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Intuition for why effective nuclear charge increases down a column
My textbook (Chemistry the Central Science, edition 12, Brown et al.) says that effective nuclear charge increases down a column. Indeed, this is consistent with the values provided by Clementi in the 1960s.
Is there intuition or some qualitative explanation for why effective nuclear charge ($Z_\mathrm{eff}$) for the outermost electrons increases down a column (as opposed to being roughly equal)?

Do you mean the effective nuclear charge ($\ce{Z_{eff}}$) for the outermost electrons?

@ShoubhikRajMaiti Yes.

One thing might be that for higher principal quantum numbers, the d, f shells are being filled, which are said to be diffuse and not good at shielding.

Initially, let's take the electrons as probability density fields instead of classical particles.
The increase on the number of protons in the nucleus increases the electric force on all electrons, but the core electrons suffer higher attraction due to higher proximity and lighter electronic blindage, so their probability density field is forced nearer the nucleus.
Reminding of Electric Force:
$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$
Consequently, the outer electrons will suffer lower blindage force from these core electrons, parallelly to the nucleus attraction, allowing them to get closer to the nucleus as well.
As effect, of course, the outer electrons also force the inner electrons towards the nucleus, but their probability densities reach a minimum repulsion energy in such a distance that electrons on the same level are forced closer to the nucleus than on the atomic element above, going down a column in the periodic table.

A qualitative way to understand this trend is that with each increase in n, going row by row down a column, the electrons of filled inner shells create a screening effect upon the charge of the nucleus. The nuclear charge grows with each added proton, but as electrons are placed into new shells (with larger n) above previous shells, they experience not only the attraction of the nucleus but the repulsion of the inner electrons.
This doesn't explain the increase in Z_eff down a column by itself though. For that, we also need to consider that the radial probability distribution (The likelihood of finding the electron a certain radial distance from the nucleus) of the (n)s, (n)p, (n-1)d, and (n-2)f electrons within a row, changes to reflect the appearance of nodes as n, the energy level, increases.
Consider the s orbitals and the alkali metals/hydrogen. The 1s orbital (Hydrogen) has no radial nodes in the probability distribution, while the 2s (Lithium) and 3s (Sodium) have one and two radial nodes, respectively. The nodes divide up the radial probability into peaks (Technically, I think it's like n - l peaks, but you asked for a qualitative explanation). As n gets larger, the most probable peak moves farther away from the nucleus (Which should hopefully make sense), but the other peaks appear deeper within the atom, closer to the nucleus. These smaller peaks of probability can actually appear deeper (Shorter radial distance) than some inner "Shielding" orbitals and allow the electron described by that probability distribution to feel an effectively less screened nuclear charge. This happens even though this electron is more probable to be found farther away than the shielding layers of electrons underneath it.
I found this image from the cited online textbook to be illustrative of this example I am talking about:

I am referencing this textbook's section on "Electron Penetration": https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07._Periodic_Properties_of_the_Elements/7.2%3A_Shielding_and_Effective_Nuclear_Charge


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Comparing Polarity of Amino Acids
I know that there are polar uncharged amino acids (serine, threonine, asparagine, glutamine, cysteine) and polar charged amino acids (the basic and acidic amino acids).
Does the charge on the acidic and basic amino acids make them more polar and hydrophilic than the uncharged polar amino acids? Moreover, cysteine is classified as an uncharged amino acid, but because it has an ionizable side chain, would it be more polar than serine, asparagine, etc.?

Does the charge on the acidic and basic amino acids make them more polar and hydrophilic than the uncharged polar amino acids?

Yes, they are less likely to be on the inside of a protein. Hydrogen bonds with water will be stronger, and interactions with ions as well.

Moreover, cysteine is classified as an uncharged amino acid, but because it has an ionizable side chain, would it be more polar than serine, asparagine, etc.?

Cysteine and tyrosine are more hydrophilic when deprotonated. So at strongly basic pH, their properties are very different from that of serine. At neutral pH, however, cysteine is less polar than serine.

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Timescales in Chemical Analysis
Has anyone seen a comparative list of typical run times of the techniques used in analytical chemistry? I am making a list of rough timescales of analytical techniques (for a slide). Assume that the sample preparation is aleady done.
For example:
Volumetric titration may take up to 10 minutes
Gravimetry can take several hours to a day
NMR- several minutes
Electron spin resonance ?
X-ray photoelectron spectroscopy: survey spectrum for surface analysis?
Auger electron spectroscopy for surface analysis
X-ray crystallography
Inductively coupled plasma: approximately a minute to read one sample
FTIR- minutes
Raman spectroscopy- typically a minute
HPLC- seconds to hours
GC/ Supercritical Fluid Chromatography/ Capillary electrophoresis
 - seconds to hours
Mass spec - less than a minute
Ion mobility spectrometry- milliseconds

There should be included 2 numbers. The device run time and analysis run time, as sample preparation can easily take 99% of time.

Often, of course, it is not the time taken for the experiment itself but waiting in the queue, or travelling to a lab where the instrument exists to get it done, that is the limiting step.

Proton NMR (1D) acquisition time can vary by orders of magnitude depending on concentration (sample amount), instrument sensitivity and relaxation times. For a standard experiment it is on the order of minutes.

On a modern XPS instrument, a non-gassy sample can be transferred from the entry airlock to the high vacuum chamber after 5-10 minutes, and a decent survey spectrum can be completed in a few minutes. Older systems will take longer. Auger analysis typically has fairly similar timescales.

Should gas chromatography (GC) also be on the list?  And what about "hyphenated" techniques like LC-MS or GC-MS?

Let us get the results for primary techniques. The speed of hyphenated technique will be dependent on the speed of GC or HPLC separations.

Proton NMR (1D) acquisition time can vary by orders of magnitude depending on concentration (sample amount), instrument sensitivity and relaxation times. For a standard experiment it is on the order of minutes.

Any ideas about the analysis timescale of Auger spectroscopy. I attended a workshop on it 8-9 years ago but don't remember how much time it takes to run a sample. I think after inserting the sample in XPS and Auger, it takes considerable time to generate the vacuum needed.

@M.Farooq I left my answer open as a community wiki, I am hoping others will contribute to the answer so we can compile a list. There are others who know more about Auger and will hopefully add some information about it.

Auger can be performed with an electron microscope, the answer would be as you suggest, a question of achieving the required vacuum.

There is a risk of the question being quite open-ended. As mentioned in earlier comments sample prep can vary enormously within and between techniques. Even in NMR depending on what you want to observe (consider for instance selective isotopic labeling).

Yes, I will edit the question so that we eliminate sample preparation. My interest for a slide was to show how much time it takes to analyze the prepared sample and get useful signals out of it. I think the fastest is ion-mobility spectrometry so far- on the order of milliseconds, the analysis results are out.

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Chemical interaction between Pterostilbene and Nicotinamide Riboside
Some supplement companies are putting a combinations of nicotinamide riboside and pterostilbene into the same capsule. Considering the stomach acid, water, body temperature, etc. I am wondering if there is any likelihood of undesired chemical/pharmacological interaction when those two are ingested at the same time.


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What is the number of stereoisomers for hex-4-en-2-ol?

For the following compound, find

Total number of stereoisomers
Number of optically active stereoisomers
Total number of fractions on fractional distillation of all stereoisomers

There is a chiral carbon marked with "*" in the compound. The compound is thus chiral and is optically active. The configuration at the chirality centre can either be R or S. There is also a double bond in the compound and cis-trans (geometrical isomerism) is possible here. So using basic permutations there are totally $2\times2=4$ stereoisomers (optical and geometrical).
Since there is a chirality centre the compound exists in right-handed and left handed systems and thus is optically active. Thus all the 4 stereoisomers are optically active.
Out of R-cis, R-trans, S-cis, S-trans isomers there are two pairs of diastereomers - R-cis, S-trans and R-trans, S-cis - so there are 2 fractions obtained due to fractional distillation.
Final Answer

Total number of stereoisomers = 4
Number of optically active stereoisomers = 4
Total number of fractions on fractional distillation of all stereoisomers = 2

Are my approach and my answers right?

The approach is right, and your answers are right, but there are a couple of things to look at in your explanation.
First, the presence of a chiral centre does not in itself make a compound chiral.
It is in this case, but a molecule is chiral if, and only if, it is not superimposable on its mirror image.  You could have a molecule with more than one chiral centre, which was superimposable on its mirror image, and was therefore not chiral.
Second, a chiral molecule is not necessarily optically active.  Its specific rotation may be zero (or at least unmeasurable), in which case it is said to exhibit cryptochirality.
Thirdly, the R- and S- form of each of the geometric isomers are enantiomers (ie non-superimposable mirror images) of one another, not diastereomers.  Diastereomers occur when a molecule has (for instance) two chiral centres and rather than inverting both (giving the enantiomer, its mirror image) you only invert one, which is actually a different compound, with different physical and chemical properties.
Personally I think using the term stereoisomer to refer to the cis and trans versions is misleading, even if strictly geometric isomers are a subset of stereoisomers.  I'd prefer to say that the R- and S- forms of each geometric isomer are enantiomers of one another, and leave it at that, rather than lumping all four together as stereoisomers.  But you don't get to choose the wording of the question, I guess.


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Redox Titration in investigating change of Iron in Avocado
I'm trying to investigate the change in iron in an avocado as it ripens. I'm planning on using redox titration but still a bit unsure of how to carry out the investigation (method). Do you have any suggestions?

You cannot do titration to determine iron content with redox methods because the amount of iron is very very small in fruits. Search for spectrophotometric methods.

@M.Farooq Your comment should probably be posted as an answer (with approx. LODs/LOQs, if possible).

You cannot do simple volumetric redox titration to determine iron content because the amount of iron is very very small in a single fruit (on the order of fraction of a milligram). Imagine what would be the buret reading? Classical methods are good for large concentrations >> 1% wt/wt
With such small quantities, UV-Vis absorption spectroscopy or atomic spectroscopy comes to our rescue. Among the common (cheap) methods, iron- 1,10 phenanthroline complex method is often used for educational purposes.


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Photoelectric effect (AP Chemistry question)

Which of the following postulates regarding the photoelectric effect is considered to be INCORRECT according to the quantum mechanical model we currently understand?
(A) Increasing the intensity of light does not change the energy per photon.
(B) The energy of a photon which ejects an electron from a metal surface must be equal to the binding energy (a.k.a. work function) of the electron plus the KE of the electron.
(C) Increasing the frequency of a photon can allow that lone photon to eject multiple electrons from a metal surface.
(D) Some electrons in a metal are easier (require less energy) to eject than others.

