text:
Why does substance that sublime not come to liquid state?
A substance that sublimates doesn't come to a liquid state at room temperature. But when we melt the substance it melts! Why does it skip the liquid state at room temperature?

Who said it melts? Some don't.

But some do. Why?

Also which don't

Also solids have a vapour pressure. And if you increase the outside pressure, all substances melt before evaporating.

There's a similar question here: https://chemistry.stackexchange.com/questions/6318/is-sublimation-caused-by-an-environmental-difference-or-an-elements-composition?rq=1 but it doesn't have answers that properly address the current question imo.

State is in P, T space. What we readily see happens normally at atmospheric P. Therw is a recent Q on sublimation. Perhaps it fixes your current one

@Alchimista  can you please give me the link

Your Q is also awkward. Does it sublimate or does it melt? Whatever happens before heating the sample is evaporation. As another said solids have vapour pressure, too.

From a thermodynamic point of view there are two competing possibilities for condensation of a vapor:
(1) vapor $\ce{->}$ liquid, with some negative $\Delta H$
(2) vapor $\ce{->}$ solid, with an absolutely larger negative $\Delta H$
In the absence of large quantum mechanical effects (see here), the second process with a more negative enthalpy change must become favorable at a low enough temperature above absolute zero; the actual temperature where this happens  depends on the material.  So when the condensation is done at a low enough temperature, it goes directly from vapor to solid, and the reverse process of evaporation then also misses the liquid.

Spot on. If the material does not sublime then below the melting point the solid is evaporating, and above the melting point the liquid is evaporating.  If material sublimes there is no liquid phase (at that pressure) and the solid will sublime (appears the same as evaporation but no chance of liquification as temperature increases, just faster sublimation).  Temperature of solid is held at the Solid/gas phase line by natural cooling due to the evaporation.

Sorry, this explains nothing. At any low enough temperature (at given pressure), the material simply either does not evaporate, or is still in thermal equillibrium with it's gas phase. Or it has a melting point. ;-)

Polar or flexible molecules tend to have a liquid phase also at low pressures, so do metals. Rigid, unpolar particles not (naphtalene, helium, iodine). The difference makes wether a not perfectly ordered phase still can have enough intermolecular forces to keep it in a condensed (=liquid) state.
A molten metal is likely still metallic. Hydrogen bonds in liquid water are still very strong. Dodecane is flexible, and methyl end groups can take up a lot of thermal energy without greatly changing the geometry. Etc.
Whereas naphtalene is too large to rotate the whole molecule in the condensed phase, but once you have enough thermal energy to break the pi-stacking, there is no other interaction present that could keep the molecule from flying away.
The better the interaction in the liquid phase, the less pressure is needed, resp. the larger the temperature range in which the liquid is the thermodynamically preferred state.


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Improvement of the thermal stability of hydrous zirconia by post-synthesis treatment with NaOH and NH4OH solutions
I read a paper about how researchers improved the thermal stability of hydrous zirconia by post-synthesis treatment with $\ce{NaOH}$ and $\ce{NH4OH}$.[1]
While the surface area of non-treated zirconia though more, but after calcinations at high temperatures the % decrease of the treated zirconias was very less as compared to non treated zirconia.
Is there any reason for this observation besides the changes the changes in pore volume and size?
Any specific reason for the morphology obtained?
G. Aguila, S. Guerrero, F. Gracia, P. Araya, Appl. Cat. A: General 2006, 305 (2), 219-232. DOI: 10.1016/j.apcata.2006.03.025

Surface area is heavily dependent on pore size. This is a general trend that when you calcine something, its surface area decreases. I think both NaOH or ammonium hydroxide promote formation of surface -OH groups and when you heat it high temperature it helps in maintaining the pore structure. Check what the paper says about pore sizes. A lot of work is done on silica and you might get a clue.

i think it will have something to do with surface hydroxylation densities ,and stabilization of alumina surface by stabilizing agents.

source: Stack Exchange

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The production of what substances has a similar environmental impact to Chlorinated paraffins?
I am currently conducting a Life Cycle Assessment (LCA) study in which the final product consists to a large part of Chlorinated paraffins. However, I cannot find Chlorinated paraffins as a process in available LCA background data.
Therefore, I am asking what other substances are likely to have a similar environmental impact during the production compared to Chlorinated paraffins (and are maybe even more commonly used so that the likelihood to find them as an existing LCA data set is high)?

Terminology might be a problem. You could try searching for chlorinated hydrocarbons or chlorinated saturated hydrocarbons or something similar.

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Is ΔV in ΔU = ΔH - pΔV always a positive number?
I know that $ΔU$ and $ΔH$ can be both positive and negative, but what about $ΔV$?
I presume $ΔV$ can be both positive and negative as $w=-pΔV$, pressure always positive, and $w$ can be positive or negative, hence $ΔV$ must be able to be either positive or negative.

See, there is only so much space around here. If something increased in volume ($\Delta V>0$), then obviously something must have decreased.

Do yourself a favour and never write "+ve" or "-ve" again.

@Loong I once had a student write "NRG" in a question response. When asked what it was, they replied that it was their abbreviation for "energy."

@Loong Out of curiosity, do you know where this cancerous trend of using "±ve"s stems from? I've never seen a textbook using these notations and when I was a student those also hasn't been used at all.

@andselisk It's definitely been circulating for at least 15 years. And I saw college freshmen doing it, so I suspect that it started in high school (or earlier).

@andselisk from a quick googling, supposedly its medical shorthand to speed up writing positive or negative, with the ve at the end being the ending of positive and negative. I don't see why + and - on their own wouldn't suffice without tacking on extra letters that could easily be confused for some sort of abbreviation, but thats all the explanation I could find.

The equation you wrote is correct only for the changes in U, H, and V between two thermodynamic equilibrium states, and only if the pressure in the initial and final thermodynamic equilibrium states is the same value.  If the pressure is not the same in the two states, then the equation should read $\Delta U=\Delta H-\Delta (pV)$.  Also, the equation $w=-p\Delta V$ is valid for calculating the work done on a system only if the only kind of work done is p-V work, and only if the pressure is constant over then entire path between the initial and final  thermodynamic equilibrium states.  With regard to $\Delta V$, the change in volume will be a positive number if the system (such as a gas) is expanded, and will be negative if the system is compressed.  That is the definition of expansion and compression.

Is there a difference between $\Delta U \text{and}\Delta E$

In many books, $\Delta E$ includes the change in kinetic energy and the change in gravitational potential energy of the system.  So, they write $$\Delta E=\Delta U+\Delta (KE)+\Delta (PE)=q+w$$In many cases, though, the changes in kinetic energy and potential energy don't contribute.


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What is a "symmetrical product" of an irreducible representation with itself?
In their classic paper outlining the Jahn–Teller theorem (Proc. R. Soc. Lond. A 1937, 161 (905), 220–235), the authors wrote "where $[\phi^2]$ denotes the representation of the symmetrical product of $\phi$ with itself". Reading this I'm thinking about a regular product of representation (scalar product with each characters).
But in the next paragraph there is this equality for the $D_\mathrm{4h}$ group:
$$[\mathrm{E_g^2}] = [\mathrm{E_u^2}] = \mathrm{A_{1g} + B_{1g} + B_{2g}}$$
Obviously something is wrong, as the direct product is a 4-dimensional representation, but the right-hand side is 3-dimensional.
I would like to add that there is a reference of Weyl 1928 unfortunately this article is in German and I cannot find anything in it.
What is the correct way to carry out this direct product?

In a (typical) direct product table there are some things in square brackets and some not in square brackets. What you want are those which aren't in square brackets. See e.g. https://chemistry.meta.stackexchange.com/questions/3435 (let the page load for a minute or so, then click on the "Desktop" link for "direct product tables" - the footnote there has more detail).

@Tom When you take the direct product of two degenerate irreducible representations, the result can subdivided into a symmetric and antisymmetric part. To obtain the symmetric part, you can use the formula from this question.

In your case you need the equations for the characters as there are symmetric and antisymmetric parts in the product. Your question gives just the symmetric part and the antisymmetric part is $A_2$ normally written as $[A_2]$ to indicate this.
The characters are found using $\chi^+ =\left([\chi(R)]^2+ \chi(R^2) \right)/2$ for the symmetrical part and $\chi^- =\left([\chi(R)]^2- \chi(R^2) \right)/2$ for the antisymmetric part. Using the point group table you need to work out the square of all the operations for $E_g$ this gives $\chi(R^2)$ (e.g. $(C_2)^2 \to E$) and then square the operations which gives $[\chi(R)]^2$, i.e. $[E]^2\to 4$. From the list of characters you then identify the symmetry species, either directly by inspection (as for the antisymmetric in this case), or by forming the irreps from the reducible representation.


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What would the chemical name be for C13H8Cl3NO
Formula

C13H8Cl3NO

SMILES

C1=C(C(=CC(=C1)Cl)Cl)N(C(C2=CC=CC=C2)=O)Cl

I found the diagram on the left in a book and drew the one on the right using
https://pubchem.ncbi.nlm.nih.gov/edit2/index.html
And got the SMILES description from that.
Any clues as to what might be an IUPAC name and the formal way to write it out?  Have I have even drawn it correctly? I could not do it exactly because the two rings kept connecting if I followed the same orientation.  I don't know enough about double ring compounds to even hazard a guess.
Is there a automatic naming engine out there?
A compound called British Impregnite found in The Scientific method by Louis F. Fieser p.137

Try Chemspider  http://www.chemspider.com/

Chemdoodle thinks it is [N-Chloro(2,4-dichlorophenyl)amino]phenylformaldehyde but there is probably a less systematic version as well.

@matt_black Wow, that's like most complicated, but very much not systematic name I've seen.

@matt_black That's Chemdoodle being Chemdoodle, I think:)

Well, let's reconstruct that starting from the very right side, where it says a $\ce{C_6H_5}$. The ring and the $\ce{CO}$ group would be a benzaldehyde if it had an $\ce{H}$ instead of an $\ce{N}$, right? Or a benzoic acid if it was $\ce{OH}$ instead of $\ce{N}$. So what would it be if it had an $\ce{NH_2}$-group? It would be a benzamide. If the $\ce{N}$ is substituted with for example a chloride we call that N-chlorobenzamide. And now we have another $\ce{N}$-centered ligand, the second phenyl ring. The ring has three positions, the one where it's connected to the rest of the molecule would be 1, so that makes a 2,4-dichlorophenyl.
Summarizing we get N-chloro-N-(2,4-dichlorophenyl)benzamide

Thank you for the steps, it gives me great comfort to know it is no more complicated than Finnish grammar. To humour an inorganic chemistry fan like me how do you know to start with the right hand side?  Is the fact that the one ring has a carbon attached the reason it is first?  Could the substance be named by starting from the other ring?

