text:
How are Uranium and such materials made
I was wondering, how are materials like uranium, plutonium and other radioctive materials made. Do they all come from the old exploded stars and just been in the mountains for millions of years. Is there getting made more of those materials, or are we gonna run out in the future?

Some radioactive materials are made by man, others have been parts of earth's rocks since they were formed
Radioactive materials are not all the same. some are man-made and some have been with us since the creation of the planet (so are indeed the product of old exploded stars).
The best way to tell which is which is to look at the half-life of the atoms involved. Half-life is a measure of how fast an isotope decays. It is the length of time it takes for half of the amount of substance to have decayed to something else. So half-lives with the same sort of size as the age of the earth are quite likely to be primordial (which basically means they were present when the earth was first formed). Other compounds with shorter half-lives would have all disappeared by now even if they were present then. They are synthesised by man.
238U (by far the most common isotope of uranium) has a half-life of ~4.5 billion years (roughly the same age as the Earth) so quite a lot of the uranium present at our planet's birth is still around. 239Pu (the isotope of plutonium used in nukes) has a half-life of ~24 thousand years so even if there was some around when the earth was formed there wouldn't be any of that left. Plutonium is created in nuclear reactors which can be designed to generate enough to make an indecent supply for the military. Having said that, a very small amount may be generated naturally in uranium containing rocks as part of the normal decay pathways of uranium or as part of natural nuclear reactors (occasionally in geological history uranium rich rocks have created the conditions for significant nuclear reactions which produce more plutonium and other materials than simple decay of isolated uranium atoms).
Other, more common, elements sometimes have radioactive isotopes. A small proportion of potassium, for example is 40K which has a half-life of ~1.2bn years (so quite a lot of the earth's original stock is still around and is the largest contributor to radiation in living things which all contains a lot of potassium).
There is little danger of running out of primordial radioactive elements as we are not using them fast enough. The man-made ones can be made in nuclear reactors if they are set up the right way (and we have way more plutonium than we know what to do with; but many rare isotopes used in medical research are made in specialty reactors where the risk is that a reactor shutdown might limit the supply).
If you want to know where all the elements originally came from, there is a good Wikipedia article with a periodic table that explains the detailed origins.


text:
Which organic compound's NMR and IR are these? Can't find out. (Data available: Melting Point ~ 116 °C)
Which organic compound's NMR and IR are these? I have figured out that the compound is a 1,4 substituted benzene. And the substituents are probably not -COOH and -OH from the spectroscopy. Kindly help me in figuring the other groups out. (Data available: Melting Point ~ 116 °C, gives sooty flame)

@Soumik Das: The presence of two clean doublets in aromatic reagion of NMR would not agree with your suggestion. It may be 4-nitrophenol, which has $113-114$ $\mathrm{^0C}$ melting point.

The very strong signal at ~1350 would fit for C-NO2

Why is there no C-H stretch band in the IR spectrum near 3000 $\ce{cm^{-1}}$ ?

@pas — When you present an NMR spectrum, you absolutely need to also give the nucleus, the frequency and the solvent. Nevertheless, I agree with Mathew on the 4-nitrophenol. The chemical shift of OH group can vary A LOT depending on the solvent, the temperature and the pH which, in turn, depends on the concentration. In this case, p-nitrophenol is acidic, so the more concentrated it is in the NMR tube, the more acidic the pH and the more the signal will move to the left.

@MathewMahindaratne The signals of 1,4-substituted benzenes form a AA'XX' system; so despite their appearance for this sample and measurement they do not qualify as simple doublets.

The integrals and shifts in the very simple NMR spectrum are a giveaway that this is a simple compound with two different types of aromatic signal, and another non-aromatic signal from a substituent, at a 2:1 peak integral ratio, hence possibly a para-disubstituted benzene ring with one proton in one substituent, none on the other.
The IR spectrum is practically identical to that in the NIST webbook for p-nitrophenol.
Some other clues:

Oop Signal at $\pu{800 cm^-1}$: para substitution
$\pu{1400-1600cm^-1}$: aromatic cc stretch, matches nicely with typical appearance for para substitution
No peak at $\pu{1700cm^-1}$: no carbonyl
Weak signals at $\pu{3000cm^-1}$: sp2 stretch, no sp3
Signals at $\pu{1350 and 1500cm^-1}$ may be nitro

Note this is a little tricky - a search would show some variation in reported spectra.
The melting point also matches reported values.
This is well in agreement with many of the comments under the original post.

source: Stack Exchange

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

sort: votes


text:
Ea is the term for activation energy, what is the mean of a in this term?
Our doctor said it is not stand for activation. it stands for something else, any help will be appreciated.

Comments are not for extended discussion; this conversation has been moved to chat.

The subscript $\mathrm a$ in $E_\mathrm a$ may refer to Arrhenius.
The IUPAC gold book states that:

activation energy (Arrhenius activation energy)
  An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$, $E_\mathrm a = RT^2 \frac{\mathrm d(\ln k)}{\mathrm dt}$,
  where $R$ is the gas constant and $T$ the thermodynamic temperature.

You haven't read any of the comments to the question, have you? :)

any one interested, the answer of our doctor is a stands for apparent or approximation value of activation energy.
thanks for everyone

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


text:
Is there a particular name for a Hill formula or sum formula omitting hydrogen?
Often, single-crystal X-ray diffraction analysis determines the non-H atoms first, and eventually puts hydrogen atoms "as riding" on idealized positions in respect to heavier atoms (e.g., C, O, N), because the latters are easier to localize in the Fourier maps.  This then considers both distance and geometry (e.g., 0.93 Å for an H attached to a sp-hybridized C) and temperature during the experiment, determined by other experiments (e.g., neutron diffraction).
Is there a particular name for a sum / Hill formula where the hydrogens are omitted, for example, when stating glucose as $\ce{C6O6}$ instead of a complete $\ce{C6H12O6}$ to describe just the backbone of the molecule?

Not an answer, but what you are describing seems like a condensed version of SMILES strings if you were to remove the bond information.

@Tyberius Not so fast, please.  In a "truncated Hill formula" $\ce{C6O6}$, constitutional information is lost.  In a reduced Smiles representation lacking bond order, CC1CCCCC1Br is (at least) a valid representative for any stereoisomer of 1-bromo-2-methylcyclohexane, or 1-bromo-2-methylbenzene; diffrerent from a short $\ce{C7Br}$.

Yeah I figure the route to what your looking for probably goes through the Hill formula rather than SMILES. I just figured it was another avenue to look into to see if someone has given a name to this type of notation. @Buttonwood

You might say it is a "united atom" representation, but I can't speak for the universality of this nomenclature.

Please note that the terminology 'sp-hybridi(s|z)ed C' is incorrect, atoms cannot be hybridised, only orbitals can be. Better options would be linear (diagonal)/ trigonal/ tetrahedral ((coordinated)) carbon (or atom in genearal). Otherwise one could write spˣ-C as an abbreviation for carbon atom with spˣ-hybridised orbitals. // P.S.: MathJax should not be used as a style element.


text:
Impediments to GC-MS characterization of petroleum products
In an attempt to characterize, and sometimes to 'match' oil on feathers of birds, a laboratory utilize gas chromatographic - mass spectrometry. Persons likely to encounter oiled birds are advise to NOT wrap the bird in plastic, as this may affect results of the analysis, rather to use foil for containment, transport, and of course, to avoid cross-contamination with other hydrocarbons.
Exclusive of issue and presumed effects of cross-contamination by other hydrocarbons, how significant is short-term (< 48 hour) storage of oiled bird(s) on the assessment of contaminated plumage?
Thank you if you can provide an answer to this question.

What are "foil" and "plastic"?

@Karl Presumably aluminium foil and clingfilm. Clingfilm contains a load of plasticiser which easily contaminates anything it touches.

You'd need to know about the specific plastic composition (as matt_black points out, the identity and concentration of leachable plasticizers) and the rate at which these might be leached. Tough question but a literature search or a control experiment(s) will probably help.

@matt_black This stuff is usually graded food-safe, and does not contain plasticisers. It's usually PE + PiB, there is definitively no need to add expensive plasticisers which also make bad PR. PVC foils do contain plasticisers, but they're more for permanent use (windows, office supplies, ...)


text:
Fragmentation involving formation of enamine
I have a problem on fragmentation which requires mechanisms involving formation of an enamine from a cyclic ketone and cyclic secondary amine. This then reacts with a cyclic aldehyde via conjugate addition, but the intermediate cannot be reached.
I've been told that I have to exchange the enamine of the ketone with the enamine of the aldehyde. What does this mean?

PROBLEM 17
Give mechanisms for these reactions, commenting on the fragmentation.

Purpose of the problem
  Revision of conjugate addition of enols, another ring expansion with an enolate as leaving group and an interesting piece of stereochemistry.
Suggested solution
  The first step is enamine formation and the second is conjugate addition. This appears to lead to a dead end as we cannot find a way to make the intermediate from the product.

The product looks like it is at the wrong oxidation level, are you sure there isn't a double bond in the [7] ring?

I believe it means you should convert the ketiminium back to a ketone and then use the pyrrolidine to make an aldiminium.

@Laura: Both Waylander and Zhe are correct in their respective comments. Your example is from the work of Hendrickson and Boeckman.1 Enamine 1 and conjugated aldehyde 2 were mixed neat forming bridged aminoketone 3. At some point, as Zhe has intimated, exchange of pyrrolidine from the ketone to the aldehyde must occur before cyclization to 3 can be accomplished. The usual explanation is that traces of water are present to catalyze the reaction. Quaternization of the amino group in 3 affords the salt 4 which cleaves with hydroxide via a Grob-like fragmentation to 5. Note the important anti arrangement of the dotted bond and the proximate vicinal bold bond. The product contains a double bond as Waylander has stated. The first example of this reaction, 6 --> 9, was reported by Stork and Landesman2 in 1956.

1) J. B. Hendrickson and R. K. Boeckman, Jr., J. Am. Chem. Soc., 1971, 93, 1307.
2) G. Stork and H. K. Landesman, J. Am. Chem. Soc., 1956, 78, 5129.


text:
Is the reaction involving a proton transfer a redox reaction?

$$
\begin{align}
\ce{H2S (aq) + NH3 (aq) &→ NH4+ (aq) + HS- (aq)}\tag{1}\\
\ce{Pb(s) + 2 FeCl3 (aq) &→ 2FeCl2 (aq) + PbCl2 (aq)}\tag{2}
\end{align}
$$
Which of the following statements correctly describes both of the equations above?
A. Both equations represent redox reactions.
  B. Equation 1 represents a redox reaction but equation 2 does not.
  C. Equation 2 represents a redox reaction but equation 1 does not.
  D. Neither equation represents a redox reaction.

