide and oxygen mixture homogeneity and stability (Litwin P. D. 2010 BMC Anesthesiol. 2010; 10: 19. doi: 10.1186/1471-2253-10-19) says:

Poynting found that the critical temperature and pressure of a vapour may be affected when it is mixed with another gas. The critical point of a substance is the temperature at above which a gas cannot be liquefied, no matter how much pressure is applied. Conversely, it is the temperature and pressure point at which condensation of the gas into liquid will commence. For nitrous oxide this critical point occurs at a temperature of +36.4°C and at a pressure of 72.45 bar (1050 psi) [12]. For a cylinder of a 50% nitrous oxide and 50% oxygen v/v mixture filled to 138 bar (2000 psi), the new critical temperature of the nitrous oxide (known as the pseudocritical temperature) decreases from +36.4°C to -6°C.

Answers to Can precipitation occur in states of matter other than liquids? tend to guide us away from thinking of the Poyting effect as being related to dissolution or solubility but my concern is that "Poynting effect" is simply a name for an observation, and by itself not a well developed scientific concept or explanation.
That Wikipedia article says:

In thermodynamics, the Poynting effect generally refers to the change in the fugacity of a liquid when a non-condensable gas is mixed with the vapor at saturated conditions.

and that linked subsection ends with:

This equation allows the fugacity to be calculated using tabulated values for saturated vapor pressure. Often the pressure is low enough for the vapor phase to be considered an ideal gas, so the fugacity coefficient is approximately equal to 1.

So if we start with tabulated basic measurements of vapor pressure and calculating fugacity f and then see an unexpected, significant increase in vapor pressure of A when introducing gas B to the vapor, we say "Oh, that's the Poynting effect!"
We might then quickly change the subject or excuse ourselves before "What's that?" or "Why does that happen exactly?" gets ask.
Question: This change has a name, but what is happening? Why does the high pressure (~72 bar) nitrous oxide's pseudocritical1 temperature drop so far below it's pure substance critical temperature (from +36.4°C to -6°C) when oxygen is add to achieve a 50% v/v mixture at ~138 bar? Is the mechanism at least analogous to dissolution in some ways? Or could it be considered a "liquification point depression" analogous to a freezing point depression?
The citation of a readily available reference or published work explaining "exactly why" this is thought to happen will be most helpful, and if in what ways the process is and is not like dissolution can be added, then that's a complete answer.

1As pointed out in comments, critical temperature is defined for a pure substance so "pseudocritical temperature" is used in the block quoted work accordingly. See Real gases; Real gas law or Interaction Parameters for Kay's Pseudocritical Temperature for some examples of the term in use.

BTW, N2O critical temperature does not drop, it is a constant, applicable for pure substances.

Ask yourself, what is dissolution? What it means thermodynamically ? When there is dissolution equilibrium from TD POV? What is the role of chemical potentials ? Why solubility depends on the solvent ? Why Pointing effect does not depend on used gas, but on the pressure only ? Note that some gas soluble in the liquid may have the opposite effect than the Pointing effect, decreasing the saturated vapour pressure by decreasing solvent molar fraction and eventually its chemical potential by chemical interaction. E.g. NH3(g+aq)+H2O(l+g).

you may start with "Does chemical potential of liquid depend on pressure ?"

The temperature change is much less striking if you think in terms of absolute temperatures instead of various customary temperature scales. The pseudocritical temperature drops from 310 K to 265 K, or in other words, by 15 %. That's still significant, but nowhere as shocking as "OMG it went from searing hot to freezing".

@TooTea but how does this drop compare to drops in other cases? Isn't this a particularly large and notable example of the Poynting effect?

@TooTea It is still on the "OMG" level, considering most people think that saturated vapor pressure is independent on presence of other gases.

