text:
What is the predominant enol tautomer of 1-phenylbutane-1,3-dione?
In 1-phenylbutane-1,3-dione (1), which enol tautomer will be predominant?

Both carbonyl groups possess α-hydrogens, so can undergo keto-enol tautomerism. In particular, the carbonyl group on the right can enolise in two ways since there are two unique sets of α-hydrogens.
The answer is supposed to be enol 2. Why can't it be either 3 or 4?

You need to retain a conjugated carbonyl group which is more stable than two enols. [Acetone exists as a ketone with little enol.] The real question is why the given answer without considering the tautomeric form of the second structure.

The question editor has modified the question, and maybe assumes that he has thought of every possible enol structure, but he didn't.

@mykhal Hmm, this is kind of two years old, but I merely copied the structures from the previous version of the question. It's in the revision history... Yes, there are missing enol forms as user55119 also suggests, but the original question did not have those either.

Two enol tautomers are possible for 1-phenylbutane-1,3-dione. Both are shown in the figure. Enol tautomer 1 is predominant due to more extended conjugation than Enol tautomer 2 which leads to more resonance structures give extra stability to 1.

The compound is highly stabilised by higher Conjugation and is thus favored over the other enolic form. Another factor is intramolecular hydrogen bonding which stabilises the enolic form even more.
It can also be explained on the basis of acidity of H in keto from.

Due to presence of 2 Electron withdrawing groups, the H atom is highly acidic and will favor the enolic form.
The reason (3) is not formed is because after the formation of (2), the H marked in the 2nd picture will not be acidic anymore, making the equillibrim highly towards (2) form. Also, formation of the allene leads to leads to loss of comjugation, since the pi orbitals are now perpendicular to each other, which makes the equillibrim even backwards (towards 2nd form).

I think this is not really answering the question. The diagrams in the question were not very clear, but I have edited it. There is a more fundamental reason why the allene 3 does not form, namely the loss of conjugation (as the pi orbitals are orthogonal). And you also don't explain why the terminal methyl group doesn't enolise, e.g. in 4. To me, it feels like you are just explaining why 1,3-dicarbonyls tend to enolise, without explicitly considering the alternative structures which OP asked about.

@orthocresol I have added explanation for (3). If the answer still doesn't answer the question, please inform me and i will delete it because i cannot answer the question any further than that.


text:
How to loosen up Tris HCl?
I have this bottle of Tris HCl that I need to use and apparently it's all hard. I don't have time to get a new bottle since this is needed for a lab for tomorrow and this is all I was given. Is there an easy way of loosening it up to be free flowing short of throwing the bottle around?

You could try warming the bottle gently in an oven or hot water bath (not on a dry hot plate!) - I would try 50 C for starters. If it is in a plastic bottle I would avoid getting it too hot. Maybe it will soften, like brown sugar, and then you could scoop it out with a spatula.
Alternatively, you could weigh the entire bottle, use the contents to prepare a stock solution of known volume, measure the tare weight of the now-empty bottle, calculate the concentration of the stock solution, and use it to prepare buffers of known TRIS concentration.

source: Stack Exchange

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text:
Why atropisomers are called conformers?
Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding) so how it is possible that according to IUPAC atropisomers are classified as conformers (conformational isomers)?
According to IUPAC:

Atropisomers - A subclass of conformers which can be isolated as
separate chemical species and which arise from restricted rotation
about a single bond.
Conformers - One of a set of stereoisomers, each of which is
characterized by a conformation corresponding to a distinct potential
energy minimum.
Conformation - The spatial arrangement of the atoms affording
distinction between stereoisomers, which can be inter-converted by
rotations about formally single bonds. Some authorities extend the
term to include inversion at trigonal pyramidal centres and other
polytopal rearrangements.

This definition of "conformation" seems vague to me, but it is often said that changing one conformation to another involves no bond-breaking and changing one configuration to another requires bond-breaking. This is the difference between them.
IUPAC gives examples of atropisomerism: e.g. (E)-cyclooctene (given in Gold Book "planar chirality" entry). This definitely should involve bond-breaking. Other IUPAC examples of atropisomerism include substituted ortho-biphenyls.

Think of the whole bond-breaking narrative in a different way: the bonds in both isomers are exactly the same.

They're conformers, but with very high energy barrier of rotation.

@IvanNeretin I don't understand? Bonds are the same in (configurational) chiral centres too, but they are oriented differently and require bond breaking as well as (E/Z)-cyclooctene which supposedly is atropoisomeric.

@Mithoron Yes, some of them, like (E)-cyclooctene requires a lot of energy to break the double bond temporarily to turn it in to (Z)-cyclooctene, but this bond-breaking is the problem I have trouble understanding. How come you can call something that involves bond-breaking atropisomerism and conformerism?

@user35715: The transformation of E- to Z-cyclooctene is a configurational change that involves bond breaking be it sigma or pi bonds. But this is not the issue. Think about the E-double bond of cyclooctene "jumping rope with the methylene chain", which interconverts enantiomers. Granted it is difficult for this ring but no bonds need to be broken.

