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Dimethylethylamine or Ethyldimethylamine?
I thought that Organic Naming was based on alphabetical for alkyl branches, so I thought it was ethyldimethylamine, however google results show me Dimethylethylamine. So which one is right? And is this an IUPAC rule?

I have reopened this. The rules that govern amines are not the same as the rules that govern substitution of alkanes. Hence not a duplicate.

There's a technicality in that the 2013 IUPAC recommendations (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book)) have slightly awkward preferred IUPAC names for simple amines like these. The preferred IUPAC name for your molecule is "N,N-dimethylethanamine" (as described by P-62.2.2.1 on p 671, but note that there is an erratum for the triethylamine example shown there).
Putting that aside, though, what you're conceptually missing is that the methyls and the ethyls in your proposed name are (implicitly) not on equal footing. The methyls are names of substituents, but ethyl is not a substituent - it is the parent chain (since it is longer than methyl). So, $\ce{CH3CH2NH2}$ is named "ethylamine", and $\ce{CH3CH2N(CH3)2}$ is named as a N,N-disubstituted form of "ethylamine", i.e. "N,N-dimethylethylamine". In common usage the "N,N" is often dropped, leaving "dimethylethylamine". I should emphasise that this is not in line with the IUPAC recommendations, but people can and will use it, and this is why you will see it on the Internet. Returning to the preferred IUPAC name in the previous paragraph, we can see that the use of "ethanamine" rather than "ethylamine" makes the difference between methyl and ethyl a lot clearer.
In fact, according to the recommendations, it is also permissible to name tertiary amines in the manner (alkyl1)(alkyl2)(alkyl3)amine (although it does not lead to the preferred name). If we were to adopt this method, then the name would be ethyl(dimethyl)amine, because now there is no notion of a parent chain and all three substituents are merely substituents - which should be listed in alphabetical order ignoring prefixes like "di". The parentheses around "dimethyl" are required (but not around "ethyl") because of a separate rule (P-62.2.2.2, p 672 - see also the discussion in the comments).
To sum up:

N,N-dimethylethanamine: Preferred IUPAC name
ethyl(dimethyl)amine: "Alternative" IUPAC name; permissible but not preferred
dimethylethylamine: Doesn't comply with 2013 IUPAC recommendations, but the logic behind the name is similar to that used in generating the preferred IUPAC name.

I believe, using the same list of errata, that the alternative name would be ethyldi(methyl)amine; see, e.g., "ethyldi(methyl)phosphane" for p. 116, P-16.5.1.3, example 4.

@LegionMammal978 I don't think we can really draw a parallel to the phosphane example. Amines have their own set of rules (P-62), and in particular P-62.2.2.2 states "the prefix(es) immediately preceding the term 'amine' is (are) enclosed in parentheses". It's not immediately clear from the text alone whether the multiplicative prefixes "di"/"tri" are supposed to be included in the parentheses, but two things suggest that they are: (1) the examples of "(triethyl)amine" and "(4-methylpent-4-en-2-yn-1-yl)(dipropyl)amine" given on p 671 and p 672 respectively;

[...] (2) the fact that if we don't include the multiplicative prefix, there can only be one prefix that immediately precedes the ending 'amine'; so the careful phrasing of P-62.2.2.2 to include plural prefixes suggests that the multiplicative prefix should be included. That said, you are right to point out that P-16.5 is not the appropriate rule to cite, and that in other cases (e.g. phosphanes) parentheses should only surround the substituent itself and not the multiplicative prefix. I have edited that - thank you.


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What is C2H33O28? (Empirical Formula to Molecular Formula)
I'm doing the final question on my homework, and got an issue with this question:

"A 76.8g sample of a liquid used as a coolant in automobiles is found to contain 3.7 g of carbon, 5.1 g of hydrogen and the rest is oxygen.  The formula weight of the compound is 60.0 g/mol.  What is its molecular formula?"

After scouring the web, youtube, and my textbook, I can't find any reason my answer doesn't make sense. My math is as follows:

Convert each of the given masses to moles, resulting in 0.30805mol of Carbon, 5.1mol of Hydrogen, & 4.25mol of Oxygen. Dividing by the small to convert to a ratio, the result is 1mol of Carbon, 16.5mol Hydrogen, and 14mol Oxygen. Finally, to get a whole number for the ratio of moles, multiply by 2, resulting in C2H33O28.
The issue with this, is that both the mass is way too large for what we are learning, and when converting this formula to a molecular formula by dividing the given mass of the molecular formula by the mass of the empirical formula results in a 0.12 multiplier. Multiplying each of the subscripts by 0.12 gives an odd result of C0.237H4O3.

I have no idea what I did wrong here besides maybe rounding, but after trying this four separate times with different rounding. If anyone can provide any insight whatsoever, I would be ecstatic, thank you so much!

Are you sure there isn't a typo somewhere in the question? Your approach seems reasonable, but the resulting formula is not a sensible molecule.

@Rocky The data is wrong. It seems the sample mass is off. Don't bother to waste your time and ask your teacher about the numbers.

Alright, I'll ask her tomorrow, thank you for the help!

Try "30.7 g of carbon".

Notice there is said liquid, not compound. In the car cooling context, it looks like water solution of ethylenglykol. From element ratio should be possible to calculate concentration. Unfortunately, M=62 g/mol.

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How many mg caffeine is 1mM caffeine?
I am finding some articles about caffeine consumption confusing, because they test in units of mM instead of mg.
https://www.ncbi.nlm.nih.gov/m/pubmed/25342885/
Wikipedia says that the molar mass of caffeine is 194 g/mol. Does that mean that 1 mol caffeine equals 194 g of caffeine? And that 1mM caffeine, in turn, equals 194 mg caffeine?

It looks to me that they are working in concentrations, not total dosage. Given that they work in cell cultures (cells in water medium), this seems sensible. Concentrations also make sense because M is a short-hand for mol/L (molarity), here apparently prefixed by m for milli.

Besides the comment above yes you are right.

1 mM stands for one millimolar: 1 mM caffeine solution has 194 milligrams of caffeine dissolved in one litre of water, or there are 194 µg of caffeine per millilitre of water.

@Aniruddha Deb. Thx for your clarification. So a 250 ml cup of coffee, containing 1 mM of caffeine solution, would then contain 194 mg divided by 4 = 48.5 mg caffeine?

@Constantthin yes

@Jan. Is 1mMol the same as 1mM?

@Jan. So you are saying that Aniruddah Deb is wrong?

This was automatically flagged for the number of comments it received; please try to not double post comments to avoid this. Comments can be edited for five minutes, please use this time to review them.

@Jan, I am confused. I would consider mM an abbreviation for millimolar, i.e mmol/L. I do not recall any abbreviation using mMol. But you write that 1 mMol refers to the amount independent of the volume. Later you agree that 1 mM = 1mMol. Your reasoning is contradictory, therefore I would say that you are wrong.

@Martin-マーチン and Constantthin , there was the confusion with abbreviations. If 1 milimol (mmol) of caffeine is 194 mg and if you put this amount in 1 liter of water, you get  a solution that is 1 milimolar (1 mM). If you take just one cup (250 mL) of caffeine solution from that 1 liter, you get 194/4 = 48.5 mg caffeine, which was already calculated correctly above.

To my understanding, M. Farooq correctly address the OP's problem. As noted in his answer:

$\pu{mMol}$ is not an acceptable notation, it better be $\pu{mmol}$ to be read as millimole. This is not the same as $\pu{mM}$, which is read as millimolar. The former is an amount unit, the latter is a concentration unit.

That's it. OP needs to understand that units $\pu{mmol}$ and $\pu{mg}$  are amount units, while $\pu{mM}$ and $\pu{M}$  are concentration units (by convention, $\pu{M}$ is equal to $\pu{mol\:L-1}$).
Let's look at OP;s problem now:

Wikipedia says that the molar mass of caffeine is $\pu{194 g/mol}$. (1) Does that mean that $\pu{1 mol}$ caffeine equals $\pu{194 g}$ of caffeine? (2) And that $\pu{1mM}$ caffeine, in turn, equals $\pu{194 mg}$ caffeine?

Answer to (1) is yes: OP must read and understand two already given answers elsewhere.
Answer to (2) is no: As M. Farooq correctly pointed out (vide supra), $\pu{mM}$ is a concentration unit (of a solution) and $\pu{mg}$ is an amount unit of a substance (e.g., it could be about substance in the solution such as caffeine in given coffee solution). They do not equal to each other ever without knowing another dimension called volume. Only way to make relationship between these two unit ($\pu{mg}$ and $\pu{mM}$, etc.) is to know how much volume we are considering.
For example, OP can ask the question correctly as: How many $\pu{mg}$ of caffeine in $\pu{1 L}$ of coffee solution, if it is $\pu{1mM}$ in caffeine?
Thus, the amount of caffeine $= \pu{1 L} \times \frac{\pu{1 mmol}}{\pu{1 L}} \times \frac{\pu{198 mg}}{\pu{1 mmol}} = \pu{198 mg}$
If the question is: How many $\pu{mg}$ of caffeine in $\pu{250 mL}$ of coffee solution, if it is $\pu{0.75 mM}$ in caffeine?
Then, the amount of caffeine $= \pu{250 mL} \times \frac{\pu{1 L}}{\pu{1000 mL}}  \times \frac{\pu{0.75 mmol}}{\pu{1 L}} \times \frac{\pu{198 mg}}{\pu{1 mmol}} = \pu{37.1 mg}$
Thus, bottom line is if you know the concentration of caffeine in any drink, you must know how much you are consuming (by volume) in order to calculate your caffeine intake (in $\pu{mg}$, etc.).
T'd like to finish this answer with some interesting data. According to Wikipedia:

Depending on the type of coffee and method of preparation, the caffeine content of a single serving can vary greatly. The caffeine content of a cup of coffee varies depending mainly on the brewing method, and also on the coffee variety. According to the USDA National Nutrient Database, an 8-ounce ($\pu{237 mL}$) cup of "coffee brewed from grounds" contains $\pu{95 mg}$ caffeine, whereas an espresso ($\pu{25 mL}$) contains $\pu{53 mg}$.

Thus, the concentration of caffeine in brewed coffee $= \pu{95 mg}  \times \frac{\pu{1 mmol}}{\pu{198 mg}} \times \frac{1}{\pu{237 mL}}\times \frac{\pu{1000 mL}}{\pu{1 L}} = \pu{2.02 mmol\:L-1}$
On the other hand, the concentration of caffeine in espresso $= \pu{53 mg}  \times \frac{\pu{1 mmol}}{\pu{198 mg}} \times \frac{1}{\pu{25 mL}}\times \frac{\pu{1000 mL}}{\pu{1 L}} = \pu{10.71 mmol\:L-1}$
That means espresso is almost 5 times stronger than brewed coffee (by volume)!

1 mole caffeine ≙ 194 g (PubChem).
1 millimole (mmol) caffeine ≙ 194 mg.
The solution of 194 mg caffeine in 1 liter of water = 1 millimolar (1 mM). (The abbreviation for molarity is M – see the list of common abbreviations in chemistry - JBC).
1 mmol caffeine (194 mg) dissolved in 1 liter of water gives 1 milimolar (1 mM) caffeine solution. 1 cup (250 mL) of such solution would contain 48.5 mg caffeine.