Dissolution(solvation) is solvation of solute molecules by molecules of solvents. This decreases chemical potential $\mu={\left(\frac{\partial G}{\partial n}\right)}_{T,p}$ of the solute, comparing it (often hypothetically ) to the chemical potential of the solute at the same concentration in gaseous phase.
Different solvents cause different chemical potential drop, leading to different solubility. As dissolution equilibrium means the equal solute chemical potentials in the solid phase and the solution.
Chemical  potential and fugacity

It [fugacity] is equal to the pressure of an ideal gas which has the same temperature and molar Gibbs free energy[=chemical potential] as the real gas.

are linked by relation $\mu=\mu_0 + RT \ln{\frac{f}{f_0}} \tag{1}$
Molecules in gaseous phase move independently on each other, unless there is a strong chemical affinity, what is not this case. Pointing effect does not depend(*) on the used gas, as there is no molecular wrapping in gas.
Pointing effect does not directly affect chemical potential of vapor. But  pressure affects the chemical potential of liquid instead, leading to higher saturated vapor pressure at the same temperature. The vapor pressure therefore increases to balance chemical potentials of the  substance in both phases.
For liquids near their critical conditions, like for the case of nitrous oxide with or without oxygen, there are big changes of state variables like molar volume or density with small change of conditions like pressure and temperature. This leads to relatively big change of saturated vapor pressure with the change of pressure of the system.
Note that for the liquid at its critical temperature $T_\mathrm{c}$ and near its critical pressure $p_\mathrm{c}$, $$\lim_{p \to p_c}{\ \left(\frac {\partial V}{\partial p}\right)}_{T }= -\infty \tag{2} $$
to get a picture what these changes are like.
As we know from the thermodynamics, for stable composition and zero non-volume work:
$$\mathrm{d}G=-S \cdot \mathrm{d}T + V. \mathrm{d}p  \tag{3}$$
Therefore $${\ \left(\frac {\partial G}{\partial p}\right)}_{T }=V \tag{4}$$
$${\ \left(\frac {\partial \mu_\mathrm{liq}}{\partial p}\right)}_{T }=V_\mathrm{m, liq} \tag{5}$$
If we assume ideal vapor behavior, we can consider $p_\mathrm{vap} = f_\mathrm{vap}$ and
$$RT \cdot {\left(\frac{ d \ln{p_\mathrm{vap}}}{\mathrm{d}p}\right)}_T =  V_\mathrm{m, liq} \tag{6}$$
$$RT \cdot \ln {\frac{p_\mathrm{vap}}{p^{\circ}_\mathrm{vap}}} =  V_\mathrm{m, liq} \cdot (p - p^{\circ}_\mathrm{vap}) \tag{7}$$
$$ p_\mathrm{vap}  =  p^{\circ}_\mathrm{vap} \cdot \exp {\left(\frac {V_\mathrm{m, liq}}{RT} \cdot (p - p^{\circ}_\mathrm{vap})\right)} \tag{7}$$
(7) is a simplified saturated vapor pressure dependency rather for "normal" liquids with $T_\mathrm{c} \gg T$, when the liquid compressibility can be neglected. Otherwise, $V_\mathrm{m,liq}=f(p)$ would have to be included to the integration, as the liquid molar volume and compressibility start steeply grow toward the critical point. E.g. water vapour and liquid densities converge to about $\rho = \pu{0.21 g/cm}$ at the water critical point.
For illustration, we can take water at the boiling point with
$V_\mathrm{m} = M/\rho = (\pu{18.02 g/mol})/(\pu{0.9584 g/cm3})=\pu{18.8 cm3/mol}$, $T=\pu{373 K}$ a $p^{\circ}_\mathrm{vap}=\pu{e5 Pa}$. It can be seen the pressure effect can be neglected for small and medium pressure.
$\begin{array}{cc}\small \\
\text{pressure [MPa]} & \text{H2O vapor pressure [kPa]} \\
0.1 & 100.0 \\
0.2 & 100.1 \\
0.5 & 100.2 \\
1.0 & 100.5 \\
2.0 & 101.2 \\
5.0 & 103 \\
10 & 106 \\
20 & 113 \\
50 & 135 \\
100 & 183 \\
\end{array}$

(*) Special cases are systems with significant chemical interactions, including gas solubility in liquid. E.g. well soluble gases cause additional effect of the opposite sign than the Pointing effect. They decrease the saturated vapour pressure by decreasing solvent molar fraction and eventually its chemical potential in both phases by chemical interaction. E.g. $\ce{NH3(g + aq),H2O(l + g)}$.

