text:
Reagent notation: meaning of dash(-) and line(|) or slash(\) in reagents
I've seen a lot of reagents which have a slash between two compounds, such as Lindlar's catalyst which is $\ce{H2}$ along with $\ce{Pd|BaSO4}$. I know that the sulphur is a poison here but in general, what does the \ or | mean? And is it supposed to be a \ or a |? Are both equivalent? Does it mean "and" or "or" or something else?
$\ce{Cr2O3-Al2O3/Mn2O3}$, Also what does the dash mean here? In?
EDIT: This in the context of the reagents mentioned above arrows in organic chemistry. If I'm not wrong,  $\ce{Cr2O3-Al2O3/Mn2O3}$ is used to convert hexane to benzene by cyclization followed by aromatization.

Generally means a reagent on a surface e.g. Pd/C or absorbed into a material e.g FeCl3/Silica Gel. The / and | are interchangeable

Have a look at What is the role of the slash (“/”) in chemical reactions?, Multiple “/” in catalysts and at the related question What is the @ symbol used for in material science?

Since you are asking about backslash and a pipeline, could you please provide resources where you found them or at least some more context, such as whether the formula is used in electrochemistry, catalysis or material science? All four symbols (-, /, , |) have specific meaning based on the area they are used and are not generally interchangeable.

You find the pipeline e.g. in electrochemistry to indicate the phase boundary of electrodes (an entry) like calomel, $\ce{Cl^-} | \ce{Hg2Cl2}\ | \ce{Hg} | \ce{Pt}$.  Next time, add more specific context to your question, please.

The '|' in $\ce{Pd|BaSO4}$ implies that the reaction is taking place in the presence of $\ce{Pd}$ as a metal catalyst. This metal provides surface area for reaction to occur.
Also '/' in $\ce{Cr2O3-Al2O3/Mn2O3}$ means that $\ce{Cr2O3}$ can be used in combination with either $\ce{Al2O3}$ or $\ce{Mn2O3}$.


text:
What is the trend in hydration entropy of alkali metal ions?
Is there a trend in alkali metal ion hydration entropy?
There's much talk of hydration enthalpy, but I have a hard time finding information about the change in entropy.
Am I right in thinking that Li$^+$ would have a more negative $\Delta S^o_{hyd}$ than Cs$^+$ because more water ligands will be ordered around it, due to it's higher charge density?

Yes, there is. The following is a plot of the hydration enthalpy versus row in the Periodic Table. Data was extracted from the Wikipedia, not the primary source (Ref 1).
Your thinking is right. The hydration entropy is evidently negative due to structuring of water (including loss of rotational degrees of freedom), becoming less negative for the larger ions (smallest loss of entropy for the cesium cation).

References

Burgess, John (1978). Metal Ions in Solution. Chichester: Ellis Horwood. ISBN 978-0-85312-027-8.

source: Stack Exchange

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

sort: votes


text:
Can I solubilize lemon essence in water somehow?
I have a water based solution (95% water, the rest are salts) to which I want to add a lemon fragrance. I bought "water based lemon essence", but upon putting a mere drop in about 500 mL, the solution turned from crystal clear to milky off-white.
I researched and found that lemon essence is apparently one of many essential oils (citral, citronellal, cintronellol). None of which are water-soluble. That's why the result is an emulsion, and it's not translucent. (It mixed very well and it's very stable, that's not my problem)
I was wondering then if it's possible to give lemon fragrance to a water solution without turning it into a milky emulsion.

Try only one drop of lemon essence in one liter water. Not more.

Some hints are provided by Wikipedia:Lemon liqueur:

To produce the Lemon liqueur requires sugar, water, lemon zest,
mixing in organic salutes  liquor, and time to mature. Lemon zest is soaked in high proof neutral spirits to extract from it the lemon oil (an essential oil). The extraction is then diluted with simple syrup.

In other words, lemon zest is drawn into water by mixing in water-soluble organic compounds, in this case sugar and (ethyl) alcohol.  If you want it to be nonalcoholic, maybe just the sugar will do.
During late autumn I make a cranberry jam by boiling the cranberries with oranges, sugar and water.  In the presence of the sugar and organic material from the oranges and cranberries and with the application of heat, orange zest is extracted into the jam.  The jam is too dark and thick to see whether the zest remains in solution when the jam is cooled, but the extraction of the zest is evident by taste.

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


text:
White growth on unopened bottle of chlorotrimethylsilane
I have a white, moldy-like substance that has been growing near the cap of an unopened glass container filled with TMSCl. It has spread to form a thin, powdery coating on all of the other containers of organic reagents it is stored with. This substance appears to have a corrosive effect on glass, but leaves plastic containers and metal containers unaffected. I'd like to know what it is before I go about cleaning it up.

Hello again! I hope you're doing well! I recently had a little safety concern about phenoxyethanol, but I think nobody active on Biology SE knows anything about it. Do you know about that substance, or know anyone who can help me with my enquiry? And feel free to drop by in the Logic chat too! =)

Chlorotrimethylsilane, like other organosilicon compounds, is easily attacked by nucleophiles and readily hydrolyzes in air yielding hydrogen chloride:
$$\ce{2 Me3SiCl(l) + H2O(l) -> Me3Si-O-SiMe3(l) + 2 HCl(g)}$$
Hydrogen chloride reacting with traces of ammonia $\ce{NH3}$ in the air builds up white solid ammonium chloride $\ce{NH4Cl}$ depositing on the nearby surfaces.
Also, the final product of hydrolysis for the majority of organosilicon compounds is silicon dioxide $\ce{SiO2}$ (sol-gel process similar to olation). Fine powder of silicon dioxide is white as well.


text:
How to retrieve excited state molecular orbital coefficients in Gaussian?
Currently I am doing TD-DFT calculations in Gaussian and want to calculate transition dipole moment for $S_1$ transition manually. However, I do not manage to retrieve molecular orbital coefficients for the excited states in Gaussian, just the ground state with population keyword and a few hours of search through the Internet hasn't helped to move forward. Is it possible to do that and if it is possible, how to achieve that? And just to make sure that I understand the output corrently: the molecular orbital coefficients are given in their casual forms (like $c_1, c_2, etc.$), not a square modulus of them?
Thanks in advance for your help!


text:
Physical interpreation of coulomb and exchange integral
When trying to solve the Schrodinger equation for the electronic hamiltonian:
$$H_{el} = \sum_{i=1}^{N} \bigg( - \frac{1}{2}\nabla_i^2 - \sum_A \frac{Z}{r_{i_A}} \bigg) + \sum_{i>j=1}^{N}\frac{1}{r_{ij}}$$
we note that two prominent integrals arise:
These are the exchange  integral:
$$  K_b(1)\chi_a(1) = \bigg[ \int d\tau_2 \chi_b(2)^* \frac{1}{r_{12}} \chi_a(2) \bigg]\chi_b(1)$$
and the coulomb integral:
$$ J_b(1)\chi_a(1) = \bigg[ \int d\tau_2 \chi_b(2)^* \frac{1}{r_{12}} \chi_b(2) \bigg]\chi_a(1)$$
$r_{12}$ is the distance between electrons 1 and 2 and the integrals are done over all space. $\chi_b$ is an orbital labelled b.
I was wondering whether someone could give me a brief physical interpretation of these two integrals.


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Helium container
Imagine a vacuum flask-like container. If we filled the main compartment with helium and the vacuum compartment with high-pressure gas would helium still escape?
Let's ignore the leakage from the valve, the valve seating, o-rings, bushings, etc.

Are you asking if helium can diffuse through solid  glass or stainless thermos bottles ?

Please give details of your questions

Partial answer , helium will not diffuse through steel . Long ago I did stress rupture tests of low alloy steel at 800 F . Tests were done to evaluate if hydrogen had an affect on rupture strength. The test specimens were thin tubes such as 0.012" ( about 0.3 mm) wall thickness. The stress was provided by 1800 psi gas pressure . Hydrogen in some and helium in the controls. When a tube was filled with helium , and the valve closed , pressure was held for roughly a month of test duration , so no helium passed through the steel. Hydrogen diffuses through steel easily so the tubes with hydrogen had a constant supply of 1800 psi. Hydrogen escaped through a bubbler and was vented ; within a couple seconds of the hydrogen valves being opened , hydrogen started venting . So helium does not diffuse through steel at 800 F ,it is unlikely at ambient temperature.

Thank you I didn't know that helium will not diffuse through steel. Do you know if the main compartment was filled with hydrogen and the outer compartment filled with gass that has higher pressure than hydrogen compartment woud it still diffuse? Im thinking maybe the pores in steel would be sealed by the gass not allowing hydrogen to diffuse?

The outer chamber / protection tube was hydrogen at very close to one atmosphere . There are no pores in steel , the hydrogen moves through the lattice; faster at elevated temperatures . At ambient temperature the moving hydrogen causes hydrogen embrittlement ( also known by a dozen other names).  At high temperatures the hydrogen diffusion is governed by the relative partial pressure of hydrogen , any other gas does not matter.


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Source for literature yields
I'm required to compare and justify the yields of my reactions against literature yields, but I cannot find any sources of literature yields for the reactions, or any reactions for that matter. Are there any resources with literature yields available?
The reactions are the formation of tetraphenylporphyrin and then complexing it with Cu and Zn, if relevant.

What databases do you have access to? A Scifinder search would find literature references with no difficulty.
A Google search finds this technical report which has the synthesis of tetraphenylporphyrin and the preparation of Fe(II) and Zn(II) complexes report here.
The Cu(II) complex is prepared in this lab report here. The references in these reports should help further.
The references in this JACS paper here should lead you to other preps


text:
Effect of specific heat capacity on heat loss
Assume we want to calculate the heat capacities of two substances with different heat capacities. We heat them both up with an immersion heater delivering a constant rate of energy (constant J/s), and we stop the immersion heater after 100 J of heat has been added. Would the lower heat capacity substance experience more or less heat loss, and would that translate into more or less error in the calculated specific heat?
I assume that less heat loss would mean less error but I am not sure as to which substance would experience less heat loss.

If anyone stumbles across this later, I am now strongly leaning towards more heat loss for lower heat capacity because heat loss is proportional to temperature difference and lower heat capacity will change temperature more easily.

If you are talking about heat loss to the environment during the test, and the heal loss is dominated by the lower heat transfer coefficient on the air side of the sample boundary, then your assessment is correct.