The answer key says that the correct answer is C. I picked A. But in equation 1, $\ce{H2S}$ donates a hydrogen to the $\ce{NH3}$ which would mean it's a redox as well. Both equations had donations and acceptances so I really don't understand why equation 1 is not a redox reaction.

Start by identifying the oxidation states of the atoms. If the oxidation states don't change, it's not redox.

Am I doing that wrong? Because I found that they did change in both I dont know what im doing wrong.

OH Nevermind I just realized I was mixing up a change in numbers with a change in state. Thank you!

Donation of a proton is not an indication of a redox reaction, rather an acid-base reaction (which may or may not be also involving a redox process). The only way to show the reduction/oxidation takes place is to demonstrate that the oxidation numbers do change:
$$
\begin{align}
\ce{\overset{+1}{H}_2\overset{-2}{S} (aq) + \overset{-3}{N}\overset{+1}{H}_3 (aq) &→ \overset{-3}{N}\overset{+1}{H}_4+(aq) + \overset{+1}{H}\overset{-2}{S}^-(aq)}\tag{1}\\
\ce{\overset{0}{Pb}(s) + 2 \overset{+3}{Fe}\overset{-1}{Cl}_3 (aq) &→ 2 \overset{+2}{Fe}\overset{-1}{Cl}_2 (aq) + \overset{+2}{Pb}\overset{-1}{Cl}_2 (aq)}\tag{2}
\end{align}
$$
As you can see, oxidation numbers only change in reaction (2), and are preserved in reaction (1).

It is C indeed.
In a REDOX reaction must exist an oxidizing agent and a reducing agent. This means that at least one atom must be reduced while another must be oxidized. In other words, the acceptance and donations must be of electrons (not of atoms), from one specie to another. In your case in the reaction B is very simple, you have:
$$\ce{Pb (s) -> Pb^2+ (aq) + 2 e-}$$
This is your oxidation reaction and therefore the $\ce{Pb}$ is your reducing agent.
While:
$$\ce{Fe^3+ (aq) + e- -> Fe^2+ (aq)}$$
The $\ce{Fe}$ is reduced from $\ce{Fe^3+ (aq)}$ to $\ce{Fe^2+ (aq)}$.
So the net reaction is:
$$\ce{Pb (s) + 2 Fe^3+ (aq) -> 2 Fe^2+ (aq) + Pb^2+ (aq)}$$
For the first reaction, there is no variation in the oxidation number of any atom, this means that there is no net electron transfer and thus no REDOX process.

Could you please explain what does "(ac)" in the net ionic equations stand for?

@andselisk Sorry. it stands for "acuoso" in spanish (my mother language) it means aqueous (aq) in english... bad habits. I just edited it.

@andselisk could you please share the link where I can find the remarks to format the equations? Thanks for the edit, I tried to do it but couldn't find how.

Ha, I didn't know that, interesting and thank you for explaining:) I edited your answer a bit: IUPAC suggests not to use indices for the states of aggregation.  Also there is MathJax to format chem and math expressions (visit this page, this page and this one), so I formatted the equations accordingly.


text:
Organic Nomenclature
10.2 It has to add 4-propyl is that correct? And 10.3 why do we choose the position 1,5 instead of 2,3?

And 10.4,, it should be included the stereochemistry of the middle double bond in front of the name as ($3E$)-3,4-diethyl...., to complete IUPAC name.

Actually, none of the suggested names is in accordance with current IUPAC recommendations.

You are correct in 10.2 and in 10.3 it should be 3-ethyl-2-methyl cyclopent-1-ene. This is to give the lowest locant to double bond.


text:
Why total heat of hydrogenation of 1,3-cyclohexadiene is more than that of benzene?
I'm told that heat of hydrogenation (HOH) is directly proportional to number of π bonds and inversely proportional to stability. So, is the aromaticity responsible for this?
Also, what is the general approach to the problems like this? Say, I encountered napthalene and some hydrocarbon which contains about 2 rings and have π bonds less than naphthalene, just like stated above – then how shall I decide?
Is there any specific rule like if a compound will be aromatic then it's HOH will be reduced by this amount or something like that?

"Heat of hydrogenation of alkenes is a measure of the stability of carbon-carbon double bonds.  All else being the same, the smaller the numerical value of heat of hydrogenation of an alkene, the more stable the double bond therein." — See https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Heat_of_Hydrogenation

Yeah I know that but how do I measure how much stable.Say I told you to give me an order of HOH of benzene, 1,3-cyclopentadiene and cyclopentene then how can I rate that whose HOH will be higher than whom.See the main thing which I'm getting stuck is that HOH of 3 π bond will be more than that of 2 π bond and so on.But see here aromaticity comes into play and so HOH of benzene is less than 1,3-cyclopentadiene but more than cyclopentene.Why HOH of benzene not lowest among all these coz of Aromaticity.I hope that you might be getting my doubt.

Yes. The relatively smaller Heat of hydrogenation (HOH) for benzene as compared to that for 1,3-cyclohexadiene is due to the aromaticity of the first. Analyzing the thermochemistry is indeed among the first and perhaps more intuitive ways to present and quantifies aromaticity itself.
A) Cycloexene HOH = -120 kJ/mol
B) 1,4-Cycloexadiene HOH = -240 kJ/mol
C) 1,3-Cyclohexadiene HOH = -232 kJ/mol
D) Benzene HOH = -208 kJ/mol
While for B the HOH is about double as compared to that of cyclohexane and thus is according to the assertion in the question, the conjugation in C does results in a "lower than expected" value.
The 8 kJ/mol difference is the energy of resonance of the two conjugated double bonds.
The situation is even more striking for D as this difference from "an expected value" amounts now to about 152 kJ/mol. This is again a measure of stability and its distinctively high value is a manifestation of  aromaticity.
The HOH of benzene is even less than that of cyclohexadienes. This shouldn't come as surprise though, as for there are not really double bonds.
In other words, while hydrogenation of individual double bond is an exothermic reaction, hydrogenation of benzene to cycloexadienes is an endothermic process.
Once you disrupt aromaticity, then you can think again in terms of individual double bonds (although delocalisation may occur to little extent, as in C). And this fact should answer the question "why HOH of benzene is higher than that of cyclopentene?", too.

Does this mean that 1,3-Cyclohexadiene is more stable than benzene?

@HarryHolmes yes. You can use heat of hydrogenation to conclude that. Ethane is more stable than ethene, the logic is the same. Just benzene is more stable than expected, and by a good amount.

Thanks for coming back here! But what did you mean by that last sentence?

@HarryHolmes it is all in the answer. Is that clear to you? Benzene is particularly stable as compared to a hypothetical localised form of it. It is not the most thermodynamically stabile beast in the universe. In term of H formation, a cyclohexene can be more stable than benzene. Still benzene is much more stable than something having three double bonds. And, in case, also than something having two, see the other answer. The exact values would depends on the tabulated data one use, but the meaning is above.

You meant cyclohexadiene, then not. But benzene is less stable then cyclohexane, thermodynamically speaking. @HarryHolmes. Let me know if you still have doubts.

Just this; have I got this straight? The absolute value of HOH is inversely proportional to stability. Cyclohexene (not 1,3-Cyclohexadiene) is therefore more stable than benzene. 1,3-Cyclohexadiene and 1,4-Cyclohexadiene are less stable than benzene. Is that it?

@HarryHolmes yes. This is however in an scale that doesn't take into account the energetic content by amount of atoms. To give you the point: a long saturated hydrocarbon is inherently more stable than a shorter one (unless entropy kicks in). But this does not really means that octane is more "chemically robust" than pentane. Same would be for a non conjugated diene respect to the alkene of corresponding length. Not to say that, as usual, thermodynamic stability and reactivity shall not be confused.

Hmmm, so 'more stable' is just in the context of heat of hydrogenation, and other factors may influence the actual stability of the species?

@HarryHolmes look at this link. Out of a defined chemical process you cannot get to much by comparing different things. It is in the meaning you want/can put into the term. Cyclohexene is more stable than benzene simply because it is more saturated, ie more energy is released upon its formation. It is quit obvious. Here is the link https://chemistry.stackexchange.com/questions/146533/what-is-the-definition-of-thermodynamic-stability

In particular my comment: Take it as an energy content. Of course in ideal case it would be better to have a "normalised stability", which in some cases is possible. For instance benzene is more stable than cyclohexane as per pi electrons. I agree that comparing different things isn't really necessary, but I have no examples in mind in which we do this comparison other than in a vague way. When it comes to quantities, we analyse reactions or transformations in which elemental composition is globally the same.

I see, thanks. So the final verdict is that cyclohexene is, considering all factors, stable? Or is no such statement really possible?

@HarryHolmes yes it is. You can isolate it in normal conditions, put it in sort of container, keep it. It is stable. Tough, its mixture with oxygen is not that stable and transform to water and CO2, soon or later. If it does it fast or not, and if it needs some initiation, that is about so called kinetic stability, ie the effective reactivity. The latter part isn't relevant to this thread here, but it can be confusing for starter readers. It is also addressed in the thread I linked, tough again is independent of the issue "what therm. stability is?".

Perhaps a diagram as to why the heat of hydrogenation (ΔHho) of benzene is less than the ΔHho of 1,3-cyclohexadiene would be helpful. The heats of formation (ΔHfo) of the relevant species except the hypothetical cyclohexatriene are available at the NIST site. The ΔHfo of cyclohexatriene is estimated as ~56 kcal/mol above the standard state. This value is obtained by tripling the ΔHho of cyclohexene to cyclohexane. One might have also averaged this ΔHho with that of 1,3-cyclohexadiene to obtain the indeterminant ΔHho of cyclohexatriene to 1,3-cyclohexadiene. Not only is the exact ΔHfo of cyclohexatriene uncertain, but, as a result, so is the resonance stabilization of benzene (-36.2 kcal/mol). The ΔHfo of benzene places it ~5 kcal/mol below that of 1,3-cyclohexadiene. Because both of these compounds are hydrogenated to cyclohexane, their difference in heat of hydrogenation is likewise ~5 kcal/mol. In other words, the ΔHho of 1,3-cyclohexadiene is greater than the ΔHho of benzene by ~5 kcal/mol. Incidently, the ΔHfo of 1,4-cyclohexadiene has been reported at NIST as ranging between 24-26 kcal/mol.

