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Basic bond strength concept
I saw C-H  bond is stronger than C-C bond... Is this because of the following reasons?
1) C-H bond is shorter than C-C bond.
2) C-H bond is more ionic than C-C bond.
I

The bond strength depends on the atomic radius of both chemical elements which take part in their bond and their difference(the chemical elements) in electronegativity.
Just for the record C-H is not ionic, it is not even polar :P
Both C and H attract the bonding pair of electrons approximately the same.

Noooo I mean atleast more ionic due to C-C as there is electronegativity difference....u must be laughing on me so hard if u think that I think C-H bond is ionic

C and H have the same electronegativity.Their difference in electronegativity is so small , which can be neglected.

Yes but atleast more than C-C right?

Electronegativity of C is 2.5 and that of H is 2.1

The main difference between the bond energies is the atomic radius of carbon being bigger than the atomic radius of hydrogen. But yes it is more than C-C since in C-C it is zero.

0.4 is a very small difference

Ok but what about overlapping. P-p overlapping is efficient than s-p

I don't what overlapping is. I know there are some bonds which are referred as σ and p but I cant tell you.Are you talking about molecular orbital theory?

Yes .... Not exactly but yes

well the shape of 1s subshell is spherical so the s-p bond would be stronger due to the fact the orbital 1s covers more area.

So more area in common between the 1s and the 2p orbital.

Its true that overlap of 2p - 2p >2s-2p > 2s - 2s due to p's directional bond. But  1s-1s >1s-2p >2p-2p , because as size inncreases extent of ovelap decreases which means here 2 opposite factors act in which size factor wins.
But the thing is whn carbon makes bond with H its orbital overlapping with it is not pure p rather sp3, sp2 or sp. Infact C always make bond with these orbitals only, so it is not right to comapre pure orbitals overlapping.
H being shorter than C, and H-C having electronegative difference unlike C-C have higher bond energy as they r able to release more energy due to these factors.Actually ectronegativity plays a important role in explaining the bond energies mostly.


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Acetone basicity
Why doesn't acetone have high basicity?

The C−O bond is polar, O becomes partially negatively charged and it should attract protons, becoming a strong base.

Related: Are hydrogen atoms in acetone molecule acidic?; Why is acetone less acidic than cyclohexanol?; Why is the pKa of a ketone higher than of water?

Andelisk im not asking about the acidity of hydrogens in acetone , just why the oxygen cant become a proton acceptor.

Inspired by (I think) comments to this question: https://chemistry.stackexchange.com/questions/119921/dna-pairs-adenine-thymine-guanine-cytosine/119924#119924

Do you have a good explanation why acetone isnt a base?

I don't know the reason, but I can describe the trend (source: http://cactus.dixie.edu/smblack/chem2310/summary_pages/pka_chart.pdf):

Protonated ketones have a pKa of about -7. Protonated ethers have a pKa of about -4 and protonated primary alcohols of about -2. Comparison to hydronium in water is difficult, but in DMSO, hydronium is more acidic by about 2 pKa units (source: comment by DavePhD below https://chemistry.stackexchange.com/a/27278/72973).
So it seems water is a little bit basic. If you replace one of the bonds to hydrogen with a bond to carbon, it becomes less basic (alcohol). If you replace the other bond to hydrogen with a bond to carbon as well, it becomes even less basic (ether and ketone).
A hand-waiving explanation would be to say the electronegativity difference between hydrogen and oxygen is higher than that between carbon and oxygen. I am not sure what to say about the difference between ethers and ketones. I think of ketones as the better hydrogen bond acceptors, and as more soluble in water (all other things being equal), so I don't have a good rationale why it would be easier for an ether to accept a proton.

And what do you mean by" protonated ketones"?

I would consider the geometry of a ketone vs an ether to perhaps explain the answer. I am pretty sure the real reason lies somewhere in the MO of the two groups (i.e. the HOMO of a ketone sits lower than an ether). However, I am not sure if it would be too rigorous to dive into that though.

@SoftwarePlayer A protonated ketone is the product of a ketone acting as a base, when a proton (or H+) attaches at the oxygen.

@SoftwarePlayer "extra oxygen atom"? Alcohols, ethers and ketones all have oxygen atoms - there is no extra oxygen.

yes you are correct i thought you meant something else

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Dipole moment of water using the definition
In many Chemistry books, the dipole moment of molecules is calculated through algebraic formulas. However, another definition of the dipole moment is
$$\vec{\mu} = \sum_i q_i \, \vec{r}_i$$
So, how can we calculate the dipole moment of water, for example using this formula? Is there a difference in interpretation of dipole moment in Physics and Chemistry?
Here is the Definition.

There is no difference in definition, that's for sure.

The document the OP references has most of the information: The overall measured dipole moment, the way bond dipoles add up to the molecule dipole, and how to calculate the bond dipole. Also, it states that a proton and an electron at a distance of 100 pm have a dipole moment of 4.80 D. From that information, we should be able to figure everything out.
Bond dipole vs. Molecular dipole
The bond dipoles add up as vectors. In the case of water, we can put a water molecule into an x-y-coordinate system so that oxygen is at the origin and the x-axis divides the H-O-H angle in half. In that orientation, the bond dipoles will add up to a vector pointing in the x-direction. To get the magnitude, we have to take the x-component of the vector. The appropriate formula is given in the textbook:
$$ \text{Total dipole moment} = \text{Bond dipole moment} \cdot 2 \cdot \cos \frac{104^\circ}{2}.$$ The term 2 comes from having two O-H bonds, and the cosine term gives the component of the vector in the direction of the x-axis (i.e. in the direction of the molecular dipole vector. The textbook has a value of 1.84 D for the molecule and 1.5 D for each bond. It is clear that they do not add up like scalars.
Relating the bond dipole moment to the molecular structure
The bond lengths in water are about 95 pm. The textbook gives the bond dipole as 1.5 debye. We can back-calculate what the charge is, given that separating a proton and an electron (with elementary charge $q_e$) by 100 pm accounts for 4.8 debye.
$$\pu{100 pm} \cdot q_\ce{e-} = \pu{4.8 D}$$
$$\pu{95 pm} \cdot q_\ce{OH} = \pu{1.5 D}$$
$$\frac{q_\ce{OH}}{q_\ce{e-}} = \frac{\pu{1.5 D}}{\pu{4.8 D}} \frac{\pu{100 pm}}{\pu{95 pm}} = 0.33$$
So the charge separation along each O-H bond is about a third of an elementary charge.
Could we calculate the charge separation?
Here, we assumed point charges at the location of the atoms. In reality, charge is distributed throughout the electron cloud. You can do quantum mechanical calculations to get the electron density, then express the charge separation as point charges. The easily accessible tool molcalc.org does this calculation (quick and inaccurately), and assigns the hydrogen atoms a partial charge of +0.21 each (so -0.42 for the oxygen atom). This is quite a bit lower than my calculation. On the other hand, the dipole moment is given as 1.88 D, close to the values given in the textbook. I am not quite sure were the discrepancy lies.

So how can we calculate dipole moment of water for example using this formula?

Yes, if you know the structure of the molecule and the partial charges (as point charges in the location of the atomic positions).

@Karsten Theis, Could you mention which textbook is this? Thanks.

@M.Farooq Now that you mention it, it is not strictly a textbook. It is Libretext supplementary material written by Mike Blaber. Here is the table of contents of that section: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties

Is there a difference in interpretation of dipole moment in Physics and Chemistry?

Addressing the above part: The definition of dipole moment is the same in physics and chemistry. However, the chemist's dipole moment has opposite directions as compared to the physicist. Which one is right? Of course, the physicists draw it correctly, which is consistent with electrostatics. There was a nice single paged article in the Journal of Chemical Education on this topic. A must read.
Misconceptions in Sign Conventions: Flipping the Electric Dipole Moment
James W. Hovick J. C. Poler
https://pubs.acs.org/doi/10.1021/ed082p889

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BOD5 respirometric method
I operate in a wastewater treatment plant laboratory and I use respirometric method for BOD5 measurement (hach direct plus). Recently I am obtaining high value for BOD (biological oxygen demand) for example BOD5 581 mg/L when the COD (chemical oxygen demand) of the same sample 895 mg/L. I did multiple assays on another sample and I obtained different values for the same sample using the same initial volume (209 mg/L and 586 mg/L while COD 630 mg/L).
I thought that some sensors are not functioning properly. I want to know if these pressure sensors can be calibrated, anyone can help please??

You might want to provide more information on what specific sensors you are talking about (make and model).

@Buck Thorn I did, thank you for notifying

Just to confirn, you mean this: https://uk.hach.com/other-instruments/bod-direct-plus/family?productCategoryId=51137362617 ?

A few simple questions just in case: 1) have you read the accompanying manual thoroughly and checked the troubleshooting section?  2) have you followed a standard protocol, if so, what is it? 3) have you checked that your sensor is not too old (past its nominal useful lifetime as provided by the manufacturer)?

@Buck Thorn, yes it is this one. I read the manual and there is nothing about this problem. All they said about the sensor's maintenance and care is to wipe them with dry cloth and that's what I was doing... I am following the procedure written in the manual. I am using the sensors from about 2 years...

I am reading the following manual: DOC342.52.90554 . It  would be good to know exactly what sensor you're using (part number etc). This one (https://www.hach.com/hq40d-portable-biochemical-oxygen-demand-bod-meter-laboratory-kit-with-luminescent-do-sensor-and-bod-bottles-1-m-cable/product?id=7640513827) for instance keeps track of how long you've used it and "remaining life".

On page 7 of the manual there are some potentially useful warnings (I don't know how experienced you are with the device).

@Buck Thorn yes the manual you are reading is the true one. I think I am experienced with the device because I am using it from about 2 years in 2 different laboratories and I am facing this problem only with one of them. I took a sample to the second lab and used the second equipment and obtained better results. I am just asking for someone who used this equipment before to see if there is any calibration method for the sensors. Anyway thank you so much for your help

That's fine, it's important to know as much as possible about the problem in question to give a proper answer. Every device should be calibrated, but to be honest that doesn't sound like the problem. The problem, as I gather from what you wrote originally, is that you observe enormous variability between measurements. That is a precision problem, not accuracy. Calibrations of the sort you might carry out with such a probe usually aim to improve accuracy.

