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XSteam IF97 tables (MS Excel version) lack of inputs
I am using XSteam IF97 to evaluate steam power plant cycles.
However, I miss some important input options such as pressure as a function of entropy and temperature ("p_st()") - when I assume isobaric heat addition (neglecting pressure any pressure losses) while reheating the steam. I end up looking at i-s diagram which is highly inefficient while optimizing cycle for different inputs.
Is there any way around or any more sophisticated software (in the best scenario, available for MS Excel). Or how do you work with steam power plant cycles?
(I intend to avoid any commercial software)

What is an i-s diagram?  Anyway, try using a p-H diagram.

@ChetMiller Enthalpy–entropy diagram (rather h-s, in my language it's also called i-s). However, my point is not to use any diagrams at all, I would like to obtain the pressure with XSteam or another software, because there is no point computing the whole cycle in Excel for fast parameters switching, while I would have to get one or two values manually.

Aren't there any other functions in if97 that are more suited for looking at this than temperature-entropy., like one where you input temperature and pressure and output enthalpy?

@ChetMiller The situation is: At a certain point, the pressure is unknown to me (for this point I also don't know enthalphy, I know only entropy). However, I can get to this point by following isobare from the point "above" (with a higher temperature), hence I need to obtain isobare based on entropy and temperature.

I'm confused.  If you know the pressure at the point "above," and it an isobar, then you know the pressure at the point in question.  What am I missing?

I'm sorry if I explain this confusingly, but I'm trying my best!
For the point "above", I know the temperature and entropy (but I don't know the pressure) and XSteam cannot output $p = f\left(s,,T\right)$. If it could, I would obtain the pressure for both points at once, and for the point "below", I could then obtain temperature and enthalpy "or whatever" (as mentioned: from the point "below" to the point "above", the path follows isobaric heating, hence temperature and entropy raise).
I hope I managed to clarify the situation a bit.

I'm not familiar with this software.  Can you please summarize what functionalities it does provide you?

@ChetMiller I won't fit the whole description in the chat here: the MS Excel version is available: http://xsteam.sourceforge.net, but I don't want to bother. It's very popular among students as it is free and does not require any SW. Just Excel - you can obtain properties of water and vapor based on two different values (in the most cases): $p,: T,: s,: v,: x,: \rho,: c_p,:c_v:$ etc.

Your inputs are the temperature and entropy, and you want to output the pressure, correct?  Do you know the reference state for the entropy?

Do you have an initial guess for p?

If you have an initial estimate for the pressure, $p_0$, then you can quickly solve iteritively for the pressure that matches the desired entropy $s=s^*$ using modified Newton's method:  $$p_{n+1}=p_n-\frac{s(p_n,T)-s*}{\left(\frac{\partial s}{\partial p}\right)_T}$$The partial derivative of s with respect to p can be obtained by finite difference at the initial guess for p.


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One-piece short distillation glassware - worth it?
Does anyone have any experience using one-piece short distillation setups similar to this: https://www.hornik.eu.com/en/3-985_distillation-bridges/8216-distilling-link-for-short-path-distillation-duran-cones-ns1423-socket-ns1423.html?
I've been struggling with a tricky distillation recently and was wondering whether something like the above would help. I wanted to hopefully get some opinions before shelling out.
Cheers

Why use such a small device ?

Define "tricky."

If tricky means you must collect only few drops than that is of immense help.

@Maurice come on, why asking?

@Alchimista I'm trying to distill off 1-2 mL of my product, which has vapours coming off at ~65C. I'm trying to heat the crude to no more than 90C to prevent possible product breakdown. Currently I'm having trouble coaxing the vapours up and into the condenser - I'm having to use plenty of tin foil to try and trap the heat in the glassware between the crude RBF and the condenser.

@Zhe See my response to Alchimista

@Technocratic yes you need a micro apparatus — milliapparatus in this case ;-) something like in the pic. Not too micro because they are expensive. Look around if you are at an institution. There should be something even to borrow from a group.

Maybe try Kugelrohr.

The complete all in one kits can be a pig to clean, much more so if a flask is incorporated. This doesn't look too bad. Are you using cotton wool or glass wool underneath the foil? I wouldn't spend out on tin foil, aluminium is cheaper

@Beerhunter I'm not using cotton/glass wool underneath the foil - would you recommend? And yes I meant to say aluminium! I've bought a one-piece short distiller similar to the one in the picture so fingers crossed it'll be worth it.

When the insulation really matters, and it can do on these smaller kits, I would use one of those options, then foil and use cable ties/wire to secure a snug fit

source: Stack Exchange

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

sort: votes


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Is iron(iii) hydroxide equivalent to hydrated iron(iii) oxide?
Are the chemicals $\ce{Fe(OH)3}$ and $\ce{Fe2O3⋅3H2O}$ equivalent?  They are both network solids and have the same empirical formula but I know that doesn't always mean they are structurally the same.

I'd rather not use long words. These two are the same.

Note that iron(III) hydroxide/hydrated oxide, or more generally all/most such substances of  transient metals, do not have well defined structure nor composition.

Our lab did over a thousand XRD patterns of corrosion deposits and these compound patterns often showed up , Technicians spent hours with the Fink Index trying to separate them  . From a practical corrosion prevention perspective , it makes no difference.

Fe(OH)3 is iron(III) hydroxide, which is a form of iron hydroxide and is different from anhydrous iron(III) oxide (which is Fe2O3) and hydrated iron(III) oxide (which is Fe2O3·nH2O, where n, in this case is 3). See the list of iron oxides and iron hydroxides.
Also see this Wikipedia article which says,

Several hydrates of Iron(III) oxide exists. When alkali is added to solutions of soluble Fe(III) salts, a red-brown gelatinous precipitate forms. This is not Fe(OH)3, but Fe2O3·H2O (also written as Fe(O)OH). Several forms of the hydrated oxide of Fe(III) exist as well.

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


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Use of TsN3, toluene at high temperature
My reasoning:
My understanding is in first step the alcohol’s O- does an Sn2 attack on the OTs containing carbon. In the second step I am confused where does the reaction happen in the triple bond or at the O-C-O Carbon?

You are correct about step 1. The azide will do 3+2 cycloaddtion with the alkyne to give a triazole, see here: https://www.organic-chemistry.org/synthesis/heterocycles/1,2,3-triazoles.shtm

There maybe a regiochemical issue in the 3+2 cycloaddition.


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Reduction of Benzamide in acidic conditions?
I am going through a mannich type reaction and ended up with my product show in the image. I compared my answer to the answer key but I noticed that the benzamide was reduced completely. I was wondering at what process during the reaction the benazmide was reduced.

You dropped a methyl group.  You did a mysterious oxidation of a benzyl group to a benzamide. The answer key is reasonable for a Mannich.

Bn is PhCH2 not PhCO

ohhhh, i see. Thank you guys for the help

Bn = benzyl. Bz = benzoyl.


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What does a filled, solid circle mean in a chemical structure?
I'm looking in a textbook at an asphaltene molecule.

The two aromatic rings next to the Nitrogen are shaded. What does that mean? I think the Nitrogen's create delocalized electrons in this configuration, so maybe the shaded circle is the range over which that free radical is delocalized?

This is not a universally accepted notation.

So it means "Ooh, I made a pretty picture!"

The open circles of course represent aromatic rings, however this is no now the accepted drawing. The accepted way is to draw out the double bonds. //  I'd guess that the solid circles are for heterocyclic aromatic rings.

Dunno, there is a hetrocylic ring with sulfur that doesnt get the solid circle. I've tried contacting the authors but this was long time ago: Gafanova O.V.,Yarranton W. (2001) The Stabilization of Water-in-Hydrocarbon Emulsions by Asphaltenes and Resins, J. Colloid Interf. Sci. 241, 469-478. .... IGNORE! I've found Professor Yarranton.

@Tunneller The sulfur ring isn't unsaturated so doesn't really count as an exception.

BTW, this may be called multisubstituted bathroom tile ;-)


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What makes the enthalpy of hydration of calcium chloride dihydrate unique?
I have recently done an experiment in an attempt to find the relationship between the water ratio in a hydrate and the enthalpy of hydration, and my results showed a correlation, but an anomaly for calcium chloride. My results were:
Sodium carbonate monohydrate: 10.03 kJ
Calcium chloride dihydrate: -19.51 kJ
Sodium acetate trihydrate: 28.98 kJ
Sodium thiosulfate pentahydrate: 47.37 kJ
Sodium sulfate decahydrate: 66.32 kJ

I have tried researching about what makes calcium chloride dihydrate unique, but I can't seem to find anything. And this is not a random error, as each of these salt compounds had 3 trials each, with consistent results.

In this group, it is unique for two reasons. It is the only species with calcium cations, and the only species with chloride anions.

Try also different hydrates of the same compound, like anhydrous, dihydrate and hexahydrate of calcium chloride.( Other hydrates exist, but are less common ).

The heats of hydration of your salts, plus some others, can be obtained by subtracting the heat of formation of the anhydrous from the heat of formation of the hydrated salt, using the numerical values from the Handbook of Chemistry and Physics, 43rd Ed., 1961, pages 1807 - 1831. The so obtained values are rather different from yours, but the tendency is the same. Here are the obtained molar heats of hydration, divided by the number of water molecules. They are strangely similar.
\begin{array}{lrl}
\text{Compound}      & E /\pu{kJ (mol \ce{H2O})-1}\\\hline
\ce{Na2CO3.H2O}      & 295.3\\
\ce{Na2CO3.10H2O}    & 295.3\\
\ce{NaCH3COO.3H2O}   & 297.6\\
\ce{Na2SO4.10H2O}    & 294.3\\
\ce{Na2S.{4.5}H2O}   & 305.0\\
\ce{Na2S.5H2O}       & 304.6\\
\ce{Na2S.9H2O}       & 301.1\\
\ce{Na2S2O3.5H2O}    & 297.2\\
\ce{Ca(NO3)2.4H2O}   & 297.7\\
\ce{Ca(CH3COO)2.H2O} & 291.2\\
\ce{CaCl2.2H2O}      & 179.6  & (!)\\
\ce{CaCl2.6H2O}      & 301.8
\end{array}
It is immediately clear that the heat of hydration of one molecule $\ce{H2O}$ in these compounds is $\pu{300\pm10 kJ (mol \ce{H2O})-1}$. But there is one odd exception: $\ce{CaCl2.2H2O}$. Has anybody an explanation?