This is a very helpful answer. It will take me some time to take it all in and to read further, but if there can be a key point, it seems to be"... pressure affects the chemical potential of liquid..." so that's where I will start. Thanks!

@uhoh Some answer elaboration...

my goal was to read thoroughly including all links, understand, leave a comment indicating how well I understand, thank the author, then accept. I haven't quite gotten there yet, but I am absolutely certain this answers my question completely so I'll click now and continue to chip away at my ignorance of thermodynamics. Thanks for your articulare and thorough answer!


text:
How easy is it to make animal horns transparent?

I never knew that HORN in HORNbook refers to animal horns, or animal horns can be made transparent? If Medievalists can make it transparent, the transparency process must've been cheap easy and quick, right?
The Horn Book | Why is it called "The Horn Book"?

Back in the sixteenth century, English monks began to make hornbooks to help their pupils learn to read. Usually a wooden paddle with an alphabet and a verse glued to the surface, hornbooks derived their name from the piece of transparent horn protecting the verse. The picture to the right shows a modern replica of a hornbook.

Hornbook - Wikipedia

A hornbook is a book that serves as primer for study. The hornbook originated in England as long ago as 1450,[1] or earlier.[2] The term has been applied to a few different study materials in different fields. In children's education, in the years before modern educational materials were used, it referred to a leaf or page displaying the alphabet, religious materials, etc., covered with a transparent sheet of horn (or mica) and attached to a frame provided with a handle.[3]

Could they have shaved off a really thin layer? If thin enough, it might be transparent without chemical treatment.

Cheap, easy, and quick do not relate at all to parchment production, for example. Yet it was made and used in medieval times. Don’t let modern ideas obscure history.

I found this description of how the horn is prepared in a post by Tammy L. Austin, at the University of Notre Dame's website (https://www3.nd.edu/~rbarger/www7/hornbook.html):

The horn of oxen and sheep were used to make the laminating structure. The horn was left in cold water for several weeks, which separated the usable part from the bone. It was then heated, first in boiling water then by fire, and pressed by plates and machines to make it smooth and transparent.


text:
Which group will migrate in this BVG oxidation?

I know that H has a higher migratory aptitude order than phenyl. But, in this case, the phenyl ring has electron donating groups attached to it which enhances its migratory aptitude. So, which group is more likely to migrate, H or phenyl?

This might answer your question: https://chemistry.stackexchange.com/questions/6269/migratory-aptitude-in-baeyer-villiger-reaction

The Ph ring will migrate. B-V oxidation of aromatic aldehydes is routinely used to prepare phenols via formates, examples https://www.researchgate.net/publication/229145649_One-pot_synthesis_of_phenols_from_aromatic_aldehydes_by_Baeyer-Villiger_oxidation_with_H2O2_using_water-tolerant_Lewis_acids_in_molecular_sieves

The reaction is a Dakin oxidation. https://en.wikipedia.org/wiki/Dakin_oxidation


text:
Does the benzene ring in styrene show a negative inductive effect on the ethenyl group?
I know that a group exerts an inductive effect on another one if there is a difference in electronegativity. In the case of styrene (1) all the atoms are sp2-hybridized. So, how could the phenyl ring show a negative inductive effect on the ethenyl group?

Negative inductive effect, no. Negative mesomeric effect, yes.

It may be meant as a -I effect, compared to an ethene hydrogen formally replaced by the phenyl. Similarly as methyl has +I effect on benzene, formally replacing a benzene hydrogen. By everyday life analogy, if 2 persons  stand on the same stair, one of them is higher than the person that originally stood 1 stair lower.