In any event, if there is a standard method to calibrate the device it would be specified in the manual or the literature, and I would not recommend using a third parties recommendations, unless you don't worry about SOPs, GLP etc. In addition you have not specified which exact probe you are using, only the readout device.

@Buck Thorn, many thanks for your help, I contacted the supplier and have been told that the sensors can't be calibrated maybe the problem is in the direct plus apparatus. Anyway I will send the equipment to the supplier and they will work to fix it. Thank you for your help.


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Primitive Technology - Production of NaOH
We need to produce NaOH with primitive Technology. This is Part 1 of my Gilligan's Island Challenge series (Part 2). You are stuck on an island with access to a beach, a forest, a river, a lake, and a surface vein of coal.
I realize it can be made from $\ce{NaHCO3 + Ca(OH)2}$, and a person in a primitive technology scenario would be able to make some soda ash, but is there another way? $\ce{NaHCO3}$ might be tough to acquire in substantial quantities without access to a trona mine.
Would adding $\ce{NaCl}$ to a highly basic solution of, say, lime cause $\ce{NaOH}$ to form? If so, could it be extracted? Assume electricity is hard to come by.

No. Why would it? CaCl2 is soluble just fine, hence no reaction.

How about sodium sulfate instead of NaCl? The wikipedia article on it shows it occurs naturally at various locations.

Ca(OH) 2 would be pretty easy to make, could that be substituted?

I like this Gilligan's Island Challenge! This is an important question: without lye, there would be no lutefisk or hominy! Lye, especially KOH, but also some NaOH, was traditionally obtained by leaching wood ashes. As stated in wiki (https://en.wikipedia.org/wiki/Lutefisk), birch ash "contains high amounts of potassium carbonate and bicarbonate".

@EdV - Wow! That is exactly the answer I was looking for.


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Why is the iron inter-system crossing non-negligible at high temperatures?
Inter-system crossing (ISC) is a radiationless process involving forbidden transitions between singlet and triplet state. These are more common with heavy atom molecules, so I understand that ISC is dependent on the atomic number.
How does the temperature influence the ISC? Is this related to activation energy? Is the ISC rate coefficient temperature-dependent?

Nothing forbidden about the ISC. The direct radiative transition from and to the triplet state is forbidden, hence its long lifetime (phosphorecence).

@Karl It is a forbidden transition in the sense that purely electronic selection rules don't allow such transitions. to OP: Intersystem crossing needs an overlap between vibrational/nuclear wavefunctions as well as the multiplets energy should match, therefore the termal occupation of the vibrational levels of the given electronic states will influence the transition probability.

@Greg Well its not even a real transition, because there is no energy transfer. ISC occurs into a vibrationally excited state, and is directly followed by lattice relaxation into the triplett ground state. The picture of a the ISC being "forbidden" and therefore having a low probability is very misleading.


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Electron Spin, and the Stern-Gerlach Experiment
I think I understand the basics of the electron spin. However, I'm trying to figure out how to tell if, when shooting a beam of atoms (the Stern-Gerlach experiment), how to tell if an element will have some atoms deflecting up and some down?
I initially thought to draw out the electron distribution of each possible element (it's a multiple choice question) and then try and figure out which ones were spinning up and which were spinning down. But this doesn't align with the answer to the question. Any guidance would be appreciated?

Your question is a tad unclear. It might help if you include the original question from the source in some way.

No worries, just added

That's a terrible question. What is upwards/downwards wrt the orientation of the field? Is there an accompanying drawing?

I'm a bit unclear myself. It's for a class, and we only touched on this experiment briefly, when we were introduced to electron spin. So I assume I must do something with the electron spin configuration of these elements? Maybe if the electron is spinning in a certain direction it impacts the deflection? But then that doesn't make too much sense because it says the whole atom is deflected

There are no such things as "spinning up", "spinning down", or "spinning in certain direction". On the other hand, the electron distribution of each element is indeed required. It will tell you which of the atoms have non-zero spin. Those will get deflected, and the rest won't.

Oh actually, I just noticed there is a picture in my lecture notes, and I think it means the positive spin means upwards deflection. But when I wrote out the Si configuration, the final shell just has two electrons spinning in the same direction? So I'm not sure how there would be variation in deflection

Okay Ivan, but the question says some atoms will get deflected UP and some DOWN, so how do you differentiate which go up and which down?

We don't. Nor does the question ask that.

Ohh, I just noticed the other elements have full outer orbitals. So is it just the fact that Si happens to have a net spin, and therefore it will be deflected either up or down? I was just confused because the question says which element will be deflected up and down, so does the spin just mean it could get deflected in general, and you can't predict the direction?

I misread the question - as Ivan points out the question is "which atoms are deflected at all"

That's right. Some atoms of Si will get deflected up and some down. The elements with zero net spin won't get deflected at all.

Thank you both, that makes a lot more sense. I think I was too focused about the direction, rather than just being deflected in general. I thought maybe some elements only deflected up or something, but that was quite silly in hind sight.

I was wondering why only electron spin and not kernel spin is considered in spin detection experiments. If kernels have odd proton or neutron number, they have nonzero spin. And even neutrons have magnetic momentum. Stable silver kernels have 47 protons and 60 or 62 neutrons.

@Poutnik The interaction between the nuclear spin and magnetic field is about a factor of 1000 weaker than the interaction between an electron spin and a magnetic field and one would need a very strong magnet to be able to observe a deflection.


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Temperature measurement using thermometer
I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different? Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this? If it is independent of this, then how thermometer works?
Please provide me the simplest answer as I am in class 9 only.
As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

The unit of temperature is Joule all right. The ambient temperature in my place is about $6.13\cdot10^{-21};\rm J$; what's yours?

Now to the first part. You seem to know that the temperature has something to do with average kinetic energy. If the temperature of two samples is the same, then so is their average kinetic energy; what's unclear about that?

Okay. Let me consider a gas which has the density of 2 mol/22.4L. Its temperature recorded by my thermometer is 60°C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is same??

I have the above doubt that temp. should vary in both these forms of gases because in first case, many gas molecules were hitting the bulb of my thermometer and when density was reduced, only few molecules hitted the bulb which should have reduced the recording.

@user36956 Temperature is the statistical parameter, implying particles are in frequent contact and mutually in equilibrium. that does not apply to 10 atoms in 22.4 L. 1 particle does not have temperature, nor do 10 particles.

So, how should i define temperature so that no such doubts come up!

Then what temperature actually is?? There is no doubt that 10 particles too have kinetic energy. Then why no temp..?

When the samples are, say, 1L and 2L, they both are infinitely big for our purposes, and hence equal. When the smaller sample is about as small as the thermometer's bulb, you'll start getting different readings.

So, if i have 10,000000 particles in 1 L and 10 particles in 1L, temperature is going to be same or different?? (I mean the readings)

Frequency of their collisions must be much greater then frequency of their collision with the bulb. For molecules of air, it is typically 10 billions collisons per second. How often collide 10 molecules in 1 L ? 1 collision in many years, just guessing.

Frequency of their collisions must be much greater then frequency of their collision with the bulb..... For what?

For temperature to have ever sense to be defined.

Hmm... Many years to collide. But the temperature will be recorded in that much time, i think!

I think i need to understand the working of thermometers i.e when those particles collide with the bulb, what happens and how temp gets recorded. I shall be thankful to you Ivan or Poutnik if you could explain me this or provide me a link.:)

It would be temperature of termometer, not temperature of the gas.

So, temp of thermometer bulb= temp of gas, only when enough particles collide with bulb to make the average K.E of bulb particles equal to average K. E of the gas particles.

I.e small difference in volumes makes no change in temp readings. The reading differs only if particles those colliding with the bulb gets reduced to a large extent.  Am i right here??

Let us continue this discussion in chat.

I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different?

The total kinetic energy is different. The average kinetic energy per particle is the same.

Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this?

In your specific example, the same number of particles hit the bulb because it just depends on the area of the bulb, as long as it is covered in water. Not only is water hitting the bulb, the atoms in the bulb are also hitting the water. After they have done this for a while, water and bulb are at the same temperature.

If it is independent of this, then how thermometer works? As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

For the example with the two gases at the same temperature but at different pressure (i.e. particle density), the situation is a bit different, but the thermometer still works. There are a different number of "hits" in this case, but if you wait long enough, the gas and the bulb will be at the same temperature.

Please provide me the simplest answer as I am in class 9 only.

I tried, but if the question addresses a complicated phenomenon, the answer can't be super-simple.

[From the comments:] So, if i have 10,000000 particles in 1 L and 10 particles in 1L, temperature is going to be same or different?? (I mean the readings)

Ideally, there should be more sample than thermometer. This means that no matter what the temperature of the thermometer was to start, it will have the temperature of the sample afterwards. With only 10 particles in the sample, the 10 particles will take on the temperature of the thermometer (and the enclosure), so you will be unable to measure the initial temperature of the sample.

[My own question] If the sample keeps hitting the thermometer, why doesn't it get hotter and hotter?

As I said above, the thermometer also keeps hitting the sample. There is some net transfer of energy while the temperatures are different (from hot to cold, warming up the colder and cooling down the hotter). Once the temperatures are the same, sample and thermometer keep "hitting each other", but the temperature of both stays constant (if the entire setup is insulated against the surrounding).

One more question- why my thermometer measures only the avg. translational kinetic energy of particles, but not rotational or vibrational one?

The short answer:
If 2 samples of water of the same temperature is considered,the average kinetic energy of water molecules of samples A nad B are the same. But, within the same water sample, water molecules have different kinetic energy with a particular statistical distribution.
Even if I had set by some magic wand the same energy for all water molecules at this moment, the energy would be redistributed back in nanoseconds to their favourite distribution.
The longer answer:
Temperature is not average kinetic energy of particles, but it is a measure of average energy per "degree of particle" freedom.
As the degree of freedom - imagine yourself it is a way the particle can move.
Imagine an oxygene molecule, consisting of 2 atoms.
It can independently move along 3 perpendicular axis x,y z -> 3 degrees of freedom.
Its chemical bond can vibrate - 1 degree of freedom.
It can rotate around 2 perpendicular axis - can vibrate - 2 degrees of freedom.
Each degree of freedom has average energy proportional to temperature(*).
The  proportionality constant in $E = k. T$ is the Boltzmann constant about $\pu{1.38 J/K}$.
If 2 macroscopic objects with the same temperature are in contact, their average energy per a degree of freedom is the same. Their particles exchange energy, but as the average is the same, the net exchange of energy is zero and objects stay at the same temperature.
(*) - It is not fully true if we consider quantum effects of quantization.  Molecules are allowed to have just particular discrete values of rotational and vibrational energy. Translational energy does not have this quantization ( at least not in steps measureable by curent methods ).
(**) - Not fully true, but for not it is.

