text:
How exactly thermal infrared (e.g. FLIR) cameras image methane leaks? Are they measuring a temperature difference or an emissivity difference or...?
The story Infrared camera shows the impact of vehicle emissions says:

In an effort to visually demonstrate the potentially fatal impacts of air pollution, FLIR Systems (Wilsonville, OR, USA; www.flir.com) has released a new video containing footage captured by FLIR infrared cameras that shows vehicle emissions and the resulting pollution issues.

[...] To accomplish this, FLIR used its GF320 infrared camera to capture video of the emissions. This camera features a 320 x 240 cooled InSb infrared detector with a spectral response of 3.2 to 3.4 μm. Designed specifically for gas leak detection and electrical inspection, the camera also embeds GPS data into the image, allowing workers to pinpoint the location of the leak or hot spot.

Wikipedia says InSb is a narrow-gap semiconductor with an energy band gap of 0.17 eV at 300 K and 0.23 eV at 80 K and a focal plane array can image light with a wavelengths in the 1 to 5 micron range. They are often cooled to reduce dark current for better sensitivity. In this case it sounds like they've added an interference filter to selectively pass 3.2 to 3.4 microns.
My question is about what it is they are really imaging here. It is true that gases such as CO2 and methane have strong IR bands, this is why they are important greenhouse gasses. Are they measuring a temperature difference or an emissivity difference or light scattering or absorption or...? How does this technique actually work?

Youtube Seeing Air Pollution with FLIR Thermal Imaging Cameras
Youtube How Dying Gas Wells Are Making One Company Rich FLIR imaging of methane leaks from gas wells

above: Image from this article.

As far as I know, the relevant IR absorption of CO$_2$ is due to the bending motion at about 600 cm$^{-1}$ - and it does overlap with the heat radiation of the Earth. For methane, it's the peak at 1300 cm$^{-1}$, the motion can be seen here

Methane does have a strong absorbance in the 3-4 micron range: https://webbook.nist.gov/cgi/cbook.cgi?ID=C74828&Type=IR-SPEC&Index=1

This paper discusses methane detection by FLIR cameras in detail. It is indeed based on absorbance of the background IR in the 3.2-3.4 micron window: https://pubs.acs.org/doi/10.1021/acs.est.6b03906

@Andrew hmm... I wonder if high absorbance also means high re-emittance of the same light? If it's at the same temperature as the surrounding material, how can it be brighter or darker?

I wouldn’t necessarily assume it’s the same temp as surroundings, but whatever the temp, the re-emission will be much broader spectrum than the absorbance, so there is still a significant decrease in observed transmission at the measured wavelengths.

Slight correction - I read more of the paper, and they say that both the emissivity of the methane and the absorbance of background are important contributors to the signal depending on the situation. A hot methane plume is visible against a cold sky because of the plume's higher emissivity, whereas a cold plume is best observed against high background emissions because of the absorbance of the plume. At least that's how I'm reading it. But you should read the paper yourself rather than trust my quick interpretation.

Did you read the paper? I think it provides much better info that you'll ever get from an answer here.

@Andrew the goal of Stack Exchange questions is to facilitate the generation of Stack Exchange answers. That's why link-only answers are so poorly received. If you think the answer is in that paper, one way you could demonstrate your confidence in that would be to post a short answer, linking to it, and explaining and quoting the bits you feel do the job.

Sorry, I should have been more clear. My intended point was that if it is important to you to have an answer to this question, you have sufficient information in the paper to write a self-answer. That seems much more effective than repeatedly bountying a question that continues to receive little interest.

@Andrew noted, but bounties are exactly the thing to use when there is "little interest", and I simply like bountying! All it takes is a single person to take an interest to write an answer.

Fair point. If you don't mind waiting for someone else to write an answer, that's your choice and I shouldn't be critical of it.

Short answer
The cameras you are asking about are called optical gas imaging (OGI) cameras, they are not thermographic cameras per se, but are a modified kind instead.
However, both use the same principles to actually detect the infrared light, the key difference is the incident radiation is filtered to allow only the wavelengths that interact with the methane gas plume (or other analyte gas) to interact with the camera sensor.
Thermographic cameras could be configured to take both kinds of images since the principle of their operation is essentially the same, but they are usually separate devices for practical reasons like cost, increasing image contrast, etc..
In the images you provided, you can see the solid objects because they emit, reflect, and absorb infrared light across the spectrum, so they can also be seen by the OGI cameras.
However, other gases are not seen because the IR light they absorb is filtered out.
Finally, a temperature difference must exist between the plume of gas being imaged and the background scenery because otherwise the amount of infrared radiation emitted will be equal to the amount absorbed by the gas, following Kirchoff's Law of thermal radiation.
Detailed Answer
Methane detection colour analogy
Cameras for visible light discriminate between colours  by placing a filter before the sensor so that only one colour (i.e., red, green, or blue) is absorbed by the sensor at any given pixel.
The sum of the RGB signal reproduces the subject's colour.
Although we cannot see them with our eyes, infrared light has different "colours" that can be discriminated by a camera sensor with the appropriate absorption range, filters, etc..
Visible light has different colours depending on the energy (i.e., wavelength or frequency) of the light:
red light has a longer wavelength (lower energy and frequency) than green, and blue, respectively.
We can see an apple is red when the green and blue light in white backlight are absorbed by its skin while the red light is reflected into our eyes (or camera).
In the same way, we can see the infrared "colour" of different objects by observing the infrared light they absorb.

Figure 1 FTIR spectrum of carbon dioxide from NIST.

Figure 2 FTIR spectrum of methane from NIST, notice the lack of overlap between this spectrum and that of carbon dioxide (above).
Each arrangement of atoms in a molecule with different nuclear charges (i.e., polar bonds) absorbs a specific wavelength of light because the vibrations are quantum mechanical in nature.
These vibrations are characteristic to each molecule, so we can identify them by observing which wavelengths of light they absorb.
See the above two Fourier-transformed infrared (FTIR) spectra of carbon dioxide and methane.
This is the basis of spectroscopy, which is out of the scope of this answer.
But, if we designing a sensor that can discriminate between certain wavelengths of infrared light, a camera can be made that can see how much of a particular wavelength (e.g., 3.5 μm for methane, 10.57 μm for sulfur hexafluoride) is being absorbed relative to the background.
Couple this principle to a system that can tell where the light is coming from, i.e. add pixels, add some optics to focus the light, and you have a camera.
Optical gas imaging (OGI)
The objects in the field of view (FOV) in OGI are illuminated by the background infrared radiation emitted from the sun and all other warm objects.
This is why OGI is also called a "passive" thermal imaging technique, in contrast to "active" techniques where the scene is illuminated with an artificial light source directed at the subject.
OGI depicts the relative decrease (or increase) in the intensity $\Delta I(\lambda)$ at a certain wavelength $\lambda$ of this background radiation $I_B(\lambda)$ due to absorption (or emission) by the analyte gas with intensity $I_G(\lambda)$.
A general OGI equation (Eq. 1) for $\Delta I$ can be derived using a radiative transfer model:
$\Delta I(\lambda) = I_b(\lambda) - I_g(\lambda) = [1 - \tau_g(\lambda)][B(T_b,\lambda) - B(T_g,\lambda)]\quad(1)$
where $\tau_g$ is the transmission coefficients of the  analyte gas, and $B(T_b,\lambda)$ and $B(T_g,\lambda)$ are the Planck law functions for blackbody radiation of the background at temperature $T_b$ and the analyte gas at temperature $T_g$, respectively.
We can obtain the transmission coeffecient of the gas plume in relation to its composition, concentration, and geometry using Beer's law (Eq. 2)
$\tau_g(\lambda) = \exp[-\alpha_g(\lambda)\bar{C}_g \ell_g]\quad\quad\quad (2)$
where $\alpha_g$ is the gas's absorption coeffecient, $\bar{C}_g$ is the average concentration of the gas through the line of sight, and $\ell_g$ is the breadth of the gas plume through the line of sight.
Now we can see that the gas plume image obtained by OGI systems requires three things to be detectable:

Absorption (or emission) of infrared radiation in a certain band of wavelengths ($\lambda$);
Plume that is big enough ($\ell_g$) or concentration that is high enough ($\bar{C}_g$) to absorb (or emit) enough infrared radiation to be detected; and
Temperature difference $\Delta T = T_b - T_g$ that is great enough between the background and the gas plume so they absorb (or emit) different amounts of infrared radiation (Eq. 3).

Each of these three factors contribute to the overall contrast observed in the image between areas where there is a gas plume and where there is not.
The final point may be mysterious to the reader, but it is due to Kirchhoff's law of thermal radiation.
Put simply, if the gas were the same (apparent) temperature as the object behind it (in the line of sight), then it would absorb as much infrared radiation as it was emitting, and would appear invisible to the OGI camera.
If the gas plume has a lower temperature ($T_b > T_g$), it will absorb more infrared radiation as it passes through it to the camera relative to the atmosphere around it, thus $\Delta I < 0$ and the gas plume will appear darker.
If the gas plume has a higher temperature ($T_b < T_g$), then $\Delta I > 0$ and the gas plume will appear brighter.
For more detail on OGI and its quantitative variant (QOGI), please see this recent dissertation by Michael Nagorski out of Waterloo.
Commercial OGI cameras
Teledyne FLIR LLC, the company you specifically are referring to in your question and who appear to make the industry standard cameras, describe the operation of their OGI and thermographic cameras and sensors in some detail in US Patent No. 9,276,161 B2 and 9,007,687 B2.
They have invented arrays of quantum wells of indium antimonide (InSb) that are specially designed to absorb radiation as close to a strong absorption band in the gas plumes that are to be imaged.
State-of-the-art camera systems include advanced optics and filters to improve the signal-to-noise ratio, thus increasing the contrast of the acquired image. Directly from US Patent No. 9,276,161 B2:

Generally, the camera system 100 is used for detecting a gaseous compound in the scene 106 when radiation from the scene 106 is received by the lens 104 and passed to the band pass filter 110 to limit the wavelength range of survey scene energy focused onto the FPA 108. In one embodiment, the wavelength range is limited to 10.3 to 10.8 μm. Other wavelength ranges are also contemplated. Each sensor element of the FPA 108 generates an analog photo current value according to a photo current responsivity profile and other factors in response to an irradiance generated by the spectrally filtered scene image formed by the lens 104 at the sensor element active surface. The analog photo current values are read out from each photo sensor element and converted to corresponding digital signal values for rendering a video image frame corresponding to the digital signal values.

In the aforementioned patents they give some details on the ways they have improved upon the principles laid out above.