Assume you add a fixed amount of heat to a substance held in an isothermal bath, so that as soon as you start heating it, the substance gives off heat to the surroundings. As heat is added the substance increases in temperature to a maximum. When the heater is turned off, the substance continues to dissipate heat until it is again in equilibrium with the surroundings.
If the rate at which heat is added is much faster than that of heat dissipation and the heat capacity is approximately constant, then the maximum temperature will be approximately
$$T_{\mathrm{max}}=T_{\mathrm{ini}} + \frac{Q}{C_p}$$
where Q is the total amount of heat added. As expected, a substance with a smaller heat capacity will become hotter.
Now consider the rate of heat dissipation. The rate of heat dissipation can be expressed with the heat equation (in one dimension) as
$$\frac{dQ}{dt} = -k \frac{\partial T}{\partial x}$$
where k is the thermal conductivity.
Therefore if k is the same for two substances, then the rate at which heat is released will differ according to the heat capacity, and the lower heat capacity substance will release heat more quickly (since temperature gradients will be greater in that substance).
This can also be argued based on the value of the thermal diffusivity $\alpha$, which is inversely proportional to the heat capacity $c_p$:
$$\alpha = \frac{k}{\rho c_p}$$
As the Wikipedia explains,

In a sense, thermal diffusivity is the measure of thermal inertia.[6] In a substance with high thermal diffusivity, heat moves rapidly through it because the substance conducts heat quickly relative to its volumetric heat capacity or 'thermal bulk'.

Since the substance with smaller $c_p$ will have a larger thermal diffusivity (assuming all else constant) it will also lose heat more quickly.
This can of course be expected to result in greater uncertainty in a measurement of the heat absorbed by the substance.

I'm not sure. Thermal effusivity, $\sqrt{\kappa \rho c_p}$, rather than thermal diffusivity, would seem to be closer to what he's looking for; but the latter expression allows one to compare two substances at the same $T$, while here the one with the lower $c_p$ would be at a higher $T$. It's an interesting problem—consider just the effect of $\rho$: OTOH, given two substances with the same $c_p$, the one with lower $\rho$ would have a greater surface area. But, on the other hand, with a lower $\rho$, the thermal energy would have to travel further, and the $T$-gradient would be less.

...I think you'd need to solve the 3D heat equation, under the assumption the two bodies have the same shape.  The math might be simpler if the both bodies were instead heated to the same temperature.

@theorist you could have a long rod insulated except at the ends that has the same exposed area and yet different $\rho$. The lengths would differ of course. That is more like what I describe with a 1d heat equation. Since the question was about $c_p$ I did not get into a discussion of the density.

Yes, the question was about calculating $c_p$ from heat loss.  But since density affects this, it must be accounted for in any solution.


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Does temperature change if change in internal energy is zero?

Find $ΔU$ for the combustion of methane in a sealed rigid adiabatic container.

Solution says since $q = 0$ and $w = 0,$ so $ΔU = 0$ by the first law of thermodynamics.
Why isn't the heat released during combustion considered?
Does the temperature of the system changes during the process? I know that
$$ΔU = nC_VΔT$$
is not valid directly as the amount of substance is changing, and so do the degrees of freedom.

The answer to you first doubt is that the temperature of the products is higher than the temperature of the reactants in adiabatic combustion.  That's where the "heat" goes.  Your 2nd doubt answers itself.

Sir if temperature is increasing , the way I am calculating ΔU is by ( Uₚ - Uᵣ ) but after writing the expression with help of U = fnRT/2 (Where f = degree of freedoms) , I am not getting ΔU = 0 , please tell where I am going wrong whether this method of calculating ΔU is wrong or something else ?

Do you know how to determine $\Delta U$ if the final temperature were equal to the initial temperature (taking into account that there have been changes in the amounts of chemical species present)?  Would you know $\Delta H$ for the case where the initial and final temperatures are equal?

I guess ΔH could be written with help of bond enthalpy , and ΔU is where I am confused ,according to my previous logics ΔU should come out to be zero.

$\Delta H$ is the heat of combustion which you look up or calculate from tabulated heats of formation.  Do you know how to do that?  If you don't know how to use this to determine $\Delta U$ at constant temperature for the reaction, you need to research that.  This is in every thermo book.  Once you have that, you need to determine what the temperature rise of the products would have to be in order for the overall $\Delta U$ to be zero.

I know ΔH can be calculated from heat of formation , and after calculating it we could find ΔU as ΔH - Δn.RT and in my question ΔU = 0 , and temperature is increasing so ΔH = Δ(nT).R , we could calculated the rise in temperature

But my doubt is why my method of calculating ΔU as ( Uₚ - Uᵣ ) , by writing individual Internal energy as fnRT/2 is wrong ??

It is wrong for this problem because it does not include the energy associated with making and breaking chemical bonds.  In problems not involving chemical reaction or molecular interactions, that does not have to be included.  But, in this problem it does.

Ok ,so it means in internal energy , we will also need to include potential energy stored in molecules ?

That is done for you already in tables of heats of formation and heats of combustion.  So you don't need to do it yourself.

Ok now i understood why my method was wrong

Sir can you kindly comment on that " Will the ΔU will be negative if the walls are made conducting instead of adiabatic as now heat is released out of the system to the surroundings" ?

Sounds correct if the surroundings are maintained at the initial temperature of the reaction mixture.

Yes, the temperature changes.  The 1st law says that the internal energy of an isolated system is constant, but it doesn't prevent it from changing form.  When you combust methane, you change the chemical composition, such that you go from bonds with a higher total potential energy to bonds with a lower total potential energy. The lost potential energy is converted to thermal energy, causing the temperature to increase.

Can you please also clarify how "1st law says that internal energy of a isolated system is constant " , doesn't the heat of reaction increase the internal energy ? {ΔU = q (from reaction) }

You're thinking q is the thermal energy generated by the reaction.   It's not.  q is the heat that flows into out of the system, as a result of the thermal energy generated by the rxn.  But if you isolate the system so heat can't flow in or out, q must be 0, thus all thermal energy changes caused by the reaction must stay inside the system, which is why the system's temperature changes.  If the reaction releases thermal energy, as in your example, T goes up.  If it absorbs thermal energy, T goes down.

Please note the following:  https://meta.stackexchange.com/help/someone-answers

I got your point that the heat of reaction is ''internal'' and we only include heat in/out from the surroundings in FLOT ; Can you tell will the ΔU will be negative if the walls are made conducting instead of adiabatic as now heat is released out of the system to the surroundings

If there is no non-pV work, then w=0, because the walls are rigid, and hence delta_U= q+w = q.  If heat flows out, q<0 and thus delta_U <0.


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Can the relationship between Gibbs (free) energy change to Equilibrium constant of a chemical reaction be derived from Maxwell-Boltzmann distribution?
Can the relation between Gibbs(free) energy change to Equilibrium constant of a chemical reaction be derived from the  Maxwell-Boltzmann distribution only ? ( Theoretically , with may be an empirical and conceptual background explanation may also be granted . )
i.e. The equation which is : ${\Delta_\text{r} G}^o=-RT\ln K_\text{eq}$

Yes, but with some restrictions such as on the number of atoms/molecules and their total energy which enables partition functions to be calculated and from these the classical thermodynamic functions G, H, S can be found. Books on Statistical Mechanics give the calculation to find partition functions.

@porphyrin please describe a bit

First you need to find out about partition functions. Most phys chem textbooks cover this. Then look for the connection between these and normal thermodynamic properties.


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Why is the concentration of hydrogen peroxide given as a percentage? For example, 3%, 6%
Conventionally, the concentration of solutions is given in moldm⁻³ or M, however, I am curious why the concentration of hydrogen peroxide isn't written in the same way.

Commercially provided liquids, like acids or alkoholic drinks,, are not provided with molarity either, but with mass or volume percentage ( unless being the precise concentration standard solution ). Like 35-37% HCl, 96-98% H2SO4 or 40% rum.

The concentration, if expressed in mol/L, has no meaning for usual people, who do not know what is a mole. That is why the concentration is expressed in percent. The percent unit is understood by everybody.


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can ozone run upwards through a silicon tube and exit against gravity?
I want to get ozone into an attic where rats are running around. For some reason, the attic is inaccessible to us.  I ordered a bubbling ozone unit. We will drill a hole into the attic and run the silicon ozone tube (without the bubbler) through a small-diameter PVC pipe and then run that pipe with the silicon tube sticking about an inch or half-inch out of the pipe up into the attic. So the ozone tube will be held in place and will exit the silicon tube straight up into the the attic air.  (The bubbler won't be attached, of course.) The ozone unit will be located inside of the house, so the actual $\ce{O3}$ will have to leave the unit and run upwards into the attic.
My question is: Can the ozone run upwards that way and exit the unit, against gravity, instead of downwards out of the unit, with gravity?
I want to get enough ozone into the attic air to cause the rats to want to leave the area.  We can't get into the attic to set traps and we would prefer not to kill the rats, rather to make the attic an undesirable space for them to occupy.  I figured we would run the unit every four hours for half an hour.  We want the ozone to exit into open space instead of from a silicon tube laying on the attic floor, so at least it will leave in air and have someplace to move around.  But there won't be a fan to move the $\ce{O3}$ around in the attic once it exits the tube. It will exit the tube into unmoving air...?

I am not at all certain that ozone is either an effective rodenticide or safe to use in any environment that is not hermetically sealed. I would absolutely not do this. Rely on professional pest controllers.

The OP explicitly prefers rather "rodentorepelent". But I would advice against this usage of ozone as well.

Ozone producers do not produce pure ozone ( if they did, they would be chemical weapons ). They provide just air somewhat enriched with ozone, with density negligibly higher than pure air.(*)
Changes of "ozone in air" temperature have much higher impact on its density than achievable ozone content. Therefore, in density context, consider the outgoing gas just as if it was air.
If the output is warmer than the targeted air, it will climb up. If it is colder, it will sink down. In any case, there would be significant mixing by diffusion or air circulation.
So in summary, expect ozone to be propagated to all directions, desired ones and not desired ones.
Much stronger effect that flow dissipation has relatively fast ozone decaying due its reactivity, disappearing in typically dozens of minutes.

(*) Note that human nose is very sensitive for ozone content, with detection threshold at ratio near 1 : 10 000 000.