Didn't get anything at all! I expected some easy basic answer.

It was easy, basic answer. :) Try this article on the Simple version of wikipedia. https://simple.wikipedia.org/wiki/Temperature    Also, try some textbooks or online courses.

Hmm... Tried but it didn't answer my question.

@user36956 I have updated the short answer.


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how is there no centre of symmetry in this case?
the question was whether the following compounds have the center of symmetry or not .the question was :

And the answer is that there is no center of symmetry .how? according to me, there should be the centre of symmetry. I have made it like this:
on asking my teacher how there is no center of symmetry he told me that since it is a Fischer projection, therefore, the horizontal lines are above the planes. therefore there will be no center of symmetry. I understand that if the horizontal lines are out of the plane then there will be no COS. but what I don't understand how come the horizontal lines are above the plane/out of the plane. please explain.

Paper is 2D. But real world is 3D. Models are like that.

obviously! but how come he know or we will know that all the groups present on the horizontal line are out of the plane and not few of them are inside and few are outside the plane?

The  angle is supposed to be tetrahedral, it is not possible on a single 2Dplane.

Problem 1: understand what a Fischer projection represents. Problem 2: if you still don't get it, build a 3D model.

One way to quickly get a visualization without building a 3D model (having a model is more instructive, though):

Go to https://chemapps.stolaf.edu/jmol/jsmol/simple.htm
Click on Load by name and type "(2S,3S)-Butane-2,3-diol"
Click on console and type "rotate branch @2 @4 180" to get a conformation that is easy to compare with the Fisher projection.


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What is a reliable way of writing significant figures in multi-step problems?
I understand the sig fig rules that pertain to addition and multiplication problems, however, in multi-step problems, I have not been able to find a systematic approach to getting the right amount. Everybody who I have asked just said "it's tricky so be careful".
If the problem is: (5.01 + 4.1) / (1.00 x 1.00)
Is the answer 9.11 or 9.1? The addition part indicates 2 sig figs because the tenths place is the last accurate decimal, while the multiplication part indicates 3 sig figs. Does it depend on whether the result inside the parentheses is added/subtracted or multiplied/divided from the other result?
What about a problem like: (2.00 x 30) + 5.01 + 2.0
Is the answer 60. or 67.0? The multiplication indicated the former, the addition indicates the latter.
Problems where the numerator has numbers being added and the denominator has numbers being multiplied like: (5.01 + 2.0 - 3) / (60.01 x 3.0) are even more confusing, especially when scientific notion is mixed in. I want to know an approach that goes step-by-step and works 100% of the time.

Related: Significant figures addition vs multiplication

First, note that there is no statistically correct way a single number can represent both a best estimate plus its error, so the rules for significant figures are very useful in practice, but they are neither perfect nor without some exceptions. For your first problem, the answer is 9.1. For your second question, is 30 an integer or is it 30. (2 sig. figs.) or is it 3. times 10 to the first power (1 sig. fig.). These kinds of ambiguities are yet another good reason to use scientific notation. Finally, forget working 100% of the time: not happening!

For physicist John Denker's 100 page discussion of "Uncertainty as Applied to Measurements and Calculations", see this: https://www.av8n.com/physics/uncertainty.htm . Personally, I think he goes way too far in his criticism of significant figures, but I post this to show that not everyone agrees about significant figures. Good for starting arguments, though. ;-)

In the second question, 30 is an integer. If I wanted to represent it as having 2 significant figures, I would write 30. At least, that's what i'm assuming you're "supposed to do" to represent 2 significant figures and hope my exam does. Should scientific notation be used in every single possibility where they can be used? Should I never write 30, but instead write 3 x 10^1?

Unfortunately, people usually write 30 (which looks like the integer) when they really mean 2 sig. figs. and hope that context will resolve ambiguity. But it is a problem. Scientific notation is best and 30. is fine if (and it is a big if) the decimal point is not mistaken for a "full stop". And notice that 30. at the end of a sentence causes the issue to pop up again. All in all, very annoying, especially if points are lost on an exam or homework.

One last thing: intregers have infinite significant figures. So if 30 in the second problem was as integer, then 30 = 30.000... = 29.999.... and the ellipsis dots means goes to infinity. So 10 (the integer) times 3. has 1 sig. fig., 30. has 2 sig. figs., 30.0 has 3 sig figs. Then on to infinity, where 30, the integer, has infinite sig. figs.

For the following, I take 3 to have one significant figure, 67 to have 2 significant figures, and 100 to have one significant figure. If you want to avoid confusion with exact integers (such as a stoichiometric coefficient), use scientific notation for numbers lacking a decimal point, as in $\pu{0.3e1}, \pu{6.7e1} \text{ and }\pu{0.1e3}$.
Pencil and paper
The step-by-step approach is to consider each step separately (observing the correct order of priority), keeping track of significant figures of the intermediate results. Then, calculate the number without any rounding of the intermediate results, and record it with the appropriate number

(5.01 + 4.1) / (1.00 x 1.00)

The sum will be significant to a tenth, so 2 significant figures.
The product will have 3 significant figures because that is the minimum of the factors
For the division, the numerator has 2 significant figures and the denominator has 3, so the result has 2 significant figures.
The arithmetic result without any rounding is 9.11, so you write down 9.1.

(2.00 x 30) + 5.01 + 2.0

The product has one significant figures, least significant figure is the tens.
Adding 5.01 will have least significant figure as the tens.
Adding 2.0 will have least significant figure as the tens.
The arithmetic result (unrounded) is 67.01, so you write down 70 or $\pu{0.7e2}$.

(5.01 + 2.0 - 3) / (60.01 x 3.0)

The sum of three numbers will be significant to the ones, giving one significant figure.
The product will have 2 significant figures because 3.0 does too.
The overall answer will have 1 significant figure because the numerator does too.
The arithmetic result (unrounded) is 0.02227406543..., so you write down 0.02.

Using a tool
I wrote a calculator in a browser that knows about significant figures (and units), described in J. Chem. Educ.2015, 9211, 1953-1955. For the first example, it gives the following output (input is:"(5.01 + 4.1) / (1.00 * 1.00)"):
$$\mathrm{result} = \dfrac{5.01 + 4.1}{1.00 \cdot 1.00}$$
$$\ \ \ =\dfrac{9.11}{1.000}$$
$$=9.1\ \ $$
The intermediate results are written with one extra digit ("guard digit") so that no information is lost (in the pencil and paper section above, I kept all digits until rounding at the very end). If you write down and then reuse intermediate results, you should follow that practice (unless your teacher tells you otherwise). The calculator is available at http://pqcalc.pythonanywhere.com/. Behind the scenes, it uses a slightly more sophisticated error propagation model, so there are times when it shows a digit more than you would expect. Also, different from most textbooks, the calculator interprets numbers without decimal points as exact integers, and numbers with a trailing decimal point (67. or 100.) as significant to the ones position.

For the second question you wrote "the product has two significant figures, least significant figure is the ones", but because 30 only has one significant figure, shouldn't the intermediate result have 1 significant figure as it is the minimum of factors? How does the number of significant figures go from 1 to 2?

@AlbertGoldman I'll fix it.

Oh that makes a lot more sense. My rounding was off. Thank you.

The valid digit number approach is simplification of evaluation of absolute and relative errors of values.
Digits
For addition/subtraction, the order (position) of last valid digit of the result keeps the higher of the orders of the last valid digits:
$$12.3 + 4.56 = 16.9$$
$$12.34 + 4.56 = 16.90$$
For multiplication/divisions, the number of valid digits is minimum of the number of valid digits
$$2.0 \times 2.3 = 4.6$$
$$2.00 \times 2.3 = 4.6$$
$$2.00 \times 2.30 = 4.60$$
Errors
The last valid digit of the result should match the order of the first valid digit of estimation of standard deviation or confidence interval.
E.g. $10.2 \pm 0.1$
Only for big number of measurements can be SD known with 2 valid digits,
like $10.23 \pm 0.45$
Errors of values follow additiveness of squares of standard deviations.
Absolute ones for additions/subtractions,
$$s^2_{A+B}=s^2_{A}+s^2_{B}$$
relative ones for multiplications/divisions.
$$s^2_{r,A+B}=s^2_{r,A}+s^2_{r,B}$$

I am not given standard deviation in pure math problems. If there is some way of determining standard deviation or "confidence level" with just math, I'd like to know it. It would be helpful if you looked at my example problems and tell me how to think about them.

You can always take 5.01 as 5.01 +- 0.01. It would have more valid digits with better precision.


text:
What is the the total heat extracted in the process?

Since this is a cyclic process:- Change in internal energy=0.
In the process C-A the work done W is zero, thus the heat Q is zero since it is an isochoric process.
Next in the process A-B, I got the heat Q to be equal to 0.5 L-atm and Work done W equal -0.5 L-atm . (Area under the graph and expansion of gas gives negative work done here)
For the process B-C its given Work done W is 1 L-atm (Positive since compression) and since change in internal energy =0, the heat must be negative of work done thus Q = -1 L-atm
Therefore the total heat exchange corresponding to each process should be -1+0.5+0=-0.5.
The absolute value being 0.5 L-atm.
But the answer gives 1.5 L-atm as the total heat exchange.
How? Please explain

The change in internal energy for each step of the process is not zero; just the overall process.  In CA, the change in internal energy is negative, as is Q.  Also, apparently in this problem, the criterion that work done on the system is considered  positive is being used.

Since the change in internal energy for the cycle is zero, if you are using the form of the 1st law written as $\Delta U=Q+W$ (where work done on the system is regarded as positive), the heat for the entire cycle must equal minus the work for the entire cycle.  The work done on the system in AB must be -(1)(1.5-1)=-0.5 L-atm.  Therefore, the work done on the system for the entire cycle is 1-0.5=0.5 L-atm.  So, the heat transferred to the system for the entire cycle must be -0.5 L-atm.