Please see question above and the teacher also supplied us with answers, so I know that answer is C.
Now, I am trying to reason why C is an answer, and A seems to be reasonable from what I read in my book. D looks somewhat good even though books keep mum about it. But I am puzzled about answer B. I looked at every AP Chemistry source I could get, and I am not sure where I can find formula for energy of photon, which would be equal binding energy + KE of electron. I can imagine this might exist outside of AP Chemistry, but within a subject? I found the equation in Wikipedia, for example:

The energy of a photon causing the photoelectric effect is found through E = hf = KE + w, where h is Planck's constant, 6.626X10^(-34) J*s, f is the frequency of the electromagnetic wave, KE is the kinetic energy of the photoelectron and w is the work function for the metal.

Any help on this matter would be highly appreciated. Desirably, reference to AP Chem book with page number. Thank you so much in advance!

I think the teacher pointed out that (c) is the correct answer because the postulate is generally regarded as incorrect according to the normal convention of photons behaving as individual packets of energy.
However what I think is interesting is that more than one electron can be ejected provided the photon is sufficiently energetic (this information will not be useful during an examination, so I suggest forgeting about this when you take an exam or do your homework, but if interested read on). For more information you can search for "multielectron ejection in auger spectroscopy" which provided references [1] and [2] below. Reference [2] opens up with the following statement:

Absorption of an energetic photon by a target should in first approximation lead to the ejection of one electron only, as this interaction is purely mono-electronic. However it is well established that sometimes several electrons can be ejected. This can happen in valence multiple ionization or after innershell ionization when secondary electrons are emitted sequentially (cascade Auger decays) or simultaneously (direct double Auger decays). Such processes and especially the ‘one-step’ or direct ones (direct double ionization or direct double Auger decay) are extremely interesting to investigate as they revealthe strongelectron correlation phenomena which are at their origin.

Not sure what Einstein would have thought of this in any case.
An Auger process does not violate the principle that intramolecular electronic transitions are quantized and therefore that the energy required to eject electrons is quantized, irrespective of the number ejected. So the first two postulates are still correct.
References

P. Lablanquie, M.A. Khalal, L. Andric, J. Palaudoux, F. Penent, J-M Bizau, D. Cubaynes, K. Jänkälä, Y. Hikosaka, K. Ito, K. Bučar, M. Žitnik, Multi-electron coincidence spectroscopy: Triple Auger decay of Ar 2p and 2s holes,
Journal of Electron Spectroscopy and Related Phenomena, Volume 220, 2017, Pages 125-132, https://doi.org/10.1016/j.elspec.2017.04.003.
https://pure.mpg.de/rest/items/item_737492/component/file_737491/content

Isn't Auger always a 1-photon -2 electron effect? One photon would knock out the signal electron and the remaining events would be dependent on the other "electron"? Spectroscopy is always interesting.

@M.Farooq I guess not: one of the references states "The Auger decay of Ar innershell holes by emission of three Auger electrons is studied." But multiple emissions can be "simultaneous" (the exact meaning of simultaneity here I am not sure about). In the context of the question I thought this more of a curiosity worth mentioning.

Answers A and D are both correct. Increasing the intensity of light just means there are more photons hitting the surface. Indeed some electrons in a metal are easier to eject than other.The band gap of the electrons of 1s subshell are harder to leave the nucleus than the electrons of a 2s subshell.C seems to be incorrect.Photons are small packets of energy absorbed/emitted by 1 electron.Also they are absorbed/emitted as a particle which makes it impossible to be emitted by 2 different electrons.

(C) Increasing the frequency of a photon can allow that lone photon to
  eject multiple electrons from a metal surface.

I am glad that you gave a second thought to your teacher's "correct C" answer. Einstein might be turning in his grave. Photoelectric effect is one photon to one electron ejection phenomenon. Answer "C" is fundamentally wrong. Otherwise, increasing the intensity of a photon of the right frequency, the photo-current would not be a linear function of light intensity.

It might have been less clear in a previous edit, but the question does ask for which of the options is incorrect.


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Help in drawing resonance structures in case of polybasic acids
While drawing resonance structure of dibasic or tribasic acid, e.g. should I take out two $\ce{H+}$ (in dibasic) or three $\ce{H+}$ (in tribasic) together or one by while drawing resonance structure?
Do dibasic acids give both $\ce{H+}$ at once or one by one?
Actually, I was drawing resonance structure of $\ce{H3PO4},$ but got confused. Should I remove all three protons at once or only one?

please forgive me for poor editing & framing of this question.Poor Editing- because i'm new.Poor Framing - because English isn't my native language & i'm a science student not literature. Sorry.

No need to apologize, showing you effort towards improving is what counts. I bet English isn't a native language for about one third of Chemistry.SE users (not for me, either), nevertheless, we do manage to communicate somehow. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Related: Most contributing resonance structure of phosphoric acid.

@andselisk my real doubt is in this part." should I take out two H+ (in dibasic) or three H+ (in tribasic) together or one while drawing resonance structure? "

It depends on what specifically the question is asking you to do.
In reality, a polyprotic acid such as $\ce{H3PO4}$ becomes deprotonated in a sequential way; the first proton is the easiest to remove (most acidic), then the second, then finally the third is hardest to remove (least acidic).
So there are four different species that the question might be asking you about: $\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$
Each of the three anionic species above has multiple resonance structures.

I almost upvoted this but the last sentence is wrong. Each of those four species has only one significantly contributing resonance structure. Note: proton shuffling (possible in all except phosphate) is not resonance but consitutional isomerism.

The most commonly taught convention for the structures of these phosphate bases involves a P=O double bond. In that case, my statement was correct and multiple resonance structures exist. If you draw them with ionized P-O single bonds, then you are correct per the structures in your example. While I think this is more realistic (AFAIK computational/spectroscopic data don't support the "expanded octet" for phosphorus) it is less commonly taught at the undergraduate level, and my money is on OP's class using P=O structures and the phosphates as examples of resonance stabilized conjugate bases.

Your bet on OP’s level and what OP got taught is probably correct. However, I refuse to use outdated and superceded concepts in my SE answers and I refuse to upvote answers that use them in the hope that they may ultimately finally die out.

Resonance structures can only be drawn for single compounds. Rephrased: if a set of resonance structures are to be valid, then all atoms (nuclei) must be at exactly the same positions.
Thus, when asked to draw resonance structures of e.g. phenol or nitric acid, you cannot deprotonate or transfer protons. All protons must be exactly at the same spot in all structures, only electrons may move—thus, the rightmost structure of nitric acid drawn in scheme 1 is not a resonance structure. Distinct from the examples above, you can draw (a different set of) resonance structures for phenolate or nitrate after removing a single proton (see also scheme 1).

Scheme 1: possible resonance structures of (top to bottom) phenol, nitric acid, phenolate, nitrate. Lesser contributors to phenol and nitric acid in brackets. For nitric acid, an isomer that is not a resonance structure is shown.
Expanding on this, if you happen to have a diprotic or triprotic acid for which you can draw resonance structures such as carbonic acid or salicylic acid, your resonance structures must have the proton connected to the identical oxygen atom for the resonance structures to be correct, compare the examples in scheme 2.

Scheme 2: Resonance structures of carbonic acid and its conjugate bases (top three) and salicylic acid and its conjugate bases (bottom three). Further possible resonance structures of salicyldiate are omitted for brevity.
Finally note that your example is poorly chosen. Phosphoric acid and its conjugate bases only have one significant resonance structure each as depicted in scheme 3.

Scheme 3: Structures of phosphoric acid, dihydrogenphosphate, hydrogenphosphate and phosphate (from left to right). No other significant resonance contributors exist.


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How to I find the equilibrium temperature in this problem?
From Castellan's Physical Chemistry, chapter 12 [1, p. 276]:

12.18 At $\pu{25 °C}$ we have for rhombic sulfur: $ΔG_\mathrm{f}^\circ = 0,$ $ΔS^\circ = \pu{31.88 ± 0.17 J K-1 mol-1};$ and for monoclinic sulfur: $ΔG_\mathrm{f}^\circ = \pu{63 J mol-1},$ $ΔS^\circ = \pu{32.55 ± 0.25 J K-1 mol-1}.$ Assuming that the entropies do not vary with temperature, sketch the value of $μ$ versus $T$ for the two forms of sulfur. From the data determine the equilibrium temperature for the transformation of rhombic sulfur to monoclinic sulfur. Compare this temperature with the experimental value, $\pu{95.4 °C},$ noting the uncertainties in the values of $ΔS^\circ.$

I have plotted the graph of $μ$ vs $T$ successfully, but unable to find the equilibrium temperature. The answer is showing 119 °C and possible range from 83 °C to 277 °C.
How do I find it mathematically? I have used the equation $\mathrm d\mu = -S\,\mathrm dT + V\,\mathrm dp,$ but lot of the things are unknown. Can anyone solve it completely with calculation?

References

Castellan, G. W. Physical Chemistry, 3rd ed.; Addison-Wesley: Reading, Mass, 1983. ISBN 978-0-201-10386-1.

Ask yourself: Is there anything here that is constant, which would allow you to simplify the equation?

I already tried to solve by making chemical potential and Equilibrium temperature as constants but still failed, so I request you please solve the problem.

That's not how it works for homework problems on this site.  We're supposed to provide guidance so that you can solve it, not solve it for you.  In this problem,  and T are not constants,  they're variables.  However, p is constant, which gives you a nice simplification of the equations you're working with.

I already considered p to be constant I have used the equation du = -SdT, on integrating both sides from limits 298K to T    I got u-u° = -S (T-298) now for both Rhombic and monoclinic sulphur u must be same in equilibrium, and also T is same , since T is equilibrium temperature,     now Gf for rhombic sulphur is zero and Gf for monoclinic sulphur is 63, both values are given so  and for rhombic S =31.88 and for monoclinic S=32.55 so using the equation u =-Sdt for both the sulphur we get       u -0= -31.88(T-298) and u - 63= -32.55(T-298) these two equations u is same so we find the equil temp

But in the above mentioned way which I did my answer is coming very weird T is coming -25°C but actual answer is 119°C

You've got it set up correctly, so your problem seems to be just with the algebra:  When I set -31.88 (T- 298) = 63 - 32.55 (T - 298), and solve for T, I get T = 392 K = 119 C.

Let us continue this discussion in chat.

Yes  thank you very much, I found it, Yeah I found out, I was so close just problem with algebra and signs, now what I realised is There was no need to post this question here when I had already solved it on my own, but since due to calculation mistake I was not sure about my approach I thought my approach is wrong, I became very scared and lost my confidence

Are the uncertainty ranges [83,277] those you calculated or as reported by the textbook?

You're welcome.  Glad I was able to help.  The last piece of this would be to use the uncertainties to find the range of different possible temperatures for phase coexistence.

Uncertainty ranges are reported by the textbook


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How to find the Cp, difference of Cp and heat of fusion from the dependence of vapor pressure on temperature?