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Using PV = nRT to solve an electrolysis problem
Question: Two electrolytic cells are connected in series (the same current passes through each cell). One cell for the electrolysis of water produces $\pu{100 cm^3}$
of oxygen, measured at $\pu{273 K}$ and
$\pu{1.01
×
10^5
 Pa}$
. The second cell contains molten lead(II) bromide,
$\ce{PbBr2}$.
Determine the mass, in $\pu{g}$, of lead produced.
My working: $$PV=nRT$$
$$n={PV\over RT}$$
$$n\pu{={(1.01 \times 10^5)(100 \times 10^{-6})\over (8.31 \times 273)}} $$
$$\pu{= 4.45 \times 10^{-3} moles}$$
$$\pu{mass = 207.2 \times ANS = 0.922 g}$$
The given solution is:

Why do I have to multiply by 2 - what is wrong with my method?

Well, because oxygen is O2.

@IvanNeretin - I don't quite follow why that is relevant. However, that has made me think: is it because...the half equation for the electrolysis of water is 2H2O --> O2 + 4H+ + 4e- and the half equation for the reduction of lead is Pb2+ + 2e- --> Pb. The ratio of electrons is 2:1...is this why?

Well, yes, and now look how much Pb you can buy with those 4e-.

And you really don't need Pv=nRT to do this since the number of moles of oxygen can be calculated directly from the volume of the oxygen (since others have already done the calculation and the volume to number of moles is the same for most gases at STP).

Both methods yield $\pu{4.45E-3}$ mol $\ce{O2}$. But to get $\pu{1 mol}$ $\ce{O2}$, you need $\pu{4 mol}$ electrons, as the half-reaction at the anode is : $$\ce{2 H2O -> 4H+ + O2 + 4 e-}$$So to get $\pu{4.45E-3 mol}$ $\ce{O2}$, the needed amount of electrons is $4$ times the amount of $\ce{O2}$, or $4$· $\pu{4.45E-3}$ = $\pu{1.78E-2}$ electrons. Now $2$ electrons are needed to produce $\ce{1 Pb}$. So $\pu{1.78E-2}$ mol electrons will produce half of it, or $\pu{8.9E-3}$ mole $\ce{Pb}$. And this amount weighs $\pu{8.9E-3} ·\pu{207.2 g}$ = $\pu{1.84 g}$


text:
In what form does the respective halides of thallium exist as?
Thallium Halides like $\ce{TlF}$ and $\ce{TlI}$ exist. However, my text says that $\ce{TlI}$ exists as $\ce{Tl+}$ and $\ce{I3-}$. This is understandable with the help of the inert pair effect. Then why does the fluoride exist as $\ce{Tl^{+3}}$ and $\ce{F-}$?

See: https://chemistry.stackexchange.com/questions/49477/why-isnt-thallium-triiodide-stable/50760?r=SearchResults

And https://chemistry.stackexchange.com/questions/69480/why-is-thalliumiii-oxide-stable-while-thalliumiii-chloride-unstable?r=SearchResults

Two factors are in play here.
First, at least in the sixth period the "inert pair" is not completely inert.  It can be pulled into chemical bonding if we use an appropriately strong oxidizing agent. Halogens lighter than iodine are better oxidizing agents than iodine itself, so are more likely to draw off that inert pair.  In the case of thallium, bromine and chlorine are capable of forming thallium(III) compounds, although these readily decompose upon warming.  Wikipedia gives a brief summary of $\ce{TlX3}$ compounds, noting the contrast in structure between the (tri)iodide and the others.
Second, as described in this answer, trihalide ions are stabilized by de-localized bonding but destabilized by the presence of counterions, which tend to break up the trihalide ion.  Triiodide is made of bulky atoms that tend to minimize this instability, whereas lighter trihalide ions are less robust and generally exist under ambient conditions only with bulky counterions such as quaternary ammonium or pyridinium cations (all the tribromide and trichloride examples given or referenced in the answer noted above involve such cations).
So the combination of a less stable trihalide ion (in the concentrated salt environment) and greater oxidizing power favors lighter halogens forming $\ce{TlX3}$ as a largely covalent compound of thallium(III) rather than a trihalide-ion salt of thallium(I).


text:
What changes are expected in chemical reaction when deuterium(D) is attached in place of hydrogen(H)
Our teachers told that deuterium (D) in place of H in organic reaction only matters when the C-H is broken in a reaction . But does it really matter only then ? Will it affect the stability of the compound so formed ?

There are whole reviews written about the deuterium effect. Here is one of them: https://pubs.acs.org/doi/10.1021/cr50004a004

Approximately 0.01% of already existing C-H bonds are actually C-D bonds. The teacher probably meant dynamic  replacement of active hydrogen, e.g. C2H5OH + D2O <=> C2O5OD + HDO, or CH3COCH2COCH3 + D2O <=> CH3COCHDCOCH3 + HDO


text:
Help to kick out the anomaly in proving $Q$ = $K_\text{eq}$
Consider a hypothetical bidirectional reaction ,  $\ce{aA + bB <=> cC + dD}$
(A , B are reactant(s) and C , D are reaction product(s)) ,
As we know the following expressions $Q=\frac{P_C^c.P_D^d}{P_A^a.P_B^b}$ (Q=Reaction Quotient , $P_x$ = Partial Pressure od $x_\text{th}$ reactant/reaction product) and while the reaction is in equilibrium , $K_\text{eq}=\frac{[C]^c.[D]^d}{[A]^a.[B]^b}$ ($K_\text{eq}$= Equilibrium constant of the reaction , [ ] = Concentration for Ideal Solution or Activity for Non-Ideal Solution) please help me to understand why at chemical equilibrium $Q=K_\text{eq}$ holds ,
although $Q$ and $K_\text{eq}$ both are dimensionless quantity , "The expression of $Q$ contains Partial Pressure and The expression of $K_\text{eq}$ contains Concentration for Ideal Solution or Activity for Non-Ideal Solution"

"As we know" is an assumption. What's the source of your equations? Is that from a book or from lecture notes? See here: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/The_Reaction_Quotient There Q is in terms of activities.

@BuckThorn , There is no mention of Partial Pressure

The partial pressure and the activity can be related as follows: $a_i=\gamma _i p_i/p^\circ$

In order to compare $K$ and $Q$, you have to define them in terms of the same measures of concentration and the same standard state. The only difference in the definition should be that in order to determine $K$, you are required to use the concentrations or activities of an equilibrium state, while $Q$ may be determined for states that are not necessarily at equilibrium.

please help me to understand why at chemical equilibrium $Q=K_{eq}$ holds

Once you have the definitions match, it becomes fairly trivial. The equilibrium constant is what you observe whenever the reaction has had enough time to go to equilibrium. The individual equilibrium concentrations might be distinct for different starting concentrations, but $K$ comes out the same (if it does not, it means the reaction is not reaching equilibrium). If $Q$ is not equal to $K$, you are not at equilibrium.
I realize this is a bit of a tautology. The explanation why the particular form of the equilibrium expression leads to a value that is a constant requires a knowledge of how chemical potentials (or $\Delta_r G$) vary with concentration. The reaction will proceed in the direction that minimizes $G$, so at equilibrium $\Delta_r G = \frac{dG}{d \xi}$ will be zero.

Please give more direct explanation . Actually , this is part of the  proof of Gibbs Free Energy Equation , for ${\Delta_\text{r} G}^o=-RT\ln Q$ is it easy to prove but ${\Delta_\text{r} G}^o=-RT\ln K_\text{eq}$ is a bit hard to establish directly .

@NilabjaSaha $\Delta_r G^\circ=-RT\ln Q$ doesn't exist... The general equation is
$\Delta_r G=-RT\ln \Big(\frac{K}{Q}\Big)$ and $\Delta_r G^\circ=-RT\ln K$ is just the special case where you plug in $Q^\circ:=1$.

@Kexanone I will not update my answer to include this but I encourage you to expand your comment into a full answer.

I've outlined the derivation of the equation, a more detailed explanation can be found in many phys. chem. and thermodynamics textbooks. Suppose, for simplicity, that the equilibrium is A(g)=B(g) and at constant temperature and pressure. The Gibbs free energy change for the reaction is $dG=(\mu_B-\mu_A)d\xi$ where $\xi$ is the extent of reaction which is $0$ for all reactants and $1$ when one mole of reactants has become product. The chemical potentials are $\mu$. At equilibrium the condition is that the change in free energy $G$ is at a minimum, $(\partial G/\partial \xi)_{T,p} = 0$.  The chemical potentials in terms of the numerical value of the partial pressure $p$ is $\mu=\mu^0+RT\ln(p)$ for each species. Therefore
$$\left(\frac{\partial G}{\partial \xi} \right)_{T,p}=\mu_B-\mu_A=\Delta G^\text{o}+RT\ln\left(\frac{p_B}{p_A}\right)$$
and this is equal to the free energy change one mole of A at partial pressure $p_A$ going into B at constant partial pressures. The term $\Delta G^\text{o}=\mu_B^0-\mu_A^0$.
It is convention to label the slope of free energy vs extent of reaction $\displaystyle \left(\frac{\partial G}{\partial \xi} \right)_{T,p}\to\Delta G'$ or something similar. The general expression is now
$$\Delta G'=\Delta G^\text{o}+RT\ln\left(\frac{p_B}{p_A}\right)$$
At equilibrium $\Delta G'=0$ and so $\displaystyle \Delta G^\text{o}=-RT\ln\left(\frac{p^e_B}{p^e_A}\right)$ where the superscript $e$ indicates the equilibrium value. Normally we write this as
$$\displaystyle \Delta G^\text{o}=-RT\ln\left(K_p\right)$$

Regarding $\mu=\mu^\circ+RT\ln(p)$. I know Atkin's and co. do this a lot, but $p$ in this case is not the partial pressure here, but a relative partial pressure (relative to the standard state). The equation in terms of partial pressure $p$ is $\mu=\mu^\circ+RT\ln\Big(\frac{p}{p^\circ}\Big)$.

According to wikipedia, the greek letter commonly used for extent of reaction is Xi, not Zeta: https://en.wikipedia.org/wiki/Extent_of_reaction

So does IUPAC: https://goldbook.iupac.org/terms/view/E02283

Ok I use $p$ as numerical value (as stated in the text) so is dimensionless, $p/p^0$ is also ok. but as  $p^0=1$ its the same. I will change $\zeta$ to $\xi$ but actually any unique symbol will do.

@porphyrin No, it's not the same. $p^\circ=10^5 Pa$. The only reason it doesn't matter at the end in your case is that the total number of particles doesn't change in your process.

