text:
Is oxalic acid dihydrate considered a primary acid standard in analytical chemistry?
Using oxalic acid dihydrate as a primary standard just seems really odd to me. I'd expect a primary standard to be oven dried. It just seems weird that a hydrate would be used.
Granted, I saw numerous references to using oxalic acid dihydrate on the web to standardize $\ce{NaOH}$. I assume that for high school and freshman labs that it is "good enough." Using student grade burettes and open air pan balances would greatly limit the possible precision.
Also I'd guess that "good enough purity" oxalic acid dihydrate can be purchased much more cheaply than potassium hydrogen phthalate (KHP), which would be my choice.

The analytical method as I remember from nearly 50 years ago...
Prepare a concentrated stock solution (4 molar?) of $\ce{NaOH}$ using distilled water. That went in a jug with a spout just above the bottom. It was capped with a dedicator tube filled with $\ce{NaOH}$ to absorb $\ce{CO2}$ from the atmosphere. It sat for a couple of days to allow sodium carbonate to settle out. ($\ce{NaOH}$ will have some carbonate.)
Dried KHP in an oven at $\pu{120 ^{\circ}C}$ for four hours and then put that in a desiccator to cool.
Boiled distilled water to remove dissolved $\ce{CO2}$ and stoppered that to cool.
Using the cooled boiled distilled water made an approximately $\pu{0.1 M}$ solution of $\ce{NaOH}$ by diluting the concentrated stock solution.
Using an analytical balance, weigh out 3 samples of KHP to nearest $\pu{0.0001 g}$ into flasks and carefully dissolved the KHP in the cooled boiled distilled water with swirling to minimize introducing bubbles into the solution.
Then using $\ce{NaOH}$ as the titrant, phenolphthalein was used as the indicator. Again careful to swirl solution, not shake, to prevent bubbles.
Using class 1 50-mL burette which was marked to $\pu{0.1 mL}$s but read to $\pu{0.01 mL}$.

Note that it is used also as a standard in manganometry.

Having one chemical that can be used in multiple ways would be good for cutting costs. But I'd expect that sodium oxalate would be a better redox standard than oxalic acid dihydrate.

Perhaps sodium oxalate, yes. Btw, we had at high school analytical scales with 0.1 mg resolution, with putting a wire on the scale shoulders, reading the pointer on enlighten scale.

DAB (Deutsches Arzneibuch) lists a total of eight "Urtitersubstanzen" (= recommended primary standards), plus a dozen more which I might (for lack of a better word) call subprime standards. KHP is among the former, oxalic acid the latter.

@MaxW My memory could be confused earlier, as we discussed it. I guess it may be more than 35 years since my last manganometric titration. Oxalic acid solution of concentration determined by primary standards is being used as a secondary, operational standard for KMnO4 solution. I remember now our high school  (focused on analytical chemistry) professor said it is almost usable as a standard, but not quite seriously.

Jander-Blasius (14. Ed., 1995) uses nonhygroscopic sodium oxalate, dried at 230-250°C (it decomposes above 250 according to wikipedia), to standardise permanganate titer solution against. They give no other recommended standard for manganometry, so I assume this is it.
I have no idea why anybody would want to use (or recommend using) the free acid instead, except perhaps to insult his first-year students' intelligence. Can't imagine it's much cheaper, in analytical grade.
For NaOH titer solution, Jander recommends using a secondary standard, e.g. HCl solution. HCl itself is standardised against freshly precipitated and dried sodium carbonate, or HgO ($\ce{+ 4 KI  + H2O -> K2[HgI4] + 2 KOH}$)  dried over conc. sulfuric acid.

Thanks! I can certainly understand standardizing HCl using NaCO3. However I'd wonder about the accuracy/precision of using HCL as a secondary standard. Just using KHP to standardize the NaOH would seem to be a better technique. KHP would only need 1 titration instead of 2.

I assume it is (or was) regarded as the better alternative, compared to using any of the widely available solid free acids. The book is a bit old school, perhaps KHP sounded too fancy for them. ;-)

My wonder in part of this was about what purity of oxalic acid dihydrate would be good enough for say a high school lab. I'm guessing that a relative standard deviation of 1% would be fine. So analytical grade might be more than is needed.

Our local supplier sells it at 20€/kg (>99%), 50€/kg for p.a. grade (>99.5%). KHP p.a. goes for 80 per kg. There are definitely simpler ways to ruin you high school chemistry budget. ;-)

At first glance, use of oxalic acid dihydrate ($\ce{H2C2O4.2H2O}$ or simply OADH) as a primary standard seems really odd to anybody including me (although I'm not a analytical chemist). I'd also expect a primary standard to be oven dried and cool it in desiccator before use as we all did in our college analytical lab using potassium hydrogen phthalate (KHP). Yet, there is a good reason why these chemist use OADH continuously.
When it comes to be a good primary standard, Low hygroscopicity is one of the features in consideration, in order to minimize weight changes due to humidity. Oxalic acid dihydrate (OADH) and potassium hydrogen phthalate (KHP) are the most accessible acids for standardization, because of their low cost and their hardly changing content per a unit (of measurement). Specifically, OADH in the crystalline solid state, represents an interesting species that is still producing many surprises since 1920s. Although there are two extra water molecules in OADH (thich is hard to loose), the alternating acid and these water molecules act both as hydrogen-bond donors and acceptors [Ref.1].  The conclusion of that reference states that:

The trends both in the metric and electronic parameters exhibited by the model clusters with increasing the number of participating oxalic acid and water molecules (Figure 1) shows clearly that the cooperative effect is the major factor in the shortening of H-bonds. This is particularly well pronounced in the optimized $R_{\ce{O··O}}$ of the carboxylic hydroxyl group bonded to the water molecules. One of the most eloquent steps in the reduction of $R_{\ce{O··O}}$ is between the hydrated single molecule and the circular model consisting of two acid and two water molecules, indicating that it is the circular motif that primarily supports cooperativity. However, both the polarization evoked by the hydration of the carbonyl groups as well as the crystal field effect also contribute notably to the shortening.

This strong H-bonding make it so stable that two water molecule won't loose in ambient conditions. That's a good reason to use it as a primary standard.
Reference:

J. Stare, D. Hadži, “Cooperativity Assisted Shortening of Hydrogen Bonds in Crystalline Oxalic Acid Dihydrate: DFT and NBO Model Studies,” J. Chem. Theory Comput. 2014, 10(4), 1817–1823 (DOI: 10.1021/ct500167n).

Interesting. I have been thinking that being hygroscopic would be much more problematic than the salt losing water. Though I know that for instance exposed copper sulfate pentahydrate tends to lose water of hydration rather than be hygroscopic. // I  wonder what the relative precision would be using both KHP and OADH.

@MaxW: I didn't see any comparison work during my survey. Interesting thoughts though.:-)


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Why does increasing the concentration of oxidizing agent decrease the E° in a galvanic cell?
In a galvanic cell, when we add ions of the oxidation agent (in this example, Zn2+) to the anode solution, it decreases the cell potential.
However, if the Zn2+ ions aren't involved in the redox reaction, then why does the extra concentration affect the cell potential? I know the Le-Chatelier and Nerst equation arguments, but I wanted a conceptual explanation.
I hypothesized that the higher Zn2+ concentration decreased the anode's affinity for solubilization, therefore decreasing the anode's desire to release electrons, thereby decreasing the cell potential, as less electrons will flow per unit of time.

$\ce{Zn^{2+} + 2e^- <=> Zn}$

If the Zn2+ ions aren't involved in the redox reaction, then why does the extra concentration affect the cell potential?

The overall reaction is
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
If this reaction were happening in a common reaction mixture (maybe with Zn and Cu powder on the bottom of a solution) we would write it as
$$\ce{Zn(s) + Cu^2+(aq) <=> Zn^2+(aq) + Cu(s)},$$
and would realize that the rate of the reverse reaction depends on the zinc ion concentration, thus affecting the equilibrium. However, the reaction at hand relies on electrons going through a single wire, and they would not go in both directions at the same time.
MaxW's comment is the key to the conundrum. There are reversible half reactions in either cell, and they determine the cell potential.
$$\ce{Zn^2+ + 2e- <=> Zn}$$
$$\ce{Cu <=> Cu^2+ + 2e-}$$
At each electrode, the half reactions are reversible, delivering electrons to the electrode and picking them up again. This back and forth establishes the potential difference that we can measure between the electrodes.

I wanted a conceptual explanation

For some, a kinetics-based answer (like the one above) counts as conceptual (or particular as opposed to macroscopic?). Here is a second way of thinking about it, using a different but related system. Let's consider a concentration cell with zinc electrodes and zinc ions in both cells. With higher concentration of zinc ions in the right cell than the left cell, the reaction would be:
$$\ce{Zn(left) + Zn^2+(right) -> Zn^2+(left) + Zn(right)}$$
Using your logic, the potential would not depend on the zinc ion concentration in the left cell because that is one of the products. Now we add some zinc ions (in the form of zinc chloride or such) to the left cell, making its concentration higher than that of the right cell. The reaction now would be:
$$\ce{Zn(left) + Zn^2+(right) <- Zn^2+(left) + Zn(right)}$$
Again using your logic, now the potential would not depend on the zinc ion concentration in the right cell because that is now the product.
The question is how the half reaction on the right would "know" to reverse direction, and to stop influencing the potential via its zinc ion concentration. I'm hoping this reductio ad absurdum (I had to look it up) is a convincing argument that zinc ion concentrations in both cells are relevant.

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Can gold cause tin or other metals to corrode?
I stumbled upon the following article while trying to identify some old RAM modules:
https://www.techrepublic.com/article/quickly-identify-ram-chips-with-these-tips/
In the article, it mentions the following about gold contacts:

A note about contact metals
Chip contacts are either gold- or tin-plated. I recommend chips with
  gold contacts because they don’t corrode. Also, never insert
  tin-plated chips into gold-plated sockets. Different metals will
  increase the likelihood of corrosion.

I don't understand how gold cannot corrode on it's own, yet cause another dissimilar metal such as tin to corrode.  Is this article correct?