The vapor pressure of zinc varies with temperature as
$$\log P(\pu{mm Hg}) = -6850/T - 0.755\log T + 11.24 \label{eqn:1}\tag{1}$$
and that of liquid zinc as
$$\log P(\pu{mm Hg}) = -6620/T - 1.255\log T + 12.34 \label{eqn:2}\tag{2}$$
Calculate:
a) boiling point of $\ce{Zn};$
b) triple point;
c) heat of evaporation at boiling point;
d) heat of fusion;
e) the difference in $C_p$ of solid and liquid zinc.

I have found out the boiling point by substituting pressure as $\pu{760 mm Hg}$ on first equation \eqref{eqn:1}, from here boiling point $T$ can be found out as $\pu{1181 K}.$
I have also found out triple point by equating the two equations because at triple point both the pressure of the two phases must be equal.
I have also found out heat of vaporization by using the equation
$$\frac{\mathrm d\ln p}{\mathrm dT} = \frac{H_\mathrm{vap}}{RT^2}.$$
But to find the heat of fusion I must know the heat of sublimation, but the heat of sublimation is not provided here. Is there any other way to know the heat of fusion (melting)?
Also, how to find the $C_p$ of both the phases and the difference of $C_p?$

Use the equation for the vapor pressure of the solid to get the heat of sublimation.  The heat capacity of the vapor minus the heat capacity of the liquid is equal to the derivative of the heat of vaporization with respect to temperature. The heat capacity of the vapor minus the heat capacity of the solid is equal to the derivative of the heat of sublimation with respect to temperature.
ADDENDUM
$$\Delta H_{sublimation}=RT^2\frac{d\ln{P}}{dT}=(\ln{10})RT^2\frac{d\log{P}}{dT}$$where log P is given by Eqn. 1.

But here vapor pressure of solid zinc is not provided, no equation for solid Zinc,  only for liquid zinc there is an equation, heat  capacity of vapor and liquid is also not provided, heat capacity of solid zinc is also not given , all are unknown , impossible to find out

When they used the words "The vapor pressure of zinc varies...," what phase did you think they meant?

If it is of solid then , we must know the sublimation Temperature and Pressure , only then we can find heat of sublimation ,heat capacity of vapor is not given, heat capacity of solid and liquid zinc is also not given,

Read my answer again.

In the first lines you said "Use the equation for the vapor pressure of the solid to get the heat of sublimation" but here I don't know neither the pressure of sublimation nor the temperature of sublimation  so both are unknown, if one of them was known we could have put in the equation of solid zinc and find out the other, and hence can calculate Heat of sublimation.

The heat of sublimation is a function of temperature, obtained in the analogous  way that you got the heat of vaporization as a function of temperature.

Yeah but for that also Pressure or temperature is required

Let us continue this discussion in chat.

BTW the question is regarding the heat of fusion but for some reason the conversation here is only about sublimation (I suppose sublimation and vaporization enthalpies combined approximate that of fusion).

@BuckThorn. It follows from Hess' law that, at the triple point, the heat of fusion is equal to the heat of vaporization minus the heat of sublimation.

Actually, the heat of fusion is equal to the heat of sublimation minus the heat of vaporization.


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Diamagnetic vs Paramagnetic
As far as I know, to know whether a substance is paramagnetic or diamagnetic, we check if that particular species has got unpaired electron or not.
But my textbook lists-
Examples of Diamagnetic Materials - Copper,Lead,Sillicon,etc
Examples of Paramagnetic Materials - Aluminium,Sodium,Calcium etc
The valence shell configuration of Copper is 3d10 4s1 . So, shouldn't it be Paramagnetic?
The valence shell configuration of Calcium is 4s2 . So, shouldn't it be Diamagnetic?

You're right that Cu0 has one unpaired electron in the 4s orbital. If you were to observe a single Cu0 atom in the gas phase, you would observe some paramagnetic behavior.
However, in metallic copper, the valence orbitals are engaged in metal-to-metal bonding. The 3d and 4s orbitals interact in constructive overlap with neighboring copper atoms to produce bonding molecular orbitals.
When two 4s orbitals (each containing 1 unpaired electron) combine to form a sigma bonding MO, the electrons in that bonding MO become paired, per the image below.
Note: this MO diagram depicts the bonding in H2, but I'm just using it to illustrate the principle of partially filled atomic orbitals coming together to form a filled bonding MO

The process of deriving the MO diagram for a metallic solid is obviously more complicated than this, but the answer to your question is that a single Cu0 atom is paramagnetic, but Cu2 would be diamagnetic. By extension, metallic copper is effectively Cu$_\infty$ and also diamagnetic.
As you can tell, the process for determining whether an atom or ion is para/diamagnetic is a very different question than whether the bulk substance (metal) is para/diamagnetic. You can't do the latter by electron counting alone; you need to know more about bonding in the substance. As another example of this, you mention that lead is diamagnetic. Lead is another example of a paramagnetic atom (6p2; the 6p orbital contains two unpaired electrons) - but lead metal is diamagnetic because those electrons become paired in the bonding MOs.

What about Sodium, Calcium and Aluminium? Shouldn't ALL these three be diamagnetic for the same reason i.e. they exist in crystals and so no electron is left unpaired? So, why are these paramagnetic?

Like I said in closing my answer, the (para)magnetic properties of bulk metals is far more complicated than my explanation and can't easily be predicted by electron counting. You get into metallic bonding, band structure, and a lot of solid-state physics that I don't understand well enough to explain. For example, look at the relative magnetic susceptabilities of Li and Na in this table: https://en.wikipedia.org/wiki/Paramagnetism#Delocalization

Why does Li have twice the magnetic susceptibility of Na, when they are both s1 in their valence shells? It has to do with the mechanism of Pauli paramagnetism, which, again, I can't explain.

So, basically I have to mug up the examples..but good explanation on your part although I don't know what Pauli Paramagnetism is .


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Is there a gauche interaction between C-3 and C-6 in cyclohexane?
To my understanding, a gauche interaction occurs between two R groups when staggered by 60 degrees in a Newman Projection. When looking down the C-1 to C-2 bond of a cyclohexane, the Newman Projection shows that C-3 and C-6 are staggered by 60 degrees. Would this still be a gauche interaction, and if so, why does it not result in an energy increase in the chair conformation? I was told this is because cyclohexane in the chair conformation acts as a long chain carbon, but I did not understand how that results in no energy increase.

I don't think Newman projections are a very good way to understand the conformations in cyclohexane. Build 3D models instead.

Yes, you're exactly right. Not only is the axial methyl gauche with respect to the methylene unit that you point out, but there are also gauche interactions of the two methylenes that are drawn in the "bridging" positions of your Newman projections.
So, why is the chair conformation so stable?
For one, gauche interactions are really not that destabilizing. You're talking about maybe a few kJ mol-1. Realistically, the axial interactions between that methyl group and other axial hydrogens is going to be more energetically unfavorable. But this is easily solved by a ring flip, which maintains the chair conformation.
Whoever mentioned that chair cyclohexane "acts like a long carbon chain" was probably referring to the fact that the bond angles are close to 109.5 degrees, which means that it is far less strained than rings that are fewer than 5 carbons or greater than 7 carbons. However, that's true for boat cyclohexane as well, so it's not that relevant to the question at hand.
For reference, boat cyclohexane is about 25 kJ mol-1 higher in energy than chair cyclohexane, so the gauche interactions you're talking about are a very minor contribution the the energy of the molecule.

Axial methylcyclohexane has two gauche interactions. One is between the CH3-C1 bond and the C2-C3 bond shown in red. The dihedral (torsional) angle is ~60o. Given the plane of symmetry passing through CH3-C1-C4, the second gauche interaction is on the opposite side of the plane (CH3-C1 bond and the C6-C5 bond). Each of these interactions is worth ~0.9 kcal/mol, a value that is reminiscent of the value for gauche n-butane.
Equatorial cyclohexane has two anti-butane interactions (CH3-C1-C2-C3 (in red) and (CH3-C1-C6-C5). As in the case of anti-n-butane, it is assigned a value of 0 kcal/mol. Thus, the axial conformation is ~1.8 kcal/mol less stable than the equatorial conformation.
Although it is true that there are gauche interactions within the ring, these interactions obviously cancel out in any case because they are the same in both conformations. It is differences in energy we are evaluating, not absolute values. It is the interactions of ring substituents with atoms in the ring that are at issue.


text:
NaCl and sour taste
This is a question from a 7th grade chemistry class.

Why does adding salt ($\ce{NaCl}$) to a cucumber soup makes it taste more sour?

My best guess is that the salt affects the dissociation levels of acids in the cucumbers - adding $\ce{NaCl}$ shifts the balance and more $\ce{H+}$ is being released, but I don't think thats the answer they are looking for. What affect does $\ce{NaCl}$ have on the sourness of food?

Salting is for enhancing taste in general. Most likely related to ( perhaps your) taste sensing and biology than chemistry.

That could be true, of course :) the question is from a 7th grade chemistry class - the teacher asked the studnets that assigment. Im not a chemist, but I work in medicine and I have no clue to be honest.

Adding NaCl will not shift the proton equilibrium at all. But on the other hand, I don’t know whether the sour taste comes from excess protons.

Note that taste is a subjective physiologic phenomena, influenced, but not determined by substance properties. While the activity of hydrogen ions may be affected by the salt presence, the acidity reporting taste sensors are affected much more.

This is a really interesting question. Mechanistically, it might all come down to flavor science/biology instead of simple chemistry.
However, I did find a really cool old paper that described how adding salts such as NaCl to acidic HCl solutions can increase the acidity (further lower the pH) of the solution.
The mechanism they suggest is that because the Na+ and Cl- ions are hydrated when dissolved (i.e., they are surrounded by a "shell" of neutral water molecules), those ions effectively end up competing with H+ (or H3O+ or however you want to depict solvated protons, which is a complicated question in itself) and therefore the acidic protons are less "protected" by the surrounding "shell" of water molecules, which increases the acidity.
I would be surprised if this is the answer that the teacher was expecting the 7th graders to come up with, though.

That's interesing, even for general knowledge. In all my teaching classes nobody metioned that, as probably it is either not know, or the outcome is so marginal.
After some thinking I concluded that the nawser to my question may be coneccted not such with acid reaction as more with osmosis. I remembered that when you make sour cabbage you salt it to get the water out of the cells. So maby more water equals more protons from the organic acid from the cucumbers. Its just a guess thew.

@Mat your last comment is probably relates to what was in the mind of the teacher. Say that perhaps a salad preparation could be more appropriate as example. Removal of water might enhance taste of food, at least if a relationship "more concentration of something → more taste" holds. Honestly as formulated I don't think it has an answer. Cucumber aren't really sour first of all, and taste isn't a straightforward linear chemical parameter. I think the teacher created an exercise out of his personal experience. Perhaps an oversalted soup made it :)

That is quite possible :)


text:
Colour in muriate of Potash?
Alkali earth metals produce white coloured salts with halogens like chlorine. But Muriate of potassium available in India generally used as a fertilizer is pink coloured(reddish white actually), why is it so?