I am not sure what is confusing the original poster. The comments were clear and the answer by Jan is clear. Let's try to walk you through the problem again.
The question asks about how many milligrams are there in $\pu{1 mM}$ caffeine solution. Did you attempt a dimensional analysis?
For mass to mole conversions (or the other way you round) you need the molecular weight. The molecular weight has units of $\pu{g/mol}$. Caffeine's molecular weight is $\pu{194.19 g/mol}$.
$\pu{1 mM}$ (read as one millimolar) of any substance $X$ means that $\pu{1E-3 mol}$ of $X$ is present in $\pu{1 L}$ of solution.
Amount of substance (unit: $\pu{mol}$) to mass (unit: $\pu{kg}$) conversion requires the knowledge of the molecular weight.
$$\pu{1E-3 mol} \times \pu{194.19 g/mol} = \pu{0.19419 g}$$
(Note that amount of substance units cancel.)
Convert grams to milligrams by recalling the definition that $\pu{1 mg} = \pu{0.001 g}$. This means that $\pu{194.19 mg}$ of caffeine are present in $\pu{1 L}$ solution. Round off to desired significant figures. Note never round off during calculations. (Hint: $\pu{1 mM}$ has only one significant figure).
Also note: mMol is not an acceptable notation, it better be $\pu{mmol}$ to be read as millimole. This is not the same as $\pu{mM}$ which is read as millimolar. The former is an amount unit, the latter is a concentration unit.

But what I don’t understand is how they determined that the molecular weight of caffeine is 194?

@Constantthin, this is a completeley different story. There are many methods to determine the molecular weight. There is an instrument that determines the molecular weight. It is called a mass spectrometer.

Caffeine is a substance whose formula is C8H10N4O2. The weight of all these atoms is the sum of their weights. As 1 atom C weighs 12 u, 8 atoms C weigh 8·12 = 96 u. As 1 atom H weighs 1, 10 atoms H weigh 10 u. As 1 atom N weighs 14 u, 4 atoms N weigh 4·14 = 56 u. As 1 atom O weigh 16 u, 2 atoms O weigh 32 u. The total is : 96 + 10 + 56 + 32 = 194 u

@Constantthin, Please start a new question. Whatever you are asking is completely different story on the determination of molecular weights and u. Readers will be lost in the comments. Also mention your grade of study.

@Constantthin, Yes this is the same dimensional analysis.

I have asked to know which grade are you in? It will help to understand your background of questions better.

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Thermal Decomposition of Sulfur Hexafluoride
What is the thermal decomposition temperature for $\ce{SF6}?$ I've heard it decomposes into $\ce{F2}$ and $\ce{SF4}$ only to decompose further into $\ce{S2}$ and $\ce{F2},$ but I haven't found much on the topic.
I became curious since I was considering using it to flush out oxygen for some reactions I was doing, due to its density and stability (plus we have a bunch lying around), but was concerned about it decomposing into more toxic compounds if the reaction or apparatus were to get too hot or if a spark were to occur.

I do not plan to actually do this. It was a silly idea I had, and I realized soon after posting this that $\ce{SF6}$ is a bad idea for these purposes since it is a greenhouse gas.

Note that SF6 is an extremely good greenhouse gas. Releasing it may be controlled or not allowed where you work.

Please do not use SF6 as an inert blanketing gas for chemistry, it is a gas which poses a great danger to the environment (Global warming). Also it is quite reactive to some things. For example it is very able to scavenge solvated electrons. It is something used in radiation chemistry to remove these free radicals. I would suggest argon (it is cheaper) or if you want a very cheap gas use nitrogen.

We have a system set up in our fume hood that recollects the gases. (I realized this was a dumb idea soon after I oosted this, ironically.) However, I'd still like to know the decomposition temperature, as this idea had sparked my curiosity.

From an IEEE paper1:

$\ce{SF6}$ begins to decompose at $\pu{300 °C}$ and the main decomposition components contain $\ce{CO2 , SO2F2 , SOF2 , SOF4 , SO2}$ and $\ce{H2S}$; [...] $\ce{H2S}$ is the special component only appears when thermal fault proceeds to some degree (above $\pu{360°C}$).

For more information about the decomposition mechanism and product yiels, see here.
References

F. Zeng et al., "Decomposition characteristics of SF6 under thermal fault for temperatures below 400°C," in IEEE Transactions on Dielectrics and Electrical Insulation, vol. 21, no. 3, pp. 995-1004, June 2014, doi: 10.1109/TDEI.2014.6832242.

As already noted the thermal decomposition s commence at $\pu{300 °C}$ and includes $\ce{CO2 , SO2F2 , SOF2 , SOF4 , SO2}$ and possibly $\ce{H2S}$.
Not mentioned relating to its stability per an Atomistry reference is:

Mixed with hydrogen it withstands a high temperature, but under the influence of powerful electric sparks formation of hydrogen sulphide and hydrogen fluoride can be effected. At a red heat copper and silver are without action on the gas, although magnesium and sodium effect its decomposition; at lower temperatures, however, it resists even these two metals and sodium can be melted unchanged in an atmosphere of the gas, as also can the hydroxides of the alkali metals. Hydrogen chloride and ammonia are unaffected by sulphur hexafluoride, but hydrogen sulphide reacts rapidly according to the equation:

$\ce{SF6 + 3 H2S = 6 HF + 4 S }$

So, with heating and also with hydrogen in the presence of electric sparks (atomic hydrogen), some hydrogen sulfide formation also. Note, electric arc generated atomic hydrogen is more powerful than the hydrogen radical ($\ce{.H}$), albeit the latter is more ubiquitously created, and I would avoid scenarios involving its potential introduction in the possible event, it also attacks SF6, speculatively (see relatedly, Eq (5) here) as follows (which could involve the extremely problematic formation of a deadly chemical weapon, disulfur decafluoride):
$\ce{SF6 + .H ->? .SF5 + HF}$
$\ce{.SF5 + .SF5 -> S2F10}$ (Source)
More definitely per above, however, H2S subsequently can further react with SF6 rapidly liberating problematic HF.
So, on the question, "concerned about it decomposing into more toxic compounds if the reaction or apparatus were to get too hot or if a spark were to occur", a definite yes from me.
Also, as a side note, SF6 reacts vigorously with Lithium metal (used to power the United States Navy's Mark 50 torpedo closed Rankine-cycle propulsion system per this source).


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How to find the lifetime of sodium atoms in the 3p using information on the spectral line width?
I am trying to find the lifetime of sodium atoms at the 3p state using the relationship between the spectral linewidth and the uncertainty principle:
$$ \Delta E \times \Delta t = h / 4\pi $$
The information I have is the full-width-half-maximum (FWHM) linewidth of the sodium atomic absorption spectrum $\lambda = \pu{9.2E-14 m}$. How would I use this equation to find the lifetime?
The atomic transition from ground 3s to excited 3p states has a wavelength $\lambda = \pu{589.0 nm}$.
I used the relationship
$$ \Delta E = \frac{h c \Delta \lambda}{\lambda^{2}} $$
Plugging in $\Delta \lambda = \pu{9.200E-14 m}$, then
$$ \Delta E = \frac{h \cdot \pu{3E8  m} \cdot \pu{9.200E-14 m}} {(\pu{589.0E-9 m})^{2}} $$
So using the uncertainty relationship, we have that
$$ \Delta t = \frac{h}{ 4 \pi \Delta E} = \frac{(\pu{589.0E-9 m})^{2}}{4 \pi (\pu{3E8 m s^-1} \cdot \pu{9.200E-14 m})} = \pu{1.000E9 s} $$
My attempt to convert the linewidth from unit of $\pu{nm}$ to $\pu{eV}$:
$$ E = \frac{hc}{\lambda} = \frac{\pu{6.626E-34 J s} \cdot \pu{3E8 ms^-1}}{\pu{9.200E-14 m}} = \pu{2.161E-12 J} = \pu{1.348E7 eV} $$

I find something in 10^-47 s. Am I wrong ? Does it make sense ?

@Maurice It is to be expected that the number will be much smaller than the literature value (17 ns), because the linewidth is smaller in the problem. However, I calculted $1 \times 10^{-9}$ s. Would you mind sharing how you calculated the time? Thank you.

The answer suggested in the comment as 10^-47 s is completely wrong. Your answer is in the right ball-park of ns. Make sure you don't round off anywhere until the end.  http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/parlif.html

@M.Farooq How do you convert the linewidth from nm to eV? I tried $E = hc/ \lambda$ and changed the unit from J to eV, but the result is orders of magnitude away from the number on that website.

Your units must be inconsistent. If you are using $\delta$E in eV your Planck's constant must be in eV as well. Please show your calculations and steps in the original post and we can start from there.

@M.Farooq I updated my calculations for the linewidth and also how I got $1 ns$ as the lifetime. Would you mind checking both? Thank you.

@ferrin, Your calculations and approach are correct. Maybe you are not getting the exact result because the 589 nm is actually a doublet consisting of 588.9950 and 589.5924 nm. Your half-width information in the textbook might be different. So the half-width of each line is actually smaller because there is a separation between the two lines. Smaller half-width in terms of energy means longer lifetime. So if you happen to find true halfwidths you should get close to literature. What is the answer in the textbook? The ballpark of nanoseconds seems to be correct.

@M.Farooq Thanks for checking! That doublet argument could be the reason. I calculated the half-width myself from the previous question that asks to plot the absorption spectrum in Lorentzian distribution. And I set the center $\lambda = 589.5 \textrm{ nm}$. I also found this literature https://steck.us/alkalidata/sodiumnumbers.pdf where they reported a lifetime of $17 \textrm{ ns}$. Maybe I didn't get the correct half-width. The problem set will be returned next week probably, so unforatunely I don't have the answer yet. Thank you for your help though! It means a lot :).

@ferrin, you half widths must be slightly off because that is the only variable you have.

Half the fwhm has to be added/subtracted  to the central wavelength and then this change in wavelength converted to an energy. Thus the limits on the wavelength are $\lambda \pm \Delta \lambda /2) $ and the energy change $\Delta E=hc/(\lambda - \Delta \lambda /2)-hc/(\lambda + \Delta \lambda /2)$ Joules with $c$ in m/s and $h$ in J s. Plugging in the numbers gives a lifetime of $\approx 10^{-9}$, i.e. $1$ ns.
As a rule of thumb, lifetimes of excited electronic states are rarely less than $10^{-12} $s and if the transition is allowed, as in this case often of the order of nanoseconds to microseconds. If transitions are forbidden then the lifetimes can extend from microseconds to seconds.

Interesting that I missed $\Delta \lambda /2$. However, I don't recall seeing this as such. If $\Delta$E is uncertainity, what is the rigorous statistical basis of dividing FWHM by 2? It is not equal to the standard deviation for a Gaussian, Voigt or Lorentz profile.

FWHM is full width at half maximum so divide by 2 and then work out energy spread. The important thing is that $\Delta \lambda$ does not directly give the change in energy. FWHM is not 1/e of width so this will not give absolutely correct answer, however using width to get lifetimes is only really true if nothing interacts with system so it is more hypothetical than real. Anyway the OP  should only use the data in the question.

