text:
How to determine if two calibration curves are parallel, within 5%
I have complex matrices and I need to make sure the matrix does not affect the measurement of my analyte. To do that, I need to prepare samples of my analyte in a solvent and of the matrix reinforced with my analyte. Both curves are going to have 5 concentration points, each done 3 times and, for sake of argument, let's assume both have comparable levels of precision. Then, I linearize these two calibration curves and check if they are parallel, which is where my question lies.
How do I prove that both curves are parallel (meaning no matrix effects are present), within a margin of 5%?
I have found the following method:

Get the y-data of the matrix data and apply them to the calibration function of the solvent and get an "apparent concentration" of the matrix.
Perform a linear regression of the concentration values of the solvent with the apparent concentrations of the matrix.

Then, I have an intercept a and a slope b. A value of b different from 1 indicates a concentration dependent bias. I have both standard deviations of the fit, stdev_a, stdev_b, so I have to determine the confidence interval (95%) of b and check if their values are withing these confidence intervals.
I have found two ways to do that in Excel and I am unsure which one is correct. For example, the slope:

Do I use the conf = CONFIDENCE.T(0.05, stdev_b, N-4), get b+-conf, and check if 1 is within this interval b+-conf?
Do I use t = T.INV(0.95,N-4), multiply that by the stdev_b, get the range b+-t*stdev_b and check if 1 is within this range?

I have used N-4 for the degrees of freedom, since it's two linear regressions, each consuming one degree of freedom, correct? In this case, N-4 == 11.
In summary:

Which function do I use to calculate the confidence intervals for this example in Excel, and
Is this method of checking parallelism correct?
Is there an easier/simpler or more correct way of doing this parallelism check?

I'm voting to close this question as off-topic because it's about statistics.

I think it's multidisciplinary. Analytical chemistry has a lot of statistics in it. If no one answers, I'll ask on the statistics board. I'm afraid the chemistry behind the question might go over their heads and derail the discussion.

If question gets closed you it might be migrated there.

Define the angular coefficients of the two resulting lines and evaluate the parallelism between them. Perform the statistical evaluation of the data.
For the parallelism test between the curves, considering the equations of the lines to be compared as: $y_1 = a_1 + b_1x + \varepsilon$ and $y_2 = a_2 + b_2x + \varepsilon$, the hypothesis for comparison of the lines slopes is given by:
$$H_0 : b_1 = b_2$$
$$H_1 : b_1 \neq b_2$$
When the null hypothesis ($H_0$) is true, the regression lines become: $$y_1 = a_1 + b_3x + \varepsilon$$
$$y_2 = a_2 + b_3x + \varepsilon$$
Considering a significance level of $5\ \%$, the null hypothesis will be true when the test returns a P-Value above $0.05$.
More information in the link:
http://www.real-statistics.com/regression/hypothesis-testing-significance-regression-line-slope/comparing-slopes-two-independent-samples/

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Here's some R code to illustrate one way to handle this situation.
First, let's generate some toy data.
require(tidyverse)
require(broom)

# concentrations
x <- c(1, 1, 1, 3, 3, 3, 10, 10, 10, 30, 30, 30)

# toy parameters
M <- 1000
Bclean <- 20
Bdirty <- -20
SD <- 2
MATRIX_EFFECT <- 0.95

# assembling data into dataframe
set.seed(0)

clean <- M*x + Bclean + rnorm(mean = 0, sd = SD, n = length(x))
dirty <- (MATRIX_EFFECT)*M*x + Bdirty + rnorm(mean = 0, sd = SD, n = length(x))

df <-
    tibble(x = x, clean = clean, dirty = dirty) %>%
    pivot_longer(-x, names_to = 'matrix', values_to = 'response') %>%
    arrange(matrix, x)

Here, x is the independent variable, presumably a concentration.  clean and dirty are names for two separate response vectors, one for a "clean" sample matrix and one for a "dirty" sample matrix.
(This way of creating the toy data assumes that you make authentic standards at the same concentrations when you make the calibration curves; but this isn't required for the statistical modeling that follows; the model will work if the two calibration curves both have totally different concentrations.)
I've assumed a value for M, B, SD, and MATRIX_EFFECT in this code, but to see how they would affect results I'd vary them and re-run the below code.  The meaning of those parameters is:

M is the slope or response coefficient for the "clean" sample matrix.
Bclean is the intercept (or offset) for the "clean" sample matrix.
Bdirty is the intercept for the "dirty" matrix.
SD is the standard deviation (in response units) of both the clean and dirty responses.  Here, I've assumed an unrealistically low value, which is helpful for being able to understand the meaning of the parameters returned by the fit.
MATRIX_EFFECT is a fraction by which M is modified in the "dirty" matrix relative to the clean matrix.

The result of this code is a "dataframe" that looks like this:
x   matrix  response
<dbl>   <chr>   <dbl>
1   clean   352.590857
1   clean   34.753328
1   clean   365.959853
3   clean   554.485864
3   clean   382.928287
3   clean   -7.990008
10  clean   814.286593
10  clean   941.055911
10  clean   998.846565
30  clean   3480.930678
30  clean   3152.718692
30  clean   2840.198150
1   dirty   -123.531402
1   dirty   48.107685
1   dirty   46.156976
3   dirty   235.697833
3   dirty   368.444690
3   dirty   139.615775
10  dirty   1147.136660
10  dirty   812.492316
10  dirty   1015.146423
30  dirty   3255.479129
30  dirty   3206.667272
30  dirty   3340.837902

We can model this data as follows:
df %>%
    lm(data=., formula = response ~ x*matrix) %>%
    tidy()

This is models the response as a linear (lm()) function of a bunch of parameters, with parameters implicitly defined by the formula.  response ~ x*matrix means, since x is a continuous variable and matrix in this case is a factor variable, to regress response on concentration interacting with matrix.  The results are:

Each row gives results, including standard errors, point estimates, and p-values that the parameter is non-zero, for a single parameter.  One subtlety is that the algorithm has implicitly assumed one (of in this case two) levels of our factor variable matrix to be the "baseline", and only parameters for the other non-baseline level (in this case "dirty") are referred to directly.  So the results mean:

(Intercept) is the model estimate of Bclean.  The estimate of 20.28 has an uncertainty of $\pm$ 0.740.  Gratifyingly, the confidence interval overlaps the "true" Bclean value we picked for this toy data above.
x is the model estimate of M
matrixdirty is the model estimate of Bclean - Bdirty
x:matrixdirty is the model estimate of M*(1-MATRIX_EFFECT) and is the parameter that you are really interested in.  Dividing by M, we see the regression says it is -49.96.  Dividing by M, in this case estimated as 1000.02839, gives an estimate for the matrix effect of 0.95004.  Thus, for these (made up) numbers, the model would say your dirty matrix has a calibration slope that is 4.996 percent of the clean matrix.  Of course, you also have error estimates here.  The std.error of 0.0658 is also in response units; after dividing by M, we can see the model thinks it's matrix effect estimate is good to the third decimal place.

These amazing results are because I've picked an SD of 2 for the simulated data, but the response varies between 100 and 30,000.  So we're assuming relative error of 2% down to 0.006%.  If you pick more realistic values for SD, the std.error of all of the parameters will be much larger.

I think that this is one of those in-between questions. Before migrating, let's define the chemistry better.
(1) Roughly what sort of precision are you expecting? What sort of instrumental precision, what sort of wet chemistry precision, and what sort of sampling precision?
Trying to figure if one error source dominates, or if all three contribute significant error.
(2)  You said "Both curves are going to have 5 concentration points, each done 3 times..." I assume you made three different solutions of the sample, then spiked 4 different aliquots each with 4 different amounts of the analyte. Thus you end up with 5 solutions for some sort of instrumental analysis. Correct?
So the 15 solutions analyzed depend on three weighings of sample. Also I assume that you're using 4 different volumes. In other words you the spike an aliquot with a one ml pipette, and the second spike is with a two ml pipette, not two shots from the one ml pipette, and so on...
So I assume that you assume that the spiking is exact and you're not worried about the spiking errors.
A point here is that you seem to accept that the intercepts of the lines will be different. So is there something in the sample matrix that causes the blank value to vary? So although you have 5 points on the line, you seem to only have three different matrix solutions.
(3) When you say you "linearize these two calibration curves", I assume you mean that you used linear least squares to compute slope and intercept of each.
(4) Exactly what do you mean by 5% error.
I assume that you mean a 5% chance of a type I error in slope. But I don't think that is what you want to test. I think you spiked the sample solution with 4 different concentrations of the analyte. So what you really want to know is the difference between the sample values at the end of the line, not variation in the slope itself.
The difference between the non-spiked values will be a lot more sensitive than test on variabilities. You have very small statistics and the variabilities are going to be huge.
The gist here is also considering a type II error and/or using multiple determinations to drive the error to 5%.

(1) I am doing HPLC at concentrations ranges of 80-120%and errors of 2%.
(2) Correct; Correct; The matrix itself has some of the analyte, at a concentration C. I have, then, these concentrations standart dilution : , 80%, 90%, 100% ,110%,120%;matrix:( standart+ feedstook) : 80% 90%, 100%, 110% 120%,  with the same dilution of standart. The intercepts will be the same.
(3) Yes, that's what I mean. But the slope is important, not the intercept, in this case.

(4)I need to know specifically the slope, because that's in a norm. It states, in full: "The line parallelism is indicative of a lack of interference by the constituents of the matrix and its determination must be realized by an adequate statistical evaluation. The significance level adopted must be 5% in the hypothesis test." So from my basic understanding of statistics, it's a 5% of a type I error in the slope.


text:
Is HDPE affected by long term exposure to methanol?
For at least several years I'be been storing various chloride solutions with methanol as a solute in Nalgene containers, which I believe are made from high density polyethylene (HDP).
But recently two of these containers had failed. On these containers the plastic seems to have hardened and cracked (see photographs below).

The other containers still seem to be pliable, in good shape. I thought that HDPE would be resilient to methanol and provide indefinite storage. Was I wrong? Or is it perhaps the methanol in combination with the particular salt that led to failure of the container? The failed containers had strontium and potassium chloride solution. Others like lithium, sodium and copper solutions are OK.

HDPE should be fairly stable to MeOH, but it crazes in the presence of UV, when subject to oxidization or to environmental stress.
Since only some of the containers failed, were those subjected to more UV (e.g. placed by windows or at the front of a shelf in a well-lit location), placed near oxidizers, such as bleaches or were they subject to higher temperature, which would have caused the methanol liquid and vapor to expand, increasing pressure and stressing the container?
My experience has also been that apparently identical polyethylene containers have different failure rates, perhaps due to subtle differences during synthesis or manufacture.
Given polyethylene's tendency to craze, it might be best to use another type of container for long-term storage.
Note, though, that ~50% MeOH/water solution used for windshield antifreeze is usually shipped in polyethylene containers. Though the water lowers the vapor pressure, perhaps the UV susceptibility is addressed with additives.