Intuition, the benzene ring + the carbonyl group directly reminded me of benzaldehyde. With an N-Cl attached it didn't become any harder as I showed above. From the other side I had to guess. A benzene ring with an N could be derived from Aniline, there are so called Benzanilides but the other way around it was much easier as it already had a name.

There is a paper [1] reporting a structural investigation of aromatic N-chloroamides.
They investigated polymorphs of similar compound they refer to as N-chloro-N-phenylbenzamide:

Chlorinated product, I suspect, is indeed is going to be named N‐chloro‐N‐(2,4‐dichlorophenyl)benzamide, as the first answer suggested.
References

Naumov, P.; Topcu, Y.; Eckert-Maksić, M.; Glasovac, Z.; Pavošević, F.; Kochunnoonny, M.; Hara, H. Photoinduced Rearrangement of Aromatic N-Chloroamides to Chloroaromatic Amides in the Solid State: Inverted $Π_\ce{N}–Σ_\ce{N}$ Occupational Stability of Amidyl Radicals. The Journal of Physical Chemistry A 2011, 115 (26), 7834–7848. https://doi.org/10.1021/jp203771c.

Thank you for your research and input.  I am not sure why there are two N's if they refer to the one nitrogen.  How does one know what to do if there were two nitrogens?

I'm glad I could help. Two N-'s reflect the fact there are two groups attached to the same nitrogen; if there were more than one nitrogen, prime(s) ' and superscripted numerical locants would be used to differentiate nitrogen atoms.


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Shouldn't the 5-membered ring expand to a six membered one in this nucleophilic addition reaction?
I am dealing with a problem involving addition of $\ce{HCl}$ to a ring:

According to what I think, the ring should expand to a six-membered following carbocation formation at the tertiary site, as a 6-membered ring will involve less strain than the current 5-membered one.
However, the correct product is:

Which is what would form if the expansion is not favored.
My guess is that the rearrangement is less likely to occur due to a tertiary carbocation, and that the rearranged product will then form in a minor concentration.

5-membered rings are perfectly alright, it is 3 and 4 that are most strained, and going from tertiary to secondary is bad.

Note also that a ring expansion would lead to a secondary carbocation. The ring strain in a 5-ring, as mentioned is not that significant, so ring expansion would actually be energetically costly here.

@Zhe Would it be feasible,then,to expand the five membered ring to a six membered one,if the double bond had a hydrogen substituent in place of the methyl group substituent as shown for the reactant?

@Zhe I agree, and that is how I reasoned the non-favourability of the expansion intermediate. Thank you for reaffirming. :)

@YUSUFHASAN Yes, if there was a hydride substituent, ring expansion would've occurred appreciably.


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Denominator in rate law?

What the possible rate laws are for the decomposition of nitramide by the following process:
$$
\begin{align}
\ce{O2NNH2(aq) &<=>[$k_1$][$k_{-1}$] O2NNH-(aq) + H+(aq)} &\quad &\text{(fast equilibrium)} \tag{1}\\
\ce{O2NNH-(aq) &<=>[$k_2$] N2O(g) + OH-(aq)} &\quad &\text{(slow)} \tag{2}\\
\ce{H+(aq) + OH-(aq)&<=>[$k_3$] H2O(l)} &\quad &\text{(fast)} \tag{3}\\
\end{align}
$$
(A) $\text{rate} = k[\ce{O2NNH-}]$
  (B) $\text{rate} = k[\ce{O2NNH2}]/[\ce{H+}]$
  (C) $\text{rate} = k[\ce{O2NNH2}][\ce{H+}]$
  (D) Both A and B

I selected A, but the answer is actually D. The answer key gives the following information:

At the first glance, you may think that Option A is the best answer due to the information available in the question stem in which details the way a rate of reaction is written. However, option B also deserves some attention. It is important to understand that $K_\mathrm{eq}$ is the equilibrium constant for step 1, as this is the fast equilibrium step
$$K_\mathrm{eq} = \frac{[\ce{O2NNH-}][\ce{H+}]}{[\ce{O2NNH2}]}$$
Rearranging for $[\ce{O2NNH-}]$, and substituting it into the rate law gives an equally valid expression which leads to Option B also being correct. The best answer is therefore Option D.

I am confused about how this can be so.
Firstly, why can we just sub the equilibrium equation in like this, and what happens to the Ksp term, it seems to just disappear.
Secondly, doesn't this mean that the overall order of the reaction is inconsistent between option A and B (i.e. in A the overall order is 1, but in B it is 1-1=0).
Thirdly, conceptually I am confused about why the concentration of the product should change the rate of the forward reaction, this does not really make sense to me, why should the reaction proceed less quickly just because there is more product around? Further, if it is the case that the product changes the speed of the forward reaction, then why don't all rate equations feature the product?
I am left very confused, perhaps someone can help me out?

(A)$ \text{rate} =K_2[\ce{O2NNH-}]$

(B)$ \text{rate} =\frac{ k_1K2}{K-1}\cdot{\frac{[\ce{O2NNH2}]}{[\ce{H+}]}}$

The $\ce{O2NNH^-}$ is formed very rapidly but reacts slowly in the second step. The rate limiting step is therefore loss of $\ce{O2NNH^-}$ in which the back reaction with $\ce{N2O}$ is considered (in the question) to be so slow as to be insignificant when writing the rate expression. Thus the rate involves just $\ce{O2NNH^-}$, however, as the fist reaction is a fast equilibrium substituting for $\ce{O2NNH^-}$ with this equilibrium is valid, provided the $k's$  are marked as being different, which they are not, and in the second reaction $k$ should be $k\times K_{eq_1} $.

$$\text{ Production rate of}~~ \ce{O2NNH-}  = \text{ Consumption rate of}~~ \ce{\ce{O2NNH-}}$$
$$k_\mathrm {1}\ce{[\ce{O2NNH2}]}= k_\mathrm{-1}[\ce{\ce{O2NNH-}}][\ce{H+}] + k_\mathrm {2} \ce{[\ce{O2NNH-}] }$$
But $k_2 << k_{-1}$ because the second step is the slow step, so the term $k_2\ce{ [\ce{O2NNH-}]}$ can be neglected  at the far right of the last equation.
$$k_1\ce{[\ce{O2NNH2}]}= k_\mathrm{-1}\ce{ [\ce{O2NNH-}][\ce{H+}] }$$
$$[\ce{\ce{O2NNH-}}] =\frac{ k_1 [\ce{O2NNH2}]} { k_\mathrm{-1}[\ce{H+}]}$$
The value of$ [\ce{\ce{O2NNH-}}]$ from the last equation can now be substituted in the reaction rate equation which represents the slow step.
$$\begin{array}{  }
\text{Rate} = k_\mathrm{2}[\ce{O2NNH-}] \\
~~~~~~~~= \frac{k_\mathrm{1}k_\mathrm{2}[\ce{O2NNH2}]}{k_\mathrm{-1}[\ce{H+}]}\\
~~~~~~~~ =\frac{k[\ce{O2NNH2}]}{[\ce{H+}]}\end{array}$$
In it we see that $k$ expresses the value of the fractional $\frac{k_\mathrm{1}k_\mathrm{2}}{k_\mathrm{-1}}$


text:
Can a long polymer chain interact with itself via van der Waals forces?
It is known that van der Waals interaction is an intermolecular forces. So is it possible that a long polymer chain interacting with itself via non-bonded interactions (such as van der Waals forces / electrostatic forces / hydrogen bonding)?

It can and does. Come to think of it, how can it not to interact? It is just the same atoms governed by the same laws. They don't have a built-in IFF system to tell them whether the other atom is from the same or different molecule.

Why these are called intermolecular interactions, they're intramolecular as well, it's just that they're secondary to covalent bonds.

You have a possible answer to your question in proteins, an example which includes some long polymer chains.
Intramolecular interactions - while not necessarily the driving force for formation of a collapsed protein globule (usually argued to be due to the hydrophobic effect, requiring intermolecular interactions) - are the basis for higher order structure in proteins and even some forms of RNA. I won't mention DNA since most commonly this forms a bimolecular double helix.
All of the interactions you mention may play a role in stabilizing a particular protein structure or "fold". A good and basic example of protein structure acquisition due to intramolecular interactions in proteins is polyalanine, which acquires alpha-helical secondary structure stabilized by intramolecular hydrogen bonds, typically between amide groups separated by 2 amino acids along the chain. Other types of structure stabilized by intramolecular interactions (H-bonds, usually) in polypeptides include other types of helices, beta sheets and various turns.


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Error with explaination of shape of weak acid strong base titration?
I have read the other posts on the same question posted on chemistry exchange such as this one: Shape of Weak-Strong Acid-Base Titration
and here : What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from?
However, I don't see why their explanation is valid. They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution. When a strong base is added to a weak acid, as more $\ce{OH-}$ is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$
However, I do not see this is useful. Isn't having the conjugate base only useful if you want to add in more $\ce{H3O+}$ but we are not! As we only add in $\ce{OH-}$ how is having buffer capacity of $\ce{CH3COO-}$ in the other direction useful (as $\ce{CH3COO-}$ and $\ce{OH}$ don't react)? I agree this is a buffer, but how does the reverse reaction help at all? This buffer doesn't seem like it would help to reduce pH changes

Just add $\ce{OH-}$ to both sides of the equation you have written.

This buffer doesn't seem like it would help to reduce pH changes

You have to be clear what you want to compare it with

Adding a strong base to pure water:
If you start with pure water, the concentration of hydroxide will be pretty much equal to the concentration of strong base you add. The pH will drop quickly.
Adding a strong base to a strong acid:
If you start with strong acid, the first few drops of strong base will be neutralized. The pH does not change much because the strong acid is still in large stoichiometric excess, and the pH is a logarithmic scale.

When a strong base is added to a weak acid, as more OH− is added this will react with the acid molecules in solutions and form more of the conjugate base, forming a buffer solution which I do agree with.

In the first approximation, the added hydroxide will be neutralized by the acetic acid, and the pH will not change. However, as you do that, the weak acid/weak base equilibrium will be disturbed, and the reaction will shift to the left, consuming some hydronium and increasing the pH. The more you change the ratio of acetic acid to acetate for a - say - milliliter of added strong base, the more the pH will change. The ratio changes fast at the beginning (where there is no acetate) and near the equivalence point (where there is almost no more acetic acid), so that is when the pH changes are largest because those are the steps where we disturb the equilibrium the most.

They typically explain the little dip at the start and the following plateau as a result of creating a buffer solution.

If you compare the slope of the dip to that of scenarios 1. and 2., you will find that it is less than in 1. (strong base added to pure water) and more than in 2. (strong base added to strong acid).