I think your notation of "Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding)" is not correct. To my knowledge, the free rotation of one single bond of an atropisomer is not restricted by connecting bond(s). I think you get confused by some literature explaining breaking a bond selectively directed to pair of atropisomers (e.g., Ref.1):

According to your post, IUPAC definition of atropisomers:

Atropisomers - A subclass of conformers which can be isolated as separate chemical species and which arise from restricted rotation about a single bond.

A variety of sources confirm this definition. For instance, Ref.2 and 3 described atropisomers as:

Atropisomers are stereoisomers (rotamers) resulting from hindered rotation about single bonds (having an energy barrier to rotation about a single $\sigma$ bond) usually due to steric hindrance. This energy barrier to rotation is high enough to allow the isolation of the conformers (rotamers).

Nonetheless, the name, coined by German biochemist Richard Kuhn in application to a theoretical concept 1n 1933 (Wikipedia) is derived from Greek (a = not and tropos = turn; thus, atropos meaning "without turn"), which supported the definition. However, the atropisomerism was first detected in 6,6'-dinitro-2,2'-diphenic acid by Cristie, et al. in 1922 (Ref.4):

It is also important to know the nomenclature of atropisomers (Wikipedia):

Determining the axial stereochemistry of biaryl atropisomers can be accomplished through the use of a Newman projection along the axis of hindered rotation. The ortho, and in some cases meta substituents are first assigned priority based on Cahn–Ingold–Prelog priority rules (Ref.5 & Ref.6). One scheme of nomenclature in based on envisioning the helicity defined by these groups (Ref.7). Starting with the substituent of highest priority in the closest ring and moving along the shortest path to the substituent of highest priority in the other ring, the absolute configuration is assigned $P$ or $\Delta$ for clockwise and $M$ or $\Lambda$ for counterclockwise. Alternately, all four groups can be ranked by Cahn–Ingold–Prelog priority rules, with overall priority given to the groups on the "front" atom of the Newman projection. The two configurations are termed $R_\mathrm{a}$ and $S_\mathrm{a}$ in analogy to the traditional $R/S$ for a traditional tetrahedral stereocenter (Ref.8}:

Atropisomers are also found in nature. The main pigment of cotton seeds, gossypol is an atropisomer, which exists in both form of $(R_\mathrm{a})$- and $(S_\mathrm{a})$-conformations. Most commercial Upland (Gossypium hirsutum) cottonseeds have an $(R_\mathrm{a})$- to $(S_\mathrm{a})$-gossypol ratio of ~$2:3$, but some Pima (Gossypium barbadense) seeds have an excess of $(R_\mathrm{a})$-gossypol. Between two isomers, $(R_\mathrm{a})$-gossypol is more toxic and exhibits significantly greater anticancer activity than its $(S_\mathrm{a})$-atropisomer:

Two fun facts about atropisomers are (KU.edu):

Atropisomers are detectable by $\mathrm{NMR}$ if half lives exceed $\pu{10^{-2} s}$.
Atropisomers are isolatable if the half-life is above $\pu{10^3 s}$.

References:

Gerhard Bringmann, Thomas Hartung, “Atropo-enantioselective biaryl synthesis by stereocontrolled cleavage of configuratively labile lactone-bridged precursors using chiral H-nucleophiles,” Tetrahedron 1993, 49(36), 7891-7902 (https://doi.org/10.1016/S0040-4020(01)88014-5).
Alan R. Katritzky, Christopher A. Ramsden, John A. Joule, Viktor V. Zhdankin, In Handbook of Heterocyclic Chemistry, Third Edition; Elsevier Limited: Amsterdam, The Netherlands, 2010 (ISBN: 978-0-08-095843-9).
Jonathan Clayden,  “Atropisomerism,” Tetrahedron 2004, 60(20), 4335 (https://doi.org/10.1016/j.tet.2004.03.002).
George Hallatt Christie, James Kenner, "LXXI.—The molecular configurations of polynuclear aromatic compounds. Part I. The resolution of $\gamma$-6 : 6′-dinitro- and 4 : 6 : 4′ : 6′-tetranitro-diphenic acids into optically active components," Journal of the Chemical Society, Transactions 1922, 121, 614–620 (https://doi.org/10.1039/CT9222100614).
R. S. Cahn, Christopher Ingold, V. Prelog, “Specification of Molecular Chirality,”  Angew. Chem. Internat. Ed. Engl. 1966, 5(4), 385-415 (https://doi.org/10.1002/anie.196603851) and Corrigendum: Angew. Chem. Internat. Ed. Engl. 1966, 5(5), 511-511 (https://doi.org/10.1002/anie.196605111).
G. P. Moss, "Basic terminology of stereochemistry: IUPAC Recomendations 1996," Pure and Applied Chemistry 1996, 68(12), 2193-2222 ()(PDF).
http://goldbook.iupac.org/terms/view/H02763: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
http://goldbook.iupac.org/terms/view/A00547: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