National Institute of Standards and Technology, U. S. Department of Commerce


text:
During formation and breaking of bonds, can the terms potential energy and bond energy be used interchangeably?
My chemistry professor uses these terms interchangeably and I was wondering if they refer to the same thing. From what I understand, bond energy is the energy required to break one mole of covalently bonded gaseous molecules. While potential energy refers to the energy gained or lost during bond formation which can be further characterised into endothermic or exothermic.

You always gain energy when you form a bond (unless you had to break other bonds in the process). What is endothermic or exothermic is a reaction where multiple bonds might be formed or broken. You can estimate whether a reaction is endothermic or exothermic by evaluating the bond energies (technical term: bond dissociation energies) of bonds made and broken.

Yes you can use them interchangeably. You may then refine the usage to be a defined in a particular way, such as bond energy. The other form of energy is kinetic energy such as in the vibrational and rotational motion of a molecule.

@KarstenTheis Hmmm, nearly the opposite of what I write here: https://chemistry.stackexchange.com/questions/108697/when-a-bond-is-formed-between-an-electophile-and-a-nucleophile-they-both-lose-e

@Try_Hard: Yes, that might have been confusing. The "You" in "You always gain energy" would be the environment, so bond formation is exothermic. Maybe "it takes energy to break bonds" is the less confusing statement.

The answer is a qualified yes. Specifically, the "bond dissociation energy" (often shortened to "bond energy") is a positive number that is the amount of energy one must add to the system to completely break a mole of equivalent bonds. A completely broken bond by definition has 0 potential energy, so bond formation results in a negative potential energy. Thus, the potential energy of the bond is equal in magnitude but opposite in sign to the bond dissociation energy. If we are only talking about the magnitude (which is often the case), they're interchangeable.
In addition to the chemical bonds, a system can have other sources of potential energy such as gravitational potential energy or electrostatic potential energy, so in that sense potential energy is not synonymous with bond energy, but generally it's pretty clear from context when we mean only the potential energy of bonding.

The potential energy  increases at the expense
of the kinetic energy during the breaking of bonds, that the kinetic energy transforms to potential energy, so the definition of bond energy as a change of internal energy,$\Delta{E} $, which associates the breaking bond.
Because of the work $\mathrm{P\Delta{V}}<<\Delta{E}$  during breaking a bond, so: bond energy equal change in enthaphy$\Delta{H}$which associates with the breaking covalent bond in the molecule at gaseous state to produce neutral particles in the gaseous state.
e.g:Bond energy of hydrogen molecule$~\ce{H-H}=436\pu{K.J/mol}~\ce{H2}$
$$\ce{H-H_\mathrm{(g)} ~-> 2H\mathrm{(g)}} \quad\left(\Delta{H}=+436\pu{K.J/mol}~\ce{H2}\right)$$
so,$\Delta{H_f^o}~\ce{H_\mathrm{(g)}}=\frac{436}{2}=218\pu{K.J}$ to produce one $\pu{mol}$ of hydrogen atoms at gaseous state $\ce{H_\mathrm{(g)}}$.
$$\text{Bond energy} =\Delta{H}= 2\times{(\text{Potential Energyof}~\ce{H_\mathrm{(g)}})}-\text{Potential Energyof}~\ce{(H-H)_\mathrm{(g)}}=+436\pu{K.J/mol}~\ce{H2}$$

@Try Hard Your comment is correct, I edited it.


text:
What is the correct number of stereo isomers in case of 3,6-dimethylpiperazine-2,5-dione?
What is the correct number of stereo isomers in case of the following compound, i.e. 3,6-dimethylpiperazine-2,5-dione?

The answer key gives three, but I think there should be four.
I recognised the two planes of symmetry, one through the methyl groups and the other through the keto groups.

The plane through methyl groups would project the keto group upon nitrogen, hence it is not a plane of symmetry. Ditto for the other one. But that's irrelevant; we are counting isomers here. What are your 4?

counting 2 nitrogen and 2 carbon as stereocentres, we hav total 4 stereocentres and the molecule does not hav plane of symmetry. but nitrogen goes inversion. so only two chiral centers so 4

Good. But there is a catch: a certain symmetry element makes the chiral centers equivalent, so RS and SR are the same thing, which leaves only 3.

The nitrogen atoms should not be considered stereocenters, not because of rapid inversion but because amide nitrogen atoms are almost planar.

You should build 3D models of the compound. If you do you will see that two of the supposed stereoisomers are the same molecule.

I just want to re-iterate @IvanNeretin's comment: There are no planes of symmetry here. There are axes of rotation for the core structure ignoring substitution, and these sometimes reduce the number of stereoisomers when the substitution is just right.

The RS compound has a center of symmetry.

Three is the correct answer.
As Ivan pointed out, there are no planes of symmetry, for example:

Ivan is also correct - the "RS" is the same as the "SR". You can prove this to yourself most easily using molecular models, but here is how to see it on the page. You just need to rotate the molecule by 180° in the plane of the page:

So, despite having chiral centres, this compound is achiral because it is superposable with its mirror image. "But wait, I was told that a molecule needs to have an internal plane of symmetry to be a meso compound" I hear you say. Unfortunately, this is an incomplete truth that many sources repeat. To be a meso compound, it can alternatively have a centre of symmetry, also known as an inversion centre. This is simply one point in space through which any atom is reflected through - ie. each atom is diametrically opposite an equivalent atom at an equal distance from this point.

The other two stereoisomers are the (R,R) and the (S,S). These are non-superposable mirror images, and hence enantiomers. Note these isomers have no plane of symmetry or centre of symmetry.

Once again, this is easier to see with physical models.

For completeness sake: meso-compound - A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members. achirality (from chirality) - [...] If the object is superposable on its mirror image the object is described as being achiral.


text:
Thermal decomposition of group 2 permanganates
Do group 2 permanganates produce manganates via thermal decomposition?

It is observed that true composition of alkaline earth metal manganate is not known. It is proposed to be a mixture of manganite-manganate and has a general formula, $\ce{R3Mn2O8.H2O}$. However the composition/formula varies according to given method of preparation. It is found on some instance that heating the permangantes of alkaline earth metal gives manganates while in other case it gives manganites.

Mangesium permanagate is prone to decomposition and it explode
violently in aqueous solution giving oxygen, ozone and the
manganate?(the reaction product is not defined properly?)
Calcium manganate is formed by decomposing calcium
permanganate.
Strontium permanaganate is also prone to decomposition and
forms either manganite or manganate depending on reaction condition
but either way, strontium manganate is formed by heating
strontium manganite at 1600 °C.
Barium permanganate also decomposes at 320 °C to form barium
manganite but if the permanganate is mixed with barium hydroxide,
barium manganate of composition $\ce{Ba3Mn2O8.H2O}$ is formed.
The reaction of beryllium permanganate decomposition is unknown to
the best of my knowledge mainly due to poor study on the compound.


text:
Why is lead(II) present in two groups of cation analysis?
Well, I found that lead cation $\ce{Pb^2+}$ is present in both group 1 and group 2 of cation analysis. But why?

Simply because lead(II) fits both analytical groups:

it forms lead(II) chloride $\ce{PbCl2}$, a poor soluble chloride, which puts it in the 1st group;
it also forms lead(II) sulfide $\ce{PbS}$, insoluble in water and diluted acids, which also puts it in the 2nd group.

One reason for putting lead in both groups is that the chloride is somewhat of a borderline case in terms of being insoluble enough to appear in Group I.  Group II represents a more complicated analysis but is more sensitive for lead.

Lead (II) chloride has enough solubility so it could slip through the cracks in Group I.  Group II where the sulfide is precipitated is more sensitive.

@OscarLanzi You are right, probably it would be worth adding that there are no sharp margins between these 6 groups and various sources add or remove cations
from one group to another arbitrarily.

@andselisk If the product of the $\text{Group I}$ test is $\ce{PbCl2}$ then how can we test for it in $\text{Group I}$?


text:
What is the difference between ΔfH and H-H° in NIST–JANAF tables?
Can someone explain the difference between $Δ_\mathrm{f}H$ and $H-H^\circ$ in NIST–JANAF tables?

As a side note, "JANAF" stands for Joint Army–Navy–Air Force; these tables were designed to provide quick access to the thermodynamic data for rocket scientists.
Here, $Δ_\mathrm{f}H^\circ$ is the standard enthalpy of formation. The difference $H-H^\circ(T_\mathrm{r})$ is required to calculate the standard enthalphy of reaction $Δ_\mathrm{r}H^\circ$ if in the given temperature range a reaction takes place.
There is a thorough explanation in Silbey's Physical Chemistry [1, p. 62]:

The JANAF tables give standard enthalpies of formation at a series of temperatures, and so these values may be used directly to calculate enthalpies of reaction. [...] Some thermodynamic tables give values of $\bar{H}^\circ_T - \bar{H}^\circ_{298}$ to assist in the calculation of $Δ_\mathrm{r}H^\circ$ for a chemical reaction or phase transition:
$$\bar{H}^\circ_T - \bar{H}^\circ_{298} = \int_\pu{298 K}^T \bar{C}^\circ_P \mathrm{d}T$$
Depending on the table, $Δ_\mathrm{r}H^\circ$ for phase transitions in the intervening temperature range may be added to the right-hand side of the equation.

NIST-JANAF tables are given for a reference temperature $T_\mathrm{r}$ and a standard state pressure $p^\circ$. Unless stated otherwise, $T_\mathrm{r} = \pu{298.15 K}$ and $p^\circ = \pu{0.1 MPa}$ [2].
References

Silbey, R. J.; Alberty, R. A.; Bawendi, M. G. Physical Chemistry, 4th ed.; Wiley: Hoboken, NJ, 2005. ISBN 978-0-471-21504-2.
NIST-JANAF Thermochemical Tables, 4th ed.; Chase, M. W. J., Ed.; American Institute of Physics: Washington, DC : New York, 1998. ISBN 978-1-56396-831-0.

Does $T_\mathrm{r}$  stand for room temperature?

@KarstenTheis - no $T_{r}$ is a reference temperature (which may be 25C but doesn't have to be). Just above the bit shown in the question there is a line stating what $T_{r}$ is for the table.

@KarstenTheis Jon Custer already commented (wow that was fast:) ), but essentially yes, the reference temperature is close to the RT. NIST usually uses $T_\mathrm{r} = \pu{298.15 K}$ (I updated the answer, too).


text:
How would you go about adding a Cl group para to a brominated cinnamic acid?

