text:
A thermodynamics problem with an interesting equilibrium condition
Usually, when we analyze reactions, we assume that the reaction takes place at constant pressure or at constant volume. But there is no reason to assume this is generally the case, and so my friend posed the following question that has a different condition:

$n=0.1$ mols of $\ce{SO3}$ are put into a container of volume $V_0 = 3\pu{L}$ at temperature $T=1100\pu{K},$ and reaches an equilibrium mixture with $\ce{SO2}$ and $\ce{O2}.$ The temperature is held constant at $T$ and the quantity $V - c\cdot P$ where $c = 20 \frac{\pu{L}}{\pu{bar}}$ is always constant during the reaction. Assume everything is ideal. Use any reference data as appropriate. What are the equilibrium concentrations of all compounds?

First, for the thermodynamic constants, it is not hard to find their values at 1100 K from the reference ones at 298 K simply by using   their relation with temperature and heat capacity. I get that $\Delta H = 95.6\pu{kJ}$ and $\Delta S = 99.3\pu{J/K}$ when one mol of $\ce{SO3}$ decomposes.
Ordinarily, in constant volume or constant pressure scenarios, we can find the equilibrium concentration simply from the equation $\Delta G = - RT \ln K$ or $\Delta A = - RT \ln K$ and then use known values of Gibbs/Helmholtz formation enthalpy to calculate the concentrations. But here, neither of these equations apply since neither volume nor pressure is constant.
From the condition of ideality it is not hard to derive a formula for the volume and pressure with respect to the extent of reaction $\xi:$
$$P_0 = n\cdot R\cdot\frac{T}{V_{0}}$$
$$V_1 = \frac{V_{0}-cP_{0}+\sqrt{\left(V_{0}-cP_{0}\right)^{2}+4RTcn\cdot\left(1+\frac{1}{2}\xi\right)}}{2}$$
$$P_1 = n\left(1+\frac{1}{2}\xi\right)\cdot R\cdot\frac{T}{V_{1}}$$
but I'm not sure how to proceed in this problem afterwards. One thought I had was to maximize total entropy, but this seems quite hard to set up.

From your initial conditions, you know that P1 = 3.049 bar, and thus that V – c P = –57.97 L (you said it was "always constant" during the reaction, which would apply to the start of the reaction as well).  Thus P, in bars, will always be (according to your constraint) 50 x (0.05797 + 0.001 V), where V is in liters.  But this reaction may not follow that constraint.  Do you know that it does?  If it does, that gives you a third equation to add to your equations for V1 and P1: V1 – c P1 = –57.97 L.  Thus you now have three (independent?) equations and three unknowns (V1, P1, and ).

@theorist The condition should be true all throughout the reaction, and I used that condition on V-cP to derive the formula for V1. It should be P0 = 3.049 bar rather than P1? So, we still only have two independent equations :/

Ah, yes, I meant P0 = 3.049 bar.   What was your friend's basis for coming up with V – c P = –57.97 L?  What was the idea behind this particular constraint?  And your friend wouldn't happen to be a professor giving you a homework problem, would it? ;)

@theorist According to him, the idea is that the reactor has a piston connected to a spring, so the force by the spring and pressure increases as the volume increases.

OK, that explanation helps.  I think the problem, as written, is presenting an underdetermined system.  I say this because if you can get the equilibrium state purely from the info. given, it means the equilibrium state is independent of the compounds reacting, which wouldn't make sense.  I.e., if you could calculate an equilibrium purely from the info. given above, that equilibrium would be the same for any reaction of the form $\ce{AB_3(g) -> AB_2(g) +1/2 B_2(g)}$, which, again, doesn't make sense. So what's missing is information specific to the chemical species involved.

@theorist Right, I guess I should've been more clear when I stated that all reference material with formation enthalpies/energies was available. I'll edit that into the question

If you have Kp at 1100 K, then you could combine that with your initial conditions to get advancement as a function of pressure.  I haven't tried it myself, but I think that would be your third indepedent equation.

@theorist Right, once we have Kp the problem is not hard to solve, so the main difficulty is in finding Kp. Unfortunately, the relation between gibbs energy change and Kp is not valid in our reactor setup here since the pressure is not constant.


text:
What does this example mean? Slow cooling of 40% Sn alloy from 800°C to 600°C: L → L and γ → L, γ, and ε → L and ε
It's in a phase diagram topic. It's about $\ce{Cu}$-$\ce{Sn}$ phase diagram.

And it has a question of:

Write the schematic diagram describing the following scenarios and identifies the regions and the phases when Slow cooling of $13.1\%$ $\ce{Sn}$ alloy from $\pu{1000^\circ C}$ to $\pu{300^\circ C}$ happens.

I have this schematic diagram that describes the following scenarios and identifies the regions and the phases at $40 \%\,\ce{Sn}$ alloy from $\pu{800 °C}$ "$\ce{L -> L}$ and $\ce{γ -> L}$, γ, and $\ce{ε  -> L}$ and ε" as an example but it's confusing to follow. However, it's just an example and it doesn't follow the $\ce{Cu}-\ce{Sn}$ phase diagram anyway.
But does the first one mean $\ce{L -> L +  γ}$?
And L , y are in different phases that manages to fall both at $40 \%\,\ce{Sn}$ alloy from $\pu{800 °C}?
Also am I correct to assume that the final one means $\ce{ε -> L + ε}$?
This is the closest image that I can give based on my assumption:

Development of microstructure in eutectic alloys (II)
At compositions between the room temperature solubility limit and the maximum solid solubility at the eutectic temperature, $\beta$ phase nucleates as the $\alpha$ solid solubility is exceeded upon crossing the solvus line

[MSE 2009: Introduction to Materials Science; Chapter 9: Phase Diagrams; Page 28]

Here's the PDF file for more information.
"$\ce{L -> L}$ and $\ce{γ -> L}$, γ, and $\ce{ε  -> L}$ and ε" is a schematic diagram for the slow cooling which means it is an equilibrium cooling Like this image given below:

Equilibrium Cooling in a $\ce{Cu-Ni}$ Binary System
Phase Diagram: $\ce{Cu-Ni}$ System.
System is:
--binary: i.e 2 components: $\ce{Cu}$ and $\ce{Ni}$.
--isomorphous: i.e., complete solubility of one component in another; a phase field extends from $0$ to $100\,\mathrm{wt}\%$
Consider $C_o = 35\,\mathrm{wt\%}$ $\ce{Ni}$

Another one is:

Ex: Cooling of $\ce{Cu-Ni}$ Alloy
Phase diagram: $\ce{Cu-Ni}$ system
Consider microstructural changes that accompany the cooling of a $C_o = \pu{35 wt\%}\, \ce{Ni}$ alloy

Cored vs Equilibrium Structures
$C_o$ changes as we solidify

These images are from this PPT file.

No. The arrows (you could label each of them with a temperature) have the highest priority, over commas and "and", which are equivalent. Homework: go down starting from 40%/800°C and describe what will happen. (Below 415°C, a further phase appears, btw.) Remember: the unlabeled areas are forbidden, you cannot cross them.

Consider the 40 at% (Sn) composition at 800 °C: it's a liquid.
When you cool it down at around 650 °C you enter a two-phase field: the solid gamma forms (Cu-enriched) and the liquid enriches in Sn. As you further cool down the gamma phase will continue to form, but with progressively higher content of Sn
You cool down at 640 °C and you enter a new different two-phase field: the liquid partially reacts with the gamma phase forming the solid epsilon phase. If you further cool down, the ε phase will continue to form from the liquid, but its composition will progressively enrich in Cu.
Just below 415 °C you enter into another two-phase field: the liquid will solidify into ε and η phase.
That's what occurs by cooling down to 300 °C, but if you further would cool down below 189 °C you will enter another two-phase field; in this case the ε and η' phases will form.
The amounts of the different phases at the different $T$ are obtained by applying the lever rule.
Take care, I described what happened from the thermodynamic point of view (that is what is represented in a phase diagram); the kinetics can remarkably change the things.

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Anti Markovnikov Product with Oxymercuration - Demercuration
Consider the Oxymercuration - Demercuration of the following compound with $\ce{Hg(OAc)2}$ followed by treatment with $\ce{NaBH4}$ in basic medium:

Here, if we go according to Markovnikov addition of water to the double bond, the $\ce{OH}$ group should get attached the carbon having the 2 methyl groups

But if go by the mechanism, we find that a more stable intermediate is formed when the partial positive charge lies on the carbon having the phenyl group as a benzyl carbocation is considered to be more stable than a secondary carbocation and thus we could very well expect an $\ce{OH}$ group on the carbon having the phenyl group.

So which one would be the major product in this case - The Markovnikov or Anti-Markovnikov product?

In my textbook, Organic chemistry by david klein it's said that the mercunium ion intemrediate involves a three memberd ring w/ resonance. This lead me to prefer benzylic position but I asked some of my friends doing o chem degrees and they told this wans't standard.

Does this answer your question? Why does water attack the more substituted carbon in oxymercuration?

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Who coined the term "gas evolution reaction"?
I found a paper from 1812 which specifically mentions the evolution of a gas [1].
There is a similar reference from 1808, except it uses the term evolution to speak of green light emitted when a material touches a green wire [2].
Is there anyone who knows for sure when the first time “gas evolution reaction” and the “gas evolved” and other such uses for the term “evolved” appeared in chemistry?
References

Henry, W. LX. Additional Experiments on the Muriatic and Oxymuriatic Acids. The Philosophical Magazine 1812, 40 (175), 337–343. DOI: 10.1080/14786441208638245.
Scholes, J. XI. On the Union of Gases. The Philosophical Magazine 1808, 31 (121), 68–70. DOI: 10.1080/14786440808563850.

There is no specific term as "gas evolution reaction" in modern chemistry but the word evolve itself conveys this meaning of emission. I am curious as to why you are interested in this terminology? Just for the sake of historical interest, the unabridged Oxford Dictionary (subscription required) shows that the term dates back to 1770s in relation to gas, heat, light etc. However there is no guarantee that the following references are the earliest. If you have more energy or you are a serious scholar of history, you might need to see German and French literature before the 1770s and what words Lavoisier used when he heated the oxide of mercury and discovered oxygen.

IV. The process of emission.

Chiefly Chemistry. The emission or release of gas, heat, light, etc.; the disengagement of a substance from combination; an instance
of such a process. Cf. evolve v. 5.

1783   Philos. Trans. (Royal Soc.) 73 350   The method put in practice
by Mr. Hutchins to settle the freezing point of quicksilver depends
entirely upon this generation, re-appearance or evolution of heat.
1800   H. Davy Res. Nitrous Oxide 425   This is on account of the
increased evolution of nitrogene from the blood.

Similarly, evolve is even older

1772   [implied in: J. MacLurg Exper. Human Bile 38   The colouring
matter of the Bile discovers also a much stronger attraction to the
mineral acids, and gives indubitable marks of a more evolved
phlogiston. (at evolved adj. 2)].
1786   Edinb. New Dispensatory i. 7   In the incipient state a
quantity of gas is still evolved, and along with it a quantity of
alkaline air.


