text:
Why is the pressure correction added in the Van der Waals equation?
Since real gas particles experience attraction to each other, that means the pressure must be less from what we would expect. So why do we have to add a term to correct it?
For example, in the correction for volume, we subtract a term to find the effective volume or the "real" volume. So, it follows that the "real" pressure must be less than the given one and so, should we not subtract a term?
This is mostly with regard to solving problems based on this law, where the values of pressure, volume, $a$ and $b$ are given

The Van der Waals  model was a crude method for adding some "fudge factors" into the ideal gas equation based on a theoretical justification. The Van der Waals model is by far not the best model for empirical data fitting. See Wikipedia article Real Gas

So you basically say that the parameter would/should always be negative, right? Well, is it?

".. real gas particles experience attraction to each other .." that is a very (too) simplistic statement. If you get the particles sufficiently close together, they always repel each other.

I was told that  the factor "a" is multiplied considering only the attraction so I framed it that way

@Karl  Interestingly, while atoms/molecules do repel each other at sufficiently close range, in gases that repulsion is entirely entropic.  I.e., neutral gas atoms/molecules are always in the attractive range when it comes to the pair potential. You can squeeze atoms/molecules sufficiently close to reach the repulsive region of the pair potential, but if you get them that close they are no longer in the gas phase.

Doesn't the "a" term make the pressure less?

You may consider the vdW equation as the deviation from the ideal gas state equation $pV_\mathrm{m} = RT$.
If you look at the vdW equation
$$\left(P+\frac a{V_m^2}\right)(V_m-b)=R T$$
you can easily see why the corrections have  their signs.

The pressure term correction has the positive sign,
as the real gas has  due molecule cohesion  smaller volume than  an ideal gas at the same pressure. Like if it was under bigger pressure than it really was.
Increasing $p$ term by a positive correction leads to lower predicted volume than for ideal gas, at the same pressure, because of cohesive force.
$$V_m=\frac {R T}{\left(P+\frac a{V_m^2}\right)}$$
The correction simulates this bigger pressure the ideal gas would need to get the same vol.

The volume term correction has the negative sign,
as the gas has at the given volume
higher pressure due molecule own volume.
Decreasing $V$ term by a negative correction leads to higher predicted pressure than for ideal gas, at the same pressure, because of own molecule volume.
$$P=\frac {R T}{(V_m-b)}$$

Smaller volume would mean more pressure than expected so the term should be subtracted?

@AnanyaPateriya  Smaller volume here means smaller atomic/molecular volume, not smaller container volume.  I.e., the particles themselves take up less of the space, so the "free volume" is greater, hence the pressure is lower.


text:
Can ionic compounds dissolved in organic solvent conduct electricity?
Ionic compounds dissolved in water, as an aqueous solution, can conduct electricity.
Can they conduct electricity if they are dissolved in organic solvent?

Yes, I have done a lot of electrolysis in MeCN using tetrabutylammonium tetrafluoroborate as the ions

That is how lithium ion batteries work. You do not want the lithium graphite intercalate to get in contact with anything aqueous.

Sure they can.  Even if most of the ions are in ion pairs or other complexes, they can still move by being transferred from one complex to another.  All you need is to get the ionic compound to dissolve, which in most cases requires a polar solvent with which the ions will not react.
A rather unexpected case involves magnesocene, whose bonding is a mixture of ionic and covalent.  When dissolved in ether or THF, magnesocene forms enough ions not only to impart significant conductivity in solution, but also to be proposed for use in batteries:

Unlike ferrocene, magnesocene displays slight dissociation and subsequent ion association in polar, electron-donating solvents (such as ether and THF).[1]
$\displaystyle {\ce {MgCp2 <=> MgCp+ + Cp-}}$
$\displaystyle {\ce {MgCp2 + MgCp+ <=> Mg2Cp3+}}$
$\displaystyle {\ce {MgCp2 + Cp- <=> MgCp3-}}$

Note the single charge on all ionic species.  This is to be expected in organic solvents with their weaker polarity and ion solvation power relative to water.
Cited reference:
1.
Schwarz, Rainer; Pejic, Marijana; Fischer, Philipp; Marinaro, Mario; Jörissen, Ludwig; Wachtler, Mario, "Magnesocene-Based Electrolytes: A New Class of Electrolytes for Magnesium Batteries," Angewandte Chemie International Edition 2016, 55(48), 14958–14962.  https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201606448

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text:
Cleaning with Vinegar - what surfaces does it react badly with? Metal window sills ok?
I'm cleaning up some mould around the home and I have some substance on my window sills which I'm not quite sure if it's dirt or mould.
I have read about using Vinegar to clean up mould, but an article I was reading awhile back said you shouldn't use Vinegar on certain things as it can eat away at it or something because it's acidic; I thought this may have been metal and/or painted substances but I may be wrong.
I know you shouldn't use it on granite, marble and natural stone, but was wondering if using it on the painted metal window sills is ok or should I use something like diluted Tea Tree Oil instead, or would that be too oily?
FYI I would be using White Vinegar.
Here is a pic of my window sills:

Why vinegar? Makes no sense except if you have limescale. Use your standard household detergent for hand dishwashing.

Because if it's mould I would like to kill it; detergent isn't going to do anything in that way.

Makes no sense, sorry. If you wash it off, it´s gone, dead or not. If you can keep it dry, it will stay clean, if it´s wet often, it will get mouldy again after some time either way.

You do realise that once you start agitating mould it can start releasing "spores" - I'd say if you're killing it off first then this mitigates this risk; why would you clean with something that doesn't kill mould when you can clean with something that does!? Sure, if the surface gets moist again, mould could come back, but not wiping the surface down with something to kill what's already there would dramatically decrease the chance of it coming back without those conditions.

Go with winegar if you want. That seems a standard window and table winegar won't do anything bad to it. No sure about the mold :)

There are commercial multi-surface cleaners based on quaternary ammonium surfactants. These surfactants are bactericides and tend to prevent reoccurrence of the infestation simply from the tiny residue after wiping off. None of the items mentioned so far are so aggressive that they will hurt your windowsills. (But you could try new agents on one sill at a time and if it begins to suffer, don't use it on all the rest.)

@JamesGaidis By "multi-surface cleaners", do you mean general cleaners in the stores like multi-purpose cleaners - same thing?

I'd add 1 part of Chlorox to 9 parts of water, apply that solution to the window sill with a worn-out or ready-to-be-discarded tooth brush, wait 30 minutes, clean the window sill with the tooth brush, wipe off the solution, rinse with water, and finally, clean up with a towel.

@Brett: The "multi-surface" cleaner I use has contents: Alkyl (C12, C14, C16) dimethylbenzyl ammonium chloride (<1% !), brand name Clorox. "Kills 99.9% of bacteria and viruses" Clorox is usually associated with NaClO, which is also good for oxidizing bugs to death, but is sometimes harsh on substrates. The multi-surface cleaner is a cationic surfactant that tends to stick to cell walls and maybe open them up. Multi-purpose could be different.

Kills 99.9% of bacteria and viruses - hmmmm.. bacteria and viruses; doesn't seem to mention anything about Mould, which is a Fungi.

Avoid aluminum alloys when working with vinegar solutions!
With time, the pH effect of acetic acid can penetrate the protective oxide coating. The exposed Al will even react with water:
2 Al (s) + 6 H2O (l) --> 2 Al(OH)3 + 3 H2 (g)
and with the presence of alloy metals, galvanic corrosion (especially if the 'dirt' is rich in salts that can serve as a good electrolyte) which apparently accelerates the attack on the Aluminum.
To quote a source:

corrosivity environment for aluminium in food industry are foodstuffs with pH 3 – 5, such as fruit juices, jams and acidic canned fruits or hot gravies, sauces as well as dressings, vegetables and fish pickled in brines with 1 –3 % salt [1]. In particular, acetic acid needs to be taken into corrosion consideration, due to its wide usage in food industry (vegetable and fish pickling), and its representative properties among acids and juices in fruits, vegetables and other organic materials that can corrode metals. Although aluminium has a good resistance to acetic acid solution at room temperature, aluminium can corrode in almost any concentration of acetic acid at any temperature if the acid is contaminated with the proper species [2].

Further comments related to the presence of select metals:

Although aluminium has a good resistance to almost all the concentrations of acetic acid at room temperature, care must be taken that the metal is free of other impurities such as iron, copper, tin and lead even in traces [2]. With increasing purity of aluminium, its resistance to acetic acid solution increases, and 99.5% aluminium can be used for the majority of engineering purposes, but components added to its alloy can increase the corrosion of aluminium in acetic acid solution [14]. Since the metal corrosion occurs via electrochemical reactions at the interface between the metal and an electrolyte solution, electrochemical techniques are ideal for the study of corrosion processes.

And importantly, the effect of pH:

It is well known that aluminium resistance is related to the thin and compact layer of naturally formed oxide on aluminium surface, but this oxide layer is stable only in pH range 4-8. Lower or higher pH values caused prominent destroying of protective layer and thus the significant metal dissolution [15].

The photo depicts a painted aluminum windows sill, not bare aluminum. So, the concern is the resistance of the paint, not that of the aluminum.

So basically the worse that can happen is it could cause corrosion on the window sill? Would you know if it likely the window sill is aluminium alloy?

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text:
Would ABS plastic degrade in a dishwasher?
I printed an object with a 3D printer and I used a roll of light green ABS plastic filament.
I would like to use the dishwasher to clean it. The temperature would be kept below 70°C, well below the glass transition temperature of about 110°C.
The dishwasher uses standard tabs, not one of the phosphate-free tabs.
What would be the effect of said cleaning on objects made of ABS plastic?

Well, what do you expect?

The plastic could be degraded, since I remember that setting the wrong water hardness can, after some time, even render glass opaque. I don't know chemistry well, obviously.

If you are planning to wash it by itself in the dishwasher, there is little risk in testing whether it is OK (at least if you can make more than one version). Try washing a model of the same material you can afford to lose and see what happens. don't use it for food or alongside things used for food. Testing is cheap and, if the risk is low, quick and easy.

According to several application guides/compatibility charts for ABS plastic (like this one), ABS is fairly resistant to the conditions present in a dishwasher (mild to strong alkaline, salts, no organic solvents, mild temperatures).
Please note that while the ABS structural backbone may be attacked by nucleophilic agents at the nitrile carbon, and that the nitrile group may be hydrolised in acidic or basic environments, the conditions required are much harsher that a common dishwasher.
For example (p.8), glacial acetic acid was found to cause significant swelling, while 25% w/w sodium hydroxide and 25% w/v HCl barely caused changes.
So, while the concern for release of toxic additives present in the formulation of the original ABS filament may or not be significant, from the chemical point of view it is almost sure it won't degrade noticeably.
Also, anecdotally, I used to wash old LEGO pieces in a dishwasher (kids are dirty), and not degradation was observed at all.