Thanks & Welcome to ChemSE! If it's absorption only, then a methane could would always appear as a simple darkening of the background behind (if thin) or big black cloud if thick. And yet, the images shown in these videos (at these times) look much more complex than that. 1, 2, 3 In the third video for example, the plume from the stack is hotter than the sky behind it, not colder. In this case it seems to be radiating more strongly than the background, not absorbing light from it.

You can just invert the image to make absorption (black) look like emission (white), that is entirely a perceptual difference. Besides, the plume should appear "hotter" than the background because the methane concentration is higher coming out of the engine than it is in the atmosphere. The intensity of the image is due to the concentration of methane which is inversely proportional to the transmission of the infrared light absorbed by methane.

I should also clarify that the image has almost nothing to do with the temperature of the gas being observed. The amount of IR light that is absorbed by the methane is mostly determined by its quantum mechanical properties, and temperature has a negligible effect on these properties. The density of the gas will change as it gets hotter which will change the number of molecules per unit volume, but it won't change how much each molecule themselves absorb. The shape of the plume will also affect the amount of absorption as there is more gas in the middle of a plume than the outside.

Okay I'm familiar with how these cameras work (inside and out) and that the displayed "temperature" is a computed value based on brightness and assumed emissivity and not a real temperature meaasurement. In the third video there is a color scale, we can see that the sky is radiating the least thermal IR and the ground the most thermal IR. The color scale is monotonic, and the intensity recorded from the plume is higher than the background sky. The plume is radiating, not absorbing. Perhaps it is just hot gas not methane. Perhaps this method does not actually specifically flag methane at all!

To avoid confusion, let's not use the word thermal when referring to these images. Thermal IR imaging measures the amount of light that is emitted per unit area due to blackbody radiation. For FLIR imaging we are representing only the amount of light absorbed by a specific wavelength that is characteristic of methane. Thermal IR images would be like a grayscale image while FLIR images would be a colour filter applied to the thermal IR image. In reality, you only collect the light absorbed by methane to increase the signal-to-noise ratio rather than filter all the light afterward.

You may be confused that you can see the earth and buildings as bright objects even though we are filtering out IR light to only "see methane". We are really only looking the absorption of 3 um light which is characteristic of methane but NOT UNIQUE. IR light is absorbed by most solids and some other molecules. But it is NOT absorbed by carbon dioxide, water, carbon monoxide, etc. So you see the solids absorbing the IR light (and others) and the methane (mostly in that band)

Okay so these "thermographic cameras" (from your answer) are not being used for thermographic observations, but this range of the EM spectrum is called "thermal infrared" to distinguish it from near infrared and far infrared/microwaves. The goal of a Stack Exchange answer is not to demonstrate one's knowledge but to answer the question as asked. I've asked "How exactly thermal infrared (e.g. FLIR) cameras image methane leaks? Are they measuring a temperature difference or an emissivity difference or...?" If answer is "No, they are measuring absorption" then the images in video #3 disagree.

You are right, it was a misnomer to call them thermographic cameras.
I've edited my answer to clarify the differences between the two and address some of your concerns about the relative contrast of different objects in the images.

Even that camera has an "Accuracy of ±1°C for temperature range 0°C to 100°C or ±2% of reading for temperature range" based on radiance within the 3.2-3.4 um band. I downloaded the three pdf files at the bottom of the page you link to, and whenever "measurement" appears, it's in reference to temperature. It seems many cameras are also recommended as optical gas imagers, and the compliance certificate says:

..."FLIR is proud to announce its GF620, GFx320, GF320, GF300, and G300a cameras have been independently tested and deemed compliant with the EPA’s NSPS 40 CFR part 60, subpart OOOOa sensitivity standard for optical gas imaging equipment." Understanding Optical Gas Imaging Quantification links to 1, 2 and...

3 and a temperature difference between the ambient environment and the leaking gas is alway essential for detection and certainly for any kind of quantitative measurement. Okay I think we're converging on an answer. The material must have some absorbance in the wavelength range of the camera (for other non-hydrocarbon gasses they use 7-8.5 or 9.5-12 μm) but that's because anything that can absorb light in a given band can also thermally radiate there. See Absorptivity

Well this is weird. I'm finding some other papers that say you can get away without a difference in temperature (like I was saying) but all the actual QOGI systems say it is essential. There must be some nuance I'm missing with the operating principle of these. I'll be back to update my answer when I have unraveled the differences.

So is this an acceptable answer now?

Sorry, while it's a great answer post, I can't say that I really know the answer to "How exactly thermal infrared (e.g. FLIR) cameras image methane leaks? Are they measuring a temperature difference or an emissivity difference or...?" yet. Thus no acceptance at this time. Specifically, I can't yet resolve the apparently contradictory messages about the importance of temperature, and https://www.tandfonline.com/doi/full/10.1080/10962247.2016.1244130?journalCode=uawm20 while long and technical-sounding, doesn't resolve the radiative transfer model (Eq. 5) wrt "RF" (Eq. 6 and on) contradiction.

Comparing to this https://www.flir.ca/discover/instruments/gas-detection/understanding-optical-gas-imaging-quantification/ which asserts the importance of a temperature difference (as does my physics intuition) between gas and background, I see two commercial companies publishing opposing views in order to sell their own products.

(I left a link the Physics SE chat room The h-bar


text:
Why does the 1H NMR spectrum of methacryloyl chloride show two singlets instead of two doublets for vinyl protons?
Based on what I have understood, for methacryloyl chloride there is a geminal coupling of the vinyl protons since they are not chemically equivalent. Why are two singlets observed for vinyl protons instead of two doublets?

If you look at "peak data" here for methacryloyl chloride ($\ce{C4H5ClO}$), you will see that there is small coupling between the two vinylic hydrogens but also weak allylic coupling to the methyl group.

The geminal $^2\!J_\ce{HH}$ coupling in terminal alkene $\ce{CH2}$ can be pretty small; often on the order of $\pu{2 Hz}$. The late Hans Reich's NMR website has a lot of data which you can check out.
(And yes, this is in stark contrast to geminal couplings in diastereotopic, $\mathrm{sp^3}$ $\ce{CH2}$ groups, which often have large magnitudes!)
So, to observe it in the spectrum, you will need to increase the resolution. The easiest way would be to perform extra zero-filling and see if that helps. The next simplest way is to process the data using a different window function / apodisation function: to increase the resolution, you can use a Gaussian window function, see e.g.

Why is the Gaussian Window Function used to enhance resolution in 1D NMR spectra?

and some practical instructions at the bottom of this answer: Multiplet shape in proton NMR of morpholines

and there are numerous resources online which you can find as well.

If none of that works, it's possible that you might need to rerun the spectrum and increase the acquisition time of the spectrum, but I'd consider that to be quite unlikely for a routine $\ce{^1H}$ spectrum.
[That said, this appears to be a spectrum you've taken from someone else's work, so... maybe all you can do in this situation is to take a magnifying glass, and hope that the spectrum was processed and the image created with sufficiently high resolution. The suggestions above obviously only make sense if you are the one who acquired the spectrum.]

source: Stack Exchange

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

sort: votes


text:
Finding the density
35.6g of NaOH is dissolved in 150g of 20% NaOH solution. What is the density od the solution?
I have tried solving this problem, but I 've come across a few obstacles.
Firstly, I found the mass of NaOH in the original solvent which is 30g. The added  NaOH weighs 35.6g, meaning there is ultimately 75.6g of NaOH, and that the mass of the solution is 185.6g. The next step would be to find the density of the solution, but this requires the volume of the solution, that is not given in the problem. I also tried to calculate the number of moles to later calculate V using the formula V=n*Vm, but I am not sure how to calculate the molar mass(M) of the solution.I am not sure if I am to use H2O or NaOH.
Can someone tell me if I am going in the right direction and if my calculations are even in the right direction, and if not, what is the right one?

You cannot determine solution density from its percentage. It is experimental parameter. For common substances, it is measured, interpolated and tabelated.

does this mean that the task cannot be solved with the given data?

Even if solution densities were linear functions of  mass percentages - but they are not - you would need at least densities of pure substances. // All depends on if you are allowed and/or supposed to use additional data sources, like density tables.

There is one possible task solution, relying on incorrect percentage usage by the task author, assuming 20% means 20 g / 100 mL of solution. Or even worse, 20 g / 100 mL of water.

This question looks a lot like this one. If you have created a new user account to circumvent a ban, you are in violation of the terms and policies of the site.

You can simplify your problems by looking at the published tables. As you have been explained in the comments, predicting densities is not an easy job and I don't know if it can even be predicted.
This link provides the densities of NaOH solution as a function of mass percentage.
https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/density-of-sodium-hydroxide/
Plot the data in Excel and generate a function. Calculate the final mass percentage of your solution.
Your first step is correct, i.e., the initial solution contains 30 g of NaOH, then you add 35.6 g to the solution. The total mass of the solution is 185.6 as you correctly calculated. What is the final mass percentage? 65.6 g NaOH/ 185.6 g solution. This is around 35% wt/wt NaOH. Roughly, interpolating from the given table, the density should be around 1.39!
Commas in German data are equivalent to decimals in English, so mass percentage of 5,5 means 5.5.

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


text:
How do I derive metallic aluminum without electricity?
I'm laying the foundation for a project called 21st Century steampunk, where I figure out what the world would look like if electricity was never discovered.
I'm wondering if it's possible to derive usable metallic aluminum from naturally occurring substances without using the Hall–Héroult process.

Recycling, maybe?

One method, which would also require a non-electrical heating source, involves reduction with carbon. Given a high enough temperature -- meaning over 2000°C -- carbon carries off the oxygen as carbon monoxide and leaves the aluminum behind. See Ref. [1](https://doi.org/10.1016/j.energy.2007.06.002), which includes the equilibrium composition calculation below.

Reference

M.Halmann, A.Frei, A.Steinfeld (2007)."Carbothermal reduction of alumina: Thermochemical equilibrium calculations and experimental investigation". Energy 32, Issue 12, December 2007, Pages 2420-2427. https://doi.org/10.1016/j.energy.2007.06.002

Thank you so much. After searching around for more than a week, this is finally the solution I needed.

This aluminium would be gaseous, and spontaneously react with oxygen should it ever encounter any...

@StianYttervik that depends on conditions. Wikipedia reports a normal boiling point of 2470°C, which is in the range where aluminum can be produced.

@Oscar the partial pressure is higher than comfortable above 1800 though. It doesn't have to boil to vanish well before it can be solidified.