The containers may have been exposed to higher than normal room temps this last summer, and correspondingly pressure, but ALL of them were. Perhaps these two were at higher pressure. But there is a marked difference in the brittleness of the failed containers compared to the ones that didn't fail which are more pliable. Which leads me to believe failure was more due to change in material property than stress itself. It also occurred to me that these containers were all reused and maybe there was something in these particular containers that led to brittleness.

In any event whether stress or material change, this gives impetus to regular, periodic inspection of ALL chemical storage containers, especially in the case of risking release of hazardous substances.

@docscience, I strongly agree: do inspect HM storage regularly.  Steel shelving can be corroded by acid vapors, bottles can crack as contents crystallize and there are many other failure modes. It is also a good opportunity to rotate usage, so oldest, most likely to be weak or contaminated, are used up or discarded.

Once stored metallic calcium in what I thought was an airtight glass container. The oxidation expanded internal volume and shattered the jar.

PE has a strong tendency for cold flow under (esp. extensional) stress. When it yields, it recrystallises, becoming more brittle. The crack position at the edge, where the highest local stress occurs, would support that this is what happened.
Different PE types (LDPE; LLDPE; HDPE; and more subtle differences in branching, comonomer content etc.), slightly different temperature, pressure, filling height, previous UV exposure, etc. made the difference for your containers. Once it's started, the crazing is likely to continue, because the anisotropic density change during recrystallisation produces even more local stress in the material.

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text:
Thermophysical properties of liquid Galden LS 230 between 25 °C and 230 °C
I'm looking for thermophysical properties of Galden LS 230, a liquid polymer (PFPE) produced by Solvay (https://www.solvay.com/en/markets-and-products/featured-products/Galden.html) at temperatures above 25 °C. Solvay only publishes properties at 25 °C.
I asked them for properties at higher temperatures, with no answer in three weeks. I also couldn't find other free online sources (this is a vapor phase soldering hobby project).
The properties published at 25 °C are:

boiling point: 230 °C
density: 1820 kg/m³
kinematic viscosity: 4.4 cSt = 4.4 × 10-6 m²/s
vapor pressure: 453.3 Pa
specific heat: 973 J/kg K
heat of evaporation at boiling point: 63 kJ/kg
thermal conductivity: 0.07 W/mK
coefficient of expansion: 0.0011 1/K
surface tension: 20 dyne/cm = 0.02 N/m
average molar weight: 1020 a.m.u
some others that I don't expect to need for thermodynamics: dielectric strength, dielectric constant, volume resistivity.

Vapor density
Using the ideal gas law to calculate the density of saturated vapor at 1 bar yields about 24.4 kg/m³, which is within the range I found in some other inofficial sources and also in a pretty hard to read plot in a solvay presentation.
Viscosity
the wikipedia page on the temperature dependence of liquid viscosity (https://en.wikipedia.org/wiki/Temperature_dependence_of_liquid_viscosity) lists a number of models that might also apply to Galden, but I'm not sure which ones would yield reasonable results. I'd be glad if someone could shed some light on this. The solvay FAQ (https://www.solvay.com/en/markets-and-products/featured-products/galden-pfpe-faq.html) shows a plot of viscosity vs. temperature for Galden HT 230, which has the same nominal boiling point as the "LS" grade, but has higher tolerances for the molecular weight distribution. The HT grade also starts with the same viscosity at 25 °C as the LS grade. If all else fails, I might use that value.
Specific heat
I'm not sure if it's reasonable to assume specific heat to remain close to constant. My gut feeling is "probably".
Thermal conductivity
Since Galden is a non-metal, I'd expect thermal conductivity to remain more or less constant, or to increase slightly. How large or how significant the change would be, I don't know.
Surface tension
I expect surface tension to decrease with increasing temperature, according to the Eötvös Rule (https://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_rule), but to really calculate values I'd need information about the critical point of this polymer (if that even exists, anyway).
A bit more technical background
In vapor phase soldering, a printed circuit board with solder paste and components is immersed into high temperature vapor (such as that of Galden). To create this vapor, the liquid has to be heated and evaporated. This must be done carefully because Galden starts to decompose at 290 to 300 °C. All the properties above are required to calculate a Nusselt number for nucleate boiling, so that I can determine the actual temperature at the heating surface, which will be higher than 230 °C.

Would it be easier if you boiled at a fixed temperature below 290C and above 230C, would that be sufficient? If you don't bug the company, they tend not to get to requests from non-corporate sources.

Unfortunately with my current hardware it's not possible to heat the Galden in such a way that I can measure the temperate at the surface where the actual vaporization takes place, which is an immersed power resistor. That's why I want to look closer at the actual heat transfer to figure out an upper limit to the heat flux.

What Nusselt number relationship are you thinking of using? If you don't get literature from the company or from journals (even data for similar molecules will be okay), you'll have to make an engineering judgement whether the worst case scenario for you is conservative or will lead to failure.

@J.Ari I thought about using correlation from Stephan and Preußer for nucleate boiling above horizontal plates, which is said to be suitable for organic fluids close to atmospheric pressure. This would be a rather bad heat transfer situation (with rather high wall temperature) when compared to, say, boiling at the wall of a horizontal cylinder, which is closer to what I have in mind.

If your elements are going to get well above 230°C, you may not have nucleate boiling but transition or vapor film boiling which will limit heat transfer to the liquid PFPE. I'm not familiar with the mentioned correlation but if I were you, I'd be thinking of doing some experiments or heating a different way. Have you considered dual stage heating?

@J.Ari I did not consider dual stage heating because I want the device to stay as simple as possible. One fact that makes this question somewhat obsolete is that I just found out that the Galden FAQ contains a section about proper heater sizing, which states that the lower end of the safe region is around 4 - 5 W/cm². If all else fails, I'll have to use that value.

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What is the mechanism for thermal decarboxylation of a generic carboxylic acid (not special cases like beta-ketocarboxylic acid or malonic acids)?
I am looking for the mechanism for thermal decarboxylation for any RCOOH, and I am guessing that its possible for the thermal energy to cleave the R-C bond homolytically leaving a R• and •COOH. Then I would guess the R• would add its electron to the H and the O-H bond would undergo homolysis as well leaving the single electrons on the C and now O to form a bond leaving RH and CO2. However I have no idea if this is what actually occurs. Thanks

Related: Decarboxylation of Salicylic acid

The decarboxylation of any carboxylic acid can basically take place in four ways:

Unimolecular free radical mechanism (homolytic fission)

However, in case of thermal decomposition there has been no evidence of this type of free radical formation. Although, free radicals are formed in electrolytic or photochemical process. There is also evidence of formation of free radical in ketonic decarboxylation.

Unimolecular heterolytic fission (carbanion formation)

This is the mechanism for most carboxylic acids.

Unimolecular heterolytic fission (carbocation or carbenium ion formation)

This has not been observed in any case.

Bimolecular decarboxylation

This occurs only in case of carboxylic acids that have high electron density on the $\ce{\alpha}$-carbon that might attract the proton from the solution. Examples include anthracene-9-carboxylic acid,2,4,6-trimethylbenzoic acid etc.
[Reference:The mechanism of thermal decarboxylation. B.R. Brown. Quarterly Reviews, Chemical Society. 1951 ; http://pubs.rsc.org/-/content/articlelanding/1951/qr/qr9510500131#!divAbstract]

Under sufficiently alkaline conditions, i.e. when the acid is deprotonated, the carboxylate can undergo an electron transfer reaction with a suitable partner (oxidant). In the course of this process, the carboxylate is oxidized to an acyloxy radical, which subsequently fragments to yield an alkyl (or alkylaryl) radical and carbon dioxide.
$$\ce{R-COO- ->[-e^-] R-COO\cdot -> R\cdot + CO2}$$
A photochemical variant of this reaction was intensively examined in the group of Axel Griesbeck at the University of Cologne, Germany. For further reading, have a look at

Acc. Chem. Res., 2007, 40, 128-140 (DOI)
J. Phys. Chem. A, 2006, 110, 3356-3363 (DOI)
Synlett, 2004, 2347-2350  (DOI)

@ShoubhikRajMaiti Nope. The photochemical step involves the generation of a sufficiently powerful oxidant, such as the excited triplet state of a cyclic imide. The electron transfer from the carboxylate might also occur to a ground state oxidant or an electrode, such as in the Kolbe reaction. The outcome might be different - dimerization of the radicals obtained from the carboxylates rather than addition to the imide radical anion - but the decarboxylation via a radical pathway is the same.


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Chemical bonding
How do we compare the strength of permanent dipole permanent dipole (PDPD) interaction? Is it by bond polarity?
Since H-X (where X is halogen) is polar, and for bond polarity (PDPD bond strength): HCl > HBr > HI, why is it that for boiling point: HCl < HBr < HI which is inversely proportional to bond strength?

You can easily Google the answer.


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Among thioester and ester which has more electrophilic carbonyl carbon?
In few reactions I’ve seen that nucleophiles preferably attack thioester’s carbonyl carbon instead of ester’s carbonyl carbon

According to me, the initiation of reaction is decided by the electrophilicity of the carbonyl compound, in esters O being more electronegative makes c=o more electrophilic, thus attack should have been at ester, but many reactions suggest that attack first occurs at thioester.
What makes thioester’s carbonyl carbon more electrophilic?

Maybe because the negative charge is more stable on $\ce{S}$ compared to $\ce{O}$.

Yes maybe that's the correct reason, but the initiation of reaction is decided by the electrophilicity of the carbonyl compound, in esters O being more electronegative makes c=o more electrophilic, thus attack should have been at ester, but many reactions suggest that attack first occurs at thioester.

Sulfur makes a better leaving group than oxygen.  That makes nucleophilic attack faster at the sulfur-bound site.

@OscarLanzi Yeah this makes sense.

Reactivity of thioesters with respect to nucleophilic attack
Why is the thioester bond weaker than a regular ester bond?
Let's consider a reaction mechanism like this:

A steady-state analysis will show that the rate of formation of product is
$$r = \frac{k_1k_2[\text{ester}][\text{Nu}]}{k_{-1} + k_2}$$

$k_1$ for the thioester is larger, because the sulfur 3p lone pair overlaps poorly with the C=O π* orbital (which is formed from 2p orbitals). Recall that esters are less electrophilic than ketones because there is an ester oxygen which has a lone pair capable of overlapping with the C=O π*. In this case, since this overlap is diminished, it turns out that thioesters are typically roughly as reactive towards nucleophilic attack as ketones.
$k_2$ for the thioester is also larger because thiolates are better leaving groups than alkoxides.