It seems that OP has full raw spectrum, he could have plotted the intensities vs. energy, and get the energy width and proceed as suggested in the answer. The key point was that FWHM/2 is not the true uncertainty in the statistical sense and division by two is an approximation. As you said it is more of a hypothetical scenario.


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Why is only monochromatic light used in polarimeter?
While using the polarimeter to determine optical rotation, why can't we use polychromatic light? What would change?

Rotation could be a function of wavelength/frequency, especially as visible wavelengths vary by 2 to 1.  Therefore, polychromatic light may give blurry results.

The reason is partly historical and partly scientific. Most of the polarimetric data exists with the sodium yellow emission at 589 nm because sodium lamps were among the most convenient ones and remember electricity is rather new. Early light sources were either flames or sunlight. It was easy to generate sodium emission by introducing salt in the flame. If you are familiar with Raman spectroscopy, it was discovered using sunlight (heliostats) not lasers in the 1920s-30s. Today nobody can think of using sunlight for spectroscopy.
At a more fundamental level, optical rotation is a function of refractive index and refractive index is a strongly dependent on wavelength. You have to see the optics of a polarimeter, the question of using polychromatic light does not arise. You will see a number of prisms are used. Impossible to use polychromatic light.


text:
Does glycerol give positive silver mirror test with Tollens’ reagent?
From my school lab manual for identification of organic compounds:

Take 2 ml of Tollens’ reagent and add a few drops of organic sample. Heat the contents on boiling water bath for 5 minutes. Silver mirror is formed. The given organic compound is glycerol.

As far as I know, the compounds with aldehyde groups give positive silver mirror test. Also, those which contain α,β-hydroxy ketone are known to react with Tollens’ reagent.
I have never heard of glycerol reacting with Tollens’ reagent. I found that on heating glycerol forms acrolein which contains aldehyde group:
$$\ce{(CH2OH)2CHOH → CH2=CHCHO + 2 H2O},$$
which I think is the reason, but I am not sure.

It is unlikely under Tollen's conditions that glycerol is converted into acrolein since both compounds are at the same oxidation level. Base is not going to effect the transformation. What ever aldehydic products are formed, they are at a higher oxidation level than acrolein. See: https://www.intechopen.com/chapters/66623 @Aniruddha Deb

Could it happen due to tautomerism? Infact, there is a dynamic equilibrium between the enol (glicerol) and keton form (see: https://tgc.ac.in/pdf/study-material/chemistry/Sem_II_Chemistry_Hons.__Tautomerism.pdf)

Your guess of formation of acrolein is a pretty good one, as indicated in this answer, but a boiling water bath will probably not give you the 160 °C temperature you are looking for to convert glycerol to acrolein.
This question is most similar to yours and has a pretty good answer by bon. The answer says:

However, from personal experience, gentle heating of a primary alcohol with Tollens’ will cause a small amount of oxidation and result in a fine black precipitate.

It is most likely that glycerol is undergoing a small amount of oxidation to any of it’s oxidation products (most probably glyceraldehyde, since that gives Tollens’ test) and this oxidized compound is giving a positive Tollens’ test.

Bon says that he is not sure, it's only his personal experience. Also I do wonder why wouldn't secondary and tertiary alcohols undergo oxidation when heated with ammoniacal silver nitrate.


text:
How do the three Miller indices (hkl) denote planes orthogonal to the reciprocal lattice vector?
The Wikipedia page for Miller indices defines Miller indices as follows:

There are two equivalent ways to define the meaning of the Miller indices: via a point in the reciprocal lattice, or as the inverse intercepts along the lattice vectors. Both definitions are given below. In either case, one needs to choose the three lattice vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$ that define the unit cell (note that the conventional unit cell may be larger than the primitive cell of the Bravais lattice, as the examples below illustrate). Given these, the three primitive reciprocal lattice vectors are also determined (denoted $\mathbf{b}_1$, $\mathbf{b}_2$, and $\mathbf{b}_3$).
Then, given the three Miller indices $h, k, \ell, (hk\ell)$ denotes planes orthogonal to the reciprocal lattice vector:
$$\mathbf{g}_{hk\ell} = h \mathbf{b}_1 + k \mathbf{b}_2 + \ell \mathbf{b}_3.$$
That is, ($hk\ell$) simply indicates a normal to the planes in the basis of the primitive reciprocal lattice vectors. Because the coordinates are integers, this normal is itself always a reciprocal lattice vector. The requirement of lowest terms means that it is the shortest reciprocal lattice vector in the given direction.

This idea of the three Miller indices ($hk\ell$) denoting planes orthogonal to the reciprocal lattice vector is something that I'm not confident that I'm understanding. My understanding is that the real (real space) lattice vector is described by the Miller indices as $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$, where the $\mathbf{a}_i$ are linearly independent basis vectors. However, if I am understanding this correctly, the Miller indices are actually not (directly) defined in terms of the real lattice vector, but are actually defined directly in terms of the reciprocal lattice vector $h{\mathbf  {b_{1}}}+k{\mathbf  {b_{2}}}+\ell {\mathbf  {b_{3}}}$, where ${\mathbf  {b_{i}}}$ are the basis of the reciprocal lattice vectors; specifically, the Miller indices are defined to denote the family of planes orthogonal to $h{\mathbf  {b_{1}}}+k{\mathbf  {b_{2}}}+\ell {\mathbf  {b_{3}}}$. This is evidenced by the fact that said orthogonal plane is not always orthogonal to the linear combination of real lattice vectors $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$ (see image below), since the reciprocal lattice vectors need not be mutually orthogonal (that is, although the $\mathbf{b}_i$ are basis vectors, they are not necessarily linearly independent, which I think is mathematically correct (?)).

(By Felix King, and posted to https://en.wikipedia.org/wiki/Miller_index)
I would appreciate it if people would please take the time to review everything that I've written so far. Is my understanding correct? Am I misunderstanding anything?
The last point that I'm having a bit of trouble with is how a plane can be defined in such a way that it is orthogonal to three basis vectors? I suspect that this is possible if and only if the basis vectors are not all linearly independent, as I alluded to in my explanation above, but I'm not sure. I think this is more of a mathematical question, so I will instead post it to math.stackexchange for a more mathematics-focused answer, but I would also appreciate it if people could please review my understanding of this and provide clarification.

The basis vectors should all be orthogonal, or you need to include additional information (such as the fourth index for hcp structures).

@JonCuster But the second paragraph of the Wikipedia article explicitly says that “the reciprocal lattice vectors need not be mutually orthogonal”?

No need for the basis vectors in the Bravais Lattice to be mutually orthogonal at all - generally they will not be.

There are two equivalent ways to define the meaning of the Miller indices: via a point in the reciprocal lattice, or as the inverse intercepts along the lattice vectors.

The reflecting plane are parallel to a plane that includes the following three points:
$$\frac{\mathbf{a}_1}{h}; \frac{\mathbf{a}_2}{k}; \frac{\mathbf{a}_3}{l}$$
To show that the plane defined by these three points is perpendicular to $h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3$, we can calculate the dot product between vectors in the plane and the supposedly perpendicular reciprocal lattice vector. To get vectors in the plane, we simply subtract pairs of points in the plane.
$$\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_2}{k}$$
$$\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_3}{l}$$
Now we multiply it with the reciprocal lattice vector:
$$(\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_2}{k}) (h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3) = \mathbf{a}_1 \mathbf{b}_1 - \mathbf{a}_2 \mathbf{b}_2 = 1 - 1 = 0$$
$$(\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_3}{l}) (h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3) = \mathbf{a}_1 \mathbf{b}_1 - \mathbf{a}_3 \mathbf{b}_3 = 1 - 1 = 0$$
The mixed terms vanish because they are orthogonal. Because the dot products are equal to zero, we have shown that the vector is perpendicular to the reflecting plane.

[OP] My understanding is that the real (real space) lattice vector is described by the Miller indices as $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$, where the $\mathbf{a}_i$ are linearly independent basis vectors.

No, the significance of $h, k, l$ in defining reflecting planes in terms of the unit cell vectors is as described in the textbook (inverse intercepts along the lattice vectors).

[OP] The last point that I'm having a bit of trouble with is how a plane can be defined in such a way that it is orthogonal to three basis vectors

That is not how it is defined. It is defined by the intersections of a plane with the three axes, i.e. through three points that are in the plane.


text:
Risks of using organic mercury compounds in analytical chemistry
I'm interested in speciation analysis. Hearing of fatal accidents with organic mercury compounds made me wonder about the risks for the analytical chemist trying to determine Hg species in foodstuff. Probably the most prominent accident is the death of Karen Wetterhahn who used dimethylmercury for NMR experiments. A few drops proved fatal. I'm assuming it was pure dimethylmercury she was using? Or is it usually diluted?
The concentration range in LC-ICP-MS analysis for quantificaton of MeHg in foodstuff ranges up to about 10 ppb from what I have seen so far. I'd expect stock solutions for preparing the calibration wouldn't exceed 1 g/L. So the consequences of spilling some drops of a stock solution are likely less severe in this case. Especially taking into account that today people are aware of how fast dimethylmercury diffuses through the usual gloves. Which is something I believe Wetterhahn wasn't aware of. Diluted or not, the handling still requires laminated gloves (silver shield), right?
I don't intend to work with MeHg, but I know someone who might in the context of analyzing foodstuff. It would be a comfort to know that the phrase "one drop can kill" applies to the handling of the pure chemical I guess.

I´m sure your friend is not the first who wants to quantify that stuff. How about some literature research? The fatal accident of KW has been extensively studied and the results published. And if there should be a need to calibrate with a sample of the actual substance, I´m sure you can buy a standard containing just a few ppb´s.

It's not like I don't do this already but I've yet to stumble upon a paper that describes what kind of safety measures were applied or what risks are involved with the chemicals used. Actually thinking about it some kind of ISO standard would possibly more helpful in this regard. Will have to look if there is any for Hg speciation. Or maybe a standard operating procedure.

https://www.osha.gov/dts/hib/hib_data/hib19980309.html and links provided therein.


text:
Simple QM9 properties for prediction
I was looking into an article (doi: 10.1063/1.5058063) which presented a histogram of radius of gyration for samples of different configurations of alanine dipeptide. This property is fairly simple to calculate using
$$
a_{\mathrm{Rg}}(x)=\sqrt{\frac{\sum_{p} m_{p}|| x_{p}-x_{\mathrm{COM}}||^{2}}{\sum_{p} m_{p}}}
$$
The sum considers all atoms $p = 1, \dots, P$ of the peptide, where $m_{p}$ and $x_{p}$ denote the mass and the coordinates of each atom, respectively. $x_{\mathrm{COM}}$ denotes the center of mass of the peptide.
In a different problem, I have a number of molecular graphs generated by a network trained using QM9 dataset and am looking for properties that are readily calculable for the newly generated molecular graphs using such packages as RDKit and OpenBable.
The properties in the QM9 dataset itself, like LUMO energy and energy of atomization, need DFT calculation, which is not something that I look forward to do. An example of a property of interest is QED, which can be easily calculated using RDKit package. But QED is more of interest in ZINC dataset.
I'd appreciate comments on any easy to compute property of interest to the community for QM9 dataset.