We recently synthesized brominated cinnamic acid in lab, and now our goal is to add a Cl para to the R group attached to the benzene ring.

There are a number of conditions that are selective for p chlorination of substituted benzenes. KCl and Oxone in MeCN at rt as described in Synthetic Communications vol 31 (2001) 2021 looks easy to use and should not disturb the other substituents details here


text:
What are methods of identifying which amino acids are in a sample of protein?
Are there any methods out there to identify which amino acids are in a sample of protein? Perhaps using NMR?

Can it be a destructive technique?

A destructive technique works fine, too.

What sample? One protein? And you'd only wish to know which rests of aminoacids are there? There a lot of methods for this, main one being just check in database. Seems you're kinda underestimating how much data about proteins is there already.

For a destructive technique that works with any mixture of proteins (i.e. to assess dietary questions), you could first hydrolyze the proteins into the amino acid building blocks. Then, you could separate them using chromatography (TLC, HPLC, GC), either as they are or after derivatizing functional groups to make them less polar.
If you are asking about a sample containing a single protein coming from an organism that has been sequenced, you could also enzymatically hydrolyze the protein (for instance with trypsin) and then subject fragments to mass spectroscopic analysis. A database search would identify the protein
For non-destructive techniques, you could start out with a natural abundance 2D NMR experiment (1H 15N HSQC), which gives you limited information (e.g. number of Asn/Gln). If you are interested in aromatic amino acids, the UV spectrum will be of interest. If you want to know whether there are free cysteins, there are coupling reagents you could use.

I'm curious whether a single crystal diffraction is ever used for amino acids identification. Or is it an overkill here?

Protein crystallography is too time consuming to use merely for amino acid identification. However, there a numerous examples where a pre-determined sequence was corrected by a medium-to-high 3D structure, especially back when DNA sequencing was less well developed and before proteins were over-expressed from a well-characterized gene. Also, amino acid modifications are sometimes detected by crystallography (mass spectrometry is usually easier and faster if you already suspect a given modification).

Another technique is Shotgun mass spectrometry. For this, protein is digested with the protease (e.g. trypsin) and peptides are identified by mass / charge. Knowing the mass of peptides, the sequence of them can be found as well as the sequence of the whole protein. The last one will tell you about the exact amino acid composition of your protein.


text:
How to calculate the standard Gibbs energy at equilibrium?

Given that $K_c = 1.7 \times 10^{-13}$, calculate $\Delta G^{\circ}$ for this equilibrium mixture at $\pu{298 K}$.
$$\ce{N2O (g) + \frac{1}{2}O2 (g) <=> 2 NO (g)}$$

I've calculated:
$$
\begin{align}
\Delta G^{\circ} &= -RT \ln K \\
  &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln (1.7 \times 10^{-13}) \\
  &= \pu{72.8 kJ mol-1}
\end{align}
$$
The result I'm expecting is $\pu{68.9 kJ mol-1}$. Where have I gone astray?

I think the author of this problem forgot to add units to $K_c$ since it's not a dimensionless entity:
$$K_c = \frac{[\ce{NO}]^2}{[\ce{N2O}][\ce{O2}]^{0.5}}$$
and should be $K_c = \pu{1.7e-13 mol^{0.5} L^{-0.5}}$. In general
$$[K_c] = \mathrm{dim}(c)^{Δn}$$
where square brackets denote the dimensions of the quantity $K_c$, $c$ is concentration and $Δn$ is the difference in the amounts between gaseous products and reactants, e.g. here
$$Δn = 2 - (1 + 0.5) = 0.5$$
On the other hand, the equilibrium constant $K$ you use for determining the standard Gibbs energy, must be dimensionless. Since we are dealing with gases only, the easiest way is to use $K_p$, which is dimensionless as required (when normalized to the standard state of pressure $p^\circ = \pu{1 bar}$):
$$K_p = \frac{\left(\frac{p(\ce{NO})}{p^\circ}\right)^2}{\left(\frac{p(\ce{N2O})}{p^\circ}\right) \left(\frac{p(\ce{O2})}{p^\circ}\right)^{0.5}}$$
so that in general
$$[K_p] = \mathrm{dim}(p)^{Δn}\cdot \mathrm{dim}(p^\circ)^{-Δn}$$
$K_p$ and $K_c$ are related (via the ideal gas law):
$$K_p = K_c (RT)^{Δn}$$
So, the equilibrium constant is
$$
\begin{align}
K &= K_p(p^\circ)^{-Δn}\\
  &= K_c(RT)^{Δn}(p^\circ)^{-Δn} \\
  &= \pu{1.7e-13 mol^{0.5} L^{-0.5}}\cdot(\pu{8.314e-2 L bar K−1 mol−1}\cdot\pu{298 K})^{0.5}(\pu{1 bar})^{-0.5} \\
  &= \pu{8.5e-13}
\end{align}
$$
Note that here I used gas constant  expressed as $\pu{8.314e-2 L bar K−1 mol−1}$ since in this case all dimensions are cancelled out and $K$ is left dimensionless. Now we can finally find the standard Gibbs energy:
$$
\begin{align}
Δ G^\circ &= -RT\ln K\\
  &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln\left(\pu{8.5e-13}\right)\\
  &= \pu{68.9 kJ mol-1}
\end{align}
$$
Here the product before the logarithm includes $R = \pu{8.314 J mol-1 K-1}$ to get answer in $\pu{kJ mol-1}$ straight away.

Yes, that definitely does the trick. It also explains why my method worked for another calculation, where $\Delta n$ happened to be $0$.

That’s right that expression is written using Kp instead of Kc so you must use the equivalence Kp = Kc(RT)^(np-nr)


text:
Where can I find reliable data for melting points of organic compounds?
I'm interested in the melting/freezing points of organic compounds, but the online literature has data that's all over the map. I can understand that especially for isomers of alkanes, pure samples can be difficult to obtain; but as basic a property as melting point ought to be well-established and easy to reference.
For example, the m.p. listed for 3-methylpentane varies from -118 °C to -163 °C depending on source. That's an enormous spread; in fact, those two numbers reoccur frequently enough that I suspect that sources are quoting older sources possibly back to the early 20th century. And that's far from the only example I could cite.
Are there any reliable sources that have experimentally confirmed the values?

CRC handbook of chemistry and physics: citable and reliable. 97th edition lists 3-methylpentane on p. 3-386 with m.p. -162.89(0.05) °C; b.p. 63.3(0.5) °C.

PubChem and Wikipedia also have almost -163 °C. They aren't 100 % reliable but usually values should be OK.

@andselisk - my 56th edition CRC did not list the melting point. Sigh - perhaps time to finally upgrade...

I think following open access melting point database will be helpful- https://pharmadigests.com/melting-point-database/

You are absolutely right about suspecting the reliability of sources, based on their value spread. I think, I found what's going on with those internet (I assumed) sources. Yet, I know everybody agree with me that CRC handbook of chemistry and physics is The Bible for Physical Constants of Organic Compounds. Here is what I found on 2005 Internet Version of CRC handbook of chemistry and physics:
3-Methylpentane (CAS #: 96-14-0): physical form: liquid; Melting Point: $\pu{-162.90 °C}$; Boiling Point: $\pu{63.27 °C}$; and Density: $\pu{0.6598 g/cm^3}$ (pp. 3-398 of Ref.1)
Perfluoro-3-methylpentane (CAS #: 865-71-4): physical form: liquid; Melting Point: $\pu{-115 °C}$; Boiling Point: $\pu{58.4 °C}$; and Density: not listed (pp. 3-458 of Ref.1)
I think some reference sources may have mistaken with values of perfluoro-3-methylpentane as those of 3-methylpentane.
Nonetheless, evem PubChem listed two other values than $\pu{-162.90 °C}$ for melting point of 3-methylpentane (Ref.2): $\pu{-162.90 °C}$ from Lide, D.R., G.W.A. Milne (eds.). Handbook of Data on Organic Compounds. Volume I; 3rd ed.; CRC Press, Inc.: Boca Raton, FL. 1994., p. V4: 3865 (most trusted value since it is PEER REVIEWED - https://toxnet.nlm.nih.gov/cgi-bin/sis/search2/r?dbs+hsdb:@term+@rn+@rel+96-14-0); $\pu{-118 °C}$ from http://www.ilo.org/dyn/icsc/showcard.display?p_version=2&p_card_id=1263 (International Chemical Safety Cards (ICSC) are data sheets intended to provide essential safety and health information on chemicals in a clear and concise way); and third value of $\pu{-245 to -148 °F}$, basically, $\pu{-153.9 to -100 °C}$, from  OSHA Occupational Chemical DB (http://www.osha.gov/chemicaldata/chemResult.html?RecNo=764). I can only speculate that they all listed wrong values because they got hits for some other substituted 3-methylpentanes with search keyword of 3-methylpentane. For example, CRC Handbook of Chemistry and Physics listed $\pu{118.27 °C}$ for boiling point of 3-ethyl-3-methylpentane (CAS #: 1067-08-9).
Yet, only CRC Handbook of Chemistry and Physics consistently listed $\pu{-162.90 °C}$ as the melting point of 3-methylpentane since 1994. Thus, I'd take that value at any time.
References:

"Section 3: Physical Constants of Organic Compounds", in CRC Handbook of Chemistry and Physics, Internet Version 2005, David R. Lide, ed., http://www.hbcpnetbase.com, CRC Press, Boca Raton, FL, 2005 (pp. 3-398).
https://pubchem.ncbi.nlm.nih.gov/compound/3-methylpentane#section=Melting-Point

It's a stretch from "many chemists regard the CRC handbook as a reliable source of data" to "I know everybody agree with me that".  Still, good research...


text:
Will fruits and vegetables be more oxidized if blended at high speed for a long time or at less rpm for a short period of time?
Will fruits and vegetables be less oxidized if blended at higher speeds for shorter periods of time, or at lower speeds for longer periods of time?
Do the oxidized molecules in vegetable juice oxidants?

What is your last sentence supposed to mean?

Oxidation occurs due to exposure to oxygen in the air (or in the case of vegetables, also due to cell damage followed by enzymatically catalyzed breakdown processes). If you moosh a banana it will oxidize faster than an unmooshed one, thus you would suppose blending it for an extended time will increase the extent of the oxidation process.