Depending on the relative rates of $k_2$ and $k_{-1}$, though, the second point may or may not be relevant. For example if $k_2 \gg k_{-1}$, then the rate equation simplifies to $r \approx k_1[\text{ester}][\text{Nu}]$, in which case the comparison of $k_2$ is no longer useful.

@ΑντώνιοςΚελεσίδης leaving group ability is quite thoroughly discussed on the internet and you will be able to find more info than I can give you in a comment. Broadly speaking the rule of thumb is the stronger the acid HX the better the leaving group X-.


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Could the inks used in Wonder Bread container have caused lead poisoning?
Decades ago I recall reading that the colorful wrapper of Wonder Bread was misprinted inside out and therefore the printed logo rubbed off on the bread itself, bread was used and kids ended up with lead poisoning. I do remember reading this but can find no further information. Is this story plausible? If so, would not some lead exposure have happened even with correctly printed packaging since it would have been touched by those taking the bread from the packaging.

It has been suggested that this would be appropriate for [Skeptics.SE].  My understanding of their scope is that it has to be a widely-held claim, and the fact that the OP remembers reading this might not be strong enough for their requirements.  I think this is fine here for the time being, but if it doesn't get an answer here after a few days, we can consider migration.

Buy a loaf of bread. Throw the bread away. Turn the bag inside out. Now look at it. Now, don't you think someone would notice if the bag was inside out? However there was a problem with bread bag reuse.

@MaxW: The kids actually liked that the bread was colored this way and no doubt their mom was unconcerned. If bag reuse is an issue, then simply touching the outside of a normal bag and then touching bread was also dangerous.

@jonsca: i would point out that books printed before 1970 or so were pulled from some children's libraries because of the lead content of the printing in them, old-fashioned printing with lead typeface. This is not that outlandish.

@Jeff I have no reason to doubt you.  It's a question of whether it's a widely enough held claim to pass muster at Skeptics.  They have a very fixed scope and are fairly unwavering about it.

@jonsca: What I would really like to find is the original article, from maybe 1970. I have of course googled it but I the search is hard since "lead" is a common verb which has nothing to do with the heavy metal. I am both convinced it really happened and that it is not even surprising given that lead was put into gasoline until the 1990s and beyond in some places and into house paint probably until around that same time. I have read that colored ink in the "funny pages" also used to have high lead content which makes the wonder bread story even more likely.


text:
Which reagents can directly oxidize ammonium ion into gaseous dinitrogen?
I have a simple question. I've looked on the internet and the only thing I found was the Anammox process - Anaerobic Ammonium Oxidation.  Wikipedia lists the reaction as:
$$\ce{NH4+} +\ce{NO2-}\rightarrow\ce{N2}+\ce{2H2O}$$
But only a few bactetria perform this process. I am looking for actual laboratory reagents that can oxidize ammonium ion to dinitrogen gas. My professor said that neither $\ce{KMnO4}$ nor $\ce{K2Cr2O7}$ can do the oxidation, and that one would have to use bleaching powder or $\ce{XeF2}$ instead. However I could NOT confirm his claim on the internet. Thus, I wish to know a list of such reagents which can do what I wish to. Thank you!

https://en.wikipedia.org/wiki/Ammonium_nitrite

@permeakra Thanks for your comment! Though it seems like a special case of the above Anammox process I mentioned above, as it can take place without those nasty bacteria, while the original reactive ions (ammonium and nitrite) remain the same.

Bromine does that. $\ce{2 NH3 + 3 Br2 -> N2 + 6 HBr}$. Because the $\ce{HBr}$ that is generated reacts with ammonia, some of the ammonia is of course not converted to nitrogen, but to ammonium bromide.

@user6376297 Wow, that's kinda surprising because among the halogens, generally fluorine is the best oxidizing agent (as it even forms $\ce{XeF6}$). Do fluorine and chlorine also give the same reaction? And what about their elder brother, iodine?

If by 'surprising' you mean 'hard to believe', I can reassure you: I tried it, and it works. If you mean 'counterintuitive', well, consider that the formation of nitrogen from ammonia probably implies the attack of one nucleophilic nitrogen atom on an electrophilic nitrogen atom, and the detailed mechanism may be very complex. Halogens will usually react with ammonia to give $\ce{X-NH2}$ compounds initially. The properties of X (and sometimes the reaction conditions) will determine what happens next. I suggest you look up the individual reactions in detail if you're interested.

@user6376297 Yes, I meant the counterintuitive part (English is not my native language :P) Thanks for detailing though!

You're welcome. One thing that I should mention too is that we don't necessarily want to use the most powerful reagent all the time. On the contrary, the finest chemistry is often the one that manages to strike that very hard to achieve delicate balance between (atom) efficiency and selectivity. In your specific example, nitrogen's highest oxidation state is not $0$, thus using a stronger oxidizing reagent may even be counterproductive, given that you want to stop at that.

Any reaction of ammonia and halogenes is not advised. Halogennitrides may form, and they are explosive. Halogeneamines are not healthy substances either.

Related: https://chemistry.stackexchange.com/questions/21817/is-there-a-simple-and-inexpensive-way-to-generate-nitrogen-gas

@permeakra : you're right, bromine+ammonia is not ideal, if the point is to produce nitrogen gas in a controlled, clean way from ammonia. I actually used this reaction to quench traces of bromine I had in some waste, using an excess of ammonia, surely NOT to make nitrogen. Sorry if my post was interpreted that way, I had understood the OP's question to be an academic/theoretical one. Have you got better/safer methods to suggest?

@NilayGhosh : indeed ( link ). Then it appears the OP is mistaken in thinking that the reaction between ammonium and nitrite can only happen in bacterial processes. Still not entirely safe; this other Wiki article claims explosions can happen in solutions of ammonium nitrite unless one is carefully controlling the stoichiometry and pH.

BTW, I'm not convinced that permanganate cannot do anything with ammonium. I know that solid ammonium bichromate and ammonium permanganate do decompose producing dinitrogen. Also, permanganate is used for oxidation of organic amines to produce nitrocompounds, so...


text:
Order of the reaction between hemoglobin and oxygen
I have the reaction:
$$\ce{Hb + O2 <=>[$k$][$k'$] HbO2}$$
where $\ce{Hb}$ is hemoglobin, $k$ is the rate constant of the forward reaction, $k'$ is the rate constant of the inverse reaction. On my book there is the rate law:
$$W = kP(\ce{O2})[\ce{Hb}] - k'[\ce{HbO2}]$$
where $W$ is in $\pu{mol m-3 s-1}$, $P(\ce{O2})$ is a partial pressure in Pascal ($\pu{kg m m-2 s-2 = kg m-1 s-2}$).
The inverse reaction is a first order reaction because:
$$\pu{mol m-3 s-1} = [k'] (\pu{mol m-3})$$
thus $[k'] = \pu{s-1}$.
I am in trouble with the forward reaction because I got a strange unit of measurement:
$$\frac{\pu{mol}}{\pu{m3 s}} = [k] \frac{\pu{kg}}{\pu{m s2}} \frac{\pu{mol}}{\pu{m3}}$$
$$[k] = \frac{\pu{m s}}{\pu{kg}}$$
Which is the order of the forward reaction?

This looks like homework, and not exactly good one. Haemoglobin is tetrameric and therefore binding of oxygen is 4 step process with 4 (significantly different) constants.

Hello @Mithoron, my reaction represents the first step.

You seem to be overthinking this a great deal. The order of a reaction is not defined by the units of the rate constant, but rather by the sum of the exponents in the empirical rate equation. From the IUPAC Gold Book:

If the macroscopic (observed, empirical or phenomenological) rate of reaction ($v$) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $k[\ce{A}]^\alpha[\ce{B}]^\beta\cdots$ [...] then the reaction is said to be of [...] (total or overall) order $n = \alpha + \beta + \cdots$

Considering that the partial pressure of $\ce{O2}$ is directly proportional to its concentration, the forward reaction has a rate equation
$$k'[\ce{Hb}][\ce{O2}]$$
where the proportionality factor between $P(\ce{O2})$ and $[\ce{O2}]$ is absorbed into the new rate constant $k'$. The order is simply 2.

Excellent reply @orthocresol! Thank you very very much!


text:
Which coordination complex does not absorb visible light?
A question was asked in one competitive exam (AIPMT 2010). It said:

Which of the following complex ions is not expected to absorb visible light?

$\ce{[Ni(CN)4]^2-}$
$\ce{[Cr(NH3)6]^3+}$
$\ce{[Fe(H2O)6]^2+}$
$\ce{[Ni(H2O)6]^2+}$

The exam is notorious for not releasing solutions of their own question papers, hence, several online sites have come up with their own versions of explanations for why the official answer is option A - $\ce{[Ni(CN)4]^2-}$, most of them copying the solution from one another!
I understand CFT and VBT (not LFT!) in terms of coordination theory. I ask this question on this community to get a definite response, as to what should be the correct answer to this question and why?

You overbroadened this. Related https://chemistry.stackexchange.com/questions/4667/why-do-transition-elements-make-colored-compounds https://chemistry.stackexchange.com/questions/62715/what-exactly-is-the-d-orbital-splitting-and-how-does-this-affect-the-colors-for and lots of others.

Absorption of visual light is associated with an energy difference between two orbitals — one occupied, one unoccupied — and electrons must be able to be excited from one to the other.
In coordination complexes, these excitations typically happen within the metal’s d subshell, so it is usually sufficient to examine that and approximately determine which excitations are possible. The main selection rules are:

the spin rule. The electron must be excitable without a spin flip
the Laporte rule. Basically, d to d transitions are forbidden in octahedral complexes

The spin rule is very strongly observed. The colour of manganese(II) whose transition is spin-forbidden is extremely faint. The Laporte rule only holds true as long as the complex is inversion-symmetric, so any asymmetric vibration is enough to make it void; thus, Laporte-forbidden transitions are typically still visible but somewhat faint.
Let’s examine the complexes:

$\ce{[Ni(CN)4]^2-}$ This is expected to be square planar and $\mathrm{d^8}$. The energy difference between the two highest orbitals — $\mathrm d_{xy}$ and $\mathrm d_{x^2 - y^2}$ — is expected to be high. The former is expected to be fully populated, the latter to be unpopulated.
$\ce{[Cr(NH3)6]^3+}$ This is a $\mathrm{d^3}$ system. It is expected to be octahedral with a standard difference between the lower and higher energy levels.
$\ce{[Fe(H2O)6]^2+}$ This is a $\mathrm d^6$ octahedral system. There is no reason to assume a low spin state. The energy difference is expected to be slightly less than in the previous case.
$\ce{[Ni(H2O)6]^2+}$ this is expected to be a high-spin $\mathrm d^8$ system and octahedral. The same expectation regarding energy levels as on the previous case applies.