While the user seems to have vanished, its worth pointing out the misconception they are having. Its not that the (E) and (Z) enantiomers of cyclooctene are atropisomers. Rather, (E)-cyclooctene has two atropisomeric conformations, which can be interconverted with out breaking the double bond. Rotation is hindered around this bond, but it can still occur due to rotation of the two surrounding single bonds. A very nice depiction of this can be found on Henry Rzepa's old website
(In case the link breaks, here is the DOI for the main hub of the site)
DOI: http://doi.org/10042/a3uy9

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text:
What studies show that d orbitals don't play an important role in covalent bonding? (> period 2 elements)
Take a look at all the following molecules: $\ce{SO3^{2-}}$, $\ce{SO2}$, $\ce{SO4^2-}$, $\ce{PF5}$ and many more you can add. In all these there is a controversy of one of these two types:

Is formal charge more important than octet rule - as to justify hypervalency- on determining Lewis structure? As a rule I'd say the answer is no. For example in $\ce{SO2}$,$\ce{SO3}$, $\ce{SO4^2-}$, Do calculations show it?
What is the central atom hybridization? For example, in $\ce{PF5}$ (see here). Does it involve d-orbitals?

In any case, both previous statements are related to the d-orbitals energy.
There is -at least- one straightforward way to answer the question:

Calculate* the Molecular Orbitals and see which of previous options fit better (the use of d-orbitals or the non use of them).
*sure some can be found in some papers.

So the question is:
Are there theoretical studies of many concrete examples as to remark that d orbitals don't play any relevant role in covalent bonding?

Sorry, can you clarify your question? What info are you looking for that is not in the related posts?

@Greg look at the title, I think the posts below show $d$ orbitals are not important in covalent bonds. I ask what is the Evidence.

Some related pub.: 1. Gillespie, R. Coord. Chem. Rev. 2002, 233-234, 53–62. DOI: 10.1016/S0010-8545(02)00102-9 2. Gillespie, R. Inorg. Chem. 1995, 34 (4), 978–979. DOI: 10.1021/ic00108a032 3. Smith, D. J. Chem. Educ. 2005, 82 (8), 1202. DOI: 10.1021/ed082p1202 4. Jensen, W. J. Chem. Educ. 2006, 83 (12), 1751. DOI: 10.1021/ed083p1751 5. Halgren, T. J. Am. Chem. Soc. 1977, 99 (21), 6793–6806. DOI: 10.1021/ja00463a002

Is this only meant to be about supposedly hypervalent compounds of main group elements?

Why don't you look for studies that do show a significant d orbital contribution to covalent bonding. Proving nonexistence is hard.

@Karl I have found 2 treating this questions, but couldn't read them completely yet, and both are quite dificult to understand. One from cooper 1994 (https://pubs.acs.org/doi/abs/10.1021/ja00089a033), the other one from magnusson 1990 (https://pubs.acs.org/doi/abs/10.1021/ja00178a014).


text:
How do I identify how many reactions are occurring when precipitating a mixture?
An example of a question is:

After dissolving $x$ grams of a mixture of $\ce{Na2SO4}$ and $\ce{NaCl}$ in water, you add an excess of $\ce{BaCl2}$ solution. You filter off the precipitate which forms, dry it, and find that it weighs $y$ grams. What is the percent by mass of $\ce{Na2SO4}$ in the original mixture?"

If I am correct, the reaction for this would be:
$$\ce{BaCl2 + Na2SO4 + NaCl -> BaSO4 + 3NaCl}$$
However, there are some questions which are of the same format but have two separate reactions:

A $x$ gram mixture of $\ce{CaCl2}$ and $\ce{RbCl}$ is dissolved in water. Addition of $\ce{AgNO3}$ causes the chloride ion to precipitate as $\ce{AgCl}$. The mass of $\ce{AgCl}$ collected is $y$. What is the percent by mass of $\ce{RbCl}$ in the original mixture?

For this one, would the reaction be two equations?
\begin{align}
\ce{CaCl2 + AgNO3 &-> Ca(NO3)2 + 2AgCl}\\
\ce{RbCl + AgNO3 &-> RbNO3 + AgCl}
\end{align}
I know how to do these types of questions past the point of deciding whether or not there are two equations. I just don't understand why I couldn't write that last one as a single equation, in the same way as I did the first example. Could anybody help me? Thanks in advance.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here.

You could: Ag+ + Cl- -> AgCl

You asked two very different questions.

(1) How do I identify how many reactions are occurring when precipitating a mixture?

There is no information in way the reactants are shown in a chemical equation as to what the products will be. That is to say that you have to know the products of the reaction.

(2) For this one, would the reaction be two equations?