Also, experimenting with a replaceable sample, as noted before, is a very practical way of getting results.

ABS isn't food safe so you may introduce toxic particles to your dishwasher and anything else in it. The combination of high temperatures and harsh chemicals could still degrade unprotected ABS despite temperatures being below glass transition. Other people have done it before and haven't had issues though, but I prefer to err on the side of caution.
I would recommend using PETG filament over ABS for food-related prints, PETG is less toxic and still handles heat well. You can also anneal PLA to make it more heat resistant but PETG would still be better than annealed PLA.
This question would probably be better answered on a dedicated 3D printing forum rather than on here.

A recommendation for a filament would indeed go to 3D print SE, but I was asking specifically about the behaviour of ABS because I wanted to keep the question in topic with the Chemistry section, since I expect more knowledge about the degradation in this community.

Although the Flyingfenix's answer goes into good detail of ABS's ability to withstand dishwasher cycles, I would warn against trusting evidence based on more ideal samples. FFF 3D printed parts are much more porous with many more microscopic defects and contaminants than injection-molded ABS, so that's something to keep in mind.

Let's not confuse food-safe with contaminant.  Uncoated ABS is not food safe, it's true, but the reason is that it's porous and may therefore hold bacteria and allow them to thrive within those pores.  However, it's a far stretch to say that it "may introduce toxic particles to your dishwasher and anything else in it."  So don't use ot to store food at all, but I doubt we shoud fear having it in the same environment with things that do store or touch food.


text:
Question about change in pressure and change in temperature for system at equilibrium
I was reading my Chemistry textbook and I noticed that it stated that for a system at equilibrium:

At constant temperature, reducing the volume of gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. Increasing the volume will shift the system in the direction of more moles of gas.
If the temperature is increased at equilibrium the system will shift in the direction to consume the heat. If the temperature is decreased, the system will shift in order to produce excess heat.

My question is why does this occur?
Please answer my questions in terms of how the molecules in the system interact, instead of just saying that what my textbook states is true.
Also, explain (if possible) using the chemical reaction of N2 (g) + 3H2 (g) ↔ 2NH3 (g), or a similar one.
Why would increasing the pressure cause it to shift in the direction of fewer moles? What, physically, makes the reaction that produces fewer moles favorable?
Also, why would increasing the temperature cause the system to shift to the more endothermic side? What, physically, makes the endothermic reaction more favorable than the exothermic reaction?

See also https://chemistry.stackexchange.com/q/125088/35806

For a general idea, see Le Chatelier's principle

When any system at equilibrium for a long period of time is subjected to change in concentration, temperature, volume, or pressure,
  (1) the system changes to a new equilibrium and
  (2) this change partly counteracts the applied change.

For the changes of $p, V$, read about equilibrium constant $K$ and Reaction quotient $Q$ and their relation.
Notice, how the Reaction quotient changes with increased pressure/decreased volume, if the reaction increases the molar amount of gases.
And how it changes if it decreases them.
For the former case, increasing pressure increases Q and the equilibrium shifts backwards to make Q  equal to K again.
For the latter case, increasing pressure decreases Q and the equilibrium shifts forwards to make Q equal to K again.

For the temperature effect, see van't Hoff equation:

$$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

Endothermic reaction direction has higher activation energy than the exothermic direction.
As the consequence,
the reaction rate of the endothermic reaction direction
raises with temperature faster
than the reaction rate of the exothermic reaction direction.
Faster endothermic then exothermic reaction means the equilibrium constant shifts at higher temperature in favour of the endothermic reaction.

See also https://chemistry.stackexchange.com/a/125159/35806


text:
What is the meaning of parentheses in "NAD(P)H"?
If someone writes "NAD(P)H is oxidized by ...", does that mean they are referring to both NADH and NADPH?
If so doesn't that lead to confusion when thinking quantitatively since the reactions may proceed at different rates, NAD may be present in a different concentration than NADPH, etc?
If not, what do the parentheses around P mean?

If an enzyme is called $\mathrm{NAD(P)H}$-dependent, that means it is using both $\mathrm{NAD^+/NADH}$ and $\mathrm{NADP^+/NADPH}$ as its cofactors. For example, most oxidoreductases are $\mathrm{NAD(P)H}$-dependent, which are able to oxidize a relevant substrate by transferring a hydride ($\ce{H−}$) group to a nicotinamide adenine dinucleotide cofactor (either $\mathrm{NAD^+}$ or $\mathrm{NADP^+}$), resulting in their reduced forms, $\mathrm{NADH}$ or $\mathrm{NADPH}$ (Ref.1):

You can find a list of $\mathrm{NAD(P)H}$-dependent oxidoreductases within Reference 1.

Reference:

Lara Sellés Vidal, Ciarán L. Kelly, Paweł M. Mordaka, John T. Heap, "Review of NAD(P)H-dependent oxidoreductases: Properties, engineering and application," Biochimica et Biophysica Acta (BBA) - Proteins and Proteomics 2018, 1866(2), 327-347 (https://doi.org/10.1016/j.bbapap.2017.11.005).

Thanks, perfect source: "NAD(P)H-dependent oxidoreductases are able to oxidize a substrate by transferring a hydride (H−) group to a nicotinamide adenine dinucleotide cofactor (either NAD+ or NADP+)".


text:
Can basic column liquid chromatography separate two extremely similar alkaloids?
Say a sample has two alkaloids, one with a molecular weight of 521.44 g/mol and one with a weight of 521.66 g/mol.
Would column chromatography be able to separate these two alkaloids?
Also, what if a sample has, say 15 different alkaloids. Will all of these separate in liquid column chromatography?
Thanks.

Similar molar mass does not say much about the alkaloid similarity, nor the alkaloid count. More information is needed, including chromatographic properties of alkaloids and the parameters of chromatography system.

TLC may give you a clue but the reality is it is very difficult to predict the exact behaviour of compounds on chromatography columns. I would think you would have a far better chance with prep HPLC.

Test it on a small scale (eg TLC) and see. It is extremely hard to predict the outcome but testing is cheap. It might be worth testing a variety of elutants or column materials. There is no easy shortcut.

Thank Matt. If they didn't separate via TLC what would be the next step? I mean, what do manufacturers use to separate alkaloids that wouldn't separate via tlc?

MS-guided prep HPLC is the current goto method

Could you recommend a company that could perform this for me?


text:
Direction of electron density in p orbitals
$\mathrm{s}$ orbitals are spherically symmetric while $\mathrm{p}$ orbitals are not. For Boron, the electronic configuration is $\mathrm{1s^2 2s^2 2p^1}$.
In which p orbital does the most energetic electron lie? Is it $\mathrm{p}_x$, $\mathrm{p}_y$ or $\mathrm{p}_z$? Since all the three are degenerate, how does the electron decide which orbital to occupy? If it occupies any of the one orbitals, is a neutral atom not spherically symmetric? Or does the electron keep changing between $\mathrm{p}_x$, $\mathrm{p}$ and $\mathrm{p}_z$ rapidly (or in a superposition of occupying all the three orbitals at once with the probability of finding in any one of the orbitals equal to $1/3$)? The same question can be asked for $\mathrm{d}$ and $\mathrm{f}$ orbitals.

Since the three p orbitals are degenerate, the electron is on all three (a linear combination of the three orbitals). The wavefunction will be spherically symmetric if you do not introduce any external field or similar that would break the symmetry. Related: https://chemistry.stackexchange.com/questions/13766/why-are-atom-spherical-in-shape and https://pubs.acs.org/doi/abs/10.1021/ed042p397


text:
Se+P White or red P?
I read this in an old chemistry book. It says:

When selenium is dropped into melted phosphorus, it dissolves rapidly, and the compound sinks through the liquid phosphorus in red streaks.

My question is whether the phosphorus is red or white?

Molten phosphorus is always P4 afaik.

Liquid phosphorus is white phosphorus. It melts at 44.1 °C.
Red phosphorus does not melt. It burns at 200 °C and sublimes in an argon atmosphere at 280 °C, without melting.

...and by sublimation, it turns into P4 anyway.


text:
Can the 5+ and 4+ oxidation state of Vanadium ion exist independently?
From my class text book:

The oxidation states of Vanadium ions 5+ and 4+ are shown in the compounds $\ce{VO2^{+}}$ and $\ce{VO^{2+}}$  respectively, whereas $\ce{V^{3+}}$ and $\ce{V^{2+}}$ are shown like this in the redox reaction (reduction half equation).

Why is this? cannot $\ce{V^{5+}}$ and $\ce{V^{4+}}$ exist indepepently as ions like $\ce{V^{3+}}$ and $\ce{V^{2+}}$ (when measuring the reduction potential)?

When an atom $M$ looses one or more electrons, it becomes positively charged and becomes the ion ${M^{z+}}$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $H_2O$, and it repells the proton of $H^+$.
If $Z$ = $1$, the ion $M^+$ attracts several molecules water around it, and it makes a Debye layer of oriented $H_2O$ molecules around the ion, and no chemical changes.
If $Z = 2$, the attraction is strong enough to repell one proton $H^+$ from one of the molecules adsorbed around the $Z^{2+}$ ion. The solution becomes a little bit acidic, and the $M^{2+}$ion is partially hydrolyzed into the ion [${M(OH)]^+}$. It is an equilibrium :$$M^{2+} + H_2O \ce{<=>}[M(OH)]^+  +  H^+$$
If $Z = 3$, the attraction between $Z^{3+}$ and water is strong enough to at least transform all $M^{3+}$ions into $[M(OH)]^{2+}$, and part of them to [$M(OH)_2]^+$ with the two reactions :
$$M^{3+} + H_2O\ce{->}   [M(OH)]^{2+}  +  H^+$$
$$[M(OH)]^{2+} + H_2O \ce{<=>} [M(OH)_2]^+  +  H^+$$
This is the case if $M^{3+}$ is $Al^{3+}$ or $Fe^{3+}$
If $Z = 4$, all ions $M^{4+}$ are at least hydrolyzed into $[M(OH)_3]^+$ and even into $M(OH)_4$, according to :  $$M^{4+} + 4 H_2O \ce{->}M(OH)_4  +  4 H^+$$It should be mentioned that $M(OH)_4$ has a tendency to loose water and produce a structure that has acidic properties. So it is often presented under the formula $H_4MO_4$ or $H_2MO_3$. That is the case with $M = Si$. Silicic acid is often presented as $H_4SiO_4$ or $H_2SiO_3$
With Vanadium $V^{4+}$, it goes another way. The molecule $V(OH)_4$ does not exist, as the ion $V^{4+}$ is too small to fix $4$ groups $OH^-$. As a consequence, $V^{4+}$ gives another positive ion $VO^{2+}$ according to :
$$V^{4+} +  H_2O \ce{->}  [VO]^{2+}+2H^+$$ Strangely enough, it looks as if the non-existent $V(OH)_4$ behaves like a basic hydroxide and gets dissociated according to : $$V(OH)_4\ce{->}[VO]^{2+}+ 2 OH^–$$
If $Z=5$, the ion $M^{5+}$ can be hydrolyzed according to $$M^{5+} + 3 H_2O \ce{->}  [MO_3]^{-}+6H^+$$$$M^{5+} + 4 H_2O \ce{->}  [MO_4]^{3-}+8H^+$$
But with Vanadium $V$, the ion $V^{5+}$ may be hydrolyzed  according to :$$V^{5+} + 2 H_2O \ce{->}  [VO_2]^{+}+4H^+$$$$V^{5+} + 3 H_2O \ce{->}  [VO_3]^{-}+6H^+$$The second reaction is favored at high pH.
This theoretical development is taken from General Chemistry, by Linus Pauling, W. F. Freemann and Co., San Francisco, 1953, Chapter XXI.