Yes, that's a pretty well-known thing in chemistry.

Yes, but how?  There has to be some kind of ion exchange between gold and tin correct?

No. But there is an electron exchange between the two, that's for sure.

Speculation: Gold has no protective oxide layer, so at least one electrode of the cell is freely accessible for any ion that comes along to discharge.

First off, read carefully:

Different metals will increase the likelihood of corrosion.

It doesn't say:

Different metals will increase the likelihood of corrosion of gold

With that in mind, it's called galvanic corrosion and occurs when contact between two (or more) dissimilar metals causes the more reactive one to corrode when it normally would not have or faster than it would on its own. This action is similar to a galvanic pile (though its only one cell) that has short circuited thus corroding the more reactive metal.
In your case it is definately the tin that will be corroded, not the gold as tin is more reactive.

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How to acetylate a thiol group in general?
I wonder (and have tried to find the answer by searching in synthetic chemistry resources) how do you acetylate thiol (R-SH) groups in general? For example the cysteine-SH group either alone to make S-acetyl-cysteine or, in the glutathione tri-peptide to make S-acetyl-glutathione.
Below I am using SMILES notation, here completely protonated. I am new to the chemistry section of stackexchange, so not sure how y'all draw structural formulas here.
So this is the reaction in general:
R-SH + X --> R-S-C(=O)-CH3

I am thinking that the best reactant should be acetic anhydride:
H3C-C(=O)-O-C(=O)-CH3

And so, I suppose the reaction is this:
R-SH + H3C-C(=O)-O-C(=O)-CH3 --> R-S-C(=O)-CH3 + H3C-C(=O)-OH

Now specifically for cysteine:
HOC(=C)-C(NH)-C(H2)-SH

We would want:
HOC(=C)-C(NH)-C(H2)-SH + H3C-C(=O)-O-C(=O)-CH3 --> HOC(=C)-C(NH)-C(H2)-S-C(=O)-CH3 + H3C-C(=O)-OH

However, how would I direct the reaction toward the thiol group and not the amino group? Difference between N-acetyl-cysteine and S-acetyl-cysteine.
Next step of the question is how this would apply to glutathione, when we try to make S-acetyl-glutathione? So glutathione
gamma-glutamyl - cysteine - glycine

where we have an available amino-group on the glutamyl part:
OC(=O)-C(H)(NH2)C(H2)C(H2)C(=O)-N(H)-C(C(H2)SH)C(=O)-N(H)-CC(=O)OH
            ^not here                       ^but here

I know people also use catalysts such as simply potassium carbonate or more complex things. Finally, I wonder, how do you get rid of the remaining acetic acid and the potassium carbonate or other catalysts from the reaction to purify the synthesized thio-ester?
I apologize, I have studied medicine not synthetic organic chemistry or pharmaceutical engineering. So, I am asking a lot in one question. I did some research but since I am no longer in academia, I don't have access to journals, such as for A simple acylation of thiols with anhydrides

For formatting see: https://chemistry.meta.stackexchange.com/q/86/23561

Replying to above comment: thanks for the link. Looks like the mhchem format is mostly good for inorganic chemistry, and linear structures. Would be nice to have a SMILES layouter. But anyway, the beauty of SMILES is they can be written free hand and read without special layout.

You can use the @ to ping members.

Welcome to Chemistry.SE! To help you become acquainted with the site, I would just like to suggest if you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and question policies. You will find StackExchange (SE) to be a different kind of Q&A website from the mainstream. To learn more visit Meta.Chemistry, Meta.SE, or chat

I searched Chem Absts. Add acetyl chloride to cysteine in trifluoroacetic acid.  The reference is  J. Am. Chem. Soc., 2014, 136 (31), pp 11050–11056 DOI: 10.1021/ja5048285 although experimental not in paper but probably online.


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What is the ANO-RCC basis set and how does it work?
On the basis set exchange, the ANO-RCC basis looks huge, even for hydrogen:
H    S
188.6144500000           0.00096385            -0.0013119              0.00242240            -0.0115701              0.01478099            -0.0212892
 28.2765960000           0.00749196            -0.0103451              0.02033817            -0.0837154              0.09403187            -0.1095596
  6.4248300000           0.03759541            -0.0504953              0.08963935            -0.4451663              0.53618016            -1.4818260
  1.8150410000           0.14339498            -0.2073855              0.44229071            -1.1462710             -0.6089639              3.0272963
  0.5910630000           0.34863630            -0.4350885              0.57571439             2.5031871             -1.1148890             -3.7630860
  0.2121490000           0.43829736            -0.0247297             -0.9802890             -1.5828490              3.4820812              3.6574131
  0.0798910000           0.16510661             0.32252599            -0.6721538              0.03096569            -3.7625390             -2.5012370
  0.0279620000           0.02102287             0.70727538             1.1417685              0.30862864             1.6766932              0.89405394
H    P
  2.3050000000           0.11279019            -0.2108688              0.75995011            -1.4427420
  0.8067500000           0.41850753            -0.5943796              0.16461590             2.3489914
  0.2823620000           0.47000773             0.08968888            -1.3710140             -1.9911520
  0.0988270000           0.18262603             0.86116340             1.0593155              0.90505601
H    D
  1.8190000000           0.27051341            -0.7938035              1.3082770
  0.7276000000           0.55101250            -0.0914252             -2.0210590
  0.2910400000           0.33108664             0.86200334             1.2498888
H    F
  0.9701090000           1.0000000
END

It looks  very different from the basis sets that I'm used to, which are Dunning's correlation consistent basis sets. Since the above example is a 4-zeta (QZ), I provide the cc-pVQZ below:
H    S
 82.6400000              0.0020060
 12.4100000              0.0153430
  2.8240000              0.0755790
  0.7977000              1.0000000
  0.2581000              1.0000000
  0.0898900              1.0000000
H    P
  2.2920000              1.0000000
  0.8380000              1.0000000
  0.2920000              1.0000000
H    D
  2.0620000              1.0000000
  0.6620000              1.0000000
H    F
  1.3970000              1.0000000
END

I do not understand why:
1) ANO-RCC has so many more exponents than cc-pVQZ at each angular momentum (2 more for S, 1 more for P, one more for D, but the same number for F!).
2) ANO-RCC has so many more contraction coefficients (3 columns for S, rather than 1 column in cc-pVQZ, and many more non-unity contraction coefficients for S, P, and D, but not F!).
It also appears that the ANO-RCC basis set has to be used in a special way. For example, do I just plug these exponents and contraction coefficients into my program? What if I want a 3-zeta ANO-RCC basis set: I know to chop off the F line, but how many lines for the other exponent types, do I chop off? Can I extrapolate to the CBS limit using 2-zeta, 3-zeta and 4-zeta?
I understand that a good place to look would be the initial papers where ANO-RCC was introduced (by Bjorn Roos?) and the follow-up papers where they reported the optimization of new basis sets. But I do not have access to journal articles since I am not enrolled at a university or working for any company.

Related: http://www.molcas.org/documentation/manual/node116.html


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Understanding Aromaticity
I was just doing a simple question and for no reason, I drew the resonating structures of the compound as shown.
I can see that some of the resonating structures seem to be aromatic while others seem to be non-aromatic
And now I wonder..

Is aromaticity a property of a coumpound or any of its resonating structure? This compound has some Aromatic and one Anti-aromatic structure. Shall we call this compound aromatic or Anti-aromatic?
To compare stabilities of resonating structures, how will we use aromaticity and what will the stability order?

There's nothing antiaromatic in there. Resonance not "resonating" structures aren't more or less stable, but are minor of major contributors to resonance hybrid,  more or less relevant to depiction of the molecule.

So, we shall say that the compound is aromatic as per Aromaticity rules and structure b contributes the least to it as it promotes Anti-aromaticity?

No. As Mithoron pointed out, structure (b) is not anti-aromatic.

Then, how will we compare the contributions to the hybrid?

Will not aromaticity be used to compare the contributions?

Structure b is not anti-aromatic but can we say that it promotes Anti-aromaticity in the resonance hybrid?

The point that confuses you is to identify some of the mesomers as aromatic (though is an almost automatic way of thinking and is understandable, even useful). But none of the mesomers you draw is per se aromatic. It is just that a) and iv) describes aromaticity, they are among the most important. Ergo the hybrid will be more or less aromatic.

The reason for the question is that OP implicitly goes a step too further while identifying "aromatic" limiting forms.
This is of course quite natural, as we treat a limiting structure that is compatible with Huckel aromaticity as aromatic, ie we all more or less consciously or not replace three double bonds with a delocalized ring of electrons.
However not one individual limiting form is aromatic, each can be at most an aromaticity descriptor. In the example given, saying that a) and iv) are aromatic, although sensible, is logically incorrect, or at least involves the reasoning I have described above.
They are certainly important limiting forms as their linear combination stabilise the molecule which will therefore be more or less aromatic. But a mesomer cannot be aromatic by definition.*
I do not enter into the correctness of the antiaromatic part of the question, see comments above pointing to the fact that there is none.
Antiaromaticity is indeed confusing as for it should be taken as a rule for what
it does not happen.
However, this means that limiting forms leading to an antiaromatic flat system will have no weight or contribute marginally.
The concomitant occurrence of sensible limiting forms leading to aromaticity or formal antiaromaticity is not possible within the same flat cyclic system, as it would require a different number of electrons, so at least this point does not need to be discussed :)
*of course in a big molecule we can treat a far away, not resonating group as aromatic, but by doing this we implicitly account for all mesomers leading to its aromaticity.

I have understood till now that aromaticity has nothing to do with mesomers. It is the extra stabilisation of a compound, not its mesomers. It is related with a resonance hybrid(the reality) against the resonance structures. Am I correct? The doubt now is, if I am correct, how will we compare the contributions of each resonance structure. How will we define which structure contributes the most, and which the least and the reason therefore?

Does 'limiting form' stand for resonance structures?

As a, i,ii,iii and iv follow huckel's rule, and they are aromatic 'descriptors', can we say that the left out b contributes least to the resonance hybrid?