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Molar conductivities of aqueous electrolytes at 1 M
In the following table from the CRC Handbook, molar conductivities of aqueous electrolytes are given for concentrations from infinite dilution to 0.1 M.  However, I am looking for values at 1 M, and I'm having a hard time finding that information.  (The following table probably stops at 0.1 M because the Debye-Huckel-Onsager equation is only accurate for very dilute solutions.)
This is a soft question, I admit, but can anyone suggest resources for finding tables of molar conductivities of aqueous electrolytes (i.e., typical ones like $KCl$, $KOH$, $H_2SO_4$) at 1 M?

Go to Google Scholar and search the title "Transport in Concentrated Solutions of 1:1 Electrolytes".


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Why does my drierite turn bright bright blue, instead of pink?
Drierite a brand name of a commonly-used, calcium sulfate-based dessicant in many chemistry labs.  The lab I work in has a dessicator box we use to store hygroscopic materials.  It isn't a a true dry box, just an air-sealed box with a big tray of dessicant at the bottom.  A cheap humidity meter we bought suggests the humidity gets down to 25% or so in there.  We store minute quantities of amine solutions in the box, and some hygroscopic dry materials.
Most drierite pellets are something like pure calcium sulfate, and are therefore a colorless white.  But some pellets are laced with cobalt chloride, a water-sensitive "indicator" of drierite freshness.  Supposedly, as the drierite absorbs moisture, the cobalt chloride turns from a sky blue into a pinkish color.  When it turns pink you know it's time to change the dessicant to a fresh dry batch.
But our drierite never turns pink.
Instead it turns into a far, far more intense, vivid blue color, much bluer and less pale than indicator drierite starts out as.  (See photo of too-blue drierite in tray, with a bottle of fresh starting indicator drierite at left.)

Why doesn't my drierite turn pink?. Why does it turn this shade of blue?  Presumably some other kind of complex or hydrate rather than the classic cobalt chloride hexahydrate is forming.  Is it an issue of pH?  Or of some volatile organic being adsorbed?

I have seen similar behavior of Drierite of acquiring a deeper blue color instead of the expected pink color in dessicators. Most likely your amines have significant vapor pressure. If you recall most amine complexes of cobalt are really deep blue like the Schweizer's reagent. You can try adding a drop of a couple of test amines in separate cobalt chloride solutions and see what happens.

OT why do you store the amine solutions in that box. Just because you manipulate them therein?

@M.Farooq Why not an answer?

@Alchimista, Because I want the OP to update the readers if his amines actually turn cobalt chloride solution blue. If they don't then this conjecture might be wrong.

They are amine solutions in DMSO, not water.  Some of the time, we use them in chemical reactions where water is detrimental.

But amines must have some vapor pressure in DMSO. This over-bluing effect takes month(s) and it is not instantaneous.

Support of M. Farooq's theory: in addition to the Drierite I put in some silica-supported P2O5, and now the Drierite does not turn this color of blue.  I guess the amine is much more preferentially attracted to the acid anhydride P2O5?


text:
Does correct bond lengths and angles imply that water is in a potential minimum?
So I come from the math community (so please disregard my ignorance on the topic).
If I have a water molecule and write the potential energy surface associated with it, my understanding is that I will find that the atoms configuration minimizes the potential energy function, so it will be a critical point of the potential energy surface.
My question is: suppose I have the building blocks of water, i.e 2 hydrogens and an oxygen, if it were the case by some magic, I added energy to the system that resulted the correct bond length and angles that correspond to water. Does that mean I have the water? That is to say ,geometrically, that I have taken a trajectory from the original potential minimum of just the original building blocks to a new potential minimum by adding energy. Since this new configuration has the correct bond length and angles, and potential energy, does that mean this atom configuration must indeed be water?
Thank you.

A water molecule (in its equilibrium geometry) has far less energy than three isolated atoms. So you should be taking energy away, not adding it. Now, if you’re saying that you hurl three atoms together and they coincidentally happen to form water, then that’s ok, but your water molecule would need to have a lot of kinetic energy to ensure that energy is conserved. In practice it would likely fall apart right after that. I know I’m not addressing your actual question, but I am slightly confused by why you’re adding energy.

hey @orthocresol thanks for answering me - I was thinking of the classical experiment where you apply a flame to hydrogen and oxygen to form water, for example. In that instance I would think you would be adding energy to the system, and for energy to be conserved, the hydrogen explodes causing water.

Ah, ok. Well, in that experiment the energy still goes somewhere – I presume it heats the water up. In that experiment, there aren't any atoms lying around: all the hydrogens are $\ce{H2}$ and all the oxygens are $\ce{O2}$, so it's a bit different from the picture that your question evokes, which is more like having two individual atoms of hydrogen, plus one individual atom of oxygen.

Now, I think, if I am understanding your question correctly, that the answer is "yes". The water molecule is defined precisely by it having three atoms in that particular geometry, so "having a molecule of water" $\Leftrightarrow$ "having two hydrogen atoms and an oxygen atom in that particular geometry". I don't know if any chemistry book would explicitly state that, but that is definitely the understanding amongst chemists.

This is exactly what I was looking for - the forward forward direction is obvious to me, but I wasn't sure if things were clean cut the other direction. Thank you so much!

It is not quite as simple. Geometry plus electron density defines the molecule; geometry alone doesn't. If your orbitals aren't set right, then the molecule with correct bond lengths and angles might be in a forced and unstable state, which will fall apart the moment you turn off the magic.

The point is that by some magic makes everything possible. But in this case you won't be able to minimise anything. It is just that you, by chance, putted the right energy to the undefined dots. If you use real atoms, no. It is a bit hard to follow your reasoning. See Ivan Neretin above, too. If they are real H and O than it can be a water so excited that it does not exist but seen in a geometrical minimum.

Hi @IvanNeretin and Alchimista, thanks a lot for adding onto the comment. Specifically, pointing to DFT as being important - it was on my radar, but knowing that a correct geometry does not necessarily imply the orbitals are correct is an extremely helpful comment to know. From your comment Ivan, does looking at this problem from a classical perspective, rather than a quantum chemistry perspective, spell a recipe for doom?

Depends on what doom means for you, but one can set up rather "classical" equations $E(angle, bondlenghts)$ that describe the hyperpotential around the energy minimum quite accurately. Of course to predict it, you need the full-blown QM approach.

@IvanNeretin Fair, but having talked about nuclear positions and electronic states, should you not also then start considering vibrational states too?

I guess, perhaps the distinction is whether you want to identify it as water (as per your last sentence), or a potential minimum (as per title). I'd still argue that the atoms in the right geometry do specify water, but I definitely agree that it isn't necessarily a minimum (in general it would be some highly unstable state that will dissociate, as mentioned by others).

Perhaps the right geometry for a time substantially longer than the vibrational frequency. I can imagine an OH molecule in a neutral atomic hydrogen beam, with a hydrogen just skimming by and being in the ‘right’ geometry the moment you look at it. Yet it is not ‘water’ in any real sense.


text:
What affects the vapor pressure of a pure liquid?
So pretend we have two samples of pure benezene in equilibrium with its vapor. One sample containing 0.5 mol and one containing 1 mol. They're also both at the same temperature. Let's pretend they're both at, say, 298 K (room temperature). Would the two samples have the same vapor pressure? I was thinking since the 1 mol sample has more molecules that could be vaporized, it would produce a greater pressure. But I was told this is wrong.

True, 1 mol of liquid has more molecules that could be vaporized, but then again, 1 mol of vapor has more molecules that could be condensed. It is a two-way street, see?

The rate of evaporation is proportional to the area of the liquid/gas boundary, not to liquid mass. And so does the condensation.

The macroscopic evaporation rate will be higher for a larger surface, as by Poutnik comment. And, as you are thinking on number of molecules, more benzene can simply fill of vapour a bigger container.

According to Raoult's Law, vapor pressure of a component in a solution is directly proportional to its mole fraction. In your case mole fraction is equal to 1(since benzene is the only component), which means that both produce the same vapor pressure.

One doesn't need ideal stuff to state that a pure liquid has a characteristic vapor pressure at given conditions. Not that the answer is wrong, but it took an unnecessary path.

Vapor pressure of pure solvent is a characteristic of the solvent which depends on the conditions and type of solvent only, Is this answer better?

Well, my comment wasn't to be pedantic. I would rather call that "vapour pressure of a pure substance...." and it is all. No one expect that melting of a ton of stuff happens at higher T than that of melting a kg. That is the point OP should have clear. Afterwards s/he could move to other cases. You can open stating the simple fact. Then go saying" it can also be seen using Raoult law with a component having x = 1.........

The vapour pressure under the circumstances of fixed T and n (amount of substance) depends on an additional constraint: either the volume of the container (if rigid) or the value of an applied external pressure.
In a rigid container (controlled fixed volume), as the volume is increased more of the liquid vaporizes but the vapor pressure remains constant until all of the substance is in the gas state. Past that point the pressure will decrease with increasing volume according to Boyle's law.
Under controlled constant pressure there will be vapor present if the applied pressure is smaller that the equilibrium vapor pressure at that temperature. At the boiling point (when applied pressure is equal to the applied pressure) one can transiently alter the applied pressure to allow increased evaporation, thereby increasing the volume for instance, but retaining the same intial and final pressures, at least until all of the substance has evaporated.
All that changing the amount of substance does is change the extent to which the volume will change when the liquid is converted entirely to gas. Between that point and the point at which there is no gas (because applied pressure is higher than vapour pressure, or because available volume is too small, which amounts to the same thing) the vapour pressure will be constant.

Does it mean that in a two phases system P and V are no longer conjugate variables of the entire system since one can control/fix both P and V? If so, how can this be consistent still with the mathematical structure of thermodynamics?

@TheQuark In a two-phase system the pressure would presumably be the same in both phases, if they are to be in mechanical balance. You could set the relative volumes of the two phases arbitrarily by adding or removing energy, provided the (T,P) values correspond to a point on the liquid-vapor coexistence line.


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Do antimicrobial abilities of copper boilers decrease over time?
As copper boilers age and develop oxide buildup, does this lessen the copper's antimicrobial abilities in killing bacteria? I was thinking since that coating would be on the copper, there would be less contact between it and the water, thus killing less bacteria.

What kind of bacteria live in the boiling water ?  Monel  ( 70 Ni : 30 Cu) ship hulls repel barnacles ,etc, indefinitely ( at least 4 years as I remember Inco reports - Inco is now gone).

Blacksmith37: This question relates particularly to boilers. The latter are notoriously for scale formation.

@blacksmith37 Why boiling ? Are not boilers usually limited to about $\pu{60 ^{\circ}C}$ ?

@blacksmith37 well, it's a espresso HX boiler which I believe the temp is around 250 F. But keep in mind the machine is not on 24/7 so the water will cool down, reheat, and cool down, etc.