This buffer doesn't seem like it would help to reduce pH changes

The way I explained it was to say the strong base reacts with the weak acid. We could have said it reacts with the hydronium ion instead. Then the shift of the weak acid/base reaction would have been the other way. In the end, we are just rationalizing the result of two equilibria, that of water autoionization (taking a back seat here because hydronium is the dominant species and hydroxide is present at very low concentration) and that of the weak acid dissociation. The shape of the titration curve is what we get experimentally, and the two equilibria explain it fully. Any explanation on top of that is just to convince ourselves.

Hm ok, so I kind of understand you. It is the excess amount of acetate ions that drives equilibrium to the left and hence makes [H+] very small. But then, why can we not say that the OH reacts with the H3O? This has been confusing me a lot with your explaination


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Why is copper(II) sulfate added drop by drop during the Biuret test?
The Biuret test is a test for proteins.  The procedure is to add a strong base, such as NaOH, to dissolve the protein.  After that, copper(II) sulfate is added drop by drop into this until a (violet) colour change is observed.  I have a few questions.

How does NaOH dissolve the protein exactly?
What is the significance of adding copper(II) sulfate drop by drop?
How would the reaction vary if copper (II) sulfate was added first and NaOH added subsequently, drop by drop?

The Biuret Test is a chemical assay that detects the presence of proteins in a sample. The test is not named after any famous scientist, but after an urea dimer called biuret $(\ce{H2NC(O)NHC(O)NH})$. However, funny thing is biuret is not a part of the test at all. The test is called Biuret Test because biuret gives a positive result to the test (See the image below).

The test relies on the characteristic color change (from blue to deep purple or violet; see the image), which confirms the presence of proteins. Biuret isn't a protein, but it gives a positive result to the test, because it has two amide bonds.
Your question (1): How does $\ce{NaOH}$ dissolve the protein exactly?
Proteins has a lot of amide (peptide) bonds, terminal amine and carboxylic acid groups, and along the $\alpha$-helix, a lot of amino acid side chains including amine groups (e.g., lysine), carboxylc acid groups (e.g., aspartic acid), thiol groups (e.g., cystine), etc. All of them can be reacted with $\ce{NaOH}$ to become ionic, which make it more soluble in water. Specifically, recall most of zwitterionic form of amino acids resist to dissolve in water. Regardless, $\ce{NaOH}$ is essential for the test, not only to make protein more soluble, but also to extract $\ce{H}$ from amide $\ce{N-H}$ bond to make $\ce{N}$ negative (Ref.1).
Your question (2): What is the significance of adding copper(II) sulfate drop by drop?
When you perform the Biuret Test in qualitative manner, you need to be careful to see color change clearly, I'd say picture perfectly. Suppose, you have very dilute protein sample. Adding the blue solution into a colorless solution is tricky, because if you are not slow enough, you won't realize there is a color change happens due to both (before and after) solutions been colored but intensities are low (see the near pipette tip of the picture). If you are doing the quantitative test, it doesn't matter how fast you add the solution. After color development, you use UV-Vis-Spectrometer to measure color intensity: at appropriate wavelength to purple color ($\pu{540 nm}$) where absorbance is zero for the $\ce{CuSO4}$ blue color, thus interference is minimal.
Your question (3): How would the reaction vary if copper(II) sulfate was added first and $\ce{NaOH}$ added subsequently, drop by drop?
When you search for the Biuret Test, you'd find out that biuret solution is already alkaline with $\ce{NaOH}$ (see the description of solution in the picture). Thus, I'd say, the order does not matter. Yet, if you want to do a crude test with aqueous $\ce{CuSO4}$ solution and $\ce{NaOH}$ solution, you should be careful, because initial $\ce{Cu(OH)2}$ precipitation might happens. If interested, you may read Ref.2 to see the effects of all solutions on the test, which gives very comprehensive review of the test.
References:

H. Sigel, R. B. Martin, “Coordinating properties of the amide bond. Stability and structure of metal ion complexes of peptides and related ligands,” Chem. Rev. 1982, 82(4), 385–426 (DOI: 10.1021/cr00050a003).
A. G. Gornall, C. J. Bardawill, M. M. David, “Determination of Serum Proteins by Means of the Biuret Reaction,” Journal of Biological Chemistry 1949, 177, 751-766 (http://www.jbc.org/content/177/2/751.short).

great explanation but one thing i can't quite understand yet. When you say "All of them can be reacted with NaOH
to become ionic" If we just want to give the NH3 group a charge why we use NaOH and not HCl? Also making the protein more polar in order to be more soluble could denature the protein. The salt bridges of the side chains whoud  definitely be affected if we put a strong base such as NaOH right?

It was the explanation for why protein dissole in the  Biuret Test solution. We have to use NaOH as necessary reagent for the test.


text:
What is the molar extinction coefficient of azorubine (carmoisine, E 122)
I'd be grateful for the molar extinction coefficient ε of azorubine at the absorption maximum (about 520 nm), with an authoritative reference.

Found the following value (see the very comprehensive reference below):
molar absorption coefficient $\pu{\epsilon = 24 000 M-1 cm-1}$ ($\pu{510 nm}$) in water
CARMOISINE (E122), CAS Number 3567-69-9, (3~{E})-4-oxo-3-[(4-sulfonaphthalen-1-yl)hydrazinylidene]naphthalene-1-sulfonic acid, SMILES(canonical): C1=CC=C2C(=C1)C(=CC=C2S(=O)(=O)O)NN=C3C=C(C4=CC=CC=C4C3=O)S(=O)(=O)O,
References
Taniguchi, M. and Lindsey, J. S. (2018), Database of Absorption and Fluorescence Spectra of >300 Common Compounds for use in PhotochemCAD. Photochem Photobiol, 94: 290-327. doi:10.1111/php.12860
https://onlinelibrary.wiley.com/doi/epdf/10.1111/php.12860
Originally from

Kamachi, S., K. Wakabayashi, K. Zaitsu and Y. Ohkura (1983)
New chromogenic substrates for the assy of esterases — acetates
and butyrates of phenolic naphthylazo compounds with sulfonic
acid group. Chem. Pharm. Bull. 31, 162–167.
Lin, J.-J., D. Liu, C. Fu and T.-Z. Li (2012) Optimization of
Fenton-like degradation conditions of Acid Red 14 azo dye under
low frequency ultrasonic irradiation. Asian J. Chem. 24, 4453–
4457.
Arvin, M., G. Dehghan, M. A. Hosseinpourfeizi and A. A. Moosavi-
Movahedi (2013) Spectroscopic and electrochemical studies on
the interaction of carmoisine food additive with native calf thymus
DNA. Spectrosc. Lett. 46, 250–256.


text:
Why has the mole been redefined for 2019?
The mole is still defined as an SI unit representing the number of constituent particles (whether they be atoms, molecules etc.) in 12 grams of carbon isotope 12 (carbon-12).
Effective from May 20 this year the definition of the mole will change to a unit representing Avogadro constant number of particles.
Since the Avogadro constant was and will be still tied to the mole (number of particles in a mole), I think this also changed the definition of the Avogadro constant, though I'm not sure. Although the old and new definitions define the constant as the number of particles contained in the mole, it seems that the definition of the mole has changed from the earlier carbon-12 definition to simply the Avogadro constant.
First of all I would like to know if anyone knows why this redefinition happened now (it was decided at the end of the last year).
Also, please correct anything wrong I've said. This is rather confusing for me and the definitions seems rather circular. The new definitions are that Avogadro's constant is the amount of particles in a mole, and a mole is defined by the Avogadro constant.

Previous definition:
  The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.
2019 definition:
  The mole, symbol mol, is the SI unit of amount of substance. One mole contains exactly 6.02214076 × 1023 elementary entities. This number is the fixed numerical value of the Avogadro constant, NA, when expressed in the unit mol−1 and is called the Avogadro number.
Avogadro constant - Wikipedia

related https://chemistry.stackexchange.com/questions/60767/effects-of-changing-avogadros-constant

If you haven't already seen it, Wikipedia has a good article on the redefinition: https://en.wikipedia.org/wiki/2019_redefinition_of_SI_base_units#Mole

Similar to the 25.4mm argument. They are not the same if we increase the precision of the measurement. The new definition effectively "rounds" the value.

You have already mentioned the last definition of the mole before the proposed change:

The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
(…)

According to this definition, the mole still depends on the definition of the kilogram. The main purpose of the proposed change is to find an independent definition. Another purpose is the possibility of linking the definition to an exact value (similar to the speed of light.)
The formal decision for the change was made during the 26th meeting (13–16 November 2018) of the General Conference on Weights and Measures (CGPM).
The motivation behind the change of the definition of the mole can, for example, be found in the resolutions of the 24th meeting (17–21 October 2011) of the General Conference on Weights and Measures (CGPM); however, the discussions and preparations started already much earlier.

On the possible future revision of the International System of Units, the SI
(…)
considering
(…)
that of the seven base units of the SI, only the kilogram is still defined in terms of a material artefact, namely, the international prototype of the kilogram (1st meeting of the CGPM, 1889, 3rd meeting of the CGPM, 1901), and that the definitions of the ampere, mole and candela depend on the kilogram,
(…)
that it is also possible to redefine the mole so that it is linked to an exact numerical value of the Avogadro constant $N_\mathrm A$, and is thus no longer dependent on the definition of the kilogram even when the kilogram is defined so that it is linked to an exact numerical value of $h$, thereby emphasizing the distinction between amount of substance and mass,
(…)

Thank you. So mainly to decouple its definition from the kilogram. I personally find this a simpler definition. I also have many other questions such as why it wasn't just defined as Avogadro's constant in the first place, but I think there are already questions about that here. Actually I don't even quite get why this was a motivating force behind the decision exactly, the separation of the kilogram to the mole, but I'm sure I'll find out with some more reading.

In the old world, the Avogadro constant was the number of particles in a mole, which was determined by experiment to be approximately some numerical value. In the new world, the number of particles in a mole is the Avogadro constant, which is some numerical value by definition. If an experiment shows tomorrow that the number of atoms in 12 grams of carbon-12 is actually more or less than that number, then we will say that that sample of carbon is more or less than 1 mol, but not that the definition of the mole or the Avogadro number has changed.

Does anyone know why Carbon was chosen (as opposed to say the more 'obvious' choice of Hydrogen)?  Perhaps gasses are too difficult to measure?

@JosephDoggie, It is a very very long story as to why C-12 was chosen. Chemists and physcists used to have a different atomic mass references. Chemists fixed ordinary oxygen as 16 (exact), physicists chose oxygen -16 isotope. Note that ordinary O is a mixture of isotopes. All the time their reported values were off and every author had to tell whether they are using chemist's scale or a physical scale. Later two persons suggested why don't we all have a common choice i.e. C as 12 exact, this will cause minimum changes in atomic masses scale. Finally the feuding parties agreed to use C-12.