Chemists use shorthand all the time. Reactions are typically netted out to the pertinent aspect under consideration. However if I were writing the reactions for the second problem I'd do it much differently that you have presented. I think I'd write:
\begin{align}
x\ce{CaCl2} + y\ce{RbCl &->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-}\\
x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} &\ce{->[excess AgNO3]} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{AgCl v}\\
\end{align}
From an analytical chemistry perspective there are two steps. First the chloride salts are dissolved and then the silver nitrate is added. You wouldn't want to add the salts to a silver nitrate solution since the silver chloride could coat the Ca and Rb salts preventing a complete reaction.
I even might write the second reaction as:
$$ (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$
I'm not a fan of one line jobs like:
$$x\ce{CaCl2} + y\ce{RbCl ->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$


text:
Why doesn't cyclopropyl methyl carbocation stabilises itself by ring expansion?
I have been taught that ring expansion stabilises smaller cyclic compounds to a great extent. So why does cyclopropyl methyl carbocation shows such type of resonance

rather than expanding its ring

thereby decreasing it's angle strain, further it will be stabilized by hyperconjugation effect and inductive effect.

perhaps the angles in the 4-ring are bad for hyperconjugation? the intermediate 4-ring is even more badly strained than the 3-ring? the resonance takes a lot of strain from the cyclic structure? (last point is definitively true)

It can't be said that the cyclobutane carbocation does not occur. March's Advanced organic chemistry, 10.C.i Neighboring-group participation by π and σ bonds: nonclassical carbocations, tries to explain the stabilization, but also cites that (chloromethyl)cyclopropane by hydrolysis provides 48% cyclopropylmethanol, 47% cyclobutanol, and 5% "allylmethanol".

I have found a similar question already answered: https://chemistry.stackexchange.com/questions/19090/what-is-the-reason-for-the-exceptional-stability-of-the-cyclopropylmethyl-carboc

I already checked this answer but this doesn't explain why ring expansion is not preferred

Question itself does - it's exceptionally stable (while cyclobutyl isn't)

Unfortunately, the premise of your question is wrong.
There is no cyclopropylcarbinyl or cyclobutyl cation. They're the same thing. Unlike cations which you might be more familiar with, for example, t-butyl cation, the cation you're looking at here, also called homoallyl, is a non-classical cation.
The open form and the two structures you drew in the second line are not distinct. They're just resonance forms of each other.
You ask does this ion undergo ring expansion? Well, yes and no. Yes, in that there is a contribution from the 4-member ring to the resonance structure. No, in that you can't expand something that doesn't exist.


text:
What is to be called as solvent?
My book says:

In a solution consisting camphor and nitrogen gas, camphor (solid) is solute whereas nitrogen gas is solvent.

Doesn't this contradict the statements that:
(1) Generally the component present in large proportion is the solvent.  I mean camphor will be in larger proportion as it is a solid. We can diffuse much nitrogen gas.
(2) Solvent determines the physical state in which solution exists. If nitrogen is solvent, then how can the solution be in gaseous state? Or if the statement is true, does that mean that the solution exists in gaseous state with camphor particles of collidial size?

Nitrogen gas can get dissolved in a condensed phase such as camphor, but not dissolve it. Unless we speak o e.g. supercritical CO2, which is not really a gas.

Regarding part 1
Camphor will readily sublime and so, in proportion to the nitrogen gas, it is the minor constituent
For part 2,

If nitrogen is solvent, then how can the solution be in gaseous state?

if the nitrogen is the solvent then of course it will be in the gaseous state as nitrogen is a gas.


text:
Colour of platinum
Why does platinum have silvery shine whereas platinum black (finely powdered platinum) is black in colour although both have the same constitution of pure platinum?
Due to similar composition, both of them will have electrons in similar kind of energy states. So we can't assume this due to charge transfer. Neither is there any scope of charge transfer.

https://chemistry.stackexchange.com/questions/45113/what-is-the-chemistry-behind-black-colour-of-platinum-black

In bulk metals, electrons in the conduction band are shared, creating a "sea" of reflective electrons. A nanoparticle, though, is so small that there is no conduction "band", just discrete levels enforced by the Pauli exclusion principle. Other metals in finely divided form, such as gold or silver, are also are black or darkly colored. Traditional silver emulsion photography relies on white silver compounds turning to black nanoparticles of silver when reduced.
Nanoparticles exhibit interesting color effects due to plasmons. Finely-divided gold particles produce ruby glass, used in stained glass windows, though thin bulk gold films appear green or blue in transmitted light.


text:
How is silver chloride more soluble in brine than water?
I heard that:

Silver chloride is more soluble in very concentrated NaCl solution (brine) than in pure water.

Is it true? I mean, won't the common ion effect operate? Or is this maybe due to some kind of complex formation as the NaCl solution is highly concentrated?

I mean won't the common ion effect operate?

The common ion effect always applies for pure substances. With the added sodium chloride, you have changed the effective solvent and changed the ions.

Or may be this is due to some kind of complex formation as NaCl solution is highly concentrated?