... because the phenyl ring has 6 sp2 atoms, and the ethylene group only two?

This can be explained by looking at the molecular orbitals. The molecular LUMO of benzene (which accepts electrons) is closer to the HOMO of ethylene (which given electrons) as compared to the large gap between HOMO of benzene and LUMO of ethylene.

Surely, the HOMO, LUMO don't exist separately and only exist as a molecular orbital. But the separation makes for easy transfer of electrons from ethylene to benzene unlike the opposite.
Because I have involved pi orbitals into the solution, I believe mesomeric effect is more apt here. To talk about inductive effect we have to involve a more complex idea (group electronegativities). Resources for such data are not easily accessible. I would appreciate if someone can find the group electronegativities of benzene and ethylene and post them here for the analysis of inductive effect as well.
Surely, if present, a soft nucleophile will it will attack the 2-carbon on ethylene side-group.

I have tried to make as good a picture as I could from what I know. Any correction to the energy levels is appreciated here. I wish I could do this over a software but I don't have access at the moment.

I'm rather curious as to how you come up with these energy levels. From Hückel theory, both benzene and ethylene have the same frontier orbital energies: $\alpha + \beta$ for the HOMO and $\alpha - \beta$ for the LUMO. I'm sure joining them together, as in styrene, will make a difference, but how do you come up with this particular arrangement?

@orthocresol I will look for an explanation for this. Even I have to verify this once before I go on telling you about the correct way to do it xD

You didn't just make it up, did you?

Lol, no. This is accurate to the best of my knowledge. Once again, I will verify this and get back. The statement I made regarding a nucleophile attacking the 2-carbon of ethylene side-group ascertains my hypothesis. There is no other place a nucleophile would attack this molecule as the pi bond must shift towards benzene ring in that case.

@Rakesh Arya You said you will come back.


text:
Which is the most stable resonating structure of naphthalene and why?
I read that the symmetric one is more stable than the other two but I didn't understand why. I am in high school so please keep that in mind

sorry, but in high school we make 3 different resonating structures and then compare their stability based on factors like charges and all. We get questions like which is more stable R.S. so that's why I required the help of someone who got what was my question. I appreciate your help though

Try mentioning the word "delocalisation" in class, and see how your teacher reacts. ;-)

Plus one just because I think too much down votes aren't deserved. It is correct to point out to novices that resonance structure are just descriptors and do not exist. As such, they cannot be called stable or unstable. Tough, one can say a resonance structure has associated  high / low energy.... This is perfectly fine.

The symmetric structure is the most stable, as it does not cause a strain of the condensed ring structure because of a longer length of a single bond, which would cause deviation from regular hexagons.
See Bond Lengths in Naphthalene. It looks, comparing the measured versus predicted length of the central bond 1+1/3, the symmetric structure is little preferred. It does not tell exactly why, though.

If we compare the compact pictures of benzene and naphthalene in the orbital representation with the cumbersome resonance representation, we might wonder why the latter picture has survived. The answer is simple: The resonance picture contains a great deal of valuable information which is often obscured in the orbital picture. As an example, suppose we wished to estimate the bond lengths in naphthalene. If we used compact picture, we might guess that they were all the same length. If, however, we chose the resonance picture of and assumed that all three structures made equal contributions to the hybrid we would see that the bond between C-1 and C-2 was double in two of the structures and single in the third. This bond is a 12/3 bond (2/3 double and 1/3 single). By contrast the C-2, C-3 bond is a 11/3 bond (1/3 double and 2/3 single). If we refer to the graph of bond order versus bond length in Fig. 4-3, we observe that a 12/3 bond is predicted to be 1.36 Å long and a 11/3 bond 1.42 Å long. The precise values found for these bonds in naphthalene are 1.364 and 1.404 Å. Although agreement is not perfect, we thus arrive by a simple process at reasonable estimates of bond lengths. Other uses of the resonance picture will become obvious as we proceed, and we shall use both resonance and molecular orbital pictures interchangeably, depending upon which one appears to illustrate a point more clearly.