It is possible that the ability to release soluble copper ions could be reduced with age.
I am assuming, for example, that a very small amount of the cuprous oxide coating could react with, say, carbonic or hypochlorous acid (from chlorinated water) to create a poorly soluble basic salt.
However, with time, to quote a source, namely Handbook of Industrial Water Treatment, Boiler Water Systems, Chapter 12 Boiler Deposits: Occurence And Control Contact:

DEPOSITS:
Common feedwater contaminants that can form boiler deposits include calcium, magnesium, iron, copper, aluminum, silica, and (to a lesser extent) silt and oil. Most deposits can be classified as one of two types (Figure 12-1)...scale that crystallized directly onto tube surfaces
sludge deposits that precipitated elsewhere and were transported to the metal surface by the flowing water
Scale is formed by salts that have limited solubility but are not totally insoluble in boiler water. These salts reach the deposit site in a soluble form and precipitate when concentrated by evaporation. The precipitates formed usually have a fairly homogeneous composition and crystal structure.

So, this growing deposit could likely reduce the concentration of copper ions that in even micro quantities has been reported as having an oligodynamic effect.
For example, per a reference a 2011 work published in Applied Environmental Microbiology, Metallic Copper as an Antimicrobial Surface, to quote:

The antimicrobial activity of copper and copper alloys is now well established, and copper has recently been registered at the U.S. Environmental Protection Agency as the first solid antimicrobial material...In connection to these new applications of copper, it is important to understand the mechanism of contact killing since it may bear on central issues...

As to possible mechanics, the article further notes:

In these enzymes, copper serves as an electron donor/acceptor by alternating between the redox states Cu(I) and Cu(II) (15)... the redox properties of copper can also cause cellular damage. A number or mechanisms have been suggested. Reactive hydroxyl radicals can be generated in a Fenton-type reaction:

$\ce{ Cu+ + H2O2 -> Cu++ + OH- + .OH}$ (1)

The extremely reactive hydroxyl radical can participate in a number of reactions detrimental to cellular molecules, such as the oxidation of proteins and lipids (45).

So at least partially, suspected Fenton-like properties of copper could contribute to its reputed antimicrobial properties. However, this requires a surface presence of either metal copper or cuprous salt, but a growing scale coating may be inhibiting to this action.

it's a espresso HX boiler which I believe the temp is around 250 F. But keep in mind the machine is not on 24/7 so the water will cool down, reheat, and cool down, etc. With the oxide build up over time, would the water still contact the copper surface enough to retain the coppers antimicrobial benefits? Or would the oxide layer be a barrier?


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isotopic composition database for carbon isotopes
I am seeking an isotopic composition database for carbon isotopes, since I would like to make a figure showing the different isotopic compositions of different materials. The best woudl be if the values are given as $\delta$ values versus the international VPDB-scale (VPDB=Vienna Pee Dee Belemnite)
I already did some research but I only coudl find figures not the raw data, and extracting the data from figures is tedious and not that accurate. Is there any data base, that lists ranges and also the literature? What I have found so far is for example, all values are on the VPDB scale.
"marine carbonates" -6‰ to 4‰
"atmos. CO_2" -8‰  to -7‰


text:
Double Equivalence Point Calculations for Oxalic Acid
Question:
Oxalic Acid, H2C2O4 has two pKa values, 1.25 and 4.27. A 0.100 M solution of oxalic acid was titrated with a 0.100 M solution of HaOH. What is the pH at the second equivalence point?
a) 1.23
b) 5.60
c) 8.40
d) 12.52
I tried to make an ICE table with, but since I don't know the amount of solution in Liters, I am not calculate the number of moles used up in the equation. I have no clue how to approach this question, therefore I can't show any work. The answer key says C is the correct answer.
A solution would be helpful.
Thank you very much.

Even if you had no clue how to solve the question, you should show your work of attempting to get the clue.

BTW, the right answer is clear just while reading the question, without any calculation.

In acid-base titrations, the pH at the equivalent point is usually taken as the average value of the last p$K\ce{_a}$ preceding the equivalence point and the pH of the final NaOH solution. Here the last $p$K$\ce{_a}$ value is $4.27$. The $\ce{NaOH}$ solution has a pH equal to $13$. Average value : $(13 + 4.27)/2 = 8.64$. It is nearly $8.40$ as proposed in solution c).


text:
How is this possible in mole concept?
L = $6.022*10^{23}$
If I have one atom of oxygen , it has exactly L value.
Now 2 atoms of a oxygen have 2L value.(Note that I mean 2O and not $O_2$here)
Well , if we take a molecule made up of 2atoms or combined i.e ($O_2$It also has L value. How is that possible ?It has bigger size and probably more atoms.
Reference :Take a look at the table

What's an L value? I don't think it is a thing at all.

@IvanNeretin Just 1st image last line

Well, Avogadro number is just a number. Admittedly, a big one, but still a number. It has no magic powers. Substitute it everywhere with "thousand", and then the confusion must vanish.

As of 2019 the Avogadro constant is no longer based on 12 g of C-12. It is now just a specific number: $6.02214076 \times 10^{23}$.

@MaxW yes.This is right

In the US the  Avogadro constant is normally denoted as $\mathrm{N_A}$. In Germany $\mathrm{L}$ seems to be the norm. See IUPAC Goldbook Avogadro constant

@MaxW Be aware there is the tiny difference between the unitless Avogadro number and the Avogadro constant with unit 1/mol.

What is the title and author of that horrible textbook, and what is the intended audience?

ICSE class 10.Viraj F dalal is the author.2019 edition.@Karl

@Karl audience are students who are preparing for national level exam of ICSE board.@Karl

https://en.wikipedia.org/wiki/Indian_Certificate_of_Secondary_Education ? Figures. Isn't India still in the Commonwealth? Can't your officials let you use British books, instead of developing crappy homemade substitutes?

Sorry for the rant. ;-)

@Karl, Many Commonwealth countries allow O-levels and A-levels where British books are used. It is equivalent to grade 10, 11-12, but it is not popular among the masses https://en.wikipedia.org/wiki/GCE_Ordinary_Level#India

Sorry to see that this is your grade 10 textbook. This also means you have no choice but to use this book for passing public exams. Please don't use L or NA as a symbol.

If I have one atom of oxygen , it has exactly L value.

No. Atoms are discrete objects i.e., countable. If you have 1 atom of oxygen, it means 1 atom of oxygen. If you said 1 mole of oxygen atoms, then it is 6.022x10$^{23}$ atoms of oxygen.
Your L comes from Loschmidt number or constant. From Science Progress in the Twentieth Century (1919-1933) , 1933, Vol. 27, pp. 634-649

Loschmidt 's number, N, is defined as the number of atoms in gram-atom
or the number of molecules in a gram-molecule.More than eighty
different experimental determinations of this number have been made
[i], and as it is a basic atomic constant its most probable value is
of great importance in atomic physics. It is, therefore, the purpose
of this article to outline the main methods by which Loschmidťs number
has been evaluated, and to give some indication of current opinions of
its most probable value
Footnote
1 This number is frequently referred to as " Avogadro's Number/' the
term " Loschmidťs Number " being then reserved for the number of mole-
cules in a cubic centimetre of a gas under standard conditions.
Unfortunately, these designations are often interchanged. Avogadro's
important hypothesis on the identity of the numbers of molecules in
equal volumes of different gases at the same pressure and temperature
was formulated in 1811, and is appropriately associated with his name
; but Avogadro made no quantitative estimate of either of the
above-mentioned constants. The first actual estimate of the number of
molecules in one cubic centimetre of a gas under standard conditions
was made in 1865 by Loschmidt, and from this the number of molecules
(atoms) in a gram molecule (atom) was later evaluated. From the
quantitative view-point it thus seems preferable to speak of "
Loschmidťs number per gram-molecule (atom)/' and of " Loschmidťs
number per cubic centimetre," as is almost invariably done in the
German scientific literature. This terminology avoids ambiguity, and
has been adopted here.

You can see here your textbook is using the symbolism of 1930s or earlier.

A mole is just a number, but you have to specify what you are counting
Aside from problematic issues about how your textbook is explaining things, the central misunderstanding arises because you need to specify what you are counting when you say there is a mole of something.
If you are counting oxygen atoms then there are two moles of oxygen atoms in one mole of dioxygen gas. There is no magic or confusion if you specify what it is you are counting.
This is not a problem unique to chemistry. If you count eggs, for example, you might use a dozen eggs (a dozen is just a signifier for 12 of anything) to make one (large) omelette. There is no confusion if you are clear about what you are counting, omelettes or eggs.
A mole is just a number, like a dozen (but a lot bigger). Indeed, you could have a mole (the number) of moles (the small burrowing mammal) but this would be enough matter to form a small planet.


text:
Are CH₃COOH and HCOOCH₃ same?
Can I write the formula of Acetic acid as HCOOCH₃ or COOHCH₃ insted of CH₃COOH would they represent the same compound with same covalent bond?

No, they wouldn't. The rules are so trivial they almost never get explained in the textbooks, but still they are not immediately obvious from the formula alone. COOHCH3 makes no sense at all, and HCOOCH3 is a different compound.

$\ce{CH3COOH}$ is an acid called acetic acid or ethanoic acid. This is the main constituent of vinegar. $\ce{HCOOCH3}$ is a rarely written formula for a volatil liquid called methyl formate, boiling at 31.5°C, which has a not too disagreeable odor and can be synthesized in the lab.

Partly related: My answer to a different question

Related (possibly duplicate): https://chemistry.stackexchange.com/questions/33135/which-one-is-the-best-notation-for-sodium-acetate/33136?r=SearchResults

$\ce{HCOOCH3}$ would likely be rendered as $\ce{H-C(O)-O-CH3}$, the lone hydrogen atom is atttached to the carbonyl group instead of the methyl group being there -- in other words, methyl formate instead of acetic acid.
I suppose that $\ce{HOCOCH3}$ could be interpreted as having the methyl group and hydroxyl group attached to the carbonyl, but you're better off sticking with the widely used ordering $\ce{CH3COOH}$.


text:
Reaction of cyclohexene oxide with LDA

The question given above appeared in the 2016 Indian National Chemistry Olympiad (INChO). What I think is that the epoxide ring is being opened by a base-catalysed mechanism ($\mathrm{S_N2}$). I proposed the following few steps for this:

The diisopropylamide ion, which is a very strong base, would not be a good leaving group. To obtain the desired products, how could the mechanism then proceed? And if not, then what would be the real mechanism? I did not find any reliable information on the Internet regarding ring opening using LDA.

For the asymmetric transformation of cyclohexene epoxide to cyclohex-2-en-1-ol, see  Tetrahedron, 2002, 58, 4669. DOI:10.1016/S0040-4020(02)00373-3. Also see, J. Am. Chem. Soc., 1970, 92, 2064 for mechanistic studies, https://doi.org/10.1021/ja00710a045

Interestingly enough, I think for cyclohexene oxide you do usually get a simple elimination reaction, but other pathways are possible and with other epoxides can lead to different major products (in particular, one can deprotonate alpha to oxygen, then alpha-elimination to give a carbenoid). The best reference I found from 5 minutes of google is this review https://www.thieme-connect.de/products/ejournals/abstract/10.1055/s-2002-33635, but there might be something more recent out there.