@StianYttervik Most likely, a cooling chamber that allows condensation before the aluminum can re-react with the CO would be involved in the process.

Prior to the electrolysis (Hall–Héroult) process, elemental aluminium was made by reducing aluminium chloride (AlCl3) with elemental sodium or potassium in a vacuum.
See the wikipedia article on "History of aluminium"

Which prompts the question: how do you get elemental sodium or potassium without electricity?

For the potassium case from the azide by heating or exposing it to ultraviolet light (https://en.wikipedia.org/wiki/Potassium_azide). The azide can be made from potassium carbonate and hydrazoic acid. I am not saying this is a good method, but you did ask! Sodium azide I assume can be made analogously, and is actually made on a n industrial scale (https://en.wikipedia.org/wiki/Sodium_azide), but that process seems ultimately to use sodium metal as an input.

@GuntramBlohm, if you prefer sodium, you can get it by heating sodium carbonate to 1100 ˚C in the presence of pure carbon.


text:
Are corrosive and reactive synonymous?
This answer on a question regarding whether water is corrosive in pure form, the author implies that "reactive" and "corrosive" are the same thing (at least with water.)
"Corrode" seems to mean irreparably damaged which also seems to imply a corrosive reaction is not reasonably reversible whereas "react" can mean several things such as the changing of an oxidation state, or bonding covalently with another element or molecule, to something as seemingly small as a hydrogen bond.
Are "corrode" and "react" truly synonymous or is corrosion just one of many types of reactions?

As not all reactions are corrosive reactions, they cannot be synonyms.  Corrosive means reactive but reactive may not mean corrosive.

Corrosion has an economic and a negative connotation e.g., corrosion scientists are interested in protecting metals/non-metals exposed to the atmosphere to reduce major loses to the city & the government.
Reaction is a necessary process for corrosion, for instance, eating away or destruction of metals by oxidation. Hence, acids are corrosive to a large number of metals. Even in geology, corrosion implies destruction of rocks mainly by water. Those who live near coastal cities know how cement is corroded (eaten away) by salty water aerosol. There is no oxidation. All senses are negative.
In my humble opinion, if something is reactive, it does not necessarily have a negative tone. Water reacts with sodium, but we would not say, water corrodes sodium because an economic loss/negative tone is not implied. Sulfur reacts with copper and so on. I would not include hydrogen bonding under a chemical reaction but this is really splitting hairs just like chemical change vs. physical change enforced upon undergraduate students.
In short, water can be corrosive or reactive depending on the context by the speaker/author. Acids are corrosive to metals (negative, warning) or aqua regia reacts with most metals (the sense maybe good or bad).


text:
Comparing the enthalpy change between the formation of water in two states

The enthalpy of the reaction $\ce{H2(g) + 1/2 O2 (g) -> H2O (g)}$ is $\Delta H_1$ and that of $\ce{H2(g) + 1/2 O2 (g) -> H2O (l)}$ is $\Delta H_2$. Then,
a) $\Delta H_1 < \Delta H_2$
b) $\Delta H_1 + \Delta H_2 = 0$
c) $\Delta H_1 > \Delta H_2$
d) $\Delta H_1 = \Delta H_2$

According to me, in a reaction forming a gaseous product the pressure would be greater than the one forming a liquid product. Therefore the enthalpy change should be greater for the first reaction. However, the correct answer is a). Where am I going wrong?

Hess' Law might prove to be your friend here. Welcome to Chem SE..

$\Delta H_2 + \Delta H_{\ce{H2O},\mathrm{evap}} = \Delta H_1$

In other words, (a) is wrong and (c) is right.  State (g) is a hypothetical ideal gas state of water vapor at 25 C and 1 bar that has not condensed.

Well in order to convert H2O into liquid we need to lower the energy of system more than H2O of gaseous state (energy of gas is more than liquids and liquids have energy more than solids).
That means more have to be released in order to make it liquid.

Since enthalpy is Q at constant pressure the PV factor can be ignored [this is for the questioner]. The difference is the heat of condensation of the vapor that is released. Therefore the reaction to give liquid water is more negative than that to give gas. I looked it up the heat of formation of water vapor is -57.8 Kcal for liquid water -68.3. Now the problem is to define what is meant by larger? the less negative number or the greater release of heat. As usual test makers do not try to answer their own questions!  [I notice that in the comments this dilemma was not resolved. altho Chet tried].  I personally deplore questions that make knowing sign conventions more important than knowing chemistry.


text:
Does water expand on freezing more than any other known substance?
Several substances expand on freezing. Gallium expands 3.1% and bismuth 3.3%. However, both of these are much less than water which expands 9%. Is there any substance that expands more than 9% on freezing?
We can generalize this question to any system at thermodynamic equilibrium and constant pressure. Density normally decreases with temperature, but sometimes increases. For example, heating plutonium increases it's density in three places: a solid-solid phase transition, a negative thermal expansion, and melting. These effects combine to make the density increase about 5% between ~320C to 640C, which is still less than cooling 4C water to -0.01C ice.
If we allow substances that dissolve/precipitate and reversible chemical reactions can we beat water? We do not allow super-cooling or super-saturation since that isn't thermodynamic equilibrium.

Silicon beats it.
Silicon has the familiar diamond-type semiconductor only when solid, and like the hydrogen-bonded structure of water ice the directional bonding adds volume to the solid phase. Melted silicon collapses to a structure more typical of a liquid metal, including its electrical conductivity.
Wikipedia reports the density of solid silicon at room temperature and the liquid at the melting point:
Solid at room temperature: 2329 kg/m³
Liquid at melting point: 2570 kg/m³
Thus the solid at room temperature has 10.3% more volume than the liquid at the melting point ... with the thermal contraction from melting point to room temperature included.


text:
What will be the product of 1 eqv Benzene reacts with 1eqv H2?

how equivalent concept is applied in the above problem, Kindly help me out with the answer and proper explanation.
As, I am getting confused in mole concept and equivalent concept here.

This is very dependent on the conditions and the catalyst used.

@Waylander I wouldn't say very dependent. It took chemists the best part of a century to find conditions that don't mostly give cyclohexane.

Those last edits make the question worse not better by leaving out a possible explanation for the confusion. We need to see what your thoughts were or it looks like we are doing your homework, which is a solid reason to close the question.

The answer makes sense (and doesn't break the law of chemical equivalence) when you consider the reactivity and stability of the intermediates in the reaction
Hydrogenation of benzene proceeds in several steps (which is intuitively obvious as a single hydrogen molecule can only add two hydrogens to a benzene ring).
But the intermediate that results from a single hydrogenation step, cyclohexadiene, is more reactive than benzene so the further hydrogenation happens faster than further hydrogenation of benzene. The same is true for the product of that reaction, cyclohexene. So, under simple hydrogenations where there is not enough hydrogen to fully hydrogenate the remaining benzene, you tend to get cyclohexane as a product and lots of benzene remains.
As described, the particular reaction shown only has enough hydrogen to turn 1/3 of the benzene into cyclohexane. Hence the mix of products: 2/3 benzene and 1/3 cyclohexane. And this, overall, obeys the law of chemical equivalence.
It is possible to use careful conditions and specific catalysts to change the mix of products. But it isn't easy. Benzene hydrogenation had been known for over 100 years before chemists observed cyclohexene as a product. The first industrial plant exploiting partial hydrogenation didn't open until 1990.

Thank you for explaining it clearly, but 1eqv of benzene + 1 eqv hydrogen gives 0.33 eqv cyclohexane + 0.66 benzene, but according to law of equivalence the each reactants with one equivalent we will give one equivalent of each of the product. But here it is 0.66 and 0.33 eqv. I think for mole concept it to be true but for equivalent concept I think something is wrong.

As, 1 eqv of hydrogen is not 1 mole of hydrogen same for benzene as they are reaction dependent, so also product dependent, can u rectify if I know something wrong....

@AmritenduBiswas The law or chemical equivalence is based on moles (or molecules if you want to count one molecule at a time). One mole of hydrogen can fully hydrogenate just 1/3 moles of benzene as it takes 3 hydrogen molecules to hydrogenate each molecule of benzene. The law holds up fine.

Hydrogenation of benzene requires high temperature and high pressure of hydrogen. An equimolar mix of benzene and hydrogen (the term "equivalent" is unfortunate here) with catalyst and appropriate temperature $T$ and pressure $p$ might form diene but as concentrations increased diene would react much faster than benzene ether by competing for remaining hydrogen or as a source of hydrogen to form the most stable product mix, cyclohexane and benzene.
A catalyst will hasten equilibrium and give thermodynamic products unless specifically fitted to a reaction as in enzyme reactions. To form the intermediate alkenes the reaction conditions must be tailored (excess benzene) and the intermediates separated immediately from the reaction, perhaps a flow system allowing only transient contact with the catalyst and with a cold trap or trapping agent downstream, say a reversible Diels–Alder.
The question, as asked, does not arouse the student's curiosity about the chemistry and that is extremely unfortunate.


text:
Why is the standard electrode potential positive for half cells that are easily reduced?
I was studying electrochemistry from my school textbook.
The cell potential in the book is defined as the difference between the electrode potentials of the cathode and the anode.$$E_\text{cell}=E_\text{right}-E_\text{left}$$
We know that electrode potential of an individual half-cell can't have an absolute value and thus we measure it taking the standard hydrogen electrode ($\ce{Pt(s) | H2(g) | H+(aq)}$, SHE) in relation that is assigned a zero potential at all temperatures.
As the electrons flow from the anode to the cathode (higher potential to lower potential), the cathode should possess a lower electrostatic potential than the anode.
That means that the electrode potential of a half cell when measured against the SHE should always be negative if the reaction is feasible, i.e., electrons flow from SHE to the cathode. Thus, the more negative the value of the electrode potential of the cathode, the more electrons will from the SHE, and the more the tendency of the substance of the cathode to get reduced.
But the values of standard electrode potentials of different half cells in my textbook and Wikipedia show that the electrode potentials of substances that get easily reduced is positive, e.g., $E^\ominus = 2.87$ V for $\ce{F_2(g) +2e^-->2F^-}$. Isn't this the complete opposite?
Shouldn't the values of electrode potentials of substances that are easily reduced be negative as we just discussed? What am I missing here?

@Poutnik I'm unable to understand what you're trying to say. I have a simple question: electrons flow from higher potential to lower potential, in our case from anode (SHE) to the cathode, i.e., cathode has a lower potential. As the potential of SHE is taken to be zero, the potential of the cathode should be negative (lower) if the reaction is feasible. But for such half cells, we see that the electrode potential is positive. Why so?

I have moved my comments to the answer.

