text:
Methods used to compare two spectra in order to determine the effects of irradiation of sample
I would like to compare two absorption spectra (or interferograms) and conclude whether between these two there are statistically significant differences at particular wavelength intervals. At the moment, I have data of two experiments that look like this:
    # A tibble: 6 x 5
      t     x1     y1     x2     y2
  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
1 3999. 0.0124 0.0132 0.0122 0.0113
2 3998. 0.0125 0.0130 0.0122 0.0116
3 3997. 0.0122 0.0131 0.0122 0.0113
4 3996. 0.0121 0.0136 0.0122 0.0114
5 3995. 0.0124 0.0139 0.0122 0.0122
6 3994. 0.0125 0.0141 0.0122 0.0129

The first column represents the wavenumber, the x columns represent the absorbance of sample and the y columns represent the absorbance of irradiated sample (before and after).
I was wondering whether I could compare these data (x and y) as time series and if so, what could be the method to quantify the differences, if any, between the samples before and after irradiation. Maybe it's already been done and there is somewhere some information as to how to compare the spectra if the wavenumber is interpreted as time ($x$ axis).
I did the t-test in R and in both experiments the null hypothesis could not be rejected, although for the second experiment (x2, y2) the $p$ value was much lower than for the first. If I average the x and y, and then plot both data, I see that there are visible differences at certain wavelength intervals. But how can I check for sure the differences between spectrums?
Here is a project with similar experiments by Zezell et al. [1]. For statistical analysis they use ANOVA and Tukey's test, but how do I do it for the vectored data?
Reference

Zezell, D. M.; Benetti, C.; Veloso, M. N.; Castro, P. A. A.; Ana, P. A. FTIR Spectroscopy Revealing the Effects of Laser and Ionizing Radiation on Biological Hard Tissues. J. Braz. Chem. Soc 2015. DOI: 10.5935/0103-5053.20150246.

Currently, your question reads like this: 1) the wavelengths are assumed to be correct (checked e.g. against a reference), 2) with sample #1, you recorded $x_1$ in function of the wavelength, 3) irradiated the sample, 4) then recorded $y_1$.  You then repeated 2 to 4 for sample #2.  Do you have multiple recordings of $x_1$ and $y_1$ to estimate their individual (and wavelength dependent) standard deviation?  Because then you test per wavelength if / where the recording were significantly different (per wavelength) with a significance level of x% for sample #1.

Equally, especially if you applied a base line correction, check the variation of the transmission / absorbance data while processing the raw data (the interferogram you should keep) for one and the same recording, processed multiple times, too.  If you do it visually by mouse click «from shoulder to shoulder» of the absorption bands displayed, variations may be large.

@Buttonwood 1) To be precise, the first column stands for wavenumber. #2 At the moment I don't have multiple recordings of the same sample. But if I have data as it is now, what can I possibly do to gauge the differences between spectra and whether they are significant? Can I do it in R?

@Buttonwood I just don't know where to look at online, there must have been some studies where the sample is studied for irradiation. What could I do, please help!

The table below illustrates a possible screen, values computed displayed are rounded to four decimals.  The suggest is to use the observations prior the intended irradiation ($x_1, x_2$) separate from the observations after the intended irradiation ($y_1, y_2$).  Per wavelength

compute for $x$ and $y$ the arithmetical mean value
determine the standard deviation of this sample (e.g., $x_1$ and $x_2$), the one pocket calculators sometimes label with $\sigma(n-1)$.  The table doesn't include it, but you may compute the confidence interval with a reasonable $t$ value.  Because there were only two realizations, the degree of freedoms ($f = n - 1$) equates to 1, and an understandably high $t_{1, 0.95} = 12.71$ to map the interval of

$$\bar{y} - \frac{t \cdot \sigma}{\sqrt{n}} \le \bar{y} \le \bar{y} + \frac{t \cdot \sigma}{\sqrt{n}}$$

subtract the arithmetical mean values from each other to determine the effect of the irradiation.  Expect positive as well as negative values to occur.
compute the standard deviation for this effect.  Because the effect is computed as a difference of $(\bar{x} - \bar{y})$, the error propagation of is a sum of the corresponding standard deviations. Or add the corresponding halfes of the confidence intervals about $x$ and $y$ instead.

I have no hands-on experience with R.

| lambda |     x1 |     y1 |     x2 |     y2 | mean_x | stdev_x | mean_y | stdev_y | mean_x - mean_y | stdev_x + stdev_y |
|--------+--------+--------+--------+--------+--------+---------+--------+---------+-----------------+-------------------|
|   3999 | 0.0124 | 0.0132 | 0.0122 | 0.0113 | 0.0123 |  0.0001 | 0.0123 |  0.0013 |          0.0000 |            0.0015 |
|   3998 | 0.0125 | 0.0130 | 0.0122 | 0.0116 | 0.0123 |  0.0002 | 0.0123 |  0.0010 |          0.0001 |            0.0012 |
|   3997 | 0.0122 | 0.0131 | 0.0122 | 0.0113 | 0.0122 |  0.0000 | 0.0122 |  0.0013 |          0.0000 |            0.0013 |
|   3996 | 0.0121 | 0.0136 | 0.0122 | 0.0114 | 0.0122 |  0.0001 | 0.0125 |  0.0016 |         -0.0003 |            0.0016 |
|   3995 | 0.0124 | 0.0139 | 0.0122 | 0.0122 | 0.0123 |  0.0001 | 0.0130 |  0.0012 |         -0.0007 |            0.0013 |
|   3994 | 0.0125 | 0.0141 | 0.0122 | 0.0129 | 0.0123 |  0.0002 | 0.0135 |  0.0008 |         -0.0011 |            0.0011 |

Hello again. Thank you for the input, but it turns out that the given x and y correspond to averages per each point before and after irradiation. In this case, I don't know which method would be usable. The relative error is 10%. I tried to plot confidence intervals ( plotted averaged data and corresponding limits per each lambda) and there where some intervals where the CI didn't overlap. But that's not the way to tell that the data is significantly different isn't it?

Yes, the safe approach is to see where the two spectra differ by $3\sigma$ or more (adding the two individual standard deviations) as strong evidence for a significant  effect at the significance level set. To assume a uniform, wavelength independent relative error of 10% of the readings obviously looses information compared to working with the non-averaged data. Without any statistics, it is the bare difference plot only, suggested by @MFarooq (https://chemistry.stackexchange.com/questions/138785/comparison-of-two-spectra-in-order-to-find-whether-the-irradiated-sample-has-sig/138858#138858)

@Buttonwood If I manage to fit the region of the spectra where the differences are seemingly different with Gaussian line, can I then compare these two lines as coming from these different distributions or am I confusing something?

@user I doubt that we are on the same page here because fitting (like in a regression) was not part of my line of thought.  Even considering your question appears pretty close to the one here, I do not understand yet the purpose of the fitting you intend.

@GianniD'Adova This now sounds more like IR mapping.  Beside clever engineering, indeed many spectra per spot are recorded to gain a S/N ratio >> 3 (e.g., https://www.jstage.jst.go.jp/article/sccj/50/3/50_209/_article, https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2732436/, https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3236568/) where the binning of the recorded sample matrix may be an issue.  A complementary check of the response surfaces of derivative spectra (https://www.whoi.edu/cms/files/derivative_spectroscopy_59633940_175744.pdf as used for UV-Vis spectra) will be a helpful tool.

@Buttonwood I read the pdf file. But there is no method I can find. There must be lots of people trying to measure the differences and the significance between differences of the spectra. If I would have the raw data then maybe ANOVA per each wavenumber could be done, but nobody would do that for each wave number, would they?

@GianniD'Adova It would require some programation (a faculty I assume R users have).  With a $t$ test, once the zero hypothesis of spectra are different along $\bar{x}$ or $\sigma$ and the critical thresholds are defined, it were up to the program repeat this testing in a for-loop over all the wavelengths recorded.  The output were a categorical yes/no binary, either permitting, or rejecting the zero hypothesis at a given wavelength, which may be color encoded.

@Buttonwood Can you explain what do you mean by "zero hypothesis  are different along $\bar{x}$ or $\sigma$"?

@GianniD'Adova Applied to the spectra problem, at a fixed wavelength, you may have five recordings prior the sample irradiation, and five recordings after the sample irradiation.  You may then test if 1) they differ significantly by the arithmetical mean value (e.g., %transmission), which I denoted with $\bar{x}$. Sometimes it is denoted by $\mu$. 2) You may check if the standard deviations (sometimes called dispersion, https://www.texasgateway.org/resource/101-two-population-means-unknown-standard-deviations) of these two populations differ significantly, or not; which I denoted by $\sigma$.

@Buttonwood Ok, thank you. But if, for example, I have multiple recordings of the same sample, what can I do, to analyse those in order to then do comparisons and what not to other groups? Apart from calculating mean of the sample. I have then sample mean, the averaged spectra, which I assume would be population mean, and raw data of few recordings of the same sample.


text:
Extent of multiple hydride shifts in a carbocation
Take this example:

Multiple hydride shifts are possible only if in each step, the formed intermediate is more stable than the previous one.
(1) to (2) converts $2^\circ$ carbocation to $3^\circ$ carbocation, thus increases stability.
Now, for (2), there are 5 hyperconjugating structures and for (3), there are 4 hyperconjugating structures. Therefore, (2) to (3) should not be favourable. There is a distant $\pu{+I}$ effect of $\pu{-Me}$ group in (3) but hyperconjugation dominates over inductive effect.
(3) to (4) hydride shift converts to the most stable carbocation possible in this scheme.
Then the final products (5) and (6) are obtained.
Is this the major product obtained from intermediate (2)?

Most probably yes because it results in the alkene with highest hyperconjugating structures and the cation won't rearrange into (3) because as you said, "...formed intermediate should be more stable than previous one."

Given a lack of stereochemistry, it is difficult to assess the number of hyperconjugating C-H bonds. A difference of 4 versus 5 such bonds may imply an equilibrium. In your scheme, why is 5 the major product and not 6? I'd opt for your last structure as the product from cation 2. Please avoid light green. Tough to read.

@user55119 (a) What can be done (stereochemistry) to make it a simple problem? (b) Are you suggesting that 4 vs. 5 hyperconjugating structure is not much to decide the stability of intermediate? (c) In 6, one bridged carbon is sp2  and the other is sp3 hybridised carbon, thus, more strain than 5? Though, 6 is most substituted product.

(a) Stereochemistry may have an effect on rates of rearrangement. (b) You  may be overcounting C-H hyperconjugation of methylenes with vicinal cations. Hard to have both C-H bonds overlapping the cation at the same time. (c) Whether 5 or 6 is more stable may depend on whether or not 5 has a cis or trans ring fusion.

source: Stack Exchange

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text:
Temperature relation for isomerism reactions
When we have an isotherm isomerism reaction
$\ce{A(g) -> A^*(g)}$  with $\ce{ΔH^0 = 0}$
is the ratio of preassures $\ce{P_A^* / P_A}$ independent from temperature?

