text:
Formation of tetrazoles from ketones in Schmidt reaction

Normally in Schmidt reaction of ketones, amides are formed however for some reason here further reaction takes place and a tetrazole is formed. Here is my try for this reaction mechanism (if my mechanism attempt is wrong please point out the mistakes and the proper mechanism):

Relating to the mechanism, I have the following doubts:

Is this a special case of reaction or are tetrazoles formed with many different ketone reactions? If not, what makes this case special?

What are the driving forces in each step? Is the aromaticity the main reason?

One particular question comes to mind. Do lactams react with HN3? You might check the stoichiometry in the Schmidt reaction.

Stoichiometry is fine I checked. It is reacting with HN3 after all. What about the mechanism I wrote? Does it seem ok? Or do you have a source with actual mechanism

Hydrazoic acid, a weak acid, is usually prepared in H2SO4 from sodium azide. The initially formed amide (lactam) forms a nitrilium ion via the enol of the amide that is protonated with loss of water. Whether the addition of hydrazoic acid is stepwise or "electrocyclic" is at issue. Take a look. here: http://organicreactions.org/index.php/Schmidt_reaction_(2)

i understand the nitrilium ion formation. but you have presented 2 pathways one is electrocyclic and other is stepwise. i dont seem to get what mechanisms you have implied here. i tried drawing a mech where the N(-) part of hydrazoic acid attacks the nitrilium ion followed by a cyclic rearrangement(after a nucleophilic attack on the N in the cycle by -NH of hydrazoic acid. which one of your mechanisms is that?  and what is the other one you think is possible.

Stepwise means a terminal nitrogen of hydrazoic acid adds to the nitrilium cation then the nitrogen of the ring adds to the other terminus to close the ring. Some double bond migration will likely be necessary. Perhaps someone will speak to the hybridization of N3 in the ring closure. You can see the problem in your third to last structure. BTW, the synthesis of losartan employs trioctyltinazide or tritylazide addition to a nitrile. Probably electrocyclic in this case.
https://patents.google.com/patent/US7915425B2/en

It is true that normally in the Schmidt reaction of ketones, amides are formed as the major product (Ref.1). However, the products from the Schmidt reaction is highly depeends on the conditions used. The other byproducts include tetrazole and urea derivatives (Ref.2).
The mechanism for tetrazole formation given by OP is acceptable but it need harsh conditions such as the presence of $\ce{NaN3}$ and $\ce{POCl3}$ as the solvent at high temperature (e.g., Ref.3). Actually, during the Schmidt reaction of ketones, the formation of terazole should be taken place even before formation of the corresponding amide. According to the mechanism (Ref.4), there should be a nitrilium ion $(\ce{R1-C#N^+-R2})$ formation, which can be considered as a N-alkylated nitrile structure:

We all knows that a nitrile reacts with hydrazoic acid $(\ce{HN3})$ in the presence of sulfuric acid to give corresponding tetrazole (Ref.5). Thus, it can be envisioned that the additional hydrazoic acid molecule may react with the nitrilium ion to produce tetrazole compound as depicted in Ref.3:

Is this a special case of reaction or are tetrazoles formed with many different ketone reactions?

It is not a special case of tetrazole formation with cyclohexanone alone. As shown in the mechanism, it is possible to prepare a tetrazole from all kind of ketones (REf.2). However, it is note worthy that Hjelte and Agback have discussed that the formation of only tetrazole with $\alpha$-tetralone even with 1:1 ketone to hydrazoic acid ratio is due to the ring size (resistance to rearrange more stable six-membered to less stable seven membered). The have isolated $45\%$ of $\alpha$-tetralone unreacted (Ref.6).
References:

Karl Friedrich Schmidt, “Process of making derivatives of hypothetical imines including amines and their substitution products,” U.S. Patent 1,564,631, 1925.
G. I. Koldobskii, V. A. Ostrovskii, and B. Z. Gidaspov, "Application of the Schmidt reaction for the preparation of tetrazoles (review)," Chemistry of Heterocyclic Compounds 1975, 11, 626–635 (DOI: https://doi.org/10.1007/BF00959947).
Rajendran Sribalan, Andiappan Lavanya, Maruthan Kirubavathi, and Vediappen Padmini, "Selective synthesis of ureas and tetrazoles from amides controlled by experimental conditions using conventional and microwave irradiation," Journal of Saudi Chemical Society 2018, 22(2), 198-207 (DOI: https://doi.org/10.1016/j.jscs.2016.03.004).
Peter A. S. Smith, “The Schmidt Reaction: Experimental Conditions and Mechanism,” J. Am. Chem. Soc. 1948, 70(1), 320–323 (DOI: https://doi.org/10.1021/ja01181a098).
Robert M. Herbst and Charles F. Froberger, “Synthesis of Iminotetrazoline Derivatives as Trichomonacidal and Fungicidal Agents,” J. Org. Chem. 1957, 22(9), 1050–1053 (DOI: https://doi.org/10.1021/jo01360a013).
Nils S. Hjelte and Tamara Agback, “Benzocycloalkanones in the Schmidt Reaction,” Acta Chem. Scand. 1964, 18(1), 191-194 (DOI: 10.3891/acta.chem.scand.18-0191)(PDF).


text:
Calcium chloride hydrolysis
I was wondering about $ \ce{CaCl2} $ hydrolysis.
The most primitive answer would have been:
$ \ce{Ca(OH)2} $ does not dilute that great, therefore it is not strong enough to have the $ \ce{Ca^2+} $s not hydrolise. Thus:
$$
\ce{Ca^2+ + H2O <<=> CaOH^+ + H+}
$$
Yet, as we can see in this answer - $ \mathrm{pK_B^{II}} $ for $ \ce{Ca(OH)2} $ is $ 1.4 $, which is not bad. Therefore, some reckon the diluted part of $ \ce{Ca(OH)2} $ to be a pretty strong base, making hydrolysis impossible. To make things worse, the cation hydrolysis is known to be harder to trace rather than anion hydrolysis.
Still, I have come across some articles on $ \ce{CaCl2} $ hydrolysis (like this one - a top Google search line; yet, this one first shows that $ \mathrm{\Delta G_{900^oC}} $ is $ \mathrm{90.951 >  0} $).
So, is it safe to say that at $ \mathrm{\approx 25^oC} $ $\ce{CaCl2}$ does not undergo hydrolysis?

You can easily construct the chart of Ca^2+/CaOH+/Ca(OH)2 distribution as the function of pH.

@Poutnik unfortunately, I do not know how. I would appreciate if you write an answer explaining how such charts are done - might be a good tip instead of the full solution.

Take a look at this. Should give you a qualitative idea. https://www.researchgate.net/figure/Molecular-form-of-the-phosphate-ion-in-chemical-balance-under-different-conditions-of_fig2_287898737

Why not to use mhchem MathJax extension \ce{Ca(OH)2} $\ce{Ca(OH)2}$ ?

Just calculating from the head by the simplifying formula $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{(\frac {[\ce{A-}]}{[\ce{HA}]})}$, $\mathrm{pH}$ of $\pu{1 M } \ce{CaCl2}$ is approximately 1/2( 12.6 - log 1)=6.3.
So concentration $\ce{[CaOH+]}$ is about million times smaller then $\ce{[Ca^2+]}$. Hydrolysis happens, but is fully negligible. Note that the real $\mathrm{pH}$ depends on impurities and dissolved $\ce{CO2}$.
About the calculation of respective fractions, we can use the formula for gradual dissociation of multiprotic acid:
If we formally suppose $$\ce{H2A <=>  HA- + H+ <=> A^2- + 2 H+}$$ then respective molar fractions for particular acid forms are
$$x_{\ce{H2A}} = x_{\ce{Ca^2+}} =\frac {[\ce{H+}]^2}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{HA-}} = x_{\ce{Ca(OH)+}} =\frac {K_\mathrm{a1} \cdot [\ce{H+}]}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{A^2-}} = x_{\ce{Ca(OH)2}} =\frac {K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
where $K_\mathrm{a1} = K_\mathrm{w} / K_{\mathrm{b1},\ce{Ca(OH)2}}$ and $K_\mathrm{a2} = K_\mathrm{w} / K_{\mathrm{b2},\ce{Ca(OH)2}}$

There is actually a somewhat unique and important aspect of anhydrous Calcium chloride hydrolysis, not yet addressed.
Namely, a very exothermic dissolution reaction that occurs with solid $\ce{CaCl2}$ from replicating the reverse reaction of its formation from Hydrogen chloride acting on Calcium oxide, at least, to a limited extent:
$$\ce{ CaO (s) + 2 HCl(g)  <=> CaCl2 (s) + H2O (l) + Energy}$$
So, where does the energy term arise to move the cited equilibrium reaction partially back to the left? The answer for solid calcium chloride apparently includes a lattice dissociation enthalpy of +2,258 kJ/mol.
Here is an example of a supporting report (also found in Concise Encyclopedia of Chemistry by DeGruyter):

Thus, when you add calcium chloride to water, the solution heats. When adding calcium chloride to water, hydrochloric acid and calcium oxide form. You must be careful when mixing the substances due to the heat of the reaction and the acid produced.

The chemical implications of a presence of some insoluble $\ce{CaO}$ also includes a possible basic salt formation (as in calcium oxychloride), or with air/CO2 exposure, $\ce{CaCO3}$. So, if ones leaves anhydrous Calcium chloride open in even a low humidity atmosphere, expect some deterioration over time to the exposed product (with possible further implications for concrete, for example).
A suggested experiment, to newly purchased $\ce{CaCl2}$ take a small amount and completely dissolve in water. Note, the heating of the solution. Take a similar amount of $\ce{CaCl2}$ and place it on a plate in a low humidity environment. In a week, try again to dissolve the salt completely.

source: Stack Exchange

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

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What is it meant by “preparing solutions on a blister”?
I was recently looking into an aspirin hydrolysis preparation lab and noticed the following:

(This preparation was taken from http://www.kii3.ntf.uni-lj.si/analchemvoc2/file.php/1/HTML/SPEKTRA/kinetics2.htm)
What is it meant by blisters, and by extension, the droplets? Does this simply indicate a ratio of added solutions? For example, to prepare 9mL of 12.5 mg/L concentrated salicylic acid solution, I would add 1mL of solution A, to 1mL of solution B, to 7mL of distilled water? Does this sound right?

Apparently it refers to a "blister" or pocket tray, with small indentations corresponding to the grid in the image, used for microanalysis.

Dr Pippik is right. Something like chemistry on a chip. The archetype of this was searching for a good solvent on a tray, for whose that know it.