Try https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/4727#4727

That is hardly an answer!

@Mithoron. I agree : it is not an answer. But I had not enough room in the "Comment" section" to insert my text and its numerous numerical values.

This is much better suited as an edit to the question. I'm sure op will appreciate this, too.

Maybe make it CW for others to put it into shape?

Am I allowed to add all my text and numerical values to Shkadov's demand ? And what is CW ?

@Maurice This is very helpful, thank you! Do you have an online source that states that the heats of hydration is equal to the differences in the heats of formation? I've tried researching it but have had no luck so far.

@shkadov. If I copy my text with numerical values and insert it in your first demand, there would be an important discrepancy between your values and mine, which is to be avoided. My values are much greater than yours. So what ?

@Maurice CW=Community Wiki. Its essentially a way of giving up ownership of post so that other users can freely edit and each receives credit (though no further rep will is earned once it is converted)

I have tried fixing up the table with MathJax (there are markdown tables now, but it was "easier" for me this way). I'd like to reiterate, this shouldn't be an answer. I see no problem with adding this to the original question though. Maybe this would have improved this whole post, leading to it to be reopened and maybe attract an answer then. In general, if you need help with the site (be it MathJax or otherwise), please ask on [meta].


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Who figured out that different isotopes have different numbers of neutrons?
I reasearched about the history of the discovery of neutrons and couldn't find who found out that isotopes are a change in the number of neutrons. Can you please tell me who it was that made this discovery?

See also History of science and mathematics SE site https://hsm.stackexchange.com/   Btw isotopes are a change in the number of neutrons is rather sloppy formulation. Because such a change is a neutron capture or ejection and not an isotope.

Frederick Soddy, I guess. But it is the other way around. He figured that the things that are chemically similar, but have different neutrons should be called something, and called them isotopes. He is the one who coined the term.

Chemistry knew about elements with different atomic masses in their atoms well before they knew neutrons were the reason for this. Neutrons were only discovered in the 1930s at which point a good explanation for the observations could be finally made. So who deserves credit is, perhaps, unclear as the observation existed long before the explanation.

@IvanNeretin Indeed it makes sense that before people could count the neutrons in an atom, they were not aware of isotopes as the difference is so subtle and isolation so difficult.

@Matt_black. Could you remember one example of elements having been weighted and found to be different from other determinations made in some other lab ?

@Maurice It isn't that different laboratories measured different values. The original experiments by Soddy and JJ Thompson in the 1910s showed that some atoms of the same element had different weights (eg Neon has atoms weighing 20 and 22 units). By 1919 better apparatus (perfected by Aston) led to the discovery of new isotopes "every week" as one book reports.

@Greg actually Deuterium was discovered the year before neutrons were so they did know about isotopes before neutrons

From Pauling's "General Chemistry" (ref. 1):

In 1921 the American chemist W.D. Harkins described nuclei as being built of protons and neutrons; he used the word "neutron" for a hypothetical particle with mass equal to that of the proton and with no electric charge. Ernest Rutherford made a similar suggestion during the same year. In 1932 the discovery of tne neutron was reported by the English physicist James Chadwick (born 1891).

The original article in which Harkins (ref. 2) reported this hypothesis is linked by the Wikipedia:
References

Linus Pauling, General Chemistry, second edition, 1970, p. 102
Harkins, William (1921). "The constitution and stability of atomic nuclei. (A contribution to the subject of inorganic evolution.)". Philos. Mag. 42 (249): 305.


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How to set up an ICE table for adding a weak acid to a strong acid
I am working on setting up an ICE table for adding 0.020 mol HCl to 1.0 L of 0.10 M propanoic acid.  Since these are both acids (propanoic acid being the weaker one and HCl being strong), how would I set this up in an ICE table to calculate the pH of the resulting acid?
Edit: I was able to find an answer by taking the negative log of 0.02 to get the pH (because HCl is a strong acid).  However, I am unaware with how to even set up the ICE table.

You should provide at least your failed trial, otherwise the question may be voted to be closed for not sufficient effort.

Thank you for your suggestion - I am unaware of how to set up the problem and was simply taking the negative log of 0.02 HCl due to the fact that it is a strong acid.

Hint: There is the universal procedure  to count with mass and charge balances and chemical equilibrium given by dissociation constants of weak acid and water. Then there is analysis what can be eventually neglected to simplify calculation.

What is the meaning of ICE ?

@Maurice  RICE chart or RICE box or ICE table.
"tabular system of keeping track of changing concentrations in an equilibrium reaction. RICE stands for reaction, initial, change, equilibrium. ".  Personally, I have never used them, or at least never heard the term until I came to CH SE site. The usage may be region or language specific.

ICE tables tabulate amounts or concentrations pertaining to an equilibrium reaction: at the initial stage (I), how they change (C), and what they are at the end or at equilibrium (E). They are an alternative to solving multiple simultaneous equations (equilibrium constant expression, mass balance), instead using a single parameter (related to the extent of reaction $\xi$).
For the problem at hand, there is one reaction that goes to completion (strong acid) and one that attains equilibrium (weak acid, both protonated and deprotonated species present at equilibrium).
The ICE table needs to have species from the second reaction only. A good estimate of the hydrogen ion concentration is the concentration of hydrochloric acid (it dissociates completely, and we are ignoring the autodissociation of water for now).
$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{HP}] & [\ce{H+}] & [\ce{P-}]\\ \hline
\text{I} & 0.1 & 0.02 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & 0.1-x & 0.02 +x & x\\ \hline \end{array}\\\\$
If you do the math, you will find that very little of the propanoic acid will dissociate, as the $\mathrm{p}K_\mathrm{a}$ is far from the initial pH.

[H+]=0.02, not 0.2

@Poutnik I should stop answering questions when I am not wearing my reading glasses!

I see, glasses are important part of my life too.


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Why does matter have spaces between them?
If you mix  sugar Crystal in a glass of water and mix it well, the level of water will not rise.The reason they say is that matter have spaces between them.If matter have spaces between them , How come that empty space is not visible to our eyes.If I look at a glass of water , I see all the molecules mixed up well.Not like there are some areas where there is no H20 molecule and some they’re are.
What does it exactly mean and how does it look like and what is happening there?
My thinking:
Is it like there are intermolecular forces between H20 molecules but they are at a separation from each other and still have the bond.Why does level inc if I put my finger.

your eyes cannot perceive distance that is that small.. Your finger is much bigger than a grain of sugar which the crystals are made up of.

Is my thinking correct or is it sth elese.

To be clear, dissolving sugar in a cup of water does cause the water level to increase, even if there is no solid left over.

Sucrose ( bulk, not loose/powdery) density is 1.59 g/mL. Volume of separate 900g water and 100g sucrose is about 900+100/1.59=963 mL . Density of 10 w/w % sucrose solution is 1.038 g/mL, so the total volume changes from 963 mL to about 1000/1.038 = 963 mL. So the total volume practically does not change.

Note that there is the law of mass and energy conservation, but there is no law about volume conservation.
Molecules of matter are in eternal motion. Molecules of gases move freely by flying between collisions. Nitrogen or oxygen molecules of air have an average speed of a supersonic fighter, colliding at rate typically 10 billions collisions per second, with the mean free flight distance typically 70 nm. Such motion creates space between them. ( Try to keep a sworm of vivid children in tight packed formation. )
Molecules of liquids and solids are held together by attractive forces. Loosely for the former, so they continuously separate and rejoin. Tightly for the latter, so they just vibrate.
Another reason for space between molecules is electrostatic repulsion of their electrons, if they get too close. If you hit a wall, you did not really touch it. The wall started to repulse you by the mighty electrostatic force, when your and it's electrons got too close.

If you mix 1 L of ethanol and 1 L of water and let it cool down ( because it warms up ), the total  volume will not be 2 L, but about 1.96 L. It is due the fact the average energy of bonds water-ethanol (via hydrogen bonds) is greater than the average energy of bonds water-water and ethanol-ethanol. This leads to shorter average distance between molecules ( fractions of nanometre ), as stronger bonds are shorter, and to the volume contraction.

Aside of this chemical reason, there is geometrical one as well, both adding to each other:
Fill a container e.g. by
peas/beans, and fill an identical container by rice. Then mix them and try to fill both containers again. You will see some volume will remain free. Rice will partially use otherwise free space between peas/beans.

The similar phenomena probably happens for the system sucrose-water, as I have no such personal experience.
Generally, such volume changes, smaller or bigger, always happen when 2 components are not indifferent with respect to each other and properties of the mixture are not a linear function of its composition.


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What plastic is this? (Raman scan)
I can't recognize the Raman spectrum of this plastic.

It's not in the common plastics raman spectrum.

It is plastic adaptor for connecting sediment filters. Anyone got idea or recognize what kind of plastic is it made of? I'm studying the high temperature resistance of it. I bought a hot water sediment filter and using it as adaptor.

I did not see any fluorinated plastics in your reference spectra. PTFE comes to mind.

I can't formally be sure but that is likely Teflon as suggested above. Eventually PVC. I have noticed that you treat Raman spectra as the way to identify materials. While it is certainly true that most of them have their Raman spectrum, Raman spectroscopy is not the best suited for tabulated data and tables. You can use it as it if would be IR, but within a sample related and an equipment/setting related frameworks.

It does't match the raman spectra of fluorinated plastics, Teflon or PVC. What materials could be pipe adaptors be made of? Is there a plastic alloy? What you mean raman spectroscopy is not the best suited for tabulated data and tables?  And what you mean one can use it as if it would be IR? What are you talking about? It's a personal raman with 532nm wavelength laser.

The fact that is a personal Raman with excitation at 532 more or less resume my comment. I mean IR (on standard configuration and with textbook samples) does have fingerprints. Not so much Raman.

Did you mean raman with 785nm or IR spectroscopy?  Most plastics I have scanned have very clear raman spectrum (all the recycle codes from 1 to 7 I have seen have them). The unknown plastic above has very clear raman too. It's so difficult to do IR spectroscopy as sample preparation difficult. With my 532 nm raman. I can take it out to bars and easily scan the contents of closed containers, etc.

