text:
Confusion in calculating $\Delta U$ from a bomb calorimeter
In the book, it is mentioned the formula for $\Delta U$ in a bomb calorimeter without any derivation:

$$\Delta U = q_v = \frac{Q\times M\times \Delta T}{m}$$
where
$$Q=\textrm{heat capacity of calorimeter,}$$
$$M=\textrm{molecular mass of sample,}$$
$$m=\textrm{mass of sample used, and}$$
$$\Delta T=\textrm{change in temperature of water in the bath}$$

I am confused regarding this formula. Can anyone give me the derivation of this formula (or a corrected formula)?
[I am 11-grader and am studying chemical thermodynamics. I can distinguish between $C$ as an extensive property and $c$ and $C_m$ as intensive properties.]
Any help would be appreciated :)
NOTE: I know that a formula is $q_v=cm\Delta T$, I want to know how the book got to the formula mentioned previously.

Does this answer your question? Calculating the heat capacity of a calorimeter

No, I know that formula and want to know how the book got to the aforementioned formula @cngzz1!

here's a link: https://chemistry.stackexchange.com/q/73656/102629

I took a very quick look at this, it might helps you: https://www.chem.tamu.edu/class/fyp/labquest/inv16.html

@Alchimista: I went through the link, but there isn't any discussion about the concerned formula.

@cngzz1: That question has nothing to do with calorimeters!

Look again there as well as here https://en.m.wikipedia.org/wiki/Calorimeter Q is a specific characteristic of each individual apparatus and must be read as a specific heat of the apparatus. It does replace Cv for the specific experiment in one-and-that apparatus. You ascribe DeltaT to the reaction of m/M moles of a substance, assume all Qreact goes to the apparatus. Ergo Qreact must be Q× DeltaT. There is no much to be derived.

Hi the previous link was the derivation for constant pressure specific heat capacity. Here's a derivation for constant volume specific heat capacity ( which explains your calorimeter experiment): https://chemistry.stackexchange.com/q/49198/102629

@Alchimista: Why is it $M/m$ instead of $m/M$? I guess number of moles is $m/M$!

@Alchimista: Here is what I think: since $q_v=C\Delta T=nQ\Delta T$, and we can substitute $n=m/M$, we get $$q_v=\frac{mQ\Delta T}{M}$$ Probably, the book has a typo, but I find the same rule in two more books!

@ultralegend5385 number of mole is m/M, right. Multiply X time M/m gives the wanted quantity per mole (1/mol). .

The formula in the book is correct.  They are trying to get the change in internal energy per mole of sample.  From the first law, for this constant volume system (no work), $$\Delta U_{\textrm{total}}=q=C\Delta T$$where C is the heat capacity of the calorimeter.  This equation assumes that the heat capacity of the water in the bath is lumped into C, and that the temperature change of other parts of the calorimeter is the same as that of the water.
The number of moles of sample is m/M.  So, $$\Delta U_{\textrm{per mole}}=\Delta U_{\textrm{total}}\frac{M}{m}=C\Delta T\frac{M}{m}$$In their notation, they use the symbol Q to represent the heat capacity of the calorimeter C.

Thank you very much! And happy new year @ChetMiller!


text:
When does Alcohol act as Solvent and as a Nucleophile?
I was recently solving problems related to Alcohol.

and its solution is

My doubt is why methanol act as a nucleophile here and when does alcohol acts as a nucleophile and as a Solvent?

On an unrelated note, the last step of the synthesis makes so little sense. You are using lithium aluminium hydride with water? LAH is pyrophoric and should never be used with water. Secondly, the production of an acid upon reduction of an ester is equally suspicious.

@TanYongBoon Actually It is a question from a book of Advanced Problems in Organic Chemistry by Himanshu Pandey for the competitive exam "JEE". I think they add LAH after removing Ester from the water but yes I didn't observe the reduction of the ester. I don't know why the author has proceeded reaction like that. I need to ask another question about it.

C5H8O3 (as does C6H10O3) has 2 degrees of unsaturation -- namely -- levulinic acid (4-oxo-pentanoic acid). The final product from LiAlH4 reduction is the starting diol, both labeled "A". The solution presented is incorrect.

@user55119 Yes. I think it should be a printing mistake as the  reaction with LAH will make the ester part to alcoholic part so that therefore we would get a diol i.e A

My main concern is that the second structure in the "solution" is not a hydroxy acid but the keto acid, levulinic acid.

@user55119 Ah! due to the action of pott. dichromate "A" oxidizes to keto acid  Am I right? that means the whole solution is wrong.

Absolutely!!! I have a corrected scheme i will post as an answer.

Welcome on Chemistry SE!
Here, you are doing an esterification, done with your carboxylic acid and methanol with $\ce{H+}$. The problem is, the yield for such reaction is quite low (an average 67% for a primary alcohol, because of the equilibrium between esterification and hydrolysis). A solution to increase it is to use an excess of reagant (preferably the cheapest one). And the simplest way to have an excess of methanol (that is really easy and cheap to get) is to use it as a solvent.
Hope this helps! And for further explanations, see Fischer-Speier esterification.

You have clarified the first part but the main thing I need is the general case.

As we’re talking about organic chemistry, it is quite hard to get a general case, it really depends on the reaction and the context... nevertheless, in general when the alcohol is the only molecule mentioned in a step of a reaction you’d better verify if your reagents can react with your alcohol (most of the time it will be the case). In any other cases, simply keep in mind that a common convention is to write the reagents above the reaction arrow and the solvent below it.

The proposed solution to this roadmap problem has several errors. I am assuming that $\ce{K2Cr2O7}$, methanol and $\ce{LiAlH4}$ are employed in excess. In the oxidation of 1 (A), chromic acid, $\ce{H2Cr2O7}$, is formed from $\ce{K2Cr2O7}$. To obtain the purported hydroxy acid 6 suggested in the solution, exclusive oxidation of the primary alcohol would be required, a pathway that leads to lactone 5 via hemiacetal 4. One could argue that hydrolysis of the lactone occurs to form 6 but, if this were the case, then hydroxy acid 6 would oxidize to levulinic acid 2 in the presence of excess oxidant. Clearly, levulinic acid 2 fits the formula $\ce{C4H8O3}$ presented in the original problem. As the problem is set forth, oxidation of the secondary alcohol is faster than that of the primary alcohol of diol 1.
The Fischer esterification of levulinic acid 2 leads to methyl levulinate 3 ($\ce{C6H10O3}$). Such an esterification is invariably conducted with excess methanol. The reduction of 3 with $\ce{LiAlH4}$ is conducted in a solvent such as ether or tetrahydrofuran (step 1) and then decomposed cautiously with water (step 2) to reform diol 1 (A) by reduction of both the ketone and ester groups of 3. Clearly, this sequence of transformations is an academic exercise and not one to be conducted in a laboratory. Hopefully, the person who provided the solution was not the same person as the one who created the question.

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sort: votes


text:
What causes little spikes in a flame?
I have a lighter at home which everytime I use it, generates little spikes (see picture below) which eventually disappear.
The Spikes are not an effect generated by the camera. I can confirm that since I made the picture.
Essentially this means the flame does not have a nice uniform flame shape, but is interrupted by the spikes.

Does someone know where this is comming from?

EDIT:
@JamesGaidis suggested to try holding the lighter sideways. It can be observed that the spot of the spike has roughly the same position when holding it sideways (left picture)

A different lighter with the same effect, but with the additional feature of beeing able to reduce the gas flow shows, that the spike is independent of the quantity of the gas (right picture).
The height of the spike (distance from the tip of the flame to the tip of the spike) is always the same, thus independent from the quantity of the gas. (In other words: the height from hole that emits the gas to the top the spike is only dependent on the size of the flame)

James did also suggest to light the lighter from an external source, and in this case, no spikes show up.
This implies that its not the gas alone generating this effect, but has rather something to do with the stone.

I don't really know if this is has to do with chemistry at all, or if it has more to do with physics. If thats the case, feel free to redirect me towards the correct SE site.

I bet it has something to do with speed profile of LPG gas injection stream. The center  of ii has the highest speed  and it's flame is longest.

Perhaps the flame follows or is carried by some particles which are scraped off the flint and carried by the rising gas, but are not sustained by the gas flow. It would be interesting to light the flame by holding the lighter sideways, or swing the lighter while lighting the flame or reduce the gas flow to minimum. Or light the gas flow without the flint - from another source, like a candle flame.

@JamesGaidis those are good ideas, I have updated the question to include your suggested experiements

How long does this jet/spike last? If it vanishes rapidly, it indicates a solid particle which came out during friction.

@M.Farooq when looking very, very closely at the flame, I observed that it never really vanishes, but gets smaller and smaller, but the significant spike that has approximately the size of that in the picture, lasts around 3 ~ 4 seconds. So it does not vanish rapidly.

Yes, this transient appearance is that of a combustible particle resulting from friction. As you noted you when you light it without friction these spikes vanish.

It would be better suited in physics.SE

@J.Doe The question is closed but I had another point to make. The image captured by your camera is the amount of light that was detected by its optical sensor for the time its shutter was open. So, the spike could be due to long enough exposure of the shutter. You should try it with a really good slow motion camera (at least hundreds of frames per second, not a cellphone camera of just 60-80 frames per second) and see if you have the same observation.

@user79161 I'm not quite sure if get you point correctly. This effect does not come from the camera. I can see the spike with my own eyes.

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text:
Comparison of evaporation rate of liquids

Evaporation: Evaporation
is a type of vaporization that occurs on the surface of a liquid as it
changes into the gas phase. When a molecule near the surface absorbs
enough energy to overcome the vapour pressure, it will escape and
enter the surrounding air as a gas.
Boiling point: The
boiling point of a substance is the temperature at which the vapour
pressure of a liquid equals the pressure surrounding the liquid and
the liquid changes into a vapour.
Volatility:
Volatility is a material quality which describes how readily a
substance vapourizes.
Flash Point: The flash
point of a volatile material is the lowest temperature at which its
vapours ignite if given an ignition source.
Autoignition
temperature:
The autoignition temperature or kindling point of a substance is the
lowest temperature in which it spontaneously ignites in a normal
atmosphere without an external source of ignition, such as a flame or
spark.
Fire point: The fire point
of a fuel is the lowest temperature at which the vapour of that fuel
will continue to burn for at least five seconds after ignition by an
open flame of standard dimension.
Smoke point: The smoke
point, also referred to as the burning point, is the temperature at
which an oil or fat begins to produce a continuous bluish smoke that
becomes clearly visible, dependent upon specific and defined
conditions.