Exactly, before your ionic equation was:
$$\ce{Ag+ (aq) + Cl- (aq) <=> AgCl (s)}$$
but with brine, the ionic equation is:
$$\ce{AgCl (s) + Cl- (aq) <=> [AgCl2]- (aq)}$$
Which has a higher solubility constant, and where as before more chloride made the $\ce{AgCl}$ less soluble, if you note that you have switched regimes and $\ce{Cl-}$ has switched from precipitating a silver ion to dissolving a silver chloride molecule, which means now with increased chloride concentration, silver chloride becomes more soluble.

The answer by A.K. is eminently reasonable, but I wondered if stability constants were available. I quote from the summary of Ref. 1 :

The solubility of silver chloride in concentrated solutions of various
chlorides was determined mainly at 25°.
The solubility is nearly doubled in going from o° to 25°, the rate of increase above and below 25° being nearly logarithmic.
The solubility is sharply proportional to integral powers of the chloride
concentration throughout considerable ranges, a fact explained by assuming the existence of the complex anions
AgCl3$^{^-2}$,  AgCl4$^{-3}$ and possibly AgCl5$^{-4}$. No evidence of the ion AgCl2$^{^-}$is found.

It seems that the simplest explanation is only the beginning of understanding the solubility increase.
References

The Solubility of silver chloride in chloride solutions and the existence of complex argentichloride ions, J. Am. Chem. Soc. 1911, 33, 12, 1937– 1946 Publ. Date: December 1, 1911 https://doi.org/10.1021/ja02225a008

Interesting, I have found earlier today AgCl2- formation constant 1.8e5 on libretexts somewhere. OTOH  [AgCl5]4- seems to me suspicious.  https://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Equilibrium_Constants/E4%3A_Complex_Ion_Formation_Constants

@Poutnik: Yes, indeed, interesting on both issues! Is the 1911 account accurate about AgCl2-, and, sure, AgCl5(4-) is unlikely, but AgCl3(2-) or AgCl4(3-) don't seem so impossible. But it could be easier to just call it AgCl2- and let it go because the solubility of the complex(es) is still so small.


text:
How to identify whether an electrophile will attack benzene ring when a deactivating group is attached to it?
I was asked to find out the major product formed when this molecule is subjected to nitration using $\ce{HNO3/H2SO4}$ mixture

Here $\ce{NO2+}$ will be the electrophile. Lets call the benzene ring which is directly attached to the carbonyl group as ring 1 (benzylic ring) and the other as ring 2 (phenolic ring). Now ring 2 is attached to an activating group as $\ce O$ will show +M and will increase the relative electron density at o-p positions. But the electrophile will preferably attack para position due to steric hindrance at ortho position.
Benzene ring 1 is attached to a carbon atom which in turn is connected to a more electronegative atom. So it is partially positively charged and will show -M effect thereby deactivating the ring. But it is also meta directing group. So will the electrophile attack meta position of ring 1?
The answer to this question says, ring 1 won't be attacked by the electrophile as it is attached to a deactivating group. The the molecule after nitration is

Then how to identify whether an electrophile will attack benzene ring when a deactivating group is attached to it?

It's not a matter of if, but of on what time scale will the reaction proceed. minutes, hours, geological eras.

I agree that the phenolic ring reacts first, but if it stops there depends on the conditions because it is possible to nitrate benzene rings with deactivating groups, after all dinitrobenzene is fairly easy to make https://www.docsity.com/en/preparation-of-dinitrobenzene-organic-chemistry-lab-manual/405022/


text:
Equalization of pressure in heat exchanger
I was solving numericals on Kinetic Theory of Gases when I came across this question

Two closed vessel of equal volume contain air at 105 kPa, 300 K and are connected through narrow tube. If one of the vessel is now maintained at 300 K and other at 400 K what will be the pressure in the vessel?

I don't want the solution of this question but I want to know why the pressure will be same in both the vessel when one is maintained 300 K and other 400 K. The molecules in  the vessel at 400 K will have greater kinetic energy and will exert greater pressure on the walls than the molecules of the other vessel.
Now one will say that some of the molecules of vessel at higher temperature will move to the other vessel so that the pressure remains same in both the vessels, but why is it so. Why can't the pressure in both the vessels be different?

The pressure can't be different precisely because the vessels are connected, and some molecules will move to the other vessel to equalize it. That's what pressure is all about.

That's what I m asking why will some molecules move to another vessel to equalise it. As motion of molecules is random, it might happen that some molecules move to vessel at higher temperature.

Of course they will. You are right, the motion of molecules is pretty random; also, there is quite a lot of them. How can we tell which way the net flow will be? We can't, unless we are able to find a measure. Luckily, some guys did that for us. The measure is called pressure.

@IvanNeretin pressure is definitely not a measure of flow. It's a measure of force per unit area.