The van't Hoff equation would predict that the equilibrium constant is not dependent on temperature, since in this case:
$$\frac{\mathrm{d}}{\mathrm{d}T}\ln K_{eq} = \frac{\Delta H^\circ}{RT^2}=0$$
Strictly speaking the equilibrium constant is a function of fugacities, but it is probably safe to assume that the pressures are also independent of T, so the answer would be yes.

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


text:
How is Pi antibonding orbital more polar than Pi bonding orbital?
While reading about the solvent effect on electronic transition in ESR spectroscopy, I found it written somewhere that the pi anti-bonding orbitals are more polar compared to pi bonding orbitals.
Can somebody explain how and why is it so?

I am also new to this theory, but I feel that $\displaystyle \pi ^_{ABMO} $ has a nodal plane in between the nuclei of atoms, and also has greater energy than normal atomic orbital whereas   $\displaystyle \pi _{BMO} $ is of lower energy  and has maximum electron density between nuclei. Thus $\displaystyle \pi ^_{ABMO} $ is more polar

We can also take into account, the high energy electronic repulsion in $\displaystyle \pi ^ *_{ABMO} $.


text:
Adding alkyl group to benzene with a deactivating group
question:
The question is asking whether we can simply do Friedel-Crafts alkylation to a benzene bonded to a deactivating group (i.e. sulfonic acid group in this problem).
(No answers to this problem)
attempt:
I am thinking this reaction cannot work. This is because:
$\ce{R-SO3H}$ strongly deactivates the benzene, making the $\ce{C=C}$ bond less nucleophilic. Therefore, it does not have enough 'power' to attack the carbocation complex ($\ce{CH3CH2-FeBr3}$), which is a limitation in Friedel-Crafts reactions.
Therefore, my attempt is:

de-sulfonation: dil. $\ce{H2SO4}$ + $\ce{H2O}$, make it to benzene
Friedel-Crafts acylation: $\ce{CH3COCl}$ + $\ce{AlCl3}$ (catalyst), make it to acetophenone
sulfonation: conc. $\ce{H2SO4}$, make it to 3-Acetylbenzenesulfonic acid
Clemmensen reaction: make it to product

question 1:
Are there any quicker ways to perform this synthesis? I am thinking is there any way to make the group less deactivating, hence allow the alkylation to happen, but no idea how.
question 2:
Although Clemmensen reaction is a good and common choice, it uses highly-toxic mercury. However, for Wolff–Kishner reduction, I am worrying that high temperature will decompose the product, plus the base will neutralize the product acid. Overall, which reaction is better for this particular reduction step?

I found a Chinese patent via Chem. Absts. that claims a 50% yield of the acetophenone by FC of benzenesulfonic acid. The conditions of a Clemmensen may effect desulfonation.

@user55119 Thanks. Can I say the -SO3H group is not that strong in the deactivation of the benzene ring in general, or just particularly in this reaction? Also, do you mean HCl used in Clemmensen will undergo desulfonation? If yes, is thioketal reduction is best choice?

Because a ring is deactivated doesn't mean it is unreactive but rather slower to react than an activated ring. I am not saying that the HCl will cause desulfonation but it may. Reduction of a thioketal with Ni(R) is an option. Also, reduce carbonyl to secondary alcohol and hydrogenolize the hydroxyl group with H2 and Pd/C.


text:
Why do bases taste bitter?
My chemistry teacher told us that acids taste sour and bases taste bitter. After reading Do acids without hydrogen taste sour? and https://en.wikipedia.org/wiki/Taste#Sourness I learned that sourness in fact is the taste of acidity. But why is it that bases generally taste bitter?

Do you have another source (except your teacher) that bases taste bitter? Because I don't think (that is generally) so.

@Karl It's in some middle/high school textbooks I think, for example https://www.classzone.com/vpg_ebooks/ml_sci_gr8/accessibility/ml_sci_gr8/page_324.pdf

Bitter compounds often contain amine groups. Amine groups are weak bases.

SAFETY - I'll point out that taste tests are absolutely not a standard test in a modern chemistry laboratory.  Tasting random chemicals is dangerous.

One should avoid tasting fruits, vegetables and suspicious plants in chemistry laboratories.

Some bases taste bitter, some do not. Between bitterness and basicity is correlation, not causation.
The strongest bases like hydroxides of alkali metals and alkali earth metals, or alkali metal carbonates, are not bitter.
Near all if not all bitter  bases are organic compounds with basic groups containing nitrogen, like aminogroups or heterocyclic nitrogen. They are often potentially poisonous alkaloids, so evolution came with unpleasant sensation for such compounds. The typical example is quite bitter caffeine.
OTOH, not all bitterness is caused by basic compounds. Grapefruits are bitter because of the glycoside naringin, which does not contain nitrogen and is not a base. Glycosides are frequently bitter, regardless of basicity.
Sensory impact is rather a topic for biology, not chemistry.

I had never heard of that. It's quite interesting!


text:
What is the liquid which has the most hydrogen atoms per volume?
What is the liquid which has the most hydrogen atoms per volume at STP ?
Is there anything better than water?
This answer does not answer this question because none of the compounds listed in it are liquids at STP, which contain more hydrogen per volume than water.

Hydrazine is slightly better.

How about liquefied methane?

Methane is not a liquid at STP.  Without this temperature and pressure restriction, the liquid Hydrogen would beat them all ;)

@GeorgeRobinson you are right

No, that's wrong. Liquid methane contains more hydrogen per volume than liquid hydrogen itself.

@ Ivan: At all pressures and temperatures where both are liquid ?

@George: there are no such conditions where both are liquid simultaneously.

Hydrogen atoms tend to get fluffy in elemental H2.  Covalent bonds with other elements or incorporation into an ionic lattice strongly favors squeezing down the hydrogen atoms, allowing higher hydrogen density even with the other element taking up space.

Does this answer your question? Most dense hydrogen containing composition

@ Matthew. No, that question and answer is not about liquids at STP.

https://chemistry.stackexchange.com/questions/67681/densest-hydrogen-storage-medium

That's too broad. Even among pure compounds there's many "better" then water, for example cyclodecane. There's no telling what's "the best" and with mixtures complexity is tremendous.

What is too broad?  Cyclodecane has a 1:2 ratio of Carbon to Hydrogen and it has a density of 0.871 g/ml.  That is a better mass ratio than the 1:2 ratio of Oxygen to Hydrogen and the density of 1.0 g/ml for water.

What mixture can beat Cyclodecane ?

Hydrazine has more hydrogen atoms per mL than water
Some simple calculations give the moles of hydrogen per mL in some possible alternative liquids. Taking into account density and molecular mass we get the following results:
Water 0.056 mol/mL 2 Hydrogens -> 0.11 mol H/mL
Hydrazine 0.0313 mol/mL 4 hydrogens -> 0.125 mol H/mL
symmetric dimethylhydrazine 0.0138 mol/mL 8 hydrogens -> 0.111 mol H/mL
methyl hydrazine 0.0191 mol/mL 6 hydrogens -> 0.115 mol H/mL
So, while some other liquid at STP hydrazine derivatives beat water, it looks like hydrazine has notably more hydrogens per unit volume. This is probably the record.

Some solid metal hydrides have more hydrogen per unit volume than either liquid or solid hydrogen.  They are candidates for storage in hydrogen-fueled processes.  A few examples of such metal hydrides are given here.

Are they liquid at STP ?

Did I not qualify the answer by identifying these materials as solid?

Did I not qualify the question by asking for identification of materials that are liquid at STP ?  Your answer refers to solid materials, thus it is off-topic.


text:
Phenethylhydrazone decomposition mechanism
I was going through a problem in the book Advanced problems in organic chemistry by Chouhan:

I figured out that since di-isopropyl amide will act as a hindered base it will take out the following hydrogens (in the form of $\ce{H+}):$

the one where the carbanion formed is in conjugation with benzene ring
the one where carbanion formed is in conjugation with nitrogen.

After that these carbanion(s) will attack respectively on their adjacent $\ce{C-C}$ σ-bonds and the nitrogen(s) will depart in the form of $\ce{N2},$ this gives me option (a) as the answer, but the solution manual proposes a different reaction mechanism:

I am not able to understand the formation of the three-membered ring, please help.
Is there a mistake in my thinking or are there some other concepts involved?
I am an undergraduate student of organic chemistry.

see https://en.wikipedia.org/wiki/Shapiro_reaction and the section on "catalytic Shapiro reaction" - that suggests that the intended starting material is the cyclopropane, which means that that your book probably has an error. Mind you, this book is also abbreviating isopropyl as "i-Ph", so I'm inclined to suggest that you deserve a better book. [edit: I meant aziridine; silly me.]

The Catalytic Shapiro Reaction: DOI:10.1021/JA951422P Answer 46(a) is correct, the aziridine ring is missing in the reactant of the question.

@orthocresol could you please propose your mechanism in detail in the answer, I am still not getting it

So, I attempted to provide a mechanism before actually reading the answer or comments. I immediately got stuck because of conservation of mass. The starting material has 30 hydrogen atoms but the products only have 28 hydrogens. There's no oxidant here, so something immediately had to be wrong.

user55119 already told you that that the given answer is correct. I gave you a link to a Wikipedia page that says the same thing. Both of us told you that your starting material in your question is wrong. I don't think there's much else to be said, to be honest. Someone else could write it up into an answer, though.