@Kexanone No: If you use  Pa or  Torr  etc  as we take the ratio vs standard state 1 bar or 1 atm or $10^5$ Pa or 760 torr, whatever we choose, the values cancel out. i.e. 0.5 atm /1 atm, $50000$ Pa /$10^5$ Pa, will be 0.5 what ever the units.


text:
Slope ≠ 1 in Henderson-Hasselbalch graph for determination of pKa

The $\mathrm{p}K_\mathrm{a}$ of indicators can be determined via plotting a graph of $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$  vs $\mathrm{pH}$ with the help of a spectrophotometer, from the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$.
From this, the graph should have a gradient of 1, and $x$-intercept = -($y$-intercept) = $\mathrm{p}K_\mathrm{a}$. However, my graph has a slope of ~0.8 which means my $x$-intercept ≠ -($y$-intercept) so which value do I use as my calculated $\mathrm{p}K_\mathrm{a}$?
From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$  axis in this situation, why is this?

What exactly did you measure with spectrophotometer, and can you post a picture of the plot?

The spectrophotometer was used to measure the absorption at 592nm of bromophenol blue at different pH's

So how do you transform the measured absorption at $\pu{592 nm}$  into two values, namely $\ce{ [HI]}$ and $\ce{[I-]}$ ?

Your term of $\log \left(\frac{\ce{I-}}{\ce{HI}}\right)$ is misleading.

I wasn't sure about your techniques. But according to the graphs and the equation, when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $.

From The Merck Index, bromophenol blue has a $\mathrm{p}K_\mathrm{a}$ equal to $4.0$. And the curve displays $4.2$

We determined log([I-]/[HI]) as follows: At a particular wavelength, the absorption of the acid form is Aa while the absorbance of the base form is Ab. The absorbance of the mixture is therefore A=FaAa + FbAb where Fa and Fb are the fractions of the indicator in the acid and base forms respectivly. It can then be shown by rearrangement and substitution that [I-]/[HI] = (A-Aa)/(Ab-A). It has been assumed that only the base form of the indicator absorbs at 592nm, thus at this λ, Aa≈0. So [I-]/[HI] = A/Ab-A

OP's question: From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ axis in this situation, why is this?

Answer for this question is easy. According to the graphs plotted using the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$ (Henderson-Hasslebalch equation), when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$ (meaning $[\ce{In-}] = [\ce{HIn}]$), then $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $. Therefore, the $x$-value where the straight-line crosses the $x$-axis is the best value for $\mathrm{p}K_\mathrm{a} $.

It is not exactly a straight line (experimental data, no information on error bars). Nice explanation why to focus on the x-intercept.


text:
Calculate the percentage in a solution
The aim of this exercise is to determine the percentage (degree) of acetic acid in vinegar.
For this purpose 10 mL of vinegar requires 8mL of sodium hydroxide of molar concentration
0.1mol/L to react with the acetic acid present.
The equation of the reaction is :
$\ce{CH3COOH + HO^- -> CH3COO^- + H2O}$
a- Find the mass of acetic acid present in 10mL vinegar.

b- Find the percentage of acetic acid in vinegar.

Given: M (CH3COOH) = 60 g/mol
Density of vinegar = 1.02g/mL
I solved question (a) in which the answer is mass = 0.048g, however I'm confused with question (b)
I think I should use this formula but I'm not sure about it.

$C = \frac{\%d}{100M} $
where d is the density and M is the Molar mass.

Instead of using a formula, think !
How much does $\pu{10 mL}$ vinegar weigh ? Answer: $\pu{10.2 g}$.
You have found that this sample contains $\pu{0.048 g}$ acetic acid. As a consequence, the percentage of acid in vinegar is : $\frac{\pu{0.048 g}}{\pu{10.2 g}} = \pu{4.7E-3} = 0.47$%
P.S. Are you sure of these numerical values ? Usual vinegars are at least $12$ times more concentrated !


text:
pKa of acetic acid in pure NH3?
Acetic acid completely dissociates in liquid ammonia and I would like to know the $\mathrm{p}K_\mathrm{a}$ value for the acid in that solvent. Can anybody point me to a reference?

I'm not sure such reference exists, but maybe... who knows. One thing is that's a bit obscure solvent, second that acetic acid is strong in something like that and it's difficult to estimate pKa of strong acids.

Is acetic acid soluble in liquid ammonia?

Yes, it is soluble as the question implies.


text:
How do I interpret the results of this chromatogram and does it tell me about the compounds' polarities?
I am reading this journal article which discusses using high-performance liquid chromatography (HPLC) to separate dyes from wine. Here are the relevant facts about the experimental setup:

Two mobile phases are used, a solvent A and a solvent B. Solvent A is is 40% methanol and solvent B is 70% methanol. Both solvents also have 25% 0.02M tetrabutyl-ammoniumhydroxide phosphate solution as well.
The stationary phase is a LiChrospher 100 RP-18e column.

Below are the chromatograms and the retention time tables that this experiment produced:

From the chromatogram, we see that amaranth has the fastest elution time of 6.60 minutes. My questions relate to what inferences can we draw from elution times:

Does the elution time correlate with the polarity of the compound itself? Here, it states that similarities in polarity correlates positively with elution time. The stationary phase used in this experiment (the LiChrospher) is a nonpolar silica column. Thus, with amaranth coming out the fastest, would it make it the most polar compound?
Further, how does the use of two solvents (A and B) affect making an inference like the one above? In the chromatogram above (specifically chromatogram "a") and the associated table, the elution time for the three compounds ranged from 6.60 minutes to 9.76 minutes. Since a solvent A and a solvent B were used, how do I know which compounds were eluted by which solvent? (If my interpretation of how the solvents A and B work is wrong, please correct me). Also, how does the use of two solvents affect understanding the relative polarity of the compounds?

You have several queries. First of let us look at the gradient method used by the authors.

The HPLC gradient for synthetic colorants was 0 min A 100 p. cent, 10
min A 0 p. cent, 14 min A 100 p. cent. The HPLC gradient for natural
carmine was 0 min A 80 p. cent, 20 min A 50 p. cent, 23 min A 30 p.
cent, 25 min A 30 p. cent

Factually, notice the trend in the mobile phase for synthetic ones. It is a linear gradient elution that they started with 40% MeOH at t=0 min, and at t=14 min, you have 70% MeOH. In the second case, it is a purely empirical "step" gradient. Try to draw a graph on a piece of paper.
Regardless, the authors use a water-rich mobile phase in the beginning and go to methanol rich mobile phase. Note I am avoiding the word polarity.

Polarity is an umbrella term in chromatography. For elementary purposes, it may be okay for serious work predicting retention on the ill-defined polarity that carries little or no meaning. You linked a GC webpage for an HPLC separation; those are different worlds.
Qualitatively, in top figure (a), it is clear that amaranth spends more time in the water-rich mobile phase, followed by azo ruby and then ponceau 4R if we use a C18 phase.

All you can say is that amaranth has a higher solubility in the water-rich mobile phase than ponceau 4R. One should not over-interpret retention behavior with this limited data.
In short, polarity is not a great terminology for describing the retention order. If you are interested, see the "Hydrophobic subtraction model" of Snyder et al. This one describes the retention in a better way of C18 molecules.
2)

In the chromatogram above (specifically chromatogram "a") and the
associated table, the elution time for the three compounds ranged from
6.60 minutes to 9.76 minutes. Since solvent A and solvent B were used, how do I know which compounds were eluted by which solvent?

This is an interesting question. As you noticed, the composition of the mobile phase changing with time, so the k, the retention factor is changing with time. People have worked out the theory of simple linear gradients. You know the retention time of elution, and you have the gradient composition chart. Correlate it, but this is a very, very rough estimate. The actual treatment is way more involved. Check "Theory of Gradient Elution Liquid Chromatography with Linear Solvent Strength: Part 1. Migration and Elution Parameters of a Solute Band" by Leonid Blumberg to see the complexity of the problem of gradient elution.
One of the best books is  "Introduction to Modern Liquid Chromatography By Lloyd R. Snyder, Joseph J. Kirkland, John W. Dolan, 2011."


text:
What happens to vaporized metal in the atmosphere when it cools down?
Imagine you vaporize a chunk of metal like an iron meteorite or plutonium or whatever in the atmosphere. What happens to that metal when it cools down? Does it form dust and drop in the nearby vicinity or does it float away in the wind or something else?

Vaporized pieces  of metals may float when cooled in the upper atmosphere for many months. It depends on the dimension of the particles. A good proof of this movement is the famous fallout problem due to the residues of the nuclear tests carried out in the years 1950 - 1964. During these years, USA and USSR were engaged in a course to more power. They developed more and more powerful atomic bombs launched from the Pacific Ocean by USA, and from Kazakhstan by USSR. The last bombs thrown in 1963 were thousand times more powerful than the first bomb in Hiroshima. The residues of these bombs was of course discharged in the upper atmosphere, and dispersed all over the world by the winds. Radioactivity was measured all over the world. And it was steadily increasing in all parts of the world. Then suddenly in 1964, Russians and Americans met in Geneva and they both decided to stop their nuclear tests. They said that it was for promoting peace everywhere. It was a great relief in the world. But the truth is that both Russians and Americans knew that the amount of radioactivity had grown up to 1/10 of the point where all food would have to be decontaminated to be eaten in the whole world, and of course at home. So they stopped nuclear tests. And slowly the radioactivity measured all over the world decreased. There was a small increase when China exploded its own bombs. And now the radioactivity of the soil has been probably washed to the oceans. On the ground, in food and people, it is back again to its level of before World War II. Today, the only measurable radioactivity is due to rare nuclides like K-40 and C-14 which are naturally present everywhere. We are used to them.


text:
Constitutional Isomers
Is there a formula to know how many constitutional isomers are there for a compound? For instance in this example, most students would miss the 6th constitutional isomer and will probably only write the first 5. So I would like to kindly ask if there's a formula or pattern for it.

Most students above the very entry level will count the 6th all right. As for the formula, well, if you want one that you'd be able to use, then it doesn't exist. Short of that condition, see https://oeis.org/A002094.

@ Ivan Neretin. I read German. But I have not been able to understand your publication, written in 1875 in German. They present a number with a cross of four numbers, somewhat like a matrix plus a cross. Have you been able to understand it ?

@Maurice I never bothered to go that deep. The OEIS page contains the formula (sort of); what's more to wish for?

@Ivan Neretin. Apparently you have been able to get this formula from OEISD. How lucky ou are ! I was not clever enough. Could you report it here ?

@Maurice Why, that's what they have in the FORMULA section.

@IvanNeretin. The formula for $\pu{A(x)}$ is simply unreadable. I copy it here: $$\pu{A(x) = 1 + \frac{1}{6} x} +\pu{(A(x)^3 + 3 A(x^2) + 2 A(x^3)}$$ It is impossible to solve it. And there is also a function $\pu{B(x)}$. What for ?

@Maurice Because A(x) is not the answer yet, but merely a step to it, and so is B(x),  and then there are more steps after that. See, I never said the formula is easy to read and implement. In fact, I said quite the opposite in my first comment.

There isn't a good overall formula that I know.
You can look at C/H ratio and easily tell a bit here.
There must be at least one of the following, but not two:

carbon ring
carbon double Bond

(1) 1-Butene

(2) 2-Butene which has two stereoisomers
cis-2-Butene
trans-2-Butene

(3) Isobutylene

(4) cyclobutane

(5) methylcyclopropane

Specifically, calculating the degree of unsaturation of the compound will tell you that either one ring or double bond exists.


text:
Meaning of the symbol “$\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$”
One of the back exercises (13(C)) from the book for practice provided at my school asked whether the following relation is true:

$\Delta_\mathrm{f}H^\circ(\ce{I2,g}) = \Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$ at $\pu{25 °C}.$

I know how to solve the question, but the answer would depend on the meaning of  $\Delta_\mathrm{sub}H^\circ[\ce{I2,s}].$ Which of the following reaction does the symbol $\Delta_\mathrm{sub}H^\circ[\ce{I2,s}]$ represent?
$$
\begin{align}
\ce{I2(g) &-> I2(s)}\tag{a}\\
\ce{I2(s) &-> I2(g)}\tag{b}
\end{align}
$$

So Chaitanya Garg's answer is correct. And the answer is (b).

Sublimation by definition refers to phase transition from solid to gas (no textbook or reference required I would think, but see: https://en.wikipedia.org/wiki/Enthalpy_of_sublimation).

@BuckThorn I'm more interested in why they used square brackets. Have you ever seen them used like this for the thermodynamic quantities? Or am I looking for the meaning where there is none?

@andselisk Rather odd to have different types of brackets on both sides of the OPs equation. Typo?

$\ ΔH_{sub}(I_2,s)$ can be interpreted as the change in enthalpy when one mole of solid $\ I_2$  converts to gaseous $\ I_2$  at a constant temperature. So the answer should be reaction (b).

Could you please provide with a reference text that says so.

@Aseem Mittal, Pg No. 50 Atkins' Physical Chemistry by Peter Atkins and also Pg. No. 165 of NCERT textbook for class 11th.


text:
Why is ruthenium dioxide a conductor while titanium dioxide is an insulator?
Ruthenium dioxide has the same crystal structure as rutile (titanium dioxide), and the metal atoms are in the same oxidation state.  Yet ruthenium dioxide is a metal-like conductor (its resistivity is given by Miao et al. [1] as $\pu{40 \mu\Omega{} cm}$ for thin films), while titanium dioxide is an insulator with a resistivity of at least $\pu{E13 \Omega{} cm}$ and a band gap of over $\pu{3 eV}.$
Why are the electrical properties of these two ceramics so different?
Reference

Miao, G. X.; Gupta, A.; Xiao, G.; Anguelouch, A. Epitaxial Growth of Ruthenium Dioxide Films by Chemical Vapor Deposition and Its Comparison with Similarly Grown Chromium Dioxide Films. Thin Solid Films 2005, 478 (1), 159–163. DOI: 10.1016/j.tsf.2004.10.032.

Well, Ru has more valence electrons.

More precisely unpaired electrons.

https://chemistry.stackexchange.com/questions/81724/how-is-solid-titaniumii-oxide-an-electrical-conductor

Note also that there are a number of factors affecting conductivity in metal oxides such as temperature, doping (impurities), and geometry (e.g. thin layers or needles vs. bulk). See http://collaborations.fz-juelich.de/ikp/cgswhp/cgswhp12/program/files_batumi/14-08-2012/Parallel_Session_5/1_Kazimierz.Conder_Batumi.pdf for a nice summary!

Diamond is a wide band gap insulator/semiconductor (depends on your point of view). Silicon is a good semiconductor, Germanium is ok with a lower band gap. Then alpha-Tin is a semi-metal. Just shows that chemistry does not uniquely translate to solid state band structure.

https://www.sciencedirect.com/science/article/pii/S1369800112000066 shows the band structure. https://onlinelibrary.wiley.com/doi/pdf/10.1002/pssa.2210690165, while older, has useful information. Bottom line - like crystal structures do not imply like electronic structure (see also C, Si, Ge, Sn above).


text:
Chemical Volcano- Experiment
In a science exhibition in 2011 , I and my  friend made a volcano that used vinegar and baking soda for the 'effects'. There was another group that made volcano too and they had brought their own supplies. But before the exhibition started, the members of other group came to us and asked us if we could share our baking soda because their baking soda wasn't 'working'. When I asked them why it wasn't working they told me that there wasn't 'salt' in it. (We were only 8-9 years old). When I tried using it it was forming bubbles but very slowly. Eventually our supplies ran out before the exhibition was over. One of my friends suggested that we should use chalk powder instead of baking soda(I don't know how he figured it out at age 9). So we ground up chalks and tried using it. In this case , it was just like the baking soda that didn't work , it was producing bubbles ,but very slowly.
What could be the issue in the two cases? Was it a matter of purity?
We had used synthetic vinegar used for food purposes(5-10%)(There was no issue with the vinegar because it worked well with our own baking soda). The chalks were ordinary chalks used in classrooms. Regarding the other group's Baking soda I do not have any other information cause it happened 10 years earlier.(And I'm no longer in contact with them)
Was reaction less fast in case 2(Chalk powder) because of the low solubility product of Calcium carbonate? But Solubility product is a thermodynamic factor and not a kinetic factor right ?

It must just have been a question of dimension of the grains. It is well known that when powders reacts, the reaction rate is proportional to the contact surface, So the contact surface is biggest when the grains are thinnest. If the baking soda had been ground with care, it must react quicker with vinegar.

It always works properly with real baking soda. No salt is needed. Calcium carbonate has poor solubility in water, e.g., white cliffs of Dover, etc., so you would need to grind it to fine powder to get high surface area contact with the vinegar. Even then, the bubbles impede contact between the vinegar and powdered chalk. Sodium bicarbonate has reasonable water solubility, and vinegar is just a dilute aqueous solution of acetic acid. So mixing them always works well. Not much more to say: use inexpensive fresh baking soda.

Baking soda is sodium bicarbonate NaHCO$_3$. It is stable in air indefinitely. Baking powder is a mix of sodium bicarbonate and a powdered acid, typically an aluminum or calcium hydrogen phosphate, which, in the absence of moisture, is stable for a long time (maybe a year? maybe more). After exposure to moisture, it will release CO$_2$ and become inactivated.
The third item, "chalk", has many definitions. Blackboard chalk is not just a stick of CaCO$_3$. A stick of CaCO$_3$ could be marble, or limestone - oh - no, you would like it to be nice and powdery but sticking together. Yep, it can be done: glue. Nah, too scratchy. Plaster of Paris! Mix with water, add some CaCO$_3$ chalk, maybe some white clay... You get the idea. Blackboard chalk is not just CaCO$_3$. Not even that, but CaCO$_3$ reacts very fast with hydrochloric acid (but that's too hazardous for grade school experiments), and its reaction with vinegar (5% CH$_3$CO$_2$H) is slower initially, and slows down markedly as the reaction progresses because the acetate ion produced inhibits the ionization of the remaining (weak acid) CH$_3$CO$_2$H.
The key lesson is that demonstrations are very temperamental, and depend on doing the same thing with the same materials many times, to see how the "same" experiment can vary so much. You may even begin to doubt that any experiment can ever be duplicated - but it can, if you keep everything constant.


text:
Hazard and precautionary statements
Is there a scientific database where one can find all the H and P statements for all chemicals? I would like to avoid using wikipedia but I can't find a database from a trusted source.

What do you mean by "all chemicals"? And what's wrong with the references usually posted at the end of every Wikipedia entry that are usually showing the origin of GHS statements?

@andselisk I meant the majority of chemicals. and the wikipedia entries aren't cited so the origin is unknown. I have just checked this with 2 examples.

There are some online sources such as Msdsonline.com, but I wouldn’t count on any of them being comprehensive. For common chemicals, though, just searching “msds [chemical name]” will often turn up something

The most comprehensive one (so not "all" chemicals) that I've seen is the Aldrich website. By law they have to supply MSDS for everything they sell. However, they may ask for a lot number or the function may not work unless you're logged in as a customer.


text:
Dissolving CNE (cresol novolac epoxy) resin covered computer chips
I intend to destroy (dissolving would be optimal) a few computer chips (NAND memory chips used eg. on SSDs to be precise).
Why? See foot note.
I am not a chemist, so of course I tried to research the topic a bit before actually attempting anything. I got to know that they are covered in a resin called CNE and that there is surprisingly little entry-level info on liquefying your electronics apart from "Get gold from scrap electronics" youtube videos - so I decided to start there.
Most common scenario described there is to use sulfuric acid heated to 90 degrees celcius which should remove the epoxy coating, then applying the nitric acid mixed with water or nitric acid mixed with hydrochloric acid (like here - link)
However: I do not intend to extract anything from these chips - I just want to have them destroyed, so that data recovery is impossible. Is there a substance (possibly available to me - just a common guy with basic protection, not a specialized lab) that could liquify the whole chip (that is CNE resin on top, and at least some of it's metal contents)? If there is such a substance - can it melt the electronic without the need of having the heat supplied from the outside?
Please do not suggest thermite or "spill gasoline all over it and light it". I am looking for a serious answer - preferably one without igniting or putting something on a heat source. I am wondering if "spill and forget" or "put it in acid bath and forget" is possible at all.

Explanation of my use case: me and my buddy work in computer security field, apart from work it is also our hobby :) The topic of removing data from hard drives is often brought in our discussions and has been widely discussed on stackexchange sites (eg. this question or this one - there's even FAQ entry on data removal). There are DoD and NIST documents mentioning melting of drives. However: while there are often mentions of melting the hard drive - nobody  is actually providing useful details.  My mistake, the documents mention melting which is of course not the same as dissolving (thanks @Mithoron).

Put in fire and the job's done. BTW melting is not dissolving.

A grinder does a good job of destroying chips. Low tech and pretty safe.

Darn, I've just made primary school level mistake (facepalm). My bad, guys - thank you for your input - however I am still interested in getting an answer for dissolving the chip, or as much from it as possible. I will fix the question.

Following Mithoron’s lead, there is always the time-honored liquid oxygen and charcoal grill method: https://youtu.be/sab2Ltm1WcM. Or a Bessemer furnace. Or a plasma cutting torch. Or a high power carbon dioxide laser.