You found your nomenclature in XIX century book? Or older?

Yes muriatic acid is an old name of HCl. Muriatic means brine, but for strange reason muriatic earth is magnesium oxide :)

If it’s sold as a fertilizer, it probably won’t be pure KCl. In addition, it may be coloured pink on purpose so it isn’t mistaken for table salt.

The color is due to iron oxides as impurities.

Related: How is Persian blue rock salt made?

If you have seen pink salt crystals from Khewra mines in Pakistan (the so-called Himalayan pink salt) sold at exorbitant rates in grocery stores, you would find out it is nothing but NaCl with a small amount of iron oxides as stated above. If you dissolve large tablespoons of such salt crystals in water, you will see a brownish precipitate sitting at the bottom on the container.
The word muriate reminded me of ancient chemistry books of the 1940s. It is a very very old name for a salt of hydrochloric acid.


text:
Why is this reaction not a redox?
My question may seem too simple. But I couldn't find why this is not a redox.
$$\ce{Hg2(NO3)2 + 2KBr -> Hg2Br2 + 2KNO3}$$
In $\ce{Hg2(NO3)2}$ compound atoms have following states:
O: -2
N: -3
Hg: +9
In $\ce{KBr}$ compound atoms have following states:
K: +1
Br: -1
In $\ce{Hg2Br2}$ compound atoms have following states:
Hg: +1
Br: -1
In $\ce{KNO3}$ compound atoms have following states:
K: +1
N: +5
O: -2
Hg changed from +9 to +1, so reduced. N changed from -3 to +5, so oxidized.

No atom ever can be +9.

@IvanNeretin Tetroxoiridium(IX): *cough cough*

Well, at least no mercury atom can hit +9 in ordinary Y chemistry (I've read that maybe there us +4, on the outside edge).  Nitrogen in nitrate ion is +5, not -3.

You got the oxidation numbers in the $\ce{NO3^-}$ anion wrong. Remember that the sum of the oxidation numbers in a species must be equal to the total charge. You claimed oxygen has a $-2$ oxidation number and nitrogen has $-3$, this would mean that you would not have an $\ce{NO3^-}$ ion but an  $\ce{NO3^9-}$, which is clearly not the case. Try fixing the oxidation numbers here and you would hopefully be able to understand the problem.

So if a compound has more than two atoms, firstly I must split it into two. One or each of them is a polyatomic ion. Each of them has suitable ions. Then find the oxidation states of each atom in the polyatomic ion. My mistake was trying to  find oxidation states of each atom directly. Am I right?

If talking about oxidation numbers then yes, you will need to find them for each atom. For ionic compounds, it is often the most helpful to do this for each ion in the system, because as I have said, the charge needs to add up for each ionic species too, and this can help in many cases. Also note that you had the same anion in the KNO3 compound and there you got it right. It is exactly the same anion there, no reason to have different oxidation numbers than before.

You got oxidation numbers mixed up. As the comments pointed out, the +9 state is possible only in very few elements and mercury is not one of those.
To my understanding in general chemistry, this problem can easily solved using Solubility Rules:

Salts containing Group I elements ($\ce{Li+, Na+, K+, Cs+, Rb+}$) and salts containing ammonium ion ($\ce{NH4+}$) are soluble in water. There are few exceptions to this rule.
Salts containing nitrate ion ($\ce{NO3-}$) are generally soluble.
Salts containing $\ce{Cl-, Br-, or I-}$ are generally soluble. Important exceptions to this rule are halide salts of $\ce{Ag+, Pb2+,}$ and $\ce{Hg2^2+}$. For example, $\ce{AgCl, PbBr2,}$ or $\ce{Hg2Cl2}$ are insoluble.
Most silver salts are insoluble. $\ce{AgNO3}$ and $\ce{Ag(C2H3O2)}$ are common soluble salts of silver; virtually all others are insoluble.
Most sulfate salts are soluble. Important exceptions to this rule include $\ce{CaSO4, BaSO4, PbSO4, Ag2SO4}$ and $\ce{SrSO4}$.
Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements ($\ce{Ca, Sr,}$ and $\ce{Ba}$) are slightly soluble. Hydroxide salts of transition metals and $\ce{Al^3+}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not soluble.
Most sulfides of transition metals are highly insoluble, including $\ce{CdS, FeS, ZnS,}$ and $\ce{Ag2S}$. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
Carbonates are frequently insoluble. Group II carbonates ($\ce{CaCO3, SrCO3,}$ and $\ce{BaCO3}$) are insoluble, as are $\ce{FeCO3}$ and $\ce{PbCO3}$.
Chromates are frequently insoluble. Examples include $\ce{PbCrO4}$ and $\ce{BaCrO4}$.
Phosphates such as $\ce{Ca3(PO4)2}$ and $\ce{Ag3PO4}$ are frequently insoluble.
Fluorides such as $\ce{BaF2, MgF2,}$ and $\ce{PbF2}$ are frequently insoluble.

Thus, according to Rules 1, 2, and 3, $\ce{KNO3}$ (Rules 1 & 2), $\ce{Hg2(NO3)2}$ (Rules 2), and $\ce{KBr}$ (Rules 1 & 3) are soluble in water: Thus, they tends to dissociate in water and exist in their ionic forms such as $\ce{K+ (aq), Br- (aq), Hg2^2+ (aq)}$ and $\ce{NO3- (aq)}$ until $\ce{Hg2^2+ (aq)}$ reacts with $\ce{Br- (aq)}$ to give $\ce{Hg2Br2}$.
According to Rule 3, $\ce{Hg2Br2}$ is insoluble in water so that it stay in solid form.
Now, look at your reaction:
$$\ce{Hg2(NO3)2 (aq) + 2 KBr(aq) -> Hg2Br2(s) + 2 KNO3(aq)}$$
Thus, in total ionic form:
$$\ce{Hg2^2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + 2 Br- (aq) -> Hg2Br2(s) + 2 K+ (aq) + 2 NO3- (aq)}$$
When remove the common ions, you got the net ionic equation:
$$\ce{Hg2^2+ (aq) + 2 Br- (aq) -> Hg2Br2(s) }$$
Can you say it is a redox reaction now? It is a simple precipitation.
Reference sited by the source:
Ralph H. Petrucci, F. Geoffrey Herring, Jeffrey D. Madura, and Carey Bissonnette, In General Chemistry: Principles and Modern Applications; 10th ed.; Pearson Education: Upper Saddle River, New Jersey, 2011.

@Oscar Lanzi: Nice edit. I appreciate your good edit to make this answer better.


text:
Why is vapour pressure not dependent on shape and size of the container?
In my textbook, it's written that vapour pressure of a liquid does not depend on shape and size of a container but won't a container that provides less surface area to the liquid will have lower vapour pressure?
for example, consider 2 containers give in the diagram below, the volume of container and liquid is same in both the diagram just the shape is changed.

wouldn't the vapor pressure be lower in the diagram on the left side? and how does changing size not affect vapor pressure?

Consider though that "vapor pressure" is really "equilibrium vapor pressure" thus ignoring dynamics. Due to smaller surface area the left system would be slower to come to equilibrium than the right system.

For a sufficiently tall container, vapor pressure does vary with height, in a gravitational field, so it is dependent on shape.

At equilibrium, both forward and reverse rate increase with increasing surface area, so the equilibrium constant is not affected. Your diagram is lacking blue molecules in the gas phase.

Changing the surface area doesn't change the equilibrium because the surface area controls both the rate of evaporating and the rate of condensation.

Rethink on what is Pressure. The same holds. And in  your sketch what changes is the time needed to reach equilibrium. It is a not clear idea about P that troubles you! Not necessarily vapor pressure.

@DrMoishePippik I am afraid your comment is confusing OP.

@Alchimista, a tall, thin container with the same volume (and even surface area) as a short, wide one will not have the same air pressure (or vapor pressure) at the top of the container, easily measured with a barometer. The statement in the text is incorrect.

@DrMoishePippik. Yes and of course. But this remark about tall containers in a gravitational field can be done all the times if the gas isn't perfect. Thus, for basically every question regarding gases. What is important is the P at the condensed ph./ gas interface, or, by definition, vapour P is function of T only. The P atop has nothing to do with vapour P has for that region is not an interface. And again the same thinking applies to P in general. Atm P at ground and at K2 are of course different. But vapour P is the same. It might be a different definition in earth science and meteo...

...because if one study the atmosphere I guess he has to deal with hydrostatic anyway.

Vapor pressure does not depend on surface area because it is derived from the thermodynamic equilibrium between the liquid and gas phases of a substance. The closer the vapor pressure is to atmospheric pressure, the closer that substance is to boiling. At its boiling point, vapor pressure equals atmospheric pressure and the molecules that comprise the liquid now have enough thermal energy to overcome the intermolecular forces that are maintaining it in a condensed phase. So it might help to think about how container geometry does not affect the boiling point of a liquid; the same reasons are true for why it does not affect the vapor pressure.
The evaporation rate in terms of mass/time is absolutely a function of the container geometry, but equilibrium vapor pressure is an intrinsic molecular property.

Evaporation of a liquid (and condensation of vapour) are physical processes. We can write the equations for a liquid material $\text{M}$ as follows:
$$\ce{M_{(l)}} \rightleftharpoons \ce{M_{(g)}}$$
By the law of mass action, we can write the expression for the equilibrium constant as follows:
$$K = \dfrac{[\ce{M_{(g)}}]}{[\ce{M_{(l)}]}}$$
We typically use partial pressures for the activity of gases and unity for the activity of liquids. Thus,
$$K_{p} = \dfrac{p_{M}}{1} = p_{M}$$
Suppose we start with the liquid $\ce{M}$ in a container with an arbitrary amount of headspace. It will evaporate till the vapour exerts a pressure equivalent to $p_{M}$ (of course, $p_{M}$ depends on the temperature).
Thus, with a change in the size of a container, there will be a difference in the amount of $\ce{M}$ in the vapour phase, but not the partial pressure exerted by the vapours of $\ce{M}$ at equilibrium (a.k.a. the vapour pressure).
The shape might affect the flux at which the dynamic exchange between vapour and liquid phase $\ce{M}$ is taking place, but nothing more.


text:
Calculation of VOCs concentration (ppm) in the headspace of a bottle
What is the right way to calculate the concentration of volatile compounds in the air in a closed container?
Suppose we have a $\pu{50 mL}$ bottle with $\pu{25 mL}$ of $100\%$ ethanol $(\ce{EtOH})$ in it.
In this case, will it be okay to use $PV = nRT?$
Doing so, it will be:
$$
\begin{align}
p(\ce{EtOH}) &= \pu{0.05872 atm}\\
V &= \pu{0.025 L}\\
R &= \pu{0.08206 L atm mol-1 K-1}\\
T &= \pu{298 K}\\
\end{align}
$$
giving us $n = \pu{6.00314E-5 mol}.$
The ppm: $\pu{1 ppm} = \pu{1 mg l-1}$
$$M(\ce{EtOH}) = \pu{46.07 g mol-1}$$
$$\pu{6.00314E-5 mol} × \pu{46.07 g mol-1} × 1000 =  \pu{2.766 mg}$$
This lead to $$[\pu{ppm}] = \frac{\pu{2.766 mg}}{\pu{0.025 l}} = \pu{110.6 ppm}$$

You have calculated the number of moles of ethanol. Do you need something else to find the concentration?