We realise that the of our complexes are average high spin octahedral complexes. For these, visible light absorption is always expected. Only one case is different. In that different case, the HOMO-LUMO difference is large. While we can still expect absorption, it seems most reasonable to assign this absorption band an ultraviolet wavelength.
Thus, $\ce{[Ni(CN)4]^2-}$ is the answer.

Hi, thanks for your detailed answer! :D Though one problem: the exam in which this question was asked did not expect its students to know "spin rule" or "laporte rule". In relation to color of complexes, we only studied that the color of complexes is only due to (1) d-d transition: takes place in species $d^1$ to $d^9$ (2) charge transfer theory: for $\ce{MnO4^-}$ $\ce{Mn}^{+2}$ a case of $d^0$ oxoanion. Can the answer be explained in terms of these rules? Thanks!

@GaurangTandon Not really, no. Leaving out the spin rule leaves you with no way to understand why manganese(II) is not coloured. But the Laporte rule is not really important.

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


text:
In a galvanic cell, do the ions from half cell move into the salt bridge?
I know that ions from salt bridge move into the half cell in order to maintain electrical neutrality. But in my book it's also given that ions move from half cell into the salt bridge.
For example, in a galvanic cell with $\ce{ZnSO4}$ and $\ce{CuSO4}$, zinc ions from zinc sulphate solution move into the salt bridge in one half cell and sulphate ions from $\ce{CuSO4}$ move into the salt bridge in the other half cell .

Of course they do. How else would the current go through the bridge?

@IvanNeretin I thought the current flowed because of the movement of electrons through the wire.

True, but you have to complete the circuit. Current can't flow to one side only.

https://chemistry.stackexchange.com/questions/19540/why-the-salts-in-a-salt-bridge


text:
Why is moist hydrogen made to react with chlorine to give hydrogen chloride?
Moist hydrogen reacts with chlorine in diffused sunlight to give out hydrogen chloride. Why is moist hydrogen used? Why not dry hydrogen?

The choice of moist rather than dry hydrogen seems to be driven by some external circumstances of which we have no idea. Dry hydrogen would react just as well.

The reaction is explosive in direct sunlight. Water acts as a catalyst.

Nice video of UV setting off the reacti0n at https://www.youtube.com/watch?v=NN82GoBG98s

H2O has h ions so it adds more + ions so that the cl- gets reacted faster towards the H+ ions so water is a catalysts or booster
I hope u understnd


text:
Is atomic weight universal?
Are we advanced enough to calculate the abundance of an element throughout the universe or is our atomic weight only based on the occurrence of an element in our planet alone?

Our planet and even on our planet isotopic variations are observed.

@MaxW I suppose you meant "even on *other planets", thank you, but in outer space and throughout the other galaxies in our known universe? Do we consider isotope occurrences  throughout there? Thanks for the comment!

Atomic weights of the elements 2013 (IUPAC Technical Report)

There is no "or".

Yes, we are advanced enough to calculate the abundance of elements throughout the universe, albeit somewhat hypothetically. (But then again, the estimates here on Earth are also largely indirect. The radius of Earth is 6400 km, and the deepest borehole pierced only 12 km.)
Yes, if you look at the Periodic table on the wall of your chemistry class, the atomic weights listed there are only based on the isotope abundances in our planet alone.

True, the "earthly" and "Universe-based" atomic weights rarely differ by much. One notable exception is argon (earthly product is mostly $\rm^{40}Ar$, cosmic version is mostly $\rm^{36}Ar$). More subtle discrepancies are numerous. They are well-known, documented, and routinely used to prove that a particular rock is a visitor from space (even if nobody saw it fall).
So it goes.

Thank you for the quick response, the conclusion that I'm arriving is: "Since we classify space rocks based on isotope occurrence. Does that mean for an universal atomic weight, we have to recalculate all the atomic weights separately as Universal Atomic Weight and  Earthly Atomic Weight

Like I said, they rarely differ by much. But if you want high precision, and if you intend to leave Earth and practice chemistry elsewhere in deep space, then yes, you'll need a whole new set of atomic weights.

Awesome exactly what I was so curious about, thanks for clarifying. Is it possible to quantify the amount by which they differ?

I believe my example with Ar is by far the most extreme in terms of weight. Then again, weight is not very telling; isotope abundance is. Say, the earthly  and cosmic fractions of $\rm^3He$ differ by an order of magnitude, but that doesn't mean much for averaged atomic weight, because both fractions are very tiny.

I'd state this a bit differently. We can calculate the mass of each isotope  much much more accurately than we can calculate the average atomic mass. The problem is that isotopic variation introduces significant error into calculating the average atomic mass. Look at iron. Wikipedia lists atomic weight as 55.845(2) but all the weights of the isotopes are known to 7 decimal places.

True, but that's a different aspect to the problem; I didn't even touch on that. See, if the mixing of isotopes within one cosmic body were perfect (which it couldn't be, of course, but let's imagine), then it would be possible to measure averaged atomic weights to 7 digits as well, but my argument would still be valid. A $\times10$ difference in $\rm^3He$ relative abundances would still translate to something like $0.01%$ difference in the atomic weights.

@MaxW @ Ivan We only know the composition of about 4% of the universe so far, so if the James Webb or Fermi Gamma-Ray Telescope manages to figure out that dark matter is some rapidly decaying isotope of extreme proportions similar to the breakdown in black holes. Will chemists decide to draft out a new Universal periodic table based on the discovery with updated atomic weights on it?

Maybe, but that's highly unlikely. For all we know, dark matter is not made of atoms.

One notable example not mentioned above involves carbon.  Carbon-14 on Earth is made continuously by the action of cosmic rays on nitrogen in the atmosphere.  We take it for granted, but the implication is that carbon-14, which per se decays with a half-life of only 5700 years, is unlikely to be seen on a body without an appreciable amount of atmospheric nitrogen (or surface nitrogen, if the cosmic rays can reach the ground).
The effect on atomic weight, however, is very small.  Carbon-14 is only a trace of Earth's naturally occurring carbon, even with all that atmospheric nitrogen floating around.

Indeed, the atomic weight of carbon in usual materials varies between 12.0096 and 12.0116. However, this is mainly caused by the variability of the isotopic abundances of C-12 and C-13.


text:
What's the most appropriate name for this reaction?
I recently wrote a test in which I was asked:

What is this reaction $\ce{CH3CH=CH2 ->[CO + H2O][H+]CH3-CH(COOH)-CH3}$ known as?

Wurtz Reaction
Koch Reaction
Clemmensen's Reduction
Kolbe's Reaction

My thoughts on this:

It's definitely not Wurtz Reaction, because that reaction eliminates terminal halogens and attaches together two alkyl groups. Also, that reaction needs sodium in dry ethereal medium.
It's definitely not Clemmensen's reduction, because that needs zinc amalgam and acid. (Although acid is present, the reduction won't take place without the zinc, right?).
Now, my options are Koch and Kolbe's reactions. As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction. However, as per Wikipedia, this reaction requires carbon dioxide and sodium phenoxide. The carbon dioxide can be taken care of, by the water gas shift reaction that converts $\ce{CO + H2O->CO2 + H2}$. But now, there's no sodium phenoxide. What to do now?

Basically, my doubt is whether there's a mistake in my procedure of eliminating the options, or if there's something I'm missing out somewhere which may lead to make a different choice?

Obviously I agree with Klaus' reply below. I'm adding my own comment because it worries me that you were looking for the phenoxide when trying to 'fit' the reaction name to the scheme you had in your test. You should have recognised that the phenoxide, in the Kolbe reaction, is not a reagent but a reactant.  What would your propene be doing there, if you were just reacting phenoxide and $\ce{CO2}$? You're clearly thorough in your searches, but you sometimes seem to take hasty shortcuts that lead you the wrong way. Which is dangerous, because details are everything in science.

As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction.

You started right, and then took the wrong turn ;-)
Under acidic conditions

the proton adds to the $\ce{C=C}$ double bond and gives a secondary cation
the cation reacts with carbon monoxide and yields an acylium cation
the latter reacts with water to result in the formation of a carboxylic acid

The whole process is known as the Koch reaction (without the Gattermann).


text:
Finding chair and boat transition states during ring-forming reactions
Reactions that form 6 membered rings most commonly go via either boat or chair transition states but the products themselves do not necessarily adopt the same conformation as the TS.
I'm attempting to model a cyclisation in which an amine attacks a pi system to form a new ring, giving rise to two diasteromers depending on which face of the pi system is attacked.
The products have been explored (MM conformational search then QM optimisation of all low energy conformers) and are both chairs (the boat being significantly high in energy as to be essentially unpopulated).
In the reaction of interest there is precedence in the literature from computational studies that the reactions can go via either boats or chairs (depending on the exact structure) and so I would ideally like to find both possible transition states leading to each diastereomer.
My initial plan was to start with the product and work backwards (the transition state is thought to be late, meaning it should resemble the product) , elongating the bond by x angstroms (optimising at each step whilst constraining the distance) and then looking for a 'maximum' from which I would then optimise to a TS however this doesn't provide  a method for me to find the boat TS's.
How does one go about finding all possible transition states? Is it valid for me to find a high energy boat conformation of the product and apply the same bond lengthening exercise?

Agree with the bounty - I think this is a great (and tricky) question, and TS searching is not my area of expertise!

As you have figured out, this is a non-trivial problem.
It's okay to find a boat conformation that might be a bit higher in energy and see if it can map to a TS. Keep in mind that for 6-membered rings, there are many conformations beyond chair and boat. A half-chair, for example, might be a useful test.
In general, for 6-membered rings, using ring-puckering searches are useful. Calculate the plane of best-fit for the 6-membered ring and displace atoms up or down -- this is similar to sampling the out-of-plane vibrational modes of the ring.
So the ideal situation for the TS would be to consider a set of conformations, that might be higher in energy than the reactant and product. After all, it's a transition state. Then use a nudged elastic band or similar method (your bond-lengthening) to look for the real TS.
Good luck!


text:
Does granite become soft as modelling clay at 2000 °C?
As Wikipedia tells us, it consists of $\ce{Al2O3}$ ($14.42\%$) (alumina) whose melting point is $\pu{2072^\circ C}$, and $\ce{SiO2}$ ($72.04\%$) (silica) whose melting point is $\pu{1713^\circ C}$. So does it soften to modelling clay (or at least partially soft, modelable via tools structure) at $\sim\pu{2000^\circ C}$, or does it stay hard and why?