As usual! Yes assigning a minor weight to b) is anyway done quickly, as for Cl carries a + charge. Basically a) & iv) are the major contributors even without considering aromaticity. The fact that they are degenerate and lead to aromaticity is even a plus. Refer to the common criteria for qualitative ordering of mesimers. Yes, lim. forms stands for structure in resonance leading to the actual, real hybrid of resonance (which, after such a long discussion, will be much similar to a) resonating with iv)).

In b) there is a -ve charge on Cl(not +ve as u said) and there is one more covalent bond which increase its contribution in hybrid. This is leads to confusion to me:-). Clarify again please.

Sorry didn't care much of the unimportant limiting forms as for I ve intended to clarify your original understandable confusion. Then b) is wrong as for it violates the octet rule.

No man, chlorine has an expandable octet, it has vacant d orbitals. This is a case of back bonding.

Correct. So what was your question at the end? The role of b does certainly result in a p-Chlorophenol to be more acidic than phenol. I would have explained this via an inductive stabilisation of ii) but b) definitely plays a role, else we couldn't explain why this base is less stronger than the fluorinated counterpart.


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Calculating pre-exponential factor using transition state theory
I would like to calculate the pre-exponential factor of a surface reaction using transition state theory. I've seen two ways to calculate it but I get different answers for each method (and I don't understand why):
$A = \frac {k_b T} {h} \frac {q^{TS}} {q^{reactant}}$ ... (1)
$A = \frac {k_b T} {h} \exp \bigg({\frac {S^{TS} - S^{reactant}} {R}}\bigg)$ ... (2)
where $q^i$ is the partition function of species i, $S^i$ is the entropy of species i, $k_b$ is the Boltzmann constant, $T$ is temperature, $h$ is Planck's constant and $R$ is the molar gas constant.
I assumed the adsorbates only have vibrational contributions to $A$ and the vibrations can be described using the Harmonic approximation.
I am using the following equations to estimate $q^i$ and $S^i$:
$q^i_{vib} = \prod_j^{n_{vib}} \frac {1} {1-\exp \big(- \frac {\Theta_j} {T}\big) }$ ... (3)
$\frac {S^i_{vib}} {R} = \sum_j^{n_{vib}} \frac {\Theta_{V,j}} {T} \frac {\exp \big({- \frac {\Theta_{V, j}}{T}}\big)} {1 - \exp\big( - \frac{\Theta_{V, j}}{T}\big)} - \ln \bigg(1-\exp\big(-\frac{\Theta_{V,j}}{T}\big)\bigg)$ ... (4)
where $n_{vib}$ is the number of vibrational modes, $\Theta_{V,j}$ is the vibrational temperature of a mode (in K).
Inputs
T = 750 K
Reactant $\Theta_V$ in K:
0.42190081
0.08952052
0.08826819
0.06520753
0.05161751
0.05077054
Transition State $\Theta_{V}$ in K:
0.21401905
0.06477819
0.05798548
0.05276798
0.04599558
0.14870034$i$ (Imaginary frequency is ignored)
Results
$A_{eq 1} = 1.61\times 10^{13} s^{-1}$
$A_{eq 2} = 1.36\times 10^{13} s^{-1}$
Intermediate values
$q^{TS} = 9.743$
$q^{reactants} = 9.430$
$\frac {S^{TS}}{R} = 4.93$
$\frac {S^{reactant}} {R} = 5.07$


text:
Can a solute be physically filtered out of a solvent?
Suppose that I have a substance dissolved in water. I then pour the solution into a filter. The filter's pores are large enough to allow the passage of water, but not the passage of the substance. In this case, would the substance come out of solution as the water goes through the filter but the substance doesn't, thus resulting in a solute being filtered out of a solvent? In this scenario, the substance is a molecular compound that does not dissociate into ions when dissolved.
In a second scenario, what if the substance is an ionic compound like salt, and the filter's pores are large enough to let one of the ions through (And of course, water molecules can also pass), but not the other? What would happen then?

Do a search on reverse osmosis for the first part of our question.

Also search under dialysis  in biochemistry (e.g., Wikipedia). Also look for ultracentrifugation if interested in separation without filtration.

Okay, but what about if a filter were to only let one of the ions that a salt dissociates into through and not the other? The reverse osmosis article on Wikipedia doesn't seem to answer that question


text:
Pressure of Gas Collected Over Water
My textbook, as well as practically every online source I have looked into, says that when you collect gas over water, the pressure of the trapped gas plus the vapor pressure of water at that given temperature is equal to the outside atmospheric pressure, or $P_g + P_v = P_a$.
However, this doesn't make much sense to me, as the weight of the column of water below the gas is being ignored (or at least that's how it seems to me). Shouldn't the external/atmospheric pressure equal the total pressure of the trapped gas(es) above the water PLUS the pressure contributed by the weight of the water below the trapped gas (just like how the force exerted by the weight of the water column is considered in a barometer or manometer)?
Thanks in advance for helping me clear this confusion up.

Are the water levels inside and outside the collection vessel equal?

Based on what I have seen online, the water levels are not the same (could just be a bad drawing). Does this make a difference either way?

If the water levels are the same inside and outside, the gas pressures above the inside and outside surfaces must be the same.  Otherwise the gas pressure inside will not match the outside pressure.  The inside gas pressure, of course, can be calculated from the hydrostatics.

It is correct. Without gas there would be vacuum, or better just Pv.. When the niche P equals atm P the liquid column does not move.

@Michael — The level zero (for the water) is when the inner and outer pressures are equal, and so water levels evens out. If you write the exact equations for such a system, you will find out that water levels (inside and outside) are actually cancelling out, so in the end only the inner vs outter pressures matter

So is the general consensus that the only way the collected gas pressure equals atmospheric pressure is if the levels of the water inside the collection tube and outside the collection tube are equal?

Yes.  I agree with that assessment.


text:
Question about liquid-phase in P-T Diagram
Currently, I have been learning about P-T diagrams and vapour pressure and I have a question that seems like it should have a simple explanation yet I am not able to understand it.
Taking the P-T diagram of water for example,

Suppose we choose a point that corresponds to the liquid region in the P-T diagram. From what I know of vapour pressure, every pure liquid in a closed container has a vapour pressure and will eventually reach a liquid-vapour equilibrium. If this is the case, what is the meaning of a "liquid region" in a P-T diagram if there must always exist a vapour-liquid equilibrium? And by extension why is there even a coexistence curve between liquid and vapour in the first place?

It is normal practice to include a reference to the figure unless it is your own :) A figure may also be copyrighted.

Wow, now that's completely misleading question title... And what you say is incorrect. You can get some water with no contact with air very easily.

The gas phase is unique in its ability to expand to fill space, so if there is space above or around a solid or liquid, molecules will eventually enter the gas phase in order to fill that space. That's the vapor-liquid equilibrium you described.
The liquid region of the phase diagram is the region where, in a closed space, some of the material will remain indefinitely in the liquid phase, even though some is also in the vapor phase. For example, in a sealed but expandable container (like a piston) maintained at an external pressure of 1 atm, the equilibrium state at 80 C has some water and some water vapor. But at 110 C, it will be all vapor.
The "coexistence curve" is a misnomer for the curve you are referring to. It does not mean that the liquid and vapor simply coexist (which as you pointed out can happen throughout the liquid region). It means that the equilibrium vapor pressure is equal to the external pressure. At that point, you cannot increase the temp of the liquid any further if the pressure remains constant. All heat added will go towards converting liquid to gas. Only after it is all gas will the temp increase. In order to get liquid at a higher temp, you have to increase the pressure. We define this as the boiling point.

Thank you for the explanation. So it would be correct to say that the P on on the y axis refers to external pressure and not vapour pressure? And it can only be interpreted as vapour pressure along the coexistence curve?  If so, then throughout the liquid region is is correct to say that there always exists a vapour-liquid equilibrium but the amount of vapour pressure differs depending on the temperature no? Then what mathematically defines the vapour pressure in the liquid region because I know that the Clausius-Clapyeron defines vapour pressure at the coexistence curve.

@rofldude188 If you draw a vertical line from the point representing current conditions to the liquid/vapor boundary line, the pressure value at that point on the line represents the equilibrium vapor pressure at that temp. So mathematically, it's given by the C-C equation even though the total pressure has moved you off the boundary line.

If the external pressure is 1 atm,, doesn't the internal pressure also have to be 1 atm?  At 80 C, the equilibrium vapor pressure of water is about 0.5 atm.  So, with an external pressure of 1 atm, the piston will not be in equilibrium.  The only way it can be in equilibrium is if there is no water vapor, and the water in the cylinder is all liquid at 1 atm.

@ChetMiller - yes, the internal pressure must also be 1 atm. I'm assuming that another gas is present so the piston is at equilibrium before the water vaporizes. As the water equilibrates with water vapor, the volume increases, but the total pressure internal remains 1 atm. The partial pressure of the water vapor will be the 0.5 atm you mentioned. The other gas will contribute the other 0.5 atm.

@Andrew If the volume of the closed container is very big (e.g. $1000$L) will all the liquid turn into gas?

@Anton - depends on how much liquid you start with

The liquid region corresponds to pure liquid at a pressure greater than the equilibrium vapor pressure at the liquid temperature.  Imagine liquid in a cylinder, with a piston pressing down on the liquid with a pressure greater than the equilibrium vapor pressure.  It is also possible to have the liquid being pressed upon by a non-condensible gas at a total pressure above the equilibrium vapor pressure; of course, in this case, at equilibrium, there is also vapor within the non-condensible gas with a partial pressure equal to the equilibrium vapor pressure.  But, as far as the bulk liquid is concerned, this gas phase in this case is essentially the same as a piston.

So it would be correct to imagine the liquid region as a liquid in a closed container with an external pressure corresponding to the P value on the y-axis of the diagram? If so, then how would one determine the vapour pressure here as it obviously exists but will be significantly less because of the greater external pressure?