@M.Farooq wouldn't C also be easier to weigh and measure than O, with it being a solid?

@JAD, no it was not a weighing issue at all. People weighed gases very very accurately long time ago. The problem was the fight between chemists and physcisits. They were not willing to change their relative atomic scales, until someone came up with a idea that if we use C-12, both chem and phys have to change their tables and the changes in their values will be minimum. It was an economic issue for chemical industries. It is an interesting story which no gen chem text talks about.


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Effect of pH on charge of jasmonic acid
It has been such a long time, that I totally forgot how it works.
I am interested in the charge of jasmonic acid at different concentrations.
More specifically, I want to know at what pH my molecule has no charge (or I should rather say jasmonic acid would occur as a zwitter ion).
Only the decaboxylic group can be deprotonated at higher pH.

Is it correct to assume that most of the time, when you have an
acidic compound it would predominantly occur in a nonionic form?
Does the $\log{D}$ say anything about this? does the maximum
$\log{D}$  value corresponds to the pH at the iso electric point?

If we would take aspirin for example
https://disco.chemaxon.com/apps/demos/logd/
at what pH should I expect the uncharged version?

Jasmonic acid has one group that is significantly ionizable, the carboxylic group:

Most molecules in the sample will be negatively charged when the $\mathrm{pH}$ is higher than the $\mathrm{p}K_\mathrm{a}$ of the molecule and an equal number will be neutral or charged when $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$. The same will hold for aspirin, which also has a carboxylic acid group. Below a $\mathrm{pH}$ equal to the $\mathrm{p}K_\mathrm{a}$ it will be mostly uncharged, and above this $\mathrm{pH}$ it will mostly be charged.
If you look up $\mathrm{p}D$, you might encounter the following equation:
$$-\mathrm{p}D = \log D_\mathrm{oct/wat} = \log\left( \frac{[C]^\mathrm{ionized}_\mathrm{octanol} + [C]^\mathrm{unionized}_\mathrm{octanol}}{[C]^\mathrm{ionized}_\mathrm{water} + [C]^\mathrm{unionized}_\mathrm{water}}\right)$$
Since the concentration $[C]$ of the ion is typically higher in water than in octanol, increased ionization at high $\mathrm{pH}$ will increase the solubility in the water phase, and $\log D_\mathrm{oct/wat}$ will decrease (become more negative) at high $\mathrm{pH}$, that is, $\mathrm{p}D$ will increase (become less negative). You can expect the solubility in octanol to continue to decrease as you near and pass the isoelectric point.
For more complex molecules with more ionizable groups you will need to consider individual contributions from different groups. Basic groups will respond in the opposite manner to acidic ones.

Thanks for clarifying this. So basically, the ionisation graph would be a more or less a mirror of the Log D graph

Hmm, that's an interesting relation. I'll have to think about that a little more.

@AnneliesRoussou I agree with your analysis. logP should be measured when the molecule is not ionized, that is low pH. So the result makes sense for aspirin, not so for jasmonic acid. It may have to do with how the programs make predictions. They are probably smart enough to predict logP at a point where the molecule is not ionized (if this is possible).

Hi Night writer, i deleted my comment because i realized when i uploaden the structure of JA in chemaxon logD predictor I made a minor error in its structure. Actually the Log D (at pH far below the pka), when the compound is unionized, does correspond to the LogP. My mistake. The chemistry part has just been such a long time ago since i switched to a different field

Yes, its completely clear now. Thanks a lot, you've been really helpfull!


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Some information about DOWXV503
I want to know about the attractor DOWXV503. Is there any scientific book or article about this material?  Any information is useful and it will be appreciated.

Please provide more information about the material ;-) - where did you find the name, give a little context etc - help us help you!

Google is your friend.
The material is Dowex V503 a polymeric absorbent for the removal of organics from humid air streams. See more  here and here


text:
Alkylation of acetophenone

When the NaOH deprotonates the hydrogen off the carbon, I thought there would be a resonance structure, with the negative charge being on both the oxygen and carbon. Since the oxygen is more stable with the negative charge, wouldn't the alkalation happen at the oxygen forming an ether? This is how i had the reaction occurring:

In theory, both can happen. The answer as to why alkylation on carbon is preferred can be quite complex, depending on how deep you want to go: a lot of textbooks teach it as being a result of hard-soft acid-base theory, which is a convenient rationalisation but not really true.

Also remember, stability of the final product plays a role as well. The proposed enol-ether is relatively (thermodynamically speaking) unstable than ketone product given above.


text:
Mixing of molecular orbitals in polyatomic molecules
I am trying to solve the following Hamiltonian,$$H=-\frac{1}{2}\nabla^{2}-\sum_{i=1}^{10}\frac{1}{|\vec{r}-\vec{R_{i}}|}$$ Here $Z=1$ . To get all ther molecular orbitals of this system. I took the following linear combination to construct the hamiltonian matrix,$$\psi(\vec{r})=\sum_{i=1}^{10}c_{i}1s(\vec{r}-\vec{R_{i}})+\sum_{i=1}^{10}a_{i}2p_{x}(\vec{r}-\vec{R_{i}})+\sum_{i=1}^{10}b_{i}2p_{y}(\vec{r}-\vec{R_{i}})+\sum_{i=1}^{10}g_{i}2p_{z}(\vec{r}-\vec{R_{i}})$$
I took this wave functions and solved for $E$ using the following equation. Usually in polyatomic molecules, one would look for symmetry to construct various molecular atomic orbitals. However, here there is no symmetry. How do I get the molecular orbitals? What combination should I have taken to solve this problem. Please suggest some references where I can read the theory of molecular orbitals for polyatomic molecules with no symmetry

The standard way is to solve for the eigenvectors of your Fock matrix. Since you have a single electron, that's your Hamiltonian. Any number of programming languages have algorithms for this implemented. Each of your coefficients would correspond to an entry in the eigenvector.


text:
Nail polish chemistry question
When I read my ingredients on my nail polish remover, it has water as its first ingredient.  Why does nail polish have water, yet the warnings still say its flammable?

I'm voting to close this question as off-topic because ingredients of healthcare (and other) products are listed in descending order of percent composition and, more to the point, mixtures can still be flammable even if water is the first component listed.

If you've ever held a flame to rum or other high proof alcoholic drink, you'll understand why.

Mixtures of water with flammable liquids can themselves be flammable, largely due to the fact that the flammable liquids generally are more volatile than water, (ie they evaporate more easily.) As a result, the air above a solution of nail polish will have a lot of molecules of the flammable liquid in it. (That's why you can smell them easily). If you have a flame near the liquid, those molecules in the air above it can combust. The heat generated will evaporate more liquid, which will fuel the fire. Some of the heat is lost to vaporizing water, which doesn't burn, but if the fraction of the total liquid that is flammable is high enough, the combustion continues.
In nail polish, the flammable liquid is usually acetone or ethyl acetate.

Flammability (technically, inflammability, but I guess people are confused by this term) is a function of some vapor. Yes, water is not flammable, but the flammable liquid, in the liquid state, also doesn't burn. It's specifically the vapor that is flammable. Usually, there is some threshold concentration, reflected in the partial pressure of the flammable gas, that is required to ignite the material. Once it ignites, the heat generated by the combustion reaction promotes combustion of the adjacent vapor and heats the liquid enough to produce more gas.
If you mix water and your flammable solvent (assuming they're miscible, which they kind of have to be), you have a solution. Raoult's law says that the partial vapor pressure of the gas above the solution is proportional to the mole faction of the liquid in the solution. In this case, it means that even if the flammable liquid is diluted with water, there's still vapor above the liquid. And though the pressure may be less, it may be high enough that it's still highly flammable.


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Redox reactions KMnO4 changing to brown
If potassium permanganate is added to a solution and it becomes brown is the substance a reducing agent?

Attribute the brown color to a compound, compare oxidation states before and after, and you get the answer.

Permanganate is a strong oxidizier so most substances will be reducing agents in comparison. Even hydrogen peroxide.

Brown in potassium permanganate reactions usually means you formed MnO2.  The cure, If you want a nicer looking reaction forming Mn(II) ion in solution, is to add acid.  If you render the balanced equations for permanganate reduction, you find you need a lot of acid to get to Mn(II).

Yes, the substance is a reducing agent which will reduce $\ce{KMnO4}$ into $\ce{MnO2 }$ in neutral or weakly basic medium.

Welcome to Chemistry.SE!! It'd be great if you could add a chemical reaction, that really adds value to an answer. It'd also be a great opportunity to learn how to use MathJax and mhchem => https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here


text:
How does a Pt and reference electrode measure the cell potential in a potentiometric titration?
I'm trying to perform a potentiometric titration to determine the redox potential of a $\ce{Fe^3+|Fe^2+}$ couple. To do this, I am using an $\ce{Ag|AgCl}$ electrode connected to an inert $\ce{Pt}$ electrode, which is submerged in a solution containing $\ce{Fe^3+}$ ions. I then titrate it with $\ce{Co^2+}$, which reduces it to $\ce{Fe^2+}$. The reference and inert electrode are connected with a standard voltmeter.
What I don't understand is why the inert electrode measures the redox potential of the $\ce{Fe^3+|Fe^2+}$ couple. As per my understanding, there is an electron flow from the reference electrode into the inert electrode, which would get accepted into the Fe solution, but now, the transfer of electrons is only between the $\ce{Co^2+}$ and $\ce{Fe^3+}$ ions, so what happens to that electron? Can someone explain exactly how the redox potential of the $\ce{Fe^3+|Fe^2+}$ couple gets measured, with reference to electron transfer between the reference and inert electrodes?

Potential measurement ideally requires no current flow. You just refer the potential at which the Pt electrode is versus a fixed one, that of the Ag/ AgCl one. Look for electrode of the third kind.

That’s what I’m confused about - how is the Pt electrode able to measure the potential of the Fe(III)/Fe(II) couple? The electrons are being transferred between Cobalt and Iron, so how does the Pt electrode pick up on the potential of that reaction?

If you have a clear picture of a Pt electrode immersed in a, e.g., Fe+++/Fe++ solution, then you see how it sense the potential of your current system.

The point is that you your are ideally working without current. This part is what a voltmeter takes care. You can also build three electrodes cells, in which current flows to the extra added working electrode, and the reference ekectride is there just to serve as reference indeed.