How can we compare the evaporation rate for water, alcohol, petrol and kerosene oil?
If we consider intermolecular forces as the criterion, then the order of evaporation would be:
water < alcohol < kerosene oil < petrol
But part of intermolecular forces still act in vapour phase...
What should be the criteria to compare evaporation rate of liquids at a given temperature?

Consider room temperature saturated partial pressure in the first place, but also  molar mass, diffusion coefficients, density and specific heat capacity. Both thermodynamic and kinetic properties play role.

Consider also molar enthalpy of evaporation, as for the same boiling point, liquid with the higher molar enthalpy of evaporation has lower room temperature saturated partial pressure.

Evaporation rates are used to compare solvents for different reasons and under various conditions. One condition would be under vacuum in a rotary evaporator, to remove a low-boiling solvent to assist crystallization of a solute. Another condition would be in a paint, where you desire a higher boiler to allow enough time to spread the paint without brush strokes.
A practical way to compare evaporation rates of solvents is to put a small amount into a dish and weigh it as it evaporates.  Pure compounds will have a fairly sharp disappearance and therefore a sharp time to evaporate; mixtures like kerosene or gasoline will evaporate fast at first, then slower and slower. This time, or the rate for 90% evaporation, can be compared with another solvent under the same conditions.
Evaporation rate is a commonly listed on Safety Data Sheets, and in other tables, generally compared to butyl acetate or ether. One table is at https://www.dispersetech.com/content/11-properties-of-common-solvents.
Many tables of solvent characteristics do not include evaporation rate, but do show vapor pressure at room temperature, e.g., https://sites.google.com/site/miller00828/in/solvent-polarity-table. Vapor pressure at some temperature can be a good indicator of evaporation rate, but a very strong factor in evaporation is wind, or blowing across the solvent surface. Water is particularly complicated because of relative humidity, but it and wind and temperature have been investigated for reservoirs: https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/97WR02405.
Chemical distributors often list all the properties of all the solvents they deal with - and sometimes of materials they don't deal with.

In that table, it is surprising that kerosene has a very low evaporation rate compare to most of the solvents (even water).

@Apurvium: Kerosene has a mixture of components, with mostly 10-16 carbons. It is distilled from petroleum over the range from 150 to 275C. https://en.wikipedia.org/wiki/Kerosene That's why it is sometimes more useful to specify an evaporation of 90% so that a difference in % of high boilers doesn't give a totally different impression of volatility.


text:
Definition of enthalpy of vaporization

The amount of heat required to vapourize one mole of a liquid at
constant temperature and under standard pressure (1 bar) is called its
standard enthalpy of vapourization.
The enthalpy of vaporization is often quoted for the normal boiling
temperature of the substance.

Suppose, some amount of water is kept inside a perfect thermally conducting container at 1 bar pressure and $\mathrm 50~^\circ \mathrm{C}$ within a thermal reservoir of $\mathrm 50~^\circ \mathrm{C}$ also. Due to evaporation, slowly liquid will convert to vapours.
After one mole of water evaporates, if we measure the heat energy received by the container from the reservoir, would it be equal to standard enthalpy of vapourization of water at $\mathrm 50~^\circ \mathrm{C}$?

Assuming that the reservoir does not absorb any heat and there is no loss of heat from the setup by:-

Radiation of heat from the reservoir,
Convection currents in the air/surrounding fluid (if any),
The entire water is at the same temperature so that heat absorbed is not used for increase in temperature,
Conduction loss from the reservoir to the ground/support it is on and air(negligible but thought I should mention)

Fulfilling these conditions ensures that the energy is only used by the water for vaporization.
So, yes, these conditions are fulfilled. But, that is never the case when experiments are conducted in labs.
Edit: The above conditions are more than just an isolated system.

Only evaporation can take place at 50 C which will be very slow... So, how enthalpy of vapourization can be defined at any other temperature than its boiling point?


text:
real meaning of osmotic and turgor pressures
I am a high school student and I am very confused in how we define osmotic pressure? Osmotic pressure is defined as the pressure required to stop the diffusion of solvent into a solution by applying external pressure over solution, which means that when the pure solvent is diffusing into the solution then it will apply pressure on the semi permeable membrane and when we have applied the same external pressure over the solution then there is no net movement of solvent occurs across the membrane and mathematically it is equal to P=cRT, where c is the concentration of solution , T is temperature in kelvin and R is the universal gas constant, if we go by this equation and calculate the osmotic pressure of 1 molar concentrated solution at room temperature it comes out to be almost 2495Pa which means 2495Newton/m^2 which is very high pressure and it doesn't seem to be true as osmosis is very slow process it surely cannot generate that much of pressure( just to imagine how large is that pressure assume an area of 1m^2 and place Mass of 249.5kg on it), so I think that it should not be that higher, so ,I am unable to understand what is actually osmotic pressure ,on which does it actually act and how do we actually calculate it?
we say that when a cell is placed in hypotonic solution then osmotic pressure is greater than Turgor pressure and water will flow inside the cell, but I have a confusion that :like turgor pressure acts on the wall/membrane of cell ,osmotic pressure(O.P)) should also act on the membrane and in opposite direction of turgor pressure, so by this logic wall should compress more as osmotic pressure on it will cause it to compress and turgor pressure will cause it to expand but Osmotic pressure is greater than turgor pressure(T.P) it should compress, but it doesn't happen? we only talk of turgor pressure(T.P) to explain the expansion but not osmotic pressure why? I know I am going wrong somewhere if Osmotic pressure would really leads the cell to compress then an animal cell would have never burst in hypotonic solution. and if osmotic pressure is also acting on the wall or membrane then net pressure on the wall is O.P-T.P (in inward direction) , and not only T.P (outward) so the wall pressure here also should not be same as turgor pressure? I think I should think of Osmotic pressure not as the pressure generated by the molecules hitting over the surface, but I should think of Osmotic pressure as the ability to take up water i.e I think we have just relate the concentration gradient in terms of actual pressure just to feel the things and to define them numerically to relate it with the actual pressure needed to stop the uptake of water.
If anyone will explain me this by giving as many examples as possible than it would be very helpful for me to understand because its my first time I came to know about these things.

2495 Pa seems to quite low, equivalent to 0.0246 atm.

Please do respect ponctuation rules when writing. Thirteen lines in one sentence is nearly unreadable.

Some languages do not have punctuation marks, and when the brain does the conversion/translation work of run on thoughts, the result is like that. :-)


text:
Synthesis of anhydrous sodium tungstate
I am interested in preparing an anhydrous sample of sodium tungstate (Na2WO4), which is normally found as the dihydrate.  Finding an example of this in literature is proving difficult, I am interested if someone can suggest a preparation method based on prior experience or difficulties I should encounter.

Is there a good reason you cannot dry dihydrate? Also, what are your starting materials and what equipment do you have access to?

I think I should be able to dry the dihydrate, previously I have normally found some literature before trying these procedures before.  I have most standard lab equipment (heaters, vacuum pump etc) and IR which I think I can use to confirm it's no longer a hydrate.  I am just unsure what conditions it would require.

Hoermann [1] prepared anhydrous sodium tungstate by melting the 1:1 mix of sodium carbonate and tungsten(VI) oxide (and, subsequently, growing single crystals); alternatively, he proposed prolonged drying of a dihydrate at 100 °C:

$\ce{Na2MoO3}$ und $\ce{Na2WO3}$ sind wasserfrei durch Zusammenschmelzen von 1 Mol $\ce{Na2CO3}$ mit 1 Mol $\ce{MoO3},$ bzw. $\ce{WO3},$ oder durch volliges Entwassern der Hydrate bei 100° zu erhalten. Die Salze zeigen ansgepragte Polymorphie.

Busey and Keller [2] obtained anhydrous sodium tungstate (reported water content 0.12%) by drying a dihydrate at 200 °C in vacuum. Purity has also been confirmed by Raman spectroscopy and powder x-ray diffraction:

The $\ce{Na2WO4}$ was prepared by heating the dihydrate at 200° in a vacuum.

References

Hoermann, F. Beitrag zur Kenntnis der Molybdate und Wolframate. Die binären Systeme: $\ce{Li3MoO4-MoO3},$ $\ce{Na2MoO4-MoO3},$ $\ce{K3MoO4-MoO3},$ $\ce{Li2WO4-WO3},$ $\ce{Na2WO4-WO3},$ $\ce{K2WO4-WO3},$ $\ce{Li2MoO4-Na2MoO4},$ $\ce{Li2WO4-Na2WO4},$ $\ce{Li2MoO4-K2MoO4}.$ Z. Anorg. Allg. Chem. 1929, 177 (1), 145–186. DOI: 10.1002/zaac.19291770117.
Busey, R. H.; Keller, O. L. Structure of the Aqueous Pertechnetate Ion by Raman and Infrared Spectroscopy. Raman and Infrared Spectra of Crystalline $\ce{KTcO4},$ $\ce{KReO4},$ $\ce{Na2MoO4},$ $\ce{Na2WO4},$ $\ce{Na2MoO4 · 2 H2O},$ and $\ce{Na2WO4 · 2 H2O}.$ The Journal of Chemical Physics 1964, 41 (1), 215–225. DOI: 10.1063/1.1725625.