If you have a gas phase over the water (including a non-condensible gas), the partial pressure of the water vapor in the gas phase is equal to the equilibrium vapor pressure of water at the system temperature.  If there is no gas phase over the liquid water, then the vapor pressure is irrelevant.


text:
Colour change when excess ammonia added to [Cu(H2O)6]2+
Since ammonia is a stronger ligand than $\ce{H2O}$ shouldn't the colour of the new solution $\ce{[Cu(NH3)4(H2O)2]}$ be lighter as the energy gap will be higher and so light with higher frequency and energy (between blue and indigo) be absorbed which means that the transmitted light is the one between blue and green which is lighter? Why does the solution turn deep blue?

Partly related: How hydrate isomerism results difference in colours?

Like the OP mentioned, the replacememt of a weak field ligand ($\ce{H2O}$) by a stronger one ($\ce{NH3}$) certainly increases the $\Delta_o$ of the $\ce{Cu(II)}$ complex.
A chemical sample appears colored when it absorbs a certain wavelength of light. What we see is the sum total of the remaining colors. The colour visible to the human eye would be the complement of the colour(i.e. wavelength) absorbed.

The increase in $\Delta_o$ leads to the absorption of a lower wavelength (or higher frequency) of light, as evident from:
$$E=h\nu= \frac {hc}\lambda $$

Note that a higher frequency would mean a shift towards the blue end.
Quantitatively speaking, the light blue colour of the initial hexaaqua complex was due to the absorption of the orange spectrum range of light.
Back to the color wheel, a shift towards the blue side (in the green direction, please) gives us a lighter shade of orange.

Why does the solution turn deep blue?

Observe the complement(the colour diametrically opposite) of orange-yellow is definitely a deeper shade of blue.

So whatever is complement to the absorbed colours in the colour wheel will be the transmitted colors(i.e. ones observed)?

Yes.
If absorbed is light red/dark orange, the solution colour is light blue. If absorption shifts to more energetic bright-orange/yellow, the solution colour shifts to violet.

Exactly. Do read the book "The Physics and Chemistry of Color - The Fifteen Causes of Color". You will benefit from the experience.

The supplied answers missed a very important point. Not only is there a shift in adsorption peak, the molar attenuation coefficient is much greater for the amine complex than it is for the aqua complex.
This image was taken from a paper "EXPERIMENT 7B FORMATION OF A COMPLEX ION" from Los Angeles City College at webpage:
https://www.lacitycollege.edu/Academic-Departments/Chemistry-Earth-Sciences/documents/E07B-complexion-(1).pdf

I agreed. Sometimes, we ignore how deeply it absorbs (make big difference). Thanks for the point.

I went through the document and I am still not clear as to why the tetraammine complex has more intense transitions. Can someone discuss how to decide the intensity of transitions in all such complexes?


text:
Whether the entropy change of the system is positive or negative
$$\ce{N2(g) + O2(g) → 2NO(g)}$$
My thinking: since gases are combined together, so there randomness or disorder gets decreased. Hence the sign of entropy change of the system should be negative.
Book Solution: the sign of entropy change is impossible to predict.
Why can't we predict the sign of entropy change for the given system?

Why would randomness get decreased? What's "together" about it? We had two gas molecules, and we still have two gas molecules.

If the reaction reaches equilibrium (and temperature constant) then entropy will have increased as there are three types of molecules present. If it goes to completion so that only NO is present then entropy decreases as there is only one type of molecule left and not two. If the reaction were A+B$\to $ C+D then the entropy would not change.

It is impossible to predict based on the simple basis covered by the book at that point, chapter or paragraph (entropy as disorder, indistinguishable particles).

The entropy of mixing is not the only factor here.  There is also the entropy change as a result of breaking and making new chemical bonds (and energetic effects associated with this).

I would think that the entropy should increase in this example to first order. The NO molecule is less symmetric, so by rough hand-waving there are more microstates in the NO system.

Could you provide a reference (source) for the stated problem?


text:
How to prepare a Gaussian input file for a binary solvent?
The UV absorption energy spectrum of a molecule has been recorded experimentally using binary solvent (DMSO:Water=9:1 v/v). I am trying to calculate the absorption energy of the same molecule using computational methods. For that I want to give the same solvent mixture (DMSO:Water=9:1 v/v) in my DFT calculation.
How can I prepare the input file for Gaussian? What are the keywords that are used in Gaussian for binary solvents?
I have calculated the dielectric constant of the binary solvent (DMSO:Water=9:1 v/v) using the following expression
$$\epsilon_\mathrm{m}= x_1 \epsilon_1 + x_2 \epsilon_2,$$
where $x_1$, $x_2$ are the mole fractions of each solvent in the binary mixture, subscripts $1$ and $2$ denote the different solvents, and $\epsilon_1$/$\epsilon_2$  their dielectric constants.

This sounds like a straightforward job for Gaussian, except for the potentially large size of the system. A basic input file might work. Look in the manual.

I am unaware that this feature exists in Gaussian. It is certainly something that stretches the possibilities of a polarisable continuum model. I am quoting (liberally) from an email to the computational chemistry list (CCL, http://www.ccl.net/), because I found no evidence, that this has changed for Gaussian 16:

Actually, there is no standard approach for solvent mixtures to you use it directly in Gaussian03/09. A good suggestion is to define the solvent parameters as a linear combination of the solvents in the proportions of the mixture.  So for toluene:water in a 4:1 mix use 4/5 of the toluene values plus 1/5 of the water values.
It is recommended to use the SCRF=SMD approach here, but that is not recommended for properties other than solvation free energy. It may be give satisfied results in the case of geometry optimization or energetic properties.  It is especially not recommended for derivative properties, because the cavity for the SCRF=SMD model has a number of continuity issues.  So for the default SCRF=IEFPCM model with the smooth cavity you mainly need the static (or zero-frequency) dielectric constant and the dynamic (or optical) dielectric constant.  You would specify EPS and EPSInf in the additional SCRF input section after the blank line which terminates the structure input.

For example:

{%chk=}
#P {Method, Basis, additional keywords} SCRF=(Solvent=Generic,Read)

{title card}

{charge, multiplicity}
{molecule specification}

EPS=(Add the value of linear combination of both solvents)
EpsInf=(Add the value of linear combination of both solvents)

See the Gaussian Manual for more details.

Another approach is to first do an optimisation and related properties calculation in both pure solvents, and then compare these results. For example the difference in the structures, or spectra.
Since you are using nine parts DMSO and one part water, it might well be that the DMSO calculation is a well enough approximation to the mixture.
Solvent calculations are tricky and often enough unreliable. One should never ever expect a quantitative agreement. It is a qualitative description you should be after.
If you need more accuracy, you have to use an explicit solvent model with MD calculations, or you can look into COSMO-RS.


text:
Would an explosion of an electrically charged material be more powerful than if it were not charged?
I was recently reading a book titled "Hitler's Suppressed and Still Secret Weapons, Science and Technology". I came across this book because I saw it referenced by someone on another site. The book's author claims that there is a variation of the typical fuel-air bomb (Where a cloud of explosive material is dispersed into the atmosphere and then detonated), whereby an explosive material is dispersed into the atmosphere, charged with electricity as much as possible, and then detonated. Supposedly, the explosion of this modified fuel-air bomb is more powerful than a conventional FA bomb. Alternatively, the explosive material is wrapped in an insulator and then charged. Now, I am very, very skeptical of this, especially since that book has been criticized as being into conspiracy theory territory. However, I am curious: If a material that acts as fuel were to be electrically charged (Like having a build up of static electricity) before reacting with oxygen, would it result in a greater release of energy than if it were not to be charged?
For instance, suppose that a cloud of coal dust happened to have a significant buildup of static electricity, and then it explodes. Would the explosion be more powerful than if the cloud did not have any static buildup?

Only if the amount of charge were significant compared to the chemical energy involved. And this is impossible as it would involve such huge voltages that the air would ionise well before a decent amount of charge were built up.

related: why sodium explodes


text:
Oxidation states of Cu in YBa2Cu3O6 based on crystal structure
How does on go about this? I understand There is a +5 charge distributed on 3 copper ions. So either we have 2+, 2+ and 1+ or 3+, 1+, and 1+. Judging from the structure there are to unequivalent types of Cu ions. But how does one determine which is which and if the formal charges are +2 and +1 or +3 and +1 respectively
Any help is appreciated

You don't, really.  Oxidation state is just a formalism.  The positive charges on copper are likely delocalized.

You're talking about *oxidation states, not formal charges...


text:
Is a discharged battery in a state of chemical equilibrium?
A reaction that has reached chemical equilibrium no longer shows changes in reactant and product concentrations, but forward and reverse reactions are still going on. I'm confused whether a completely discharged battery is in a state of equilibrium.
Argument that an equilibrium has been reached
If I measure the cell potential, it will be zero. That means that the Gibbs free energy of reaction is zero, which is a property of a chemical equilibrium. When I discharge a battery by shorting it, there are concentration changes in the battery that cease when the battery is discharged, which looks like a reaction going toward and then attaining equilibrium.
Argument that this is not an equilibrium state
One property of a chemical equilibrium is that it is dynamic, i.e. the forward and reverse reactions are ongoing but occur at the same rate. For the battery, the reaction does not happen unless I close the circuit by electrically connecting the electrodes (e.g. shorting the battery with a wire). Even if I connect the electrodes with a wire, no current will flow because the battery is dead. So I don't see forward and reverse reactions going on. This would be like placing two beakers with 1 M aqueous salt solutions next to each other and saying they are at equilibrium. However, if the solutions were both contacting a dialysis membrane, ions would flow in either direction (at the same rate) so it would be fair to call this a state of dynamic equilibrium.
Which argument wins?
Is the discharged battery at a state of equilibrium or not? Does it matter whether the electrodes are connected electrically to decide on the answer?  Does it make sense to discuss a state of equilibrium in the context of electrochemistry in the first place?

Yes a discharged battery, with the two terminals of the battery connected, is at a state of equilibrium.

Post-script: After recently encountering a related post I re-read my answer here and realized I could have provided a much clearer and more useful definition of equilibrium (rather than my somewhat philosophical earlier answer) that encompasses the different cases you present: in thermodynamics, a system is in an equilibrium state when no lower free energy states are accessible.
Therefore, the two beakers, just like the exhausted battery, are at equilibrium, because there are no accessible lower energy states under the defined conditions.