Suggestion: Add a bit about your infrastructure, what program do you typically use, do you work on your own or on a larger computer cluster?  Equally as a suggestion: please add a link to the publication you refer to, e.g. the doi, because  there are already some posts here including histograms about properties in data sets.

Thanks for the suggestion @Buttonwood. I added doi for the paper, but I'm afraid that the program that I use and my infrastructure would be irrelevant to the question. Not that it is a well kept secret, but it won't help with the problem.

I found 3 properties:

The molecular weight (MolWt)
Wildman-Crippen partition coefficient (LogP)
Topological Polar Surface Area (TPSA)


text:
Maximum Value of Monolayer Capacity
What is the possible maximum value of the monolayer capacity? Is it possible for the monolayer capacity to exceed more than $(100)\frac{mol}{g}$ or even $(1000)\frac{mol}{g}$? Any help is highly appreciated. Thank You.


text:
Why did life end up with the big 4 elements, and specifically nitrogen?
This question is mostly about the human body, but it's also about life in general.
It's not difficult to notice a big different between the abundance of elements on Earth & that in the body. I've been pondering why. Such a huge question, but here I choose to focus on the big 4. Below is my current understanding, please correct me if I'm wrong or add things that I miss.
An umbrella reason for why $\ce{O}$ & $\ce{H}$ are prevalent is because $\ce{H2O}$ is such a wonderful solvent. Life generally depends on reactions to function, and they happen best when immersed in a liquid. Water is good and copious, hence life use it.

Another reason for $\ce{O}$ to be the most common element in our body is that it can oxidize things, which can become handy.
For $\ce{H}$ is that it's basically a proton, thus can be a "filler" in most situations.
$\ce{C}$ is the backbone of life because it's the element with the best bond force. In other words, not a goldilock but the GOLDilock. Life loves to handle such flexible tool, thus it uses carbon.
That leaves the question for $\ce{N}$. Why do we even use nitrogen? It's not anywhere close to the top of abundance in the Earth's crust. It does not have attractive characteristics like $\ce{O, C or H}$. One may say that it's dominant in the air, but just because there's a lot of $\ce{N}$ doesn't mean we evolve to use it. We can always eat dirt to get some hypothetical elementary replacement for $\ce{N}$, for example. Besides, life began in the ocean and there's little $\ce{N}$ there. Do fish have different big 4? I haven't looked into it, but I doubt it.

Both $\ce{H2O}$ and $\ce{C}$ have something "best" in their belt. I guess $\ce{N}$ must have some stuff of highest quality, too, but I have no clue what it is. So, even though my knowledge of the other 3 elements is crude, I'm bugged by nitrogen the most.

Basicity. Amines are probably the best proton acceptors that can exist in water solutions. Also, nitrogen is much better in forming metallocomplexes, again, one of the best donor atoms in water solutions, second only to cyanide-ion, which simply cannot be easily tied to an organic skeleton.

Your first point may not be very relevant. $\ce{Cl}$ and $\ce{F}$ can also oxidize more things than $\ce{O}$, yet they are not as prominent as  $\ce{O}$. Also, one of the reasons of using $\ce{N}$ is it's very high bond enthalpy: N can form very strong (mostly triple) bonds with stuff, which is something none of the other 3 can do.

Without going through your points by point too deep, roughly I find the only one relevant is that about carbon. Then basicity as stated above for N. Consider that nitrogen is abundant in the atmosphere and it is fixed from there, like C. Basically what we need as implant comes from air. We means living organisms, of course no soil no bones.

There is no "replacement". P has very different properties and it only get worse for others.

Note that the composition of living things is very close to the composition of non-living things (though it only slowly adapts to changes in availability, so there may be things we've been stuck with for billions of years). You could build better biochemistry with different proportions, but unless the advantage is great enough to offset the lower bioavailability, life is shaped by the resources available. If the readily available resource proportions were different, so would we. It needs to be especially useless (or dangerous) not to be used.

@Luaan The composition of living things is very far from the composition of non-living things in most cases. Otherwise we'd be made of silica.

Although mostly we are made up of the elements you list, it is the organisation of molecules and the possibilities for synthesis of new molecules that make life possible. Life would not be possible without many metals, e.g. Mg in chlorophyll, Fe in blood , Calcium ions in signalling as well as sodium/potassium ions. Many other elements  are involved such as phosphorous in ATP/ADP and DNA and sulphur in amino acids. In fact life has found use somewhere for most elements.

I'm not at all sure why this has been closed as an "opinion-based" question. It may not be possible to provide a definitive single answer but it is clearly possible to bring many important chemical facts to bear on the question. Closing as opinion based should be reserved for questions where there are no facts and there are only opinions.

@matt_black I broadly agree with you. At the same time, I'd point out that 2/4 people voted to close as "opinion-based" and 2/4 voted to close as "needs more detail". SE insists on hiding this, despite it being one of the more popular feature requests for almost a decade. Mind you, I don't entirely agree with either of those close reasons... Anyway, back to the question itself: it has 2 reopen votes, so there's a reasonable chance it will be reopened soon; but if it doesn't then to meta it goes...

Because chemistry built on carbon, hydrogen and oxygen would be very dull and building complex structures from modular components would be impossible
The key point to remember is that life depends on complex chemistry. Living things are, by definition, things that can replicate and things than can build structures that can do complex chemistry. Metabolism is the process of taking simple things from the environment and making them into complex things. This requires complex machinery, built from simpler chemicals available in the environment.
While it might, in principle, be possible to build very complex carbon-oxygen compounds that could metabillise and reproduce themselves, there is no easy way to do so from simple components. But throw nitrogen into the mix and simpler routes to complexity are available.
Take the molecules that make up most of the complex machinery of all living things: proteins. Proteins make up most of the complex machines that do the job of metabolism and they also constitute many of the structural components that make up living things. But they are built from a small number (about 20) simpler chemicals: amino acids. Amino acids are fairly simple chemicals that might be thought of as the equivalent of Lego bricks for building an incredible variety of living structures. Most consist of only carbon, hydrogen, oxygen and nitrogen. All have the core structure $\ce{NH2CH(R)CO2H}$ (where R can be a variety of organic chunks) which constitutes the equivalent of the standardised bricks in Lego allowing all pieces to fit together so more complex structures can be built. In the case of proteins, that core functionality allows a variety of different amino acids to be "strung" together end-to-end (the amino end reacts with the carboxylic acid end on each acid).
The variety of different structures than can be created by changing the sequence of amino acids is staggeringly, unimaginably large. Many proteins contain hundreds of amino acids and you only need about 60-70 for the possible number of different sequences to exceed the number of atoms in the universe. This means that, by exploiting combinations of simple chemicals (which are not that hard to create with simple chemistry), life can suddenly explore an unimaginably wide range of possible complex structures. Perhaps such structures could be built from fewer elements, but not in a modular way that uses a small range of available components and a single way of joining them together.
The availability of this level of complexity from things that (mostly) consist of C,H, N and O is why the use of nitrogen is key to life. The constraint isn't the availability of the elements (though simple nitrogen-containing compounds are fairly available), it is the accessibility of complex structures from simple components.

Even when we can accept that C,H,O combinations are rather poor, there's no reason to specifically choose N as the last piece. Why not, say, C-H-O-S? Or if complexity is the biggest goal, why not add 2-3 elements to the party? Why only N?

@longtry: Nitrogen is a light atom and for that reason very abundant. Sulfur is not exactly rare, but not nearly as common. Still, there are common amino acids that contain sulfur. That's why decomposing biological matter has that particular odor - those are all volatile sulfur compounds we smell, and we evolved those sulfur receptors exactly because sulfur is so biologically common.

@longtry Abundance isn't the reason. The point is that the chemistry of nitrogen-containing carbon compounds like amino acids opens up a unique variety of new structures. No other element does that. And sulphur is used just not in anything like the same amount as nitrogen. As is phosphorus.

I kind of doubt the statement "No other element does that" unless you can help me understand why N is uniquely so. See, in the structure of the amino acid you mentioned, NH2CH(R)CO2H, what's the most interesting and 'opens up a number of variety some 30 orders of magnitude more than our universe's total atoms' is that "R" part, and it connects to a carbon atom. What I'm trying to say is somehow, (maybe by chance?) life came up with that nitrogen in that structure, thus we say "amino acid". It can be whatever else. How about ClH2CH(R)CO2H?

@longtry The key point about amino acids isn't the R, it is the core structure that allows long chains to be built with a variety of Rs. If it could easily be done in other ways using other elements, then we would have examples from known chemistry (chemists are pretty creative). We don't.

Hmm, makes sense... but why? I tried searching "nitrogen role in amino acids" but didn't get any leads. I get your point about the structure, but then what does N do within it? How does that single N molecule help with the molecule's ability to connect with so many different Rs?

@longtry It is the easy condensation reaction between amino acid amines and carboxylic acids that is the key.

I feel that this answer, which is very good already, may be improved by a few words regarding why $\alpha$-hydoxy acids (think lactate) could not do the job of nitrogen in amino acids. Proline being the glaring exception. matt_black mentioned the condensation, but esters are rather stable, too.

"Sulfur is not exactly rare, but not nearly as common (as nitrogen)".  Actually they are fairly close, see my answer.  Nitrogen has the advantage of being lighter and requiring fewer steps, but sulfur forms an especially stable even-even nucleus and thus tends to accumulate.

@TAR86 polymerization of alpha-hydroxy acids, which require no nitrogen, may have aided the formation of early life, see here. Possibly once life forms capable of fixing nitrogen developed, amino acids which have a better nucleophile for the polymerization reaction took over later.

My take is a little bit different.  Organisms evolved to adapt to the chemistry that is there.  And what's there, primarily, are hydrogen, carbon, nitrogen and oxygen.  Plus, as we shall see, sulfur, which cannot be ignored in proteins.
Here we find a list of the most common elements in the Universe.  Although everything else is made from fusion of hydrogen, the abundance order is not based on atomic number alone.  Nature tends to favor certain elements that form especially stable nuclei as the fusion processes proceed:

Ten most common elements in the Milky Way Galaxy estimated spectroscopically[1]

$$\begin{array}{|c|c|c|} \hline
\text{Z} & \text{Element} & \text{Mass fraction (ppm)} \\ \hline
1 & \text{Hydrogen} & 739,000 \\
2 & \text{Helium}   &        240,000\\
8 & \text{Oxygen}    &       10,400\\
6 & \text{Carbon}   &        4,600\\
10 &    \text{Neon}   &      1,340\\
26 &    \text{Iron} &          1,090\\
7 & \text{Nitrogen}  &       960\\
14 &    \text{Silicon}     &     650\\
12  & \text{Magnesium} &    580\\
16 &    \text{Sulfur}     &      440\\
\hline
\end{array}$$

These are also most common in the Solar System, accounting for 99.9% of the total (the ppm numbers quoted add up to 999,060).  Of these ten, helium and neon are chemically almost inert; and magnesium, silicon and iron do not form fluid or volatile compounds under typical conditions in space and thus do not lend themselves to reactions that form complex "organic" molecules (where they do appear, it's only as very minor constituents in the presence of plenty of the other, more volatile-compound forming elements).  This leaves hydrogen, carbon, nitrogen, oxygen and sulfur, the last of which also has some importance in proteins.  Thus, plausibly, life found what was there and adapted the chemistry of those elements for its purposes.
An additional note: If life were going to have carbon and oxygen to work with at all, inevitably it would find nitrogen around in good abundance too. Nitrogen can be interchanged with ixygen through a common type of stellar process, the CNO cycle. See this answer.
Cited reference:
1.
Croswell, Ken (February 1996). Alchemy of the Heavens. Anchor. ISBN 0-385-47214-5. Archived from the original on 2011-05-13.

