text:
Liquid Versus Aqueous States?
I am studying ionic equations and have encountered the statement of the phase of matter in the equation since ionic substances tend to react differently based on their phase.
I know that
(l) state is when the substance itself is a liquid
(aq) state is when we put the substance in an aqueous solutions, most commonly water.
But I don't know
1) Why and when would you use one state over the other. (Answered)
2) How do you define the concentration when you say (aq) state? Is it predefined/can be computed based on the substance or reaction or is it irrelevant?

An aqueous solution is water based. You might have other liquids like acetic acid or ethanol in the solution too, so water doesn't have to be the only liquid. Does this answer you question?

Almost. See question 2.

One could theorize that the liquid phase must be at least 50 mole % water to be an aqueous phase, but that certainly oversimplifies the situation. I'd say that if the reactions are what you'd expect in a water solution then it is an aqueous phase.

Perhaps I can be a little more helpful, but only by introducing another factor. If we're using a aqueous Ksp for example we'll typically use concentrations, but it is really activities of the ions that should be used for the equation. So for an aqueous solution I'd typically assume the concentration simplification. So for a 40 mole % solution of ethanol, I wouldn't expect good agreement with the Ksp equation even though the solution is more than 50 mole % water. I'd expect that you'd have to use activities which would greatly complicate the mathematics.

Thanks! Could you please make your comments into an answer so that I can accept it?

2) How do you define the concentration when you say (aq) state? Is it predefined/can be computed based on the substance or reaction or is it irrelevant?

When a reaction equation is written down, one usually has a specific process (initial and final states including concentrations) in mind. When this information is not provided, and when it is not say the objective of a (e.g. textbook) question to identify the concentrations, it is usually safe (sometimes even necessary) to assume that the change is between ideal reference states.
Tabulated thermodynamic data such as heats of formation usually refer to ideal standard states, which in the case of a heat of formation of an aqueous solution corresponds to a pressure of 1 bar, and an ideal solution (with the solute having the same properties as in an infinitely dilute solution) at either 1 molal or 1 molar concentration of the solute.
So for instance the process (taken from another answer - thanks @Farooq)
$$\ce{HCl (g) -> H+ (aq) + Cl- (aq)}$$
refers to transfer of a gas molecule into a 1 M or 1 molal solution, if the process refers to standard states (as tabulated data usually does).
During the transfer, the properties of the bulk solvent do not change, and so the solvent does not appear explicitly in the equation.
It can be hard to understand how a 1 molar or 1 molal solution can simultaneously have the properties of an infinitely dilute solution. That's because we extrapolate from the data for an infinitely dilute state to a finite (1 M or 1 molal) concentration, using Henry's or Raoult's law.

Is it predefined/can be computed based on the substance or is it
  irrelevant?

The state symbols, l, s, g, (aq), as they are called are nothing but qualitative descriptors. They have nothing to do with concentrations. It is a good point when do we distinguish (l) and (aq)? You have already answered it. When a given component in the equation is a pure liquid and it is participating in a reaction, you would prefer to write "l". If a reactant or a product is dissolved in water, it is better to write (aq). Consider hydrogen chloride gas being dissolved in pure water to yield hydrochloric acid solution.
e.g. HCl (g) + H2O (l) -> H3O+ (aq) + Cl- (aq)

So if you gave me an equation with one or more aqueous reactants, are you saying I would be able to perform the reaction with any concentration of aq since the reaction does not depend on it?

How does a gas dissolve in water? That seems quite interesting and weird.

Yes, you are right. (aq) is just a medium. Reaction certainly depends on the nature of a medium. If gases such as oxygen did not dissolve in water, all fishes would be dead in the oceans and seas. Don't forget Coke which has CO2 dissolved in sugary water. HCl gas is extremely soluble in water.


text:
How do I get irreducible representations?
I'm currently trying to do a problem where it asks me to "show that the irreducible representations spanned by the atomic x-, y-, and z- coordinates is independent of the choice of local coordinate system. Use the diagrams below."

I know the general process is to determine the irreps for the separate coordinates. So, when I attempted to do the first one, by doing the symmetry operations to the x vectors from the D3h character table. However, it did not work as well. So now I don't know where to go with it. Please help and thank you.

Have a look at the method given in this answer. In your example subject each vector on each atom to the operation in the point group, e.g rotation, reflection etc. Count 0 if the vector is moved else count 1 if it remains unchanged and -1 if it is unchanged but just points in the other direction. Now you have the reducible representation. Use the method suggested above to get the irreps.
If the angles are not 180 then it is necessary to use a rotation matrix and determine the trace (sum of diagonal elements). (This is explained in a good symmetry text book such as R. Carter, 'Molecular Symmetry and Group Theory'. Of course this has all been worked out and so it is necessary only to use a formula for $C_n$ and $S_n$ operations.
These are  $\displaystyle \chi(C_n)=1+2\cos\left(\frac{2\pi}{n}\right) $ for $n$ fold rotations and $\displaystyle \chi(S_n)=-1+2\cos\left(\frac{2\pi}{n}\right)  $ for rotation reflections, e.g. $\chi(S_3)=-2\;; \chi(C_5)=(1+\sqrt{5})/2$,
Excepting rotation, for  each atom each operation contributes an amount as shown below.
$$\begin{array}{ccc}
\text{operation} & E & \sigma & i  \\
\chi(n) & 3 & 1 & -3
\end{array}$$

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Spectroscopic constants - explanation
I'm a beginner in spectroscopy and I have troubles understanding papers describing values of spectroscopic constants. This paper, for example, describes constants $T_e, R_e, \omega_e, \omega_ex_e, B_e, \alpha_e, D_0$ and $D_e$.
I can see, that $R_e$ is an internuclear distance, but I have no idea, what are the remaining constants.
Could you provide some brief explanation? Preferably with picture of some potential energy curve(s), if possible, as I find it much easier to understand such concepts from schemes.

You can find plots and further explanations online (as I easily did). This is a homework-type question, yet I do feel that the information request for diatomic molecular spectroscopic constants warrants at least this partial answer. See this writeup for the "Vibronic Absorption Spectrum of Molecular Iodine" for more details and illustrations.
From NIST:
T$_e$ minimum electronic energy (cm$^{-1}$)
R$_e$ internuclear distance (Å)
ω$_e$ vibrational constant – first term (cm$^{-1}$)
ω$_e$x$_e$  vibrational constant – second term (cm$^{-1}$)
B$_e$ rotational constant in equilibrium position (cm$^{-1}$)
α$_e$ rotational constant – first term (cm$^{-1}$)
D$_e$ centrifugal distortion constant (cm$^{-1}$)
β$_e$ rotational constant – first term, centrifugal force (cm$^{-1}$)
From Wikipedia:
D$_0$ dissociation energy (cm$^{-1}$)

The subscript $e$ means measurements relating to the minimum of the internuclear separation, i.e. at the bottom of the potential energy. Many textbooks do not use the subscript $e$ at all but the meaning is the same.
The $D_0$ by contrast is the dissociation energy measured from the lowest vibrational energy level ($n=0$), some books also quote $D_e$ which is the energy from the lowest point on the potential energy and so this is $D_0$ plus the zero point energy. However, $D_e$ may also be the centrifugal distortion constant and has the same units as $B_e$. The context in which it is used will determine what it means, i.e. if the use is in determining the rotational energy $D_e$ (or just $D$) will be the centrifugal distortion constant.
The term $x_e$ is the anharmonicity constant and is dimensionless and $\alpha_e$ is the vibration-rotation coupling constant, and has the same units as $B_e$. (It describes the fact that as the molecule has a different bond length in vibrational level $n=1$ than in $n=0$, and that in $n=2$ from $n=1$ etc, consequently the rotational constant $B$ has to be modified). The other terms are described in the first answer.
However, there is a wrinkle, the fact that $\omega$ is used and not $v$ for frequency means that angular frequencies are being used. They are related as $\omega =2\pi v$. The frequency $\omega $ has units $\mathrm{radian\,s  ^{-1}}$ and $v$ units of $\mathrm{\,s^{-1}}$. Some books use $v$ some $\omega$; it is really quite messy you just have to be careful.

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By how many pH units does rain vs distilled water change a soil pH test?
Sources state that the concentration of $\ce{H2CO3}$ in rain is of a few µmol; I don't know if that's per ml. The buffering effect of most soils is >1000 µmol $\ce{H+}$/kg.
Is the difference from low acid rain water versus lab distilled water on a pH test for soil less than 0.1 in pH values?

Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about
$$\begin{align}
\mathrm{pH}=-\log[\ce{H+}]&=5.65\\
-\log[\ce{HCO3-}]&=5.65\\
-\log[\ce{CO3^2-}]&=10.3\\
-\log[\ce{H2CO3^*}]&=5.0\\
-\log[\ce{CO2}]&=5.0\\
-\log[\ce{H2CO3}]&=7.8\\
\end{align}$$
(The calculation can be found in: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 159–161.)
Note that $[\ce{H2CO3^*}]$ is the total analytical activity of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$. $[\ce{H2CO3}]$ is the concentration of true $\ce{H2CO3}$.
Thus, $[\ce{H2CO3^*}]=10^{-5}$ and $c(\ce{H2CO3^*})=10^{-5}\ \mathrm{mol/l}$; i.e. a few µmol/l.


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Why do we assume 100% ionization for acetic acid (a weak acid)?
This is about a titration. The base titrant used is sodium hydroxide.

You do not assume 100% ionization of acetic acid during titration. Acetic acid is a weak acid. Look at this equilibrium
$\ce{CH3COOH <=> H+   + CH3COO-}$
The base, $\ce{NaOH}$, keeps consuming $\ce{H+}$ during a titration, as a result acetic acids keeps on providing $\ce{H+}$ to maintain the equilibrium in water until all the acetic acid is converted into acetate ion.


text:
Nitration of N-phenylbenzamide
Why during nitration of N-phenylbenzamide the $\ce{NO2}$ group is directed at para-position of the ring attached to the nitrogen atom?
Okay, I understand that this is due to the para directing effect of the $\ce{>NH}$ group. But carbonyl $\ce{>C=O}$ group has a meta directing effect, then why is $\ce{NO2}$ not directed to the meta-position of the benzene ring attached to carbonyl group?

In the following reaction

(1) $\ce{NO2}$ group at meta position w.r.t. Ring 2
(2) $\ce{NO2}$ group at para position w.r.t. Ring 1
(3) $\ce{NO2}$ group at para position w.r.t. Ring 2
(4) $\ce{NO2}$ group at meta position w.r.t. Ring 1

The carbonyl attached to ring 2 is meta directing but deactivating.