Also, I will fix it soon, but you need to be more careful with the number of hydrogens on your nitrogen. If you look back at your scheme, you consistently have too many. Negatively charged nitrogen should only have two bonds, neutral nitrogen should have three. In some places you even have nitrogen with five bonds.

@orthocresol, thanks for pointing out the mistake in my scheme, and for the review, it seems to be very interesting.

Because of the steric hindrance of the two iPr groups, LDA is a very poor nucleophile. It is, as you note, a very strong base. The two products observed both arise from deprotonation of cyclohexene oxide.
Deprotonation of one of the methylenes alpha to the epoxide ring opens it to give the allylic alcohol.
Removal of one of the protons on the epoxide ring carbons opens the ring to the enol, which on quench gives cyclohexanone.


text:
What has been the accepted value for the Avogadro constant in the "CRC Handbook of Chemistry and Physics" over the years?
Some questions about the constant made me curious as to exactly how much the accepted value has changed over the years. I am pretty sure that I learned a rounded value of $6.023\times10^{23}$ about 1970 which would have been around the 50th edition. I know that sometime after that the accepted value rounded to 4 significant figures was $6.022\times10^{23}$. Of course since 2019 the accepted value has been fixed at exactly $6.02214076\times10^{23}$
It would seem that the CRC handbook is a reasonable source for the accepted value of the constant since it has been updated almost yearly.
Please update the table in the answer in the community wiki page with other values

You can find the value in past CODATA publications. This is probably the most reliable and easily found historical record.

See Table 1 in Becker, P. History and Progress in the Accurate Determination of the Avogadro Constant. Rep. Prog. Phys. 2001, 64 (12), 1945–2008. DOI: 10.1088/0034-4885/64/12/206. It covers year up to the year 2001 and is going to save a lot of time.

I dislike offsite image links since they disappear over time. So below is Table 1 from andselisk's comment on the question.
The paper itself is behind a firewall and I can't see it.

See Table 1 in Becker, P. History and Progress in the Accurate Determination of the Avogadro Constant. Rep. Prog. Phys. 2001, 64 (12), 1945–2008. DOI: 10.1088/0034-4885/64/12/206. It covers year up to the year 2001 and is going to save a lot of time. – andselisk

Couple of things about the table. (1) It shows individual values determined by experimenters with improving exactness. The CRC data and CODATA show weighted values. (2) the basis for the AMU changed in 1961 from oxygen-16 to carbon-12. I don't think that the table corrects for that. (3) I'm assuming that $\mathrm{u_{rel}}$ is an absolute value not a % value. // The particular article is behind a firewall that I can see.

Please edit the table to include missing data
Avogadro constant in the "CRC Handbook of Chemistry and Physics"

Edition
Years(s)
page
value
CODATA?

1  (Link)
1913
--
Not Given
--

2
1914

No

3 (Link)
1914
123
6.42(31) × 1023
No

4
1915

6.42(31) × 1023 (note)
No

5
1917

6.42(31) × 1023 (note)
No

6 (Link)
1917
286
6.42(31) × 1023
No

7 (Link)
1918
369
6.42(31) × 1023
No

8 (Link)
1920
523
6.06 × 1023
No

9 (Link)
1922
605
6.06 × 1023
No

10 (Link)
1924
678
6.06 × 1023
No

11 (Link)
1926
784
6.06 × 1023
No

12
1927
884
6.06 × 1023
No

13
1928
986
6.06 × 1023
No

14
1929
1151
6.06 × 1023
No

15
1930

No

16
1931
1244
6.064(6) × 1023
No

17
1932

6.064(6) × 1023 (Note)
No

18
1933
1488
6.064(6) × 1023
No

19
1934
1890
6.064(6) × 1023
No

20
1935

6.064(6) × 1023 (Note)
No

21
1936
1980
6.064(6) × 1023
No

22
1937
2026
6.064(6) × 1023
No

23
1939
2177
6.064(6) × 1023
No

24
1940
2519
6.064(6) × 1023
No

25
1941
2454
6.064(6) × 1023
No

26
1942
2403
6.0228(11) × 1023
No

27
1943
2499
6.0228(11) × 1023
No

28
1944
2517
6.0228(11) × 1023
No

29
1945
2585
6.0228(11) × 1023
No

30 (Link)
1946
2631
6.0228(11) × 1023
No

30(Rev)
1947
2631
6.0228(11) × 1023
No

31 (Link)
1949
2683
6.0228(11) × 1023
No

32
1950
2811
6.0228(11) × 1023
No

33
1951
2833
6.0228(11) × 1023
No

34
1952

No

35
1953
3092
6.0228(11) × 1023
No

36
1954
3104
6.025 44(11) × 1023
No

37
1955
3080
6.0228(11) × 1023
No

38
1956
3136
6.0247(2) × 1023
No

39 (Link)
1957
3134
6.0228(11) × 1023
No

40
1958
3376
6.022 8(11) × 1023
No

41
1959
3301
6.022 8(11) × 1023
No

42
1960
3399
6.022 8(11) × 1023
No

43
1961

No

44
1962

No

45
1964
F-95
6.022 57(9) × 1023
No

46
1965

No

47
1966-1967
F-138
6.022 52(28) × 1023
No

48
1967-1968

No

49
1968–1969
F-167
6.022 52(28) × 1023
No

50
1969-1970
F-168
6.022 52(28) × 1023
No

51
1970-1971

52 (Link)
1971-1972
F-189
6.022 169(40)  × 1023
No

53
1972-1973
F195
6.022 169(40) × 1023

54
1973-1974

55 (Link)
1974-1975
F-222
6.022 169(40)  × 1023
No

56
1975–1976

6.022 045(31) × 1023

57 (Link)
1976–1977
F-242
6.022 045(31) × 1023
Yes, bulletin #11

58
1977-1978
F-242
6.022 045(31) × 1023
Yes, bulletin #11

59
1978-1979
F-250
6.022 045(31) × 1023
Yes, bulletin #11

60
1979-1980
F-244
6.022 045(31) × 1023
Yes, bulletin #11

61
1980-1981

62
1981-1982

63
1982-1983

64 (Link)
1983-1984
F-198
6.022 045(31) × 1023
Yes

65 (Link)
1984-1985
F-192
6.022 045(31) × 1023
Yes

66
1985-1986

67
1986-1987

68
1987-1988

69
1988-1989

70
1989-1990
F-215 & F-217
6.022 1367(36) × 1023
Yes, 1986

71 (Link)
1990-1991
1-1
6.022 1367(36) × 1023
Yes, 1986

72 (Link)
1991-1992
1-1
6.022 1367(36) × 1023
Yes, 1986

73
1992-1993

74
1993-1994

75
1994-1995

76
1995-1996

77
1996-1997

78
1997-1998

79
1998-1999

80
1999-2000

81 Link
2000-2001
1-7
6.022 141 99(47) × 1023
Yes, 1998

82
2001–2002
1-7
6.022 141 99(47) × 1023
Yes, 1998

83
2002-2003

6.022 141 99(47) × 1023 (Note)
(Yes, 1998)

84
2003–2004
1-7
6.022 141 99(47) × 1023
Yes, 1998

85
2004-2005
1-2
6.022 141 5(10) × 1023
Yes, 2002

86
2005–2006
1-5
6.022 141 5(10) × 1023
Yes, 2002

87
2006–2007
1-5
6.022 141 5(10) × 1023
Yes, 2002

88
2007–2008
1-1
6.022 141 5(10) × 1023
Yes, 2002

89 (Link)
2008-2009
1-1
6.022 141 79(30) × 1023
Yes, 2006

90
2009-2010
1-1
6.022 141 79(30) × 1023
Yes, 2006

91
2010-2011

6.022 141 79(30) × 1023 (Note)
(Yes, 2006)

92
2011–2012
1-1
6.022 141 79(30) × 1023
Yes, 2006

93
2012–2013
1-2
6.022 141 29(27) × 1023
Yes, 2010

94
2013–2014
1-2
6.022 141 29(27) × 1023
Yes, 2010

95
2014–2015
1-2
6.022 141 29(27) × 1023
Yes, 2010

96
2015–2016
1-2
6.022 141 29(27) × 1023
Yes, 2010

97 (Link)
2016–2017
1-1
6.022 140 857(74) × 1023
Yes, 2014

98
2017-2018

6.022 140 857(74) × 1023 (Note)
(Yes, 2014)

99
2019-2020

6.022 140 857(74) × 1023 (Note)
(Yes, 2014)

100
2020-2021

6.022 140 857(74) × 1023 (Note)
(Yes, 2014)

>101?
2021-2022
?
6.022 140 76 × 1023 (exact)
Yes, 2018

NOTES FOR TABLE

Early editions are actually titled "Handbook of Chemistry and Physics". Not sure when the title formally switched to the "CRC Handbook of Chemistry and Physics".

Some early editions are available online at HathiTrust. The "Limited search" editions can only be searched for a hit. There is no page number or preview for the found text.

The first edition that I found to hit on the phrase "Avogadro number" was the 16th edition. It would seem to that phrase until at least the 31st edition.

By 84th edition the phrase was changed to "Avogadro constant"

A CODATA newsletter had a list of publications which listed: No. 11 (Dec 1973) 8 pp, Recommended Consistent Values of the Fundamental Physical Constants, 1973 (Report of the CODATA Task Group on Fundamental Constants, August 1973).

1st Edition
The link is to the facsimile edition.
3rd edition (c)1914
This version is online at HathiTrust. An old edit of the Wikipedia article  CRC Handbook of Chemistry and Physics indicates the third edition has 322 pages.
3rd thru 12th Editions
In at least the 3rd through 12th editions of the CRC Handbook in have a "Heat" category in the Table of Contents where there is an entry for "Number of molecules in a molecule-gram" which points to a table with three values. Note also that none of these three values are in Becker's paper. (Only seen 1st, 3rd, and 6th through 12th editions)

My calculation is that the mean = $6.42(31)\times 10^{23}$
Millikan 1910 would seem to be from rounding $6.18^\times10^{23}$ value in  Phil.Mag. Feb,1910,209

4th and 5th edition
See likely that these editions are same as 3rd and 6th.
8th through ? edition
In at least the 8th through 12th editions of the CRC Handbook the Table of Contents contains a category "Miscellaneous Tables" which has a link to "Atomic and Molecular Constants" which references the "Smithsonian Tables" and gives a value of $6.06\times10^{23}$ for the "Number of molecules in a gram molecule".
The reference to the Smithsonian Physical Tablesdoesn't give any more indication as to which tables those are. This seems to be the 2nd reprint of 6th rev. ed. Prepared by Frederick E. Fowle. in 1916
13th through and 30th(?) Editions
There is another version of the table Number of Molecules in a Gram-Molecule which has four values.