I mean that is difficult to compare Raman sp as it can be done with IR ones. You are right, can't be a fully substituted molecule as it seems you have both aliphatic and aromatic C-H str. Still for such a screw I would expect Teflon (now excluded), PVC, perhaps PS.

it doesn't match the exact spectrum of PS.. see https://imagizer.imageshack.com/img923/3127/SH9znP.jpg, nor PVC, see   https://imagizer.imageshack.com/img924/7128/IEUj01.png   what wavenumber is the alipatic or aromatic C-H str? What's C-H str?


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Nature of materials
Many a time I become confused when materials (and substances) are described with the following words: granular, crystalline, pellet.
Worse still, the same material can be described with a combination of the words (e.g. sodium chloride granules vs sodium chloride crystals).
My question then is the exact difference between these three words.

A crystal is the generic term for solids, having at atomic/molecular level regular structure, what macroscopically manifests in formation of regular geometric 3D crystallic objects. The size and shape can be natural, formed by crystallization from solutions or melted solids, or mechanically broken, e.g. in production mills.
A pellet or a granule refer rather to a particular shape than to nature of the material or way of its forming. They are not particular crystal forms, but they rather relate to technology that produce them. They do not follow the natural geometric patterns of crystals and are usually polycrystallic. Solids are forced by the technological process to take a specific shape. Material of pellets/granules may not be crystalline at all.
Pellets as somewhat flatted shapes are often result of pressure binding and cutting of solid microcrystallic or amorph matter. But it could be prepared from originally liquid, or liquid+solid forms as below.
Granules as more spherical-like shapes than pellets. Granules of homogenous inorganic or organic substances are often produced by fast cooling of melted solid or by fast evaporation of hot saturated/oversaturated solutions. Alternative option is using a solidifying binder, if granules do not have homogeneous composition - e.g. like pet feed.


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Extracting other mineral from salt in evaporated seawater
I recently stumbled upon this news:

Seawater contains 47 minerals and metals. Starting with the most abundant and proceeding to the least abundant, these are chloride, with a concentration of $\pu{18 980 ppm}$ (parts per million) in seawater, sodium ($\pu{10 561 ppm}$), magnesium ($\pu{1 272 ppm}$), sulfur $(\pu{884 ppm})$, calcium ($\pu{400 ppm}$), potassium ($\pu{380 ppm}$), bromine ($\pu{65 ppm}$), inorganic carbon ($\pu{28 ppm}$) and strontium ($\pu{13 ppm}$). Then follow boron ($\pu{4.6 ppm}$), silicon $(\pu{4 ppm})$, organic carbon ($\pu{3 ppm}$), aluminum ($\pu{1.9 ppm}$), fluorine ($\pu{1.4 ppm}$), nitrogen in the form of nitrate ($\pu{0.7 ppm}$), organic nitrogen ($\pu{0.2 ppm}$), rubidium ($\pu{0.2 ppm}$), lithium ($\pu{0.1 ppm}$), phosphorous in the form of phosphate ($\pu{0.1 ppm}$), copper $(\pu{0.09 ppm})$, barium ($\pu{0.05 ppm}$), iodine (also $\pu{0.05 ppm}$), nitrogen in the form of nitrite (also $\pu{0.05 ppm}$) and nitrogen in the form of ammonia (once more $\pu{0.05 ppm}$). Thereafter, we have arsenic ($\pu{0.024 ppm}$), iron ($\pu{0.02 ppm}$), organic phosphorous $(\pu{0.016 ppm})$, zinc $(\pu{0.014 ppm})$, manganese ($\pu{0.01 ppm}$), lead ($\pu{0.005 ppm}$), selenium ($\pu{0.004 ppm}$), tin $(\pu{0.003 ppm})$, caesium ($\pu{0.002 ppm}$), molybdenum (also $\pu{0.002 ppm}$) and uranium ($\pu{0.0016 ppm}$). Then come gallium ($\pu{0.0005 ppm}$), nickel (also $\pu{0.0005 ppm}$), thorium (also $\pu{0.0005 ppm}$), cerium ($\pu{0.0004 ppm}$), vanadium $(\pu{0.0003 ppm})$, lanthanum (also $\pu{0.0003 ppm}$), yttrium (also $\pu{0.0003 ppm}$), mercury (once more $\pu{0.0003 ppm}$), silver (also $\pu{0.0003 ppm}$), bismuth ($\pu{0.0002 ppm}$), cobalt $(\pu{0.0001 ppm})$ and,finally, gold ($\pu{0.000008 ppm}$). Altogether, there are some 50 quadrillion tons (that is, $\pu{50 000 000 000 000 000 tons}$) of minerals and metals dissolved in all the world’s seas and oceans.

Which is quite surprising for me (thought those white crystal was only $\ce{NaCl}$)
Assuming I already have TONS of general salts (mixed) from simple seawater evaporation, is there any way to separate each mineral? Let's say by pouring the salts in magnet (for material that is having magnetic properties), etc

(lithium salt pond, which seen exactly same as $\ce{NaCl}$ pond)
Apologize for shallow understanding of the topic. Any suggestion is highly appreciated since I can't find the answer elsewhere.
P.S: This question have been posted earlier in Physics

Many such extractions are technically available, but economically would wait, until prices of such minerals from other sources raise enough to be profitable.

I commented in the Engineering entry of this question that magnesium is only material economically produced from sea water ( other  than the obvious salt ).

I think bromine can be extracted economically as well, though this may depend on the source of the sea water.

One material extracted from seawater is magnesium.

In the United States, magnesium is obtained principally with the Dow process, by electrolysis of fused magnesium chloride from brine and sea water [emphasis added]. A saline solution containing $\ce{Mg^{2+}}$ ions is first treated with lime (calcium oxide)[sic; calcium hydroxide would seem more likely] and the precipitated magnesium hydroxide is collected:

$\ce{Mg^{2+}(aq) + CaO(s) + H2O → Ca^{2+}(aq) + Mg(OH)2(s)}$

The hydroxide is then converted to a partial hydrate of magnesium chloride by treating the hydroxide with hydrochloric acid and heating of the product:

$\ce{Mg(OH)2(s) + 2 HCl → MgCl2(aq) + 2H2O(l)}$

The salt is then electrolyzed in the molten state. At the cathode, the Mg2+
 ion is reduced by two electrons to magnesium metal:

$\ce{Mg^{2+} + 2 e^− → Mg}$

I would say the preferred method are magnezite/dolomite quarries, unless not available.


text:
Why is the concentration term in van der Waals correction for ideal gas squared?
In van der Waals correction for ideal gas equation, the pressure term has a correction where concentration term is squared. I saw derivations for this and they say that the concentration is squared because the number of particles at surface and the force exerted by other particles on the particles at surface, both are dependent on the concentration. I don't get this, could someone please explain why this is so mathematically?
Edit- upon reading more, I found that van der waal gave credit to Laplace for coming up with the pressure reduction term being directly proportional to the square of the density. Could you also explain this?

Intuitively, the rate of molecular interactions leading to vdW cohesion pressure formally follows kinetics of the 2nd order, so the volume is squared.

Could you please elaborate?

Try to simulate a real gas as mixture of ideal monomer and dimer gases in equilibrium, shifted strongly toward the monomer. Then write for the mixture the state equation f(p,V)=n(pure monomer)RT,  taking into account the chemical equilibrium.

By other words, one V is for the absolute deviation being proportional to 1/V, And the other V is for the relative deviation being proportional to 1/V. Twice as much of dimer stored in half volume causes 4 times greater pressure deviation

See:  https://chemistry.stackexchange.com/questions/70616/why-is-the-pressure-correction-in-the-van-der-waals-equation-proportional-to-n/126525#126525

Okay I get the explanation, but how does Laplace fall into this picture?

The links explanation of the cohesive pressure is better, even if I think that formally it is equivalent to the dimer model.


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What is the difference between fire and warm body?
A burning body has flames (Candle flame)whereas a hot or warm body (warm utensil or body being hot but doesn’t burn )is just very hot. Keeping hand near a burning body makes you feel hot but not exactly near a hot body.A burning body has flames which are yellow,then blue and purple for some outside in colour.
What exactly is a burning body and how does it differ from a warm-body?
Please tell how can I improve my question if not suitable ?

Define a "burning body"...what are you specifically referring to? A burning wooden log? Candle flame?

Yes I am referring to that.

Please also tell me how did you not get my question at first instance .What is the information that is missing ? It helps me improve my posts.Thank you

I feel that you should clarify what burning body you are referring to? If it is a candle flame, write candle flame. Be more specific please.

Being hot is a state. Being burnt is a process of ongoing chemical reaction, usually with aerial oxygen. If there are evolving flameble gases or vapours, a flame is formed with light emitting ions, radicals and hot solid particles. If nonvolatile solid is burning, like some metals, flame does not occur, heat concentrated on the surface, reaching very high temperature with bright light, like magnesium or termite. Material can be burnt without oxidation, just by thermal decomposition.

You're really asking about the difference between a physical change and a chemical change.*
In a physical change, the chemical identity of the substance doesn't change.  In a chemical change, it does.
Suppose you take cube of sucrose (table sugar) at room temperature.  If you heat it up, it will become warmer.  But it will still be, chemically, sucrose.  It will just be sucrose at a higher temperature.  Since the chemical identity did not change, it is a physical change:
$$\ce{C12H22O11(s, room\,T)  -> C12H22O11(s, warm\,T)}$$
[$\ce{C12H22O11}$ is the chemical formula for sucrose; "s" means solid, "l" means liquid, "g" means gas.]
If you warm it more (to $170 \,^\circ \text{C}$) it will melt.  Melting is a physical change, because after the cube melts, it still consists of sucrose molecules; though these will now be in the liquid state:
$$\ce{C12H22O11(s)  -> C12H22O11(l)}$$
But suppose you instead flow enough energy into it to break the bonds that hold the sucrose together, causing it to combust, in the present of oxygen, into water and carbon dioxide.  That is what happens when it burns.  This is a chemical change because, as it is happening, the sucrose and oxygen are being changed into new chemicals: water and carbon dioxide:
$$\ce{C12H22O11(s) + 12 O2(g) -> 12CO2(g) + 11 H2O(l)}$$
This chemical change releases energy, which (in this case) takes the form of heat and light (which is why it feels hot, and you can see a flame).
In sum, a warming body is undergoing a physical change.  A burning body is undergoing a chemical change.
[*Yes, some might take issue with this distinction, but it is used in all first-year chemistry textbooks I've seen, indicating there is a consensus among educators in this field that this is a useful distinction for beginning students to learn.]