Exactly, thats what I coclude as well! But the given answer is +1.5 L-atm

There are multiple issues with this problem statement.  Our answer is correct and the "given answer" is incorrect.


text:
Primitive Technology - Extract Silica and Sulfur from ash
Given some plant ash containing silica and sulfur, we need to find a way to extract the silica and sulfur using only resources available to a primitive person stuck on an island, with access to a forest, lake, and coal vein. (This is part 2 of my Primitive Technology series. Part 1)
I've imagined a way to extract the silica, but I am not sure if it would work. Can I get this reaction confirmed possible?
$$
\begin{align}
\ce{SiO2(ash) + NaOH &->[Heat] Na2SiO3 + H2O}\tag{1}\\
\ce{Na2SiO3 + 2 HCl &->[Heat] 2 NaCl + SiO2 + H2O}\tag{2}
\end{align}
$$
I am unsure whether the sulfur content could be feasibly extracted in this scenario.
Notes

Some plant parts, such as rice husk, have high silica content (over 90%).
Coal and sea water typically have substantial sulfur content. Sulfur is a by-product of coke production.

For making sodium sulfate, heat gypsum hot enough to drive off all the waters of hydration and sulfur trioxide, leaving behind CaO. Of course, the sulfur trioxide plus water yield sulfuric acid, which can then neutralize NaOH to give sodium sulfate. This would be hard to do, for a number of reasons. Still, compared to the discovery of phosphorus, it does not look so bad.

Thanks for your reply about using gypsum. I guess maybe it's not feasible to extract the small amount of sulfur contained in plant ash.


text:
Provide a condensed structure for the following compound

I keep getting, (CH3)3CCH2CH2CHNH2CHCHCH3. However, the homework system keeps counting it as incorrect, and I don't know what part is incorrect. Thanks for the help!

Well, i can see at least one other option, but as yours is also correct, that's rather exercise in futility.

You are probably missing a set of parentheses. For compounds like this, the NH2 group should be in parentheses to show that it is not part of the chain.

$\ce{(CH3)3CCH2CH2CH(NH2)CHCHCH3}$.
However, Mithoron's comment makes a valid point. There are lots of ways to do this. If your homework system only codes one as correct, then that is frustrating. Better to talk to your instructor about the specific question.

Thanks for the help! I tried the parentheses and it still marked it wrong. I'll go ask my instructor on Monday.


text:
is 6-benzylhexanoic acid the same as 7-phenylheptanoic acid?
is 6-benzylhexanoic acid the same as 7-phenylheptanoic acid? If the two compounds stated are equal, what's the more preferred IUPAC name?

Draw their structure for yourself and you will know, if they are equal.

The two molecules below are the same, with just an alternative naming:
$$\ce{(Ph-)-CH2-CH2-CH2-CH2-CH2-CH2-COOH}$$
$$\ce{(Ph-CH2-{})-CH2-CH2-CH2-CH2-CH2-COOH}$$
The former is preferred, as the benzyl substituent does not have sense to be used, if attached to the last carbon of the main chain.


text:
Is cyclo[18]carbon aromatic?
How does this compound even exist? Having such a massive ring strain.
Is it even conjugated? As it's resonance structure will have all double bonds.
Since it was reported to be found, there must be some stabilizing factor. I was wondering whether that might be aromaticity, since it does have 4n+2 electrons.

Usually, when you're reading on bleeding edge chemistry, it's safer to assume the compounds are not stable at all. If you read the article more closely, you'd see that it speaks of interstellar space and very low temperatures, and only a few of the molecule were ever spotted.

The compound doesn't quite exist. Then again, the ring strain isn't all that massive (think cyclopropane).

https://chemistry.stackexchange.com/questions/119283/why-does-the-image-of-cyclo18carbon-look-like-a-nonagon


text:
Is it important to use arrows as bonds in writing coordination compounds?
Why are the coordination compounds representated by arrows? Is it necessary to write every coordination compounds with the arrow side of the bond facing the metal ion and the other part the ligand?

Why dont you check what your textbook or one of IUPACs colourful books dictate on the matter?

Related: Is there any way of representing a dative/coordinate covalent bond?

Possible duplicate of Is there any way of representing a dative/coordinate covalent bond?

Historically, coordination compounds have been depicted in various ways.
According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), bonds representing coordination from one atom to a single other atom should be represented as normal plain single bonds. The use of dative bonds (i.e. arrows pointing to the central atom) to represent coordination is not acceptable.


text:
Basic concept of the first rule of thermodynamics
There is a cylinder-piston system with some gas in it, and let's say this system is a closed system.
Say that the system is submerged in a large beaker of water whose temperature is $25\ \mathrm{^\circ C}$. A spark is used to trigger a reaction in the system, and after the reaction is completed, the temperature of the water bath increased to $28\ \mathrm{^\circ C}$, and the piston went inside the cylinder.
By instinct, I think that heat is lost by the system since the surrounding(water)'s temperature increased, and work has been done to the system. However, when I consider the thermal equilibrium, the system's temperature must be $28\ \mathrm{^\circ C}$ as well. In this case, is heat added to the system or not? Furthermore, in the equation $\Delta E = q+w$, what are the signs (plus or minus) of the three terms?

You have a system and a water bath, both at 25°C before and 28°C after. I would definitively say q is negative, wouldnt you. And you have done no work on the outside, on the contrary, the outside has pushed in the cylinder, so w is positive. Whats your guess on the sign of deltaE?

@Karl Maybe negative, but we don't know for sure unless the actual numbers are goven, right?

Thats it, exactly.

Chemical energy of system has been converted to thermal energy of both  the system and the surrounding water.
Additionally, water did work on system, as the system shrinked down.
$$\Delta E_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}-p \cdot \Delta V_\mathrm{system}$$
For $H=E + p \cdot V$,
$\Delta H=\Delta  E + p \cdot \Delta V + V \cdot \Delta p$
As p is constant,
$\Delta H=\Delta  E + p \cdot \Delta V $
$\Delta H_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}$
where $H$ means enthalpy, energy at the constant pressure.
If we considered other thermodynamic quantities like Gibbs energy or entropy, we would need to examine the underwent reaction and system composition. But if the system energy is considered, its change is related just to exchanged thermal energy and work.
$q$ in the equation $\Delta E = q+w$ means heat transfers to the system, so it is negative. $w$ is work done on system and is positive.  $\Delta E$ is very probably negative, as work value would be much less than transferred heat.

In my judgment, this answer is not 100 % correct and, in addition, does not discuss an important feature of the reaction.  See my answer.

I always leave space for other answers.

The first law of thermodynamics tells us that, for this process at constant pressure, $$\Delta E_\text{system}=q_\text{system}-p\,\Delta V\tag{1}$$and $$MC_\mathrm s\,\Delta T_\mathrm w=-q_\text{system}\tag{2}$$where $\Delta V$ is the change in volume of the system.  Eqn. 1 and 2, in terms of system enthalpy, can also be combined and written as $$\Delta H_\text{system}=q_\text{system}=-MC_\mathrm s\,\Delta T_\mathrm w\tag{3}$$So there is no $p\,\Delta V$ in this equation.
What is happening in this system is that (1) the reaction is exothermic so that heat is transferred from the system to the water and the water temperature rises and (2) the total number of moles of gas decreases in the reaction so that the volume of the system decreases and the piston moves inward.

So, what about the signs $\Delta E$, $p$, and $w$ of OP's equation in this instance?

@Mathew Mahindaratne The op is using the convention that w represents work done on the system.

I would like to address one of the OP's questions about heat transfer directly because the existing answers did not (yet):

[OP:] I think that heat is lost by the system since the surrounding(water)'s temperature increased, and work has been done to the system. However, when I consider the thermal equilibrium, the system's temperature must be 28 ∘C as well. In this case, is heat added to the system or not?

The short answer is: no heat is added to the system; the system gives off heat.
Heat is a transfer of energy from one place to another. So you can't "add heat" somewhere without "losing heat" somewhere else. In a heat transfer across a boundary, $q$ is positive for one side of the boundary and negative (but of equal magnitude) for the other.
The surrounding (in this case the water bath) is often chosen to be "boring" (no chemical reactions, for sure), while the system can have all kinds of stuff going on (a spark, a chemical reaction, large fluctuations in temperature). This way, you can measure the heat transfer via temperature changes in the water bath.
In the case at hand, both system and surrounding show an increase in temperature after things have calmed down. That tells us that some other form of energy (potential energy or specifically "chemical" energy in this case) has been transformed into thermal energy. This took place in the system (the surrounding is too "boring" for that).
To get back to the question I wanted to address:

... is heat added to the system or not?

The thermal energy of the system increases because of the exothermic chemical reaction. Some of that thermal energy is lost to the surrounding through heat transfer, warming up the surrounding. You could imaging that the temperature of the system first went up a lot (if the reaction is rapid), and then cooled down to 28 $^\circ$C, heating up the surrounding from 25$^\circ$C to 28 $^\circ$C. Or you could skip the details and just compare the initial and final state and not worry what happened in between, in the spirit of thermodynamics. The conclusion is that heat is transferred from the system to the surrounding. Nevertheless, the temperature of the system increased because of the exothermic reaction, which generated more thermal energy than the system gave off. In other words, no heat is added to the system; the system gives off heat.
In this explanation, I have set aside the work done on the system. The OP is vague about whether the process runs at constant pressure or not, and we don't know the magnitude of the work compared to the heat transfer; for my answer, you should assume that the magnitude of the work is negligible compared to that of the heat transfer, otherwise some of the statements might become inaccurate (such as "the exothermic reaction, which generated more thermal energy than the system gave off").