There is no granite as a compound; it's chemical composition deviates quite a lot, and so do the thermal properties. This reminds me of the average hospital temperature of 36.6 °C (averaged from the patients with fever, 40 °C, and morgue, 4 °C)

Granite at this temp. is magma, not exactly modelling stuff...

One of the reasons why granite looks the way it does (some large crystals of feldspars embedded in a matrix of smaller crystals of quartz or mica) is because the components of granite don't crystallise at the same temperatures. Also granite is formed by slow cooling deep in the earth not fast cooling on the surface.

Mixtures of different compounds do not usually work like that.  They generally melt at lower temperatures than the constituent pure compounds.  So I would expect your granite to be not like modeling clay, but molten even before getting to the melting point of silica.
Relevant:  https://en.wikipedia.org/wiki/Eutectic_system.
If you want modeling clay, try 1300 °C where the Earth's lithosphere gives way to the asthenosphere.  See the Characteristics section of https://en.wikipedia.org/wiki/Asthenosphere.

Granite is not composed of the oxides listed, they are just a way of expressing the weight percentage of elements in the sample. Granite (and other rocks) are actually composed of minerals.
All granites contain quartz + alkali feldspar + plagioclase feldspar. Biotite is also pretty common and a variety of other minerals can also be less commonly present.

This is a phase diagram of the quartz - alkali feldspar - plagioclase feldspar system at 5 kbar pressure. The first melt forms in this system at the eutectic point at around 650 °C, but complete melting would not occur until a much higher temperature. I expect the last thing to melt would be left over quartz although it would first transform to tridymite and then cristobalite before melting finally melting.

So... when it gets to be "soft" so that one could model with it like with clay (*having tools that do not melt at that temp)?

@Blender Rocks never get that soft. They melt before they get to that stage.


text:
Refining Tin (Sn)
I am a Copper Cookware Tinner & have a problem in reusing much of the tin that I end up sweeping off the floor etc. Prob about 5kg over the year & it would be great to save this?
Now, my workshop will definitely have copper dust around the place & possibly other metals?
What is happening when I re melt all the sweepings is that I am getting a casting of tin that is not quite all tin I believe? Instead of nice flat crystalline structure, I am getting long needle like crystals. This I am told is due to some copper being percent in the finished metal? I have heard this can be cleaned/removed using sulfur?
As you can probably tell from the terms I am using, I am no scientist, but I would be very grateful in learning the practical aspects of getting this tin back to near 999 grade as I originally receive it

No expert in refining metals, but I can't imagine that the extra equipment, chemicals, and time to refine the tin would be worth it financially. Also you'd need some way in the end to analyze the tin to be sure of its purity.

Thanks for the reply MaxW. Was kind of hoping the sulfur method was simply a case of adding it to the pot and this smoking off the copper or something like that????
The problem is that the tin with probably less than 1% copper in it is not the same crystalline structure as 999 tin & will not be non stick as tin is.
Will keep checking here & possibly its not too hard to accomplish this?
All the best, Steve

Even if you could "smoke off" the copper as you suggest, it seems like a poor idea. Smelting is plain nasty. Putting a metal in the atmosphere should give you pause. You would also get sulfur, sulfur oxides and who knows what else. So you'd really need an exotic setup to do something like that safely.

Cheers Max. Although I am still looking for the practice instruction on how to do this. Its nothing new and has been easily accomplished for thousands of years with only the most basic of equipment.
I figure a Volcano pumps a fair bit of these type of things out every now and then!?


text:
What kind of chemical used in a chromatography run could be denoted with the letters D.H.?
A fellow translator is having trouble translating (into Russian) the mysterious D.H. abbreviation in the following passage:

Slight variations of the ratio of the mobile phase constituents or adjustments of the mobile phase flow-rate can be made occasionally, in order to provide a suitable elution times for Methyl p-hydroxybenzoate and D.H., and to meet the requirements of the system suitability tests.

If by some chance it's a widely known compound, or at least there are several chemicals that could be abbreviated thus, I'd be grateful for answers. A crapshoot, I know, but I became curious.
It could be some rare chemical; in that case it would be futile to try figuring it out.

Do you have a link to the original article? Some context might be helpful!

@NotEvans. - Sadly, no - only a link to the translators' forum thread. The discussion is in Russian, of course.

Дэ.Хэ. and call it a day:D

Recently I had to read up on HPLC-MS technique and I encountered D.H./DH several times as an acronym for dehydrogenase, also, like in the quoted section, in the context of parabens analysis with HPLC.
For example, DH is explicitly defined as an acronym for dehydrogenase in Kastner's Protein liquid chromatography [1, p. 5] or in the paper by Zimmerling et al. [2].
Other examples for the exact dehydrogenase composition include acronyms such as ADH (alcohol dehydrogenase), FDH (formate dehydrogenase), $\small\text{D}$-LDH ($\small\text{D}$-lactate dehydrogenase), 11βDH (11β-hydroxysteroid dehydrogenase) and many more.
References

Protein Liquid Chromatography; Kastner, M., Ed.; Journal of chromatography library; Elsevier: Amsterdam; New York, 2000; Vol. 61. ISBN 978-0-444-50210-0.
Zimmerling, J.; Tischler, D.; Großmann, C.; Schlömann, M.; Oelschlägel, M. Characterization of Aldehyde Dehydrogenases Applying an Enzyme Assay with In Situ Formation of Phenylacetaldehydes. Appl Biochem Biotechnol 2017, 182 (3), 1095–1107. DOI: 10.1007/s12010-016-2384-1.

Since the context is about chromatography and elution times it seems probably that D.H. actually stands for an abbreviation of DHBA. Dihydroxybenzoic acids are family of phenolic acids (as it can be seen here) similar to methyl 4-hydroxybenzoate.
Another possibility is that D.H. stands also for Dorset-Henley liquid which is a synthetic medium for the culture of tubercoline but this seems improbable.


text:
Stability of Halide Salts of Different Metals

Between $\ce{EuCl3}$, $\ce{RuCl2}$, and $\ce{VCl}$, order in
  increasing stability.

I don't know if this is a dumb question and if the answer is blatantly obvious, but I don't understand how to "solve" this question. Is there a rule that determines the stability of chloride salts / any salt in general that can be applied by just referring to a periodic table?
(My rationale for this problem was that Eu and Ru have 3+ and 2+ charges, respectively, and since Eu's 3+ oxidation state would empty both its s-orbital and d-orbitals and Ru's 2+ state would empty its s-orbitals, the complex would be able to form (Just to show that I've done work) but that honestly seems obviously wrong as well.)
This is not an HW question.

Vanadium doesn't seem to exist in the +1 oxidation state.

Unless it's in some sort of organometallic complex.  Definitely not a simple chloride salt.

"Stability" needs to be defined, it seems it's one of more abused words.

I guess it would be like "be able to exist at room temperature without rapidly decomposing" or something like that, I don't know. The answer is that EuCl3 and RuCl2 are both "stable" but VCl isn't. The question is why is this so and how are we able to tell so quickly.


text:
Would the concentration of oxygen around fire increase as we get close to the fire (from the outside)
Since fire require oxygen, wouldn't a fire, like those in the American forest along west coast, constantly draws in oxygen from around? And if that is so, would the concentration of oxygen around the fire higher than farther away?

A fire has no way to directly "draw in" oxygen for it to burn. It consumes what oxygen is in the immediate vicinity of the flame, depleting the air in oxygen compared to the concentration of oxygen in air further away. This sets up a concentration gradient such that oxygen will diffuse into the depleted region near the fire.
An even more important process than diffusion of oxygen into depleted air is convection driven by the heat of the fire. Oxygen-poor air that has been in the fire gets hot and rises, drawing fresh air into the base of the fire. Note that this still isn't oxygen being drawn to the fire, but a mass of fresh air flowing into the lower pressure area at the fire resulting from convection.
The bottom line is that a fire consumes oxygen, so the region around the fire will be slightly depleted in oxygen, not enriched as you suggest.

Due to convection it draws in air, which happens to contain oxygen.  It doesn't draw in oxygen specifically.

Since fire require oxygen, wouldn't a fire, like those in the American forest along west coast, constantly draws in oxygen from around?

That doesn't make sense. You are reversing cause and effect; you are saying that since the presence of oxygen (along with other factors) causes fires, that this somehow mean that fires will cause the presence of oxygen. That would be like saying that putting solar panels on your roof causes your roof to get more sunlight.

No, it'd be more like the water level around a drain.
Sure the presence of a drain causes the water to flow in that direction, but the water flows in that direction specifically because that region's still water-deficient despite water flowing to it:
 -source
Same deal with the fire and oxygen.  Oxygen's net flow rate will tend to be toward the fire, but it'll be toward the fire because that area's oxygen-deficient, i.e. lower oxygen levels.  If the fire for any reason stopped having lower oxygen levels, oxygen'd stop having a net positive flow toward it.
Phenomena like this are often modeled with Fick's laws of diffusion.  Typically, the result looks like the concentration profile of the diffusing species (in this case, oxygen) decreasing from some bulk concentration (in this case, atmospheric concentration) to the concentration at the sink (in this case, the fire's vicinity).
  -source
In the above, if the oxygen concentration around the fire $C_1$ were to become higher than the atmospheric oxygen concentration $C_0$, then oxygen would start flowing away from the fire.

Kinda funny to think about a candle's flickering flame due to the dynamic conditions at the flame, kinda like how the whirlpool around the drain also flickers.


text:
Preventing oxidation of copper(I)
By aerial oxidation $\ce{Cu+}$ can easily oxidise to $\ce{Cu^2+}$. Which of the following should be added to $\ce{Cu+}$ to prevent it from oxidation: $\ce{F-}$, $\ce{Mg^2+}$, $\ce{Mg}$ or $\ce{F2}$?
$$
\begin{align}
E(\ce{Cu^2+/Cu+}) &= \pu{0.15 V}\\
E(\ce{F2/F-}) &= \pu{2.87 V}\\
E(\ce{Mg^2+/Mg}) &= \pu{-2.37 V}
\end{align}
$$
Why it isn't $\ce{Mg^2+}$?

What would Mg2+ do to prevent oxidation of Cu+?

Teacher asking this question should get fired. One of the given would indeed prevent Cu+ from getting oxidised, but that's not the only thing it would do.

Ditch all the choices and go with thiourea. It works beautifully, try adding it to a copper(II) sulfate solution (color disappears) or better yet acopper(II) chloride solution (color disappears and a hite precipitate of copper(I) chloride appears which rediscover with more thiourea).

