text:
What is the smallest molar volume?
I wondered how small a volume a mole of material could occupy, so I started with carbon, which would need 12 grams. That's 60 carats, and there happens to be a famous 60 carat diamond.

If my density/atomic mass calculations are correct, the substances with the smallest molar volume would be nickel, carbon, beryllium, and boron, with boron being the smallest. But I had trouble finding images of large-enough beryllium or boron crystals.
Are there any compounds of sufficient density and low enough atomic mass to beat boron? Can a maximal density boron crystal be made that large?
What substance has the smallest molar volume?
EDIT -- The Noor-ul-Ain diamond is closer to being 12 grams.

Apparently you are insisting on only single crystals? A mole is just a particular (large) number.

A small number of crystals would be fine, so long as it made for a good image.

If you're asking about molar volume in standard conditions, you should call it.

You're really asking about the smallest molar volume, not the smallest moles per se.

As pointed else where, diamond has the smallest molar volume ($\pu {3.42 cm^3/mol}$). The other 4 of 5 smallest molar volumes are: Boron ($\pu {4.39 cm^3/mol}$) < Beryllium ($\pu {4.85 cm^3/mol}$) < Carbon ($\pu {5.29 cm^3/mol}$) < Nickel ($\pu {6.59 cm^3/mol}$) < Cobalt ($\pu {6.67 cm^3/mol}$), according to Elements' Handbook. Iron and Copper are not far behind with molar volumes of $\pu {7.09 cm^3/mol}$ and $\pu {7.11 cm^3/mol}$, respectively.

Boron is a covalent solid with high melting point, like diamond (though not quite), and hence its crystals are hard to make. Unlike diamond crystals, they are not nice and probably wouldn't make a great display.
The table on http://periodictable.com/Properties/A/MolarVolume.v.log.html seems to corroborate your findings about boron molar volume being the smallest among all elements. Pity it's wrong, and so are you. (Or rather, it is technically right, but in a way that conveys a wrong impression.) Some elements just tend to have multiple polymorphs (called allotropes in this case), and carbon is one of them. All data in the standard tables are for the standard polymorph, which is graphite. But diamond at $3.5\;\ce{g/cm^3}$ is much denser, and decisively beats boron in the contest for the smallest molar volume.
Sometimes it takes walking around the world to realize that the aim of your quest has been in your pocket all along. The picture of the "smallest mole" is the one you brought here.
So it goes.

...but only barely

For the record: https://www.facebook.com/ed.pegg/posts/2733734803333970

A mole of neutrons in a neutron star would take up about $10^{-20}$ m$^3$. And in a black hole, they would be even smaller.

For completeness, metallic hydrogen is predicted to have a metastable state that would exist at standard temp and pressure (after forming at very high pressure) and would occupy a smaller volume per mole than diamond.

@Andrew: At STP? Really? That's the first I've heard of such a claim, and Googling it turns up this question on physics.SE with a rather skeptical answer.

Not that a black hole or neutron star is STP, either.  While Ice VII might beat diamond (on an atomic density basis) on Earth by being in a high pressure matrix (see another answer), diamond is probably the best that's strictly at STP.

To address you concern about boron, there is a cubic diamond form of boron nitride $\ce{c-BN}$, ICSD #182731 [1], which possesses $V_\mathrm{cell} = \pu{7.99 Å3}$, $Z = 2$ and the molar volume
$$V_\mathrm{m} = \frac{N_\mathrm{A}V_\mathrm{cell}}{Z} = \frac{(\pu{6.022E23 mol-1})(\pu{7.99 Å3})}{2} \approx \pu{2.406E-6 m3 mol-1},$$
which is about $30\%$ less than that for the diamond.
The only drawback is that this form of boron nitride is predicted to be stable above $\pu{11 Mbar}$.

Figure 1. Unit cell of $\ce{c-BN}$. Color code: $\color{#FFB5B5}{\Large\bullet}~\ce{B}$; $\color{#3050F8}{\Large\bullet}~\ce{N}$.
References

Qiu, S. L.; Marcus, P. M. Structure and Stability under Pressure of Cubic and Hexagonal Diamond Crystals of C, BN and Si from First Principles. J. Phys. Condens. Matter 2011, 23 (21), 215501. DOI: 10.1088/0953-8984/23/21/215501.

The "cubic" structure to which you refer looks tetragonal to me, unless the given cell is contained in a larger cubic structure because $c/a=\sqrt{2}$.  Also in a compound, is the molar volume based on atoms or "molecules"?

The cubic diamond phase stems, it seems, from the arrangement of B- and N-networks, separately resembling the diamond phase; and the elongation is caused by the shift between the both. And yes, you are right, to me it also looks like a tetragonal crystal system, although it's been deposited as triclinic for some reason. The molar volume is determined for the formula unit.

The abstract in the reference tells me that the dense phase os not the usual cubic boron nitride.  It's a collapsed phase that forms from the c-BN phase at 11 Mbar (and from cubic diamond at 13 Mbar).  We have to check the actual cell structure for the condensed phase.  I suspect the actual unit cell has only one formula unit so it's half as dense as the claim.

@OscarLanzi That's why I uploaded the init cell content. There are $(1 + 8\cdot\frac{1}{8}) = 2$ B atoms and $4\cdot\frac{1}{2} = 2$ N atoms, which makes up to formula unit BN and $Z = 2$. As for "collapsed" phase, I wasn't able to find the reported crystal structure, and the search among boron nitrides in ICSD reveals structures with higher molar volumes.

If you allow a mole of atoms, then some compounds come to the fore.  Like water.
Ordinarily, liquid water occupies $6.0\text{ cm}^3/\text{mol atoms}$.  Freezing this to ordinary ice (Ice $\text{ I}_h$) increases this volume slightly as water expands upon freezing.  But there are high pressure ice phases that are denser and thus give diamond a run for its money ... or maybe more.
Ice $\text{ VII}$ has been found on Earth as inclusions in diamonds. According to Wikipedia this phase has a density of $1.65\text{ g/cm}^3$, which translates to about $3.6\text{ cm}^3/\text{mol atoms}$.  But that is just at the minimum pressure for this phase, $2.5\text{ GPa}$.  At higher pressures, which can be maintained internally within the diamond lattice, this phase is fairly compressible because the hydrogen bonds can be squeezed towards a symmetric bonding arrangement (at which point we would have Ice $\text{ X}$).  So the densest arrangement of atoms naturally occurring on Earth might be not diamond per se, but Ice $\text{ VII}$ included within it.


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How does a change in the surface area of the electrodes affect the voltage produced by a galvanic cell?
I did a lab where I increased the surface area of the electrodes in a galvanic cell, that resulted in an increase in voltage. However, I have been reading and that is not supposed to happen. I am a bit confused, if you increase the surface area of the electrodes, the rate of reaction as well as the current increases, therefore, if Ohm's law ($V = IR$) is taken into account, the voltage should also increase. But according to some sources, the voltage should remain unchanged.
Could anyone please tell me if an increase in surface area will or will not increase the voltage produced by the cell, and why?

@NightWriter Please don't use \mathrm{...} where it shouldn't be used (I noticed you really like to put it everywhere in your edits). Please (re)visit this page, this page, this one and Math.SE Meta post on how to format your future posts better with MathJax and Markdown.

Ok, thanks @andselisk! Yeah, I have no idea when it should be used, I just go by appearances, like most people...

The standard electrode potential, $E^\circ$, in volts, does not depend on the surface area of the electrodes. However the standard electrode potential is measured with an infinitesimal current flow. In the simplest model you can imagine a cell, galvanic or electrolytic, as having an internal resistance. So as the current of the cell increases the voltage decreases. Larger electrodes would effectively reduce the internal resistance of the cell.

But don´t current and voltage have a positive relationship, meaning that if current increases, so does voltage? According to V=IR that is how it is, but again, other sources say what you are saying but I still cannot understand how that can happen.
Thanks for your help.

No, in the simple model $\mathrm{V}{obs} = i(R{battery} + R_{load})$

May I know what Vobs is?

Sorry, didn't mean to be obtuse -- Volts observed.

source: Stack Exchange

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

sort: votes


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Radius with Bohr model
$$r_n = \left(\frac{h^2}{4\pi^2me^2}\right)\times\frac{n^2}{Z} $$
Would I be able to calculate the radius of sodium ion ($\ce{Na+}$) with the help of the above Bohr atomic model formula?

This formula is for single electron systems, and $\ce{Na+}$ has 10 electrons. You could do $\ce{Na^10+}$, though.

The atom is the chemical equivalent of the three body gravity problem in physics. There is no exact solution only numerical ones when you have more than one electron.

To follow-up on @MaxW's point, I believe the system is also chaotic, so even having a nice, closed form solution is kind of useless because very small changes to initial conditions can lead to wildly different behavior of the system.

The hydrogen atom wavefunctions can be useful for multielectron atoms as a means of looking up their size (by means of a parameterization). The hydrogen atom wavefunctions form one possible basis for the definition of effective nuclear charges, see for instance here, where  Hartree–Fock orbitals are used to compute effective charges:
$$ Z_{eff} = \frac{<r>_H}{<r>_Z}$$
Here $<r>_H$ and $<r>_Z$ are the mean hydrogenic and Hartree-Fock radii (for nuclear charge Z).
The mean hydrogenic radius for ground state hydrogen is related to the Bohr radius $a_0$ as
 $$<r>_H = \frac{3}{2}\frac{a_o}{Z}$$
Expressions for $<r>_H$ are available for other orbitals, allowing $<r>_Z$ to be computed from the tabulated values of $ Z_{eff}$ as
$$ <r>_Z = \frac{<r>_H}{Z_{eff}}$$
So using the hydrogen wavefunctions in this way is a little more complicated than using the equation for the Bohr model, but just the s-orbitals can already be useful to give you an idea about the size of atoms.

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


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why is methane gas solubility in water higher than carbon monoxide solubility in water?

The above graph depicts solubility in water at various temps.
If we are dissolving in polar water, why is methane more soluble than a polar molecule like carbon monoxide?

So, given the strong dependence on temperature, this implies that the solubility is driven more by entropic considerations than enthalpic considerations, as the latter should be relatively constant over different temperatures. That means that your question may be flawed in that polarity is not the driving factor of solubility for gases.

Carbon monoxide is much less polar than it might seem.

If CO were nonpolar wouldn't it experience greater dispersion forces due to its higher mass relative to methane? My book states that gas solubility should increase with mass.

Well, it's molar solubility so value is already divided by molar mass... Weight percent for CO is higher then for CH4.

Polarity get trumped by hydrogen bonds?


text:
What is the molecular weight of Tris(ethylenediamine)cobalt(III) iodide pentahydrate?
What is the molecular weight of
Tris(ethylenediamine)cobalt(III) iodide pentahydrate ?
The chem formula is: (Co(en)3(I)3*5(H2O)
According to my calculation, it is: 928.77 g/mol.
But I am uncertain as to whether this is correct or not.