I just want to verify that it is the right way to calculate concentration of volatile liquids in the air? Thank you!

Just a couple more steps and you're done.

Thank you, made them, is it right? As this calculator giving me completely different answer http://madur.com/index.php?page=/partial_pressure

Don't you think approx.0.06 atm should lead to approx.60000 ppm v/v, even without any calculation ?

Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) <> 1 ppm (V/V) <> 1 ppm (n/n) <> 1 ppm (w/V).
It is recommended to use explicit units, like e.g mg/L.
Expecting ideal gas behaviour $pV = nRT$ is not reasonable.   Ideal gases are not in equilibrium with their liquid phase.
There should be rather used a real gas equation.The easiest to use is the van Der Waals equation, but even that has significant error near condensation state.
Additionally, the sample(!) temperature  has to be strictly controlled, otherwise $n$ would be variable.
Why not rather to use small volume of ethanol to be known to fully evaporate in given volume ? Or perhaps other solvent, unless ethanol is waterless.
Note that ethanol has significant deviation from the Raoult's law.

I prefer $R = \pu{8.31446 J K-1 mol-1}$
Note that by May 2019 redefinition,
 It's exact value is $\pu{8.31446261815324 J K-1 mol-1}$

Thank you for your answer! I will follow your advises to try to calculate a more reasonable number. Could you please also advise on the ppm calculation in that case? Thank you!

@VGranin ppm is ambiguous: w/w, V/V, n/n ?

Pure liquid water would be 1,000,000 ppm w/w, V/V, n/n, and m/V. Pure steam at normal pressure and a temperature a bit above the normal boiling point on the other hand would be 1,000,000 ppm w/w, V/V, n/n, but ~800 w/V because the density of steam is much lower than that of liquid water.

Using ppm is a recipe for disaster...

Engineer here. Using the vapour pressure and the ideal gas law (IGL) will usually give a sufficiently accurate solution for gases for which $\dfrac{R\cdot T}{P} > 20 \frac{L}{g\ mol}$.
That said, if you want to correlate vapour-liquid equilibrium, using the IGL is not a good idea.


text:
How is the ln k term in the linearised Arrhenius equation acceptable?
In my syllabus we only study about rate laws under the topic "Rates of reactions", but as I was interested in it, I started to search about that topic. And then I found the Arrhenius equation
$$\ln k = \ln A - \frac{E}{RT}.$$
But according to this equation rate constant $k$ doesn't have units because we are taking the natural logarithm of it. According to rate laws, rate constants have units and they depend on the reaction.
Are we talking about two different rate constants here? I searched this on internet and all the books I could refer but I couldn't find an answer.

Rewrite as $$\ln \frac{k}{A} = - \frac{E_A}{RT}$$ and $k$ is allowed to have the same units as $A$.

I saw this equation in this format too but my problem occurred when I saw the above equation was used to draw a graph between lnK and T.

There are two problems: The y-axis will be ln k, and you are supposed to read off ln A from the y-intercept. You circumvent the problem by plotting ln (k / units), where "units" are the appropriate units for k. The y-intercept will be ln(A / units).

The graph would be ln k vs. 1/T to get a linear relationship (and -Ea / R as slope)

The reason is because of sloppy notation. This sort of logarithmic plot is widespread in chemistry, though, and isn't limited to the Arrhenius equation, so do not expect it to change very much in the near future. You will have to get used to it.
With the Arrhenius equation
$$k = A \exp\left(-\frac{E_\mathrm a}{RT}\right)$$
you cannot directly take logarithms on both sides, because of the inherent units in $k$ and $A$. First you would have to divide through by a constant which we could call $k^\circ$:
$$\frac{k}{k^\circ} = \frac{A}{k^\circ} \, \exp\left(-\frac{E_\mathrm a}{RT}\right)$$
Now we can write
$$\ln \left( \frac{k}{k^\circ} \right) = \ln \left( \frac{A}{k^\circ} \right) - \frac{E_\mathrm a}{RT}$$
and as long as $k$, $k^\circ$, and $A$ are expressed in the same units, the resulting arguments of the logarithms will not have units. Note that the actual value of $k^\circ$ does not matter: you will of course get a different y-intercept, but the final values of $A$ and $E_\mathrm a$ that you get out of the Arrhenius plot will not be affected.
The division by $k^\circ$ is often implicit. When people write $\ln k$, what they usually mean is $\ln(k/k^\circ)$ where $k^\circ$ has the numerical value of $1$ and the units of whatever $k$ is in. Once you have the y-intercept, you take the exponential of that and tack the units back on to get $A$.
Although it is not rigorous, from a practical standpoint, this generally does not lead to much confusion as long as the author is consistent with their usage of units. Another reason for omission is that the actual value of $k^\circ$ has no physical effect, as previously discussed, and an overly pedantic focus on writing $k/k^\circ$ and $A/k^\circ$ can distract one from the chemical principles behind these equations, especially for more complicated equations which have many more symbols.

The above mentioned approach is described in detail in a nice article "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions" in the Journal of Chemical Education. I think that this is rather a new invention, or to justify the "dimensionless-ness" of transcendental functions. The original authors never bothered to define their functions that way. They were only interested in the numerical values. DImensional analysis is rather a later invention, much later than the work of Arrhenius and Sorenson.

This is the same thing what confuses some people with the equilibrium constant $K$.  A lot of undergraduate textbooks define $K$ as having units depending in the stoichiometry of the reaction in question. Then they move to Gibbs free energy of the reaction through $ \Delta G = - RT ln(K)$ and just kind of forget that they spend an awful amount of time on the units of $K$ seven lectures ago.

@Ezze, it’s not exactly the same situation. K is supposed to be dimensionless. The rate constant k is not supposed to be dimensionless.

$$
\begin{align}
k &= A \exp\left(-\frac{E_\mathrm a}{RT}\right) \label{eq:1}\tag{1}\\
\ln k &= \ln A -\frac{E_\mathrm a}{RT} \tag{2}
\end{align}
$$
If you use the equation \eqref{eq:1} to get the units for $A,$ you can get the units for $A$ as the exponential term is unitless.
The units of $A$ are the same as those of the rate constant since $A$ carries the units for $k.$ The units for $A$ vary according to the order of the reaction^

The first order: $\pu{s^-1}$
The second order: $\pu{M^-1 s^-1}$


text:
Could dimerization account for weak IR signals?
How would you interpret the IR spectrum of your product if it's peaks match the literature spectrum, but the signals are weak? Could the argument that the sample was very dilute and lots of water was in it work as a viable explanation?
What about in a scenario where the experimental IR spectrum has all necessary peaks but is missing a key peak in, for example, the 2000-3000 region?
I synthesized acetylglycine and am trying to see why I am missing a peak.
For the purposes of this lab, it's not necessary to have absolute purity of compound but for my own curiousity, i'm trying to figure it out.
I'm trying to consider all the physical reasons for why my IR may not match the literature, while my melting point ranges are sufficiently accurate, meaning my product shouldn't have too many impurities.
I believe a dimerization could have taken place where for example, a carboxylic acid terminal of one molecule hydrogen bonds with the carboxylic acid terminal of another. Carbonyl oxygen to hydroxyl group, hydroxyl to carbonyl oxygen.
This could lead to a suppression of the hydroxyl (-OH) signal.

Could you show the data? Is the key peak strong or weak in the literature spectrum? Impurities would not explain a missing peak. What is the expected concentration in your sample (let's say you assume 100% yield), and how does it compare to the concentration in the literature spectrum? Are you using a solvent correction and do your solvents absorb in the 2000-3000 cm(-1) region?

Intensity is (in general) proportional to sample concentration, so yes, if you are performing IR in solution, then an overly dilute sample would have less intense peaks. But solution IR is somewhat uncommon; I suspect you are either doing IR on your sample in neat form (using an ATR stage) or in a pressed KBr plate? If the latter, you may not have used enough sample.
Regarding your missing peak: sometimes peak intensities can change, depending on your sample prep. For example, if the peak in question corresponds to proton capable of hydrogen bonding (e.g., -COO-H, -NR2-H, -O-H, etc.) then the sharpness and intensity of the peak will depend on the extent of H-bonding your sample, which, in turn, can depend on things like sample concentration, the presence of water, etc. Sometimes these peaks get broadened to the point of being invisible if your sample is not very concentrated.
If your missing peak corresponds to a C-H, C-C, C-O, C-N, or some other non-labile bond, then there really isn't an explanation other than your compound not being what you think it is...

i was just speculating but in actuality i was using a solid sample and i was trying to see if i placed too little of the 'dust' on the IR spot. but your answer is good and helps me think about the possibilities. Thank you!

I think all modern teaching labs use attenuated total reflectance (ATR)-FTIR systems... you place some crystals and then collect the spectrum right away. ATR is least sensitive technique because the IR penetrates only only a few nanometers of the sample. In good old school times, one would dry the sample-literally bone dry, then oven-dried KBr was ground with the sample and a pellet was made under intense pressure. The resulting spectrum was a high quality spectrum with reasonable transmission values. Water and remaining solvents can often ruin the IR spectrum. There is always a comprise in life in terms of quality vs. quantity. ATR is fast but spectrum quality is not that great.

Good answer. I just want to point out that ATR has many uses and pros, otherwise it seems that it only allows for quick analysis.


text:
What element has good conductivity but is also brittle?
I'm confused by what element(s) can be identified to have these properties.
The following is text from my 2019-2020 Kaplan MCAT General Chemistry book under the metalloids section

Silicon (Si) has a metallic luster but is brittle and a poor conductor

So onto the MCAT Concept Check 2.2:

Identify a few elements that  likely possess the following  properties

Good conductivity but brittle

I flip to the answer key and it says:

Metalloids exhibit brittleness but good conductivity

So I'm confused how this makes sense because Si was the main example they provided in that passage, and it seems to contrast the two points they make.
So what is an actual example of good conductivity but brittle??

Graphite, an allotrope of carbon, is a decent conductor and brittle

Sorry what I meant specifically is what elements on a periodic table, not allotropes, isoforms, different configurations, etc

What is the cutoff value for "good conductivity"? The Wikipedia article on metalloids gives conductivity of arsenic as 3.9 × 10⁴ S•cm⁻¹ and the conductivity of antimony as 3.1 × 10⁴ S•cm⁻¹. This looks like it might make them qualify as "good conductors".