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


text:
Understanding transition state theory
Let's say we have standard chemical reaction given by
$$A-B + C \leftrightharpoons [A\cdot\cdot B\cdot\cdot C]^{\dagger} \rightarrow A +B-C$$
I want to estimate the pre-exponential factor of this reaction using Transition State Theory.
From Fundamentals of Chemical Reaction Engineering by Davis and Davis, I see that
$$\text{rate} = \nu _i C_{\dagger} = \nu _i \mathcal{K} C_{AB} C_{C}$$
Where $\mathcal{K} = C_{\dagger}/(C_{AB}C_C)$, and $C_i$ is the concentration of species $i$.
Then we redefine $\mathcal{K}$ as
$$\mathcal{K}_C = \frac{C_{\dagger}}{C_{AB}C_C} = \frac{Q'_{\dagger}}{Q'_{AB}Q'_C}$$ where $Q'_i$ is the partition function per unit volume.
We redefine energies to lowest ground state, so
$$\text{rate} = \nu _i C_{\dagger} = \nu _i \mathcal{K}e^{-E/k_BT}$$
We know that $Q = Q_tQ_rQ_{\nu}Q_{el}$, where $t$ is for translational, $r$ is for rotational, $\nu$ is for vibrational, $el$ is for electronic partition function.
We also know that $Q'_{t} = \frac{(2\pi mk_B T)^{3/2}}{h^3}, Q_{el} \approx 1$.
Everything good so far.
My question lies with the rotational and vibrational partition function.
$$Q_r = \frac{8\pi}{h^2 \sigma} (2\pi k_B T)^{3/2} (I_1I_2...)$$ where $\sigma$ is the rotational symmetry number. My question is, what is the rotational symmetry number? How do I find it? How do I find moment of inertia for molecules? More importantly, does a single atom like $C$ even have a rotational partition function?  How do I calculate $Q_{v}$ for $A-B$? and $C$? The formula says
$$Q_v = \prod_{s=1}^n (1-e^{-hv/k_BT})^{-1}$$
The end result says
$$r = \frac{k_B T}{h} \cdot \frac{Q'_{\dagger}}{Q'_{A}Q'_{B}} e^{-E/k_BT}C_{AB}C_C$$
The final question would be, how do I simply something like this? With the $\frac{k_B T}{h}$, what exactly got simplified?
Edit: I figured out I can get moments of inertia from Calculated moments of inertia for molecules are listed in the NIST Computational Chemistry Comparison and Benchmark Database: http://cccbdb.nist.gov/).

Most phys. chem. textbooks (MqQuarrie & Simon for example) describe how to calculate the vibrational and rotational partition functions including any symmetry factors.

So rotational spectroscopy is something one can spend a lot of time talking about so I'll try to keep this brief. Although this isn't a complete answer, it was too long for a comment, but hopefully this sheds some light on the topic.
For rotational transitions, one of the selection rules tells us that in order for EMR to induce a transition in the rotational energy levels, a permanent dipole moment must exist in the molecule (i.e. a single atom won't be observed in rotational spectra, nor would a molecule such as $\ce{H_2}$).
The rotational symmetry number is a variable that is used to account for the amount of rotations one can perform on a molecule that would make it indistinguishable from the starting point (i.e. a 90 degree rotational of $\ce{PCl_5}$ along the principal axis can be performed three times before returning each atom to its original position in space, however, each rotation produces an identical picture of the molecule each time).
Moment of inertia can be calculated from the rotational constant, however, in practice, one would usually calculate the moment of inertia with computational software and then use that to approximate the rotational constants which are employed to predict the energies and allowed transitions for the rotations of a molecule.
This is a link to an appendix with the equations used for calculating the moment of inertia in molecules based on the geometry (e.g. symmetric rotor, linear)
Here is an article that further discusses symmetry numbers and partition functions.
This one talks about rotational partition functions of diatomic gases.
This article talks about transition moment integrals and selection rules.


text:
What is the reason for decrease in electron donating tendency with increase in no. Of bonds
The order of electron donating tendency with polar protic solvent goes as
$$\ce{RCH2 > RCH=CH > RC=C-}$$
However, I think that with the increase in the number of bonds the electronegativity between $\ce{R}$ and $\ce{C}$ increased and the leaving group should rather be able to leave easily. Where am I going wrong here?

Okay so the first thing to note is that the "electron donating" tendency follows the given order because the electronegativity of "carbon" increases. You could explain this with something as simple as the hybridisation theory. Notice the hybridisation difference between the carbon atoms in the first and second compound. $sp^2$ hybridised carbon atoms have more electronegativity than $sp^3$ ones(Because of more percentage of s character). Therefore carbon atoms with $sp^3$ hybridisation more easily share electrons than those with $sp^2$. Now between the last two, what do you mean?
If you meant $\ce{RC#C-}$ which should most likely be the case since a -1 charge is present in what you have drawn. Well here the carbon which would act as a "donor" has a $sp$ hybridisation and you could use similar arguments as stated above.
This is the reason why the order is what it is. You most probably were not wrong has you have got all the facts about electronegativity right, but looks like you have arrived at a slightly wrong result while connecting the facts and arriving at an explanation.
Also, please elaborate and explain your problem and attempt better next time.
Hope this helps, in case you find anything unanswered, feel free to comment.

i was thinking in a way that if electronegativity increse then two species have a greater difference in the electronegativity and bond should becomes more ionic but I see it becomes more covalent


text:
The importance of the cold chain in the food and the pharmaceutical industry
There are these everyday things that one should know as a scientist and especially as a chemist, but which never come to light in an academic curriculum — at least not in mine.
One such thing is the purpose of imperative non-disruption of the cold chain. It is clear that there are plenty of reasons why it is crucial to transport food and also pharmaceuticals in a refrigerated condition. On the one hand, it reduces the risk of bacteria proliferating significantly, but it also prevents labile substances from decomposing or reacting with other ingredients. So far, so good. But why shouldn't you cool down products again once the cold chain has been interrupted?
I have read, for example, in the package insert of a pharmaceutical product that the drug should always be stored below 25 °C. In pharmacies, the product is kept in the refrigerator, and I, as the consumer, should continue storage in the refrigerator if possible. If this is not feasible for me, e.g., due to lack of space, the storage temperature should not exceed 25 °C. It is also pointed out that I should not keep the drug in the refrigerator after storage at room temperature. This reminded me of the wise words of my grandmother, who said that fish, for example, should only be frozen once because otherwise, it will go bad. With many other foods, similar claims are made. Since these indications can be found across countries and cultures, I assume that there are known scientific reasons for this, but I am not aware of them. I can't figure out why it should be bad to refreeze a defrosted piece of meat or why I shouldn't put my medicine in the refrigerator on hot days anymore after storing it at room temperature for a while.
Maybe a few food chemists and/or pharmacists can shed some light on this.

Edit: Poutnik and Julian, I wish I could accept both of your answers. I find it difficult to decide which answer is "the best", as they both address a different aspect of my question which they explain in a very understandable way. I have decided to mark Julian's answer as accepted since he — despite his long membership — has a much lower reputation than Poutnik.

In general, pharma companies can and will only guarantee that a product will behave properly when treated within certain bounds which they are legally expected to test and have tested. If you handle the product outside these bounds then you cannot expect legal compensation if the product does not behave as advertised (including causing harm). For specific food and pharmaceuticals the reasons are bound to be more specific.

Answer here from a quality manager in the pharmaceutical field in Europe.
Pharmaceutical companies are obliged to perform stability tests for their products according to the relevant pharmaceutical (GMP) agency in your market (FDA/EMA/etc) and the international agreed guidelines like ICH Q1A-F and the WHO).
These tests are carried out in stability chambers with specific temperature and humidity according to your market. A product intended to be stored at room temperature in Europe and the US will need a study for minimum 12 month up to 5 years at 25 °C and 60% relative humidity (RH) as well as an intermediate study at 30 °C and 65% RH for 6 month to 5 years and a an accelerated test at 40 °C and 75% RH for at least 6 month. Cool stored products will have the long time storage at 5 °C and the accelerated test at 25 °C and 60% RH and there is a whole lot of other stability studies you can (or can be forced to) test.
Based on a risk assessment and your stability protocol you will analyze your drug after defined times (like 1, 3, 6, 9, 12, 18, 24, … month).
For each pharmaceutical product there must be a defined specification. This might includes the assay, impurities like degradation products, metal residues, solvent residues, color, taste, microbiological contamination etc.
Now, after each time point you test your defined specification if everything is still within the specification. The scope is based again (as nowadays everything in the pharmaceutical environment) by a risk assessment.
Using this information you can define your storage conditions and shelf life of your drug. If you find out you fulfill the specification at 25 °C for the whole duration but not at 30 °C, you will define the storage condition to be below 25 °C to be on the safe/legal side. Same with the storage at 5 °C. If the tests at 25 °C fail after 2 month, you can e.g. define a storage of max 2 weeks at room temperature to be on the safe side.

I remember doing such stability tests by quantitative TLC by CAMAG scanning device.

@Julian Specifically about tablets, because of the formulation of the active ingredient into a matrix (including @Poutnik's comment) one may read this like as to probe if the chemicals did not undergo unwanted chemical reactions during storage where covalent bonds altered.  The complementary question, do these tests include checks of the phase purity, i.e. as a probe for crystallographic phase transitions which may differ e.g., by  solubilisation rates as the original API (and may be under a different patent than the intended API)?

@Buttonwood If this is in the specification, sure. Usually if there are phase requirements, these are in the specification for the product and should be analyzed. I just never had such a product for testing.

@buttonwood  This is IMHO controlled indirectly, as there are performed simulated dissolutions at standardized conditions. So matrix or phase alternation effects would affect the content of active or controlled compounds in the resulting solution.

@Poutnik My speculation was this were characterizing the powdered tablets by PXRD equally allowing to check for identity and quantification, perhaps complemented by IR spectroscopy.  Your suggest to perform standardized dissolution experiments obviously complements this set which possibly already is known by recording the ADMET data.

@Buttonwood It was not suggestion, it is done routinely anyway within stability studies.

Avoiding cooling again after keeping it (for noncritical products) at room temperature is mainly to prevent forgetting it was not cold all the time and that it may not last as long as expected.
Taking food from the fridge, you may not remember after a week or two that it was outside longer than it should have been. It can violate the safety of the growth of microbes, which does not fully stop when it gets cold again.
For drugs or similar critical products, it violates stability and guarantee of usability, as thoroughly described by @Julian answer. I did related analysis within drug stability studies in past as well.
Food freezing is another case. Industrial freezing of food is done by shock freezing using liquid nitrogen. This leads to the quick forming of small ice crystals that do not puncture the cell walls, so after melting food before usage, it is OK.
But when refrozen back e.g in a freezer, freezing is slow, leading to the growth of large ice crystals puncturing the plant/animal cells. After another melting, cell liquids leak out, cells lose their tonus and the food gets an unpleasant mechanical and sensory state, affecting the final taste. Additionally, what was told above about the fridge applies here as well.
The presence of free ice along with frozen vegetables is a sign the content was refrozen.
Slow melting of previously shock-frozen stuff leads to the growth of bigger crystals, similar to recrystallization, as bigger crystals are thermodynamically preferred due to lower surface energy. So some cell wall tearing and subsequent deterioration of the content happens even during melting and continues until the content is frozen again. Also, shock freezing still induces some stress on the product, even if lower than slow freezing.
So the final effect of

fast freezing -> slow melting -> slow freezing -> melting

can be much worse than just

slow freezing -> melting

as the former has more time and opportunity to cause damage.