You should not be looking at $\ce{-NH-CO\bond{-}}$ as two different functional groups i.e. as $\ce{-NH-}$ and $\ce{-CO-}$.
Since as in $\ce{-(NH-CO)\bond{-}}$: $\ce{N}$'s lone pair are actually in resonance with the carbonyl group.
The left handed phenyl ring is more activated than the right handed phenyl ring. Hence the electrophile would preferentially attack on the left handed phenyl ring.
The amide group is too weak base to be protonated by the acid and therefore it does not direct the electrophile to meta position.
Ortho position is hindered due to steric hindrance of the another phenyl ring.
And therefore

 the attack takes place at para position of the left handed phenyl ring

so answer should be option B.

@andselisk : Could you suggest me some perfect and to the point editing tips

Feel free to visit this page, this page and this one on MathJax and Markdown formatting. Also there is a glorified MathJax reference from Math.SE.


text:
Change in H+ ion concentration of water on boiling
I know that on boiling water, pH scale decreases but water remains neutral. At 100°C the pH=6.14 is neutral.  And at this temperature also, the H+ ion concentration and OH- ion concentration are equal.
My question is what happens to the  H+ ion concentration on boiling  relative to the concentration at 25°C? Does it increase or decrease or remain same?

It seems like you answered your own question: as you said, at 100 °C the $\mathrm{pH} = 6.14$, and $[\ce{H+}] = 10^{-\mathrm{pH}}$.

@andselisk You mean [H+] increases, but then [OH-] too increases equally. So in this case how is it that decrease in pH implies increase in [H+] as now the pH scale itself has changed?

I'm afraid I don't understand what you are trying to imply. The pH scale itself is a superficial term (e.g. typical claims that pH can only take 0 - 14 values and pH 7 is always neutral) to the point it's almost a meme.

So according to your previous comment, [H+] increases as pH=6.14 . But I don't want to get that one word answer, I would rather like to have an explanation for my question i.e., how does [H+] changes with change in 'pH scale'(not pH I repeat)?

It is an interesting question, however there is a misconception. You are giving too much attention to the word neutral. The word neutral is just a qualitative version. Numbers and equations are more accurate way to discuss science.
You have to see how water's $\pu{pH}$ at 100 °C is calculated. How do you get those numbers? Lets us do it algebraically, so that the word neutral does not matter.
Autoionization of water at temperature T:$\ce{ H2O <=> H+  + OH-}$, if the conc. of $\ce{H+}$ is x, the quantity of $\ce{OH-}$ is also x, if you have accepted the ionic product of water is
$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = x^2=
51.3\times{10^{-14}}$
Solve for $x = [\ce{H+}] => -log [\ce{H+}] = 6.14$
So the numbers prove that $\pu{pH}$ has changed. Water has become slightly acidic. If you associate the word neutral = 7.0 at 25 °C. However, if you use the word neutral with respect to water temperature, then $\pu{pH}$ of 6.14 is neutral at 100 oC. Ultimately, the numbers Kw and pH matter, nobody will quibble about the word neutral in serious research.
Water is the most amazing solvent in Nature. People are still exploring its properties and it is yet to be fully understood. For example, I just heard in a conference that someone was trying to understand what happens on the surface of micron sized water droplets. "Pure" water showed very odd chemical effects which have never been shown before.
Read this well written Wikipedia article: https://en.wikipedia.org/wiki/PH#cite_note-2

https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know


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Primary kinetic isotope effect in the E1cB reaction
How is the primary kinetic isotope effect (KIE) observed in the E1cB reaction when breaking of the $\ce{C}$-$\ce{H}$ bond is not the rate determining step of this reaction?

There is a section on Wikipedia describing KIEs in the E1cb reaction in quite some detail: https://en.wikipedia.org/wiki/E1cB-elimination_reaction#Kinetic_isotope_effect Have you already seen that? Is there something specific you don't understand about it?

Is it like....1) A Deuterium 'Exchange' indicates that an E1CB REVERSIBLE reaction has taken place, & ...2) A beta deuterium substituted reactant showing primary kinetic isotopic effect indicates that an E1CB IRREVERSIBLE reaction has taken place? I'm confused. Please help.


text:
How to decolorize blue fabric whitener?
After adding 4 drops of fabric whitener in water, what should I add to decolorize the solution and turn it clear like water?

Try a very mild solution of bleach such a that sold under the trade name of 'Milton'

Can Dettol (Chloroxylenol) work?

This is probably better as an antiseptic than an oxidising agent.


text:
Which of these chlorides is more reactive?
We do learn that for $\mathrm{S_N1}$ reaction the rates are tertiary > secondary > primary > methyl. The reasons mentioned are usually inductive and hyper-conjugative effect.
But which of the effects is more dominant? So, let's take two chlorides, ethyl chloride and isobutyl chloride, which will be more reactive to $\mathrm{S_N1}$? If we refer to inductive effect, then isobutyl chloride should be more reactive. But if we refer to hyper-conjugation, then ethyl chloride should be more reactive.
Which is more reactive actually?
Note: Which carbocation is more stable, the ethyl- or 1-propyl-carbocation? is not the same as this as I am not satisfied with the answer there. That question compares the stability of carbocation while I am comparing the reaction rates which may depend on kinetic factors too. Actual data could be helpful.
Also, the isobutyl cation rearranges to a tertiary cation. Does the stability of rearranged carbocation affect the rate of reactivity?

I’m have been told that it follows this order , aromaticity>Electromeric effect>resonance >hyperconjugation>Inductive effect

I fully expect to receive some complaints about closing a question as a duplicate of another question without answers, but I see no point in keeping two copies of exactly the same question. If anybody still wants to complain, please bring it to meta.

@orthocresol my question was asked 1 hour before that


text:
Sulphonation of benzene
Will nitrobenzene and benzene have the same rate towards sulphonation reaction, since the formation of sigma complex is not the RDS of this reaction?

Nitrobenzene and benzene would NOT have the same rate towards any type of electrophilic substitution reaction as strength of nucleophile - (nitro)benzene does matter.
Nitro group as in nitrobenzene is highly deactivating group due its electron withdrawing nature and therefore decreases the nucleophilicity of benzene ring.

IJOART_document

But how shall it matter in "SULPHONATION" if the attack of electrophile, and thus the formation of sigma complex, is NOT the rds of sulphonation?

@Anonymous : Read the document... It is very well written and presented

@Mithoron : Should I not capitalise compound names in order to make them stand out and increase the readers' attention towards them?

I don't think it should be used for emphasis, also using emphasising in multiple ways at once tends to be too much.

@Mithoron : Your reputations speak up for your experience in answering on this interface. I'll try to follow your advice since it seems (highly) logical.


text:
Is isobutyl chloride or ethyl chloride more reactive in an SN1 reaction?

Compare the rate of $\mathrm{S_N1}$ reaction in the following compounds:

We see that both compounds form primary carbocations. However, there is more hyperconjugation in the second compound as compared to the first one.
Also as the step in which the leaving group leaves is rate determining step (slowest step), which is before the rearrangement of the carbocation, I concluded that the second compound must have a greater rate of $\mathrm{S_N1}$ as compared to the first one.
But the answer says that the first has a greater reactivity than the second. Is the answer correct? If yes, then why?

I don't know the answer without either looking into the literature or (maybe) running some calculations (unlikely unless I'm v bored), but what I want to point out is that the ionisation (i.e. loss of chloride) and the rearrangement (alkyl / hydride migration) are not necessarily separate: they can occur in a concerted fashion.

The inductive effect of the methyl groups should help stabilize the carbocation formed from 1.
The answer may be a bit clearer if you add equilibrium to your thinking.

The initial dissociation of the 1 to form carbocation A is the rate determining step.  However the rearrangement of A to B should be relatively quick, and will drive the initial dissociation forward by Le Châtelier's Principle.
In contrast, 2 has less stabilization by the inductive effect, and there is no possibility of a rearrangement.


text:
Ring contraction in benzilic rearrangement
Wikipedia says that,

Benzilic acid rearrangement occurs with ring contraction when used on cyclic diketones.

But gave no explanation, can you please elaborate on this with some mechanism. One such reaction I found here, why ring contraction taking place?

Ultimately because it is energetically favourable and the resulting carboxylic acid can deprotonate giving a stable species that disfavours the reverse reaction.

@Waylander: Could you explain, why is the formation of a five-membered ring energetically favourable when it's more strained than a six-membered ring? Thanks!


text:
Which of the following carbanion is most stable?

The carbanion which is most stable is

The answer is 1. But why won't the answer be 2 because it shows resonance, whereas 1 only shows I effect and resonance effect is more dominant?

did you miss out  (3).. carbanion is most stabilised in option 3 due to presence of  phenyl ring

now that you've pointed out I'm confused between (2) and (3) but the answer is (1)

I am not able to make out why answer is 1... but will surely let you know once I am able to find out some excerpt on it... that is the reason I commented it and not answered ;)

Think of the pKa, as the answer by glucose has suggested. One way is to think about which bases it reacts with. Acetylene is deprotonated by even bases like ammoniacal silver nitrate (a.k.a. Tollen's reagent) Which is not a very strong base. But toluene is very hardy and can't be deprotonated even by very strong bases like sodamide.

Some points to check while comparing stability of a carbanion:

Inductive effect: preferentially $\ce{-I}$ effect
Resonance: preferential $\ce{-M}$ effect
Hybridization: preferentially attached to carbon atom having more s-character

... many more: stability
Now the problem is option 1 and 3 are in itself very 'nice' (w.r.t. stability) possible cases of Hybridization-stabilization and Resonance-stabilization respectively.
So the natural (what I thought) way out here is comparing $\ce{pKa}$ 's of acetylene and toluene respectively for option 1 and 3.
from here :

$\ce{pKa}$ of acetylene is about 25.
$\ce{pKa}$ of toluene is about 43.

Less $\ce{pKa}$ denotes how (much more) feasible is for the acid to dissociate by deprotonating itself.
So, acetylene carbanion is more stable than toluene carbanion

You can read up this link and widen your concept. https://chemistry.stackexchange.com/a/10703/73527
I liked the fact what @glucose stated ,so my answer is just  a continuation.
You can opt out option 2& 4

4 option

You should know that it has a +I effect

3 option

because the no.of congujating structures is just two ..
Not enough for stability.
 right!
 Now coming to the real answer
Why pka of toluene is more than acetylene?
pka facts shows that acetylene can deprononate quite easily but why?
Why hybridisation precedes resonance?
With toluene , the methyl group is electron-donating onto the benzene ring, so toluene extends the p orbital system all throughout the system, but sometimes the electrons are outside of the ring. Since the conjugation extends outside of the ring, not all of its resonance structures contribute to a stabilization of the ring itself.
So we can see why hybridisation precedes. Greater the electronegativity
more it can carry negative charge stability

 whereas 1 only shows I effect

No, it doesn't show only inductive effect
but hybridisation hence electronegativity also play a role.
Greater the s character greater the electronegativity and hence greater the charge stability.


text:
Hartree Fock for an atom (Beryllium)?
I'm working on solution of Hartree-Fock equation for Beryllium in configuration $1s^2 2s^2$, expanding the orbital with STO-3G basis
$$\phi_k=\sum_{p=0}^3 C_{kp}e^{-\alpha_{kp}r^2}$$
I know the values of $\alpha_k$ both for 1s and 2s.
All books speak about solving molecules and the only case with atom is the helium case, so I have some doubts. In molecular case they expand the molecular orbital as
$$\phi_k=\sum_{p} C_{kp}\chi_p$$
without index $k$ on $\chi_p$ and so also all matrices I need to solve the eigenvalue problem. For this reason I can't understand the difference between the basis set used in molecular case and atom case and consequently how to solve the Fock equation in the case of an atom, and in particular of Beryllium.