Ok, granted this is actually not a trivial question, as food scientists would surely note. In general you can slow the rate of oxidation by keeping things cold (as one posted answer indicates), and mechanical rupture with a blender will heat your fruit, elevating the temperature. My guess is the oxidation is very fast, on the order of seconds, so that if you want to achieve a certain level of homogeneity you're better off doing it cold and slow (but not too slow). You can also add lemon juice or vitamin-C to impede oxidation.

The breakdown due to enzymatic activity is more difficult to interfere with and I'd have too look into this more. In the lab all sorts of enzyme inhibitors are available to interfere with this.

The higher the RPM and the longer the time, the faster your fruits will be oxidized
The oxidation is a chemical reaction that occurs when something reacts whit oxygen (a molecule, an atom, etc...) and as any chemical reaction the more divided the reactants are, then the faster the reaction will be.
That’s why usually the detergent for the washing machine comes as a powder, so it can be easily dissolved, while the soap are in a solid and robust bulk, so it can be used in a controlled way for a longer period of time. In your case I’m sure that the higher the rpms for blending the fruit, the smaller the pieces will be. And if in addition you let them exposed to the air for a long period of time, then they’ll surely oxidize really fast. So I will recommend you to blend them at slow rpm and once they are small pieces, protect them, keeping them inside a closed recipient in the fridge. The important thing here is that if it’s possible don’t turn the fruit into puree, if you still can use them as small pieces and is very important that the fruit does not gets oxidized, then I suggest you to use a knife or something similar. If oxidation is not critical, then just blend them and use them as quick as you can, you won’t notice the difference.


text:
Why degree of dissociation/ionisation affected by dilution?
What happens when we dilute, say, an acid which is weak, and then add water to it? What motivates or pushes it to dissociate more?
We are just adding some molecules of water, right? So, what exactly happens to a weak acid like $\ce{CH3COOH}$ so it starts to dissociate more?
Is it because of the fact that water being a polar solvent having high dielectric constant starts to split the $\ce{O-H}$ bond in $\ce{CH3COOH}$, or something else?

This is not a homework question 1st of all.And I wasn't able to grasp the concept that why Ostwald's dilution law comes so.

Thanks @andselisk for improving the question while retaining the ideas of the OP.

@AbnerAlfredThompson The idea of a "homework question" on this site is a question which doesn't show the person's efforts to solve the problem or express why they think some given answer is incorrect.

Well then now it does, doesn't it?So,can you please put it off hold or I need to state something more to convince you......

What motivates or pushes it to dissociate more?

Perhaps surprisingly, if you could look at an individual acid molecule, it does not (or hardly) changes its behavior upon dilution. It just dissociates with a certain rate, yielding the conjugate base and a hydrogen ion. There is no "extra motivation" to dissociate, and no "tapping on the shoulder" pushing it to react.

Why is the degree of dissociation/ionisation affected by dilution?

The reverse reaction, where a dissociated acid (the conjugated base) reacts with a hydrogen ion to make the acid again, is affected by dilution. Here, two molecules have to come together, and this takes longer when the solution is more dilute.
This description is for the simplest case, where the net reaction not only describes the stoichiometry but also the mechanism of the reaction. However, even if there are intermediates, at equilibrium the forward and reverse rates are equal, and when you dilute solution, rates of the reverse steps would be significantly lowered, leading to a higher degree of dissociation.

Why so?Why when we are diluting it,it is taking longer to come together.I would guess that the opposite should happen as there are more water molecules so the chances that say a CH3COO- ion attracting an H+ from it's surrounding water should increase shouldn't it.Also,can this be the possible explanation that since we are adding more water to it and since water has high dielectric constant so it'll ionise CH3COOH more.Also,since water is more so it'll hydrate the broken ion and surround it,make it harder for them to recombine.

Try adding your argument why Ostwald’s law is counterintuitive to you to your original question. If you do, the question might be reopened.

I started a new question with two specific scenarios to give others a chance to address the ideas the OP shared in this question and in these comments.


text:
Why are pi antibonding orbitals lower in energy than sigma antibonding orbitals?
I know when a carbonyl group is attacked by a nucleophile, the electrons of the nucleophile enter the antibonding π orbital of the $\ce{C=O}$ bond, which breaks the π bond. My textbook says that it enters the antibonding π orbital instead of the antibonding σ orbital, because the antibonding π orbital is lower in energy. This is my point of confusion: why is it lower in energy?

Related: How can antibonding orbitals be more antibonding than bonding orbitals are bonding?

When two atoms form a bond, if two atomic orbitals mix, then two molecular orbitals are created, one of lower and one of higher energy than that of the atomic orbitals.
Now, just how much the energy changes is dictated by how well the atomic orbitals overlap. Overlapping depends on three things:

Orbital energy
Distance between atoms
Orbital symmetry (they must be "aligned")

Now, assuming that the atomic orbitals are similar in energy and the symmetries are correct, which is the case for C and O.
The sigma bond is a result of two orbitals very close together, directly facing each other frontally. The overlap is great. The pi bond emerges from two orbitals overlapping side-by-side, which isn't as efficient because it happens at a greater distance.
Here is a pic to exemplify:

Thus, not only is the bonding combination of a sigma bond lower in energy, but also the antibonding combination is higher in energy.

Correction: s orbitals could bond with f orbitals, if they are aligned right. // 'assuming that the atoms are similar in energy' I assume you mean the orbitals of the atoms? // This is a very simplified viewpoint. Sigma orbitals are lower in energy, because the maximum of the electron density is aligned on the bonding axis, which is not the case for pi, delta, phi, ... orbitals (where it is exactly 0).

@Martin Thanks for the corrections. Indeed it is a very simple explanation. I felt the need to write the answer in a very basic, introductory way. Once OP understands the core concept he can perfect it by reading more about the topic.


text:
What are nanoparticles made of?
No, really. I can't seem to find anything that properly explains this concept. What are nanoparticles? Are they charged metal ions held together by bridging ligands, or...?

Nanoparticles can be made of pretty much anything, like ordinary particles.

Nano- implies an average size of  $10^{-9} m$

I was referring more specifically to colloidal metals (i.e. silver, gold). What are those made of?

You are still asking a wrong question. The literal answer is bound to be useless. Particles of colloidal metals are made of metals; nothing more general can be said.

The prefix nano is about size not the chemical makeup of the particles
Nanoparticles are particles of anything. As long as that thing can be made in a form where the particles are of the order of nano-meters (10-9m). Since atoms have sizes of around 100pm or 0.1 nm(well, 20-200pm) this implies that very crudely nano particles consist of blocks of atoms with about 10 atoms per side or ~1,000 atoms in total.
Some nano-particles are made from molecules, some from single elements and some from clusters of multiple elements. They are interesting as they often have different properties to the bulk material (suspended gold clusters are red and used in colouring some glasses).
Nano particles are often stabilised by having some surface-active molecule on the outside to prevent the agglomeration of the small units. But there are a variety of ways to make and stabilise them. It is a big subject, but the important thing is that the definition relates only to size not to what the particles are made of.


text:
Amount of enantiomers

I got a question about this molecule, and I'm not sure how to answer it. I know that this one has 4 chiral C-atoms, but I can't seem to get the number of enantiomers right. I know that the formula for the amount of enantiomers is $2^{n-1}$, with $n$ being the amount of chiral centers, which is $4$. So that makes an amount of $8$ enantiomers. But that's not what it says in the answer.
Am I doing something wrong here?

Try making combinations of R and S nomenclature on each carbon atom and see the distinct cases

The formula is useless if you don't know when to use it. And that formula only works if there some level of symmetry in a linear molecule. Frankly, you will find formulas of this type to be less helpful in organic chemistry.

The compound in question is called D-galactose in carbohydrate chemistry. It has 4 chiral centers (C2, C3, C4, & C5): Thus, stereochemistry of them are $(2R,3S,4S,5R)$. The rule is if a compound has $n$ chiral centers, then it has maximum total of $2^n$ possible stereoisomers, if it does not have any symmetry. D-galactose does not have any symmetry within the molecule (e.g., It would have symmetry if aldehyde group, $\ce{-CHO}$ has been reduced to $\ce{-CH2OH}$). Thus, four chiral centers of D-galactose are accounted for a total of 16 stereoisomers ($2^4$). If you draw them all, you would find they are collection of 8 pairs of enantiomers as:

{$(2R,3S,4S,5R)$, $(2S,3R,4R,5S)$}
{$(2R,3R,4S,5R)$, $(2S,3S,4R,5S)$}
{$(2R,3S,4R,5R)$, $(2S,3R,4S,5S)$}
{$(2R,3S,4S,5S)$, $(2S,3R,4R,5R)$}
{$(2S,3S,4S,5R)$, $(2R,3R,4R,5S)$}
{$(2R,3R,4R,5R)$, $(2S,3S,4S,5S)$}
{$(2S,3R,4S,5R)$, $(2R,3S,4R,5S)$}
{$(2R,3R,4S,5S)$, $(2S,3S,4R,5R)$}.

That make maximum total of 8 pairs of enantiomers, meaning $2^3 = 2^{4-1} = 2^{n-1}$, where $n$ is number of chiral centers. Therefore, your formula should have been for pair of enanthiomers.

That was very helpful, I understand now.

@Guido If you are happy with the answer, feel free to accept it:)


text:
Is there any difference between isotonic and isosmotic solution?
Is there any difference between isotonic and isosmotic solution? Can we say isotonic solution do not show osmosis because they have same osmotic pressure?

I think often you can use those terms interchangeably but they aquire different meanings in some contexts. According to the current entry in the wikipedia, an isotonic solution does not alter the solvent concentration in a second solution, the standard example being the interior of a red blood cell. In contrast an iso-osmotic solution may do so.

Yes, there is a difference. Isotonicity implies a biological compatibility, whereas isoosmoticity implies similarity of chemical and/or physical composition. Solutions that are isoosmotic to biological fluids/blood are not necessarily isotonic as tonicity refers to a given cell membrane [1, p. 230]:

The 0.9% sodium chloride solution is said to be iso-osmotic with physiological fluids. In medicine, the term isotonic, meaning equal tone, is commonly used interchangeably with iso-osmotic. However, terms such as isotonic and tonicity should be used only with reference to a physiological fluid. Iso-osmotic actually is a physical term that compares the osmotic pressure (or another colligative property, such as freezing-point depression) of two liquids, neither of which may be a physiological fluid, or which may be a physiological fluid only under certain circumstances. For example, a solution of boric acid that is iso-osmotic with both blood and lacrimal fluid is isotonic only with the lacrimal fluid. This solution causes hemolysis of red blood cells because molecules of boric acid pass freely through the erythrocyte membrane regardless of concentration. Thus, isotonicity infers a sense of physiological compatibility where iso-osmoticity need not. As another example, a chemically defined elemental diet or enteral nutritional fluid can be iso-osmotic with the contents of the GI tract, but would not be considered a physiological fluid, or suitable for parenteral use.