Most likely, vanadium cannot exist as a 4+ or 5+ ion except as part of a covalently bound species such as $\ce{VO^{2+}}$ or $\ce{VO2^+}$.  However, its heavier congeners niobium and tantalum form minerals whose structure indicates at least partial ionic bonding in the +5 oxidation state.  One such mineral is Schiavinatoite, a niobium-tantalum borate in which the niobium or tantalum (in the +5 oxidation state) has a high coordinate number (8) with all neighbors at essentially equal distances, which indicates a strong ionic contribution to the bonding.  The mindat.org site referenced above gives an original reference 1, accessible here.

(Illustration from mindat.org. Blue = Nb/Ta, pink = B, red = O.)
Reference
1.
Demartin, F., Diella, V., Gramaccioli, C. M., Pezzotta, F. (2001) "Schiavinatoite, (Nb,Ta)BO4, the Nb analogue of behierite."  European Journal of Mineralogy: 13: 159-165.


text:
Chemical structure of Turquoise
I know nothing about chemistry, I managed not to take it in correspondence high school. I don’t know how to read or make sense of this: $\ce{CuAl6(PO4)4(OH)8.4H2O}$ the chemical formula of turquoise. From looking at a children’s periodic table I figured out some of the letters like Copper and aluminum, but I don’t know what the parenthesis or numbers mean. Also, is the PO polonium?

PO4 is phosphate ion - Phosphorus plus 4 oxygens with an overall charge of 3-, the 4 outside the parenthesis means there are 4 phosphate ions. The OH is hydroxide ion (oxygen plus hydrogen bearing a -1 charge).

PO in that formula is not polonium. It is phosphate ion.

I don’t know how to read or make sense of this: $\ce{CuAl6(PO4)4(OH)8.4H2O}$ the chemical formula of turquoise.

It means for every atom of copper (Cu), there are 6 atoms of aluminum, 4 phosphate groups (which each contain a phosphate atom and four oxygen atoms), 4 hydroxide groups (which each contain an oxygen and a hydrogen atom) and 4 waters (which each contain two hydrogen atoms and one oxygen atom). So the parentheses are for grouping, and the numbers are to count how many atoms or groups there are.
The structure is a crystal, so it goes on and on in all directions. That is why I was careful to say "for every atom of copper...". For other substance like carbon dioxide that are found as molecules, it is less complicated (there, the formula actually tells you the number and kind of atoms in a single molecule).

Chemical structure of Turquoise

Some of the components in turquoise are positively charged (copper and aluminum) and some are negatively charged (hydroxide and phosphate). The building blocks are arranged that positive and negative come close together, and like charges are further apart. The entire structure is very complicated.
The picture below does not show all the atoms, but instead shows copper, aluminum and phosphate as 3D shapes (the metals and the phosphorus would be in the center of each of the shapes). The shapes have common corners occupied by oxygen atoms (in the formula, these appear as part of phosphate, hydroxide and water). The hydrogen atoms are shown at tiny spheres.
Source: http://rruff.info/uploads/MM64_905.pdf

Thank you! That’s all very helpful! My interest is in finding out if there’s any scientific support for the claims that gems and crystals do anything like heal or keep bad dreams away. I don’t believe they have ANY spiritual properties and can’t do things like bring wealth or protect from harm, but since they are minerals and minerals in general work with the human body, I though some claims might be reasonable. So I decided to start learning just because it interests me. So thank you very much for helping!


text:
Colour of VO3^-?
To aquire the yellow colour of vanadium ion, it must be present in a 5+ oxdiation state.
This can be acheived (according to my textbook) by adding dilute sulfuric acid to $\ce{NH4VO3}$, then followed by zinc.
$$\ce{VO3- (aq) + 2H+ (aq) -> VO2^{+} (aq)  + H2O (l)}$$
where $\ce{VO2^{+}}$ gives the yellow colour from the $\ce{V^{5+}}$ present in the compound
However, is not $\ce{VO3-}$(aq) yellow in colour?
If it is, then why is $\ce{NH4VO3}$ not yellow in colour but white?
Side question: what is the purpose of zinc in this reaction explained above - not including reduction to lower oxidation states?

Ammonium vanadate is white but it can turn to yellow owing to impurities in the form of $\ce{V2O5}$.


text:
Gibbs energy: the term −TdS. Where does the reversible heat go?
When we have an exothermic reaction as well as a decrease of the entropy in the system after the reaction, why does the decrease in the entropy of the system diminishes the amount of useful work we can extract from the system?
In my opinion, if we have at the final state less accessible microstates than at the  initial state, the kinetic energy (at STP most of them as translational energy) of the system must be more concentrated and therefore have a higher capability of doing useful work. Thus, it gains reversible heat from its system, "stores" it as kinetic energy and could do work.
Where is my mistake? Could you please help me, maybe with a depiction?

Does this answer help? https://chemistry.stackexchange.com/a/112958

The way to think about microstates is as alternative configurations of a system which satisfy constraints consistent with the observed macroscopic properties of the system. The collection of such microstates forms an ensemble.
The misconception lies in the idea of "concentrating" kinetic energy, and in thinking that reducing the number of available microstates changes the available energy. It doesn't necessarily do so. It is necessary to be careful, since the idea that "concentrating kinetic energy reduces the entropy" is not generally correct. In general you should consider all available degrees of freedom. An ideal gas only has translational degrees of freedom (kinetic energy) so in this case this restriction to kinetic energy applies.
In addition, during an isothermal process performed on an ideal gas, the internal energy does not change, and neither does the average kinetic energy ($K=\frac32 N \mathrm{k_B}T$), but the number of available microstates may change (for instance with an expansion), as reflected in the dependence of entropy on volume. The change in entropy reflects a change in the probability distribution over the ensemble.
To illustrate what is meant with "dispersion" (and by extension entropy) envision a maze in which the system wanders. As the maze gets larger the system gets more lost in that maze, it has more choices where to go.
If you use a particle in a box model, assuming quantum like behavior, then the gaps between energy levels become smaller as you increase the size of the container, that is, with expansion. This results in spreading probability of occupation (dispersion) over more states. Now, irrespective of how the isothermal expansion is performed - reversibly or irreversibly (assume a free expansion) - the entropy cost for the system is the same. However, in the case of the spontaneous expansion the surroundings remains at constant entropy, whereas its entropy is reduced in the reversible case. In addition (and this is a key point), no work was performed in the free expansion.

I appreciate your text. For your example we have constant internal energy and the same average kinetic energy, respectively, throughout the expansion. The system has now more accessible microstates and can therefore disperses its energy more ergo (in my opinion) "looses" some energy per Kelvin. How can this work in terms of the molecular view. The more spread out thermal energy is less useful now, but why

I think the idea of dispersion is causing confusion. Hope my edits help.

I like that maze comparison. Now where the system has more choices what happens to the energy quanta. I always read that the entropy increases with every irreversible heat transfer and becomes less useful to us due to its reduced concentration though the quantity stays the same. Where is the connection to the changes probability distribution.

What energy quanta? And in the second sentence, which quantity stays the same?

I have meant with "quantity" the amount of energy and energy quanta, respectively. And to your first question I have learned that the internal energy is quantized and is therefore the sum of energy quanta which are distributed over the microstates. Do you know what i mean. I apologize for my english. I am not a native speaker.

I think this is a difficult problem to wrap ones head around but maybe my further edit helps. Its useful to keep account separately of the entropy change for the system and that of the surroundings.


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What is the main difference between computational methods such as Hartree–Fock and tight binding?
I was reviewing many electronic computational methods recently and got a bit confused about relation between different approaches. In particular, what is the essential difference between Hartree–Fock (HF) and tight binding (TB)? I am not that deeply familiar with theory. And just for note I am physicist more than chemist.
I know that HF is more general method not essentially relying on LCAO approximation. But say we mainly use LCAO for HF anyway in most cases. In TB the Hamiltonian matrix elements $H_{ij}$ are mainly used from some tabulated values or fitting with experimental data. This is good for speedy calculations.
But can we just use LCAO orbitals and calculate matrix elements analytically using general many-body Hamiltonian? Will it not become HF method essentially?

Not too familiar with solid state stuff myself but I got an impression that TB is essentially a semi-empirical method used in solid state physics. If so, it is indeed nothing more than a simplification of HF with certain integrals precomputed 'for all cases' and included in the model.

There is also the so-called 'density functional tight-binding' which similarly uses some precomputed elements of the Hamiltonian but within the DFT, not HF framework. A step further in some way, and considered generally better than conventional semi-empirical methods for molecules (like PM6 or AM1).

Maybe this question could be migrated to mattermodeling.stackexchange.com.