It is a very good question. Your question has nothing to do with current being drawn or not or a reference electrode etc. Imagine a single Pt electrode dipped in an Fe(II) solution and forget about the presence of cobalt. Imagine that a metal is a source (sea) of electrons.
The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).
As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.
Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


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Which one is more acidic – 3,5-dinitrophenol or 4-nitrophenol
I’m confused because in 3,5-dinitrophenol, the inductive effect is dominating whereas in 4-nitrophenol the mesomeric effect dominates. So which one should we consider for acidity?

Part of the answer depends on why you're asking. If you want to know the answer for which is acidic, you just look it up or measure it. You can't always determine the relative magnitude of competing effects in chemistry (unless you've got amazing intuition and/or experience), so you might not be able to determine the answer just by thinking about it.

As Zhe's comment points out, intuition does not count for much here, unless you have enough actual experience working with these systems to know what is more acidic. There are some things which cannot be predicted a priori using simplified models.
The only way to find out is to look up the data. The acidity constants of some of these phenols are tabulated in J. Org. Chem. 1998, 63 (19), 6680–6688. 4-Nitrophenol has a $\mathrm pK_\mathrm a$ of $7.14$, and 3,5-dinitrophenol a $\mathrm pK_\mathrm a$ of $6.73$, in aqueous solution.

I have used MarvinSketch 17.28 to estimate the acidic strengths of substituted nitrophenols, as measured by their $\mathrm pK_\mathrm a$ values.
$$\begin{array}{cc} \hline
\text{Substitution} & \mathrm pK_\mathrm a \\ \hline
\text{2,4,6-trinitro} & 2.344 \\
\text{2,4,5-trinitro} & 3.224 \\
\text{3,5-dinitro} & 6.238 \\
\text{4-nitro} & 7.075 \\
\text{3-nitro} & 7.892 \\ \hline
\end{array}$$
It is evident that 3,5-dinitrophenol is more acidic than 4-nitrophenol.
The relative acidities of the meta- and para-monosubstituted nitrophenols are in line with predictions based on resonance arguments.

MarvinSketch seems to get most of those pKas fairly accurately, but I'm surprised at how bad the number for picric acid is.

Picric acid has pka about 0.7


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Why do reference electrodes have porous frits or membranes as a salt bridge?
From what I understand, they act as a salt bridge between the reference electrode electrolyte and the analyte solution (in a potentiometric titration). But why do we use partially permeable membranes or frits instead of making a normal salt bridge for the reference half cell?

The answer is here on ResearchGate https://www.researchgate.net/post/What_is_the_function_of_porous_frit_for_reference_electrode

Hi there, the link is not opening - it redirects me to the researchgate.com/topics page.... I think the post may have been deleted. Could you perhaps clarify my question on this thread?

You can register there as a user. The discussion is pretty long.


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Relationship between structure and shock sensitivity for branched vs straight hydrocarbons and ammonium ions
In p. 137 of John D. Clark's Ignition! (in the new edition), Clark reports an interesting observation regarding the shock sensitivity of different isomers of alkyls and ammonium salts.
Specifically, he says that he compared the sensitivity of tetramethyl ammonium nitrate ("Tallulah") in $\ce{N2O4}$ with the isomeric chain ion $\ce{NH3+CH2CH2CH2CH3}$, with the latter being far more sensitive (58 cards vs 8 cards, on his card-gap test setup, which he explains in p. 69).
And, analogously, he compared neopentane with pentane (i.e. the same structures, swapping out $\ce{N+}$ for $\ce C$, again in $\ce{N2O4}$), and again got a fifty-card difference between the two, with pentane testing at 100 cards and neopentane much more stable at 50 cards:

Again a difference of 50 cards, which meant that the ratio of the critical shock wave pressures due to structure was the same in both systems. I was fascinated by this coincidence, but never had the chance to carry the work any further. It's recommended to the attention of some future investigator.

Has this observation been followed up on? If so, what explains the coincidence?

@Mithoron Which aspects of the question would benefit from bibliographic references beyond the one that's already provided?

Ah, there's only a parallel, still I'm curious is the test for hydrocarbons in mix with N2O4 too?

@Mithoron Identical molar mixtures with dinitrogen tetroxide, yes.


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Why the chemical shift value of a proton be negative in proton NMR spectroscopy where TMS is used as standard reference?
If there's any possibility of chemical shift value of a proton be negative.And if yes...why..when I am using TMS as a standard reference.

Yes, it's possible. What's so wrong about a negative number, though?

@Anirban it would just mean the protons are more shielded than those in TMS.

Related: https://chemistry.stackexchange.com/questions/7198/chemical-shift-value-and-proton-nmr-spectroscopy

Is it normal to get negative chemical shifts in $\pu{ppm}$ for NMR analysis?
In general, negative chemical shifts in NMR is possible when you locked it with tetramethylsilane (TMS) as your internal reference, but it is not quite normal with NMR spectra of our day to day chemicals though. If you are not expecting a negative resonance for your compound, then you must calibrate (or make a request to calibrate) your NMR instrument because it may be a calibration problem. In any given case, when you run a NMR spectrum known or unknown compound, you should have a signal that you can identify (such as TMS for example) and whose chemical shift you know. By definition, chemical shift of methyl signal of TMS is exactly to $\pu{0.00 ppm}$. Alternatively, if your solution does not contain TMS, you could set the residual proton signal of the deuterated solvent to its accepted chemical shift (e.g., residual $\ce{CHCl3}$ in $\ce{CDCl3}$ to $\pu{7.28 ppm}$).
Chemical bonds are regions of high electron density and therefore can create local magnetic fields. The best example for that is benzene ring (See the following scheme):

The induced field by circulating electrons is opposed to the applied external field $\mathrm{B_\circ}$ (Lenz’s law). Electrons within $\sigma$-bonds are not usually able to circulate freely and so the chemical shift will depend on the orientation of the nucleus with respect to the bond. As a result, higher frequency chemical shifts can be obtained for protons outside the ring and lower frequency chemical shifts can be obtained for protons inside the ring. The perfect example is illustrated in following scheme (From Ref.1):

As you see in the example, inner methyl groups giving negative resonances and outer $\ce{t-Bu-}$ groups having higher field than normal. Hope this simple version make you understand.
References:

M. Jacquet, F. Lafolet, S. Cobo, F. Loiseau, A. Bakkar, M. Boggio-Pasqua, E. Saint-Aman, G. Royal, “Efficient Photoswitch System Combining a Dimethyldihydropyrene Pyridinium Core and Ruthenium(II) Bis-Terpyridine Entities,” Inorg. Chem. 2017, 56(8), 4357–4368 (DOI: 10.1021/acs.inorgchem.6b02861).

AFAIK modern spectrometers use the solvent deuterium resonance to lock. TMS is just for referencing, not locking

@orthocresol: Thank you for the comments. I agree. That's what I wanted to say but didn't put right words. :-) It was Friday night and I was rushing to go home. Well done!


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Chemical that won't freeze below melting point
I have an organic chemical (can't say due to an NDA) that has a nominal melting point of 28 °C. When I raise it above this temperature, it melts, but then it will stay a liquid indefinitely at lab temperature (≈22°C), and in the fridge (4 °C) for at least a week. When put in the freezer (-20 °C), it will freeze, and then stay a solid at lab temperature indefinitely.
Is there an explanation for this phenomenon? I'm assuming this may have something to do with the kinetics of the crystallization, but I've never seen something with this level of hysteresis.

Supercooling is a pretty common thing with organics.

We even exploit this to make hand warmers. Sodium acetate is easy to supercool and needs a "kick" to start crystallisation. Small sealed pouches are sold as hand warmers. They contain a flexible metal disk that when clicked kicks off crystallisation and this releases heat. The solid push can be reused by warming it and melting the solid. It stays liquid when cooled.


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What is it in the composition of natural fabrics that cause them to retain smoke odors versus synthetics?
Just an observation I notice when around smoker friends or when I visit Las Vegas - polyester clothing doesn't seem to stay smelling of smoke for the rest of the day like my cotton shirts do (or also my hair for that matter).

You would probably need to [edit] and reframe this question as something practical, like how to select odor resistant textiles.

Adsorption is a very complicated phenomenon which involves so many interactions. Smoke has thousands of chemicals too, therefore one cannot pin-point a single cause. It is just like the concept of stains. Some stains on cotton never come off easily, which means that molecular interaction is very strong between the dye in the stain and cotton fiber. As a rough estimate, the -OH groups on cellulose may be providing the hydrogen bonding sites to nitrogen containing aromatic (in a chemical sense) compounds, hence stick for a long time. I am drawing an crude analogy because silica which has plenty of silanols (Si-OH) groups bind amines too strongly.


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How do water models fit into force field methods?
What I Understand:
In molecular mechanics, it is recognised that molecules tend to be composed of units that are structurally similar in a vast array of environments so atom types are defined from which molecular models can be built.
A force field is defined that describes the energy required to distort the molecule from equilibrium.[1]
$$E_\mathrm{FF} = E_\mathrm{str} + E_\mathrm{bend} + E_\mathrm{tors} + E_\mathrm{vdw} + E_\mathrm{el} + E_\mathrm{cross}$$
Once a functional form has been decided on, the field is parameterised based on empirical results.
What I Don't:
A large array of force fields have been developed, each different in their functional form or parameterisation. Numerous "water models" exist, and these appear to be force fields in their own right, often with far fewer energy terms.
The SPC model, for example, appears to contain only Coloumbic and Lennard-Jones terms, as described in Berendsen.[2]

We developed a three—point charge model (on hydrogen and oxygen
  positions) with a Lennard— Jones 6—12 potential on the oxygen
  positions only.

What is the purpose of these water models?
Question:
Why are standalone water models required, and how are they incorporated into more general forcefields like MM2?

References

F. Jensen, Introduction to Computational Chemistry, John Wiley &  Sons, Chichester, 2007.
Berendsen, H. J. C., Postma, J. P. M., van Gunsteren, W. F.,
and Hermans, J., Interaction Models for Water in Relation to Protein Hydration, in: Intermolecular Forces: Proceedings of the Fourteenth Jerusalem Symposium on Quantum Chemistry and Biochemistry Held in Jerusalem, Israel, April 13--16, 1981. DOI: 10.1007/978-94-015-7658-1_21; pdf mirror

Are you familiar with continuum models and their shortcomings?

I've read that a solvent can be primitively modelled by a medium of constant dielectric constant, but I wouldn't know how to apply that in the context of my question...

The reason I ask is because you wondered why standalone models were required, and I assumed this could mean, why not use something simpler, like a continuum. Perhaps you meant why are force field parameters from other molecules not used?

I suppose I'm looking for the latter. Why isn't water taken care of in, for example, the MM2 forcefield, like every other molecule in the simulation. Why does it require its own forcefield? And then, how does that run alongside the MM2 part?

