text:
How is Raoult's law a special case of Henry's law and not vice versa?
I know that Raoult's law can be a special case of Henry's law when the value of the Henry's constant becomes equal to the vapour pressure of the pure state of the liquid. Why can't we say the opposite, i.e. Henry's law is a special case of Raoult's law?

Raoult's law is dealing with the equilibrium between the gas phase and a liquid phase of two liquids. Henry's law is dealing with a gas phase in equilibrium with gas dissolved in a liquid. I don't see how either is a special case of the other.

If we formally consider the Raoult scenario as gas(vapour ) dissolving in liquid, then these mixed solvents are special case of more general solutions the Henry law applies to.

@MaxW That's what confused me, because they seem unrelated to me. But it's an actual subsection in my textbook, where the both are compared and Raoult's law is considered a special case if the above mentioned criteria gets satisfied.

@Poutnik Can you please clarify what kind of general solutions would Henry's law apply to?

@Nandini Rather all solutions where Henry law is valid.

@Poutnik - Thanks for the correction.

Related: What are the key differences between Raoult’s Law and Henry’s Law

Raoult's law is a proposition regarding the relationship between the vapor pressure of a solvent (component 1 below) in its pure form ($p_1^\circ$) and when a solute (component 2) is present at a mole fraction $\chi_2=1-\chi_1$ (the vapor pressure then $p_1$):
$$\chi_1 = \frac{p_1}{p_1^\circ} $$
It is generally true for very dilute solutions because we know independently that
$$\frac{p}{p^\circ} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$
For some ideal binary systems Raoult's law is applicable over a broad range of composition. In fact, sometimes the solute also observes Raoult's law in dilute solutions:
$$\chi_2 = \frac{p_2}{p_2^\circ} $$
This however is exceptional behavior. It is more common that the solute observes the general Henry's law
$$\chi_2 = \frac{p_2}{K_H} $$
where $K_H$ is a proportionality constant for the solute-solvent pair. It should be evident that when $K_H=p_2^\circ$ the solute displays Raoult's law behavior even in dilute solution.
Henry's law is a general statement observed by solutes with a measurable vapor pressure $p_2$, since the following limiting behavior is expected as solute-solvent interactions far exceed solute-solute interactions in importance in the limit of zero solute concentration:
$$\frac{p_2}{K_H} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$
However, there is no principle that guarantees that $K_H=p_2^\circ$. Again, for a substance to follow Raoult's law at low concentrations is regarded as exceptionally ideal behavior.

Hi, I do not understand why $\frac{p_2}{K_H}\to 1$ as $\chi_1\to 1$. Shouldn't we, by Henry’ Law, have $\frac{p_2}{K_H}\to 0$ because $\chi_2\to 0$?

@insipidintegrator $\chi_1\to 1$ and $\chi_2\to 0$ are equivalent for a two component solution since $\chi_1+\chi_2=1$

No wait so we should have $\frac{p_2}{K_H}\to 0$ right? I understand they  are equivalent, that’s why I asked. Because $p_2\propto \chi_2$


text:
Is bleaching powder same as Sodium Hypochlorite?
i want to know is Sodium Hypochlorite is same as bleaching powder.
can some one please let me know.

The main ingredient in any bleach is hypochlorite. So, sodium hypochlorite, potassium hypochlorite, calcium hypochlorite all are bleach.

can e use bleaching powder for covid if not getting Sodium Hypochlorite ?

Please don't do anything like that..

Your comment doesn't make sense. Didn't you understand my comment above? Check the composition of your "bleaching powder". If it has hypochlorite, it can be used as bleach.

The hypochlorite is the bit with the bleaching power, the metal ion is not important

Historically, 'Bleaching Powder' has referred to Calcium hypochlorite, Ca(OCl)2. To quote from Wikipedia:

Bleaching powder (formerly known as "chlorinated lime"), usually a mixture of calcium hypochlorite (Ca(ClO)2), calcium hydroxide (lime, Ca(OH)2), and calcium chloride (CaCl2) in variable amounts.[16] Sold as a white powder or in tablets, is used in many of the same applications as sodium hypochlorite, but is more stable and contains more available chlorine.

It was prepared by the action of chlorine on moist lime, hence the products listed.
$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl }$
$\ce{Ca(OH)2 + 2 HOCl <=> Ca(OCl)2 + 2 H2O }$
$\ce{Ca(OH)2 + 2 HCl -> CaCl2 + 2 H2O }$

Complement to what Ajkoer states. The only known solid substance containing Hypochlorite ions is Calcium Hypochlorite. Sodium hypochlorite can not exist in the solid state. It only exists in solution. If you try to evaporate it, it gets decomposed.

can we use Bleaching powder instead of  sodium hypochlorite for covid @Maurice

@SMG, the question is where do you want to use it?

yes  for this only can i do it @Maurice

for sanitation  @Maurice

@SMG, the question is where do you want to use it?

for sanitation in my housing

@SMG, Please be more specific. Housing can include everything. Very dilute bleach is only safe in the laundry and washrooms and sometimes mopping the floor, if the floors are compatible. There is no other use. Do not spray, do not use on any other stuff and avoid skin and eye contact.

You may use calcium hypochlorite for use indoors. It does not make any difference whether you use sodium or calcium hypochlorite. The only advantage of sodium salt is the fact that sodium hypochlorite can be easily washed away with plenty of water. Calcium hypochlorite contains always some chalk that is not soluble in water. This will let white spots on the surfaces after washing. On the other hand, calcium hypochlorite is more concentrated as it is solid. You will need a smaller volume of calcium hypochlorite, with respect to bleach.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Forming AgCl from electroless Ag
We are looking to design flexible substrates with electroless plated silver on them (for use in sensor applications).  We need convert the surface of the silver to silver-chloride (AgCl) for conduction to hydrogel/hydrocolloid.  What are simple ways to do this.  The material will be on rolls and we need a simple production method that can be applied to the rolls.  Thank you.

We can't give you a one paragraph magical answer. You need to hire an industrial chemist or firm, to guide you through whatever you're trying to do.

@MaxW No one is expecting a single "magic" answer.  But they are looking for a discussion of the issues.  That is the whole point of StackExchange.

It has been my experience  that inventors don't want to share pertinent details of their process which are needed to solve the problem unless a NDA has been signed. Just throwing answers on the wall and hoping one sticks is futile.

I would recommend this patent which generates AgCl via the electrolytic chlorination of sheet silver, to quote:

The present invention relates to a method for chlorination of silver materials in the forms of wire, thick film paste and metallic sheet/disc. The method is environmentally friendly since the process does not produce hazardous chemicals or flammable gas. The electrochemical setup includes position the electrode containing silver materials in the anode and a platinum wire at the cathode. A DC voltage supply between 1 to 5 volts oxidizes the silver surface which receives chloride from imidazolium chloride solution to produce silver chloride on silver wire, film or disc. The imidazolium chloride reagent can be in aqueous solution or polar organic solvents or naturally exists in ionic liquid form. The imidazolium radical cation decomposes to produce a stabilized organic radical and an imidazole molecule. The stabilized radical recombines to produce volatile organic compound that can be recovered by simple distillation. The chlorination reagent can be regenerated...

In place of platinum wire perhaps a diamond electrode.
Also, some pertinent comment from atomistry, to quote:

The rose-red to purple-brown mixtures of silver subchloride and silver chloride are called by Carey Lea " photochlorides." They are formed by the interaction of colloidal silver and a suspension of silver chloride; by the action of organic reducers and of light on silver chloride; and by the electrolytic chlorination of sheet silver. The photochlorides are characterized by their great sensitiveness to light, blue light producing a blue coloration, and red light a red coloration. This action is reversible, the blue coloration being transformed into red by the action of red light, and so on. This phenomenon is inapplicable to the production of colour-photographs, for white light causes a darkening in colour.

I really don't think that pulling such a answer out of thin air can be useful. There are just a zillion pertinent unanswered questions about the OP's process.

On the contrary, the above response is quite helpful.

@AJKOER  Thank you for the excellent answer!  This is exactly why StackExchange is so powerful!

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
The IR/ H-NMR analysis of benzophenone?
I think I have figured out the IR spectrum for benzophenone. Would it be that the sharp peak in the middle (1700 cm^-1) is for the ketone and the little peaks pass 2900 cm^-1 rang are the sp2/sp3 carbons from the benzene rings?

Also, I have an H-NMR spectrum for the same compound and I know there are only three peaks because the compound is symmetrical. But I am confused on how to determine the J values for each hydrogen.

Can someone kindly assist?

If you print expanded portion with the resonance in Hz, you can easily calculate J values.

Can you explain what you mean?

Your peaks in the main spectrum is in $\pu{ppm}$. You have to print it again in $\pu{Hz}$ (there is a option to do so in printing software). Then expand it to be more clear. It is hard to read and isolate in compact mode.

Oh...I had to screenshot this photo because I was unable to save it correctly especially since it was given to me as a cropped photo. Because of this, I was wondering if I can figure the J values another way

Do you know the power of spectrometer? Such as whether it is $\pu{400 MHz}$ or higher or lower, but exact number?  If so, each $\Delta\delta$ in $\pu{ppm}$ can be multiply by that numerical value (e.g., $400$) to get approximate $J$ in $\pu{Hz}$. This calculation is based on this: For instance, for a spectrum in $\pu{400 MHz}$, $\pu{1 ppm} = \pu{400 Hz}$.

I'll ask because that picture was all I was given.

Why do you want the J couplings? The assignment of all peaks is totally clear without, and the two triplets are higher order peaks anyway, so analysing the coupling is not trivial.

In the ir there is a CO, and many CC and CH vibrations and of these the CH might be expected to have the highest frequency.


text:
Determining amount (in mols) of oxygen needed to fully combust coal
I'm trying to calculate the amount of oxygen needed to fully combust some coal.
As an example, assume I have $\pu{1 kg}$ of coal which is, by mass fraction, 60% carbon ($\ce{C}$) 40% hydrogen ($\ce{H}$). I know this is unrealistic but it's just for the sake of the example.
There is then $\pu{0.6 kg}$ of $\ce{C}$ and $\pu{0.4 kg}$ of $\ce{H}$. I can convert to $\pu{mol}$ by using the MW of both species, $\pu{0.012 kg/mol}$ for $\ce{C}$ and $\pu{0.001 kg/mol}$ for $\ce{H}$. Which means I have $\pu{50 mol}$ of $\ce{C}$ and $\pu{400 mol}$ of $\ce{H}$.
The reaction stoic for the oxidation of both species is as follows:
$\ce{C + O2 -> CO2}$
$\ce{4H + O2 -> 2H2O}$
Would it then be correct to say that I need $\pu{100 mol}$ of $\ce{O2}$ to combust the $\ce{H}$ and $\pu{50 mol}$ of $\ce{O2}$ to combust the $\ce{C}$?