In the 13th, 14th, 16th, 18th, 19th, 23rd, 25th, 26th and 27th editions the table is listed in the table of contents in the Heat section.

In the 29th, 30th and 30th (rev) editions the table is listed in the table of contents in the Miscellaneous section.

The table is gone in the 31st edition (and presumably in any subsequent editions).

16th, 18th, 19th, 21st, 23rd, 24th, 25th and 27th Editions
These editions have identical entries, listing it as "Avogadro's number" in a table named "Probable Values of the General Physical Constants" which is in the "Miscellaneous Tables" section of the Table of Contents. The tables cite Raymond T. Birge: The Physical Review Supplement, Vol. 1, No. 1, July, 1929, as the source for the entire "Probable Values of the General Physical Constants" table. Note however that the value given here doesn't match the one attributed to R T Birge in 1929 in Becker, P. "History and Progress in the Accurate Determination of the Avogadro Constants"
17th, 19th, 20th, and 22nd Editions
These editions are book ended by editions which have the same value. These editions are assumed to have this value too.
28th, 29th, 30th, 30th (rev), 31st, 32nd and 35th Editions
The 30th revised edition has "Revised" above "Thirtieth Edition" on the title page
Listed as "Avogadro number (chemical scale)" in "New Table of Values of the General Physical Constants." The table is listed in the "Miscellaneous" section of the Table of Contents.
Cites: Raymond T. Birge: The Physical Review Supplement, Vol. 13, 233, 1941 as the source for the entire table. (This is the value listed for R T Birge in 1941 in Becker, P. "History and Progress in the Accurate Determination of the Avogadro Constants". With the quoted error the extra 3 is unnecessary.)

The reference is better known as: "A New Table of Values of the General Physical Constants (as of August, 1941)", Raymond T. Birge
Rev. Mod. Phys. 13, 233 (1941) DOI: 10.1103/RevModPhys.13.233

36th edition
Value found in table of "Atomic Constants" which references James W. M. DuMond and E. Richard Cohen, Physical Review 82:555, 1951
38th edition
Found in table of "Atomic Constants" which references "A survey of Atomic Constants (1955) by J. A. Bearden and John H. Thomsen, The John Hopkins University"
40th, 41st and 42nd Editions
This cites Raymond T. Birge: Reviews of Modern Physics 13, 233, 1941. This matches the value attributed to Birge in the same year by Becker. (Because of the error the extra 3 is superfluous so I left it off.)
45th Edition
"Avogadro's Number" in table "Values of the General Physical Constants" which simply states "Furnished by R. D. Huntoon."
50th Edition
Cites source for data in Table named "Physical Constants" as "National Bureau of Standards Technical News Bulletin, October 1963."
52th, 53rd and 55th editions
References Rev. Mod. Phys 41, 375 (1969)
56th edition
From user  theorist's comment on this post.
57th, 58th, 59th, and 60th Editions
Avogadro Constant in table Recommended Consistent Values of the Fundamental Physical Constants.
Table cites: Data from CODATA Bulletin No. 11, ICSU CODATA Central Office, CODATA Secretariat: 19 Westendstrasse, 6 Frankfurt/Main, German Federal Republic (copies of this bulletin are available at no cost from this office).
64th and 65th editions
Avogadro Constant in table titled Recommended Consistent Values of the Fundamental Physical Constants.
Table cites: Data from CODATA Bulletin No. 11, ICSU CODATA Central Office, CODATA Secretariat: 51 Boulevard de Montmorency, 75016 Paris, France (copies of this bulletin are available at no cost from this office).
70th through 72nd Editions
Cites 1986 CODATA values as published in E. Richard Cohen and Barry N. Taylor, Journal of Research of the National Bureau of Standards, 92, 85, 1987.
83rd edition
Based on pattern of CODATA being used for 4 years.
84th edition
calculated page number from on-line pdf file
85th edition
Internet edition
85th edition
Got value via "Look Inside" at Amazon
90th edition
Based on the 2006 CODATA recommended values
91st edition
Surely same as 90th and 92nd data. Pattern of using CODATA for four Years.
92nd edition
From print copy (MaxW)
98th through 100th Editions
Based on pattern of CODATA being used for four years.

The 56th edition (1975-1976) has it as 6.022045(31) x 10^23. Since the uncertainty gives the estimated value an upper bound of 6.022076 x 10^23, it seems they underestimated the uncertainty. The uncertainty they'd need to encompass the value at the redefiniton is 6.022045(97) x 10^23.

I chose the CRC handbook because it has been published for about as long as scientists have been able to determine a reasonably accurate value, and it is updated yearly. Given the CODATA values it seems worth noting CRC's reliance on that data, so added a column to track that.

@theorist - Typically the error number is the standard deviation. So  (6.022140 - 6.022045)/0.000031 =  3.06 standard deviations which doesn't seem horrible.

3.06 stand deviations means that, according to their predicted uncertainty, there was only a 1:500 chance that the actual value would be that far from their predicted value.  If you subsequently asked those who provided that value about this, I think they themselves would acknowledge they significantly underestimated the uncertainty.

The Committee on Data of the International Science Council (CODATA) was established as ICSU Committee on Data for Science and Technology in 1966.
Avogadro constant, CODATA recommended values

Year(s)
page
value

1969
477
6.022 169(40) × 1023

1973
718
6.022045(31) × 1023

1986
6
6.022 136 7(36) × 1023

1998
6
6.022 141 99(47) × 1023

2002
6
6.022 141 5(10) × 1023

2006
7
6.022 141 79(30) × 1023

2010
5
6.022 141 29(27) × 1023

2014
5
6.022 140 857(74) × 1023

2018
5
6.022 140 76 × 1023 (exact)

NOTES FOR TABLE
2018 - exact by new definition of constant

I wonder if it would be helpful to note that the definition of mole was standardized to a new definition in 1971: https://en.wikipedia.org/wiki/Avogadro_constant#SI_definition_of_1971

@Zhe - The values do need to be corrected. In 1961 the amu was based on 1/12 of a $\ce{^{12}C}$ atom, so 1 amu was $1.0\bar0$ g/mol. With the 2018 suggested values now accepted, the molar mass constant is now 0.999 999 999 65(30) g/mol. So if I figured this correctly the values from 1969 to 2014 need to be multiplied by 0.999 999 999 65.

@Zhe - pragmatically the 0.999 999 999 65 correction only changes the value if there are at least 9 decimal places in the value.

Oh, yes, practically, there's no difference, but it just felt weird to juxtapose values with (slightly) different meanings.

Here are plots of the historical CODATA and CRC values for $N_{\text {A}}$ listed in MaxW's posts.  Plots were made in Mathematica.
I've provided plots with successively later lower bounds for the dates, and correspondingly smaller $N_{\text {A}}$ ranges, so that the shrinking uncertainties in later years can be visualized.
N.B.: When the CRC value had a two-year range, I averaged the years (e.g., 1975-76 became 1975.5). When the CRC values were repeated in successive editions, I plotted only the earlier value.
The last three of the four CRC values shown each correspond to the CODATA value from 3–5 years past.  This suprised me—especially with its emphasis on yearly updates, I would have thought the CRC would be kept more up-to-date than that.  I wonder if this is generally the case for data in the CRC—that data updates stop a few years before the publication date, to give time for proofreading, typesetting, etc.
1998 CODATA => 2003–04 CRC
2002 CODATA => 2005 CRC
2006 CODATA => 2010 CRC


text:
Why isoelectronic species give similar spectral lines?
In my perspective those species should give similar spectra those which have same energy of different energy levels as the transition would yeild same energy photon
But still the answer to this question is always the isoelectronic species
Not the ions of the element

?? The spectral lines of $\ce{Na+}$ and $\ce{Ne}$ do not have the same energies.

But Mg+2 and Al+3 gives series of similar spectral lines (this was the question of my exam i.e.NEET)

RE: comment

Isoelectronic species will not have an identical matching spectrum" ok, but will they be similar but why the spectrum of Mg+2 not similar to Mg and why to Al+3 or Na+

This should really be another question be here goes anyway...
The answer is complicated.
Non-isoelectronic species
$\ce{Mg^{2+}}$ and $\ce{Mg}$ are not isoelectronic. Hopefully it will suffice to say that in quantum mechanics there are selections rules about what electron transitions between orbitals are "allowed" and which are "forbidden." The forbidden transitions do occur but with a very much reduced probability compared to the allowed transitions.
The gist is that because of the selection rules the pattern of lines for non-isoelectronic species will be different.
Isoelectronic species
Because of the pattern of allowed transitions isoelectronic species will have a similar parrern of spectral lines but at slightly different energies due to the difference in atomic number. Thus since they are isoelectronic  $\ce{Na+}$, $\ce{Mg^{2+}}$, and $\ce{Al^{3+}}$ have a similar pattern of spectral lines but at different energies.

There is also another factor in all of this. Slater's Rules give a rudimentary method for calculating shielding between orbitals. However quantum mechanics shows that the actual shielding mechanism is more complicated because the spatial distribution of the orbitals depends not only on the electron orbitals which are filled, but also on the atomic number of the nucleus.

You are completely forgetting the nuclear charge of isoelectronic series. Their valence shell configuration is the same but the electrons experience a different attraction in each. As you stated

$\ce{Mg^{2+}}$ and $\ce{Al^{3+}}$ give series of similar spectral lines
(this was the question of my exam i.e.NEET)

Yes the spectrum pattern is similar but not identical. Atomic spectrum and that of ionized atoms is like a fingerprint, otherwise identifying the elements in the stars will become difficult and perhaps impossible. There is certainly some overlap in atomic spectrum but not identical as your question title suggests. A quick search of Google shows that this non-sensical question is quite popular in local exams. It is wrong. Isoelectronic species will not have an identical matching spectrum. Even the wavelengths are different.
Search this title in Google Scholar to see the wavelengths of isoelectronic series, where it available free of charge. None match with neutral sodium spectrum.
Wavelengths and levels of the Na I isoelectronic sequence from K IX through Mn XV
Journal of the Optical Society of America Vol. 66, Issue 9, pp. 899-904 (1976)

Yes, similar $\ne$ same...

" Isoelectronic species will not have an identical matching spectrum" ok, but will they be similar but why the spectrum of Mg+2 not sumilar to Mg and why to          Al+3 or Na+

This must have a deeper quantum mechanical explanation. I do not know. Keep in mind, after hydrogen multielectron atoms are very challenging to deal with theoretically.


text:
A trick to find resultant pH
Consider the following problem:

Find the $\mathrm{pH}$ of the solution formed by the mixing of two solutions of $\mathrm{pH}$(s) $2$ and $3$ of equal volumes.

The normal way:
Since the resulting solution will have twice the volume, the new $\mathrm{pH}$ will be:
$$-\log\bigg(\frac{10^{-2}+10^{-3}}{2}\bigg)=-\log\bigg(0.0055\bigg)=\boxed{2.26}$$
Interestingly, my book mentions a trick:

Find the average $\mathrm{pH}$ of the two solutions and subtract $0.24$ to get the actual $\mathrm{pH}$.