Apart from the dehydration method mentioned by @andselisk, there are also 2 chemical methods where you can obtain the anhydrous salt:

by the fusing together equivalent quantities of tungstic anhydride
and sodium hydroxide or sodium carbonate, the resulting mass being
taken up with water and allowed to crystallize. (@andselisk) Tungstic acid or hydrated
tungsten oxide also works in this case.
Fusion of alkali on mineral wolframite, $\ce{(Fe,Mn)WO4}$

For further information regarding the compound, see here

I have actually, to my surprise and that of others, successfully dissolved the filament from an incandescent oven bulb (see before and after pictures here, click to expand on the final picture), which is composed of, per a reference, to quote:

Incandescent light bulbs consist of an air-tight glass enclosure (the envelope, or bulb) with a filament of tungsten wire inside the bulb, through which an electric current is passed.

in a galvanic cell setting. The piece of tungsten metal serves as the anode and carbon (as graphite in my experiment) can act as the cathode. In the current case, I would recommend an electrolyte of NaOH plus 6% NaOCl (which relates to the so-called Bleach battery cell).
Heat in a microwave replenishing the electrolyte as needed. When dissolution is evident and Sodium hydroxide exhausted, remove electrodes and evaporate down the solution to effect salt separation.
As to mechanics, the NaOCl is expected to result in the anodic oxidation of the tungsten metal (as generally reported in the noted Bleach Battery reference) as follows:
$\ce{W + 3 NaOCl -> WO3 + 3 NaCl}$
Then, per Wikipedia, to quote:

$\ce{Na2WO4 + 2 HCl → WO3 + 2 NaCl + H2O}$

$\ce{Na2WO4 + 2 HCl → WO3.H2O + 2 NaCl}$

This reaction can be reversed using aqueous sodium hydroxide.

Hence, the presence of added Sodium hydroxide acting on $\ce{WO3}$ is expected to result in Sodium tungstate.
Now, why a microwave approach? To quote a 2013 paper: Microwave-Specific Enhancement of the Carbon−Carbon Dioxide (Boudouard) Reaction:

This changes the position of the equilibrium so that the temperature at which CO becomes the major product drops from 643 °C in the conventional thermal reaction to 213 °C in the microwave. The observed reduction in the apparent enthalpy of the microwave driven reaction, compared to what is determined for the thermal reaction from standard heats of formation, can be thought of as arising from additional energy being put into the carbon by the microwaves, effectively increasing its apparent standard enthalpy. Mechanistically, it is hypothesized that the enhanced reactivity arises from the interaction of CO2 with the steady-state concentration of electron−hole pairs that are present at the surface of the carbon due to the space-charge mechanism, by which microwaves are known to heat carbon. Such a mechanism is unique to microwave-induced heating...

Also, interesting cited chemistry on the microwave action on carbon reputedly creating some hydroxyl radicals (${.OH}$). Reference: see per this 2007 paper, Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances. Note: this is not unexpected, as per the prior reference especially in alkaline conditions, the action of an electron-hole on ${OH-}$ should produce ${.OH}$.
Note: The presence of radicals is expected to accelerate the oxidation of the tungsten metal.
Bottom line, this microwave-assisted electrosynthesis is relatively safe, effective, and avoids high temperatures. It requires only generally available inexpensive reagents.
Note, one can use alcohol to source an anhydrous Sodium tungstate as per Wikipedia, to quote: "insoluble in alcohol", but any Sodium chloride presence is sparingly soluble in methanol (14.9 g/L).

Sorry, but I fail to see what all of this has to do with the question. I find many of your answers pretty ingenious, but there is also a bunch that for some reason reminds me of  Kissinger's shuttle diplomacy. This one leans towards the latter, I'm afraid.

Well, yes, I did forget to add the part of employing alcohol to create anhydrous crystals! However, given the general reluctance of old school chemists to ever consider a galvanic cell synthesis, albeit as my added pictures clearly demonstrate its successful feasibility here, I still expect significant dismissal. That, however, does not change pictures reflecting reality, a reality increasingly important for those without access to their labs! For those in want of a reagent,  try expanding your horizon to encompass electrosynthesis and  may I even further stray, with photochemical synthesis?

At this point I'm no longer sure whether you are serious or just trolling. Neither your post nor any of the references don't have any mentions of anhydrous sodium tungstate, let alone a straightforward synthetic procedure which OP asked about. No one denies the credibility of the literature sources, it's that they are completely irrelevant for what's being asked.

To be completely honest, I did not separate out the sodium tungstate. I suspect using alcohol here would enable a salt separation leading to the anhydrous salt. Hence, a further edit to my answer.

Again, there is no mentioning of sodium tungstate in your post or linked content whatsoever. Please post a clear and concise procedure of preparing anhydrous sodium tungstate $\ce{Na2WO4}$ backed up by relevant reputable sources and analytical data. Currently I don't see how any of this information makes any sense in the context of the question.

True, I have added my understanding of the anode oxidation of W via NaOCl including a cited Wikipedia reference as to the effect of a NaOH presence citing the formation of sodium tungstate. Hopefully, a much more clear presentation as to how sodium tungstate is, indeed, formed here.

Please don't use edit statements; it makes the post harder to read, and the intention harder to understand. If anyone is interested in how the text came to be, an edit history is linked below.

Martin: At this point, I agree it is appropriate to remove edit citations (which I have and will do in the future).


text:
Negative coefficients for DIIS in SCF procedure
I am trying to implement the direct inversion of the iterative subspace (DIIS) algorithm in my simple HF program. After some problems it now seems to work. I can calculate the correct energies in roughly the same number of steps as in the literature.
My problem is that when generating the Fock matrix from previous Fock matrices I sometimes have negative expansion coefficients. This leads to expansion coefficients for the newest Fock matrix larger than one. The coefficient still sum to one.
Did anyone else encounter this problem?

I just ran an SCF calculation with DIIS method and I can positively confirm that Fock matrix elements are sometimes greater than 1 or less than -1.  Additionally, MO coefficients can certainly be negative.

After some further research I came to the conclusion that the DIIS doesnt restrict coefficients to $[0,1]$. This is only true for EDIIS. http://dx.doi.org/10.1063/1.4740249


text:
Interpretation of Coloumb operator in Hartree Fock equation
I have read in a textbook (Modern Quantum Chemistry Szabo and Ostlund) that the Coulomb operator of the form
\begin{equation}
\mathcal{J}_{j}\left(\mathbf{x}_{1}\right)=\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2}r_{12}^{-1}
\end{equation}
describes the average local potential at $\mathbf{x}_{1}$ arising from the charge distribution from an electron in $\chi_{j}$. And therefore the term $\sum_{j \neq i}\left[\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2} r_{12}^{-1}\right] \chi_{i}\left(\mathbf{x}_{1}\right)$ for an N-electron system gives the total averaged potential of $N-1$ electrons in other spin orbitals on the electron in $\chi_{i}$.
However, if one expands the sum
\begin{equation}
\sum_{j \neq i}\left[\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2} r_{12}^{-1}\right] \chi_{i}\left(\mathbf{x}_{1}\right) = \left(\int d \mathbf{x}_{2}\left|\chi_{1}\left(\mathbf{x}_{2}\right)\right|^{2}r_{12}^{-1} + \int d \mathbf{x}_{2}\left|\chi_{2}\left(\mathbf{x}_{2}\right)\right|^{2}r_{12}^{-1} +...\right) \chi_{i}\left(\mathbf{x}_{1}\right)
\end{equation}
it is indeed a sum over the spin orbital but not of the electrons in the system. My interpretation for the sum is as follows: the total averaged potential acting on electron 1 due to the charge distribution of electron 2 in N-1 spin orbitals . I think it has something to do with the fact that electrons are indistinguishable so one cannot assign an electron to an orbital. However, I think the Hartree Fock integro-differential equation
\begin{equation}
h\left(\mathbf{x}_{1}\right) \chi_{i}\left(\mathbf{x}_{1}\right)+\sum_{j \neq i}\left[\int d \mathbf{x}_{2}\left|\chi_{j}\left(\mathbf{x}_{2}\right)\right|^{2} r_{12}^{-1}\right] \chi_{i}\left(\mathbf{x}_{1}\right)-\sum_{j \neq i}\left[\int d \mathbf{x}_{2} \chi_{j}^{*}\left(\mathbf{x}_{2}\right) \chi_{i}\left(\mathbf{x}_{2}\right) r_{12}^{-1}\right] \chi_{j}\left(\mathbf{x}_{1}\right)=\epsilon_{i} \chi_{i}\left(\mathbf{x}_{1}\right)
\end{equation}
was derived for hydrogen molecule. Is it the same for an N-electron system?

From the labels ($i,j$), I am guessing that $\chi$ is a molecular orbital (MO). When formulating the equations for molecular orbitals, one assumes that there are as many (spin-)orbitals as there are electrons. In any event, when writing down the Coulomb operator in the MO representation for the ground state, only occupied orbitals appear.
Therefore, I believe that the equations you are showing here are for the general problem, not just $\ce{H2}$. In some sense, the equations are the same for either, it is just a question of how many summands there are. Finally, note that the last equation you showed is valid for any system, but that it is actually part of a set of such equations, one for each MO.


text:
How to explain regioselectivity in nucleophilic aromatic substitution
I’m trying to rationalise the major regioisomeric product of the following reaction of an arylchloride in the presence of sodium amide:

I have predicted that the para-substituted product is formed as the major product because the negative charge in the meta position in the carbanion intermediate is more strongly stabilised by the electron-withdrawing CO2NMe2 group:

Is this observation based on inductive effects correct? Can the carbanion be stabilised mesomerically by delocalizing the negative charge through the π-bond framework?

Are you sure this reaction proceeds through a benzyne mechanism?

In the benzyne mechanism, mesomeric effects are not involved as when protons are removed, the electrons are situated in orbitals perpendicular to the pi-system, in non-bonding orbitals (which makes sense since the electrons come from a sigma bond). Therefore the electrons left over are not delocalised, which is why such a strong base is required. This also means regioselectivity is explained by inductive effect, which acts through the sigma bonds. The amide is not involved in any -M stabilisation in this case.
In the first part of the mechanism, the most acidic proton is abstracted, which will be either of the identical protons either side of the Cl, as this provides the most stabilisation by inductive effect. This is correctly shown in your diagram. Deprotonation at the carbons adjacent to the amide is less favourable and leads to no reaction due to lack of leaving group anyway.
The point where the alkyne is attacked is determined by both inductive and steric effects; and optimisation of both simultaneously usually occurs in this mechanism. What I mean by this is that in the major product above, the negative charge is situated as close to two -I substituents as possible, so is stabilised, and at the same time the added NH2 is furthest from the other substituent, reducing steric strain.
In the minor product, the negative charge has less -I stabilisation, and more steric strain since the substituents are now closer together.
Of course this only applies if it progresses through the benzyne mechanism. As @Waylander hinted at, a likely possibility is aromatic nucleophilic addition-elimination since you have an electron withdrawing substituent para to the leaving group. This just so happens to give a product identical to the major product of the benzyne route.