References

Gennaro, A. R. Remington: The Science And Practice Of Pharmacy, 21st ed.; Lippincott Williams & Wilkins: Philadelphia, 2005. ISBN 978-0-7817-6378-3.


text:
Why do colligative properties depend only on number of solute particles?
Colligative properties depend solely on the number of even though the interactive forces are different for different solute-solvent pairs. So why is the dependence only on the number of solute?

I'm a little uncertain what you are asking about specifically. But, in a solution (taking a per mole of solution basis for the thermodynamics), if there is a concentration $x$ of solute, there is a concentration $1-x$ of solvent. You only need one number ($x$) to specify the composition of the solution.

A very simple, qualitative explanation:
After your solute has dissolved, there are no more enthalpic effects to take into account. The solvation enthalpy is converted to a temperature change, and that's it. For colligative properties, the solute ideally has no significant own vapour pressure, does not precipitate, it just stays in solution.
The solute molecule moves around (Brownian motion), the solvent molecules around it exchange places, but are all the same, and that's that.
Now everything in solution is about entropy. In a solution, there is only orientational and translational entropy. (Molecular) Vibrational excitation is only relevant at high temperatures, rotation is practically imposible due to the many interactions.
That's why every solute particle, irrespective of its size, has the same contribution to the entropy.
If the solute concentration inceases, its particles start interacting. Their coming close to each other and detaching again makes local energy fluctuations, which add entropy, but depend strongly on the kind of interaction. That's when the identity of the solute starts to matter.

So, basically the statement valid as long as the entropy changes due to solute-solute interactions are negligible?

solute-solute interactions do not lead to a violation of colligative property expressions, unless their occurrence leads to a violation of Raoult's law behavior for the solvent.

It's a lie. Colligative properties do depend on the chemical nature of the solute and solvent - their interaction. The trick is: an ideal solution, in which the solute-solute intermolecular forces and solute-solvent interactions are more or less of the same character, has the property that colligative properties are dictated by composition given only in terms of amount of moles. However, any sufficiently diluted solution will behave somehow ideally, that is where the statement goes: for diluted solutions, colligative properties depend only on the number of moles of solute.

This explanation is no better than just believing in the definition. If it's about the nature of the interaction, why isn't the property proportional to at least the effective surface area (whatever that exactly is) of the particle? Why $n$, no matter if the particle is a Li ion or a 100kg/mol polymer molecule?

You are correct. A more rigorous explanation would require a very deep look into the thermodynamics and why it is usually formulated in terms of molar fractions instead of mass. One would have to analyse very carefully the derivations of equations such as Raoul's law to understand why molar fractions show up - we can get a hint from the simple ideal gas equation pV = nRT, which uses number of moles and not mass (or surface area like you said). By working with Raoul's law it is easy to deduce that boiling point rise for example of a nonvolatile solute depends on solvent molar fraction.

Colligative properties are defined as solvent properties that display a linear dependence on the mole fraction $x_2$ of solute in solution, but are otherwise blind to the specific nature of intermolecular interactions. The key condition for such properties to be observed is that Raoult's law (one definition of an ideal solution) holds. Colligative behavior is purely statistical. For colligative properties - including freezing point depression, boiling point elevation, and osmotic pressure - the change in the property of the solvent can be expressed as $\Delta J=K_J x_2$, where $K_J$ is regarded as a constant for the solvent at the given temperature and pressure, independently of the identity of the solute. The relations assume explicitly that the solute concentration is low ($x_2\lt\lt1$). In practice, the requirement of dilute solutions is often necessary to ensure that Raoult's law is observed.
Raoult's law implies that the chemical potential of the solvent, $\mu_1(l)$, can be written as
$\mu_1 (l) = \mu_1^*(l)+RTlnx_1$
where $\mu_1^*$ is the chemical potential of the pure solvent at the same temperature and pressure. This equation is the basis for deriving the colligative property equations.
It should be emphasized that it is necessary to ensure ideal solution behavior - that Raoult's law is observed - in order to implement the mathematical formalism associated with colligative properties. In that sense this is a functional definition: one searches for systems that match the definition and allow application of the formalism.
When are these conditions met?
Quoting Atkins' physical chemistry textbook (Freeman, 4th Ed.):

some mixtures obey Raoult's law very well, especially when the
  components are chemically similar. [emphasis mine]

And while I'd rather you took my word for it, you can always check another tertiary source, the wikipedia:

In chemistry, colligative properties are properties of solutions that
  depend on the ratio of the number of solute particles to the number of
  solvent molecules in a solution, and not on the nature of the chemical
  species present.1 The number ratio can be related to the various
  units for concentration of solutions. The assumption that solution
  properties are independent of nature of solute particles is only exact
  for ideal solutions, and is approximate for dilute real solutions. In
  other words, colligative properties are a set of solution properties
  that can be reasonably approximated by assuming that the solution is
  ideal.

From what I read at https://chemistry.stackexchange.com/questions/33896/difference-between-ideal-ideal-dilute-solution, it would be sufficient to consider an ideal-dilute solution.

Raoult's law sometimes holds in the dilute regime even when it doesn't over the entire range of mixing. That is why I emphasize the dilute part, even though the ideal part is also important.


text:
Is there an element with no liquid state?
I know this question might be dumb but, is there an element that will go directly from solid to gas, with almost no visible or no liquid state? Just a random curious question.
I am asking if there is an element that sublimes at normal pressure and temperature.

@Danii well $\ce{CO2}$ sublimes if the pressure is below 5.1 atm.

Wait? Temperature? You're getting it even more muddled. You can pick only one variable here otherwise sublimation would need to be in this exact point. Carbon heated at 1 atm. to over 4000 K sublimes not melts. Arsenic does so at over 600 deg. C

Even though this may not be the OP's intention, I hold that is question is very relevant. I'm thinking about what's the requirement upon the interaction potential to allow a liquid state; from a thermodynamics viewpoint, this is barely a trivial question.

The answer to your question is no, there is no element with no liquid state. The state of an element (or molecule) depends on the temperature and the pressure. All elements can be in any of the three states (solid, liquid or gas) with the right temperature and pressure. We say that oxygen is a gas because at standard temperature and pressure it is in gas state, but if the temperature is low enough it will condense into a liquid and if temperature is even lower it will become a solid bulk. The opposite happens when increasing temperature but of course from solid to gas.
There is a phenomenon known as sublimation, this is when a substance (carbon dioxide for example) goes directly from solid to gas, leaving no trace behind. This doesn’t mean that you cannot obtain it as a liquid, it means that at our normal pressure and temperature it will pass directly to gas, but if the conditions are different then you could have it as a liquid. I know only two elements (pure elements) that can sublimate: iodine and arsenic. But again it does not mean that they don’t have liquid state

The answer is a bit misleading. Solid iodine will sublime as $\ce{I2}$ at "normal pressure and temperature," but at normal pressure arsenic must be heated to around 600 C. // user Mithoron also mentioned that carbon will sublime. Not sure what allotrope or at what temperature.

Technically, carbon sublimes at atmospheric pressure.  If you can get it roughly as hot as the surface of the Sun.

Also there is the point about pressure and the liquid phase. To my knowledge no element has any allotrope for which the critical temperature (gas to liquid limit) is lower than the melting point (solid to liquid). Thus there is some temperature and pressure at which every element will form a liquid.

What about sulfur?


text:
Why does coke foam up when I add it to (cold) black coffee (no sugar, no milk)?
Album of the different gifs detailing this situation: Album of coke-coffee related gifs
I thought that the answer would be similar to the ice-cream float situation, but I don't see what in the coffee would cause the soda to float.
Could it be some reaction with the coffee oils and the coke?

Above is a shot of my materials:

Can of Coke Zero (ingredients: Carbonated water, caramel color,
phosphoric acid, aspartame, potassium benzoate, natural flavors,
potassium citrate, acesulfame potatssium, caffeine)
Nespresso coffee    (ingredients: Roast and ground coffee, hot water)

Above is after the coffee cooled.
Experiments:
Here is a link to a gif of me pouring the coffee:
Gif of me pouring coke into the coffee
To address one of the comments, here is a gif of me pouring coke into an empty bottle: Gif of me pouring coke into an empty bottle, comparison
To address another comment, link to me pouring coke into a mug of tea (instead of coffee): Gif of me pouring coke into a mug of tea

Doesn't coke fizz when you add it do just about anything else, be it pure water or an empty glass?

I should try it; it looks foamier than usual. I can add a pic when I do it again.

I have to wait for the coffee to cool off. I'm going to put it in the refrigerator for about 15 minutes

Actually, I've heard there are some recipes among the students to prepare some sort of a wake-up cocktail with various ratios of both drinks, and the key factor is to get as much foam as possible, which, according to the myth, is what gives the caffeine kick. I can confirm that foam on the surface of the cola-coffee mix is denser and more stable and there is more of it. Interesting question:)

Coffee is inhomogeneous  - reasons for "more fizz" are exactly the same as in  adding mentos, ice-cream of pretty much whatever else.

@Mithoron I think you are on the right track. Even if you don't add anything but just shake the coke before opening the can, the bubbles on the side of the can act as nucleation sites. Maybe the coffee has some solids in it or some oils that can act as nucleation sites. The equilibrium of coke is on the side of flat, i.e. carbon dioxide in the atmosphere rather than in the solution, and foaming is a kinetic phenomenon.

@Mithoron here's a link to me adding coke to tea: Celestial Mandarin Orange spice using a Keurig https://imgur.com/G4Ef88W

@IvanNeretin please look at the following gif of me pouring coffee into tea: https://imgur.com/G4Ef88W

@Mithoron I'm finding that just doing a google search on "is coffee a homgeneous mixture" yields results saying that it is a homogeneous mixture. What do you mean by inhomogeneous?

@Nonsingular Exactly what it means. Even instant coffee leaves  some solid particles behind and even colloidal particles can act as nucleation sites.