A large number electrochemicals misconceptions will be solved if we take the electrode potential sign as the electrostatic sign of the galvanic cell with respect to the hydrogen electrode. Thus this sign will not be affected by how the half cell is written. This was the European understanding of the leading 20th century top-electrochemists. The Germans decided to set the hydrogen electrode at 0 V, the reference point. The Germans also decided to set the hydrogen electrode at 0 V, as the reference point. This is the take of many leading modern electrochemists including A.J. Bard.
Now let us take an electrode such as copper. Its half-cell electrode potential is +0.34 V vs. SHE. Now interpret the sign this way: If we make a galvanic cell of copper and SHE, the copper electrode will be *positively charged" = the cathode, and the hydrogen cell will behave as an anode.
In the same way, take the fluorine half-cell (hypothetical), and if we make a galvanic cell of fluorine and SHE, the fluorine electrode will be *positively charged" = the cathode, because its half-cell electrode is + 2.87 V.
In a galvanic cell, the current is from A(node) -> C(athode) i.e., A to C.

One nagging detail that has bothered me:  shouldn't the Faraday constant (and by extension the elementary charge of the electron) be negative?  It would seem easier to visualize this distinction under that circumstance, but I can't recall seeing the Faraday constant written that way.

@MikeSerfas, Good question. Please post a new question. I have not historically explored this but will search about it.

@AChem Does that mean that in actuality, the cathode has a lower potential than SHE (thus negative), but just to make it easier to understand, we use the positive sign for feasible reactions?

No, this will be the actual electrostatic sign.

@AChem Do you mean that the electrostatic potential at the cathode will be higher than at the anode?

@Silica19, Yes.

@AChem How is that possible? Don't electrons move from higher to lower potential?

From the modern electronic current point of view, electrons flow from "lower" potential to "higher potential. Positive current flow (so called conventional current) flows from higher to lower potential.

Blame Ben Franklin or physicists for not knowing what + and - are and having terms like current flow! We chemists know better and think only of electrons altho we do lapse into positive ion flow in Li ion batteries and in acids. No one is perfect.
Oxidation = Loss of electrons;  Reduction = Gain of electrons
Oxidation is at the ANODE; Electrons LEAVE the anode into the wire. Anions move to the anode.
Reduction is at the CATHODE; Electrons enter the cathode from the wire. Cations move to the cathode.
Those are the simple rules. Unfortunately the physicists still get a say! They defined volts backwards and we are stuck with it [but some references are unstuck so it still can be confusing, so be careful]. A positive voltage means a reaction proceeds as written and its deltaG is negative. Electrode potentials are written [usually] as reduction half reactions. To appease the physicists a positive voltage means the reaction proceeds as written; a negative voltage means it proceeds in reverse. These half reactions are for a reaction under standard conditions with a hydrogen electrode. To get any reaction simply reverse one of the half reactions with its sign and add the equations and the voltages, then correct for differences from standard conditions. If the voltage is positive the reaction goes as written; if negative reverse the equation and change the sign. [or else change the concentrations and the cell voltage. If the cell voltage is close to zero and the electrode reactions reversible the direction of reaction or the cell voltage is easily manipulated.]
The pertinent equation is: DeltaG = -EnF = -E[0]nF +RTlnQ  Q is the reaction quotient with the form of the equilibrium constant Keq. At equilibrium deltaG = 0 so
E[0]nF = RTlnKeq = -deltaG[0].  As can be seen all these + and - signs can get confusing. [LOOK THIS UP IN A PCHEM TEXT or two] [The font this site uses substitutes a lower case o for the numeral zero.]
To sum it up a positive cell voltage corresponds to an equilibrium constant greater than 1. and a negative deltaG. The reaction proceeds as written. How far it goes depends on the voltage and the amount of material [the size of the battery]. At equilibrium the cell voltage is zero and the various concentrations satisfy the equilibrium constant
To react lithium and fluorine: galvanic cell: Li = Li+ + e- E= +3v; F- = 1/F2 +e- E= -3. We have two reduction reactions and need one oxidation and one reduction so must reverse one. Lets do Lithium   e-+ Li+  = Li E= -3v. Put the two half reactions together Li+  + F- = Li + 1/2F2 E = -6V. Wow it is going nowhere. So we reverse the reaction.  Li + 1/2F2  = Li+ +  F-; E = 6V! an explosion!! Not so Good so we set up a cell with an anode, Li, a wire to a motor then to a cathode [who knows what, it can't react with F2], surrounded with F2 and a magic electrolyte to carry the Li+ and F- ions formed remotely and desperately trying to get to the cathode and anode respectively. To reverse this a potential greater than 6V is applied in reverse. I do not think that this particular cell has been made not even by accident.
Work out some simple cells such as a hydrogen ion concentration cell or electrolytic deposition of copper to get an idea what happens to cell voltages as concentrations change.  If you get stuck on conventions or signs remember Li is the strongest reductant and F2 is [almost] the strongest oxidant use these two to determine if a different convention is being used for signs.

Easily reduced redox systems "suck" electrons from an electrode, charging it to the more positive equilibrium potential, at which the electron transfer rate is zero.
And vice versa for easily oxidized redox systems: They are pushing electrons to an electrode, charging it to the more negative equilibrium potential.
In galvanic cells, cathodes have higher potential than anodes, in contrary to electrolytic cells. By other words, when switching the cell mode, electrode names switch places, as the current switches direction.
E.g. Imagine the notoriously known Daniell cell:
$\ce{Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)}$
If we perform electrolysis on it, with the negative contact attached to $\ce{Zn}$, the cathode is the left, more negative electrode, as electrons come to $\ce{Zn}$ and reduce $\ce{Zn^2+(aq)}$.
If we use it as the galvanic cell, the cathode is the right, more positive electrode, as electrons come to $\ce{Cu}$ and reduce $\ce{Cu^2+(aq)}$.

SHE has the absolute alectrode potential $\pu{+4.44 \pm 0.02 V}$.
For Ox/Red redox pair and a formal reaction $\ce{Ox + n/2 H2 -> n H+ + Red}$, there is direct relation of the standard reaction Gibbs energy and the respective standard electrode potential $\Delta_\mathrm{r} G^{\circ}=−nFE^{\circ}$.

This is because of Gibbs free energy. The equation
$$
E^{\circ}=-\frac{\Delta_{\text{r}}G^{\circ}}{nF}
$$
makes it clear that for spontaneous reactions $(\Delta_{\text{r}}G^{\circ}<0)$, the electrode potential is positive. Highly electronegative elements such as fluorine undergoes reduction very spontaneously and hence, the positive value. In contrast s- and p-block elements have negative electrode potentials as they tend to oxidize easily.
The reference of SHE allows one to determine if the given element can reduce $\ce{H+}$ to hydrogen gas.


text:
If I knead my dough long and fast enough, can it be cooked itself?
May be Seasoned Advice question, but I thought that it's more of a scientific question. Correct me if I was wrong.
Kneading generates friction between dough chemicals(gluten, starch, etc.), thus generating heat. That's why we constantly chill pastry dough because the heat originated from the dough can ruin butter-dough layers AFAIK.
So... will that heat eventually cook the dough? Will it be heated enough to cook itself?

This is pretty much the same as "can you cook a chicken by slapping it". Answer is... not really.

@orthocresol Good point. I thought that if a chicken is slapped many times it can also be counted as some sort of... liquidy... doughy... goo, and that if I knead or blend that "dough" long enough, it can also be cooked by friction heat, if we can still call it a chicken. Dough does not have that concrete form, so dough will still be dough when deformed and then heated.

All depends on scenario. If you used n kW electric engine, thermally isolated space and properly geared mixing device, the dough would get surely baked, with mechanical power converted to thermal power. If the same mechanical power is used as thermal power needed to keep baking temperature (for the same arrangement), mechanical baking is possible. // Practical use is another question.

Also, long before the dough would be cooked, it would be reduced to a dried-out crumbling mass. However, friction stir welding is effective: https://en.wikipedia.org/wiki/Friction_stir_welding . So, you could weld the stirrer to the metal bowl...

Good comment, as intensive stirring would cause intensive drying out, unless there is closed space and limited air volume.

No, if you consider heat transfer the bread will never get warm enough to actually heat up/bake.


text:
comparing basic strength of toluidine isomers

I would like to compare the basicity of ortho and para isomers of toluidine.After comparing the pKb values online the order of basic strength is:
p—toluidine> m—toluidine> aniline> o—toluidine
My reasoning: according to me ortho form should have been the most basic because while comparing there is no SIR, SIP, localised lone pair, proton sponge or mesomeric effect and in case of ortho and para forms the hyperconjugation is till same extent so we finally come down to inductive effect due to which the clearly the ortho form should be more basic. But it is not so, Why?
I saw a similar question before posting my question on stackexchange but it talked about SIR and SIP effect but I was thought that in case of small groups both these effects are not applicable and that was the sole purpose of posting this question.
Thanks!!

Ever heard of the ortho effect? https://en.wikipedia.org/wiki/Ortho_effect

ya but it is due to SIR and here as NH2 is not very big it won't be sterically hindered

The article I linked specifically discusses the toluidines case as an example of the ortho effect. Do you think you know better?

@Waylander does that article mean that irrespective of the fact that group is small or big, any group on the ortho position increases the acidity for a substituted benzoic acid and decreases the basicity of substituted aniline. And also do we consider SIR applicable in presence of small groups like (-OH,-CN,-NH2)? cuz my professor told us that ortho effect is due to SIR which is not applicable when small groups are present.

Read the answers to this question https://chemistry.stackexchange.com/questions/87461/is-steric-inhibition-of-resonance-or-steric-inhibition-of-protonation-dominant-i

Any group in the ortho position is to have SIP. In the cases of OH,CN & NH2 you also have H-bonding complicating the situation. I think your professor's understanding of this is flawed.

@Waylander Thankyou after your article and jan's answer I am clear with this question.


text:
Is anodizing of mixed metals (aluminium and cooper) possible?
I'd like to blacken my Noctua NH-P1 passive cooler, made from an aluminium fins and copper heatpipes. Since thermal dissipation is key in my build, I thought about anodizing it, instead of spray paint coating.
The tutorial for aluminium anodization I found on Wiki-How looks quite straight forward, but since my knowledge in chemistry is quite basic, I do not know if and how the heatpipes would interfere the process.
In my case, not coloring the cooper at all, would not be a problem, but damaging the heatpipes (which are not solid, but filled with some liquid for ensuring an even better heat transfer) or breaking the soldering between the heatpipes and the cooling fins or even risking to do so, would be a total deal-breaker.