This trick surprisingly works for many problems with a remarkable accuracy. I'm curious to find why this method works.

Answered here in detail: https://chemistry.stackexchange.com/questions/7051/why-does-this-shortcut-work-for-calculating-ph-proof-required

@NicolauSakerNeto thank you. It solved my query. Unfortunately I was not able to find this question before.

The hidden part of this trick is that the volumes of solution A and B must be equal. Just be careful when you apply it.

The question, in the way it is formulated, has infinite number of possible solutions in near the whole pH range 2-3. I guess it implies, what should be stated explicitly, that both solutions are diluted strong acids.

This "trick" seems to me to be something that might be useful in a Chemistry Olympiad competition but useless in real laboratory work.

@MaxW of course. Students dont really get access to log tables in exams

@newbie105 - LOL, I did have log tables. When I started chemistry there were no such things as hand-held calculators. We used a sheet of log tables or in my case I was fortunate enough to have a Post Versalog sliderule in high school.

The "trick" is valid accidentally for particular pH difference equal to 1 for the same volumes of diluted strong acid solutions. It is mathematical consequence and has no direct relation to chemistry. It has no practical value.


text:
Does electron mass decrease when it changes its orbit?
I have studied  this in my chapter atomic structure that when an electron changes its orbit from lower energy to higher energy state , it does not state in my book that it moves there but that it disappears.It just kind of reaches that orbit without crossing the path.We don’t know whether it moves there or not.
So , I also read about the fundamental properties
of electron in which it said that when an electron moves or is in motion , it’s mass changes.
If we see , We know that electron only absorbs certain wavelength of light. If we can find that when it changes its orbit , if it’s mass decreased or not during that interval.Is it that we can find that whether  electron moves and changes orbit or it just disappears kind of.
Yes , I am only talking electron in an hydrogen atom here.
Please do give your answer in reference to other atoms if it is helpful.

What is called an "electron orbit" is not something like a planetary orbit but a quantum state. Quantum states mostly escape human intuition.

Ohk.what I did write about is neils Bohr theory of atomic structure.We also know that his theory did came right without much explanation but with experiments.

Note that an electron need not to change its location to change its quantum state.  The wave functions aka orbitals  with different quantum numbers spatially overlap.

See https://physics.stackexchange.com/questions/149744/does-the-mass-of-an-electron-change-with-its-energy-state

I'd like to copy the answer by John Rennie to a similar question at Physics.SE, since it's much better than current answers here IMHO. Although it mostly speaks about binding energy, the same principle applies to the quanta of excitation energy.

The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron together then they come to $1.67353272 \times 10^{-27}$ kg. The difference is about 13.6eV, which is the ionisation energy of hydrogen (though note that the experimental error in the masses isn't much less than the difference so this is only approximate).
This shouldn't surprise you because you have to add energy (in the form of a 13.6eV photon) to dissociate a hydrogen atom into a free proton and electron, and this increases the mass in accordance with Einstein's famous equation $E = mc^2$. So this is a direct example of the sort of mass increase you describe.
However you can't say this is an increase of mass of the electron or the proton. It's an increase in mass of the combined system. The invariant masses of the electron and proton are constants and not affected by whether they're in atoms or roaming freely. The change in mass is coming from a change in the binding energy of the system.

Yes, this is an important point, the masses of the particles don't really change. If they did, then fundamental particles like electrons would cease to be indistinguishable (you could distinguish them by their different masses!), and this would wreak incredible havoc in their quantum mechanical description.

Wanted to post the same link as Ruslan. Comment is to stress that this are confusing/difficult concept and this answer picked up a nice one that I recommend to OP

Does electron mass decrease when it changes its orbit?

Essentially yes. If you add the mass of a free proton and a free electron you'll get a greater mass than that of a hydrogen atom. The mass difference will be equivalent to 13.6 eV which is the ionization energy of hydrogen.
Now for any "practical" chemistry experiment the assumption is that masses don't change. The reason is that the mass change is so small that it is undetectable with ordinary chemical laboratory procedures.
There is also another weird effect here. For the very heavy elements the binding energy, and the orbital so small that the particle model of the electron has the electron traveling at relativistic velocities. This increases the apparent mass of the electron.

I disagree on purely formal aspects. The system as such surely changes energy, but I don't see any reason why or how to attribute it to the mass of the electron. Nitpicking, but has to been said.

It is a change in kinetic energy of the electron, too, so we actually know that the mass is changing if we consider relativistic effects.

@Greg your comment use the relativistic mass idea which is confusing. Better to stick to system atom (nucleus and electrons) and say energy input - - > increase in rest mass, as it is for a bottle of gas.

In spite of my comment @Greg, essentially the energy goes to the electron because its mass is small. However the answer is nice as it is about proton and electron. It says "essentially"...

@Alchimista There is also a tiny difference between invariant and rest mass, as the former is not necessarily the latter.

@Poutnik not clear what you mean if not switching between single particle view - system. The latter case is what I have meant. Invariant and rest mass coincide for the bottle of gas at rest.

It would be prudent to clarify on the definition of "mass", in this particular answer, as mentioned in the physics forum. Is it "rest mass", "inertial mass", or "gravitational mass", or something else?

The Bohr Model tried, quite successfully for its time, to model the energy states of an electron.
This model has turned out inadequate, as it cannot answer question like yours.
There have been more refined models (where your question isn't possible), but they all have their drawbacks.
Not an answer to what you were asking, but some kind matching your question.

In my younger days I was fooled be thinking of hydrogen atoms as of planets orbited by  a moon. This would have given them a disk-like 2-dimensional shape with really strange properties.

We have quantum and Newtonian mechanics.Is it that my question is of point of view from Newtonian and not quantum mechanics.Conclusion : Bohr model is actually wrong and so answer does not exactly become valid since it is mostly on the basis of only Neils Bohr theory

@srijannahar The question as in the title still makes sense in context of the Special Relativity and relativistic wave equations.

@Poutnik yes, but with very different aspects.

@GyroGearloose Sure. In context of non-relativistic quantum mechanics, Schroedinger wave equation and classical Newtonian mechanics, an electron mass is constant.

The emphases is on “for its time”. I wish schools stop teaching old theories without context, just because they want to put two more pages in their textbooks to show off.


text:
Why do different substances containing saturated hydrocarbons burns with different flame?
LPG and Candle wax both contain saturated hydrocarbon, yet LPG gives a clean white flame but a candle gives yellow flame with lots of smoke. Why so?
I have read some answers on web which says that it is due to the insufficient supply of oxygen, so I tried burning candle beside LPG ( so there is no difference in supply of oxygen for both candle and LPG ) , but  still candle gives a yellow flame. Why?

Don't know for sure, but LPG, as a gas, is less prone to have traces of Na with it. See https://en.wikipedia.org/wiki/Flame_test.

I did upvoted all the three answer to date as for each point to different points all at play here...

Candle flames need a wick. Probably made of cotton, containing at least small amounts of Na, enough to give a yellow shine.

The yellow color of the candle flame is absolutely unrelated to sodium (Na).

The whole point of a candle is to emit light. So the burning system (ie a particular wax plus a wick) is designed to burn "poorly" and emit a good dose of soot which heats up in the flame and emits a lot of light. An LPG flame (or a butane cigarette lighter) is not designed for light, but heat. There there design mixes oxygen and the hydrocarbon well giving complete efficient combustion and very little light or soot.

LPG is propane, butane or a mixture of both. Paraffin candle wax is $(CH_2)_n$

typically $C_{31}H_{64}$ (other, nonparaffin, waxes are also used in candles, like stearin, beeswax, etc.)
However, LPG is a gas from the start and can mix with air (i.e., oxygen) before it begins to burn, while candle wax has to be heated by the flame before it can begin to burn, so the outer periphery of candle wax vapor (the part of the vaporized candle not yet heated up to ignition) is very small - a dark zone surrounding the wick. The possibility of contact with oxygen is limited to this area.
You will notice a blue zone around the wick also - perhaps obscuring the dark zone closer to the wick. This is a hydrogen-rich flame; under ideal conditions, you can burn off all the hydrogen and leave the carbon (put a glass into the flame: it will get coated with black soot). But if you don't cool the flame with a glass, the carbon which would otherwise deposit will burn yellow.
So, "insufficient supply of oxygen" is an OK answer, but maybe a better one is "scanty mixing of candle wax vapor with oxygen". I suspect that if you boiled candle wax rapidly in a closed container with a vent, and ignited the gases coming out of the vent, they would burn much more like LPG and less yellow.
The colors are dependent not only on line spectra of the atoms, but also on black body radiation from solid carbon. In the candle flame, by the time the carbon get access to oxygen, its temperature is not as high as in a LPG flame, and the possibility of solid carbon particles burning increases the amount of yellowish light emitted. Kind of like charcoal burning, but on a much smaller scale.

James, I don't think there is such a notion as hydrogen rich flame in candle flames. The blue color of candle flame is from small diatomic carbon based species.

"Hydrogen burns with a pale blue flame that is almost invisible" https://h2tools.org/bestpractices/hydrogen-flames. However, the idea of diatomic carbon forming, then burning before agglomerating to soot (nanotubes or graphene) is also quite interesting! Another link https://candles.org/candle-science/ actually calls the blue flame oxygen rich (!) because that's where the hydrogen burns off. I would compromise and call it the first-burn zone, but that's where the hydrogen gets first dibs on the oxygen; carbon is next in line and yellow.

Yes, it is true that hydrogen burns with a blue or sometimes near invisible flame but as I stated earlier, the blue color of a hydrocarbon flame is not due to hydrogen or hydrogen rich zone but due to small diatomic carbon species. Those candle.org type website are not reliable for scientific purposes.

if you boiled candle wax rapidly in a closed container with a vent, and ignited the gases coming out of the vent — This is more or less what steam boilers designed to burn heavy fuel oil do. I don't know what those flames look like, but they do manage to get closer to complete combustion that way.

The existence of C2 molecules in the blue flame has been proven by Herzberg, when he analyzed the line spectrum of this blue light. The found lines which are separated by exactly the vibrational energy of a C2 molecule.

I have read some answers on web which says that it is due to the insufficient supply of oxygen, so I tried burning candle beside LPG ( so there is no difference in supply of oxygen for both candle and LPG ) , but still candle gives a yellow flame. Why?

You cannot conduct this experiment by yourself without using a proper apparatus. Otherwise comparing the combustion of a solid wax which is slowly melting by the heat of the flame vs. LPG, is a comparison of apple and oranges.
What matters is the proper fuel to air ratio to determine the color of the flame. In atomic spectroscopy, one uses acetylene $\ce{C2H2}$, a small hydrocarbon. It burns with a beautiful blue flame when acetylene is pre-mixed with a proper ratio of air (i.e., air flow is higher). Sometimes we need a yellow flame, and all that is needed is to reduce the air flow. Most hydrocarbons should burn with a blue flame in an adequate amount of air.