I'm not sure how far you've come down this road. Maybe the following is useful to review: http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/molmech-lecture.pdf

I feel like that review pretty much covers all I know. It introduces force fields, the functional forms of each component (stretching, bending, etc.), and atom types. But I'm still not comfortable with the combination of force fields within a simulation.  If water needed to be treated differently, why not just define new atom types specific to the water environment?

I don't know the history of mm2 or force-field dev to give you a detailed answer, but I know water is hard to simulate, because the point is to replicate properties. Capturing properties over a wide range of T/P for water, and including also behavior in a range of environments, is hard because you want to minimize computational costs. TIP3P, SPC/E etc all vary in accuracy under different conditions. Parameters were optimized for selected conditions. Most MD packages allow choice of different water models.

Dunno why mm2 doesn't have explicit water parameters, not that familiar with it. It would be possible in principle to incorporate but the package was not intended for water (I suppose), given other focus at the time.

Using a special parameterization optimized for a selected molecule can be done by providing a parameter labels/files that the program understands, and if the program has rules for mixing molecules with different parameters (particularly LJ).

I guess for starters, try finding a torsion angle on a 3 point molecule.
Is it fast enough, accurate enough?
The real issue is why do some force-fields have intra-molecular bonds and others do not.
Including bond and angle forces in water models make their computation more expensive. There are classes of flexible forcefields for water. There are even more indepth forcefields that are polarizable.
Why don't we use the most sophisticated forcefield models for water?
The more sophisticated the model, the longer our simulations take. Often, rigid water models are good enough for certain properties. So, we use the simplest model that gives us the right answer for the property we are after. This is just a general good rule.
Parameters
Force-fields use parameters, and proteins use ALOT of parameters. These parameters are often fit for a specific water model. Hence, you are kind of stuck using that water model if you are using that Force-Field for the protein/solute molecule.
Complexity
As our interests move towards more involved calculations that involve water it does become important to include more effects such as flexibility and polarizability. This is increasingly the case in calculating Free Energies, whether it is the Free Energy of hydration or the Free Energy of binding a ligand to a protein. Another example is modelling supercritical water - only polarizable models have been successful. Rigid models such as SPC/E and TIP3P simply are not good enough for many properties under many conditions. You can of course fudge parameters to make anything fit a particular problem, but we want to use generalizable force-fields as much as possible. In this case, models with more parameters (flexible, polarisable) give us additional accuracy for the difficult problems.
Bandwagon
Another reason many people use one model more than another is because their colleagues use it, so they use it because other people use it and it likely won't be questioned since other people use it. This is not a good reason for using a certain model! usually the first people to use a certain model had a reason, it solved their problem. However often the following bandwagon users don't read the strength/weaknesses of the model but instead just use it because they see it used lots for seemingly similar areas of research. Welcome to molecular modelling.
Mixing Rules
Generally atoms have dispersion and electrostatic terms. In the case of dispersion, when the force or energy between two atoms is being calculated a single parameter for each term is required for the calculation. Take the well known Lennard-Jones for instance
$$U(r_{ij}) = 4 \epsilon_{ij} \left[ \left( \frac{\sigma_{ij}}{r_{ij}}\right)^{12} - \left( \frac{\sigma_{ij}}{r_{ij}}\right)^6 \right]$$
different forcefields may have different numeric values for $\epsilon_i$, $\epsilon_j$, $\sigma_i$,$\sigma_j$ but they will all have a value. You need to use mixing rules to take the respective $i$ and $j$ parameters and create an $\epsilon_{ij}$ and $\sigma_{ij}$. We call these mixing rules a.k.a. combining rules. Wikipedia lists several of the common ones. https://en.wikipedia.org/wiki/Combining_rules
For instance for the Lorentz-Berthelot
$$\epsilon_{ij} = \sqrt{\epsilon_{i} \epsilon_{j} }$$
and
$$\sigma_{ij} = \frac{\sigma_i + \sigma_j}{2}$$
Finally, in SPC and SPC/E the hydrogens do not have Lennard-Jones parameters. It is not uncommon for hydrogens to have no dispersion parameters. This can lead to problems though, since dispersion parameters keep other atoms/particles from getting too close.
So a specific example of an SPC water molecule interacting with a Sodium ion, the electrostatics just needs to know that the Oxygen has a negative charge of $q_o =−0.82$ and the sodium has a charge of $q_{N_a} =+1$. A naive approach would be to treat this as a bare coulomb interaction which is just
$$ U(r_{ij})^{\rm electrostatic} = k \frac{q_o q_{N_a}} {r_{ij}}$$
where
$$k = \frac{e^2}{4 \pi \beta}$$
$e$ is the charge of an electron and $\beta$ is the permittivity and a global and constant value for the simulation.
A better approach is to use EWALD's summation for electrostatics. It is more complicated, but the only charge specific parameters it needs to know is still just $q_o$ and $q_{N_a}$. The other parameters it uses for damping can be considered global parameters. Note: also do this for each of the hydrogen charges interacting with the Sodium ion.
For Lennard-Jones you would need to use mixing rules for the oxygen (there is no LJ parameters for the hydrogens)
$$\sigma_{(o-N_a)} = \frac{\sigma_o + \sigma_{N_a}}{2}$$
and
$$\epsilon_{(o-N_a)} = \sqrt{\epsilon_o \epsilon_{Na}}$$
then do the standard LJ calculation for force and or potential energy.

This is a good overview of why a particular forcefield might be used, but what I'd like to know next is how are they integrated? For example, if I were to simulate an organic molecule solvated in water, where would each force field apply. Presumably the water-water interactions would be handled by SCP, and solute-solute interactions would be handled by MM2/CHARMM/..., but how are water-solute interactions handled?

Ahh, very perceptive! For the dispersion forces (typically Lennard jones) you use mixing rules. For electrostatic charge is charge.... There is no mixing rules. I will add this to my response when I have a keyboard rather than my mobile :)

In short you either take the average of the two parameters or you multiply them and take the square root.

I would also recommend using modern forcefields, not MM2. Check out Allen & Tildesley or Frenkel & Smit for references.


text:
Solubility of a tribasic weak acid
I have a question that reads:

$\ce{H3A}$ is a tribasic acid. The three deprotonations can be written as
$\ce{H3A <=>[pKa1 = 2.9] H2A- <=>[pKa2 = 4.2] HA^{2-} <=>[pKa3 = 5.2] A^{3-}}$
What is the predominant species in a 0.10M solution of $\ce{H3A}$ at pH 5?
a) $\ce{H3A}$
b) $\ce{H2A-}$
c) $\ce{HA^{2-}}$
d) $\ce{A^{3-}}$

I worked this out as
$\ce{Ka1 = \frac{[H+][H2A-]}{[H3A]}}, \ce{Ka2 = \frac{[H+][HA^{2-}]}{[H2A-]}}, \ce{Ka3 = \frac{[H+][A^{3-}]}{[HA^{2-}]}}$
At pH 5, $\ce{[H+] = 10^{-5}M}$
So,
$\ce{[H+] = 10^{-5} = Ka1\frac{[H2A-]}{[H3A]} = 10^{-2.9} \frac{[H2A-]}{[H3A]}}\ce{\implies \frac{[H2A-]}{[H3A]} = 10^{2.1} }$
and similarly,
$\ce{\frac{[HA2-]}{[H2A-]} = 10^{0.8}}$
$\ce{\frac{[A^{3-}]}{[HA^{2-}]} = 10^{-0.2}}$
Since $\ce{[HA^{2-}] > [H3A], [HA^{2-}]>[H2A-] and [A^{3-}]<[HA^{2-}]}$, the predominant species is $\ce{HA^{2-}}$, so c).
This is the correct answer according to the solutions, but I'm not sure if my logic is correct, or if there's an easier way to arrive at the same conclusion (it's significant working out for a MCQ).

You can also use the Henderson-Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{\ce{[A-]}}{\ce{[HA]}}$$
rearranges to:
$$\mathrm{pH} - \mathrm{p}K_{\mathrm{a}} = \log \frac{\ce{[A-]}}{\ce{[HA]}}$$
$\log \frac{\ce{[A-]}}{\ce{[HA]}} > 0$ for conjugate base dominating.
$\log \frac{\ce{[A-]}}{\ce{[HA]}} < 0$ for acid dominating.
$\log \frac{\ce{[A-]}}{\ce{[HA]}} = 0$ if they're present in the same concentrations.
The equilibrium of interest is the third dissociation.
Here, the LHS equaluates to $-0.2$ indicating that the acid in that equation ($\ce{HA^{2-}}$) dominates over the conjugate base ($\ce{A^{3-}}$). You should be able to extend this for each equilibrium.


text:
Electrical resistivity
I have read that in non-metals solids the electric resistivity is reduced if we increase temperature.But this isn't logical.I mean if temperature is increased ok the electrons get extra energy but as well the vibration of the atoms is increased so electric resistivy is increased.Temperatures cant make electrons overcome the band gap easily.There should be an increase till the temperature is high enough so with little energy the valence electrons would get ionized and then electric resistivity to be dicreased. Where am I wrong?Help appreciated

You are right. But till the temperature is that high, there should be an increase, and so it is.

Yeah but why tables show a negative temperature coefficient of resistivity for some materials then?

Well, just like that: as T grows, more electrons make it across the band gap, hence more charge carriers, hence more conductivity, even though the situation for each particular electron grows worse.

What you write doesnt make any sense . 1 hour ago you said that the resistivity is increased

Then I might have misunderstood your original post, sorry about that. No, it is the conductivity that increases, and I just explained why.

Wait so not all electrons have the same energy or absorb the same energy? What is happening?

Of course not. Wait, didn't you use the term "band gap" a while before? What is that gap and why is it important? I'll tell you why: most electrons just sit below the gap and do not matter. But a few of them have energies that put them above the gap. They are allowed to roam freely around the crystal, hence the conductivity.

Well band gap is the energy required for a  valence electron to leave its orbital and enter the conduction band.Non-metals usually have a relative medium band gap.

Yeah, that's right, and that's pretty much what I just said, isn't it?

They must have this energy to overcome the band gap . If one valence electron has the energy to overcome this energy barrier then all valence electrons of similar atoms will overcome this energy barrier.

There is no such thing as few of valence electrons  will pass the energy barriers and some wont.

I appreciate your contribution despite you having not answered my question yet.

OK, I got it. You think that since all atoms are identical (which they are), all their orbitals are identical, and so the electrons that reside on these orbitals must have the same energy. Well, that's about right, and that's why once an electron receives a handful of energy from whatever source, it ceases to occupy that orbital.

Yes all atoms of the same element are identical , aren't they?

They are, but look again at my previous comment.

Yes I mean there should be a huge change in resistivity if valence electrons have found the energy to overcome the band gap: and before the valence electrons get that energy the resistivity is increased.