There's no contradiction there.  Silicon, in and of itself, is not a good conductor, as the passage says.  Silicon in semiconductors is doped with a number of elements to introduce "holes" for electrons.

@user3665690 Different allotropes of elements have different properties including conductivity so it makes no sense to ask for the properties of "an element". Graphite and diamond are not the same but are both carbon.

Whether a "metalloid" is both brittle and a good conductor depends on what you mean by "metalloid", which varies in literature, what is a "good" level of (electrical) conductivity, and even what exactly you mean by "brittle" versus "ductile".  Silicon, for instance, is only a semiconductor, with much less electrical conductivity than metals, when pure.
Wikipedia identifies beryllium and magnesium as brittle metals, although the latter can be made ductile with minor alloying.  Zinc is called "slightly brittle".


text:
Mutually immiscible liquids
How many mutually immiscible liquids can be stored in a glass bottle at S.T.P.?
According to

PATTLE, R. E. (1950). Systems of Mutually Immiscible Liquid Layers. Nature, 165(4188), 203–204.

a system of 6 immiscible phases can be formed (including mercury), but the ammonia layer disappears after a month. They suggest gallium could add another layer, giving:

ammonia solution
isobutyl alcohol
liquid paraffin
potassium carbonate solution
methylene iodide
gallium
mercury

Could silicone oil and fluorocarbons create additional stable layers, and have there been any other systems suggested since 1950? I saw a 2006 reference with allegedly 8 phases.

I guess they could. Temporarily there cold be much more layers even from miscible stuff.

Better than gallium, could the alloy Galistan be used? It’s a mixture of Gallium, indium and tin. Other liquids like Bromine, Cesium and NaK might have some potential but may just react explosively.

If one allows miscible liquids, there could be layers with just two solvents, with differing amounts of solutes added to make one layer denser than another. The layers might stay separated for months or years, unless shaken or stirred.
By adding denser solutes, and by increasing the amount dissolved, the series could be extended indefinitely, with alternating miscible layers separated by a non-miscible layer.
For starters, a thought experiment:

Top layer octanes, 0.7 g/ml
Ammonia water, 0.9 g/ml
Octanes + a little bromoethane, 0.95 g/ml
Water, 1.0 g/ml
Octanes + more bromoethane, 1.1 g/ml
Water + $\ce{MgSO4}$, 1.2 g/ml
Octanes + yet more bromoethane, 1.3 g/ml
Water + $\ce{BiCl3 + HCl}$, 1.4 g/ml
usw.

Actual calculations are left to experimenter ;-)
And as for shaking and stirring, it would be interesting to see everything merge into just two layers.


text:
Can neutrons be ejected from an atom?
In a neutral atom, can neutrons be ejected, and if so, by what (nuclear) process or reaction? I have read Neutron emission and Can a proton be ejected from an atom, but I don't understand how and if that would work for removing neutrons from a heavy atom such as lead-208. Can neutrons be ejected from lead-208 so that it is lead-197, and if not, why?

https://en.m.wikipedia.org/wiki/Neutron_emission

@Tyberius How would that work for lead?

Neutron spallation is one way to remove neutrons from atoms. Bring in a high energy proton beam and cause (p,n) reactions.

The typical binding energy per nucleon $(E_\mathrm B/A)$ of most nuclides is about $5{-}8\ \mathrm{MeV}$. Such values are higher than the released energy of a typical radioactive decay. Therefore, emission of neutrons doesn’t occur during most usual radioactive decay processes.
In order to make a neutron source, you have to find a suitable target (not $^{208}\mathrm{Pb}$) and a corresponding nuclear reaction that releases a neutron and that is also energetically favourable, for example
$$^9\mathrm{Be}\ (\alpha,\mathrm n)\ ^{12}\mathrm C$$
$$^9\mathrm{Be}\ (\gamma,\mathrm n)\ 2\ ^{4}\mathrm{He}$$
Note that $^{12}\mathrm C$ and $^{4}\mathrm{He}$ are very stable nuclides with a relatively high binding energy per nucleon so that these reactions can release a large amount of energy, but still you need to irradiate the beryllium targets with high-energy radiation (high-energy $\gamma$ radiation of more than $E=1.665\ \mathrm{MeV}$ e.g. of $^{124}\mathrm{Sb}$ or $\alpha$ radiation) to release neutrons.
Nevertheless, neutron emission can occur during radioactive decay if the neutron excess is large and the beta decay energy is high enough, i.e. at the edge of the valley of beta stability. At very high values, spontaneous emission of neutrons becomes possible.
More important, however, are so-called neutron precursors, e.g. $^{87}\mathrm{Br}$ or $^{137}\mathrm{I}$, which have a very high decay energy so that the product nuclides ($^{87}\mathrm{Kr}$ and $^{137}\mathrm{Xe}$, respectively) can get enough energy to make neutron emission possible in the next step.
$${}^{87}\mathrm{Br}\xrightarrow[{E_\mathrm{max}=6.83\ \mathrm{MeV}}]{\beta^-}{}^{87}\mathrm{Kr}^*\xrightarrow[\quad]{}{}^{86}\mathrm{Kr}+\mathrm n$$
$${}^{137}\mathrm{I}\xrightarrow[{E_\mathrm{max}=5.89\ \mathrm{MeV}}]{\beta^-}{}^{137}\mathrm{Xe}^*\xrightarrow[\quad]{}{}^{136}\mathrm{Xe}+\mathrm n$$
Such reactions are responsible for the so-called delayed neutrons, which are very important for most nuclear reactors – without this effect, controlling a nuclear chain reaction would be difficult.
Of course, the most important neutron sources are nuclear reactors, which generate neutrons (about 2–4 neutrons per fission) in a controlled nuclear chain reaction.


text:
Basic sites on Indene ion
Answer key says that A is the most basic site but I don’t understand the concept behind that.
To assess the basic site, I’d normally look at the molecule as is and assess the current stability of the atom in question. The least stable would be my strongest base.
Looking at this question both the atoms properties are the same (C with lone pair) so resonance is the next thing I would look at. Both of these rings have resonance which moves (delocalizes) the negative charge around. So I would think since the volume of resonance is greater in the bigger ring, A is more stable thus making B the stronger base.
What am I misunderstanding or what concept am I not considering?

I believe the concept assessed here is the idea of aromaticity. You must first be able to recognise that the lone pair held by each carbon atom with the negative formal charge is in a different type of orbital. In the carbon atom of the 6-membered ring, it is in an $\ce {sp^2}$ hybrid orbital that is in a plane perpendicular to the $\pi$ electron cloud of the indene molecule. This lone pair is not delocalised into the indene $\pi$ system. On the other hand, in the carbon atom of the 5-membered ring, it is in an unhybridised p orbital that is parallel to the other p orbitals of the indene molecule. This lone pair is, in fact, delocalised into the indene $\pi$ system. As such, it is much less available for donation to  an acid.
Protonation at sites C and D would also be unfavourable due to destruction of aromatic character. Between sites A and B, I have explained above that the lone pair at A would be much more available for donation due to lack of delocalisation in to the $\pi$ system. Hence, the preferred site of protonation would be site A.


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What is the reaction between piperine and concentrated sulfuric acid?
Piperine was chemically characterized by reaction with acids. This was done by dropping concentrated $\ce{HNO3},$ $\ce{HCl},$ and $\ce{H2SO4}$ to a sample of piperine. Both $\ce{HNO3}$ and $\ce{HCl}$ produced yellow solutions while sulfuric acid produced a blood-red solution.
What happens to piperine as it reacted with $\ce{H2SO4}$ and why it turned blood-red? Also, what is the difference of that reaction with the reaction of piperine with $\ce{HNO3}$ and $\ce{HCl}?$
Edit: Late addition for viewer's convenience:
The piperine structure is shown below:

Would help if you gave the structure of Piperine

Interesting question... The reactions with the acids have likely changed the conjugated system, thus resulting in different wavelengths of light being absorbed by the resultant compounds. Nitration may occur with conc. nitric acid but as to why both nitric acid and hydrochloric acid both give a yellow solution, it is indeed puzzling.

The methylenedioxyphenyl system is very electron rich so protonation seems a possibility

@Waylander: I have added the structure for your convenience.

I think OP is confused with the colorimetric method of detection of piperine, which uses the concentrated sulfuric acid as one of reactant with chromotropic acid, 1,8-dihydroxynaphthalene-3,6-disulfonic acid (Chromotropic Acid Test; Ref.1 & 2). Accordingly, this colorimetric method, developed in 1965, relied on hydrolyzing the methylenedioxy group of piperine by chromotropic acid in the presence of concentrated sulfuric acid to formaldehyde, upon which a definite color was developed with chromotropic acid (Ref.3). Color intensity is proportional to the amount of piperine present. Water inhibits color development, but a small amount of methyl alcohol is necessary. The abstract of Ref.3 tells them all:

A highly stable reagent consisting of para-dimethylaminobenzaldehyde (I) or para-hydroxybenzaldehyde (II) or other cylic aldehydes, plus thioureau in concentrated sulfuric acid, has been developed for use in determining piperine. The slightly yellow reagent, if stored under refrigeration in a brown bottle, maintains a constant chromogenic capacity for 4–6 weeks. Piperine when heated with this reagent for 15 min at 100°C develops a red color (I max. 490 mμ) or purple color (II max. 570 mμ) whose intensity is proportional to the amount of piperine present. Water inhibits color development, but a small amount of methyl alcohol is necessary. Ethyl alcohol cannot be used, since it develops color with the reagent. Critical factors for reproducible, quantitative assay are the concentration of acid and thiourea in the reagent and the time and temperature of heating the reaction mixture. For I, the most sensitive of the aldehydes used, good precision, with a standard deviation of it ± 3470, was attained. Between the levels of 0.01 and 0.06μM of piperine per ml of reagent, there is a linear relation which may be described by the least-squares equation.

The theory behind the test is concentrated $\ce{H2SO4}$ would knocked off formaldehyde from its cyclic acetal form in piperine and formaldehyde then react with formaldehyde specific reagent to form the red color. What happen with the other two acids, $\ce{HCl}$ and $\ce{HNO3}$, is they can also remove the cyclic acetal protective group from piperine forming free catechol moiety. In the absence of formaldehyde specific reagent in the solution, the highly conjugated catechol moiety predominate to give the supposed color.
Note: It is good to know that chromotropic acid test is very sensitive to presence of formaldehyde and detect them in very low concentrations (Ref.4). It is also fact that chromotropic acid is specific to formaldehyde (Ref.2).
References:

Leila Gorgani, Maedeh Mohammadi, Ghasem D. Najafpour, Maryam Nikzad, “Piperine—The Bioactive Compound of Black Pepper: From Isolation to Medicinal Formulations,” Comprehensive Reviews in Food Science & Food Safety 2017, 16(1), 124-140 (https://doi.org/10.1111/1541-4337.12246).
Clark E. Bricker, Hilding R. Johnson, "Spectrophotometric Method for Determining Formaldehyde," Ind. Eng. Chem. Anal. Ed. 1945, 17(6), 400–402 (https://doi.org/10.1021/i560142a021).
Horace D. Graham, “Quantitative Determination of Piperine. I. The Komarowsky Reaction,” J. Food Sci. 1965, 30(4), 644–650 (https://doi.org/10.1111/j.1365-2621.1965.tb01818.x).
Jingping Zhang, David Thickett, Lorna Green, "Two Tests for the Detection of Volatile Organic Acids and Formaldehyde," Journal of the American Institute for Conservation 1994, 33(1), 47–53 (doi:10.2307/3179669).