The only problem with this view is that the most common elements in the universe are not the most common on earth or, more relevantly, the early earth. If "common on the earth's surface" was the criterion, we'd be based on silicon.

As mentioned, the elements must be available for reaction under conditions where life can evolve.  That's where the "big five" (I include sulfur) win, at least for life as we know it.  Might silicon become available and silicones come alive at higher temperatures?  That is another question.

Availability is part of the explanation, perhaps, but it is deeply insufficient to explain why current life uses C, H, N and O. Also, relatively complex silicon chemicals (clay minerals) are widely available available on the earth's surface and one theory of the origin of self-replication starts with clay minerals and only has other compounds taking over later as they can do a better job and can form more complex structures.

@matt_black Silicon doesn't readily dissolve in water, though. It wouldn't be available at the same conditions as early life (that is, excluding the clay minerals which are an enticing possible mechanism of the first replicators). Modern plants (especially grasses) do use silicon, but it's exactly because of how awful it is - the hard grains of silicon dioxide take great toll on grazers. Availability is a big part of the explanation, as are potential alternate biochemistries (obviously N has a huge role in our biochemistry). It doesn't explain everything, of course (e.g. iron).

Iron can be explained in context: in the presence of oxygen and water, it will form iron oxides. Those are ionic in nature, and relatively simple as a result. Solubility is limited, as is bioavailability, but plants and animals do use iron for oxygen-related chemistry.

Dusting off my geochemist hat here - In addition to elemental abundance, the matter of temperature & pressure is also key. The "solid" planet that we call home is undergoing solution chemistry on a vast scale, with most minerals differentiated mainly by local T/P conditions, rather than chemical composition ($\ce{SiO_4}$-based minerals being most common by far in the crust and mantle).
The narrow range of temperature & pressure conditions on Earth's surface at the cosmically "current" moment just happens to intersect with the T/P zone in which a few crust-abundant elements are able to form stable, reproducible, but also potentially-variable compounds. In our case, that means the protein-in-water "soup" that forms the basis for all life on Earth.
But it's not unreasonable to posit that a different "sweet spot" of solution chemistry and geochemical/astrophysical conditions could have occurred (or might occur elsewhere/elsewhen).

It might not sound unreasonable to posit alternative chemistries of life using different elements, but it is unreasonable given what we know about chemistry. The nearest anyone has come to positing a coherent theory of self-reproducing systems not based on carbon is the Cairns-Smith idea that replication started with complex silicates. But even he realised that, to get beyond a certain level of complexity, you have to switch to C H N O (and a few others) as the components.

It's a bit curious to me why clay didn't break through.  Silicon does not have to bond with oxygen to make extended structures; we know of such structures in $\ce{SiC, Si3N4, SiS2}$.  Alas, or not, the full possibilities of bonding did not materialize in clays at least on Earth.

@OscarLanzi Silicon is much harder to work with, given the ambient conditions on Earth. It also doesn't dissolve in water. Clay minerals are an interesting possibility that might explain one of the first replicators, but they didn't allow for much interesting stuff to happen - they would form the early membranes, separating the insides of the replicator from the outsides. But even if clay was enough, its chemistry would be glacial compared to sugars. As soon as first replicator "found" sugars, it would completely take over (just like chemes made way for genes, and memes overtake genes).

My point was that most of our explorations into potentially life-forming chemistry all work in the frame of reference of Earth's conditions. But even the comparatively alien/hostile conditions of 3+Byo are not the most extreme possible when considering exoplanets. What is and isn't most efficient, or even feasible, changes considerably when we open up the T/P scale beyond Earth-norms.

@MandisaW Science fiction writers are free to speculate that complex systems can emerge under very different physical conditions to earth. But none of them have produced convincing examples that don't violate what we know about chemistry. Hence the widely held view that life can only exist in a very narrow range of conditions. It isn't just us being parochial.

@matt_black More astrophysicists & exogeologists than sci-fi writers, but in any case it's hard to produce examples when so much of this is unobservable in our current frame of reference. How exactly, for instance, to prove the impact on solution chemistry under the influence of supermassive gravity wells, or intense, prolonged, high-energy radiation? We can theorize that all our understanding holds equally true, but proving it definitively is a high bar.

@MandisaW Yes, complete proof of a negative is hard. But the onus is on the speculators to demonstrate that complex alternative chemistry is possible in radically different environment. Just claiming ignorance is a very weak argument and largely untrue since we do, for example, know a lot about geological chemistry which involves extreme heat and pressure.

@matt_black A - Geological chemistry on Earth / within the Solar System is still not that "extreme", if we're talking about extra-Solar planets. Our home is in a thankfully "tame" system.

B - What we know of Earth's geology has been refined over decades (centuries, in some aspects) of Earth-based observation from multiple disciplines/channels. Unfortunately we have vanishingly few observation channels for studying other planetary systems, so we have a lot of blanks to fill-in, some of which may be flawed. "Life-supporting" requires an intersection of many variables/zones. My point isn't to challenge all of known chemistry, just to point out that relying on an Earth-based framework may well be fixing our field of view too narrowly.

Shouldn't it be $SiO_2$?

@Liam Clink - Nope, https://en.m.wikipedia.org/wiki/Silicon%E2%80%93oxygen_tetrahedron. It's an anionic tetrahedral structure that forms all sorts of linked superstructures, depending on the cation(s) involved, and temperature/pressure conditions (as well as the presence of water, in some notable cases).

According to the RNA-hypothesis, life did not start with DNA and proteins (separate molecules for catalysis and storing sequence information) but RNA instead (one molecule to store sequence information, act as polymerase and catalyze other reactions).
While that hypothesis is realistic in terms of the awesome properties of RNA, it requires a lot of complexity and five different elements: C, H, N, O, P. There is a hypothesis of warm little ponds (WLPs) being fed nucleobases from meteorites and polymerizing as the years go by: https://www.pnas.org/content/114/43/11327.
The classic hypothesis of atmospheric chemistry creating proteins (Miller-Urey) has fallen out of favor because apparently the atmosphere did not contain much ammonia in most models (see e.g. Trainer (2013)).
Of course, I am going backwards in discussing these hypotheses, assuming that we need RNA. The OP was going the other way - what chemistry is possible with what was most abundant or most likely to make interesting life-supporting polymers.

Both H2O and C have something "best" in their belt. I guess N must have some stuff of highest quality, too, but I have no clue what it is.

In terms of nucleic acid chemistry, nitrogen plays the role of hydrogen bond acceptor and donor. The donor role is a unique one - there is no good way to position a hydrogen on oxygen in a certain direction, the hydroxyl group almost always rotates. In the other hand, imine functional groups (or amides or heterocycles with nitrogen) have the hydrogen pointing in a fixed direction.
Examples are Watson-Crick base pairs in nucleic acids and main chain hydrogen bonds in proteins. In both cases, all hydrogen bond donors are N-H rather than O-H.

That last added part is very interesting. Do you know why that imine can get the hydroxyl group into order, something apparently no element other than N can do? And what's so important of a non-rotating -OH group?

@longtry I added some examples to illustrate what I meant by non-rotating hydrogen bond donor. Also, -OH always comes with an acceptor while the nitrogen functional groups can be pure donor, making for greater specificity.

Because life uses whatever it finds useful. And organic compounds with nitrogen can be easily synthesised, metabolized and may have many useful properties. And also, various stable organic compounds with nitrogen will just come into into existence in abiogenic conditions.
Why though? Well, nitrogen-carbon bonds are fairly strong and unlikely to decompose under temperature or through reaction with other compounds. Also, nitrogen can bind to one, two or three other atoms, so nitrogen allows the compounds for more complex structure. Nitrogen atoms also can form organic aromatic compounds which are very stable.
So what about other elements? Metals (and semimetals for all pratical purposes) and noble gases go straight out of the window as they can't form stable bonds with carbon. We're mostly left with nonmetals.
The fluorides can for fairly stable bonds with carbon, but are much more likely to exist in form of anions and it's hard to oxidize those anions so they can bond with carbon. Also, they can only bond with one carbon so they can't allow for more complex compound.
The sulfur can actually form fairly stable bonds with carbon and we indeed see that sulfur is fairly common in live organisms. Selenium is similiar in that regard. Carbon-phosporus bond is not very stable and phosphorus is mostly found in organisms in form of inorganic anions.
Most other nonmetals are either too rare or simply don't form stable bonds with carbon.

Thanks. So why did N get the edge over S & Se? I feel it has something to do with Permeakra's comment right under my question, but his/her is quite advanced and do not go into the details as your answer.

N did notventirely get the edge over S as the latter is also in some amino acids.  Among the three elements in the previous comment, abundances are given by N > S > Se.

Abudance is not so important here. More so the fact that C-S bond is just much less stable than C-N bond. Silimiliary with Se, but in this cas the abudance is actually visibly much lower than S.


text:
Lab goggle recommendations?
I'm taking my first chemistry class with a laboratory component this quarter. My first laboratory session was earlier this evening (observing simple metathesis reactions). While I found the lab work to be fairly straightforward, the fact that the interior surface of my safety goggles accumulated a considerable amount of condensation proved to be a moderate hinderance.
I'm wondering if anybody has recommendations for lab goggles? Particularly ones which may avoid this problem?
Note: it is school policy that we wear lab goggles with side shields, not lab glasses.
Any insights here would be appreciated!

A suggestion would be to not get too close when boiling stuff in the lab.

@AniruddhaDeb although this is a good point, I was referring to condensation on the inside of the goggles, not on the outside (we were not boiling anything in this particular experiment). I'll edit my original question to clarify :)

I am continually amazed that the lab does not provide such vital safety equipment. We had some that allowed for some ventilation: https://www.draeger.com/de_de/Applications/Products/Head-and-Eye-Protection/Protective-Eyewear/X-pect-8100

@TAR86 worry not, lab goggles are indeed supplied by the school, that said I'm seeking a higher quality set to use which may avoid condensation (although based on some of the responses to my original question, this sounds like wishful thinking)

OT of course the first lab session has to be a metathesis reaction, :)

I don't have specific product recommendations (and I think they're discouraged in general on SE), but these are the general principles:
You'll only get condensation on goggles (or anything else) if they're cooler than the dewpoint of the air contacting them. If your goggles are cold, you'll get condensation.
If you're sweaty, it'll be worse.
If the goggles are glass or another high-thermal-mass material, they'll stay cold longer, making the problem worse.
If the goggles are ventilated, humid air can escape from the space between your goggles and your face, making the problem better.
Does the problem only arise after you've been working in the lab for a while, or does it happen as soon as you put the goggles on? If the latter, you might be able to fix it by just putting the goggles in a warm interior pocket for a few minutes before you start work.
There are also spray-on anti-fog products for goggles. If you can't find one of those, try rinsing the goggle interior surfaces with soapy water, then not rinsing the soap off. The soapy film left on the goggles won't prevent condensation, but it will slow the formation of droplets, which will slow the fogging effect.