"The point where the alkyne is attacked is determined by both inductive and steric effects" Alkyne? :-) More seriously, sterics are really not a consideration for 1,3- vs 1,4. In both cases, they're nowhere near each other. 1,2-disubstituted rings are different; for those you do need to start considering sterics.


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Molecular dynamics simulation of a protein in acidic medium
I want to perform an MD simulation of a protein under acidic solvent conditions. A quick literature search seems to indicate that people are more interested in the protonation of protein side chains (which is important, of course), but seem to ignore the abundance of $\ce{H+}$ ions (i.e. $\ce{H3O+})$ in water surrounding the protein.
Is there a standard practice for performing MD simulations of proteins in acidic environments? Should I account for $\ce{H+}$ at low pH?

I may just be making a silly mistake in a back of the envelope calculation, but wouldn't the concentration of water around the protein dwarf the concentration of $\ce{H3O+}$ by several orders of magnitude? A pH of 1 (very acidic for a biological setting) would correspond to an $\ce{H3O+}$ concentration of 0.1 M. Pure water has a concentration of 55 M. While this concentration would be reduced somewhat in an acidic solution, it still suggests that there is on the order of 100 times more waters than $\ce{H3O+}$. This ratio gets an order of magnitude larger when you increase the pH by 1 unit.

I think you're onto something. Just continuing in your train of thought, at pH = 1, an area of 10 angstrom^3, would just have 0.06 H3O+ molecules. At 100 angstrom ^3 (still pH = 1), you get 60 H3O+ molecules, which might become non-negligible. But rarely, do biomolecules function at pH=1. At pH = ~6, the number of H3O+ is still negligible at 100 angstrom^3.
I wonder if the effects of H3O+ in biology (and hence simulation) would have a large effect, despite its small number...

@Mithoron but that's my point. The concentration of H3O+ is probably never large enough where one would actually need to alter the solvent itself for the simulation. I would think this would only be necessary in a solution that was so concentrated (e.g. negative pH) that H3O+ ions made up a significant portion of the solvent.

I think you have largely answered the question yourself, and it's not an entirely trivial answer, it depends on what you are probing and the conditions you are trying to simulate and the size of your simulation. If you are interested in the fate of the hydronium ions diffusing about a protein, then it is obviously a good idea to introduce them into the system. You would not need to even replicate the nominal concentration, but obviously have at least one ion. If you are interested in the effect of counterions in general at physiological pH then you may want to skip this.


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Will CO2 bubbles ever fully dissolve in water?
Background:
Many aquarists keep freshwater planted tanks add $\ce{CO2}$ from a pressurized canister to their aquaria to promote plant growth. The injection is at a slow rate to get dissolved $\ce{CO2}$ in the neighborhood of 30 ppm. Methods vary, but ceramic discs to release a very fine mist of bubbles are popular, as are $\ce{CO2}$ reactors which trap the bubbles in a bell or canister and stir them around with a water pump.
In any case, within the hobby there is an odd fixation on getting the gas bubbles to dissolve completely, I suppose on the reasoning that a bubble that pops at the surface is $\ce{CO2}$ that escaped to air rather than being dissolved in the water.
I'm skeptical of this logic -- I'd think while the bubble is dissolving, water vapor and whatever other dissolved gasses will be diffusing into the bubble. At some point, the bubble becomes just humid air in equilibrium with the surrounding water, and no amount of stirring will get more $\ce{CO2}$ out of it. This is just my guess, based on casual observations where the bubbles will diminish to less than 10% of their initial volume within 30 seconds, and thereafter they don't seem to get any smaller.
Questions:
Is that correct, and does it mean the aquarist's quest for a "100% efficient" $\ce{CO2}$ bubbler that never lets bubbles pop at the surface is doomed to eventual failure?
And if so, what are the basic principles that describe this equilibrium and how can we figure how much $\ce{CO2}$ can be extracted from a bubble, and how big the bubble will be after it's become just humid air?
If not, why does it seem the bubbles will rapidly dissolve in the water at first, but never fully dissolve even though the solution is far from saturated?

There is a limit as to how much $\ce{CO2}$ will dissolve into water. I'd wonder about aggressively adding $\ce{CO2}$ in an aquarium with fish or other oxygen dependent animals. The excess $\ce{CO2}$ will sparge dissolved oxygen from the water.

@MaxW The rule of thumb in the hobby is to aim for 30 ppm, though it can vary by species. The $\ce{O2}$ released from photosynthesis is impressive, so there's plenty of oxygen if the tank is running well. I'm no biologist but I always figured excess $\ce{CO2}$ will kill the livestock directly by acidosis before oxygen deprivation does.

@Mithoron Thanks, I'm here to learn! Perhaps you could put that in an answer?

Maybe, if post won't get closed and no one answers it anytime soon. It's just that a bubble with other gasses could also dissolve, unless they are too close to saturation point - so you may be not wrong after all if that's the case.

@MaxW i believe that too. Oxygen has very less solubility with respect to othr gases. So the aquarium must have no fish

They equilibium achieved naturally is more probable that s why aquatic life exist

Questions: (1) Fresh water or salt water? (2) Are there fish or other animals that depend on dissolved oxygen in the aquarium? (3) Is the pH adjusted regularily in the aquarium? (4) To what value is pH adjusted?

I just think of an installing the plants only phase, no other living organisms inside. What you say is true, there could be in and out passage for all gases. However, a big bubble escape without having the possibility to get to equilibrium, while fine bubbles should be more efficient. When we have to flush a solution, we do that with the smallest bubbles as possibile, without being aquarist. And that it is not only due to preventing spill, it works exactly like dissolving a solid. Finer the powder, faster the equilibrium is reached.

@MaxW (1) fresh. (2) yes, but I am certain they aren't harmed by this technique. It's a very common technique, and the target $\ce{CO2}$ levels are too low to displace a significant amount of oxygen, and the plants more than replenish what little oxygen is lost by generating oxygen through photosynthesis. (3) pH is not adjusted -- there's no need. Minor fluctuations in pH don't harm the livestock, and the carbonates in my tap water buffer the pH. With the added $\ce{CO2}$ the pH ends up pretty close to 7.

@Alchimista Fine bubbles and agitation are good and I certainly see how they can help, but does that mean if the bubbles are fine enough or the solution is stirred enough, the bubbles will eventually disappear entirely? I know I can put some solid salt in some water, stir it up, and eventually there will be zero solid salt left. But stirring CO2 bubbles in water, they get smaller but never seem to go away entirely. Why is that?

@PhilFrost well I the liquid is saturated they certainly escape. They escape nevertheless as they move to the surface. That is why finer bubbles "dissolve better". If you imagine the smallest bubble you can have, it will enter the liquid basically already dissolve, in a limit case. Still, what you say is in principle true, the inside of the bubble will contain some oxygen and water, as they have vapour pressure. Bubble can certainly disappear entirely, tough unless the inlet is in the depth, it will be difficult to see that. Finally, a bubble of water can't form, it is a droplet among the othe

@Alchimista that sounds more like an answer than a comment

@PhilFrost and for your question why a trapped bubble does not dissolve in spite the saturation isn't reached, is rather the diffusion that is hindered. See the bubble as a container. The interior is at hydrostatic pressure, there is an extra force due the the water surface tension to be overcame. No answer because I can't quantify, especially this latter aspect.

Situation at equilibrium
If you have pure water in contact with air in a closed system (like a capped bottle filled halfway with water), the $\ce{CO2}$ concentration will reach an equilibrium.
$$\ce{CO2(g) <=> CO2(aq)}$$
"(aq)" refers to aqueous concentration, not aquarium.
Currently, our outside air contains about 400 molecules of $\ce{CO2}$ in a million molecules total (400 ppm or 0.04% by volume or by number of particles). At room temperature, the concentration of $\ce{CO2}$ in the water will be about $\pu{13E-6 mol/L}$, corresponding to 0.6 mg of $\ce{CO2}$ per liter.
If you let water stand in pure $\ce{CO2}$ at atmospheric pressure (or bubble pure $\ce{CO2}$ through pure water that is not in contact with air), the concentration of $\ce{CO2}$ at equilibrium will be about 2500-times higher, so $\pu{0.033 mol/L}$. This corresponds to 1400 mg of $\ce{CO2}$ per liter.
Getting "30 ppm"
In the aquarium, the goal is to have 30 mg of $\ce{CO2}$ per liter (confusingly described as 30 ppm, which is different from 30 ppm in the gas phase). This is 50-times higher than the equilibrium concentration you get with air, and about 50-times lower than you get in an atmosphere of pure $\ce{CO2}$.
So what happens in the aquarium is that $\ce{CO2}$ enters the water from the $\ce{CO2}$ bubbles, and exits when water has contact with the air. In a covered aquarium, the rate of $\ce{CO2}$ removal is a bit slower because the air has more $\ce{CO2}$ than fresh air. There are also biological processes (plants, fish, microbes) that produce or use up $\ce{CO2}$. For these reasons, you monitor the $\ce{CO2}$ content and adjust the $\ce{CO2}$ flow (input) to maintain the desired concentration.
The $\ce{CO2}$ concentration is measured with a "drop checker" via the pH of a test solution of defined carbonate hardness (4 dHK, or degrees of carbonate hardness). As the $\ce{CO2}$ concentration in the aquarium changes, the $\ce{CO2}$ partial pressure in the air gap between aquarium water and test solution changes, eventually changing the $\ce{CO2}$ concentration in the test solution, and with it its pH. This takes a while, so the measurement lags by minutes or hours.
Your questions
Update I edited this after Poutnik posted a comment

I'm skeptical of this logic -- I'd think while the bubble is dissolving, water vapor and whatever other dissolved gasses will be diffusing into the bubble. At some point, the bubble becomes just humid air in equilibrium with the surrounding water, and no amount of stirring will get more $\ce{CO2}$ out of it.