We need a better reducing agent (that itself oxidises) than $\ce {Cu^+}$.
So choose the ones which have greater reduction potential than $\ce {Cu^+}$.
It isn't $\ce {Mg^2+}$ as it cannot oxidise any further hence it has a very low(negative) reduction potential, so it cannot stop $\ce {Cu^+}$ from oxidising.

In one sense none of the options really works.  Metallic magnesium certainly prevents the oxidation of $\ce{Cu^+}$ to $\ce{Cu^{2+}}$, but also reduces $\ce{Cu^+}$ to elemental copper.  To stabilize $\ce{Cu^+}$ for real use a soft-base complexing agent like thiourea.


text:
How to make sure molded silicone rubber adheres to the ABS mold?
For one of my experiments I need to coat a 3D printed ABS part with a thin layer of silicon rubber:

Basically I want to make sure the molded silicon will adhere to one mold (the negative side in the picture above) but not the other. I'm using V-SURE SILICONE NL Standaard from VOSSCHEMIE BENELUX :

To ensure the positive side will be realised after rubberisation I use the realise agent V-SURE TRENNSPRAY 400 GR from the same company:

I was wondering if there is anything similar for adhesion. Something like a spray glue or something I can smear with a brush. Thanks for your help in advance.
P.S.

I have absolutely no idea what are the products I'm using. I just bought them in a local shop for sculpting. I tried translating the website or googling the names with no good results. It would be also a great help if you could help me know what is the official/scientific names of the mentioned products.
I'm not able to find the right tags for this post. I would appreciate if the moderators could guide me with that.
I have also posted this question on Reddit (molding, modelmakers and ResinCasting)
Maybe there are other moldable ruberry materials which have a better adherence to ABS? Any other type of flexible, wear resistance and strong rubber will do. For example I have seen lots of tinkers all over the internet using Caulk mixed with some moist absorbent powder (e.g. cornstarch) and oil, or soaked in water+soap.
Surprisingly the silicon sticks firmly to the glass beaker. I can't even remove it with a cutter!

No such thing, at least not one easily accessible to a customer. Just clean the part properly, it'll stick imo. But a form with parallel walls is not going to work. You'll never get anything out of it.

Thanks for the comment. No such thing, at least not one easily accessible to a customer If there is anything I might be able to order it. If you give me any names?  Just clean the part properly, it'll stick imo. Well, I did my best to be sure the surface is clean. but it doesn't. I'm thinking maybe I should change the morphology so there will be more roughness. a form with parallel walls is not going to work. You'll never get anything out of it. I do not understand this. would you please elaborate?

What you have is prepolymerised PDMS with acetyl endgroups. Those hydrolyse with water from the air to form acetic acid and the remaining reactice chain ends connect and crosslink. Unless VOSSCHEMIE has developed sth. revolutionarily new. ;-)

@Karl So in my publication should I name the material as PDMS or silicon?

I don't know exactly what could prepare the surface so it sticks, but it'll involve actual chemistry, and you probably shouldn't mess with it. ;-) ABS is perhaps not ideal, try PLA?

I have no choice for PLA over ABS. We have this one 3D printer and it works with ABS :(


text:
Does the diiodide(1−) anion exist?
With respect to chemical bonding does the $\ce{I2-}$ ion exist?

Probably not, is assume that the equilibrium is entirely on the right side: $\ce{2I2- <=> I- + I3-}$

It seems like it does (+refs 1-7 therein).

I found a paper$\ce{^{[1]}}$ regarding the formation of diiodide anion species($\ce{I2^-}$). It is assumed to be an unstable species which forms during the formation of triiodide anion (excess of iodide in solution which makes it brown). This is a part of the paper which describes the formation of the species. For more information, read the full paper. Also, you can check the abstract which @andselisk linked.

The $\ce{I2^-}$ species is formed by the 248-nm laser photolysis of
iodide through the following reactions:
$$\ce{I- + hν -> I + e_s-}$$
$$\ce{I + I- ->[k2]I2^-}$$
$$\ce{I2- + I2- ->[k3]I3- + I-}$$
As discussed later, we found that photodetachment of electrons from
$\ce{I-}$ ions produces solvated electrons ($\ce{e_s-}$) in ionic
liquids as well as in molecular solvents. After photodetachment,
iodine atoms react with $\ce{I-}$ to form diiodide anion radicals. In
aqueous solution, the transient absorption maxima of $\ce{I2-}$ are
located around 400 and 720 nm. The extinction coefficients of
$\ce{I2-}$ in aqueous solution at 385 and 725 nm are 10000 and 2560
M-1cm-1, respectively.[...]

$\ce{^{[1]}}$: J. Phys. Chem. B, 2007, 111 (18), pp 4807–4811


text:
Equivalent weight of the anhydrous acid

The cupric salt of a monobasic acid contains $3$ molecules of water of hydration per atom of $\ce{Cu}$. $\pu{1 g}$ of hydrated salt yielding on strong heating $\pu{0.3306 g}$ of $\ce{CuO}$. What is the equivalent weight of anhydrous acid?

My approach:
Let the formula of the acid be $\ce{HE}$ (where $\ce{E}$ is some element). So, the formula of the salt is $\ce{CuE2*3H2O}$
Don't know how to proceed from here. Any help would be appreciated.

I replaced C with E in your notations. C is "reserved" carbon and using this letter for any element would cause a lot of confusion.

Let the anhydrous acid be $\ce{HA}$ and assume its Molar Mass to be $X$. There is no a priori reason to assume $\ce{A}$ is a mere element.
The copper salt in question is $\ce{CuA2.3H2O}$.
Its Molar Mass is:
$$63.55+2\times(X-1)+3\times 18=115.55+2X$$
Assume reasonably and without losses, that:
$$\ce{CuA2.3H2O(s) -> CuO(s)}$$
So that $1\ \mathrm{mol}$ of the hydrated salt decomposes to $1\ \mathrm{mol}$ of cupric oxide.
The Molar Mass of $\ce{CuO}$ is $79.545$, so $0.3306\ \mathrm{g}$ of it is:
$$\frac{0.3306}{79.545}=4.156\times 10^{-3}\ \mathrm {mol}$$
So now we know that $1\ \mathrm{g}$ of the hydrated salt is $4.156\times 10^{-3}\ \mathrm {mol}$, so its Molar Mass is:
$$\frac{1}{4.156\times 10^{-3}}=240.6\ \mathrm{g/mol}$$
And with the above:
$$115.55+2X=240.6$$
Solve for $X$.
As the acid is monobasic, its Molar Mass is also its Equivalent Mass.


text:
Why doesn't cottage cheese melt?
Most kinds of cheese melt but cottage cheese does not at usual cooking temperatures.
During the preparation of various kinds of cheese, I have heard of a coagulant named ‘rennett’ being added along with the acidifying agent. For cottage cheese it is just lemon juice mostly (acidifying agent). Does this have anything to do with the difference in properties?

Well, it's valid question, but you should elaborate a bit.

You essentially make cottage cheese in the process of making cheddar cheese. But cottage cheese is about 80% water. The "cheese" solids are soluble in oil, but not in water. So cheddar cheese will "melt" but cottage cheese won't.

I do not get how solubility and melting are connected..

I don't think the cheese solids "melt" as much as they dissolve in the oil present in the cheese. So if you "melt" the cheese once and separate the oil, then you wouldn't be able to get the "cheese" which is left to melt again.

It melts but only if it is in a proper medium. I discovered that shredded cottage cheese melts in my pasta sauce.
I do not know if it is the acidity or the presence of extra ingredients that does this but any macro-physical trace of it is gone, just like mozzarella.

“Macro physical” . The usage of the term itself conveys the ambiguity. Melting is definitely not defined as the transit of the solid state of a material is from macro to something less.. please rethink and modify your answer..


text:
Is blending 15% methanol with petrol a good idea?
I just came to know that the Government of India is going to implement a scheme under which 15 percent methanol will be blended with petrol and which will reduce the cost of the fuel by 10 percent.
I want to know that how would this practically affect all the machines where petrol is being used as fuel? Will those machines have to be reengineered to work with the new blended petrol or would those machines work the same way they worked with normal petrol? How would their efficiency be affected?

I think one reason methanol hasn't been used as an oxygenated fuel additive in other countries is that it can attack engine components like gaskets. It will also raise the vapor pressure of the fuel and could result in higher VOC emissions at the pump and venting from the tank. It could help reduce some unwanted emissions, but I don't know why it would reduce the cost of the fuel by so much.

https://en.wikipedia.org/wiki/Methanol_fuel

There are two reasons why methanol may require caution: it is toxic to people (so you need to take care not to spill it) and it is much more corrosive to the chemicals used in gaskets (so you may have to redesign many engine parts). combustion-wise it isn't a problem, but many countries have avoided it because of those other problems and have chosen ethanol instead. But this often fails to be "greener" as the production of ethanol isn't usually green and diverts source material from food production.

I think it's important to first understand why methanol is added to petrol to begin with, outside of what I will discuss later on as some of the benefits to atmospheric emissions.
For internal combustion engines, petrol should be of a high "octane rating". This is a measure of the compression ratio of the fuel, and it needs to be high in order for the engine to work efficiently. If the petrol has too low of an octane rating, then something called knocking can occur, which over time can cause very bad engine damage.
In order to improve the octane rating of petrol then, a variety of compounds can be added that boost the overall octane rating (appropriately named octane boosters). In the past this has included tetraethyl lead, which when burned during combustion lead to the accumulation of lead in the atmosphere, which was enormously damaging to the environment and to human health. Methanol, and other alcohols like ethanol, are one such example of an octane booster.
Before I discuss the effects on engine performance, I will say that the use of using methanol blended fuel does have positive environmental impacts over the use of diesel or just straight petrol. Combustion efficiency is improved, thus leading to lower emissions of $\ce{NO}$ and $\ce{NO2}$ (collectively called $\ce{NOx}$ gases), which has positive implications for the atmosphere overall (lower $\ce{NOx}$ generally leads to lower ground level ozone, but this is dependent on VOC emissions of course!), as well as to lower particulate emissions. They also do not contain sulfur, and thus blended mixes have a lower sulfur load - that means less sulfur dioxide emissions.
In terms of redesigning engines on the basic, fundamental level - no change is required. Any internal combustion engine can use blended fuels, it just has implications of the lifetime of the engine. However, methanol can act as a corrosive on certain components of the engine. As blended fuels are increasingly seen as a good way for governments to reduce their $\ce{NOx}$ emissions - particularly after the problem with diesel emissions over the last decade - it seems likely that manufacturers will respond to this by redesigning engines to resist the corrosive effect of methanol within the blend. For now, though, having a small ratio of methanol to petrol - and it seems in this case, India has designed for 15% methanol blend, which should have little impact on engine damage through corrosion - is sufficient.
One last thing to note, however, that may not have come up. Some people might wonder why we don't just a 100% methanol fuel once we design engines that don't corrode due to it. The simple answer is - it's not as fuel efficient as petrol!


text:
Does increasing the entropy of a system cause it to absorb heat?
This is my first time posting here, so bear with me. It was stated (in this video: https://youtu.be/7jT5rbE69ho?t=325) (Watch about the next fifteen seconds, until 5:40) that an increase in entropy leads to a decrease in heat (or temperature or absorbs energy - unsure of the correct terminology). (If I'm understanding correctly.) Now, I've never taken a thermodynamics class or anything past high school chemistry, but this is a little confusing to me and I'm hoping to get an explanation. What follows below is my best attempt at understanding why. It is most likely mis-guided, so feel free to correct at will.