There are two separate ways to color copper and aluminum chemically.
Aluminum and titanium are easily anodized in an acidic water bath. Once anodized, the surface aluminum oxide becomes porous and can be dyed any color, or it can be colored by interference of light through the anodizing process.
Pure copper cannot be anodized, though some Al-Cu alloys can be. Instead, copper can be colored by oxidizing (torching) in air to produce interference colors. However, unlike the very hard aluminum oxide ($\ce{Al2O3}$) coating of anodized aluminum, copper oxides are soft and absorbent, so that touching the unsealed surface leaves fingerprints. The video link also describes sealing the oxidized surface.
Copper can also be colorized through chemical baths and other processes. Sulfides can produce browns and blacks, for example. Note that some of the chemicals used, such as "liver of sulfur", are toxic and are not suggested for use at home. Again, a coating to protect the surface is needed.
Since this is to be done to a heat transfer device, though, consider that any coating will reduce cooling efficiency. The thicker coating needed on copper might be more problematic. If there is sufficient extra cooling capacity, though it might work.


text:
What do the positive and negative signs on the atomic orbitals diagram imply?
I was reading up on wave functions and electrons behaving like a wave when they are bound in an atom. In the orbital diagram, the + and - signs don't make any sense to me, and the reasoning that one side of the nucleus taken as positive displacement is probably wrong. Can someone explain it?

Not only charges have signs.

What do +ve and -ve mean in the first place? It seems to occur only in posts of India users and I have never seen an explanation of the symbolics. Does it mean positive and negative partial charge in multiples of the elementary charge? But I have seen it even out of charge context.

@Poutnik +ve and -ve are short hand for positive and negative and are used in any context where those words are used (not just chemistry - for example quite common in medicine for positive test results). Possibly more widely used in chemistry in India, but certainly not specific to there. I see it often here in the US.

@Andrew - Thanks. Interesting. Like being selfexpanding character macros.

@Poutnik The ve might mean "valence electron" I suppose. I was recently confused by this notation.

@BuckThorn So did I get very confused, thinking all the time about scientific notation I may not be familiar with. Seeing larger and larger scope of usage, I got more and more confused.

@Poutnik actually upon further thought the "v" probably represents a greek "$\nu$" meant to indicate a stoichiometric coefficient. Maybe there is a question on usage or notation worth posting in there.

The sign on atomic orbital diagram represents the sign of the respective wave function, (a solution of the Schroedinger equation). It is analogical to sign of sin(x) function with positive and negative sections.
There are 2 major implications:

For adjacent atoms, overlapping of orbital parts with the same/opposite sign causes constructive/destructive wave interference, known from classical physics. 2 electrons with the same quantum numbers in the same atom would lead to their wave function mutual cancellation and zero probability.
Regions of the given orbital with the opposite signs are separated by 3D "node surfaces" with $|\Psi|^2=0$ and zero probability density of electron occurrence.

An orbital with the main quantum number n has n-1 of such node surfaces. E.g. the orbital 3s contains 2 inner spherical 3D surfaces with zero probability of the electron occurance, having 3 radial probability maxima.


text:
How is a Barrierless Reaction Confirmed in Quantum Chemistry?
Background:
Some quantum chemistry papers explore potential energy surfaces by characterizing critical points with an ab initio method. Reactants, products, intermediates, and transition states are found through optimization calculations, and then confirmed using frequency calculations. Intermediates connected by a transition state can be confirmed by intrinsic reaction coordinate (IRC) or some other minimal energy path (MEP) calculation. Here's an example potential energy surface from a paper online:

Question:
For any given intermediate and reactant/product pair, how does one confirm (1)
a barrierless connection or (2) a lack of connection? A connection with a barrier is simple: a transition state must exist and connect back to the two species. That's like TS3 connecting H + SO$_2$ and HSO$_2$ above. But it's not clear how something like SO--HO and OH + SO can be confirmed, or the lack of connection between HOSO and OH + SO can be confirmed.
My thoughts:
(1) Intuitively, an optimization that does very small steps and always goes downhill from the reactant/product to the intermediate of interest may be sufficient to confirm the barrierless connection. At the same time, this feels like a bit of a slippery slope.. If there is a downhill route going from OH+SO to TS', then this could be used to confirm that OH+SO can barrierlessly form HOSO (see crude cyan-drawn path below).

(2) Intuitively, confirming a lack of connection would require some sort of constrained optimization starting from the intermediate which shows that all possible degrees of freedom lead to other intermediates or products.
((Here, I am just assuming we stay on a single, simple, adiabatic potential energy surface. Here is a similar question with smaller scope: What is a barrier-less reaction in Quantum Chemistry?))

The absence of TS is sufficient to confirm a barrierless path. On the scheme the transition between SO...OH and SO + OH is barrierless because SO+OH is not a minimum on PES. This configuration lies on the slope near the SO...OH minimum or corresponds to a big plate area on the PES*, depending on particular orientation and distance between two molecules, so there is no saddle point between them.
*because there are a lot of possible configurations for SO+OH system with relatively equal energies. Slope corresponds to interacting SO+OH (Van der Waals), plate area corresponds to nearly not interacting molecules
In my opinion it is not correct to consider any existing path as locked on the one side, a system just may have lower probability to follow it starting from a "bad" side. This situation may correspond to a high barrier or very narrow path which is not good because it requires concentrating energy only on (for example) one vibrational degree of freedom. If I understood correctly, here transition SO+OH -> HOSO considered as locked, which in my opinion is not correct, I see two possibilities for this transformation:

stage by stage go backward from SO+OH (slope or plate area) -> SO..HO (minimum) -> TS' (saddle point) -> HOSO (minimum), with barrier because we have TS on this path
directly from SO+OH to HOSO, which in my opinion should correspond to narrow path case because requires quick, relatively hard collision O atom of OH to S atom of SO, because  molecules prefer another orientation (partial charges are bigger on H (OH) and O(SO) and corresponding molecular orbitals’ energies should be closer), so this situation may be possible only if two molecules directly collide and rotation of molecules  is freezed (which doesn't seems to be highly probable).

P.S.
I may not be right, I will be glad to discuss my point of view

I agree with your opinion, all around! I am interested in a more rigorous criteria, though. For instance, (1) how do you confirm or prove the "absence of TS" in the first case, and (2) how can you compare or quantify the "probability" of a path or how "narrow" the path is? These two don't seem well-defined compared to something simple like calculating the frequencies of a geometry after optimization.


text:
Does nitrous oxide decolorize potassium permanganate?
My teacher says that nitrous oxide can't form any complex with manganese in aqueous solution, so it can't be oxidised. However, shouldn't it get oxidised as it is present in the intermediate oxidation state of nitrogen?

Your teacher is partially right. Potassium permanganate solution is used for cleaning/scrubbing nitrous oxide for medical uses. If the nitrous oxide contains nitrogen oxide, NO, as an impurity it will certainly decolorize the solution but if pure nitrous oxide is being bubbled through potassium permanganate solution then the color will stay as it is. This is one of the ways to do chemical analysis of this gas. It is not that nitrous oxide cannot be oxidized or reduced under any conditions because it is an intermediate oxidation state. This is a fallacy and there are no free manganese ions in potassium permanganate.


text:
How does adding oxalic acid to water change the volume of the solution?
Does $\pu{87.4 mL}$ of $\ce{H2O}$ and $\pu{12.6 g}$ of $\ce{HOOC-COOH.2H2O}$ create $\pu{100 mL}$ of a solution?
So in our practical class our teacher wanted to make $\pu{100 mL}$ $\pu{1 M}$ aqueous solution of $\ce{HOOC-COOH.2H2O}$ [Oxalic Acid Dihydrate]. So first he measured $\pu{12.6g}$ $\pu{(0.1 mol)}$ of $\ce{HOOC-COOH.2H2O}$, then he measured $\pu{(100-12.6) mL} = \pu{87.4 mL}$ of $\ce{H2O}$ on a beaker. Then he mixed the  $\pu{12.6g}$ $\pu{(0.1 mol)}$ of solute with the $\pu{87.4 mL}$ of solvent.
I have some confusion about this. Because usually for preparing such a solution I would first add $\pu{12.6 g}$ of solute on a beaker and then keep adding water until the volume reaches $\pu{100ml}$. But I have never seen such a process for creating a solution by measuring a specific volume of solvent and a specific volume of solute and mixing them to get a specific volume of solution. Like: $a$ volume of solvent + $b$ volume of solute = $a+b$ volume of solution.
Can someone explain the theory behind this process?

Guess your instructor was just having a bad day.. Check this Chem SE post out.. It might help.

There are 2 false assumptions in the proposed procedure:

Density of the solute is the same as density of water.

Most solids are significantly denser than water. Density of oxalic acid dihydrate(OAD) is $\pu{1.653 g cm-3}$

The total volume does not change when the solute is being dissolved.

The total volume can shrink or expand. This  factor is usually less significant than solute density, but e.g. mixing equal volumes of ethanol and water leads to about 5% volume contraction.

These assumptions may be acceptable if the desired molarity is intended to be just orientational and the volume deviation is acceptable.
The initial total volume of water + OAD, before dissolving, would be:
$$V_\mathrm{tot} = \pu{87.4 mL} + \frac{\pu{12.6 g} }{ \pu{1.653 g mL-1}} = 9\pu{5.0 mL}$$
Generally, there is no way to predict the final solution volume (or total volume change) with the given solute mass and solvent volume without knowledge of density of the final solution. If such data is not available, the initial volume can be taken as approximate estimation.
For solution of the given molarity, the standard procedure is proper solute dissolving in such a solvent volume that total volume is somewhat less than final total volume. When dissolved, the volume is then topped to the desired nominal value by the solvent.
Alternatives need to know the final solution density:

The procedure calculates the needed solvent volume (or mass) for the target nominal volume.
The procedure calculates the solute mass for given solvent volume, leading to not a well rounded final volume.


text:
Solubility comparison for para-nitrophenol and phenol
During lecture with our chemistry teacher, he randomly asked us to compare solubility of phenol and p-nitrophenol, we expected p-nitrophenol to have greater solubility than phenol due to better dipole moment (5.43 D which is 1.70 D for phenol) and better hydrogen bonding but as it turns out phenol is way more soluble (84 g/L) than p-nitrophenol (10 g/L). Teacher was also confused, he didn’t resolve the problem that day and came to class next day saying he did some research and has found the reason but he hasn’t disclosed it yet and has asked us to do research ourself to get the reason.
I think the reason behind this abnormal behaviour is the strong attractions already existing between p-nitrophenol molecules is a lot more stronger than interactions between water molecules and interactions with water molecules are weaker, I am not sure of my answer and would like a better explanation / ideas to think about solubility. What do you guys think should be the reason for this abnormality.