I think this might be indeed useful for the OP, that certainly is a beginner.

Fire or flame is a luminous chemical reaction in the gas phase, i.e, such a chemical reaction is generating heat + light. Not all flames are very luminous e.g., hydrogen oxygen flame is very very light blue...hard to see.
Candle flames are yellow because of glowing carbon particles. The blue/purple color comes from small molecules formed by the decomposition of hydrocarbons emitting blue light. If you are burning wood etc., the yellow color can also come from sodium. The flames keep moving because the hot air rises around the flame.
A warm body is relatively more simple. There is no chemical reaction, that is you do not see multiple colors. A very hot body (again no chemical reaction) is a blackbody radiator. It emits a continuous spectrum. Flame spectrum is not continuous.

Fire is also produced when stones are rubbed hard and their spark if put on grass , generates fire.Similarly a lighter.How does that fire get formed . So how did fire actually come ?

How did that chemical reaction happen and why did it?.

When you rub two stones and see sparks, that is not fire. This is a glowing piece of stone. As I said, fire is a chemical reaction in the gas phase.

Also read about Flint https://en.wikipedia.org/wiki/Flint

@robertpatrick Stone sparks are not fire, but electrostatic discharge of high voltage, because of quartz piezoelectric phenomena. BTW, think thoroughly at least an hour about each your question, and search for possible answers, before posting such a question. Writing answers about topics described in textbooks and on internet many times is wasting human resources .

A hot or warm body stays without any change after the end of the experiment. It can be reheated over and over again. A burning body is changed into ashes at the end of the experiment. It cannot burn twice.
Edit : Answer to a comment by Robert patrick.  I don't understand the following question in your comment : How is that burn produced ? Why is it yellow ?  For me a burn is a mark made by a cigarette or a similar hot piece. It is brown or black. It cannot be yellow. Sorry. I don't understand. Maybe you would speak of the color of the flame. Well. The flames are not always yellow. The candle flame is yellow indeed, and this color is due to the presence of solid carbon in the middle of the flame. And carbon, as every substance heated to more than 800°C is incandescent, or white-hot, producing a yellow light. If you pass a spoon or any solid  object quickly across the flame, it will be soon covered by soot (charcoal), which shows that there is some charcoal in the center of the candle flame.

How is that burn produced? Why is it yellow?Why does the flame move a bit(You know like it moves a bit , for example the candle flame . It isn’t that it stays constant.

I hope you get my question .@Maurice.


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What is this white layer on brass?
I have an object made of what I assume to be brass, i.e. an alloy of copper and zinc.
The curved parts have both a black layer and a white layer (also note the verdigris (I guess) on the left top, but this is the only spot on the whole piece) :

If I understand correctly, the black layer is tarnish, or patina, and due to oxidation from sulfur dioxide in the air.
But I can't find reliable reports of white corrosion on brass.  Could it be wet storage stain, a.k.a. "white rust" ?
Hints on how to remove it are welcome.

As brass contains zinc and as most of zinc compounds are rather white or "colourless", it could be some zinc oxide, hydrated oxide, basic carbonate.. like that. Or some external dust ,or abrasive from past cleaning.

The "white rust" in your reference refers to zinc or Zamak /die cast alloys; nothing to do with brass. I think the white ,black coatings are more related to paint or dirt.

Judging from the location and colour, it is most likely residue from liquid polish that was not properly rubbed/polished off. Quite a few types of liquid polish for brass turn white when they dry, and any folds and creases in the design is where they would be most likely to accumulate if the person doing the polishing is not very thorough.
Depending on how long its been there, you may be able to shift it with just warm water and a soft brush or cloth, leaving the patina mostly in place.  Otherwise you may need to use brass polish and polish the whole thing.
You could try asking advice from some of the better antique dealers or ideally someone who specialises in conserving antique brasses if the piece is a valuable antique.

I just noticed a flat spot of the same white depot on a (non-metal) mineral part, so you must be right about it being some residuals of a cleaning product.  And it's indeed surprisingly easy to remove, a damp cloth seems to be enough.


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Name of proton directly bonded to a functional group e.g. carbonyl
For example, a carbon bonded to a carbonyl group is the alpha carbon, and has alpha protons. Following that there are beta and gamma carbons and protons.
Is there a specific name for the proton in an aldehyde that is bonded to the carbonyl, outside of "aldehyde proton" etc?

Welcome to ChemSE. You may use "aldehyde hydrogen", also. The problem is that it is not a proton (H+) although the usage is common.

@user55119 thanks. Do you know what the distinction between calling an H bonded to a carbon a hydrogen or a proton is? In all my lectures we have always said "proton"

As I said, the term "proton" is often used loosely.

it is a hydrogen atom, a proton  + electron, that bonds to a carbon, atom, not just  proton.

One can see it as a proton bound due to electrons.... Still I've upvoted Poutnik comment. Origin of the loose use of the term proton might be in HNMR. But in the latter case the term has a physical ground.


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Can you do SERS without a Raman microscope?
Surface-enhanced Raman spectroscopy or surface-enhanced Raman scattering (SERS) is a surface-sensitive technique that enhances Raman scattering by molecules adsorbed on rough metal surfaces or by nanostructures such as plasmonic-magnetic silica nanotubes.
Does this require a Raman microscope? Remember you are not scanning on a large object or liquid in vials but molecules in the rough metal surfaces. If you just use a normal Raman spectrometer. Will it work on the SERS sample?

SERS is not a sample, it is a phenomenon in Raman spectroscopy. SERS does not need a Raman microscope. Any ordinary Raman spectrometer can be used to study SERS.

But SERS sample uses molecules that are absorbed in the rough metal surface. How can the laser reach this depth without the Raman microscope also focusing down the laser to the very smaller nanometer scale?

I used to teach SERS on an educational Raman spectrometer with an integration time of 1 min (Au nanoparticles). You do not need an extra-powerful laser or a Raman microscope.

I only have a 532nm Raman spectrometer. I plan to buy an Au SERS substrate. Should I also buy a sample stage? what I have are only attachments in front like this https://imagizer.imageshack.com/img924/4927/5D1GEw.jpg

We never used a sample stage and I am not familiar with your Raman spectrometer. What is your analyte to begin with?

Just water for a start. How do you put the slide of glass in front of the horizontal Raman device? You just tape it?

It was not horizontal. The configuration was such that the beam would hit the surface vertically, so you could put the sample on the sample. Water will not show SERS effect.

so need to use this vertical sample holder to use SERS? https://imagizer.imageshack.com/img922/3048/cRWs7K.jpg  can't horizontal position work? what other substance in house can show SERS effect when used with a silver substrate? I just want to learn about SERS.,

First study the theory of SERS. If you wish to spend money, this vertical stand would work. Just curious, if this is a home research project or an academic one?

A home research project. If I won't buy the vertical stand. Horizonal setup can work if I know where is the focal point and just taping the glass slide to the front, isn't it?

SERS does not care about the orientation of the sample.

ok. last question. What cells or tissues in the human body will show SERS effect using 532nm Raman? I know colored ones can't be used because of fluorescence. But SERS is supposed to overcome fluorescence. So can red blood be used as analyte? or red tissues?

No experience in this domain.

Indeed study spectroscopy in general first. Because I think it is not the best to discuss (here) polymers identitification, sers, fluorescence microscopy without knowing same basic first. This said, congratulations on the experimental part. Your spectra look great.

Btw Farooq. Why wont water show any SERS effect? I read it can detect contaminants in water in one part of millions. One reason id buy expensive SERS substrate is to study contaminants of water. Tnx


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What is the purpose of this glassware?
I have seen these small test tube looking doohickeys being supplied from many chinese suppliers, but never with any description as to what they are for. They almost invariably have the description "Automatic Filling Tube Liquid Test Tube With Spout" which didn't lead me anywhere.
With most glassware, even relatively complex pieces, I can usually discern what it is for and why it is designed in that way but this has me stumped. My best guess so far is that it is meant to collect liquid from a sealed flask where the long stem passes through a bung to the bottom of the flask and gas pressure pushes some liquid up and into the tube. So maybe it is for volumetric reagent dispensing? If you had a squeeze bulb on the bottle it would push liquid up into the tube, then suck back and reduce the level to a consistent volume, but you would have to tip the whole bottle to pour it out so idk?
If anyone is familiar with what these are for and how they are used it would be much appreciated because this has been bothering me for a while.

I had several of them and I never figure out. I think what you and they say might be the purpose. Nothing too volumetric, though. Just I don't see the type of sealing that the long stem can easily and practically go through without loosing the sealing itself.

@Alchimista. I agree. This technique is not perfectly volumetric. It is used when  an order of magnitude on the result is considered as satisfying. It is often the case in waste water treatment plants. And if the punched rubber stopper is leaking, it is replaced.

Ah, thanks. I wonder if they were used with some sort of air pump like a squeezable bulb on a separate hole on the bung, or if there is there some way to tip the whole bottle so that it fills? With the little angled tube inside it looks like you could get accuracy around 0.5mL if it sucks back the excess. Either way it's a clever method of getting rough volumetric dispensing without moving parts.

It is an apparatus for taking a sample of a liquid from a big container. Let's suppose you have at your disposal a big bottle (at least one liter) which has two holes on a level with one another at middle height. Now imagine that a flow of a small river is entering the bottle from the upper neck, at such a rate that the level of the water inside is always on a level with the holes. The water gets out through the lateral holes. Now you put a punched rubber stopper in one of these holes. And the long thin pipe of your glassware is horizontally inserted into the stopper hole. Your glassware can then be turned and placed in such a way that the broader tube is vertical with its neck above. With this equipment, the broad tube is all the time half-filled with water. If the composition of the water slowly changes, a sample of its composition may be taken by turning the broad tube so as to empty it upside down, then putting it vertically again.

Indeed I've found them among dismissed glassware at a water waste treatment plant.