Macroscopics
Imagine the tube is initially closed. Then one vessel is heated up. As $V$ and $m$ are constant:
$$p=k\,T$$
So if $T$ is increased, $p$ is increased. As $p$ is increased, there will be a net flow of mass, until they are balanced. That's what we observe. Now, we can think of it microscopically.
Microscopics
Initially, there are the same number of particles in each vessel. When heating up there is an increase in velocity. So, in the high-temperature vessel the same number of molecules are with higher velocity. Imagine a plane in the middle of the narrow tube, the number of collisions per time, because of velocity, will be higher in $400\ \mathrm K \to 300\ \mathrm K$ direction. It indicates a net flow of mass. This flow of mass will equal pressures on both sides.

I can't follow, so what if the collision rate is greater in the vessel at higher temperature? I want to know why will the pressure will be same in both the vessels

@santimirandarp Your second to last sentence is very, very misleading. It seems to draw an equivalence between the gas flow and the velocity of the molecules - this is very much not the case though! You should really be clearer as to how the greater velocity of the molecules exerts a greater force, which causes gas flow.

@santimirandarp Why does a higher number of collisions increase the flow?

This experiment divides the container so that pressure (a "flow" of gas molecules against the walls) can be equalized thru the narrow tube but net flow of heat from hot molecules passing thru the narrow tube is seriously impeded (essentially does not occur because each section is maintained at 300 K or 400 K).
This non-equilibrium condition is held static by the external devices maintaining the set temperatures and the narrow tube which connects the  pressures in the two "systems" but disconnects the heat flow.
To see that the pressures will eventually be the same, imagine connecting the two vessels at 300 K by the narrow tube, then quickly heating one to 400 K. The pressure will increase in the heated tube, and depending on the narrowness of the tube, will decrease as gas flows into the cooler tube (where the pressure will increase because there will be more molecules there, even tho still at 300 K). After some time, the pressures will equalize.

I'm going to answer this question in two parts: First, why will some of the molecules of the vessel at higher temperature move to the other vessel?
Pressure is simply the force per unit area that the gas exerts. When a gas is sitting in a vessel, it exerts a force on the walls - and the walls in turn exert an equal an opposite force. Thus, nothing moves (on a macroscopic level) in the system.
If you open a tube to another vessel, then the gas in each vessel will be exerting a force on the gas in the tube. When both vessels are at equal temperatures and pressures, they exert equal and opposite forces, and the gas remains static (on a macroscopic level), and there is no net flow.
Heating up one vessel would increase the pressure due to the ideal gas law:
$$pV=Nk_\mathrm BT.$$
Since the higher temperature vessel will have a higher pressure, it will exert more force on the gas in the tube, and the gas has to move away from it.
Secondly, why can't the pressure in both the vessels be different? The answer is that the system is closed, and the number of molecules is conserved - molecules cannot enter from outside the two vessels. Thus, molecules can't move from the hotter vessel to the cooler vessel forever - the hotter vessel would run out of molecules.
The critical part of the ideal gas law to keep in mind is the $N$, the number of molecules. In the hotter vessel, molecules will travel out, and $N$ will be reduced, as long as the pressure is greater than the cooler vessel. Since volume and temperature are constant, pressure must drop until the pressures are equal.


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Boiling of a liquid
Consider water as an example. I m really confused when does boiling takes place. When it's vapour pressure equals atmospheric pressure or when it's saturation atmospheric pressure equals atmospheric pressure.
Wait this is even more confusing. Either of two doesn't make sense. Consider this image

Here the water isn't covered by lid so there won't be any vapour pressure, as all the vapours will escape. This is the case where vapour pressure will exert
vapour pressure.

But in the first case irrespective of the vessel covered by a lid, boiling takes place. Neither vapour pressure equals atmospheric pressure nor saturation vapour pressure equals atmospheric pressure.

Related: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

Saturated atmospheric pressure is a phrase meaning that the vapor pressure is equal to 1 atmosphere. So it isn't just "atmospheric pressure" but a pressure of exactly one atmosphere. // For a open pot boiling imagine a thin layer above the liquid which is the pure vapor of the liquid. The vapor then diffuses into the atmosphere.

@IvanNeretin reading your post makes me even more confused. Firstly why are bubbles forming at the bottom of the liquid. Secondly, as you wrote that the bubbles that are formed at the bottom contains pure water vapour so why does boiling takes place when vapour pressure equals atmospheric pressure, because at the bottom there will also be and additional pressure of $h\rho g$ which would prevent the formation of bubble

$\rho gh$ will be there all right; I simply ignored it.

The red heat of the burner makes the bottom of the pot a little hotter and causes the highest vapor pressure. If "bubbles"  developed at the very top surface of the liquid, they might be invisible.

It's the first of the two options given. Vapour pressure is a property of the liquid, not of the vapour, and it is 1 bar at 100°C!

Water boils because the vapor pressure IN THE LIQUD reaches a level where the molecules can overcome the external pressure of the surrounding atmosphere and form bubbles. This level can be reached either by raising the temperature (the molecules gets more energy) or lowering the surrounding pressure (like in the mountains).
Actually, the pressure a water molecule has to overcome to turn into vapor, is the sum of the pressure from the water above plus the atmosphere. That is why water in hydrothermal vents can be several hundred degrees without boiling.
It makes absolutely no difference whether the atmospheric pressure comes from a bone dry gas or from pure water vapor.