Also intact cell walls inhibit microbial growth rather effectively. Fresh apples or pumpkins stay good on a cool shelve for months, even tomatoes, but not if there is the smallest dent in their skin. When thawed a second time, food should be used immediately, because all their cells are broken. +1

Thanks for your answer! The growth of differently sized ice crystals is very interesting. Why is that?

Also, the answer explains why deep-frozen products should not be re-frozen. But often we also freeze leftover food, or freeze the meat after visiting the butcher. So far I have never had problems with the meat tasting odd, even though it was frozen slowly and therefore these larger ice crystals must have formed.

@Sam below the freezing point, the solid phase is more energetically favorable than the liquid phase, but the solid/liquid boundary is still unfavorable. Small ice crystals have very high surface area / volume, so they are unstable. Close to the freezing point, freezing will mostly proceed by increasing the size of the few crystals that manage to grow large enough to be stable. So average size of crystals is large. At temperatures well below the freezing point, the energetic preference for the solid phase is greater, so far more crystals survive, and the average size is smaller.


text:
Calculating heat of combustion in a bomb calorimeter
In a constant volume bomb calorimeter, we measure the change in temperature of the calorimeter and find out it's change in  internal energy. We then convert it to change in enthalpy.

To determine $\Delta H_c$ we write the balanced chemical equation to find $\Delta n_g$ and using the value of $\Delta U$ obtained from calorimeter, we apply $$\Delta H = \Delta U + \Delta n_g RT$$

The above equation is only valid for constant pressure and temperature. How can we apply this equation on calorimeter, where the temperature change is observed?

What do you mean by how can we apply the equation? if you have the numbers then all you have to is plug them in

The main requirement of the equation is that temperature should be constant throughout the reaction. That's how it is derived. But temperature changes in a calorimeter. So this equation should be invalid here.

I'm not sure from where you've studied it but the formula is valid in a constant volume bomb calorimeter. There is no particular conditions on pressure and temperature.. where did you get that from?

From the basic definiton, $\Delta H = \Delta U + \Delta (pV)$. If we consider ideal behavior, then $\Delta (pV) = \Delta (nRT)$. This will become $\Delta n_g RT$ only when $T$ is constant.

In a constant volume bomb-calorimeter, we can make chemical reactions take place. If we let chemical reactions freely take place in it, then the total internal energy change is zero. This is due to the fact there is no expansion work since the system is a constant volume by definition and the heat transferred is zero since the bomb calorimeter is well insulated. The heat released in a reaction is automatically absorbed by the bomb calorimeter device. Hence the total internal energy change is zero.
Now, consider another path of the reaction. We start with reactants and turn them into products under constant volume and constant temperature conditions (*) and then these products we raise the temperature as to reach the final temperature we got in the previous path.
The energy associated with moving from reactants to products isothermally is $\Delta U_{rcxtn}$ and the energy change for raising up the products to the temperature of first path is $ \Delta U_{heat}$
Now, we can write:
$$ \Delta U_{heat} = C_{v}^{cal} \Delta T$$
And,
$$ \Delta U_{rcxtn} + \Delta  U_{heat} = 0$$
Or,
$$ \Delta U_{rcxtn} = - C_{v}^{cal} \Delta T$$
Now, consider the enthalpy change of reaction:
$$ \Delta H_{rcxtn} = \Delta U_{rcxtn} + \Delta (PV)$$
Now, if we were to assume our reactants as ideal gases a, then $PV=nRT$ and also considering constant temperature, we get:
$$ \Delta (PV) = \Delta (nRT) = RT \Delta n$$
Using the above result and expression for internal energy, we finally arrive at the following equation:
$$ \Delta H_{rcxtn} =  - C_{v}^{cal} \Delta T + RT \Delta n$$

Refernces:
Around 42:49 of this video
Page-2 of these notes


text:
nodes and probability density
The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.
The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.
so if probability density cannot be zero, then what about nodes? Arent they regions where probabilty density is zero?

What is a "node" here?

A node is in the orbital context generally a 3D surface with the zero $|\Psi|^2$ value, what has interpretation as zero probability of an electron occurance.

The text is simplifying, as what it says is valid for and only for 1s orbital.
The text author had probably in mind that there is no sharp end of electron density as misinterpretation of orbital shapes may suggest. There is an asymptotic exponential-like ceasing of the wave function values toward infinity.
All higher s orbitals with n>1 have n-1 radial nodes, having zero $|\Psi|^2$ value of wave function at n-1 different radii.
Similarly, 2p orbitals have the node plane passing the center between the two lobes and upper p orbitals have additional node surfaces. Similar for d and f orbitals.


text:
Vapour density dependence on temperature
Does vapour density change with temperature or is it solely dependent on molecular mass
as per the formula VD=molar mass/2

Vapour density is the density of gas with respect to the density of hydrogen at same temperature and pressure. You measure density of the gas at, say $p$ and at temperature $T$ and divide it by the density of Hydrogen at the same temperature and pressure.
The formula you used was only valid for an ideal gas.
$$\rho_{gas} = \frac{pM_{gas}}{RT}$$
$$\rho_{\ce H_2} = \frac{pM_{\ce H_2}}{RT}$$
Thus $$\frac{\rho_{gas}}{\rho_{\ce H_2}} = \frac{M_{gas}}{M_{\ce H_2}}$$
And we know that $M_{\ce H_2}$ is $2\pu{g/mol}$ $$\frac{\rho_{gas}}{\rho_{\ce H_2}} = \frac{M_{gas}}{2}$$ So vapour density of an ideal gas will be independent of temperature.
But for real gases after if you set the pressure of the gas and Hydrogen gas in such a way that after van Der Waal correction, the ideal pressure of both comes out to be the same, then your formula will ve valid for real gases as well.

I see a clash of definitions, or rather their terminology for the relative definition of vapour density versus the absolute density definition with the unit kg/m3 in sense of "density of vapour". As it is easy not to be aware of "vapour density" is not "density of vapour" but an independent term. Personally, I have never met the term and meaning "vapour density" yet.

"Mass of a certain volume of a substance divided by mass of same volume of hydrogen". This is just another way to define it. Wiki has defined it by stating that volume of both gases is same. But it still is the ratio of densities. It is defined in such a way because it is easier to calculate masses of the two gases in similar container than to calculate the absolute densities of the two gases in container of different volume.

Sure, I have linked the Wiki article as well, and modified my answer. I do not have any objection about the term nature, but the term itself.  What I wanted to say is that the choice of term "vapour density" is unlucky, leading to confusion with "substance density" as mass per volume. As non-native, learning English after chemistry, it is easier for me to fall into a terminology trap.

Yes, they did mess up the terminology.

In fact, it has both meanings, depending if you consider "vapour" as adjective or as a part on the term.

For ideal gas vapour behaviour, the vapour density is the ratio of its density and of the density of hydrogen at the same conditions
$$\frac { \frac {pM}{RT}}{ \frac {pM_{\ce{H2}}}{RT}} \simeq \frac M2 $$.
There for assumption of vapour ideal gas begaviour, "vapour density" does not depend either on temperature either on pressure.

I see a clash of definitions, or rather their terminology for the relative definition of vapour density versus the absolute density definition with the unit kg/m3 in sense of "density of vapour" or substance density.
It is easy not to be aware of "vapour density" is not "density of vapour". In fact, lack of context gives the reader the confusing choice if vapour is an adjective to density or a part of the term.

so it is dependent?

Is y in y=2x dependent on x ?

BTW, when you ask a question, always write down also the context or background. It prevents the need of clarifications and helps to provide relevant answers. If constex/bckground is omitted, it often leads to clarification ping-pong and/or helper may blindly shoot useless answers as they do not know the needed context. Last but not least, provide you resolution effort first, otherwise the questions may get closed.

understood my bad , what kind of context should i have added?

At least what scenario you consider. And it is expected you search through available resources before asking.

@JustJohan See the update, apologizing for confusion.


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Dissociation constant for Indicators
So I came across a question which asked to calculate the $\mathrm{pH}$ at which an indicator of given $\mathrm{K_{a}}$ would change colour. The solution was given in the following way
Since when the color change occurs $\ce{[HIn]=[In^{-}]}$,
As $\mathrm{K_a}= \left(\frac{[\ce{H^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$,
At $\mathrm{K_a}=[\ce{H+}]$ the colour change would occur.
And consequently from the $\ce{H+}$ concentration the $\mathrm{pH}$ was calculated, the solution imposes that during the colour change the equilibrium is attained, but how can that be since to observe colour change you add an acid or base which disturbs the initial equilibrium for which the $\mathrm{K_{a}}$ was defined and hence the now won't hold true.

Hint: Indicators change their color not at Ka but rather pKa+1 and pKa-1. Sort this out by searching more on the web about the indicators. The dissociation constant is indeed a constant, it does not change with pH. It is only affected by temperature and pressure.

I totally get that and that isn't my question whether or not the Ka changes but is rather why is the  author taking the condition of equilibrium to plug the values of HIn and In-.

Ka value has to remain the same, regardless of the pH. The system is at equilibrium all the time no matter whether you add acid or base to an indicator solution. The rate of reaction is instantaneous.

Note that indicator range is much dependent of the color transition and related eye sensitivity. Reported intervals are AFAIK always narrower than 2 pH units, like 1.3 ( methyl orange 3.1-4.4) - 1.8 ( phenolphthalein 8.2-10.0). The perceived center pH is biased toward the less intense colour.

First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is why there is a range of colour change(look at these pH charts), with the upper and lower bounds having a $\mathrm{pH}$ difference of $2$. As an example, here's a simplified reaction for Phenolphthalein ($\ce{Ph}$ is the phenolphthalein anion)
$$\ce{\underset{\text{colourless}}{H2Ph} <=> 2H+ + \underset{\text{violet}}{Ph^2-}}$$
If you take a simpler, weak monobasic acid indicator $\ce{HIn <=> H+ + In-}$, the $\mathrm{K_a}$ is
$$\mathrm{K_a} = \frac{\ce{[H+][In^-]}}{\ce{[HIn]}}$$
The relative concentration of the indicator and it's conjugate base is only determined by the concentration of the $\ce{H+}$ ion, as $\mathrm{K_a}$ is a constant.
Answering your second query, the author takes the condition of equilibrium, simply because the system is in equilibrium, that is $\mathrm{Q_a = K_a}$. It is assumed that the reaction is instantaneous or that titration is performed slowly enough so that the system is in equilibrium at every single instant.

It still doesn't clear my confusion, I know that the system is assumed to go through the process slowly and that causes the system to be in equilibrium all the time but the Ka was defined for the concentration of the species that were a result of dissociation of the indicator only. When the pH changed due to adding of extra acid or base the concentration of the species also changed so how can you take Qa=Ka.