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Electrophilic Aromatic substitution reactions of aromatic compounds
Benzene and substituted benzene (phenols, aromatic ethers, benzaldehyde, aromatic ketones like acetophenone, benzoic acid, aniline and nitrobenzene) generally can undergo electrophilic Aromatic substitution reactions.
Now among all of these only benzene and aromatic ethers can undergo Friedel-Crafts alkylation reactions (or I'm wrong?).
My question is do all of the Aromatic compounds undergo halogenation,nitration and sulphonation cause these reactions as far as I know fall under electrophilic Aromatic substitution reactions.

You are asking far too general. You know the mechanisms of FC and electrophilic substitution. They always work, unless there is a reaction with a substituent instead of the aromatic core. That you have to decide for each case individually.

Btw. you do know the difference between activated and desactivated aromats? (You usually wouldn't if you haven't learned about FC-Acylation yet. In that case just wait a week until yout teacher gets to it.)

I do know about it..there are some groups that activate the ring and there are some that deactivate the ring.like OH,NH2 OCH3 etc these groups...they are electron donating and are activating..in addition are ortho para directing...and NO2,CHO,COOH etc ..these groups are electron withdrawing and are deactivating and meta directing...but halogens are an exception..cause they are electron withdrawing and still are ortho para directing and there inductive effect is stronger than resonance effect.hope I'm right.

With the right conditions you can nitrate pretty much any aromatic unless there is a substituent that is unstable to the reation conditions.


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Calculating the density of a magnesium sulphate solution
I am trying to calculate the density of a magnesium sulphate solution and came across this paper which determines an equation from experiment to do that. However, I am having trouble understanding the paper, as parameters A, B, C and D are mentioned but I don't understand how to calculate these from Table 2 in the paper. Can someone please explain how this works?
EDIT:
I am adding the data from the above paper:

The paper mentions the equation
to calculate the relative density where the parameters A, B, C and D are said to come from the table. m stands for the molality (I assume mol/kg) t is temperature (°C) and d is in g/cm^3. As you can see there is no mention of A, B, C or D in the table, leading to the confusion.

Unfortunately behind the paywall...

Sorry I didn't notice but I still seem to have some kind of access from my university. But there are ways around the paywall

I did not get a clear answer on the SE policies when googling but I'd also prefer not to post potentially illegal links or copyrighted material.

Note that it seems there is an error in the exponent in the formula ( 2/3 versus 3/2  ). Rather check it within the paper context and/or the calculated values.

Mistake from my side. corrected

For eventual writing and formatting of chemical/mathematical formulas or equations, see  MathJax helper reference page

Each of parameters A-D is a polynomial function of t:
$$A = \text{"m"} + \text{"mt"}.t + \text{"mt2"}.t^2 + \text{"mt3"}.t^3 + \text{" mt4"}.t^4$$
$$B = \text{"m3/2“} + \text{"m3/2t"}.t + \text{"m3/2t2"}.t^2$$
$$C = \text{"m2“} + \text{"m2t"}.t$$
$$D= \text{"m5/2"}$$
The terms in quote means the equally denoted numerical values from the provided table.
The complete final function $1000 \cdot \Delta \rho=f(m,t)$ has for each term $a_{b,c} \cdot m^a \cdot t^b$ the value for $a_{b,c}$ equal to the particular table value named "$m^bt^c$".
So one can the complicated function use in 2 ways:

The full blown function $1000 \cdot \Delta \rho=f(m,t) = \sum {\left( a_{b,c} \cdot m^a \cdot t^b \right)}$ without the need of coefficients A-D.
Temperature specific $1000 \cdot \Delta \rho=A \cdot m + B \cdot m^{3/2} + C \cdot m^2 + D \cdot m^{5/2}$ with precalculated A-D values for given temperature.

BTW, it looks like a very interesting numerical approximation. Maybe even better could be rational functions like Padé approximant, that have smaller deviations and better  extrapolation behaviour, compared to polynomials.

Feedback for the Padé approximation:
For approximating of analytical functions, see e.g Math SE site - how-to-compute-the-pade-approximation. For approximation of experimental discrete data, the best approach is probably numerical mathematics, the least square method combined with function minimum search algorithm. They probably used the same technique for their numerical model with half-integer powers.
Using half-integer power model is unusual. It may work better than polynomials in many cases, but just for positive numbers. Fortunately, they do not calculate with negative Celsius temperature :-) ) Note that Math SE site has even tag pade-approximation . Also, it is included in MATLAB libraries, if applies. See Octave as both online/offline free functional clone of MATLAB.
See also my answer in this Q: Is there a quick way to approximate a logarithm in Nernst equation?.

Interesting. Do you have some resources on how to use Padé approximation for curve fitting? Aside from that, what is your guess on how they arrived at this kind of equation? The paper doesn't touch on this.

The answer was updated by Padé appr. feedback.


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Spin orbital coupling and total angular momentum
The Sodium emission spectrum has 2 bright yellow lines called the Na-D lines. Superficially I understand the cause; there's spin-orbit coupling and the 3p orbital is split into 3p(3/2) and 3p(1/2).
But I don't understand why those numbers (3/2 and 1/2)?
Are these the total angular momentum values or the orbital angular momentum values?
And how does one work them out?

J is the total angular momentum. L is the orbital angular momentum and S is the intrinsic total electron spin angular momentum. J = L + S
L = $\sqrt{L\left(L+1\right)}\frac{h}{2\pi }$
--> L can be n-1 where n is the Principal quantum number
S = $\sqrt{S\left(S+1\right)}\frac{h}{2\pi }$
--> S is an integer or half an odd integer, depending on whether the number of electrons is even or odd
J = $\sqrt{J\left(J+1\right)}\frac{h}{2\pi }$
J can be L+S ... L-S
For the  Na-D lines the J = 3/2 is caused by L=1 and S=1/2 for the 3/2 value and L=1 and S=-1/2
The values for $J,L,S$ have a range of values e.g. $-X, -X+1,\cdots X-1, X $, e.g. if $L=5/2$ the range is $-5/2,-3/2,3/2,5/2$


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Why is 1-Ethyl-1-methylcyclohexane achiral?
I was wondering why the 1-Ethyl-1-methylcyclohexane molecule is achiral.
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

Because of absence of chiral center* & a better reason, you can easily cut out plane of symmetry in this molecule.

The functional groups are on the same carbon, so either configuration can be rotated into the other. If the functional groups were on different carbons, then depending on their relative locations the molecule could be chiral.

Second sentence is not correct.  Cis and trans 1-ethyl-4-methycyclohexane are achiral.

Should be OK now.

Oversight on my part, thank you for the correction!


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Finding wavelength of the centre of Stokes Ro-vibrational spectrum
I have the following rovibrational Raman spectrum for $N_2$:

and I am trying to calculate the "wavelength where the centre of the Stokes ro-vibrational spectrum would be observed" if the $N_2$ is excited with a 532nm Nd:YAG laser. I am very unsure of how to do this. I think I need to use the fact that Stokes transitions are given by $\tilde v = \tilde v_0 - 4BJ-6B$ ($B=$ rotational constant, $J=$ rotational quantum number and $\tilde v_0=$ excitation wavenumber) and that the separation between Stokes and Anti-Stokes lines is approx. $4B$. I don't really know how to make use of this information though nor what exactly is meant by the "centre" of the Stokes ro-vibrational spectrum. I'd appreciate it if someone could help me work through this.

You should draw out the rotational energy levels and transitions for this vibrational level, then you will understand how these transitions arise. Usually these are also shown in a textbook, if not for Raman then for IR and you could use these as a starting point. The principle is the same but the transition energy gaps are different in Raman vs  IR. You have all the info you need in the spectrum.


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Is pressure extensive or intensive?
I've learnt that pressure is an intensive measure,
but I have also learnt that pressure can be added (partial pressures), which doesn't make sense since intensive variables are non-additive.
Where is the flaw in my understanding?

It is a the constituting feature of partial pressures that they add up to the total pressure. You have no choice about it. And you cannot change one p.p. independently of the others, you always also have to readjust the absolute pressure.

Btw. other intensive variables can be added, too. Concentrations, for example, but if you actually, physically mix the two concentrations, you get their average. That's the point, imho, extensive properties add up when you physically combine objects in a single act.

@Karl thank you, your most recent comment cleared the doubt.

You are right that intensive properties are not additive like the properties which depend on the extent of the system. Pressure is indeed an intensive property because it is force divided by an extensive quantity (area). Now partial pressure is an interesting property. It is a hypothetical construct which cannot be directly or "experimentally" measured in a mixture of gases yet it can be calculated for individual gases in that mixture.
The basis of calculation of partial pressure is the number of particles in the container of each gas (hence an partial pressure originates from the number of particles of gas-A and gas-B).  Greater the number of particles of each gas in a container of fixed volume, greater is the calculated partial pressure of each component. The number of particles in a container determine the partial pressure which is an extensive property and hence partial pressures are additive.
Wikipedia explains the limitations of classifying quantities as extensive or intensive.
Consider reading the Journal of Chemical Education Article (free download from Google Scholar)
Redlich, "Intensive and Extensive Properties" J. Chem. Educ., 1970, 47 (2), 154–156. Redlich says that this classification is an incredible source of confusion in thermodynamics. I feel that he is not wrong.

The general validity of the division of physical properties into extensive and intensive kinds has been addressed in the course of science. Redlich noted that, although physical properties and especially thermodynamic properties are most conveniently defined as either intensive or extensive, these two categories are not all-inclusive and some well-defined physical properties conform to neither definition. Redlich also provides examples of mathematical functions that alter the strict additivity relationship for extensive systems, such as the square or square root of volume, which may occur in some contexts, albeit rarely used.

> Other systems, for which standard definitions do not provide a simple
> answer, are systems in which the subsystems interact when combined.
> Redlich pointed out that the assignment of some properties as
> intensive or extensive may depend on the way subsystems are arranged.
> For example, if two identical galvanic cells are connected in
> parallel, the voltage of the system is equal to the voltage of each
> cell, while the electric charge transferred (or the electric current)
> is extensive. However, if the same cells are connected in series, the
> charge becomes intensive and the voltage extensive. The IUPAC
> definitions do not consider such cases.
>
> Some intensive properties do not apply at very small sizes. For
> example, viscosity is a macroscopic quantity and is not relevant for
> extremely small systems. Likewise, at a very small scale color is not
> independent of size, as shown by quantum dots, whose color depends on
> the size of the "dot".