This was super helpful, thank you! I will consult my Professor and see which of these solutions may be most appropriate in our particular laboratory.

Most goggles with side shields should have some ventilation source on them. I'm assuming your goggles look somewhat like this:

The vents on the side should facilitate enough ventilation and prevent defogging.
If you're using a full face goggle with a shield, It should look something like this:

The black things on the side are covered vents: they prevent accumulation of water vapour inside while also not allowing chemicals to get in directly.
Try to obtain a pair of goggles with these features in them. If you have already bought a pair, a (slightly unsafe) workaround would be to simply drill a few small holes in the side shield to allow the water vapour to exit.

the goggles provided by the school are actually comparable to the ones in the lower photo. That said, they are presumably fairly old, it may be that the vents are clogged, preventing the escape of water vapor. It might be worth getting a new set for myself to see if that makes a difference.

The upper pictures shows a pair of lab goggles. I do not know what those in the lower picture are supposed to be used for specifically, but they are thoroughly unsuitable for normal lab work, for exactly that reason the OP experienced. They surely look cheaper. :/


text:
How to determine the composition of a gas mixture after a combustion?
The problem is as follows:

A mixture of gases consisting of $\ce{CH4}$ and $\ce{C2H4}$ is let to pass over red hot
  $\ce{CuO}$. Then $\pu{0.6g}$ of $\ce{H2O}$ is collected along $\pu{1.185g}$ of
  $\ce{CO2}$. What will be the composition of the mixture if it is known
  that the combustion was complete?.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.& \ce{CH4}=51.8\%\,and\,\ce{C2H4}=48.2\,\%\\
2.&\ce{CH4}=50\%\,and\,\ce{C2H4}=50\,\%\\
3.&\ce{CH4}=33.3\%\,and\,\ce{C2H4}=66.6\,\%\\
4.&\ce{CH4}=38.8\%\,and\,\ce{C2H4}=64.2\,\%\\
5.&\ce{CH4}=36.3\%\,and\,\ce{C2H4}=63.7\,\%\\
\end{array}$
I'm not sure exactly how to proceed with this question. Should it be treated as if it is an analysis by combustion?.
What I did to solve this was to treat each gas in the mixture as reacting independently for the oxygen from interacting with the red hot $\ce{CuO}$ as follows:
$\ce{ CH4 + 2 O2 -> CO2 + 2 H2O}$
With: $x=\ce{ CH4}$
$x\,g\times\frac{1\,mol\,\ce{ CH4}}{16\,g\,\ce{ CH4}}\times \frac{2\,\,mol\,\ce{ H2O}}{1\,\,mol\,\ce{ CH4}}\times\frac{18\,g}{1\,mol\,\ce{H2O}}=\frac{18}{8}x\,g\,\ce{H2O}$
$x\,g\times\frac{1\,mol\,\ce{CH4}}{16\,g\,\ce{CH4}}\times \frac{1\,\,mol\,\ce{CO2}}{1\,\,mol\,\ce{CH4}}\times\frac{44\,g}{1\,mol\,\ce{CO2}}=\frac{44}{16}x\,g\,\ce{CO2}$
For:
$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O}$
With $y=\ce{C2H4}$
$y\,g\times\frac{1\,mol\,\ce{C2H4}}{28\,g\,\ce{C2H4}}\times \frac{2\,\,mol\,\ce{H2O}}{1\,\,mol\,\ce{C2H4}}\times\frac{18\,g}{1\,mol\,\ce{H2O}}=\frac{36}{28}y\,g\,\ce{H2O}$
$y\,g\times\frac{1\,mol\,\ce{C2H4}}{28\,g\,\ce{C2H4}}\times \frac{2\,\,mol\,\ce{CO2}}{1\,\,mol\,\ce{C2H4}}\times\frac{44\,g}{1\,mol\,\ce{CO2}}=\frac{88}{28}x\,g\,\ce{CO2}$
This is reduced to a system of equations as follows:
$\frac{44}{16}x+\frac{44}{14}y=1.185\,g\,\ce{CO2}$
$\frac{18}{8}x+\frac{18}{14}y=0.6\,g\,\ce{H2O}$
Solving this system yields:
$x=0.1024\,g\,\ce{CH4}$
$y=0.2874\,g\,\ce{C2H4}$
Then to calculate the composition of the mixture I would use the molar fraction:
$n_{\ce{CH4}}=\frac{0.1024}{16}=0.0064\,mol$
$n_{\ce{C2H4}}=\frac{0.2874}{28}=0.01026\,mol$
Then the percentage for composition for $\ce{CH4}$ would be:
$\frac{n_{\ce{CH4}}}{n_{\ce{CH4}}+n_{\ce{C2H4}}}=\frac{0.0064}{0.0064+0.01026}=0.3841$
For $\ce{C2H4}$:
$\frac{\ce{C2H4}}{n_{\ce{CH4}}+n_{\ce{C2H4}}}=\frac{0.01026}{0.0064+0.01026}=0.6159$
To which correspond each one as roughly $38.41\%$ and $61.59\%$ respectively for $\ce{CH4}$ and $\ce{C2H4}$. But none of this seem to check with any of the answers.
Could it be that my method was wrong or did I overlooked something?.
The only answer which is closer to mine is the forth alternative but the second number is off by a big margin thus I'm not very convinced on my result. What would be the recommended procedure in this case?
Can somebody help me here?.

Let's put it this way: can a mixture consist of 90% of A and 90% of B?

BTW, just in case it makes you feel better, my numeric answer is similar to yours.

@IvanNeretin I'm still confused at what were you trying to say with the first comment?. It doesn't make sense that a mixture can be simultaneously 90% of two components. Wouldn't it be 90% of A and 10% of B or viceversa?. It's a relief that it seems that you arrived to the same answer as me. But I'm left pondering how did you got it there?. Mind sharing your method please?. :)

Why, the method is trivial, same as yours: from grams to moles and then solving a couple of linear equations. (Except that my x and y were the moles of CH4 and C2H4, since we never need their masses at all.) Now to the point: OK, indeed 90%+90% does not make any sense. What about 38.8%+64.2% (I am referring to the answer #4 from the book), does that make any more sense?

@IvanNeretin I see. Of course it doesn't. That was why it bothered me and I began asking myself if my procedure was wrong or did I made some sort of miscalculation. Now that I think more into it, I just catch up with what you mentioned regarding moles, that would have skipped some steps from my method.

@IvanNeretin $4$ Looks suspiciously similar to $1$ and $4$ was rounded from $3$ which if you look the decimal from 38.8 in the percentage would make $8$ similar to $3$. So I suspect there was some error in the printing or typing the answer. Who knows?.

Well, books are written by humans, and humans make errors. An answer option where the percents do not add up to 100 is not necessarily bad per se (maybe the intention was to teach you to notice just that?), but the absence of the right answer certainly is. So it goes.

@IvanNeretin Your comments have greatly helped me, but I feel it doesn't seem right that the question itself is left without an answer.

You can self-answer the question, this is allowed and even encouraged.

Although the book answer is clearly wrong, I'll quibble with the OP's analysis. The OP came up with 4 significant figures after solving the simultaneous equations. The OP should have been using MW's to at least 4 significant figures. In general I like to use two extra digits in intermediate calculations. That isn't any problem with today's calculators.

This is clearly a homework question. Yet, since OP has given good effort to solve the question, I'd like to give OP some insight to solve these kinds of questions, taking current question as an example.
Suppose the mixture contains $x\;\pu {g}$ of $\ce{CH4}$ and $y\;\pu {g}$ of $\ce{C2H4}$. The combustion reactions of $\ce{CH4}$ and $\ce{C2H4}$ can be given as:
$$\ce{CH4 + 2 O2 -> CO2 + 2H2O} \tag{1}$$
$$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O} \tag{2}$$
Thus, the amount of $\ce{CO2}$ produced from the reaction $(1)$ is:
$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{1 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{11}{4}x\,\pu{g} \,\ce{CO2}$$
And, the amount of $\ce{H2O}$ produced from the reaction $(1)$ is:
$$x\,\pu{g}\,\ce{CH4} \times\frac{\pu{1 mol}\,\ce{CH4}}{\pu{16 g}\,\ce{CH4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{CH4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{4}x\,\pu{g} \,\ce{H2O}$$
Similarly, the amount of $\ce{CO2}$ produced from the reaction $(2)$ is:
$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{CO2}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{44 g} \,\ce{CO2}}{\pu{1 mol}\,\ce{CO2}}=\frac{22}{7}y\,\pu{g} \,\ce{CO2}$$
And, the amount of $\ce{H2O}$ produced from the reaction $(2)$ is:
$$y\,\pu{g}\,\ce{C2H4} \times\frac{\pu{1 mol}\,\ce{C2H4}}{\pu{28 g}\,\ce{C2H4}} \times \frac{\pu{2 mol}\,\ce{H2O}}{\pu{1 mol}\,\ce{C2H4}}\times \frac{\pu{18 g} \,\ce{H2O}}{\pu{1 mol}\,\ce{H2O}}=\frac{9}{7}y\,\pu{g} \,\ce{H2O}$$
Since the mixture produced $\pu{1.185 g}$ of $\ce{CO2}$ and $\pu{0.6 g}$ of $\ce{H2O}$, we can derive two equations with two variables, $x$ and $y$:
$$\frac{11}{4}x+\frac{22}{7}y=1.185$$
and,
$$\frac{9}{4}x+\frac{9}{7}y=0.6$$
Up to this point OP has taken correct path. Now, let's consider these two questions. These equations can be simplify as:
$$77 x + 88y=1.185\times 28$$
or
$$ 7x + 8y=\frac{1.185\times 28}{11}=3.016 \tag {3}$$
and,
$$ 7x + 4y=\frac{0.6\times 28}{9} = 1.867 \tag {4}$$
Thus, $(3)-(4)$ gives you:
$$4y = 3.016 - 1.867 = 1.149 $$
$$\therefore \; y = \frac{1.149}{4} = \pu{0.2873 g}\,\ce{C2H4}$$
And $(4)\times 2-(3)$ gives you:
$$7x = 1.867 \times 2 - 3.016 = \pu{0.718 g}\,\ce{CH4}$$
$$\therefore \; x = \frac{0.718}{7} = \pu{0.1026 g}\,\ce{CH4}$$
Thus, it is clear that you were still in the correct path on solving the same two equations for $x$ and $y$.
According to my solution, mass percentage of $\ce{CH4}$ is:
$$\frac{0.1026}{0.1026 + 0.2873} \times 100 = 26.31\% \; (w/w)$$
And mass percentage of $\ce{C2H4}$ is:
$$\frac{0.2873}{0.1026 + 0.2873} \times 100 = 73.69\% \; (w/w)$$
Since this is not the one of the given answers, let's check for mol% like what you have done:
$$n_{\ce{CH4}}=\frac{0.1026}{16}= \pu{0.0064 mol}$$
$$n_{\ce{C2H4}}=\frac{0.2873}{28}= \pu{0.0103 mol}$$
Thus, the percentage of composition for $\ce{CH4}$ would be:
$$\frac{0.0064}{0.0064+0.0103}\times 100 = 38.32\%\: \pu{mol}$$
Similarly, the percentage of composition for $\ce{C2H4}$ would be:
$$\frac{0.0103}{0.0064+0.0103}\times 100 = 61.68\%\: \pu{mol}$$
Accordingly, the most closest answer is (5), although both given values are $\approx 2\%$ off from calculated values. It is noteworthy that OP's calculations are correct to the end.
Note: This question is not well written. That's why I have to go through two calculations to see which $\%$ is they are talking about. Also, the sum of compositions given in answer choice (4) is not $100\%$.