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why does adding a non volatile solute not decrease the boiling point?
If I add a solute that decreases the bond strength of the pure solvent then wouldn't the solvent molecules evaporate more easily as it has lesser force holding it down now?But in my textbook, it's written that BP always increases when we add a non-volatile solute.

At least some solvent interacts strongly with the solute, to use your picture.

Decrease of the solvent activity due its decreased molar fraction is bigger than the solvent activity increase due eventual solvent solute interaction.
So the pressure versus solute molar fraction curve may not be linear as by the Raoult law, but is monotonous.


text:
Will there be some ethanol evaporating before the boiling point of azeotrope? And will it bring some water with it because of the hydrogen bond?
Before the specific temperature, even if the temperature doesn't reach the boiling point of azeotrope or ethanol or water, we know that there are still some ethanol evaporate. In this time(before the boiling point of azeotrope(because the boiling point of azeotrope is the smallest)), will there be some water come out with ethanol BECAUSE OF hydrogen bond? I'm confused that if the hydrogen bond will count before the boiling point.

Some water will evaporate, but not because of the hydrogen bond. Also, the azeotrope is not exactly formed due to hydrogen bond.

some liquid will evaporate, but when the temperature is below Tb it will be due to the thermodynamic effect of vapor pressure. This is why you can smell ethanol at room temperature and why laundry dries at 25C.

Your question is (in)valid at whatever ratio, not only for the azeotrope! See @Ivan Neretin comment.

Azeotropes exist for miscible liquids with mixing enthalpy sufficiently different to zero, what leads to existence of either minimum or maximum of total vapour pressure for particular molar fractions.
It also leads to partial pressure not being linear function of the molar fraction, being a deviation from the Raoult's law.
Hydrogen bonds themselves do not lead to the azeotrope. The important thing is the change in the average enthalpy of intermolecular bonding after mixing.
Mixture of liquids forming an azeotrope  acts as if the azeotrope mixture were the 3rd liquid and if there were evaporating the azeotrope and just one ( for binary azeotropes ) of the pure solvents, depending on the composition.
If the mixture is more ethanol rich than the ethanol-water azeotrope, there is evaporating azeotrope plus ethanol. And vice versa.
So at any temperature, there is always some partial vapour pressure of all volatile components, no matter if they form an azeotrope or not, or if there are hydrogen bonds or not.

I understand some parts of it, but not everything because I haven't learnt some concepts. And also because I'm still learning English  Thanks anyway!!:)


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Sodium + water reaction, noxious fumes?
I was reacting pure sodium and tap water, and at one point I put some splattered sodium back into the water. I observed fumes that reacted with my respiratory system, producing a burning sensation.
I read that it is possible to produce sodium oxide fumes from sodium hydroxide at high temperatures (the sodium did burn at the end, producing an orange flame).
My question is what could be the chemical composition of these fumes?

This video shows what happened in 1947 when the U.S. Army dumped 20 thousand pounds of sodium (in 3500 pound containers) into frozen alkaline Lake Lenore, WA, USA:    https://youtu.be/OBm8fM8sV0w  .Not sure how alkaline the lake was prior to the sodium addition, but the clouds rising thousands of feet contain droplets of relatively concentrated NaOH solution. And, of course, very hot water vapor.

I was expecting a snarky comment by Ivan Neretin but that video practically tops any comment he could ever have given!

Related: Vaporized sodium hydroxide?

I agree.  This is a dangerous reaction and should not be tried without proper equipment, including a lab hood and fire extinguishing equipment.
The fumes could be "just" droplets of sodium hydroxide solution.  I've seen similar fumes involving hydrochloric acid, just by adding water to rinse out traces of concentrated acid (in a hood, of course).  They may be carried by increased evaporation even without boiling and without need for a further "chemical reaction".

Well sodium hydroxide(NaOH) is solid at room temperature but at high temperatures it can be converted to gases.This happens due to the fact that the molecules of NaOH gain so much energy that the intermolecular bonds (dipole-dipole) between 2 sodium hydroxide molecules are broken.The reaction of sodium with water is exothermic and produces lot of heat and  sodium hydroxide may have absorbed some of the heat.
What you did was very dangerous and I wouldnt recommend anyone to try it at home without laboratory equipment.

Sodium hydroxide is an ionic compound.

NaOH is still a molecule which contains ions Na+ and OH-.

No, NaOH is not a molecule at normal temperatures and pressure. Maybe there are some NaOH molecules in high temperature flames and such. But that is not relevant to the OP’s question.

Ed V it is an ionic molecule.

No. Look up “sodium hydroxide” in wikipedia. It is an ionic crystalline solid, just as it says there. Also shown is the crystal structure.

Each NaOH is an ionic molecule. In metals each metal cation is an ionic molecule.

Please look at the wikipedia article. I will have no further interaction on this.


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How do I explain the existence of these 2 peaks in this mass spectrometry?
Question about HRMS:Phosphatidylcholines and phosphatidylethanolamines are common types of glycerophospholipids found in biological membranes. Because of the high amount of heteroatom content, electron impact mass spectrometry is a bad idea because the ions fragment too fast. For biological molecules, electrospray ionization is the most common method. A compound in the glycerophospholipid class was analyzed by low-res electrospray ionization. It produced peaks: 773 Th(100 percent abundance), and 795 Th(about 20% abundance). The question says these are not isotope peaks. The same compound was analyzed by high-rest EI mass-spec to have peaks at 772.5856 Th and 794.5680 Th. The question says to explain the 2 peaks’ presences in the high-resolution mass spectrometry.
For question #6, I came up with the molecular formula for C40H80NO8P for both classes of lipids but I was unsure of how to use that to come up with the actual molecular formula. The first lipid I found was dipalmitoylphosphatidylechloline with a molar weight of 734. But it's smaller than either of the 2 fragments. Then I added 6 methylene groups to come up with a molecular formula of C46H92NO8P. But I'm not sure how to get the molecular formula for both fragments(or if we actually need them). Can you help with this question or let me know of what I'm missing?

I don't understand...

Never mind, I thought the mass was different by 2, but it is different by 22.

So the second species could have two carbons more (+24) and two hydrogens less (-2). Making a fatty acid chain longer by two would add two carbon atoms and four hydrogen atoms. For each single bond turned into a double bond, you would get to remove two hydrogen atoms.

But how do you figure out the molecular formula for the first and second species?

You already had a reasonable molecular formula, C40H80NO8P, with a molar mass of 734 g/mol. It ionizes by adding an H+, so 735 g/mol. That is 38 g/mol off from the 773 g/mol you need. You mentioned methylene groups at 14 g/mol. Three of those (not six) brings us to an additional 42 g/mol (overshooting a bit), so make two C-C bonds into C=C double bonds to take away 4 g/mol worth of hydrogen and you are there.

Okay, but then how do I use the molecular formula for both the first and second species in order to come up with a molecular formula for the whole molecule?

Oh, same compound, so it has to be something brought along non-covalently. If there is a Na+ ion instead of an H+ ion, you get +22.

So you're saying that if I add an Na+ ion instead of H+ ion to the molecular formula of the first fragment, I can then get the molecular formula of the second fragment and that molecular formula for the second fragment would be the molecular formula for the whole compound?

In electrospray ionization, a molecule can end up with an extra proton or an extra cation. Either one of these +1 charged species is accelerated and detected in the mass spectrometer (negative charged particles or neutral species are not; if there are multiple cations, they would supposedly repel each other once the solvent is gone, and not fly together).
So a molecular formula that fits the data is $\ce{C43H82NO8P}$. The lighter species detected in the spectrometer would be $\ce{C43H82NO8PH+}$, and the heavier species would be $\ce{C43H82NO8PNa+}$. The masses given by https://www.sisweb.com/referenc/tools/exactmass.htm for these species match the given values pretty well. The difference in mass is not due to fragmentation, but to a cation piggy-backing with the molecule of interest (instead of the usual proton).
The data does not distinguish between phosphatidylcholines and phosphatidylethanolamines. You can, however, infer the number of C=C double bonds in the fatty acids attached to the glycerol.
I had to look up the unit "Th" on wikipedia:

The thomson (symbol: Th) is a unit that has appeared infrequently in scientific literature relating to the field of mass spectrometry as a unit of mass-to-charge ratio. The unit was proposed by Cooks and Rockwood naming it in honour of J. J. Thomson who measured the mass-to-charge ratio of electrons and ions.

HRMS stands for high resolution mass spectroscopy, I think.

Hey, it would be great if you could answer this question: So you're saying that if I add an Na+ ion instead of H+ ion to the molecular formula of the first fragment, I can then get the molecular formula of the second fragment and that molecular formula for the second fragment would be the molecular formula for the whole compound?

I edited my answer accordingly. The covalent compound is $\ce{C43H82NO8P}$ and it forms two distinct positively charged species in the gas phase, one with a covalently attached H+ and one with a sodium ion (not covalently bound).

Why did you not use the values from the high-resolution mass spectroscopy?


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Can concentrated sulfuric acid oxidize iron(II) to iron(III)?
Can concentrated (18 M) $\ce{H2SO4}$ oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}?$ Or does it actually block the oxidation? I read in some forum somewhere that in order to prevent $\ce{Fe^2+}$ from being oxidized, one should acidify the solution. I don't know if that's the reason why we added $\ce{H2SO4}$ in our $\ce{Fe^2+}$ solution. But then again, isn't iron(II) unstable in the presence of acid and water? And will it oxidize?

Welcome to our site!  Please note that Chemistry.SE is a formal place and informal language is usually frowned on, especially the use of expletives.

Please control your language. It does not contribute anything. There are good teachers and bad teachers in the whole world.

@M.Farooq Complete edits are appreciated better!

@M.Farooq I'm really sorry I'm new to this site I'll mind my wordings next time thank you so much for pointing that out

Which reaction are you talking about, i.e. what is the oxidation agent? Sulfuric acid is usually not classified as oxidizing acid (nitric acid, in contrast, is).