I asked my teacher and also this is what i suspected , para nitro phenol is less soluble in water than phenol as para nitro phenol is way more polar than water and so the hydrogen bonding and other interactions formed by interaction between para nitro phenol and water are weaker than formed by interaction between molecules of para nitro phenol because of huge polarity para nitro phenol as compared to water but as dipole moments of water and phenol are almost same so they bond better than being alone and hence making phenol more soluble .

The nitro group does not seem to cause water solubility in anything! Nitrobenzene is insoluble, nitromethane and nitroethane are slightly soluble. It must have something to do with the diffuseness of the negative charge about the O atoms in the nitro group and the nitrate ion. Any overt dipole or hydrogen bonding attraction will involve disruption of the resonance lowered molecular orbitals. Add to that the entropy decrease from association and there is no increased intermolecular attraction.


text:
How to identify unknown, insoluble solid with limited resources?
Looking for suggestions on how to identify an unknown white solid we received as a donation. It was originally labeled as $\text{MgSO}_4$, but clearly is not. We suspect it might be $\text{CaSO}_4$. So far, I can say that it is practically insoluble in water. We are a DIYbio lab and so were aren't equipped for complex chemical analyses. We also don't have a fume hood so we can't utilize any reactions that might produce noxious gases. Any ideas would be greatly appreciated if only so I don't have to see the bag sitting on my desk anymore.

Either $\ce{CaSO4}$ (~55 USD per 1 kg) or $\ce{MgSO4}$  ((~85 USD per 500 g) is not seriously expensive in fairly pure form. Your chemical is in very crude form judging by the color (probably in tech grade). So why waste money to analyze?

You can take advantage of the difference in pH required to precipitate calcium and magnesium hydroxides for this one. I would acidify a solution of it, then slowly add in a 0.1 molar solution of KOH while recording pH, until you see a precipitate of the hydroxide salt. Off-hand, I cannot recall which precipitates first as pH increases but this will give you an idea if you are dealing with Ca or Mg sulfate.

@Todd, CaSO4 is not soluble in acids, it is very hard to dissolve. However, insolubility does not prove that it is CaSO4. It is best to buy the material from a reliable source.

A good reason not to accept “donations” of unidentified materials…

@AChem - I disagree. See this for example.

A donation of a big plastic bag of white stuff of unknown composition? How in the world does this make any sense at all?

@everyone Did you not see the part about DIYbio makerspace? The bag was originally labelled, but incorrectly. MgSO4 is very useful for PCR, hence the accepted donation. Regardless, it was before my time as lab manager.

Since you don't know what it might be contaminated with, it might need to be disposed as hazardous waste. It could be contaminated with biological material, or even with radioactive wastes, which would not show on any but the most sensitive chemical assay.
As the old riddle goes:
What do you call 1,000 liters of sewage mixed with 1 ml of Chateau Mouton-Rothschild 1945? Sewage.
What do you call 1,000 liters of Chateau Lafite 1869 mixed with 1 ml of sewage? It's still sewage.

This isn't really an answer, just moralism. However, I think you are correct that it's easier just to handle it as potentially hazardous waste since it is definitely not the compound of interest.


text:
Meaning of equivalents in organic synthesis
The use of "equivalents" as used in titrations/chemical analysis is an obsolete concept except in some countries. The original meaning of equivalent weights is the weight of a compound that combines with 16 g O or 1 g H and it was developed during the time when electrons did not "exist". With this definition it is easy to determine the equivalent weights, hence equivalents can be calculated directly or indirectly for most of inorganic simple compounds provided a chemical equation is available.
I still occassionally see equivalents in organic synthesis works. For example, one abstract has
"Deprotonation of pyrazine, pyridazine, pyrimidine, and quinoxaline using an in situ mixture of ZnCl$_2$·TMEDA (0.5 equiv) and LiTMP (1.5 equiv) was studied."
This is another paper title
"Why Do Catalytic Quantities of Lewis Acid Generally Yield More Product than 1.1 Equiv in the Intramolecular Diels−Alder Reaction with a Furan Diene? "
It appears that the organic chemist's equivalent is quite different from old-shool's analytical chemist's equivalent. I see this has been asked before, but the answers are not relevant to organic synthesis and it is hard to find a relevant definition in any chemical dictionary or even textbooks with reference to organic compounds. Is there a solid reference which defines how equivalents are defined/used in organic chemistry?

https://books.google.com/ngrams/graph?content=stoichiometric+equivalent&year_start=1500&year_end=2019&corpus=26&smoothing=3&direct_url=t1%3B%2Cstoichiometric%20equivalent%3B%2Cc0#t1%3B%2Cstoichiometric%20equivalent%3B%2Cc0

Karsten, I think this reaction equivalent in J.Chem.Ed. is quite different from what orthocresol is explaining. The article has a nice solved equation for an inorganic reaction but the term reaction equivalent there is not "molar equivalents" as explained in your first link and this answer. Also it will not fit in the examples I quoted. The book is also related to inorganic compounds. They are apparently using equivalents in a classical way (the analytical chemist's way).

“ The use of "equivalents" as used in titrations/chemical analysis is an obsolete concept except in some countries.” What???We don’t use equivalent weight internationally anymore?

@insipidintegrator, Equivalent weights are longer taught in a majority of Western universities.

Wow. @AChem and here, (won’t specify country, but it is to the east of Greece and to the west of New Zealand) we literally have a whole chapter called “Redox Reactions “ in school which focuses on redox titrations and where the main task is to find out the n-factor and equivalent weights of a variety of reactions.

I have studied those as a student but in the US, most of them will unaware of these terms even PhDs in analytical chemistry. For the sake of records, normality is still used in analytical procedures in industry.

@AChem Here is how equivalents is used on StackExchange (well, N=1, single example): https://chemistry.stackexchange.com/a/167376

Is there a solid reference which defines how equivalents are defined/used in organic chemistry?

I'm not sure about a reference per se; but from my (limited) experience as an organic chemist, the usage of 'equivalents' in this context invariably refers to molar equivalents. It's most certainly not based on the concept of 'equivalent weights' or normality, which as you noted is obsolete.
Typically this is measured with respect to the substrate of the reaction, so in the case you quoted,

Deprotonation of pyrazine, pyridazine, pyrimidine, and quinoxaline using an in situ mixture of ZnCl2·TMEDA (0.5 equiv) and LiTMP (1.5 equiv) was studied

would refer to the addition of $\pu{0.5 mol}$ of $\ce{ZnCl2.TMEDA}$ and $\pu{1.5 mol}$ of $\ce{LiTMP}$ per $\pu{1 mol}$ of pyrazine (for example).
As for the second quote, the authors are probably comparing the Diels–Alder reaction using catalytic Lewis acid versus (super)stoichiometric Lewis acid. In this case, the Lewis acid is also measured in terms of molar equivalent, relative to the substrate.

Molar equivalents make perfect sense in reaction schemees but organic chemistry books are quiet about calculating or quoting equivalents. I will search this term in Google Books. It is clear now that this equivalent is not the analytical / inorganic chemist's equivalent.

Molar equivalents is how synthetic organic chemists apply the term. If substrate A is either expensive or valuable then one would use 1.1 molar equivalents of inexpensive or readily available substrate B to form bond A-B.


text:
How to determine if the ring compound has optical isomers?
The first column and fourth column of the table below says that the compound has the plane of symmetry and is not optically active. However, I noticed that each side has different arrangement of the H and Cl atoms. Does it still count as being symmetrical?

When you say “each side” which molecule(s) are you referring to?

Is there a plane of symmetry, if the sides are not mutual mirror images ?

Are you concerned about the chlorine in the lower right in structure 1 not lying on the line? It is on the line. It is where it is so that you can see it in the structure.


text:
Are all humectants sticky?
While using a topical lotion based on ethanol, water and propylene glycol, I noticed that it becomes sticky while drying. Propylene glycol is a humectant and is used for this purpose in pharmaceuticals and cosmetics. This made me wonder: Is stickiness a general property of all humectants?

Hmm, humectants share 2 essential properties: 1/high hydrophilicity 2/ low volatility  // Both can be achieved e.g. by many hydrophilic groups like OH  ( glycols, glycerol, polyvinylalcohol PVA).  Not sure if PVA is sticky as well, but I would guess it may be. // It is quite possible stickiness and water retention are 2 aspects of the same quality here.// What may play role is the type of emultion ( oil in water or water in oil )


text:
Is hydrochloric acid or sulfuric acid more corrosive to titanium?
Which acid is more corrosive to titanium at 20 to 30 °C and similar concentrations: hydrochloric acid $\ce{HCl}$ or sulfuric acid $\ce{H2SO4}$?
I found some info on AZoMaterials — Titanium - Corrosion by Acids, but it was not exactly apples to apples.
Also, if the acid concentrations are very low (say, less than 0.01%), which one will have worse long term effects on the metal?


text:
Electronic configuration in d-block
According to Bohr-Bury rule, 4f and 5d orbital come after 6s orbital. But if I search for the electronic configuration for Os or any  atom in d block for that matter, they give the electronic configuration as 4f 5d 6s instead of 6s 4f 5d
Why does this happen ? why is S orbital written in the end even though it has less energy
Can someone please explain ?

What is "Bohr-Bury rule"? This looks pretty much like Aufbau approximation known by a plethora of other names, but Bohr-Bury rule is something new (at least to me). Anyway, https://chemistry.stackexchange.com/questions/141708 seems to be related.

Thanks, but my question is not on the order of filling of orbitals per say, but the order with which each orbital is portrayed on the internet. For example if you search for the electronic configuration of Ir (Iridium) in the internet or any reliable source,  you get 4f(14) 5d(7) 6s(2) which is right, but why not 6s(2) 4f(14) 5d(7), same answer but this one is ordered in such a way that it follows the order increasing energy of orbitals. Energy wise 5d has more energy than 4f and 6s respectively so why do they have to mix  the order up can't they just follow the order of increasing energy ?

It doesn't matter the way you write it. 4f 5d 6s is the same as 6s 4f 5d. The first one is ordered by energy and the second by the way the orbitals are filled. Even though, you must note that the 5d${^1}$ fills first in the element lanthanum [Xe] 6s2 4f0 5d1 ^[1] https://en.wikipedia.org/wiki/Electron_configuration

Thanks, so both the answers are right then, just 2 different ways of representing the same configuration.