@Zhe: You are correct. I immediately realized there are two double bonds in the products while SM has only one. So, as as orthocresol said, aziridine has to be there to begin with. Also note that in the solution, it starts with exyra double bond (terminal one). That confirms the sloppiness of authors (poor proofreading).


text:
Comparison of two spectra in order to find whether the irradiated sample has significantly different resulting spectra
I would like to compare two absorption spectra ( or interferograms) and conclude whether between these two there are statistically significant differences at particular wavelength intervals. At the moment, I have data of two experiments that look like this:
    # A tibble: 6 x 5
      t     x1     y1     x2     y2
  <dbl>  <dbl>  <dbl>  <dbl>  <dbl>
1 3999. 0.0124 0.0132 0.0122 0.0113
2 3998. 0.0125 0.0130 0.0122 0.0116
3 3997. 0.0122 0.0131 0.0122 0.0113
4 3996. 0.0121 0.0136 0.0122 0.0114
5 3995. 0.0124 0.0139 0.0122 0.0122
6 3994. 0.0125 0.0141 0.0122 0.0129

The first column represents the wavenumber, the x columns represent the absorbance of sample and the y columns represent the absorbance of irradiated sample ( before and after). I was wondering whether I could compare these data ( x and y) as time series and if so, what could be the method to quantify the differences, if any, between the samples before and after irradiation. Maybe it's already been done and there is somewhere some information as to how to compare the spectra if the wavenumber is interpreted as time ( x axis). I did the t test in R and in both experiments the null hypothesis could not be rejected, although for the second experiment (x2, y2) the p value was much lower than for the first.
If I average the x and y, and then plot both data, I see that there are visible differences at certain wavelength intervals. In R, I used IRISSeismic package and function crossSpectrum, that gave me such an output:
   freq        spec1        spec2 coh        phase             Pxx             Pyy
1   0.0002666667 2.121935e+01 2.152532e+01   1  0.010220252 1.856693e+01+0i 1.883465e+01+0i
2   0.0005333333 7.011069e+00 6.869078e+00   1  0.008984730 6.134686e+00+0i 6.010443e+00+0i
3   0.0008000000 8.385363e+00 8.197039e+00   1  0.011999039 7.337193e+00+0i 7.172409e+00+0i
4   0.0010666667 7.483070e+00 7.272319e+00   1  0.015419031 6.547686e+00+0i 6.363279e+00+0i
5   0.0013333333 4.086251e+00 3.899095e+00   1  0.017551350 3.575469e+00+0i 3.411708e+00+0i
6   0.0016000000 1.537709e+00 1.405562e+00   1  0.008722742 1.345496e+00+0i 1.229867e+00+0i

For all data points the coh index is given as 1. But obviously that doesn't help me to determine, whether the difference is substantial between two data sets.
I tried to average data points between groups (x1, y1 and x2, y2) at each wave number and plotted the graph, it looks like the difference might be significant at certain wavenumber interval. Maybe I could try to express one line ( averaged data points after irradiation of the sample) as a function of the other? But how?
Here is a project with similar experiments: https://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532015001202571 , for statistical analysis they use ANOVA and Tukey's test, but how do I do it for the vectored data?
I cannot find online anything. Here is something of the sort I am looking for: https://www.youtube.com/watch?v=gjKSfILE9nM He compares two spectra in Matlab, but he is using self made toolkit which I can not obtain. How can I do something like this in R? I've read something about modelling the data, but here, I don't have any variables just data of two spectra.

The first thing to do is to know how these measurements are reproducible. If you repeat the first measurement of $\ce{x_1}$ , do you always get $0.0124$ ? This absorbance is extremely weak. Or do you obtain from time to time  $0.0122$, then $0.0125$ and $0.0127$ or $ 0.0123$ ? Are these measured values reliable ? Maybe the six first measurements of $\ce{x_1}$ and $\ce{x_2}$  are all equal. Who knows ?

@Maurice There are intervals that have much larger absorbances. I guess, we could assume that the x2 outputs would be the second given measurement of the same control sample. So, in that case can we produce estimated line for both x and y and then graph error areas to see, whether they overlap? In that case we only have two data points to estimate the approximate point, I guess that's too small. But even so, is this method even legitimate?

@ User. How were the wavelengths chosen ? Why stop at 3994 ? This is the first time and the only time where an effect can be accepted. Why didn't you not measure at 3993, 3992, then 3980, 3950 ? What is the unit of these wavelengths ? Is it in visible, UV, IR ?

@Maurice Sorry, that I didn't clarify. It's only beginning of the spectra. The spectra is taken from 4000 - 400 cm^-1 . I just put the first few rows, it's IR. The measurements are taken per each wavenumber approximately, the difference between each measurement ~ 0.964

Sorry, stupid question: What is your background? From your question I gather you know your statistics, but have not had your basic training in physical chemistry. Right?

This is already posted at the CV stack exchange: https://stats.stackexchange.com/q/482560/247352.

The question reads very similar to this one here on ChemSE (https://chemistry.stackexchange.com/questions/138770/methods-used-to-compare-two-spectra-in-order-to-determine-the-effects-of-irradia?noredirect=1#comment273089_138770), with one approach suggested.  Maybe a cross-post, maybe the two OP's do know each other?

Does this answer your question? Methods used to compare two spectra in order to determine the effects of irradiation of sample

Why don't you start with the simplest possible approach. I am sure there are very fancy spectral matching statistical algorithms, but simplicity has its own beauty. Comparing an inteferogram (not many features there) might be difficult but what about absorbance rather?
Look at the absorbance difference spectrum: (Spectrum after irradiation)-(Original sample spectrum). How does the plot look like? What is the signal which you are collecting? Sometimes the human eye can save a lot of time rather than doing significance testing on data points and forming null hypotheses.
The features which remain the same will be zero now and the features which change will be visible in the the difference spectrum.
Alternatively, you can take the first derivative of both spectra and smooth it with Savitsky-Golay or study the difference of "first derivatives". Noise will be an issue but I guess you can take care of that.

After I have smoothed out both derived spectra, what are the methods to compare for significant differences?

Could share the pictures of the simple absorbance spectrum and the derivative subtracted spectrum.


text:
What surface treatment can turn nickel black?
Questions:

The detailed surface treatment to turn nickel plated steel black.
The reaction mechanism of the process.

Methods:

(preferred) Anodizing, oxidizing : The only black nickel oxide is $\ce{Ni2O3}$ but I'm not sure if it's suitable for surface protection. I do not know the color or other Ni(II) or Ni(III) compound. There are nickel blackening solution for sale, but I don't know their ingredients.

(not preferred) Electroplating onto the existing nickel: US military have a standard to electroplate black nickel (MIL-P-18317). However this process is quite a fuss.

Sodium thiocyanate (?) is not an uncommon chemical. https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19760021316.pdf (just my 10ct after 90 sec. on google)

See https://www.finishing.com/330/81.shtml . Also, ferrous metal blackeners that contain alkali selenides or selenous acid (selenium dioxide solution) might work. N.B. All are fairly toxic.

The article provided by Karl has answered the question with electroplating method, which is not preferred but this is the only answer I have.

The method is based on MIL-P-18317. The electroplating solution consists of $\ce{NiSO4, Ni(NH4)2(SO4)2, ZnSO4}$ and $\ce{NaSCN}$

The black nickle coating are $\ce{NiS}$ and $\ce{ZnS}$.
$\ce{NiS}$ and $\ce{ZnS}$ are deposited by $\ce{M^2+ + H2S -> MS + 2 H+}$.
$\ce{H2S}$ is generated by $\ce{SCN- + 3H+ + 2e- -> H2S + HCN}$

note: Is reducing thiocyanate a standard method to generate sulfide in electroplating ? I know soluble sulfide salts smells horrible, but it is worth the trouble ? not mentioning the cyanide byproduct, in acidic solution.

Metal $\ce{Ni}$ and $\ce{Zn}$ are also deposited $\ce{M(NH3)2^2+ + 2e- -> M + 2NH3}$.
The role of $\ce{NH4+}$ is to generate $\ce{M(NH3)2^2+}$ prior to metal deposition.

You may want to review those chemical formulas.  You have nickel(I) sulfate and a double sulfate with nickel(0). ???

You can use PVD or CVD method. You will get a nano coating of Ni oxides. In PVD chamber you can heat Ni steel with $\ce{N2 + O2}$ or $\ce{Ar + O2}.$


text:
Finding percent contamination of sulfur dioxide in air by redox reaction with iodine
I would really like some help with this problem. My work is at the bottom. I've tried multiple ways and have gotten the same answer for both, but for some reason the computer program won't accept the answer as it is. Any help is appreciated!

An environmental engineer analyzes a sample of air contaminated with sulfur dioxide. To a $\pu{500 ml}$ sample at $\pu{700 torr}$ and $\pu{38 ^\circ C}$, she adds $\pu{15.50 ml}$ of $\pu{0.01017 M}$ aqueous iodine, which reacts as follows (equation 1 is unbalanced):
$$\ce{SO2(g) + I2(aq) + H2O(l) → HSO4−(aq) + I−(aq) + H+(aq)}\tag1$$
Excess $\ce{I2}$ reacts with $\pu{11.37 ml}$ of $\pu{0.0105 M}$ sodium thiosulfate (equation 2 is unbalanced):
$$\ce{I2(aq) + S2O3^2−(aq) -> I−(aq) + S4O6^2−(aq)}\tag2$$
What is the volume % of $\ce{SO2}$ in the air sample?

Here's my work and answer:
Find the number of moles of $\ce{I2}$:
$$  0.01550 \times 0.01017 =\pu{1.576 \times 10^{-4}mol} $$
number of moles of $\ce{S2O3^2-}$:
$$0.01137 \times 0.0105=\pu{1.1938x10^-4 mol}$$
excess $\ce{I2}$:
$$1.1938 \times 10^{-4} \times \frac{1}{2}=\pu{5.969x10^-5 mol}$$
number of moles of $\ce{I2}$ in reaction with $\ce{SO2}$:
$$1.576\times10^{-4} - 5.969 \times 10^{-5} = \pu{9.791x10^-5 mol}$$
$$9.791\times10^{-5} \times \frac{2}{1} = \pu{1.9582x10^-4 mol} $$
$$V=\frac{(1.9582 \times 10^{-4})(0.082)(311.15)}{0.921}= \pu{5.425 ml}$$
$$\frac{\pu{5.425 ml}\,\ \ce{SO2}}{\pu{500 ml}\, \text{air}}\cdot (100\%) = 1.085= \pu{1.09 vol.\% }$$

Welcome to Chemistry! For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

@Safdar - Thanks for making the problem readable. Very nice edit!

@ User. Why do you show an equation that is unbalanced ? What is the meaning of the words "the computer program won't accept the answer" ? Which computer program ? Is it to understand that the program obtains another result ? How far away from yours ?

Please try to find a descriptive title for your question. While being as concise as possible, it shall indicate, without ambiguity, the subject matter of the question in such a way as to distinguish it from that of other questions, without going into unnecessary detail.

It is good practice to stay with general algebraic expressions as long as possible, plugging in the numbers in the last minute. This way it is much easier for authors and readers to either verify the correctness either spot errors or mistakes.

Your problem is you didn't balance any equation. You correctly use mole ratio for second equation. But, mole ratio for first is 1:1 (if you balance it).


text:
Analysis of post-HF wavefunctions
Hartree-Fock method introduces electron (spin)orbitals and they are commonly used for qualitative rationalization of many molecular properties. However, MOs have meaning only if we ignore electron correlation. Post-HF non-DFT methods build correlated wave-function, but due to correlation extracting independent orbitals from them is not possible.
Electron density still is a somewhat informative concept even with no orbitals involved, but we like to talk about $\pi$-systems, sigma-bonds and the like and relate electron transitions to molecular features. So, I'm interested in existing approaches to analysis of post-HF wavefunctions (for example, obtained from Coupled Cluster method).
Are there any well-known approaches to analysis of correlated wavefunctions that do not rely on concept of orbitals that allow to relate some molecular properties to local structural features?

As a starting point: the quantum theory of atoms in molecules comes to mind, and possibly the electron localisation function. If I recall correctly, they could be applied directly to only the density (even a measured one).