In a nutshell
The tight-binding model relies on the assumption that we can forgo the GLOBAL antisymmetry of fermions and instead purely work with a LOCALLY antisymmetric basis. In that sense, the tight-binding model is an approximation to the total wavefunction that uses site-specific antisymmetric wavefunctions to form basis states for the total wave-function.
Details
The tight-binding Hamiltonian takes the form:
$$ H = \sum_{i} E_i | i \rangle \langle i | + \sum_{i} V_{ij} | i \rangle \langle j |  $$
Where $E_i$ is the energy of the particle (usually either an electron or exciton) on each site and $V_{ij}$ is the coupling between sites. Often this model and it's parameters are selected purely empirically, but the tight-binding Hamiltonian, and the formula for it's matrix elements, can be derived from first principles.
To do so we realize what the tight binding Hamiltonian is: it is the exact many-electron Hamiltonian projected into some basis $ |i \rangle $, where each basis state $ |i \rangle $ is a tensor-product of individual site wavefunctions. To give a concrete example consider a lattice of organic molecules, each individual molecule representing a site in my tight-binding model. In such a case the basis states of my tight binding model become:
$$ |i \rangle = | \psi^{(1)}_i \rangle \prod_{j \neq i} | \psi^{(0)}_j \rangle $$
That is my basis state is a tensor-product of molecule $i$ in the $1$ state and all other molecules $j$ in the $0$ state. For electron transfer, this would mean molecule $i$ is in the charged site, while other molecules $j$ are in the neutral state. For exciton transfer $i$ is in the excited site, while $j$ are in the ground state. Note that such a basis cannot be antisymmetric, a tensor product of antisymmetric wavefunctions is NOT itself antisymmetric.
From here, with a given model for the site wavefunction (Hartree-Fock, DFT, Configuration Interaction, etc.) one can derive the matrix elements of the tight-binding Hamiltonian. The derivation in the case of CI wavefunctions was first given by Davydov in his seminal paper on molecular excitons: http://dx.doi.org/10.1070/PU1964v007n02ABEH003659
The question that remains is why such a basis is a good choice for the total material wavefunction when we know it does not even obey the Pauli Exclusion Principle. It all comes back to the idea that that the electrons must somehow be localized. Consider the HF exchange integral:
$$ K_{ij} = \int  \psi^*_i (r_1) \psi_j (r_2) \frac{1}{r_{12}} \psi^*_j (r_1) \psi_i (r_2)  $$
Note that this integral is 0 when there is no spatial overlap between orbitals "i" and "j", e.g. $\psi^*_i (r_1) \psi_j (r_2)=0$. When the exchange integral is 0, the Hamiltonian matrix element of the Slater Determinant and Tensor-Product wavefunctions become identical. So, in essence, when electrons can be spatially localized, they are no longer really indistinguishable, and it becomes OK to violate antisymmetry.
This also means you come come across people using tight-binding Hamiltonians for systems that really aren't tight-binding (say for atoms in an individual quantum dot), you should be extremely wary.

This is quite a specialist subject, and I don't know enough to judge the correctness of your answer, hence I refrain from voting; but regardless, welcome to Chemistry SE!


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Reactions involving carboxylic acids
The following question was given in FIITJEE study material:

I guess (i) step is de-esterification, which gives the compound of option (A).
In next step, I wouldn't expect it to decarboxylate, instead I guess it'll dehydrate. So, according to me, final product is:

which I can't see in the options. Then, what should be the final product?

 The correct answer given was: (C)

The conditions for the first step specify base but not water present, so you are not going to get hydrolysis which means the product is not A . What might a sufficiently strong base do to the starting material?

@Waylander: Ok, so this is Claisen condensation. And, then this is the product, which converts into acid on hydrolysis and finally decarboxylate on ∆. Thanks for helping.

Correct. Depending on how strong the acid is, how long you heat it and to what temperature.

The answer is B. You use NaOEt as the base to avoid trans-esterification.

@joehua: I agree with your expl. But, I think we get (B) in the first step itself, what about the second one?

Now, I see that when acid is added, the solution is also heated. So, clearly, this is hydrolysis then decarboxylation. The answer should be C.


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Why does 2+ oxidation state become more stable relative to 3+ oxidation state for first row of transition metals?
I viewed an image showing all the possible oxidation states of each element in the first row of transition metals, and the main oxidation states highlighted in a different colour.
I noticed all elements Sc to Cr had in common a   3+ main oxidation state,
and from Mn to  Zn 2+ was the common main oxidation state.
The reason stated for this in my good textbook was 2+ oxidation state become more stable relative to 3+ oxidation state, I can't quite understand why.
So  Q : why does 2+ oxidation state become more stable relative to 3+ oxidation state  for first row of transition metals?
I had drawn the electron - in -boxes configuartion for all elements in the first row, and looked at Hund's rule and partially and fully filled orbtals, but I have not really seen a clear, fundamental explanation of why, for this question.

While question may be more specific, answer covers whole this topic precisely^

It is due to the difference in their second and third ionisation enthalpies. Due to the large difference in the second and third ionisation enthalpies arising from the disturbal of the stable half filled electronic configuration, +2 oxidation state is more favourable.

(-1) It is true that half-filled subshells are more stable. It is not true that this special stability contributes in any significant way to the overall trend. See the linked question & answer in the comments above...


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What in salt makes ice unable to refreeze?
I know that salt makes ice melt, but why does salt make the ice not refreeze?
I have tried all different kinds of experiments, but I still cannot figure it out.
The chloride acid in the salt makes the ice melt. What makes then the ice colder than it was before we put the salt on the ice?

It doesn't make it colder, but it does make it more difficult to crystallize

@Tyberius Table salt is famous about simultaneously melting ice and getting  the mixture colder down to about -20 Deg C, what is freezing point of saturated solution. Calcium hydrate can go down to -50 Deg C

@Poutnik you are right, I was too focused on the crystallization process and didn't think about the thermodynamic impact

what thermodynamic impact

It might be worth looking as the answers here https://chemistry.stackexchange.com/questions/116302/where-does-the-energy-come-from-to-lower-the-temperature-of-a-brine-solution/116316#116316

There is no such thing as "a salt that makes ice unable to refreeze". Well, sort of depending of the conditions.
What happens is that salts added to pure water ice form a solution (of the salt in the melt water) that has a much lower point of freezing that pure water ice.
This is known as the freezing-point depression (Wiki for the basics).
For example, adding common salt (sodium chloride) to pure water ice will depress the freezing point of water to about -21°C, so ice will melt until the solution will reach this temperature. Then, you'll have an equilibrium of -21°C salt water and -21°C solid ice.
If kept at the freezing temperature of pure water, the salt water will never re-freeze on its own.
If you take this salt water (now at -21°C) and lower even more its temperature, it will then freeze, forming salt-water ice.
Other salts (like sodium acetate used in airports, for example) will give you different temperatures, but all of them will re-freeze if the temperature is lowered below some definite temperature dependent on the salt and concentration.


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Can you explain DFT and TDDFT functioning (math aside)?
I have been recently reading a lot on the quantum mechanical theory regarding Density Functional Theory, DFT, and Time Dependent Density Functional Theory, TDDFT (Oscillatory and Rotatory Strengths in particular) in order to understand how first principle calculations fundamental theory work.
I get that DFT is used to calculate the ground-state configuration of a system and TDDFT gets you the excited states useful for spectra determination and that there have been developed several algorithms in order to make calculations more efficient (timewise).
But I still can't answer myself in a short way how does each one works (math aside).
If you were to explain them to a person that is not necessarily familiar with quantum mechanics, how would you do it?

I do not have access, but if memory serves, this may help you: https://pubs.acs.org/doi/abs/10.1021/cr0505627

@TAR86 Well then, http://iopenshell.usc.edu/chem545/lectures2014/HEAD_GORDON_REV.pdf

Thank you both, that is indeed a great review on ab initio methodologies, still, a simple how and why does it works answer is still yet to be found

I'm closing this question because it has been asked and answered on Physics Stack Exchange. Please note that Cross-Posting is highly discouraged on the network.

Dear @Martin-マーチン I think that a different answer can be given by a physics and/or a chemist approach. Both of them can potentially be useful as the fundamental explanations from each approach can give more information on what is happening.

While this might be true, it does not change the fact that the question was cross-posted, which is highly discouraged. Also the audience for these questions on our network will be very similar, hence the interested users will be redirected towards the question, which has the most information. If the question can be reshaped to highlight the different perspectives of the different communities, then the users may be persuaded to reopen this question. If you seek further clarification, please ask on [meta].


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What is "salpotricon"?
I read in my old Polish history handbook about the black powder recipe that was taken from some medieval manuscript, if I remember correctly.
One of the ingredients was called “salpotricon”, but years passed and I still have no clue what could it be.
So, I'm asking the internet to solve this mystery. Maybe it's a Latin word, I'm not sure.

Could this be Saltpetre? Otherwise known as Potassium Nitrate.

Jeżeli chcesz sporządzić proch silny, weź cztery funty czystej saletry, jeden funt siarki, jeden funt węgla, jedną uncję salpotricon* i jedną uncję... Saltpetre is already in the list of ingredients.

Using google translate (so take this with a handful of Sodium chloride) I get "If you want to make strong gunpowder, take four pounds of pure saltpeter, one pound of sulfur, one pound of coal, one ounce of salpotricon * and one ounce" which implies Saltpetre and salpotricon are different things. In that case I have no idea what the later may be, and a very, very quick scan of the wikipedia article on gunpowder leaves me no wiser.

Where has Karsten Theis found this unreadable text ?

If you read Polish I suspect it is fairly clear. Just google "salpotricon", it's one of the very few hits. Unfortunately it seems to be hidden behind some sort of login

You mean this link https://forum.vmc.org.pl/viewtopic.php?t=1023&start=125 unfortunately there is no answer for this question

OK, I asked one my Research Associates who is half-Polish. She didn't know, so she asked her Polish mother, who is also trained as a translator. She didn't know either, saying "It sounds rather old and yes, nobody seems to know. I've even checked my old printed encyclopaedia, but nothing there either.
It definitely is not an originally Polish word, it seems verbatim ("c" never appears in that kind of an ending in Polish)  taken from a different language." So it is quite a puzzle - and I'm beginning to suspect a languages group might be better than this chemistry one. The on suggests Greek ...

@IanBush The weird thing is that the word has an asterisk after it, so I suspected that there was some explanation in the question.


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What gas is evolved when using vinegar, hydrogen peroxide, and salt to etch copper, and why?
The question of how vinegar + hydrogen peroxide + table salt can be used to etch copper has been asked before (Full equation when using vinegar, hydrogen peroxide, and salt to etch copper). However, the accepted answer seems to miss a key fact when observing this reaction, so I wanted to ask for clarification in the hope that someone can explain the unaddressed fact.
When the vinegar + hydrogen peroxide + table salt solution is used to etch copper, there is a gas that vigorously forms on the surface being etched. However, none of the reaction products in the accepted answer for the full equation for this reaction are volatile. So if none of the reaction products are volatile, how can a gas be released?  Clearly, there is something missing from the accepted answer. Does anyone have any ideas?

Make a bold guess! ;)

My guess would be hydrogen, but if so, the accepted equation for this reaction is wrong, so I am hoping someone can address both points at once.

OK, it´s not hydrogen. Hydrogen evolves when base metals react with water (or acid). Next guess, please. (The accepted answer of that other question is correct as far as I see, but it does in fact not tell you what gas comes out. There is another reaction.)

OK, so it is probably oxygen, then. So what is the missing link in the reaction process?

Correct. The point is that you add a rather large excess of peroxide. It just bubbles away.

So it is just direct decomposition of H2O2?  Why does it happen only at the copper surface?  I have been using this chemistry to etch PC boards, and the bubbles form very vigorously at the Cu surface, but hardly at all in the masked areas, or on the surface of the tank.