It should be indeed "just for the sake of the example," because your example is not realistic ($\ce{CH8}$ empirical formula is not real). Nonetheless, your calculation is acceptable.

Why don't you use reasonable examples ?

It would be too tedious to type out the 10 different mass fractions of the real sample and list all of the reactions happening (plus I bet no one would read it). I just wanted to check my understanding really quick. I don't do this stuff very often.

Your calculations look broadly correct. But your assumption for what coal is looks way off. and you calculations also assume complete combustion which is rarely true for coal except under ideal conditions. Coal fires, for example, produce a lot of CO from incomplete combustion.


text:
How does the purification until analytical grade work?
What are some industrial methods used for puryfing organic solutions, like toluen, until it is analytical grade? Im not looking for details just some names and somewhere to keep reading

Multiple distillations


text:
Resonance of diazomethane
I understand the diazomethane resonance forms that have two double bonds between $\ce{C=N}$ and $\ce{N=N}$ in the structure and also the form that has one single between $\ce{C-N}$ and triple between the two nitrogens ($\ce{N#N}$) in the structure.
Why is not possible for diazomethane to have a double bond between $\ce{C}$ and $\ce{N}$ as well as a triple bond between $\ce{N}$ and $\ce{N}$ as a resonance form?

I guess you could write it, but I would be cautious with suggesting hypercoordination.

Are you suggesting a resonance structure with 10 electrons on a nitrogen atom?

As OP suggested there are three resonance structures of diazomethane contributing to the structure:

If you want to neutralize the formal charges of last resonance structure, you can put a double bond on negatively charged carbon and adjacent positively charged nitrogen to get the fourt resonance structure. However, traditionally we don't do that because it make the middle nitrogen hypervalent (or pentavalent nitrogen). On other word, it gave that nitrogen 10 valance electrons in its outer cell, which can't have in a 2nd row element.
It's just like resonance structures of nitrate ion, in which central nitrogen contains a full positive charge in all three resonance structures and two of three oxygen atoms contain full negative charges. So far, every one accept those resonance structures without questioning hypervalency of the central nitrogen atom until recently.
In 2015, Durrant has published his work on quantum mechanical calculations, suggesting possibility of hyper valency in diazomethane and some other possible molecules (Ref.1). According to its abstract:

From the inception of Lewis' theory of chemical bonding, hypervalency has remained a point of difficulty that has not been fully resolved by the currently accepted qualitative definition of this term. Therefore, in this work, a quantitative measure of hypervalency has been developed. The only required input is the atomic charge map, which can be obtained from either quantum calculations or from experiment. Using this definition, it is found that well-known species such as $\ce{O3, \color{blue}{CH2N2}}$ and $\ce{ClO4−}$, are indeed hypervalent, whilst others such as $\ce{XeF4, PCl5}$ and $\ce{SO4^2−}$, are not. Quantitative analysis of known species of general formulae $\ce{XF_n^m−, XCl_n^m−}$, and $\ce{XO_n^m−}$ shows that there are no fundamental differences in chemical bonding for hypervalent and non-hypervalent species. Nevertheless, hypervalency is associated with chemical instability, as well as a high degree of covalent rather than ionic bonding. The implications for accepted Lewis structure conventions are discussed.

The experimentally determined atomic charges from electron densities obtained by X-ray crystallography have been compared with theoretical values obtained by QTAIM analysis of the output from quantum calculations. In order to calculate the overall electron count at the $\ce{X}$ atom, the author has defined a parameter called the valence electron equivalent, $\gamma$, as ‘the formal shared electron count at a given atom, obtained by any combination of valid ionic and covalent resonance forms that reproduces the observed charge distribution’. Mathematically:
$$Q(\ce{X}) = \displaystyle \sum_{i=1}^n{a_iq_i(\ce{X})}; \ \text{and then, } \ \gamma(\ce{X}) = \displaystyle \sum_{i=1}^n{a_i\gamma_i(\ce{X})}$$
where $Q(\ce{X})$ and $\gamma(\ce{X})$ are the overall charge and valence electron equivalent of atom $\ce{X}$, $q_i(\ce{X})$ and $\gamma_i(\ce{X})$ are the formal charge and electron count of each contributing resonance form, and $a_i$ are the coefficients required to reproduce $Q(\ce{X})$.
According to the paper:

It follows that for any given atom $\ce{X}$, if $\gamma_i(\ce{X}) = 8$, the atom obeys the original Lewis octet rule. If $\gamma_i(\ce{X}) \lt 8$, the atom obeys the ‘modified octet rule’, but if $\gamma_i(\ce{X}) \gt 8$, neither form of the octet rule is obeyed and the atom is hypervalent. In such a case, no combination of formally charged Lewis octet-compliant structures can reproduce the observed charge distribution and it is necessary to include a contribution from an ‘expanded octet’ structure.

In the case of central nitrogen atom of diazomethane, $\gamma_i(\ce{N}) = 10$ ($ \gt 8$), thus it is considered to be hypervalent (Note: for nitrate ion: $\gamma_i(\ce{N}) = 8.30$ and hypervalent as well). The author has compared the results of $\gamma$ calculations on a variety of species including diazomethane, using quantum method 5 plus the QTAIM, Natural Bond Orbital (NBO), Hirshfeld, and Mulliken charge models. All of these methods agree with the findings, giving value of $\gamma_i(\ce{N})$ of diazomethane $10, 9.91, 9.78,$ and $9.56$, respectively.
References:

M. C. Durrant, "A quantitative definition of hypervalency," Chemical Science 2015, 6, 6614-6623 (https://doi.org/10.1039/C5SC02076J).


text:
The connection of the "R" in the ideal gas law and osmotic pressure
I am studying chemistry and biology at the same time, and have encountered a problem. In the ideal gas law, the $R$ is the ideal gas constant, and its value is $\pu{0.0821 L atm mol-1 K-1}.$
However, in the osmotic pressure formula $\Psi_s = -iCRT,$ the $R$ is called the pressure constant, and its value is $\pu{0.0831 L bar mol-1 K-1}.$ Is it a coincidence that their values are identical (after adjusting for the $\pu{atm}$ and $\pu{bar}$ conversion)?

For R in all sorts of units see https://en.wikipedia.org/wiki/Gas_constant

No coincidence, and you’ll encounter it in other contexts as well. A bit like pi and e in maths.

R exists because we measure temperature in different units than energy. If you measure temperature in joules/mole, R disappears.

It's not a coincidence at all!  If you do an online search for "derivation of osmotic pressure", you'll see how $R$ enters into the derivation.
Indeed, that's one of the beauties of the van 't Hoff equation for osmotic pressure – it reveals that (under the modest simplifying assumptions of the van 't Hoff equation derivation) the osmotic pressure created by a dilute solute is equal to the pressure that solute would exert if it instead acted as an ideal gas!:
$$\text{van 't Hoff equation:  } \Pi_i = \frac{n_i}{V} R T = C_i R T$$
$$\text{ideal gas equation:   } p_i = \frac{n_i}{V} R T = C_i R T$$
Here the "$i$" refers to species $i$.


text:
if we mix bleaching powder in water will this be converted to Sodium Hypochlorite?
if we mix bleaching powder in water will this be converted to Sodium Hypochlorite ?

by the way, how does a mixed salt of calcium hypochlorite, calcium chlorite, calcium chloride and water ( as quoted by NCERT, but generally hypochlorite being the actual active component), which doesn't have any sodium, suddenly become sodium hypochlorite?

actually, both are oxidising agents which provide nascent oxygen atoms. so, both have similar action and hence, bleaching powder is sometimes also used instead of sodium hypochlorite, just that, I think, sodium hypochlorite is a bit more active than bleaching powder when both are taken in the same weights.

Bleaching powder is made of Calcium Hypochlorite mixed with some other Calcium compounds in various proportion. It does not contain any Sodium compounds. So in water it will not be converted to Sodium Hypochlorite.


text:
Meaning of hydrogen bonds for life
Why are hydrogen bonds so important for life? Or more generally asked: Why is water so important for life / biological systems? Or maybe again in other words: Why does especially water provide this special property? And not a single other substitute(?) ?
I can read what the hydrogen bonds cause to happen but I would rather like to know: Why is that in detail or what makes so water so much different from any other (similar?) liquid?
I know this touches heavily biology (or also physics) but I think I will find the reason in an chemical explanation?

This is a huge question without a simple answer.  H bonds help to hold DNA/RNA and proteins together and so give them structure. Also it may be that it is the polar nature of liquid water (as a large dielectric constant) that is important as  this forces many types of molecules to form bilayers (rather than dissolving) and hence vesicles and cells which compartmentalise different molecules.

Hydrogen bonds sure do make supramolecular chemistry more accessible, and life is all about supramolecular chemistry.

Water is (almost) everywhere on earth in copious amounts. It is liquid at earth conditions (H-bridges!). It's rather non-toxic (or organisms were able to evolve to tolerate water well). It is a good solvent for the molecules involved in energy metabolism and biostructure ana- and catabolism. Water has neutral pH, it happily donates and accepts protons which makes it ideal for participation in metabolism.
I'd broaden the question, asking for other molecules that could support life as we know it in a similar way like water does it and also check other liquid molecules which are available in large quantities in the universe (e.g. ammonia, nitrogen, hydrogen) if they could support other forms of life that depend on analogous chemical processes, or even go beyond that (silicon- and aluminum oxides, etc.)

The idea that other liquids could support "other forms of life" is science fiction not chemistry as nobody has ever showed how any non-carbon, non-water based chemistry could come remotely close to the complexity required for life.


text:
Chemical shift difference methyl groups in 4-hydroxy-2,5-dimethyl-3-hexanone
Can someone explain why the NMR Spectrum of 4-hydroxy-2,5-dimethyl-3-hexanone shows a chemical shift difference for the methyl groups at the end? Why aren't the other carbons have a difference?
Below you see the NMR spectrum:

There are four methyl groups and four relevant peaks. What is the problem?

C-2 is a chiral centre, so both C-5's are diastereotopic, likewise with both C-6's. https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-08-symmetry.htm

Thanks, the website really helped to understand the issue.


text:
Are SN1 reactions pseudo unimolecular first order reactions?
I believe the the Solvent (present in excess) helps in ionisation of R-X to carbocation in Rate Determining step so shouldn't the Rate Determining step be a pseudo-unimolecular reaction as 2 molecules are involved but solvent species has no role in determining the rate?

Actually your concept of molecularity is applicable to only simple elementary reactions where the substrate is not so complex. This fact, hence can't be determined theoretically, as far as my knowledge goes( I'm still an undergraduate ) and hence the experimentally determined result, the rate of reaction is directly proportional to concentration of substrate should be accepted, in my opinion.

Actually your conception of solvent presence influencing the rate of reaction is more of an SN2 reaction concept where the solvent forms a complex with the cationic part of the attacking compound and helps it ionise ( as the solvent is polar aprotic solvent in SN2 reaction kinetics). But this exactly may not be the reason why it is bimolecular.

The mechanistic approach, in my opinion should be used only for reactions with simple substrates, when you'd like to determine the molecularity of a reaction.

@user93057 I am not talking about the solvent influence on rate by freeing the nucleophile like in SN2 rather in SN1 solvent doesn't play any role in rate of reaction but its involved in RDS (polarising the R-X bond further as shown in the picture) so I was wondering whether it would be correct to term SN1 as a pseudo-unimolecular reaction as there are many molecules playing a role in RDS but still a first order reaction.

this link may help: https://www.quora.com/Chemical-Kinetics-Can-SN1-reactions-be-considered-as-pseudo-unimolecular-reactions


text:
In a double displacement reaction, why don’t the two products reform after being dissolved in water?
Let’s say I have $\ce{AB (aq) + CD (aq) —> AD + CB}$. When AB and CD are dissolved in water, they get dissociated into their component ions. But why don’t they reform as AB and CD, if it is already known they (A and B, C and D) can form a chemical compound with each other? Why must the ions from each compound combine with ion(s) from the other compound?
By nature, dissociation in water causes ions from within an ionic compound to separate. When the ions are separated from their respective compounds, they are surrounded by water molecules to prevent them from recombining. But I’m guessing this blockage somehow doesn’t apply to to the relationship between the ions of different compounds.

https://chemistry.stackexchange.com/questions/76147/what-are-some-ways-to-predict-that-a-double-replacement-reaction-will-occur

If one dissolves both AB and CD in water, they are split into their ions i.e. A+, B-, C+ and D-. As long as all ions remain in aqueous solution, thus no precipitation, they don't really exist as AD + BC in the solution. The solution is just a homogeneous mixture of ions at this stage. The difference between reactants and products comes after the species are removed from the solution, e.g. by precipitation.
Then, why is AD and BC formed on precipitation, rather than the reactants?
This is because AD and BC are more stable than their reactant counterparts. There are several reasons why this could be the case. One possible explanation is HSAB, Hard Soft Acid Base Theory. In general, the larger of the anions pairs with the larger of the cations and vice versa in this method. However, bear in mind that reactions very rarely give 100% yield, and there is likely to be at least traces of the original reactants left over at the end.

I see, that really helps clarify things a bit. (Again, sorry for the late response). Thank you so much!

If the four substances AB, CD, AD and CB are all soluble in water, there is no such reaction in water. If you dissolve AB and CD, or independently AD and CB, you will obtain a solution containing the ions A+, B-, C+ and D-. That is all you can say. There are no pure substances any more. And it does not depend on the original choice of the substances, either AB + CD, or AD + CB. The sense of the reaction can be known when evaporating the solution, because AB, CD, AD and BC are not equally soluble. The least soluble of the four will first crystallize, and there is no way of predicting which one it is. The solubilities cannot be calculated. It may be one of the reagents, or one of the products. The result may be a chemical reaction, or not. Is it what you wanted to know ?

Goodness my reply comes so late... so if there is a reaction, one of the products will crystallize?

For example, if you dissolve some KBr and NaCl in water, there are no more KCl or NaCl in the solution. There

For example, if you dissolve some KBr and NaCl in water, you will obtain a clear solution and there are no KBr and no NaCl any more in solution. There are only Na+, K+, Cl- and Br- ions.  And the same solution could have been done by dissolving NaBr and KCl and mixing. But, if one of the four possible compounds is insoluble, it will produce a deposit at the bottom of the flask, and this would be a chemical reaction. This would happen if you replace NaCl by NaClO4. When mixing NaClO4 solution with a KBr solution, a precipitate will happen, because KClO4 is sparingly soluble in water.

I didn’t know that crystallization could also refer to precipitation in a solution. Many thanks for your response!


text:
Reaction Mechanisms: Free Radical Halogenation
In Free Radical halogenation using chlorine and sunlight, why does the chlorine-free radical get attached to the allylic carbon?
Will it be correct to say that it follows the anti-Markovnikov rule?
But if so, there is no $\ce{H2O2}$ present...
The substrate can be considered as but-2-ene.

What's the substrate for free radical halogenation?

@zenix but-2-ene

This is not $\ce{HX}$ addition to an alkene. Therefore, Markovnikov rule does not apply here. You should change your question accordingly.

More specifically, this is not HBr addition in the presence of peroxide (Kharasch reaction). HCl and HI do not work in the Kharasch reaction.

H. C. Brown has suggested that the halogen atom does add but, at high temperatures, is expelled before the second step of free radical addition can occur$^1$.

Addition of halogen atom gives this radical, which falls apart to regenerate the starting material if the temperature is high or if it does not soon encounter a halogen molecule to complete the addition. Hence, low concentration of halogen can be used with high temperature to favour substitution over addition$^1$.
The allylic radical, on the other hand, once formed, has little option but to wait for a halogen molecule, however low the halogen concentration$^1$.

For the formation of minor product you may look into this question.
To follow the Markovnikov's or anti-Markovnikov's rule, there should be difference of  electronegativity between the two atoms. Therefore, I do not understand what you are trying to ask.
Reference

Organic Chemistry- Morrison and Boyd (6E); Page 389-390

The first step of the allylic mechanism should have the half arrow going to the chlorine radical.

I would appreciate a reference to H. C. Brown's suggestion.

@Zhe It's done now. Can you please verify the linked question?

@Apurvium Sorry, but I don't see a fix, and the linked question has the same problem. The product from step 1 does not correspond with the arrows (the product based on arrows are the allylic radical, hydrogen radical, and chlorine radical, with no new bond formed between chlorine and hydrogen).


text:
Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?

Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?

I know that dipole moment depends upon the strength of dipole and angle between the dipoles.
1,1-Dichloroethylene has two chlorine atoms with 120° angle between them because of $\mathrm{sp^2}$ hybridization of carbon atom.
cis-1,2-Dichloroethylene has the same with the angle between chlorine atoms as 60° (or else the dipole moment of trans form would not have been zero).
So, the dipole moment of cis should be greater than that of 1,1-dichloroethylene, but the answer says the opposite (μ(1,1-dichloroethylene) > μ(cis-1,2-dichloroethylene)). Can anyone help me?

The 1,1 compound has the larger dipole. The dipole is charge times distance, so as well as calculating the component of the charge along each axis you also have to consider the distance apart of the charges. The total is the sum of each part.

Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:

For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$.
Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.
Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.
Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.

+1 I like that you actually showed the vectors broken down into perpendicular components and did the math.

Yes, you are correct. And I even checked the values for confirmation, but the book states the reason that u11>u12 to be resonance. They state that in u11 due to resonating structures, there will be charge involved in dipole moment, whereas not that effective in u12

cis-1,2-Dichloroethene has the larger dipole.
cis-1,2-Dichloroethene,  1.9 D
The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 1.9 D.

1,1-dichloroethylene, 1.3 D
The charge separation is on one carbon atoms with ability to get electrons from double bond and distribute some of the charge to the other carbon atom. Vector addition gives the net dipole of 1.3 D.

trans-1,2-Dichloroethene,  0 D
The charge separation is on two different carbon atoms. Vector addition gives the net dipole of 0 D.


text:
Symmetry Adapted molecular orbitals of Formaldehyde
I am currently trying to find the Symmetry Adapted linear Combinations (SALCs) of Formaldehyde $\ce{H2C=O}$, and use the  1s orbitals on the hydrogen atoms, and 2s, 2px, 2py and 2pz orbitals on the carbon and oxygen atoms to construct molecular orbitals. However I am finding some issues.
I understand how to work out the SALCs of the hydrogen S orbitals, which is $A_1 = H_{1s} + H_{2s}$ and $B_1 = H_{1s} - H_{2s}$. However, I don't understand how figure out the SALCs of the orbitals belonging to oxygen and carbon.
I can only guess the combinations by drawing them, and not through calculation, like I can with the hydrogen S orbitals as these can interchange. This isn't the case with the oxygen and carbon orbitals.
Is there a way of finding this through calculation? Or do I just draw out all the possible combinations?

SALCs are only relevant for atoms that are symmetry-related. The two H atoms in CH2O are related by symmetry (interchangeable via C2 or a mirror plane). The C and the O aren't related at all.

@orthocresol Thank you so much! I realise there isnt a delete button for my obvious question

No worries. It's not stupid at all. You can most certainly leave it up.

There is an example calculation here https://chemistry.stackexchange.com/questions/118342/molecular-orbitals-only-pi-system-of-allyl-system-using-group-theory/118346#118346

If you want to know how the individual orbitals transform, the "linear" and "quadratic" part of your character table can help. For example, for the O orbital that forms the pi bond (the $2p_y$), the $C_{2v}$ character table tells you that it is $B_2$. Similarly the C $2p_y$ orbital is $B_2$. The $2p_z$ orbitals are $A_1$ and the $2p_x$ are $B_1$. You can use these symmetry labels to determine which orbitals may mix to form molecular orbitals.


text:
Why does the amount of CO2 produced decrease with more baking soda?
I'm in chemistry class and we have an assignment and they gave us a video and I took down the data they gave us. They fist measured out 10mL of vinegar for each trial and used different amounts of baking soda in all 8 trials. They put the baking soda in a beaker then poured the vinegar into the beaker and capped it with a one hole stopper and a syringe in the hole. It was about a 1-1.5 second delay between poring the vinegar and capping it. I don't understand why the amount of CO2 produced wouldn't increase at the beginning then flat-line when the vinegar becomes the limiting reactant. I posted a graph of the data we received. All trials had the same amount of vinegar.

Well, you know on what it depends if any gas will evolve is you add some acid?

@ Logan. I agree with your surprise. Your results are difficult to understand

Not really, if there's too little acid, then pH won't be low enough for any gas evolution.

@Maurice There shouldn't be any space after @ for pinging, it wasn't needed this time though https://chemistry.meta.stackexchange.com/questions/3889/pinging-in-comment-to-suggest-answering-question

@Mithoron. First : What do you mean by "pinging"?  Second, there is never "too little acid". Logan states that they always use the same amount of acid.

@Maurice Is what you just did to bring my attention back here. Just read the post I linked. And think how pH is related to decomposition of carbonates.

@Mithoron. I understand pinging is the way of using the @ sign if somebody wants to ask specifically the author of one of the previous messages. Second I know how carbonates are decomposed vs. pH. But here it is not a carbonate. It is a hydrogenocarbonate : NaHCO3.

This graph is a very good example of experimental errors which may give rise to a pseudo-phenomenon i.e., an observation which apparently looks like a scientific phenomenon but it is an artifact of the experimental design.
First of all work out the moles of acetic acid (5% acetic acid) and moles of NaHCO3. As you can see vinegar has become a limiting agent. And it should have become a flat curve only if your system was a closed system.
So where is the culprit in the experimental design?
As you stated "It was about a 1-1.5 second delay between poring the vinegar and capping it."
By the time, someone puts a syringe, a large volume of gas has already escaped when acetic acid was poured on a heap of sodium bicarbonate. How big is the heap of 3 g NaHCO3!
There is a nice trick to avoid these problems. You put the sodium bicarbonate in an open plastic pouch so that acid does not immediately encounter the salt. You can stopper and then shake to mix (quite well).
So your hypothesis is right for a flat curve, sadly there is an experimental design issue.


text:
Why do azeotropes boil at constant temperature?
As given in the Wikipedia page,

An azeotrope (/əˈziːəˌtroʊp/) or a constant boiling point mixture is a mixture of two or more liquids whose proportions cannot be altered or changed by simple distillation.

It further goes on to give the reason;

This happens because when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture.

But why? Why do they have this special property? Why don't substances normally have the same composition in solid and liquid phase?

It is a strange property, that nobody can really explain. Azeotropic mixtures behave like pure compounds. But they are not pure. The composition of azeotropic mixtures change with pressure.

At the azeotropic point, the solution components are boiling off at the same rate proportional to their concentration. I.E. the composition of the vapor is the same as the composition of the solution.

Functionally it's no different than pipetting out an aliquot of solution from the initial bulk stock - boiling just uses a phase change to achieve the same result. Does the concentration of a bulk solution change when you partition out an aliquot? Certainly not. So too does the concentration not change when you effectively do the same thing via vaporization.

Your answer is correct. But it is not related to the question. The question is why ?Usually when heating mixtures of volatil liquids, the vapors do not have the same composition as the liquid. Azeotrops are exceptions. The question is : why ?


text:
What is the degree of unsaturation of oxalate anion?
I was practicing a DPP for JEE Mains & Advanced exam. There is a question in organic chemistry, a reaction sequence was given whose answer was oxalate anion.
But the next question was what is the degree of unsaturation of the compound formed in the previous question .
I simply did it 2 because there are 2 C=O bond but the answer was given 3. Please help me in making out, why it is so.

Please be more specific how you have calculated the degree of unsaturation. in that way we can see your basis for the problem.

The Degree of Unsaturation can be considered as the number of Pairs of Hydrogen Atoms that could be added to the structure. One can simplify the structure to consist of two carbon-oxygen double bonds and two carbon-oxygen single bonds in one resonance contributor. Note that the single bonded oxygens have a negative charge and thus could accept a hydrogen each. This accounts for 1 degree of unsaturation. There are also two double bonds, and a pair of hydrogens could be added across these also. This gives two degrees of unsaturation. Hence there are 3 degrees of unsaturation in total (as the book says).
In order to calculate the degree of unsaturation for any compound, one can use the formula:
((2* Number of Carbons) + 2 - (Number of Hydrogens and halides) + (number of nitrogens)) / 2.
In this case, the formula gives 3, which is the degree of unsaturation.

Then you mean any anion would atleast have a degree of unsaturation ?

In principle, for any negative charge there is half a degree of unsaturation. So a double negative charge is one degree of unsaturation. If it was a single negative charge, that would be half a degree of unsaturation. However I think it is mostly applied to organic compounds.

What is a pair of hydrogen atoms in this context?  You can't really add three pairs of hydrogen atoms; to get the (theoretical) saturated compound $\ce{C2H6O4}$ two of the hydrogn atoms must be cations.  That shouldn't count since those do not bring in electrons.   Seems more realistic to require maintaining the charge, so the saturated species would be $\ce{C2H4O4^{2-}}$ and the degree of saturation would be two.


text:
Why can't two 3° free radicals combine?

In Kolbe Electrolysis, my teacher told that when a 1° or 2° free radical is used, we get alkanes through dimerisation. Why wont dimerisation occur for a 3° one??

Why have you drawn ester? Kolbe electrolysis is of sodium salt of an acid. I guess, we no longer use "°", for primary, secondary, tertiary radicals/other intermediates..

Question is: Why can't two $3^\circ$ free radicals combine (during Kolbe Electrolysis)?

The simplest answer is there is no reason why not. Here is the counter example to show it is possible and achieved as far back as 1959 Ref.1). I attached the available PDF file for convenience. There are several trisubstituted succinic acid derivatives, which are achieved by the dimerization of corresponding disubstituted malonic acid derivatives after decarboxylation during Kolbe Electrolysis.
Although the intermediate in this reaction is not technically a $3^\circ$ free radical but $2^\circ$ free radical with really bulky group ($\ce{CH3O2C-CH^.-CH2Si(CH3)3}$), I think this paper (Ref.2) is worth considering. Yet, some examples in similar vain (e.g., $\ce{CH3O2C-CH^.-CH2C(CH3)3}$) are also discussed in Ref.1. Eberson has continue his work with Kolbe Electrolysis to present some other example of tetraalkylsuccinc acid derivatives (Ref.3).
These examples are thoroughly reviewed by Schäfer in 1990 (Ref.4).
Other than Kolbe Electrolysis, there are some other examples of dimerization of $3^\circ$ free radicals have been reported. For instance, synthesis of 2,2,3,3-tetramethysbutane has been undergone with similar mechanism. The compound can be obtained by reaction of Grignard reagent, tert-butylmagnesium bromide with ethyl bromide, or of ethylmagnesium bromide with tert-butyl bromide in the presence of manganese(II) ions (Ref.6). The mechanism of this transformation is believed to be undergoing through the dimerization of two tert-butyl radicals, which are generated by decomposition of the organomanganese compounds generated in situ such as:
$$\ce{(CH3)3C-MgBr + MnCl2 -> (CH3)3C-MnCl + MgBrCl}$$
References:

Lennart Eberson, “Studies on Succinic Acids I. Preparation of $\alpha,\alpha'$-Dialkyl- and Tetraalkyl-Substituted Succinic Acids by Kolbe Electrolysis,”  Acta Chem. Scand. 1959, 13, 40-49 (DOI: 10.3891/acta.chem.scand.13-0040) (PDF).
Lennart Eberson, “The Synthesis of Some Aliphatic Organosilicon Dicarboxylic Acids. III,”  Acta Chem. Scand. 1956, 10, 629-632 (DOI: 10.3891/acta.chem.scand.10-0629) (PDF).
Lennart Eberson, “Studies on Succinic Acids V. The Preparation and Properties of Diastereoisomers of Tetraalkylsuccinic Acids,”  Acta Chem. Scand. 1960, 14, 641-649 (DOI: 10.3891/acta.chem.scand.14-0641) (PDF).
Hans-Jürgen Schäfer, “Recent Contributions of Kolbe Electrolysis to Organic Synthesis,” Topics in Current Chemistry: Electrochemistry IV; Vol. 152, Eberhard Steckhan, Editor; Springer-Verlag: Berlin, Germany, 1990, pp. 91-159 (PDF).
M. S. Kharasch, J. W. Hancock, W. Nudenberg, P. O. Tawney, “Factors Influencing the Course and Mechanism of Grignard Reactions. XXII. The Reaction of Grignard Reagents with Alkyl Halides and Ketones in the Presence of Manganous Salts,” J. Org. Chem. 1956, 21(3), 322-327 (https://doi.org/10.1021/jo01109a016).


text:
Equilibrium involving both gas phase and aqueous phase
If I have been given an equation that involves a gas-aqueous phase equilibria, how do I write the equilibrium constant?
I have come across several questions that give equations having BOTH gas and the aqueous phase and I couldn't comprehend how to attempt them.
Should we find the concentrations of the gas too and use it for the equilibria?
But in several questions I was given partial pressure of the gas at equilibrium but I was never given the temperature to find concentration.
What is a proper way of representing the equilibrium constant?
Here is an example:
$$\ce{RNH2(g) + H2O(l) <=> RNH3+(aq) + OH-(aq)}$$
How do I write the equilibrium constant for this, if I have been given concentrations of $\ce{RNH3+(aq)}$ and $\ce{OH-(aq)}$ and the partial pressure of $\ce{RNH2}$ at equilibrium?

It would help if you would give an example of a specific problem.

Yes I have edited my question now

I dare say most equilibrium constant don't cite specific units because the appropriate units are implied implicitly. In the case of the reaction:
$$\ce{R-NH2(g) <=>[aqueous] R-NH3+ + OH-}$$
I'd write the mathematical equilibrium equation as:
$$K_{eq} = \dfrac{\ce{[R-NH3+][OH-]}}{p_\ce{R-NH2}}$$
and give the constant the appropriate units: $\pu{mol^2\cdot L^{-2} \cdot kPa^-1}$. (I just picked kPa as one of the zillion units for pressure. Use whatever units you were given the vapor pressure in: atm, torr, mm Hg, lbs/in^2, slugs/furlong^2 ... whatever)
This is obviously a "overall" equilibrium. According to Henry's law there should be an equilibrium between the vapor pressure of $\ce{R-NH2}$ and the concentration of $\ce{R-NH2}$ in solution. Since you have no way of calculating the concentration of $\ce{R-NH2}$ in solution you can't break the overall reaction into the two different equilibria.

There are no units in any equilibrium constant under any circumstances. The reason is that the activity coefficient is what truly defines the equilibrium constant, and activity is unitless. Typically, we just hide the fact that the pressures are measured with respect to a standard pressure of 1 atm and a standard concentration of 1 M.

So there are different ways to express equilibrium there is the Kc value which, as the c denotes, is based on the concentration. In a question that was giving you moles or molarity of a substance that is a gas, we can find the concentration and using the coefficients to express a Kc.
Kc = [products]/[reactants] - take note of the bracket which means it is concentration.
However, in questions when you are given partial pressure, you must express the equilibrium expression as a value equating to Kp. The K constant is particular to a given temperature which is why increasing or decreasing the temperature will affect your Keq, however, you do not need the value of temperature to write an equilibrium expression.
Kp = (Pa)^a(Pb)^b - these are pressures and must be written with a parenthesis.
The third kind of equilibrium is dealing with substances in aqueous, for example, ions. In this case, the K value is Ksp or solubility of products and similarly, you are finding the concentration of the ions that dissociate.
In most cases however dealing with Kc, gases, and aqueous solutions are treated the same meaning we can directly put them in the Kc equation for equilibrium. It is only pure solids and liquids that are not expressed as their concentration does not change.
I hope this helps!


text:
Way to create acidic/HCl gel?
Is there a way to turn azeotropic hydrochloric acid into a gel?
I am trying to remove barnacles from the bottom of a boat and normal hydrochloric acid works, but it is tedious because in liquid form it just runs off, so I have to keep applying it over and over, kind of wearing the barnacle away.
If I could turn the HCl into a gel then it could just be applied for a few minutes and remove the whole barnacle.

Just use proper (and standard) cleaning agents. I've never heard before that someone use conc. HCl for cleaning (and I heard of way more original methods).

It turns out somebody already answered the question in a different post:

Xanthan Gum can handle hydrochloric / muriatic acid (HCl) for a while,
  depending on how concentrated the acid is. Here is how I made an HCl
  gel to clean an old toilet that was heavily coated with lime scale...
I first dissolved a level teaspoon of xanthan gum into 1/4 cup of
  rapidly stirring water. After a few hours of letting it sit, with
  occasional stirring, I had a very thick, viscous goo. To make the HCl
  gel, I mixed one part of this goo with 1.5 parts of 31% hydrocholoric
  acid, by volume. This gave a gel of a suitable viscosity to get the
  lime dissolved off the toilet. I don't know how long this gel would
  last before the HCl broke down the gum, but my gel lasted several
  hours, which was enough to get the toilet clean.

If you are quoting another post it would be nice if you would include a link as reference.


text:
How to Visualize the crystal planes of calcite?
I've been struggling to visualize the structure of calcite. Most papers say it grows in a rhombohedron capped by (104) faces. However, I don't see why it would do this when looking at it using CrystalMaker

Can one, by looking at the unit cell, discern which crystal faces will be expressed? How?

Why crystals grow the way that they do is still a mystery. Calcite crystals of visual size have grown into more than 300 different forms of crystal structures.


text:
Why does HgCl2 react with HCO3^- but not with CO3^2-?
I have started learning Qualitative Salt analysis recently. When I was reading about tests to distinguish $\ce{CO3^2-}$ and $\ce{HCO3-}$, I came across a line in a book which states

$\ce{HgCl2}$ forms a reddish-brown precipitate of $\ce{3HgO.HgCO3}$ with $\ce{HCO3-}$ but no reaction is reported in case of $\ce{CO3^2-}$.

I am not able to understand why there will be no reaction of $\ce{HgCl2}$ with $\ce{CO3^2-}$.
It would be better if I can get a sound logic for this so that I can remember it better.

The problem is $\ce{CO3^{2-}}$ only exists in very basic solution where the predominate reaction is: $$\ce{Hg^{2+} + 2OH- <=>[aq] HgO + H2O}$$

Mercuric compounds do react with carbonate ions, but the precipitate does not contain any carbonate anion. The equation is _$$\ce{3Hg^{2+} + 2 Cl^- + 2 CO_3^{2-} -> 2 CO_2 + Hg_3O_2CO_3}$$ The precipitate $\ce{Hg_3O_2CO_3}$ is reddish brown.

@Maurice I am confused, how does the precipitate contain no carbonate ion and then you render it $\ce{Hg3O2CO3}$?

@MaxW As far as what I understand from your answer,all reactions of CO32- is accompanied by a competitive reaction with OH-.Then,why do certain reactions of CO32- occur even though they exist in very basic solution
For example: I have learnt that CO32- can react with AgNO3 to form a white precipitate of Ag2CO3.Here,why isn't Ag2O reported as the major product?Why is there a difference in product observed even though both Ag2O and HgO are both oxides that lie above the ΔG=0 reference line in the Ellingham Diagram?

@Maurice Do you mean the product is 2HgO.Hg2CO3?May I know the source from which you got this information?Any snapshot regarding this or link to an authentic source or page number of a standard book where this is mentioned would be very helpful.

Anirudh - I should have written "My guess is that ...". Obviously forming $\ce{HgO}$ in a carbonate solution would be a reaction of some sort. // I'd agree with your suspicion that $\ce{Hg^{2+}}$  doesn't react at all, in any way, with a carbonate solution as being wrong.

@Anirudh The formula I gave for the precipitate is $\ce{Hg_3O_2CO3}$ this can be considered as $\ce{ 2 HgO + HgCO_3}$. See Farooq's answer for a reference. I also found this data in the old Treadwell treatise "Analytical Chemistry"

@Maurice Will it exist as a double salt?

@Chem-Learner. Usually compounds like $\ce{Hg_3O_2CO_3}$ are not considered as double slt, because in the past oxides were not considered as salts. That is why they are simply described as basic salt. In the past a salt was defined as the result of a reaction between an acid and an oxide (or a hydroxide). Today a salt is often the result of an anion joined with a cation. In this cas the basic mercury carbonate may be considered as a double salt.

Anirudh, which book is teaching this stuff? Sorry, I can only find this discussion in 1890s and the latest in my favorite Mellor's Treatise from 1930s.
Coming to 2020:
After doing a simple acid test for carbonates, a nice safe & sound method to distinguish carbonate vs. bicarbonate salt is
(a) to react it with a neutral barium chloride solution. If the solution is turbid, carbonate is present, otherwise bicarbonate. Of course the methods fail with a mixture of the two.
(b) Sodium Carbonate-Phenolphthalein Test: phenolphthalein is turned pink by soluble carbonates and but remains colourless by soluble bicarbonates
Your textbook is not completely correct when it states that bicarbonate does not react with $\ce{HgCl2}$. For your reference from A treatise on chemistry, Volume 2
by Henry Enfield Roscoe, Carl Schorlemmer, 1913, pg 699

Is there a way to carry out test for HCO3- and CO32- when they are present as a mixture?Will CaCl2 followed by NH3 addition work?

Yes, you will have to do a acid-base titration with methyl orange as indicator. Which textbook are you using?

Oh Okay,I was given that if the white precipitate of CaCO3 is formed,then it contains HCO3- along with CO32- in my lecture notes.Can you elaborate on why the further acid-base titration is required?
I am using Concise Inorganic Chemistry authored by JD Lee 4th Edition along with my notes as my main reference.Can you suggest me any book where I can learn Inorganic chemistry with concrete logic and not just filled with reactions?

For qualitative analysis, A. I. Vogel A Textbook of Qualitative Analysis is the best https://www.amazon.com/Vogels-Qualitative-Inorganic-Analysis-7th/dp/0582218667. There is more more German book Lehrbuch der analytischen und präparativen
anorganischen Chemie which has a lot of good schemes for qualitative analysis. One can read relevant sections with the help of machine translation.

If you are willing to spend some money and find a lot of details, one of best books is Holleman and Wiberg "Inorganic Chemistry" in English. It will help you as a handy reference for life. More than > 2000 pages and highly highly detailed. It is the only book which has survived for 100 of years as a textbook from Germany.

Looked for Holleman and Wiberg "Inorganic Chemistry" in English -- Couldn't find it anywhere, even as a used book.

Check the WorldCat catalogues too see libraries near you https://www.worldcat.org/title/inorganic-chemistry/oclc/48056955. Unfortunately, out of stock on Amazon: https://www.amazon.com/Holleman-Wibergs-Inorganic-Chemistry-Nils-Wiberg/dp/0123526515/ref=sr_1_1?dchild=1&keywords=wiberg+inorganic+chemistry&qid=1589064998&sr=8-1


text:
What can create black stains on metal after putting HCl on a tap full of limestale?
After putting HCl on a tap part that was really attacked by limescale, lots of gas began to form and limescale progressively disappeared, as expected.
However, the reaction left behind black soot-like deposits, on the cup holding the reaction but also on the sink (when washing away), that don't go away easily.
So, what could be this black deposit made of?
Also, the chromium-plated parts of tap began turning a bit red after exposure to HCl and the non chromium-plated parts turned black and slowly began to be covered with some kind of white powder (see the picture below).
What is the material of the tap that would explain this behavior?

I’m very interested in what created the blue and orange fringes of the object’s shadow!

The red is copper; the knob is likely brass , possibly bronze depending on age. Your knob is not zinc or aluminum or the HCl would have done more damage. Difficult to determine what is black and what is shadow. It is less likely to cause damage in the future if you use vinegar and steel wool to clean-off carbonates.

Actually the knob itself was damaged, on non-chromium plated places

Also, what could explain the black deposit/stains?


text:
Linear molecule that is liquid at room temp. and pressure?
I am trying to find a linear molecule that is a liquid at room temperature and pressure.
Do you know of any?
By linear I mean that all the bonds are co-linear (ie. simple alkane carbon chains don't count). Non hazardous materials are preferable.

https://en.wikipedia.org/wiki/Dicyanoacetylene

Bromine, B.P. 58.8 °C

Dichloroacetylene is an oily pyrophoric liquid BP 33 °C - explodes.

Also dibromoacetylene and, who knows, maybe even a hundred others - don't ask such open-ended questions!

@Mithoron, Hmm may I ask why? I am actually trying to figure out a suitable molecule for a Raman spectroscopy experiment in my lab and the answers to this question helped me out a lot (and it would've helped me if someone had asked it in the past). Should I close it? Does it go against the rules?

Well, I put a "too broad" close vote, it's up to others if they agree. This is a bit of a grey area, but unfortunately questions "are they any" tend to lead to a list of examples, which isn't a favourable outcome.

Well, that is a typical example of an xy problem: You are asking about y without the context of x. If you had described the purpose you want it for, and possibly given some tighter criteria of what you are looking for, this would be a less broad question, be of more use for further guests, possibly have a wider audience, and will most likely have been perceived better. I generally agree with Mithoron about it being too broad.

@Martin-マーチン alright, I don't see the harm of having this question up nonetheless... I already found what I wanted. Should I delete it? Am I harming the stack exchange by keeping it up?

You cannot delete it; and I never even implied that this was the way forward. I would encourage you to improve the question by [edit]ing it. You might be surprised to find even more answers eventually, and that would be a great service to future users of this site, just liked you would have found it helpful if someone had asked a question similar to this.

@Gyromagnetic: Does 2-butyne count? Its melting point is $\pu{-32 ^\circ C}$ and boiling point is $\pu{27 ^\circ C}$.

@MathewMahindaratne Nah, he wants all atoms in line.

Technically liquid, but close to the boiling point and a bit nasty to work with, is carbon disulfide.  It melts at about -112°C and boils at +46°C.


text:
Do acids stay acid forever?
I mean their pH value never change unless we dilute with water etc?

An acid is a molecule or ion capable of donating a proton (hydrogen
  ion H+) (a Brønsted–Lowry acid), or, alternatively, capable of forming
  a covalent bond with an electron pair.

So acids stay acids forever?

Related: https://chemistry.stackexchange.com/questions/20039/do-chemicals-have-an-expiry-date

So acids stay acids forever?

You know a company markets that diamonds are forever. Well may be not, a diamond is not thermodynamically stable. It may wish to turn back into graphite if you somehow provide the right activation energy.
So there is nothing special about acids. Acids are like other molecules some may remain stable for a long time and some may not.
An example is that of sulfuric acid. If someone leaves concentrated acid in an open beaker, in period of few months its volume will increase. How? It absorbs moisture from air.
If you leave conc. HCl bottle open, HCl gas will escape into air and the acid concentration will decrease.
Moral of the story: Nothing is permanent but change.


text:
Atoms and their vibration
We all know that the atoms in any solid matter vibrate about their mean position but why do they do so?? What will happen if they stop vibrating??

I don't know who downvoted you. But I can say that your question is not clear enough. What do you mean with "any matter" ? Solid state ? Gaseous state ? Furthermore, there is no "mean position" for a Helium atom in the gaseous state.

By 'matter' I presume you mean ' molecule'  In that case the motion is explained by the Heisenberg Uncertainty Principle, according to which is is not possible at the same time to  measure exactly position and momentum. The limits imposed mean that there is a zero point energy in all vibrations.  (By 'mean position' you should use 'centre of mass'). Bonds cannot stop vibrating. If there was a molecule with no zero point energy, we would know that quantum mechanics had failed. So far quantum mechanics has passed all the tests.

I think atoms of perfectly crystalline solids will stop vibrating at absolute zero.

What will happen then??

@Zenix, nothing stops vibration at absolute zero. Vibration continues, it is called zero point energy.

@M.Farooq I didn't know that, thanks. I just guessed...

They vibrate because they have kinetic energy yet are confined locally within a potential well resulting from attractive and repulsive interactions with neighboring atoms. At absolute zero quantum mechanical principles suggest that they should just keep vibrating. For more on this you might want to ask at physics SE or philosophy SE.

Doesn't that kinetic energy ever lost or released or transformed ??

I will interpret your question to mean: why must all particles vibrate? This is explained best using quantum mechanics. One of the postulates of quantum mechanics is the Heisenberg Uncertainty Principle. It means that one cannot measure both the position and momentum of a particle to arbitrary precision. This means that if one measures the momentum of a particle to greater levels of precision, the position of the particle becomes less certain and vice versa. The particle must vibrate since if it was perfectly stationary, it would be possible to determine position and momentum to greater levels possible than that allowed by the Heisenberg Uncertainty Principle.


text:
Can aqua regia/royal water be produced with sources of chloride and nitrate other than hydrochloric acid and nitric acid?
According to Professor Poliakov and a question on Chemistry Stackexchange here, gold dissolves in aqua regia because the nitrate ion oxidizes the gold into $\ce{Au^3+}$, and the chloride bonds to the gold to form tetrachloroaurate III ions ($\ce{AuCl4−}$). This means that the actual acidity of the compound plays no role in the actual oxidization and dissolution of gold. Therefore, if I just provide nitrate ions and chloride ions, the reaction should still proceed (ex: use $\ce{NaNO_3}$ for the source of nitrate ions and $\ce{NaCl}$ for the source of chloride ions). According to the equations given by an answer to the Stackexchange question earlier, aqua regia can react with gold in two ways: $$\ce{Au + 3HNO3 + 4HCl <=>> AuCl4- +  3NO2 + H3O+ + 2H2O}$$
$$\ce{Au + HNO3 + 4HCl <=>> AuCl4- + NO + H3O+ + H2O}$$
And so therefore, by my supposed substitute for the dangerous acids, they can react with gold with the following reactions:
$$\ce{Au + 3NaNO3 + 4NaCl +3H2O<=>> AuCl4- +  3NO2 + 6NaOH + Na+}$$
$$\ce{Au + NaNO3 + 4NaCl + 2H2O<=>> AuCl4- + NO + 4 NaOH + Na+}$$
The sodium hydroxide that is produced by the reaction could be a problem, but we can then add a weak buffer like $\ce{HCO3-}$ or $\ce{H2CO3}$ afterward to deal with it.
Am I correct, or is something else at play in this reaction?

Intersting questen, badatchemistry! From the title "Does aqua regia only work with nitric acid and hydrochloric acid?" I'd say yes, as Aqua Regia is the mixture of of HCl and HNO3. But the underlying thoughts make up a good question. The answer probably is in the mechanism, as if e.g sodium nitrate and sodium chloride would work, they would be used for the commercial process, as they are much safer to handle than Aqua Regia. What would happen if one added sulphuric acid? Would that make Aqua Regia in situ?

@imalipusram Thanks for that. I have fixed the title to make it more relevant. Also thanks for the insight. I might one day do an experiment and write a paper on this (I am currently a high school student studying chemistry).

The clue is probably the redox potential of tetrachloroaurate. Au(III) easily is reduced to Au(0).

@Poutnik then what is special about aqua regia so that it can dissolve gold? Does that mean the hydrogen ions appear in the reaction mechanism but acts as an intermediate or a catalyst?

@Poutnik Sorry if I got some basic concepts wrong but this is my first year in chemistry. I have am unsure about the last statement, because the oxidation number of chlorine is the same before and after the reaction (-1).

Yes chlorine works as desinfectant. Otherwise, see my answer.

$\ce{HNO_3}$ or $\ce{HCl}$ or nitrate ions are not able to dissolve Gold. Only chlorine molecule $\ce{Cl_2}$ does it. All my students have succeeded in dissolving a tiny piece of gold sheet suspended in $2$ or $3$ mL water where some chlorine gas is slowly bubbling. It can be done in one minute or two at room temperature in any high school lab. And in aqua regis, Chlorine is produced as shown by Pouting.

@Poutnik. Sorry, sorry. I should have re-read my message before sending it. I don't understand how I could have written your name so poorly. But I thank you for the description of its meaning.

Not even zinc would react with neutral nitrate + chloride, why should gold ?
Aqua regia must be strongly acidic for nitrates to have oxidative properties for oxidation of chlorides to chlorine and to oxidize gold to trace amounts of $\ce{Au^3+}$(what redox potential allows)
Mixing of acids leads to these reactions:
$$\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 H2O}$$
$$\ce{2 NOCl -> 2 NO + Cl2}$$
$$\ce{2 NO + O2 -> NO2 }$$
That also means aqua regia dissolution ability decreases with time.
The major oxidative reagent is free chlorine.  Hydrochloric acid saturated by chlorine dissolves gold as well.   Chlorides additionally help gold dissolution  by reaction $\ce{Au^3+ + 4 Cl- <=>> AuCl4^-}$, this decreases the redox potential $\ce{Au/Au^3+}$, making gold dissolution by chlorine and even by nitric acid easier.
$$\ce{2 Au + 3 Cl2 + 2 Cl- -> 2 AuCl4-}$$
$$\ce{Au + HNO3 + 4 HCl -> HAuCl4 + NO + 2 H2O}$$
$$\ce{Au + 3 HNO3 + 4 HCl -> HAuCl4 + 3 NO2 + 3 H2O}$$
See also https://en.wikipedia.org/wiki/Aqua_regia

Gold dissolves in aqua regia because there is a redox reaction: The nitrate ion oxidizes the gold in the presence of $\ce{Cl-}$ ion into $\ce{Au^3+}$, which combine with the chloride ions to form $\ce{Au(Cl)4-}$ ion. In this reaction, the presence of $\ce{Cl-}$ ion is important because it reduces the $\ce{Au/Au^3+}$ couple's potential as follows:
$$\ce{Au <=> Au^3+ + 3e-} \qquad E^\circ = \pu{-1.498 V} \tag{1}$$
However, in the presence of $\ce{Cl-}$ ion, this half-reaction becomes:
$$\ce{Au + 4Cl- <=> AuCl4^- + 3e-} \qquad E^\circ = \pu{-1.002 V} \tag{2}$$
The nitrate reduction half-reaction is:
$$\ce{NO3- + 4 H+ + 3e- <=> NO + 2H2O } \qquad E^\circ = \pu{0.957 V} \tag{3}$$
Therefore, the complete redox reaction is:
$$\ce{Au + 4Cl- + NO3- + 4 H+ <=> AuCl4^- + NO + 2H2O} \qquad E_\mathrm{rxn}^\circ = \pu{-0.045 V} \tag{4}$$
Although, $E_\mathrm{rxn}^\circ$ for this reaction is $\pu{-0.045 V}$ under standard conditions ($\pu{1 M}$ and $\pu{1 atm}$), at working conditions (e.g., concentrated $\ce{HCl}$ is at least $\pu{12 M}$), the corresponding $E_\mathrm{rxn}$ becomes positive and forward reaction proceeds (Recall Nernst equation).
According to the above half-reactions, having strong acidic conditions is essential for this reaction. Therefore, OP's suggestion of having just nitrates and chlorides (e.g., $\ce{NaNO3}$ and $\ce{NaCl}$) will not work as it in aqua regia. For example, reduction half reaction for $\ce{NO3-}$ is as follows:
$$\ce{NO3- + H2O + 2e- <=> NO2- + 2OH- } \qquad E^\circ = \pu{0.01 V} \tag{5}$$
Sum of $(2) \times 2$  and $(5) \times 3$ gives:
$$\ce{2Au + 8Cl- + 3NO3- + 3 H2O <=> 2AuCl4^- + 3NO2- + 6OH-} \qquad E_\mathrm{rxn}^\circ = \pu{-1.001 V} \tag{6}$$
This $E_\mathrm{rxn}^\circ$ value is too much to overcome.