If the bubble is pure $\ce{CO2}$, it should dissolve completely given sufficient pure water. However, the aquarium water is not pure water, and some oxygen, nitrogen and water will be transferred from the liquid into the gas bubble while $\ce{CO2}$ dissolves in the water. I would guess that in the vicinity of the $\ce{CO2}$ bubbler, the concentration of $\ce{CO2}$ in the water is already near its maximum, so the $\ce{CO2}$ concentration in the bubble does not go to zero. On the other hand, the partial pressure of oxygen and nitrogen will increase in the bubble until it matches the vapor pressures of the solutes. What the final volume of the bubble will be depends on the kinetics, but if you prevent it from escaping to the surface, it will not disappear completely.

[...] does it mean the aquarist's quest for a "100% efficient" $\ce{CO2}$ bubbler that never lets bubbles pop at the surface is doomed to eventual failure?

In pure water, if the bubble is small and there is plenty of time before it reaches the surface, it should be possible for bubbles to completely dissolve.

And if so, what are the basic principles that describe this equilibrium and how can we figure how much $\ce{CO2}$ can be extracted from a bubble, and how big the bubble will be after it's become just humid air?

It is not an equilibrium situation. If you let the system come to equilibrium, the composition of the bubbles (near the surface, i.e. at ambient pressure) will be the same as the air on the surface. However, as explained in the first section, the system is never at equilibrium if you bubble in $\ce{CO2}$.

If not, why does it seem the bubbles will rapidly dissolve in the water at first, but never fully dissolve even though the solution is far from saturated?

I suspect the solution surrounding the bubble becomes saturated very quickly, and then the $\ce{CO2}$ has the diffuse away from the bubble before more can dissolve. Also, once nitrogen and oxygen have entered, they will maintain the bubble as long the water is saturated with nitrogen and oxygen.

Hm, until CO2 strips out O2 and N2, bubbles would not dissolve completely, not being CO2 bubbles. Water is in equilibrium with 0.78 atm N2 and possibly oversaturated by O2 wrt  0.21 atm O2. Dissolved O2 and N2 would rush into bubbles with pure CO2 and no O2 nor N2.

@Poutnik I guess it is a question of kinetics. If a bubble has the same composition as the bulk gas the solution is in contact with, it is stable. If it has any components with higher or lower partial pressure than in the bulk gas, the component will go into solution or come out of solution. I will have to edit my answer, thanks for the comment.

@Poutnik There might also be an effect similar to Ostwald ripening so that the bubbles eventually disappear (in favor of the biggest "bubble", the water-air interface at the surface of the aquarium), see foam ripening.

See also https://www.researchgate.net/publication/325821502_Ostwald_ripening_for_air_bubbles_and_decompression_illness_phenomenological_aspects

To improve your CO2 intake and not lower dissolved oxygen levels, an apparent solution is to make the water in your fish tank more alkaline and also consider increasing the surface turbulence.
The logic is that there is an equilibrium based dissociation of CO2 in water creating Carbonic acid. However, in the presence of OH- ions, the formation of an insoluble carbonate (like CaCO3) will continue to drive the reaction to the right.
On the effect of turbulence from an air pump agitation of the water or perhaps a waterfall creating mist, will capture the carbon dioxide.
So, as solutions consider adding alkaline stones. There is also a supporting article: Limestone Rocks in Aquariums and Water pH, to quote:

Have you set up an aquarium and then noticed that the pH of the water continues to rise? Could rocks or gravel that were added as decor in the aquarium affect the water pH? Yes. If your rocks are actually limestone, they are the cause for the pH elevation in your aquarium water. Limestone is calcareous (contains calcium) and is known for its ability to both harden the water and increase the pH. In fact, if someone asks how to raise water pH, one method we suggest is to place crushed limestone, coral, oyster shell or any highly calcareous material in the filter.

On the waterfall idea a reference, to quote:

Speaking of pond health and waterfalls, they're a great way to aerate your pond, but you might not want to overestimate how much oxygen is getting into your pond from the waterfall. Specifically, the larger your pond is, the higher the waterfall would need to be to effectively aerate it alone.

Hope this helps.

This is interesting, but it doesn't really answer the question, "Will CO2 bubbles ever fully dissolve in water?"

On a practical note, increasing the pH and the hardness of the water interferes with nutrient uptake for the tropical species commonly used in freshwater planted aquaria which are adapted to soft, acidic water; and I don't think carbonate is bioavailable for photosynthesis. Oxygen saturation is really not a concern: photosynthesis converts $\ce{CO2}$ and water into carbohydrates and $\ce{O2}$, which is so abundant it bubbles off the plants. It's not difficult to get oxygen saturation over 100%: the concern for livestock is the toxicity of $\ce{CO2}$ but 30 ppm isn't harmful for most species.

Phil getting CO2 gas to dissolve becomes increasingly difficult when it has been effectively 'salted out'  (because of other dissolved compounds) or due to elevated water temperatures.  However, as the chemistry discussed in my answer notes, no matter how reduced this level, once dissolved in an alkaline solution, it is removed (via, for example, CaCO3)  and more CO2 can be dissolved. The release of acidic products from organisms in the fish tank can release this stored CO2.

That makes sense, but my question isn't about the total amount of CO2 in the water, but rather the fate of an individual bubble. My observation is that if I take a fresh tank, which could absorb many, many grams of CO2, and trap just a single CO2 bubble under a bell in that tank, the bubble will be mostly dissolved in about 30 seconds and then get no smaller. Surely this isn't because the solution is saturated, because adding a second bubble the same thing happens: rapidly mostly dissolved, then no further visible process. Why is that?

In the frame of the question posed by OP, this is waste of CO2.

Way back when the experiment was done with air all the O2, was removed, then all the N2 and there was still air left; conclusion: the AIR remaining wasn't air and argon was discovered. So either you have reached equilibrium maybe unlikely because added CO2 still partially dissolves, or the gas remaining in your bubble is NOT CO2. It really helps to control all variables in an experiment and to be aware that there might be some you weren't aware of.


text:
When do lone pairs affect dipole moments?
My textbook says that NH3 has more dipole moment than NF3 despite F being more electronegative than H, because the dipole moment due to the lone pair on N points in the same direction as the ones due to the H atoms, but not the F atoms. But this is the only time it talks about a dipole moment being caused by a lone pair.
In multiple examples and questions involving calculating dipoles with H20, HF, etc (that contain lone pairs), they are just ignored. It's the same with other websites. Are lone pairs negligible when calculating dipoles if they are on O, F, etc atoms?

I think you are missing the very point of it, and the lone pairs are not the issue. Think of a molecule where N is connected to N, that is, N2. N is more electronegative than H and less so than F, so the dipole moment should be somewhere between those of NH3 and NF3, shouldn't it? Is it?

Otherwise stated when you think of the actual electronegativity of a fragment the distribution of electrons, either in the other bonds or in lone pairs is important, too. Also the electrical dipole moment is a vector. It can be that it sums to zero. So it is not you neglect lone pairs, just the effect is cancelled by some other(s).


text:
When to include the mass of the dissolved substance in calorimetry problems?
In some calorimetry problems, in which a substance is dissolved in water and we want the heat of the reaction, the mass of the dissolved substance is included in the mass plugged into mcat but other times it isn't. When should I include the mass of the substance?
For example, in this problem the mass of $\ce{NH4NO3}$ is not added to the mass of water to find the enthalpy:

$\pu{5.44 g}$ of ammonium nitrate was added to $\pu{150.0 mL}$ of water in a coffee-cup calorimeter. The temperature changed from $\pu{18.6 °C}$ to $\ce{16.2 °C}.$ Calculate the enthalpy change for dissolving $\ce{NH4NO3(s)}$ in water in $\pu{kJ mol^-1}.$ Assume the solution has a specific heat capacity of $\pu{4.18 J g^-1 K^-1}.$

But in this problem the mass of $\ce{CaCl2}$ is added to the mass of water:

In a coffee cup calorimeter, $\pu{2.6 g}$ $\ce{CaCl2(s)}$ was dissolved in $\pu{260 g}$ of water at a combined initial temperature of $\pu{23 °C}.$ The final temperature was $\pu{26.4 °C}.$ Calculate the enthalpy change of the reaction.

I would really appreciate if someone could explain to me when to include the mass of the dissolved substance.

There are multiple issues here, and the answer is: it depends.

How much of your substance is there relative to the solution? If the mass is, say, 1% of total and the specific heat is comparable, then your answer may be off by 1%, which is probably smaller than the systematic measurement error from heat loss in the calorimeter.

What is the specific heat of your substance? If it's very small, then it will hardly require any heat to raise the temperature of your substance.

Frequently, you have to make assumptions. Note that in your first example, it asked you to make assumptions about the specific heat of the solution. This is the specific heat for a mixture of your substance and water. That basically addresses both (1) and (2) above. Note that in this case, since it's the total solution, you should use the total mass.
In the second question, the total mass of calcium chloride is very small, so whether or not you include it is almost irrelevant.


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pH Dependency of IR Stretching Frequency in CO-binded Heme Protein
If one consumes CO, then CO binds with Heme Protein forming a Fe-CO bond. My question is if there will be any pH dependency on IR Stretching Frequency, i.e. ν(CO) and ν(Fe-CO). And if there is, then what will be the result?


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How to identify a blow-out pipette?
Is my pipette a blow-out pipette?
I suspect that it is a blow-out pipette, though the double lines aren't rings. My guess is it is a TC blow-out pipette. It's a big 50 mL pipette, longer than my arm.
I wonder, how am I supposed to know? I also have another one with a single dark band on the top.
EDIT: This video explains how blowing out a pipette is actually done.