The problem is that in your last wondering, i.e. "wouldn't the system be at a higher temperature because it's 28 C?" is that you're confusing the system and its environment. The system here is the cylinder full of explosive chemical, the environment is the surrounding water. Note that technically, though, there is no reason that you can't call the surrounding water a "system" and the cylinder an "environment" - it sounds weird to our ears but it works out in theory, it's just that usually an "environment" is much larger/more complex than we can describe in detail in real cases. The real trick is in being consistent in your identification of such and such throughout.
In the thermodynamic equation
$$\Delta E = Q + W$$
the $E$ referred to here is the energy of the system, its internal energy (sometimes also denoted by $U$). Hence, $Q$ is the heat input to the system, and likewise, $W$ is the work input: to see this, consider the case where one is zero and the other nonzero, to leave only one to consider. Now, if $\Delta E$ is positive, then the nonzero one must be positive. Moreover, since this represents an increase in the system energy, that implies that both $Q$ and $W$ measure inputs, and their sign comes to be thus.
The energy change in the environment, e.g. $\Delta E_\mathrm{env}$, is, then, just the negative of this (alternatively, just exchange the roles of system and environment as mentioned before), in order to make it balance.
Hence, in your particular case, since the system both deposited heat into the environment and did work on it, for the system, $Q$ and $W$ are both negative: negative heat/work "received by" the system is equivalent to such done by the system. But in the case of the environment-as-system, its $Q_\mathrm{env}$ and $W_\mathrm{env}$ will then be positive, since it is receiving both from the cylinder.


text:
Units of the rate constant for reactions whose order is greater than 1
In many textbooks, it is written that:

The value of rate constant depends on the nature of the reactants, temperature and catalyst. It is independent of the concentration of the reactants.

However, the unit of the rate constant is
$$\left(\frac{\mathrm{mol}}{\mathrm{l}}\right)^{1-n} \mathrm{s}^{-1}$$
where $n$ is the order of reaction.
Therefore, except for first order reactions, the unit of the rate constant will have a concentration term $\displaystyle\frac{\textrm{mol}}{\textrm{l}}$ for all other reactions. Does this contradict the statement given in books?

So what if it will have a concentration term?

Think of it like the equilibrium coefficient, the units don't realllly matter. Still use them to check your units and stuff.

@EashaanGodbole I think units matter. However, you can't tell from the units what a quantity depends on. Let's say I drive with a constant speed of 100 km/h on the freeway. Just because the units have the length-unit km in them does not mean my speed depends on how far I've already driven (or how long, for that matter). The idea of constant speed is that the ratio of distance and time is constant, even as time passes and you move along.

@Ivan Neretin So, rate constant does not depend on the concentration of reactants for all type of reactions including pseudo first order reactions?

@KarstenTheis I wasn't saying units don't matter in general--Units are very important and non-neglectable for most cases. I'm just saying they may not hold as much importance in this specific scenario.

As an example if the reaction is $\ce{A + B -> C}$ the rate expression for the appearance of $\ce{C}$ is
$$\frac{\mathrm d[\ce{C}]}{\mathrm dt} = +k[\ce{A}][\ce{B}].$$
The units have to be the same on both sides of this equation, these are concentration/time on the left and to be the same on the right the rate constant $k$ has to have units $(1/\text{concentration})(1/\text{time})$ usually expressed as $\mathrm{dm^3\,mol^{-1}\,s^{-1}}.$
It is true that the numerical value of the rate constant usually depends of several things such as temperature, but the units remain the same for each type of rate constant, 0th,  1st, 2nd, 3rd simply because the rate equation units have to balance.

No, units does not remain the same for each type of reaction. Here you have assumed the order of the reaction w.r.t. both 'A' and 'B' as 1. So net order n= 1+1= 2.

@Apurvium You have probably mis-understood, I wrote that units have to be the same on both sides the the equation because the equation has to balance. By type of rate constant it is clear from what I wrote that all first order have the same units etc. etc.

But you have mentioned that "the units remain the same for each type of rate constant, 0th, 1st, 2nd, 3rd simply because the rate equation units have to balance", so I assumed that 0, 1, 2, 3 refers to order. This makes the statement incorrect!

No, all reactions of 1 st order have the same units for the rate constant  and similarly for other orders, which is what 'same for each type of rate constant' means.

OK. But my original query remains unanswered. Which is " Does rate constant depends on concentration of reactants for any type of reaction?"

No, you can see from the equation I give that $k$ is a constant independent of any concentration. The same is true for all other rate expressions excluding the pseudo rate constant that is sometimes used in special cases.

OK, so for pseudo first order reaction, the value of rate constant will depend on the concentration of reactant(s).

yes and only because one concentration is vastly in excess of the other and does not change by any realistic amount while the reaction occurs.

Suppose, in the given reaction by you, the reactant 'B' is in excess. Therefore, the rate law for this pseudo first order reaction will be dx/dt = k[A], neglecting  [B] since it will remain unchanged during the reaction. So, why the value of 'k' will depend on [B] ?

In a bimolecular reaction where the rate is $k[A][B] $ if B has such a high concentration compared to A that A can react without affecting B to any great extent, then $k[B]$ is effectively constant and we can write rate as $k[A][B] \equiv  k'[A]$ where $k'$ is a pseudo rate constant. All this will be in your textbook.


text:
How to convert P2O5 concentration to H3PO4 concentration?
Why are some elements such as phosphorus reported as $\ce{P2O5}$ concentrations when the materials have no $\ce{P2O5}$ in them. I would like an explanation of how the conversion is done between real life and this idealized state.
For example. the Wikipedia page on phosphoric acid states that a $\ce{P2O5}$ concentration of 70% is equivalent to nearly 100% $\ce{H3PO4}$. Can someone explain this conversion?

What's to explain? P2O5 reacts with water. Do you know what the product is? Can you balance the reaction?

okay I didnt realize the concentration was in reference to water because its often discussed as purity etc. makes sense thanks.

@IvanNeretin Im reading a paper which states energy used to process phosphoric acid is "6500 kwh to make a ton of phosphoric acid (P2O5)". I  want to express this in kWh/ton H3PO4. So I get 4710 kWh/ton H3PO4. Agreed? The paper is: A Technical Review of the Improved Hard
Process

It is a common practice to express content of the component A in the equivalent content of the component B. There is an older unit of the water "hardness" dGH ( degree of General/German Hardness ) where 1 dGH means equivalent of 10 mg CaO / 1 L. But there is no CaO in water. Similarly, in fertilizers, there is AFAIK historical custom to use content of oxides, even if there are none.

P2O5 + 3H2O --> 2H3PO4. Using molecular weights  P2O5/2H3PO4 = 142/198 = 0.71 = 71%

Your question is a valid analytical chemistry question which has deep roots in history. No, the reason is not that $\ce{P2O5}$ reacts with water to from phosphoric acid and that is why $\ce{P2O5}$ values are quoted as suggested in the comments below the question. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $\ce{K2O}$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $\ce{P2O5}$ but the convention to report elements as oxides continues.
Second part: I will let you solve the rest of the problem: Start from here
1 mol $\ce{P2O5}$ = 2 mol $\ce{H3PO4}$
and do wt percentage conversions to moles. A 70% $\ce{P2O5}$ solution in water means 70 g $\ce{P2O5}$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.

Thanks for the answer I believe I solved it based on first poster response as follows (in 1 kg). 700g P2O5 x 1 mol/ 142 g = 4.9 mol P2O5. 300 g H2O is 16.7 mol H2O.
4.9 mol P2O5 + 17.78 mol H2O --> 9.8 mol H3PO4
Mass of 9.8 mol H3PO4 * 98 g mol = 960.4 g H3PO4

Seems right, when dealing with percentage concentrations, it is better to use 100g as the unit. So using your numbers, it is 96% w/w phosphoric acid.

$\ce{P2O5}$ content is useful as it is the phosphorous content that a scientist often cares about when using phosphoric acids. This might seem silly given that as you pointed out anything from $0-70\% \ \ce{P2O5}$ could just be expressed as a percentage of $\ce{H3PO4}$ but there are times when a higher $\ce{P2O5}$ concentration is desired. For example polyphosphoric acids are useful reagents that have more phosphorous content that phosphoric acid. It would be ambiguous to state the concentration as $115\%\ \ce{H3PO4}$ (what does 115% mean) so instead procedures are reported in terms of $\ce{P2O5}$ concentration with the balance assumed to be water (for those in the know).


text:
Cycloalkanes and saturation
I thought molecules that have a ring have an added degree of unsaturation, but this explanation has confused me:

How can something like cyclohexane which has degree of unsaturation of 1, be considered to be "saturated"? Are there different meanings of saturation?

Saturated is an old term in organic chemistry, dating back when the atomic structures and electrons did not exist. The meaning of saturated, even today is "chemically combined to the maximum extent, having no ability to combine further." The carbon atoms in cycloalkane are saturated in the sense that all the four valencies of each carbon atom are satisfied by four individual atoms.
The degree of unsaturation is a related concept: It tells us how many rings or double bonds are in the molecule. Don't mix these two terms.


text:
How there are 7 enolizable hydrogens in 2‐methylcyclohex‐2‐en‐1‐one?

49. How many enolizable hydrogens are there in the following compound?

(a) 2
(b) 4
(c) 5
(d) 7

I know that enolate ion is the anion formed when an α-hydrogen in the molecule of an aldehyde or ketone is removed as a hydrogen ion.
In the above question I find only two α-hydrogens which can participate (denoted with red color):

But the answer is (d) 7.
Where am I getting wrong?

How about the protons on the Me group?

I don't see how the protons in the methyl group could be involved.

@J_B892 That's what TAR86 mentioned (γ not para position as ring isn't aromatic).

Enolizable with what? With a base such as hydroxide, the answer is 4. Two α- hydrogens and two vinylogous hydrogens at the γ-position.

I think there is a mistake in the question; the structure should be 3‐methylcyclohex‐2‐en‐1‐one.

Actually there are exactly 7 alpha hydrogen in the given structure.  Just look that there is also a double bond and it is also considered as functional group.  So, you should count alpha hydrogen w.r.t  double bond and they are 5 .
Finally 2 + 5 you get the required answer = 7

Look, that's just wrong. You can't deprotonate the methyl group and delocalise the negative charge onto oxygen.


text:
Concentrating sulfuric acid
Since calcium oxalate is 3000 times less soluble than calcium sulfate, would it be possible to concentrate sulfuric acid from let's say $\pu{50\%}$ solution using following steps? :

adding calcium carbonate to cold dilute $\ce{H2SO4}$
filtering off resulting $\ce{CaSO4}$
mixing $\ce{CaSO4}$ with oxalic acid in 1:1 molar ratio
adding just enough water to submerge the solids
vacuum filtering off the now concentrated (hopefully) $\ce{H2SO4}$

My reasoning behind this is that even though at any given time there's only a small amount of $\ce{CaSO4}$ dissolved, the reaction producing sulfuric acid would eventually go to completion with high yield due to it's the massive kinetic favourability. The only problem I see is that the amount of water needed would limit the final concentration. Would adding merely enough water to wet all the solid still work, improving the final concentration?