There are! First off, you're correct that Hartree-Fock and DFT produce effective one-particle states $\phi_i$ which are the molecular orbitals (MOs) in the case of HF or the Kohn-Sham (KS) orbitals in the case of DFT (I'll focus on Hartree-Fock since I am more familiar with it than DFT). The MOs are defined via expansion coefficients ($C_{\mu i}$) in the basis set $\{\chi_\mu\}$ such that $\phi_i = \sum_\mu C_{\mu i} \chi_{\mu}$. The more physically meaningful quantity one works with is the one-particle reduced density matrix (1PRDM) defined as $\Gamma_{pq} = \sum_{\mu \nu} C^*_{\mu p} P_{\mu \nu} C_{\nu q}$ where $P_{\mu \nu} = \sum_i C_{\mu i}^*C_{\nu i}$. Diagonalization of the 1PRDM produces what are known as natural atomic orbitals (NAOs) $\phi^{\rm{NAO}}_i = \sum_p a_{p i} \phi_p$ where the coefficients $a_{p i}$ are defined by the eigenvalue relation $\Gamma_{p q} a_{q i} = \rho_i a_{p i}$ where $\rho_i$ is the occupation number corresponding to th $i^{\rm{th}}$ NAO. Since $\Gamma$ is Hermitian, the NAOs are orthonormal and form a perfectly suitable basis to perform any quantum chemistry calculation, however, their value lies in the fact that NAOs actually show you where the electron density is so it's a very useful visualization tool. Once you have NAOs, all of the analysis you want is available. There are things called natural bond orbitals (NBOs) which rotate NAOs in such a way that they maximize the electron density between 2 atoms, thus simulating the notion of a chemical bond. Here, you find the orbitals describing your $\sigma$- and $\pi$-bonds. The point I'm trying to make is that what's important for this kind of analysis is having $\Gamma_{pq}$.
So the answer to your question depends on whether or not you calculate $\Gamma_{pq}$ for a variety of post-HF methods, such as configuration interaction (CI), coupled-cluster (CC), and others. If we denote the Hartree-Fock determinant as $|\Phi\rangle$, the 1PRDM from before is simply $\Gamma_{pq} = \langle \Phi | a^+_p a_q | \Phi \rangle$. The wavefunction originating from post-HF calculations is $|\Psi_\mu\rangle = (1 + C_\mu)|\Phi\rangle$ in the case of CI and $|\Psi_\mu\rangle = R_\mu e^T |\Phi\rangle$ in the case of equation-of-motion (EOM) CC. The analogous expression of the 1PRDM for the $\mu^{\rm{th}}$ excited state ($\mu = 0$ is the ground state) is
$$\Gamma^{\mu}_{pq} = \langle \Psi_\mu | a^+_p a_q | \Psi_\mu \rangle$$
You can calculate this object directly for the various types of wavefunction that come out of correlated electronic structure calculations. It's not light work for sure, but if you know your way around many-body algebra (e.g. Slater's Rules, diagrammatics, Wick's Theorem), you can calculate this object and diagonalize it to find correlated bonding orbitals just as you would in Hartree-Fock. Also note that in the case of CC, the bra state $\langle \Psi_\mu | \neq [|\Psi_\mu\rangle]^+$ and instead we use the biorthogonal left-CC parameterization $\langle \Psi_\mu | = \langle \Phi | L_\mu e^{-T}$ where $L_\mu$ is a linear de-excitation operator complementary to $R_\mu$. These and other technicalities are of course well-documented in the CC literature. The key observation here is that we never explicitly build a correlated $|\Psi_\mu\rangle$ defined via Slater determinants directly. In the case of CC wavefunctions, you'd blow up your computer since $e^T$ is an object as large as the full CI wavefunction for any level of truncation. So the way things are done is you obtain $|\Psi_\mu\rangle$ indirectly though a set of CI coefficients or CC cluster amplitudes and calculate the 1PRDM, again, completely defined through this sequence of finite coefficients. It may still be a single-particle orbital, but it has many-body information embedded within it.
That being said, these calculations are available in most electronic structure programs.

The point I'm trying to make is that what's important for this kind of analysis is having $Γ_{pq}$. || Is there no analysis tools without resorting to this representation?

I mean, as far as I can read, your answer states "well, if we use some trickery, we can kinda extract something that looks like MOs if we squint hard enough" Not that it isn't an answer, but it isn't answer I hoped for.

It's not trickery at all! The reason the density matrices are central is because any observable in a $k$-body theory can be expressed using up to the $k$-body RDM (there are entire theories built on this principle - DFT is one of them, in an incomplete way). So as far as any physical theory is concerned you don't need any more. Quantum chemistry is governed by a 2-body interaction so $\Gamma_{pq}$ is important and so is $\Gamma_{pqrs}$. But that's all there is to the theory. I emphasized $\Gamma_{pq}$ since I read your question as being charge density-oriented, presumably for visualization.

On one hand, you are right (in mathematical sense). On the other hand... It seems, I have to invest more into how various responses are calculated and what they actually mean local-wide and molecule-wide.


text:
Could an extremely strong reducing/oxidising agent attack nitrogen molecules?
Could a very strong reducing agent like lithium aluminium hydride or a very strong oxidizing agent like manganese heptoxide react with $\ce{N2}$? Usually when people talk about reactive substances they emphasize either strong electron acceptors or donors. I am wondering if using such compounds could be employed in a brute force method of dissociating $\ce{N2}$ molecules. I know there exist industrial methods for that already, but i'm poking around to see if maybe there's a way to do it under atmospheric pressure without tailored molecules like enzymes or organometallic complexes.

"since nitrogen can both give and take electrons to achieve a more stable configuration" Huh?

What's the point of this question exactly? Agents you mention don't react with molecular nitrogen in standard STP. Still bacteria can reduce it enzymatically and ammonia is produced industrially in via reduction with H2. And I agree with Zhe that your last sentence there isn't OK at all.

ok, ill remove that part,i wanted to know if a brute force method could work in destroying the N2 molecule

It is hard to make a full answer to questions where the answer is simply: "no". There is a reason why it is a billion-dollar problem, and you may not find the answer to this problem on SE.

You mean like this?

I know there exist industrial methods for that already, but im poking
around to see if maybe there's a way to do it under atmospheric
pressure without tailored molecules like enzymes or metalorgnic
complexes

Though Haber process which was used to industrially produce ammonia was well-known, some researchers have also attempted to split nitrogen molecules using metal complexes:

Chirik et.al (Nature, volume 427, pages 527–530(2004)) attempted to split nitrogen molecules by using a zirconium complex containing two bulky chemical groups called methylated cyclopentadienyl. The complex was the right size to snuggle up to both nitrogen atoms and give an electron to each atom. This double donation take place in an organic solvent at 100 °C and atmospheric pressure, disrupts the triple bond. Further reactions in the same solution then finish the job of splitting the nitrogen molecules and adding hydrogen atoms to make ammonia. (source)
Carbon monoxide and a molecule containing hafnium to break apart nitrogen and create a fertilizer called oxamide. (source)
John Mark P. Martirez and Emily A. Carter at Princeton University propose a gold-molybdenum catalyst that could split the nitrogen triple bond at room temperature using energy from visible light. That dissociation step is the primary limit on the reaction rate in the Haber-Bosch process. (source)

Brute force would be really good if it were also clever. An article on using a large spark generator (Ref 1) discussed controlled experiments and used to data to estimate the amount of nitrogen oxides produced by lightning. I suppose cleverness could be added if a proper surface (i.e., catalyst) could be found, or if a proper reaction could be designed.
The trouble is, brute force is pretty random, and destroys as much of the product (even more!) than it converts of the starting material (N2). Oxidation is usually considered for brute force work. However, the Haber process uses heat and an iron + KOH catalyst at 400-450 degrees C to make ammonia. The heat is to make it fast enough, and the catalyst needs the high temperature to become active.
The equilibrium is favored to produce more NH3 at lower temperatures: $$N2 + 3H2 --> 2NH3$$ is exothermic. (Ref 2)
Now I speculate: the Haber approach relies on the dissociative adsorption of N2 upon the metal (iron is cheapest, but there are others that work well too). The actual mechanism appears to be multi-step, with dissociation of nitrogen, and hydrogen, and stepwise addition of H to N to form NH3. Suppose you took a mixture of N2 and H2 and dissociated the hydrogen with a spark; I wonder if the H atoms would combine with N2 and start the process of dissociating the N2 molecule before recombining to H2.
So, a spark at low temperature (and perhaps at moderate pressure) might initiate a reaction in N2 + H2 like a spark in a mixture of H2 + O2. Oh no, it couldn't be that easy!
Ref 1: https://agupubs.onlinelibrary.wiley.com/doi/pdfdirect/10.1029/1999JD901138
Ref 2: https://en.wikipedia.org/wiki/Haber_process


text:
Why is black-body radiation curve smooth without a sharp cutoff?
Planck's law is able to predict a graph that is consistent with experimental observation:

In essence, unlike Rayleigh-Jeans law that assumes equipartition theorem to hold (that each mode of motion shares equal energy at thermal equilibrium, so all modes excited at thermal equilibrium ⇒ ultraviolet catastrophe), Planck's hypothesis proposed that oscillators of frequency v will be excited only if they can acquire energy of at least hv , where h is Planck constant.1
So based on this argument, I am not sure why the shaded region of the graph will slope in a smooth manner when wavelength is being decreased. Say if you are only able to supply energy of hv, then all oscillators with frequency > v should not be excited and hence will not contribute to energy density ⇒ ρ = 0:

I expect the behaviour to be same as that in photoelectric effect: if frequency of incidence wave is below threshold frequency, you will never observe photoemission. But the actual observed graph is the first one. Where did I go wrong? I am not a physicist so a not-too-complicated explanation will be appreciated.

1 Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 247.

Long story short, not all molecules in a sample have the same energy.

Consider the thermal energy of molecules at given temperature does not have a sharp cut off either, so there is always some nonzero probability the energy hν can be obtained. In contrary, photons in photoemission contexts have the particular energy level, so there is the sharp cut-off edge.

What made you think there's just one single way to radiate for any substance? Your argument would make sense only perhaps in very close vicinity of absolute zero!

The key to understanding the curve is thinking about how energy is distributed among components of the system
The mathematics that produce the curve involves some not-very-simple statistical mechanics some of which Planck didn't understand when he first developed his theory.
But it isn't that hard to get the intuitive idea. Consider the case of a gas where molecules have some kinetic energy. The individual molecules don't all have the same energy: some are moving faster than others. The temperature of the gas is function of the average kinetic energy of the molecules. But individual molecules are exchanging energy very frequently as they bang into each other. But those random collisions exchange random amounts of the kinetic energy between the molecules. Sometimes one molecule will gain a great deal of energy so it ends up with far more than the average amount. But the probability of a series of collisions giving a single molecule a very, very large kinetic energy is low and the higher that energy, the lower the probability. There is no sharp cut-off just an exponentially decreasing probability of getting a higher and higher energy.
Take that picture and apply the appropriate statistical probability theory and you get the overall Planck distribution: a curve where the average kinetic energy is given by the temperature but where individual molecules have some probability of having much lower energy (with slowly decreasing probability) and others have some probability of having much higher energy (but with a sharply decreasing probability). There is no sharp cutoff at high energies just sharply decreasing probability of reaching those levels.
In reality the picture is far more complex as molecules have vibrational and electronic energy as well as kinetic energy, but these details don't matter much for the intuitive picture.