I understand that entropy can be partially understood as disorder. When the Helium-3 enters the Helium-4, that constitutes an increase in entropy.
The change in entropy is defined as Heat/Temperature (see below)

Rudolf Clausius invented the idea of entropy in such a way that the change in entropy is the ratio of the heat exchanged in any process and the absolute temperature at which that heat is exchanged. That is, he defined the change in entropy DS of an object which either absorbs or gives off heat Q at some temperature T as simply the ratio Q/T.

I then figured that maybe the Q-T ratio (entropy) increased by the total energy (of the helium system) increasing, requiring it to pull heat from somewhere

Miscellaneous thought: My above reasoning confused me because I thought it required energy to decrease entropy locally (add order to a system), so it seems strange that a system increasing in entropy
(like my helium example) would absorb energy.

I consider watching any video a waste of time, so I'll be judging from your words alone. (Anyway, your question is essentially self-consistent, which is good.)

Yes, entropy is a measure of disorder (sort of).
No, $dS={\delta Q\over T}$ only for reversible processes.
And no, entropy is not the Q-T ratio for two reasons. First, because the above formula deals with infinitesimal increments and not with finite quantities. Second, because like I just said, that formula is not universally applicable.

All in all, the increase of entropy alone is not necessarily accompanied by the heat flow to the system. It may be so, but it may be otherwise. You may imagine pretty much any kind of possible outcome:

You dissolve some $\ce{NH4NO3}$ in water (which surely increases entropy), and the beaker gets palpably cold. The system starts sucking heat from the surroundings, just like you anticipated.
You mix your $\rm^3He$ with $\rm^4He$, and... nothing happens (except the increase in entropy, of course, but that's a thing you can't see or touch). There are no thermal effects to speak about, and hence no heat flow.
You burn a candle in a closed vessel (supposedly, containing enough air), and guess what? Entropy increased, but heat is flowing outside.

As for your last remark: true, it takes some energy to decrease entropy locally (say, in your fridge). But in doing so, we inevitably increase entropy elsewhere, so the overall balance is positive.
So it goes.

Good answer. Just to avoid confusion, the second point in the latter list is approximately true.  A small heat of dilutation (endothermic) exists, and has been utilised since the 1960s. doi: 10.1007/978-1-4899-6443-4_133

Very interesting. I appreciate the quick response. I am giving the video the benefit of the doubt, just because I'm certainly no expert on it. So, why is it that the increase in entropy causes a drop in temperature? Again, thanks - very helpful

Increase in entropy does not necessarily cause a drop in temperature (see my examples 2, 3). In the cases when it seemingly does (like example 1), this is due to some endothermic reaction, which is made spontaneous by the positive entropy gain ($\Delta H>0$, but $\Delta G=\Delta H-T\Delta S<0$).


text:
Is there a substance that will eat through and completely dissolve carbon but not damage engine cylinder, piston nor rings?
I'm afraid that when I was instructed to be specific in the title of my post, I kind of said it all... Here is the issue:
I have a 2001 Isuzu Rodeo. It has been going through excessive amounts of oil. No blue smoke, no drips whatsoever.
Through an Isuzu forum I find this is a common problem for model years up to and including mine as oil drain holes for the cylinders that were made in the pistons clog with carbon.
In that the vehicle is a 2001, I can well suspect those little holes are well caked and packed.
Though some folks state that they remove the spark plugs, pour a commercial internal auto engine cleaner called B12 into the cylinders and allow an over-nite soak, other auto-techs state that the only way to clear the drain holes is to tear apart the engine and drill the holes in the pistons out or replace them altogether.
Is there a chemical that will eat away and dissolve the carbon whereas after allowing whatever sufficient soak time is suggested I could then change the oil and have a hope for cleared out ports without damage to the metal engine parts?
Thank-you for your thoughts!

I don't think that's possible. Carbon is pretty durable. In fact, it is much easier to dissolve the whole engine while keeping the carbon.

If it's pure carbon, then the only way to remove it I can think of is by rigorous oxidation (e.g. by burning in oxygen flow or by reaction with strong oxidants (e.g. conc. $\ce{H2SO4}$, $\ce{HNO3}$)), but this is going to damage the metal parts either, as Ivan said.
However, if the clog is caused by soot and resinous residue from the incompletely burnt fuel, then most likely hot toluene or xylene can dissolve the residue and liberate the carbon particles. Both are widely available solvents which won't damage the metal parts (but avoid contact with rubber tubing) and are also used to remove clogs caused by paraffin wax.
P. S. Both are highly flammable liquids, so do not warm them up using an open fire (e.g. use an improvised water bath)!

To the OP: Definitely mind any safety warnings and/or get a hold of an MSDS when using either organic solvent.

@brucemc777 No prob:)

Thank-you, very much! I have to believe that the temperature of the engine at the time would quickly influence the temperature of whichever solvent I was working with. Therefor, instead of pre-heating the solvent, I would think it best to simply get the engine to an acceptable temperature. The problem is, just what now defines "acceptable temperature"... In that I would kind of like to avoid pouring the solvent into a cylinder and having my arm blown off, would either of the afore cited fluids be stable across all normal engine operating ranges?

@brucemc777 Please don't even think of using these solvents with the working engine, or the one that is still hot, this may cause severe injuries. Take off the parts that need cleaning, and apply warm solvent to them (or just soak if you have enough solvent).

You might of just saved my life :)


text:
Are the breaking of bonds simultaneously endothermic and exothermic?
I understand that energy is NEEDED to BREAK bonds. But don't chemical bonds already have some form of potential energy in them already? So wouldn't it make sense that even though energy is required to break them, their potential energy is also RELEASED once they are broken?
In other words, shouldn't the bond breaking process be considered simultaneously endothermic AND exothermic?
Side note: I'm studying grade 12 level chemistry right now, so I don't have too much knowledge on the nitty gritty details of this concept. That being said, It would be greatly appreciated if anyone could explain this in very simple terms.

A reaction overall cannot be simultaneously endo- and exothermic, the values would cancel out such that the reaction would be overall endothermic or overall exothermic, taking into account all of the bonds being made/broken.

Think of "exothermic" and "endothermic" as representing the difference in energy from start to finish. Yes, energy may need to be added to trigger a reaction (e.g. lighting thermite, which takes a lot), but if, at the end, more energy was produced than added, it is exothermic.

Bond breaking in itself does not release energy!!

Bond breaking does NOT release energy!!
In the course of a chemical reaction, you typically break some bonds and form some new bonds. The bond breaking is always endothermic. The formation of new bonds is exothermic, so depending on whether the old bonds or the new bonds were stronger, the reaction overall can be either endothermic or exothermic.
When somebody says chemical bonds “contain potential energy” it is a HIGHLY misleading statement. You justifiably interpreted this as: when you break bonds, they release the potential energy that was stored within. This is, unfortunately, wrong. That phrase should honestly be banned, but somehow it has always persisted.
What that phrase really means is more nuanced. It means that some chemical bonds can be broken in a reaction which subsequently produces more energy via bond formation.
Bond breaking is always endothermic!
For further reading see: Is Bond Formation "Strictly" Exothermic?

https://en.wikipedia.org/wiki/Helium_dimer#Molecular_ions ?

Impressive counterexample. “Always” should therefore be replaced with “almost always”, then.

Typical case when the original question was not well put, here had this misunderstood/sloppy "potential energy" thing in it. Great how this makes you think about long learned stuff again. :-)

A bond is always some local energy minimum. You definitely need to add some energy to get out of there, but when you separate the remaining parts to infinity, you might get more energy back than you put into it. Certainly not if you split water or sodium chloride.
So you're principally correct, but the two (or more, breaking a bond will usually induce further rearrangement in the parts) processes cannot be separated. Can't break a bond but not remove the parts from each other.

While I understand what you mean, I think it doesn’t hit the nail on the head. If you decompose an azide the bond breaking is still endothermic. Subsequent bond formation may well drive the reaction (in this case the formation of the N-N triple bond) but OP is only asking about bond breaking, not a reaction as a whole. The problem with OP is that they have been taught something about how chemical bonds “contain potential energy”, which is obviously not true because of your first sentence.

@orthocresol The azide example might be wrong, and definitely too complex, right. My thinking was this: Two particles at infite distance are at energy level zero. You bring them closer, and they will first repel each other (unless it's an anion and cation), so E > 0. When (if) a bond forms, the energy level goes down again, but not necessarily below zero.

(I notice I ignored London etc. interaction. Hm.)


text:
USNCO Question - Determining solute using spectrophotometer
Below is a question from the 2017 USNCO:
The concentration of which approximately 0.01M solution could be most accurately determined by a visible spectrophotometer (or colorimeter)?
(A) $\ce{Mn(NO3)2}$
(B) $\ce{Co(NO3)2}$
(C) $\ce{Zn(NO3)2}$
(D) $\ce{Pb(NO3)2}$
The answer sheet states B is the correct answer.
I researched the colours of the ions in solution, and found:

Mn is pink
Co is pink
Zn is clear
Pb is clear

I was unsure how to progress from there.
How should I approach this question in an exam if there was no reference book?

Can you find any information about the faintness/brightness of the color of given ions? I don't know if there is a standard scale for this.

This really strikes me as a question where you must just know the answer rather than being able to rationalize it.

This is a poor question, and can only be answered properly if only one species is coloured in the visible part of the spectrum (also not defined in the question). How can one determine which can be 'most accurately' determined without knowing something about the instrument to be used. Just because one species may absorb more strongly than another does not mean that it can be measured more accurately: solutions can absorb too much as well as too little.

An aqueous solution of $\ce{Mn^2+}$ is only faintly pink, whereas that of $\ce{Co^2+}$ is much more colored:

Thank you. Is there any way of determining this in an exam situation?