Well, the Cu surface is probably the most coarse of all, and then it´s somewhat catalytic. Why exactly? Tried if other metals do the same?

Thanks for the helpful discussion. I'll have to see what happens with other metals...


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Why does Pb have a higher electronegativity than Sn?
I recently learnt that electronegativity generally decreases as I move down a group and from right to left within a period. However, according to the table below, Pb has an electronegativity of $2.33$, which is higher than $1.96$ of Sn. There are also a lot other elements, such as Au, Hg, and W, that does not follow the general rule of electronegativity.

Can someone offer an explanation of why some elements have a higher electronegativity than the elements directly above them?

Not necessarily, value depends on source https://en.wikipedia.org/wiki/Electronegativities_of_the_elements_(data_page)

The explanation is actually well-known.

Elements in Period 6 get f orbitals for the first time. 4f subshell has relatively poor shielding effects, thus it's harder for the atoms to lose the outer electrons, increasing electronegativity. (similar reasons also account for some chemical properties of Period 4 elements, like $\ce{H2SeO4}$'s oxidizability)
Relativistic effects also contribute. For a very informal (and actually incorrect) understanding, just imagine that the electrons in these heavy atoms move fast and their masses increase and their energy decreases.

Electronegativity, symbol χ, is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.
Lead has an electronic configuration of [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
Inorder to achieve half filled electronic configuration which stabilises the lead atom, it attracts on the shared pair of electrons with greater force making it more electronegative than Tin.

But Sn is also one electron away from half filled 5p. How would you explain that?


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What is the priority order of the four groups?

I'm trying to figure out whether this is in R/S configuration....I thought it would be R since the groups seem to be decreasing in priority in the clockwise direction (since F has the largest atomic number). However, the answers say that it is actually in the S configuration....could someone explain this to me? Have I incorrectly assigned priority?

So what if F has the largest atomic number? It is not like we look at the whole group and count the largest atomic numbers. No, we look at the first atom alone, and think of it for a while. If that's indecisive, we look further...

Oh ok....so does that mean that the MeO group has highest priority followed by CF3...?

Looks like so.$,$

The priority is MeO > CF3 > COOH. I think the configuration is R.


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Flexibility of long-chain aliphatic compounds
PubChem is unable to generate the conformers of Z-6-Heneicosen-11-one? The website says that the molecule is too flexible
What does flexibility of a molecule mean in this context? I understand that it is a long chain compound, but I assumed that the bonds are rigid and do not flex? In layman’s terms, how does one quantify flexibility? Is it flexible like a skipping rope or like something else?

The bonds are rigid, but the torsion angles aren't.

Related: https://chemistry.stackexchange.com/q/125574/5017

Rather than to flexible it want to say too much atoms /bonds

@Alchimista - it's not that it has too many atoms or bonds. A rigid molecule with the same number of atoms will appear (e.g, cholesterol is okay).

PubChem 3D has limits on the number of 'rotatable bonds'
I recently answered a related question about PubChem 3D:

Beyond the limits of MMFF94 and MMFF94s themselves, PubChem3D had several limits indicated in the accompanying manuscript: Bolton et. al. "PubChem3D: a new resource for scientists" J Cheminf. (2011) v. 3, art. 32)

Not too large (with ≤ 50 non-hydrogen atoms).
Not too flexible (with ≤ 15 rotatable bonds).
Consists of only supported elements (H, C, N, O, F, Si, P, S, Cl, Br, and I).
Has only a single covalent unit (i.e., not a salt or a mixture).
Contains only atom types recognized by the MMFF94s force field.
Has fewer than six undefined atom or bond stereo centers.

Your molecule falls into the "more than 15 rotatable bonds" category. I count about 18 $\ce{sp^3}-\ce{sp^3}$ single bonds.
Basically, the problem is that the search space for conformers with that many rotatable bonds is too big. In principal, it's $\approx3^n$ where $n$ is the number of rotatable bonds.


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Surface energy for different crystallographic planes
For a given crystal structure, say body centred cubic, on what factors does surface energy of each plane depend?
A basic approach considers the atomic density on each crystallographic plane, the number of bonds broken if the surface is cleaved along that plane and combines these two.
Is this a realistic and reasonable approach if I want to compare the surface energies of two crystallographic planes, say (100) and (111)? What are the other factors, if any, that should be included in the comparison?
Edit : Background : My problem related to the wettability of a crystal surface by a liquid. The contact angle that the liquid makes with the crystal surface is determined by the interfacial energies between solid-liquid, liquid-vapour and solid-vapour. Keeping all other factors, I only change the nature of the solid crystal surface, by presenting a different crystallographic plane. I would like to theoretically estimate which crystallographic plane would have the highest energy, and which would have the lowest, so that a prediction can be made about the extent of wettability of the surface.
Thank you!

As you speak of a solid/liquid interface, obviously you must take into account the liquid phase as well. But I think you have overgeneralised your problem. Be more specific!

Thanks @Karl, I have removed the bit about the interface, and now am asking only about the surface energy.

Background info is generally good! What I meant was that you should be more specific about your actual problem. You ask a very general question, those rarely have a clear answer.

Thanks @Karl, now I understood. Hopefully with the background, the problem is now more specific. :)

You can think about this problem from the definitions of surface energy.
The surface energy quantifies the disruption of inter-molecular bonds that occurs when a surface is created

The composition or structure of a surface depends on how the bulk crystal has been cut.
The electronic structure (or number of dangling bonds) depends on surface composition
The reactivity of a surface depends on the electronic structure.

Imagine a battenberg cake:

The cake has been cut horizontally and the surface is comprised of tasty red and yellow squares.
If however we had cut the cake vertically straight down the middle, you would only see a red line next to a yellow line (Cut 1).
If you cut vertically and diagonally across the cake, you might only see either a purely red or purely yellow surface. In the case of the example image, you would only see a red surface (Cut 2).

I hope this example shows that (by relating the coloured squares to atoms), different cuts result in different surfaces with different properties.
When calculating the surface energy of a system:

You need to calculate the bulk crystal energy.
You need to calculate the energy of the surface slab. Make sure your model is converged with respect to slab thickness. The central layers should be representative of the bulk crystal. Look at magnetic change, charge difference, degree of ionic relaxation.
You then calculate the difference between the slab energy and the bulk crystal energy.

Note: In periodic calculations, a surface slab is constructed by cutting the bulk in a specified plane and introducing a vacuum gap. This vacuum gaps should be sufficient that the surface does not interact with its periodic mirror images. Around 15 Angstroms.
There are a number of codes that can help in the construction of surface slabs (METADISE, Materials Studio, Vesta)
If you wish to calculate the energy of a surface (and assuming you are using a periodic DFT code), I would highly recommend the paper by Wenhao Sun and Gerbrand Ceder (https://www.sciencedirect.com/science/article/pii/S003960281300160X)
Note: be careful of magnetic influences.

Thanks @Wychh, you explained this beautifully. Now, I'm not familiar with DFT packages, so I'm thinking of using a simplified broked-bond model, such as that described here (https://www.sciencedirect.com/science/article/abs/pii/S0022369799004151). Do you think this is a reasonable approximation, or will I make major deviations if I dn't do an exact DFT calculation?


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How to explain that an increase in temperature favours an endothermic reaction?
I'm trying to explain how a change in temperature favours an endothermic/exothermic reaction from a collision theory point of view, but I'm not really familiar with the concept just yet. For an increase in temperature, can I say that:

An increase in temperature leads to a greater frequency of collisions between reactants and products, leading to an increased rate of reaction. Since the rate of the endothermic reaction increases more than the exothermic reaction, the equilibrium shifts towards the reactants/products.

If this is right, I'm struggling to see why the endothermic reaction rate will increase more with this logic. The exothermic reaction has a lower activation energy, so shouldn't its rate of reaction increase more than endothermic since it has a lower energy requirement to complete the reaction? Unless the increase in particles with enough energy for the exothermic reaction is comparatively lower to the increase in particles with enough energy for the endothermic reaction (aka proportion)?

You cannot just explain the kinetics with thermodynamics and vice versa. Your proposal seems to work qualitatively in one direction, but it leaves out e.g. activation energy, so it can never give you a correct picture of the kinetics.

I believe that collision theory in itself wouldn't be enough to explain this behavior. You can explain it by having Arrhenius equation and the activation energy in mind. What you got wrong was the effect of temperature on reaction rates depending on the activation energies. Actually, the speed constant of a reaction with higher activation energy will be more significantly affected by variation in temperature. This can be verified easily by manipulation of the Arrhenius equation. Apply ln to both sides to obtain:
$\ln{k}=\ln{A}-\frac{E_a}{RT}$
The graph of ln(k) as a function of (1/T) is a straight line with slope's magnitude directly proportional to activation energy.
Now, assuming that the endothermic reaction has higher activation energy than its inverse, it will be more affected by, say, an increase in temperature. While both reactions will become faster, the speed constant of the endothermic reaction will increase to a greater degree and that will shift the equilibrium favoring the products of the endothermic reaction.
If the temperature is decreased, the endothermic reaction will be slowed down to a greater degree, favoring the products of the exothermic reaction.

If you look at the graph of the reactivity with relation to temperature it increases proportionally until it reaches its activation energy and then it proceeds to decrease. For exothermic reactions since the activation energy is met early the reactivity decreases when more temperature is provided. While for endothermic reactions since they require a higher activation energy combined with the fact that they absorb some of the energy for the reaction to take place, we can draw the conclusion that increase in temperature favours endothermic reactions.

I think I kinda of understand what you're trying to say. Do you have a link with a graph perhaps that I can look at? I tried looking at the Arrhenius equation on wikipedia, but that just shows an exponential graph. Not sure if that's the right path to go down.

Here is an answer to that question, according to a final exam's marking guidelines at a particular school:

As the temperature increases, the average kinetic energy of the particles increases. This means that an increasing proportion of the particles has energy equal to, or greater than, the activation energy for the reaction. In terms of collision theory, a change in temperature has a greater impact on the rate of a reaction with a large $E_\mathrm{a}$ than on one with a small $E_\mathrm{a}$. As an example, raising the temperature from $\pu{20 ^\circ C}$ to $\pu{30 ^\circ C}$ for a reaction with an activation energy of $\pu{60 kJ mol-1}$ increases the fraction of collisions with enough energy to react by 97%. The same increase in temperature for a reaction with an activation energy of only $\pu{42 kJ mol-1}$ produces only a 75% increase in the fraction of collisions with enough energy to react. For an endothermic reaction, the activation energy is greater for the forward reaction than for the reverse exothermic reaction. The rate of the endothermic reaction is increased more than the rate of the backward reaction in response to a change in temperature.  Hence, an increase in temperature will always favour the reaction that cools the reaction vessel (the endothermic reaction).