A propane molecule $\ce{C3H8}$ in the gas phase must find $7$ molecules $\ce{O2}$ to burn totally. A wax molecule $\ce{C_{31}H_{64}}$ must find $63$ molecules $\ce{O2}$ to burn completely.
During its trip through the flame (having maybe $10 - 20$ cm length), the propane molecule will have enough time to meet $7$ molecules $\ce{O2}$. The wax molecule may not be lucky enough to meet enough molecules $\ce{O2}$ and also simultaneously  to get rid of the $32$ $\ce{H2O}$ molecules (and $31~\ce{CO2}$ ) formed in the reaction to burn completely during the trip through the flame. As a consequence, the first atoms of the wax molecule to react with $\ce{O2}$ are the outer atoms, which are mainly Hydrogen atoms. The probability of producing long chains of carbon atoms in the flame between the lower part and the upper part of the flame is high. These chains of carbon atoms are hot but solid and they emit light as every heated solid.

+1 I'm really surprised that no one has mentioned the bunsen burner, which provides a simple demonstration of how restricting the oxygen supply can cause even methane to burn with a yellow flame.

A flame light is emitted by two mechanisms:

An electron transitions in free molecules, giving off a linear spectrum with rather large optical depth. The intensity and the grouping of these lines in the usual flames is perceived as blu-ish light and the large optical depth makes the flame transparent.
A blackbody-ish radiation of soot. (Other possible solid particles, like magnesium oxide in magnesium flame behave more or less the same). It is usually red, orange or yellow, depending on the temperature of the flame. It usually has small optical depth, meaning the flame is bright and opaque.

You always get the first mechanism. Depending on the conditions, you may or may not have soot in the flame or you may have soot only in some parts of the flame. If you make a flame without soot, it is blue and transparent. Perfect if you want heat and no light. When you have both mechanisms in action, the second one usually gives way more light and you don't see the blue color.
How to make a flame without soot: mix the fuel well with enough air or oxygen before it gets in the flame. That's how a LPG-powered stoves work. It doesn't really matter what type of fuel you use, gasoline or diesel or even wax vapors can burn with a blue flame as well. Most modern internal combustion engines do just that, because light emission is pretty much unwanted in the combustion camera.
How to make a sooty flame: light a jet of a gaseous fuel and let the flame itself mix the fuel with air. It usually doesnt do the best job. In a candle, the jet is the vaporized wax around the wick. The flame forms zones rich in fuel that burns only partially. The unburnt carbon forms soot. Again, it doesn't really matter what type of fuel you use. Propane can burn with a candle-like flame pretty well, just block the air inlets below the stove.
How to form a perfect candle or kerosene lamp flame: you want soot only in a part of the flame, it should burn completely as it advances up in the flame. Too much exposed wick and some soot will leave the flame without burning completely. The flame will smoke. Too less exposed wick and you get too little soot and too little light. In candles, the wick tickness and material is choosen in a way that it burns off at near the right length. In a kerosene lamp, you have a wheel that regulates the exposed wick surface.
How some substances don't make yellow flame when burning (e.g. alcohol): a combination of a fast diffusion, comparatively low need of oxygen for a unit of fuel and a chemical mechanism of burning that doesn't favor free carbon and soot formation. But in specially aranged conditions it is still possible to make alcohol burning somewhat yellow. Other fuels (like, hydrogen gas) simply don't contain carbon at all so no soot there, no matter of the burning conditions.

It's not just the light emission that is unwanted in a typical internal combustion engine. The soot itself is something you really want to avoid, as it would coat everything increasing friction (bad), foul the spark plugs/valve seats/injectors/oil/exhaust system (bad), pollute the environment (bad) and massively decrease efficiency by leaving lots of stuff unburned (bad as well). ;-)

@TooTea pretty much, but goes somewhat offtopic. My point is, light emission in the combustion chamber is an important efficiency loss mechanism even if there is no soot at the end of the combustion.

The color that we see in a flame is due to transitions of electrons between different energy "states" in hot molecules. The nature of the molecule determines the energy states and the temperature determines which highly energetic states are occupied. Which molecules are present in a flame depends on the fuel molecules, the fuel/air ratio, and some other aspects of the environment, as mentioned by others. In many flames, Polycyclic Aromatic Hydrocarbons (PAH, think hydrogenated fragments of graphite or the benzene molecules dehydrogenated and stuck together: multiple planar rings of carbon atoms with hydrogens mostly only on the outer edge) happen to have transitions that produce yellow light. In air, a small molecule like propane is likely to completely burn up before it can produce a PAH. Of course, propane isn't big enough to produce a PAH by itself; they can add to one another to produce them; it is just not likely under the flame conditions considered. Something like a long-chain wax molecule is relatively more likely to react with itself to cyclize and produce a PAH (or benzene-like ring molecules which go on to produce PAH's which go on to produce soot...) as it is 'easy' for a reactive part of the molecule to "bite" some other part of the molecule to produce a ring. While 'smoke' can sometimes be due to condensing water vapor, PAH's go on to produce soot or soot precursors which can be visible as smoke. Yellow flames and smoke are due to PAHs and soot and these are present in candle flames because they are easily formed by large chain molecules like those in waxes.


text:
Heat of reaction for the decomposition of AgCl means?

Certain sunglasses have small crystals of $\ce{AgCl}$ incorporated into the lenses, on exposure to light of appropriate wavelength produces a grey colour to reduce the glare following the reaction:
$$\ce{
  \underset{(colourless)}{AgCl}
  ->[h\nu]
  \underset{(grey)}{Ag} + Cl
}$$

If the heat of reaction for the decomposition of $\ce{AgCl}$ is $\pu{248 kJ/mol}$, what maximum wavelength is needed to induce the desired process?
Does this mean that by absorbing only a certain wavelength, electrons move for a lower to higher energy state and react?
Why does it only emit grey light, but does not any colours of the spectrum in order?
The question asks about the "heat of decomposition".  Does this mean the energy required to excite an electron and cause a reaction?
When a molecule absorb a photon, it is breaking the atoms apart by  moving electrons?
Please help clear up any misunderstandings in this question.

HINT "Heat" is just energy...

Does this answer your question? Why is the dissolution of AgCl considered endothermic?

This has nothing to do with sunglasses: that is the irrelevant pseudo-pretext stuff. The question is this: “What wavelength of light, in nm units, is such that a mole of photons of that wavelength would have a total energy of 248 kJ?”

The intent of the question is that the reaction:
$$ \ce{Ag+} + \ce{Cl-} \rightarrow \ce{Ag} + \ce{Cl}$$
requires absorbing exactly one photon. That photon must have at least the energy required to perform this reaction. The grey colour produced is caused by the metallic silver ($\ce{Ag^0}$) formed by the reaction.
Because a photon's energy is inversely proportional to its wavelength, $E_{photon}=\frac{h c}{\lambda}$, in order to have the required energy or more to cause the reaction, the photon must have a required wavelength or less.  So there is a maximum wavelength that the photon may have in order to cause this reaction.
You're told that the energy required to decompose 1 mole of $\ce{AgCl}$ is 248 kJ/mol, so you have to figure out the energy per atom/photon and convert that energy to a wavelength.
This is an example of the photoelectric effect - that light comes in quantised particles (set amounts of energy) called photons, and certain processes require absorbing a single photon at a time with enough energy, rather than absorbing more lower energy light over time, as if the light was only a continuous wave.

This is actually very similar to a previous question you asked and has a similar answer:
https://chemistry.stackexchange.com/questions/144039/what-happens-if-we-continuously-hit-an-atom-with-photons


text:
Why is it called N-Methylmethanamine?
The IUPAC name of $(\text{CH}_3)_2\text{NH}$ is N-Methylmethanamine according to PubChem. Shouldn't the locant N- be dropped, given that there is no ambiguity over where the substitutent methyl may be located? (If it were attached to the carbon atom, the IUPAC name of the compound would be Ethanamine).

Related: My answer to a different question

I think we should always keep questions even if they seem related. If the question is duplicate then the community will mark it. If it's not duplicate I would humbly suggest that it shouldn't be deleted after being answered. This way firstly contributors earn rewards, secondly the people trying to search about related topics on the internet would be able to find an answer quickly.

@Desai That is my intuition too. However, it is beneficial for the community if questions have accepted answers. My question is more along the lines of, would Loong write an answer here or should I answer my own question with a link pointing to Loong's older answer.

You can do it yourself, just make sure that you credit @Loong properly with a link to the answer and a mention!

Does this answer your question? IUPAC rules on omission of locants in carboxylic acid

It is according to IUPAC rules that while naming alkyl substituted amines, we add a prefix of N before it. In this case also, it is the same... nevertheless, your reasoning of no ambiguity over dropping the prefix N is correct, but the name would not be Methylmethanamine (incorrect), instead it would be dimethylamine which is also correct.

Thank you for your answer. I know that dimethylamine is the common name for the compound, but my question was specifically why Methylmethanamine is the wrong IUPAC name.


text:
Number of lone pairs as a measure of basicity
$\underline{\text{Consider the following question:}}$

The first difference I noticed in these compounds was the large number of oxygen atoms in (IV). This definitely means that there is a larger number of lone pairs in this compound. So the obvious conclusion is that compound (IV) is the most basic or the least acidic one. However none of the options seem to have (IV) as the least acidic compound.

Where have I gone wrong?

You should better consider that (I)  is a diketone that can enolize. (II) is a lactone. (III) is an ether + ketone. (IV) is an ester + diketone.

@Maurice I understand. However I'm interested in knowing why this reasoning fails here opposed to it being the sole property in comparing the basicity of species like diamines and triamines

You have it all backwards. Basicity of a compound is unrelated to its acidity, and both are unrelated to the number of electron pairs. Say, hydrazine N2H4 has two of them, while ammonia has one; which of the two is a stronger base?

Related: https://chemistry.stackexchange.com/questions/128120/strength-of-weak-acids-and-bases-is-methane-better-base-than-ammonia/128141#128141

@IvanNeretin I see. There seems to be some intermingling of different definitions of base- Lewis base, Bronsted base.

Is it 2<3<1<4...? The definition you have to go with is the lewis one.


text:
Neutral FeCl3 Test
What are the specific conditions for a compound to have in order to show positive $\ce{FeCl3}$ test? I got this doubt, when I took 2 different test papers having different key for $\ce{HCOOH}$ and $\ce{CH3COOH}$. I never found anywhere, except for the classical example of $\ce{-OH}$ group attached to $\ce{sp^2}$ carbon, which gives $\ce{FeCl3}$ test.

What do you call the $\ce{FeCl3}$ test ? Is it the colored reaction with phenols ? Is it the red reaction with thiocyanate ions ?

For the ferric chloride test the gist is that the subject molecule has to form a complex with $\ce{Fe^{3+}}$.

Thanks, but could you give me common feature of all the compounds which give (+)ve FeCl3 test? Like, why that compound forms a complex with Fe3+ ion...