Your definitions of $\phi_k$ and $\chi_p$ are unclear, but maybe that's part of your confusion? If your first equation is correct, then $\phi_k$ are the AOs. But the second equations looks like $\phi_k$ are MOs and $\chi_p$ are AOs. That doesn't add up. Please double check what those are supposed to be.

Exactly! That's the part of my confusion! In the case of a molecule I expand my MOs as combinations of AOs and for closed shells I find Roothaan equation (that's what books treat and what I understood). But in the case of a simple atom, like beryllium, what happens?

Formally, the very same thing. But the result may be that the MO coefficient matrix becomes the identity matrix, so MOs equal AOs. But this depends on your AO basis set of course.

The problem is that I have no idea of the form of the matrices in the case of beryllium. In fact my book expand each MOs with a given basis and then create the Fock and overlap matrix. In my case I have an expansion of each AO with ITS OWN STO-3G basis and I don't know how to use the expression for the Fock and overlap matrix given.

The matrices in HF (Overlap, Fock etc.) are in the basis of AOs, not the actual basis functions. Whether you choose STO-3G, STO-6G, or maybe even actual Slater type orbitals does not matter for those matrices. This will affect how the AOs look like, and therefore indirectly the numbers in those matrices. But the matrices itself only depend AOs and do not care about the basis function the AOs are expaned in. If you only have 1s and 2s orbitals, then these matrices will be of dimensions $2\times 2$ (using a minimal basis, no p orbitals and spatial orbitals).

Ok, maybe now it's clear. So in the case of beryllium I have the following expansion in HF $\phi_k=\sum_p C_{kp}\chi_p$, where the index $p$ stands for $1s$ or $2s$. This expansion does not care about the fact that both 1s and 2s AOs (that are represented by $\chi_p$) are expanded into STO-3G basis. So my matrices in Fock equations are in the beginning $2 x 2$, since I have two AOs, but the single elements in those $2 x 2$ matrices are in reality $3 x 3$ blocks due to the expansion of single orbitals in STO-3G basis. Am I right?

Yes, except for the $3\times 3$ blocks part. When you perform the integrals of the AOs, then you get 3  primitive Gaussians in both, bra and ket vector. Expanding this product of sums yields 3*3=9 primitive products, which all are summed into the final $2\times 2$ matrix element. You may however first calculate all the primitive Integrals and keep them in a $6 \times 6$ matrix, which you then transform to the $2\times 2$ one. See https://chemistry.stackexchange.com/questions/76323/how-to-compute-2-electron-integral-for-hartree-fock-code/76460


text:
Question on deflection of cathode rays in a magnetic field
My book states that :

When only electric field is applied, the electrons deviate from their
  path and hit the cathode ray tube at point A. Similarly when only
  magnetic field is applied , the electron strikes the cathode ray tube
  at point C.

I tried to apply the Fleming's Left hand rule used to find the direction of force on a current carrying wire where the direction of current is taken opposite to the direction of flow of electrons.
The rule states that :

Stretch the thumb, forefinger and middle finger of your left hand such
  that they are mutually perpendicular. If the first finger points in
  the direction of magnetic field and the second finger in the direction
  of current, then the thumb will point in the direction of motion or
  the force acting on the conductor.

Applying this rule I found that the electrons should hit the cathode ray tube at A in a magnetic field . But in the book it is given that they strike it at C.
Please explain me where I went wrong . Thank you.

I'm voting to close this question as off-topic because it's not about chemistry.

Ok so should I ask it in Physics SE.

@Chemist I don't see any physical chemistry in here. If the question had been how to make an electron beam, OK. Or how the flourescent screen works.

Cross posted on Physics SE

1)  for individual charges Lorentz force F =q[v*B], so curl your right hand from velocity vector of positive charge towards magnetic field so that your thumb points at the force acting on the charge. Here we have considered positive charge
2)It is better use this mnemonic to find out the direction of force to avoid confusion
Stretch the thumb, forefinger and middle finger of your right hand such that they are mutually perpendicular

fore finger denote direction of electron moving
middle finger denotes the direction of magnetic field (north to south on outside the magnet)
finally force direction can be found out using your the direction where your thumb is pointing

The book is correct
Tip- If current is mentioned then Use your left hand, to avoid twisting your hand in weird ways! .

Ok with the right hand rule you stated I am getting it right. But when using the left hand rule I pointed the direction of current opposite to the direction of cathode  ray and I am still getting the opposite answer that they should hit at A.

Chemistry and physics cannot be separated by a boundary. Everything is grey beyond textbooks. We should stop teaching atomic structure because that is also physics.

I want to use my LEFT hand for finding the deflection of  cathode ray and even after using the direction of current as opposite to that of cathode ray I am not getting the right answer.

@AshokSharma You are using N-->S for the magnetic field lines, right? ... and you are right, and your book is wrong. If they wanted the beam to hit in C, they should have but the magnet the other way round. ;-)

@Chemist  In the picture in your answer you have shown the green line pointing north to south as middle finger . But in the left hand the middle finger shows the direction of current and not of magnetic field.Thanks for your efforts

@Chemist  you  said that in left hand the forefinger points in the direction of current.  No it points in the direction of MAGNETIC FIELD and MIDDLE FINGER POINTS IN THE DIRECTION OF CURRENT.

But this is the well known FLEMING'S LEFT HAND RULE. You can google it and see it for yourself.

I think Fleming’s rule is for current carrying conductor.


text:
ApexTrack: peak tail looks like a shoulder
I'm trying to implement ApexTrack algorithm in my software by following the description written by Waters. But many chromatograms that I see have noticeable tailing of peaks: .
No matter what Touchdown% threshold I set (even 0) - it won't be able to go past the bump in the tail because the slope becomes positive for a bit.
Are there modifications to ApexTrack that could join such tail to the peak? And at the same time keep recognizing true shoulders (which you see to the left of the largest peak).

Stanislav, the terminology you are using is not correct. The arrow shows peak tailing. Front side is the early eluting side. This would help in communication with analytical chemists. I think you can significantly get rid of this tailing by simply modifying your chromatography methods. This will always cause trouble in experimentation regardless of any software. Could you mention if this is a single component chromatogram or two components which are co-eluting? Secondly have you used PeakFit?

Oops. Corrected the wording, thanks! The tail and the peak seem to be the same compound. No, I haven't tried PeakFit, should I? :)

Yes, please. It will make your life much easier! PeakFit can easily fit exponentially modified Gaussians after smoothing. You did not mention, if you have a single compound or a two compounds? If it were a single compound, and you see a shoulder, this is an indication of a clogged or damaged column.


text:
What happens when mixing elemental mercury with sodium hypochlorite solution
Say you spilled elemental mercury from a thermometer and used household bleach on it, what would be the reaction if any?

More importantly, this is a very poor way to decontaminate anything where mercury has spilled.

There are several redox reactions can be taken place when you add household beach on elemental mercury metal. However, this question is more like a homework question. Thus, I give you insight to the answer and you may read a bit and find out what's happening. This reaction is studies and results have been published (Reference 1), the abstract of which states that:

The absorption of elemental $\ce{Hg}$ vapor into aqueous hypochlorite was measured in a stirred tank reactor at $25$ and $\pu{55 ^{\circ}C}$. $\ce{NaOCl}$ strongly absorbs $\ce{Hg}$ even at high pH. Low $\mathrm{pH}$, high $\ce{Cl-}$ and high-temperature favor mercury absorption.  Aqueous free $\ce{Cl2}$ was the active species that reacted with mercury. However, chlorine desorption was evident at high $\ce{Cl-}$ and $\mathrm{pH}<9$. Overall second-order reaction was observed between mercury and chlorine with an apparent second-order rate constant of $\pu{1.7\times 10^{15} M^{-1} s^{-1}}$ at $\pu{25 ^{\circ}C}$ and $\pu{1.4\times 10^{17} M^{-1} s^{-1}}$ at $\pu{55 ^{\circ}C}$. Gas-phase reaction was observed between $\ce{Hg}$  and $\ce{Cl2}$ on apparatus surfaces. Strong mercury absorption in water was also detected with $\ce{Cl2}$ present. Results indicate that the chlorine concentration, moisture, and surface area contribute positively to mercury removal.

I hope you will figured it out when you consider the standard reduction potentials of $\ce{Hg}$ and $\ce{OCl-}$ in any electrochemical series. For example, see Reference 2.
References:

L. L. Zhao, G. T. Rochelle, “Mercury absorption in aqueous hypochlorite,” Chemical Engineering Science 1999, 54(5), 655-662 (https://doi.org/10.1016/S0009-2509(98)00263-2).
S. G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).

As noted by @Mathew, sodium hypochlorite can be used to remove elemental mercury. It is a redox reaction involving oxidation of mercury to $\ce{Hg^2+}$. There are various papers which discussed about the reaction of sodium hypochlorite and mercury and its role in removing of mercury. This paper1 discusses about the dissolution of mercury in aqueous solution of sodium hypochlorite at pH of 5.9–8.5 and the mechanism of oxidation of mercury to $\ce{Hg^2+}$. There is also another paper[2] which discuss about the role of sodium hypochlorite in the removal of mercury by wet-scrubbing method.

The important step for increasing gaseous elemental mercury
  removal in wet scrubber systems is altering the chemical form of the
  Hg to a water‐soluble oxidized species. This work focuses on the
  removal of elemental mercury from simulated flue gas by aqueous sodium
  chlorite in a bubble reactor. The effects of initial oxidizing
  solution concentration, reaction temperature, pH and mercury
  concentration in the inlet of the reactor on mercury oxidative
  absorption in sodium chlorite were investigated. The results indicate
  that higher concentrations of sodium chlorite favor Hg removal, with
  a greater efficiency observed in acidic than in alkaline solution.
  High temperature inhibits Hg absorption in aqueous sorbent when the
  reaction temperature is lower than ca. 40 °C, and the removal
  efficiency increases when the temperature is higher than that value.
  In conclusion, the major influencing factors on the levels of Hg
  removal are pH and chlorite concentration in solution.