OK, now look at it this way: there is no "before". As soon as the temperature is above 0 K, some electrons (maybe very few) are above the gap. Say, at a given temperature about $1\over 1,000,000,000$ of all electrons get enough energy to overcome the gap; that may seem like not much, but give it a little heat, and you'll find 10 times that much, which is still not much, but hey, it's a 10 times increase.

Between 273 K and 500 K the intrinsic carrier concentration in Si increases by about 5 orders of magnitude. Compare with the decrease in conductivity of a metal over that range. Increase in carriers far outweighs the increased scattering of carriers.


text:
What does the reaction of hydrogen and iodine to hydrogen iodide look like?
The reaction, $\ce{H2 + I2 <=> HI}$ is often used as an example for equilibrium reactions. It is however a difficult reaction to show in a school laboratory: we are not equipped to work with flammable gas at $\pu{450 ^\circ C}$ and high pressure.
It would be great if I could show my students what this process looks like in real life. I have a description, with drawings, by Max Bodenstein, from 1896 -- slightly dated.
Does anyone have pictures or videos of the gases involved and of the equipment used to make $\ce{HI}$ from the elements and to measure reactants and products?

Synthesis from elements is an industrial method. In the lab you probably want to use hydrolysis of $\ce{PI3}$ instead.

What's the point? It won't make a great display anyway.

andselisk: Thanks, but I want to show this particular equilibrium reaction. I don't need the hydrogen iodide.

Ivan: Students will encounter this reaction in textbooks and/or exams. I think it's important to show that it's a real thing that happens to actual substances, rather than just theoretical juggling with symbols. "Great display:" there's a purple gas that turns transparent, and there must be cool gear involved.

To show an example for equilibrium reactions to your students, you can use a reaction which has visual change as shifting the equilibrium position such as color. My best bet is, use of cobalt(II) chloride equilibrium with water:
$$\ce{[CoCl4]^2-(aq) + 6H2O(l) <=> [Co(H2O)6]^2+(aq) + 4Cl-(aq)}$$
The beauty her is The ion, $\ce{[Co(H2O)6]^2-(aq)}$ is pink while $\ce{[CoCl4]^2-(aq)}$ is blue.  According to the Le Chatelier’s principle, at relatively low chloride concentrations (e.g., diluting the solution or adding $\ce{AgNO3(aq)}$ to remove $\ce{Cl-}$ ions), the equilibrium shifted to the right and, therefore, the solution becomes pink. However, if there is a large concentration of excess chloride (e.g., adding $\ce{Cl-}$ ions by means of adding $\pu{12 M}~\ce{HCl}$ drop wise or adding $\ce{CaCl2}$ beads), the equilibrium tends to shift to the left, making the solution blue.
In addition, similar to its sensitivity to the concentration of solutes, the equilibrium is also sensitive to temperature as well (See picture below). At colder temperatures, the equilibrium tends to shift to the right(pink color at LHS of the picture) while at warmer temperatures, it lies to the left (blue color at RHS of the picture):

Reading:

http://www.rsc.org/learn-chemistry/resource/res00000001/the-equilibrium-between-two-coloured-cobalt-species?cmpid=CMP00005957
https://eic.rsc.org/cpd/understanding-equilibrium-a-delicate-balance/2000012.article
http://research.chem.psu.edu/mallouk/ilab/Equilibrium_and_Le_Chatelier.htm

Thanks Mathew, this is nice! I'll try that sometime.
I like to show [Fe(SCN)2]+ -- might be less toxic? Or starch with iodine.

Well you could create extra pressure by adding both gases in a container close the container with a moving top . As long as you pressure the top and the container's volume is decreased according to the law of thermodynamics (Boyle) you can increase the pressure at a significant level .

In theory, yes. In a school lab? Er, well...


text:
Requesting clarification on specific gravity
I am unclear why the ratio of molecular weights will yield the specific gravity of a substance relative to the standard substance (usually air or water at a specific T and P). I always took specific gravity to be a function of temperature since densities are temperature dependent, so I would find density data at approximate conditions to calculate specific gravity whenever I needed to.

I always took specific gravity to be a function of temperature since densities are temperature dependent, so I would find density data at approximate conditions to calculate specific gravity whenever I needed to.

You are correct. Densities, and therefore specific gravity values, are temperature dependent.

I am unclear why the ratio of molecular weights will yield the specific gravity of a substance relative to the standard substance (usually air or water at a specific T and P).

This requires a bit of unraveling. I'll point out that specific gravity is unitless being the ratio of the density of the substance in question to a particular standard.
For gases:
For the specific gravity of a gas at sea level, air is assumed as the standard. However I remember gas values expressed mostly as densities g/L or kg per cubic meter.
Assuming ideal gas behavior and STP as defined by IUPAC, the molar volume of any gas is 22.7 liters. So the relative densities of ideal gases would be determined exactly by the relative molecular masses.
For liquids and solids:
The standard would be water which has a value of about 1 gram per ml. For solids and liquids atmospheric pressure fluctuations can generally be ignored, but temperature must be taken into account.
The relative molecular masses would, in general, have a poor correlation with specific gravity though there would might be good correlations with some series of similar chemical molecules (for instance the monohalo benzenes).

Under the "For gases" header, do you mean to say "...gas at sea level, air is assumed as the standard"? Why would we have to make an assumption about the volumes of the gases being compared, isn't density an intensive property?

@J.Ari - Thank you for the correction. You are correct that I didn't include "air" in the sentence. // I assumed ideal gas behavior because then using PV=nRT at STP the volume of 1 mole of every gas would be exactly 22.7 liters. But of course not all gases conform to the ideal gas model exactly. It is really a matter of the precision of the measurement. To two significant figures I can't imagine any difference. To three maybe a few gases would differ. To six significant figures then a lot will vary from ideal behavior at STP.


text:
Equilibrium constant for the neutralization of weak acid by strong base

Acetic acid has a $K_\mathrm{a}$ of $\pu{1.8e-5}$. What is the equilibrium constant for the neutralization of this acid with $\ce{NaOH}$?

Given acetic acid
$$\ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} \qquad K_\mathrm{a} = \pu{1.8e-5}$$
$$\ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$
So, if we do $K_\mathrm{w} = K_\mathrm{a}K_\mathrm{b}$, then we get $K_\mathrm{b} = \pu{5.55e-10}$. How do I use this to find $K_\mathrm{eq}$?
I know how to find $K_\mathrm{eq}$ using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?

Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant $K$ and unravel a tangle from there:
$$\ce{HOAc + OH- <=> OAc- + H2O}$$
$$K' = \frac{[\ce{OAc-}][\ce{H2O}]}{[\ce{HOAc}][\ce{OH-}]}$$
Since $[\ce{H2O}] = \text{const}$ (reaction medium), $K'[\ce{H2O}] = K = \text{const}$:
$$K = \frac{[\ce{OAc-}]}{[\ce{HOAc}][\ce{OH-}]}$$
By multiplying both numerator and denominator by $[\ce{H+}]$, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant $K_\mathrm{a}$ and ionic product of water $K_\mathrm{w}$:
$$K = \frac{\color{red}{[\ce{OAc-}][\ce{H+}]}}{\color{red}{[\ce{HOAc}]}[\ce{OH-}][\ce{H+}]} = \frac{\color{red}{K_\mathrm{a}}}{K_\mathrm{w}}$$
For acetic acid:
$$K = \frac{\pu{1.8e-5}}{\pu{1e-14}} = \pu{1.8e9}$$

wait so is keq always equal to ka/kw and would the same apply for kb/kw?  From my understanding we setup the reaction kb and saw that it is similar to keq but is only lacking H+.  So by multiplying in our H+ we see that it is also has our ka equation so we then can say ka/kw?  Thank you for your help, truly I am so confused in this chapter and its hard stringing all these k's together.

No, Keq should be determined from the equilibrium for each given situation. Remembering some formulas won't help, or at least you should understand on what premise they've been derived. That's why I edited the title of your question to cover the cases where this formula works: weak monobasic acids neutralized by strong alkali.

Oh I think I understand now, cuz keq=oh*h3o and that is equal to ka and then when we solve for the equation to isolate the k constant then we see its ka/keq.  So I see how you thought about manipulating the variables and now it makes sense a lot more to me now.  Thank you, you are the GOAT!


text:
Pressure at which graphite and diamond are in equilibrium

The standard state Gibbs free energies of formation of graphite and diamond at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.
The conversion of graphite to diamond reduces its volume by $\pu{2e-6 m3 mol-1}$.
If graphite is converted to diamond isothermally at $T = \pu{298 K}$, the pressure at which graphite is in equilibrium with diamond, is
(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$

I applied
$$\Delta G_{(p,T)} =\Delta_\mathrm{f}G^\circ + \int_{p_1}^{p_2}V\,\mathrm dp,$$
and since the system is at equilibrium,
$$\Delta_\mathrm{f}G^\circ = -\int_{p_1}^{p_2}V\,\mathrm dp.$$
Now I am stuck. I have not been given any relation between pressure and volume. Is there any assumption I have to make to solve this integral?

Forget integrals, just use $p\Delta V$.

I just came to know the equation $$ dG = VdP - SdT$$ is only applicable when no reaction is taking place , so my expression is incorrect.

My comment still applies.

Allowing for the volune change to be $2.0×10^{-6}$ in the units given, we really have only two significant digits and thus the one bar ambient pressure is not significant. It should not be included in (a), (b) or (d).

For each phase $i$ (graphite or diamond) you can show that
$$\mathrm{d}\mu _i = V_i \mathrm{d}p - S_i \mathrm{d}T$$
or after integration
$$ \mu_i = \mu_i^\circ+\int_{p^\circ}^{p} V_{i} dp $$
at constant T (note that each phase consists of a pure substance and $V_{i}$ refers to the molar volume of phase $i$; below the subscript $m$ is used instead to refer to the molar volume of a pure phase).
We are asked to find the pressure $p=p_{eq}$ at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that
$$ \mu(diamond) = \mu(graphite)$$
This leads to
$$\Delta \mu^{\circ}=-\int_{p^\circ}^{p_{eq}}\Delta V_mdP$$
and, since $\Delta \mu^{\circ}= \Delta_f G_m^{\circ}$, ultimately to the expression you provided:
$$\Delta_f G_m^{\circ}=-\int_{p^\circ}^{p_{eq}}\Delta V_mdP$$
The approximation you are allowed to make at this point is that the solids are incompressible, such that their $V_m$ are constant with change of pressure. It follows that
$$\Delta_f G_m^{\circ}=-\Delta V_m\int_{p^\circ}^{p_{eq}}dp=-\Delta V_m(p_{eq}-p^\circ)$$
which can be solved for $p_{eq}$:
$$p_{eq}=-\frac{\Delta_f G_m^{\circ}}{\Delta V_m}+p^\circ$$

The answers to(including yours) : https://chemistry.stackexchange.com/questions/138563/entropy-paradox-and-confusion-with-gibbs-energy. suggest That the general equation for dG is: dG=vdP-sdT +sum(UdN). However your first equation seems to only consider dG=vdP. Why then, did you end up with a correct answer?