@Ed V Thank you for your answer, but since his remark I have clarified my question - I am looking for a way to dissolve not melt :)

I know, but it is extremely unlikely that there is a way to dissolve the pesky CNE. Total destruction, however, is always possible! Good luck!

Damn, I didn't know this stuff is so resistant. Thank you!


text:
What would a (gas) chromatogram look like if two compounds have the same retention times?
In gas chromatography, the number of peaks represents the number of compounds in an unknown sample, as retention times for each component differ.
But what would happen if two components have the same exact retention time? Would you see two separate peaks or one?

Please avoid clickbait terms in titles (ie "urgent")

Why would you get two?

] Complete overlap (= identical retention) is a regular feature of experimental chromatography with very different compounds and complex mixtures. It just happens when the selectivity happens to be equal to 1.00. You can have one, two, or three different compounds hidden under a single peak and this usually happens in early eluting peaks in all modes of chromatography of a complex mixture.
Peak overlap is undesirable, but it happens all the time. All you need to do is to change the method.
People have spent a portion of their life in something called as peak deconvolution to find out what is hidden in a single peak. Note how the area adds up.

Have a look at this one from reference (1) [my highlight]:

In the simple case, the use of retention data for identification purposes involves
merely a comparison between the retention time of an unknown and that of a series
of authentic compounds. Since the essential ability of chromatography is to separate
substances, a negative conclusion - these two compounds can be separated, thus
they are different - can be reached with certainty. A positive identification, on the
other hand cannot; there are just too many compounds which may have the same
retention time. The odds can be improved very simply, by just measuring the
column efficiency for the unknown and for the mixture containing both the
unknown and the suspected compound in the same amounts. If the measured
efficiency is the same and if its precision is about 108, the compounds could not be
resolved with a resolution unity, by a column about 10 times more efficient than the
column used. We must emphasize again that a positive result for this test is not a
proof that the two compounds are identical.

I firmly recommend at least two reliable methods of identification, ideally more, not just gas chromatography.
Sources

Guiochon, G. and Guillemin, C.L. (1988) Chapter 11. Qualitative analysis by gas chromatography: The use of retention data., in Quantitative gas chromatography, J. Chromatography Library 22:481-529. DOI: 10.1016/S0301-4770(08)70083-6

Strictly same retention time means exact same signal in GC (i.e."apparently" same compound). This means you'd have one peak for the mixture as if it was a single compound. On an unknown sample, you might miss the fact that there are two compounds coeluting.
To avoid the issue, you would always inject the sample on (at least) two columns with different polarities. If your GC is set up properly, you can do this in parallel without additional effort, it is definitely good practice. I have been trained to use gas chromatography that way.

The signals add up. If compound A and B peaks merge into a single peak, their areas should sum up. It is a good idea to test two orthogonal columns.

Or you could use GC-MS to determine peak purity.


text:
Variation of battery potential with current density
It is found experimentally that the potential of an electrochemical cell with applied current density varies as follows:

To produce the passage of current through a cell it is necessary to apply a potential difference that exceeds the equilibrium f.e.m by at least the overpotential of the cell. Furthermore, the potential of the cell will decrease as it generates current, because under such conditions it no longer works reversibly and therefore does less work than the maximum. When the current is high and close to the limiting value of one of the electrodes, a sharp drop in the work potential is observed.
I know this is what is observed, but I can't explain why. Why does this happen?

What is the meaning of "potencial de salida" and "potential de la pila" ? Other question ? Are you speaking of an electrolysis or of a cell producing the currant ?

The "potencial de la pila" is the voltage measured during the experiment with a multimeter. The "potencial de salida" is simply the product of the current and the voltage. I am referring to the case of a fuel cell, such as a methanol cell. @Maurice

The voltage that you measure is equal to the redox potential $E$ if the current is zero. If the current, the measured voltage is equal to ${E - rI}$ where $r$ is the internal resistance. In principle this internal resistance should be a constant. But in your reported data, it is obviously not the case. The internal resistance is about $\pu{1 ohm}$ in the beginning, and decreases when the current is increasing. This is surprising, and I don't know its reason.


text:
Calcium chloride as a drying agent for solutions
I'm aware of the reaction Calcium chloride has with water vapour and how it can therefore be used as a good and inexpensive drying agent but could that be applicable for solutions?
For example if there is an Ethanol, water mixture which you want to dehydrate could you add Calcium chloride so it forms a complex with the water molecules and then destill the solution to obtain dehydrated ethanol?

You can do that with some solvents, but not with ethanol.

CaCl2 binds with  6 H2O or  4 CH3OH or 3 CH3CH3OH, therefore not applicable for drying ethanol

Yes, See: https://chemistry.stackexchange.com/questions/54878/methanol-adducts-with-cacl2

IF you want to get really good drying for small alcohols, the better choice is the right molecular sieve. Some (type 3A sieves for example) have fine-tuned cavities in their structure where water molecules can fit but ethanol, for example, cannot fit. They are the preferred choice of drying agents for several small molecule solvents in the lab. CaCl is a far worse drying agent for those solvents.


text:
Small scale gas-to-liquids via Fischer–Tropsch — is it even possible?
I'm really interested in the Fischer–Tropsch gas-to-liquids (GTL) process. A lot of people are using wood or other biomass feedstock to make syngas on small scales (farms, shops, etc.).
Practically, what would be required to build a small scale GTL reactor to run a Fischer–Tropsch process on a small scale in a shop? Is that even possible to do safely?

Josh Velson wrote about it on Quora. Take a look at this and this.

It probably isn't practical to do Fischer-Tropsch on a small scale
Gasification of biomass is easy. And produces syngas, a mixture of carbon dioxide, carbon monoxide, hydrogen and other compounds.
But the Fischer-Tropsch process to convert that mixture into useful liquids (like diesel) is not easy. The process requires carefully designed reactors, controlled heat and specific catalysts which are poisoned by many of the typical minor constituents of biomass originated syngas (especially any containing sulfur). So industrial reactors have to clean the syngas so the catalysts are not poisoned.
It is unlikely that you will be able to build a carefully controlled reactor that guarantees the right conditions (moderate but controlled temperature and pressure, complex geometry, specific catalysts, clean syngas). Even less likely that the result will be remotely economical on a small scale given the required cost and energy consumption. Heck, major chemical and oil firms have been trying for decades and few of their efforts have been  economically viable.
And there is another reason why this is futile on a small scale. Syngas itself is a usable fuel without further processing. It has been used to power engines (when diesel or petrol/gasoline are scarce as they were during WW2). And variants of syngas were once common as a piped domestic fuel for heating and cooking (before natural gas which is mainly methane became far cheaper and less toxic for most customers). So why not just burn the syngas?


text:
Chemical difference between dexedrine and methamphetamine
I am a newbie at chemistry and wondering what the key difference between dexedrine and methamphetamine HCL is. I know that methamphetamine has an extra methyl group compared to Adderall. But what about dexedrine?

Read this Wikipedia article.

I have read these articles....I want an intuitive explanation.

Adderall
Adderall is a drug(trade name) that contains a combination of four salts of amphetamine (equal parts racemic amphetamine and dextroamphetamine, which produces a (3:1) ratio between dextroamphetamine and levoamphetamine, the two enantiomers of amphetamine).
Racemic amphetamine: levoamphetamine and dextroamphetamine respectively. Notice the dash-wedge notation. The methyl group is pointing away from the reader in the former structure and is pointing towards the reader in the latter structure.

Enantiomeric amphetamine: chiral molecules that have a mirror orientation of one another. See this question.

Dexedrine
Dexedrine is the trade name for dextroamphetamine
Methamphetamine
It is a drug similar in structure but has an extra methyl group (So, primary amine becomes secondary amine).

Also, do note that a simple change in the drug structure can change the potency.

The difference is that it is not the same molecule (group $ \ce {NH_2} $ for one, group $ \ce {NH-CH_3} $ for the other) so not the same properties.

The N-Methyl is important because it changes the metabolism profile

@Waylander: So would you say that methamphetamine and adderall are more similar to each other compared to Dexedrine and Methamphetamine?

No - Adderall contains the 2 enantiomers of ampthetamine, dexedrine is a single isomer. Neither contain methampthetamine.

@Waylander: But methamphetamine and adderall feel.subjectively similar.

A general principal in medicinal chemistry - a small change can make a huge difference


text:
Propagating uncertainty with the slope of a tangent line
I have done an experiment that found the initial reaction rate for a reaction that created gas in $\pu{kPa s^-1}$ by finding the slope of the tangent line at time = 0. How do I propagate uncertainty for this? Is there any uncertainty propagation that has to be done?
$$\text{slope} = \frac{\pu{(99.726 ± 0.01) kPa} - \pu{(99.71753 ± 0.01) kPa}}{\pu{0.076 s} - \pu{0.061 s}} = \pu{0.56 ± 0.02 kPa s^-1}$$
This is what I have so far. My teacher said that the error for the time was so small that it is negligible (but not a perfect number). I do not think using percentage uncertainty would make sense here because you could measure the slope with however big or small numbers you wanted, depending on what two points you chose to use.
I got Vernier Logger Pro to do the tangent and slope calculations for me which is why there are more decimal places than uncertainty. Not rounding until the end. Is that wrong?
These are the instructions I was given:
only a simple treatment is required. For functions such as addition and subtraction, absolute uncertainties can be added; for multiplication, division and powers, percentage uncertainties can be added. If one uncertainty is much larger than the others, the approximate uncertainty in the calculated result can be taken as due to that quantity alone.

You have not propagated your errors properly. Note that the sum in your nominator evaluates to a small number with an error of roughly the same size (99.73±0.01 - 99.72±0.01 = 0.01±0.01). Also, error propagation cannot be done by simply adding up the errors.
Instead of calculating the error propagation manually (cumbersome, and prone to errors, hehe ;-) I think it's better to use suitable software.  Both R and Python (both are freely available, commonly used tools) can handle error propagation.
Here's how to do it with R (the code below should be reproducible, assuming you have installed R).
install.packages("errors")
library(errors)
options(errors.notation="plus-minus", errors.digits=1)
y1 <- set_errors(99.726, 0.01)
y2 <- set_errors(99.71753, 0.01)
t1 <- set_errors(0.076, 0)
t2 <- set_errors(0.061, 0)
(y1 - y2) / (t1 - t2)
# 0.6 ± 0.9

The value of the slope is 0.56 kPa/s, but since the error is 0.9 kPa/s, we should report just a single significant digit, so that makes it 0.6 ± 0.9 kPa/s.
And not rounding till the end is not wrong, in fact, it's good practice. As long as you handle your uncertainties correctly, too many significant digits (due to intermediate calculation steps or whatnot) become practically irrelevant, since your value cannot be more precise than the size of your uncertainty anyway.
Good luck!