References

I. P. SizenevaV. A. Val’tsiferV. N. Strel’nikov "A Study of Mercury Dissolution in Aqueous Solutions of Sodium Hypochlorite" Russian Journal of Applied Chemistry April 2005, Volume 78, Issue 4, pp 546–548 (link)
Y. Zhao  S.T. Liu  C.M. Chen  X.Y. Ma "Removal of Elemental Mercury by Sodium Chlorite Solution" 28 February 2008 (https://doi.org/10.1002/ceat.200700381)
https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/98JD02304


text:
Statistical model of ligand substitution
Recently, I was told that in case of a particular step of a generic ligand substitution reaction:
$$\ce{M(OH2)_{$N - n$}L_{n} + L <=> M(OH2)_{$N - n - 1$}L_{$n + 1$} + H2O}$$
The probability of the forward reaction and by extension, the equilibrium constant of this step, $K_n$ would be proportional to
$$\frac{N - n}{n + 1}$$
by a purely statistical analysis. Now I have thought about this for quite a bit, but I can't understand the mathematical reasoning behind arriving at this expression. I suspect it has something to do with the numbers of the ligand being replaced and the ligand which is replacing the other one. Can anyone explain the process of arriving at this expression using simple (if possible) reasoning?

I may add a more complete answer later, but for now, read this page, in particular example 2 and the remaining paragraphs in that section. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Complex-Ion_Equilibria

I think the answer may just come down to a simple counting of available sites. The equilibrium constant for a reaction is equal to the ratio of the forward and reverse reaction rates. For the forward reaction, there are $N-n$ sites available at which a ligand can replace an $\ce{H2O}$. Conversely, for the reverse reaction, there are $n+1$ ligand sites at which a water molecules can replace it. If we assume that in each case the reaction rate with $m$ sites available is equal to $m$ times the reaction rate with $1$ site available, we obtain an equilibrium constant that is proportional to the ratio of sites available for the forward and reverse reactions.
$$K=\frac{k_{f,N-n}}{k_{r,n+1}}\approx\frac{(N-n)k_{f,1}}{(n+1)k_{r,1}}\propto\frac{(N-n)}{(n+1)}$$

I think the easy way out is to invoke $S_\mathrm m = R \ln \Omega$. If we assume that for a generic complex $\ce{MA_{n}B_{$N-n$}}$,
$$\Omega = {N \choose n} = \frac{N!}{n!(N-n)!} \quad \left[ = {N \choose N-n} \right]$$
and that for the individual molecules $\ce{A}$ and $\ce{B}$, $\Omega = 1$, then the equilibrium constant $K$ for
$$\ce{MA_{n}B_{$N-n$} + A <=> MA_{n + 1}B_{$N-n-1$} + B}$$
is given by
$$\begin{align}
K &= \exp\left(\frac{-\Delta_\mathrm r G}{RT}\right) \\
&= \exp\left(\frac{\Delta_\mathrm r S}{R}\right) \\
&= \exp\left(\frac{S_\mathrm{m}(\ce{MA_{n + 1}B_{$N-n-1$}}) + S_\mathrm{m}(\ce{B}) - S_\mathrm{m}(\ce{MA_{n}B_{$N-n$}}) - S_\mathrm{m}(\ce{A})}{R}\right) \\
&= \exp[\ln\Omega(\ce{MA_{n + 1}B_{$N-n-1$}}) + \ln\Omega(\ce{B}) - \ln\Omega(\ce{MA_{n}B_{$N-n$}}) - \ln\Omega(\ce{A})] \\
&= \exp\left[\ln\left(\frac{\Omega(\ce{MA_{n + 1}B_{$N-n-1$}})\Omega(\ce{B})}{\Omega(\ce{MA_{n}B_{$N-n$}})\Omega(\ce{A})}\right)\right] \\
&= \frac{\Omega(\ce{MA_{n + 1}B_{$N-n-1$}})\Omega(\ce{B})}{\Omega(\ce{MA_{n}B_{$N-n$}})\Omega(\ce{A})} \\
&= \frac{N!}{(n+1)!(N-n-1)!} \cdot \frac{n!(N-n)!}{N!} \\
&= \frac{N-n}{n+1}
\end{align}$$
The reason for ignoring $\Delta_\mathrm r H$ is because we are only interested in statistical effects, i.e. entropy, and we don't care about the actual stability of the complex or the strength of the M–L bonds. However, the exact justification for assuming this form for $\Omega$ still eludes me. It makes intuitive sense (that there are $N!/(n!(N-n)!)$ ways to arrange $n$ different ligands in $N$ different coordination sites), but I can't convince myself (and don't want to attempt to convince you) that it's entirely rigorous. In particular, I feel like symmetry should play a role here; maybe it is simply that the effects of any symmetry eventually cancel out.


text:
Sodium Sulfate vs Sodium Sulfite
I recently took a test and one of the questions was:

Which procedures will allow a student to differentiate between solid sodium sulfate and solid sodium sulfite?
I. Make solutions of each and look for a precipitate when added to $0.10$ M barium nitrate.
II. Add crystal of each to $0.10$ M HCl and watch for bubbles.
III. Make solutions of each and test with a pH indicator.

The answer is 2 and 3, however, I incorrectly answered 2 only. I have actually thought a long time about this question.
Why is III also correct, if sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color?

The last statement

…sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color

is true only for $\ce{Na2SO4}$ as it's formed by both strong base and strong acid and won't noticeably affect pH.
$\ce{Na2SO3}$, on the other hand, undergoes hydrolysis:
$$
\begin{align}
\ce{Na2SO3 + H2O &<=> NaOH + NaHSO3} \\
\ce{2Na+ + SO3^2- + H2O &<=> Na+ + OH- + Na+ + HSO3-} \\
\ce{SO3^2- + H2O &<=> HSO3- + OH-}
\end{align}
$$
resulting in elevated pH (basic solution) up to 9 (according to Wikipedia).

But sulfate is a weak base too, right? (HSO4- is a weak acid)

The questions asks you to differentiate Na2SO4 and Na2SO3. There are two main concepts involved here:
H2SO3 is a very weak acid (better to write SO2 (aq)) and H2SO4 is an extremely strong acid. Certainly, HCl would not be able to displace H2SO4. Another point which you should remember is that salts of strong acids and strong bases are pH neutral. However, once there is a disbalance in the strength of acids or bases, their corresponding salts solutions are either weakly basic or acidic. Salts or weak bases and weak acids are quite complicated in terms of pH calculations.
A solution of Na2SO3, formed from a strong base and weak acid, would show alkaline behavior.


text:
How to interpret parts-per notations? Is 1 ppb equal to 1 μg/kg?
In an article I recently submitted, a reviewer asked that I provide a concentration in μg/kg instead of ppb (parts per billion), and mentions that the later is not correct. I am not a chemist, and I thought that 1 μg/kg = 1 ppb.
Is 1 ppb equal to 1 μg/kg ? What is a reason to consider ppb as incorrect?

Just using ppb is ambiguous unless you state the basis (eg w/w for mass %). ppb could also be referring to molar proportion or volume proportion.

You are correct suggesting that 1 μg/kg implies 1 ppb, however the reverse is not true. For instance, 1 ppb can also be 1 nmol/mol, and the reader will never have a chance to deduce which one is it unless you explicitly define the usage of the "parts per something" in the text.
This clutters the manuscript with redundant notes and causes overall confusion.
Recommendations from authoritative sources
IUPAC
IUPAC's “Green Book” lists ppb and related symbols (ppm, ppt, ppb etc.) as deprecated [1, p. 98]:

Although ppm, ppb, ppt and alike are widely used in various applications of analytical and environmental chemistry, it is suggested to abandon completely their use because of the ambiguities involved.
These units are unnecessary and can be easily replaced by SI-compatible quantities such as pmol/mol (picomole per mole), which are unambiguous.
The last column contains suggested replacements (similar replacements can be formulated as mg/g, μg/g, pg/g etc.).
$$
\begin{array}{lllll}
\hline
\text{Name} & \text{Symbol} & \text{Value} & \text{Examples} & \text{Replacement} \\
\hline
\ldots & & & & \\
\text{part per billion} & \text{ppb} & 10^{-9} & \text{The air quality standard for ozone is a} & \pu{nmol/mol} \\
  & & & \text{volume fraction of}~\varphi = 120~\text{ppb} & \\
\ldots & & & & \\
\hline
\end{array}
$$

ISO-80000-1
ISO-80000-1, section 6.5.5 The unit one [2, p.20] stresses out the necessity for the coherent usage of “per” units in general, and prohibits the usage of “parts-per” units:

Abbreviations such as ppm, pphm, ppb and ppt are language-dependent and ambiguous and shall not be used.
Instead, the use of powers of 10 is recommended.

European Pharmacopoeia
According to European Pharmacopoeia, section 1.2. OTHER PROVISIONS APPLYING TO GENERAL CHAPTERS AND MONOGRAPHS [3, p. 4], in the absence of context ppm implies mass fraction:

The expression ‘parts per million’ (or ppm) refers to mass in mass, unless otherwise specified.

Related incident
In 2020-02-26 there was a major accident with both engines stalling on the Airbus aircraft resulting in emergency landing due to the overdose of a biocide fuel additive.
A base engineer was not familiar with the term “ppm”, and since the term was not defined in the maintenance manual, he incorrectly used an online calculator, which resulted in overshooting the correct biocide-to-fuel ratio 38 times.
The fact that the instructions for two different biocides were not only extremely brief, but also referred to different ppm values by mass and by volume also greatly increased confusion:

Unit standardization
The dosage of biocide required was previously provided in ppm by weight for Biobor® JF and by ppm by volume for Kathon® FP 1.5.
There is now industry-wide agreement to use mL/L unit for both Biobor® JF and Kathon® FP 1.5 biocides.
Airbus maintenance procedures for all Airbus aircraft, the AMM for
A300/A310/A320/A330/A340/A380, MP for A350 aircraft and AMP for A220
aircraft, were updated accordingly by removing the use of the ppm unit.

Luckily, the plane managed to land safely and there were no human casualties.
This accident made the aircraft manufacturer to take action against using ppm notation in the future versions of technical documentation, and include transparent definitions for the “parts-per” notations still used in the current service manuals [4, p. 105].
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. (PDF)
ISO 80000-2:2009 Quantities and Units – Part 1: General; International Organization for Standardization: Geneva, CH, 2009.
Council of Europe; European Directorate for the Quality of Medicines & Healthcare. European Pharmacopoeia, 10th ed.; Council of Europe: Strasbourg, 2019. ISBN 978-92-871-8912-7.
Air Accidents Investigation Branch. Report on the Serious Incident to Airbus A321-211, Registration G-POWN at London Gatwick Airport on 26 February 2020; United Kingdom, 2021. (PDF)

This difficulty with "parts-per" notations should probably be extended to percentages as well (given that, really, 'percent' is just 'parts per hundred'). I can't tell you how many times I've struggled to identify what kind of percentage is meant: % w/w? % w/v? at. %? % v/v? For % v/v, what are the bases? Is 10% one volume A for every 9 volumes of B? For every 10 volumes of B? For 10 volumes of final mixture/solution?  Headaches, headaches, headaches, every dang time.