Suppose a hypothetical reaction A + B <--> C + D, where left to right is exothermic and right to left is endothermic.
To understand why endothermic rate is more impacted with increase in temperature you must consider the Arrhenius equation:
k = Ae^(-Ea/RT)

k = rate constant
A = frequency factor (or per-exponential factor) --> this factor varies only slightly with temeprature.
e^(-Ea/RT) = fraction of molecules that have energies equal to or more than the activation energy.

To understand why temperature favours endothermic reaction you need to graph y = e^(-Ea/RT).

The x represents Temperature. The y represents frequency of molecules with energies equal to or above the activation energy. The green curve is for exothermic (forward) reaction where Ea is small, and the blue curve represents the endothermic (reverse) reaction where Ea is high.
You can see that as temperature increases the green curve plateaus the blue curve continues to rise. Eventually the blue curve and green curve will be almost on top of one another. This indicates that given a temperature high enough both the forward and reverse rates will have similar values for their rate constant k.
Whats interesting to note here is that for the exothermic curve (green) the rate constant rises very rapidly with a small temperature increase and then any increase beyond that has no significant impact. For the endothermic reaction (blue), however, the rate constant rises steadily and the higher the temperature goes the more it increases. Because both the curves eventually plateau as you extrapolate out the exothermic rate will never be less than the endothermic rate, they will just become really close to one another.
From a molecular standpoint A and B already have a good proportion of their molecules close to the forward Ea, and so any further increase in temperature won't make a significant difference to the amount of molecules that can reach the transition state. For C and D the majority of the molecules have low energy compared to the reverse Ea, and so increasing temperature will drastically increase the proportion of molecules that can reach the transition state.
Effect on Equilibrium Constant Keq:
So we've established why increasing temperature increase the endothermic reaction more than it increase the exothermic reaction rate. This means we will now have more A and B forming than we did at the lower temperature and more C and D disappearing.
Keq = [C][D] / [A][B]
This will shift Keq to a lower value. If we decrease the temperature the endothermic rate (reverse rate) will decrease faster than the exothermic rate (forward rate) and so [C] and [D] will rapidly form and [A] and [B] will be used up, shifting Keq back to a higher value.
I know that was a long answer but hopefully it helps understand why reactions behave the way they do under temperature change.

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(A = B + energy) is the total equality for a chemical reaction; Mass-Energy is conserved. At equilibrium the chemical potentials of products and reactants are equal. Since the equilibrium is dynamic this means that the rates of the forward and reverse reactions, when the products, reactants and energy content are contained, are equal. kf [A] = kr [B] [energy], [B][Energy]/[A] =K[equilibrium]= kf/kr. Keq expressed thusly is independent of temperature. [Normal Keq and rate constants ignore energy and are functions of temperature.] This means an increase of heat content, an increase in temperature, means a decrease in [products] and increase in the [reactants]. An increase in temperature changes the equilibrium in the endothermic direction.
This explanation uses the First Law and makes the combination of concentration and inherent energy content of the reactants and products paramount. The kinetic theory of matter is implicit; the chemical activity of the molecules depends on their concentrations and their energy content. It makes no mention of activation energy, which is good, because the activation energy concept depends on mechanism; this approach by starting at equilibrium is independent of mechanism. A second apparent contradiction is a phase change at constant pressure of a pure substance. Here both activities are defined as 1.; a change in energy will change the potential of both phases. The result is phase change to eliminate the change in energy by changing the amounts of each phase maintaining constant chemical activities, T, and P.  The reaction in one direction is energy driven in the other entropy driven.


text:
Why does changing the salt of the salt bridge effect the voltage produced in a galvanic cell?
My experiments results showed that the voltage decreases down $\pu{Group 1}$: $\ce{Cl-}$ salts in the salt bridge. All other variables were kept constant. $\ce{Zn}$ and $\ce{Cu}$ rods were used. Concentrations of the salt in the salt bridge were $\pu{1M}$.
I dont understand this result. What could the reason be behind this?

The voltage measured in a Zn/Cu cell does not depend on the nature of the salt bridge. Could you explain exactly what you have done and measured ? And what is a "Group 1 chlorine salt" ? Is it not a mistake ? Is it not "chloride" that you like to mention ?

@Maurice I believe Group 1 chlorine salt is Alkali metal chlorine salt eg NaCl, KCl etc. The voltage is probably decreasing as the mobility of the ions in salt bridge increases i.e. $\ce{K+}$ has more mobility than $\ce{Li+}$


text:
Coordination number of chromium(III) oxalate complex
The coordination number refers to the number of binders that interact with the central metal ion. So, in the complex $\ce{[Cr(C2O4)3]^3-}$ chromium is bound to three oxalate ions.
Why its coordination number is considered 6? So, is it an incorrect definition of coordination number?
Can you help me?

Each oxalate ion has two binders.

A better way to think of it is that the coordination number is the number of points of contact between ligands and the metal. If a ligand like oxalate touches the metal at two places, it adds 2 to the coordination number instead of 1.

This seems to be the case of "show, don't tell".
IUPAC's definition of coordination number (C.N.) which applies to inorganic complexes:

In an inorganic coordination entity, the number of σ-bonds between ligands and the central atom. π-bonds are not considered in determining the coordination number.

Let's have a look at the one of the first determined structures of trioxalatochromate(III) inorganic salts [1]:

Figure 1. Crystal structure of $\ce{[Cr(C2O4)3]^3-}$ anion (ICSD-109607). Color scheme: $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#8A99C7}{\Large\bullet}~\ce{Cr}$.
Here, there are three oxalate $\ce{C2O4^2-}$ ligands arranged in distorted octahedral environment (depicted with semi-transparent coordination polyhedron).
Each oxalate is a bidentate chelating ligand, resulting in C.N. 6 of chromium center.
References

van Niekerk, J. N.; Schoening, F. R. L. The Crystal Structure of Ammonium Trioxalatochromate(III), $\ce{(NH4)3[Cr(C2O4)3].2H2O}.$ Acta Cryst 1952, 5 (4), 499–505. DOI: 10.1107/S0365110X52001428.

As Ivan Neretin told it, each group oxalate has two groups $\ce{-COO^-}$. So it may be bound to a central metal by two oxygen atoms, one per group $\ce{-COO^-}$.

So, the binders aren't the whole molecule or ion, but the atoms bound to the metallic ion, I'm right?


text:
What are computed energies in DFT and FF methods relative to?
I am not talking about zero-point energy.
I wish to understand how the energy of a system is calculated relative to a zero baseline. This is best explained with an example.
From DFT, the energy could be:
Isolated H = -13.6 eV,
H2 = -31.8 eV

What are these numbers related to? What would be a system at 0 eV? Or what is the zero baseline from which these number are calculated from?
I have also noticed that force-field energies tend to be positive. What is the zero baseline here?
In summary:

What is the DFT baseline?
What is the Forcefield baseline?

Start by looking at the Hamiltonian for a molecular system
\begin{equation} \label{eq:coulomb_hamiltonian}
    \hat{H}
    = - \sum\limits_{α=1}^{ν} \frac{1}{2 m_{α}} \nabla_{α}^{2}
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{α=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{α}}{r_{αi}}
    + \sum\limits_{α=1}^{ν} \sum\limits_{β > α} \frac{Z_{α} Z_{β}}{r_{αβ}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
\end{equation}
where $m_{α}$ is the rest mass of nucleus $α$ and $Z_{α}$ is its atomic number, $r_{αi} = |\vec{r}_{α} - \vec{r}_{i}|$ is the distance between nucleus $α$ and electron $i$, $r_{αβ} = |\vec{r}_{α} - \vec{r}_{β}|$ is the distance between two nuclei $α$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.
Now, thinking classically, the total energy is equal to zero when all the terms in the Hamiltonian are equal to zero, so that the zero of the total energy corresponds to the case when, (again) classically speaking, particles are infinitely far away from each other and not moving. Obviously, the total energy of any stable molecular system should be lower than that, i.e. negative, for otherwise the system will be unstable with respect to dissociation towards particles. That is the way it works for any method of computational chemistry that solves the Schrödinger equation or its relativistic counterpart is some way or another (HF, post-HF, DFF, etc.).
But force field methods and energies are completely different beasts, because the meaning and the zero of energy in this realm depends on the functional form of potential energy and parameter set used in a particular force field. So, in molecular mechanics in general absolute energies have no meaning, only relative energies (energy differences) should be used for any kind of analysis.

Shouldn't an isolated H be about -13.6 eV relative to the zero state?

@IvanNeretin, yes, it should be close to -0.5 Hartree in calculations. And I have no idea why OP's number is so far away from that value.

@Wildcat. I apologise for the original numbers. These were chosen at random as I had no computational software, I have now updated to reflect computational values.


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Some substances appear white when crushed while others do not. Why?
Calcite crystals are semi-transparent but turn white when crushed. I used to think that it has something to do with the scattering of all wavelengths of light from small particles. But that cannot be the whole story. By that explanation crushed carbon should also appear white (which is not the case).
So my question is why do some substances appear white when crushed while others do not.

If you downvote, please leave a comment.

You probably mean calcite as crystallic calcium carbonate. Calcium is a soft alkali earth metal, reacting with water and forming gaseous hydrogen and calcium hydroxide.

@mithusengupta123, ignore the downvotes, some people have no clue why they are doing it! The problem with anonymous downvotes is that we don't know whether it is Einstein or a high school drop out who has cast a vote. I am sure this phenomena has to do with random scattering/reflection process. For solids you look at the diffuse reflectance spectrum. The they reflect all visible wavelengths, they appear white.

Read about  Kubelka-Munk theory developed for powder's color.

Specimens of hematite can be black, red, brown, or silver in color and occur in a wide variety of habits; however, all specimens of hematite produce a streak with a reddish color. The "streak test" shows a lot of different coloration from rocks: https://geology.com/minerals/streak-test.shtml

Nice answer below. It is always the net balance between reflectivity and absorbivity, modulated also by density and shape. Think of beer foam, seen in both pale pales and stouts

It depends if the substance in bulk is transparent, highly reflective (metallic) or highly absorptive.
Carbon in its graphite allotrope is highly absorptive, and even finely divided is still black.
Carbon as diamond is transparent. Crushed diamonds are white -- look at a diamond file or abrasive disk. What happens is that incident light is scattered as it encounters particle after particle, refracted or reflected repeatedly, based on particle size, shape and index of refraction (nD). A higher nD increases the amount of light reflected from particles. For that reason, titanium dioxide, $\ce{TiO2}$, which has a comparatively high index of refraction (~2.5), makes an excellent white pigment.
Metals reflect light because the "cloud" of electrons on the surface return a reflected wave out of phase with the incident wave. However, finely-divided metals have fewer electrons to share, and electrons start behaving as discrete, localized charges. Finely divided silver appears black, for example. This is the basis for traditional silver halide photography. Powdered $\ce{AgCl}$ appears white, but where hit by light, it yields metallic silver nanoparticles, which appear black (as a negative image).
If your interested, look up the effect of surface plasmons on metallic reflectivity, which can yield surprising colors, such as gold in ruby glass.