I feel it is more useful as a descriptive, or experimental support. It is more a useful view, than a view that reveals something about the universe in itself. Extensive properties are very often those we can measure with reasonable efficiency and precision. We can then use the maxwell relations to transpose to intensive variables, which for some reason tends to tell us more about what is actually going on.

I think voltage or current should not be regarded as intensive or extensive. Charge is clearly an extensive property. Generally imho the classification ex/intensive is an educationally useful concept, but it is not fundamental, and one must be (and make the pupils aware of its) limited applicability, i.e. it belongs into basic thermodynamics, and that's it.

Yes, Karl. This is the indeed a good summary.

The last paragraph above is a bit weird (it's from Redlich, right?). Almost none of the classic intensive thermodynamic properties makes sense on a microscopic scale, I only mention temperature. Even most extensive properties fail, like entropy, or at least become illdefined. The only two which are beyond doubt are charge and spin, and mass (let's ignore relativistic nonsense ;-)). Viscosity however seems to make no sense on a microscopic scale, however you can predict the behaviour of small particles in a fluid phase very well if you know the macroscopic viscosity. Same goes for T. ;)

Yes, very true. The quoted paras are from Wikipedia.

I wouldn't be so quick to dismiss the utility of the concepts of intensive and extensive properties. Yes, there are corner cases to which they don't apply.  But that doesn't mean that, outside of those, they can't be defined precisely, and used formally.  See Chapter 1 of Callen's Thermodynamics and an Introduction to Thermostatistics, as well as my answer here: https://chemistry.stackexchange.com/questions/142206/if-x-and-y-are-2-extensive-properties-of-a-system-then-is-x-y-always-intensive/142228#142228 ...

... Indeed, I'd argue you need to understand the concept of intensive vs. extensive properties to understand why at least one of the independent variables in the total differential of a thermodynamic state function must be extensive. E.g., if you start with U = U(S, V, N), you can Legendre transform it twice to get G = G (T, p, N).  But if you Legendre transform it a third time to get some function X = X(T, p, $\mu$), X will not be a state function.


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Will there be a temp change in free expansion
I did a little thought experiment, where I place an infinitely long beaker filled with some gas, which has a moveable piston at the open part. So if I place this beaker in vaccume, there will obviously be no work done on the system and hence no loss of energy due to the work done by the system (because the int pressure of the energy will obviously push the piston upwards).
So here's my main question:
If in the Beginning, let's say we have a certain amount of gas placed in the beaker such that the piston is just 1 cm above the base. But after infinite time the lid will probably be at infinity. So if a t=0, the gas occupies a limited space, won't the collisions be high and hence kinetic energy be high, which would mean the temperature of the system is high? But at t= infinity, the gas would occupy an infinite amount of volume which would mean that there are less collisions, so wouldn't this reduce the velocity of the molecules, hence reduce the kinetic energy, hence reduce the temperature?
I feel something is wrong over here because mathematically, I know there shouldn't be any change in int energy, therefore the temperature. So how do we justify this experiment?

Internal energy is a function of both temperature and volume for a real gas.

But can you explain this wrt ideal gases? Cus we have been taught temp of ideal gas is the only factor which when changes can be said to change the internal energy of the system

Why should be kinetic energy high because collision rate is high ? Compare it to the expanded gas at the same temperature, where collision rate is lower. Imagine a sworm of drunken birds unable to avoid each other. If swarm is packed, they collide more frequently. Does that mean they fly faster ?

When a  "lid" opens, it is just open. You want a piston!

Hehe, it should be a flock, not a sworm.

In a real gas, the internal energy is the sum of the kinetic energy of the molecules and  their potential energy of interaction.  An ideal gas is the limit of real gas behavior in the limit of large molar volumes in which the potential energy of interaction of the molecules approaches zero.  The kinetic energy of the molecules depends only on temperature, but the potential energy of interaction of the molecules depends on both temperature and specific volume.  The temperature of an ideal gas does not change when subjected to free expansion, but it does change for a real gas.

In a real gas, the internal energy is the sum of the kinetic energy of the molecules and their potential energy of interaction. An ideal gas is the limit of real gas behavior in the limit of large molar volumes in which the potential energy of interaction of the molecules approaches zero. The kinetic energy of the molecules depends only on temperature, but the potential energy of interaction of the molecules depends on both temperature and specific volume. The temperature of an ideal gas does not change when subjected to free expansion, but it does change for a real gas.  So the deviation from ideal gas behavior caused by potential energy of interaction of the molecules is what causes the temperature of a real gas to change in free expansion.


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Is it accurate to describe these radiation measurement units in terms of where-it's-measured and relationship-to-time?
I'm reading about measuring radiation and trying to come up with a more intuitive retelling for an audience familiar with math but not necessarily chemistry or any physical science. Is any or all of the following inaccurate or oversimplified?

Radioactivity (Ci, Bq) - Radiation from the material's point of view. Measured as a rate of disintegration.

Exposure (R, C/kg) - Radiation from the air's point of view. Measured as a sum of radiation that's passed through.

Absorbed dose (rad, Gy) - Radiation from the absorbing target's point of view. Measured as a sum of energy deposited.

Dose equivalent (rem, Sv) - Radiation also in terms of sum energy-absorption, but adjusted for a medical point of view.

As you can tell, the 2 pieces I want to highlight are (1) where-it's-measured and (2) time dimensionality (rate versus summable quantity).
My objective question is, where have I got it wrong? What has potential to cause confusion?
But of course, interested in complete reformulations.

Exposure rather as  Measured as an air ionization effect of passed radiation. Dose equivalent as adjusted for biological sensitivity

Maybe some numerical values may help.

Radioactivity. The activity is the number of disintegrations occurring per second in a given source. The unit is the Becquerel : $1 Bq = 1 s^{-1}$. For example, the experience shows that $3.7·10^7$ nucleus are decomposed per second in $1$ g $\ce{Ra-226}$. The activity of $1$ g $\ce{Ra-226}$ is $3.7·10^7 Bq$ which was previously called $1$ Curie ($1$ $Ci$). One Curie is a huge activity for usual measurements in the laboratory.

Exposure dose. If a gamma-ray radioactive source is in air, the gamma rays ionize the air around. As a consequence, they deposited some energy in a small volume of air situated at a distance from the source. If the small air sample is far away from the source, the amount of energy is smaller than if it is near the source. The amount of energy so absorbed by the air sample is called the dose (or exposure dose) and is determined with dosimeters, and it is measured in $J/kg$ air. The unit of dose is röntgen ($1~r$), equal to $0.01 J/kg$.

The experience shows that $1 Ci$ of a gamma-ray emitter delivers a dose of $1$ röntgen per hour ($1 r/h$) in air at $1 m$ distance from the source. This value is  somewhat dependent on the nature of the radioactive source. Let's forget about this difficulty for the present purpose. Most important, the dose decreases with the square of the distance. So, to be consistent with the units, the dose delivered by a gamma-ray source is $1 r·m^2 Ci^{-1} h^{-1}$. A small volume of air (some milliliters) situated at $1 m$ distance from a $1 Ci$ gamma source, absorbs in one day ($24 h$) a dose equal to $(1~r·m^2·Ci^{-1} h^{-1})·1 Ci·24 h/(1 m)^2 = 24~ r = 0.24 J/kg $. This energy is finally transformed into heat.

If the irradiated matter is not air, but another substance, like water, the amount of energy is called absorbed dose and is expressed in Gray, abbr. $Gy$ ($1 Gy = 1 J/kg$). The experience shows that an exposure dose of $1$ röntgen delivers an absorbed dose of $0.01 Gy$ in water, like in air.

Living tissues are mostly made of water. But they don't react the same way to a given absorbed dose. Some molecules like ADN are much more sensitive to a given adsorbed dose than water. To compare the biological effects of a given dose expressed in Gray, it is necessary to define a new parameter, the effective dose, which is expressed in Sievert, abbr. $Sv$. $1 Sv = 1 Gy$ in water. But $1 Sv$ = up to $20~Gy$ in sensitive tissues.

The present explanation is just a first approach of the domain of dosimetry and radioprotection. As quickly mentioned in the text, the energy deposited in a target depends also on the energy of the gamma ray, which was forgot here. And we have not spoken about alpha and beta rays, whose dosimetry is not similar to gamma rays. Dealing with these problems should take hours, and cannot be done in the present message.


text:
Why is enthalpy change at constant pressure not only internal energy change?
Okay so this might seem a bit Weird but the formula for change in enthalpy has been given the following way-
∆H=∆U+∆PV
so in this case when we say ∆V is zero we get that ∆PV  is also zero and thus ∆H=∆U. But when we say pressure is constant, my teacher simply takes out the pressure term outside the delta function because it's constant and writes,
∆H=∆U+P∆V
Why does he do this? Why can't this also result in the expression we get from constant volume? Plz explain

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

It can be thought of as H = U + PV, and operating with delta on all terms may be treated as differentiating the whole equation. So for the “PV” term, you need to use the product rule.
You end up P(dV) + V(dP) and if pressure is constant (which it generally is, so it’s a fair assumption), dP=0 and you’re left with P(dV).
(And to your point, if volume is constant, dV=0 and that just means that at constant pressure and constant volume, the change in enthalpy is equal to the change in the internal energy of the system.)


text:
Will entropy increase in adiabatic expansion?
The formula for entropy change states that change in entropy is not possible if heat change is nil.
∆S=∆Q/T
But what if we adiabatically expand a gas? Won't the molecules get more space to vibrate in, hence increasing the randomness of the system, therefore increasing the entropy?

The thinking and searching comment from your other question applies here as well. A good start is e.g. reading your textbooks and relevant articles on Wikipedia or educationnal sites and following their links.

The formula you provide is applicable during isothermal (constant T) processes. In addition, it can be applied only when the process is carried out reversibly, that is, q is the heat exchanged during a reversible process.

Related : What is the difference between reversible and irreversible adiabatic expansion?