I revisited my calculations by hand two times and I get to the same answer obtained before. Can you check your equation? I'm referring to $3$ and $4$. I've even compared with Wolfram Alpha's solving for equations and get the same value as in my post here When you multiply by $28$ you forgot to multiply in the numerator for both equations by $4$ and $7$!. Can you please recheck this?.

I did rechecked my work twice so I appreciate you could check this again. I agree that not all of the alternatives check in the sense that they do not add up to $100%$ and this could be a hint of an error. But if your procedure validates mine, then it seems that the important aspect of this problem I was right isn't it?.

Your calculations are correct to the end. Thank you for finding my mental error.

I'm glad that the error was not by my side. Can you please check this topic in solubility? Perhaps does it exist another way to solve the problem more intuitively?.


text:
How can every atom have unique spectral lines?
Why does every element have unique spectral lines?
I have studied that spectral lines help us to identify the properties of an element. This is because after excitation when the excited electron(s) fall back to their original energy level(s), photons of certain frequency corresponding to the difference between the energy levels are emitted.
Now I have a question,  every substance consists of atoms, and each atom consists of electrons, electrons are fuzzy clouds and are described by wave-functions given by the solutions of the Schrodinger equation, so what makes the spectral lines different although every substance has the same thing?
I searched on the site and discussed with various members of the community and some of the conclusions I drew are mentioned below:
Electron-nucleus attraction and the electron-electron repulsion.
Hydrogen is a special case because it has only a single electron so there is no electron-electron repulsion.
Helium has two electrons so now we have some e-e repulsion, Lithium has three electrons so there is even more e-e repulsion and so on...
Every atom has a different number of electrons and a different nuclear charge, so the balance of nuclear attraction and e-e repulsion is different. So every atom has its own set of atomic orbitals that are specific to it and are different from every other atom.
The spectrum comes from transitions between orbitals, and since the orbitals are different for every atom the spectrum is different for every atom.
Are my conclusions reasonable or is there something else I missed to note?
But this answer creates one more question in my mind,
Do all noble gases or alkaline earth metals have similar spectral lines considering the above points I mentioned?

Well, in the question I'd replace atom with element. I'd just use Rydberg equation and point out the effect that Z has on the series of lines. Shielding is another factor, electron-electron repulsion a smaller one, and nuclear mass (different isotopes) an even smaller one.

No offence, but can you try to put some more effort into the grammar and logical structure of your text? As is, this is very hard to read.

@Yuvraj Singh, Did you notice that Raman spectrum has nothing to do with atomic spectrum. And there is nothing such as a fuzzy cloud of electrons. It is just a mathematical construct so do not take it in a literal sense.

@M.Farooq I will edit will you open my answer again!

I think one can only vote to reopen after you re-edit. I do not have moderator rights. Delete this question and make a more focused question after double checking what you are trying to say. Hopefully you will get better answers. Raman had nothing to with atomic spectra. It is based on vibration of molecules or in other words movement of the nuclei.

@M.farooq I update my question, but you all guys taken my question in a wrong way, because my main question is written in last regarding Nobel gases for that I have written how this question arises in my mind hope it is fine now!

Please re-post the question and delete this one. Simply go to edit, copy and paste into a new question box. The problem with public forums is that the poster and the other commentators are complete strangers. They cannot read how sincere your question is. All they can read is what is written online. I wanted to add some explanation but when I saw Raman effect connected with atomic emission and "fizzy electrons", I stopped.

@M.Farooq they are saying that I need to wait two day more to ask a new question.

Let us continue this discussion in chat.

@M.Farooq Please do not give the advice of deleting and reposting a question. Question on [chemistry.se] should always be edited to be improved; this one doesn't even have a negative score. You can always vote to reopen a closed question, as long as you have not already done it (this might include previous attempts).

I voted to reopen after the OP improved the question.

Do all noble gases or alkaline earth metals have similar spectral
  lines considering the above points?

The question is interesting after you modified it. The basic set of reasoning you provided is the main story. Each element has a different nuclear charge and the outermost electron(s) is responsible for the atomic emission spectrum. Since the nuclear charge is different, those outermost electrons experience a different potential energy. Their kinetic energy is also different from element to element.
The key question is what is meant by similarity? The atomic spectra of all the elements is visualized as bright lines on a dark background. The reason they appear as lines is just because of the instrument used to observe the atomic spectrum. There is nothing fundamental in the "line"spectrum. The atomic emission appears as lines because the slit in the monochromator is shaped like a very narrow rectangle. This is the image of the slit. If I made a very narrow circular opening, the images will appear as bright points rather than lines. Consider that line spectrum is the conventional way of looking at the atomic spectrum.
For the visible region, let us look at the visible spectrum of alkali and alkaline earth metals.

or look at the visible spectrum of noble gases

Are they similar in any way? Visually there is no similarity at all in the spectrum. Look at the spectrum. The reason we cannot see any similarity is because the dispersion of the lines is on the "wavelength" scale. However, very bright, I mean really genius spectroscopists of the 18th and the 19th century were able to find out a patterns (or call it as mathematical series). They figured out that each set of lines forms a pattern as Sharp series, Principal series, Diffuse Series and Fundamental series and there they found similarities in patterns-in terms of mathematical series. In short, our eyes cannot interpret those series. Another complication arises because we are just looking at the visible region. There is a ultraviolet region and then there is an infrared region. We cannot see it without using more sophisticated instruments.

I should like to clarify a couple of points for the OP. When Fraunhofer first observed to Sun's spectrum it was in absorption so that the lines appear black, the ones  illustrated are in emission so are bright. Either method, absorption or emission can be used. Secondly, if the spectrograph slits are narrow enough the spectral profile of the atomic transition is observed, and not the width of the slit, because a transition has an inherent width that is not determined by the slit width of the spectrograph. This spectral line-shape is observed if the experiment is done properly.

Thanks sir, I am not too much known to this   topic . and learning is good that was my topic.

@porphyrin, I think you misunderstood my point. The wanted the OP to know the origin of the word line spectrum is because the slit image is projected on a photographic plate. If we replace slit with a pin hole, while avoiding the diffraction limit,, we will see "points"  rather than typical lines. Then historically, atomic spectrum would be a point spectrum rather than line spectrum.


text:
Why does the ortho effect not work here?
The basicity order of aniline and 2,6-tertiary butyl aniline is the latter greater than the former. Why does the ortho effect not make it the other way around? Or am I missing something?

Is your first sentence a statement or a question? Providing pKb's would be helpful.

First thing to correct is that ortho effect is  applicable only when u have -COOH..ie..comparing acidity or stuff for substituted benzoic acid.

Here u have to consider SIP effect ....

I don't know where to find pKbs, sorry :(

Ortho effect is present in anilines too

https://chemistry.stackexchange.com/questions/15412/why-can-the-lone-pair-not-align-with-the-phenyl-moiety-in-2-6-xylidine https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases

https://chemistry.stackexchange.com/questions/14767/steric-hindrance-and-br%c3%b8nsted-basicity https://chemistry.stackexchange.com/questions/87461/is-steric-inhibition-of-resonance-or-steric-inhibition-of-protonation-dominant-i

Thank you! I don't understand all of the answers very much, but I think I caught the gist of it. Does it mean that m-xylene with NH2 meta to both is less basic that with the NH2 between the CH3's? Even with the double ortho effect?


text:
Why H3PO2 does not form acid salt?
Why $\ce{H3PO2}$ does not form an acid salt? I didn't understand the language.

Because it is a monoprotic acid, I guess.

It is a OK question

Does this answer your question? Ionization of hydrogens in hypophosphorous acid

Hypochlorous acid has 1 hydrogen attached with oxygen so it is monobasic acid so why can't it make its salt.

but the statement , hypochlorous acid does not form acid salt is correct. how?


text:
Why for compounds of elements in a group with larger anions have nearly constant lattice enthalpy?
Why do compounds of elements in a group with larger anions have nearly constant lattice enthalpy?
In the reasons for the decrease of the solubility of sulphates of alkaline earth metals, it is given that:

"the size of anions being much larger compared to cations the lattice enthalpy will remain almost constant within a particular group"

I am unable to understand the reasons behind it.
When I was searching I also found that larger cations stablise larger anions and small cations stabilise small anions. I didn't understand this.
Please explain.

If we assume that the lattices have the same structure, the lattice enthalpy is dominated by the charge of the ions and the distance between them. Since we are dealing here with cations from the same group, the charges are the same, so the remaining variable is the distance between them.
With moderately sized or small anions, a smaller cation can get closer, resulting in a shorter overall distance and a larger magnitude lattice enthalpy. In the case of very large anions, however, the anions are already packed as closely together as possible, so if a smaller cation moves closer to one sulfate, it moves farther from another. The equilbrium position of the cation is thus determined solely by the distance between the sulfates and does not vary with cation size.
For the second part of your question, I'm guessing that your statement that "larger cations stablise larger anions and small cations stabilise small anions" might be made with respect to solubility?
Solubility is a function of both the lattice enthalpy and the solvation enthalpy, which are in competition with each other. Generally speaking, small ions have higher solvation enthalpy in water. However, they also can have greater magnitude lattice enthalpy if they are able to pack more closely together.
Empirically, it has been observed that if the cation and anion are both small, the lattice enthalpy "wins" over the solvation enthalpy, and the salts tend to have low solubility in water.
If both ions are large, both the lattice enthalpy and the solvation enthalpy have smaller magnitude, and the empirical observation is again that the lattice enthalpy "wins".
When the sizes are mismatched, however, we find that the solvation enthalpy "wins" and the salts are generally soluble. We can see why in the extreme example from the first part of your question, where a smaller cation does not increase the magnitude of the lattice enthalpy at all, but will definitely increase the magnitude of the solvation enthalpy relative to a similar salt with a larger cation.


text:
Comparison of the dipole moments of CHCl3 and CBrCl3

Consider compounds $\ce{CHCl3}$ and $\ce{CBrCl3}$: Which compound has the larger dipole moment? Explain your choice.

Okay, so I've been trying to figure it out and I thought dipole moment had to do with electronegativity but now I'm not sure.