@buddy001 Because the extra acid reacted with the indicator and attained equilibrium, so $\mathrm{Q_a = K_a}$


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Acorrect range of concentrations for the reactants in the equilibrium reaction

What would be a correct range of concentrations for the reactants in the equilibrium reaction, $\ce{Fe^3+(aq) + SCN-(aq) <=> FeSCN^2+}$?

For a Chemistry project I am measuring the effects of changing the concentration of $\ce{Fe^3+(aq)}$ on the yield of $\ce{FeSCN^2+(aq)}$ produced. I'm going to use a spectrophotometer to calculate the absorbance of the products.
I was wondering what amounts of concentrations of $\ce{Fe^3+(aq)}$ and $\ce{SCN-(aq)}$ would be suitable.
My research suggests that I should set the spectrophotometer to around $\pu{450 nm }$and that the domain of the concentration of the product $\ce{FeSCN^2+(aq)}$ should be between $0 \le [\ce{FeSCN^2+(aq)}] \le 2 \times 10^{-4}$. Since I don't know the equilibrium constant, I would need to look at literature to find out what concentrations would be ideal since when absorbance > 1, the relationship is no longer linear.
I'm only varying the concentration of $\ce{Fe^3+(aq)}$ in 10 intervals (currently $\pu{0.1-1 mol/dl^3}$), the concentration of the $\ce{SCN-(aq)}$ will be kept constant.
What would be the ideal concentration range for $\ce{Fe^3+(aq)}$ and the ideal concentration for $\ce{SCN-(aq)}$?
ps. I am adding $\ce{HCl}$ as a neutralizer to the $\ce{Fe(OH)-}$ formed since it distorts the colour of the solution.
I am also using $\pu{20 ml}$ of $\ce{Fe^3+(aq)}$ solution and $\pu{20 ml}$ of $\ce{SCN-(aq)}$ and $\pu{10 ml}$ of $\ce{HCl}$.

Is not 0.1-1 M Fe^3+ too much ? Absorbance would be IMHO out of range. The reaction is very sensitive, used to quantitative Fe determination. So, rather consider [Fe^3+] the same as the final [FeSCN^2+]

Why do you think that the absorbance is not linear when it is less than 1 ? Linearity goes from absorbance of 0 to approx 2.5 the upper value being limited by the quality of the instrument used and not for any fundamental reason based on Beer-Lambert law. From memory the FeSCN has a max extinction coefficient of approx 450 $\mathrm{dm^3,mol^{-1},cm^{-1}}$ so you will need dilute solutions in the range you give.

@porphyrin, I made a typo, it should be A > 1. My teacher recommended that I try to measure values of absorbance less than 1.


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About spontaneous combustion of methane
I heard that spontaneous reaction happens if $\Delta G=\Delta H-T\Delta S$ is negative.
For combustion of methane, according to Chemguide:

$\Delta H=\pu{-891.1 kJ K^{-1} mol^{-1}}$
$\Delta S=\pu{-0.2422 kJ K^{-1} mol^{-1}}$

Thus, I would deduce that at normal temperature, $\Delta G<0$, combustion of methane is spontaneous, while at very high temperatures, it will stop.
But, from another reference, I deduce the opposite conclusion: if we consider the diagram below (taken from https://fr.wikipedia.org/wiki/R%C3%A9action_chimique),

I would think that in order to be able to go to the intermediate state, one needs to provide energy, thus that it could not be spontaneous for this reason. One needs to provide energy, for example by heating, that is increasing temperature.
How is that possible? Any comment?

Practically every spontaneous reaction requires activation energy. This comes from thermal motion of molecules, thermal, mechanical or electric shock, light, breaking crystals or from already ongoing reaction. Or, spontaneous reaction is not ongoing, if needed activation energy is too high. We then say reactants are thermodynamically unstable, but kinetically stable. Typical example for that are diamonds.

Thermodynamics and kinetics are two different things.

Does this answer your question? Difference between thermodynamic and kinetic stability

A catalyst can reduce the activation energy, but that "bump" still must be passed to "roll downhill", even though the reaction is thermodynamically favorable. For example, methanol won't ignite at room temperature, but a Pt catalyst can set it on fire. Look up "catalytic lighter".

Thank you all. Wikipedia saus something similar : "Because spontaneous processes are characterized by a decrease in the system's free energy, they do not need to be driven by an outside source of energy.". But so the word spontaneous is a bit clumsy

But then, at very high temperature, does the reaction stop because $\Delta G$ becomes positive ?

@MathieuKrisztian Not really, because also methane will decompose at some point, and water too. At such high temperature, you have a different reaction. There is not p,T region where burning methane is an equillibrium reaction that you can push forwards and backwards by changing pressure and temperature a bit.

Thank you. It helps me a lot in the understanding of the concepts

For starters, above 800°C, you start getting CO instead of CO2. The Bouduard equillibrium.

The combustion of methane is spontaneous at ordinary temperature. But it is is extremely slow.

@Maurice That is a bit of a dubious statement. This is a multi-step radical chain reaction, and I doubt it has an finite reaction rate at RT in the dark.

"Spontaneous" means different things in different contexts
Your penultimate paragraph captures a key idea. The explanation for why this is right requires a recognition of the context of the term "spontaneous".
The context of the statement at the start of the question $\Delta G=\Delta H-T\Delta S$ is negative is thermodynamic stability. But this is somewhat at variance with the more natural use of the term which implies "things happen without being pushed". This idea is closer to the idea of kinetic stability in chemistry.
You correctly identify the need to add energy to get the reaction past the transition state. Even if the reaction overall releases energy (thermodynamically spontaneous) the reaction won't just happen if there is a huge barrier to getting over the transition state. There are big kinetic barriers that stop the reaction "just happening". Oxygen and gasoline will react to release energy but this doesn't happen without the push given by the spark plug in the engine of a car.
That barrier is so low in some cases a compound will react with nearly everything with little excuse (chlorine trifluoride will set fire to asbestos). That is a spontaneous reaction in any context. Luckily, few thermodynamically spontaneous reactions are also kinetically spontaneous or humans would catch fire in air.
So when you see the term "spontaneous" ask what is the context: thermodynamic or kinetic? And don't confuse them.


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Reactivity Dilemma of elements
While learning Periodic Properties, I stumbled upon a doubt regarding reactivity of elements of period 2 and period 3.
(I am denoting reactivity, of any $X$ element as $R_{X}$.)
For reactivity of $s-block$ elements, I saw $R_{Li}<R_{Na}$ . Acording to me this is so because since the metallic character increases down the group, (Am i correct $?$)
But as we go towards right side, things become complicated. For boron group, I saw this: $R_{B}<R_{Al}$. I  Similarly for other groups I saw following:
$R_{N}<R_{P}$;
$R_{O}>R_{S}$;
$R_{F}>R_{Cl}$;
(According to me reactivity of third period elements should be greater than that of second period but i don't know why it doesn't follow for oxygen and halogen groups)
Please tell reasons why there is such unusual recativity trend for the elements I mentioned, and thanks in advance $!!$
(I saw on internet having divided opinions, which was not helpful).

$\ce{Li}$ vs $\ce{Na}$, $\ce{Be}$ vs $\ce{Mg}$, $\ce{B}$ vs $\ce{Al}$:
Rather, increasing metallic character and reactivity have the common cause, decreasing the ionization energy due nucleus charge shielding and greater electron distance from a nucleus.
$\ce{N}$ vs $\ce{P}$, $\ce{O}$ vs $\ce{S}$, $\ce{F}$ vs $\ce{Cl}$:
The same reason as above leads to decreasing of electron affinity of their atoms, what has the opposite effect - decreasing of reactivity.
Note that atomic nitrogen is more reactive than atomic phosphorus, so it follows the general trend. But molecular nitrogen is much less reactive than molecular phosphorus due its much stronger bond.


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What are the advantages of using ellipsometry over AFM to determine film thickness?
I am trying to understand why one would chose one over the other and I am struggling to understand this since both are non destructive. I know AFM can only go at 5 nm but other than that I don't see anything.
I am also struggling to wrap my head around the concept of ellipsometry so any suggestions or insights into how this technique works would be amazing.


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What should be the nomenclature of this organic compound?
I am confused with my teacher's explanation for the nomenclature of this organic compound.

He said that we should start numbering from the sulphonic acid at the bottom of the structure. But one of the students asked if we should start numbering from the other sulphonic acid and with this method, the double bond will be on number $1$ (as shown in the image below). But my teacher didn't give a reason for that.

So isn't the student saying correct ?

Although this isn't disputed by anybody here, let's start by noting that in general, suffixes have greater "priority" over double and triple bonds in numbering. The rules are given in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);
[...]
(e) saturation/unsaturation

This means that the sulfonic acid suffixes should obtain the locants 1 and 2, which are the lowest possible for two adjacent groups. There are two possible ways of doing this.

With the sulfonic acid groups sorted, we can now turn to the thorny issue of the alkene. It is more complicated than I thought, actually. First, let's consider how to refer to the double bonds in the names. Again, the answer is provided by the Blue Book.

P-31.1.1.1
[...] Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

On the left, this is a "typical" double bond where the locants (2 and 3) differ by 1, and so only the lower locant is cited.
On the right, the locants (1 and 6) differ by  more than 1, so both locants are cited with the higher one in parentheses.

Lastly, we have to make a choice between these two. We turn to the last rule for today:

P-31.1.4.2 If there is a choice of names and numbering, the following criteria are considered in order until a decision is reached:
(1) a minimum number of compound locants. A compound locant is used for a double bond if the locants of the atoms at each end of the bond do not differ by a value of one. When a compound locant is required, the higher locant is cited in parentheses. [...]

Since the option on the left-hand side has no compound locants, and the one on the right has one compound locant, the left choice is preferred. Therefore, the preferred IUPAC name for the molecule is cyclohex-2-ene-1,2-disulfonic acid.

There is a convention we follow while numbering cycloalkenes, numbering is always done through the double bond, meaning that the double bonded carbon atoms must always get consecutive numbers.
This rules out the student's way of assigning numbers.
So, now we have two options, and using basic rules of assigning numbers, you would easily come to the conclusion that your teacher's answer is the correct one.
Reference :
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/03%3A_Organic_Nomenclature/3.04%3A_Alkenes_Cycloalkenes_and_Alkadienes
See the paragraph corresponding to the naming of cycloalkenes.

Please see Answering nomenclature questions and cite the rule explicitly. Note that chem.libretexts.org is not an authoritative source for chemical nomenclature.

That's not true; there are a few (fairly esoteric, admittedly) examples where the doubly bonded carbon atoms do not receive consecutive numbers. See https://i.stack.imgur.com/3mHyT.jpg for an example where this is forced (2013 Blue Book, P-31.1.1.1, page 374). This is not the case here, so your answer turns out to be right! But not for the right reason, unfortunately.


text:
How many dichlorinated products including stereoisomers can be isolated when (S)-2-chlorobutane reacts with Cl2 /hv?