Edit: A simple overview for the correct calculation for anyone who might stumble upon this question in the future:
[Co = 58.933 g/mol]
+ [(en)*3 = 180.30 g/mol]
+ [I*3 = 380.712 g/mol]
+ [H * 10 = 10.08 g/mol]
+ [O * 5 = 80 g/mol]

= approximately 710.03 g/mol

Seriously, if you don't trust your own calculations, get a molecular weight calculator. Here is one by Matthew Monroe that is quite trustworthy and mature.

Since you asked, here is the ChemDraw figure with analysis window for Tris(ethylenediamine)cobalt(III) iodide only. To have a pentahydrate, you may add $\pu{18.02 g/mol} \times 5 = 90.1$ to the $619.95$, which gives a total of $\pu{710.05 g/mol}$ for the formula weight. Thus, your answer, $\pu{928.78 g/mol}$, is way beyond this value.

Thank you for showing me the correct answer with the clear explanation! @Mathew Mahindaratne


text:
Equilibrium - Pressure Vs. Concentration
Does increasing the pressure increase the concentration of reactants as well?
If:
$$\ce{A(g) + B(g)<=>C(g) + D(g)}$$
Assuming all reactants and products are gases.
According to Le Chatelier's Principle, increasing the pressure will cause the system to shift wherever has the least amount of gaseous molecules. However, in this case, the molar ratios are 1:1; thus, increasing the pressure will not cause a shift. That being said, when the pressure is increased, does the concentration of reactants and products both go up? Or do they maintain the same.
Thanks!

There are some cases where le chateliers fails if you follow typical wisdom. It is better to approach it quantitatively using differential sensitivity analysis. This is probably out of the scope of the question, but it works every time.

See the following for a prime example of why you can't just follow Le Chateliers principle blindly...  https://www.researchgate.net/publication/290591065_Exceptions_to_the_Le_Chatelier_principle

when the pressure is increased, does the concentration of reactants and products both go up? Or do they maintain the same?

Le Chatelier's principle in its most general form makes statements about what happens to a reaction that used to be at equilibrium when changes are made to concentrations, temperature or pressure.
To keep things simple, let's say the temperature stays constant, but we are changing the overall pressure of the reaction mix by decreasing the volume. As a result, all concentrations (or partial pressures) will increase by the same factor. If the sum of the stoichiometric factors for reactants in the gas phase is equal to that of the products, the reaction quotient Q will not change (all factors cancel out) and the system stays at equilibrium. If this is not the case, the reaction will shift to re-establish equilibrium.
On the other hand, if you change the pressure at constant volume by changing the temperature, the concentrations (partial pressures) will stay the same, but you are out of equilibrium anyway because the equilibrium constant is temperature dependent.
In both cases, you can use Le Chatelier's rules to predict in which direction the reaction has to shift to re-establish equilibrium.


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How can I prevent my spin-coated silica thin film from cracking when I heat it up?
I am spin-coating a porous silica solution on a glass microscope coverslip, and then heating the resulting thin film to $\pu{350 °C}$, in an attempt to create a thin film of porous silica sol gel. I am trying to replicate only the porous sol gel portion of the work of Yantasee et al.:
 Nanoengineered electrochemical sensor based on mesoporous silica thin-film functionalized with thiol-terminated monolayer
The problem I am encountering is that my heated sol gel is cracking and partially delaminating. This is unacceptable. Can anyone offer advice on how to prevent this cracking when heating to $\pu{350 °C}$?
Relevant Information

The recipe for the silica precursor solution is:
2.578 g quantity of Pluronic F-127 surfactant was dissolved in 3.430 g of deionized water and 8.846 g of ethyl alcohol. Then 0.108 g of concentrated $\ce{HNO3}$ was added into the mixture. Finally, a 5.0 g quantity of TEOS was added to the solution.
 The solution was sonicated for 8 hours and then spin-coated at 3000 rpm onto the surface of a 12 mm glass microscope cover slip.

 Film thickness: ~10 μm.

Possible Causes
 Perhaps I am using the wrong type of glass for the substrate, and the respective thermal expansions are causing cracking at the interface. If so, what relatively cheap glass could I use?

 Maybe I need to rigorously prepare the glass substrates by cleaning them with a solution of acid (or base) and solvent. Could an ill-prepared substrate contain contaminants acting as nucleation sites for crack formation when heated? If so, what best method is used by these groups to prepare their slides? I can't seemed to find a paper on sol gel thin films where they describe their substrate preparation.

I suggest trying the thermal treatment without placing the glass slide in direct contact with the hottest walls of the furnace and focus on a T ramp. These kind of problems are normal and rarely mentioned in experimental parts as for their description and work around are generally looking childish ;) as for most of the users are from disparate fields of chem and phys. You can also contact an author. Normally s/he will answer telling you how no to crack the film.  I colleague of mine struggled with TiOx, I suspect he was just making so many films that he did find some of them usable.

The reference you sited has used gold surface to make the thin film. If the method you wanted is the one in that reference, folloeing is how they descibed it in experimental section:

The modification of a gold electrode array on microchip with a SH-FMS thin film consisted of two steps. First a mesoporous silica film was deposited onto the electrode surface by the spin-coating technique. Then the thiol functional groups were immobilized inside the mesopores of the film by self-assembly chemistry. To prepare the silica film solution, a 2.578 g quantity of Pluronic F-127 surfactant was dissolved in 3.430 g of de-ionized water and 8.846 g of ethyl alcohol. Then 0.108 g of concentrated HNO3 was added into the mixture. Finally, a 5.0 g quantity of TEOS was added to the solution. The molar ratio of TEOS : water : ethanol : HNO3 : Pluronic F-127 was 1 : 8 : 8 : 0.05 : 0.008. The solution was shaken for 24 h, then spin-coated at the speed of 3000 rpm onto the bare surface of a gold microelectrode array on microchip (part No. M1450110, Microsensor System, Inc.) for 30 s. Fig. 1a shows the schematic of the microelectrode array. The specific electrode array had 50 finger pairs (15 microns finger width and finger spacing) with gold as the electrode material and quartz as the substrate. The electrical connection for the array was covered with plastic tape before spin coating with the film solution. After the spin coating, the tape was removed and the film-coated electrode was calcined on a hot plate at 350 °C for 8 min in air.

Hope this is helpful.

I have encountered similar issue while spin coating thin film of NiO. In my case, following steps resolved the problem:

Preheating the substrate for 1 min at 100°C
After the solution is coated on the substrate, drying it in pre heated oven for 10 min at 120 °C (My solvent was 2 methoxy ethanol)
Then calcining in furnace at required temp. The ramp rate should be taken into account.

Thanks and Regards


text:
Pure Hydrogen and Dry Hydrogen Difference
Do pure and dry hydrogen are the same? What does dry hydrogen actually means. And what does pure hydrogen means.

This is not really about chemistry, rather meaning of the words: dry vs. pure.

dry means without water, pure means only the relevant species are present.

It depends what you (or your field) is considering what an impurity is for a given substance. For example, if you want really pure methane, ethane will became an impurity. However, if you want gaseous alkane(s), there is no real difference between methane and ethane, so ethane is no longer an impurity.
On the other side, moisture can be your enemy, so pure methane (without ethane) can be considered bad if it contains 3% water whereas a mixture of methane and ethane (50/50) could be considered better because it contains only 0.1% water.
In your case, "pure hydrogen" would mean that "it is composed almost entirely of hydrogen gas" (but it can contain water) whereas "dry hydrogen" would rather mean that the gas does not contain water (but it can contain other molecules as well, but not too much)

Well, I wouldn't consider a wet gas pure unless water is not considered an impurity....


text:
Rate of change of reaction rate of 0th, 1st, 2nd order reaction?
Does the rate of change of the reaction rate for a 1st order reaction change linearly over time? My teacher says it does, but my textbook seems to contradict this. Could anyone clarify?

The logarithm of the concentration of one of the components will vary linearly. However, the concentration does not vary linearly with time, only its logarithm does.
Reaction Rate $= -\frac{d[A]}{dt} = kA = $ constant * [concentration] for a first order reaction.
If you integrate this using separation of variables you end up with
$\ln \frac{[A]_t}{[A]_0} = -kt$
if you plot $\ln[A]_t$ verse time the slope will be constant.
If you plot concentration as a function of time it will not be constant because we can rearrange the above equation to get
$[A]_t = [A]_0 \exp(-kt)$
which is clearly exponential and not linear.
So yes, for a first order reaction, the reaction rate is constant because it depends linearly on the reaction rate constant which is... constant because it is the slope of the logarithm of concentration verse time. Here is a useful link:
www.chem.libretexts.org/reaction_rates
Here is a short answer to a similar question that is useful: chemistry.stackexchange.com

Thanks Charlie. I a still a little confused, how can it be that "for a first order reaction, the reaction rate is constant" when the rate changes with concentration?

"the reaction rate is constant because it is the slope of the logarithm of concentration verse time." $\text{the reaction rate constant is the slope of the logarithm of concentration verse time.}$

yes Adnan, the reaction rate constant, is constant as all derivatives of linear functions are.

what I mean @user2351418 is that after plotting the logarithm of concentration verse time THEN you calculate the slope, which is the reaction rate constant. It is a single number.

The rate constant does not change with concentration, but surely the reaction rate must? I thought the definition of a 0th order was that rate does not change with concentration, and of a 1st order is that rate changes linearly with concentration, and of a 2nd order that rate changes with the square of concentration. As depicted here https://chem.libretexts.org/@api/deki/files/15966/14.16.jpg?revision=1&size=bestfit&width=962&height=943

Thank you Charlie! I understand, you are saying the rate constant does not change with time


text:
Ranking basicity from a structure
I just came across this question in my revision.

I selected (A) but the correct answer is (D).
Could someone explain where I am going wrong?

I is least stable --> most reactive --> most basic
Cl is more electronegative than F so III will have a stronger inductive stabilization effect -->  III is more stable than II --> III is less reactive than II --> III is less basic than II
Hence, rank order: I, II, III

You might want to review the electronegativities of F and Cl.

From an instructive perspective, it would be nice to see an answer including the pKa all three compounds determined under shared conditions (for the the same solvent); especially for the comparison II vs. III, though.

Solution: I had the electronegativity of F vs Cl the wrong way round.


text:
Preference for basic conditions in ester hydrolysis
It has occurred to me that it is very common see the use of basic conditions for ester hydrolysis in various syntheses. One such sequence commonly seen in the case of the malonic ester synthesis is:

$\ce { OH^-,H2O}$
$\ce {H^+, heat}$

The first step is for the hydrolysis while the second step is for the decarboxylation. It is strange that there is a first step with basic conditions since ester hydrolysis and decarboxylation can both take place in acidic solution. Are there practical reasons for why ester hydrolysis is done with aqueous base?