Textbooks like Atkins and others do a severe injustice to science by re-writing fictitious history as if it were a nice and smooth story. You can see below in the quote that the thought process of Planck is an "act of desperation" as he called it himself. He had to derive a formula that will fit the experimental black body curve by all means even if it required violating classical physics rules. The original derivation in Planck's paper used very advanced mathematics and statistics (and I do not fully follow his arguments as a lowly chemist). You can roughly think at that you are heating a body which is glowing white and there is a distribution of frequencies of oscillators (hence advanced statistics). I remember reading that Planck was walking with his son and told him that I have discovered something which will be as important as Newtons. It was a feat!
In his time, electrons or modern atomic structure was not fully known and hence photoelectric effect was another story. It was Einstein who explained the photoelectric effect which earned a a Nobel Prize. In photoelectric effect there is no distribution of energies of incident photons. Roughly, you can say that is why the photoelectric effect is abrupt.
See Planck, the Quantum, and the Historians* by Clayton A. Gearhart Phys. Perspect. 4 (2002) 170–215.

letter from Planck to the American physicist Robert W. Wood, October 7, 1931:
In this letter, Planck responded to Wood’s request for a description of the
‘‘considerations which had led me to propose the hypothesis of energy quanta.’’
Planck spoke of his work as an ‘‘act of desperation,’’ and said:
I also knew the formula that expresses the energy distribution in the normal
spectrum. A theoretical interpretation therefore had to be found at any cost,
no matter how high. It was clear to me that classical physics could offer no
solution to this problem, and would have meant that all energy would
eventually transfer from matter to radiation. …This approach was opened to
me by maintaining the two laws of thermodynamics. The two laws, it seems to
me, must be upheld under all circumstances. For the rest, I was ready to
sacrifice every one of my previous convictions about physical laws. …[One]
finds that the continuous loss of energy into radiation can be prevented by
assuming that energy is forced at the outset to remain together in certain
quanta. This was purely a formal assumption and I really did not give it much
thought except that no matter what the cost, I must bring about a positive
result.


text:
Molecular Structure
What's the meaning of the triangle and striped line in this structure of $\ce{H3O+}$:

Can someone suggest me a site to learn about this?
(I'm quite a novice in chemistry.)
Thanks

The part of this answer already fully adresses your question, and this comment provides references. Alternatively, look up wedge and dash notation in any organic chemistry textbook or in the internet.

@andselisk IMO, this question is a duplicate of that.

@NilayGhosh That was also my initial thought, but I refrained from single-handedly closing this one as a duplicate since in What are the meanings of dotted and wavy lines in structural formulas? OP explicitly mentions that they are already aware of the wedge and dash notation. Let the community decide.

Does this answer your question? What are the meanings of dotted and wavy lines in structural formulas?

@Mithoron Oh yes

@Shub For future reference: Often, it is good to substantiate questions and answers on ChemSE with an image - and you rightfully did so here, too.  However, it is better if you insert these them directly in your question.  A link toward a reference external to ChemSE may break (e.g. for change of the address of a site, merger / closure of a business, etc.) and may represent an unnecessary obstacle to grasp the content of your question fully.

@Buttonwood I tried to do so but failed (both in the case of uploading from the URL and from the device). I think I lack the required reputation for that.

@Shub I'm sorry but I forgot if the inclusion of a picture requires a minimal reputation, or not.  In the present case, I accessed the reference site given by you, basically took a screen photo which then was cropped and scaled to a lesser dimension in an editor (https://www.pinta-project.com/) and then uploaded the resulting .png.

I would suggest you Dr. Wayne Breslyn's YouTube channel. The image you got from this video's thumbnail by Dr. Wayne Breslyn https://youtu.be/Jm0kFK5IiGo

It is the way how to display the 3D structure of molecules on the 2D plane.
Imagine that

Atomic bonds represented by a normal line are  placed on the plane of  a paper/screen.
Atomic bonds represented by triangles point toward you from the plane.
Atomic bonds represented by stripes point behind the plane.

It is frequently used in organic chemistry, where the particular 3D structure matters, especially for optically active molecules (Their mirror image is not the identical molecule).
See the andselisk comment above.

But how will I show the double and triple bonds then?

Arrange them to be in the plane :-) E.g. both ethene and ethyne are planar.


text:
How can we say that Helium (He) is liquid at "absolute zero"?
According to Wikipedia, Helium can never be solidified at $\pu{1 atm}$ ($\pu{101325 Pa}$) pressure, not even at absolute zero. A minimum of $\pu{25 atm}$ pressure is required to make Helium solid. But absolute zero ($\pu{0 K}$) is not attainable, how can we say that Helium is liquid at absolute zero with $\pu{1 atm}$ pressure? It can never be proved. Is it a theory?

This is an extrapolation, based on our understanding of its nature.

It is similar as if mathemations say $$\lim_{x\to 0}{\left(\frac {\sin x}{x}\right)}=1$$

Because the zero-point energy exceeds binding forces. https://en.wikipedia.org/wiki/Zero-point_energy Are you certain about that position, Herr Heisenberg?


text:
Does mercury(II) cyanate exist?
Recently, I have answered a question "Comparing explosive properties of mercury(II) cyanate and mercury(II) fulminate" where stability of cyanate vs fulminate was discussed. While I was answering that question, I found out that fulminate vs cyanate stability was highlighted when chemical isomerism was discovered by Liebig and Wohler. They both reported independently that cyanic acid and fulminic acid that apparently had the same composition had very different characteristics and so they worked on the corresponding silver salts. These two compounds had the same chemical composition, yet are chemically different. Silver(I) fulminate is explosive, while silver(I) cyanate is a stable compound.
My thought process is that since silver(I) cyanate and silver(I) fulminate were extensively discussed to explain chemical isomerism, could this be applied for corresponding mercury compounds i.e mercury(II) cyanate and mercury(II) fulminate. Mercury fulminate is well known in explosive industry and is used commercially. OTOH, mercury(II) cyanate is not known(?). I could hardly find any information of this compound, So, I assume (incorrectly?) that mercury(II) cyanate doesn't exist and any attempts to isolate the compound would decompose explosively. I also found mercury(II) cyanate as one of the decomposition intermediate of mercury(II) fulminate:

$$\ce{4 Hg(CNO)2 → 2 CO2 + N2 + HgO + 3 Hg(OCN)CN} \\
\ce{Hg(CNO)2 → 2 CO + N2 + Hg}   \\
\ce{Hg(CNO)2 → :Hg(OCN)2} \\
\ce{2 Hg(CNO)2 → 2 CO2 + N2 + Hg + Hg(CN)2} $$

Questions:

Does mercury(II) cyanate exist?
Fulminates are considered unstable and sensitive than cyanates and thus difficult to isolate. So, in this case mercury(II) fulminate should be more unstable and sensitive than mercury(II) cyanate and yet the latter one hasn't been isolated yet. Why?

Reference:

Mercury study report to Congress, DIANE Publishing
Accounting for Resources, 2: The Life Cycle of Materials by Robert U. Ayres, Leslie Ayres, Leslie W. Ayres, Edward Elgar Publishing, 25-Nov-1999, Nature, 380 pages

Sadly no answer to your question but as you did some research on the topic, do you know if something like silver cyanate + mercury(II) chloride has been tested and studied so far?

"mercury(II) cyanate should be more stable and less sensitive than mercury(II) fulminate" and therefore so useless that it's totally overshadowed by better known isomer.

"The Chemistry of Mercury" by Charles Andrew McAuliffe says only what @Justanotherchemist said + that reaction's in methanol. So, it exists, but people generally don't are about it.


text:
Thermodynamic and Transport properties input files for Chemkin
I am using Chemkin pro 2009 version. I want to model a decomposition+combustion reaction on my catalyst in a plug flow reactor.So I'll need to create my own Thermodynamic and Transport Properties file for my own material.
Can you anyone tell me what's the format for the various input files for Chemkin Pro? I have thermodynamic and some diffusion data for my system. Just need to know how to arrange that data in a text file.
Any help will be appreciated

I don't know the program, nor their handbook (e.g., [a senior version ?] https://www3.nd.edu/~powers/ame.60636/chemkin2000.pdf, p. 17). A limited search yields tutorials about this program on youtube (note, for example the section past 1:36 min:ss of https://www.youtube.com/watch?v=5lq0mbHuABY stating «let's use the pencil icon to see the [chemkin] files' content»), and a page by MIT about modeling combustion including reaction mechanism, thermodynamic properties, and transport data in three chemkin files shared with the public http://web.mit.edu/anish/www/MITcomb.html, too as a reference.

Thanks a lot .I found thermo file format and mechanism file format ,still searching for transport properties file format

According to a pdf (https://personal.ems.psu.edu/~radovic/ChemKin_Tutorial_2-3-7.pdf, p. 2/274), «To request technical support ... support@ReactionDesign.com».  MIT's group around professor Howard no longer exists, yet their reaction engineering continues (https://cheme.mit.edu/research/catalysis-and-reaction-engineering/). I speculate the departmental chair (https://cheme.mit.edu/people/faculty/) may help to identify those continuing to use this program possibly licensed for the department rather than just by one group.  Good luck.


text:
Diazo coupling of α-naphthol

Why does coupling of α-naphthol take place at δ-position (para to $\ce{-OH}$) rather than the β-position (ortho to $\ce{-OH}$)?
In the case of β-naphthol, we consider the aromaticity of the entire compound to decide the position.
Aromaticity is not disturbed even when the attack takes place at the ortho position. So I don't find any reason for the diazo compound to not attack there.

Does this answer your question? Regioselectivity in coupling reactions of α-naphthol

In his picture, the OP says that ortho position is sterically hindered. I don't find any reason why.

OH can cause steric hinderance. Also, there is much greater -I effect on ortho compared to para..

We dont consider ortho effect in phenols because of the small size of OH. Why here then?

Take second case.. not first


text:
Karl Fischer Titration simple explanation
I'm looking for a simple explanation how the Karl Fischer method to determine the content of water works. I'm not a chemist but an engineer from an electronic-physical field and my chemistry courses are a while ago, meanwhile. That's why I can't understand what's going on but I don't really need a detail explanation, just low-level enough that I can say "Yep, makes sense" and high-level enough, that I can understand it.
Hence, could someone please describe roughly what's going on?

Wikipedia article does a good job https://en.wikipedia.org/wiki/Karl_Fischer_titrationm maybe after reading this one, you can ask a specific question.