This is my attempt at solving this problem. I get 6 products. However my book says the answer is 5 products.

III and IV are the same. You can rotate III to make IV

I have to rotate the structure III in the plane of paper. How do you get IV from III by rotating it ? I don't get it. If I rotate III in the plane of paper 180 degrees, the structure remains the same, it doesn't change and it surely doesn't become IV.

Turn the paper over and look through it

Is rotation out of the plane of paper allowed in Fischer projections?

Are we assuming only one chlorination?

Actually, the 4th structure is wrong, you can't get that as C3 must have S configuration. But the structure you made has R configuration on both C2 and C3.

Elaborating on what Waylander said, The C-2 carbon chlorine should be in S configuration so both (III) and (IV) structures can't be correct and only one with S configuration(IV) will be considered and rotating that will result in the other structure.


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Solving detergent powder in a solution
I used a detergent powder mixed with water to clean my espresso machine. Some micro particles of powder are stuck into machine's inner tank. So when I pull the hot water out of the machine's nozzle, it contains micro powders and it tastes powdery.
What solution can I use instead of water to wash out all those sticky powders?
A harmless solution to break bond between powders and solve the powder into itself.

Water until everything is gone. Everything means both dust and soapy taste.

Some coffee pot makers recommend using white vinegar; boil it thru the system, then boil some water thru the system to "rinse" it. Perhaps some coffee oils are holding some non-soluble material from the detergent powder.

I think using acetone will be helpful. Acetone will evaporate fastly and it is polar and can form hydrogen bonds to the detergent powder like water. Though you need to use much acetone to wash all detergent powders and after that, wait for some minutes for acetone to vaporize from your machine.

Acetone may be a disaster, if there is any plastic.


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How does this rotation minimize the dipole moment?
A paper I'm reading says:

The low energy conformation of 3 minimizes the overall dipole moment of the compound and thus rotates the exocyclic amide 180° from desired conformation, as can been seen in structure 4.

Why does this rotation minimize the overall dipole moment?

Isn't simple enough?

Hello
At first, let me tell you the reason for the rotation. As I have mentioned in the picture, the lone pair on Nitrogen can contribute to two different resonance structures. One of the resonances leads to the aromaticity of the ring which I have mentioned with number 2. This aromaticity makes the compound more stable ( lowers the energy of the molecule ). But the fact is that both resonance structures contribute to hybrid structure.
Thermodynamically, it is more likely for the compound to only have the resonance which makes the ring number 2 aromatic, Though ring number 1 rotates to prevent the lone pair on Nitrogen from having the resonance form which I have drawn with blue color. The red resonance can't be done, because after rotating, ring 1 and ring 2 are not in the same plan anymore. So because of the rotation, the more likely resonance structure will contribute to the hybrid structure more than before.
Now your question. Why does this rotation minimize the overall dipole moment?
As a result of the resonance structure ((the red one )), there will be a negative charge on Oxygen. This negative charge causes a dipole moment which is directed toward the Oxygen. Though according to picture 4, the molecule will have a smaller overall dipole moment.

This does not address the question at all. It’s about rotation of the exocyclic amide, which is to the right of the rings in the drawn structure


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How to calculate planar density of a face /plane of crystal with cation and anions?
Planar density is given by numer of atoms in plane /plane area  .
Suppose I want to calculate
planar density of 100, 101 and 111 planes of Nacl. Radius of Na+ and Cl- ions is 0.086 nm and 0.126 nm.Na+ are present at octahedral voids and cl- at FCC but the problem which I am facing is that I cannot add simply the number of atoms in given face/plane  for calulation of numer of atoms(zeff) as sizes of cation and anion  will be diffrent .What should I do??


text:
How many monochloro products including stereoisomers of the given compound are possible?

I tried solving problems and I got 2 products.

However according to my book the answer is 3 products.

(II) has a stereoisomer. I also had to look twice, no sweat. ;)

Your second J molecule has got two stereoisomers, one (S) and one (R)


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how to derive equilibrium concentrations from initial conditions for mass action reactions?
I am reading a tutorial on biochemical reactions and mass action kinetics (https://www.math.utah.edu/~keener/books/control.pdf, pp. 1-2) and would like to derive an analytic solution to confirm simulation results. This is NOT a homework question, I am simply trying to understand things.
The tutorial considers a simple reaction where A and B combine, reversibly, to produce C:
$A + B \rightarrow^{f} B$
$C \rightarrow^{r} A + B$
where $f, r$ are the forward/reverse rate constants. It is clear that the change in [C] over time is:
(Eq. 1) $\displaystyle\frac{d[C]}{dt} = f[A][B] - r[C]$
It's simple to then show that the equilibrium constant $K_{eq}$ is:
(Eq. 2) $K_{eq} = \displaystyle\frac{r}{f} = \frac{[A]_{eq}[B]_{eq}}{[C]_{eq}}$.
The question: how can we use this to derive the equilibrium concentration of one of the species (like $[C]_{eq}$) as a function of the initial concentrations of A and B, $A_0, B_0$ and rate constants $f, r$? Simulation of Eq. 1 for $A_0 = 200, B_0 = 100, f = 0.0001, r = 0.001$ shows that $[C]_{eq}$ is around $90$. How can this be confirmed analytically?
Attempt at solution: try to rewrite Eq. 2 in terms of initial concentrations. We can use the fact that $[A]_{eq}$ and $[B]_{eq}$ can each be rewritten in terms of $A_0, B_0, [C]_{eq}$:
$[A]_{eq} = A_0 - [C]_{eq}$
$[B]_{eq} = B_0 - [C]_{eq}$
this is because the equilibrium concentration of "pure" A (A that wasn't used with B to make C) has to the total amount of A we started with, minus the amount of A that went into producing C. Same argument for B.
But it's not clear if this helps get a solution? Plugging these quantities into Eq. 2 we get:
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} - K_{eq} = 0$
which seems too messy to be correct. I was expecting a simpler quadratic equation for such a simple problem. Guidance on solution or references to derivations will be great.

Here is the expression from the question:
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$
Get rid of the fraction:
$(A_0 - [C]_{eq})(B_0 - [C]_{eq}) = K_{eq} {[C]_{eq}}$
Distribute the sums in the product:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0)  + [C]_{eq}^2 = K_{eq} {[C]_{eq}}$
Sort the terms and you have your quadratic equation:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0 - K_{eq})  + [C]_{eq}^2 =  0$


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How can you tell if a metal ion will hydrolyze to form an acidic solution?
I saw somewhere that all metal ions except for the alkali metals and alkaline earth metals will hydrolyze to increase acidity. If I'm interpreting things correctly, sodium ions, for example, will not hydrolyze because sodium hydroxide is strong, thus sodium ions will not react with water to form sodium hydroxide and hydrogen ions. But I was told that magnesium will not hydrolyze, but magnesium isn't on the strong bases list that was given. By the way, I'm a grade 12 student, so I would appreciate if you kept things down to earth. Can someone clarify please?

So the ability of a metal cation to hydrolyze water molecules to form acidic solutions is based on the charge of the ion and the size of the atom. The larger atoms will stabilize charge better (i.e. it spreads it out over a larger space). Unstable cations form hydroxides or oxides with water molecules and release hydronium ions.
Estimations of the acidity for metal cations can be found with the equation below,
$pka = 15.14 - 88.16(\frac{Z^2}{r})$
Z being the charge of the ion and r being the atomic radius. I bold the word estimation cause this isn’t a perfect system and I want to make that clear. However, better equations do exist.
Electronegativity does play a role too. The more electronegative a metal cation is, the more acidic it will be.
I would recommend taking a look at this link for more information

Thank you for your answer. To clarify, group one and two metal ions have low charges --> a low Z value --> a low pKa value, thus they don't hydrolyze?

And would you also mind confirming that magnesium ions will not hydrolyze?

@SamLiu See https://chemistry.stackexchange.com/a/41678/43057 for an example with magnesium

@ManRow ahh ok thank you, that cleared it up


text:
Good Hess's Law experiment
I am doing an Internal Assessment for my IB chemistry class, and I want to do something on Hess's Law, and I'll be doing it simulation based, using this website. With the solutions and equipment on that simulation, what is a good experiment I can do? I want to avoid doing something relatively  easy like 'finding the enthalpy change of this reaction' and proceed to do so in under 5 pages, I want something a bit more advanced, but still able to be done on this simulation.

You could possibly do a formation of a salt using an acid/base reaction, evaporate it, then calculate the heat released when it enters solution. Extra points if you can explain heat lost to the solution in terms of the thermal stability of water.
This would give you a more advanced look into the thermal processes of chemistry, its the kind of lab we would do in Analytical Chemistry. Multiple steps of synthesis and measurement.


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conceptual doubt on change in entropy
we have learnt in thermo that delta S is zero in reversible process
Now, when a solid melts reversibly is delta s zero since reversible or positive since the particles are more energetic in liquid phase
which one do I follow?
Thanks in advance.

delta S total is zero in reversible process. For solid melts reversibly, delta S of system increases and delta S of surrounding decreases by same amount, so delta S total is zero.

how can delta total be zero should it not  always increase?

Note that "more energetic", especially in the conte of the question, is a vague assertion as for energy might refer to different state functions. Besides, the comment above should answer.

@Alchimista yep ,but can delta total be zero?

delta zero total increase in spontaneous process; delta zero total = 0 during reversible process.

@JustJohan yes it can be zero if the process of reversible.

For a reversible process, the entropy change of system plus surroundings is zero.  But neither of them has to be zero individually, and one can be positive while the other is negative.
For an irreversible process, the entropy change of system plus surroundings must be positive.  But, here, one of them can be negative while the other is more positive.


text:
Doubt about Van Der Waals equation
I got how the real pressure would be less than ideal pressure due to net force of attraction between gas molecules acting towards the center.
But, why is the real volume greater than ideal volume? I thought the real volume would be less than ideal one since the ideal volume is the volume of container (volume of gas particles negligible) while the real volume is the available volume for gases to move excluding their individual volumes i.e. Vi = Vr + volume of gas particles

Available volume for gases to move is not a part of the picture at all. Both volumes are those of container. Now, imagine the molecules moving about the same way in both cases, but in case of real gas they also have a volume of their own. Who would need a bigger container?

Does this answer your question? How can we replace volume in ideal gas equation with V - nb?

In the ideal gas equation $$p_iV_i = nRT$$ $p_i$ is the pressure if there was no intermolecular forces between the molecules and $V_i$ is the free volume available for the molecules to do motion in, which is the volume of the container.
In real gases however, the pressure that you observe (real pressure) is less than the expected pressure. This is because of the intermolecular forces. So you convert the $p_r$ to a pressure which the gas would exert if there were no intermolecular forces.  $$p + a\frac{n^2}{V^2}$$
The free volume available to real gas is less than the volume considered in ideal gas equation. That is because Kinetic theory of gases made an assumption that the volume occupied by molecules of an ideal gas is negligible. But in real gases, the molecules are of considerable size. Thus, the full container will not be free for motion, some volume of the container will be occupied by the molecules. Let the volume occupied by one mole gas molecules be $b$, for n mole gas this is $nb$, so free volume available to gas is less than the volume of container $$V_i =  V_{\text container} - nb$$
Edit: In your question, the $V_r$ they're talking about is volume of the container. And $V_i$ is actual free volume available to the real gas.