Thank you! I am thinking that that might be a bit too advanced for me... The instructions I was given is " Only a simple treatment is required. For functions such as addition and subtraction, absolute uncertainties can be added; for multiplication, division and powers, percentage uncertainties can be added. If one uncertainty is much larger than the others, the approximate uncertainty in the calculated result can be taken as due to that quantity alone" I forgot that it's more complicated irl. I realize I probably should have put that in my original question, oops! Does this change your answer?

Thanks for the clarification. Assuming the errors you reported were absolute errors, then no, the answer is the same (mostly).
The simplified error propagation method you suggested is probably fine for homework, but would underestimate the final error in this case.
First we need to determine the relative error in the nominator: 0.01 / 0.0085 = 118%.
Then we can perform the division while adding the relative errors: 0.0085±118% / 0.015±0% = 0.565±118%.
So using the simplified method, the final value becomes 0.6±0.7 kPa/s, by my reckoning.

I feel like using R for a calculation as small as this is a bit of an overkill.


text:
Disparity between LeChatelier's Principle and Solutions
Let's say that there is a solution
$\ce{Na+(aq) + Cl-(aq) <=> NaCl(s)}$
If the above solution is at equilibrium we can say that it is saturated and hence no more Sodium Chloride will dissolve. However, according to LeChatelier's Principle, if we add more Sodium Chloride it will just dissolve. Is there another rule that governs how this kind of dissolution behaves?

You misinterpret the principle. It does not say if you add more salt, more salt dissolves. More generally, the principle focuses on changes of intensive properties, that do not change with the system scaling.
The principle says if the thermodynamic activity of the solid had increased by adding more salt, more salt would have dissolved to decrease this activity. But by adding more salt, the salt activity remains constant.
You can also consider  chemical potential $\mu = \frac{dG}{dn}$  instead of activity, as they are closely related.
By other words, thinking more salt dissolves by adding more salt to saturated solution is equally wrong as thinking more salt crystalizes by adding more of saturated solution. As the salt activity in saturated solution or in solid state does not depend on their amount, which is extensive property.

What do you mean by "activity"

See the answer update.


text:
Why does nuclear waste contain plutonium?
According to this page on Wikipedia:

Reactor-grade plutonium (RGPu) is the isotopic grade of plutonium that is found in spent nuclear fuel after the uranium-235 primary fuel that a nuclear power reactor uses has burnt up.

Since plutonium is not found naturally, how does it end up at nuclear waste from nuclear fission reactors?

Because that is the way how 239Pu is created. ( 238U + n ->  239U -> 239Np -> 239Pu ). What is the desired intermediate product in 239Pu production, is a waste in 235U fission based power plants.

@z1273 Thanks! Also, the tour and help center links are for the biology stackexchange :)

Wikipedia:

The fissioning of an atom of uranium-235 in the reactor of a nuclear
power plant produces two to three neutrons, and these neutrons can be
absorbed by uranium-238 to produce plutonium-239 and other isotopes.
Plutonium-239 can also absorb neutrons and fission along with the
uranium-235 in a reactor.

As noted by @IanBush, Pu-239 has a half-life of 24110 years and Pu-240 has a half-life of 6560 years, so they will stay as it is for years before fissioning to any other isotopes.

It's worth nothing Pu-239 has a half life of ~24,000 years, and Pu-240 ~6500 years, so once they are produced by the above mechanism they are going to hang around for a while, and not just disappear in minutes


text:
Predict the valence configuration of this element using the first five ionization energies
If the first five ionization energies of an element are, respectively: $\pu{1.09 kJ/mol}$,  $\pu{2.35 kJ/mol}$, $\pu{4.62 kJ/mol}$, $\pu{6.22 kJ/mol}$ and $\pu{37.83 kJ/mol}$, to which group of the periodic table does this element belong? Predict the valence configuration of this element. Justify your reasoning
I don't understand how to solve this. How do I relate ionization energy to finding the element, then predicting the valence configuration?

Those ionization values are too small. Did you copy them correctly?

You expect ionization energies to increase as you remove more electrons from an atom, because the charge of the nucleus remains the same (and therefore its attraction to the remaining electrons) but repulsion from other electrons is reduced with removal of each additional electron. However, in addition to this simple intuitive trend, quantum mechanics throws in a twist, namely the existence of electron "shells" about an atom, sharing similar energy (the same principal quantum number). Electrons in the same shell repel each other with the same intensity. This is the justification for Slater's Rules.
There is a third way to solve this: look up values in a table of ionization energies. It helps ensure accuracy to have "second and third opinions", so this is a good exercise. From the Wikipedia  (in $\pu{kJ/mol}$), the values are 1086.5, 2352.6, 4620.5, 6222.7     37831, 47277.0.


text:
Chemical safety information: Where to look to find out how poisonous poisons are?
Could someone point me toward a primer on how to read the toxicity warnings on standard chemical data sheets? Or alternatively, where one should look for more precise information than is contained on such sheets? Specifically, when I see that "Poison" skull and crossbones, I don't know how to tell when it means "Don't drink a quart of it" as versus "If it splashes on your finger, chop it off immediately."
I looked at a data sheet for rosmarinic acid, a component of rosemary, and the Caymen Chemical data sheet it said it was acutely toxic if ingested. So I looked at a data sheet for curcumin, which I have been taking three grams a day of for ten years, and the ThermoFisher data sheet said to seek medical attention immediately if you get any on your skin. That is clearly not right, or I would be dead. Which makes me wonder, in general, where to look for safe levels of exposure, and specifically, how to tell toxic rosemarinic acid (which has been reported to increase the average lifespan of c. elegans by 63 percent*) is to humans.

Biogerontology 12(4): 329–347. doi:10.1007/s10522-011-9334-7.  Biofactors 45(5): 774–787. doi: 10.1002/biof.1536.

Indeed reading the safety data sheets is a bit confusing. NaCl is quite surprising as well. One important point - I guess - is that they serve as regulatory in industrial environment. So I am not surprised to read about NaCl, but then I am confused while reading about chemicals which are exotic even for a chemist....


text:
Why are complexes of type MA3B3 not optical active?
Complexes of type $\ce{MA3B3}$ have two geometrical isomers, namely:

fac-isomer
mer-isomer

If we look closely at the mer-isomer, it has a plane of symmetry, so it is optically inactive.
But the fac-isomer has no plane of symmetry. Still, it is optically inactive, why?
Being a 12th grader, I wonder where am I wrong in my understanding of optical isomerism in coordination complexes.

Because of three planes of symmetry, to begin with.

There is no plane of symmetry in fac-isomer of type ${MA_3B_3}$

Your assertiveness is not quite justified.

there is a plane of symmetry in the fac- along the axis AMB and between two A and two B atoms. see the image here https://courses.lumenlearning.com/introchem/chapter/isomers-in-coordination-compounds/#:~:text=In%20a%20fac%20isomer%2C%20any,cis%20pairs%20of%20identical%20ligands.

@IvanNeretin I understood the point . I missed the Plane of symmetries along AMB and 2 As and Bs. That response was quite irresponsible.

Build a model of your two isomers. There are planes of symmetry in both, but they can be hard to see if you don't have a 3D model.

https://www.chemtube3d.com/symfecl3br3fac/

@ShoubhikRMaiti That helped a lot !

Orient the fac- isomer so one A atom is facing towards you and a space between two B atoms is below it.  Then the left half of this view is a reflection of the right half so you are actually looking down a mirror plane.
Because you could have chosen any of three A atoms in your view there are three mirror planes, which pass through a common rotation axis.  The axis passes through the face that has all A atoms at its corners and the opposite face which has all B atoms.  This symmetry is called $C_{\mathrm{3v}}$ in the language of point groups.


text:
Why is it that protons and electrons undergo the same amount of deflection in an electric field if they have the same energy?
The question I have is with respect to this diagram. Which depicts a stream of protons and electrons entering a proton field with equal energy.

Why is it that in the case where the energy of protons and electrons are equal, their angle of deflection is exactly the same? However when speed is kept the same, the deflection of electrons is greater. Why is this so?

a better quality image would be appreciated and appreciated! In an electrostatic field, the force undergone is worth $\vec{F} = q \vec{E} $ therefore the force depends only on the charge of the particle

I’m voting to close this question because it's not about chemistry.

@Mithoron it's about mass spectrometry, isn't it?

@LevelRiverSt: It's a question about basic electromagnetism, which'd commonly be accepted as a straightforward Physics topic.

@Nat it had not escaped my notice that was about basic electromagnetism. I was pointing out that OP had presumably posted it due to its application in mass spectrometry, an important technique in analytical chemistry. If so, I think it has a place here.

Because they have equal (but opposite) charges. That is all.

Now that's a mildly non-trivial observation. Why would they be equal, really?
Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the acceleration $a={qE\over m}$. By the time the particle reaches the right side of the picture, it gets deflected by $y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$.
That pretty much sums it up. $E$ and $L$ are the same for everyone, $q$ is the same in absolute value for electron and proton, so the only difference is $mv^2$. Now, considering your two situations:

Same speed: same $v$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
Same energy: same $mv^2$, same deflection (just to the opposite sides).

Things get hairy in relativity, but that's another story.
So it goes.


text:
Out of KCl, CaCl2, MgCl2, NaCl, which causes the greatest corrosion rate of iron and why?
I know that salts act as an electrolyte in a redox process and rusting of iron, where iron loses electrons and oxidizes, and oxygen gains electrons and reduces. I did an experiment with KCl, NaCl, MgCl2 and CaCl2 salts dissolved in water, then dipped iron nails in the separate solutions to see which nail would rust more over a period of time. My result was MgCl2>CaCl2>KCl>NaCl. For my analysis I'm looking for an explanation as to why MgCl2 solution had the most rust, the CaCl2 the second most ect. I wondered about periodic trends and how I could look at electron affinity to see which salt could accept electrons the easiest, but I'm not sure if it's right. I just want help on this chemical process and an in depth explanation for my results. Thanks!

It would be nice to know how you pre-treated surface of the nails, what steel they are made of, what temperature and concentrations of the salts were used, and how you estimated rusting quantitatively.