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Oxidation state in transition metals
Why does scandium has +2 oxidation state unstable? It only has +3 oxidation state stable.
Scandium has outermost electronic configuration as 3d¹4s². Why it does not lose two 4s e- only instead of all three?

Because then it only has one single electon in all its outer orbitals.  Doesn't sound very desireable to me. Also https://en.m.wikipedia.org/wiki/Scandium#Uncommon_oxidation_states

Scandium actually does have oxidation states lower than +3.  Wikipedia reports a double chloride $\ce{CsScCl3}$ where scandium is clearly in the oxidation state +2.
The stability of oxidation states in transition metals depends on the balance between ionization energy on the one hand, and binding energy due to either ionic or covalent bonds on the other.   In the case of scandium the third ionization energy is low because all three valence electrons are held rather loosely, being in diffuse orbitals that are shielded from most of the nuclear charge by the argon core.  Thus it's easy to get from the +2 to the +3 oxidation state because of the greater binding energy enabled by the latter.  Later first-row transition elements like iron have more tightly bound $3d$ electrons and thus higher third ionization energies, making the +2 oxidation state more favorable.


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Reaction of nitrobenzene with sodium metal in ethanol
The reactant is nitrobenzene(only) and reagent is Na metal in solvent liquid ammonia along with ethanol (which i am assuming is playing the role of proton donor please correct me if i am wrong).
The first thing that comes in mind is "alkyne" but there is not any alkyne in nitrobenzene( not even in its resonance structures).How this reaction will proceed?
In answer, the DBE is just reduced by one.I was also trying what if it was toluene?Can i have a explanation based on mechanism?

Birch Reduction will occur with Nitrobenzene.

The ethanol makes little sense, unless it is added in the end only, to destroy remaining Na metal.

Benzene Rings or Substituted Phenyl rings undergoes reduction in presence of $\ce{Na}$ in liq. $\ce{NH3}$ in presence of $\ce{EtOH}$ solvent, and generally forms substituted Cyclohexa-$1,4$-dienes.
This reduction generally follows Single Electron Transfer (SET) Mechanism.The single unpaired electron of Sodium is transferred to the Benzene ring forming a radical which subsequently forms a Carbanion. This carbanion eventually takes the Proton from $\ce{EtOH}$ and becomes Cyclohexa-$1,4$-diene.
But in the first step, the transfer of electron by $\ce{Na}$ occurs in that place, where the intermidiate formed after this step gets maximum stabilisation. In case of Electron-withdrawing group, such as $\ce{-NO2, - COOH, -SO3H, -CHO, -COOR, -COR }$ etc., that transfer occurs para to that group, as the negative charge generated is highly stabilised by that electron-withdrawing group. The mechanism thus in case of Nitrobenzene is as follows,

But if the group is electron donating such as $\ce{-CH3, -OH, -OMe, -NMe2 }$ etc. The transfer will occur in such a way to stabilise the generated negative charge by keeping it in meta to the group (where the effect of electron donation by the group is least). For those compounds, the mechanism and corresponding product will be as follows,

Thus, in your case, the product will be $\text{3-nitrocyclohexa-1,4-diene}$ as described in the first mechanism.


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where is sulphate going here? electrolysis of magnesium sulphate
I am electrolyzing a water + magnesium sulphate solution using copper for both electrodes. I am getting a light-blue precipitate forming which I believe to be a mixture of copper hydroxide and magnesium hydroxide, which are both poorly soluble. The solution is remaining clear which tells me that basically no copper sulphate is forming.
My question is, where is the sulphur from the sulphate going?
The mixture is not becoming acidic as if sulphuric acid was forming, and I do not detect any smell of sulphur is if some sulphur oxides were being released as gas.

It stays in solution?

I wasn't aware that could happen. Could the sulphate ion just sit in solution by itself? I thought it would have to be in equilibrium with a positive ion as the copper sulfate would be?

Are you sure the solution is not becoming acidic? Precipitation of Mg(OH)2 indicates that the MgSO4 has been split into Mg(OH)2 + H2SO4.  If no mixing occurs, the cathode region will become alkaline while the anode region will become acidic (and will dissolve some copper).

I'm pretty sure. I don't have pH indicators but when I add a base to the mixture, there is no reaction. And when I add an acid(vinegar), almost all of the precipitate disappears as it reacts with the acid to form soluble salts (magnesium and copper acetate in this case). If any acid was formed, wouldn't it just immediately react with the hydroxides and it would go back to mostly magnesium sulphate again? There is a really huge amount of white precipitate that forms very quickly when electrolysis starts.