Writing 6s orbital at end of electronic configuration indicates that electron will first be removed from 6s orbital then 5d then 4f.
d blocks are exceptional casses. That's why they behave like that.
Example  for scandium [Ar] 3d¹4s² indicates that electron will be removed from 4s subshell.

Both form of EC are correct but 4f5d6s gives us idea about from which shell electrons will be removed


text:
Negative or fractional order of reaction
Let $\ce{A}$ be the reactant and $\ce{P}$ the product at the imaginary elementary reaction $\ce{A->P}$.
Could it possibly have a net total order of $-1?$
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^{-1} \tag{1}$$
Or even more, a reaction whose rate depends on the concentration of the product as it follows:
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{P}] \tag{2}$$
Is there any example of an actual reaction that behaves like that?
On the other hand, can we think about non-integer orders of reaction like $1.5$, $0.5$, $-0.5$, $2.368$, $\mathrm e$ or $\pi$ for instance? Any examples?

Orders of reaction are experimental parameters. So they cannot be predicted in advance. They can be 0, 1, 2 or 3. But a reaction may have no order at all, like the reaction H2 + Br2 -> 2 HBr, which rate is a complicated fraction containing the three concentrations. In principle, it is not absolutely impossible to have reactions with an order different from 0. 1, 2 and 3. But I must admit, I have never seen any such reaction.

Wikipedia has you covered for both parts of the question in their article on Rate equation.
They give an example of a negative order of reaction:
$$\ce{2 O3(g) -> 3 O2(g)}$$
Under certain conditions, this reaction is of -1 order in $[\ce{O2}]$. This covers your question about a reaction being dependent on the product concentration (equation 2 in OP's question) as well as having a negative order (equation 1 in OP's question).
They also give an example of a fractional order of reaction:
$$\ce{COCl2(g) -> CO(g) + Cl2(g)}$$
This has an order of 0.5 in chlorine.
Elementary or one-step reaction?
If the reaction is elementary, the reaction orders are equal to the the stoichiometric coefficients of the reactants. I'm not sure what is meant with a one-step reaction. For the examples given in this answer, there are radical intermediates in multi-step mechanisms (What does a fractional order of reaction mean for the mechanism?) that explain the unusual kinetic features.

Ok, but about the question 1 I meant negative order for a product.

A starting point for searching for a negative order for the product would be the term "product inhibition". This is fairly common for enzyme-catalyzed reactions where the product, looking similar to a reactant, clogs up the active site: https://en.wikipedia.org/wiki/Product_inhibition


text:
Can alkyl halides react with guanidine/guanidino groups?
I wonder how the π delocalization of guanidino groups affects the reactivity of the terminal nitrogens with respect to alkyl halides compared to, say, primary amines?

Doesn't delocalization prevent reactivity with alkyl halides? For instance once amides are formed from acyl halides, why are there no further substitution on the newly introduced NH2 if there are halides nearby?

Well, yes, they do react. I guess imino position is activated, while amine groups deactivated. That clearly shows in guanidine's very high basicity.

Yes they alkylate, example  https://pubs.acs.org/doi/abs/10.1021/jo0265535

@Waylander in this paper they use protected guanidines so they are able to deprotonate an acidic N-carbamate hydrogen; I assume that same approach wouldn't work with a naked (basic) guanidine?

@Mowgli  I don't know, but if you can access this review it may tell you https://www.thieme-connect.com/products/ejournals/abstract/10.1055/s-2006-926325


text:
Is formation of cyclic carbonate from trimethyl orthoformate and vicinal aromatic 1,2-diol possible?

Hello, my question is, "is it possible cyclic carbonate formation between trimethyl orthoformate and  vicinal aromatic 1,2-diol"

Trimethyl orthoformate is at the wrong oxidation level to give a cyclic carbonate. You need phosgene or a dialkyl carbonate. The efficient preparation of catechol carbonate from catechol (benzene-1,2-diol) and dimethyl carbonate is described in this paper


text:
where could I find the standard atomic orbital energy of all elements?
I am trying to find the standard atomic orbital energy of all elements.
Here is the data I found in some old books:

I am looking for some more modernized data, I tried to look for the data in the CRC book, but didn't find anything.
My question is:
Where could I find the standard atomic orbital energy data?

It's difficult to answer this until you define "standard atomic orbital energy" Hartree-Fock? Something else?

Also units are always important. Looks like electron volts. It would be good to specify

How old are the books you've already looked at? It's hard to 'update' something if you don't know how old your previous information is...

NIST is a reasonable place to start. They've posted data from electronic structure calculations using DFT online (Ref 1) . However it looks like you might need to scrape the site to extract the desired information and assemble it into a table, or find a secondary reference or website that has done that for you already. Note also the link within the reference does not exist but will forward you to the current page.
Reference

Kotochigova, S., Levine, Z.H., Shirley, E.L., Stiles, M.D., and Clark, C.W. (2003), Atomic Reference Data for Electronic Structure Calculations (version 1.3). [Online] Available: http://physics.nist.gov/DFTdata [year, month day]. National Institute of Standards and Technology, Gaithersburg, MD.

There is a periodic table of the results at https://www.nist.gov/pml/atomic-reference-data-electronic-structure-calculations/atomic-reference-data-electronic-7 But note these results will not agree with the table above, for instance the LDA value for the 1s eval of H is given as -0.233471 Hartree = -6.35 eV presumably due to the lack of self-interaction correction. We need to know what approximation is meant by "standard atomic orbital energy", especially for elements with Z>1 where orbitals don't really exist.

@IanBush Regardless of what level of theory the OP chooses as a reference, a problem with the NIST data may be that it only lists energies for occupied orbitals whereas the table in the question lists excited ones as well.

Is there any experimental data for orbital energy of single atoms? Like in this page http://www.graylark.com/eve/orbital-energies-table.html

@Jack Well you can't use straightforward PES as given on that site to measure an energy to ascribe to unoccupied orbitals. But remember that orbitals do not exist when Z>1, they are only an approximation. And there are many, many approximations, both experimental and theoretical so to answer your question you need to  tell us which one is "standard"

Sorry, let's do this right. Orbitals do not exist for number of electrons > 1

@Jack It might help to know more about how you intend to use the information or why you need the information. At present it seems like you want experimental data on the energy of the highest occupied atomic orbitals, ionization energies, or something similar, but originally it seemed you wanted tabulated data on atomic orbital energies computed theoretically.

Here is my intention: Like explained in http://www.graylark.com/eve/orbital-energies-table.html      What are they?
These are the energies of valence atomic orbitals in the elements. They were found using Photoelectron Spectroscopy, which measures the energy needed to ionize electrons from atoms in the gas phase.
What are they good for?
Usually we use these to make MO diagrams. They tell us the relative energies of the AOs that we start with. You can also look at the periodic trends, etc. The lowest value is the same as the first ionization energy.


text:
Why are the products of Grignard reaction on an alpha-chiral ketone diastereomers rather than a racemate?

Comment on stereochemistry of products.

This is a Grignard reaction and the product has two chiral carbons. I think it should be a racemic mixture.
However, my teacher says the correct answer is diastereomers. He says if the reactant has no chiral carbon and product is chiral, then it would be a racemic mixture. But why?

If a reaction scrambles an existing single stereo-center, you get a racemic mixture. If it creates a new stereo-center without affecting the existing one, and the reactant was chiral (not a racemic mixture), you can't get a pair of mirror images. The new stereo-center will be a mix of R and S, and the existing one will be just R or just S, so it is a mixture of diastereomers.

The reaction introduces a new chiral center into your mix, that chiral carbon being the one that has the tertiary alcohol on it (let’s call this #2).
The product will produce a mixture of those chiral centers being flipped between R and S, but you need to remember that there is a preexisting chiral center (let’s call it #1) that we have before creating this new one.
Recall:

enantiomers occur when all chiral centers are flipped
diastereomers occur when one, but not all, chiral centers are flipped.

This reaction can have a mixture of #2 being R or S - but it has #1's chirality the same (s). This means not all chiral centers were flipped, creating diastereomer mixtures.

He says if the reactant has no chiral carbon and product formed is chiral then it would be a racemic mixture. But why?

So that occurs whenever we transition that sp2 carbon to sp3, turning the carbonyl into an alcohol (reduction), but that "rule" he gave says if there's no chiral carbon — we do actually have chiral carbons before the reaction starts — so the auto answer of "racemic mixture" doesn't quite apply here.

First of all, welcome to ChemSE on your first post. Allow me to expand on the Comment of Karsten Theis and provide you with an illustration. If a compound is racemic and bears a single stereocenter, then it is properly represented without stereochemistry as shown in structure 1. The example you provide is a single enantiomer, (S)-4-methylhexan-3-one, 1-S. [If you are unfamiliar with R,S-nomenclature (CIP-rules), go here.]
Let's presume that our Grignard reagent is methyl magnesium bromide. Without addressing Felkin-Ahn issues, addition of the Grignard reagent to the carbonyl group of 1-S from the front (next to the methyl group) will afford stereoisomer 3 while addition from the rear of the carbonyl group produces stereoisomer 4. Both of these diastereomers are normally produced and invariably in unequal amounts. Why, you might ask. Felkin-Anh considerations notwithstanding, the reagent "sees" a methyl group from the frontside approach and a smaller hydrogen from the rear. There is difference in the energy of the two transition states. For the sake of argument, let's say that the ratio 2/3 is 3/1. Had you started with enantiomer 1-R, the ratio of the diastereomers 4/5 [red box] would also be 3/1; they are the enantiomers of 2/3, respectively. Enantiomers 2-5 are optically active.
Conducting the reaction on the racemate 1 will still give the same ratio 3/1 but the products will be racemic and not optically active.


text:
Wittig reaction without ketone or aldehyde?

In the reaction below the ethylene glycol would react with the ketone in the first step turning it into a protecting group. How would the ylide react, as I thought it only reacts with ketones and aldehydes to form alkenes?

Phosphorous ylides are good nucleophiles. They react with alkyl halides to give alkylation

The catalyst in the first reaction should be $\ce{H+}$ and not $\ce{H3O+}$ because one is trying to remove $\ce{H2O}$ during the ketalization.

The catalyst is not H+ but dry HCl in alcohol solution, so possibly protonated alcohol molecules.  If possible, the water is removed azeotropically with benzene or toluene.


text:
How does bis(glycinato) Nickle exhibit optical isomerism?
I'm a medical practitioner but sometimes I find time to discuss few topics of chemistry with my son who has just passed out his 12th standard exams. Regarding optical isomerism I have learned that a chiral center is an essential requirement as is there in "Lactic Acid" {CH4-CH(OH)-COOH}. But in co-ordinate compounds like "bis(glycinato)Nickle(II)" I find same ligand (glycinato) bound on either side of a tetrahedral Nickle with two atoms each of Oxygen & Nitrogen occupying the four corners of tetrahedron.
Then where is the chiral center?