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Why is concentration used as a proxy for activity instead of mole fraction?
I am trying to understand how the Gibbs energy change of a chemical reaction is computed and I do not understand why concentrations (mol/L) are used as a proxy for activity rather than mole fractions (mol/total mol).
From the Wikipedia article on the chemical potential, in an ideal solution, the chemical potential of the chemical species $i$ is expressed as
$$\mu_i \approx \mu_i^\circ + RT \ln x_i,$$
where $\mu_i^\circ$ is the chemical potential of species $i$ in standard state, $R$ is the gas constant, $T$ is the temperature and $x_i$ is the mole fraction of $i$ in the system. Because real solutions are not ideal, the activity of $i$ ($a_i = \gamma_i x_i$) is used rather than its mole fraction $x_i$ in the calculation of $\mu_i$.
When $T$ and $P$ are constant, the Gibbs energy change of a reaction is formulated as
$$\Delta_r G = \sum_i \mu_i dn_i,$$
where $dn_i$ is the change in the number of particles of $i$ caused by the reaction (its stoichiometric coefficient). Consequently;
$$\Delta_r G = \sum_i \mu_i^\circ \cdot dn_i + \sum_i dn_i \cdot RT \ln \gamma_i x_i.$$
$\sum_i \mu_i^\circ \cdot dn_i$ is equivalent to $\Delta_r G^\circ$, while $\sum_i dn_i \cdot RT \ln \gamma_i x_i$ is equivalent to $RT \ln Q$ in the more usual formulation $\Delta_r G = \Delta_rG^\circ + RT \ln Q$.
Molar concentration (mol of species per volume of solution) are often considered as a proxy for the activity of solute species, while partial pressures are used for gas species (which gives the "mass action ratio" $\Gamma$ instead of the reaction quotient $Q$). While partial pressures are a concept similar to mol fractions (they are fractions of a total), concentrations and mol fractions are not the same quantity and I do not think they scale the same way. So why are $\Delta_r G$ quantities often computed using concentrations instead of mole ratios?
EDIT: Thinking about it, I think I found a simple justification for the use of concentration as a proxy for activity.
Let's consider a fictive reaction $A \leftrightharpoons B$. The chemical potential change associated with this reaction is
$$\Delta_r \mu = \Delta \mu_B - \Delta \mu_A$$
$$\Delta_r \mu = \Delta \mu^\circ_B - \Delta \mu^\circ_A + RT \ln x_B - RT \ln x_A$$
$$\Delta_r \mu = \Delta \mu^\circ + RT \ln \frac{x_B}{x_A}$$
$$\Delta_r \mu = \Delta \mu^\circ + RT \ln Q$$
mole fractions $x_A$ and $x_B$ are respectively $\frac{n_A}{n_{total}}$ and $\frac{n_A}{n_{total}}$ with $n_{total} = n_A + n_B$, so $Q = \frac{n_B}{n_A}$ ($n_{total}$ cancels). Obviously $Q$ has the same value if concentrations ($\frac{n}{V}$) are used since the volume $V$ is the same for both species and then cancels itself. This justification is obvious for a 1:1 stoichiometry. However, with other stoichiometries, either $n_{total}$ or $V$ remains as a factor in $Q$ (to the power $\sum_i \nu_i$, where $\nu_i$ is the signed stoichiometric coefficient of species $i$). I am pretty sure that $n_{total}^{\sum_i \nu_i}$ is not always equal to $V^{\sum_i \nu_i}$; this is where the difference may happen.

I have often thought like you without having ever been brave enough to ask it. On the other hand, there has been a recent publication in the Journal of Chemical Education, saying that the pH values at relatively high acidic concentration is better described by - log x  than by - log[H+]. The trouble is that I have not noted the reference of this article, which must be a couple of years old.

The posted notation seems unorthodox if not incorrect.
For the Gibbs free energy of reaction one usually writes
$$\Delta_r G = \left(\frac{d \Delta G}{d \xi}\right) = \sum_i \nu_i\mu_i^\circ  + RT\sum_i  \nu_i \ln \gamma_i x_i$$
$$ = \Delta _r G^\circ + RT \ln Q$$
where $ \xi$ is the reaction progress, $Q=\sum_i  \nu_i \ln \gamma_i x_i$ is the reaction quotient, and one can equate $\Delta _r \mu= \sum_i  \nu_i\mu_i $ with $\Delta_r G$. For the reaction $A \leftrightharpoons B$ one would write $\Delta \mu = \mu_B - \mu_A$.
Changing concentration units is ok but a set of units usually implies a choice of thermodynamic reference state. The reference state generally defines the limit in which $\gamma \rightarrow 1$ (the limit of ideal behavior). If you decide to express the reaction quotient Q in terms of molar concentration rather than mole fraction but want to retain the same general form of the above equations, then it becomes necessary to modify both Q and $\Delta _r G^\circ$. Some reference states are better choices than others under certain circumstances, and it is not accurate to say that mole fraction is not used, it is, just not always. For solids or solvents a mole fraction unit is often convenient. When dealing with solutions, and particularly with solutes as a minor component, it is generally more convenient to deal in molal and molar units (the prior being favored), in combination with Henry's law. In addition, at small concentrations of solute the mole fraction, molarity and molality are linearly proportional, differing by a multiplicative constant.  For instance, at low concentration we can write the mole fraction of solute as
$$\chi _2 = \frac{n_2}{n_1 + \sum_i n_i} \approx \frac{n_2}{n_1} = \frac{c_2}{c_1} = \frac{c_2}{1000\delta_1/M_1} \approx \frac{M_1}{1000 \delta_1 ^\circ}c_2$$
where 1 and 2 label solvent and solute and $\delta_1 ^\circ$ is the mass density of the pure solvent, $M_1$ its molar mass.


text:
In vivo, does it take energy to move double bonds from cis to trans long chain unsaturated fatty acids
In vitro, long chain fatty acids may yield more energy when bonds break, but in vivo, I understood that energy is burned before unsaturated fatty acids can be oxidized because the side groups with double bonds are crowded, and bunched on one side in the cis position, as all unsaturated FA are, and the molecules of oxidation cannot get in between - they cannot access these bonds to oxidize them.
So I thought I read that ATP is used (how much?) to first get those side double bond groups into the trans position so the oxidation enzymes can reach them.  Let me know if this is not true. Or if there is a thread here that really explains this well.

This question might be a better fit for Biology stackexchange...

Look up fatty acid cis trans isomerase. They typically do not require energy.


text:
Reimer-Tiemann reaction on indole
According to me net reaction should be at $\ce{C3}$ when Reimer-Tiemann reaction is done on indole (see image), since in the resonance structure, negative charge is on this carbon.

But, the answer given is on $\ce{C2}$, and the solution given is

In pyrolle, the most electrophilic site is $\ce{C2}$. Think in a similar way.

But, I can't either find fallacy in my reasoning, not do I convince with the solution.
TL;DR Where will dichlorocarbene attack on indole (am not interested in ring expansion).

The paper by Order and Lindwall [1] agrees with you that the R-T reaction gives indole-3-carboxaldehyde.

Since its isolation in 1903 by Hopkins and Cole (1), 3-indole aldehyde has been investigated very little. It was first prepared by Ellinger (2) from indole through the use of the Reimer-Tiemann reaction. This method was improved on later by Boyd and Robson (3).

References

Order, R. B. V.; Lindwall, H. G. 3-Indole Aldehyde and Certain of Its Condensation Products. J. Org. Chem. 1945, 10 (2), 128–133. DOI: 10.1021/jo01178a006.

What about the explanation given in the solution?

Indole is not pyrrole

Also note that the answer refers to C2 as "most electrophilic", which is true. But in the first step of the R-T reaction, the aromatic ring acts as a nucleophile, so the reaction happens at the most nucleophilic position (C3) rather than the electrophilic position (C2).


text:
On the origin of Gibbs Free Energy
It is intuitive to me that for a spontaneous process involving any system i.e. one occurring without constant human interaction $$ \text{d}S \geq 0 $$ as I can observe increasing energy dispersal in everyday life, for example my tea turning cold.
What is not intuitive to me is that for all systems $$ \text{d}S \geq 0  ⇔\text{d}G = \text{d}H-T\text{d}S-S\text{d}T  \leq 0 $$
Why is this? I imagine that the explanation requires an understanding behind the origin of the definition of Gibbs free energy, G = H - TS.

have a look at this answer as well as that below from Maurice. https://chemistry.stackexchange.com/questions/122162/what-is-the-true-meaning-of-gibbs-energy-and-chemical-potential/122170#122170

It's the entropy change of the universe (not the system) that determines spontaneity and, at at constant T and p, we can determine the change in the entropy of the universe directly from the change in the Gibbs free energy of the system.  I.e., at constant T and p, the change in the Gibbs free energy of the system serves as a surrogate for the change in the entropy of the universe.  See the bottom of my answer here:  https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/124470#124470

The Gibbs energy has not been derived this way. It came from a thought process. About 100 years ago, Gibbs was puzzled by the very existence of spontaneous endothermic reactions. In mechanics, objects always fall down. They never "fall up" spontaneously. They never gets more energy spontaneously. In exothermic reaction, the matter looses spontaneously energy, like a falling object. In endothermic reactions, the atoms and molecules receive energy from the surroundings. They increase their energy. It is surprising. So Internal energy U and Enthalpy H are not equivalent to the gravitational energy. A spontaneous reaction can happen in chemistry with $\Delta$H <$0$ ou >$0$. As a consequence, H or U are not the good criteria to foresee whether a chemical reaction will spontaneously happen.
So Gibbs was thinking : What is the energy that always gets out of a system of molecules and atoms in a spontaneous reaction, even when the reaction is endothermic ? The answer is of course : the electric energy. A galvanic cell can only produce energy, whatever its heat effet. A working galvanic cell may become hotter or cooler, but it can only deliver electric energy. The amount of electric energy delivered by a cell is defined by the difference of a new potential energy, called Gibbs energy $G$, and defined so that $\Delta$$G$ is always negative in a cell. $\Delta$$G$ is obtained form the expression : $\Delta$$G$ = - $zEF$, where $E$ is the voltage of the cell, $F$ is the Faraday, and $z$ is the number of electrons exchanged in the reaction inside the cell. Furthermore, $\Delta$$G$ is defined and calculated in the same way as $\Delta$$H$. Gibbs energies of formation exist like enthalpies of formation.
For example, one may write : $\Delta$$G_{react} = \Sigma$$G_{fin}$-$\Sigma$$G_{in}$ and so on. You probably know the rest to the theory. And for example the relationship between Gibbs energy and the equilibrium constant.
Now if you plot $\Delta$$H$ and $\Delta$$G$ of a given chemical reaction versus $T$. you will notice that $\Delta$$H$ practically does not change much with $T$. But the voltage $E$ and of course $\Delta$$G$ do change much with $T$. The line $H$ vs. $T$ is nearly horizontal. But the line $G$ vs $T$ is inclined. The strange result of this drawing is the fact that both lines are joining one another at $T$ = $0$ $K$. So the difference of these energies is of course : $H-G$ = $T\Delta S$, where $\Delta S$ is the slope of the $\Delta G$ curve.


text:
Reactivity of carbonyl derivatives with a nucleophile
As observed, the reactivity of carbonyl derivatives with a nucleophile goes like:
Acid halide > acid anhydride > acid > amide > ketone > aldehyde. So is nucleophile a code word to nucleophilic substitution (nucleophilic acyl substitution in this place)?
My professor gave the explanation by saying that look out of $\delta^+$ charge on the carbonyl carbon of each compound, the one that has the most is the one which reacts most with a nucleophile.
And, if we consider that, then the order according to professor is:
Acid halide > aldehyde > ketone > acid anhydride
Is there anyway this order may be correct?
I'm a student btw, so I got no knowledge about the lab and practical stuff.