Good to see classical analytical chemistry queries. These graduated pipettes or Mohr pipettes they are called, confuse even the relatively advanced users. First you have to check where the graduations end. If the pipette was meant to deliver 10 mL or 50 mL you should not see 10 or 50 mL mark near the tip of the pipette. That is one hint. Add a picture of the tip as well. Secondly, a double ring indicates that all liquid has to be expelled.
Notice that when you are doing very high precision work, one has to know the full drainage time as well after the liquid has been delivered. For example, after delivering 10 mL you need to wait for 5 s before blowing out the last drop. Very high quality glassware use to mention this time on the glassware.
In your blurred picture one can se Ex 5s, which means wait for 5 s before blowing out the last drop.
Volumetric pipettes should never ever be blown out, but they also have a drainage time, you have to wait for a certain time before dislodging the last drop purely from surface tension effects.

Can't I just "help" blowing it out with the bulb? I could blow air into the pipette with the bulb.

If you require high precision, shouldn't you avoid graduated pipettes all together?

Yes, this pipette is a blow out pipette because the 50 mL mark is missing and it has double rings.

@M.Farooq: I guess you need to wait so all liquid, even from the top, drains into a container. You're not supposed to "help" it go down with the bulb?

@user1095108 I was actually asking more general. If I needed a volume of 50 mL; shouldn't I use a more precise approach? I also did not claim that volumetric pipettes have infinite accuracy, just more than what you'll be able to do with a graduated one.

@Martin-マーチン: Oh, my situation is like this: I prepared a standard solution of an acid and stored it in a big 2000 mL glass bottle (I used glass so the bottle wouldn't react with the acid). Since the bottle is so large I wanted a long glass pipette, accuracy is not critical, so I went for the graduated glass pipette. I'm still astounded by the accuracy, is +/- 0.1 mL so bad?

@user1095108, No you are supposed to let the gravity work while the liquid is draining.

@user1095108 We are digressing into a discussion I am not comfortable having, one that doesn't benefit this answer or the question. From a picture of a pipette I cannot possibly judge whatever you're trying to accomplish and whatever you would need for that. I have not directed the question at you or your setup, I have simply asked another question related to it; one that remains unanswered and I do now regret asking it in the first place.

Elsewhere here, M. Farooq has given excellent answer for this analytical chemistry question. I like to explore this answer by explaining what are these lines means and some confusions arose from M. Farooq's answer.
The pipette is an important tool in many fields of science. There are few types of them:

Volumetric pipettes: This type of pipettes is used to deliver a single specific volume of liquid (e.g., $\pu{1, 10, 25, 50 mL}$, etc.). This type of pipettes are reminiscent of big rolling pin with a belly. When emptying its content, the liquid is allowed to drain-out by gravity, should never be blown-out. After emptied the content, a small amount of liquid should remain in the tip of the pipettes.
Measuring pipettes: This type of pipettes is used to measure various amounts of liquids using the same pipette. This type of pipettes are straight glass or plastic tubes with one tapering end. The straight tube is calibrated into small sections so that different amount of volumes can be measured (e.g., see images attached to the question). Measuring pipettes can be divided into two other types: Mohr pipettes and Serological pipettes.
Mohr pipettes: The graduation of this type of pipettes is end before the tip (maximum volume does not include the volume in the tip).
Serological pipettes: The graduation of this type of pipettes is continued to the tip (maximum volume includes the volume in the tip).

There are a few specifications on Measuring pipettes. Closed to the tapered end, some pipettes contain two circles, which indicate the pipette is a blow-out one (see below):

After that, there are calibration information, which contains the maximum volume and the temperature where the measurement is most accurate. For example, if it reads "$\pu{10 mL}$ in $1/10$ TD at $\pu{20 ^\circ C}$" that means maximum volume is$\pu{10 mL}$ and minimum volume is $\pu{0.1 mL}$ at $\pu{20 ^\circ C}$, and pipette is to deliver (TD). If it is TC, then it for to contain (see above image). Some manufacturers use a color coded ring to indicate the maximum volume. For example, Eppendorf uses yellow, green, blue, orange, red, and violet colors to indicate $1, 2, 5, 10, 25,$ and $\pu{50 mL}$ Serological pipettes.
Reference:
Karen Guzman, "Pipetting: A Practical Guide," The American Biology Teacher 2001, 63(2), 128-131.

+1 Sadly, all these basic skills and classical calibration concepts are no longer taught except in universities with strong analytical chemistry programs. Even if you ask a PhD student, they will not be aware of these things, let alone undergraduates.

The reason for me asking this question is this video and the Wikipedia article, which says: "Less commonly, some TD pipettes are made "to contain" as per manufacturer and made to be blown out". As you can see, my pipette has EX specified, but is a TC pipette, so saying this is "less common" is wrong IMO. It seems to be the norm for serological pipettes.

In the Wiki link you provided, I this is wrong  " The delivery time is described as the duration of time that the meniscus reaches the end of the tip starting from the topmost volume, which is specified as 5 seconds for Class AS bulb and graduated-pipette volumetric equipment.". I think Ex+5s means to deliver plus 5s wait before you blow out the last last drop. I don't think it will take 5 s seconfs to empty the full 50 mL pipette as claimed here.

Where does your pipette mention TC?

I think M. Farooq  is correct here. The article I mentioned didn't explain 'Ex' but it make sense it to be for extra.

I think Ex means to deliver.

Yes, the pipette is calibrated to deliver. But I didn't heard of 'Ex' before.

In the top photo by the OP, Ex+5s is written just above 20 C.


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How to compare electrical conductivity of ionic liquids and aqueous electrolytes
I am trying to compare the electrical conductivity of selected ionic liquids (ILs) and aqueous electrolytes.
By ionic liquid (IL), I mean a pure IL -- no water. Electrical conductivity of ILs is measured in units of Siemens per length. For example, the following table from Leys et al., J. Chem. Phys., 2008 gives the electrical conductivity of selected ILs in units of S/m:

On the other hand, aqueous electrolytes are obviously salts in water. I think, then, electrical conductivity of aqueous electrolytes are concentration-dependent. The following table in the CRC Handbook gives the electrical conductivity of selected aqueous electrolytes in units of S cm^2 / mol:

How can I compare the electrical conductivities of ionic liquids and aqueous electrolytes?
In other words, is there a way to relate ionic ($\sigma$ in S/m) and molar ($\Lambda$ in S cm^2 / mol) conductivities?

No, it is an apple to orange comparison. They are changing the frequencies and in the molar conductivity measurements you do not know what was the current frequency.

What do you want to learn from this comparison?

@Karl I hope to show that electrical conductivities of aqueous electrolytes are higher than those of ionic liquids (or vice versa).

Yes, aqueous electrolytes must be better because acids and bases have the highest molar conductivity known. IL are bulky, viscous, I don't think they are great conductors.

@M.Farooq How do you think I could quantify that, from results available in the literature?

No, it is an apple to orange comparison. Ley et al. are changing the frequencies and in the molar conductivity measurements you do not know what was the current frequency. It is usually measured at 1 kHz if I remember correctly. Similarly, S/m cannot be changed to Scm^2/mol. You can indeed calculate the molarity of pure ionic liquids. The molarity of pure water is ~ 55 M. You should know the moles and the volume of the IL, but there is no point in calculating it.
You need to some heavy literature search. People have actually measured molar conductivity of ILs. Also note that many ILs are not water soluble so you have narrow down your choice and goals. Why are you comparing the conductivities to begin with? See this paper "On the concept of ionicity in ionic liquids" in Phys. Chem. Chem. Phys., 2009, 11, 4962–4967.
Also Walden plot might be useful for you.

Why not dividing the conductivity of ionic liquid by their molar concentration in the pure state (similar to the famous 55 M for water) ? The unit will be the same as in the equivalent conductivity of electrolytes.

@Maurice Then you have the same unit, the comparison still makes no sense.


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How to determine if a base is bulky
I remember my teacher saying that during E2 (elimination bimolecular) reactions of alkanes, if the strong base that is used as a reagent is a bulky one, it has a higher probability of giving a Hofmann (less stable) product as the major product even if there are possibilities for a Saytzeff (or Zaitsev) product to form.
How do I determine whether a base is bulky or not?
Some bases like t-butoxide ion are pretty obvious. I usually just guess it is based on how much steric-clouded it seems when I draw a representation on paper. Is there a better way to figure this out?

Related: Is the -OCH3 group bulky?

https://chemistry.stackexchange.com/questions/104019/how-does-base-size-affect-the-ratio-of-kinetic-thermodynamic-enolate

user55119's answer has the details, but the underlying point is that it's empirical. There isn't some theoretical cutoff where you perform some calculations and then it gives you a bulk coefficient $\xi$ where if $\xi > 4$ it's bulky and if $\xi \leq 4$ it's not. Well, maybe you can calculate the size, but then one should ask, exactly what size is considered "bulky"? which brings us back to square one.

This question about the effect of the bulkiness of bases on the mode of E2 elimination has been studied and was published in 1956 by Brown, et al.[1] The elimination of 2-bromo-2-methylbutane was examined using solutions of potassium ethoxide, potassium t-butoxide, potassium t-amyloxide and potassium t-heptoxide in their respective alcohols. The clear result was that increased branching of the alkoxide increased the yield of the Hofmann product over the Zaitsev product. The bulkier the base the less selective for attacking methylene hydrogens over methyl hydrogens. The numbers in parentheses are the selectivities corrected for the number of hydrogens available in each elimination.

H. C. Brown, I. Moritani and Y. Okamoto, J. Am. Chem. Soc., 1956, 78, 2193.


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Using RDKit to count "RR, the number of rigid single or fused ring systems in the molecule"
I am a computer scientist, not a chemist, working with RDKit. I need to compute something described to me as "RR, the number of rigid single or fused ring systems in the molecule."
I know I can access the rings of a molecule with AtomRings and BondRings. However, I don't know how to select only the "rigid" ones. Perhaps all ring systems are "rigid" and the word is just there for emphasis and redundancy? If not, how can I tell?