There are two likely contributing factors here. What you're seeing is rapid release of carbon dioxide from solution, indicating that its solubility has decreased and that the rate of re-equilibrating to that new lower solubility is quite fast. The solubility of carbon dioxide is a factor of
1) the partial pressure of CO2 in the air above the liquid (higher pressure forces more into solution),
2) the pH of the liquid (because higher pH leads to formation of more $\ce{HCO3-}$ and $\ce{CO3^2-}$ thus increasing the total CO2-related species in solution), and
3) temperature of the solution (like most gases, CO2 decreases in solubility as temp increases).
In your case, (1) and (3) are probably most important. [EDIT: I overlooked that the coffee was cold, so (3) is actually not likely to be relevant] The can before opening is pressurized with a high CO2 atmosphere, so it is overloaded with CO2 relative to just about any ambient condition. In order to equilibrate with ambient conditions, it needs to lose some CO2. On the other hand, you've diluted it into the coffee, so there is a slight offsetting effect there. The coffee is slightly acidic, but is not likely to be very well buffered, and probably is not much different than diluting with water.
In addition to these thermodynamic effects that change the equilibrium dissolved CO2 concentration, there is also the kinetic issue of how quickly the CO2 escapes in order to return to equilibrium. The CO2 can only leave solution at the boundary points where the liquid meets air. The stirring effect of pouring the liquid (even into an empty glass) creates more surface area of the liquid/air boundary and also keeps the CO2 well distributed throughout the solution so that the areas near the boundary are continuously resupplied with CO2 as it is lost to the air.
To determine if coffee specifically has any effect, I would try the same thing with water instead of coffee. I suspect the result will be very similar.
UPDATE based on new data:
I was wrong about the coffee itself not being important. The coffee mixture clearly foams more than the tea. There are two possibilities I can think of (not mutually exclusive):
1) the rate of CO2 evolution is greater in coffee. As the other answer postulates, that could be due to faster nucleation by particles in the coffee or, if the pH goes up very quickly, buffering by acids in the coffee once the pH reaches 5 or so (from the Coke starting point ~ pH 2).
2) the same amount of CO2 is released, but the coffee mixture foams more. Studies on sparkling wines such as the one described here indicate that proteins, especially ones with carbohydrates attached, contribute to foaming by acting as surfactants (that means like a detergent molecule, essentially making soap bubbles). It is likely that coffee contains similar molecules. I now think this is the likely explanation, though I also like the idea that there are more nucleation sites from the dissolved solids.
The way to test some of these hypotheses would be to filter the coffee with varying molecular weight cut-off filters, but I suspect you don't have access to that sort of equipment?

Hi I appreciate the chemistry you're laying down on me! To note: I cooled my coffee so that at best it was lukewarm. Usually I add coke when my coffee is stone cold and I do see the same effects.

I can repeat the experiment above with hot water to address your point.

Just use water at whatever temp your coffee usually is in order to see if there's something in the coffee that is having an effect. I just assumed the coffee was hot even though I now see that in your title you said it was cold. I should read more carefully.

I added https://imgur.com/G4Ef88W and https://imgur.com/a/QFasLFg to compare the different solutions

Coffee contains organic acids, notably 3,4-Dihydroxy-cinnamic acid, in German Kaffeesäure. (same name in English, https://en.wikipedia.org/wiki/Caffeic_acid)
Acids make coke foam, but then coke already is extremely sour, so adding coffee probably doesn't make a difference.
If you leave a cup of coffee overnight, and then slowly pour it in the sink, you will see it contains a lot of small solid particles. Those act as very efficient nucleating agent for bubble formation.

Typically pH of coffee according to this site :https://www.sheltondentistry.com/patient-information/ph-values-common-drinks/ is well above the pH of the coke itself. And the coke has so much carbonation that it is likely much better buffered than the coffee, so adding coffee will have little effect on pH.

@Andrew Yea, coke is sour as hell. I don't agree on the buffering, because I believe it contains a lot more acid than CO2.

@Karl I added a gif: https://imgur.com/G4Ef88W showing me pouring coffee into tea. The same argument for nucleation should work, however, experiment shows otherwise

@Nonsingular Coffee into tea? That sure tastes disgusting, but what's in there that could foam? :-))

Sorry, that was a typo: it should read "coke into tea"

Any other solutions that could help with whether particulates in a solution causes the foam that I'm recording?

@Nonsingular Your tea contains a different amount of differently shaped particles, you cannot expect the same effect as with coffee. Also coffee can make a stable foam, caffè crema. Tea not really.

You bring up a good point and brings up a few questions. What causes the nice foaming "crema" in espresso? Do other stable foaming solutions provide the same effect as coffee when coke is mixed with it?

Coke (or your high pressure espresso machine) delivers the gas for the foam, the oils etc. in coffee are the chemicals that deliver stable walls for the bubbles!


text:
Transferring suspended substance from isopropyl alcohol to water?
I have this UV-reactive invisible ink (a clear liquid.) According to the data sheet, it is suspended in an isopropyl alcohol base. Unfortunately, this is terrible for my particular application for two reasons: (1) it dries too fast (I need it to stay wet on a rubber stamp for ~15 minutes before stamping.) (2) It is too "thin" and thus bleeds into the porous material a little too much.
I've used regular black ink refills that (I believe) are water-based and they are perfect (except for the visible, non-UV-reactive part.)
If I understand this stuff correctly, the UV ink would be UV-reactive particles suspended in isopropyl alcohol base. If I were to mix this with water to get a workable liquid, I would likely dilute the ink too much.
Is there a way to transfer the suspended particles from the isopropyl-alcohol to water? Perhaps by mixing them, letting them separate and then let the alcohol evaporate off the top?
Alternatively, to reduce dilution I've considered finding a clear viscous liquid (think simple syrup without the stickiness) or even a clear gel as a thickening agent. Unfortunately, I don't know of a good viscous liquid or gel that would evaporate without a visible/sticky residue. Possibly one of the inactive ingredients in a hand sanitizer gel?

I suspect that suspended is the wrong term. To chemists suspended means that solid particles are "floating" in the liquid. I'd rather suspect that the compound was dissolved in the isopropanol.

isopropyl alcohol is miscible with water, so a mixture of water and isopropyl alcohol won't separate, which is what I think you're suggesting. however, if the ink is soluble in water or a water/alcohol mix, you might be able to evaporate off a good bit of the alcohol and restore the volume by adding water. On a separate point, glycerol and propanediol (1,2 or 1,3) are used as thickeners in hand sanitizers, but they would all leave a residue on paper.

After asking this question, I researched the ingredients in hand sanitizer gel and recognized glycerin as a possibility. I tried a series of tests, mixing with water, mixing with hand sanitizer gel and mixing with glycerin (vegetable glycerin, according to the package) and all tests failed miserably except for the glycerin, which worked admirably well.

Slowly drive off the alcohol in a vacuum or by gentle heating. Replace the lost alcohol with distilled water. Experiment with small volumes first to ensure that heat does not degrade the ink.


text:
Can i breathe pure oxygen gas?
Sorry for the irrelevant tag because I was unable to find An appropriate one .Suppose there is a vaccum chamber and i filled completely with O2 without any traces of any other gases . Can a normal person breathe in it comfortably. I suspect it as pure O2 may act as oxidiser but since we need it for breathing . Can our organs manage that.

Can you? Yes. Comfortably? No, to the point that you'd probably get oxygen poisoning and die. Then again, pressure is important.

Like everything in chemistry, the amount and the proportions are very important, it is often the difference between the poison and the antidote. Too much of anything is usually bad. It is true that you can drink water but if you drink too much it could drop your body pressure and be harmful for you. So yes, you can breathe, the real question is for how long?

https://en.wikipedia.org/wiki/Oxygen#Toxicity, last paragraph

Breathing pure oxygen is potentially toxic. At reduced pressures, pure oxygen can be inhaled for days, or weeks. It is used to treat individuals suffering from stroke, heart attack, etc. and can be comfortably used for up to 24 hours. The body can be overwhelmed by oxygen during long periods or higher pressures, free radicals in the oxygen cause damage through oxidative stress which can result in tissue damage and death.


text:
Properties of Water
Water is a substance characterised by properties such as high boiling point and its ability to act as a universal solvent. So why do we say that “the smallest unit of water is a water molecule”? Single molecules cannot possess such properties. Am I missing something here?

You need a lot more water molecules until you have water. A drop of 10,000 will show a huge difference e.g. in vapour pressure. Even a billion would still be quite strongly influenced by it's surface, and not show the ideal bulk water properties.

Not all water is defined the same. "Water" defines a molecule - $H_2O$ - but in our everyday parlance also a bulk substance which is a collection of such molecules. Perhaps it would help to say "the smallest unit of a volume of water molecules is a single water molecule". Not very enlightening perhaps.

Water is a substance. Would it perhaps be more correct to say that a water molecule is the smallest unit that defines the properties of the substance water?

@J.Smith I don't think that's going to solve your question, because you still need to define "properties". The properties in question can depend on how many water molecules you have.

The problem here is that conventional usage of "the smallest unit of water is a water molecule" leaves out a lot of qualifications. The sentence ignores the qualification that some properties are intrinsic to the molecule but others only appear when there are a lot of molecules together. Every molecule in bulk water has the same mass, geometry and electric dipole but many of the bulk properties only appear when there are millions of the molecules (since they depend on how the molecules interact).

No, you aren't missing anything. Indeed, one molecule doesn't have any boiling point, neither high nor low. In fact, you've just stumbled upon one of the holes deliberately left in the early definitions. It is a very common thing in the textbooks on natural sciences: first we give the kids a simplified definition, so they don't have to read a ton of prerequisites before they start reading the book, and then they kinda already know what a molecule is, so why bother returning to it.
If we were to invent a more-or-less realistic definition of a molecule in simple terms, it would go like this: "the smallest unit of matter such that the matter retains its properties when split to these units and then allowed to settle down".
Indeed, say we have a glass of water, and we magically split it to individual molecules and then step aside. The molecules will assemble back into liquid water; probably it will be all over the place, but who cares, it is liquid water all the same, with unchanged boiling point and stuff.
Now what if we magically split a sample of some matter to smaller units, like atoms? Well, as soon as we step aside, those will assemble back to molecules, but not necessarily the same molecules. They may well be different, and so will be the physical properties of the sample.
This definition is not without its flaws either, but that's another story.

Many molecules won't survive being taken into vacuum. ;-)  Imo trying to define real world stuff like molecules in terms of abstract entities like "smallest unit of matter" is generally a bit of a philosophical exercise in futility. You get to statements which are often flawed, true in best case, but all pretty useless in gaining further understanding.