Real Volume: Volume of the container
Ideal volume: Free space available
Real pressure: Observed pressure
Ideal pressure: Pressure if the gas didn't have any intermolecular forces.


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Student Question regarding molecular van der Waals volume and Polarizability
I would like to request a crosscheck of my understandings regarding molecular van der Waals volume and polarizability. Please have a look through my statements about these 2 concepts below and point out any mistakes or inadequacies in my understanding.
Kindly please also provide citable authoritative research or review articles to any mistakes or inadequacies in my understanding. I do not have good access to books so articles it has to be.
Statement1
Van der Waals (vdW) bonds are the same thing (synonym) as vdW interaction.
vdW bond/interactions are molecular attractions resulting from dipole-dipole attraction, which means permanent polarity. And London dispersion forces, which are temporary polarity from either random fluctuation of the electron cloud around the atom nucleus or from an external electric field, e.g. a nearby polar molecule).
As such, vdW interaction occurs when the electron clouds of two molecules just touch each other.
Statement2
Van der Waals (vdW) volume of a molecule is (roughly) the volume of the electron cloud of that molecule. And is approximately the sum of the collective vdW volume of the constituent atoms of that molecule.
Statement3
vdW volume of an atom is calculated from the vdW radius of that atom, as a result, I can say that vdW volume of a molecule is calculated from the collective vdW radius of its constituent atoms.
Statement4
Polarizability is the ease by which the shape of the electron cloud is disrupted by either random fluctuations or external electrical influence. The higher the polarizability value the, easier for a molecule’s electron cloud shape to be disrupted.
Statement5
When a symmetric (non-polar) electron cloud is disrupted (being polarizable), the molecule becomes polarized and gains temporary polarity, meaning exists one positive and one negative end on the molecule (like a magnet).
Statment6
Polarizability value becomes higher when the molecular weight increases, meaning polarizability is positively correlated with molecular weight.
Statment7
And both Polarizability and molecular weight is positively correlated with vdW volume, meaning these 3 molecular parameters are positively correlated with one another, an increase of the value of any one increases the other 2. That is because, the greater the molecular mass, the greater the number of electrons. As a result, the electron cloud’s volume (vdW volume) becomes larger and easier to disrupt (high polarizability) since the nucleus has to influence a large number of electrons and electrons that are on higher orbitals with greater distance to the nucleus.
Statment8
High vdW volume molecules more readily form vdW bonds with other molecules because the larger electron cloud volume allows vdW interactions at a greater distance. And because to the larger electron cloud volume means greater polarizability and thus making it easier for London dispersion forces to arise, imbuing the molecule with temporary polarity.
Statment9
High vdW volume and high polarizability CAN indicate higher solubility in water. Although this is not always true and depends on the specific solution and solvent situation.
In a pro water solubility scenario,
The higher polarizability (and by correlation higher vdW volume) and resulting higher London dispersion forces imbuing non-polar molecules with temporary polarity allow non-polar molecules to dissolve in water, though not nearly as well as polar molecules.
Higher dispersion forces can also contribute further to the solubility of polar molecules.
Higher polarizability allows nearby polar molecules to induce polarity(dipoles) in non-polar molecules, thus making otherwise poorly soluble molecules become soluble in water. E.g. the Iron cation in hemoglobin polarizes non-polar oxygen, allowing it to dissolve in blood.
In an anti-water solubility scenario,
The high polarizability results in strong vdW bonding of non-polar molecules with each other, forming aggregates. In order for such molecules to dissolve in water, the vdW bonds need to be broken before allowing the formation of new vdW bonds with water molecules. This process
requires energy and thus making such molecules harder to dissolve in water.
These are the statements I request a cross-check. Aside from this, I would also like to request scientific authoritative link to tables of vdW volumes and polarizability values for the 20 canonical amino acids.
The table should have a legend explaining what a high numerical value for vdW volume and polarizability mean.
E.g. I know that vdW volume is expressed in cubic Angstrom (Å^3), therefore, a high numerical value e.g. 1000 Å^3 necessarily means a greater volume.
However, regarding polarizability, I found they can be expressed in many different ways and I don’t have the level of knowledge to say a high number means higher polarizability (ease to disrupt the electron cloud shape). Thus I need the reference to clearly state that high value means high polarizability.
Thanks for your attention and awaiting your kindness.


text:
Conversion reaction - Organic chemistry

I tried solving this problem, but my answer doesn't match with any of the given options. Here's my solution.

You seem to be asking a lot of questions from a particular book (MS Chouhan). Do you know that there is a solution manual provided with the book?

Yes. But this is the NEET version. I don't think it comes with a solution manual :(

What else can happen after the first Grignard is made?

Your final product will decompose to form cyclopropane.
$$\ce{Br-(CH2)_3-Br + 2 Mg ->[ether] BrMg-(CH2)_3-MgBr -> (\Delta)  + other side product}$$
See this answer for more details.


text:
Melting point of objects and food
What is the melting point of egg yolk?
Well, generally, when we break an egg, the yolk is in liquid state. After we pour it on a pan, the egg starts to cook and it hardens. My question is, can an egg melt/boil ?
What are the conditions to be imposed?

Most foods never melt. Instead, they decompose.

@IvanNeretin that does mean that the egg does not has a definite melting point? since it'll split up?

That's right, it doesn't melt at all.

Defined melting point requires pure substance or in some cases well defined mixture of few components, which do not decompose at melting temperature. Egg yolk is not such a case.

Egg yolk is a complex mixture including a lot of proteins. Heating proteins does not melt them, it causes them to denature or unfold. Once denatured, the proteins interact to form a gel-like solid. If you continue to heat up the solid, it will burn.

As Poutnik said, the melting point is well defined for pure substances. In the case of egg yolk, it is not a pure substance and also when you heat it, some molecules decompose. So we can't measure or even define a melting point for it. Also, the temperature is not constant, when the egg is melting.
A substances' phase doesn't get influenced only by temperature. Maybe it is strange that heating the egg yolk in a pan, makes it to behave like a solid. Actually, some complex changes in intermolecular forces are getting done.
If you want to cause an egg yolk to melt, I think that the only work to do is to heat it under the normal pressure.


text:
Ratio of rates of diffusion of gases
I stumbled upon a question, A mixture of 50 mL of Hydrogen gas and 50mL of Oxygen gas is allowed to effuse through an effusimeter. After some time it is observed that the residual gas occupies 90mL. Find the composition of (1) Effused gas and (2) Residual gas.
Since there was no information given about the pressure, I took it to be constant.
Acccording to the Graham's Law of Diffusion, $\frac {R1}{R2} $ = $\frac{P1}{P2} \sqrt{\frac{M2}{M1}} $ , where M1 and M2 are the molar masses of the gases, P1 and P2 are the partial pressures of the gases and R1 and R2 are the rates of diffusion of the gases.
Now, some of my friends took the ratio of the Partial pressures $ \frac{P1}{P2}$ as a constant and calculated the ratio of rates of diffusion of Hydrogen to Oxygen to be 4 and reached the result composition as: Effused gas- H2=8mL, O2=2mL, Residual gas- H2=42mL, O2=48mL.
I personally think that the ratio of partial pressures will change as the gases effuse out, but I'm not sure how to approach the problem and get the answer with the lowest approximation possible. Please help me out. Is there any way we can solve this problem by not taking the difference in partial pressure as zero?

It is rather mathematical than chemical problem.You have to solve a differential equation.

As @Poutnik indicated, this will be a differential equation problem. This is a high school problem and what your friends actually calculated was the initial rate of diffusion.

Should I post it on the Maths Stack Exchange then?


text:
Understand Hexabromocyclododecane CAS numbers
Hexabromocyclododecane has 16 possible stereo-isomers. So far so good.
I found that:

134237-50-6 is the CAS of alpha-Hexabromocyclododecan

134237-51-7 is the CAS of beta-Hexabromocyclododecan

134237-52-8 is the CAS of gamma-Hexabromocyclododecan

3194-55-6 is the CAS of a mixture of different Hexabromocyclododecan

Now, I have two more CAS which are a bit opaque to me:

678970-15-5

Pubchem says it's the same as 134237-50-6
However, ChemIDplus says it's alpha-Hexabromocyclododecane, (-)-

138257-18-8

Pubchem says it's the same as 134237-51-7
However, ChemIDplus says it's beta-Hexabromocyclododecane, (-)-

Question: Describe each of the CAS different chemicals or are 678970-15-5 and 138257-18-8 just synonyms. If they are different chemicals, what does (-)- mean here?

See questions regarding multiple CAS numbers: FeOOH, potassium sorbate

The (−)- in the chemical name refers to the optical activity of the compound, that is to tilt the plane of linerarly polarized light to the left (levo-rotatory), or to the right (dextro-rotatory) as determined with a polarimeter.  It may be either (−)- (minus sign) or (+)- (plus sign), as seen, for example, for two of the three forms of tartaric acid.  Compounds with stereogenic centers which do not tilt the plane of the polarization (e.g., meso-compounds) may be labeled by (±)-, too.
Similar to this, and equally used for molecules with stereogenic centers − especially carbohydrates  − is the the use of (D)-, (L)- or (DL)- which however relates to the spatial arrangement around the sterogenic atom of highest priority (there are rules how to determine this just from reading structure formula).  It may be used independently as well as simultaneously to the former system of (−)- and (+)-; but note the knowledge what to write within former system does not translate 1:1 into what to use in the later system.
As for the CAS number, this is a book-keeping number (see here); contrasting to a chemical name, it does not convey structural or even chemical meaning.  CAS for example systematically attributes more than one registry number to the same molecule if there is evidence that it crystallizes in different polymorphs.  This «already» happens with inorganic molecules like $\ce{TiO2}$ which may crystallize in the form of the mineral anatase (CAS 1317-70-0), or rutile (CAS 1317-80-2).  Thus, if your literature search is about the isolated molecule alone (say, molecular mass), it does not matter which entry you follow further.  Yet if you are interested in say the melting point of your compound or crystal structure, you have to treat the CAS numbers separate, like referring to different compounds.  Thus, the relationship between CAS number and a structure is not bijective.
(Occasionally, e.g. for some of the more substituted fullerenes, you may encounter many CAS numbers for one and the same compound in a database like Scifinder or Reaxys.  Possibly the difficulty to draw their formulae contributed to create whole sets of CAS numbers.)


text:
Is it possible to make a solution of zinc oxide in a high school chemistry lab?
As part of my IB Chemistry curriculum, we are required to conduct our own experiment researching our own question. I'm in the preliminary research phase and wanted to conduct an experiment looking at the effectiveness of zinc oxide (the active ingredient in many physical sunscreens) in absorbing UV light at different pHs ranging from 4 to 7.
I've found some procedures similar to the experiment I want to conduct, but all of them seem to dissolve a commercial sunscreen in ethanol and run it through a spectrophotometer. I want to isolate the active ingredient itself, and as such, I wanted to make a solution of zinc oxide -- though the molecule is not soluble in water, might there be another way to make a solution (perhaps dissolving in 2-propanol?). Then, I would ideally change the pH of the solutions using a buffer.
Does this seem viable, or is it likely I will need to purchase a zinc oxide suspension or perhaps look at a different chemical? Thank you for your help.