You need to be more specific as to the nucleophile.

No it didn't ...I also asked if my professors order could be somehow correct. And I know the reason behind the order but what I'm asking is reactivity with a nucleophile a code word to nucleophilic substitution?

It's from one of the questions where they say that check reactivity with a nucleophile...so the nucleophile remains the same (any) with carbonyl derivatives changing.

The reactivity order in you list is different from that in Mithoron's reference. Also, you have amide>ketone>aldehyde. I'd think the reverse is true. You said "as observed", what reaction was studied in the observation?


text:
How to determine the chemical state of a reactant or product in a chemical equation?
I am asked this question:

Write a balanced equation, including state symbols, for this reaction:
$\ce{calcium + hydrochloric acid -> calcium chloride + hydrogen}$

I answered this:

$\ce{Ca(s) + 2HCl(aq) -> CaCl2(s) + H2}$

This was the correct answer:

$\ce{Ca(s) + 2HCl(aq) -> CaCl2(aq) + H2}$

I thought that the $\ce{CaCl2}$ was a solid precipitate (ppt.) instead of aqueous(aq).
Here are the known chemical states:

solid(s)
liquid(l)
gas(g)
aqueous solution(aq)

Is there any method to work out the chemical state of a reactant or product in a chemical equation, if I know the chemical states of all of the other reactants?

According to these basic solubility rules:

Calcium chloride (CaCl2) is soluble. Hence, you write CaCl2(aq) in the equation.

Generalised tables like this should be taken with a pinch of salt. Sodium Bismuthate, for one, is an insoluble sodium salt.

You must first learn two things before hoping to solve any chemical problem:

You must learn how to write a balanced equation, with the correct numbers of atoms, before guessing about their physical state. For example, in your first equation, there is one $\ce{Ca}$ on the left-hand side and $\ce{2 Ca}$ at the right-hand side. This is not balanced.
You should realize that $\ce{CaCl}$ and $\ce{CaCl2}$ are not the same substance. When you say that you "typed $\ce{CaCl(s)}$ instead of $\ce{CaCl(aq)}$" you still are wrong. You should have said, "instead of $\ce{CaCl2(aq)}$".

Now to answer your question, there is no way of obtaining the physical state from the periodic table.
Simply speaking, you should here know that pure $\ce{HCl}$ is a gas at room temperature, and it is very soluble in water. In the pure state, $\ce{HCl}$ is called hydrogen chloride. Dissolved in water, $\ce{HCl}$ is called hydrochloric acid.

Thank you for trying to help me. I’m still unclear about this issue and would be grateful if you could help me understand the rest of my question, which is, how to work out the chemical state of CaCl when I know the state of the reactants and products ?

CaCl does not exist.

I’m sorry, I meant CaCl2. So my question is, how to work out the chemical state of CaCl2 when I know the state of the reactants and products?

You cannot. Not from the basic principles. You can use as a guidance the provided table,but it covers just a minor part of substances and there are exceptions. Additionally, the state may depend on quantitative circumstance. There is lot of empirical knowledge in chemistry.


text:
Why does a mentos make soda explode?
So if you put a mint mentos in a cup of soda it will explode everywhere. The mint in the mentos makes the soda have a chemical reaction which will then make the soda explode right? Well what would happen if we put a mento in different types of soda would that make a difference in how much chemical reation there will be? I already put mentos in cream soda and it made a small explosion. So I wondering if the different sodas had different effects on  how big the explosions are.

Don’t think it’s a chemical reaction. More so that it creates nucleation sites that drive CO2 out of solution I believe.

When you open the soda bottle, the CO2 present in solution must get out, as the pressure is suddenlyvery  low. But to get out, nucleation sites are needed, that is bumps or rough edges of something, around which the bubbles can be formed.  The pill do have a lot. So the bubbles are formed all of a sudden around the pill. It looks like an explosion, but it is not.
The only trouble is that nobody has ever explained why this phenomena is so important with a Mentos pill in Coca light. Other pills or other sodas
do not work so violently.

@Maurice must be that the structure of mentos leads to a larger surface plus, perhaps, a special wax or so more affine with CO2 but just the right amount. Logic says that others similar products should do the same with the same violence.

Comparison between sodas. Unfortunately the surface is fixed. https://www.compoundchem.com/2017/05/02/coke-mentos/amp/

@Alchimista. Why is the explosion stronger with Coca Light than with regular Coca ? And why does it not work so strongly with Pepsi or other sodas ? Or Mineral water ? Nobody knows !

what do you mean the surface is fixed

@puppylover the "catalitic" surface, being the experiment in the link just using mentos.

@Maurice that is why I've linked the poster. From that page one finds ref to an experiment using different sodas and (trying) linking composition to fountain height.


text:
Euler's Theorem and Extensive properties
I read in a textbook (Mathematics for Physical Chemistry by D. McQuarrie) an example of Euler's Theorem used in Thermodynamics. The author states "The [variables] are all extensive [quantities], and so [Euler's theorem is applicable]" (I'll provide more details below)
My question: why do variables being extensive indicate that this theorem is relevant?
The theorem starts by stating that a function is homogeneous to degree $N$ in some set of variables if those variables always form terms such that their powers sum to $N$. For example:
$$f(x,y) = yx^2 + y^2x + \frac{y^4}{x} + x^3$$
that would be homogeneous degree 3 for $x$ and $y$.
The theorem concludes with:
If $$f(\lambda x,\lambda y) = \lambda^N f(x,y)$$
Then $$Nf(x,y) = x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}$$

This
is the wikipedia page for the same theorem.
The example uses the above theorem to derive:
$$U = S(\frac{\partial U}{\partial S})_{_{V,n}}+V(\frac{\partial U}{\partial V})_{_{S,n}}+n(\frac{\partial U}{\partial n})_{_{S,V}}$$
The author's solution immediately states: "The entropy, volume, and number of moles are all extensive thermodynamic quantities, and so $U(\lambda S,\lambda V,\lambda n) = \lambda U(S,V,n)$." After that statement the solution is simple.
I know that extensive variables means that they scale with the size of the system, but the relation to this theorem is unclear to me. I suppose that if the system is "twice as big" ($\lambda$ is doubled), then energy, entropy, volume, and amount are doubled. That still seems vague to me if its even a good way to think about this.

see https://chem.libretexts.org/Courses/New_York_University/CHEM-UA_652%3A_Thermodynamics_and_Kinetics/Lecture_11%3A_Physical_significance_of_free_energy%2C_Euler%27s_theorem%2C_Maxwell_relations

Thanks, this part was especially relevant: "In fact, any extensive thermodynamic quantity must be a homogeneous function of degree 1 in its extensive arguments."

It really is that simple. Stating that a thermodynamic system observes Euler's Theorem can be considered axiomatic if the geometry of the system is Cartesian: it reflects how extensive variables of the system scale with size. In a homogeneous system with a Cartesian geometry all extensive properties scale identically and so any extensive property can be written as a homogeneous function of degree 1 of the remaining properties.


text:
Difference between a species and a phase in a solution?
Overview:
As of my knowledge from metallurgical thermodynamics a phase is a thermodynamically stable and homogeneous entity made up of different species of a given concentration.
e.g = 'Austenite' is a phase of steel with a given (Fe, C) concentration (%wt). So, we can call the Iron and carbon to be species.
But we may observe that 'Austenite' is nothing but Fe&C atoms arranged in a particular structure, which happens to be FCC (Face centered cubic). In that sense, a phase is nothing but a homogeneous aggregate of species at a different length scale(a larger length scale).
My questions related to the fluid system are.

What's the difference between a species and a phase in a solution (in a fluid)?
Given a mixture of water+salt+sugar, with the conditions that 'salt'
is homogeneously mixed in the water, while 'sugar' is inhomogeneously
mixed in water. May you please define the phase and species
respectively for this system?
And if what does the transport of species in a fluid phase mean?

Air, fresh water, sea water, whiskey are fluid phases, aside of solid phases.
A phase is a space region of homogenous properties of matter, possibly with gradient of theses properties (like sea water profile, or solution of statically dissolved solid), with strong discontinuity of these properties at the region boundary.
The gradient can be at special cases questioned, as it can be property of the phase boundary, if much stronger then elsewhere. The phase interior has minimal gradient of the considered gradient.
By other words, if you underlayer water by a thick sugar solution, you will get 2 liquid phases, with more or less sharp boundary and with the property gradient becoming suddenly steep. If you let it diffuse statically, or stirr it just a little, the solution properties with get about stable gradient and the liquid can be considered a single phase.
By even other words, there are 2 phases, if you can distinguish the boundary between them. If once sharp boundary becomes diffuse, occupying both once separated phases, it forms just a single gradient phase.

Nitrogen, oxygen, argon, water, sodium chloride, calcium bucarbonate, sugar are species. Species are chemical elements or compounds, consisting of the same kind of their smallest parts - atoms, molecules, ions.

If concentration of species within a phase is not homogenous, they can be  subject of redistribution via diffusion controlled by Fick's laws and convection ( thermal or mechanically forced )

Is a phase-only made up of its corresponding species? Like saline, water is a phase, and its species are water and salt.
Now if we somehow remove the water, does this forms a new phase that's only made up of salt

@Anshuman Sinha Yes, It is and it does. If you evaporate enough water from saline, the is the phase of saturated salt solution and the phase of the solid salt.