I guess it all boils down to "how much chloride".


text:
What is the correct Lewis structure of difluorophosphate?
I can draw two possible Lewis structures for $\ce{PO2F2-}$ anion:

What is the correct structure and why is the other one wrong?

I would prefer one with plus on P and minus on each O (if the matter were of any significance, that is).

@lvan but wouldn't the increased amount of charges in a molecule would make it unstable ( - repells- )

If anything, given that O is much more electronegative than P, my structure is slightly more realistic in terms of charge distribution. Also, it correctly reflects the fact that both O atoms are identical.

@ivan neretin there’s a difference between the most accurate representation of the molecule and the conventional Lewis structure. Lewis structure conventions prioritize minimal formal charge over adherence to octet rule for third row elements (but the reverse for second row). I

@Andrew so in your opinion what is the correct structure in above Lewis structures?

Definitely structure 2


text:
Aromatic Substitution in acidic medium

Find product A.
I tried solving but why isn't the answer this
The answer is given this

Three parts to this answer:

You have two electro-philes, the tertiary carbo-cation formed by protonating the alcohol and breaking off a water molecule and then the bromine.  Looks like you got this part.

Assume the two electro-philes attack consecutively.  The methoxy group is more strongly activating and ortho/para directing than the alk-yl group for the second attack.  So the bromine and the newly formed carbon-carbon bond will be meta to each other.

The sterically favored positions for each are as shown in the given answer.  The carbo-cation favors the position para to the methoxy group.  The bromine picks the position ortho to the methoxy group that's away from the alk-yl function.

Please excuse my use of hyphens.  Device auto-correct was putting in wrong words and I had to defeat it.

I think you will also get some bromination in the other position next to the -OMe

Probably so @waylander, and maybe even the carbo-cation could do the same.  But that position is sterically less favorable and presumably the most favored product is wanted.

@Oscar_Lanzi  Murphy's Law applies. I suspect that "A" will be a mixture of at least 4 compounds in the real world.

One of Murphy's Laws is everything in textbooks is much simplified compared with the real world.

Never was a truer word spoken!

The group $\ce{O-Me}$ is an o/p orientator for the second substitution not meta as you answer

But i want to know the mechanism

the attack of $ \ce {H ^+} $ leads to cyclization

$ \ce {Br_2, FeBr_3} $ forms the electrophile $ \ce {Br ^+} $ which can a priori be substituted for $ \ce {H} $ in the ortho or para position of the benzene ring except that here the para position is not allowed because of the cycle formed in step 1. Write the Wheland intermediates that form when $ \ce {Br^+} $ arrives on the ring: you will see that the ortho position is favored because the substituent $ \ce {O-Me} $ already in place stabilizes the intermediary.

For the choice of the 2 possible ortho positions which are favored, one is sterically less favorable


text:
Question regarding chemical equilibrium when adding a salt to acid solution
Having read the following was wondering something:
$$\mathrm {p}K_\mathrm{a} = -\log\frac{[\ce{H+}]_\mathrm{eq}[\ce{A-}]_\mathrm{eq}}{[\ce{HA}]_\mathrm{eq}} $$
$$\ce{HA <=> H+ + A-}$$
$$\mathrm{pH = p}K\mathrm{_{a}} +\mathrm{log\frac{[\ce{A-}]}{[\ce{HA}]}} $$
This rmgd
$$ \mathrm{[\ce{A-}] = [salt] }$$
when acid is weak.
Now my question is like this: take a beaker of $\ce{CH3COOH}$. Now add $\ce{CH3COONa}$. Now here using the a over formula as salt concentration increases $\mathrm{pH}$ should decrease when take in $\ce{A-}$ as salt concentration. However this same $\ce{A-}$ was there in the $\mathrm{p}K_\mathrm{a}$ formula so shouldn't the $\mathrm{p}K_\mathrm{a}$ change to compensate? If so then there will be no $\mathrm{pH}$ change but I have read that there is $\mathrm{pH}$ change on adding $\ce{CH3COONa}$ in a solution. What is wrong with this?

The first equation is wrong : it’s Ka or not pKa

It is now editted .

The first equation only works if the system is in equilibrium. I added to subscripts to indicate that.

indeed @KarstenTheis, I corrected my answer as well

the second relation is wrong to : $pK_{a} = -\log\frac{[\ce{H+}]_{eq}[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}=-\log[\ce{H+}]_{eq} -\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$ so $pH=pK_a+\log\frac{[\ce{A-}]_{eq}}{[\ce{HA}]_{eq}}$
the acidity constant is the relation which links the concentrations of the species present (and not the reverse): if we add the salt $ \ce {A^-} $, then according to the relation you see that the pH increases. You can also see it by reasoning thus: you add a base so the pH can only increase.


text:
How is the selective complexometric titration testing equation simplified?
When there is $\ce{N^m+}$ and $\ce{M^n+}$  metal ions in a solution with respective m and n charge,
$\ce{N^{m+}, M^{n+}}$ reacts with EDTA($\ce{Y}$) as follows:
$$\ce{N^m+ + Y^4- <=>  NY^{m-4}}$$
$$\ce{M^n+ + Y^4- <=>  MY^{n-4}}$$
Formation constant for each chemical reaction,
$$\ce{K_{NY^{m-4}}} = \frac{\ce{[NY^{m-4}]}}{\ce{[N^{m+}][Y^{4-}]}}$$
$$\ce{K_{MY^{n-4}}} = \frac{\ce{[MY^{n-4}]}}{\ce{[M^{n+}][Y^{4-}]}}$$
where as $\ce{N^m+}$ has the higher value for formation constent KNYm-4 > KMYn-4
Fraction of each ion exsists ${α}$, $\ce{[N_{total}][M_{total}][Y_{total}]}$ are the total concentraion of the ion in any form.
$\ce{α_{Y^{4-}}}$ = $\frac{\ce{[Y^4-]}}{\ce{[Y_{total}]}}$
$\ce{α_{N^{m+}}}$ = $\frac{\ce{[N^m+]}}{\ce{[N_{total}]}}$
$\ce{α_{M^{n+}}}$ = $\frac{\ce{[M^n+]}}{\ce{[M_{total}]}}$
After $\ce{N^m+}$ has reacted with EDTA,the remaining amount EDTA is used for $\ce{M^n+}$
$$\ce{[Y^{'}_{total}] = [Y_{total}] + [NY^{m-4}]}$$
Where as $\ce{α_Y^{'}}$ is,
$$\ce{α_{Y^{'}}} = \frac{\ce{[Y_{total}]}}{\ce{[Y^{'}_{total}]}}$$
These are all the definition constants but I have no clue how this equation came
$$\ce{K''_{MY^{n-4}} = K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}} - α_Y^{'}}$$
To perform a selective complexometric titration this condition need to be satisfied,
$$\ce{K''_{MY^{n-4}} = K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}} - α_Y^{'} \geq 10^7}$$
where αY' is given below,
$$\ce{α_{Y'}} = \frac{1}{\ce{(1 + K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$
Using EDTA as for an example Y4-
How did this equation came with the assumptions (The simplified equation)
$$\ce{K''_{MY^{n-4}}} = \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}]}$$
One of the assumptions would be
$$\ce{ K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] \gg 1}$$
Addition of one can be neglected.
What are the assumption do need to use to get the simplified expression
Can someone help me just using basic mathematics did not help I tried it by myself but no luck I am pretty sure I have missed some assumptions

Can you reference the book?

What are all those variables K'', K, K, Y', $N_{total}$ etc.? Also, please visit this page, this page and this one on how to format your posts competently with MathJax and Markdown.

I can define all the constants and respective chemical equations then the question will be very long?

Defined the constants now but $\ce{K''}$ is unknown

You have not involved acidobasic conditions, affecting [Y^4-]/[EDTA total] and pH changes due titration.

I finally found the answer to the question I have made a huge mistake in the testing equation
In order for a titration to happen the Formation constant $\ce{K}$ should be greater than or equal to $\ce{10^7}$
What I have written in my note book is

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}-α_{Y^{'}}}$

Which is incorrect it,
Should be like this

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}α_{Y^{'}}}$

The conditional formation constant of metal $\ce{M}$ should be multiplied by $\ce{α_{Y^{'}}}$ not subtracted
Substituting the equation for $\ce{α_{Y^{'}}}$ will give the simplified equation
$$\ce{α_{Y^{′}}=  \frac{1}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$
$$\ce{K''_{MY^{n-4}} =  \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}} $$
But since
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] >> 1 }$$
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] + 1 \approx K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] }$$
Which gives the following final simpified experssion
$$\ce{K''_{MY^{n-4}} =  \frac{K_{MY^{n-4}}α_{M^{n+}}}{K_{NY^{m-4}}α_{N^{m+}}[N_{total}]}} \geq 10^7 $$
Finally the testing equation is sloved for selective titrations


text:
Explain the mechanism of glycerol solubility in ether.?
I need some guidance regarding the issue of solubility, as you can see in my question i have glycerol and ether.
What factors play a role in the solubilization process between these two compounds?
I know that glycerol has polar character and intermolecular dipole-dipole forces, while ether (ethyl ether, i imagine) has less polar character than glycerol and Hydrogen bonding type intermolecular forces. But how can this type of solubility originate taking into account these aspects?
Excuse me for my English.
Thanks a lot...

Well water is soluble in diethyl ether (6.05g per 100ml at 25C)

According to a 1951 article in J. Am. Chem. Soc., water is soluble in diethyl ether to the extent of 1.468%, although other investigators had found between 1.19 and 2.66%. https://pubs.acs.org/doi/abs/10.1021/ja01150a531 And the solubility of ether in water is about the same.
Water and glycerol are completely miscible - no insolubility at all.
According to the CRC Handbook, glycerol and diethyl ether are immiscible. Of course, water and ether are also immiscible, but there is some slight solubility, so you might expect similar solubility between ether and glycerol. Rather than explain tiny solubilities, the immiscibility reaction suggests that the two compounds are just too different to be compatible.
While water is a good example of hydrogen-bonded molecules rejecting the less polar ether molecules, I see glycerol, a viscous group of hydrogen-bonded molecules just not opening up to let an ether molecule in. And ether is so non-polar that it doesn't even stick to itself very well (i.e., it's quite volatile), so instead of associating with glycerol, it vaporizes and leaves the system quickly.

Thank you so much.
Everything makes sense now, your explanation was very helpful to me to understand this case.

I love the downvote! The answer is acceptable to the OP, but not to some browser. I guess Einstein was right: "All problems should be solved as simply as possible, but not simpler." (Unless maybe you are a beginner.)