The decisive factor is if the mirror image of the complex is or is not identical to original. E.g. meso-tartaric acid has 2 chiral centers that mutually cancel, the molecule has plane of symmetry and mirrored images are identical, so not optically active.

There are octahedral derivatives of this compound (and they can be chiral as all bis(bidentate-ligand) octahedral complexes are. If it really is tetrahedral, it won't be chiral.

This question/answer is somewhat related.


text:
How to calculate corrected Gas Volume?
We are working on a project called 'Electronic Gas Volume Corrector'. The device reads pressure and temperature from sensors mounted on a gas pipe and gas volume from a gas meter. These are the flowing condition parameters : $p_f$, $T_f$ and $V_f$. A computer code then uses the following equation to calculate gas volume in "base" conditions :
$V_b = V_f \times P_f/P_b \times Z_b/Z_f \times T_b/T_f$
This equation tells us how gas volume would be if it was measured in billing conditions for the same number of moles n as in the flowing conditions.
These are the billing conditions values: $T_b = \pu{15°C}, P_b = \pu{1 atm}, Z_b =0.978$. The gas in question is propane. I found Zb in this website.
The only parameter left to determine is $Z_f$, and we do not know how to calculate it, especially since it depends on sensors readings of flowing conditions $P_f$ and $T_f$. I tried using $PV=nZRT$ but the number of moles is always unknown since it depends on gas consumption, and $V/n$ which is the molar volume and also depends on $T$ and $p$ according to my research, and we do not know how to calculate it.
What are we missing ? Is there any other equation that helps in calculating $Z_f$ ?

There should be tables of empirical formulas to get compressibility factor of the real gas  Z as function of p and T.

What is the meaning of the indexes $b$ and $f$ ?

@Poutnik we found some old documents about Volume correction but they use graphs (Z(P) for a few values of T ) instead of equations. If you can share links/titles it will be much appreciated

@Maurice b is for billing conditions and f for flowing (real) conditions
Two consumers might get different bills even if they consume the same energy because of different T and P, that's why we need to convert flowing Volume to a base Volume calculated in the billing conditions

One possible way is calculating V or Z from van der Waals' state equation for real gases.

See Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics for graphs of z as a function of $T/T_c$, $P/P_c$, and ascentric factor $\omega$, where $T_c$ and $P_c$ are the critical temperature and pressure and $\omega$ is a tabulated (constant) physical property characteristic of the specific gas under consideration.

Poutnik's suggestion, about using van der Waals equation, is the best, and the simplest, as it needs only two correcting factors.

I found this equation :$ Z = \frac{V_m}{V_m-b} - \frac{a}{V_m R T} $ but the problem the molar volume figures in it. Is it always constant and equals 22.4 l/mol ?

I found something interesting in this website (https://www.e-education.psu.edu/png301/node/840#:~:text=Gas%20Super%2DCompressibility%20Factor%2C%20Z,pressure%20and%20high%20temperature%20conditions.) Equation 3.66 seems to be exactly what I was looking for. I don't why they call Z supercompressibility factor. So what do you all think  ?

@IssamH Vm=f(p,T)


text:
Is there any way to measure or otherwise quantify the extent of caramelization experimentally?
I've been trying to design an experiment where I want to compare the efficacy of caramelization of a monosaccharide under different pH conditions (the sugar would be placed in buffers of varying pH). The only problem is that I'm really not too sure how I would go about quantifying my findings. Perhaps I could use absorbance as an indicator, whereby I can conclude that the darker the solution becomes, the greater the extent of caramelization. However, I'm not too sure if that would be the most valid approach and I'm wondering if there's any better alternative.

Do you actually have to be able to carry out the experiment, or is this a "thought experiment"? Have you thought about the reaction(s) occurring during caramelization, and what particular analyte you might target as an indicator of the extent of caramelization? Perhaps caramelization is complex, resulting in many different products, and you would like to obtain a particular distribution of these? Also, do you have a budget? Access to a lab? If so what kind of instrumentation? There are many spectroscopic tools that you could use to solve this.

Typically, quality of caramelization is measured by taste ;)

I would use semiquantitative subjective parameters, like discrete visual and taste scales of intensity of brown  colour and caramel taste.

Eventually also additional parameter for unpleasant taste presence.

Get a high quality digital camera with controlled exposure, color temperature color space etc.  and controlled lighting and substrate background. Find camera settings and lighting that discriminates.  Possibly a diode array uv-vis spectrometer Try Ocean Optics or Agilent They might sell or loan a used or returned instrument cheap.


text:
Can flaxseed (linseed) oil undergo hydrogenation?
I'm curious if it's possible to chemically hydrogenate flaxseed (linseed) oil. I've never seen it in hydrogenated form. Given its quality as an excellent wood protectant, I figure it would be easier to apply as a paste. In fact, there's a popular product on the market which has a paste-like consistency called Odie's oil. The MSDS doesn't list its ingredients since they're considered non-toxic. Interestingly, it hints it may contain linseed oil, since it warns the product can spontaneously combust.
Could it be the hydrogenation process would negatively impact its ability to harden when cured?

Linseed oil hydrogenation would cancel its nutrient qualities.

It would be possible to hydrogenate linseed oil, if you were to expose the oil to hydrogen gas while it is in contact with a cataylsis such as palladium on carbon or finely divided nickel then the carbon carbon double bonds in the fatty acid residues in the oil will adsorb hydrogen gas and become carbon carbon single bonds.
Here is a diagram of a unsaturated fatty acid like one of the ones in linseed oil.

The physical properties of the oil would change, normally if you hydrogenate vegtable oils their melting points increase and they become harder. This would tend to convert the oil from a liquid into a solid.
While it might be easier in some conditions to apply a paste of hydrogenated oil to an object like a wooden surface. The oil would not perform the same tasks as the oil normally does after the hydrogenation.
When exposed to oxygen gas an unsaturated oil like linseed oil will undergoe an oxidation reaction whcih forms organic peroxides. The funny thing is that the polymerisation reaction of alkenes is normally inhibited by oxygen. The breakdown of the organic peroxides will form free radicals which then make the oil turn into a polymer. A common use of linseed oil is to treat wood, you paint it onto the surface of bare wood. The oil enters the wood where it will be exposed to less oxygen than it is on the surface. It will then undergoe the polymerisation reaction and turn the wood into a composite material which has better properties than the wood.
What can happen is that the free radical can remove (abstract) a hydrogen atom from the side chain of the fatty acid. This forms a radical which is stablaized by resonance.

The radical can then react with oxygen to form a hydroperoxide which on heating can form alkoxy and hydroxyl radicals. This process can result in the number of radicals in the system increasing.

Instead of reacting with an oxygen molecule it is possible for the carbon centred radical to react with another alkene to form a new carbon carbon bond. This can result in the polymerisation of the oil.
It reminds me of a paper I saw years ago where someone impregnated wood with styrene before irradating the wood with either gamma rays or high speed electrons. This treatment formed a wood / polystyrene composite material which had good mechanical properties and also was resistant to wood boring creatures.
I think that the hydrogenated oil would be unable to enter the wood, also the polymerisation of the oil would be greatly reduced if it was hydrogenated. If we were to take a saturated oil such as a typical diesel fuel and soak wood in it then we will not improve the properties of the wood.
The diesel fuel would not undergoe a polymerisation reaction in the wood, a few years ago in Sweden an old diesel train used on a branch line had a fire. The train was partly made from wood. The wood had become contaminated with diesel over the years as a result of spills. The wood turned into a gaint version of a fire lighter. When the train caught fire the whole thing went up in smoke very quickly. So I think that hydrogenation of linseed oil would make it useless as a thing for treating wood surfaces.
One comment made after the original post was made was to point out that hydrogen peroxide can be used to oxidize the oil. This will form epoxides, the epoxide group is a three memebered ring with an oxygen in it. It is a special ether which is much more reactive towards nucleophiles such as amines. If the oil was treated with hydrogen peroxide then it would be possible to make a polyepoxide. If the polyepoxide was reacted with either ammonia or an amine. Then it would be possible to form a polymer. I am sure that there have been papers written on this subject.

Awesome explanation! Now you have my brain spinning in strange directions.  Would hydrogen peroxide enhance linseed oil?

Here's. an interesting article: http://article.sapub.org/10.5923.j.chemistry.20180805.01.html


text:
Why acidity of hydrogen hallide increase down the group 17
I just wandering when going down group 17 the electronegativity decreases, and acidity increase with an increase in electronegativity. Then theoritically it should have decrease its acidity of hydrogen hallide when going down the group, but why the result show that the acidity increase down the group instead?

The same happens in group 16, it is actually related to the fact that the hydrogen-halogen bond energy decreases as we go down the group and thus losing H+ ion becomes easier and therefore down the group H+ donating capability increases and so does the acidity.

So that mean when going down the group, the effect of bond energy overweight the effect of electronegativity ,right?

We always see bond energy, you might have a case in which electronegativity plays the major role in deciding bond energy but it is not that case here.


text:
Mechanism of Grignard reaction with 2-(thiiran-2-yl)oxirane

What will be the final major product?

The answer is 2-(methoxymethyl)-3-(substituted methyl)thiirane:

I think $\ce{R}$ of $\ce{RMgX}$ attacks thiirane ring because negative charge on sulfur is more stable than that of oxygen.
But why doesn't the negative sulfur subsequently attack on less hindered carbon of the ring? Instead, it attacks the β-carbon which is more hindered and opens the ring so that methyl of $\ce{CH3I}$ can make bond with negative oxygen atom. Why does the reaction proceed that way?

Sulfur attacking the less-hindered end of the epoxide would generate a [4] ring. This is more strained than a [3] ring and thus energetically less favourable. The attack is reversible, so the more energetically favourable (thermodynamic) product will be the outcome which is trapped by methylation in the final step.

Actually cyclopropane is a bit more strained than cyclobutane.  There is an entropic effect favoring three- over 4-ring formation. See here.

One other aspect is after the initial opening of the sulfur-bearing ring, the sulfur ends up closer to the bridgehead carbon than to any other part of the oxirane ring. (You can see this in a molecular model of the intermediate anion, in which the other two atoms are tied back from the sulfur by the small ring size.) So electronic overlap and thus bond rearrangement occurs more readily through the bridegehead.