The word "water" can be used to mean different things. In everyday language we usually use "water" to mean a large collection of water molecules, which we associate with such phenomena as boiling, freezing, solubilization of salts, etc. We can conjure a different definition of "water", namely, a single water molecule, with molecular formula $\ce{H2O}$. These are not the same thing. In a way, one is a subset of the other.
The solvation capabilities and boiling point of water are thermodynamic properties, usually (but not only) defined for macroscopic ("bulk") quantities of water under equilibrium conditions. The boiling point defines a temperature at which a transition occurs between the liquid and the gas states of a substance. The boiling point is a property of the ensemble, the entire collection of molecules. It is important to emphasize that when chemists discuss properties such as boiling points they are pretty strict. Such properties depend on the pressure, amongst other things. Even the geometry of the water volume can influence the experimentally observed boiling point. Chemical thermodynamic properties belong by definition to a collection of molecules under well defined equilibrium (constant) conditions. Thermodynamicists observed that properties like solubility and phase transition temperatures reach limiting values provided you have sufficient molecules, that is, above a limit dealing with bulk matter. In fact, a thermodynamic limit can be defined by observing how properties change with the amount of a substance.
It doesn't make much sense to attribute a bulk property such as a boiling point to a single water molecule, but you can say that a single molecule of water has intrinsic molecular properties that lead reproducibly, when the molecule aggregates with others of its kind, to specific macroscopic properties.
On a historical note, properties such as the boiling point were defined quantitatively as the foundations of thermodynamics were layed down and a formal molecular theory of matter began to be developed, the latter involving amongst other things statement of Avogadro's hypothesis and, later, advances in spectroscopy.

Thanks for the response Try Hard. Does that mean that we cannot define a single molecule of water to be the substance “water”? Since water is characterised by the aforementioned thermodynamic properties.

Water is divisible down to the molecular level. It exhibits properties (thermodynamic or possibly other) that may depend on the number of such molecules and their interactions.

Then let me ask this question: What properties characterise a substance? Are they thermodynamic properties or something else? What am I missing here?

The word "substance" typically implies a macroscopic ensemble, a large collection of molecules. A thermodynamic description might be adequate, assuming the substance is in equilibrium, which for a pure homogeneous liquid such as water would mean stating what temperature and pressure it is at. Then you can proceed to list all of the properties you might be interested, extensive (such as volume, enthalpy of formation, etc) or intensive (molar properties such as molar heat capacities, thermal expansion coefficient, etc).

With thermodynamic properties you would not be speaking of one water molecule, but of the properties of the ensemble of molecules. But of course properties can also include spectroscopic properties that are impossible to explain without an atomic/molecular/QM description of matter. When you start dealing with certain spectroscopic properties, you move away from the aggregate properties and shift into the realm of the molecule. The aggregate properties become "perturbations" on molecular properties that are similar in an ensemble and in a single molecule. Often, but not always.

the smallest unit of water is a water molecule

To say that a molecule is the smallest unit of a substance that shares its properties is not a good definition, and has the same problems as defining an atom as the smallest unit that has all the properties of a given element.
For the definition of an atom, you can salvage that definition by specifying exactly what properties are retained. For example, the electronic transitions involving inner electrons will be the same for an isolate atom and for different allotropes of the element (and you can use that to identify atoms from their spectrum even if they are extraterrestrial).
Similarly, you can specify a small set of properties that a bulk sample of a pure molecular substance has in common with a single one of the molecules it contains: Composition, bond lengths and angles (with some minor variations), electronic spectra (more or less).
Of course, if your bulk sample is in the gas phase, there will be more properties that are the same, so pure water in the gas phase or water in air will share most properties with a single isolated water molecule.


text:
Nomenclature of carbohydrates
In the chapter on Biomolecules in my book, the following points are given regarding carbohydrates.

Numbering the C-chain with consecutive rank
The C-atom present in the aldehyde group of aldoses and the terminal C-atom nearest to the keto-group of the ketoses are numbered as the first carbon or C1.
D- and L- Configuration
i) The D- and L- symbols refer to the configuration of monosaccharides. These do not indicate the dextrorotatory or laevorotatory specific rotations.
ii) In case of a monosaccharide, if the asymmetric C-atom of the highest rank is similar to the asymmetric carbon of D-glyceraldehyde (H-atom on the right,-OH group on the left) then the monosaccharide is of D-configuration. If the H-atom is on the left and -OH group is on the right, then the monosaccharide is of L-configuration.
iii) As per convention, if in a monosaccharide unit (expressed by Fischer projection formula), the -OH group falls on the right of the highest ranked asymmetric carbon, then the monosaccharide has D-configuration and if the hydroxyl group falls on the left, it has L-configuration.
Use of + and - sign
D- and L- symbols are followed by '+' and '-' in brackets. These repesent the dextro- and laevo- specific rotation. There is no relation between optical activity and nature of rotation. Example: Both glucose and fructose have D-configuration, but their optical activities are positive (dextrorotatory) and negative (laevorotatory) respectively.

Points ii and iii seem to be contradictory to me. Can someone please shed some light on this. I noticed that point iii is talking about Fischer projection. May be point ii is talking about some about mode of representation. Can someone please explain the paragraph under Use of + and - sign

It's really strange and contradictory info, I don't know what the ii paragraph is talking about. Referencing the source as “my book” makes is hard to verify and e.g. check if there is some illustratory graphics.
To the last paragraph: The cited information would lead to e.g. “D(+)-glucose”, which is not consistent with typographical details of IUPAC nomenclature of carbohydrates (or IUPAC nomenclature of organic chemistry). The correct name would be (+)-ᴅ-glucose.[1]

2-Carb-4.5.  Optical rotation
If the sign of the optical rotation under specified conditions is to be indicated, this is done by adding (+)- or (−)- before the configurational prefix. Racemic forms are indicated by (±)-.
Examples:
ᴅ-Glucose or (+)-ᴅ-glucose
  ᴅ-Fructose or (−)-ᴅ-fructose
  ᴅʟ-Glucose or (±)-glucose

Fructose is an example of laevorotatory ᴅ- sugar (therefore is also called levulose).
The stereochemical configuration (especially when it is of one of many chiral atoms, and in sugars, sometimes it's the one that forms the acetal group of the major cyclic form, in some sugars not) is not directly related to the physical optical rotation.
References:

Nomenclature of Carbohydrates (Recommendations 1996), 4.5. Optical rotation

The '+' or '-' sign is written after 'D-' or 'L-' as per my knowledge. All the articles I have read online follow this.

@MrAP Yes, the mentioned notation is used quite often, though it's not the recommended one.

In the last paragraph, I was facing problem in understanding this line: "There is no relation between optical activity and nature of rotation".

@MrAP I tried to address that – there are many chiral atoms in sugars, but the D/L notation is related to single one only. You did not mention if you are aware of "optical rotation" term, it's a physical unit, rotation of polarized light passing through the sample.

For background, chiral molecules affect plane-polarized light such that the angle of polarization increases or decreases from the reference point. Historically, this was described as rightward or leftward rotation. From the Latin words for right ("dexter") and left ("laevus") we get dextrorotary (D) and laevorotary (L). But when carbohyrate structures were first being elucidated, there was no way to tell which enantiomer went with which rotation, so conventions were established to designate enantiomers as D or L without knowing the actual rotation associated with each.

When modern techniques enabled determination of absolute stereochemistry, the (+) and (-) were added to indicate the actual direction of rotation associated with each structure.


text:
What is ultimate decomposion product of Teflon (PTFE) in environment
There is a lot information available regarding thermal decomposition e.g. burning, but what happens to PTFE in ordinary environmental circumstances? Or if time for degradation is short enough, what happens in households (indoor air) which contains some PTFE impregnated materials? As far as I understood fluorotelomers might be also involved here somehow, but this is maybe due e.g. some dirt repellent product product which contains those fluorotelomers among PTFE.
One answer can be that PTFE is inert and stable and so fort, yes but I assume that some degradation happens still?

PTFE starts degrading above 260°C but I am not aware of any decomposition in the environment. Common plastics take centuries to be degraded and PTFE in much much more stable.

We are yet to see the small planet Pluto to complete its first orbit since its discovery. Likewise, we are yet to see the first sample of PTFE to decompose in natural environment.

@IvanNeretin I'm not sure that is strictly true. It surely decomposes very slowly as do other, more easily dispersed, fluorocarbons. But many other apparently inert plastics decompose more quickly than originally expected.

It is important to distinguish PTFE from other fluorocarbons. The polymer is very inert and won't spread around without considerable deliberate effort. But many fluorocarbon stain relevant chemicals are much more volatile and will spread. But they are usually so inert this isn't a big worry, though some manufacturers have cut down on their use because they do accumulate in the environment.

related https://chemistry.stackexchange.com/questions/34190/is-there-any-chemical-that-can-destroy-ptfe-or-teflon

Carotene, much less stable than PTFE, has been found in shale ~52,000,000 years old. https://www.sciencedirect.com/science/article/pii/0016703787902328


text:
Reaction between NaOH and Carbonic acid
I was presented with the following reaction when we were just learning about acids and bases:
$$\ce{2NaOH + H2CO3 -> H2O + Na2CO3}$$
however I had thought it might've been something like
$$\ce{NaOH + H2CO3 -> H2O + NaHCO3}$$
They both seem to fit the notion of a neutralization reaction creating a salt so how do I know that the first one is correct? The second reaction looks fine to me (is it that bicarbonate is too reactive?) and more intuitive as the carbonic acid just gives a proton away ending up as an ion to bond with $Na$, so how do I distinguish between these kind of cases?

You're right, both of these reactions take place.
From Wikipedia, the Free Encyclopedia

In aqueous solution, carbonate, bicarbonate, carbon dioxide, and carbonic acid exist together in a dynamic equilibrium.

Now, the answer to the question depends on context. If you're asked to do this reaction in a mildly acidic/weakly basic medium, then the answer would be the bicarbonate, while in a strongly basic medium, the answer would be the carbonate.
Interesting for further reading:
This $\ce{pH}$ dependent equilibrium between carbonate and bicarbonate is one of the ways in which the blood $\ce{pH}$, and the $\ce{pH}$ of the mouth is maintained. It also helps aquatic life to survive in ocean waters. (Also see urea cycle and carbon cycle)

@ AbhigyanC What about $\pu{pH}$ at the second equivalence point, more or less than 7?

@AdnanAL-Amleh $\ce{pK_{a1}}$ and $\ce{pK_{a2}}$ of carbonic acid are $4.5\times10^{-7}$ and $4.7\times10^{-11}$, respectively. So it happens at $\ce{pH}=8.34$, using the formula for pH of a amphiprotic anion is the average of the two pK values. See http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt