text:
Hydration, Coordinate Bonding, and Reduction Potential in Groups 1 and 2 Elements
Whenever an alkali metal such as Lithium is hydrated/solvated, coordinate bonding occurs and it is the high hydration enthalpy of lithium which makes it the strongest reducing agent in aqueous solution. My question is, if lithium is hydrated and becomes surrounded by water molecules in coordinate bonding, then how does the overall reaction of adding lithium to water produce LiOH and H2(gas) ?  Is the coordinated ion an intermediate product?
I also am wondering what happens to the "lost" electrons from the metal in coordinate bonding.  Take Beryllium dissolving in water as an example. As it is hydrated it will lose the 2 valence electrons in its outer orbital and then it hybridizes its orbitals into 4 empty orbitals so that 4 molecules of water, supplying 1 lone pair each, can bond with the positive beryllium ion. Where did the original 2 electrons go?  How do these electrons vanish in order to allow the hybridization of the 4 empty orbitals for water molecules to bond with?
And if Lithium is the strongest reducing agent in aqueous solution I assume it means it is reducing water itself, in order to create lithium hydroxide and hydrogen gas, but again, how does that come about from the coordinate bonding and the hydrated lithium? Coordinate bonds preserve the charge of the cation but I am not quite sure how this is so.
I have included an image below from a website that attempts to explain it, but I am confused about to where the original electrons go.
The website from which I took the image is:  https://www.chemguide.co.uk/inorganic/group2/beryllium.html#top

This answer addresses your question regarding the reaction
$$\ce{2Li}(s) + \ce{2H2O}(l) \rightarrow \ce{H2}(g) + 2\ce{Li+}(aq) + 2\ce{OH-}(aq)$$

Whenever an alkali metal such as Lithium is hydrated/solvated,
  coordinate bonding occurs and it is the high hydration enthalpy of
  lithium which makes it the strongest reducing agent in aqueous
  solution. My question is, if lithium is hydrated and becomes
  surrounded by water molecules in coordinate bonding, then how does the
  overall reaction of adding lithium to water produce LiOH and H2(gas) ?
  Is the coordinated ion an intermediate product?

The reaction mechanism for this reaction involves some radical chemistry:

Lithium hydroxide is soluble in water, and will exist as separate ions as long as the solution is not saturated. As soon as the ions are generated, they will be solvated by nearby water molecules. The solvated ions are not intermediates - they might better be viewed as products.
The lithium metal is very reactive due to its free electron, and will react quickly. Besides, Li metal is neutral, and will not immediately form any strong interactions with the polar water molecules, I would think.

And if Lithium is the strongest reducing agent in aqueous solution I
  assume it means it is reducing water itself, in order to create
  lithium hydroxide and hydrogen gas, but again, how does that come
  about from the coordinate bonding and the hydrated lithium? Coordinate
  bonds preserve the charge of the cation but I am not quite sure how
  this is so.

Indeed, in the course of the reaction water is reduced and Li is oxidized. But I don't think the Li metal (neutral atomic Li) is solvated by water. The reaction with water takes place extremely quickly. And as mentioned above, it is not intuitive to me why water would want to solvate a Li atom in the first place, due to it being non-polar.

If we recall though that in the alkali and alkaline earth metals the ratio of charge to ionic radius becomes very important in influencing the behavior, especially in the elements near the top of the respective groups, such as Lithium and Beryllium. The positive charge of lithium in such a small atom strongly attracts polar water and permits coordinate bonding, where 4 water molecules surround lithium, thus hydrating it. So, would the hydration occur only after it has already reduced other water molecules to produce lithium hydroxide and hydrogen gas?

I was certain that the hydration of lithium had to occur first, because the hydration enthalpy is what actually permits, thermodynamically, for lithium to so easily give up its outer electron, being that it is so positive. The ionization energy decreases as you go down the alkali metal group, meaning it is easier to remove an electron, but lithium is a special case at the top which defies this trend, due to its hydration enthalpy and ratio of charge to size. This makes it the strongest reducing agent in water. So does hydration actually occur first? then the reaction you demonstrated above?

Can you give a reference for the high hydration enthalpy of the lithium atom (not the lithium cation)? I'm trying to find something about this, but I cannot. The link you provided for Be talk about the hydration energy for the Be cation, not the Be atom.

You are correct in that it is the lithium cation and not the neutral lithium atom which has this very high enthalpy of hydration, but this is also where some of the confusion arises for me. Lithium is the strongest reducing agent in aqueous solution because of the high hydration enthalpy can compensate for its high ionization energy. Lithium will not give up its outer lone electron, it is too electronegative to do so despite being an alkali metal, but since the cation can be hydrated so thoroughly and efficiently, it will give up the electron very readily, having highest reduction potential

I am wondering how we end up with LiOH and H2(gas) though in the very end. Does the hydration happen at the same time as the radical reaction mechanism which you have already shown? We have hydration of lithium cation occurring, but we also have lithium reducing water to form lithium cation, hydroxide ions, and hydrogen gas. How does this all unfold? I want to understand step by step what is occurring.

The lithium atom has a low ionization energy, among the lowest in the periodic table. Its electronegativity is also quite low. It is counter-intuitive to me that Li, when placed in water, wants to hold on to its electron. Its tendency to give away its electron is the origin of the high reductive power of Li. The large hydration energy of the lithium cation can be rationalized by considering the large charge density of the cation.

As for the reaction mechanism, of course many things take place simultaneously. I'd imagine that $\ce{Li+}$ does not wait around to be hydrated - that happens immediately after the cation is formed. For two H atoms to form $\ce{H2}$, they also need to be near each other and actually collide. LiOH is soluble in water, so it is more correct to say we get the $\ce{Li+}$ and $\ce{OH-}$. To best understand the order of events as detailed as you want, you may need to run ab initio molecular dynamics simulations. But I think our intuition would be fairly accurate. I'll update the answer later today.

Lithium in fact has the highest ionization energy and the highest electronegativity of all the alkali metals. If you look up trends in ionization energy you will see this is the case.  Lithium has the highest ionization energy but in water it also the lowest reduction potential. It easily gives up electron very easily despite high ionization energy because of its high hydration enthalpy.  The hydration enthalpy is what resolves this seeming contradiction. Energy released in hydration compensates for high ionization energy.  Without hydration it could not lose its electron in aqueous solution.

Yes, relative to other alkali metals you are correct. But still, in the grand scheme of things, Li's ionization potential is not very high, and Li's electronegativity is not very high. It is not a contradiction to me, because the properties of the other alkali metals are irrelevant for the question you want answered.

I'm assuming that lithium is hydrated which allows it to give up its electron which will then reduce water.  In the beryllium example I provided its the same mechanism where the loss of an electron allows the cation to hybridized orbitals so that it can create coordinate bonds with water.  But they never specify where that electron went. I'm just wondering if the electron it loses to reduce water makes way for the empty orbitals that can be hybridized to allow hydration.  Sorry if I've not been clear


text:
How is hair tissue mineral analysis performed?
This might be a mixed chemistry / physics question, but I've always wondered what methods were used to precisely determine the levels down to trace amounts of all minerals in human tissues. I know there's a tissue preparation phase, in which I am interested though less than in the details of the actual final testing methods and processes to determine those trace amounts of all minerals.

The analytical technique is called inductively coupled plasma optical emission spectroscopy. It was invented by a chemist V. A. Fassel, although it is pure physics. The specimen is digested in an acid, and a fine spray is created. That "spray" is heated to a temperature, which is higher than the surface of the Sun. Most of the elements of the periodic table emit specific wavelengths in that hot plasma- the characteristic emission of each element is just like a fingerprint. Not all elements can be detected easily but it is very good for metals. This is the mineral analysis. Organic analysis is even more complicated and it is a separate story.

Thanks for your answer! :) Is this used to detect all the different mineral elements in the sample or only some of them? Do you you know what acid it is digested in? HCl? Or maybe a combination of acids?

Almost all elements can be detected by ICP-OES with a few exceptions like noble gases, some non-metals etc. There is no general recipe to dissolve samples for ICP analysis. It is very much like the art of cooking. The recipe to make soups is dependent on the cook. It also depends on sample to sample.

I'd point out aqua regia digestion as one of the most common sample preparation techniques for mineral analysis (which is also a DIN norm, german industry standard) as a concrete example, specially in combination with microwave treatment.

@LoschmidtsSchnitzel Thanks a lot for your answer!

Inductively Coupled Plasma (ICP) based techniques are very common for elemental analysis. The goal of such analyses is to quantify the elements present in a sample. The neat thing about this, in my opinion, is that the integrity of the molecular compounds does not need to be maintained throughout the analysis, since we only care about the elements that constitute the molecular compounds.
Common acids for decomposing samples include HCl, $\ce{HNO3}$, and HF. For certain samples, combinations of acids are used. The acids decompose by different mechanisms; for example, HCl is said to be a "complex binder", and $\ce{HNO3}$ is an oxidizing acid.
The extreme temperature of an Argon plasma leads to the introduced sample becoming 1) vaporized, 2) atomized, and 3) ionized. The final ions can be detected in a number of different ways, and I think the two most common techniques are Optical Emission Spectroscopy (OES) and Mass Spectroscopy (MS).
ICP-OES is based on detecting the characteristic emission from the ionized atoms. Very high-resolution detector systems are used in order to differentiate between all the different emission lines. Usually a few common detection lines are used for each ion, carefully chosen to minimize any overlap with nearby emission lines.
ICP-MS is different, and many different MS techniques exist. A few common detection methods used in elemental analysis is "magnetic sector" and "quadropole mass-filter". Regardless of the method used, the goal is to physically separate the ions based on their mass-to-charge ratio (m/z). For example, with magnetic sector instruments, the ions are guided into a magnetic field. Depending on the m/z ratio, their trajectory will be affected to different extents, and the ions end up in different detectors based on the offset of their trajectories. Here, you must be aware of any polyatomic ions that happen to have the same m/z ratio as the ion you are interested in, as that will interfere with your measurements. For example, a common interference to $^{107}\ce{Ag+}$ is the polyatomic $[^{91}\ce{Zn}^{16}\ce{O}]^+$. As you may see, having an idea of what is in your sample may help you selecting the best way to quantify whatever you are interested in.
So these techniques are very likely to be used in the type of analysis you mentioned, especially if extremely low detection limits are needed (down to parts-per-billion or even parts-per-trillion).

Neutron activation analysis is a technique that can be used for trace mineral quantification in a sample.
You start by irradiating your sample in a small experimental nuclear reactor. These were more common in the 1960's and '70's than they are today. Slow neutrons are absorbed by many different isotopes, converting them to radioactive species. The sample is pulled out of the reactor and then put in front of a high-resolution gamma-ray spectrometer.
From Wikipedia:

NAA can detect up to 74 elements depending upon the experimental procedure, with minimum detection limits ranging from 0.1 to 1x106 ng g−1 depending on element under investigation. Heavier elements have larger nuclei, therefore they have a larger neutron capture cross-section and are more likely to be activated. Some nuclei can capture a number of neutrons and remain relatively stable, not undergoing transmutation or decay for many months or even years. Other nuclei decay instantaneously or form only stable isotopes and can only be identified by PGNAA.

Do operators not risk exposing themselves to minute amounts of radiation with such equipment? (naïve question popping up in my head) I'm also wondering how the sensitivity would roughly translate in ppbs of the sample (at common densities found in human biological samples).

@Veritas it's of course not the method one would choose to use today. I added this for historical reference. As far as risk is concerned, everything around us already exposes us to "minute amounts of radiation", even the concrete of the building itself. Standard procedures involve checking samples with meters...

source: Stack Exchange

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

sort: votes


text:
Difference between oxo and formyl
What is the difference between oxo and formyl? both represent aldehyde when used as a substituent(correct me if I am wrong). But when should I use formyl and when oxo?
This answers my problem and also says that oxo is also used for aldehyde.

ALDEHYDES:
If they get priority in the name they get the suffix ‘al’ or ‘carbaldehyde’
As substituent prefixes ‘oxo’ and ‘formyl’ are used. The prefix ‘formyl’ is used in preferred IUPAC names, except for a -CHO group at the end of an acyclic chain.

However he did not give any sources.

No, oxo means ketone.

@IvanNeretin its almost certainly right. Oxo is also used for aldehyde.https://the-educational-blog.quora.com/IUPAC-nomenclature-for-organic-compounds-Part-VI-Aldehydes-and-ketones.

Partly related: https://chemistry.stackexchange.com/q/58339/7951 and https://chemistry.stackexchange.com/q/19868/7951

Oxo is the "O" of a ketone or aldehyde functional group as in 4-oxocyclohexane-1-carboxylic acid or 6-oxohexanoic acid.

Oxo is the substituent name for the $\ce{=O}$ group of an aldehyde or a ketone if that group is found on the longest continuous carbon chain that contains a higher priority group.

e.g., $\ce{OHC-CH2-CH2-CH2-CH2-CO2H =}$ 6-oxohexanoic acid
If the aldehyde carbon is not part of the longest continuous chain, then it becomes a formyl substituent.
e.g., $\ce{CH3-CH2-CH(-CHO)-CH2-CH2-CO2H =}$ 4-formylhexanoic acid

Welcome to our site! Could you perhaps add a few referenced (preferably from the IUPAC Blue Book), to back your answer up? Nomenclature systems are defined as standards, hence the requirement of references.

"Oxo" refers to a substitution of a single oxygen atom on a carbon atom by a double bond.  "Formyl" refers to the substitution of a three atom HCO group onto a different carbon or other atom.
I'll leave it to others to explain the differences in IUPAC nomenclature.
So for example 3-oxopentanal has five carbons just like pentanal, but it has one extra oxygen atom, double bonded to carbon #3.  (This means it has two less hydrogens, a fact that the nomenclature elides.) This is a case where "oxo" refers to a ketone, not an aldehyde.
"Formyl" is sometimes used as a substituent, but not always.  "Formyl benzene" is a incorrect name for benzaldehyde, but 4-formylbenzoic acid does exist and is used in practice as the name of chemical.

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


text:
Carbon dioxide concentration from pressure changes
Is there a way of determining the concentration of CO2 if we know the pressure in the container before the addition of CO2 and the pressure in the container after the addition of CO2 if we have no way of measuring the amount of CO2 delivered to the container (like with a CO2 controller with a CO2 monitor)? For instance, if we have a sealed container with air in it at 298 K, whose pressure is measured at 1 atm, and we introduce carbon dioxide gas causing the pressure to increase to 1.2 atm... can we use the ideal gas equation and assume that the 0.2 atm increase is due to addition of CO2 and is the partial pressure of CO2 and just use 0.2 atm/(0.0821 × 298) = 0.008 m/L (approx. 352 ppm) as the concentration of the CO2 in the chamber? Or, is there more to it?

The m/L is correct, but how did you get that 352 ppm?

I was working on the molecular mass of carbon dioxide being about 44g/mol and 1M = 44g/L and so 44000 ppm, so 44000 x 0.008 = 352... is that correct?

That would be the grams per liter.  The mole fraction of CO2 is 1/6, and the mole fraction of air is 5/6.  So, on a molecule basis, the ppm CO2 is 167000.  What is it on a mass basis?

I found this: "To convert from ppm by mass to ppm by volume, divide by the density of the particles." so I am assuming I will need to determine the density of the carbon dioxide and will also need to know the volume of the container.

Not really.  I know that it may not make sense, but in common terminology, ppm volume means 1 million times the mole fraction.  So you won't need to know the volume of the container.

If you know the partial pressures of the gases, then you can calculate the mass fractions.  In the present case, assume you have 1 mole of gas.  Then the number of moles of CO2 is 0.167 moles and the number of moles of air is 0.833.  The mass of CO2 is (44)(0.167)=7.33 grams and the mass of air is (29)(0.833)=24.2 grams.  So the mass fraction of CO2 is $$\frac{7.33}{(7.33+24.2)}=0.233$$
So the ppm CO2 by volume = 167000 and the ppm CO2 by mass = 233000.

That's great! Thank you, Chet. This makes perfect sense and I will be able to apply this to the problem I am trying to solve!


text:
Why is Libethenite (Cu2(OH)PO4) color green?
It is not a complex compound and the copper atom is not oxidized so I'm wondering how does its color change. The book mentioned that at favorable conditions, adding water in Copper phosphate, which is blue, will result in the formation of Libethenite.

First of all, why do you think your copper is not oxidized? Each copper has a charge of +II while $\ce{OH^-}$ and $\ce{PO_4^3-}$ add up to -4.
For all further discussions we need to look deeper into the crystal structure. The compound crystallizes with two crystallographically destinguishable copper ions. One forms a $\ce{[Cu(PO_4)_4(OH)_2]^12-}$ octahedron while the other one is, at least to me only fivefold coordinated to form a trigonal bipyramid $\ce{[Cu(PO_4)_4(OH)]^11-}$.
Of course those are no simple $\ce{OH}$ units as there is further bridging and connections happening but for a simple view this should be okay. So for the color there are, among other effects, still $\ce{d-d}$ transitions happening.

EDIT:
To add something to the color change in the reaction itself. In your copper phosphate you have copper polyhedra that only feature phosphate as ligands, for the $\ce{[Cu_3(PO_4)_2]}$ these would be trigonal bipyramid and square plane.So you have the addition of $\ce{OH^-}$ to these polyhedra as well as some substitution. And introducing a different ligand will also change your splitting energy and therefore change the color.

can you please elaborate how the copper is oxidized in the chemical reaction?

It's not oxidized in the reaction, the copper ions in both compounds, your copper phosphate (I assume it's copper(II)) and your product both have the same oxidation state of +II. The phosphate ion has a charge of -III while the hydroxide has a charge of -I. Together is adds up to -IV. So your cations need to be +IV to make a neutral compound. As there are two copper ions this will give +II for both of them.

so if not for the oxidation of copper, what does causes the change of color of the crystal?

I added something to my original answer. I did not read your original question well enough. I though you were interested in the color itself and not in the color change.

Thank you. So to verify, the reason of the color change is just because the OH group is a weak field ligand?

I don't know how strong phosphate is but yes OH is more or less weak. But you are also changing the geometry for one of the polyhedra from square plane to octahedral. You are also changing the crystal structure so you'd have to check how the two are finally related. Changing the structure also influences things like the band gap that can play a role for the absorption of the compound.

Does the change on geometry, from square planar to octahedral, contributes to the change of color?

The ligand coordination, and thus geometry, affects the crystal field splitting of the d-orbitals, and therefore affects the d-d transition frequencies, and therefore (possibly) the color.

Disclaimer: I realize that I am not really answering the question of color change, but perhaps this perspective adds something nonetheless.

To pinpoint the origin of a mineral's color is not an easy task, I think. Some minerals "inherently have color", in the sense that the pure mineral has a color. Other minerals rely on impurities in their crystal structure in order to have color - without these impurities the mineral would be colorless.
According to Mindat.org$^1$, a common impurity in libethenite is arsenic (As). Whether this impurity is responsible for the color, I do not know. Mindat also reports that the most common color is a range of colors, which I presume is an indication that varying impurities do affect the color of libethenite.
The apparent color of the mineral also depends on the lighting source. So for argument's sake, lets assume that the mineral is illuminated by white light from the Sun.
Quantum mechanically, all light scattering involves absorption and emission of light. It is the electrons that light interacts with. If the light frequency is far from any resonance transition of the material, then the light is emitted very quickly without loss of energy. However, if the incoming frequencies match any transition frequencies, then electrons will be excited to higher electronic states. Due to non-radiative pathways that compete with radiative pathways back to the ground state, the emitted light has (for almost all materials) a shorter wavelength than the incoming light$^2$.
Light from the Sun is "white", in the sense that all visible frequencies are represented. Therefore, a bunch of different transition frequencies are likely to be satisfied when libethenite is illuminated. The color we observe is the net effect of the interaction of all emitted wavelengths.
I'll make the claim that the origin of color in materials under "normal conditions", originate from electronic transitions. The transitions take place either due to absorption of electromagnetic light (fluorescence/phosphorescence) or due to thermal excitation (bio- and chemiluminescence). Regardless of whether the material is a semi-conductor or an organic molecule, ultimately we are talking about electronic transitions.
They can be d-d transition (i.e. effects due to ligand coordination), or valence band -> conduction band transitions, but both are electronic transitions.
Incidentally, the band gap of libethenite is around 3.4 eV$^3$, which is the same as about 364 nm. This band gap is not prohibetively large for electronic transitions that may lead to emissions in the visible region - but some energy needs to be lost for that to happen. It might be that this contributes to the observed color. If you have a piece of libethenite and an UV flash light, try to illuminate it in a completely dark room and see what happens.
So why exactly is libethenite green? I don't know, but it is related to the electronic structure of libethenite. Perhaps it can be rationalized qualitatively (why wavelength of emitted light decreases after the reaction) with crystal field theory, but such pen-and-paper arguments always contain a lot of assumptions that may or may not hold for the particular case under study.
$^1$ Libethenite on Mindat.org
$^2$ This is related to Kasha's rule and Stoke's shift.
$^3$ Li et al (2014), ChemElectroChem 1(3), pp. 663-672

That is also true, you are dealing with solid state compounds. That means you have to look into band theory and band gaps. Compounds like CdS for example are only colored due to their band gap, somewhat comparable to how charge-transfer would work in a complex. And then there is also, like you mentioned physical color origins like scattering for example. So things like d-d-transitions remain and are origin of many colors but then further effects can happen once you go from solvated systems to solid states.

There is a really good summary on these things but it's in German unfortunately. But there are multiple reasons for color origins in pigments, minerals, etc. Some are radicals like in Lapis Lazuli, dd-transition like in cobalt pigments or your case where copper is used. There are band gaps like in CdS. There is scattering and many more things that can happen. In your case I'd say green-blue is something I expect from copper so it's mostly coordination chemistry.

I happen to read a theory that briefly explains the relationship of the size and pigment color. The theory was called Kubelka-Monk Theory.


text:
Most dense hydrogen containing composition
I am searching for a composition which could host the most hydrogen atoms in a specific volume. Note that this density is sought for justifiable conditions in terms of pressure, temperature, ... and should be realizable in a reasonably equipped lab in quantities of a few kilograms. As for hydrogen itself, besides its very low melting point (which is a problem in certain respects), the density itself is 0.086 g.cm-3 which is lower that water (nearly 0.110 g.cm-3), so that it is evident that a composition should be sought after. Any help would be appreciated.

I think you need to be more specific about what you want. Are you looking for compounds that can release hydrogen easily? Like, for example, palladium. And BTW you have understated the density of water bay a factor of 10: it is close to 1g/mL not 0.1g/mL.

What is the context?  Are you trying to generate hydrogen gas?  For what purpose?

@matt_black Note that the effective density of hydrogen mass is of importance in my problem which for water is about 2/18 of the original density.

@SeyedMohsenAyyoubzadeh, then you should have written: "which is lower than the density of the hydrogen in water (nearly 0.110 g.cm-3)." Have you consulted https://en.wikipedia.org/wiki/Hydrogen_storage?

@DrMoishePippik Yes, you are correct but that could be clearly inferred from the question itself.

Obviously not, see matt_black comment. Check your writing.

A few binary metal hydride candidates:
Magnesium hydride gives about 0.11 grams of hydrogen per cubic centimeter, and unlike water it can release its hydrogen leaving the element behibd on simple heating.  It does, however, require 287°C to decompose, limiting reversibility.
Lithium hydride offers nearly the same hydrogen density as the magnesium compound but decomposes at a much higher temperature, limiting its utility.
Beryllium hydride approaches approaching 0.12 grams hydrogen per cubic centimeter.  But it has several drawbacks; beryllium is expensive, it's a serious health hazard and the element does not react directly with hydrogen, making an element/hydride cycle impossible.
Aluminum hydride, with a favorable stoichiometry, comes inat nearly 0.15 grams hydrogen per cubic centimeter and also releases its hydrogen at 150°C.  Like magnesium hydride it has been considered for hydrogen storage.  However, reversible cycling between the element and the hydride is again inhibited because aluminum and hydrogen do not react directly except at very high pressure (10GPa for aluminum hydride versus 20 MPa for a direct synthesis of magnesium hydride).  As with beryllium hydride, getting the aluminum back to hydride form requires indirect methods with other hydride sources.
A ternary compound with 0.15 grams of hydrogen per cubic centimeter is $\ce{Mg2FeH6}$, which can be made from the elements at 2-12 MPa pressure after compressing the metals into pellets.  This compound can also release its hydrogen thermally and is one of many ternary hydride candidates for hydrigen storage.

Thanks, is there a theoretical reason why the hydrogen density is in such a narrow range for these different kind of materials, e.g. water and Magnesium hydride? In other words, is 120 mgH/cm3 some kind of upper bound related to STP conditions?

Found one with nearly 150, see edit.  But yes, there is a limit to how much any atom can be squeezed by chemical bonding unless assisted by high pressure.

Please add Mg2[FeH6] https://pubs.acs.org/doi/abs/10.1021/ic00181a032

Done (with some delay b/c my first edit somehow got lost).


text:
What are the isomers of a complex having a structure of M(AB)2C2?
Our instructor have us answer an isomerism problem of a crystal structure that has a form of $\ce{M(AB)2C2}$. I keep getting 9 isomers, which includes the optically inactive isomer, and my peers and I have different answers.

I thought of several wrong ways to count isomers, and still failed to get 5. If you want us to tell which of your isomers is wrong or which one is missing, then we really need to know more.

After more fiddling I succeeded in getting 5. If your peers are getting more, they must be counting enantiomers.

Sorry, but I have noticed that 4 of my isomers have D and L forms so instead of 5, the isomers that I have is 9 including the one which is the optically inactive isomer.

Great. (Or maybe not.) Now you are ahead of me, because I'm getting 8, of which two are optically inactive.

How did you form another optically inactive isomer? I am new to this topic that's why I can't trust my intuitions.

By morphing together two of your supposedly optically active isomers which are in fact the same, that's how. Look again at those which have their C ligands in trans position to each other.

I see what you mean but from what i see, there is no plane of symmetry in the complex.

Yes there is. It passes through all A's and B's.

Is it possible? The tutorial video that i've watched on youtube said that the ligands that passes through all A's and B's can't have a mirror plane since they are not in the same plane.

In my book they are. Do we really have the same picture before our eyes? Look again at the isomer(s) in question. Which pairs of ligands are opposite?

You said that I should look at the C ligands that are trans with each other so I did. There is only one complex that I have that has C ligands trans with each other, not counting the complex that has C on the plane of the octahedral geometry. Sidenote, sir may I ask what is the name of the book you are using that contains isomers for a complex with M(AB)2C2 form? Thank you.

I'm not using any book, it was a figure of speech. OK, maybe you'll add some drawings to your question.

I see now what you mean sir, thank you! It is indeed the same molecule.

Great! Now I guess you may write an answer yourself, or just leave it at that and call it a day.

I will answer my own question and would mention you sir for the help.

Referring to the replies of Mr. Ivan Neretin, complexes that has a form of M(AB)2C2 posses 8 stereoisomers, 2 of which are optically inactive. The other 6 isomers can be interchanged to their respective enantiomers.

Without some pictures this answer is hardly helpful for other people struggling with the question. It doesn't even specify the coordination of M.

I tried to at least put a picture of the complex for easier visualization but I failed to generate one.

Do you have a pen and paper and a camera? This should work fine for starters, someone else might use the chance to produce higher quality images if necessary.

I don't have any gadgets with me except for this PC that I am using.


text:
Ionization enthalpy for group 13 elements

The ionization enthalpy for elements along a group generally reduces.
But there is an exception for group 13 elements and the order is not
uniform.
The order is:
B>Tl>Ga>Al>In

According to my textbook, Xam idea Class 11 Chemistry Book for CBS, on page 153, the reason for this is the weak screening effect of electrons in d and f orbitals.
What does this mean?
The following is an excerpt from my textbook:

The ionization enthalpy values as expected
from the general trends do not decrease
smoothly down the group. The decrease from
B to Al is associated with an increase in size. The
observed discontinuity in the ionization
enthalpy values between Al and Ga, and
between In and Tl are due to the inability of d- and
f-electrons,which have a low screening effect, to
compensate for the increase in nuclear charge.

Source

Except for boron, the differences between the ionization enthalpies in G13 seem quite small, no?


text:
Procedure for determining the levels of Acrylamide from a cup of coffee
For experimental purposes, I want to determine the level of Acrylamide in cups of coffee. Is there a method that doesn't use spectrometric analysis to determine it as I do not have access to it in the lab. Is Acrylamide a Heterocyclic Amine?

Acrylamide is a $\alpha,\beta$-unsaturated amide (https://en.wikipedia.org/wiki/Acrylamide). Also, your question is not clear. You want to analyze acrylamide in coffee or acrylamide in a plastic coffee-cup?

I want to analyse it in coffee...

Analysis of Acrylamide in Coffee


text:
What kind of isomerism does Ni(PPh3)2Cl2 have and what is their IUPAC name?
I know that there are 2 isomers that $\ce{Ni(PPh3)2Cl2}$ have and they exist in equilibrium. I am tempted to name them cis- and trans- but it does not sound right.

Given that one isomer is tetrahedral and one is square planar, cis and trans would not be correct as there is no cis/trans in a tetrahedron. See also: https://en.wikipedia.org/wiki/Dichlorobis(triphenylphosphine)nickel(II)

Answer for the question is  here https://chemistry.stackexchange.com/questions/115553/why-is-nipph%e2%82%83%e2%82%82cl%e2%82%82-tetrahedral/115563#115563..One of the structures is square planar ,while the other is  tetrahedral.The square planar complex can both be cis and trans.

A metal phosphine complex, dichlorobis(triphenylphosphine)nickel(II) exists in two isomeric forms (blue and red) based on how you prepared it (Wikipedia). The blue isomer, which is paramagnetic can be prepared by treating hydrated nickel chloride ($\ce{NiCl2.6H2O}$) with triphenylphosphine in alcohols or glacial acetic acid. When allowed to crystallize from chlorinated solvents (e.g., dicloroethane (Ref.1), dicloromethane (Ref.2)), the tetrahedral isomer converts to the square planar red isomer, which is trans and diamagnetic:

References:

Benedetto Corain, Bruno Longato, Roberto Angeletti, Giovanni Valle, “trans-[Dichlorobis(triphenylphosphine)nickel(II)]•$\ce{(C2H4Cl2)2}$: a clathrate of the allogon of venanzi's tetrahedral complex,” Inorganica Chimica Acta 1985, 104(1), 15–18 (https://doi.org/10.1016/S0020-1693(00)83780-9).
A. S. Batsanov,  J. A. K. Howard, “trans-Dichlorobis(triphenylphosphine)nickel(II) bis(dichloromethane) solvate: redetermination at $\pu{120 K}$,” Acta Cryst. 2001, E57, m308–m309 (https://doi.org/10.1107/S1600536801008741).


text:
Solubility: Is the common ion effect proportional to individual solubility values?
I'm trying to grasp how solubility works in real applications. Is it so that in a solution with only one salt, it's solubility will change upon the addition of another salt with a common ion in a manner strictly proportional to the latter salt's solubility in water alone? (i.e. that highly soluble salts will decrease the solubility of common ion salts the most and vice versa) If so, is the solubility of each salt in a combined solution a ratio of their respective solubility values when alone in the solution?

It is more complicated than that. Do you know what is a solubility product?

Not yet! I'll read into it. Thanks for the hint!

There are basically 3 effects.

The solubility is controlled by the solubility product, the constant being the multiplication of ion activities
$$K_\mathrm{sp}={a_\mathrm{M}}^m\cdot  {a_\mathrm{X}}^n$$
for a salt $\ce{M_{m}X_{n}}$
For low concentration of ions, it can be approximated by concentrations, related to activities via activity coefficients $\gamma$, that are supposed to be equal to 1.
$$K_\mathrm{sp}={c_\mathrm{M}}^m\cdot  {c_\mathrm{X}}^n$$
where $a_\mathrm{M}=\gamma_\mathrm{M} \cdot c_\mathrm{M}$, the same for $\text{X}$
Therefore the presence of other salt with the common ion decreases the salt solubility, as the common ion contributes to the rate of precipitation.

The second effect is general effect of dissolved non common ions, given by ionic strength
$$I=\sum_i {c_\mathrm{i}\cdot {z_\mathrm{i}}^2}$$
and its effect on the ion activity, given for diluted solutions approximately by Debye-Hueckel equation
$$\log \gamma=-0.5 \cdot z_+ \cdot z_- \cdot \sqrt I$$
where $\gamma$ is the mean activity coefficient for given salt and $z$ are respective ion charges. The coefficient $0.5$ is rounded up.
For a particular ion, it is often written as
$$\log \gamma=-0.5 \cdot {z}^2 \cdot \sqrt I$$
For less diluted solutions, it is often modified as
$$\log \gamma=-0.5 \cdot {z}^2 \cdot \frac {\sqrt I}{c + \sqrt I}$$
where $c$ is ion dependent constant close to 1.
Generally, the higher is overall ion concentration, the higher is salt solubility. But the effect is smaller than effect of common ions.

For high concentration of general salts, ion activity coefficients revert their decreasing trend and start raising. For very high concentration to values much higher than 1.
As result, the solubility decreases.


text:
Does dichlorobis(triphenylphosphine)nickel(II) have a d⁰ electron count?
I am currently in dilemma because I'm trying to construct an octahedral splitting diagram for the $\ce{NiCl2(PPh3)2}$ complex. If the complex is $\mathrm{d^0}$ system, how would I determine whether the isomer is paramagnetic or diamagnetic?

Work out the nickel oxidation state as +2.  Can you count the $d$ electrons from there?

Oh, and this is not octahedral, unless there is some recent mathematical discovery of an octahedron with four rather than six vertices.

Considering the advised you gave, the d-count of the Ni2+ would be 8.


text:
Can antimony pentafluoride be made by adding hydrogen fluoride to antimony trifluoride?
I was looking at how the strongest acid was made and I found something I did not understand. On the Wikipedia page for antimony trifluoride it shows the production of $\ce{SbF3}$ using $\ce{HF}$:
$$\ce{Sb2O3 + 6 HF -> 2 SbF3 + 3 H2O}$$
The page also shows production of $\ce{SbF5}$ using $\ce{F2}$:
$$\ce{SbF3 + F2 -> SbF5}$$
However on the page for fluoroantimonic acid it shows production of:
$$\ce{SbF5 + 2 HF -> SbF6- + H2F+}$$
So my question is why the production of $\ce{SbF5}$ seems to require fluorine insted of hydrogen fluoride? And is it possible to use hydrogen fluoride to make $\ce{SbF5}$?

It can be prepared using hydrogen fluoride, in a double displacement reaction:

$\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$

However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$.
Consider the reverse of your proposed reaction, which is preferred energetically:

$\ce{SbF5 + H2 -> SbF3 + 2HF}$

@EdV, sorry, I don't have that info. And, as you state, it's not something I'd try to measure. Does Derek Lowe have an article on SbF5, similar to this in his series Things I Won't Work With https://blogs.sciencemag.org/pipeline/archives/2010/02/23/things_i_wont_work_with_dioxygen_difluoride?

@EdV, your welcome, but I have Wikipedia to thank.

so my takeaway is keep your Antimony pentafluoride away from your hydrogen.


text:
Why does benzoyl peroxide homolytically cleave to give phenyl radical

Benzoyl Peroxide cleavese rapidly to give out $\ce{CO2}$ and phenyl radical.

What I could not understand was why phenyl radical is formed because it is not at all resonance stabilized and nor any electronic effects are acting on it that could help in its stabilization then why is it being formed ?? Wouldn't it be too unstable ??
I mean after first cleavage, (Before $\ce{CO2}$ leaves) it should be more stable due to resonance (and that too equivalent resonating structures)

I guess not! It is more stable than the homolysis product with a manageable activation barrier.

See this related mechanism of Kharash effect

The mistake you're making (which is an easy one to make) is that you're neglecting to consider the carbon dioxide product. Carbon dioxide is a stable molecule, and its release also increases the entropy of the system.
So even if the phenyl radical is less stable than the benzoyl radical intermediate, the combination of phenyl radical + $\ce{CO2}$ can still have an overall lower free energy than the benzoyl radical.


text:
Why does pH of a buffer solution change according to Henderson–Hasselbalch equation?
If I add a little amount of acid or base to a buffer solution, the pH of the buffer won't change. That's what the diagramatic mechanism tells us.
But according to Henderson–Hasselbalch equation, the $\mathrm{pH}$ changes!
For example, a buffer solution of $\ce{CH3COOH}$ and $\ce{CH3COONa}$:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)$$
Now I add a little amount of acid $(\pu{0.01M},\;\pu{5 mL}).$ It disturbs $\ce{CH3COONa}$. Reaction:
$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$
It is noticed here that a small amount of $\ce{CH3COOH}$ is produced and a little amount of $\ce{CH3COONa}$ is spared. So simply $[\ce{CH3COOH}]$ and $[\ce{CH3COONa}]$ go high and low, respectively. As a result, the ratio between $\ce{CH3COONa}$ and $\ce{CH3COOH}$ decreases. As the $\mathrm{pH}$ depends on the ratio, the $\mathrm{pH}$ changes as well.

First, Henderson–Hasselbalch equation is: $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\text{base}]}{[\text{acid}]}\right)$. Second, $\frac{[\text{base}]}{[\text{acid}]}$ ratio is in log scale so numerical value of change is minimal at $\mathrm{pH} $ around $\mathrm{p}K_\mathrm{a}$.

so Log ([base]/[acid]) is minimal but not 0.  Still a little  change at pH but according to mechanism there shouldn't be any change in pH, shouldn't it? Don't these two ideas contradict each other?

Do your calculations with suggested change (adding $\ce{HCl}$), and see what happens.

Yes I did before asking this question and got a little change around 1.8*10^(-3). If add more acid, the change is more

Your premise re: "If I add a little amount of acid or base to a buffer solution, the pH of the buffer won't change." is wrong. The pH does change, just not by much. The gist is that the pH of a buffer solution changes much less than the pH of pure water would. Also look up "buffer capacity."

There is a worked examples in this answer to https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566

Suppose you have a $\pu{1.0 L}$ of acetate buffer solution made by $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COONa}$ solution adding to $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COOH}$ solution. Thus, $[\ce{CH3COOH}] = [\ce{CH3COONa}] = \pu{0.5 M}$. Therefore, according to Henderson–hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.75$ ($\mathrm{p}K_\mathrm{a} = 4.75$ for acetic acid).
Now you add a little amount of acid ($\pu{0.01M}, \; \pu{5 mL}$). It disturbs the equilibrium and change both $[\ce{CH3COONa}]$ and $[\ce{CH3COOH}]$, according to the reaction:
$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$
Now, the amount of $[\ce{CH3COONa}]$ is reduced by $\pu{0.01M} \times \pu{0.005 L} = \pu{0.00005 mol}$ and, similarly, $[\ce{CH3COOH}]$ is increased by $\pu{0.00005 mol}$.
Hence, new $[\ce{CH3COONa}]$ is $\pu{0.49995 mol/L}$ and $[\ce{CH3COOH}]$ is $\pu{0.50005 mol/L}$ (neglecting the volume change for convenience). Thus, new $\mathrm{pH}$:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)= 4.75 + \log \left(\frac{\pu{0.49995 mol/L}}{\pu{0.50005 mol/L}}\right) = 4.7499 \approx 4.75$$
You see, it is a very minimal change. It is true that if you add more acid or base, change could be significant. However, all buffers have buffer capacity (usually, $\mathrm{p}K_\mathrm{a} \pm 1$). Nobody ever said $\mathrm{pH}$ of buffer is constant. It fluctuate withing the capacity. If you add too much acid or base, its $\mathrm{pH}$ would drastically change.

You have just avoided my question. I acknowledge it's a minimal change but not 0 . The mechanism shows all H+ added get neutralised , no extra  H+ left. Then how and why does this minimal change happen?

See my last paragraph.

The mechanism also includes that the system has reached a new equilibrium state in line with the  minimal change in pH

I am a bit confused at this point that "How does the mechanism include minimal change in pH?"

Your buffer has $\ce{CH3CO2H}$ and  $\ce{CH3CO2Na}$ in equilibrium with water. Their concentrations will adjust according to the changes following  Le Chatelier principle.

After adding acid, a little amount of CH3CO2H is produced and its concentration gets high. That's why dissociation of CH3CO2H happen to gain equilibrium state according to Le Chatelier principle.     Is that the reason of minimal change at pH?


text:
Do "random" "2D" polymers constrained to a box actually form "space filling" curves?
Background: The self avoiding walk model is (allegedly) a reasonable model for polymers, at least with respect to certain "universal" (i.e. lattice independent) constants -- see section 1.1. in this survey.
Consider an $n \times n$ square grid, with lower left corner marked $a$, and upper right corner marked $b$. Let $S_n$ denote the set of self avoiding walks in the grid from $a$ to $b$.
There is a family of probability distributions on $S_n$, parametrized by a single number called the "fugacity." (Allgedely, a measurement of pressure.) In particular, for $ x > 0$, define $\nu_x$ as the probability distribution on $S_n$ that awards a walk $\omega \in S_n$ mass proportional to $x^{|\omega|}$, where $|\omega|$ is the length of the walk.
This family of probability distributions is known to have a phase transition around certain values of $x$. In particular, for $x > 1 / \mu$, where $\mu$ is the connective constant of the square lattice, a walk is "space filling" with high probability. (My usage of space filling here is heuristic, it is made precise in the linked article.)
Mathematically, this is all very interesting. I'm wondering if there is empirical evidence that complements these results. This leads to my imprecise question, but first some caveats.
Caveat one: These results I mentioned concern $2D$ self avoiding walks. The closest analogue in our 3D world is presumably polymer systems under tight constraints, such as in this. I'll call these 2D-ish in the question.
Caveat two: I ask this question as a mathematician extremely ignorant of chemistry. Please take that into account in your answers, if possible.
Question: Is there a physical process that produces a "random" (that is, $\nu_x$ distributed) polymer chain in a 2D-ish medium or 3D medium, and if so, is it "space filling" for sufficiently high pressure / fugacity $x$?
Totally naive thoughts:
I debated whether to include these -- I decided to err on the side of sounding ignorant but at least communicating a better picture of what I'm curious about, and where the boundaries of my knowledge are.

As far as making the process, what if just chucked some monomers in a solvent along with a polymer strand and waited for it to grow by random rerouting attachments? Is there any incentive for a long polymer path to replace one of the monomers in our starting strand? I guess one hard part is insuring that you get one chain, and not one chain and many polymer loops. Maybe there are physical reasons why the loops would like to merge into the chain, at least under high pressure? Or maybe it's not a problem that there are extra polymer loops, as long as we can somehow point to our favorite polymer chain.
Taking inspiration from the algorithm in that last link, one process I can imagine is sticking a long polymer into solvent in a 3D container, shaking it up (very gently), and then squeezing down one of the sides of the container. It seems hard to ensure that this can be squeezed thinly enough to make criss-crossing of the polymer impossible. Maybe instead of physically squeezing the polymer to one side, we can squeeze it using some other force (magnets???) ? Or maybe criss-crossing is inevitable if you try to do something like this?

Very interesting question. This is largely beyond what I can answer, but I should mention that restricting a "normal" linear 1D polymer to a 2D plane may not be so difficult. For a century we've known a marvellous method for producing layers a single molecule thick: placing hydrophobic molecules on the surface of a hydrophillic liquid, allowing it to spread, then carefully squeezing the surface layer together. This is the principle of the Langmuir-Blodgett trough. I'm sure this has been applied to the study of polymer packing.

Hm. The self-avoiding walk is a model which is correct (=rather good) for polymers in a good solvent. Actually that it's self-avoiding doesn't make a big difference in that case. However random walk (and self-avoiding walk even less) does not fill space. The larger the molecule gets, the lower the space-filling. I'm rather convinced this is the same in any dimensionality. So, the question is: What is your question?

@Karl The mathematical model for self avoiding walks *constrained to lie in an $n \times n$ grid * is space filling at supercritical fugacity. My question is whether this is an empirical fact as well.

Is that an artefact? Obviously a self-avoiding walk of $n^2$ steps must fill an $n \times n$ rectangle: But the last steps would be unnaturally (unphysically) constrained.

@Karl There is no constraint on the length of the walk, just that it is drawn from the distribution $\nu_x$. Not every such walk is space filling, it's just that they are "space filling" with very high probability. (And my question is whether $\nu_x$ is an empirically supported model.)


text:
Reaction enthalpy using bond energy in resonating molecules
To find out and enthalpy of a reaction using bond energy we use the formula
$$\Delta H = \sum\Delta_\mathrm{bond}H_\mathrm{reactants} - \sum\Delta_\mathrm{bond}H_\mathrm{products}$$
since we use all the individual $\Delta H$ of the bonds to calculate the enthalpy, e.g.
$$\Delta H_\ce{C2H6} = 2\Delta_\mathrm{bond}H_\ce{C-C} + 6\Delta_\mathrm{bond}H_\ce{C-H}$$
But if a molecule is in resonance, how is the bond enthalpy taken? For example,

Is bond enthalpy the sum of the bond energies of that canonical structure which is the most stable or the average of all canonical structures or something else?

To every resonance form you can assign those value. But none is there. The hybrid structure has less energy. This question suggests 1) you took values as physical constants, and they are not;  2) the concept of resonance isn't clear to you.


text:
Regarding enthalpy of a reaction
Enthalpy of a reaction $$\ce{2H2 (g) + O2 (g)-> 2H2O(g)} \Delta H_1$$
$$\ce{2H2 (g) + O2 (g)-> 2H2O(l)} \Delta H_2$$
Now I was asked to compare the enthalpy.
If I subtract reaction 2 from 1, I get $$\ce{2H2O(g)-> 2H2O(l)} \Delta H_2 - \Delta H_1$$
Since this is exothermic enthalpy is negative and $$ \Delta H_2 - \Delta H_1 < 0 $$ or $$ \Delta H_2 < \Delta H_1 $$
But if I consider
Enthalpy of $\ce{2H2O(l)}$ is less than enthalpy of $\ce{2H2O(g)}$ therefore more energy will be released in the formation of $\ce{2H2O(l)}$ and thus $$ \Delta H_2 > \Delta H_1 $$
So which one is it, where am I going wrong ??

Negative enthalpy means energy released to surroundings, positive enthalpy means surroundings transfers heat to system. You might be confusing the sign and magnitude of the enthalpies. A smaller absolute (unsigned) number is not generally the same as a more negative number.

You should be careful to distinguish between the magnitude of enthalpy (or some other change in energy) and the sign of the property:

the magnitude tells you how large the change was (yes, self-evidently)
the sign tells you in which direction the change occurred:
a) $\text{system}\rightarrow \text{surroundings}$, $\Delta H<0$, or exothermic
b) $\text{surroundings}\rightarrow \text{system}$, $\Delta H>0$, or endothermic

This is the typical convention used (not the only one).
In the case of the condensation of a gas, the magnitude can vary depending on the strength of intermolecular interactions in the liquid compared to the gas. The sign is negative because condensation at a constant temperature requires dissipation of heat to the surroundings.

So which one is it, where am I going wrong ??

That the enthalpy of $\ce{H2O(l)}$ is less than the enthalpy of $\ce{H2O(g)}$ means in the scenario you present that
$$\Delta H_2 < \Delta H_1$$
not the other way around. You could also start from the fact that condensation is exothermic (or vaporization endothermic) so that $H_g-H_l=-\Delta_{cond}H =\Delta_{vap}H>0$$^\dagger$, from which $$\begin{align}H_g&>H_l\\ H_g-H_{reagents}&>H_l-H_{reagents}\\ \Delta H_1&>\Delta H_2\end{align}$$

$\dagger$ (Writing "$H_g$" is a bit "dangerous" because it suggests there is an absolute enthalpy scale (rather than relative) but you can assume that here you are using a common reference state for water).


text:
Which of acid aqueous solution 1) to 4) is non-volatile one?
I am working on my scholarship exam practice assuming high school chemistry knowledge and struggling at how I can rationalize this question.

Which of acid aqueous solution 1) to 4) is non-volatile one?
  1) $\ce{CH3COOH}$
  2) $\ce{H2SO4}$
  3) $\ce{HCl}$
  4) $\ce{HF}$

The answer key provided is 2) $\ce{H2SO4}$ but I wonder, in the exam condition, how should I solve this rationally? What thought should I have on this to determine the volatility of the compound? Please help.li

Chemistry is not quite like math. You can't expect to deduce everything rationally. That said, I'd think of the bonds holding these molecules together (what are they, BTW?), and compare the compounds along these lines.

I found this question very broad since the exam does not specify the scope of content and it is from other country (Japan) which has different curriculum so I tried to find some rules of thumb to tackle such problems. Anyway, I think yes, $H_2SO_4$ has two hydrogen bonds with two additional polar $O$'s while others have less. For example, $CH_3COOH$ and $HF$ only contain a hydrogen bond while $HCl$ not. Perhaps molecular weight should be taken into account too.

Forget molecular weight; hydrogen bonds are quite enough of a reason. See, despite what I said in the first comment, it is possible to solve this one logically after all, and you just did that yourself.

@TreyAnupong Please have a look on https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here, with mhchem-syntax (equally available for LaTeX) input and rendering of chemical equation and formulae is greatly improved and may be equally used in questions, answers and commentaries.  About content of your question, at grammar school-level we had little booklets with math / phys / chem relevant tables repeating textbook content, including some about typical chemicals and their properties (molar mass, melting / boiling point).  Didn't you too?


text:
Fuel cell potential for the electrolysis of ammonia
I am trying to calculate the fuel cell potential for the electrolysis of ammonia with the following equation:
E = Gibbs free energy for the reaction/(# electrons X Faraday's Constant)

The half reactions are as follows:
$$2NH_3 + 6OH^- -> N_2 + 6H_2O + 6e (Anode) E^0 = -0.77 V $$
$$ 6H_2O + 6e -> 3H_2 + 6OH^- (Cathode) E^0 = -0.83 V $$
I obtained the potencial values from the following paper and (double checked online as well): https://www.researchgate.net/publication/327875102_Review-Ammonia_Oxidation_Electrocatalysis_for_Hydrogen_Generation_and_Fuel_Cells
Equations (4a,4b and 4c)
The overall reaction is then:
$$2NH_3 -> N2 + 3H_2 $$ $$E^0 = +0.06 V $$ ** in the paper they seem to imply that the cell voltage is $$ + 0.06 V$$ which I don't understand why, since the oxidation takes place at the anode, therefore for the anode the potential should be $$ E^0 = +0.77 V$$ from my understanding.
I have doubled check my standard enthalpies of formation and entropies of formation. However, when I calculate the Gibbs free energy of the reaction I keep getting $$ E^0 = -0.06 V$$ for the overall reaction.
My Gibbs free energy is positive and the temperature I am using is 298 K, also NH3 is in a gaseous state. I have tried calculating the $$ E^0 $$ for $$ H_2 + 0.5 O_2 -> H_2O $$  and I obtain a cell voltage of $$ E^0 = 1.23 V $$ which corresponds to the reaction Gibbs free energy of $$ - 237 kJ/mol $$ , I just don't know where I am getting confused in the ammonia reaction.
In the paper they mention that the reaction is thermodynamically favorable, but I don't see how when the Gibbs free energy for the reaction is positive.

You can see my detailed response on the sign of electrode potentials here:
Still taught to reverse oxidation half cells in electrochemistry?
To summarize, electrochemists all over the world have decided to quote all electrode potentials as reduction potentials. This tug of war of signs between the so-called European convention and American convention of electrochemistry for more than a century, resulted in this decision. Basically, the current convention is the same as that of Ostwald.


text:
When H2NCH2CH2PH2 acts a monodentate ligand, it is the phosphine end that chelates. Why is that?
In some occasions, if the ligand acts on Pd it do so as a monodentate ligand. I am attempting to explain this phenomenon but came into a block.

I believe there must be cases when it works with NH2 end alone, just not on Pd. Long story short, think of HSAB theory.

So basically, you're comparing a soft base and a hard base. In HSAB theory, a soft acid like Pd2+, would preferably react with a soft base like R-PH2. Soft acids and bases are easily polarized because of their size. Although, the possibility of the ligand acting as a bidentate can't be ignored, the P to Pd2+ interaction is thermodynamically favored than N to Pd2+ interaction.


text:
Enthalpy of combustion of 1-butanol
1-butanol $\ce{C4H10O}$
So I have absolutely no idea how to find the enthalpy and I did some looking on the internet with no use since I don't understand.
I came here hoping someone generous could do the sample calculations so I can understand and do the rest my calculations and test for my experiment.

That's the data for one the mixture of alcohol.
How would I figure out enthalpy for test 1?
(maybe I could cut out test 2, 3 and average in the picture since its irrelevant in figuring out the enthalpy of test 1, and causes confusion)
For temperature before and after. The solution was used to heat up 200 ml of water. By burning the mixture, igniting it.

Welcome to Chemistry Stack Exchange. Can you clarify two things about your question? You want to figure out "enthalpy", but determining the absolute enthalpy is challenging. I assume you are trying to calculate the enthalpy change. Can you verify that is correct? Also, can you tell us a little bit more about this experiment? What procedure are you following? That will help us provide a better answer that is more specific to your situation.

Yes, I did 5 different mixture of alcohol and water. and we are seeing if the enthalpy changes. That is all. there is 4 more table like this for different mixtures and that is it. I will compare the enthalpy of each solution and talk about it.

45 ml of butan-1-ol + 15 ml of water do not have a mass of 156.3–156.5 g. Does the mass include the container? (It also cannot include the additional 200 ml of water that are heated during the experiment.)

Answering your question as it stands is impossible, as we readers do not have enough information about your actual experiment. What do you mean with 'enthalpy of 1-butanol'? You can only measure changes in enthalpy, so you need to specifiy what 'change' you're looking at.
You could mean the enthalpy of evaporation or the reaction enthalpy of some chemical reaction or hundreds of other possibilities. It's essential to clarify what process you're referring to, so one can associate the corresponding starting (e.g. liquid) and end state (-> gas) to this energy change.
However, you stated that 200 ml of solution were heated up by whatever your process was. At constant pressure, the enthalpy required to change the temperature by $\Delta T$ is
$$\Delta H = m c_p \Delta T$$
where $c_p$ is the isobaric heat capacity. Knowing the value of $c_p$, you can calculate this energy change. Now where did that energy come from? Presumably, from your process. At this point, assume that there are no energy losses (f.e. neglect the heat lost to the surrounding air). Then by conservation of energy, the enthalpy increase measured by the temperature increase must be equal in size to the energy released by the process.
Finally, note that by convention we say if a process releases energy, it's enthalpy change is negative (we could've just done it the other way around; think of it as the same idea as defining a sea level).

I ignited the solution made a fire that heated up the water. the mass of the fuel was before and after.

Yes so what you are trying to measure is a enthalpy of reaction (combustion energy). Calculate the energy change as outlined, and then divide that by the mass of fuel that has burnt and you obtain a value of energy released per gram burnt. Best practice would be to calculate this for each test seperately, then average only the final values for the energy content.


text:
Calculating Sensitivity in Mass Spectrometry
My mass spectrometry textbook gives the following example:

Calculating sensitivity Magnetic sector instruments are specified to have a sensitivity of about $4 \times 10^{-7} C \ \mu g^{-1}$ for the molecular ion of methylstearate, $m/z \ 298$, at $R = 1000$ in $70 \ eV$ EI mode. One microgram of methylstearate is equivalent to $3.4 \times 10^{-9}  mol$ or $2.0 \times 10^{15}$ molecules. The charge of $4 \times 10^{-7} C$ corresponds to $2.5 \times 10^{12}$ electron charges of $1.6 \times 10^{-19} C$ each. Vice versa, in dividing the number of molecules per microgram by the number of charges at the detector we may conclude that only one out of $800$ molecules is finally detected.

I have three questions:

What is $R = 1000$ supposed to be? The author never mentioned such a quantity at any previous point in the textbook.
How did the author calculate that the charge of $4 \times 10^{-7} C$ corresponds to $2.5 \times 10^{12}$ electron charges of $1.6 \times 10^{-19} C$ each? Or is this just a fact that he has knowledge of?
How did the author calculate the last part:

Vice versa, in dividing the number of molecules per microgram by the number of charges at the detector we may conclude that only one out of $800$ molecules is finally detected.

?
I would greatly appreciate it if people could please take the time to clarify this.

@EdV Yes, but I'm referring to the converse of the calculation; that's the point of the author's statement in the second question. We start with the information that the instruments are specified to have a sensitivity of about $4 \times 10^{-7} C \ \mu g^{-1}$. The author then decomposes this into $2.5 \times 10^{12}$ electron charges of $1.6 \times 10^{-19} C$ each. The point of question 2 is, starting with the value of $4 \times 10^{-7} C \ \mu g^{-1}$, how did the author induce that this is $2.5 \times 10^{12}$ electron charges of $1.6 \times 10^{-19} C$ each? ...

... After all, we could have used any other combination of $x$ and $y$ such that $x + y = 4 \times 10^{-7} C$ - not necessarily $2.5 \times 10^{12}$ and $1.6 \times 10^{-19} C$. So, the question becomes, how did the author get the specific values $2.5 \times 10^{12}$ and $1.6 \times 10^{-19} C$?

@EdV Ok, I understand now. Thank you very much for the clarification. You should have posted it as an answer so that I could accept it, haha. And yes, it seems that the quantity $R$ might remain a mystery.


text:
Freezing point vs Intermolecular forces
Water has a higher boiling point (100°C) than cyclohexane (81°C). This is probably because of stronger intermolecular forces between water molecules as compared to cyclohexane molecules. Then, why is the freezing point of cyclohexane higher than that of water? By the previous logic, water should be having a higher freezing point as well.
This is one among many examples. I am pretty sure than boiling point depends primarily on intermolecular forces while freezing point depends on both intermolecular forces and packing efficiency in the solid state. I am sure that both factors support a higher freezing point for water. So, what other factors govern the freezing point of a liquid substance?

Note that water is a bad substance to compare to, since it is an exception in so many cases due to its hydrogen bond network. Examples are boiling and melting points compared to $\ce{H2S}$, $\ce{H2Se}$, density anomaly at 4 °C and many others.

Freezing is tricky, it can't be justified as easily as boiling.

This is a tough question to answer because the intermolecular distances are similar in the  solid to liquid transition unlike those in the liquid to gas phase transition.
In the case of the elements there is a correlation between the Debye temperature and the melting temperature. The Debye temperature is that temperature at which the atoms gain their full degrees of freedom, and the correlation, while not prefect  provides evidence for a connection between melting point and inter-atomic vibrations.
The idea (originally by Lindemann ) is that when the temperature reaches a certain value, the potential energy of the (harmonic) intermolecular vibration has now increased by a  sufficient amount that atoms become independent of one another and the solid melts. Thus the energy profile of a solid is like an egg box with many depressions in which the atoms reside and have some limited freedom, given a bit more energy they become free. An approximate calculation shows that only a 5% increase in intermolecular spacing is enough to cause melting, which to me seems very small. Clearly this is an over simple model but does give some insight into the problem.
In your example it is far more difficult to give any quantitative answer and many examples should be studied to understand the general picture. If we follow this simple model we must assume that the vibrations are of whole CX or water molecules moving in their intermolecular potential.  Hopefully someone who is an expert in this area can provide you with a more satisfying answer.


text:
Shift of Equilibrium
A friend of mine stumbled upon this question while revising equilibrium. Apparently the answer is A.

The answer obviously cannot include adding a catalyst. Furthermore, according to le Chatelier's principle, as temperature increases, the side with the endothermic reaction is favoured, in this case 2NO(g). Also, if there is an increase  in pressure, the side with fewer moles of gas will be favoured as it consumes more moles of gas. However, as both sides of the reaction have an equal amount of moles of gas, I would believe this to not have any effect and thus answer C. Hence, how come the answer is A?

By mistake, that's how. Your reasoning is fine.

You cannot apply Le Châtelier's principle to temperature. The effect is correct, but the reasoning is wrong. https://chemistry.stackexchange.com/questions/62852/le-ch%c3%a2teliers-principle-and-heat

@Zhe I stated that as temperature increases, the side with the endothermic reaction is favoured. I did not provide any reasoning for this.

It is really disappointing how many  exam papers contain errors.

Application of the Le Châtelier's Principle for a temperature effect is evidence of the incorrect reasoning. It gives the correct answer but for the wrong reason.

@Zhe So my textbook is incorrect?

If your textbook is using Le Châtelier's Principle to talk about the temperature dependence of equilibra, then yes, it's wrong.


text:
Why do foraminifera use less magnesium in their shells when the water is cold?
When scientists study ancient seabed cores, they say they can judge the temperature of the water in the past by analyzing the magnesium to calcium ratio in foraminifera shells (both in the form of carbonate, presumably).  Supposedly, they added less magnesium when the water was colder.
Why?  Is it because the lighter magnesium was more likely to be locked up in ice, like Oxygen-16 vis-a-vis O-18?
Also, why the focus on foraminifera?  Why not use diatom shells?

Might it have to do with the solubility of magnesium and calcium carbonates with respect to temperature?

Good question.... I will look into it...  But still no answers from anybody :(


text:
Relation between activation energy and rate constant

Consider two types of reaction. One has more activation than other. If we increasing temperature by same amount in both, rate constant of which reaction will increase more?

My attempt: as we know that $k = A\mathrm e^{(-E/(RT))},$ if temperature is same in both, then the one having more activation energy will have lesser rate constant. Am I correct?

Note that the question asks which one increases  most, not which is the greatest. You have given the equation by which to work this out.

So shall i differentiate the equation

This is a poorly worded question, it does not state whether you should compare the change in rate in absolute or relative terms (is it a factor or a difference you should compute?)

I think it is asking to calculate relative change(whose rate will increase more).

$$k_\mathrm{high} = A e^{(−E_a/RT_\mathrm{high})}$$
$$k_\mathrm{low} = A e^{(−E_a/RT_\mathrm{low})}$$
$$\frac{k_\mathrm{high}}{k_\mathrm{low}} = e^{-E_a \cdot \beta}$$
If the activation energy is zero, the ratio of rate constants will be one, i.e. there is no temperature dependency of the rate. The higher the activation energy, the larger the exponent, the larger the ratio of the rate constants ($\beta$ is negative, so the exponent is positive).
It follows that if you are interested in the ratio of rates, it will be larger the higher the activation energy is.
The answers that come to the opposite conclusion are also correct. This is because the question is ill-defined and it is not clear what is being asked.

This can be answered both conceptually and mathematically.
Conceptually: Let's call the two reactions 1 and 2 with reaction 1 having the larger activation energy. Here is a plot of the arbitrary reactions.

It is clear from this graph that it is "easier" to get over the potential barrier (activation energy) for reaction 2. Thus if we increase temperature, the reaction would get faster for reaction 2 before it gets faster for reaction 1. Therefore, when increasing the temperature by the same amount, the rate constant for the reaction with the lower activation energy will increase more.
Mathematically: You can understand how the rate constant changes with temperature by simply graphing. Here is a very quick and very rough graph on how the rate constant changes with temperature (plotting the equation $k=Ae^{-E/RT}$).

Notice how increasing the temperature causes a quicker increase in the reaction with the lower activation energy $(E_{a2})$. However, this is only an a priori expectation.
You mentioned in a comment above about differentiating. You can use that method as well. Taking the derivative of the equation for $k$ with respect to temperature while assuming $A$ and $E_a$ do not depend on temperature yields
$$\frac{dk}{dT} = \frac{AE_a}{RT^2}e^{-E_a/RT}$$
Edit: As porphyrin pointed out in the comments below, this equation elucidates the true behavior of the temperature dependence. Plotting this function for the two energies, we see that the derivative for the reaction with a smaller activation energy increases faster for lower temperatures. But for higher temperatures, there is a shift in the change of the rate constant. Therefore, the process is temperature dependent. At low temperatures, the rate constant for the reaction with low activation energy changes is more sensitive to temperature change while at high temperatures, the reaction with larger activation energy is more sensitive to temperature change.

As a side note: at very large (infinite) temperatures, both functions approach the same value and there will be an equivalent distribution among the reactions' products.
From comments below: Prove that if we increase temperature in an endothermic reaction, equilibrium proceeds in forward direction.
There are a couple ways to do this. As this is now an equilibrium question, I will keep the regime to equilibrium instead of rate constants. First, let's use fundamental chemistry to get an expected result then use equations to show what the result would be quantitatively.
Fundamentals: Say you have a simple reversible reaction
$$A\rightleftharpoons B$$
Given this reaction is endothermic, energy is required to drive the reaction. Because of this, we can write energy as a "reactant"
$$A + Energy \rightleftharpoons B$$
According to Le'Chatelier, if you increase temperature then the equilibrium must shift towards the products, thereby answering the question.
The Maths: Now, let's apply some math as overkill. The equilibrium constant is dependent on the standard free energy of the reaction given by the following equation
$$\Delta G^o = -RT\ln(K)$$
If we want to understand how the equilibrium constant will change we can rearrange this equation to solve for $K$ and plot as a function of temperature. Doing this and making the substitution for the free energy $\Delta G^o = \Delta H^o - T\Delta S^o$ gives
$$K = e^{-\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}}$$
Since the reaction is endothermic, $\Delta H^o > 0$. This gives the same monotonically increasing profile that we see for the Arrhenius equation (plotted above). Therefore $K$ must always increase, and therefore the reaction proceeds in the forward direction, for increasing temperature. Of course, this is making the assumption that $\Delta H^o$ does not change with temperature.

Thanks for this wonderful answer.But now i have another query.We know that in an endothermic reaction,by increasing temperature reaction proceeds in forward direction.But by going through this way,we know change in enthalpy is positive.Therefore activation energy of forward reaction is greater than that of backward.Therefore rate constant of backward reaction will increase more than rate constant of forward reaction,which means that equilibrium proceeds in backward direction.These two things are contradicting.Help me with this concept.

The conclusion you draw is not general. If the ratio of slope $dk/dT$ is plotted vs $T$ and different $E_a$ (large $E_a$ over small) then the ratio is  $\lt 1$ at low $T$ but exceeds $1$ at some finite temperature. So the answer depends on where the initial temperature is set and the relative activation energies..

@Aastik You are correct. So was your question actually about forward and reverse rate constants? Or was it about rate constants for two different reactions?

@porphyrin That is absolutely right.

Actually my question is stated like this-:Prove that if we increase temperature in an endothermic reaction,equilibrium proceeds in forward direction.

@Aastik Okay. I answered your question in the answer above because it merited some explaining.

So we know that
$$\ln\frac{k_{T_2}}{k_{T_1}} = \frac{E_\mathrm a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$
If we consider both were raised from $T_1$ to $T_2$, $$\frac{1}{T_1}-\frac{1}{T_2} = \text{constant}$$
So
$$\frac{\ln\frac{k_{T_2}}{k_{T_1}}}{E_\mathrm a} = \text{constant}$$
So the one with higher $E_\mathrm a$ has a higher value for  $\ln\frac{k_{T_2}}{k_{T_1}}$
Thus the fractional change of a reaction with higher $E_\mathrm a$ is more than the one with a lower $E_\mathrm a$.

The activation energy is the kinetic energy required from the reactants to form the products.Now lets go to your second part.Since the first reaction has a bigger activation energy than the second reaction , if we add energy to the system the first reaction has better chances of happening than the second  reaction because more molecules from the first reaction will be converted in products than in the second case.To actually understand it you have to take quantum mechanics of every molecule in both cases and you will  see it.

This answer could be significantly improved by adding details and examples or even sample calculations.

Using math in some cases creates dangers for the understanding of science and chemistry.


text:
Issue with evaporation rate of water

The evaporation rate of water (kg/s) is given by:
$$g = \frac{P}{P_o} AΘ$$
Where $P$ is given in Pascals and $A$ is the surface area (in $m^2$) and $Θ$ is the empirical evaporation coefficient:
$$Θ = 12v + 16$$
where $v$ the wind speed (perpendicular to the water's surface).
a) Calculate the evaporated mass of water in one hour when $v = 0$.
b) Calculate the enthalpy and entropy difference within $1$ hour when $v = 0$.

What I have tried:
a) I understand $P$ to be the partial pressure of water vapor in moist air and $P_o$ to be the atmospheric pressure of moist air.
I am not given neither the values for these pressures nor the surface area... How am I supposed to calculate the evaporated mass without such data? (It's stated that all the necessary information is provided)
b)
$$\Delta H = \Delta U + P\Delta V$$
Holding fixed the number of molecules:
$$dH = TdS+ VdP$$
But again I don't have enough information...
Are there flaws in this exercise or am I missing something?

You are correct, you cannot calculate the answer in absolute terms, e.g. kg, so you must answer the question in terms of the pressures, areas and the constant 16 (since v drops out). How does theta become gamma, though? Typo??

Yes, that is a typo. I'll fix it.

What kind of exercise are they assigning! Basically this asks for "how many second are in one hour" in a convoluted way.

Yeah, this exercise is not really useful. I am interested in doing exercises in vapor-gas equilibrium (saturated vapor), which is described by the Clausius–Clapeyron relation: https://en.wikipedia.org/wiki/Evaporation#Factors_influencing_the_rate_of_evaporation

So if anyone has a set of exercises on it, could you please share it? I have more exercises on it but are confusing like above.


text:
Electric glow discharge of last 2 noble gases
Radon is an extremely unstable noble gas. Due to this, putting it in a gas discharge lamp for a prolonged period of time is extremely hard, as pure radon is needed to give a true color, but radon decays, contaminating the lamp. But suppose we were to find a way to not let radon decay,  concentrate it in a gas discharge lamp, and turn the lamp on. What color would the radon gas glow?
Sources I find are contradictory. images-of-elements.com says

Radon is said to glow red in discharge tubes

and the image that it gives

is only an illustration, not radon itself.

Wikipedia says that it "glows green or red in discharge tubes", but it doesn't cite any sources there.
The spectral lines of radon are listed in physics.nist.gov/PhysRefData/Handbook/Tables. According to this source, radon gives probably somewhat reddish due to the strong lines at 705 and 745 nm.
reddit: what_color_does_radon_glow_when_electricity_is says that

Radon glows violet/purple when discharged. The reason why is simply we can infer it off of the patterns of the periodic table. You see, radon is at the very bottom to the right, in the noble gases column. Noble gases are already stable (for the most part) and do not want to react with other particles; thus requiring a lot of energy to react. As radon is the largest natural noble gas element, we humans can mess with it a little bit more than its other noble gas brethren. Now since we know it requires a lot of energy to react, we know it is related to light with an insanely short wavelength (shorter wavelength=higher energy). If we look at the light spectrum we can see the shortest wavelength and highest energy wavelength is purple/violet/indigo. Thus making radon that color when discharged.

And based on the emission spectrum from chemistry.bd.psu.edu, it appears that radon would fluoresce with a deep bluish-violet color.

Same question for oganesson. Although Wikipedia says it should be a solid due to relativistic forces, what color would oganesson glow in a gas discharge lamp?

The reddit comment seems just bizarre, given the prevalence of neon lights (particularly in Vegas, although most lighting seems to be LED there now).

Interesting question! While I don't have an answer, there are stars that emit lines from technetium... for a relatively short time. ;-)  https://en.wikipedia.org/wiki/R_Geminorum . See also the Capitol Steps song, Everything Glows :https://www.deseretnews.com/article/338958/CAPITOL-STEPS-BRING-MUSICAL-MOCKERY-TO-SL.html

As there would be glowing, melting or vaporization due very intense alpha bombardment, such a question is rather funny.

@Buttonwood Rn VIS emission line intensities are provided in the NIST link in the question.

According to Rutherford, a discharge of Radon (radium emanation) had a bluish color when he measured it. He also notes that the color of the discharge changed to 'pale rose' when the radon was condensed using liquid air.

@Poutnik Paschen's law tells us that a glow discharge will work best in a pretty low pressure gas. Would a thin glass tube with say 10 Torr of radon inside be melting or vaporizing, or visibly glowing under self-ionization or incandescence? Maybe so, that would be worth finding out!

@uhoh It is quite clear, but even low pressure with half-life 3.6 days of 222Rn means enormous radioactivity, ionization and thermal effects, not speaking about oganesson.

@uhoh  I did some quick calculation and unless I made error, decaying of 1 mol of radon gives power about 1.2 MW.  If we take 1 L at 10 torr, it gives about 700 W. It look like radon plasma  even without external voltage.

@Poutnik Yikes! I see what you mean. Subconsciously I was probably thinking of Rn as an everyday household substance and most houses don't glow from it, but yes those partial pressures are a dozen orders of magnitude or more lower than our glow discharge tube. Thanks!

The Royal Society of Chemistry has a Twitter account (here) and they have shared the spectrum of radium emanation as it was known in the beginning. It is a beautiful picture showing the comparison of helium with the newly discovered gas.
In photographic spectroscopy, darkness of the lines indicated intensity. One can see the intense lines at 416 and 418 nm and around 460 nm. So the emission must be appearing violet blue in a discharge tube. Like most elements, the ground state transitions, hence the strongest, are in the UV region.
The reasoning at reddit and other unauthentic sources of information are not reliable and close to gibberish.
Of course, NIST is the most reliable source today, however the line 745 nm is on the very edge of visible deep deep red. It is very hard to see, just like potassium, one cannot easily see the deep red lines, hence the flame appears violet/lilac to us.

beautiful!

The argument here is that the behavior of Ra and K generalize to other elements, in particular Rn. However, since you selected radium, not radon, as an example to illustrate your point, and the names are potentially confusing in being so similar, not to mention that the figure compares Ra to He, a noble gas like Rn, you might want to clarify the point. Or is the point to potentially confuse?

Please read the label carefully, it is "Radium emanation" not radium, which is an old name for Radon.

Thank you for that clarification, I learned something new. I should also have read your answer more carefully, "...as radon was known in the beginning..."

@BuckThorn, I like chem history, that is why it was a familiar term. Emanation was common name for a radioactive gas from actinide elements too. Those turned out to be isotopes of Rn.


text:
How do I do stoichiometry for 2 limiting reactants

The Drano reaction produces sodium aluminate along with hydrogen gas. If a typical container of Drano weighs 500 g which is 35% by weight by weight sodium (l) hydroxide and 2.5% by weight aluminum, what volume hydrogen gas would be produced if (excess) water was added to an entire can of crystal Drano at 22 °C and 1.2 atm?

My equation:
$$\ce{4 Al + 4 Na(OH) + 12 H2O -> 4 Na[Al(OH)4] + 6 H2}$$
If my balancing correct, I am trying to get the excess for water. Would I need to do the moles left over? Or would 500 x 62.5% (2.5% and 35% added together) suffice?
If I do need to do moles left over, how do I manage with two limiting reactants?

Richard, there can be only one limiting reactant. You cannot have two by definition. The one which is present in the least amount in a stoichiometric sense, is the limiting reactant!

Besides the fact there is only one limiting reagent, the term "moles left over" is more applicable to the population of moles, the animals. In chemistry you probably want to use "extensive amount" instead.

You can simplify the equation to $\ce{2Al + 2NaOH + 6H2O -> 2Na[Al(OH)4] + 3H2}$.