We're going to need to know what "rigid" means to help here.  One strict possibility is it means only aromatic rings.  Another is that any ring that isn't too flexible should count, so maybe 6 and possibly 7 and maybe (?) 8-membered aliphatic rings would be fine.  But then cyclooctyne isn't very flexible even though it's all aliphatic.

You can try on the RDKit mailing list. We've thought about the same kind of thing and .. it's hard to come up with a good definition. Small rings are rigid because they can't flex (e.g. 3-membered rings). Some rings qualify because they have double or aromatic rings.

I think the easiest thing would be to go through the SSSR (smallest set of smallest rings) and first see if any of the bonds are in another ring system (so fused) and then see if any of the atoms could be considered flexible. The best definition I can come up with for a potentially flexible ring atom seems to be "both neighboring atoms only have single bonds and are not connected to each other." I'm not sure how well that heuristic holds up - "flexibility" of ring atoms mostly goes back to hand-waving around "I can move it."

My group has thought about this as we pursue better methods for finding conformers which would ideally include flexing rings.
I'm not aware of a current descriptor for "rigid" rings because it's one of those concepts that chemists discuss but is more of a heuristic for "I can move some of the atoms." (Which would imply you need some sort of molecular dynamics to determine - solvable but not fast to compute.)
Let's try something like this:

To get the number of fused ring systems, you'd go through the SSSR (smallest set of smallest rings) and see if a ring bond is also a member of another ring.  If so, that's a fused ring system. You'd then need to do graph traversal to find all ring bonds in the fused ring system (i.e., in cholesterol there's only one fused ring system).

To get "rigid single rings" you'd need to go around the ring atoms and see if there is at least one "flexible" ring atom. The best heuristic I've derived for a potentially flexible ring atom seems to be "both neighboring atoms only have single bonds and are not connected to each other."

This last heuristic would ignore 3-membered rings (e.g., neighboring atoms are connected), benzene (i.e., atoms have double/aromatic bonds), .. seems to get most things right.
IMHO, even fused ring systems can flex, which is why my group has centered on item #2 - finding ring atoms that can move. But if you're implementing someone else's heuristic from a paper, you'll need to carefully do the ring traversal in #1 to accurately count the fused ring systems.


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Question on Hydrogen Balloon
I tried to extract hydrogen for my kid's birthday, by mixing caustic soda with water and aluminum foil (mixed in a glass bottle). A lot of heat and (hopefully)hydrogen was released.
However, within 2-3 hours, the balloons started losing altitude and began falling down. I supposed it was caused by losing the gas slowly.
But when I tried to burn the ballon the next day it did not create the pop sound for the hydrogen test.
Please excuse me if this is a naive question. I'm curious since this is the first time I'm dealing with hydrogen and balloons. I attempted this for almost 4-5 time. And every time the balloon stopped flying within 2-3 hours. All other balloons filled with air did not lose gas.
I want to know what happens to the gas. Does it go out? Is it some sort of diffusion?
Thanks.

Friend first of all things using hydrogen is very dangerous and especially in ballons. Be precautious. I have already met with an Hydrogen accident like this one done (faultof the party organizer)

Thanks, I will be, It was outdoor and from where I am, Helium is not easy.

Thanks, It's weird why air ballons did not deflate while Hydrogen ones deflated fast. Also if I understand correctly H2 is inflammable highly, When I ignited the balloon the next day, There was no loud pop sound, just a regular balloon pop sound.

He ballons you can buy on fairs ec. are usually made from some metallised plastic instead of rubber, which makes them keep the He longer. H2 is similarly good in diffusing through thin, biaxially stretched rubber.

Hydrogen diffuses through rubber faster than anything else. Helium is also fast which is why most commercial helium balloons are not made of rubber but other less-permeable plastics which are also usually metallised to further reduce diffusion.

Hydrogen diffuses through steel at 400 F , it is difficult to hold.

I noticed the same thing and I'm not sure why.  Not only are the smaller hydrogen molecules permeating through the latex and exiting the balloon over time but, somehow, air is entering.  I blew the gas into an inverted test tube from a H2 balloon that was no longer floating and, lo and behold, a spark did nothing!  It was 100% hydrogen when I inflated it.

Hydrogen, having very small and fast moving molecules,  diffuses faster and through smaller holes than air. Your balloons seem to be permeable for hydrogen. Try a different brand, if it helps. But as others have mentioned, rubber is not the best material to keep hydrogen. Better is a metallised plastic.
Pure hydrogen just burns in a flame.  If you have ever  watched the film footage (1937)  of the catastrophic accident of the airship Hindenburg , it was just a huge scale of burning in a flame.
What makes an explosive "bang"is a hydrogen mixture with air, even better with oxygen. Because  hydrogen does not explode, until is contains >6% v/v of oxygen.
Any binary chemical explosion of gases requires a self-sustaining chain reaction. That does not happen if a mixture contains too little of one of mutually reacting gases.

Agree with the first part, which is the answer. About the second part, if you mean that pure hydrogen doesn't burn and so it does not pop, fine. But otherwise of course it pops.  I won't compare Hindenburg whit a bottle. Why the Hinderburg did not exploded was likely its pressure and size being big enough, or trapped gas sacks popping in sequence. Nor the balloon of OP contained pure hydrogen.

@Alchimista How do you interpret "It just burns." ? It was not clear or I have missed it that it did not contain pure/nearly pure hydrogen. Hydrogen does not explode, until is contains >6% v/v of oxygen.

I have missed the just burned part. I wasn't discussing the "intensity of the stuff". I simply understood that hydrogen wasn't enough to observe its burning, or perhaps the balloon was much opened for the test. Anyway these are details which are difficult to quantify here.

A curiosity about diffusion. Working with a vacuum/argon line, we observed some vacuum forming overnight if the system wasn't operated. It was due part of the piping .... The material let some of the argon flow out. I wonder how much vacuum one can get using just selective membrane...

@Alchimista The passing gas will diffuse until its partial pressures are equal. If there is 1 L of passing gas at 1 ATM, connected via a semipermeable membrane to 100 L of not passing gas at 1 ATM, the residual pressure at the 1 L side would by approx 0.01 atm

that is quite a "high" vacuum just for that. @Poutnik

What vacuum ? :-) Partial pressures are equal then.

vacuum with respect to P on the other side

Let us continue this discussion in chat.


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Using RDKit to determine "nonring, nonterminal atoms"
I am a non-chemist using RDKit to determine the number of "nonring, nonterminal atoms" in a molecule.
I can't find anything in the RDKit documentation that helps with "nonterminal". Can it be used for that?
I can using RDKit's AtomRings function to identify atoms in rings, but perhaps there is a more direct way?

Welcome! I'm not 100% sure I know what you mean by "non-terminal" but I guess you mean that an atom has more than one bond. I'd suggest using SMARTS - seems like you want something like [*R0!D1]. It's often helpful to use SMARTSview to test these patterns.

I think you want to use SMARTS patterns to perform searches. These are something like "regular expression" patterns for molecular graphs.
At a rough guess, your query converts to [*R0!D1]:

* matches any atom (i.e., you don't care if it's a carbon or an oxygen)
R0 matches an atom in a ring with size 0 (i.e., not a ring atom)
D1 matches an atom with explicitly one bond (so !D1 matches atoms with more than one bond).

It's often helpful to use SMARTSview to test patterns.
More on RDKit's SMARTS support can be found in the documentation


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Dehalogenation of vicinal dihalo alkanes
Which mechanism will the following reaction follow?

Will it follow E2 mechanism?
If yes, then how would 2,3-dibromobutane undergo this reaction?
Does it form both cis and trans butene or is the product butyne?
Here is a link to a website that has mentioned it is E2 mechanism but has low stereospecificity.  So does Zn act as the base donating the electrons? If this happens then why isn't hydrogen leaving the substrate resulting in dehydrohalogenation and formation of alkyne? Why are two halogens leaving the substrate to give an alkene instead of alkyne

A repost of a closed question will be closed as well. The proper way is to address the closure reason by editing of the original question and asking to reopen it.

Now it depends on reagents you use. You need to provide a point where you feel yourself to be unsatisfied. Now there are multiple ways using a bulky and a strong base, using NaI acetone, Using Alcoholic KOH. So you need to specify the reagents also as different reagents follow different mechanisms

I had provided a whole explanation on what I was unsatisfied.
I even requested that an elaborate question was necessary for the contributors to understand what doubt I had.

It was taken down stating that it was too broad of a question and needed specific focus. I am not happy with the moderators for the first time. They seem to be focusing more on the 'how to post a question' than explaining a doubt. I understand serious offenses can be taken down, I don't understand why moderators cannot just suggest edits if they feel the question is so easily explainable in lesser magnitude of words. @Poutnik

@Anonymous The reagent was Zn
Can you just tell me what reaction mechanism it follows? I'll try to continue it myself from there. My teacher had taken it as an example for E2 but i doubt it.

Well, moderators of any moderated sites are not very happy with users posting questions before  getting familiar with the site policy. It seems to me you have confused too broad with too deep, so instead of narrowing it,  you have ended with shallowing it. Site policies are for advantage of the whole SE community, which aims for the high-quality of answers AND questions. Without it, it would be like dozens of other topical sites ,mostly forums, with lots of garbage.

BTW, they were not necessarily moderators, they could be as well senior users with Q close voting privilege. 5 such votes, with chosen reason, is enough for the Q to be closed.

@Desai If there is a metal than the most common use of metal is to give electrons. Now it depends on you how and where you feel that your substance should give electrons(it is on you how judiciously you use the electrons)

Also please add the reagent in this question so that other community members can help you too

I found an article on ChemLibreTexts which states that the reaction proceeds by E2 mechanism but has low stereospecificity.

@Desai provide a link to that in your question and also provide your point of concern in that

related: https://chemistry.stackexchange.com/q/57928/102629


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What is the bond energy of H-O? Wikipedia disagrees with itself
What is the bond energy of H-O? Wikipedia disagrees with itself:

On English Wikipedia:
467 kJ/mol

On French Wikipedia:
366 kJ/mol

What is the right value?