Dipole moments are related to electronegativity.
Draw the structures of $\ce{CHCl3}$ and $\ce{CBrCl3}$; in both cases, carbon is the central atom connected to either H, Cl, and/or Br. Then consider the electronegativities of H, Cl, and Br as compared to C. How do they pull on electrons? Does one molecule lead to a more uneven pulling of electrons compared to the other?


text:
Is there any chemical that can catalyze "reverse combustion?"
I am trying to build a machine that converts carbon dioxide and water into gasoline (i.e. octane). Because the energy needs are so large, is there any catalyst chemical that can cause "reverse combustion" at room temperature?

Catalyst or no catalyst, the energy needs are still as large as they were.

If we had that catalyst, we won't have greenhouse effect any more. Yet, there were some publications on turning $\ce{CO2}$ into methanol (not exactly hydrocarbon and oxygen). You can find these info here. As we all know and as sighted in this paper, Turning $\ce{CO2}$ into fuels is exactly what photosynthetic organisms have been doing for billions of years, although their fuels tend to be foods, like sugars (here, your catalyst is chlorophyll).

As above, this is somehow a field. Point is that the energy must be of "green" type. But even running your own PV panel thermodynamics is indeed challenging. I would say proibitive, but I am not a particularly optimistic and "visionary" person so let's hope one day....

By principle, all catalysts accelerate both forward and backward reactions
in the same extent, so the equilibrium with and without the catalyst is the same.
So the catalyst catalysing the "reversed combustion" would catalyze the normal combustion as well. The equilibrium would be  far at the combustion products side, as if the catalyst was not used.
If it had not been so, one could have constructed a chemical perpetuum mobile. It could generate energy by shifting equilibrium forward and backward just by adding and removing the catalyst.

but is there a catalyst or not?

As platinum is a catalyst of combustion, it is a catalyst of reverse combustion as well. It helps you reach the equilibrium, which is fully on the side of H2O and CO2. :-) Without provided energy, like attempts to use first steps of photosynthesis, catalysts  are useless.

thanks! but is there anything cheaper?

Yes, there is plant photosynthesis. If there was a cheap, easy and revolutionary solution, it would be already in place.


text:
Measure ethanol volume fraction in a beverage
Given a sample of an alcoholic beverage (be it beer, wine, etc..), what are possible methods to find out the ABV (alcohol by volume) of it with moderate accuracy?
As fas as I know, some methods exist already but are not really suitable in my condition:

Distillation. Impractical because of the apparatus needed
Boiling temperature based. Not accurate if exact condition of sugar is not known
Estimation by measure of specific gravity before and after fermentation. I've found it to be inaccurate with the data I have access to

I was thinking about a process similar to the tritation method used to find water hardness, but I'm not sure what kind of reactions would be involved for this kind of test, nor the nature of the titrant that would be required.

What, pray tell, is "estimation"? ;) Anyway, you have missed the most commonly used methods used in the production of wine, beer, etc. You should read a bit more.

"Estimation" meaning use of a function of density before and after the fermentation (https://en.wikipedia.org/wiki/Alcohol_by_volume#Practical_estimation_of_alcohol_content). However, I've found them to be quite inaccurate. If you say I've missed a method, would you be so kind pointing me to it, so that I can research it further?

OK, but "estimation" just means "a method that gives a rough first guess". Density is one, refractive index another, and optical activity tells you how much sugar you have (if you know what sugar it is). I guess you can use that to correct the values from density and refractivity measurement.

Your question: Given a sample of an alcoholic beverage (be it beer, wine, etc..), what are possible methods to find out the ABV (alcohol by volume) of it with moderate accuracy?

There are a few method used in winery industry as  Karl mentioned a comment in elsewhere. I assume you want to find out better methods to find the alcohol content:

The best technique is the gas chromatography method, which is the most accurate method for determining alcohol content. Gas chromatography has the added benefit of providing additional information about the wine, beer, or any other liquor product (e.g., Ref.1).
Seemingly most popular technique is the determination of ethanol in alcoholic beverages by titrimetric and/or spectrophotometric dichromate methods (e.g., Ref.2-Ref.4).
Another method is enzymatic assay using spectrophotometry: Ethanol, in an alkaline environment, is oxidized through an enzymatic reaction and converted in acetaldehyde in presence of $\mathrm{NAD^+}$, which in turns produces $\mathrm{NADH}$. The increase of the absorbance, measured at $\pu{366 nm}$ (end-point), is proportional to the concentration of ethanol in the sample (e.g., Ref.5).
In addition, you can also use NMR technology to determine the ethanol content in alcoholic beverages (e.g., Ref.6).

References:

M.-L. Wang, Y.-M. Choong, N.-W. Su, and M.-H. Lee, “A rapid method for determination of ethanol in alcoholic beverages using capillary gas chromatography,” Journal of Food and Drug Analysis 2003, 11(2), 133–140.
G. J. Pilone, “Determination of ethanol in wine by titrimetric and spectrophotometric dichromate methods: collaborative study,” J. Assoc. Off. Anal. Chem. 1985, 68(2), 188-190 (https://www.ncbi.nlm.nih.gov/pubmed/3988696).
Arthur Caputi, Masao Ueda, Thomas Brown, "Spectrophotometric Determination of Ethanol in Wine," Am. J. Enol. Vitic. 1968, 19(3), 160-165 (https://www.ajevonline.org/content/19/3/160).
Malinee Sriariyanun, Parita Mutrakulcharoen, Surapun Tepaamorndech, Kraipat Cheenkachorn, Kittipong Rattanaporn, "A Rapid Spectrophotometric Method for Quantitative Determination of Ethanol in Fermentation Products," Oriental Journal of Chemistry 2019, 35(2), 744-750 (http://dx.doi.org/10.13005/ojc/350234).
Amy A. Caudy, "Spectrophotometric Analysis of Ethanol and Glucose Concentrations in Yeast Culture Media," Cold Spring Harbor Protocol 2017, (9), 735-741 (doi: 10.1101/pdb.prot089102).
Meden F. Isaac-Lam, Thomas Brown, "Determination of Alcohol Content in Alcoholic Beverages Using 45 MHz Benchtop NMR Spectrometer," International Journal of Spectroscopy 2016,  Article ID 2526946, 8 pages (https://doi.org/10.1155/2016/2526946).


text:
What is the empirical formula of a compound where the mass ratio of sulphur and oxygen is 2:3?
This is how I tried to solve it:
$$\frac{2}{3} = \frac{0.67}{1}$$
$$\frac{32.06}{16} = \frac{2.00}{1}$$
$$\frac{0.66}{2.00} = \frac{0.33}{1}$$
I assumed by the result that if for every $\pu{1g}$ of $\ce{O}$ there are $\pu{0.33g}$ of $\ce{S}$ then if we have roughly $\pu{1g}$ of $\ce{S}$ there must be $\ce{3 O}$ atoms.
Is this result acceptable or do you know easier, clearer methods for this kind of example?

This is pretty much how I would do it.  And indeed the compound is $\ce{SO3}$.


text:
Is it safe to touch salts given to you for analysis?
In our chemistry practical paper (High school level), Inorganic salt analysis is one of the questions asked. We are given a petri dish with a sample of salt and are asked to find the cation and anion present in it.
I noticed quite a few of my classmates touching the salt (pinching it between their fingers and sprinkling it, similar to what you would do with table salt). Is this safe? Note that we do not have much in the way of PPE (No gloves, eyewear, only a lab coat). For reference, the cations may be any one of the following:

$\ce{NH4+}$
$\ce{Pb^2+}$
$\ce{Cu^2+}$
$\ce{Al^3+}$
$\ce{Zn^2+}$
$\ce{Sr^2+, Ba^2+, Ca^2+}$

The anions are $\ce{CO3^2-}$, $\ce{Cl^-}$, $\ce{NO3-}$, $\ce{SO4^2-}$ and $\ce{CH3COO-}$ (I don't believe these pose much of a health hazard)
The first thought is that this is very unsafe as the salt may be a lead/strontium/barium salt and you might end up ingesting a few milligrams if you don't wash your hands. Otherwise, I cannot see any further hazards here.
Is there anything that I have missed in my analysis? Any further thoughts on the safety of this procedure are welcome.

The students should never ever touch salts with their bare hands. There is no need to. Skin readily absorbs many things. In short it is a very bad practice and the teachers should keep their eyes open in the labs. One can understand that some Indian colleges and high schools are overpopulated with 40-60 students in lab session. Still if you see your class fellows doing this tell them not to do so.

@DrMoishePippik is there supposed to be a link in your comment?

@AniruddhaDeb Don't you have spatulae? Avoid touching any salts.

You won't get dramatic consequences in a normal scenario. But there is no need to do so, first of all. You might soon pass from qualitative to quantitative. Or the assistant can put something wrong in the Petri, and so on.

Downvoters, please clarify the reason for the downvote. It seems like a valid question to me :/

Pb, Ba (and Hg) are particularly nasty. Read the cautionary obituary, https://www.mentalfloss.com/article/501147/retrobituaries-karen-wetterhahn-chemist-whose-poisoning-death-changed-safety . N.B. She had on gloves!

The comments more or less say it all.  Wearing gloves is a good idea with just about any chemical, not only to protect you but also to protect the chemical from contamination.  For instance, if you find chloride in your sample you might want to be sure said chloride is not from salt on your skin.


text:
What's a non-toxic flavorless water-soluable chemical I could use for freezing point depression when making ice cream?
I'd like to make a savory version of ice cream. So I can't use sugar or any artificial sweeteners for the freezing point depression needed to control ice crystal sizes. What can I use?

What gives you the idea that freezing point depression is responsible for (I assume small?) crystallite size?

I've heard that at various parts of the internet. It doesn't really make sense to me but I'm not knowledgeable enough regarding making ice cream to dispute it.

http://icecreamscience.com/sugar-in-ice-cream   http://icecreamscience.com/fiber-in-ice-cream/#211_effect_of_increased_viscosity_on_ice_crystal_size   https://foodcrumbles.com/secret-of-ice-cream-freezing-point-depression/   These gave me the idea of that. Maybe I need to look into the fiber one more...


text:
Reactivity of anhydride vs aldehyde with nucleophile
It is known that order of reactivity of carbonyl derivatives towards nucleophiles follow the order as:

Acyl halide> acid anhydride>ester> amide> amide derivatives

(Source: "A Guidebook to Mechanism in Organic Chemistry" by Peter Sykes M.Sc., PhD., F.R.S.C., C.Chm. Fellow and Vice-Master, Christ’s College, Cambridge.)
(Here amide derivatives are NRR' only)
It is difficult to find out the order in books and online as well, if we also include aldehyde and ketones.
Well there is one such debated question on Chemistry Stackexchange, but the fact that anhydride is greater in reactivity than an aldehyde confused me as  why is the acid anhydride's direct conjugation (even though if it's cross conjugation) less effective in stabilising the positive charge on carbonyl than the aldehydes hyperconjugation?

Could you cite any sources that got you convinced that anhydrides are more reactive than aldehydes towards nucleophilic attack? From the data that you cite, there's no mention whatsoever about ketones or aldehydes.

https://chemistry.stackexchange.com/questions/14618/rationalising-the-order-of-reactivity-of-carbonyl-compounds-towards-nucleophiles

Oh sorry, I was actually thinking you found it in some other book or article or on the internet.