Note that H3O+ activity coefficient in concentrated HCl is >10, so activity wise, the reaction (4) is pushed even more than the concentration suggests. Overall, the great answer, I have intentionally kept the level
of my answer rather low.


text:
Naming substituted bicyclo compounds

I have to name the above compound. I am getting 5-bromo-2-chloro-7-fluorobicyclo[2.2.2]octane.
But my teacher is telling me that the answer should be 2-bromo-5-chloro-8-fluorobicyclo[2.2.2]octane.
His reasoning is that the bromo substituent should get a lower number as it comes first in alphabetical order. Which is correct?

You have already correctly identified the parent structure of the compound as bicyclo[2.2.2]octane.
The most important simplified criteria for the numbering for such structures are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given in the question doesn’t have any principal characteristic group that is expressed as a suffix. Thus, Rule (c) is not relevant in this case.
The compound also doesn’t have any multiple bonds. Thus, also Rule (e) is not relevant in this case.
Next, low locants are assigned to the substituents that are expressed as prefixes according to Rule (f). Note that all substituents are considered together in a series of increasing numerical order. Therefore, the correct name is 5-bromo-2-chloro-7-fluorobicyclo[2.2.2]octane rather than ‘2-bromo-5-chloro-8-fluorobicyclo[2.2.2]octane’ since the locant set ‘2,5,7’ is lower than ‘2,5,8’.

Only if there is still a choice, a low locant would be given first to the bromo substituent in accordance with Rule (g).

A doubt : i thought we should count the bridge carbons at the end? Is that right?

@Aditya suresh: Yes, but now, new bridge is $\ce{-CHF-CH2 -}$!

@Loong: I modified your structure for viewers' benefit (see Aditya suresh's comment) and gave appropriate numbering. Please feel free to change it back if you don't like it.


text:
Convoluted explanation of different types of Raman scattering
I am currently studying the textbook Surface-Enhanced Raman Spectroscopy -- Bioanalytical, Biomolecular and Medical Applications by Marek Procházka.
I just read this passage from chapter 2:

Rayleigh scattering arises when the energy of the molecular system both before and after the scattering process is the same ($\hbar \omega_{\text{s}} = \hbar \omega_{\text{inc}}$). The occurrence of inelastic light scattering can be explained by classical theory in terms of a modulation of the incident radiation $E_{\text{loc}}(\omega_{\text{inc}})$ by a vibrating molecule with frequency $\omega_{\text{vib}}$. Raman frequency can be smaller $\omega_{\text{inc}} - \omega_{\text{vib}}$ or larger $\omega_{\text{inc}} + \omega_{\text{vib}}$ than that of the incoming photon $\omega_{\text{inc}}$. When the incident radiation is shifted to a lower frequency (lower energy), the scattered light is called Stokes shift. In this type of scattering the energy is transferred to the vibrational mode of the molecule which ends up in a higher vibrational energy state. Similarly, when the incident radiation is shifted to a higher frequency (higher energy), the scattered light is called anti-Stokes shift. In this type of scattering the energy is transferred from a vibrationally excited molecule, which ends up in a lower vibrational state after the scattering event. The scattering process when an intermediate “virtual state” does not correspond to a real energy level of the molecule is called normal Raman scattering (NRS)$^2$.
$^2$ The term “RS” is often used for RS under nonresonant (normal) conditions, term “normal RS” (or sometimes classical RS) is used to distinguish it from resonance RS under resonant conditions. Note that there is also the case of preresonant RS.

In particular, it is this part that I don't understand:

$^2$ The term “RS” is often used for RS under nonresonant (normal) conditions, term “normal RS” (or sometimes classical RS) is used to distinguish it from resonance RS under resonant conditions. Note that there is also the case of preresonant RS.

This is written so convolutedly that I do not at all understand what it is saying, nor can I distinguish between the different types of RS (Raman scattering) that it is referring to. I would greatly appreciate it if people would please take the time to clarify this.

The author does not seem to have a good command of the English language. There are obvious semantic errors in the more simple parts (check the upside-down logic of the first senctence), and he fails as soon as it gets complicated. I would return that book.

The only different is that the exciting light is close to an electronic transition in resonance Raman Scattering . Look at Wikipedia for Resonant Raman Scattering. This explains this clearly compared to 'ordinary' or conventional Raman Scattering.

@Karl Hmm, the first sentence seems to make sense to me; what, specifically, are you referring to? The earlier parts of the textbook seem to be fine; it's just this part that I thought was confusing. Do you happen to have experience a different surface-enhanced Raman spectroscopy textbook that you think is superior?

@porphyrin Can you please be more specific in what parts of the statement you are referring to? Like I said, it's very convoluted, so I'm having difficulty mapping what you said onto it.

I would ignore the book and look at the Wikipedia pages then at another textbook or review article.

@porphyrin So you also think that I should return this textbook?

yes, certainly do that. There is bound to be another that will clarify things for you or look in the literature for a review article.

@porphyrin Ok, will do. Thank you to you both for the advice.

The word "arises" is simply wrong. If you have an "afterwards" state, then the scattering has already occured. If the two states have the same energy, you call it a Rayleigh scatering. You can say no problem, it´s clear what he means, but as soon as you want to expain sth. beyond the obvious, sloppy semantics are fatal.


text:
Which double bond is more electrophilic : carbon-carbon or carbon-oxygen?
Question

Attempt
It seems to me that the carbon-oxygen bond would be more electrophilic due to the higher electronegativity of oxygen. Hence I believe $\ce{CH3-}$ should attack the carbonyl site. But this doesn't seem to be the case.
Is it because that the intermediate carbanion can undergo resonance with oxygen, had $\ce{CH3-}$ attacked carbon- carbon double bond?
What can be a valid reason?

Google for gilmans reagenthttps://www.masterorganicchemistry.com/2016/02/05/gilman-reagents-organocuprates-what-theyre-used-for/

Your title doesn't match your question. You should have said "In a $\alpha,\beta$-unsaturated system.." An isolated $\ce{C=C}$ bond is not electrophilic.

This may assist  https://en.wikipedia.org/wiki/Reactions_of_organocopper_reagents

of course, C=C bond is less electrophilic when compared to C=O bond, but the CH3 group seems to attack the C=C bond due to 1,4-addition. detailed explanation below.

Here is useful info on Gilman reagents and conjgate addition: http://www-oc.chemie.uni-regensburg.de/OCP/ch/chb/oc5/Cu_Zn_Organometallics.pdf

step-1: cleavage of C=O bond:

the C=O bond undergoes heterolytic cleavage to give positive charge on carbon and negative charge on oxygen just like almost all other organometallic reagents react with a ketone, for example a grignard reagent (not exactly, but similar to it).

step-2:(the real step which caused deviation from your answer)-resonance of intermediate:

the positive charge on the carbon of the erstwhile C=O group now undergoes resonance with the double bond at the allylic position to give positive charge at carbon no. 3 instead of carbon no. 1
in this sort of reaction is 1,4 addition occurs instead of 1,2 addition. For more information, you can visit this link: https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/1%2C4-Addition

step 3:(as usual):-attack of CH3 group:

now on as usual, the CH3 group from our organo copper reagent ( called as gilmann reagent, used in Corey-house synthesis), just like a grignard reagent, attacks the new carbocation formed at carbon no.3 and the water molecule in the second step donates it's proton to the remnant oxygen of our old C=O group and it becomes an enol. tautomerisation takes place and we bet 3-methyl cyclohexanone as a result.


text:
How to find the number of valence electrons with the help of quantum numbers?
I'm bit confused on how to find the number of valence electrons. For example, Boron ($\ce{B) has the electronic configuration as: $\mathrm{1s^2 2s^2 2p^1}$. So what would be its number of valence electrons : 1 or 3. According to me the answer should the 1 as the last orbital ($\mathrm{p}$ orbital) has 1 electron. Am I wrong? Please guide me if I am wrong.

Any electrons present in the electron configuration of the element in question, but not present in the nearest noble gas with an atomic number below that of the element, is a valence electron.

Not quite correct. For example  manganese (Mn) has configuration $\ce{[Ar]\ 4s^2\ 3d^5}$ and hence 7 valence electrons.

Fair point @MaxW, I will edit.

I'd use the KISS principle here. // The valence electrons are the electrons that can participate in bonding. For n=2 there can be $\ce{sp^3}$ hybridization so all the 2s and 2p electrons are valence electrons. // If you want to elaborate more certainly some paragraphs would help.


text:
Effect of charge density on electrolysis of dilute sulphuric acid
I am told that if electrolysis of dilute sulfuric acid is performed under high charge density, then the following reaction takes place at anode instead of evolving oxygen:
$$\ce{2 HSO4- -> H2S2O8 + 2 e-}$$
Why is that? And can a general conclusion be drawn?

High current densities mean high enough applied voltage so the major reaction current is saturated and higher/lower electrode potential is able to perform the next available electrode reaction.

As the voltage increases it may exceed the decomposition potential for other reactions to occur. If these reactions are kinetically favorable they can compete with the initial low potential themodynamically favorable one, see selectivity in parallel https://youtu.be/oGbht7Jryfo You can calculate Decomposition potentials using the Nernst Equation

Related: Reaction of sulfate ion in copper sulfate electrolysis


text:
Extent of Hydrogen bonding in H2O2
Several books that I have read say that in $\ce{H2O2}$ the degree or extent of H-bonding is 6 as each H atom would form 1 and O atom would form 2 H-bonds respectively. However if we look at HF the no.of H-bonds formed by it is only 2 (because in overall mixture of molecules the extra lone pairs present in F would out no. the no. of available H atoms with partial positive charge)
My question is that going with the same logic even $\ce{H2O2}$ should form only 4 hydrogen bonds as the no. of lone pairs in O would outnumber the H atoms available then why is the degree of H bonding in $\ce{H2O2}$ 6 but not 4?

Avoid do over usage of markup, $\LaTeX$, don't use $\LaTeX$ in titles. Also I suspect your reasoning for lesser number of hydrogen bonds for HF (compared to $\ce{H2O}$).

@Zenix So what would be your reaaon for that?

I think not one but there are many co-existing factors that act. Availability of number of hydrogens is an important factor, but, F is also, ig, too small to accommodate 2/3 hydrogen bonds. Availability and directionality of lp must also play their roles.


text:
What explains the stability of aromatic compounds?
We are taught in High school that aromatic compounds are highly stable. Why is that true?
Also,
Why are Antiaromatic compounds highly unstable?
Aromatic compounds have 4n+2 (n is a whole number) electrons whereas anti-aromatic compounds have 4n electrons (n is a natural number)
The difference in number of electrons in aromatic and anti-aromatic compounds is just of two, right? So what role do those two electrons play in making the compound so stable?
Please explain in detail.