Note that resonating structures are just artefacts of the resonance bond model and do not really exist. following more advanced theory of molecular orbitals, all bonds have character of about 1.5 bond, as $\pi$ electrons are distributed across all the naphthalene structure, not being localized each between just 2 carbon atoms. A question should rather be, if the density of $\pi$ electrons is even, or if some bonds are preferred.


text:
how many calories per unit of sweetness between table sugar and aspartame?
How many calories per unit of sweetness between table sugar and aspartame?

Welcome to Chemistry Stack Exchange. This is a question that could easily be answered by using your favor internet search for "sucrose calories" and "aspartame calories". Can you give us a little bit more information on how this information is hard for you to find or interpret?

This is the wrong question to ask. The right question is "how many calories per unit of sweetness". Even if the calories per gram are the same, the calories to give a specific amount of sweetness (the primary use of the molecules) is all that matters in practice.

Aspartame 4 calories per gram source
Sucrose 3.9 calories per gram source
But you need far less aspartame than sucrose to get the same level of perceived sweetness

I believe in giving good answers to questions even if the question is not quite there yet. In this case, it is clear what the question is, even if it is not clearly formulated. It would be nice to address the lack of "denominator" in the question, i.e. calories per what.


text:
What is best to clean mold off PVC plastic window frames?
To remove black mold on window sills right next to the glass pane (as a result of condensation), I usually use a combination of water and dish soap or diluted bleach. Is there another chemical that would work better? Would Rubbing Alchohol (isopropyl) hurt the PVC? Does the bleach hurt the PVC? Thank you for any answer you can give.

Contractors frequently use water with a commercial cleaner (surfactant, alkalizer/chelator like NTA) and added bleach (about 10-20%) to pressure wash vinyl siding on houses. The bleach lasts only several hours because it oxidizes the surfactant, so the mixture must be made fresh each day.
For home use, a little dishwashing surfactant in water plus a little bleach should serve quite well, without doing any harm. Just wipe up with a sponge or rag and don't leave puddles on the windowsill.
Condensation frequently occurs indoors on windowframes in winter when humidity is high. Some people like the humidity for easier breathing. Aluminum window frames are particularly effective at condensation; vinyl can be better, but not quite perfect.


text:
Is there such a thing as a temperature-controlled reflux column?
You know the deal - you're doing a distillation in the lab, but want a fraction that comes over at a very specific temperature. I find it very hard to control the temperature in my air-cooled reflux column. For anyone that's ever had to wrap their reflux column in aluminum foil, you know what I'm talking about.
Is there a piece of laboratory equipment that controls the temperature of the reflux column within a degree or so of an arbitrary setpoint? I'm thinking something like a condenser column (Liebig, Graham, etc) where the water (or other working fluid) that circulates through the jacket also passes through some sort of temperature controller so that the water/fluid keeps the walls of the condenser at a tightly controlled temperature. Any tips on where to find this kind of device?
I know of vacuum-jacketed reflux columns, but I'd prefer something where I didn't need to use a very long tube in order to separate close fractions.

Back in the day spinning band distillation stills had controls for the pot, column and head.

Yes, there are (were) electrical patches or bands, similar to thermal blankets in operating principle.


text:
Chemical analysis of soapstone
Soapstone is a metamorphic mineral with a number of constituent compounds including principally:

silicon dioxide
magnesium oxide
iron oxide
aluminum oxide
calcium oxide
water/carbon dioxide

What would be a typical procedure to analyze a sample of soapstone and determine the proportions of its constituents?

Questions without elaboration effort are not very welcome, and may be closed. Always think about and search for possible answers thoroughly before posting a question.
Written effort attracts elaborated answers.
Include eventually all relevant  circumstances and background to prevent wrong assumptions and requests for clarifications.
Chemistry SE site expects to explicitly elaborate the question by at least basic textbook/online search, writing down what you already know/thought about, what you found/understood or not/tried/failed.