You can ask yourself: if the universe is expanding (that is, if it is in a dynamic state), is thermodynamics truly ever applicable? The answer is that in science we need to take approximations, during measurement or when extracting information from data. So it is in thermodynamics, which works well enough even though on long enough scales the approximations fail (on timescales of the "in the long run we are all dead" sort).
This is of course a tricky point. We often need to distinguish between "real equilibrium states" and "kinetically trapped or metastable states". However it is useful to invoke a "duck-typing" rule applicable to thermodynamics problems: if the measured properties of the system remain constant and homogeneous during the interval of the experiment, then the system can be considered to be at equilibrium during that interval. This definition is particularly valuable since thermodynamics is an empirical science. Caveat: that doesn't mean you can be sure that the system is in the state you meant it to be in (!), since if you wait long enough you might end up in another state (which might be the one you actually wanted the system to be in!). Perhaps more important is reproducibility: can you put the system into the desired state, reproducibly?
To summarize: the concept of equilibrium, like that of reversibility, is subject to the practical condition that no variations be observed across the system or over time during the course of the experiment, when the system is in the proposed state.
As regards a battery, if you don't see a change in the potential, you can regard the system as being "at equilibrium", at least temporarily! Again, the definition of equilibrium is not as strict as we'd like to think. The materials in the battery, left exposed to the environment long enough, will rust and fall apart, leading eventually to  other "equilibrium" states.


text:
Adding weak acid to strong base, find pH
30mL 3M acetic acid (pka 4.7) is added to 50mL 2M NaOH. what is the resulting pH
a) <2
b) between 2 and 7
c) 7
d) >7
ans =d
can someone explain how to actually solve for pH? the solution describes doing this as a limiting reagent problem in which you solve for the moles of acetic acid and sodium hydroxide.
I'm just a little lost because doesn't acetic acid form its own equilibrium? So doesn't acetic acid form an equilibrium, which means that the actual acid in solution is not that which you calculate from simply multiplying the initial concentration by the initial volume?

HINT - How many millimoles of acid and base are there?

It does but it only raises pH further - you'd need to treat the solution as buffer if there would be less NaOH added then the amount of acid .

It's not really it's own equilibrium in the sense that all of the acid-base equilibria are coupled via the exchange of hydronium ions.

As MaxW intimated, you are adding more millimoles of a strong base to fewer millimoles of a weak acid. Intuition, if nothing else, tells one that you will have an excess of strong base remaining and therefore the final solution will be basic; that is >7.


text:
Why is N-acetyl alanine methyl amide considered a dipeptide?
I'm going through an Amber Molecular Dynamics tutorial here and am a bit confused about why in step 8, we create a sequence as:
foo = seq { ACE ALA NME }
and this is called a dipeptide. I think there's only one amino acid, alanine, isn't there?

I would consider this a monopeptide, but there is a non-zero probability that I am missing some other salient point about why this particular system was chosen for modelling.

I miss seeing why this was closed, even if the posed question does not reveal a remarkable amount of effort. Basically the poster says: the peptide contains one aminoacid, so why is it called a dipeptide? Not sure what else the critics were looking for.

I agree, there’s really no need for closure.

Plus, more factually, it wasn't a homework question but one that I asked after discussing with other computational folks in the lab I work in. Given the responses denote some ambiguity in the interpretation of counting peptides, this seems like a helpful reference to keep around.

The nomenclature here is based on counting the number of backbone peptide bonds rather than the number of aminoacids. This does not apply the usual naming convention.
If you look at the wikipedia page on "peptide (amide) bond", it states

A peptide bond is an amide type of covalent chemical bond linking two consecutive alpha-amino acids from C1 (carbon number one) of one alpha-amino acid and N2 (nitrogen number two) of another along a peptide or protein chain.
It can also be called an eupeptide bond to separate it from an isopeptide bond, a different type of amide bond between two amino acids.

Furthermore,

When two amino acids form a dipeptide through a peptide bond it is type of condensation reaction.
[...] The two joined amino acids are called a dipeptide.

Note that you can have aminoacids connected through their side chains. Glutathione (GSH) comes to mind as a good example of this:

It contains three aminoacids (Gly, Cys and Glu). Would you call this a "tripeptide"?
The wikipedia does:

It is a tripeptide with a gamma peptide linkage between the carboxyl group of the glutamate side chain and the amine group of cysteine, and the carboxyl group of cysteine is attached by normal peptide linkage to a glycine.

What about the bonds in GSH? The wikipedia goes on:

[...] the tripeptide glutathione is synthesized in two steps from free amino acids, by two enzymes: glutamate–cysteine ligase (forms an isopeptide bond [emphasis mine], which is not a peptide bond) and glutathione synthetase (forms a peptide bond).

So, here they call GSH a tripeptide based on the number of aminoacids, not the number or even type of "peptide bond".
Hmm, pretty confusing. Perhaps then it is best to go to the original literature on the subject, linked in the wikipedia:

"Nomenclature and Symbolism for Amino Acids and Peptides. Recommendations 1983". European Journal of Biochemistry. 138 (1): 9–37. 1984.

"Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)". Pure and Applied Chemistry. 66 (5): 1077–1184.

According to reference 1:

Part 1, Section C : PEPTIDE NOMENCLATURE
3AA-11. DEFINITION OF PEPTIDES
A peptide is any compound produced by amide formation between a carboxyl group of one amino acid and an amino group
of another. The amide bonds in peptides may be called peptide bonds. The word peptide usually applies to compounds whose
amide bonds are formed between C-1 of one amino acid and N-2 of another (sometimes called eupeptide bonds), but it includes
compounds with residues linked by other amide bonds (sometimes called isopeptide bonds). Peptides with fewer than about 10-
20 residues may also be called oligopeptides ; those with more, polypeptides. Polypeptides of specific sequence of more than about
50 residues are usually known as proteins, but authors differ greatly on where they start using this term.

There is some degree of ambiguity regarding the numerical prefixes to the term "peptide", which are not discussed explicitly. But the above article provides a solid guideline about what you should call a peptide, namely two or more aminoacids linked through amide bonds.
So what about AcAlaNHMe (the capped alanine peptide in question)? One reason it is called a dipeptide (and not a peptide or tripeptide) is to emphasize that it contains two peptide (or amide) plains with intervening $\phi$ and $\psi$ dihedral angles which confer the potential conformational flexibility of a regular tripeptide. This is discussed for instance in this article.

In your last sentence, do you mean it confers the conformational flexibility of a regular dipeptide?

@MarteriaGirl I actually meant a tripeptide, a regular linear peptide consisting of three aminoacids linked by two peptide bonds, with the central aminoacid and the peptide planes defining psi and phi dihedral angles. The phi/psi conformational space available to the Ala dipeptide you show can be argued to approximate that available to a tripeptide.

If the peptide were made exclusively out of amino acids, you would simply count the amino acids to get the name. One amino acid is an amino acid, two amino acids connected by a single peptide bond is a dipeptide, three amino acids connected by two peptide bonds is a tripeptide and so on.
For this case, you have two peptide bonds, so you might call it a tripeptide. Or you argue that there is one amino acid only, and you call it an amino acid with two protecting groups. Or, and this is a strange way of thinking about it in my mind, you say that there are two amino groups and two carboxylic acid groups upon hydrolysis so it is - a dipeptide.
I would not dare say this if I hadn't found this way of counting elsewhere, in a tutorial of another software.


text:
Is heat just a change in temperature?
According to my chemistry textbook, heat is defined as "Energy that is exchanged because of a difference in temperature or a change in phase." My textbook also says, "Heat is a form of energy. Temperature, on the other hand, is a measure of how quickly the molecules (or atoms) in a substance are moving," implying that temperature is not a form of energy, just a measurement of a physical condition.
This seems all wrong to me. If heat is just a change in temperature, but temperature isn't energy, doesn't that imply that energy is created when two objects of different temperature interact, or is there potential heat energy in the temperature of an object?
Also, the textbook says that the reason we feel hot is because of heat from temperature change, and not temperature itself, but this doesn't explain why chemical reactions occur faster at different temperatures. Nor does it explain why there are different states of matter at different temperatures, or even why our bodies need to stay at a certain temperature at all. Where is the energy from heat and other chemical processes coming from, exactly?
Other sources online also say that heat is a change in temperature and temperature is not energy, only a measurement, so I don't think it's just my textbook. I feel like I'm missing something.

[...] is there potential heat energy in the temperature of an object?

The energy associated with temperature is called thermal energy. If you know the heat capacity of an object, you can figure out how much energy is required to change the temperature by a certain amount.

If heat is just a change in temperature, but temperature isn't energy, doesn't that imply that energy is created when two objects of different temperature interact [...]?

Heat is energy transferred from one object to another. When heat is transferred from a hotter to a colder object, the hotter object loses thermal energy and the colder object gains it. No energy is created.

Where is the energy from heat and other chemical processes coming from, exactly?

A chemical reaction can give off heat because electrons can be in higher states (weak bonds) and in lower states (strong bonds). Potential energy can turn into thermal energy, and vice versa. It takes energy to melt ice, so melting ice cools down the surroundings. An explosive reaction releases energy, so it increases the temperature of the surroundings by transferring heat to it.

I really like your answer. I think it addresses the major puzzlements of the OP without dumbing down the concepts so that erroneous ideas are created. Temperature and heat are surprisingly hard to define.