@ andselisk, I am slightly confused by this new recommendation. My question is related to the denominator in nmol/mols. Suppose I prepare 100 ppb NaCl in water. Do I have 100 nmol of NaCl in 1 mol of water? How we practically prepare it. What if my solvent is a mixture such as air. How will we calculate moles of air? This new definition does not seem to be any better than the previous ones.

@M.Farooq I'm not sure I understand, and this is not a novel directive, part-per-somehing has never been a favored/preferred notation as long as I remember myself. ppb is just a fraction, use whatever quantities of the same dimensions; you have to find out the amounts to calculate ppb anyways, why is this suddenly a problem?

@andselisk, I am actually wondering about different scenarios. There is a very subtle difference in using nmol/mol vs. ug/kg. The typical definition of ppb is ug X in a kg of solution not the solvent. 100 ppb NaCl in water mean 100 ug NaCl "in" 1 liter of water solution. The total volume is 1 L. The analyte is a part of the solvent weight. With 100 ppm NaCl as 100 nmol NaCl dissolved in 1 mol water. In this case, NaCl is not a part of solvent. Similarly if I have a mixture as a solvent, whose moles should we use? The 1 L and kg definition do not care about the identity of the solvent system.

@M.Farooq There are commonly used unambiguous quantities such as molar concentration for solutions, volume fraction for gas mixtures, molality if you want to use only the mass of the solvent etc. ppb and Co. is ambiguous, harder to use in calculations with units and always requires extra annotations. What's the point of making own life's harder? I'm sorry, but maybe you might want to ask a new question as I'm not getting what the problem here really is.


text:
Making hard water
Does anyone know how to make hard water with a hardness of 200+ ppm in terms of $\ce{CaCO3}$?
I live in Seattle and we have very soft water and we need hard water for testing purposes. Obviously you cannot dissolve $\ce{CaCO3}$ in water. I've read many "recipes" even ones that claimed you can dissolve $\ce{CaCO3}$ with a soda stream pop maker. Best I could do was a 60 ppm.
I've tried baking soda and Epsom salt and was able to make 130 ppm.
I've tried $\ce{HCl}$ to dissolve it in water, but the resulting pH was not suitable and if I tried to raise the pH, the $\ce{CaCO3}$ precipitated back out of the water.
I'm at a loss on a "recipe" that actually works. Am I trying to replicate the possible? Any ideas?

You might want to read Sources of hardness on Wikipedia. It's not about just carbonates, you actually want to have soluble bicarbonates (and other anions such as sulfates and chlorides, for completeness) in solution.

Andselisk,Thank you for taking the time to respond. Yes I've read about sources of hardness. Allow me to explain what I'm attempting to do. I am trying to replicate the formation of hard scale. I have a clear understanding of how to get CaCo3 to precipitate out of solution. (Pressure, pH or raise in temperature) what I'm lacking is a recipe to create a solution that will precipitate CaCo3 in the form of hard scale. Only thing I can think of trying is Calcium hydroxide and bubble Co2 through it.

What about some calcium bicarbonate rich mineral water ( composition is usually provided ), diluted to desired concentration ?

@JasonRusk, what andselisk says is correct, but I'm not sure you understood correctly. "Hard scale" as you call it does not just consist of carbonates (and calcium), it is also composed of magnesium salts and both with chlorides, sulfates, etc. So, you need to make a mixture. What you are referring to or trying to get around sounds like just calcium carbonate solubility. I'm unsure why you are using CaCO3 - are you referring to the equivalent unit 'mg CaCO3/L'? That is not to be taken literally...best to convert that to mg/L Ca and Mg and choose a few different salts from those.

I think hard scale might occur only slowly. Fast precipitation tends to make smaller crystals and many of them.

When analytical chemists talk about hardness of water as ppm CaCO3, it is just a statement of "titratable" calcium / magnesium by ethylenediamminetetraacetate (EDTA). From your comments, you would like to replicate hard water perhaps for a project. Keep in mind that hardness are of two types temporary and permanent hardness. Permanent hardness can replicated with a soluble calcium salt.
Basically, you need a soluble calcium salt, but most water soluble salts of calcium are hygroscopic and nobody uses them as primary standard. Fortunately, CaCO3 is actually a primary standard. The way a Ca standard is prepared is by dissolving pure CaCO3 is smallest possible amount of acid. If this is a school project, you can take 200 mg solid CaCO3 in a 1 L volumetric flask. Add a very small amount of dropwise HCl until it dissolves. Test the pH and neutralize the acid if needed with NaOH. Dilute with distilled water. This would be titratable 200 ppm CaCO3 in water and would produce "scum" if you add soap solution (not the liquid ones).


text:
Thermal decomposition of ester
I have a clear liquid with a faint sweet smell. I know this compound contains an aromatic group and an ester. However, something strange happens when I try to distill it - it forms a white solid. I believe this solid to be the result of thermal decomposition of the ester - the ester reacting with oxygen in the air. I have searched around, but can't find much written about thermal decompositions of esters. What reaction is this? What is the product? Could it be a carboxylic acid?

Well, you are right. Esters decomposes at higher temperatures to form several small carboxylic acids and ketones and at even higher temperature, these carboxylic acids decomposes into carbon monoxide, carbon dioxide, carbon, oxygen and/or water. A paper[1] discusses about the thermal decomposition of ethyl formate, ethyl acetate and ethyl propanate at 1248-1638 K, pressure around 1.5 atm at a chamber filled with argon at a concentration of 2000 ppm.
After the experiment, it was found out that 90% of oxygen has been released as oxygen gas and the pyrolysis mechanism was determined to be a six-centered decomposition producing equimolar ethylene and corresponding acid and at drastic conditions, it was known to have a dehydration reaction releasing water and a decarboxylation reaction to form carbon dioxide.
It was found out that EF gave formic acid, EA gave acetic acid and EP gave methyl ketene as the main products which subsequently decomposed into various smaller products at higher temperatures and the concentration ratios were noted.
Carboxylic acid can form white solid as noted in case of formic acid and acetic acid.
References

https://sutherland.che.utah.edu//USCI2013/PAPERS/1A07-070RK-0061.pdf


text:
Is it possible to boil a liquid by just mixing many immiscible liquids together?
In open air, when vapour pressure reaches 1 atm, boiling takes place.
I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to $p = p^*_A + p^*_b$, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.
If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that  $p = p^*_A + p^*_B + ... > \pu{1atm}$ , will boiling results at room temperature?
I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make $p>\pu{1atm}$), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (e.g. rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.

This brings interesting question how many non miscible liquid phases can coexist in equilibrium.

I guess a good answer here should make clear what exactly is going to happen, and how we can predict the final state of such a system.

@Karl IMHO AKs answer describes the situation pretty well. The OP is apparently confused about when something will boil, as AK explains. The steam distillation technique does not rely on changes in boiling points, that is a misunderstanding, but if the volatile compounds are soluble in and  form an azeotrope with water then the boiling point will be lowered. However the question asks about immiscible solvents, so azeotropes are essentially irrelevant. Steam distillation relies on transport of the volatile compound, not changes in b.p.

@BuckThorn Steam distillation is done with nonmixing azeotropes. Some high-boiling unpolar organic substance and water. Toluene and water would be a perfectly good example. since the partial pressure of each of the phases is less than ambient, no boiling will occur. is of course the correct answer here, no doubt. But what exactly will happen?

Yes they will boil all right. Sure, there might be some kinetic impediment to it if you let the liquids to settle in layers, but if you stir them so as to expose their surfaces, they will boil†. This is what steam distillation is all about.
As for the first law, it will hold just fine. You burn your firewood, you get the heat, but it is not for free: the firewood is gone. Same thing here. Your liquids are gone. What you have now is a vapor mixture. Sure, you may cool it down, and it will separate back into the liquids, which you may then heat up and repeat... Congratulations, you've just invented the steam engine. Pity it's been done before by one James Watt. Also, you don't really need two liquids for it.
So it goes.

(†) There seem to be some ambiguity as to what exactly "boiling" entails.

Will vapor bubbles form in the whole bulk of both liquids? No.
Will both liquids eventually evaporate completely, if left in a cylinder under a piston and maintained at these pressure and temperature until they reach equilibrium? Yes.

For a single pure liquid these two questions are equivalent, hence the confusion.

Found a link where the same is explained from a different perspective: https://en.wikipedia.org/wiki/Azeotrope#Homogeneous_and_heterogeneous_azeotropes

What happens when one of the components has evaporated completely?

Then the other will stop evaporating too.

Vapor pressure of A and B, obviously.

Those are saturated vapor pressures. As for the vapor pressures at specific conditions, these can be anywhere from there to 0.

As to your story: with 10 ml of A evaporated, we'll have (supposedly) vapor pressure of A greater than its saturated vapor pressure, so no, that won't happen.

"Saturated" vapour pressure is a pleonasm. What you mean is partial pressure. Won't happen, right, but I don't follow your reasoning.

Both A and B will evaporate until they reach their respective saturated pressures. There is nothing ambiguous about it.

Hm, water will not boil just because I replace air by hydrocarbon vapour. In fact, both liquids may not be in contact at all. Both will evaporate as usually with the total vapour pressure higher than atmospheric one.

True, hence my post-scriptum about the exact meaning of "boiling".

Nothing special would happen, immiscible liquids would just form layers. As for the expression, $$p_T=\Sigma p^o_i$$
I suggest you read this answer. Quoting Ivan Neretin:

This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.

It would take infinite time for that expression to prove itself correct.
In speaking within an average man's lifespan, the liquids would remain in separate layers, with the pressure above liquid surface would remain approximately the pure liquid vapour pressure ($p^o$) of the uppermost layer if left undisturbed (according to your question: no energy transactions).
Don't worry about the first law: it is rarely violated these days.

This "equilibrium" can be attained sooner by agitation (stirring) of the mixture, but mechanical stirring certainly counts as work done.
From a chemguide page:

Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapour pressure you measure will simply be the vapour pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapour. The top one is sealing it in.
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapour pressures are bound to reach the external pressure before the vapour pressure of either of the individual components get there.

In other words, although less evident, even if you do get the mixture to boil and do work, it would be because you had done work on it earlier during agitation.

I object to the last point. Work done during agitation can easily be made negligible. For example, we may put the liquids into two separate open compartments, so that their vapors mix, but the liquids themselves don't. Then they both will have open surface without any work.

@IvanNeretin I wish to respectfully mention that I do not agree. Could I get back to you in a day? I have an exam tomorrow and am short on time today. I will certainly participate in a fruitful discussion.

Looking forward to hear your arguments.

Ok, well, as A.K. mentions in his answer, boiling happens when the vapour pressure of the phase is greater than ambient pressure. Keeping the liquids in separate partitions will not create a single phase.

Imagine two open beakers standing side by side. The liquids in them don't mix, but their vapors do.

I am unable to see how that helps the problem here. That would be no different from a pure liquid and its vapour pressure contained in a closed vessel and an inert gas inserted.

What problem? You seemed to agree that an agitated mixture will boil below the boiling points of both liquids, and that because of exposed surface. I just proposed a setup that lets you have exposed surface of both liquids without any agitation.