Given its heat resistance to thermal decomposition and the high chemical inertness of soapstone, as noted in this source, for example, to quote:

Soapstone is chemically inert, resistant to heat and durable, yet also malleable...As a metamorphic rock, it is very resistant to heat. In fact, soapstone is often used for casting metals, such as lead, pewter and silver...Because soapstone is primarily talc, which is chemically inert, it is highly resistant to etching and chemical reactions that may mar it’s beauty.

I would suggest placing a small amount of finely powdered soapstone in a vessel with highly alkaline aqueous mix of Sodium hydroxide (NaOH).  Apply conventional heating or, given an iron oxide presence, microwave heating. This is an attempted relatively safe synthesis of, for example, Sodium aluminate.
The proposed reaction is a hot alkaline attack of any of the several metal oxides present, including, for example, aluminum oxide, which could proceed as follows:
$$\ce{Al2O3 (s) + 2 NaOH (s) + 3 H2O (l) -> 2 Na[Al(OH)4] (aq) }$$
Support is provided per this reference, to quote:

All the solutions removed some $\ce{SiO2, Al2O3, and Fe2O3}$ from the samples, but the quantities were small. Sodium hydroxide attacked the kaolin group minerals more strongly...

I would then neutralize the soluble salts with $\ce{CO2}$, creating corresponding carbonate (or oxide, in the case of Aluminum ions) precipitates.
Please read the cited reference for more information on procedures (albeit, as alluded to per this 1971 source).

but what about non-destructive methods? Some spectroscopic tools work well for elemental analysis in minerals.

Matt_black: Good comment. I have since edited to add the date of my cited reference (1971) and do agree, more possible paths likely do now exist.


text:
Difference in titration curves
In my titration curve for HCl being titrated with NaOH, the initial pH-value was plotted as 2.1 according to a pH-meter. When I found the equivalence point and calculated the concentration of HCl, it was 0.075 M. I calculated how much the initial pH-value should be from that by -lg(0.075) which is around 1.12. That is way off from the original pH-value I got.
Why did this happen? Is it because the risk of me calibrating the pH-meter incorrectly? If that is the case, will the x-value of the equilibrium point also be wrong?? For the latter, I feel like the answer would be no, because what is wrong is that if the pH-meter was calibrated wrong, then all of the values would be 0.9 pH higher than what they should be, so the curve would only differ in the y-direction, but would be exactly the same in the x-direction. Is that correct?
Thanks in advance!

Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of $\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. Just for the fun try this experimentation : Take your $\ce{HCl}$ solution with an electrode showing  pH $2.1$. Add an equal volume of saturated $\ce{NaCl}$ solution (which is neither acidic nor basic). You will not believe your eyes by reading your electrode : The pH values goes down to $1$, as if diluting the initial solution produces an increase of concentration !
The activity is sometimes described as the amount of $\ce{H3O+}$ ions divided by the volume of the so-called "free water" (and not by the volume of the solution). In concentrated solutions the "free water" is this small amount of water that is free to move, and not attracted and fixed around the ions. One Liter $1 M$ $\ce{HCl}$, should contain much less that $1$ Liter free water, maybe about $0.1$ Liter free water, or less.

Intreresting! I did not know about that. Thank you!


text:
Acidity of butyric acid vs. thiophenol
I am being asked to compare the acidities of various compounds for an assignment, but I am confused about why butyric acid is a stronger acid than thiophenol/benzenethiol.
From what I was taught, I assumed the element effect (for elements in the same group on the periodic table) would take effect. Since sulfur is larger than oxygen and has a larger electron cloud to stabilize a negative charge, it has a more stable conjugate base and thus should be the stronger acid.
However, according to PubChem, $\mathrm{p}K_\mathrm{a} = 4.82$ for butyric acid, and $\mathrm{p}K_\mathrm{a} = 6.62$ for thiophenol.
What contributes to increased acidity of butyric acid? And why does it overrule the element effect?