There are two concepts here, temperature and heat.  The temperature of a 'system' is only that property that determines whether or not a system is in thermal equilibrium with another system.
This accords with our experience that we can distinguish by touching between something that is hot and that which is cold but have little feeling if it is at the same temperature as we are.
We can measure temperature by many different means, such as pressure of a gas, electrical  resistance, expansion of a liquid, magnetic susceptibility of a paramagnetic salt and by radiant emission, which is how we can determine the temperature of a furnace or of the surface of a star. The reference of zero is then some property such as the triple point of water for the celsius scale.
A molecular interpretation of temperature indicates that in a molecule with an infinite number of energy levels, as the temperature is raised more and more energy levels  become populated but any given level always has less population than the one immediately below it. This is in accord with the Boltzmann distribution. Thermodynamically this is expressed as the rate of change of internal energy with entropy, i.e the slope of a graph of internal energy vs entropy (at constant volume).
( In systems where there are a finite number of energy levels, such as an arrangement of spins, (for example in some types of nmr experiments) then negative temperatures are possible, i.e. larger population in upper than lower levels. This does not mean a temperature below absolute zero.)
Heat is internal energy in transit, it flows from one part of a system to another, or between two systems both by virtue of a temperature difference,   and can only be quantified when the transfer has finished. It is incorrect to refer to 'the heat in a body' just as it is incorrect to refer to 'the work in a body'. The heat and work are ways in which the internal energy of a body is changed. Put another way, it is impossible to divide the internal energy into some amount of heat and another amount of work. We have no direct knowledge of heat from our senses (or instruments) and heat is quite distinct  to 'hotness'.
The first law defines the change in internal energy $\Delta U$ as the sum of heat $Q$ and work $W$, $\Delta U = Q+W$.
This means that we can define heat as that energy transfer brought about by non-mechanical means, and is equal to the internal energy change less the work done when the system is at a different temperature to it surroundings.
The first law has three features, (a) it is based on conservation of energy, (b) to satisfy (a) it introduces the idea of internal energy and (c) it defines heat as energy in transit by virtue of the temperature difference.
That heat is energy was first determined quantitatively by J. Joule in the mid 1800's in elegant experiments where the increase in temperature of water agitated by a paddle wheel, rotated by the lowering of a weight under gravity, was measured.

This seems all wrong to me. If heat is just a change in temperature, but temperature isn't energy, doesn't that imply that energy is created when two objects of different temperature interact, or is there potential heat energy in the temperature of an object?

One example of how heat and temperature are related is when describing mixing cold water and hot water. Take for example a cup of water at room temperature. The water molecules will have an average kinetic energy that leads to the "room temp" feeling. If you add "heat" to this system, the temperature will rise, adding kinetic energy to the molecules of water in the system. When talking about heat, you're talking about the energy of the water molecules, as well as the amount of energy you could add to the water molecules to raise them to another temperature, so it's important to be specific when saying "room temperature water has been heated from absolute zero using xxxxxx joules of heat". This would allow you to keep a mental pointer of the amount of heat that any specific sample of water has.
I recommend looking at PhET's States of Matter Simulator, click states, then water, then heat it from 0 K and watch what happens with the water molecules. (remembering that the heat in that container is constant so there is no loss in temperature when you let go of "heating".

Also, the textbook says that the reason we feel hot is because of heat from temperature change, and not temperature itself, but this doesn't explain why chemical reactions occur faster at different temperatures. Nor does it explain why there are different states of matter at different temperatures, or even why our bodies need to stay at a certain temperature at all. Where is the energy from heat and other chemical processes coming from, exactly?

Different states of matter at different temperatures is a difficult and expansive topic in chemistry that is mostly understood to be governed by the individual intermolecular forces that dominate in each chemical that you study. For the case of water, it is liquid around room temperature because the average kinetic energy that the water molecules have at room temperature is around the same energy as the intermolecular interactions that bind the water together, allowing water molecules to move around each other but still have forces that hold them together. As you heat higher and higher, the average kinetic energy of the water molecules will increase, till it is great enough to break all of the intermolecular bonding (i.e.) at 100 degrees Celsius and 1atm, you will boil water because all of the hydrogen bonding is broken.

Other sources online also say that heat is a change in temperature and temperature is not energy, only a measurement, so I don't think it's just my textbook. I feel like I'm missing something.

Another important thing to think about is the fact that temperature is an intensive property, so if you take a tank of hot water, divide that water in half, and none of the heat is lost, then both tanks will be the same temperature hot water, meaning that their molecules will still have the same average kinetic energy, but they can only give a certain amount of total heat to another substance using the formula, where Q is the amount of heat transferred between two samples at different temperatures (the difference in energy between hot water and cold water), m is the total mass of both samples when mixed, c is the specific heat capacity which for water is $4.184\frac{J}{g  ^\text{o}C}$. This means that 1 gram of water takes exactly 4.184 Joules of heat to raise that sample 1 degree Celsius in temperature.
$$ Q=mc\Delta T$$
I hope that my previous answers also help this question, but if you're looking for more help, I would suggest looking up Khan Academy videos on Youtube of heat transfer problems.
TL:DR (Here's a YouTube video of someone explaining the simulator and much of what I just explained) Heat vs. Temperature - Chem Academy

Heat is energy transferred from one 'body' to another, you cannot say that a substance has 'an amount of heat ' as you do in your first paragraph, any more than you can say that a substance has an amount of work.

At the risk of confusing the issue, there is the concept of latent heat.

Heat is a form of energy. According to the kinetic theory of matter, particles in solids vibrate about their mean position with mean kinetic energy. Particles in fluid travel with a mean velocity. In either case, the kinetic energy is proportional to the temperature. If there is a medium possessing a temperature gradient, particles at a higher temperature level will possess higher kinetic energy. Upon collision with the adjacent particle with lower temperature and hence lower kinetic energy, some kinetic energy is transferred from the higher to the lower kinetic energy particle causing it to move further than its mean position. The kinetic energy of the lower temperature particle increases while that of the higher energy particles decreases. With the increase in the kinetic energy of the lower temperature particle, there is an increase in the temperature of the particle and hence change in its mean position. In this manner, heat is transmitted from one point to another and this gives the microscopic description of the heat conduction process.
            Generally, in liquids and electrically non-conducting solids, heat transmission is caused by longitudinal oscillation of the lattice structure. In metals, thermal conduction is as a result of free electrons while in gases, conduction is as a result of the elastic collision of molecules. Particles are more densely packed in solids than in liquids and in liquids than in gases. Liquids have shorter mean free path than gases. Thus, the probability of collision occurring in solids is higher than in liquids which in turn is higher than in gases. Consequently, solids conduct heat faster than liquids which in turn conduct heat faster than in gases.
I hope this might help you to understand your current and probable future doubts.
Please read the following links for better understanding:
https://nptel.ac.in/courses/112108149/pdf/M1/Student_Slides_M1.pdf
Also kindly take a look at the following Link as well:
https://sciencing.com/role-heat-play-chemical-reactions-13455.html
Please ask for any further clarifications or suggestions.
Keep learning and have fun. :)


text:
Effect of inert gas on Q
Adding an inert gas will not affect Q if that reaction is held at a constant volume. But, if the reaction is held at constant pressure, then adding inert gas will change Q bc the partial pressures of the gases will change.
I don't understand this. If the reaction is held at constant pressure, this means that you changed the volume when adding the inert gas. Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

If the number of gas particles is the same for reactants and products, then Q would not change. However, if you have a reaction like
$$\ce{2N2O(g) <=> N2O4(g)}$$
even though the reactant and product partial pressure changes by the same factor, the changes don't cancel out because the coefficients are different.
$$ Q = \frac{[\ce{N2O4}]}{[\ce{N2O}]^2}$$

Say PN2O = 2 and PN2O4=4. Say the volume doubles as a result of adding an inert gas at constant pressure. So now PN2O = 1 and PN2O4=2. Won't Q be the same?

No because $P_\ce{N2O}$ gets squared.

Couldn't you make the same argument for adding an inert gas at constant volume? Wouldn't any change be magnified for the squared reactant?

At constant volume, the partial pressures don't change at all.

But, if you add an inert gas, won't that increase the total pressure in the system, so the reactant and product partial pressures will increase, and due to the squared reactant in the Q expression, the N2O partial pressure will increase more than that of N2O2?

If you apply the ideal gas law, you will realize that when  temperature, amount and volume are constant, the partial pressure is constant as well. The additional pressure is from the inert gas, but the amount of N2O in a given volume (the "concentration") is still the same.


text:
More stable carbocation, tertiary or benzylic?
This is a very confusing question that which is more stable, benzylic or tertiary butyl carbocation?
Some sources say benzylic due to resonance some say tertiary due to +i effect and 9 hyperconjugative structures.
I also found that benzylic has almost 3 kcal/mol more energy than the other making it less stable. But some very good examinations has approved benzylic.
So finally whats the concrete answer?


text:
What is the correlation between electronegativity and the sharing of electrons?
For example in a C-O bond, the electronegativity difference is 1.2. does this mean that the Oxygen has 1.2 of the shared electrons and Carbon 0.8 (or some other numbers)? Is there a specific equation that can give us how the bonding electron pair is shared?

The Wikipedia page should answers your Q. https://en.m.wikipedia.org/wiki/Electronegativity

Is there a specific equation that can give us how the bonding electron pair is shared?

No. The electronegativity scale is based on bond strength, not on partial charges. There are quantum-chemical methods to calculate the electron density in a molecule, and you can derive partial charges from that.

For example in a C-O bond, the electronegativity difference is 1.2. does this mean that the Oxygen has 1.2 of the shared electrons and Carbon 0.8 (or some other numbers)?

No, the electronegativity difference does not allow you to make a quantitative statement like that. It is useful, however, to predict how a bond would hydrolyze. It also gives you an estimate about how polar a bond is.

I just want to add that the way you calculate partial charges is not unique, i.e. you have different models, and they don't necessarily give the same answer, or the same values. As Karsten said, I would take the kind of information that you get from electronegativity only as qualitative, not as quantitative. The same goes for calculated partial charges.


text:
Tautomerism in alkenes
Why doesn't but-2-ene exhibit tautomerism like alkenols?
By transfer of 1 hydrogen from terminal CH₃ to the middle carbon and forming a double bond between the two?

Getting a proton off an oxygen is easy peasy. Getting it off a carbon is not usually quite so easy.

In order for tautomerism to occur, the enol must be deprotonated, so as to allow the negative charge to come down, kick the pi bond of the C=C bond, which will bond to a proton. Hydrogen atoms on oxygen are much more acidic than hydrogens on carbons because of the electronegativity differences between O and H. Thus, the deprotonation does not happen in alkenes.


text:
Electrons in methane
Why does a molecule of methane have 10 electrons?
I used to think that the number of electrons in methane were 8 in number but it turns out that there are 10 in number but i dont know why.
Please help
Regards

The core 1s electrons of carbon aren't depicted in the dot cross diagram.

Methane, CH₄ has an atom of C and 4 atoms of H.
Looking at the structure of CH₄, there are 4 sigma bonds present. A sigma bond is a type of covalent bond. Each covalent bond represents a pair of electrons shared between two atoms.
This gives us 8 electrons; however one must realize that there is a pair of electrons from 1s² orbital (of the Carbon atom) which is not participating in any form of bonding. Hence the total 10 electrons present in a molecule of CH₄.
Another simple way to know the total number of electrons is to know the atomic numbers of the individual atoms (6 for carbon, and 1 for each hydrogen), so 6 + 1*4 = 10 electrons.
Now check the formal charge present on the molecule (in this case it is 0, CH₄ is electrically neutral - if a positive charge 'x' is found, subtract 'x' from 10 & if a negative charge 'y' is found, add 'y' to 10. Again, a total of 10 electrons is found.

I regret writing this down lol. @user198885 do see the tour: https://chemistry.stackexchange.com/tour


text:
Why oxidation state decreases
oxides of group 1 elements are Li2O , Na2O2 , KO2 , RbO2
Oxygen’s oxidation state in each compound is -2 , -1 , -1/2 ,-1/2 respectively I.e oxide-> peroxide-> superoxide
my question
The oxidation state of the oxygen Is increasing down the group why is that ?
My attempt
the tendency to donate electron increases down the group , by this statement the oxidation state of oxygen in the oxides down the group should decrease as oxygen gets easily reduced.
What am I missing here?

What about the size of the cations?

Firstly, the oxides you have listed are not the only ones that exist. It's that on reacting with excess oxygen, Li forms monoxide, Na forms peroxide, and the elements further down form superoxide. These are not necessarily the strongest oxide form too, infact, according to this link:- K2O2; the most stable oxide of K is peroxide despite being able to form a superoxide. (Also, Li, Na can also make superoxides (though not very stable) and K, Rb can also make monoxides.)
So I think the question changes to why are the elements down the group more able to form oxygen compounds with decreasing oxidation state of oxygen, and that is due to the size of the metal atoms; the larger the metal atom, the higher number of oxygen atom containing species it can (stably) accommodate around it.


text:
What effect on boiling and melting points does intramolecular hydrogen bonding have?
I know that intermolecular forces increase the boiling and melting point of a compound, but what is the effect of intramolecular forces on boiling and melting points?
I found contradicting answers while searching on the web.

Why would there be an effect? There is none.

@IshanJawale As Ivan says, there shouldn't be an effect of intramolecular forces. Changing from gas to liquid to solid is only affected by the interactions between molecules, not within them.

@WilliamR.Ebenezer There is a general problem with these unspecific, short titles. They will almost always lead to confusion and sometimes would appear to be not fitting at all. The mix-up in the other post did not make this better. If you are suggesting these kind of edits, please double check if it (partially) invalidates an answer (as was the case in the linked one).

@Martin-マーチン, yes, I did notice that earlier, before you pointed out, but my edit had already been petitioned. But I judged that it would be better, as a question with a contradicting title is worse than a few edits made to the answers mentioning the inadvertent error(the site must edge towards the perfection!). The point being, this question would not be posted had this error been non-existent, as the same question had been addressed satisfactorily earlier, under the same title.

@Tyberius I agree with that, too close.

@WilliamR.Ebenezer Please excuse my brevity earlier, it was supposed to be  a friendly reminder to watch out for these things, and maybe suggest another edit/put a flag on the answer to fix these occasions. I agree with your stance, which is why I accepted your edit and modified the answer. It also was already mentioned in the comments there. (This question is now in the close queue, but I wouldn't want to hammer it shut, I believe the community will do that without my help.)

@Martin-マーチン I will make sure to look out for such connected edits in the future.. Thank you.

The effect of intramolecular forces versus intermolecular forces on boiling and melting points is sometimes seen when we compare the boiling points of ortho- and para-nitrophenol (or similar compounds).
Para-nitrophenol shows intermolecular hydrogen bonding, which causes an increase in boiling point as different molecules bond better with each other, as compared to ortho-nitrophenol, where there is intramolecular hydrogen bonding, which does not help in increasing the boiling point, hence p-nitrophenol has a higher boiling point than o-nitrophenol.
However, while comparing intramolecular bonding with no hydrogen bonding, no difference can be expected unless the intramolecular bonding affects the intermolecular bonding in some way, for instance by changing size and thus packing.


text:
Can buffer solutions be used to control the pH of ethanol?
For my experiment, I want to investigate the effect of differing pH on the solubilities of a sparingly soluble drug. However, the solubility is too low for me to conduct a reliable study on the solubility (0.003 mg/ml in water). In ethanol, the solubility increases to 10 mg/ml. Can I simply add buffer solutions and acids into ethanol to change its pH as I would do with water? Are there alternatives to adjusting the pH of ethanol?

Your basic notion seems wrong. pH aside, solubility in ethanol isn't the same as solubility in water.

Although pH scale is rather strictly used for aqueous solutions, the concept of pH in organic or non-aqueous solutions is not trivial. It is called apparent pH or operational pH when organic solvents are present. There is a practical problem here as well, do you think your buffers will dissolve in ethanol? Once you figure this out, you can read a section in  "pH measurements in Nonaqueous Solvents" by Laitinen & Harris, “Chemical Analysis ─ An Advanced Text and Reference". It is a very useful book.


text:
Why doesn't H₄O²⁺ exist?
I know this question has been asked previously but I cannot find a satisfactory explanation as to why is it so difficult for $\ce{H4O^2+}$ to exist. There are explanations that it is so because of $+2$ charge, but if only that was the reason then existence of species like $\ce{SO4^2-}$ should not have been possible.
So, what is exactly the reason that makes $\ce{H4O^2+}$ so unstable?

Note that even $\ce{H3O+}$ is questionable, as some sources prefer existence of $\ce{H9O4+ }$, where each of 3 hydrogen atoms is bound to another $\ce{H2O}$.

Yes but isn't it kind of like hydration, where H2O molecules surround the ions , whereas in H3O+ a covalent bond exists between O and H?

Well, it can be,  it could be considered as coordination bonds of 3 H2O ligands. The idea was, that not only the naked $\ce{H+}$ does not exist, but neither the naked $\ce{H3O+}$.   https://pubs.acs.org/doi/abs/10.1021/ic00126a005    A Crystalline [H9O4]+ Hydronium Ion Salt with a Weakly Coordinating Anion

I do not doubt $\ce{H3O+}$ exists, I have implied the liquid water context.

Define "exist." If it "exists" in interstellar medium as a few ions, does that count? And if so, how would you ever detect that?

Related question, though the asker may already be aware.

https://chemistry.stackexchange.com/questions/24342/what-is-the-ka-of-oh-and-kb-of-h3o

@Zhe "Exists" as in "can exist" (in space, in an argon matrix, in theory).

@Karl I wasn't looking for an answer from you. I was challenging the OP to define / state their question better. Exists is not the same thing as stable (relative to an as yet undefined reference).

@Zhe I mainly wanted to ask what makes it so unstable rather than does it exist or not.

I myself was always confused why $\ce{H3O^+}$ is so well-known and yet almost nobody talks of $\ce{H4O^2+}$. I mean, $\ce{H3O^+}$ still has a lone pair, right? Why can't another proton just latch onto that? Adding to the confusion, $\ce{H4O^2+}$ is very similar to $\ce{NH4+}$, which again is extremely well-known. Even further, the methanium cation $\ce{CH5+}$ exists (admittedly not something you'll find on a shelf), and that doesn't even have an available lone pair!
It is very useful to rephrase the question "why is $\ce{H4O^2+}$ so rare?" into "why won't $\ce{H3O^+}$ accept another proton?". Now we can think of this in terms of an acid-base reaction:
$$\ce{H3O^+ + H+ -> H4O^2+}$$
Yes, that's right. In this reaction $\ce{H3O^+}$ is the base, and $\ce{H^+}$ is the acid. Because solvents can strongly influence the acidity of basicity of dissolved compounds, and because inclusion of solvent makes calculations tremendously more complicated, we will restrict ourselves to the gas phase (hence $\ce{(g)}$ next to all the formulas). This means we will be talking about proton affinities.
Before we get to business, though, let's start with something more familiar:
$$\ce{H2O(g) + H+(g) -> H3O^+(g)}$$
Because this is in the gas phase, we can visualise the process very simply. We start with a lone water molecule in a perfect vacuum. Then, from a very large distance away, a lone proton begins its approach. We can calculate the potential energy of the whole system as a function of the distance between the oxygen atom and the distant proton. We get a graph that looks something like this:

For convenience, we can set the potential energy of the system at 0 when the distance is infinite. At very large distances, the lone proton only very slightly tugs the electrons of the $\ce{H2O}$ molecule, but they attract and the system is slightly stabilised. The attraction gets stronger as the lone proton approaches. However, there is also a repulsive interaction, between the lone proton and the nuclei of the other atoms in the $\ce{H2O}$ molecule. At large distances, the attraction is stronger than the repulsion, but this flips around if the distance is too short. The happy medium is where the extra proton is close enough to dive into the molecule's electron cloud, but not close enough to experience severe repulsions with the other nuclei.
In short, a lone proton from infinity is attracted to a water molecule, and the potential energy decreases up to a critical value, the bond length. The amount of energy lost is the proton affinity: in this scenario, a mole of water molecules reacting with a mole of protons would release approximately $\mathrm{697\ kJ\ mol^{-1}}$ (values from this table). This reaction is highly exothermic
Alright, now for the next step:
$$\ce{H3O^+(g) + H+(g) -> H4O^2+(g)}$$
This should be similar, right? Actually, no. There is a very important difference between this reaction and the previous one; the reagents now both have a net positive charge. This means there is now a strong additional repulsive force between the two. In fact, the graph above changes completely. Starting from zero potential at infinity, instead of a slow decrease in potential energy, the lone proton has to climb uphill, fighting a net electrostatic repulsion. However, even more interestingly, if the proton does manage to get close enough, the electron cloud can abruptly envelop the additional proton and create a net attraction. The resulting graph now looks more like this:

Very interestingly, the bottom of the "pocket" on the left of the graph (the potential well) can have a higher potential energy than if the lone proton was infinitely far away. This means the reaction is endothermic, but with enough effort, an extra proton can be pushed into the molecule, and it gets trapped in the pocket. Indeed, according to Olah et al., J. Am. Chem. Soc. 1986,  108 (5), pp 1032-1035, the formation of $\ce{H4O^2+}$ in the gas phase was calculated to be endothermic by $\mathrm{248\ kJ\ mol^{-1}}$ (that is, the proton affinity of $\ce{H3O^+}$ is $\mathrm{-248\ kJ\ mol^{-1}}$), but once formed, it has a barrier towards decomposition (the activation energy towards release of a proton) of $\mathrm{184\ kJ\ mol^{-1}}$ (the potential well has a maximum depth of $\mathrm{184\ kJ\ mol^{-1}}$).
Due to the fact that $\ce{H4O^2+}$ was calculated to form a potential well, it can in principle exist. However, since it is the product of a highly endothermic reaction, unsurprisingly it is very hard to find. The reality in solution phase is more complicated, but its existence has been physically verified (if indirectly).
But why stop here? What about $\ce{H5O^3+}$?
$$\ce{H4O^2+(g) + H+(g) -> H5O^3+(g)}$$
I've run a rough calculation myself using computational chemistry software, and here it seems we really do reach a wall. It appears that $\ce{H5O^3+}$ is an unbound system, which is to say that its potential energy curve has no pocket like the ones above. $\ce{H5O^3+}$ could only ever be made transiently, and it would immediately spit out at least one proton. The reason here really is the massive amount of electrical repulsion, combined with the fact that the electron cloud can't reach out to the distance necessary to accommodate another atom.
You can make your own potential energy graphs here. Note how depending on the combination of parameters, the potential well can lie at negative potential energies (an exothermic reaction) or positive potential energies (an endothermic reaction). Alternatively, the pocket may not exist at all - these are the unbound systems.
EDIT: I've done some calculations of proton affinities/stabilities on several other simple molecules, for comparison. I do not claim the results to be quantitatively correct.

$$
\begin{array}{lllll}
\text{Species} & \ce{CH4}  & \ce{CH5+}    & \ce{CH6^2+}     & \ce{CH7^3+}     & \ce{CH8^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 556 & -246 & -1020   & N/A   & N/A  \\
\end{array}
$$
Notes:

Even without a lone pair, methane ($\ce{CH4}$) protonates very exothermically in the gas phase. This is a testament to the enormous reactivity of a bare proton, and the huge difference it makes to not have push a proton into an already positively-charged ion.
For most of the seemingly hypercoordinate species in these tables (more than four bonds), the excess hydrogen atoms "pair up" such that it can be viewed as a $\ce{H2}$ molecule binding sideways to the central atom. See the methanium link at the start.

$$
\begin{array}{lllll}
\text{Species} & \ce{NH3}  & \ce{NH4+}    & \ce{NH5^2+}     & \ce{NH6^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 896 & -410 & N/A   & N/A    \\
\end{array}
$$
Notes:

Even though the first protonation is easier relative to $\ce{CH4}$, the second one is harder. This is likely because increasing the electronegativity of the central atom makes the electron cloud "stiffer", and less accommodating to all those extra protons.
The $\ce{NH5^{2+}}$ ion, unlike other ions listed here with more than four hydrogens, appears to be a true hypercoordinate species.  Del Bene et al. indicate a five-coordinate square pyramidal structure with delocalized nitrogen-hydrogen bonds.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2O}  & \ce{H3O+}    & \ce{H4O^2+}     & \ce{H5O^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 722 & -236 & N/A   & N/A    \\
\end{array}
$$
Notes:

The first series which does not accommodate proton hypercoordination.
$\ce{H3O+}$ is easier to protonate than $\ce{NH4+}$, even though oxygen is more electronegative. This is because the $\ce{H4O^2+}$ nicely accommodates all protons, while one of the protons in  $\ce{NH5^2+}$ has to fight for its space.

$$
\begin{array}{lllll}
\text{Species} & \ce{HF}  & \ce{H2F+}    & \ce{H3F^2+}     & \ce{H4F^3+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 501 & -459 & N/A   & N/A    \\
\end{array}
$$
Notes:

Even though $\ce{H3F^2+}$ still formally has a lone pair, its electron cloud is now so stiff that it cannot reach out to another proton even at normal bonding distance.

$$
\begin{array}{lllll}
\text{Species} & \ce{Ne}  & \ce{NeH+}    & \ce{NeH2^2+}    \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{No}   \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 204 & N/A   & N/A    \\
\end{array}
$$
Notes:

$\ce{Ne}$ is a notoriously unreactive noble gas, but it too will react exothermically with a bare proton in the gas phase.
Depending on the definition of electronegativity used, it is possible to determine an electronegativity for $\ce{Ne}$, which turns out to be even higher than $\ce{F}$. Accordingly, its electron cloud is even stiffer.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2S}  & \ce{H3S+}    & \ce{H4S^2+}     & \ce{H5S^3+}     & \ce{H6S^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 752 & -121 & -1080   & N/A   & N/A  \\
\end{array}
$$
Notes:

The lower electronegativity and larger size of $\ce{S}$ means its electrons can reach out further and accommodate protons at a larger distance, while reducing repulsions between the nuclei. Thus, in the gas phase, $\ce{H2S}$ is a stronger base than $\ce{H2O}$. The situation is inverted in aqueous solution due to uniquely strong intermolecular interactions (hydrogen bonding) which are much more important for $\ce{H2O}$.
$\ce{H3S+}$ also has an endothermic proton affinity, but it is lower than for $\ce{H3O+}$, and therefore $\ce{H4S^2+}$ is easier to make. Accordingly, $\ce{H4S^2+}$ has been detected in milder (though still superacidic!) conditions than $\ce{H4O^2+}$.
The larger size and lower electronegativity of $\ce{S}$ once again are shown to be important; the hypercoodinate $\ce{H5S^3+}$ appears to exist, while the oxygen analogue doesn't.

@santimirandarp I don't have good graphing software on me right now, but I made something on Desmos. I also put a link to graphs that curious readers may play around with.

Unstable by 248 kJ/mol! I smell a monopropellent.

I've rarely seen this level of investigation in an SE answer. This answer could be turned into a manuscript and published in a reputable journal.

The "pairing up" of excess hydrogen atoms (beyond 4) may not be entirely accurate.  This reference renders the NH5 cation as a square pyramid with multiple delocalized bonds.

@OscarLanzi Huh, interestingly enough that's also what my calculations found, and I overlooked it. It seems to be the only truly hypercoordinate structure out of all the ones I calculated. I was planning on linking the geometries for each structure, but I'll have to do it later. Thanks for the tip, though!

Edited in the exceptional structure of the NH5 dication.

What tool did you use to create those diagrams?

In chemistry ask "why" only after you ask "if".
Given a sufficiently strong superacidic medium, $\ce{H3O^+}$ can be protonated to $\ce{H4O^{2+}}$.  Evidence for this reaction, by studying isotopic exchange in a $\ce{HF + SbF5 +SO2}$ solvent, is given here[1].
$\ce{H4O^{2+}}$ is, of course, a powerful protic acid, and it would be leveled to something weaker, meaning $\ce{H3O^+}$ and maybe other products, except in superacids as in the above reference.
A subtle difference between $\ce{H4O^{2+}}$ and $\ce{SO4^{2-}}$ is the presence of atoms with odd atomic number in the former.  An odd atomic number means an odd number of protons, so $\ce{H4O^{2+}}$ can split off or transfer a component ($\ce{H^{+}}$) having a single charge and an odd number of odd-proton atoms, without having to break up any electron pairs (no need to form radicals).  Hence, unless the environment is suffuciently acidic/nonbasic, this ion could break up relatively easily to singly charged products.  $\ce{SO4^{2-}}$ lacks this option; as this ion has only even atomic number atoms it cannot break up into singly charged species without the unfavorable formation of radicals.
We see a similar phenomenon in the autoionization of solvents like water, ammonia or even dinitrogen tetroxide.  Solvents are more likely to autoionize if they have odd atomic number atoms (like hydrogen or nitrogen) with which singly charged, nonradical species can form.
Reference
1.
George A. Olah, G. K. Surya Prakash, Mario Barzaghi, Koop Lammertsma, Paul von R. Schleyer and John A. Pople. " Protonated Hydronium Dication, H4O2+: Hydrogen–Deuterium Exchange of D2H17O+ in HF:SbF5 and DH217O+ in DF:SbF5 and Theoretical Calculations", World Scientific Series in 20th Century ChemistryAcross Conventional Lines, pp. 837-840 (2003).

Good point. If you can't break a molecule without making radicals or moving more than one atom (or subunit), you have a good chance it's at least metastable.

$\ce{H4O^{2+}}$ would be much smaller than $\ce{SO4^2-}$. The charge density is too high much higher.
The oxygen in water already becomes electron-deficient when it forms $\ce{H3O+}$. In your (not completely, see below) hypothetical $\ce{H4O^{2+}}$, the additional proton barely finds any electron density to make a covalent-ish bond with.
Whereas for the sulfate ion, you just have to distribute some extra charge over a relatively large molecule.
So $\ce{H4O^{2+}}$ exists, at least in theory, but it is extremely unstable. It's properties have been detailed (with links to scientific papers) here: What is the Ka of OH- and Kb of H3O+?

It is quite a valid species, OP is just wondering why it's very difficult to make, even transiently.

@Mithoron Are you implying that this particle can exist, at least theoretically and with a very short lifetime, in a bound state? I find that hard to believe, unless of course you have some evidence? (Ok, found it under the link you gave above https://pubs.acs.org/doi/pdf/10.1021/ja00265a031)

Yeah, it seems it's repulsive, but metastable.