Look, both links are Wikipedia pages containing a table each. The table headers have superscript numbers seven $(^7$ and $^{[7]}),$ which refer to the entries from the list of references on the bottom of each page. These are the sources you should compare, not Wikipedia pages.

Sure, but for the french wikipedia, the link is broken, so I could not compare. For english wikipedia, the link is given the same as information in english wikipedia.

Then I'd prefer English Wikipedia — at least you can look up the source of information. Also, it's closer to the accepted value.

Plus 1,not really for the question that perhaps is a bit naive, but for having spotted the difference. I think this link should clarify why the pages show different values. Out of calculation, one should identify what O-H bond mean. https://en.m.wikipedia.org/wiki/Bond-dissociation_energy

thank you. Apparently, there are several values, depending on the environnement : O−H  in water  497
O−H  in methanol  440

I think generally it's a prudent recommendation to avoid Wikipedia as a reference for precise scientific values.  The CRC handbook is a gold standard for a reason.

According to the 97th edition of the CRC Handbook of Chemistry and Physics, table 1 Bond Dissociation Energies in Diatomic Molecules [1, p. 9-76]:
$$D^\circ_{298}(\ce{H-O}) = \pu{(429.74\pm 0.03) kJ mol-1}.$$
The value has been taken from the article by Ruscic et al. [2]. Note that this value is strictly for the diatomic molecule $\ce{OH}$ in gas phase and is determined as the standard enthalpy change for the fission process
$$\ce{O-H → O + H}.$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Ruscic, B.; Feller, D.; Peterson, K. A. Active Thermochemical Tables: Dissociation Energies of Several Homonuclear First-Row Diatomics and Related Thermochemical Values. Theor Chem Acc 2014, 133 (1), 1415. DOI: 10.1007/s00214-013-1415-z.


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Optical activity of tartaric acid
The meso-form of tartaric acid is optically inactive due to the plane of symmetry. But how are these forms able to show optical activity when the σ-bonds can freely rotate, which can change the configuration of the atoms?

Bond rotation doesn't change the fact that the molecule is still overall chiral and lacking in symmetry.

In other words, bond rotation can't change the configuration of the atoms.

@Nivedita Chowdhury: You are confused with conformation with configuration. There are no bond breaking with free rotation to change the configuration. yet, each isomer has infinite conformers due to rotation.

Consider the R,R-acid. Cut the central C-C bond with your magical chemical scissors. Now you have two tetrahedral objects that are identical in all respects. The fact that they happen to "dock" together is immaterial. After all, they are objects before they are chemical entities. The same holds for the S,S-acid.

https://chemistry.stackexchange.com/a/38989/9961


text:
Is it possible to heat up water using chemicals? (Meaning no outside source of heat like electricity, or gas)
I am asking in terms of heating something small like a 10oz cup of water with it also being safe to consume. It would have only need to heat the water for around 1-2 hour(s).

Not sure what you mean exactly. There are heat packs involved to heating meals for military personnel (“MREs”) and chemical heat warmers for use in boots, etc. But adding a chemical substance to water, in order to heat the water, and then having the hot water be potable, is another matter. I did not downvote, by the way.

It is very easy to heat up water with chemicals. It is impossible to do so and still have drinkable water.

@matt_black  I assume by "drinkable" you meant "safe to consume", since you were responding to the OP, who asked about the latter.  If so, it's not impossible.  Please see my answer below.

Does this answer your question? Can you heat water with additives?

Heat by how much?

Technically, you could heat water by adding chemicals to it, and have something safe to consume. But there's two issues:
(1) In so doing, you wouldn't have pure water any more—you'd have water plus those chemicals.
(2) It's wildly impractical.
But here's how you'd do it:
(1) Add pure concentrated HCl to water. This is exothermic (i.e., thermal energy is released).
(2) Add just enough pure concentrated NaOH to neutralize the HCl. This is also exothermic. The result would be water with dissolved NaCl (table salt), which is safe to drink.
But the amount of HCl and NaOH you'd need to add to obtain significant heating might make the water too salty.
Also, once it's heated, just 10 oz. of water would not stay hot for 1-2 hours without an insulated container.

Here we calculate the amounts and concentrations that would be needed.
$$\Delta \text{H}_{solvation} \text{, HCl} \approx -57 \text{ kJ/mol}$$
$$\Delta \text{H}_{neutralization} \text{, HCl and NaOH} \approx -75 \text{ kJ/mol}$$
Thus for every mole of HCl that we add, and then neutralize with NaOH, about 132 kJ of thermal energy is released.
10 oz. water = 296 mL ~ 296 g
Now suppose we want to heat it from 21 C to 71 C (70 F to 160 F).  The thermal energy needed is:
296 g x 4.184 J/(g C) x 50 C = 62 kJ
Thus we need .5 moles of HCl and .5 moles NaOH.  To avoid the need for special handling, let's use lower concentrations of HCl and NaOH—say 6 M for each (6 M HCl is commonly seen in freshman chemistry labs).  The quantity we would need of each would be:
(.5 moles/(6 moles/L))(1000 mL/L) = 83 mL, for a combined total of 166 mL of NaOH and HCl.
Since our goal is 296 mL (= 10 oz.) of hot water, we would add 83 mL of 6 M HCl and 83 mL of 6 M NaOH (both also assumed to be at 21 C) to 296 mL – 166 mL = 130 mL of water.
[Here, for the purposes of this rough calculation, I've assumed the volumetric heat capacity of salt water is about the same as that of pure water, i.e., ~4.2 J/(mL C).]
This would give a salt concentration of .5 moles/0.296 L = 1.7 M.  That's about 3 x the 0.6 M salt concentration of seawater, and about 2/3 the 2.6 M salt concentration of soy sauce.  So unless you plan to use it as a condiment, you would probably find it unpalatable.

Hm, safety of drinking NaCl solution is  conditional. If you are under conditions to control your daily sodium intake, for medical or dietary reasons....

This particular problem is amenable to a “back of the envelope” thermo calculation, especially for a theorist! :-) So, how much concentrated HCl and solid NaOH would have to be added? If the final NaCl concentration rivals seawater, then I would say it is closer to survivable than potable. Feel free to make any reasonable simplifying assumptions and Happy Holidays! :-)

How about using the "chemicals" dihydrogen and dioxygen, and letting them react?

@EdV  Yes, I was thinking I should do that very calculation and add it to my answer.  Hopefully I'll have time later this weekend.

Great! Then as a loophole cheat, reverse osmosis to get rid of the salt! Just needs high pressure for that and maybe get that as hydrostatic pressure from a largish dam.

@KarstenTheis  Ah, very clever. And even more wildly impractical than my approach. :) Since the OP prohibits "outside sources", it would need to be something added to the water.  So how would that work?  Could the OP could bring along tanks of compressed H2 and O2, and bubble them into the water in the presence of a Pt catalyst?  But then how do you control the reaction to avoid H2O2 formation?

Silver metal catalyzes the decomposition of hydrogen peroxide, as does platinum. So no major problem there

@EdV  Though you'd need a lot of surface area to ensure it decomposes fast enough; I recall it took a few hours for the Pt catalyst to remove the H2O2 from the disinfecting solution for my contacts.


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Reducing olefinic bond in conjugation with C=O bond using LiAlH4
I'm studying the synthesis of alcohols. One of the methods given is "Reduction with metallic hydrides like $\text{LiAlH}_4$". Under this title the following reaction has been listed [1, p.635]:

$$\text{CH}_{3}-\text{CH=CH}-\text{CHO}+\text{LiAlH}_{4}\rightarrow \text{CH}_{3}-\text{CH=CH}-\text{CH}_{2}\text{OH}$$

Also on [1, p.298], the following proposition is made which seems to contradict with obtained results of the above reaction.

In certain scenarios, these reagents may reduce double bonds in conjugation with $\text{C=O}$ as well as reduce the $\text{C=O}$ bond.

It would be great if someone could clarify exactly under what cases and conditions would the double bond reduce as well. Thanks.
Reference:

Shahi, R. Essential Organic Chemistry for JEE Main & Advanced; Arihant  Prakashan: New Delhi. ISBN: 978-93-13192-10-7.

Conjugate reduction with NaBH4 in protic medium is an option because protonation of the resultant enolate will occur and the ketone will reduce to cyclopentanol. Conjugate reduction with LiAlH4 in aprotic solvent leaves a non-reducible enolate. If protic workup protons the enolate faster than excess LiAlH4 is destroyed, then cyclopentanol can form.  Compare with: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not/87195#87195


text:
Phase rule for a pure substance in vapor-liquid region
I'm confused with how the Gibbs Phase Rule works for pure substances in liquid-vapor mixtures. According to the phase rule, there should only be one degree of freedom for a two-phase mixture with a pure substance. For example, if we pick a temperature for two phases to exist, there will only be one pressure that satisfies this condition. But how is the amount of vapor in the mixture, defined?
Looking at the figure below, it appears as if the pressure and specific volume are both independent of the two-phase region. Does this contradict the phase rule? If I wanted to obtain a vessel containing a mixture of water with 50% vapor and 50% liquid, could it be done by changing the pressure and temperature alone?

Note that the phase rule is qualitative. It says nothing about the particular amounts. It simply enumerates the theoretical number of independent parameters needed to determine amounts and compositions for the given number of components and phases.
The definition of a system, like total amounts of components, does not count among degrees of freedom.
The number of degrees of freedom of a system is the number of independent parameters, that are needed for the given system to determine its state.
For the given container volume and the substance amount, a given temperature determines the equilibrium pressure and vice versa. So there is 1 degree of freedom.
Amounts of particular  phases are determined by the temperature as the  single degree of freedom.

What determines the amounts of each phase are the physical constraints. There simply must be enough liquid to produce gas at the vapor pressure. If not enough liquid the system is below the liquid-gas equilibrium line.  If the pressure of the gas is increased [with no inert gas present ie. a pure substance] Gas will condense until the gas phase vanishes.  The phase rule describes what happens at equilibrium and isn't concerned about how to attain equilibrium that is the chemists job. The graph shown describes situations where both phases exist except at high P below Tc

@jimchmst Sure, I was addressing the qualitative rule aspect. I should include note about this too.