Precedent - why change conditions that others have shown work well?
Control - it is much easier to react 1 eq of base/nucleophile under mild conditions, whereas acid hydroysis generally requires excess reagent and higher temperatures.
Ease of operation - base hydrolysis gives a carboxylic salt as product which frequently extracts into aq, giving an easy workup and purification. Acid hydrolysis requires neutralisation of excess reagent as a first step.


text:
Is it possible to dissolve Raspberry ketone in warm water?
I poured a small amount of RK in cold water and it clumped together so I warmed it in the microwave and it disappeared in the water meaning it dissolved, but i was under the assumption that RK is insoluble in water, so what happened ?

According to PubChem, 4-(4-hydroxyphenyl)-2-butanone (Raspberry ketone) is estimated to be partially soluble in water, ca. 25.1 g/L @ 25 celcius.
https://pubchem.ncbi.nlm.nih.gov/compound/4-4-Hydroxyphenyl-2-butanone#section=Experimental-Properties
My initial thought is what would one need with Raspberry ketones? Their health effects seem to lack any clinical evidence.

The question is not about 25 °C, rather what happens with solubility at elevated temperatures, how microwave radiation affects it, and whether hydrolysis or any other reaction takes place.

@andselisk I can agree somewhat with your comment and perhaps my answer lacked a breath of information. I will remove it shortly because I haven't the time to expand upon it but wish you to see this reply.

No, it gives some useful information, it's just that it has questionable relevance. Probably leave it for now and edit later when you will have some more time.

I am not in favour of this question/answer as for the same can happen to everything if the premise are trivially wrong. OP supposed the stuff much less soluble but it turns out that at RT is already soluble for a good 25 g/l. Heating up would speed up dissolution and likely increase solubility. I didn't get why to ask in first place. The answer seems trivial  but correct.   At LigninPauling and @andselisk

@Alchimista According to IFA's GESTIS database,  it's "Practically insoluble in water" which is in agreement with what OP said. However, what would happen after microwave treatment puzzles me as there is still no definitive answer. Besides, PubChem refers to ToxNet, ToxNet to EPA homepage for TSCA. I fail to see a solid reference for this solubility data.

@Andselisk I am just failing to see the ground for all this. Some grams per litre are enough to answer the question. If I would do what OP describes, I would rather say RK is (sparingly, readily, ecc) soluble in warm water. If RK is listed as much more insoluble than I agree. By Googling I found the same value 25. No polemic at all. Just failing see the relevance of the Q and thus of the related feedbacks.

@Alchimista Yeah, sometimes science is not about "why?", rather "why not?" It seems to be the case here:)


text:
Effect of temperature on equilibrium constant in terms of entropy change
This question was asked in the JEE examination of 2017 :

[A]  option is incorrect since the system is losing heat and hence $$ \Delta S\lt0$$
[B]   Lets assume we have an endothermic reaction  $$\ce { A <=> B - Heat} $$
Now if we increase the temperature, the reaction will shift to the product side and hence the entropy of the surroundings will decrease more as compared to the initial conditions. Therefore the Unfavourable change in entropy(decrease in entropy) should increase.
[C]  Though the reaction is endothermic, since we are supplying heat to the system, the net change in entropy of the system is positive.
[D]  Lets assume another exothermic reaction $$ \ce{A<=>B + Heat}$$.
When we increase the temperature, the reaction will shift to the left side and hence less heat will be given off to the surroundings. Therefore, favourable change in entropy(increase in entropy) decreases.
Hence, according to me, the correct answer should only be [D], but [B] option is also given correct. What is the error in my explanation?

The key point is that the entropy of the surroundings changes, not that of the system (at least for a small change in $T$), so only [B] and [D] are relevant, as you correctly determined.
You can write
$$\log(K)=-\frac{\Delta H}{RT}+\frac{\Delta S}R$$
This of course is one form of van 't Hoff's equation. It tells you, assuming $\Delta H$ and $\Delta S$ are constant, that the temperature dependence is due to $\Delta H-$ the heat transferred to/from the surroundings (entropy increase/loss of the surroundings):
$$\log(K)\propto-\frac{\Delta H}{RT}=\frac{\Delta S_\text{surr}}R$$
Now increasing $T$ makes the change in the entropy of the surroundings smaller.
When $\Delta H\gt0$ (endothermic), ie $S_\text{surr}\lt0$, this makes the reaction more likely to happen. When $\Delta H\lt0$ (exothermic), ie $S_\text{surr}\gt0$, this makes the reaction less likely to happen.
So both [B] and [D] are correct.


text:
Why is the volume related to the probability of finding an electron?
thank you for your time.
I was wondering about the correlation between volume and probability of finding an electron.
we know that if we move away from the nucleus, we are going to get less probability of finding an electron. this happens with a 1s sub-level.
But when i want to know ( graphically ) the probability of finding an electrons in a 2s sub-level, then the volume comes up.
my professor during the lessons said "The PROBABILITY increases because the PROBABILITY DENSITY decreases, but the VOLUME increases."
he also said that this is true to a certain point, since after that it (volume) starts to decrease.
he then said that the DENSITY OF PROBABILITY is 0 in then nucleus.
what i wrote above it's probably messy. i'm just looking for a better explanation of this concept since i didn't find anything online that talks about density of probability and volume.

Outside of context, what your professor said makes zero sense, or maybe even less (that is, it destroys whatever little sense we could have had previously). The probability increases, he says. Probability of what?

there are lots of answers on this site addressing this eg: https://chemistry.stackexchange.com/questions/92244/what-is-the-difference-between-%cf%88-%cf%88%c2%b2-radial-probability-and-radial-distribut?rq=1

also: https://chemistry.stackexchange.com/questions/84726/most-probable-point-for-finding-an-electron-in-the-1s-orbital-of-a-hydrogen-atom

This part is wrong: "he then said that the DENSITY OF PROBABILITY is 0 in then nucleus."

There is a straightforward relationship between g(r) - the radial distribution function of the electron (probability per unit radius) - and $\rho(r)$ - the electron density (probability per unit volume).
You can write that
$$\mathrm{g(r) =  4\pi r^2\rho(r)}$$
Here all that's been done is to integrate $\rho(r)$  over a concentric area $4\pi r^2$ surrounding the origin, removing the angular dependence of the density. This works for an s-orbital function with no angular variations in the density.
The $r^2$ factor means that even when the density becomes small away from a nucleus, the total radial probability can be substantial (and reaches a maximum). Inversely, you can have a very high volume density but very low radial probability, for instance at the origin of an s-orbital, where $r^2$ goes to 0.
This is shown schematically (not intended to depict a real H-atom) in the following:

where the red curve illustrates the behavior of g(r), the blue curve of the density.


text:
Amber Force Field Improper Parameterization Huge Kb Problem
I am trying to parameterize a molecule. I know dihedrals and impropers are the same for Amber Force Field. (Source: http://archive.ambermd.org/201004/0334.html)
 So, I did PET scan for improper and I found this chart:

According to this chart and Amber7 manual Appendix-C I should define impropers like this in my mini-force field:
C-O-B-C 40.5 180.0 2.0

But in gaff and gaff2 the biggest improper values are not greater than 15
( here is gaff2: https://github.com/choderalab/ambermini/blob/master/share/amber/dat/leap/parm/gaff2.dat and here is gaff: https://github.com/choderalab/ambermini/blob/master/share/amber/dat/leap/parm/gaff2.dat)
And to compare my improper values with gaff, I scanned C-C-C-C impropers and I obtained almost the same chart as above but in gaff, this improper is defined like this:
C-C-C-C 1.1 180.0 2

(line 6693 : https://github.com/choderalab/ambermini/blob/master/share/amber/dat/leap/parm/gaff2.dat#L6693)
so it is obvious gaff doesn't use PES scan values for improper but it seems unreasonable because dihedral values fits with PES scan.
So what should I do? Should I use values which I obtained from PES scan ( C-O-B-C 40.5 180.0 2.0) or what should I do?

Amber has a mailing list for these types of questions where the creators of GAFF will be answering your question. Also, Amber7 is old. They are at least at Amber18 now.

@charlie it has nothing to  do with ambertools version and gaff. from amber 4 to amber 18 force field definition hasn't changed.

There are a few reasons I don't use windows 95 anymore

"force field parameterization" is not a software. it is a basic rule like f=m*a

in amber7 and amber16 manuals, parameter definitions are the same one to one. in amber16 manual page 230s and in amber7 manual appendix-C. and other amber manual page bla bla about parameters are the same. something never changes. (also amber has no windows version) .


text:
Measuring a high Michaelis constant using fluorescence
We have the task of measuring kinetic parameters of an oxidase reaction that has a $K_M$ of about $2 \,\text{mM}$. For that, we want to use a fluorescence assay based on Amplex Red. The latter is a fluorogenic substance of an upper detection limit of $5 \, \mu \text{M}$. At higher concentrations, the inner filter effect causes non-linear behaviour.
Usually, we would use substrate and fluorophore concentrations of a similar range of the $K_M$ value. Obviously, we cannot do that here, as the fluorophore concentration would be way above the upper detection limit. This led us to the idea of reducing the substrate, fluorophore, and enzyme concentration by the same factor to do measurements in a micromolar range. Later we could simply scale up the results.
We doubt that this method is possible as the Michaelis-Menthen kinetics never look at the ratio between substrates and enzyme, but only regard absolute substrate concentrations.
So our question is if the proposed method is in fact not applicable and if so, what your suggestions would be?

What is the purpose of this kinetics? Knowing that would be helpful to suggest something else.

Based on your description, it sounds like you are measuring the production of hydrogen peroxide by an oxidase enzyme. In that case, it is not necessary for the Amplex Red to be at the same concentration as the substrate. You only need enough reagent to determine the initial rate of reaction at each substrate concentration, ie when only a small portion of the substrate has been consumed.
If $\pu{5 μM}$ is not sufficient for that purpose, you can determine the resorufin (produced by reaction of Amplex Red with $\ce{H2O2})$ by absorbance rather than fluorescence for lower sensitivity and higher maximum. According to the Thermo Fisher assay kit info, resorufin has an extinction coefficient of $\pu{58000 M^-1 cm^-1},$ so with a standard $\pu{1 cm}$ path length cuvette, your linear range would probably go up to around $\pu{50 μM}$ depending on your spectrophotometer. With a short path cuvette (these are available with $\pu{0.2 cm}$ and $\pu{0.1 cm}$ path lengths), you could go quite a bit higher, certainly more than sufficient for your needs.
As for why you cannot just scale the concentrations down, consider the limiting case where the Michaelis constant is equal to the $K_\mathrm{d}$ of the substrate. We have that
$$K_\mathrm{d} = \frac{[\ce{E}][\ce{S}]}{[\ce{ES}]},$$
so $[\ce{S}] \approx K_\mathrm{d}$ when $[\ce{E}] = [\ce{ES}]$ (assuming $[\ce{S}]_\mathrm{tot} \gg [\ce{ES}]).$ Thus, the absolute concentration of $\ce{E}$ is not a factor, as long as it is small enough that the amount of $\ce{S}$ bound does not substantially change the concentration of free $\ce{S}.$ You should get the same result for $K_\mathrm{M}$ regardless of what concentration of enzyme you use.