It's probably a good idea to: 1. Cite the source of the image; 2. Explain the intensity of the color based on electron transfer of the complexes.

@GeorgeTian, you might be able to say that $\ce{Zn^2+}$ should be colorless because it has filled orbitals, but I'm not aware of a simple way of differentiating between the colors of the other options.

@andselisk could you please elaborate on the electron transfer?

@GeorgeTian Various electronic transitions require unequal energy and can therefore be observed at different wavelengths and, what's more important here, have different values of the molar absorption coefficient.

@andselisk, right now I have the image source in what I think is a markdown field for the image description (you'll see it if you try to edit my post), but I'm not sure if it is accessible directly. Do you suggest putting the source explicitly underneath the image? Regarding your other point, I'm not qualified to provide such an explanation, unfortunately. At the USNCO level this is simply a matter of memorization.

@a-cyclohexane-molecule Oh, my bad, I didn't know that, disregard my comment:)

Unfortunately some sort of unjustified rationalization  is often used to explain facts.


text:
What are the following equations called?
Our teacher was writing out the following ways of expressing redox and she used a particular term for what these equations are called. When we write out the following what is this called? Its a way to show what atoms lose and gain electrons, but I forgot what these are called:
$$
\begin{align}
\ce{Li &-> Li+ + e-}\\
\ce{F + e- &-> F-}
\end{align}
$$

Those are called half-reactions because they show how electrons associate or dissociate, but don’t show the full reaction. I believe that’s the term you’re looking for!

@user54747 If you are happy with the answer, don't forget to click "Accept" (looks like "v" on the left) button – this is your way of expressing gratitude on SX sites:)

@andselisk Dont see it.

@user54747 Then see the tour, please.


text:
Why can't we rotate a carbon while checking for chirality?
If I were to rotate the front carbon clockwise by $60^\circ$, it would become achiral. But my text book says that it's chiral. Is it that rotating an individual atom about a sigma bond changes its conformation is that why we can't rotate it?

Related: Is (2R,3S)-butane-2,3-diol chiral?

We can rotate whatever rotates freely (or nearly so) in the wild. Whether or not it changes some conformations is irrelevant. Also, this particular molecule would never become achiral, rotate it all you want.

I meant about the sigma bond

Yes, I got that. I meant the same.

Wait, this is (2R,3S) thing, and hence a meso form. I take the last part back. You were right, it is nominally chiral and will become achiral if rotated 60°. Then again, actual molecules rotate around the sigma bond all the time, and you can't separate the "0° isomer" from the "60° isomer".

So chirality is for a specific conformation and not a molecule as a whole?

Chirality is there when you can separate and isolate the enantiomeric forms. In this case you can't.

Wait. Wait. Wait. We need to be much more clear about what we're talking about. The front carbon is a chiral center regardless of rotation. In speaking of the chirality of a molecule, we generally do not limit ourselves to any one conformation, but rather all accessible ones. Most of the comments and answers assume that you're speaking about the molecule, but it's not 100% clear to me that that's what the book is also referring to.

@Zhe it was that question (I edited the question)

Your first structure (labeled II) is equal to structure I in the problem set. Is this an error or a result of copying the structure from somewhere else?

@Terje my bad that's a copying error

@Zhe No, let’s really be clear. There is not really such a thing as a chiral carbon (even though the term keeps popping up in chemical literature). Molecules and centres can be chiral, only molecules can be achiral and carbon atoms (or other atoms) are typically asymmetric. So both carbons are asymmetric but the molecule is also achiral.

If I vote to close this as a duplicate, it will become unclear. I’m not going to vtc in this case.

It simply depends on what is the subject. The book asks for the displayed conformation, which we agree is chiral.
The question would have been more tricky to answer if it were about optical activity of a sample.
In this case we know the conformer can not be isolated at room conditions but this is not a general rule.
And one can always cool down, at least in principle.


text:
Potential Energy Diagrams and the Effect of Catalysts
Why does a potential energy diagram showing the effect of a catalyst on activation energy not move left on the reaction pathway scale (compared to uncatalysed reaction) if a catalyst speeds up reactions?

You might be imagining that we all have the same textbook as you, opened on the same page, and are looking at the same picture as you. I hate to break it to you, but this is not so.

You should provide the diagram. To do this, take a clear photo of the picture, and then upload it by using the image upload option that appears on the top bar when editing or posting the question.

Hi, sorry, this is the first time I've used this sight. I am not a chemistry expert by any means; I'm just sitting higher. I have got the answer now though.

Because the reaction coordinate does not indicate speed.
It represents a certain state of the reaction (e.g. 50 % completed) but it does not have any information about time in it.
Therefore the higher reaction speed is not represented in those diagrams.


text:
Does calcium hydroxide react with citric acid in orange juice?
I noticed that calcium-fortified orange juice on the shelf at my local grocery store contains calcium in the form of calcium hydroxide.
And I also noticed a couple other ingredients that ordinary orange juice doesn't list explicitly:  citric acid and malic acid.  Are these meant to neutralize the calcium hydroxide to form calcium citrate (commonly listed on calcium supplement pill bottles)?
I'm just curious. I want to know what final chemical(s) will end up in my stomach if I swallow this concoction.

Most probably, yes.

calcium citrate

Don't know if this is the brand but here is such a label

Therefore calcium hydroxide is an actual additive. In this case advertised along with vitamin D (calcium fortification).
As for what gets to your stomach, there are reasons why ingredients, not compounds produced via cooking are listed. Cooking unlike chemical synthesis isn't supposed to create specific compounds, but food in which all compounds are mixed and every possible equilibrium happens at once.
In the juice there's just more calcium cations and pH may be slightly affected, if it was significantly changed you'd find it by your taste.

Looked today and the store label does read like this even though the formatting has changed a bit.

What is this supposed to be? Doesn't look like answer, I'm afraid.

@Mithoron -  paracetamol had given an answer saying that he didn't think that any food label would list calcium hydroxide. I was just trying to show that the OP wasn't nuts. Such labeling does exist.

Well, then do sth so it shouldn't be comment-answer, it was flagged as VLQ.

I changed it to community wiki. Edit it however you want...

I'm not a food chemist, nor am I associated with any regulatory bodies for food/beverages, but here's my take on this

I notice the calcium-fortified orange juice on the shelf at my local grocery store contains calcium in the form of calcium hydroxide.

I haven't seen the mineral content of a beverage expressed that way, i.e- you don't see labels talking about the added compound, rather, they'll just mention the mineral cation and the concentration in mg per 100 or 1000ml (or in some percentage, it's the manufacturer's call).

Food regulatory rules are pretty the only reason for the uniformity you see in labels from licensed manufacturers.
One possibility is that the juice you have at hand was produced locally by a small-scale manufacturer (and meant to be distributed locally as well); that would easily explain the unorthodox labeling. (edit: MaxW's answer would indicate that such a label can also be found in online documentations for some well known products as well; I've never seen it in print myself)
I'm not 100% sure why you have calcium hydroxide in orange juice, but from a cursory spot of Googling, it would appear as though calcium fortification is pretty much the only reason it's there (you even mention it's calcium fortified).
Apparently, the usual way of fortifying foodstuff with calcium is by addition of calcium citrate malate. However, the whole process appears to bump up manufacturing costs.

For example, calcium chloride can give a salty aftertaste; calcium hydroxide and calcium oxide give a 'flat' taste; while calcium lactate can leave a bitter aftertaste and tricalcium phosphate a metallic aftertaste.
 Source: an article on a Tropicana patent from a couple of years ago

Going by the description in that excerpt, for a small-scale manufacturer, calcium hydroxide (which is easily available in 'food-grade') would be the cheapest option (that doesn't impact the taste as negatively as other compounds).

And I also noticed a couple other ingredients that ordinary orange juice doesn't list explicitly: citric acid and malic acid. Are these meant to neutralize the calcium hydroxide to form calcium citrate?

Well, calcium hydroxide has poor solubility in water (1.73 g/L according to the Wikipedia page), so the amount of calcium citrate, calcium malate and calcium citrate malate formed is tiny. In all likelihood, your juice will still be acidic.

I think the OP was correct - regardless of the solubility of $\ce{Ca(OH)2}$, that the calcium hydroxide, citric acid, and malic acid are added to generate the citrate malate through a neutralization reaction - with water as the side product.

@Geoff Ah, now that you bring it up, I suppose that might be the case (I still hope to find an authoritative source that validates it). At any rate, I'll taking out that line on neutralization, thanks for the heads up! :-)


text:
If you introduce a 3rd group into O- chlorotoluene, which position will the 3rd group prefer?
I read that Cl is ring deactivating group even if it is O,P- directing. Which group among the methyl and chloro controls the orientation of the 3rd substituent. What are the products formed?

It depends on many things.

The nitration of o-chlorotoluene gives all four possible isomers according to Wibaut [1].
Reference

Wibaut, J. P. Recherches Quantitatives Sur La Nitration Des Chlorotoluènes. Recueil des Travaux Chimiques des Pays-Bas et de la Belgique 1913, 32 (10), 244–320. https://doi.org/10.1002/recl.19130321002.


text:
What is the reaction mechanism for the polymerisation of formaldehyde?
Clayden et al. (2012) mention that formaldehyde is rather prone to polymerisation. In an aqueous solution containing formaldehyde, it is known that the polymeric "paraformaldehyde" is formed. However, the book does not did not display any information regarding the mechanism of polymerisation. Upon searching the Internet, I was only able to find this image which shows a plausible mechanism for polymerisation:

However, this mechanism seems to be for polymerisation under acidic conditions. I was wondering if there is a general scheme for the polymerisation of formaldehyde.
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Related: https://chemistry.stackexchange.com/questions/83084/can-chloride-attack-the-carbonyl-bond-in-formaldehyde-in-the-reaction-with-hydro

And: https://chemistry.stackexchange.com/questions/53226/how-is-naoh-able-to-clear-a-solution-of-paraformaldehyde


text:
Mechanism of Biosynthesis of Terpenes
There is a specific part of the biosynthesis of terpenes that makes me confused:

According to this mechanism, the double bond of the isopentenyl pyrophosphate attacks the primary carbocation, when it would make more sense to attack the tertiary carbocation that could be formed by resonance. Am I missing something here? Clearly, this is not a concerted process if the mechanism is displayed in this way, and an intramolecular resonance conversion would be faster than a nucleophilic attack.

There is only one carbocation! There is no "resonance conversion" between two distinct carbocations, one primary and one tertiary. Resonance forms do not interconvert! Anyway, the regioselectivity is probably enzyme-controlled.

Yeah, sorry about that. I was trying to make a point though.

Enzyme-controlled and helped by sterics (it’s easier to attack the primary position than the tertiary one).