Calcium oxalate solubility depends on acidity, as the weak oxalic acid is protonated in highly acidic sulphuric acid.
Additionally, oxalic acid would be dehydrated if sulphuric acid was concentrated enough.
If we for convenience mark oxalate anion $\ce{(COO)2^2-}$ as $\ce{Ox^2-}$, then
$$\ce{CaOx <<=> Ca^2+ + Ox^2-}$$
$$\ce{Ox^2- + H+ <=>> HOx^-}$$
$$\ce{ HOx^- + H+ <=>> H2Ox}$$
$$\ce{  H2Ox ->[H2SO4] H2O + CO ^ + CO2 ^}$$
Therefore oxalic acid has no chance to create concentrated sulphuric acid.


text:
distilled water and copper(II)chloride
In class we did a lab of adding blue crystals of copper(II)chloride into distilled water, the water turned blueish green,were the changes chemical or physical?

Next time, you should write down your previous effort, what have you done and failed, to answer your question. If the question does not express your own effort, others may not be willing to answer, and may even vote for closing the question.

https://en.wikipedia.org/wiki/Copper%28II%29_chloride

Aqueous solution prepared from copper(II) chloride contain a range of copper(II) complexes depending on concentration, temperature, and the presence of additional chloride ions. These species include blue color of $\ce{[Cu(H2O)6]^2+}$ and yellow or red color of the halide complexes of the formula $\ce{[CuCl_{2+x}]^{x−}}$

On the Wikipedia page, you can see a photo of 3 solutions with different colour.


text:
How are there three possible tautomers of 2,2,4-trimethylheptane-3,5-dione?

The answer for the above question is three. But I am able to draw four structures:

Where am I getting it wrong?

The last structure does not really exist in slightly acidified solution of the given compound.

Coming to this question won't even cis and trans also contribute to Number of tautomers?

yes I was confused whether last structure is correct or not. but don't know the reason to eliminate it. why it doesn't exist in acidified solution? also if we make the cis and trans structure as well then we will have more than 4 structures which are not in the option. that means the question is just asking the standard structures.

Once you do a single tautomerism there's a hydrogen bonding (6 membered ring) formed between -OH and the oxygen. This gives stability to the compound and prefers not to further tautomerise

I consider your structures correct. I agree that there should be E/Z isomerism considered, too, even though equilibration will probably be fast.

In your attempt the 4th compound is not reported to have significant percentage in solution (acidic or basic). This is because once the tautomerisation occurs for example let's say the 1st compound, there is a tautomer having a six-membered conjugated ring involving H-bonding. This thereby increases the stability of the compound and consequently increasing the activation energy for the tautomerisation you proposed in compound 4. This is why compound 4 is not formed.
Also keep in mind that stereo-chemistry is also important while finding number of possible compounds.

This is a typical example of a exercise asked in the wrong way. It is obvious that all the structures OP wrote are correct, therefore the question itself must ask an entirely different thing. If it were stated that high energy conformers are to be excluded, then your answer would (at least partially) be correct. Your reasoning should exclude structure 2, too, because it is a very high energy structure compared to 1 and 3. (Please don't use tag lines in your answers.)

"Not reported" by who?

thank u, sir ... I understood ur point

can u please look at this question as well:https://chemistry.stackexchange.com/questions/120080/how-there-are-7-enolizable-hydrogens

When a tautomer is formed from a diketone, a six-membered pesudo-ring structure is formed that stabilizes the structure through H-bonding. This structure is responsible for not letting any tautomerization of the second carbonyl.

Note: the structure I've shown on the top is made from MS Paint so its quite messy but you'll get the point.

If one day you get tired of apologizing for using MS Paint, there is a post covering some programs for drawing chemical structures. Among those, MarvinSketch is probably the sanest option (check whether their licensing applies to you). Also, check out a robust tutorial from Wikipedia.

thank u, sir ... I understood ur point

can u please look at this question as well:https://chemistry.stackexchange.com/questions/120080/how-there-are-7-enolizable-hydrogens


text:
Spontaneity of an exothermic chemical reaction
When the process is exothermic (ΔHsystem<0), and the entropy of the system increases (ΔSsystem>0), the sign of ΔGsystem is negative at all temperatures. Thus, the process is always spontaneous.
It is easy to prove this using the formula of Gibbs free energy, but how can we prove it by logical reasoning?
I was thinking that in an exothermic process, the system loses heat. Shouldn't "the loss in heat" normally cause more ordering of the system and eventually a decrease in entropy? Might be a stupid question, but what am I missing here?

In an exothermic process, the system produces heat. Whether it loses that heat or keeps it to itself is an unrelated question.

Aha okay that clarified everything. Thank you

It's a good idea to work out the assumptions implicit in your question. Since you are considering use of the Gibbs free energy as a criterion for spontaneity you are presumably concerned with a process carried out at constant T and P:
$$dG \leq 0 ~~~(\textrm{const. T, P)}$$
The second assumption is that $\Delta S$ and $\Delta H$, associated with the macroscopic process you are investigating, are independent of T and P over the range of interest. At constant T and P you are then allowed to write
$$ \Delta G = \Delta H - T \Delta S$$
for the entire range of interest, over which, again, $\Delta S$ and $\Delta H$ are assumed constant. Under these conditions, the above equation suffices to show that what you state in your first paragraph:

When the process is exothermic (ΔHsystem<0), and the entropy of the system increases (ΔSsystem>0), the sign of ΔGsystem is negative at all temperatures. Thus, the process is always spontaneous.

is true. The logical reasoning is as follows: a negative change in enthalpy corresponds at constant P and T to heat being released by the system. Of course this is what is understood as "exothermic". That heat increases the entropy of the surroundings. On the other hand, $\Delta S>0$ means the entropy of the system increases. Overall we then have that $$\Delta S_{sys}+\Delta S_{surr}>0$$ which guarantees spontaneity.
Regarding your second question, you should distinguish between changes in the entropy of the system at constant T and P, described by $\Delta S$, and changes in the entropy of the surroundings, described by (or proportional to) the heat emitted, or $\Delta H$. This is how you analyze, for instance, a reaction at constant T and P.
Consider instead a closed adiabatic system at constant P. Then
$$\left(\frac{\partial S}{\partial T}\right)_p = \frac{C_p}{T}$$
which means, since $C_p>0$ that, yes, as you state the entropy goes down if the temperature drops, all else being equal. But then T is not constant. When a reaction occurs or a piston moves, all else is not equal, and other sources of entropy production/depletion must be considered.

Not all reaction is like this but, exothermic reactions are usually spontaneous because molecules always want to be more stable. So by reacting to another molecule and releasing energy, they can lower their own energy thus making them more stable and unreactive.

My question was that why should being exothermic be accompanied by an increase in entropy and not a decrease in order for a rxn to be spontaneous (for all temperatures)


text:
Distinguishing between octahedral and tetrahedral holes
I know the question sounds a bit silly and there are many similar answered questions but my 3D visualization (imagination) is very bad. There are some images of octahedral and tetrahedral holes but still there's no easy explanation. Can someone help me?

Well, the tetrahedral holes are tetrahedral, and the octahedral ones are octahedral. It is not going to be any other way. What is your question, really?

@Ivan Neretin Nah the problem is there's no picture that the holes are tetrahedral or octahedral. They're just round holes and many balls surround it. If I see the holes from top, will be like seeing pyramid from top or what?

I know tetrahedral but tetrahedral hole .. Someone should enlarge the hole and shows it is really that tetrahedral.

If you are in an empty space with 4 atoms as closest neighbors, it is a tetrahedral hole. If 6 atoms a close by, instead, it is a octahedral hole. Now we just need a small 360 $^\circ$ camera to fly through crystals to visualize it.

The shape of the holes, taking atoms to be spherical, is not that of a tetrahedron or octahedron. The geometry of closest neighbors is what is described by the terms.

An "octahedral" hole is shaped like a curvy cube, not an octahedron. See the pictures below and my comment there.

To visualize holes in closest packing of spheres, you need 6 balls (marbles, table tennis balls, etc) of equal size. 3D closest packing can be constructed from 2D layers of closest packed spheres as such:

The light blue layer is on the bottom and the dark blue layer is on top. The holes, tetrahedral or octahedral, are in between layers.
To find a tetrahedral hole, take three balls and arrange them in a triangle to form part of the lower layer. Then place one sphere in the center on top (you have to keep the lower level together somehow or the structure will collapse). Your tetrahedral hole is in the center of the tetrahedron formed by the 4 balls (marked T in the picture).
To find an octahedral hole, you need a helper or glue. Take three balls and arrange them in a triangle. Take another three balls and arrange them in a triangle. Place the two triangles on top of each other, twisted against each other by 60 degrees. This is a octahedron on its side, and in the center is the octahdral hole.
In the picture above, these positions are marked by T and O. You just have to imaging that they are in between the layers, and then you might be able to count nearest spheres. The picture below shows the arrangement of the nearest spheres, only.

If you don't have 6 balls handy, but have two hands handy, you can also do the following. Place thumb, pointer and middle finger together so that the finger tips form a triangle. Point at the center of the triangle with the pointer of the other hand. That's the tetrahedral hole. Now, form a triangle with the finger tips of your other hand. Bring the 6 finger tips together with a little twist so that they "lock in". At the center of the 6 finger tips, you have the octahedral hole.
So, whether you are a visual or a haptic learner (not that that would be a thing), this will do the trick, I hope.

There are some images of octahedral and tetrahedral holes

Actually, I only found images of the spheres representing the atoms, no images of the shape of the space in between. If you squeeze some blue play dough between two layers of marbles, here is what you get (see also https://leancrew.com/all-this/2019/06/cheesegraters-and-crystals/):

Within a layer, each indentation is surrounded by six others. The arrows point at the six locations with the largest voids (in the center of triangles formed by the sphere and two of its nearest neighbors in that layer). If you look closely, each indentation has three holes, the contacts with the layer below. The voids take on different shapes depending what is underneath. If a sphere is directly underneath, it is a tetrahedral hole. The shape of this is a tetrahedron with "dented in" faces (far right). If a center of a triangle is directly underneath, it is an octahedral hole. The shape of this is a cube with "dented in" faces (near right; cube and octahedron are complementary archimedian solids: the centers of the faces coincide with the other's vertices and vice versa. A tetrahedron is complementary to itself).
If you shrink the spheres around the atoms to points, you get a space-filling arrangement of tetrehedra and octahedra called a tetrahedral-octahedral honey comb.
Teaching this material
Stefanie Lenzer, Bernd Smarsly and Nicole Graulich recently published a paper in J Chem Ed on using clear spheres to explore crystal packing and voids.