The answer to your question is yes and no. You are correct in supposing that as an ideal gas expands the entropy will increase as it has more space to occupy and so the number of ways the molecules can be placed in the total space has increased. (Try to not think of entropy in terms of randomness but in the number of ways molecules can be positioned in space, or occupy different energy levels). However, accompanying expansion is cooling and if the adiabatic process is carried out reversibly the decrease in entropy due to cooling exactly matches that increase due to expansion, and so the entropy change is zero.
If the expansion is irreversible and adiabatic, the gas does less work, looses less internal energy and so there is entropy increase as the cooling is now not as large as when expansion is adiabatic and reversible, and so the two process, entropy increase due to expansion and decrease due cooling no longer match.


text:
Why do molecules showing a pure rotational spectra require a permanent dipole?
I understand that pure rotational spectra only exist for molecules which have a permanent dipole moment.  The common explanation is "so that they can interact with the E-field of the incoming photons", but I don't think it is clear why photons can only interact with dipoles.
I'm also aware that the answer probably has something to do with the matrix element $\langle\psi_i|x|\psi_j\rangle$ where $\psi_i$ and $\psi_j$ are the initial and final states of the molecule, and $x$ comes from the dipole moment operator.  I'm just not sure why we need this matrix element and not something else.  And as an extension, why does Raman spectroscopy depend on the polarisability of a molecule, whose operator behaves like $x^2$?

In semi-classical terms think of the the electromagnetic wave (ir /microwave , visible uv etc)  as an oscillating electric field, this can interact with the oscillating electric field of a rotating molecule caused by its permanent dipole, hence dependence on $x$. When the energies are the same, i.e. in resonance, energy can be absorbed from the EM wave into the molecule and vice versa.  Raman is scattering, the polarisability is a volume, $x^2$ is due to the cross section (area) seen by the radiation.

The matrix element for rotational spectra comes from https://en.wikipedia.org/wiki/Fermi%27s_golden_rule. The relevant Hamiltonian is proportional to $\hat{x}$, hence your expression. (Well, I think it is $\propto \hat{x}$ as long as you define your coordinate system carefully; but in free space, we have the freedom to do that.) Not sure about Raman, though.


text:
Help determining normalisation constant in SALC of methane
We have been told that a symmetry adapted combination of four hydrogen orbitals is used to create localised hydrogen orbitals for methane. These orthonormal combinations are denoted ψa1, ψt’2, ψt’’2 and ψt’’’2.
It then asks to calculate the normalisation constant for the localised orbital Ψ = ψa1 – ψt’2 – ψt’’2 – ψt’’’2.
What I have done is reduce the symmetry representation of the 4 H s orbitals to Γ = A1 + T2. From here I have drawn an MO diagram and shown that there is overlap between the C 2s orbital and the A1 combination of H s orbitals, and also overlap between the C 2p orbitals and T2 combination of H s orbitals. So I am assuming that the combinations denoted above are these respective molecular orbitals. Therefore, the orbital for which I am asked to determine the normalisation constant is just the A1 MO subtracting the 3 T2 MOs.
From here I am stuck as to how to normalise this wavefunction. I know that N = 1/sqrt(Σci^2) but I cannot determine what the ci coefficients are?
Surely they aren't just the coefficients of ψa1, ψt’2, ψt’’2 and ψt’’’2? Which would give N = 1/2 - seems far too straightforward.
I know that for Ψ to be normalised we require the integral of its square to be equal to 1 which requires each of the components' square integrals be equal to 1/4. This would suggest the normalisation constant is sqrt(1/4) = 1/2?

Indeed, it is that straightforward. :-) By the way, the individual square integrals need not be all $1/4$. As long as they add up to $1$, that's fine.

Thank you. So just to clarify, the normalisation constant for Ψ = ψa1 – ψt’2 – ψt’’2 – ψt’’’2 is simply 1/2?

Yes, that's right. As long as the individual $\psi$'s are normalised, of course. But you said that already.


text:
AlCl3(aq) reaction with NaHCO₃ to form alumina Al2O3
In a demonstration, aluminum foil was add to a solution of $\ce{HCl}$ and allowed to react, the reaction was presented as
$\ce{2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)↑}$
After filtration, the solution appeared optically clear.
Next, baking soda, i.e. sodium carbonate $\ce{NaHCO3(s)}$, was added, and the solution then turned whitish opaque. After a while a white precipitation could be observed. In the presentation it was claimed that alumina, $\ce{Al2O3(s)}$, was precipitating out of the solution.
What reaction takes place when $\ce{NaHCO3(s)}$ is added to the solution? Is it correct to claim that $\ce{AlCl3(aq)}$ exists in the solution?

Aluminum cation is an acid in water solution.  2) Acids react with bicarbonates to release carbon dioxide.

$\ce{Al^{3+} + 3 HCO3^- -> Al(OH)3  +  3 CO2 }$ then $\ce{2Al(OH)3 -> Al2O3 + 3 H2O}$

$\ce{AlCl3}$ does not really exist in solution. it is dissociated into ions $\ce{Al^{3+}}$ and $\ce{Cl-}$, which are themselves hydrated

In the reaction formula written by @Maurice it looks like Al(OH)3 is an intermediate. Would it be possible to control the reaction and stop it at this stage, i.e. before any Al2O3 precipitates.

$\ce{Al(OH)3}$ and $\ce{Al2O3}$ are both precipitates, and they are both hydrated. I don't know whether it is possible to distinguish between them.


text:
Which has more internal energy, a bucket of hot water or a bucket of cold water
Online answer I found :
The bucket of hot water has more internal energy.
Internal energy depends on the amount (mass) of substance and the kinetic energy of the molecules of the substance.
My thinking to it :
Since it is written dependant on mass, then swimming pool has more mass than bucket of hot water.Since it has even higher K.E , then how can Bucket of hot water have more Internal energy than swimming pool?

One thing is total amount of energy in J, other thing is specific energy in J/kg or molar energy in J/mol. The latter two depend on temperature, the former one on temperature and amount of substance. So hot water in the bucket has higher specific energy in J/kg, the pool higher energy in J.

Ok thnx.I understood

If you thought and searched about it little longer, you would not propably post your question at all.


text:
How does heat transfer between molecules happen in deep?
Inside the container is cold water and outside the container is hot water. B is the microscopic view of container walls .W is water
And A is hot water.
What I have shown is the microscopic view of container walls(Small part )
A is the surrounding and B is the microscopic view of the walls of container ( Since there is some thickness of container walls and we can say that there are molecules present there ).
As you can see , when cold water molecules apply force on B (walls of container at left side) , B walls also apply force on water molecules and same has happened in the part where there is hot water.This is from the Newton’s third law of motion.
Average Kinetic Energy of water molecules will be less since it is cold water.So less force applied by water molecules on the wall.
Average kinetic energy of hot water will be more than cold water.So it applies more force on walls of container than by cold water.
I hope I have written the things clear till here.
Q1 Is it that the walls container will be pushed by B towards right side and not left side ? Because there is more force by Hot water and not cold water.
Q2 Does heat get transferred here because the Average kinetic Energy of the walls of container will increase due to force of A on B? And those will put the molecules inside B into motion which will further put water molecules  into movement and therefore making heat transfer ?
Q3 Does heat transfer happen only because of this displacement?
HAVE I WRITTEN THE MEHTOD OF HEAT TRANSFER?IF NOT, THEN PLEASE HELP IN CORRECTING ME.
I am trying to make my self understand heat transfer in details.
Please tell if how can I improve the quality of my question if not good.

Using photos/screenshots of ( even handwritten) text instead of typed text itself is strongly discouraged. It is impossible to index/search/reuse it in answers referring to it. In a  case  of handwritten text, it puts extra burden on responders to properly decipher it.
That all may lead to the question being ignored or even closed.
Consider copy/paste or retyping and using eventually  MathJax for expressions and formulas.

Ok.I have described everything above needed.Only diagram is needed to see that’s why I put this

I have edited it and made the text gone.

You need to be more clear and specific when you explain the problem. That greatly helps yourself correctly understanding it. Example: "K.E of water will be less since it cold water" 1.) Spell out K.E. 2.) it's not kinetic energy, it is thermal energy, or average kinetic energy of water molecules 3.) that's not the conclusion you want to draw, but a prerequisite: not "will be less", but "is smaller" 4.) the second part has no verb, "it is cold water". All this might sound petty to you now, but it is crucial so the reader, including yourself, can concentrate on the issue.

This text is not understandable. The container contains cold water. But what is outside the container ? In the beginning it was hot air. At the end it is hot water. So what is what ? How can one understand that a rectangle represents "inside the container walls" ? Further more the last picture mentions that A is hot water, and W is water. Where is this W water ? Is it hot or cold ? It is not on the picture .

Ok.I am very sorry for this. W is cold water.A is hot water only.The rectangle represent the view inside the container walls (Since it will have some thickness).I am sorry for not making it very easy to understand.

@Maurice I have edited my question and written below exactly what is what.

@Karl Editing is done

Please help.I really thought a lot about it.

OK, rework the rest of your question in the same manner. Complete sentences, grammar, the lot. Your question will suddenly get upvotes, and perhaps you don't even need an answer then. Tthinking hard how to correctly put the problem often solves the problem itself.

Yes @Karl you are right.Is it fine now ? Or need some work to be done on it?Done bit more changes now.

Forces are useful to explain the pressure. They are no use for explaining heat transfer.

Ok. I wrote it because force will the displacement of walls which in turn will send the Average kinetic energy of molecules into vibration.That is what I know that $When molecules vibrates or move fastly , their temperature increase $From there I thought of this.If this is wrong, then pls tell how does heat transfer actually happen at molecular level.Please.@Maurice

@Poutnik Do take a look at it for now.

Note that liquid temperature and pressure are independent parameters, so as they are independent for gases.

@Poutnik How does this heat transfer happen then?

Have you properly searched and read about it ? As writing what was already written many times is not very productive.

I have searched a lot but couldn’t find the way I have written or similar to it.

A1: There is no displacement of the wall, forces are in equilibrium with zero net force. There are initiated stronger vibrations on warmer wall side by energetic hot air molecules. And weaker vibrations on colder wall side by energy leaching cold water molecules .
A2a: If the wall initial temperature was the cold water temperature, then yes. Otherwise no, as the mean kinetic energy of wall molecules can generally goes both ways, depending on initial temperatures of air, wall and water, weighted by air and water thermal capacity a conductivity.
A2b: Yes
A3: No, as there is no displacement. Hot air with in average more energetic molecules causes stronger vibrations of wall molecules, which cause faster motion of water molecules.
Imagine the whole scenario in more familiar macro world. 2 boxes of ideally elastic balls in gravityless ISS station, separated by a wall of balls interconnected by elastic strings holding all of them at their place.
Now, balls in one of boxes move faster. What will happen during the time ?