A nonfluorescent derivative of dihydroresorufin, 10-acetyl-3,7-dihydroxyphenoxazine has been marketed under the name Amplex® Red (AR). AR is regarded as the best fluorogenic substrate for peroxidase because it is highly specific and stable. It has been used to detect Reactive Oxygen Species (ROS) such as $\ce{H2O2}$ in various experimental enzymatic systems (e.g., $\ce{H2O2}$ is produced by many enzymatic reactions). During these assays, nonfluorescent AR is oxidized to highly red fluorescent resorufin by ROS (excitation/emission maxima: $\lambda_{ex} = \pu{563 nm}$ and $\lambda_{em} = \pu{587 nm}$) in a peroxidase-catalyzed reaction with $\ce{H2O2}$ (Ref.1). The abstract of that original reference states that:

The enzymatic determination of hydrogen peroxide can be accomplished with high sensitivity and specificity using N-acetyl-3,7-dihydroxyphenoxazine (Amplex Red), a highly sensitive and chemically stable fluorogenic probe for the enzymatic determination of $\ce{H2O2}$. Enzyme-catalyzed oxidation of Amplex Red, which is a colorless and nonfluorescent derivative of dihydroresorufin, produces highly fluorescent resorufin, which has an excitation maximum at $\pu{563 nm}$ and emission maximum at $\pu{587 nm}$. The reaction stoichiometry of Amplex Red and $\ce{H2O2}$ was determined to be 1:1. This probe allows detection of $\pu{5 pmol}$ $\ce{H2O2}$ in a 96-well fluorescence microplate assay. When applied to the measurement of NADPH oxidase activation, the Amplex Red assay can detect $\ce{H2O2}$ release from as few as 2000 phorbol myristate acetate-stimulated neutrophils with a sensitivity 5- to 20-fold greater than that attained in the scopoletin assay under the same experimental conditions. Furthermore, the oxidase-catalyzed assay using Amplex Red results in anincreasein fluorescence on oxidation rather than adecreasein fluorescence as in the scopoletin assay. In comparison with other fluorometric and spectrophotometric assays for the detection of monoamine oxidase and glucose oxidase, this probe is also found to be more sensitive. Given its high sensitivity and specificity, Amplex Red should have a broad application for the measurement of $\ce{H2O2}$ in a variety of oxidase-mediated reactions and very low levels of $\ce{H2O2}$ in food, environmental waters, and consumer products.

As you have described it, your enzyme is better suited for designed UV–vis-based assay. Yet, adapting a UV–vis-based coupled enzyme assay to a fluorescence-based assay is not unprecedented since it has been done before. For example, Gutheil and coworkers have developed high-sensitivity fluorescent assays for D-Alanine, which use a microtiter plate format based on D-aminoacid oxidase/horseradish peroxidase (DAO/HRP)-coupled reactions (Ref.2). They have also taken the standard UV–vis-based coupled enzyme assay for D-Alanine using o-phenylenediamine (OPD) and adapted it to microtiter plates for comparison purposes. The fluorescent assays allow the sensitivity of the assay to be increased from $\pu{2 nmol}$ (chromogenic (UV–vis) assay) to $\pu{2 pmol}$, a 1000-fold increase in sensitivity. One of the two fluorogenic HRP substrates tested in this assay was AR (which is commercially available that has given a lower limit of sensitivity of $\pu{2 pmol}$ D-Ala, and was linear up to $\pu{400 pmol}$. During the assay, AR is converted to fluorescent resorufin upon oxidation in HRP-coupled reactions. Resorufin was then detected with excitation at $\pu{546 nm}$ and emission at $\pu{595 nm}$.
You may follow their design for your enzyme as well.
References:

M. Zhou, Z. Diwu, N. Panchuk-Voloshina, R. P. Haugland, “A Stable Nonfluorescent Derivative of Resorufin for the Fluorometric Determination of Trace Hydrogen Peroxide: Applications in Detecting the Activity of Phagocyte NADPH Oxidase and Other Oxidases,” Analytical Biochemistry 1997, 253(2), 162-168 (https://doi.org/10.1006/abio.1997.2391).
W. G.Gutheil, M. E. Stefanova, R. A. Nicholas, “Fluorescent Coupled Enzyme Assays for D-Alanine: Application to Penicillin-Binding Protein and Vancomycin Activity Assays,” Analytical Biochemistry 2000, 287(2), 196-202 (https://doi.org/10.1006/abio.2000.4835).


text:
what is actually supercritical fluid?
let’s take an example: if we have a glove-box filled with helium at 5 bars at room temperature which is far beyond its critical point, what actually it would be like ?
would it be like a normal gas ? would it be like a liquid ? or it will be like a condensed vapors ? can this supercritical fluid dissolve salts for example ? would it quench hot metals like water ?
if I put may hands inside this fluid would it feels like putting my hands inside a water ? or it feel like a normal air ?
is supercritical fluid a special types of a liquid ? if no, why then some SCFs used as a solvent ?
I searched a lot but really can’t find any clear answer explains what they ar exactly.
Note: this is not duplicate, I asking about the nature of SCFs while the "duplicated" question was asking about room temperature SCFs which is clearly different.

Just to make it clearer: are you familiar with "phase diagrams" or not at all?

If you search supercritical carbon dioxide on Youtube, you will see very nice videos of supercritical carbon dioxide in transparent cells. I work with supercritical CO2 for some purposes but never saw it in real.

5 bars of He at room temperature is a gas. Not even close to a critical point. The question is unclear.

my question is not duplicate !! I see that video about supercritical CO2 and it seems like a free moving liquid which make me confused.
by definition critical temperature is the temperature at which gas can not liquefied no matter what how much pressure is applied , so 5 bars at room temperature exceeding critical point .

@JonCuster Technically it's supercritical, because He has low crit. pressure, but yeah, in practice it's just a gas.

@Mithoron  so, SCFs is similar to gases?

https://chemistry.stackexchange.com/questions/22185/does-the-density-of-a-liquid-and-gas-phase-of-a-substancenitrogen-or-water-con

this not answer my question, I will try to clear it more.
if  I have a cup of a liquid and put it inside a large pressure chamber under pressures exceeding its critical pressure, then I apply a heat until reaching the critical point and surpassing it. would my liquid expanding and fill the chamber like a gas ? or it will stay in the cup like a liquid ?

The name supercritical fluid is fancy but don't be confused by thinking that it this would a very odd looking state of material. Supercritical CO2 is just a state of CO2 at a certain T and P, so it looks very much like gaseous CO2 i.e., it is a transparent clear "fluid" whose density and viscosity is higher than a gas but lesser than that of the liquid. If you are able to see an interface in those transparent cells (e.g. here https://www.youtube.com/watch?v=GEr3NxsPTOA), it means that the liquid phase and the gas phases exist in the system. We are able to see the interface (boundary) when the refractive index is different from the container walls, and the surroundings. In short do not expect a new form of matter in supercritical state. It is just like thinking what does a plasma look like? It very much appears a like an ordinary flame which is extremely hot.
Response to the comment "is supercritical fluid a special types of a liquid ? if no, why then some SCFs used as a solvent ? what if a small pot of liquid heated inside a large pressure chamber to its critical point? would it expanding fill the whole chamber like a gas or stay in its container like a liquid ?"
I am afraid you are mixing several concepts here. The key question is, how does one distinguish a liquid phase and a gas phase? Supercritical fluid is not a liquid because its density and viscosity are not close to liquids, nor it has a definite boiling or melting point (a liquid does, right?). Think of supercritical fluid as a state or condition of a gas where it is above its critical temperature and pressure. That is all. We cannot say supercritical carbon dioxide has a boiling point of x Celsius or a freezing point of y Celsius.
Coming to your second part on SFC extraction: First, let us talk about gases. Can the gases act as a solvent or carrier of molecules? The answer is yes, and this is the very basis of gas chromatography. How do you detect odors? The gas molecules are carrying the odor causing molecules from a far away place to your nose i.e., gases also have some capability of "dissolving" substances. In the same vein, if I use a temperature of Tc, and a pressure Pc for carbon dioxide (or any gas), we can use it for extraction. It just turns out that supercritical carbon dioxide has higher dissolving power than gaseous carbon dioxide and certainly less than liquid carbon dioxide.
Third point: Let us do an imaginary experiment: If you have a sealed container of fixed volume containing nothing but carbon dioxide at very low temperature. Also assume a pressure gauge is also attached. You have filled in enough carbon dioxide in the chamber that the gauge shows a value of 85 atm. Now you start heating the chamber until the temperature of the chamber reaches 35 Celsius. Imagine you can see the insides of that chamber with a camera: You will see nothing because there will be no liquid in that chamber. It will look empty to your eyes. Given the fact you know that an empty glass is not empty, it is filled with gases (nitrogen, oxygen etc). In the same way, our chamber is also filled with a carbon dioxide which our eyes cannot see. We will still call this carbon dioxide as supercritical carbon dioxide because we know from our experimental settings that we are above critical temperature and pressure of carbon dioxide. You can go higher temperature to 45 Celsius, still you are in supercritical state. Thus in a phase diagram, there is a supercritical region.
Don't let the word fluid confuse you. Since we cannot call a gas above its critical temperature and pressure a typical gas, nor we can call it a liquid, it is best to call it a fluid. Hope that clarifies your confusion.

that video is actually what make me confused. when CO2 reaching critical point it appears like an expanding liquid or a condesed vapor which rise a question : is supercritical fluid a special types of a liquid ? if no, why then some SCFs used as a solvent ? what if a small pot of liquid heated inside a large pressure chamber to its critical point? would it expanding fill the whole chamber like a gas or stay in its container like a liquid ?

See the edited answer above.

thank you for explanation. but it is not yet clearly understanded. when I watched this video:https://www.youtube.com/watch?v=QHcqyFm0i9M , the supercritical CO2 seems like an expanded liquid or a liquid that act like a gas . it appears like any liquid !!
I know that SCF is not a true liquid and it is a fluid combining liquid and gas properties but the video show it is a liquid-like fluid which expanding like gases.

Could you mention what grade you are in-this will help in explaining? At 2:35 min of the video, this is the supercritical form of CO2. Let me ask a question. Why do you see a glass filled with water? The reason is refractive index RI of water is different from glass and air. You are able to see liquid CO2 b/c its RI is different from the container walls. Now they heat it up. I can see some ripples after 2:35 min of the video, this is due to refractive index changes in the supercritical state. Have you seen stars twinkling? This is also due to refractive index changes of the air in the sky.

I studying pharmaceuticals.I did not know a lot about refractive index but I think if the supercritical CO2 differ from normal CO2 it should also differ in refractive index . I think also that filling glass with a the fluid after transition is not related to refractive index . CO2 expanded as its density decreased by increasing temperature, until it filled the glass.
I did’nt say supercritical fluid is a liquid, but the video shows that it appears like a liquid sharing some properties of gases "i.e. expanding" .


text:
Why does beryllium lose two electrons when it’s orbitals are already full?
I have a rudimentary understanding of orbitals, as in what they are, the shapes ($l$, I think) and of the principles.
So I was doing the electronic configuration for beryllium. It has $4$ electrons, so the electronic configuration is 1s2, 2s2. Why does berylium lose two electrons to complete its “inner duplet”? I thought that atoms want to gain, lose, or share electrons so their orbitals are complete, and thus don’t react.
Its orbitals are complete, is it something to do with the promotion of an electron to a higher energy level, but won’t that be temporary change? Or perhaps there is some fundamental misunderstanding on my part.
(I’m in my IGCSE, this isn’t in our syllabus, just trying to understand, so I hope my ignorance is forgiven)
Edit: I am not familiar with hybridisation but I don’t think it applies to this, unless you are giving a bonding example.

Complete shell not "orbitals" 2) Actually Be is tetracoordinate and it's bonds quite covalent.

@Mithoron Oh, I see it forms covalent bonds due to its strong pull on its electrons. One example being BeCl2 with chlorine’s dative bond. But aren’t orbitals involved? And your comment answers my question to some extent that it isn’t ionic (thanks for that), but it’s s orbital is complete what reason does it have to form covalent compounds?

It's p orbitals are accessible so it can just try to get octet (even if sometimes incomplete). It's bonding may be more ionic or more covalent depending on compound, but it's very much not binary thing, but continuum.

https://chemistry.stackexchange.com/questions/64683/what-does-the-molecular-orbital-scheme-of-beryllium-chloride-and-hydride-look-li https://chemistry.stackexchange.com/questions/31817/what-is-the-bonding-type-in-hydrides-of-group-2-elements

Electrons exist in the outer spaces of elements, away from the nucleus. Orbitals are mathematical descriptions of places in these outer spaces where electrons seem to be when they are there; they can be mathematically manipulated and reorganized but are a way to help you imagine what is really going on.
Manipulating the electrons is done experimentally, not mathematically. For example, an element can be ionized and the energy required can be measured. Beryllium can suffer the loss of up to four electrons: the first requires $\pu{9.3 eV}$, the second requires $\pu{18.2 eV}$, the third requires $\pu{153.9 eV}$, and the fourth requires $\pu{217.7 eV}$. The energy required to remove the first two electrons is in the range of other metals that lose two electrons: For comparison, the first and second ionization energies of $\ce{Mg}$ are $7.6$ and $\pu{15.0 eV}$ while those of $\ce{Ca}$ are $6.1$ and $\pu{11.9 eV}$, respectively. So, $\ce{Be}$ can lose 2 electrons with a relatively small amount of energy (which will be supplied by something like empty orbitals and ionic or covalent bonding energy to other atoms). Losing those last two is a big jump and essentially doesn't happen in chemistry (it hypothetically happens in physics and mathematics!).
The empty orbitals on $\ce{Be}$ are quite high in energy, i.e., if you stuck an electron into a $\mathrm{p}$ orbital on $\ce{Be}$, it would be floating so far out from the nucleus that it would be looking for a better place. The idea of orbitals gives you a mental picture of what might happen - whether it really happens or not depends on whether that mathematical picture is true and is the most stable (lowest energy) condition available.
Completing shells or orbitals is a qualitative way of looking at the issue, but numerically, ionization potentials gives you the data you need to develop the picture, and the orbitals (as calculated for the hydrogen atom) are a reasonable way to begin.

I have added few words to make it better. If you don't like what I have done, please feel free to roll back.

I don't remember what it was before, but it sounds good now, so, Thanks!

All main group elements follow the octet rule, which is to bond in such a way as to achieve 8 electrons in their valence shells.
I've never heard of the term "inner duplet", but perhaps this is referring to its desire to complete its valence shell with 8 electrons. Valence electrons are the electrons in the outermost shell of an atom.
For example, Beryllium can bond with oxygen and donates those 2 electrons to oxygen to combine with oxygen, which has 6 valence electrons. After this, Be will have a charge of 2+ and oxygen will have a charge of 2-. These ions will be attracted to each other and bond. This results in them both having completed valence shells with 8 electrons each.
Here is a brief explanation of how the octet rule affects metals:
https://socratic.org/questions/how-does-the-octet-rule-affect-metals
Explanation of valence electrons:
https://socratic.org/chemistry/the-periodic-table/valence-electrons-and-the-periodic-table
I hope this answers your question and I didn't just re-explain something that you already knew :)

I already know that, beryllium loses two electrons so it has completed the duplet of the first shell and then subsequent attraction to the anion. But when you draw out the orbitals, it doesn’t need to lose 2 electrons since the 2s orbital is complete. I always thought that atoms complete their orbitals to achieve stability which is inertness.

You seem confused by the usage of inner duplet, beryllium has 4 electrons. Effectively if it were to lose two electrons it’s first shells duplet would be complete, I used air quotes since the 2s orbital is already complete, so it doesn’t need to lose electrons (which I learnt now it forms covalent bonds). But even so it’s s orbitals are complete, I simple can’t understand why it would react? My knowledge on orbitals isn’t very extensive, half knowledge is dangerous they say

Right, I understand what you are saying about both s orbitals being full. I suppose the only reason it would lose electrons would be to achieve the the state of a noble gas by following the octet rule.

After the electron configuration, the last shell of the beryllium atom has two electrons. In this case, both the valence and valence electrons of beryllium are 2. We know the details about this. The elements that have 1, 2, or 3 electrons in the last shell donate the electrons in the last shell during bond formation.
The elements that form bonds by donating electrons are called cations and beryllium is a cation element. Beryllium donates the electrons of the last shell to form bonds and turns into beryllium ions:
$$\ce{Be^0 – 2e– -> Be^2+} \ \text{or} \ \ce{Be -> Be^2+ + 2e–}$$
The electron configuration of beryllium ion $(\ce{Be^2+})$ is $\mathrm{1s^2}$, which shows that beryllium ion, $\ce{Be^2+}$ has only one shell and that shell has total of two electrons. The electron configuration of $\ce{Be^2+}$ also shows that the beryllium atom has acquired the electron configuration of helium. That is, in this case, the valence of the beryllium-ion is $+2$. Since the last shell of a beryllium-ion has two electrons, the valence electrons of a beryllium-ion are two. Source- https://valenceelectrons.com/category/valence-electrons-of-all-elements/


text:
What are the pi-MOs in anisole like?
The $\pi$-orbitals of benzene are well known and given in any standard undergraduate textbook. When considering anisole, phenol, etc, a lone pair in a p orbital of the O can delocalise into the $\pi$-system of the aromatic ring. How would this affect the $\pi$-orbitals of the cyclic system? Is it possible to account for the effect of this delocalisation by starting with the benzene $\pi$-orbitals and then interact those with the p-orbital, or does the change in point group disallow this - surely if we used this method, we would mix each benzene MO in turn with the p-orbital to yield 12 MOs, which is clearly not the case?

You don't get 6*2 = 12 MOs, you get 6+1 = 7 MOs.

I understand that this is what I should get, however I am unable to actually get this. What I have been doing is considering each pi MO in turn for the benzene molecule, then adding a p-orbital on a C next to a ring carbon and looking at its overlap with the MOs. My issue is, the p-orbital can overlap constructively or destructively with any one benzene MO, which would give a cap of the p-orbital perturbing only one benzene MO to give rise to 2 more - the 6+1 situation. However, based on symmetry the p-orbital should be able to interact with 4 different benzene pi MOs (6*2 situation)?

Symmetry isn't the problem here; there's no way you can put 7 things in (6 benzene MOs + 1 p-orbital on oxygen) and get 12 things out.

A first-order answer to answer is relatively easy to compute with some basic linear algebra. Trying to qualitatively sketch out the MOs and get the relatively energies correct may be highly prone to error.

@orthocresol, in that case are you suggesting that the p orbital can only interact with one benzene MO?

@Zhe, I have tried a preliminary calculation using basic group theory, which gave me 5 b1 and 2 a2 orbitals in c2v. However I found constructing the form of all these orbitals difficult.

@K.P. no, not at all. You can have $m$ orbitals from fragment A interacting with $n$ orbitals with fragment B and you will only get $m+n$ MOs from that – no more and no less, and it doesn't depend on whether they are related by symmetry or not.

@K.P. The calculation does not require group theory at all, you just need to diagonalize a matrix and find it's eigenvalues and eigenvectors.

Specifically, I assume that it's just a carbon $p$ orbital interacting with the ring. You can perturb the system with an oxygen orbital: https://www.wolframalpha.com/input/?i=Eigenvectors(%5B%5Ba,b,0,0,0,0,0%5D,%5Bb,a,b,0,0,0,b%5D,+%5B0,b,a,b,0,0,0%5D,+%5B0,0,b,a,b,0,0%5D,%5B0,0,0,b,a,b,0%5D,%5B0,0,0,0,b,a,b%5D,%5B0,b,0,0,0,b,a%5D%5D)

@Zhe for anisole, I would suggest a smaller absolute $\beta$ value and a larger absolute $\alpha$ value in the Hückel method. Maybe $0.8\beta$ and $1.2\alpha$?

@TAR86 Probably. I picked those values just so that I could get a faster answer out. :)


text:
Why is the ROHF spin density of the propenyl radical not symmetric?
I was preparing some material for my students and in order to illustrate the concept of spin density and how the ROHF method gives a purer description of this property than the UHF method. I prepared a simple calculation for the $\ce{CH3}$ and $\ce{CH2CHCH2}$ radicals.
The calculations for both radicals performed properly. Given the symmetry of the molecules, I expected the distributions to be symmetric. That expectation was borne by the results of the calculations for the methyl radical and the UHF method of the allyl radical. However, in the case of the allyl radical the spin density surface is asymmetric as shown in the following picture obtained with GaussView 5.

The canonical MO have the proper symmetry (see below)

Can anybody point to an explanation of these behaviour or a mistake in the calculation?
P.D. Find enclose a copy of the Gaussian input field I used in the calculation.
%chk=c3h5_rohf.chk
# opt=tight rohf/6-31++g(d,p) guess=mix

ROHF

0 2
 C
 H                  1    1.06999999
 C                  1    rCC    2  ACCH
 H                  3    rCH    1  AHCC    2    0.0
 H                  3    rCH    1  AHCC    2  180.0
 C                  1    rCC    2  ACCH    3  180.0
 H                  6    rCH    1  AHCC    3  180.0
 H                  6    rCH    1  AHCC    3    0.0

 rCC 1.40140000
 rCH 1.0700000
 ACCH 120.0
 AHCC 120.0

I believe your wave function is not the one you would expect in ROHF. The reason for this is guess=mix which will try to construct a broken symmetry wavefunction from the HOMO and LUMO. Why the SCF converges to that solution, and why this is stable (-116.460412224 ha), I don't know, and I have currently not the time to investigate (I will try to come back to that), but remove the keyword and you should obtain the symmetric correct solution (-116.457980797 ha). Inspect your output file, and you will notice, that the calculation is run in Cs symmetry, not C2v. (Also HF might not be the best method to show delocalisation.)

Thanks. Repeated the calculations without the guess=mix keyword and know the results of the calculation are consistent with what is expected. The wavefunction has C2v symmetry and the spin density distribution has the right symmetry,

My suspicion based on the higher energy of the symmetric solution is that the radical is actually a multireference problem. A two config calculation might shed light on it ( or cisd). The correct solution is closer to a superposition of two localized determinants ( for the $\pi$ bond as well as the SOMO), i.e., this is a case in which mesomers are a better description than conjugation.


text:
How to measure the electrical conductivity of a substance?
Suppose you are given these substances:
$\ce{NaOH(aq)}$
$\ce{HCl(aq)  }$
$\ce{C6H12O6(aq)  }$
$\ce{NH3(l)}$
Are there rules to figure out how conductive each compound is? Is it based on intermolecular forces or polarity?

A lot of work was done in the 19th and 20th century on determining the conductivity of aqueous solutions. It was a huge field called conductometry (check Wikipedia). You experimentally measure the resistance of a solution by placing the desired liquid between two inert electrodes and pass high frequency alternating current to prevent electrolysis. In solutions, electrical current is carried by ions. If your solution has ions or if it is ionizable, it will conduct electricity. Now you can determine which solution has ions based on ionization, e.g. you can write $\ce{HCl (aq) -> H+ + Cl-}$ and can predict that it must be a conductive solution. In fact, the aqueous $\ce{H+}$ has the highest conductivity known in water. $\ce{OH-}$ is next.

But how did you come to that conclusion? I understand it is based on the ions but oh is o2+ + H-, why is that less than HCl?

It is actually a circular argument, people first measured conductivities of various solutions and then they classified that certain substances are good conductors (like NaCl solutions) and some of them are poor conductors (high resistance like acetic acid solutions). OH is not O2+H(-), it is a unit made of oxygen and hydrogen atoms. Check your textbook for electrolytes.


text:
How to measure concentration of arsenite and arsenate in a water sample
A colleague and I are developing a transgenic bacterial strain that contains an arsenite oxidase gene. The bacteria can oxidize arsenite into arsenate autotrophically. We need a method to measure the concentration of arsenite and arsenate over time in a water sample. First we thought of using a pH meter, but we need to measure the arsenite and arsenate concentration over a 48 hour period, and we are worried that the pH meter wouldn't hold up. Could there possibly be a chemical that reacts with arsenite to change the color, and we could use a spectrophotometer or a colorimeter. If there are any ideas for measuring concentrations of arsenite (around 5 mM), please let me know.

Speciation analysis is very challenging because you would like to separate the species first before detection. In order times electrochemical detection was done by now ion chromatography is the way to go for arsenic speciation. You will need ion chromatography system with inductively coupled plasma with mass spectrometry. Have a look at this paper: Determination of Anionic, Neutral, and Cationic Species of Arsenic by Ion Chromatography with ICPMS Detection in Environmental Samples, Anal. Chem., 1998, 70 (17), pp 3649–3655 The good thing about Analytical Chemistry papers is that most of the methods are reliable.

Thank you for your help. My equipment, however, is quite limited. My local university does not have such high-end equipment, so it would be okay if there are other methods that would be less precise, but more cost-effective. I have access to a pH meter, colorimeter, spectrophotometer, and maybe some other equipment. Are there any other ideas?

That is the main drawback of speciation analyses (too expensive!). Check this spectrophotometric method: Spectrophotometric determination of arsenite, arsenate, and phosphate in natural waters, Analytica Chimica Acta, 1972, 58, 289-299. https://doi.org/10.1016/S0003-2670(72)80005-9


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Good software/module in python to determine nuclear attraction integral in Slater type orbital basis sets
I am trying to solve a Hamiltonian of one electron and 10 stationary nuclear centers. here, the electron is treated quantum mechanically and nuclear centers are treated classically objects. The Hamilton is as follows,
$$H=-\frac{1}{2}\nabla^{2}-\sum_{i=1}^{10}\frac{1}{|\vec{r}-\vec{R_{i}}|}$$. Charge on each nuclear center is one. For the given set of locations of nuclear centers I tried to solve the Hamiltonian using MO theory by mixing  $1s$, $2px$,$2py$,$2pz$ of all the 10 centers making a total basis of 40 with arbitrary coefficients to be determined.
Now I have used STO -3G to compute the one electron attraction integral by using the existing algorithm and wrote my own code. My question is whether there any software package to compute all the integrals (one electron integral) using slater type orbital in python. Because STO-3G is not an accurate representation of slater type orbital. Do I need to go for FORTRAN for slater type orbital basis set calculations?
Also is there any reference books or papers where I can find the algorithm for slater type orbitals one-electron integrals?

I would argue that in order to get accurate results, using several GTOs on each center will probably get you farther than one fixed STO. Nonetheless, https://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid_state_physics_software lists DFTB+ as having STOs. The integral code is likely to be Fortran still.

Thank you so much.What GTO basis set would be appropriate for the above Hamiltonian. I already used STO-3G to solve the above Hamiltonian. How do I verify my answers? and how do I know how much, the answers are deviating from the right answers? @TAR86

Any sufficiently sized basis set will do, check https://bse.pnl.gov/bse/portal for candidates. Given that you have a single electron, Hartree-Fock is exact in the given basis set. You could run your calculation in e.g. ORCA to verify your code. To assess the quality of the solution, observe the rate of convergence when using subsequently larger sets, such as cc-pVXZ, where X = D, T, Q, 5 ...

@TAR86. Thank you so much again. Is there any better references or lecture notes or any review articles about theoretical chemistry which has all the pieces of information you said about Hartree Fock

I learned from Wikipedia that ORCA is supported only in C++. Do we have a similar software package in python to verify the answers and observe the rate of convergence @TAR86

The write-up on wikipedia is not entirely bad, for detailed derivation and mathematics have a look at Szabo, mentioned on wiki. ORCA is closed source, but would do all that I think you try to do here. The point is that HF is the same and exact for one electron in any quantum chemistry package - so why program it (unless that's the point).

I am completely new to Quantum chemistry. Thank you so much for explaining patiently. All I need is the eigenfunctions of the above mentioned Hamiltonian. If I get it from ORCA, then that is fine. Thank you so much once again. Please suggest some good reference or tutorials about ORCA@TAR86

There is a google site: https://sites.google.com/site/orcainputlibrary/home and the manual is not bad (I wrote parts of it). You would set up a "molecule" consisting of 10 hydrogen, multiplicity 2, charge +9. Choose HF, a basis set and let it rip.

Let us continue this discussion in chat.

Thanks sir, In this software, is it possible to get excited states for the above Hamiltonian ? Does this give a all the excited states of this Hamiltonian mentioned above ?@TAR86


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Is it possible to build a non-degrading battery?
I am not a chemist, nor do I know anything about chemistry. What I am interested in are batteries. I don't really care about the whole "Charge in minutes, last for days" types of batteries (the famously hyped graphene battery) But is it possible to have a battery that will never degrade and always keep its performance without upkeep/maintenance?
I just didn't know if there were some types of laws in chemistry that prevents this from happening. I would ask in electrical engineering but this sounds to me more of a chemistry question.

"Nothing is static. Even the Mona Lisa is falling apart."

There is not a big principle. But  this would require all materials to be in their most stable state  and this is not possible during operation. Moreover in real batteries one can have deformation, leakage, ecc. But in theory a battery is not required to degrade by its working principle.

But is it possible to have a battery that will never degrade and always keep its performance without upkeep/maintenance?

Well, consider the dry pile powering the Oxford Electric Bell, which has been running since 1840. However, that is an example of a nonrechargeable pile.
The question is, can a rechargeable battery be designed that retains it's charging/discharging current profile, reflecting a conserved structural integrity, during an unlimited number of cycles?
While in practice there are many reasons this is difficult to do (loss of electrolyte, occurrence of irreversible passivating reactions, etc) in principle there seems to be no reason why this should not be the case, at least over time on the order of a human lifetime. We manufacture many things which last an awful long time, and batteries would last even longer if designed or treated in an exceptional way, shielded from the environment and charged/discharged in a highly regulated manner. Some battery designs such as thin film lithium-ion allow an exceptional amount of cycling.

I just didn't know if there were some types of laws in chemistry that prevents this from happening.

The second law of thermodynamics places limits on what you can expect to achieve given the constraints of a particular choice of battery chemistry. This reveals itself for instance in Peukert's law, particularly applicable to lead-acid batteries operating at constant temperature, which says that the capacity of a battery is restricted by its discharge rate. Discharging quickly reduces the total work you can do with the battery.
The second law however does not say we cannot use work to recharge a battery to the same original state a very large number of times. But it does make it very hard! It begs the question, "is such a  battery useful?" Probably not, except theoretically. The design and operation of a battery involves tradeoffs between charge density, discharge and recharging rate, and preservation of structural integrity during charging/discharging cycles.
By the way, you could generalize the question to include other portable high-capacity sources of electrical power. Think of probes that are sent into outer space and continue to operate after, well, a long time. Or consider the quantum battery.

But is it possible to have a battery that will never degrade and
  always keep its performance without upkeep/maintenance

Nothing is permanent but change. As stated above, there is a battery which is almost working for more than 100 years at Oxford; it is a Zamboni pile. I was looking for the original work for some time and it turned out to be in Italian.
Zamboni G., 1812, Della pila elettrica a secco. Dissertazione, Dionigio Ramanzini, Verona
It is working for ages because it draws very very tiny amount of current. The moment you start drawing a current from anybattery, chemical changes will happen which will finally lead to thermodynamically favorable state of all elements in the battery. For example, if you have a copper zinc battery, Nature would like to keep the zinc in Zn2+ state and copper as metallic Cu. In short, no chemical battery can last forever.

Nothing is permanent but change, at least until entropic heat death of the universe, assuming that theory is correct. - Some guy reading this.


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What happens if water is treated to AC power supply of 12 kV and 14 kHz?
Generally water splits to oxygen and hydrogen at $\pu{1.23 V}$. What happens if we treat water at several $\pu{kV}$ with frequency range of $\pu{10 kHz}$ to $\pu{15 kHz}$ AC supply.

Short circuiting, fuse blowing or even buying a new power supply is going to happen.

Well, you can try it and find out, if you wanna risk your life, that is. BTW 1.23 V isn't enough to make the reaction go with and significant pace.

Using up to 30 kilovolts is not uncommon with aqueous systems in a technique called capillary electrophoresis. Similarly, applying kilohertz current is also common in a technique called conductometry to measure the resistance of aqueous solutions.
In general, when you pass high frequency current, electrolysis does not occur.
You have to check the dielectric breakdown voltage of water, perhaps with high voltage and high frequency current, severe Joule heating would be observed, if there is no dielectric breakdown (which means internal sparking).


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Should the tip of the burette of an automatic titrator be immersed in the analyte?
In all the tutorials I found online, the tip of the burette (containing the titrant - NaOH) was immersed into the analyte solution. My question is: shouldn't the tip be out of the solution? Isn't there a possibility for the analyte to crawl up the tip and react with the NaOH inside the burette that hasn't been dispensed yet?.
Everyone does that... Even in the official tutorials of the company

I would think that the tip of the burette should never be in the liquid being analyzed. If it is then an indeterminate amount of the titrant in the tip can diffuse into the liquid.

I quess it's a question of the diameter and lenght of the burette tip, and the speed in which the automatic system can finish the titration. If the former is thin, and the latter is fast, then you can probably neglect the diffusive backflow, or calibrate and deduct it from the result.

In classical titrimetry you would avoid dipping the burette into the sample. However, for autotitrators, the electronic dispenser tip (electronic burette) should be dipped in the sample. This is to avoid any error due to the a drop clinging to the tip. In manual titration one would wash it. Most autotitrators are like this design including Karl Fischer system. Your concern can be alleviated by the fact that you need pressure to cause a back flow of the sample into the tip of the dispensor.

Just to quibble you wouldn't wash the tip of the burette in classic titrimetry. You touch the tip to the side of the flask, and then wash down the side of the flask.

Chemists are in general Sceptical Chymists !

I have tried either inside or outside and they both worked fine. Since I deliver very small volumes at a time (20 μL), I validated the dispensed volume using a balance and dipping the tip inside the analyte was more precise. My system dispenses through a small micropipette tip (jerry-rigged).

This is the way to go as an analytical chemist! Calibration with 20 micoliters is not so trivial. Both deliveries should have been identical, however, there are tons of weighing errors with small weights such as evaporation issues and buoyancy errors. Once you dipped the tip in solution, evaporation error is minimized.


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What is a good resource for learning about the intermolecular bonding in water beyond high school?
I'm writing a paper for school on boiling water at various temperatures and pressures, but only noting cases wherein high temperatures were achieved without boiling under low pressure (metastable states). I haven't been able to find a good explanation for this phenomenon other then the chaotic nature of boiling coming from heterogeneities. I believe it is because of the high surface tension, but I have not been able to find any in depth resources for intermolecular bonding (in water) other than at the high school level (hydrogen bonding). I'm just looking for some guidance, thank you.

So you wanna an analysis of overheating or hydrogen bonding? You may think one can get from one to another, but it's rather overoptimistic.

Well that was my initial thoughts as the surface tension of water is uniform (I think I've never found anything at all so this is just brainstorming) so it becomes hard for imperfections to appear on the surface until the surface tension varies enough for air pockets to form beneath them. I'm not sure this is just guessing, but I'm trying to find any concrete papers on this. Or just metastable states in general.

It is an interesting topic which I am not knowledgeable on but you may want to look into nucleation theory as well. https://en.wikipedia.org/wiki/Classical_nucleation_theory

Google Scholar is very helpful when you need to read advanced level of work. The key word you can search is Superheating of liquids (water is one of the common examples). Anyone who has been burnt by microwaved water knows the pain of superheated water. There are some excellent reviews on the phenomenon of superheating.
This is pretty good overview:
http://www.lassp.cornell.edu/sethna/Nucleation/
Starting from the original work in 17th Century to recent ones...
An experience to prove, that water, when agitated by fire, is infinitely more elastic than air in the same circumstances
Phil. Trans., 41 (1) (1739), pp. 162-166
Experimental superheating of water and aqueous solutions
https://www.sciencedirect.com/science/article/pii/S0016703709000994
Maximum Superheating of Water as a Measure of Negative Pressure
https://aip.scitation.org/doi/abs/10.1063/1.1722122

Thank you, I'd heard of arXiv, Research Gate, and Google Scholar but I think I was just using incorrect terminology trying to search for papers or chapters from textbooks on the subject.


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Action of H2S and OH- on an alkyl halide?
What would be the action of $\ce{OH-}$ and $\ce{H2S}$ on 2,6 dichloroheptane? I came across this question in my organic chemistry workbook but I have simply never seen a question of this type before. I think a cyclic compound may be formed, but I am not sure which.
I thought really hard with help from the comments under my question, I came up with this solution.  However in my book it says that there should be only one sulphur atom in the cyclic compound. How does this make sense as there are two halogens in the original haloalkane? Is my mechanism incorrect?

It rather depends on the reaction conditions, but the formation of 5 and 6-member rings is a favourable process and likely to occur, particularly with sulfide as a nucleophile

I assumed you mean $\ce{OH-}$ (like alkaline $\ce{H2S}$?)

@Waylander thank you! But why would there be a sulphide?

@Mathew Mahindaratne Yes.

H2S plus OH- gives HS- which will react readily with the haloalkane to give RSH, this will be further deprotonated by OH- to give RS- which will react internally to give the cyclic species.

@Waylander oh okay I get it! Thanks very much!

Please why has this question been down voted? I did research and could find no explanation on this forum.  Even on the general net, i could not find any explanation for the mechanism of this . Thank you.

Can this question please be put off hold? I have edited it.

It is unlikely you will get a cyclic disulfide as you have drawn. The second chloride is far more likely to be displaced by the internal nucleophile, that is the RS- already in the molecule. Redraw you mechanism with stepwise addition of sulfide.


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How to actually calculate work from force times distance?
I understand that work = force $\cdot$ distance, but I'm not sure of the units and conversions to use.
What I know is that I have a piston head of area $\pu{28cm^2}$ (~$\pu{4.4in^2}$) traveling a distance of $\pu{6cm}$ (2.4") perpendicular to an opposing force of $40,000$ psi . How much work is being done? I'm guessing I am looking for an answer in joules?
My thinking so far is that the answer to this question is as follows:
work in $\pu{Joules}$ = $\pu{Pa * m^3}$ displacement.
Displacement = area of piston head $\cdot$ distance travelled $=  \pu{28cm^2} \cdot \pu{6cm} = \pu{168cm^3} = \pu{1.68\cdot 10^{-3} m^3}$. Force in $\pu{Pa}$ is $\pu{40000psi} \cdot 6895 = \pu{275.8MPa}$.
So force $\cdot$ displacement = $\pu{46,334 Joules}$.
I'm not sure if this is correct and am confused by some of the conversions.

A question being put on hold does not stop people from commenting. Therefore, I don’t agree with the claim that “had [we] put the question on hold before he had the opportunity to answer [you] would still be in the dark”. Based on your edit I have reopened the question (that is the point of closing things; it is to encourage people to edit their questions). I removed your meta commentary; if you want to discuss site policies, please consider posting on [meta]. Lastly, if you don’t want your question to be misconstrued as schoolwork, then try not to make it look like schoolwork - eg context...

Comments are not for extended discussion; this conversation has been moved to chat.

I know: (The Displacement = The distance traveled), but : (area of piston head $\cdot$ distance traveled = $\Delta{V}$)

Adnan, according to Merriam webster at least, displacement is a volume, (area x distance), not merely the distance travelled, which is what is calculated above.

Orthocresol.  I think your comment that it is my responsibility to "not make my question look like homework" is silly, frankly.  If the site was called Chemistry Homework Stack Exchange that would be something else.  Maybe you should ask rather than presume?

@user1023110 People here volunteer their time to answer questions. What you wrote initially was a problem that very easily could have been pulled directly from an introductory physics assignment. There are dozens of people per day on this site who simply copy and paste an assignment and hope someone will do it for them. Its much easier to have the default position be to close these questions and reopen them when the user clarifies the post then to leave them open to clutter the site.

You might want to familiarize yourself first with the correct use of quantities, and then with the correct use of the corresponding units. In particular, $40\,000\ \mathrm{psi}$ is the value of a pressure and not of a force. Furthermore, you should avoid confusing quantities, names of quantities, and units in equations as in “work in joules = Pa * m^3 displacement”. You might want to use proper quantity calculus instead.  You should also avoid the wrong use of SI prefixes as in “40000 × 6895 = 275.8M”.
Your initial assumption is correct. Work is force times distance, assuming the movement is in the direction of the force. In quantity calculus, this is expressed as the following quantity equation (for simplicity’s sake ignoring the vectorial character of the quantities):
$$W=F\cdot x\tag1$$
where
$W$ is work,
$F$ is force, and
$x$ is distance.
You do not know the value of the force $F$; however, you know the value of the pressure $p$ and the value of the surface area $A$. And you know that pressure is defined as
$$p=\frac FA\tag2$$
Thus, the force $F$ acting on the piston head is
$$F=p\cdot A\tag3$$
You could use this equation to calculate an intermediate value for the force $F$; however, it is generally preferable to insert such intermediate equations into the total equation for the final result; i.e. in this case, insert Equation (3) into Equation (1), which yields:
$$W=p\cdot A\cdot x\tag4$$
Now you have only a single equation for the work $W$ that only depends on known independent quantities (the given pressure $p$, area $A$, and distance $x$).
It is only now that you should insert actual numerical values for the quantities. In order to avoid the need for conversion factors, you might want to convert the given values to a coherent system of units first, preferably SI base units. Thus
$p=40\,000\ \mathrm{psi}=2.7579\times10^8\ \mathrm{Pa}$,
$A=28\ \mathrm{cm^2}=0.0028\ \mathrm{m^2}$, and
$x=6\ \mathrm{cm}=0.06\ \mathrm m$.
Inserting these values into Equation (4) yields
$$\begin{align}W&=p\cdot A\cdot x\\&=2.7579\times10^8\ \mathrm{Pa}\times0.0028\ \mathrm{m^2}\times0.06\ \mathrm m\\&=46333\ \mathrm J\\&\approx5\times10^4\ \mathrm J\end{align}$$
(Note that $1\ \mathrm{Pa}=1\ \mathrm{kg\ m^{-1}\ s^{-2}}$ and that $1\ \mathrm J=1\ \mathrm{kg\ m^2\ s^{-2}}$.)

Thank you.  , conceptually I think I understand all of this, suggested by the fact that we come up with the same answer,  although not sure I understand  what is wrong with saying “work in joules = Pa * m^3 displacement”, though perhaps better if I said "work in joules = pressure in Pa * m^3 volume displaced"?  From what I can see this appears to be a true statement.

$\mathrm{Pa}\cdot\mathrm{m^3} =\frac{\text{newton}}{\mathrm{m^2}}\cdot\mathrm{m^3}=\text{newton}\cdot\mathrm{m}=\text{Joule}$


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store alcohol tinctured herbs in polypropylene for extended time
Can I store alcohol tinctured medicinal herbs in the 2 oz. food grade polypropylene cups  (PP 5) for about a week?  I would like to have several "doses" ready to go because they are taken several times a day, and I would like to have several days ready to use at all times.  Thanks

Keeping alcohol in plastic is usually better avoided. Hard to tell on individual basis.

The alcohols should not be an issue, it is used for storing often.  Some oils will affect it.
Here is a random guide that includes a few common oils.


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Mechanism with hypohalite in haloform reaction
The haloform reaction is usually done in a solution of NaOH and $\ce{I2}$ (or indeed, any halogen). Now the first part of the reaction mechanism in the wikipedia page and in the most organic books describes the reaction between the enolate and a halogen molecule ($\ce{X2}$).

[Picture source: Wikipedia, URL:https://upload.wikimedia.org/wikipedia/commons/8/8d/Haloform_Schritt_1.svg ]
Now, one of the things that I read in the wikipedia page is that the haloform reaction can also be done with sodium hypochlorite($\ce{NaOCl}$). We also know that hypohalite ions are produced in a solution of base and halogen, which is what we use in case of haloform reaction. So, I was wondering, is there any way to draw the mechanism with hypohalite only, without bringing in the halogen molecule? What would that mechanism look like?

I'd expect same thing, just with OH- leaving instead of halide anion.

The mechanism of haloform reaction will also be almost similar in the case if we use hypohalites (here, it is $\ce{NaOCl}$).
In the first step, instead of $\ce{OH-}$, the acidic $\alpha$ proton will be taken up by $\ce{OCl-}$ (Remember, $\ce{HOCl}$ is a weak acid (but not as weak as water), so it's conjugate base i.e $\ce{OCl-}$ is a moderately strong base to take up a proton), and $\ce{HOCl }$ will be formed, which will be attacked by the enolate formed in the previous step, and $\ce{OH-}$ will leave to form the $\alpha$-haloketone. The reaction will proceed further two steps to form the tri-halo substituted ketone, and correspondingly $\ce{OH-}$ will be generated. These generated $\ce{OH-}$ can then attack the electrophilic carbonyl carbon give the haloform and the salt of the organic acid .
The mechanism is shown below,

This is quite interseting but is there any way to experimentally prove that this is the mechanism? Do you know any research paper on this?

@ShoubhikRajMaiti This might be helpful https://pubs.acs.org/doi/pdf/10.1021/cr60052a001 . Although you might need to search through the references a lot.

@SoumikDas  in the second step of the mechanism, its OH group to be attacked and not the halide atom, because halide groups have a more electron withdrawing power than hydroxyl groups, and so, in ClOH, OH is more electrophillic and hence its attacked.

@HonBisasoIsmail Seriously?? Think about it once again what you are saying

yes we all know that oxygen is more electronegative than chlorine, but my argument is that despite that fact, also when it come to functional groups, halides are more electron withdrawing than hydroxyl group

More simply, starting with Cl2 + NaOH implies:
$$
 Cl2 + NaOH = NaCl + NaOCl
$$
with the above reaction being the basis for the commercial production of aqueous chlorine bleach.
So, equivalently, we should be able to start with commercial bleach (which is actually not pure NaOCl, but sodium hypochlorite plus NaCl and even possibly Na2CO3).
Interestingly, if there is an electro/photochemical redox aspect to the haloform reaction (leading to radical and non-radical associated reactive halogen species), then the presence of sodium chloride, which is an excellent electrolyte, may be significant as could be carbonate radicals.
Working with freshly prepared AgOCl (stable in the presence of Ag2O) in place of NaOCl/NaCl would be one way to test, as the formation of insoluble AgCl and Ag2CO3 removes chloride and carbonate ions.

From another vantage, departing from the classic organic chemistry depiction, I start with  2012 article in Atmospheric Chemistry and Physics 12, pages 6237–6271 (full paper ) regarding the haloform reaction in atmospheric conditions (and a path variation below for natural waters), to quote:
“Halocarbons can then be formed through the reaction of HOCl, HOBr and HOI with the dissolved organic matter through the haloform Reaction (R6, written for X = Cl), or less likely, through addition Reaction (R7).
RCOCH3 + 3 HOCl → RCOOH + CHCl3 + 2 H2O (R6) “
Based on the following comment on the reactivity of HOCl contained in this 2017 compilation by ScienceDirect at to quote:
“The reactivity of HO-Cl mimics those of the two-electron oxidation reaction exhibited by peroxynitrite (HO–NO2) or hydrogen peroxide (HO–OH) whereby the homolytic cleavage of O–X (X=N, O, or Cl) to form the HO and Cl pair could impart a diversity of chemical reactions with a variety of biomolecules.”
Based on the above and other mentioned radical pathways, I suggest the following largely radical pathway, letting G = RCO for convenience, assuming the presence of sunlight (or lab light) containing UV, as part of a hypochlorous acid activation path:
HOCl + hv → •OH + •Cl
Reference: See Scheme I and also a link to a free pdf download of the full paper, which I recommend, as this advanced work is replete with references to rRHS (radical Reactive Halogen Species) attacking organics.
Or, an alternate route per other sources including  to quote “and HOCl can react in chloride containing aerosols to form Cl2. Cl2 can then escape to the gas phase and photolyze, releasing Cl atoms which may then reform HOCl” , and a similar observation in this full paper at, which also cites a role for natural organic matter (NOM) and HCO3- (contained also in some chlorine bleaches with added Na2CO3) acting as a catalyst.
Then, the proposed  action of the hydroxyl radical (as frequently employed in environmental remediation, see as an example here ) together with the chlorine radical could continue as follows:
GCH3 +  •OH + •Cl → •G + •CH2Cl  + H2O
HOCl + hv → •OH + •Cl
•CH2Cl +  •OH + •Cl → •CHCl2 + H2O
HOCl + hv → •OH + •Cl
•G + •OH → GOH
•CHCl2  + •Cl → CHCl3
Resulting in the cited net reaction for the haloform reaction in an assisted UV reaction (via  possibly free liberated chlorine, NOM, HCO3-,..., per above references):
$$
GCH3 + 3 HOCl + hv → GOH + CHCl3 + 2 H2O
$$
Hypochlorous acid can also apparently be radicalized in the dark in the presence of transition metals occurring in natural waters in a Fenton-type reaction also per this reference:
$$
Fe(2+)/Cu(+) + HOCl → Fe(3+) + •ClOH-
$$
As such more generally, in the presence of a source of electrons  (e.g. , Fe(2+) →  Fe(3+) + e- ) and a medium that supports an electron’s half-life (like organics including humic acid and ethanol):
$$
HOCl + e- (aq)  → •ClOH-
$$
With the created chlorine hydroxyl radical anion (above) displaying a pH dependence, with some authors assuming, contextually, an appropriate pH condition, citing different Fenton-type reaction products via hypochlorous, more explicitly:
At pH > 5 (Source: Scheme 1 reference above to quote: "Above pH >5 ketones and aldehydes are halogenated by electrophilic substitution at the α carbon"):
•ClOH- → •OH + Cl-  (k = 6.1×10^9)
Source: See Reaction 6 at Supplement Table S1 at and click on Supplement pdf to download)
For completeness, at pH < 5:
$$
•ClOH- + H+ → H2O + •Cl    (k = 21×10^9)
$$
Interestingly, the presence of chloride ion, can also be impactful, albeit slow, per the reaction:
$$
•ClOH- + Cl- → •Cl2- +  OH-  (k = 1.0 ×10^4)
$$
Where the dichloride radical anion (•Cl2-), while being less chemically reactive is, however, more stable and provides a source of •Cl (which slowing reacts with water liberating the hydroxyl radical at pH >5):
$$
•Cl + H2O → •ClOH- + H+   (k = 2.5×10^5)
 $$
The Science Direct compilation also relatedly postulates a radical argument pathway with HOCl, to quote: “Reaction of HOCl with thiols yields RSOH via the radical–radical coupling reaction of the intermediates, RS• and HO•” displayed in Equations (2.77) to (2.79).


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Identify KNO3 and KH2PO4 at home
I am not expert in chemistry. I need to find a way to identify a $\ce{KNO3}$ and $\ce{KH2PO4}$ powder at home. I suspect the seller gave me the wrong powder. What is a simple way to test between these compounds at home?

KNO3 is much better soluble in water.

Given the number of votes and views gained in such a short time, I hope that you will edit this question to be more clear, As-is I don't know how to answer it.

In the absence of other chemicals you can utilize the fact that nitrates(V) are much stronger oxidizing agents than phosphates(V).
You can soak a piece of newspaper in a saturated solution of the salt, let it dry well in a ventilated place and try to ignite it.
If the salt was $\ce{KNO3}$, it will burn vigorously and fast (that's why kids are often using paper soaked in sodium nitrate as primitive fuse for home-made firecrackers and rockets).
$\ce{KH2PO4}$ doesn't promote combustion of paper, so it will burn normally or even slower.

Okay. The burning technique is surely simple for detecting $\ce{KNO3}$. Thanks!

@sanch21 It looks like you used an unregistered account to ask this question. I just want to point out that you can request to have your accounts merged and be able to accept this answer.

Well, given that one is a weak acid and the other is the salt of a strong base and a strong acid, the pH of a solution of the powder should tell you which it is. If it's $\ce{KNO3}$ it should neutral, and if it's $\ce{KH2PO4}$ it should be acidic. Just need some pH paper, which you could probably buy at a pool supply store.

This could work, but KH2PO4 solution would have pH somewhere around 4.7 - water used at home might be as acidic as this, I guess.

@Mithoron so be sure to use the pH-paper to test regular water too.

Touch medium samples of both salts with a glowing ember. No noticable effect would be observed on contact with $\ce{KH2PO4}$. Contact with $\ce{KNO3}$ shall produce noticeable flame.
Video https://www.youtube.com/watch?v=dJRqJ145dyM


text:
Calculate the empirical formula of the compound containing nitrogen and oxygen

Calculate the empirical formulas of the compounds containing: nitrogen 47% and oxygen 53%.

My guess was to get the molar mass of N and O and divide that by 47 and 53 and continue from there because that was my teachers example but it doesn’t seem right

Given is amounts of $\ce{N}$ and $\ce{O}$ in $\pu{g}$. To find amounts of $\ce{N}$ and $\ce{O}$ in $\pu{mol}$, divide $\pu{g}$ amounts by relevant molar masses. Rest is pure mathematics.

Empirical formula reflects lowest ratio of the amounts of the elements expressed in natural numbers $\mathbb{N}$.
Amount $n_i$ of $i$th element is the ratio between mass $m_i$ and molar mass $M_i$; mass $m_i$ can be found via mass fraction $ω_i$ and total mass $m$:
$$n_i = \frac{m_i}{M_i} = \frac{ω_im}{M_i}$$
In general, finding empirical formula $\ce{A_aB_bC_c}$ is just a matter of finding the ratio
$$a:b:c = n(\ce{A}) : n(\ce{B}): n(\ce{C}) = \frac{ω(\ce{A})}{M(\ce{A})} : \frac{ω(\ce{B})}{M(\ce{B})} : \frac{ω(\ce{C})}{M(\ce{C})}$$
where $a,b,c\in\mathbb{N}$ ($m$ is cancelled out as it's a constant).
For $\ce{N_xO_y}$
$$x : y = \frac{ω(\ce{N})}{M(\ce{N})} : \frac{ω(\ce{O})}{M(\ce{O})} = \frac{0.47}{14.01} : \frac{0.53}{16.00} = \pu{3.35e-2} : \pu{3.31e-2} \approx 1:1$$
resulting in empirical formula $\ce{NO}$.