As a non-chemist, I have to accept that the volume between 6 spheres is called an "octahedral hole", but the notation strikes me as odd. The shape left between 6 spheres is much more like a cube. See the play-doh model above. The octahedron referred to is the dual of the cube. You get a wire-frame of this octahedron by joining the centers of touching pairs of spheres by rods. Maybe this is why the OP claims to have "very bad 3D visualization". Maybe it's actually good, and OP can't understand why this cube is being called an octahedron!

@AdamChalcraft the volume between the spheres is indeed like a squashed cube. But if you draw the lines between the centres of of nearest atoms (a point and stick model) you see they form the edges of an octahedron. Similarly, a small atom in the hole will touch its 6 nearest neighours at the corners of an octahedron. See for example the "octahedral" environment of the sulfur atom in sulfur hexafluoride (which is a molecular substance rather than a lattice, but you can see the analogy.) https://en.wikipedia.org/wiki/Sulfur_hexafluoride

@AdamChalcraft Thanks for reintroducing me to the technical term dual. And you are right, octahedral hole is jargon; I suppose it should be "space for atoms that would have octahedral coordination", i.e. six nearest neighbors. Until this question, I had never thought about the exact shape of the available space. Making the layer model in play-dough is instructive but kind of challenging. Taking a small ball of play-dough and squeezing it between 4 or 6 marbles, on the other hand, is very easy and gives quick results, so it would be a fun class room activity.

@AdamChalcraft I was having the same problem digesting that this squashed cube is named octahedral. This is a tetrahedral void placed on another tetrahedral void with 60° rotation. It is more like a 'hexahedron'.


text:
How to compute the number of ATP molecules that can be produced from breaking down an unsaturated free fatty acid?
From studying $\beta$-oxidation of free fatty acids, I've learned how to compute for the ATP equivalence of even-chain and odd-chain free fatty acids.
What really bothers me now is how to compute for the ATP equivalence of an unsaturated fatty acid. Since one of the major difference of the process is in unsaturated fatty acids, normal $\beta$-oxidation rounds will occur until the unsaturation is encountered. The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of FADH2.
Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

Is the computation same and I just need to subtract a molecule of FADH2 per unsaturation (double bond)?

Definitely subtract a molecule of $\ce{FADH2}$. After all, you can skip the first of the the four steps in each cycle of removing two carbons (formation of double bond, adding water across the double bond, oxidizing the double bond to a carbonyl group, oxidative decarboxylation). You would also have to check whether any of the steps are coupled to ATP hydrolysis (e.g. the formation of the link to the acyl carrier protein, or transport steps) if you want to get an accurate number of ATP molecules synthesized per fatty acid turned into carbon dioxide.

The cis-isomer will be converted to trans-isomer completely skipping the step where the fatty acyl-CoA molecule is oxidized to the trans-form with the release of $\ce{FADH2}$.

Most unsaturated fatty acids in unprocessed food are in the cis-form. To turn them into the requisite $\ce{\Delta^2}$ trans-double bond, you need one or two additional enzymes, see https://www.ncbi.nlm.nih.gov/books/NBK22387/. For turning cis- into trans- and, if necessary, moving the double bond into the correct position, an isomerase is used. If there are two double bonds, one is reduced first by a reductase.


text:
How do I find pH of a mixture of weak acids?
Suppose that we are given two weak bases $\ce{HA_1}$ and $\ce{HA_2}$. We are also given the $K_\ce{HA_1}$ and $K_\ce{HA_2}$. How can we calculate the $\mathrm{pH}$ of the mixture?

It depends... The "full" treatment would consider both HA species and the $K_w$ of water. Depending on the values of $K_{HA_1}$ and $K_{HA_2}$ and the concentrations various approximations may be appropriate.

@Mithoron No. That question is different. In that question we are given the pH of both acids whereas in this question we are given some other information. Only the title looks quite similar.

@AdityaJain I answered your question here.


text:
What materials are resistant to hot chlorine?
I'm looking to do an experiment involving hot chlorine gas (900 Celsius), however at the moment the only material I'm aware of that could cope with this is Hastelloy's C-22 alloy. What other alternatives exist?

How much water is this chlorine going to have with it? Wet chlorine and dry chlorine behave differently.

The system will be entirely flooded with nitrogen so there shouldn't be any HCl

I'm not sure simply flooding the piping volume with N2 will prevent HCl formation if there's significant amounts of water in the Cl supply or in the piping but make sure to discuss the wet case with the materials specialists.

Short answer — none. Hastelloy C276 or C22 are the best alloys, also considering availability, presumably cost is no object. At that high temperature there is no readily available data. At 500 °C for a limited time experiment, other nickel alloys like 625 or 201 and even 316 SS have some limited life. But for 900 °C, I expect the Hastelloy C family was recommended because they have the best life at about 500 °C. Your conditions are where it would be best to get a manufacturer of high alloys or ceramics to work with you.

You are dealing with nasty stuff. For example , i saw liquid bromine ( less active than chlorine ) spontaneously  ignite a piece of titanium at room temperature. The titanium had been in a Hastelloy C holder which partly melted during the incident.

Alloy 22 has been specifically tested with chlorine which is why I was using it as my example.


text:
Bond Dissociation Energy in Water
Why do the bond dissociation energies of the O−H bonds in water differ? Is it related to an increase in lone pair-bond pair repulsions upon the breaking of the first O−H bond (decrease in stability, so lesser energy is required to break the bond)?
Bond dissociation energy of the first O−H bond: 502 kJ/mol
Bond dissociation energy of the second O−H bond: 427 kJ/mol

This has been asked here: https://chemistry.stackexchange.com/questions/107378/bond-enthalpy-of-hydroxide/


text:
Collision that governs kinetic model of gas
From Atkins' Physical Chemistry 10th, it states that

The kinetic model is based on three assumptions:

The gas consists of molecules of mass $m$ in ceaseless random motion obeying the laws of classical mechanics.
The size of the molecules is negligible, in the sense that their diameters are much smaller than the average distance travelled between collisions.
The molecules interact only through brief elastic collisions.

I quite can't get the second assumption. The intermolecular distance between the molecules are greater than the diameter of the molecules, so we can neglect the volume. Given Boyle's law apply (as omitting size and volume of molecules is a property that ideal gas has), there will be many molecules collide with the walls and the molecules will hit another molecules after its collision with wall (right?), thus the pressure of the system should be low. How often the collision of what kind happens under what condition?
If the separation doesn't result any influence on each molecule (because third assumption said elastic collision, so any interaction (hydrogen bonding, van der Waals, so on) higher than the collision energy should be omitted), will the collision come purely from their collision with walls, not collision between each molecule?
Wikipedia states also the assumption that

The number of molecules is so large that statistical treatment can be applied.

What's the question, really? There can be no walls and even if there are, something like you imagine would invalidate model, I'm afraid.

@Mithoron The main question is what kind of collision in the kinetic model under the assumptions. If there's no walls, is it possible to relate the collision with Boyle's law?

The molecules interact only through brief elastic collisions (with each other and perhaps with other stuff).

@Mithoron Interesting, so that brief collision is happened with high pressure?

In any pressure whatsoever - that how gas works! This theory is supposed to be somewhat realistic...

@Mithoron I didn't know the model can be thought to be simple like that. The book put the context in chapter of perfect gas so the book overstates the theory, I think. Thank you.

The key requirements are (1) thermal equilibrium (2) a continuous energy distribution (not discrete particle counts) and (3) negligible intermolecular energy. There is a lot of room between "enough particles" to satisfy these conditions and "high pressure". You can simply scale up the volume if you want to increase the number of particles and ensure smoother statistics.

The measured pressure of a gas in a vessel (ie the pressure the gas exerts on the walls of the vessel) can be calculated from the properties of the gas molecules (like their average momentum) assuming elastic collisions with the walls. in short, collisions of molecules with walls is the pressure.

So I've no idea where the "thus the pressure of the system should be low" comes from.

There will be many molecules collide with the walls and the molecules will hit another molecules after its collision with wall (right?), thus the pressure of the system should be low. How often the collision of what kind happens under what condition?

The collision condition doesn't need to be strictly in Boyle's law condition, that the elastic collision is happened between anything and under any pressure condition. The requirement of large number of molecules but only collision interactions are applied implies that it is just not applicable to apply Boyle's law in kinetic model of gas.


text:
Why is it ill advised to eat the solid aspirine formed from salicylic acid + acetic anhydride
In the laboratory notes it is indicated that it is ill advised to eat the solid formed from this experiment. An esterification reaction, in which salicylic acid is treated with acetic anhydride.
If it is for the acid added as catalyst either sulfuric acid or phosphoric acid. Why isn't it possible to neutralize it with a strong basic?
The addition of water should be enough to quench (destroy) any remaining acetic anhydride and turn it into acetic acid, and this latter one is more soluble than aspirin so it should be possible to crystallize aspirin without any acid.
The question appears in this lab notes http://www2.xula.edu/chemistry/crs-orgleclab/Organic_Lab_1_2230L_Web_Files/23_Aspirin_notes.pdf
I think it is self-evident that your basement shouldn't become a pharmacy. I think it has more to do with the acids used in the synthesis. Does the industrial process differ from the synthesis presented in the article with regards to this aspect (acid catalyst and their neutralization after the process)?

Uhm, quality control. You may not know what you are eating.

It is ill advised to eat pretty much anything in the lab.

The last paragraph puts you on the right track - you need an extensive purification procedure, quality control of both reactants and products, which involves a lot of spectroscopy and chromatography. How many hoops do regular suppliers of medicine jump through before you are allowed to buy their products?

if you think just adding water detroys all the acetic anhydride then you have not worked with it.

Why would you want to eat anything that is formed in the laboratory?

Pharmaceuticals that are sold for human consumption must go through extensive purity and quality testing, and must be made under carefully controlled conditions. See: GMP - Wikipedia

I don't think this is a personal medical question: it is about good practice in a lab which is a mainstream chemistry question. I'm voting to reopen.