Attempting an almost non-chemical description, you need to know that the concentration of a sample, $c$, equates the number of moles $n$ per unit of volume $V$; expressed as $c = n/V$.
For once, a known volume $V$ of your liquid to be analyzed for its concentration $c$ of water into is injected into a chamber equipped with a stirrer and electrodes.  There is a stock of reagent solution for which you know ahead of the analysis its concentration. You equally know ahead how many molecules of the reagent will react with how many water molecules to yield a balanced reaction equation (stoichiometry).  The reagent solution shall react with water only.
The addition of the reagent solution to the analyte is monitored; for one, you know the volume of the reagent solution added.  Because of this known volume added, and the known concentration of the stock of reagent solution, you know how many molecules of the reagent were added.  The mixing and stirring ensures that each molecule of the reagent quickly encounters a molecule of water to react with.
For two, during the addition of the reagent solution to the analyte, the electric potential between the electrodes dipped into the analyte is monitored. Initially, the addition of a little volume of reagent solution changes this potential only a little.  However, continuing the addition of small volumes of reagent solution, there is one point where this change of potential suddenly is large.  This is the point when 1) basically all the water molecules reacted with the reagent, 2) the electrodes now monitor an excess of the reagent in respect of the water molecules, and 3) the titration may be stopped.
Because you know the volume of the reagent (of known concentration) added and the balanced reaction equations used in this analysis, you may calculate the number $n$ of water molecules consumed to reach this point of sudden change of the potential.  However, because you know both the number of water molecules consumed as well as the volume $V$ of analyte initially injected, you may compute the concentration of water molecules in the analyte, too ($c = n/V$).

The Karl-Fischer analysis offers a rapid characterization of organic liquids (minute scale), provided they do not react differently with the reagent.  Samples containing alcohol, for example, may differ in the underlying reaction equations and stoichiometries from samples containing water.  Generally speaking, to work well, typical samples submitted do not contain more than 1 mol/L of water (about 18 g/L).

Awesome, that was the type of explanation I was looking for! I think I got the idea, I just wonder how it can be that precise as I have to add liquid/material continously. Furthermore, I wonder how it can be that fast as these added ingredients have to react (resp. reach an equilibrium?!)?

#1 Karl-Fischer titration is a batch-wise analysis (e.g. publicity video by Metrohm, https://www.youtube.com/watch?v=W7wxPpADsRw); one-time addition of the analyte, step-wise incremental addition of the reagent, e.g. by a piston driven by a stepper motor such the reactions still advance much faster than the addition. #2 Energetically speaking, the underlying reactions run this much downhill (like an apple falling from a table) that even if the redox reactions could be reversed, the typical reaction conditions render the backward reactions (reformation of water and reagent) insignificant.


text:
Is benzene a polymer of ethyne?
In one of my classes, my teacher told me that benzene can be formed by cyclic polymerization of ethyne. But I think that benzene isn't a polymer because only 3 ethyne molecules are combining to form benzene whereas, thousands of monomers are combined to form a polymer in a polymerization reaction. So, after all, can benzene be called a polymer of ethyne?

To be precise, benzene is a cyclic trimer of ethyne. Trimer is also a polymer which has 3 constituent units or monomer. It is not necessary that polymer should contain thousand of monomer.

This is essentially a question of "how many stones form a heap". To me, three is not many, hence a trimer is not a polymer.

@IvanNeretin trimer can be technically called a polymer whose degree of polymerization is 3. Now, it is a matter of convenience whether one wants to consider trimer a polymer.

Oligo (few) perhaps; poly (many), I don't think so. Definitely trimer.

@NilayGhosh But a minimum of several dozen! Oligomers become polymers when the chemical properties of neighbouring $n$-mers are indistinguishable, partly because they become also unseparable. A trimer is a distinct chemical compound, not a polymer!

The scientist who coined the term polymer, Jöns Jacob Berzelius, used the word to refer to different substances which had the same empirical formula. In this sense, benzene is a polymer of ethyne (acetylene), because benzene $\ce{C6H6}$ and ethyne $\ce{C2H2}$ have the same empirical formula, namely $\ce{CH}$. In other words, all substances with molecular formula $\ce{(CH)_x}$ were polymers of the smallest unit in the family (in this case, $\ce{C2H2}$, because $\ce{CH}$ molecules are only known as extremely fleeting species on Earth, though they are abundant in space).
However, that was back in the 1830s, when we had no clue about atomic and molecular structure (in 1808, one of the eminent chemists of the time, John Dalton, defended $\ce{HO}$ as the formula for water). The meaning of the term has evolved over time to accommodate our increase in knowledge. The most current IUPAC recommendations on polymer nomenclature (Compendium of Polymer Terminology and Nomenclature - IUPAC Recommendations 2008) states the definition of polymer as:

A polymer is a substance composed of macromolecules.

And, fortunately, they also define a macromolecule as:

Molecule of high relative molecular mass, the structure of which essentially comprises the multiple  repetition  of  units  derived,  actually  or  conceptually,  from  molecules  of  low  relative molecular mass.

Note  1:  In  many  cases,  especially  for  synthetic  polymers,  a  molecule  can  be  regarded  as  having  a  high  relative  molecular  mass  if  the  addition  or  removal  of  one  or  a  few  of  the  units has a negligible effect on the molecular properties. This statement fails in the case of certain properties of macromolecules which may be critically dependent on fine details of the molecular structure, e.g., the enzymatic properties of polypeptides.

Note  2:  If  a  part  or  the  whole  of  the  molecule  has  a  high  relative  molecular  mass  and  essentially  comprises  the  multiple  repetition  of  units  derived,  actually  or  conceptually,  from   molecules   of   low   relative   molecular   mass,   it   may   be   described   as   either   macromolecular or polymeric, or by polymer used adjectivally.

Note  3:  In  most  cases,  the  polymer  can  actually  be  made  by  direct  polymerization of  its  parent  monomer  but  in  other  cases,  e.g.,  poly(vinyl  alcohol),  the  description  ‘conceptual’  denotes that an indirect route is used because the nominal monomer does not exist.

As you can see, there is quite a bit of detailing to try to keep things consistent. Observe that the definition speaks of high relative molecular mass, and this point is expanded upon in Note 1. With these considerations, benzene is not a polymer - it is certainly a very small molecule, and adding or removing a portion of it (presumably cyclooctatetraene $\ce{C8H8}$ and cyclobutadiene $\ce{C4H4}$, respectively) would result in a substance with drastically different properties.
In some sense, Berzelius wasn't really wrong to call benzene a polymer, because for all he knew back then, benzene could well have the formula $\ce{C20000H20000}$ (actually, it was not widely believed that "large" molecules could exist at all, with some resistance to the idea existing as late as the 1920s-1930s). He did the best with what knowledge was available to him. But now we can do better, and it just is more convenient and self-consistent to not label benzene as a polymer.
This isn't to say that, however, that benzene and ethyne are completely chemically unrelated. There is in fact a close relationship, in that ethyne can efficiently undergo a [2+2+2] cycloaddition reaction to form benzene. This is not exploited industrially, however, as there are much cheaper (and safer) ways of making it.

Here comes the question how small can a macromolecule of  polymer be  ? What is the least number of ethene molecules to create a macromolecule of polymer ? Like Ivan's number of stones for a heap.

Ah yes, no matter where one draws the line, there will be arguments about what makes the cut and what doesn't. It's best to keep the boundary relatively soft and adjust as necessary depending on context. Regardless, this case is relatively clear in just about any reasonable context, so making a definitive statement is not that risky.

So, if benzene isn't a polymer, how can this reaction be called a polymerization reaction?

@Habib It is a very specific trimerisation reaction. Calling it a polymersiation is simply a misnomer.

Short answer: Yes and No, depending if we consider polymer(1) or polymer(2).
Long answer:
The first step in decision "Is X A or not ?" must be clarification what we mean by "A".
Benzene is a cyclic trimer of ethyne. Trimers are a special case of polymers(1) as repeated monomer structures, being a superset of oligomers and polymers(2). Benzene is a polymer(1) with n=3.
OTOH, we consider polymers(2) as multiple repetitions of monomer unit, where repetition count $n \gg 1$. Such polymers(2) occur rather with a range of $n$ and no particular $n$ has a special, privileged status. Like a polymer plastic material may have $n$ in range 2000-4000.
In such a sense, benzene is not a polymer(2). Neither we say $\ce{N2O4}$ as the dimer of $\ce{NO2}$ is polymer(2), but we can still say it is a polymer(1).

The IUPAC Gold Book defines "polymer: A substance composed of macromolecules." Nothing else. Your definition (1) is very unusual.
.

IUPAC definition is polymer(2) definition. Definition of macromolecule is very vague.

Of course, but your definition (1) should not be used. It is a misnomer perpetuated by a few highschool teachers.

It is not definition (1) versus (2). It is (1) and (2), a wider and a narrower sense. How do you call the superset of oligomers and polymers(2), what must be a useful term, as the boundary between the 2 subsets is very vague ? Perhaps polymer(1) has already assigned other name I am not aware of.

There is no name for the superset (dimer-trimer-oligomer-polymer), or parts of it. I can also see no real use for such a term. Especially there is no use for an umbrella term that includes dimers/trimers and polymers. Polymers have end groups, specific n-mers do not.

That is quite funny,as it's definition would be much sharper than definition of polymer(2) or of an oligomer.

So dimer of ethene with end groups is polymer, right ? Cyclical trimers/tetramers do not have end groups. Linear ones do.

Your polymerisation could yield 1-x-4-y-butane as a side-product. You can also directly synthesise such a molecule. It is not a polymer.

Let us continue this discussion in chat.

A useful criterion I learned in organic chemistry: It is a polymer if removal of one or so monomers from the end does not change the properties, like viscosity significantly. You can use this rule of thumb to distinguish between oligomers and polymers in general.
So by this account, benzene is definitely not a polymer.

This goes to the core without indulging in semantic and nomenclature. Conjugated polymers will be again special here... As far their electronic molecular properties are concerned the switch from oligomers to polymer will be clear as well. Perhaps it won't coincide with a rheological one, but in all cases there is physical significance. Plus one.


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why is NaClO 'liquid chlorine'?

Sodium hypochlorite. $\ce{NaOCl}$.
Appears to be: one part sodium, one part chlorine, one part oxygen.
Five questions:

Couldn't it just as easily be 'liquid sodium' or 'liquid oxygen'?
It's not really any of these though, is it?
It's a compound of all 3, right?
So, doesn't the new molecule have it's own properties?
Or is a compound/molecule(s) properties simply derived from it's constituent elements?

It is just a sale trick.

yes perhaps, but the questions it led me to are real

which of my assumptipns are correct? which are not?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic of the question by doing at least basic own topic review, writing what he/she has found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

I've once met that $\ce{H2O2}$ was called as "hydrogen"... unfortunately, this crap attitude is quite common from company PRs.

@voices For future reference: wikipedia contains many articles about chemicals, and for many of the inorganic ones (like NaClO), the search box permits both a search by chemical name, or alternatively by sum formula (try for example NaOCl2, or NaOCl3, or NaClO4). There equally is a site (http://www.cheminfo.org/wikipedia/) to retrieve 17k wikipedia entries based by a (sub)structure, especially helpful for organic compounds, too.