Related for phenols: https://chemistry.stackexchange.com/questions/43187/what-is-the-product-of-the-chemical-reaction-between-phenol-and-ferric-chloride .... https://chemistry.stackexchange.com/questions/61926/which-phenols-give-the-colours-blue-green-or-violet-in-the-neutral-fecl3-test

The test used to detect carboxylic acid is called neutral $\ce{FeCl3}$ test. It is called neutral because the pH value should be maintained at 7. This is achieved by adding very dilute ammonia solution drop by drop while shaking the solution. If the characteristic odor of ammonia becomes predominant, the solution has become basic (you can also check with litmus paper or phenolphthalein indicator). It can be made neutral by gently heating to remove excess ammonia (the odor will disappear). Cool the solution and then add few drops of ferric chloride solution to get different colors of precipitates. The reactions are:
$$\ce{RCOOH + \underset{aq. ammonia}{NH4OH} -> \underset{neutral}{RCOONH4} + H2O}$$
$$\ce{3RCOONH4 + FeCl3 -> \underset{colored salt}{(RCOO)3Fe}  + 3NH4Cl}$$

Why reaction condition should be neutral/Why pH should be at 7?

If the solution is basic:

Presence of excess ammonia leads to formation of brown-orange precipitate of ferric hydroxide which interferes with the color of the ferric carboxylate salt resulting in a false result.
$$\ce{FeCl3 + \underset{excess aq. ammnonia}{3NH4OH}  -> \underset{brown orange ppt.}{Fe(OH)3} \downarrow+ 3NH4Cl}$$
$$\ce{FeCl3 + 3RCOOH ->[excess NH4OH] (RCOO)3Fe + Fe(OH)3 \downarrow}$$

If the solution is acidic

If the solution is acidic (little ammonia is added), colorless complexes can form between the acid and ferric ions leading to a false negative result.
$$\ce{FeCl3 + 3RCOOH ->[acidic medium] (RCOO)3Fe + other colorless complexes}$$

Neutral pH

At right pH, you will get colored precipitate. For formic and acetic acid, you will get red colored solution indicating positive results.
$$\ce{3RCOOH + FeCl3 -> \underset{red}{(RCOO)3Fe} + 3HCl~~~~~R= H, CH3}$$

Do formic acid and acetic acid both give FeCl3 test ? since this test is used to distinguish both of them, it doesn't make sense to use the test if both are giving positive result...

@Math_Whiz Yes both give positive FeCl3 test. If you do want to distinguish between them, try Tollen's test. Formic acid give positive result while acetic acid do not. See: https://www.vedantu.com/question-answer/give-one-chemical-test-to-distinguish-between-class-12-chemistry-jee-main-5f62645101faef2daa4e25e0


text:
Naming quaternary alkylammonium cations
Which is the correct name, (1) or (2), for the following cation?

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-62.6.1),

Salts of tetravalent nitrogen $\ce{R4N+X-}$ (where one $\ce{R}$ group represents the parent hydride of the amine or imine and the other groups are hydrogen atoms or substituent groups) are named by one of the following methods:
(1) by adding the suffix 'ium' to the name of the amine or imine, with elision of the terminal letter, 'e' if present, substituent groups being cited as prefixes, and the name of the anion added as a separate word;
(2) by substituting the parent hydride 'azanium', $\ce{NH4+}$;
(3) by substituting the parent hydride 'ammonium', $\ce{NH4+}$, for quaternary salts only.
Method (1) leads to preferred IUPAC names.

The form of the names you suggested use the third of these: that is, the names are based on the parent hydride "ammonium" with a number of substituents. Since the cation in question is quaternary (i.e. doesn't have a free $\ce{NH}$), this is an acceptable method of naming the cation, although it's not the preferred name.
So, time to address the question proper. As given in P-14.5, substituents are cited in "alphanumerical order":

Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

The prefixes "butyl" and "ethyl" are already preferred prefixes (P-29.3.2.1), so we are good on that front. Since "butyl" comes before "ethyl" alphabetically, "butyl" is cited first in the name. The "tri" prefix indicating three ethyls is not taken into consideration, leading to the correct name butyltriethylammonium.
The Blue Book, in fact, further specifies that "ethyl" should be placed within brackets (P-16.5.1.3):

Parentheses are placed around prefixes denoting simple substituent groups in front of parent hydrides when no locants are necessary and around prefixes defining simple substituent groups qualified by locants. A minimum of parentheses must be used. Enclosing marks are never used around the name of the first cited simple substituent group.

This leads to a "more correct" name of butyltri(ethyl)ammonium. However, the parentheses are only mandatory in preferred IUPAC names (P-16.5), and as established above, names based on "ammonium" are not preferred IUPAC names:

Enclosing marks must not be omitted from preferred IUPAC names. In general nomenclature, when there is no possible ambiguity, enclosing marks may be omitted to simplify a name.

In this case, I don't think that there is any ambiguity introduced by dropping the parentheses, so the original name butyltriethylammonium is still acceptable, albeit not preferred.
The preferred IUPAC name would be derived using method (1) in the first quote. There are a few more rules to look up, but I will not quote them here. Application of these leads to the preferred name N,N,N-triethylbutan-1-aminium.

If you name two groups that are the same in priority, name it alphabetically. Since b comes before e, butyl is before ethyl. The first answer is correct.


text:
Naming phosphonium cations in ionic liquids
In Rennie et al., J. Phys. Chem. Lett., 2013, the following cation and anion structures are given in Figure 1.  Figure 1(f) is captioned as butyltrimethylphosphonium, which seems to match the given structure.  However, in the caption, the abbreviation $\text{P}_{2225}$ is given, which seems strange:

Rennie et al. repeat this in their Supporting Information:

But, other authors, such as Tsuchida et al., J. Mol. Liq., 2018, use $\text{P}_{2225}$ as an abbreviation for triethyl-n-pentylphosphonium, which makes sense if the subscript numbers represent the number of carbon atoms in the substituent alkyl groups:

Does it seem to you that there is a typo in the Rennie et al. paper? Or, what are the standard abbreviations for phosphonium cations?

Skepticism is an integral part of a good scientist. Remember the motto "Paper never refused ink" when reading papers and its Supporting Information. You will find incorrect references (very frequent), occasionally wrong calculations, wrong formulae, or wrong names as you read more and more papers just because scientists, like normal human beings, make mistakes.
The strategy in such cases is not to stop at two papers, because the odds are 50:50, but find at least 6-7 papers to confirm the truth.
Your notation is correct for $\ce{P2225}$.

Reference: Another paper from RSC Two phosphonium ionic liquids with high Li+ transport number, Phys. Chem. Chem. Phys., 2015,17, 23041-23051
The odds are in your favor now.


text:
Does the decay of Carbon-14 effect health in biological systems?
Please note that this is not fundamentally a Biology question. It relates to the properties of molecules containing Carbon and Nitrogen.
Carbon atoms are everywhere in living systems from sugars to proteins etc.
Carbon-14 is present in living systems in a certain percentage compared to Carbon-12.
This Carbon-14 is decaying to Nitrogen-14 even when the system is alive.
Does this effect the health and functioning of the living system at all ?
I mean take a glucose molecule $\ce{C6H12O6}$, suddenly one of the Carbon atoms randomly decays to a Nitrogen atom. Does this molecule continue to function as a glucose molecule ? This question extends to every molecule in the body that contains Carbon atoms.

Some relevant information from Wikipedia: "Since many sources of human food are ultimately derived from terrestrial plants, the relative concentration of carbon-14 in our bodies is nearly identical to the relative concentration in the atmosphere. The rates of disintegration of potassium-40 and carbon-14 in the normal adult body are comparable (a few thousand disintegrated nuclei per second)."

See https://xkcd.com/radiation/

I have a feeling that this was asked here before... but it's just a feeling.

Surface water also contains tritium, $\ce{^3H}$. So there are multiple isotopes to which humans have been exposed before we ever starting fiddling with nuclear materials.

See also the radiation hormesis hypothesis: https://en.wikipedia.org/wiki/Radiation_hormesis

@Martin-マーチン https://chemistry.stackexchange.com/q/43616/102629

@Martin-マーチン Not rather https://chemistry.stackexchange.com/questions/65809/what-happens-to-a-radioactive-carbon-dioxide-molecule-when-its-carbon-14-atom-de

I mean take a glucose molecule C6H12O6, suddenly one of the Carbon atoms randomly decays to a Nitrogen atom. Does this molecule continue to function as a glucose molecule ? This question extends to every molecule in the body that contains Carbon atoms.

To answer this part of the question, obviously a glucose molecule which has had a carbon-14 atom decay isn't a glucose molecule anymore. What the glucose molecule turns into is anybody's guess, but no doubt there would be a wide range of molecular fragments from the multiple decays of glucose molecules. Some of the molecular fragments would be highly reactive and cause unusual reactions within our bodies.
Worse, the cabon-14 emits a beta particle when it decays. It would be sort of like a bullet ripping through the body. Thus the beta particle would disrupt the bonding in other molecules, not just in the original glucose molecule.

You do know that only about one carbon in a trillion is C-14 and that during a typical human lifetime, only about 1% of all the C-14 we have decays.  Right?

I think it's important to note that even though each beta decay event potentially destroys multiple biological molecules (including important ones such as segments of DNA strands), the events themselves are so comparatively rare that even something as trivial as body-temperature spontaneous thermal degradation likely poses a greater metabolic challenge/health hazard to the cell.

See also https://en.wikipedia.org/wiki/Radiation_hormesis


text:
What is the viscous component in viscoelastic materials?
All real solid materials are in certain way viscoelastic, because an ideal fully elastic material doesn't exist as well as a fully viscous solid material. According to rheologic models it is possible to define approximately the deformation of the material under stress.
I was searching information, but I don't know yet what is the cause of the viscosity.
For example this is the rheologic Tuckett model, which clearly shows the viscoelasticity of a real material (specifically a polymeric material)

An ideal elastic material after stress release would return to its original shape, but in a viscoelastic one, energy dissipation occurs and the material has not the sufficient energy to come back to its original shape (plastic deformation is resulted)
Here is the example of a ideal elastic material, where all the energy returns and the material takes back its shape and note, that the deformation response to the stress is immediate. On the other hand the viscoelastic materials have a late reaction to stress (why?).

The issue at stake is:

what does it mean "when a material is more viscous than another one" and what is the cause of the viscous component in solid material?
Viscosity in liquids is caused by friction of the particles during movement. Surely it is not possible to apply the same answer to the viscosity in viscoelastic materials.

Thank you in advance for helpfulness

The Wiki section on viscoelastic materials says "Viscoelastic materials have elements of both of these properties and, as such, exhibit time-dependent strain. Whereas elasticity is usually the result of bond stretching along crystallographic planes in an ordered solid, viscosity is the result of the diffusion of atoms or molecules inside an amorphous material"

Does this answer your question? Are viscoelastic materials to be considered solids or fluids?

Check "rheologic mastercurve".