I'm surprised at the downvotes. Just not knowing something doesn't mean the OP deserves downvotes. And the question will be helpful for future audiences with little or no experience in practical chemistry.

Unless you worked to Pharma industry Good Manufacturing Practice you shouldn't trust your product
The simple reason why you should not trust the product is that that you don't know it is safe unless you have followed rigorous quality control processes. The drug industry follows carefully regulated quality-control processes known, in Europe, as Good Manufacturing Practice. This is far from easy to meet and even major drug firms sometimes have their factories closed because they fail to ensure it.
For your particular reaction there are many possible things you would need to check before knowing the product was safe to consume. You really don't want to guess whether all the acetic anhydride or acetic acid has been removed: even a small amount will be bad for you (you can't rely on intuition that it should all have been removed). And are you sure there are no nasty side products? Have you checked? Have you checked the purity to make sure? Have you had a check by an independent party?
If you can't say yes to all of these (and more: go read the GMP standards for aspirin manufacture), don't take the product.


text:
Reaction of getting jarosite from pyrite
I am doing a PhD about archaeometry in order to study the contaminations and alterations that occur to the ceramics which are buried in a marine environment.
In my research, I have found pyrite in the cracks of my ceramics. Now, I am trying to find jarosite.
Well, I have found bibliography about pyrite's oxidation reaction to give jarosite (Secco et al. [1]). Now, trying to understand the redox reaction of pyrite to jarosite, I am getting blocked. The image of the reaction is the next:

The presence of jarosite and absence of pyrite in the outermost part of the margin in the Olbia potsherds is due to the oxidation of pyrite and reaction with dissolved alkali (Faure 1991), according to the following reactions (Faure 1991):
$$
\begin{align}
\ce{FeS2(s) + 7/2 O2 + H2O &-> Fe^2+ + 2 SO4^2- + 2 H+};\tag{18}\\
\ce{Fe^2+ + 1/4 O2 + H+ &-> Fe^3+ + 1/2 H2O};\tag{19}\\
\ce{Fe^3+(aq) + 3 H2O &-> Fe(OH)3(s) + 3 H+};\tag{20}\\
\ce{(K+{,}Na+) + 3 Fe(OH)3 + 2 SO4^2- + 3 H+ &-> (K{,}Na)Fe3(SO4)2(OH)6 + 3 H2O}.\tag{21}
\end{align}
$$

I would like to understand the half redox reactions (oxidation and reduction). I have tried to solve this, but I have not been able to. I identify the reducing agent (the S, which oxidizes from S(1-) to S(6+)), and the oxidizing agent (the oxygen, which reduces from O(0) to O(2-). However, I think I am unable to write the half reactions properly, because I do not obtain the same as the equation (18) of the bibliography:

What I obtain is that the second half reaction also seems an oxidation. I can't solve it like a reduction.
I will appreciate a lot your help, maybe you can shed a little bit of light on this.
References

Secco, M.; Maritan, L.; Mazzoli, C.; Lampronti, G. I.; Zorzi, F.; Nodari, L.; Russo, U.; Mattioli, S. P. Alteration Processes of Pottery in Lagoon-like Environments. Archaeometry 2011, 53 (4), 809–829. https://doi.org/10/cp6gkh.

Oxygen is the oxidant

Yes, thank you! I am trying to solve the reaction of the reduction, but I can't. What I get is as if the second reaction was another oxidation.

The second reaction is also an oxidation with oxygen as the oxidant

So, I understand that it is not a redox reaction? Only oxidation? Sorry, I think I am missing some concepts. And how can I obtain the same as equation (18)? Thank you

What I don't understand is the statement "due to the oxidation of pyrite and reaction with dissolved alkali" when none of the processes quoted have OH- in the equation. The process requires only water and oxygen.

quation (18) is a redox equation. We can never have an isolated reduction or isolated oxidation.
$\ce{FeS2(s) + 7/2 O2 + H2O -> Fe^2+ + 2 SO4^2- + 2 H+}$
You can see here iron (II) is just an spectator ion. It is sulfur (the disulfide S) which is getting oxidized to a $\ce{SO4^2-}$. As a result oxygen has to gain electrons. So your identification of the redox species is correct but the blue hand-written equation is wrong.


text:
Favoured position in nitration of pyrrole
I try to figure out the preferred position of nitration (determine whether compound A or B is the major product) by looking at the resonance structures of the intermediate cation.

The solution then showed that for compound A, there is no intermediate 2. I wonder why. Thanks

Not strictly accurate as pyrrole nitrates 2:3 in approx 4:1 ratio

Your first set of resonance structures are wrong.


text:
Smoky lights at events - Piped in or a chemistry reaction of some kind?
Was at Dreamhack MTL this weekend and realised that in the whole venue there seems to be just a little bit of smoke in the spotlights and other light sources.
You also see that kind of smoke in shows and conventions. So I was wondering is that smoke something that gets piped in for that volumic effect on the lights, or is it just something that tends to happen when you shove 3000+ human beings in an enclosed space?
And if so, what is happening here, on a chemistry level, to cause that smoke?


text:
Protocol during antechamber heating
I recently setup a heating apparatus for the large antechamber in my glovebox (i.e., large amounts of heating tape). Eventually, I want to do bake-outs for my moisture-sensitive chemistries (e.g., battery electrolytes).
What is the standard procedure for bake out in an antechamber? I know that many commercial entities like mBraun sell heating add ons for their boxes, so I'm sure that best practice exists.
Is it best that I put the chamber under slight vacuum so that pressure doesn't build up? Is it best that I keep argon in the chamber to help with heat transfer? What are other considerations I should have when running a bake?


text:
Why do some compounds without carbonyl group give positive 2,4-DNPH test?
I was performing some experiments in Chemistry Lab today in which I had to find the functional groups present in a compound.
It being a very basic undergraduate course introducing us to the Chemistry lab, the functional groups were quite basic, and we just worked with mono-functional group compounds, including Carboxylic Acids, Amines, Aldehydes, Ketones and Phenols.
When performing the 24DNPH test on a white powdery substance (which I think had the carboxylic acid group), I got an yellowish-orange precipitate which was further characterized when I added dil. $\ce{HCl}$ to the test tube.
However, the sample gave negative Fehling's, Tollen's and Legal's tests (Legal's test is the sodium nitroprusside test).
Later, I even performed a test on another substance which gave positive primary amine test, as it gave me a scarlet dye in an azodye coupling reaction with $\beta$-naphthol. (This has stained my hand with reddish spots :D).
Point is, even this substance gave me a yellowish precipitate on adding 24DNPH, but it wasn't the normal orange/red precipitate which we see on reacting actual carbonyl compounds with 24DNPH.
This left me wondering, what compounds are these yellow precipitates that I had got?
I'd be very happy if someone can answer this as it is really intriguing me for a long time.

2,4-DNP is a nucleophile so will react with other groups than  just carbonyls.

So the product formed may be R-CO-NH-NH-{2,4(Ph(NO2)2)}? (I know it's not really a proper formula writing form, but I'm just asking to know)

It could be. A species like that would likely be coloured. I can't quite see how it would form from the free acid without some kind of coupling agent present. It could be just the salt of the acid (2,4-DNP being basic). With respect to the second (amine positive) material, what was the process for the azo dye forming reaction?

@Waylander The azodye process involved adding a little bit of $\ce{NaNO2}$ and $\ce{HCl}$ to the unknown substance placed in an ice bath, followed by the addition of β-naphthol


text:
Four-electron integrals over GTOs
I was looking through the following presentation:
http://folk.uio.no/helgaker/talks/SostrupIntegrals_10.pdf
Looking at the section on 4-electron integrals, I couldn't help but notice that the slides wrote that the 4-electron terms (Kij and Jij) were Gabcd x Dcd. [see slide 25]
I realised that Dcd was referring to the dipole-moment integral. Maybe I just missed something here, but what does the dipole moment integral have got to do with the 4-electron integrals?
This is particularly confusing for me since I notice that the guide here (which uses the same notation afaik) uses Dij^2 as the auxiliary kinetic-energy integral.
https://joshuagoings.com/2017/04/28/integrals/

$J_{ij}$, $K_{ij}$ are certainly not four-electron quantities (which probably are a thing in extended, explicitly correlated theories). We could debate whether those are one- or two-electron quantities. On slide 25 of the first link, $D$ refers to the density matrix (often called $P$ in other literature), compare slide 30. This is also clear from the context (Fock matrix $F$ from $J,K$ and density matrix). The $g_{abcd}$ on slide 25 are two-electron electron integrals, which are typically four-center.


text:
How is the Hessian evaluated?
I was reading about the various ways to optimise the geometry of the system, and found that none of the sources mention how the Hessian matrix is obtained.
How does one obtain the Hessian matrix? I’m thinking it’s either some analytical formula or just numerical differentiation.

Both analytically and via numerical derivatives is possible and has been used extensively. However, for straight geometry optimization, one typically uses approximate forms of the Hessian (a pseudo-Hessian, if you will, built over several geometry optimization cycles). For transition state search, one usually requires the true Hessian.

What method are you talking about (ab initio electronic structure, atomistic MD)? What software?

I'm referring to ab initio electronic structure.

Are you looking for a detailed mathematical description or a more qualitative description of how the Hessian is estimated?

The documentation for the Gaussian software is actually pretty complete, even if qualitative. On the subject of optimizations, I recommend you look up the freq and opt keywords.
The manual explains the procedure followed by the Berny optimization as involving construction of an initial analytical estimate of the Hessian using a simplified force field, with subsequent optimization steps improving on that estimate using computed gradients. A number of options are provided for improving the initial estimate of the force constants where the defaults derived from the estimated Hessian are not good enough.
For frequency calculations, accuracy requires a good estimate of the full Hessian, better than estimates that might suffice for geometry optimizations. Gaussian allows for analytical computation of force constants for some of the methods (e.g. RHF) and uses analytical gradients to compute a numerical Hessian with some other methods, or if requested. In optimization cases where convergence is an issue, Gaussian facilitates convergence to a real minimum by employing the Opt=CalcAll keyword, which results in analytical (where possible) computation of the full Hessian (at some cost though).

Thanks! I'll certainly look it up. Could you also point me towards a mathematical explanation, if possible?

@ANZGCFlyingFalcon I can add a little info if it helps, when I get a chance. A numerical methods textbook that covers optimization methods should discuss this btw.