Let us continue this discussion in chat.

Is it possible to boil a liquid by just mixing many immiscible liquids together?

No*, boiling is when the vapor pressure of a phase is greater than the ambient pressure. You might create a total pressure of all of the partial pressures in excess of ambient, but it wont cause boiling since the partial pressure of each of the phases is less than ambient, no boiling will occur.
To get boiling by mixing you need two mixable substances that form a positive azeotrope when mixed and are at a sufficiently high temperature that the mixture can boil. Take for example methanol and chloroform system in the phase diagrams below. At $\pu{330K}$ neither pure methanol nor pure chloroform boil, but if you mixed $\pu{2 mol}$ of chloroform at $\pu{330K}$ with $\pu{1 mol}$ of methanol at $\pu{330K}$  together the phase would boil upon mixing as the vapor pressure of the phase is now above ambient.

Okay I said no, but that technically isn't correct. Grease fires are a (too) common example of mixing immiscible liquids and causing boiling as the oil temperature is well above the boiling temperature of water. This is why you can't throw water on a grease fire. For perspective on thanksgiving of 2016 there were about 1570 fires in the US  related to this phenomenon.

Is steam distillation not a thing at all, then?


text:
Why does the carbonyl of carboxylic acids get protonated and not the hydroxyl group?

Is the O of the C=O group more electronegative or is it for the sake of getting from product A to B that the carbonyl is protonated rather than the hydroxyl?

I suppose it because $\ce{R-C^{+}(OH)2}$ is more preferred. I think protonation of both oxygens is possible, but the symmetric structures is more stable.

Because the protonated acid is resonance-stabilized when the carbonyl is protonated, but isn't if the hydroxyl is protonated:

So the first protonated form is much more energetically favourable than the second.
The same resonance in the non-protonated acid means that the lone pair on the alkoxide has reduced basicity, meaning that the carbonyl lone pair is the more basic since it isn't involved in any energy-lowering orbital interactions(/resonances).


text:
Are precipitations exothermic and/or endothermic?
Should be an easy one. I'm fumbling a concept. I've read precipitations are exothermic. Is this accurate? Why would there be no endothermic precipitation reactions?

Do you mean chemical reactions where precipitation occurs, or precipitation alone ?

There are plenty of exothermic dissolution reactions, so if by precipitation you mean the reverse of dissolution, there are plenty of endothermic precipitations.

No, absolutely not.
Precipitation reactions can be either endothermic and exothermic.
Table 1. Thermodynamic data of precipitation for some salts
\begin{array}{cccccc}
\hline
\text{Salt} & \Delta G_\text{ppt}^\circ & \Delta H_\mathrm{ppt}^\circ & -T\Delta S_\mathrm{ppt}^\circ(\pu{25 °C}) \\
\hline
\ce{Be(OH)2} & -121 & -31 & -90 \\
\ce{Mg(OH)2} & -63 & -3 & -61 \\
\ce{Ca(OH)2} & -28 & 16 & -44 \\
\ce{Li2CO3} & -17 & 18 & -34 \\
\ce{MgCO3} & -45 & 28 & -74 \\
\ce{CaCO3} & -48 & 10 & -57 \\
\ce{SrCO3} & -52 & 3 & -56 \\
\ce{BaCO3} & -47 & -4 & -43 \\
\ce{FePO4} & -102 & 78 & -180 \\
\hline
\end{array}
It is more than clear that all precipitations are not exothermic.
Also definitely noticable is the fact that $\Delta S$ is a positive quantity.
But that is unexpected, as an increase in the disorder of the system upon precipitation may seem counterintuitive to many.
What is often overlooked is that several other factors are at work.
Paving the way
Ions can be of two types:

Electrostatic structure breakers
Electrostatic structure makers

This classification is based on how ions interact with water.
Electrostatic structure breakers
Sometimes referred to as chaotropic ions, these disrupt the hydrogen bonding interactions between water molecules. Examples include $\ce{NH4+}$, $\ce{Cs+}$, $\ce{Br-}$, $\ce{I-}$.
It is tempting to assume that large cations with higher possible coordination numbers would attach and order water molecules more effectively than small cations, but this is not experimentally observed.
Smaller the radius and/or greater the charge (the more acidic or basic the ion), the greater its structure-making properties in aqueous solution. It is hence clear that the above mentioned ions are structure breakers rather than makers, evidently having very low values of $Z/r^2$.
It is easy to deduce that upon their introduction into aqua, they will correspond to an increase in entropy.
Electrostatic structure makers
AKA kosmotropic ions, these increase the ordering of water. $\ce{Fe^2+}$, $\ce{Al^3+}$, $\ce{Mg^2+}$, are members of this club.
These ions have large charges and relatively smaller radii, resulting in a higher $Z/r^2$ value.
Via hydrogen bonding these ions create relatively larger shells of hydration than their low $Z/r^2$ counterparts.

The differences in hydration for the two types of ions is visible in the following table.
Table 2. Hydration data for some cations

Uh..What does this have to do with the question?
Everything.
The reason that most acidic cations and basic anions combine to give precipitates is due to the disorder that results from the release of large number of water molecules from the well structured hydration spheres.
The second law favours this.

For an example, let us consider the precipitation of $\ce{MgCO3}$
$$\ce{Mg(H2O)36^2+ + CO3(H2O)28^2- <=> MgCO3(s) + 64 H2O}$$
The number of water molecules are derived from Table 2.
The release of $64$ molecules of water per molecule of magnesium carbonate precipitated is very indicative of the positive entropy of the process.
Now for the enthalpy.
Rather than the Gibbs–Helmholtz, let's return to the good old Hess law.
According to Hess' law,
$$\Delta H_\mathrm{ppt}(\ce{M_yX_m)}) = -y\Delta H_\mathrm{hyd}(\ce{M^m+})- m\Delta H_\mathrm{hyd}(\ce{X^y-}) + U(\ce{M_yX_m})$$
The electrostatic structure makers (type 1) would have higher negative
$$y\Delta H_\mathrm{hyd}(\ce{M^m+})+ m\Delta H_\mathrm{hyd}(\ce{X^y-})$$
leading to an overall positive heat of precipitation. e.g. $\ce{FePO4}$, $\Delta H_\mathrm{ppt} = \pu{+78 kJ/mol}$
Vice-versa for type 1: the structure breakers.
I hope I managed to enlighten you on this matter.
P. S. This was a great question. Pardon the short conclusions, I am currently travelling. Members of the community may feel free to edit and make a more dramatic exit.

References:
Hanzlik, R. , Inorganic Aspects of Biological and Organic Chemistry 1976, II, 26-30
https://www.wou.edu/las/physci/ch412/ppt2.htm
https://www.wou.edu/las/physci/ch412/pptrxn.html

Thank you very, very much. Wasn't expecting such a thorough yet concise response, especially considering entropy can increase which was also a counterintuitive point as you said. Above and beyond answer.


text:
About the percentage yield and purity
I was wondering why all of the percentages like mass percent composition, percentage yield and percentage purity are calculated using masses.
Wouldn't it make more sense to use number of moles of the substance since we're counting the number of atoms in the molecules?

It is for the sake of convenience and practical purposes. Analytical balances or any other instrument cannot count the number of atoms/molecules, neither they are calibrated in terms of moles. All we can do is to measure its mass directly. Secondly, "measuring" moles assume that we know the exact molecular mass of that material. What if you make a new unknown product in a synthetic reaction? All you can do is to measure its mass directly and later characterize it.

Oh I see, thank you for your helpful answer!


text:
Why is there a lone pair in thionyl fluoride?
Why is there a lone pair in $\ce{SOF2}$? I drew its structure, which according to me should look like this:

Why is there a lone pair on sulfur? Isn't its octet complete? If yes, why should it expand its octet and gain more electrons?

How many valence electrons did S have in the first place?

@Ivan Oh you're right! S has 6.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

When you want to write a Lewis structure, I suggest you start from considering how many valence electrons each atom has. In your case, you would have:
S = 6
F = 7; total = 14
O = 6
The final total would be 14 + 6 + 6 = 26. The structure you guessed for is correct, sulfur is the central atom. You start by drawing a single bond per each atom, getting something like this:
F
|
S–O
|
F

Then, you can draw a double bond for oxygen:
F
|
S=O
|
F

If you do the math, by counting the electrons that you put in the previous structure, you would have 2 + 2 + 4 (two S-F bonds and a S=O) = 8.
Then we will add in the lone pairs: three on each halogen, two for oxygen, and one for sulfur. Therefore we add 3 × 2 × 2 = 12 electrons (three lone pairs on each fluorine) plus 4 electrons on O (the two lone pairs), and we have a total of 24. The final 2 , adding up to 26, come from the lone pair on sulfur.
This makes sense because sulfur is in period 3, so it is possible for it to have more than 8 electrons: in other words, the octet rule applies not only for 8 electrons, if that makes sense.

A much, much better description would use partial charges at the sulfur and oxygen and a single bond instead. A double bond would require d-orbitals from sulfur, a theory that hat been disproved. This has been discussed on our platform a couple of times. In any case, following the octet rule is usually safer than expecting 'octet expansion' (because the latter is wrong).

You are right, thank you for catching that! Do you have any references about what you said? :)

See for example https://chemistry.stackexchange.com/questions/29101/why-is-the-bond-order-in-the-so%e2%82%83-molecule-1-33-and-not-2


text:
Why doesn't open-chain glucose react with water readily in aqueous solution?
Though the exact number given varies from 0.02% to 3%, it's well agreed that glucose spends a very small percentage of its time in aqueous solution as a linear aldehyde. Glucose's overwhelming preference as a cyclic pyranose is commonly explained by the cyclization being intramolecular; while the aldehyde group in its linear form can react with other molecules like protein and lipids during rare encounters, it far more readily reacts with an alcohol group that is always around by virtue of being on the same molecule.
My problem with that explanation is that there is indeed another molecule that is always around in aqueous solution: water. All sources I could find said that the common instability of hemiacetals and geminal diols are both explained by oxygen repulsion. Therefore, I would expect there to be a glucose-geminal diol concentration comparable to the cyclic hemiacetal concentration in aqueous solution, yet I cannot find any mention of a glucose-geminal diol, as if it didn't exist. What am I missing?

Why doesn't water react with the aldehyde to form a linear geminal diol?

It does, and depending on the nature of the aldehyde, this will be the major species. You are right in saying that the high concentration of water helps to form the diol. In other solvents, the diol form will be present at lower fraction, even when they contain a bit of water. The Wikipedia article on geminal diols has some equilibrium constants for various aldehydes and ketones in aqueous solution: https://en.wikipedia.org/wiki/Geminal_diol#Reactions

Glucose's overwhelming preference as a cyclic pyranose is commonly explained by the cyclization being intramolecular; while the aldehyde group in its linear form can react with other molecules like protein and lipids during rare encounters, it far more readily reacts with an alcohol group that is always around by virtue of being on the same molecule.