Resonance, resonance. It matters a lot! If you want to compare apples with apples, then you should look at thiophenol (6.6) and phenol (~10), in which case your argument works.

I'd hate to be in the lab unless there's a very effective hood. ;-) Then there's selenophenol... https://blogs.sciencemag.org/pipeline/archives/2012/05/15/things_i_wont_work_with_selenophenol

“The biggest stinker I have run across. . .Imagine 6 skunks wrapped in rubber innertubes and the whole thing is set ablaze. That might approach the metaphysical stench of this material.”

Butyric ac. is a real carboxylic one.

Similarly, to compare pears with pears, compare butyric acid pKa=4.82 with acetic acid pKa=4.75 or formic acid pKa=3.75. Where is the increased acidity ? Carboxyls are acidic, as its carbonyl carbon atom has partial positive charge . It polarizes the $\ce{O-H}$ bond, making its braking to ions easier, and stabilizes the formed anion by charge delocalization over the whole carboxylate group. That is all.


text:
How can one determine if all reagents will react in the reaction?

Is it correct that all the reagents above the arrow will participate to form an intermediate product, to which the reagents below the arrow are then added, and they react to form the final product?
Thank you in advance.

Welcome to ChemSE! Your question will likely be considered as homework. You need to make some effort to answer the question yourself . Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange. The reagents are there because they are critical to the success of the transformation.

Thank you... No, it isn’t homework question. And am not asking to get my question solved. I just need to understand when reagents will be considering for reaction. How to determine if a reagent take place within the reaction. It is just a general question. But i tried to upload this picture to explain my question and make it clear @user55119

One very important aspect about organic chemical reactions is their very detailed descriptions about reagents. Some of them are catalysts or inhibitors. Some of them are solvents. Some of them are acids or bases which provide an optimal medium for the reaction to happen. In your case, in the first step you perform the oxymercuration first and then the demercuration with sodium borohydride. Methanol is the polar solvent we need in this reaction. I suggest you read "Robert B. Grossman - The Art of Writing Reasonable Organic Reaction Mechanisms".

Is there a list that one can memorize to differentiate between solvents which are part of the reaction and mechanism, and the ones that don’t affect the reactions at all? @TheRelentlessNucleophile

I doubt such a comprehensive guide exists. Yes, there are lists of solvents commonly used throughout organic chemistry (https://organicchemistrydata.org/solvents/) and other useful reagents and their purpose (https://en.wikipedia.org/wiki/List_of_reagents). But predicting their exact behaviour and how they are going to impact the reaction is someting the chemist has to train themselves, through consistent exercise and studying the literature.

Chemistry is neither a set of easy rules, neither set of empirical knowledge to memorize. It is both, intertwined in a wild mixture of both extremes. Most non chemists are too impatient to recognize many patterns that are hybrids of rules and evidence. Chemistry is projection of very complex quantum interatomic relations to our macro world.

That is correct, the two numbered reagent and solvent pairs refer to two reactions, that will be done one after the other.
There is an intermediate product that can be isolated, or at least identified.


text:
Smallest unit of Sodium Chloride
Since ionic solids are not molecular, can a single sodium chloride unit exist and if so, what is such a unit called? I am aware of the concept of "formula units", but as far as I can tell, this is a notional term and does not describe any physical structure. Is "unit cell" the correct term and can a single unit cell exist?
Just to be sure people understand the question: I am talking about solid sodium chloride and not an aqueous solution of it.

Suppose you have a crystal of sodium chloride. What does the surface of the crystal look like? Now make the crystal smaller.

A “formula unit” would be closer to the answer than a “unit cell”. A unit cell is used in crystallography to define the smallest repeating unit of the crystal that has all the symmetries of the overall crystal structure. For NaCl, there are a variety of different ways to visualize the unit cell (2 interpenetrating FCC lattices, FCC with a 2-atom basis, etc. ), but regardless, there are four formula units of NaCl in the NaCl unit cell (4 Na and 4 Cl atoms).