The explanation given to you is rather incorrect. Dilute sulfuric acid is added to iron(II) or even iron(III) salts is added to prevent hydrolysis of the salts. What happens when you don't add an acid to iron(II) solution in water. With time, you will see that a brown precipitate is forming and settling to the bottom. Iron forms hydroxides which are very quickly oxidized by dissolved oxygen (not by dilute acid) to a brown rust colored product.
As a result, the concentration is no longer the correct concentration. The same job can be done adding a small amount of HCl so there is nothing special about sulfuric acid. Chloride salts are even more prone to hydrolysis that's why iron(II) chloride or iron(III) chloride are never used for solution preparation.


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Relating collision theory to rate of reaction
I want to relate the macroscopic advancement of a reaction, the rate of the reaction, with the microscopic collisions of the molecules. This is the type I have developed so far:
$$u = A^2 \times \frac{n_1}{m^2} \times\frac{n_2}{m^2}\times \frac{(m_1u_1^2+m_2u_2^2)}{E_\mathrm{act}}\times 1/t $$
where $u$ is the rate of reaction in $\pu{mol2 sec-1}$.
$A =$ the common area of the substances
$n_1/m^2 =$ the moles per square metre of the first substance
$n_2/m^2 =$ the moles per square meter of the second substance
$u_1$ and $u_2 =$ average velocity of molecules of each substance
$E_\mathrm{act} =$ activation energy of the reaction
Now the molecules of the first substance must bump to the molecules of the second substance with the appropriate orientation. How can I add this to my type?

Please use mathJax formatting for variables and equations. I formatted your equation, but unsure of the content you intended to. Please correct if I'm wrong.

Reaction velocity cannot be $\pu{mol \cdot s}$ anyway. So, I corrected accordingly. If you are not comfortable with my edit, you may change it as you want, but I warn you, this would be questioned by others. :-)

It is ok.I will add a constant which will give the correct units of measure.

@BrightFuture To my present knowledge, despite the time-dependent advancement of a reaction of a reaction is subject of (chemical) kinetics, the English term you search is either «rate of reaction» (http://goldbook.iupac.org/terms/view/R05156) or «reaction rate». On  the other hand, your thought of line is very close to the concept of collision theory, the (effective) cross-section of moving molecules is an important parameter.  See, for example, https://en.wikipedia.org/wiki/Collision_theory.  This may justfiy rewording the title of the question.

I know what collision theory is.

This is an interesting question and many experimental method exist to study it. For more information have a look at 'Chemical Kinetics and Dynamics' by Steinfeld, Francisco & Hase.

It is interesting to see you creating your own approach. The comments above provide good responses. I would recommend looking at Buttonwoods suggestion regarding collision theory and cross-sections. However, it appears none have answered your question directly, so I will provide my response to it here.
Question: "Now the molecules of the first substance must bump to the molecules of the second substance with the appropriate orientation. How can I add this to my type?"
My Response: You recognize that orientation is important when it comes to chemical reactions. There are many ways to handle this, I will provide three short suggestions and you can consider them or use them to come up with something else.
Arrhenius (constant) approach: The pre-exponential factor, $\alpha$, in the following Arrhenius equation absorbs two important concepts of chemical reactivity - 1) the frequency of collisions, 2) the orientation of the molecules.
$$u = \alpha e^{-E_a/RT}$$
Note, I changed from the conventional notation in order to match and not confuse with your own. In fact, $\alpha$ is defined as the frequency of properly oriented collisions. So a simple constant multiplier that accounts for orientation can help modify your equation.
Boolean logic: You can possibly simplify things further by using a simple selection function that tells you whether the molecules are in proper orientation. Here is the simplest you could use:
$$f(x) =
\begin{cases}
1 \text{ if proper orientation}\\
0 \text{ if not proper orientation}
\end{cases}$$
Functional dependence: Last mentioned here would be the most creative approach to come up with a function that describes the orientation of your molecules. You will have to come up with the functional dependencies you need (i.e. a reaction coordinate) to help solve your problem. Here is an example of a reaction between atomic hydrogen with molecular hydrogen to make the same species using sinusoidal function where the independent variable (reaction coordinate) is the angle of the H-H bond relative to the hydrogen atom.
$$\text{Reaction:    } H-H + H^{\boldsymbol{\cdot}} \to H^{\boldsymbol{\cdot}} + H-H$$
$$\text{Function:    } f(\theta) = \frac{1}{2}\left[\cos\left(\frac{\theta}{2}\right)+1\right]$$

The interpretation of such a function as this would be 100% likely to react if the $H$ atom is in the plane of the bond with $H_2$ while 0% likely to react if $H_2$ is vertical. This isn't an exact reality, but just an example.


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Can some one please explain alkyl shifts for me

Provide a mechanism for the following reaction. Include all intermediates, resonance structures, charges, and electron pushing arrows to obtain full credit.

I’m stuck with this problem. I got as far as step 2 of the mechanism. I know the double bond acts as a nucleophile and attacks one of the hydrogen on sulfuric acid thus creating its conjugated base $\ce{HSO4-}$ leaving a carbocation on carbon 1.
In previous problems I have done the carbocation will rearrange to form a more stable cation. I’m told it’s done by either a hydride shift (which I understand) or an alkyl shift. I’m having a hard time grasping the concept of a six-membered ring going into a five-membered ring. And essentially understanding why there is a shift.

#1) In an analogy: Are you able to number non-H atoms in your starting material in a pattern like a shoelace (this time with definitive start and end), and may you place this on the product?  Don't feel bad if you need to draw such numbering patterns for you, at the beginning you probably need multiple attempts to find your way through.  #2) You probably recognize $\ce{H2SO4}$ as an acid, capable to donate protons.  What could be sites in the organic molecule accepting protons -- even if only in an equilibrium?  #3) Sketch how these new molecule could serve as starting materials.

#4) Probably there are multiple pathways found; which one is more reasonable, than others?  (It is not saying that the reaction definitively works this or that way, but finding what makes more sense.) This site won't provide you right on the sleeves the answer.  But if you document what you already tried to tackle the question, it is easier to recognize where you need help.

Thank you for you feedback. I was not clear with my question.

I seriously question the product in this reaction. The carbon bearing two methoxy groups and one hydroxyl group is the intermediate in ester hydrolysis. Orthoesters are acceptable but not this 2/3-orthoester.

@user55119 You are right, the targeted compound (the «2/3-orthoester») is unknown to Elsevier's Reaxys database; regardless if searched as drawn here, or allowing further than methox-substition as only $\ce{-C(OMe)2OH}$.

@Buttonwood: I'd like to hear from Lromero2 as to whether or not 8 --> 9 is the real problem. The problem as presented is patently wrong.

The reaction of the ketal 1 with aqueous H2SO4 is more likely to follow the pathway 1 --> 2 --> 3 to form α,β-unsaturated ketone 3. This is a typical ketal hydrolysis in acid medium. For the proposed reaction pathway 1 --> 4 --> 5 --> 6, species 6 is unstable. Indeed, if it were formed as an intermediate, it would afford ester 7a and ultimately the carboxylic acid 7b. Is there a possibility that the actual reactant doesn't bear methoxy groups but methyl groups? If this is the case, then the reaction 8 --> 9 is more reasonable.

Many organic reactions' steps may be described by equilibria.  In other words, at least on microscopic scale, both «forward» and «backward reaction» is possible.  So a homework for you is to become fluent in this reversibility.  As an example relevant here:

I'm not aware about an better approach than to learn a handfull of rearrangement reactions till you recognize «typical pattern», e.g. molecular substrates which are more likely than others to undergo this type of chemistry.  Here, the formation of the product bears similarity to the Pinacol rearrarangement.  As already identified by yourself, an intermediate carbocation is needed to trigger the cascade like,

and to memorize larger alkyly and phenyl groups are more likely to move, than methyl groups (e.g., Beckmann rearrangement).  In early exercises it helped me to recall that each single dash not only represents to carbon atoms connected with each other, but equally a pair of two electrons forming a bond between these two atoms in questions.  Depending on teacher and school, a drawing like the following may be permissible until creating enough «muscle memory»:

Probably soon you will learn that drawing a bent arrow anyway symbolizes the movement of two electrons (contrasting to the fish hooks to symbolize moves of single electrons) which renders drawing to dots for the two electrons in question frequently redundant.
The formation of six- and five-membered carbocyles is thermodynamically favourable.  If a six-membered ring interconverts into a five-membered carbocycle might be of lesser importance for your current level of study, than to recognize patterns typical for this kind of reaction.  It depends both on thermodynamics as well as kinetics, and over time you develop a reasoning when one, or the other favours the starting materials, or the products.  In some way, it is like learning grammar of a foreign language (e.g., I speak, you speak, s/he speaks, we speak, you speak, they speak).  As such, keep speaking Chemistry.


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Home made dispersion/solution with detectable >0.1μm particles
Sorry for the possibly wrong terminology. I'm not a chemist and I'm not a native English speaker.
Please edit the question as needed. Thank you.
Is it possible to buy or prepare a dispersion with the following properties?

the particles should be >0.1μm in size, for example 0.11μm - 0.2μm
there must be a way to detect them - even in a small amount (colour, taste, chemical reaction with something else, ...)
it mustn't be a "glue"; it must be possible to flush them completely away with water or something else readily available
it mustn't be a toxic or corrosive substance
the dispersion (or components to prepare it) should be readily available (drugstore, grocery etc.)
the total price shouldn't be higher than $20 (I know this difficult with prices different in every country - this is just a guideline)

Why do I need it?
There's a very slight chance that a Sawyer Micro Squeeze Water Filter might have been exposed to a below zero temperature and might have been damaged by frozen residual water.
The manufacturer says there's no way to test the filter functionality in that case and a new filter should be bought instead. I wonder if this is really needed or there's a way to test the filter at home.

Thanks for the answer & all confirming upvotes. As stated above, I'm not a chemist and don't know how to reword the question to be acceptable for this SE site. Is it possible to move it to https://outdoors.stackexchange.com/ with the last paragraph only and the accepted answer? (This is probably the right SE site for this question.)

I think this question was ok, perhaps added additional tags would be good, but otherwise it's prompted a really good answer that addresses a common problem which non-chemists would otherwise not have thought a good solution to.

Try to get some clay and shake it vigorously with water. Let it settle for a several hours. Test your filter with the supernatant water and collect the filtrate in a very clean glass tumbler. Colloids have an interesting property of scattering light. In a dark room, try to shine light (ordinary flashlight might work or perhaps an ordinary pointer used in presentations/offices). Be careful, never ever look at the pointer directly. Look at the filtrate at $90^o$ (right angles to the light beam), if you see a light beam travelling in water, it means that colloidal particles are seeping through the filtrate and the filter is damaged.
Read more about the Tyndall Effect https://www.sciencesource.com/archive/Tyndall-Effect-SS2423500.html or on Wikipedia before doing anything.