Yes, that is the special thing about glucose: the aldehyde can form an intramolecular hemiacetal without strain. It can also form a geminal diol, but that is no different from other aldehydes, so it is usually not mentioned in that context.
Is the geminal diol ever important for the reactivity?
There is an enzyme called aryl sulfatase which has an aldehyde (a cysteine modified post-translationally to a formyl glycine) in its active site [review]. X-ray crystal structures show that the main form is the hydrated species, i.e. the geminal diol. In the suggested mechanism, one hydroxyl attacks a sulfate ester, displacing the alcohol. Then, the second hydryoxyl group reforms the aldehyde and releases the sulfate in an elimination reaction.
Are there other instances where we don't explicitly mention reaction with water?
The hydrogen ion occurs in aqueous solution in various hydrated states. Carbon dioxide in aqueous solution occurs either hydrated (carbonic acid) or not. In either case, it is easier to lump the species into one than to talk about them separately.
What is the equilibrium constant for the hydration of the aldehyde in glucose
I did not find data for glucose, but did find some for glyeraldehyde 3-phosphate (Petersson and Petersson 1999). [In the HTML version of the paper, some equations and schemes don't show, but the PDF is an improvement.] In aqueous solution, only about 3% of this phosphorylated sugar is present in the aldehyde form. In the reaction they were studying, only the aldehyde form reacts in the enzymatic reaction, so there is a slow non-enzymatic first step to dehydrate the diol and a fast second step.
Wikipedia has an article on glycoaldehyde, with 4% of it in the aldehyde form.
If glucose behaves in a similar manner, most of the linear form would be in the diol form. Equilibrium constants given in the literature probably use the total concentration of linear glucose (hydrated or not) in the equilibrium constant expression rather than the actual concentration of aldehyde.

I did read that geminal diol stability varies and why (oxygen repulsion). However, I read the exact same reasons for hemiacetal instability. So I figured I should find hard numbers like equilibrium constants. The only source I can find showing equilibrium constants for a wide variety of carbonyl species is a page of lecture notes that cites two old sources, one for hemiacetals and one for hydrates, and it suggests that the more readily an aldehyde reacts with water, the more readily it reacts with methanol. I can't find a single mention of a glucose-geminal-diol, like it didn't exist.

You are referring to this source

Yes, exactly. Not great because the methanol is the most similar to water in size and shape of all the alcohols, but it was the only source I could find with both hemiacetal and hydrate reactions for a wide variety of species that wasn't behind a paywall.

@BatWannaBe I added two references (still not exactly what you were asking, but relevant).


text:
Negative charges in acid-catalyzed reactions
Whenever my professor draws a mechanism in an acid-catalyzed reaction, he always says you can't form negative charges. Why is that?

Well, with all those greedy $\rm H^+$ skulking around, any negative charge won't last long.

Similarly, reactions in alkalic environment do not produce $\ce{H+}$

The acidic environment would not allow creation of anions but anions of strong acids, that are too weak bases to be significantly protonated.

The reason behind your professor's statement is that when you have an acidic solution, the H+ is always around, and therefore it quickly reacts with any negative charges that it meets. That's why you can't have any negative charges, they won't last long!
Remember that proton exchange is among the fastest chemical reactions too. I hope this helps!


text:
Why does the reaction of the primary alcohol cyclopentylmethanol with hydrogen bromide follow SN1 mechanism?

Example 2: Identify all the products formed in the following reaction and indicate the major product.

Solution: Three products will be formed in the given reaction as shown below:

In this reaction the primary alcohol undergoes SN1 mechanism by forming a carbocation and then undergoing ring expansion, but primary alcohols prefer SN2 mechanism. So why is this primary alcohol undergoing SN1 mechanism?

The Sn1 mechanism should be considered relatively unreasonable here. You might consider a Sn1-like mechanism instead where ring expansion is the thing that displaces water, i.e., the C-C bond is the nucleophile.

But the carbocation is formed in Sn1 mechanism and ring expansion can only take place if carbocation is formed,also in this reaction the product formed by Sn1 mechamism is major product

"ring expansion can only take place if carbocation is formed" That is incorrect. The ring expansion could displace the leaving group.

The bottom line here is the the provide solution is wrong.

Is the given answer to the question regarding major product 1-bromo cyclohexane?

Actually i think what happens is Sn2 in beginning and then once that happened alchol slowly rearranges(you cannot stop the reaction , i mean formation of carbocation in acidic medium) , so final product same as due to Sn1 mechanism

It does not. The proposed mechanism is wrong.
The conversion of cyclopentylmethanol to 1-bromo-1-methylcyclopentane is not reported anywhere in the literature. If you could somehow generate the primary carbocation, then one would expect a rapid 1,2-hydride shift to generate the more stable tertiary carbocation, giving rise to 1-bromo-1-methylcyclopentane.
The ring-expansion product could be rationalized by a concerted 1,2-alkyl shift displacing $\ce{OH2^+}$ without violating first-principles.


text:
What's the purpose of stirring the mixture at 0 degree and dding NaOH to the organic reaction mixture?
I used to follow the lab protocol without thinking, but I now hope to understand why certain procedures are done.
I am synthesizing dimethoxyacetic acid from methyl dimethoxy acetate(so hydrolysis from ester to COOH).

Add lithium hydroxide monohydrate (89.5 mmol, 3.75 g,1.2 equiv) to a solution of methyl dimethoxyacetate (74.6 mmol, 10 g, 1 equiv) in dioxane (45 mL) and water (45 mL) at 0 °C.
Stir the mixture for 1 h at 0 °C and one additional hour at room temperature.
Add sodium hydroxide (1M aq soln, 75 mL) to the mixture.
Extract the mixture with diethyl ether (375 mL).
Acidify the recovered aqueous layer with hydrogen chloride (6N aq soln) until a pH of 1-2.
Extract the organics with diethyl ether (375 mL).
Dry the combined organic layers with magnesium sulfate and filter.
Evaporate the solvent slowly under reduced pressure to obtain 2,2-dimethoxyacetic acid.

I thought the reason why you acidify the aq. layer was to convert -COONa or -COOLi (ester salts) in the layer to acid (-COOH). Then doesn't it mean that the product I want ( dimethoxyacetic acid ) will be in the aqueous layer? Why do I need to care about the organic layer?
Would appreciate any insight&help!

For 6 - don't you mean extract the aq phase with ether? Also 7 seems to suggest that you combine the ether phase from 4 with the ether phase from 6.

Your description does not make sense at all! You say first that you are making the acid, then in 8, you say you get the "acid chloride" ?

@SteffX yeah you are right, amended step 8. it was a mere typo

@Waylander so after extraction with diethyl ether, I thought the organic layer would be consisted of : (1) some unreacted ester (starting material) & (2) diethyl ether, and the aq. layer would be of (1)sodium acetate (2) dioxane (3) water obviously & (4) dioxane and possibly LiOMe. What I don't get is the next instruction... they say I need to dry the organic layer to retrieve the product, but isn't it the aq. layer where my desired acid will be located?

Instruction 6 is misleading. Once the aq layer is acidified in 5 your product will extract into ether. I would extract the acidified aq phase with ether at least 3 times to make sure you get all the product. You then combine these ether extractions and dry them.

@Waylander you are amazing, thank you


text:
Standard Solutions for UV-Spectrophotometry
Can one use a mixture of compounds when making a standard solution for UV-Spectrophotometry. For instance, one is attempting to find the casein concentration of milk using UV-spec and they have milk protein which has casein protein in, but not pure casein protein. Would it be feasible to make the standard solutions using the milk protein, knowing how much casein is in it, even if there are other proteins in there or if it is not pure casein?

In general I would say yes, if you are sure that the absorption spectra do not overlap. Besides, even in that case other interferences might happen, for instance if a coloring reaction must be performed on the "standard" and on the analyte.

@Alchimista's caveat that the peaks don't overlap is a big assumption. UV-VIS peaks are very broad. With something as complex as milk there must be numerous compounds that would respond in the UV-VIS range. You'd also have to consider scattering.

In theory, I would agree with Alchimista but in practice, how can you determine exactly the real amount of casein you have? I mean, a "standard solution" is the base for determining the purity/concentration of a compound and that requires a very pure compound. Here it seems you are trying to stretch the definition to suit your needs, but it won't work

To all. I a not saying you assume anything. You must be sure! As far as I know, I do not see proteins as a good  UV-Vis analyte at all (except some biochem procedure I've forget or I never was aware of). To me they are  big ensemble of chromophors for which I am already in trouble finding cuvette and solvent.

@MaxW. Agree.  But this is a problem mostly of the OP sample rather than of my general comment. The latter is anyway meant to circumvent lacking of a pure reference.

One problem with spectroscopic techniques for colloids is that there is not only absorption but scattering as well which can confound the results.

@Alchimista - Sorry my comment above wasn't worded well. I agree with your first comment. I was just trying to point out to the OP that the broad peaks in UV-VIs mean that peak overlap is a common rather than rare occurrence when there are multiple UV-Vis sensitive species.


text:
Why does calcium phosphate dissolve at low pH
Why does calcium phosphate dissolve at low $\mathrm{pH}$? What makes its solubility  $\mathrm{pH}$ dependent? This question arises after reading about the calcium phosphate bridges in casein micelles and how they 'break apart' at a $\mathrm{pH}$ below 6, such that the micelle is destabilised and falls apart.

Calcium (and other divalent cations) are extremely good at bridging phosphate moieties together. I suspect (but it's only my opinion) the following: when lowering the pH, you will partially protonate phosphates and that lets less chances for them to be linked together by calcium.

It is because the phosphate ion $PO_4^{-3}$ is a weak base. In general, ionic substance with anions that are weak bases dissolve better in acidic solution.