@satan29 In this problem T is held constant.

but what about the UdN term?

There is no $\mu dN$ term. Think of it as "phase equilibrium" between graphite and diamond. You keep both phases at constant composition and look for the point where their chemical potential is equal.

@BuckThorn can you please give the source of derivation of this u=u°+integral(VmdP)

@SarveshMaheshwari For each phase you can show that $\mathrm{d}\mu _i = V_i \mathrm{d}p + S_i \mathrm{d}T$.

@BuckThorn There is no $\mu dN$ term but doesn't  moles of diamond and graphite change as reaction proceeds?

@Lalit I made a correction to my previous comment. In any case, at equilibrium there is no reaction. However, if you shift the amount of each phase reversibly (which is presumably what you mean) then the chemical potentials remain constant at constant T and p.


text:
Why does electron flow from anode?

The above is a picture of Galvanic cell. Zinc anode is placed in Zinc sulfate solution so we have ions as Zn2+ and SO42-. On the cathode side we have the same situation we have Cu2+ and SO42-. My question is why the zinc rod is getting converted into zinc ions and copper ions are getting converted into atoms and get deposited on copper rod. Why is electron flowing from anode to cathode?

The semi-reaction of reduction for Zn is :
$Zn^{2+} +2e^- \rightarrow  Zn \quad E^\circ =-0.763$
The semi-reaction of reduction for Cu is :
$Cu^{2+} +2e^- \rightarrow  Cu \quad E^\circ= 0.337$
The standard redox potential for Cu is more positive thus it is reduced and Zn is oxidized.
Electrons flow from a negative to a positive potential. They flow towards the cathode because their potential V, at the anode is more negative, thus they have an higher potential energy: $U_e = V*e$
note that $e$ the charge of the electron is negative as well as V.

Thank you so much. Your answer is very nice.

I want to ask that if I’m right in saying this:- the ionisation potential (that is electric potential created when the atom gets converted to ion) of Zinc ion is 9.39 and copper ion is 7.72. Since there is a potential difference therefore electrons would flow from higher potential to lower potential.


text:
Why is electrospray ionization a soft technique in comparison to electron ionization?
I know that electrospray ionization (ESI) is a soft ionization technique and that electron ionization (EI) is not, because it causes a great amount of fragments in a fast time due to the fact that a single electron is accelerated up to 70 V. ESI MS can cause fragments, but slower and not as good. But why is this effect less strong than in EI?
Thank you for your help!

The title of the Nobel Lecture of the person who invented electrospray ionization was "Electrospray Wings for Molecular Elephants". By that he meant that this technique is good for very large molecules and it does not break them. The interesting part is nobody fully understands how the big molecules get a charge. There are two models and who can explain it better than John Fenn, the inventor. Have a look at his lecture: https://www.nobelprize.org/prizes/chemistry/2002/fenn/lecture/
Coming to the electron impact ionization, it is a physical contact (don't take it literally) of a very high energy electron beam with the molecule, the energy of this process is sufficient enough to knock out an electron and break other bonds too in the gas phase. This breakage is understood in terms of molecule's cross section. Note that in ESI, no free electrons are used to ionize molecules. The ionization process is a mix of solution phase chemistry (i.e., what ions were present, pH, etc) and the molecule's own properties. Not all can be ionized by ESI. Certain ions will suppress electrospray phenomenon like trifluoroacetic acid.


text:
How much of Ag has formed on the Cu tile?

$\pu{9.547 g}$ $\ce{Cu}$ tile is added to an $\ce{AgNO3}$ solution. After some time, $\ce{Cu}$ was taken out of the solution, washed, dried and weighed. The mass appeared to be $\pu{9.983 g}$. How much of $\ce{Ag}$ has formed on the $\ce{Cu}$ tile?

The right answer to be found is supposed to be: $\pu{1.08 g}$.
My imaginary equations (with unlikely scenarios):
\begin{align}
\ce{Cu + 2AgNO3 &-> Cu(NO3)2 + 2Ag}\\
\ce{Cu + AgNO3 &-> CuNO3 + 2Ag}
\end{align}
I cannot solve this problem. There are a few others, but this is bugging me the most. How does one solve it? I believe this is something I wasn't taught before or there is again a mistake in the references given.
I've tried calculating in ways I thought was right and anyhow, I cannot find the answer.
After discussions, I have come to the conclusion that I am taking $\pu{0.618 g}$ as the solution to this problem and will look forward into it tomorrow with the teacher as well.
This is the only closest answer I seem to arrive at by myself and others, as we get $\pu{0.618 g}$ when using the $\Delta m_\mathrm{p}=\pu{0.436 g}$ and $\Delta m_\mathrm{t}=\pu{22.853 g}$ differences. However, the problem regarding the other exercises still persists, which are of similar nature.
I am starting to question my teacher's logic and abilities.
There is a close answer to the referenced answer, which assumes the $\ce{Cu}$ having a charge of $+1$, thus $\ce{Cu + Ag+ -> Cu+ + Ag}$ and $\Delta m_\mathrm{p}=\pu{0.436 g}$, $\Delta m_\mathrm{t}=\pu{16.2 g}$.
However, I assume that to be a very unlikely scenario considering the current circumstances. This scenario lets you acquire \pu{1.0617 g} as the answer.

Could you please edit what you've tried so far into the question?

To me it seems like there is a typo somewhere. For some reason this question is all over Runet (Russian internet) and the answer is 0.62 g (which to me is the correct one).

Could you provide a link for that? I can't seen to find anything regarding this anywhere

@Alchimista This is my first post on stack, I got extremely surprised by the hold put on the topic, as well as it being treated as off-topic. I even put the homework tag beforehand, yet it was removed by the moderator

I have reopened the question, but I'd still think it would have been a much better question, if the actual calculation that you did were included. In its current form it only discusses results, not the way to get them, therefore it doesn't point to the actual question you have. You should also explain the abbreviations you use. This question was probably closed, because there was a certain lack of attempt in previous versions. the edits then have not reached the critical number to reopen. (Please note that the homework tag is deprecated, it says that in all caps in its description.)

In this experiment the tile lose mass due to oxidation and dissolution as Cu dication (Cu++) and acquire mass due to the reduction and deposition of Ag.
This is what is measured by weighting before and after.
This difference in mass correspond to 2n moles of Ag minus n moles of Cu so is just matter of solving a first degree equation.
Quickly, without being consistent in digits and truncation.,
solving it for the current data gave about 0.6 g of Ag.
All others outcomes seems to me to assume an unlike stoichiometry or refuses perpetuating around.
A way to get into an "effective stoichiometry" of 1:1 is to consider formation of Cu+ followed by disproportionation AND assuming that all metal copper finely precipitates without depositing back to the tile from which originally came from. While, partially occurring, this scenario would explain discrepancies found performing a real experiment, it seems unlikely that is intended as to give a fixed result as for it  requires the assumption in italic above. Doing that the amount of silver is now about 1.06 g.

Thank you for the answer. Can you possibly provide how you got the answer and why 1.08g is not a viable answer? I wish to discuss this with the teacher whether she got the wrong answer herself

You get it exactly as I wrote assuming  that tge  reaction is Cu and 2 Ag+ giving 2 Ag and Cu++. It seems the likely one. You can play with 1 to 1 stoichiometry but I guess you did and does not give the answer. If not retry following the same reasoning as in my answer. It would be funny if it turns to be the wanted 1.08 g :)) but what I and and others have proposed (Cu++) is the one that makes sense.

It seems given by an author that thought of that.  But it could be a mistake. You could add another exercise of that sort to see if it is the same problem. Read again my answer as I added an end. But my opinion is unchanged.

Another exercise: 73g of Zn put in NiSO4 solution, which has a mass of 240g. After some time, the Zn tile appeared to be 71.8g. Find ZnSO4 mass percent in the solution (Answer: 13.35%)

Thank you. If you can, could you provide the steps if you manage to get the answer? (possibly in chat or somewhere) As well as for the previous exercise where you get 1.06g, as that seems to be the most accurate answer I've yet to see and never gotten, as I am completely frustrated with just these exercises

To solve this you use exactly the same rationale. One would need the exact wording but like this isn't clear what is what.  At least here you are sure about stoichiometry. You solve for n and then you know how many miles deposited how many & went in solution. You know also what happened to the mass of the solution so you know what the w/w % conc is.

Well, let's work backwards... Given:
9.547g Cu tile
9.983g tile + Ag
1.08 g Ag
then the Cu that is left on the tile is
9.983 - 1.080 = 8.903 g
and the Cu that dissolved must be
9.547 - 8.903 = 0.644 g
Using the atomic masses:
AM(Cu) = 63.546
AM(Ag) = 107.8682
we can calculate the moles of Ag
1.08/107.8682 = 0.010012
and the moles of Cu
0.644/63.546 = 0.010134
So the poser of the question was assuming the chemical reaction:
$\ce{Cu + Ag^+ -> Cu^+ + Ag}$

Copper(I) in aqueous solution? No way.

Didn't say that I agreed, just what poser of the question must have done...

Yep, looks plausible (although I arrive at 1.06 g for this imaginary $\ce{Cu + Ag+ → Cu+ + Ag}$ reaction), but then this textbook has negative value.

I wonder if the poser was assuming in solution that $$\ce{Cu^{2+} + Ag^+ -> Ag v + Cu^+}$$ so that only half the Ag deposited on the Cu plate.

Charge balance is fubar for that one.

All in all I don't know how the poser messed the problem up. Old problems get reworked to form new problems. Perhaps in the original problem there was some sort of metal complexes that gave the overall 1:1 reaction.

Cu+ 2Ag+ --> 2Ag Cu2+

This seems a bit closer to what I am looking for, except I need to find the 1.08g by myself, and not use it in the equations. The answer is given for reference whether you did it right or wrong.
This problem was provided by my teacher, and I translated it into English, so sorry for those who have a hard time understanding it

In case for others who might be trying to find the answer, I assumed the reaction to be Cu↓+2AgNO3⟶2Ag↓+Cu(NO3)2

@I'mPatrick - Lacking any other information I'd expect an acidic solution and that the copper would be oxidized to $\ce{Cu^{2+}}$. However to get the given answer you have to reason that the copper goes to $\ce{Cu^+}$ and assume some small rounding errors.