Pure water won't dissolve $\ce{ZnO2}$, nor the typical organic solvents used in the lab (EtOH, EtOAc, ether, etc.).  Similar with $\ce{TiO2}$ in toothpaste, $\ce{ZnO2}$ in sun creme will be provided as a suspension of small particles in an emulsion.  Thus searching for a dispersion at a chemical supplier (e.g., https://www.sigmaaldrich.com/catalog/search?term=Zinc+oxide%2C+dispersion&interface=Product%20Name&N=0+&mode=mode%20matchpartialmax&lang=en&region=US&focus=productN=0%20220003048%20219853286%20219853101) may be a way to go.  Seek for similar experiments, e.g. in J. Chem. Educ., too.

Note it is ZnO, not ZnO2, use of the later in sunscreens would not be a very good idea, see https://en.wikipedia.org/wiki/Zinc_peroxide

https://en.wikipedia.org/wiki/Zinc_oxide#Laboratory_synthesis

@IanBush Your interjection is correct, I meant zinc oxide.  The suggest to look into the 34 articles in Journal of Chemical Education with both concepts of «ZnO» and «UV» present (https://pubs.acs.org/action/doSearch?AllField=%22ZnO%22+%22UV%22&startPage=&SeriesKey=jceda8) shouldn't invalidate this.

@Poutnik Your are correct.

Zinc oxide is made commercially by boiling zinc and admitting air in a controlled manor and collecting in a fabric bag. Zinc boils about 1670F. Substantial tonnages of ZnO go into rubber.

If you want to analyze the amount of $\ce{ZnO}$ in  a commercial mixture, you may dissolve the zinc oxide in $\ce{HCl}$ and determine the amount of $\ce{Zn^{2+}}$ in the obtained solution by complexometric titration with EDTA. It is quite easy.

Zinc oxide is not soluble. It does dissolve at very acidic or alkaline pH, but it is a chemical reaction, not true dissolution.
The role of zinc oxide ( or titanium dioxide ) is mainly to cause UV light dispersion, effectively prolonging a lot the ray paths to allow sufficient absorption.  Dissolving the sunscreen formula would create a thin suspension with very different UV properties, which would have tendency for sedimentation.
I see a sense rather in measurement of passing UV through a thin film of a cream with zinc oxide, prepared by some controlled way, e.g. controlling its mass per area.

You have to do some sunscreen solubility tests first.
The used formula would be an emulsion either "oil in water", either "water in oil". The solvent should dissolve well both.  I would perhaps dissolve sunscreen in acetone, performing filtration/centrifugation or sedimentation, whatever applicable, to get rid of eventual $\ce{ZnO}$ or $\ce{TiO2}$ particles.
Then I would evaporate the aceton and add defined $\ce{ZnO}$ amount.
Or, you can try your own sunscreen formula.

Thanks for your help! Since my experiment is based around sunscreen, do you think simply dissolving (or maybe a better word is just mixing?) sunscreen with an active ingredient of zinc oxide into EtOH would work, or is it better to avoid a commercial sunscreen in a chemical investigation such as this? Otherwise, I probably will look at a zinc oxide dispersion online.

@ibchemstudent See the A update.


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How is dipole moment of ammonia less than sulphur dioxide?
Sulphur dioxide has only two bonds at 180º so shouldn't the dipole moment of molecule be 0 and as ammonia has 1 lone pair so it's dipole can't be 0 so how Sulphur dioxide has more dipole moment than ammonia?

Sulphur dioxide isn't linear - also has a lone pair on central atom.

I suggest you visit this page. Sulfur dioxide is almost always in this form. Two Oxygens are more electronegative than a single Nitrogen and also S-O bond is longer than N-H bond, though the dipole moment of S-O is more than N-H and despite the lone pair on S, SO2 has a greater dipole moment than NH3.


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Why does [Ni(gly)2] show optical isomerism despite having no chiral carbon?
For optical isomerism chiral carbon should be present i.e. carbon with four different groups but this compound doesn't have one then why does it show optical isomerism?

For optical isomerism, a molecule should be different from its mirror image; that's all. Whether it contains a chiral carbon (or any carbon at all, for that matter) is irrelevant.

@Maurice Glycine is the achiral amino acid since there is no carbon center with 4 different substituents (R = H).

@AY The key here is to ask how the glycines are arranged around the central nickel atom. If the arrangement breaks enough symmetry, then the complex is not superimposable on its mirror image.

Is it optical isomerism, though? I can see how it could have cis/trans, but not optical.

@permeakra There's not enough information to know for sure, but I could imagine a chiral complex if the glycines were bidentate on a tetrahedral complex.

Dear anonymous editor, please stop submitting edits that attempt to change gly into a sum formula. For one, gly is easier to read and understand. For two, the sum formula you are suggesting is outright wrong. For three, MathJax should not be used in titles anyway (but that only applies to the question).

@Jan (or any one else): Please flag if the vandalism continues. We might have some rather harsh options at our disposal, but we might need to use them if this continues.

Could you all please explain to me why somebody else has an access to edit my question because continuously somebody is editing it , actually I am very new to this platform , please tell if there is anything that I can do to help.

On this site we aim to have high quality questions and answers. Every user has the ability to edit posts, or suggest such edits. In the latter case they will be reviewed by more experienced users. Editing can have multiple reasons, some are introducing better formatting, correcting spelling, adding tags, giving a better title, etc.. Please take the [tour] to find out more about this site and if you have more questions related to that, visit [meta]. Unfortunately, you question has been targeted by an anonymous editor for a while. I hope that is over now.

Your assumption that $\ce{[Ni(gly)2]}$ shows optical isomerism is incorrect: the complex is square planar in nature and has a plane of symmetry (the plane passing through all the atoms). It is thus achiral and does not have any optical isomers.
It does show geometrical isomerism though, as the ligands are not symmetric with reference to each other.


text:
interconverting Van Der Waal's equation for real gases from number of molecules to density
I want to ask a question about Van Der Waal's equation for real gases.
I was shown the following formula for real gases (VDW) which accounts for the liquification of water vapour which is not possible with an ideal gas model.
$$P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2} = \frac{\rho RT}{1-\rho b} - a \rho^2$$
which can be expressed using densities by making the denominators equal first:
$$P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2} = \frac{NkT}{V-Nb}\left(\frac{V}{V}\right) - \frac{aN^2}{V^2} = \frac{\frac{NkT}{V}}{\frac{V-Nb}{V}} - \frac{aN^2}{V^2} $$
but I appear to yield the result
$$P = \frac{\rho kT}{1-\rho b} - a \rho^2$$
where Boltzmann's constant $k$ has been switched for the gas constant, $R$. I can't find any sources that explain this derivation apart from "Molecular Driving Forces" by Dill, and he doesn't do a thorough conversion either.
How do I convert the numerator with Boltzmann's constant $k$ to the gas constant, $\rho$ which expresses VDW's equation involving densities?

Your analysis is correct in terms of number density.  But let's see how it plays out in terms of molar density.
Let n be the number of moles and A be Avagadro's number.  Then N=nA.  If we substitute this into your first equation, then I get $$P=\frac{n(Ak)T}{V-nAb}-\frac{aA^2n^2}{V^2}$$But, since Ak=R, we obtain:$$P=\frac{nRT}{V-nb'}-\frac{a'n^2}{V^2}$$where $a'=aA^2$ $b'=bA$.  In terms of molar density $\rho=n/A$, this becomes:$$P=\frac{\rho RT}{1-\rho b'}-a'\rho^2$$

Oh, so you're saying that the density in question is molar density, not number density, correct?

It’s not clear how their parameters are defined.

your answer works and matches, so i'll take your answer as the right one. Thanks very much!


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Why is the ionization energy of radium higher than barium?
The ionization energy of radium is higher than barium. The reason provided is that occurs fue to lanthanide contraction. But, Radium isnt a member of lanthanide series. How come we say that lanthanide contraction occurs in it?


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How do I get the m/z = 101 peak for 3-methylbutyl propanoate?

(reference, second source here)
I've found the bigger peaks like m/z = 57, and 70 but how do I get 101? I've tried everything I could think of, like inductive cleavage or McLafferty rearrangement. Since it's 101, using rule of 13 I get the most likely formula to be C5H9O2. But the alcohol-chain has 11 hydrogens, how do I take 2 hydrogens away and end up with 5 carbons?

M/z=101 is ~ 4 times less intense than m/z=43. Minor fragmentation between C2-C3 of the alcohol moiety gives these two cations CH3CH2CO2CH2CH2+ (101) and CH3CHCH3+ (43).

I should add that m/z=43 is more intense because it is a secondary cation rather than primary (m/z=101).


text:
does NH4OH lose concentration when it is open?
I get a lot of variation on NH4OH titration and I am curious if anyone knows if NH4OH at concentration greater than 20% will evaporate the NH3 and decrease in concentration if it is open and at room temp. 20c

Of course it will. That's how you feel its smell, BTW.

Avoid using $\ce{NH4OH}$ formula, it is the old and wrong way to describe  the ammonia solution. It does not exist in water solutions either in non dissociated form, either ( in majority ) as  ions $\ce{NH4+(aq)}$ and $\ce{OH-(aq)}$. $\ce{OH-}$ is a strong base, while $\ce{NH4+}$ is a weak acid, mutually reacting as $\ce{NH4+(aq) + OH-(aq) <=>>NH3(aq) + H2O(aq)}$.

NH4OH does not exist.

My PI when I was doing my PhD told this hilarious story from his time as a student when another student was tricked into getting solid $\ce{NH4OH}$ from another lab. The lab assistant sent her, not thinking she would do it (it was for her own experiment), then quickly called the person he sent her to, telling him to give her the correct chemical but label it $\ce{NH4OH}$ (solid). By the time she came back, he had supplied everyone else with beer and declared the lab day over.