The solubility of calcium salts is highly dependent on $\mathrm{pH}$. For example, let’s look at tricalcium phosphate, which presents the solubility equilibrium:
$$\ce{Ca3(PO4)2(s) + H2O <=> 3Ca^2+(aq) + 2PO4^3-(aq)}$$
The solubility product constant for this equilibrium $(K_\mathrm{sp})$ is vastly varied from source to source. Some listed it as $\pu{2.0 × 10^–29 mol^5 L{–5}}$, and another listed it as $\pu{2.07 × 10^−33 mol^5 L^{–5}}$ at $\pu{25 ^{\circ}C}$ and $\mathrm{pH}~7.00$. However, I found a reliable $K_\mathrm{sp}$ value in literature, which reported that $K_\mathrm{sp}$ of $\beta$-$\ce{Ca3(PO4)2}$ in the $\mathrm{pH}$ range $6.0-7.5$ is $\pu{1.2 × 10^–29 mol^5 L^{–5}}$ at $\pu{25 ^{\circ}C}$ and $\pu{2.83 × 10^–30 mol^5 L^{–5}}$ at $\pu{37 ^{\circ}C}$ (a negative thermal coefficient of solubility; Ref.1). Therefore, the quantitative analysis of $\ce{Ca3(PO4)2}$ is near impossible.
Still, we can qualitatively look at the problem: If you decrease the $\mathrm{pH}$ of the solution in above equilibrium by adding acid to the solution, some of the phosphate ions would protonate and transform into $\ce{HPO4^2–}$ ions, as what happens in phosphate buffer solutions:
$$\ce{PO4^3-(aq) + H3O+ (aq) <=> HPO4^2-(aq) + H2O}$$
As a result, the concentration of $\ce{PO4^3-}$ ion would be reduced. The system would respond to this reduction by producing more $\ce{PO4^3-}$ ions, according to the Le Chatelier principle. To do so, some solid $\ce{Ca3(PO4)2}$ would dissolve in first equilibrium, and the equilibrium will be shifted to the right.
Keep in mind that in first equilibrium, $\ce{Ca/P}$ ratio is 1.5. The solubility of calcium phosphates is also depend on $\ce{Ca/P}$ (Ref.2). For example, both monocalcium phosphate $(\ce{Ca(H2PO4)2})$ and monocalcium phosphate monohydrate $(\ce{Ca(H2PO4)2.H2O})$ are highly soluble in water at $\pu{25 ^{\circ}C}$ ($\ce{Ca/P}$ is 0.5). On the other hand, $K_\mathrm{sp}$s of dicalcium phosphate ($\ce{CaHPO4}$) and dicalcium phosphate dihydrate $(\ce{CaHPO4.2H2O})$ are $1.26 × 10^{–7}$ and $\pu{2.57 × 10^{–7} mol^2 L^{–2}}$, respectively at $\pu{25 ^{\circ}C}$ (Ref.2). Based on these values, when $\mathrm{pH}$ is gradually reduced to a point such that all calcium phosphates would dissolve.
References:

T. M. Gregory, E. C. Moreno, J. M. Patel, W. E. Brown, “Solubility of $\beta$-$\ce{Ca3(PO4)2}$ in the system $\ce{Ca(OH)2-H3PO4-H2O}$ at $5, 15, 25,$ and $\pu{37 ^{\circ}C}$,” Journal of Research of the National Bureau of Standards–A. Physics and Chemistry 1974, 78A(6), 667–674 (http://dx.doi.org/10.6028/jres.078A.042).
L.C. Chow, “Solubility of Calcium Phosphate,” In Octacalcium Phosplrate; L. C. Chow, E. D. Eanes, Vol. Eds., Volume 18 of Monographs in Oral Science (G. M. Whitford, Ed.) Karger Publishers: Basel, Switzerland, 2001, pp. 94-111 (DOI:10.1159/000061650).


text:
Precipitating silver(I) salts from the solution of barium(II) cyanate and iodide

Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$.

The answer was only $\ce{AgCN}$ will precipitate, but I don't understand why $\ce{AgI}$ wouldn't precipitate as well since there is more than enough excess $\ce{AgNO3}$ available to precipitate with both $\ce{I-}$ and $\ce{CN-}$?

"There is more than enough." How did you determine that with out any quantitative calculations?

I did calculate that max CN- that could be precipitated as AgCN is 2 x 10-12 mol. This leaves 3.498 x10-9 mol Ag+ remaining to react with I-

Only if the concentrations are such that the solubility product exceeds $K_{\mathrm{sp}}$.

Consider a $\pu{10.0 mL}$ solution containing $\pu{1.0e-10 M}$ each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$. If $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, will any precipitate(s) form? If yes, what compound(s) will precipitate?
$K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0e-17}$; $K_\mathrm{sp}(\ce{AgI}) = \pu{8.5e-17}$.

Assuming that $\ce{Ba(CN)2}$ and $\ce{BaI2}$ dissociate completely.
$\ce{[CN-]_i = [I-]_i =} 2\cdot10^{-10}$ molar
Neglecting any volume change of solution the initial concentration of $\ce{Ag+}$ will be
$\ce{[Ag+]_i} = \dfrac{3.5\cdot10^{-9}\pu{mol}}{0.010\pu{L}} = 3.5\cdot10^{-7}\pu{M}$
Now if both the $\ce{CN-}$ and $\ce{I-}$ are quantitatively removed then the same amount of $\ce{Ag+}$ must be removed.
$\ce{[CN-]_i + [I-]_i =} 4\cdot10^{-10}$ molar
$\ce{[Ag+]_f} = 3.5\cdot10^{-7}\pu{M} - 4\cdot10^{-10}\pu{M} \approx 3.5\cdot10^{-7}\pu{M}$
So the final concentration of $\ce{Ag+}$ is essentially the same as the initial concentration. The concentration of $\ce{Ag+}$ with the Ksp's can now be used to calculated how much of the two anions can remain in solution.
The final concentration of $\ce{CN-}$ is
$\ce{[CN-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{6.0\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{1.7e-10}$
The the final concentration of $\ce{I-}$ is
$\ce{[I-]_f} = \dfrac{K_{sp}}{\ce{[Ag+]_f}} = \dfrac{8.5\cdot10^{-17}}{3.5\cdot10^{-7}} = \pu{2.4e-10}$
Conclusion:
Since $\ce{[CN-]_i > [CN-]_f}$ some $\ce{AgCN}$ will ppt.
Since $\ce{[I-]_i < [I-]_f}$ no $\ce{AgI}$ will ppt.

Alternative method to MaxW method:
Assume that an initial $\pu{10.0 mL}$ solution of $\pu{1.0e-10 M}$ in each of $\ce{Ba(CN)2}$ and $\ce{BaI2}$ is clear (homogeneous). That means $\ce{Ba(CN)2}$ and $\ce{BaI2}$ have dissociated completely. Thus concentrations of ions are as follows:
$$\ce{[CN-]_i = [I-]_i} = \pu{2\cdot10^{-10} mol \! L^{-1}}$$
Suppose when $\pu{3.5e-9 mol}$ of $\ce{AgNO3(s)}$ is added to this solution, no volume change has occured. Thus, the initial concentration of added ions in the solution will be:
$$\ce{[Ag+]_i = [NO3-]_i} = \dfrac{\pu{3.5\cdot10^{-9} mol}}{\pu{0.010 L}} = \pu{3.5\cdot10^{-7} mol \! L^{-1}}$$
For precipitation of $\ce{AgCN(s)}$:
$$Q_\mathrm{sp} = \ce{[Ag+]_i}\cdot \ce{[CN-]_i} = (\pu{3.5\cdot10^{-7} mol \! L^{-1}})(\pu{2\cdot10^{-10} mol \! L^{-1}}) \\ = \pu{7.0\cdot10^{-17} mol^2 \! L^{-2}} \gt K_\mathrm{sp}(\ce{AgCN}) = \pu{6.0\cdot10^{-17} mol^2 \! L^{-2}} $$
Therefore, $\ce{AgCN(s)}$ will precipitate.
For precipitation of $\ce{AgI(s)}$:
$$Q_\mathrm{sp} = \ce{[Ag+]_i}\cdot \ce{[I-]_i} = (\pu{3.5\cdot10^{-7} mol \! L^{-1}})(\pu{2\cdot10^{-10} mol \! L^{-1}}) \\ = \pu{7.0\cdot10^{-17} mol^2 \! L^{-2}} \lt K_\mathrm{sp}(\ce{AgCN})=\pu{8.5\cdot10^{-17} mol^2 \! L^{-2}} $$
Therefore, $\ce{AgI(s)}$ will not precipitate in this condition.


text:
Can glucose and insulin exist in the same container without reacting?
Someone asked if glucose and insulin could be mixed in the same container and used as an injection to provide a quick energy boost. I'm not interested about the medical part of this question, but just if glucose and insulin (which is a protein) could exist in the same container without reacting with each other without forming some sort of glycoprotein. Is it possible to make a reasonable prediction what would/not happen in the mixed glucose/insulin solution in the container?
I know that insulin and glucose do not react with in the body, I want to know if in aqueous solution they would react.

Glycoproteins come in O-linked and N-linked forms. For O-linked sugars, a glycosidic bond forms between the sugar and the hydroxyl of a serine or threonine side chain. For N-linked sugars, a glycosidic bond forms between the sugar and the amide of asparagine. The equilibrium for all these reactions lies on the side of hydrolysis in aqueous solution.

Is it possible to make a reasonable prediction what would/not happen in the mixed glucose/insulin solution in the container?

Yes, in aqueous solution there is no expectation that a glycoprotein forms between insulin and glucose. In vivo, glycoproteins form in a coupled reaction involving the carbohydrate (usually an oligosaccharide rather than a single glucose), the protein and a nucleotide triphosphate.
However, if an individual has high blood sugar levels (i.e. high concentrations of glucose in the blood) for extended periods of time, hemoglobin (specifically, the N-terminal amino group and the sidechain amino group of lysine) will react with the sugar and become glycalated. This is used in the HbA1c diagnostic test for checking if someone has diabetes. After the sugar binds (reversibly), there is a further reaction (Amadori rearrangement) that leads to a long-lasting product.

OK. Is there any enzyme required for the formation of glycoproteins in vitro ?

It would not form because the equilibrium lies on the side of the reactants.


text:
How fast is the interconversion triplet → singlet in organic molecules?
When computational studies are performed on transition state geometries, the "thermodynamic" (that is, with no kinetic considerations) energy of the analyzed state is obtained.
I was wondering: if a (multistep) reaction begins in a biradical state, the kinetics of the interconversion from triplet to singlet is so slow that the reaction is likely to proceed in a radical fashion, or if a lower transition state exists for the ionic (singlet) counterpart, then it is likely to proceed in a ionic way, because the rate of interconversion is extremely fast?
I am sorry if the question is dumb, but all I could find was about transition metals, and I don't know if the same applies.

It depends on the organic molecule.
When spin-orbit coupling facilitates intersystem crossing, it can be fast: 100 femtoseconds for nitronapthalene (viz. Chem. Eur. J. 2018, 24, 5379 – 5387).
The Wikipedia article on intersystem crossing cites a timescale for radiative decay (phosphorescence) of triplet to singlet of 10−8 to 10−3 s.

Spin-orbit coupling will cause the spin flip needed. This is always present but becomes larger the heavier the atom are, iodine, xenon for example present either in solvent (such as iodobenzene) or as part of the molecule. Paramagnetic species also work in the same way, e.g. O2, NO as does an external magnetic field because, as the molecule tumbles, the apparent changing magnetic field the molecule experiences causes T-S crossing.
How fast intersystem crossing is will depend largely on the energy gap between singlet and triplet; the larger the gap the slower the process, and  as the strength of spin-orbit coupling increases so does the rate constant. This 'energy gap law' has been confirmed by many experiments on excited states and of aromatic molecules and metal complexes and has the form $k=Ae^{-\beta\Delta E}$ where $A$ depends on the strength of the interaction and $\beta$ on the properties of the molecule, this varies as $\ln(\Delta E)$ so is almost constant as $\Delta E$ changes, typical values are 3 to 5.