A chemist would not refer to sodium hypochlorite solution as ‘liquid chlorine’ – the term is instead reserved for chlorine that has been cooled to below $\pu{35^\circ C}$ and has thus condensed to a yellow liquid. The easiest way to prepare it is probably to cool a container to $\pu{-78^\circ C}$ using dry ice and acetone and pass chlorine gas through said container.
Analogously, liquid X to a chemist means compound X in liquid state. Liquid nitrogen is probably one of the most well-known and most-used compounds where the state is explicitly mentioned – it does make a difference whether you are talking about a supercooled liquid at $\pu{-196^\circ C}$ or an inert gas at room temperature.
Thus, liquid oxygen is a blueish paramagnetic liquid that boils at $\pu{-183^\circ C}$. It is always on a chemist’s mind if they’re using a vacuum pump with a cooling trap cooled by liquid nitrogen as liquid nitrogen is cold enough to condense oxygen and any organic compounds can react vigorously even at very low temperatures with liquid oxygen.
Somewhat differently, liquid sodium requires elevated temperatures of $\pu{98^\circ C}$ – and very likely an inert gas atmosphere as it can react exothermically with ambient oxygen or water (vapour).

As you correctly assume, sodium hypochlorite is a compound with the formula $\ce{NaOCl}$ and like other multi-element compounds its properties have little in common with the elements that make it up. In this case, elemental sodium is a metal while elemental oxygen and chlorine are (molecular) gases but the compound is ionic with a sodium cation and a hypochlorite anion (which itself is a molecular ion made up of an oxygen and a chlorine atom). But that leaves the question why it is referred to as liquid chlorine by the manufacturer?
The main uses of sodium hypochlorite are bleaching and disinfecting and both of these derive from the same property. The hypochlorite anion is inherently unstable and tends to decompose depending on the conditions it is subjected to. The decomposition usually leads to chlorine gas being liberated which remains dissolved. Thus, you get chlorinated water. As you can smell chlorine from a hypochlorite solution and since the solution is typically produced by dissolving chlorine gas in a base (see Poutnik’s answer for details), you have essentially ‘captured’ the chlorine ‘in liquid form’ and thus an arguably acceptable common name would be: ‘liquid chlorine’.

Short answer:
No. No. Yes. Yes. No.
Next time, try searching before posting.
Longer answer:
It is just a sale trick. Chemistry community has no direct influence on used trademarks of commercial products. Some may call whiskey "liquid gold", but nobody expects there is any gold there.
The product is called "liquid chlorine" just because it shares some of characteristics of chlorine dissolved in water/water based solutions. As in both cases , there are some equilibrium concentrations of hypochlorous acid, free chlorine and atomic oxygen.
The "liquid chlorine" is basically a solution of sodium hypochlorite, sodium chloride and sodium hydroxide. There is the answer directly on the photo: "130 g/L available chlorine(Cl) present as sodium hypochlorite". I.e. chlorine is not present in its free form, but in chemically alternative form of hypochlorite, that can release chlorine at different than storing conditions.
Sodium hypochlorite is produced by dissolution of chlorine in solution of sodium hydroxide:
$\ce{Cl2(g) + 2 NaOH(aq) -> H2O + NaCl(aq) + NaClO(aq)}\tag{1}$
If acidified, e.g. by sulphuric acid, the reaction (1) is reversed:
$\ce{H2SO4(aq) + NaCl(aq) + NaClO(aq) -> Na2SO4(aq) + Cl2(g) + H2O}\tag{2}$
If dissolved by about neutral solutions, reaction (2) happens in lower extent, as there occurs also
$$\ce{ClO-(aq) + H2O <=> OH-(aq) + HClO(aq)\tag{3a}}$$
$$\ce{HClO(aq) -> H+(aq) + Cl-(aq) + O(aq)\tag{3b}}$$
The atomic oxygen is reportedly the true desinfecting agens, rather then chlorine itself.


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Adding a base and a salt to a hydrated complex
The question goes like this:

$\ce{Fe(NO3)3}$ V= $\pu{1 liter}$ and $\ce{[Fe(NO3)3]} =\pu{0.1 moles/liter}$.
Calculate the $\mathrm{pH}$ of this solution using the values of $K$ given.
We add a solid of $\pu{0.2 moles}$ of $\ce{NaOH}$. Give the reactions and the $\mathrm{pH}$. How many precipitation is formed?
After this we add $\pu{0.5 moles}$ of $\ce{KCN}$. Give the reactions, the $\mathrm{pH}$ and the end concentrations in equilibrium.
Given:

$K_\mathrm{st}\,\ce{[Fe(CN)6]^{3-}} = 1\times10^{42}$
$K_\mathrm{a}\, \ce{HCN} = 4\times10^{-10}$
$K_\mathrm{a_1}\,\ce{Fe^{3+}} = 6.3\times10^{-3}$
$K_\mathrm{s}\,\ce{Fe(OH)3} = 1\times10^{-38}$

Up to and including the addition of $\ce{NaOH}$ I understand what is expected and how I should solve for this. However, as soon as the $\ce{KCN}$ is added, I don't know if I should let the $\ce{Fe^3+}$ react first with the $\ce{CN-}$, with the $\ce{OH-}$, or if I should establish a new equilibrium starting from the previous one?
My first thought would be to let the cyanide react completely with the iron-ion, and only form the iron hydroxide with the remaining ions [because $K_\mathrm{st}\,\ce{[Fe(CN)6]^{3-}} >> K_\mathrm{s} \ce{Fe(OH)3}$]. However, in some provided solutions of similar exercises, they seem to let the hydroxide react first with acid, not caring about a higher solubility constant.

$\ce{Fe_3(NO_3)_3}$ does not exist ! Only $\ce{Fe(NO_3)_3}$ does exist !

What is the meaning of $\ce{K_{st}}$ ?

@Maurice $\ce{K_{st}}$ = $\ce{K_{form}}$ = $\ce{1/K_{diss}}$, it's the product of all the stability constants inbetween $\ce{Fe3+}$ ... $\ce{Fe3(CN)3}$ ... $\ce{Fe3(CN)6^{3-}}$

Well ! $\ce{Fe_3(CN)_3}$ and $\ce{Fe_3(CN)_6^{3-}}$ do not exist

If you know how to solve this, you should be able to figure out. After you add the $\ce{NaOH}$, the pH will be still acidic, and some $\ce{Fe(OH)3}$ will precipitate. If you disregard this amount (assuming it wont be solving back), then after adding the $\ce{KCN}$ you will end up with a basic solution, for which you need to solve a system of equations to get the right answer. You don't need to worry about which will react first, since they will create and equilibrium.

Whats the definition of $\ce{K_{a1}Fe^{3+}}$? Obviously it is not about hydroxides, then what is is about? The first proton dissociation from the coordination sphere of the hexaaqua complex? You also mentioned a hydrate complex in the title but you did not mention any water complexes in the post, nor any constant for it.

@AndrewKovács The $K_\mathrm{a_1},\ce{Fe^{3+}}$ goes for the following reaction $$\ce{Fe(H2O)6^{3+} + H2O <=> Fe(H2O)5(OH)^{2+} + H3O+}$$There are no other constants given, which I think suggests the next acidic reaction of the iron complex is negligible.


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Alphanumeric preference between hexyl and heptyl
Our chemistry teacher taught us that hexyl (five letters) is preferred to heptyl (six letters) as hexyl has lesser number of letters. After searching many sources I could not find any such rule.
Which one would be preferred while writing IUPAC names?
For example, will it be 8-heptyl-10-hexylheptadecane or 10-heptyl-8-hexylheptadecane?

Hexyl comes from the Greek name "hexa* which means "six". "Heptyl" comes from the Greek name "hepta" which means seven. It has nothing to do with the number of letters. The next radical, after "heptyl", is "octyl", coming from the Greek, since "octa" means "eight" in Greek.

@Maurice Nit picking, but I believe it is "octo."

@ Zhe. OK. You may be right. It does not change the reasoning.

Please read the example I gave,I wanted to ask-between hexyl and heptyl which is given more preference according to the Alphanumeric order of IUPAC.

@James I think you would like to review your examples.

@Habib it was just a random one to explain my point.Btw, thank you for answering

According to the IUPAC nomenclature, preference for placing prefixes is given in accordance with alphabetical order (just like in a standard English dictionary). It has nothing to do with the number of letters in a prefix. Since "P" in heptyl comes before "X" in hexyl, heptyl will be placed before hexyl in IUPAC nomenclature.
Hence, the correct name will be the one where heptyl is placed before hexyl (8-heptyl-10-hexylheptadecane).

Every time you back up your answer referring to "IUPAC nomenclature", you also need to provide a citation. This is not only considered a good practice in general, but also makes your post more future-proof since IUPAC rules are often a subject to change.

Is there any such rule in IUPAC of counting the number of letters to decide priority between two substituents?

As far as I know, there isn't, at least not for determining the order of prefixes in an organic compound.


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More Anti-bonding or less bonding electrons?
Is a molecule more stable with less bonding electrons or more anti-bonding electrons? This question arose when I was asked the stability comparison between $\ce{N2^+}$ and $\ce{N2^-}$.

Both (N2)+ and (N2)- have same bond order of 2.5 but (N2)+ would be more stable as it has lesser number of electrons leading to lower inter-electronic repulsions.It can also be said that (N2)- has more antibonding electrons leading to a lower stability than (N2)+

Elaborating a bit- intro texts often show bonding and anti bonding orbitals equidistant from the starting atomic orbitals, but when you do the math, bonding orbitals are stabilized by a smaller amount than antibonding orbitals are destabilized

@Andrew, Is that really true? I don't know much about this topic but I always thought that energy is partitioned equally such that the potential lost by BMO is same as that gained by ABMO.

Yes. The equal separation is a very useful approximation, but not quite quantititatively correct. The extra destabilization of the antibonding orbitals is why, for example, He2 is not as stable as He.


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Do solid insoluble ions react with soluble ions in aqueous solution?
Do solid and insoluble ions (like aluminium phosphate) react with soluble ions in aqueous solution (like potassium sulfate)?  When would a solid ion react with an ion in aqueous solution in general?

Surface equilibria happen all the time, but what are you imagining with this example?

And @Maurice here should read https://en.wikipedia.org/wiki/Lattice_energy

@ Mithoron. The mentioned publication does not explain the difference between the solubilities of $\ce{CaF_2}$ and $\ce{CaCl2}$

@Mithoron, I thought AlPO4 (insoluble) and K2SO4 (water-soluble) were a good example for this question. I was wondering whether they would swap ions in a precipitation reaction.  In purely basic information about double displacement reactions I've found online, both reactants are always initially in aqueous solution, never in solid or gas phase. I can't find anything about the process solid ionic substances undergo if they are placed in a solution.

What you seem to be asking about is a two phase system, a solid phase in equilibrium with ions in solution. For such a system there are two reactions occurring at the same rate. Thus some of the solid is constantly dissolving and some of the ions from solution depositing at the same rate.
There is a named process, Oswald ripening, that describes the interaction in such a solution. The overall effect is that smaller particles are more likely to dissolve so the particle size tends to grow as the process continues since larger particles have a smaller surface area to volume ratio.