Your balance equation is correct except for not simplifying it. Simplified equation should be:
 $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$
I also think you're getting a little bit confused about your equation. You probably think reacting $\ce{Al}$ and $\ce{NaOH}$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are absolutely wrong. As M. Farooq commented elsewhere, the one reactant, which is present in the least amount in a stoichiometric sense, is the limiting reactant. Let's see your question in hand:
The amount of $\ce{NaOH}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{35 g}\; \ce{NaOH}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{40 g}\; \ce{NaOH}}  = \pu{4.38 mol}$
The amount of $\ce{Al}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{2.5 g}\; \ce{Al}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{27 g}\; \ce{Al}}  = \pu{0.463 mol}$
Since $\ce{NaOH}$ and $\ce{Al}$ react with 1:1 ratio according to the equation, $\ce{Al}$ is clearly the limiting reagent.
Hence, $\pu{0.463 mol}$ of $\ce{Al}$ reacts with $\pu{0.463 mol}$ of $\ce{NaOH}$ if you added enough water to $\pu{500 g}$ of Drano container. The leftover amount of $\ce{NaOH}$ would be: $\pu{(4.38 - 0.463) mol} = \pu{3.92 mol} = \pu{3.92 mol} \times \frac{\pu{40 g}\; \ce{NaOH}}{\pu{1 mol}} = \pu{157 g}$
Now, how much water you need? You need more than $\pu{0.463 mol}\times \frac{6}{2} = \pu{1.39 mol}$ of water, which is more than $\pu{25.0 g}$ (to be excess).
Note: Similar calculations would appied to find out amount of $\ce{H2}$ released by the reaction.

I think you're getting a bit confused with your equation. It's probably better to write two distinct reactions, one with $\ce{NaOH}$ and one with $\ce{Al}$.
35% of $\pu{500g}$ is $\pu{165g}$ $\ce{NaOH}$, and 2.5% of $\pu{500g}$ is $\pu{12.5g}$ $\ce{Al}$.


text:
How to oxidize tantalum?
I am interested in learning how to oxidize tantalum. I have seen some tantalum rings that the sellers say is a permanent smokey blackish color. But I've spoken to some people that say tantalum cannot be oxidized, and the rings I saw must have been coated.
So anyone know if and how tantalum can be oxidized?

Any metal can be oxidized. Although Ta is rather resistant, there's lots of ways. As for the black color, it might be tantalum nitride, but this question falls under both unclear and too broad, IMO.

The smoky black stuff is probably not tantalum oxide.  The only such oxide that's well characterized is $\ce{Ta2O5}$, which is white.  Tantalum itself is a rare element, so if the jewelry contains tantalum at all it's probably a thin coating or cladding on a base metal substrate.


text:
Amphoterism and the Metal and Non-Metal Oxides/Hydroxides. Why does amphoterism occur in these?
I am having trouble understanding why Beryllium Oxide and Aluminum Oxide are amphoteric. Also, while they are not oxides, Aluminum and Beryllium Chloride as well.
Most of the things I have been taught or have read is that the partial covalent character of the bonds in these compounds is what allows them to be amphoteric. By not having to behave as strictly ionic allows them to be either an acid or a base depending on what type of solution they are in and what they are being reacted with. Is this correct?  If so, why does the slight change in bonding contribute to the amphoteric behavior?
Covalent bonding seems to be a common theme in amphoteric substances. Why is this so?
I've been staring at reaction of Beryllium Hydroxide and Aluminum Oxide and can't grasp why they alternate between acid and base behavior, depending on what they are reacted with, and how this phenomenon is related to the electron arrangement and bonding. Can someone elucidate this in some detail?


text:
Reactivity towards nucleophillic addition

Phenyl ethanal or ethanal

I thought phenyl methanal is less reactive towards nucleophilic attack due to presence of bulky group. But answer in my book suggest ethanal is less reactive.

Para-methoxybenzaldehyde vs para-methylbenzaldehyde.

in this case due to the prescence of $\ce{OCH3}$ $\ce{O}$ will act as an electron withdrawing group as well as electron donating (resonace) whereas $\ce{CH3}$ group will only donate it's electron (resonance) thus para-methoxybenzaldehyde  should be more reactive. Again, the answer in my book states para-methylbenzaldehyde is more reactive.

The resonance effect of the methoxy group (donation of electrons) is more important than its inductive effect and the resonance effect of the methoxy group overwhelms the donation effect of the methyl group. Thus, your book is correct. [BTW: By phenyl methanal do you mean benzaldehyde? Ethanal (acetaldehyde) should be more reactive toward nucleophiles than benzaldehyde owing to resonance contribution from the phenyl ring.]

i meant phenyl  ethanl benzene with ch2-cho

I think you mean phenylacetaldehyde. As to my previous comments, you might look at: https://chemistry.stackexchange.com/questions/110641/nucleophilic-addition-propanal-vs-benzaldehyde?rq=1


text:
Spatial Wavefunction Symmetries and Identical Particles
I was reading this and it mentions in the 3-electron section, that for a spacial wave function to be symmetric under fermion swapping, it must be a function of even parity. Similarly for anti-symmetry under fermion swapping, it must be a function of odd parity.
It is not immediately obvious to me why parity symmetry $(-1)^l$ has anything to do with the bosonic or fermionic properties of the spatial wave function. So I suppose my questions are:
Is it true that spatial inversion is the same symmetry as swapping?
And if so, why?

Spatial inversion is generally not the same as swapping. As for why the latter is intimately connected with spin properties: it is very non-obvious and deep, so for now just take it as given.

When we talk about exchanging electron $i$ with electron $j$, we are actually changing the wavefunction according to
$$\Psi(..., x_i, ..., x_j, ...) \to \Psi(..., x_j, ..., x_i, ...).$$
The operation is taken by the parity operator $P$. Applying it twice would
return the wavefunction to its original form. So the following eigenvalue equation is satisfied
$$ P^2 \Psi = 1 \Psi. $$
Then the $\lambda$ eigenvalue of the $P$ operator ($P\Psi = \lambda \Psi$)
also satisfies an analogous equation
$$ \lambda^2 = 1. $$
$\lambda$ can only be either $\pm 1$. If it is positive the wavefunction is symmetric under swapping and if the eigenvalue is negative the wavefunction is antisymmetric under swapping. It can also be said that the symmetric (antisymmetric) wavefunction has even (odd) parity but do not confuse it with spatial symmetry.
It is assumed that fermions must be described by antisymmetric wavefunctions
and bosons by symmetric wavefunctions. Many have argumented this fact alluding
to the spin-statistics theorem but others have, as well, stated that such justification is not satisfactory. Besides, it is not completely clear why.


text:
Changing conditions of gaseous mixture

A gaseous mixture of $\ce{NO2}$ and $\ce{N2O4}$ is sealed in 4 identical vessels with movable pistons. When the conditions are modified as described in 1) to 4), in which condition does the molar ratio of $\ce{NO2}$ to $\ce{N2O4}$ remain unchanged after a long time enough to reach equilibrium?
  1) Total pressure is increased by moving the piston at a constant temperature.
  2) Temperature is increased while keeping the total pressure constant by moving the piston.
  3) Additional $\ce{NO2}$ is introduced from outside while keeping the total pressure constant by moving the piston at a constant temperature.
  4) Additional $\ce{N2}$ is introduced from outside while keeping the total pressure constant by moving the piston at a constant temperature.

The answer provided is (3).
My attempt: I think when we add $\ce{NO2}$, the equilibrium will shift to the right so the concentration of those two will changed. I answered (4) because $\ce{N2}$ is unreactive, making the concentration of both compounds the same. Please help why the answer 3 is correct and others are wrong.

Why do you think the equilibrium shifts in the answer 3 ?

Because when we balance the equation we have $\ce{2NO2}$ => $\ce{N2O4}$. Although the pressure remains constant, since choice (3) introduced $\ce{NO2}$ (increase in concentration), the system will counteract the change by shifting equilibrium to the right.

Try again.. :-)

Hint: Enumerate equilibrium conditions via equilibrium constant before and after addition.

Thank you for your hint. I have looked closer again and yes, both before and after addition will give the same equilibrium constant, meaning the molar ratio of those two compounds is equal so (3) is correct. I am also interested in the choices that are wrong. The equilibrium constant will change only when the temperature changes if I understand it correctly so number (2) is indeed incorrect. However, I am still struggling why number (1) and (4) are incorrect at a constant temperature. Like in number (4), adding inert $\ce{N2}$ would not affect partial pressure of those two compounds.

Hint2: At the same temperature and equilibrium constant, is partial pressure ratio of compounds independent on partial pressures ?

Hint3: With total pressure constant, how can be partial pressures constant after addition of nitrogen ?


text:
Can I compare DFT calculations with different grids?
When doing DFT calculations, some integrations are commonly done numerically on grids. [In fact, more than a single grid may be used at the same time for different integrals, e.g. approximations such as the resolution of identity (RIJCOSX, RIJK, etc., see e.g.  J. Chem. Phys. 118, 9136 (2003)) use grid schemes too (I believe some programs call this approximation density fitting).]
There's the downside that those grid schemes introduce a source of error. In fact, I find that only by increasing the quality of the grid I can remove some imaginary frequencies. Can I rigorously compare energy values among calculations that used slightly different grid schemes?
I think yes and I reasoned as follows. Since I'm interested in energy differences, let's imagine two structures with true energies $E^*_1$ and $E^*_2$. Both energies are approximated by the calculated energies $E_i = E^*_i + \epsilon_i$, where $\epsilon_i$ is the error introduced by calculations (including grids, etc.). Now the energy difference $\Delta E^* = E^*_2 - E^*_1$ is approximated by $\Delta E = E_2 - E_1 = \Delta E^* + \Delta \epsilon$, where $\Delta \epsilon = \epsilon_2 - \epsilon_1$. Since grid schemes introduce errors that are smaller than other sources of error (e.g. implicit solvation), everything should be fine as long as $\Delta \epsilon$ is acceptably small for the particular application at hand.
Does this seem reasonable? If not, what's wrong with the above?

Yes, your error analysis is valid for energy differences. However, I believe it is also valid for the absolute error $\epsilon_i$ of any calculated quantity, not just the error $\Delta\epsilon$ of an energy difference.
In any numerical computation, the key thing for situations like this is to ensure that these "structural" sources of error have been reduced to a magnitude that don't affect your results. In this case, this is achieved when you reach a grid quality $Q$ where the numerical results no longer change when you further increase to $Q+\delta Q$.
To be clear: in this case elimination of the imaginary frequencies is not a good criterion for a sufficiently large $Q$.  The proper stopping point for refinement of $Q$ is when the results no longer change appreciably.
Note that, as you allude to implicitly in the question, this situation is different than the one of trying to compare, e.g., absolute energies from computations run with different methods or at different levels of theory.  In this case, there are fundamental/theoretical reasons why those comparisons cannot be meaningfully made.