Whenever you read the term "something is stable" you should ask, stable with respect to what? Aromatic compounds are stable but with respect to what? If I say someone is tall, your natural question would be "compared with whom?"

@M.Farooq My bad! I wanted to ask, why are aromatic compounds more stable than non aromatic and anti aromatic compounds?

Still you will have to think, stable in terms of what? There is no such comparison. When you are talking about stability you would think that aromatic compounds are more stable towards chemical reactions as compared to alkenes. That is the proper way to think about it. Now you can read here https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.01%3A_Aromatic_Compounds_Are_Unusually_Stable

AFAIK electrons going to BMO/ABMO/NBMO determine there relative stability.

LOL - When chemists see the phrase highly unstable they think explosive...


text:
Thermodynamic model of inclusion process of alpha-cyclodextrin
I have a question about thermodynamic model of inclusion process of CD, published by Tabushi (DOI: 10.1021/ja00471a043). Here, in 3rd step, action is described as: "An apolar solute (guest molecule) is transferred from water to the ideal gaseous state leaving a structured cavity behind. The cavity then collapses with redistribution of the water molecules." I don´t understand this part where guest molecule is transferred to the gaseous state (why this state). Can somebody try to explain this in some simple way if it is possible, please? Thanks for all the answers!

As the Buck Thorn's comment suggests this is not a mechanism for reaction (although it is drawn that way and is clearly silly in that respect) but a way of calculating the thermodynamic properties. (in thermodynamics a state function is independent of the path so that any set of paths for which data is available can be used as long as they starts and ends in the right palce)

This is a variation on the application of Hess's Law (or looks like it, not having read the article). The idea is that the individual steps add up to the complete reaction
cyclodextrin with included water + hydrated guest $\rightarrow$ hydrated cyclodextrin with included guest
According to Hess's Law you can break up this process into a sum of intermediate steps. One of those steps is dehydrating the guest molecules and shoving the waters into the bulk solvent. That last step is what is meant by collapse of the cavity  with redistribution of the water molecules..


text:
Interference of Ca-EDTA with fluoride delivery in mouthwash
The goal of fluoride mouthwash is to deliver $\ce{F-}$ to the surface of the teeth to turn hydroxyapatite in the dental enamel into fluorapatite.
Typically sodium fluoride or stannous fluoride are used.
ACT fluoride mouthwash also contains $\ce{Ca}$ in the form of $\ce{Ca-EDTA}$.

$\ce{CaF2}$ has a very low solubiliy product of $3.9 × 10^{−11}$, I would expect the $\ce{F-}$ to precipitate as $\ce{CaF2}$ and become unavailable.
This is in agreement with the literature where $\ce{EDTA}$ has been used in the determination of $\ce{F-}$ through backtitration of $\ce{Ca^2+}$. This shows that the complex is less stable than the fluoride.
http://www.sciencedirect.com/science/article/pii/S0003267000876369
Note:
It was suggested in the comments to use 10x the concentration of $\ce{Ca-EDTA}$. I don't see how this could work, all it would do is depress the concentration of available fluoride even more.
I suspect $\ce{EDTA}$ is used as an anti biofilm agent or preservative, and the $\ce{Ca}$ is used so it doesn't chelate dental $\ce{Ca}$. I assume this use is unrelated to the process of fluoride delivery.

Good observation but don't surprised by the fact a lot of food items have EDTA. A mouth-wash is no major exception. You will have read the patent in order to get the idea why EDTA is beneficial in oral rinses https://patents.google.com/patent/WO1988000044A1/en

Secondly you concern of NaF is not a . Ca-EDTA must be present in excess.

EDTA suppresses biofilm formation and it also has antimicrobial properties. I assume this is why it as added, or maybe it is used as a preservative. Na-EDTA will scavenge Ca from the teeth, so presumably the Ca-EDTA will leave the Ca in the teeth. What I don't understand is why it does not react with the fluoride.

think about the concetration. If Ca-EDTA is 10-fold (let us assume) higher conc. than NaF, what do you think about the equilibrium?

If [Ca-EDTA] is 10x[F-], would not all F- be precipitated as CaF? The other way around we won't have Ca protection against chelation of the Ca in the teeth. If that is the reason they use the Ca salt in the first place.

There is no $\ce{CaF}$. Presumably you mean $\ce{CaF2}$

@BuckThorn calcium subfluoride does exist but that isn't relevant here ; ).

@NilayGhosh Thanks. You learn something new every day, sometimes two things.

@ Buck Thorn I meant CaF2, fixed the typo. Thanks!


text:
Can Diels-Alder reaction occur using reagents of Robinson annulation?
Robinson annulation seemed somewhat like the Diels-Alder reaction to me, if all ketonic groups were to be represented as enol then cyclohexanone would be like diene and the Michael acceptor as a dienophile.
On trying out this thought, I got pyran derivatives. The doubt is attached as the following image.
"Heathcock, C. H., Ellis, J. E., McMurry, J. E., & Coppolino, A. (1971). Acid-catalyzed Robinson Annelations. Tetrahedron Letters, 12(52), 4995–4996. doi:10.1016/s0040-4039(01)97609-9 " report that Robinson annulation can be achieved using sulphuric acid instead of a base. Though no mechanism is reported in it, I think enol is formed as an intermediate.
I request you to clear this doubt of mine whether if there is any slightest possibility in Robinson annulation that basic/acidic medium would enolize the ketone then undergo diels-alder reaction on providing heat.
Also, if it does then would it follow Woodward-Hoffmann rules as Robinson annulation looked like a thermally allowed cycloaddition reaction
The following paper reports such possible diels-Alder reaction with enols though not directly for reagents of Robinson annulation: "Akbulut, N., Hartsough, D., Kim, J. I., & Schuster, G. B. (1989). The triplex Diels-Alder reaction of 1,3-dienes with enol, alkene, and acetylenic dienophiles: scope and utility. The Journal of Organic Chemistry, 54(11), 2549–2556. doi:10.1021/jo00272a017 "
.

( Reference: https://en.wikipedia.org/wiki/Oxo-Diels–Alder_reaction and https://en.wikipedia.org/wiki/Inverse_electron-demand_Diels–Alder_reaction )

Those classes of reactions are known, but generally not via the enol. More common is via a silyl enol ether or something like Danishevsky's diene or possibly with a Lewis acid as catalyst.

Thank you for your insightful comment. Is there report of diels-alder reaction between ketone and enol or enol- dienol ?

Another intriguing question about the C-C bond formation in the annulation is does the carbon of the enolate add 1,4 to the enone or does the oxygen of the enolate add 1,2- to the enone and then undergo an alkoxy (hydroxy) Claisen rearrangement. There is no evidence to my knowledge that the reaction proceeds this way so there is no need to invoke it.  Nonetheless, it is an interesting thought experiment.

I don't remember ever seeing a cycloaddition via a naked enol.


text:
Why bridged bicyclic compound isn't a preferred product over fused bicylic compound in Robinson Annulation reaction?
Why a doubly-fused cyclohexanone is formed instead of bridged cyclohexanone like product which would be more flexible (entropically-favoured)?
Also, the OH group not being at the bridgehead carbon unlike former compound, shouldn't it have a better chance of forming stable alkenone after dehydration as the preferred product formed in Robinson annulation?  The following image elaborates my doubt.

 (source : https://en.wikipedia.org/wiki/Baldwin%27s_rules#Rules_for_enolates)
As per the Baldwin rule for enolate, the enol-endo reaction and enol-exo reaction is favoured for both exo-tet and exo-trig for six/seven-membered chain (from the excerpt is provided as above image).
Both the alternatives follow six-membered exo-trig which is favoured according to above Baldwin rule. (Refer to the image where the carbon chain of both the alternatives is numbered.)
The bridged-bicycle compound has 3 faces, where each face has a 6-membered ring when it was visualized through ChemDraw 3D.
So, it seems both the alternatives as the product of Robinson annulation is favoured according to modified Baldwin rules.

What is "better dehydration?"

more preferable dehydration mode', hope that helps

First of all, the reaction is usually conducted in protic solvent with base. The alkoxides all get protonated. The fused beta-hydroxy ketone gives facile beta-elimination. The protonated, bridged beta-hydroxy ketone cannot eliminate in base (Bredt's Rule). Acid is needed to form the bridged unsaturated ketone. Baldwin's Rules apply to competitive reactions, i.e., which pathway is kinetically faster.

One issue is that the drawing on the left does not depict the correct location of the oxygen. It should be in an axial position. Therefore, successive protonation would lead to very facile elimination.


text:
Strains in conformations of cyclohexane
I know some the basic kind of strains that can exist in molecules namely Torsional strain (Vanderwaals' strain), steric strain and angle strain.
Is it possible to associate a particular type of strain that causes the majority (and the strain that causes least) of the instability to the molecule eventhough all of the types of strains could possibly exist in that conformation of cyclohexane?

For example: I was taught that in the twist boat conformation of cyclohexane, the torsional strain is approximately zero but angle strain is huge making this form unstable.

What would be corresponding most and least contributor of strain for chair, half-chair and boat conformations of cyclohexane?

Why is this question not been answered yet?What extra information should I give?
Is the question not clear?


text:
Explaining allene IR stretching frequency
So I was comparing the IR stretch frequency of ethyne, ethene and allene. Of course triple bond is stiffer than double bond so ethyne has a higher stretch frequency. The problem then is where allene sits when compared to these two. I guess it should be the lowest because I just think of the cumulated double bonds as serial spring, which should give lower spring stiffness than only one double bond in ethene. Obviously, I'm wrong... It turns out allene IR stretch frequency is in the middle of ethene and ethyne.  I wonder how might we explain that. Thanks!

There are two normal modes for the along-axis stretch in allene. Are you thinking about the right one?

Can you expand on that a bit? I’m not sure which two mode you are talking about

The symmetrical stretching mode won't participate in IR spectroscopy just like that of $CO_2$ but the asymmetrical stretching mode would.

@SurajS okay, but I’m asking about how to explain the frequency

@SurajS we don’t cover stretching mode in the intro org class

@Macrophage $$\bar{\nu} = \frac{1}{200\pi c}\sqrt{\frac{k}{\mu}} $$ Above is the formula based on Newtonian mechanics for calculating stretching frequency. So, we can't come to a conclusion just by comparing the spring constant $k$. We should also consider the reduced mass $\mu$ for a linear three-body system in case of an allene which I'll let you know if I get it. Alkene and alkyne can be compared just by looking at the spring constant since reduced mass is the same in both the cases but the situation of allene is a bit different.

Thank you, I will be very interested it if you can explain allene's frequency with just that formula, which we did learn in our organic chemistry course.