The mean Kinetic energy of molecules is proportional to absolute temperature. Their mean speed is then proportional to the square root of absolute temperature and momentum to the absolute temperature.
Pressure on the walls is proportional to momentum change rate, what is proportional to molecule momentum and the rate of collisions. For gases in constant volume, both is proportional to square root of absolute temperature, so pressure is proportional to absolute temperature of ( ideal ) gas.
But pressure is not essential for molecular mechanism of heat transfer.
Molecules of matter at given temperature do not have the same speed and kinetic energy. During their collisions, it fluctuates all the time. What is constant is their mean value.
For monoatomic ideal gas, the mean kinetic energy of 1 mole of the gas is $E_\mathrm{k}=3/2RT$, where R is the universal gas constant about 8.314 J/K/mol. The speed distribution follows the Maxwell-Boltzmann distribution.
It means that for given kinetic energy and speed of a molecule, we cannot say what is the temperature of matter the molecule belongs.
When 2 different regions of matter have different temperature, they exchange heat, what means molecular kinetic energy.
Molecules from adjacent regions mutually exchange kinetic energy by random way by collisions. As by random collisions, they can both gain or lose the energy.
Molecules from both colder and warmer regions mutually both transfer and accept the energy. If temperatures are equal, their mean kinetic energy (*) is equal and the rates of energy flow are equal in both directions. If one region is warmer, the rate of passing energy is greater than the rate of accepting energy.

(*) It is more complicated, as more exactly, it means the mean kinetic energy per each way the molecule can move, called degree of freedom. But the above is still valid for monoatomic gases.

Do you mean to say what I have written is right or wrong @Poutnik.You have told the bit same thing but I don’t know what are you trying to say from my question.If you can also give a diagram.It will be really helpful.

I do appreciate your efforts.That coordination with my question.Telling me where am I exactly wrong from the lines would help a lot to understood this concept.

Ok Thank you so much.I will keep it in mind.

For atoms , displacement is not necessary but vibration can also happen instead of displacement of walls.

Yes, as written.


text:
Are there any sulfides, sulfates, carbonates that cannot be dissolved by aqua regia?
I was looking into these specific minerals: arsenopyrite, pyrite, chalcopyrite, barite, calcite, dolomite. I investigate issues with tellurides not showing expected gold concentrations.

What do you mean when you say : " I investigate issues with tellurides not showing expected gold concentrations"  ? Issues ? What sort of issues ?  What is an "expected gold concentration " ? Has it any relationship with the previous sentence ? Has it any relationship with the aqua regia from the title ? Explain !

There is gold predicted at the site where these samples are. We tryed to prove the presence of gold through aqua regia tests, but the resulting gold concentrations are low. My suspicion is that the majority of the gold resides as an inclusion in these minerals, and the aqua regia test won't be able to show that because of its inability to dissolve the host.

If gold is included in inclusions, why don't you grind your minerals to a powder ?

It was ground, but the particle sizes are probably smaller then the smallest grinding size.

Osmium or iridium minerals are not soluble in aqua regia. The arsenide and
sulphide of platinum are slightly soluble in aqua regia but the
rate of solubility is very slow. Sperrylite(PtAs2) and Cooperite(PtS) is insoluble in aqua regia. (http://www.minsocam.org/ammin/AM22/AM22_1016)

Also see: https://actlabs.com/geochemistry/exploration-geochemistry/aqua-regia-partial-digestion/

Chemical Book reports that barium sulfate (barite, or baryte) is

slightly soluble in concentrated sulfuric acid and soluble in carbonate alkali metal solution in which it is converted to barium carbonate; it is insoluble in other kinds of acid or base.

It sounds like the barite is your culprit.  If you try grinding it with an eye towards physically separating the denser gold from it, the Mohs hardness of the mineral is 3-3.5, so there appears to be no issue with the physical grinding process.

Thank you for your input. It is likely that the grinding sizes are insufficient as well.


text:
IUPAC name of aldehyde with two functional groups
I am confused as to what suffix to use for the following compound:

Should it be carbaldehyde or just -al, like should it be propane-1,2-dicarbaldehyde or should it be propan-1,2-dial?
Or maybe it's 1-methylethane-1,2-dial? I am pretty sure its either of these three or maybe something else.

You have correctly identified the principal characteristic group as aldehyde.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple mono- and dialdehydes are named substitutively using the suffix ‘al’.

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

For example: pentanal and pentanedial, respectively. Thus, the compound that is given in the question is a butanedial.
The terminal locants of aldehydes, i.e. ‘1,4’ in case of butanedial, are not cited.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Also the locant ‘2’ of 2-methylbutanedial is not cited.

P-14.3.4.3 The locant is omitted in monosubstituted symmetrical parent hydrides or parent compounds where there is only one kind of substitutable hydrogen.

Therefore, the preferred IUPAC name is methylbutanedial.

By way of comparison, the Blue Book contained the similar example ‘2-chlorobutanedioic acid’, which was later corrected to ‘chlorobutanedioic acid’.


text:
For a flat, planar molecule, how do you determine which side is up? Or down?
I was just reading about alpha and beta linkages in certain cyclic molecules, and it said that the alpha configuration was the one in which a hydrogen was sticking out above the plane, and an OH moiety was below the plane.  A beta linkage for such a molecule had the hydroxyl group upwards, and the lone H below...
But, I cannot find out how you determine which side is 'up' or 'above' and which is 'down' or 'below'...

See https://en.wikipedia.org/wiki/Skeletal_formula

For sugars: https://chemistry.stackexchange.com/questions/136694/why-is-the-1-6-linkage-in-amylopentin-an-alpha-link-and-not-a-beta-link/136707#136707. For steroids: beta is above the plane of the molecule; alpha is below given the standard way of drawing steroids.

Sugars ain’t flat.

And even if something is flat, there is still an up and down. This is anyway relevant only in relation to another objects, for instance while fusing rings etc. Nomenclature rules fixes this aspect.


text:
Are all 'dextrorotatory' sugars in biology actually 'd' or '+' in chirality?
When papers or articles say that all proteinogenic amino acids are 'levorotatory' or 'L', they often make a point of saying that only half of them are truly, optically levorotatory.  All of them (except glycine) are called 'L' because of their 'opposing' configuration to glyceraldehyde, the original molecule deemed to be dextrorotatory...
Well, all sugars used by living things are supposedly 'R', or so I keep reading....
Are they similar in configuration to glyceraldehyde, or actually dextrorotatory to light?
EDIT:  What I really wanted to know, in posting this question, is whether less than half, or most, or all organic sugars and carbohydrates are ACTUALLY optically active, in the sense of lowercase d or +,....
Remember, I noted that only 10 of 20 proteinogenic amino acids are actually levorotatory to polarized light....

Levorotatory is lower case "l" or a minus sign, not "L".  All natural amino acids, save, glycine, are of the Fischer/Rosanoff designation L. Dextrorotatory means "d or +". Most, if not all, natural sugars are of the Fischer/Rosanoff designation D. The carbon that gives this assignment happens to be R. D and L are configurations; d and l are optical rotations. Search "carbohydrate configurations" in Search on Chemistry at the top of the page.

So, the answer to your question is "yes". Dextrorotatory sugars are "d" or "+".  But a sugar has no chirality. Chirality is not a property of a molecule. Only carbon atoms have a chirality, and this chirality is R or S.

I would say chirality is a property of the molecule rather than just a specific center. A molecule is chiral if it has no improper rotation symmetry elements, which can be caused by point center of chirality like a carbon atom, but also from axial chirality (e.g. a helical structure). I guess you mean we can't assign R or S labels to a whole molecule, but I would refer to these labels as "configurations" rather than "chiralities".

Dextrorotatory = d = (+). Levorotatory = l = (-). These terms describe the absolute sense of rotation in an optically active substance.
D and L are older terms for the sense of chirality that can be traced back to dextro- or levorotation of glyeraldehyde. The actual compound may have the same or opposite sense of rotation, but the assignment is determined by analogy. (Cf. Fisher/Rosanoff).
R and S describe absolute stereochemistry of a center, so a specific molecule with multiple chiral centers cannot be sufficiently described with a single R or S, but require one for each stereocenter. This designation is part of the more modern (and widely accepted) Cahn-Ingold-Prelog system, which has largely been adopted by IUPAC.
Note that while D and L might be sufficient to describe a sugar, anything more complicated than a 3-carbon aldose or 4-carbon ketose is not likely to be properly described with a single R or S because there are multiple stereocenters.


text:
Is there a word for "damage done by fire" of a material?
If a house had a wooden frame and was caught in a fire, there'd be nothing much left at the end. The damage would be huge. If a house had an iron frame and was caught in a fire, it probably wouldn't be too badly off at the end. Is there a scientific measure of "how things fare in a fire" please?

It does not seem to be about chemistry.

I’m voting to close this question because this question is not about chemistry.

'Damage done by fire' is also found in  insurance parlance linked to insured loss claims. It is not even necessarily a direct fire loss, as could occur with an insurance policy (namely, a business interruption policy) covering loss of sales as a consequence of a covered fire event (for example, fires from illegal activities or civil unrest may NOT be  covered events).

It totally depends on the material.  Example , steel surface will oxidize but the nature of the oxide will depend on the time/temperature. The steel strength may be reduced or increased depending  on time/temperature and the original strength and composition.

In English at least, there is no single word that captures this quality, for the simple reason that fire can affect materials in many different ways.
For example, an organic material may actually catch on fire itself and be combusted. "Combustibility" and "flammability" refer to slightly different descriptions of the tendency of materials to do this.
Metallic materials, on the other hand, may simply melt in the heat of the fire without combusting or may undergo microscopic structural changes due to the heat. The result can be a significant change in the properties of the material even after it has cooled back to ambient temperature, even though the material is not combustible. The effect is dependent on the temperature of the fire, which varies widely depending on fuel and conditions.
As a third example, a inflammable material that contains volatile components (like the water in a silica gel) might lose that volatile component in high heat, again permanently altering the properties of the material without combustion or melting.