Thanks. I think this points me to what I am trying to understand.


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In Na extraction, is Ca2+ of CaCl2 reduced in addition to Na+ of NaCl in electrolysis
In Na extraction from $\ce{NaCl}$, $\ce{CaCl2}$ is added in electrolysis to lower the melting point of the mixture to about $\pu{750 ^\circ C}$. But doesn't $\ce{Ca^2+}$ cation compete with the $\ce{Na+}$ ion to be reduced at the cathode? If so, since $\ce{Ca^2+}$ is right next to the electrochemical series, should we not have a considerable amount of $\ce{Ca}$ as impurity in the $\ce{Na}$ that we have extracted? Where do the inconsistencies lie?

Electrochemical series is for aqueous solution only!

@Mithoron, Got it, thanks. But in this case, how will one decide which cation will be reduced without the help of the electrochemical series?

Well, there are data for various solvents and melts, not nearly as commonly found or thorough, though.


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Why did my synthesis of calcium peroxide fail?
I dissolved calcium acetate in about $\pu{80 mL}$ of water and added $\pu{30 mL}$ of 60% peroxide. I should have obtained calcium peroxide which is insoluble in water, but i saw no precipitate. I read somewhere a synthesis where very dilute peroxide was used, so i later tried the synthesis at a  maximal concentration of $\ce{H2O2}$ as 6% and still without result.
Should I have used a different salt of calcium? Is it possible the stabilizers in $\ce{H2O2}$ interfered?

According to the English edition of Wikipedia (https://en.wikipedia.org/wiki/Calcium_peroxide), $\ce{CaO2}$  is described both with the low solubility in water mentioned in your question, as well as decomposing in water (see property box).  Perhaps the product decomposed before you were able to isolate it?  Side note: 60% $\ce{H2O2}$ is about twice as concentrated typically seen in the chemistry lab (stay safe!).  On the other hand, if your group has access to this high concentration, possibly a database like Reaxys listing multiple protocols of synthesis may be accessible for you, too.

See also https://chemistry.stackexchange.com/questions/119807/neutralization-between-caoh2-h2so4-in-a-30-hydrogen-peroxide-solution

Wikipedia states that the industrial preparation of calcium peroxide is done with calcium hydroxide, not the acetate or a neutral salt.  Compare this choice with those used commercially for magnesium peroxide and lithium peroxide.
You want to do these syntheses with strong bases because the metal peroxide itself is most stable in basic conditions.  Neutral or acidic water would promote hydrolysis to hydrogen peroxide and thence decomposition.

One error is the presence of an acid source which results in moving the below equilibrium (written in the reverse of the formation reaction) to the right which occurs slowly with water for many peroxides:
$\ce{ CaO2 + H2O <=> CaO + H2O2}$
Even if you performed the indicated reaction above with H2O2 acting on CaO, you still should check the pH of the hydrogen peroxide, as it is likely acidic. Apparently, phosphoric acid, for example, is frequently employed stabilizer for H2O2 as neutral to basic H2O2 has a much reduced shelf-life.
Further, any presence of a transition metal (Wikipedia on Magnesium peroxide cites an issue, for example, with iron) could also move the equilibrium to the right. Select transition metal oxides are known to produce a cyclic decomposition reaction with H2O2, again effecting the above equilibrium.
My success making a peroxide also entailed working with cold solutions, albeit, in the presence of a stabilizer (like sodium silicate), one may work at warmer temperatures.

This brings up another advantage of using a basic source of the metal.  You neutralize the added acid and also precipitate transition metals.  Magnesia, lime and lithia all do that.


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Why does formation of bond releases energy and where does that energy come from?
I am having a hard time to imagine why does bond formation releases energy and where does that energy comes from?? It is easy to understand that breaking a bond requires energy because it can be viewed in many ways like breaking a wood into pieces requires energy, ripping your shirt into half requires energy, converting liquid water into gaseous water requires energy. These tangible concepts make it easy to understand why bond breaking requires energy. From these examples it answers the question why does bond breaking requires energy and where does that energy comes from (surroundings/thermal energy/mechanical force applied by the surroundings). But bond formation have so little comparisons and most of the examples associated with it are vague and hard to communicate with non science major. Thanks for the help.

Bond formation is like the same processes in reverse. True, this doesn't happen in our everyday world. Well, on atomic level it does.

Well, electrons have a lot of energy and can lose a bit. While why they lose it is hard to explain, the source is kinda simple.

If you want to keep it simple, you can say that an electron (negative charge) that can interact with two nuclei (positive charges) has a lower energy than an electron near a single nucleus. Of course, this would mean that all atom-atom interactions would be covalent, which is not true. To understand this, you will need to get more sophisticated: orbitals, Pauli exclusion principle, exchange interaction...

The shirt and the wood are not the best analogies. Two magnets are - take a look at Snatoms, https://www.youtube.com/watch?v=He30D8M5fNc

I think the shirt and the wood are good analogies because it is what ultimately felt at a macro scale. Just keep into account that some of the bonds might be not covalent. Nice question. Plus 1.

It's a matter of perspective. You say you understand why bond breaking requires energy. In other words:
$$\ce{A2 + energy -> 2A}$$
But consider the reverse reaction:
$$\ce{2A -> A2 + energy}$$
This is also valid and can be understood in the following way: Bonds stabilize the electrons. In the unbound or atomic state, the electrons have a bit more energy. You can consider that atoms have some potential energy, like a ball on a hill. By forming a bond, the atoms "go down the hill" into a more stable state and thus lose this potential energy. This energy must go somewhere, and it is thus released in one form or another (heat, light, etc.) To break the bond is to push the atoms "up the hill" again, which requires energy.


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Artificial foods — Could you make them? And would they compare to their "organic" counterpart?
This is partly just a hypothetical, but basically.
If you were to take all the chemicals in an apple and put them together, could you make an apple? (Probably closer to apple sauce though.)
If this is possible, How would my almost-apple paste compare to a normal apple when it comes to nutrition?
I imagine it would not be nearly as healthy as a normal apple. As from my research, supplements to a vitamin do not have the effect that the vitamin would have if consumed naturally.

I know apples contain Alpha-linolenic acid, Ferulic acid, Isoquercetin, Lutein, Neoxanthin, Niacin, Thiamine, Riboflavin, Pantothenic acid, Vitamin B6, Folate, Vitamin C, Vitamin K, Carbohydrates, Sugar, Calcium, Iron, Magnesium, Manganese, Phosphorus, Potassium, Zinc, Asparagine, D-Categin, Farnese, P-Coumaric-Acid, Avicularin, Quercitin, Rutin, Ursolic-Acid, and Protocatechuic-Acid.

This is presumably not a comprehensive list by any means. I also only really used apples as an example. If there's another food you feel would be easier to "build" then please use that.

Thank you for giving this a read! And I apologize if I posted in the wrong place, I'm new to stackexchange.

No, it is not currently possible to create whole foods (like an apple) (as opposed to processed foods, like sugar or vinegar*) from their component chemicals.  The reason is simple:  All whole foods are composed of tissues, which are in turn made of cells, and we aren't close to being able to synthesize cells, especially eukaryotic cells, from their component chemicals. Their structures are far too complex. [https://en.wikipedia.org/wiki/Artificial_cell]
*In this sense I would also consider honey a processed food, though it is processed by bees instead of humans.
The closest we can come to creating artificial tissues is to clone existing cells; but that wasn't your question.
So then the question becomes whether we could fully characterize all the components of a whole food (again, let's use the apple as an example), and create a mixture containing all those components—and whether, in so doing, we would have something nutritionally equivalent.
The answer to that is not known, and I would say the burden of proof lies with those who wish to claim it would be nutritionally equivalent.  Even if we could characterize every possible micronutrient in an apple, we still wouldn't necessarily know, much less be able to recreate, the states in which they exist (in particular, their states of association with other compounds) within the apple.
Certainly there is a difference between a micronutrient that is complexed with other micronutrients, and then contained within a structure and then contained within another structure, and one that is mixed within a paste and thus complexed with compunds with which it might not be naturally associated in the intact apple. Of course, we do "paste" the apple ourselves when we chew it, but that process likely does not lead directly to the type of paste you would have if you tried simply mixing all the apple's constituent compounds together.
And it's not just how we directly would react to the difference.  There's also how the organisms of our microbiome would react.  It's certainly possible that consuming intact apples would favor one species of microorganism, while consuming synthetic apple paste would favor another, thus leading to changes in the balance of our microbiome.
Then we have to combine all the above uncertainty with the enormous added uncertainty that exists surrounding human physiology and nutrition.

Your question seems broad.  Still, in short, a tentative answer is no, however...

Currently, it is not possible to mix dry chemicals from the shelf in a round bottom flask to produce a steak.

However, you may heard of tissue engineering for example to multiply human skin cells to help patients with burns.  It consists of the isolation of healthy skin cells which are grown outside the human body to larger patches eventually transplanted again to the human body (example, example):

(reference)

(open access reference)

There is interest to produce meat without recurring to animals as cultured meat.  Beside plausible reasons based on religion or animal welfare, this is appealing because of it may become easier to scale than to care for growing flocks of cattle, may represent a more efficient use of resources (think e.g., about use of water water, soy) with less footprint (e.g., less waste water and methane emission).


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Why don't dicarboxylic acids show hydrogen bonding in spite of the fact that carboxylic acids exist usually as dimers in polar aprotic solvent?
I have seen several examples:

The fact that maleic acid is more soluble in water than fumaric acid in spite of the fact that it must be able to show intramolecular hydrogen bonding.
trans-Cyclopentane-1,2-dicarboxylic acid is more stable than the cis form (cis form should be able to show hydrogen bonding).
o-Nitrobenzoic acid is more acidic than the para isomer.

I tried linking point 1 to acidic strength (maleic acid > fumaric acid because of hydrogen bonding in conjugate base) and point 3 to acidic strength (o-nitrophenol < p-nitrophenol)
What am I missing? Something related to the size of the ring?

Refer to these papers: here, here and here

This is a prime example of question that's both badly worded and too broad.

I would like to object the statement dicarboxylic acids would not engage hydrogen bonding.  Terephthalic acids ($\ce{HO2C-C6H4-CO2H}$) are a prominent example since early attempts to formalize the description of intra- and intermolecular hydrogen bonding in the solid state. Consider the pairing of benzoïc acid and $p$-terephthalic acid, sharing the same motif of hydrogen donors and acceptors, described as $R_2^2(8)$ in Etter's notation:

(Etter, Acc. Chem. Res. 1990, 23, 120-126,  doi 10.1021/ar00172a005)
But there is no limit to $p$-substitution to find this motif again, e.g.:

(Holy et al., Collect. Czech. Chem. Commun. 2006, 71, 139-154, doi 10.1135/cccc20060139, researchgate.net)

But why exactly isn't hydrogen bonding predominant in explaining the 3 specific cases I mentioned in my question.  Would really appreciate your help and support.