As by the comment by Ivan Neretin «of course it [i.e., an aqueous solution of $\ce{NH4OH}$] will [lose concentration if exposed openly to air]».  Because an aqueous solution of ammonia mostly is ammonia physically soluted in water (example) just as observed with other gases like oxygen, nitrogen, carbon dioxide, too.
Of course you may write the chemical equilibrium of
$$ \ce{NH3 + H2O <<=> NH4+ + OH-}$$
however, as already indicated by the arrow, the corresponding equilibrium constant is small. According to wikipedia
$$K_b = \frac{[\ce{NH4+}] [\ce{OH−}]} {[\ce{NH3}]} = 1.8 × 10^{−5}$$
corresponding to about 0.42% of ammonium in a 1M aqueous solution of ammonia.  Thus, it is only a matter of time of exposure to air and the temperature of the solution until ammonia is evaporated.
A second reason why aqueous solutions of ammonia become less concentrated over time is their reaction with $\ce{CO2}$ of air. If the concentration of ammonia is high enough, you may see this next to the mouth of the bottle, too.  Because it can build up a solid deposit literally baking the stop cocks of glass bottles like seen for lye like $\ce{NaOH}$, on occasion you may see rubber plugs being used instead of stop cocks made of glass to close flasks of $\pu{250 mL}$ volume and smaller.  But compared to $\ce{NaOH}$, this issue is of smaller importance.


text:
Question on the formula to calculate enthalpy change from calometric data
I have a sample question here.
"In an experiment, a student placed 20cm^3 of a solution containing 0.04 mol of HCl in a plastic cup and added 20cm^3 of solution containing 0.04 mol of Na2CO3. The highest temperature rise is 15K. If the heat capacity per unit volume of the final solution is 4.2JK^-1cm^-3, what is the enthalpy change of reaction?"
And what I calculated was:
Assuming that the density of the final solution is equal to the density of pure water, 1.00gcm^-1
∆H = (mass of final solution x heat capacity of final solution x maximum ∆Temperature)/ number of moles of the limiting reagent = (40 x 1 x 4.2 x 15)/ 0.04 = 63kJmol^-1
However, the correct answer was 126kJmol^-1.
Am I using the formula correctly here? Or is there something wrong in my calculations?

Isn't that they ask you for H of reaction? What you have calculated is the actual H change.

Sorry, it's enthalpy change of reaction to be more specific. So what I'm asking is if there's anything wrong with the formula I've used or something about my calculations.

Since you have the capacity per unit vilume, you don't need to use the density of the solution. Your answer is true. The reaction is (( carbonate + HCl ----> Hydrogencarbonate + cholorid )). Since, all HCl has consumed, the reaction won't preceed more (( I mean that Hydrogencarbonat won't react with HCl, because there is no HCl left )).
I think that the guy who computed the correct answer, has 0.02 mole as the moles of limiting reagent


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Using the mass of calcium carbonate, what volume of carbon dioxide could theoretically be obtained if the reaction progressed to completion?

Using the mass of calcium carbonate $m(\ce{CaCO3}) = \pu{0.500 g}$, what volume of carbon dioxide $V(\ce{CO2})$ (in $\pu{mL})$ could theoretically be obtained if the reaction progressed to completion? Assume that the pressure was $\pu{101.3 kPa}$ and the temperature was $\pu{22.5 ^\circ C}$ and that the gas was behaving ideally $(R = \pu{8.314 kPa L mol-1 K-1})$.

I know that I will work with
$$pV = nRT$$
and to bring the $n$ (amount of substance), I have to divide the mass by the molar mass and I assumed that $\ce{CO2}$ mass will be $\pu{0.5 g}$ as well because $\ce{CaCO3}$ will give $\ce{CaO + CO_2}$ so the ratio is $1:1$ so $\ce{CO2}$ mass is $\pu{0.5 g}$, but my problem is that when I bring the amount of substance it should be for $\ce{CO_2}$ or $\ce{CaCO3}$? Because it will differ, $\frac{0.5}{44.01}$ or $\frac{0.5}{100.09}$, that's what I am stuck on.

It seems that you perform a little calculation without thinking on its meaning. You are interested on the number of mole of CO2. There is a 1:1 correspondence between CO2 and the carbonate, ie one mole of the latter gives out one mole of the gas when it totally decomposes. Thus you are ultimately interested on how many moles of carbonate you are starting with. So you need to plug in the molar mass of the carbonate,because it is what you are weighing at start.

so you mean in order to bring the volume, when i calculate the n (moles number) so i should divide the mass (0.5) by the molar mass of the co2 not caco3?

it is getting puzzling

well it was an experiment and the experiment ends with 73ml of co2 but i donot think that it is the answer because the question says calulate it theroitically

Consider question titles to be like book titles that provide rather sufficient content hints than story summaries. E.g. "Computation of CO2 volume evolved from carbonates"

@Poutnik I personally prefer question titles to be actual questions; in any case, they should summarise the query not explicitly state every detail.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

@Martin-マーチン  I had in mind mostly cases where the title is formulated in such a way it almost ( or even literally ) does not need the question body. It may be a question, if if is compact enough, otherwise I guess a summarised topic is better.

You are confused about chemical equations. The stoichiometric coefficient (the numbers before a compound) do not indicate the mass of the compound, they indicate the amount of substance. For example
$$\ce{CaCO3 -> CaO + CO2}.$$
This equation means that $\pu{1 mol}$ of calcium carbonate will decompose to give $\pu{1 mol}$ of $\ce{CaO}$ and $\pu{1 mol}$ of $\ce{CO2}$.
It does not mean that if 1 gram calcium carbonate were to decompose, 1 gram of calcium oxide and 1 gram of carbon dioxide will be obtained.
In your question the mass of $\ce{CaCO3}$ is given. From it you can calculate the moles of $\ce{CaCO3}$ involved. Same number of moles of $\ce{CO2}$ will be obtained and you can put that in the ideal gas equation.
Just so you know,
$$n_{\ce{CO_2}} = n_{\ce{CaCO_3}} = \frac{\pu{0.5 g}}{\pu{100.09 g/mol}}.$$
I recommend you study the chapter about the mole concept again.
As a side note: You mentioned in your question that the ratio of products is $1 : 1$. That's actually the ratio of the amounts of substances, not the mass ratio.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

I'd upvote that, but I'll do that once it is complete, because I'd still think that OP will struggle with the completeness of this answer, too.

@Martin I used the word number of moles because some people, including me, confuse amount of susbtance  with mass of the compound

@Eyyboss which is indeed reasonable

I don't find this reasonable at all. This is a technical term with a proper definition. This term is ancient, and now deprecated for some time. From a purely scientific view it is incorrect. I admit that in the English language this term is quite cumbersome; nevertheless it is the correct terminology. But I will not edit it out again, it is your choice after all, and it is the other user's choices to do differently and to vote on this answer. And all of this aside, in my personal view, this answer is still incomplete.

I am not saying that it is correct. I am saying that it is a reasonable mistake at least in four languages. @Martin.....

I used this link for another Q. But it is somehow related to the point raised by some comments on this thread. https://www.degruyter.com/configurable/contentpage/journals$002fci$002f41$002f3$002farticle-p50.xml

You are mentioning $0.500$ g $\ce{CaCO3}$. Whatever the use of this substance later on, it contains $\ce{0.500 g/(100 g/mol CaCO3) = 5.00·10^{-3} mol CaCO3}$. Of course, if this sample reacts with an acid later on, it will produce $\ce{5.00·10^{-3}}$ mol $\ce{CO2}$. And these $\ce{5.00·10^{-3}}$ mol $\ce{CO2}$  occupies a volume $V$ = $\ce{5.00·10^{-3}} RT/p$.

yeah i know this but thats not my question my question is when i want to bring the n i shoud bring the n by dividing the mass by the molar mass my question is which molar mass i should put the molar mass of co2 or caco3 ?

@lizzy you need to understand that the molar mass is irrelevant to the idea used in your answer.. the number of moles in this reaction is what is formed in the ratio given above.

If you know how to calculate it, this answer is easy to understand. If you are struggling with the concept of what react to what and which molar masses to use when, then this answer is indeed impossible to understand. @Safdar Obviously the molar mass is needed to calculate the initial amount of substance, so it is relevant to know which one to use.


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In Energy-Dispersive X-ray Spectroscopy, do higher atomic elements ever generate just K lines without L or M lines?
The heavier elements only show L or M lines. Is that because the critical ionisation energy is too high to get K lines in higher atomic numbers? Can you ever get just the K lines, or just the M lines or L lines?

With a tuned X-ray source, e.g. from a synchrotron, it should, at least in theory, be possible to excite any specific line, or cause absorption at a specific wavelength, as in X-ray absorption spectroscopy.


text:
Weak acids compared to strong ones in pH > 7 solution
In the book of Molecular Cell Biology (Big Alberts) by Alberts, at page 46, it is stated that

Acids—especially weak acids—will give up their protons more readily if the concentration of > H3O+ in solution is low and will tend to receive them back if the concentration in solution is high.

Does this mean that, in a solution with high pH, the weaker acids will behave like a strong acids in water, in the sense that they will be more inclided to give up their protons compared to strong acids?

When the ratio of conjugate base concenteration to conjugate acid concenteration is more than 1000 , we  say that the  dissociation of acid is complete.
According to Henderson–Hasselbalch equation the ratio of conjugate base concenteration to conjugate acid concenteration gets more than 1000 , when PH-PKa is more than 3.
For strong acids, this ratio is more than 1000 in PH>2 ((  actually it is true for most strong acids )) , but for weak acids , there is no specefic PH . Actually, higher PH , higher dissociation ratio.
So ,when the PH gets higher, the dissiciation ratio gets greater and more poroton will be given up.
Actually, when PH-PKa gets more than 3 for weak acids , they behave like strong acids and allmost all of the acid will dissociate.


text:
How much D2O is in Vienna Standard Mean Ocean Water?
I'm curious how much $\ce{D2O}$ (heavy water) is in Vienna Standard Mean Ocean Water. I am interested in $\ce{D2O}$ molecules and not $\ce{HDO}$/$\ce{D+}$/ $\ce{DH2O+}$/$\ce{D2HO+}$/$\ce{D3O+}$ or any other deuterium species or ions. Vienna Standard Mean Ocean Water is pure water extracted from some ocean with a specific isotopic composition. Sweet water has slightly different isotopic composition.
A suitable unit for the result could be mol/kg or a mol fraction.
I'm not sure if temperature or pressure make a difference. I suggest $\pu{20 ^\circ C}$ (or any other temperature, please specify, if possible) and $\pu{101325 Pa}$, but other conditions are ok, too. Liquid water makes most sense, I guess.

Allright, I took some effort as suggested and used my brain ...
2H/1H = 155.76 ± 0.1 ppm (a ratio of 1 part per approximately 6420 parts)1
So if an HDO autoprotolyzed ...
HDO --> D+ + OH- or
HDO --> H+ + OD-
and then D+ or OD- reformed water with OH- or H+, then it would have a 1 in 6420 chance of forming D2O, and so D2O would be 1 in 6420^2=41,216,400 parts.
Sounds good?
1 https://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water

Not too much, $\approx$ 1:36000000 .

This question makes no sense. Please elaborate, and show your own effort.

The question makes sense, butI agree with the lack of effort while the asnwer has the obvious way to obtain.