In electrolysis of aqueous solutions sulfate ions just sit there.  Protic hydrogen (or the metal in the cation if this metal is not highly electropositive like magnesium) is reduced more easily at the cathode, and water is more easily oxidized at the anode.
Sulfate ion (actually, bisulfate ion) can be oxidized electrolytically at the anode if you use sulfuric acid instead of water as the solvent.  See the preparation methods for potassium peroxydisulfate.

What is "protic hydrogen" ?


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Are 3-center-2 electron bonds and and banana bonds same?
While reading about diborane, I came to know that the two bridging hydrogen atoms in diborane form 3 center 2 electron bonds. On account of repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are drifted away from each other giving the shape of a banana. Such bonds are called banana bonds. In diborane 3 center 2 electron bonds and banana bonds are same.
What if a compound has only one bridging atom which forms 3 center 2 electron bond. As there is only one bridging atom, it would not face repulsion so it won't form a banana bond.

In that case, it won't dimerise/polymerise to form 3C-2e- bonds(such type of delocalizations of electrons won't be possible).

@thewitness not really, https://en.wikipedia.org/wiki/Trihydrogen_cation

The phrase "banana bond" is usually reserved for situations where steric conditions prevent the orbitals of two atoms from aligning properly so that the bond appears "bent" and is weaker than an optimally aligned bond would be.  Cyclopropane is the poster child for such bonds, diborane is not because this description of banana involves only two atomic centers.
It's a different concept from three-center two-electron bonds, where typically the bonds are well aligned but the three atoms bonded together are not always collinear.  This is what diborane does.  The atoms in the diborane bonds are not collinear because we need two such bonds, not because either bond suffers from imperfect overlap.


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Does Cyclooctatetraene decolourise Br2 water and why?

I am thinking yes because it is not aromatic according to Huckle's rule. However, why in terms of orbitals etc?
I am a first year and so I am new to these concepts.


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Reading Stereographs
On an old paper, I found an interesting representation of some symmetry operations: stereographs.
Online I found a set of symmetry operations and related stereographs (see: http://newton.ex.ac.uk/research/qsystems/people/goss/symmetry/Stereographs.html), yet it's  not clear to me how the representation is related to the symmetry operations (or: how are the representations obtained?).
Some samples from the website linked above:

C4

C5h

D6

I would like to understand the geometrical (and symbolical) meaning of a stereograph, in order to relate non-mnemonically it to its symmetry operation

Just to be clear: You are asking about the difference between things like $\oplus$ and $+$?

Not only that. Apparently, although I couldn't find a complete description, stereographs are obtained via "mathematical" projection over a sphere of points derived from the symmetry of the object. Then, the location of the points gives the symbol in the stereographs. I think that every strange symbol in a stereograph could be understood in those terms.
I am sorry if the description is terrible, but I couldn't really find anything "complete" about this notation.
If someone could suggest, for lack of a description, a book which explains it, I could make a search in my university's library

I vaguely remember a book with all the space groups (regarding symmetry in solid state chemistry), which had similar pictures. I just don't remember its name. But maybe this can be of help: http://img.chem.ucl.ac.uk/sgp/misc/symbols.htm

The books ' The Basics of Crystallography and Diffraction' by C. Hammond, IUCR texts on Crystallography, publ OUP, and McKie & McKie, 'Essentials of Crystallography, publ Blackwell Scientific, both contain chapters on these projections.

The solid elements are the symmetry operations. The plus and open circle symbols show the positions of atoms and their symmetry mates. The center of the circle is the z-axis. The edge of the circle are directions perpendicular to the z-axis. In the examples in the question, all axes are either along z or perpendicular to z. It gets more complex in cubic symmetry, which also has axes in other directions.

Here, the triangles are three-fold axis along the body diagonals. The symbols are further explained at http://pd.chem.ucl.ac.uk/pdnn/symm2/pntgrp1.htm.


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Link1 and multiple output files in Gaussian: doable?
I have recently started learning Gaussian in my Computational Chemistry class, and I have recently learned how to use the "Link1" option, in order to run multiple calculations from a single input file.
What I was wondering is: is there an option that allows the creation of multiple output files (one for each "Link1" section), thus allowing an easier interpretation of the results for each calculation?
I couldn't find anything about the problem on the manual or online.

Can't guarantee because I haven't tried this myself, but I believe what you are looking for is "punch=all". The one issue this might have is I think it only produces a single fort.7 file, so it probably requires an additional option or a separate program to ensure later link1's are sent to different outputs.

That's indeed an useful option (better than nothing, thank you!)
Apparently, as far as I see, there are no options to change the filename, but this might be a start :)


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Reactivity and its meaning
When we say that “potassium” is reactive, are we referring to a potassium atom or bulk potassium as in the solid metal? In other words does “potassium” refer to a type of atom or an agglomeration of potassium atoms?

Typically it refers to the bulk, but you can never be sure. For this and other reasons, the word "reactive" is more confusing than meaningful, and hence is better avoided.

Reactivity is a vague concept in chemistry. It is rather confusing than useful. Reactivity involves both thermodynamic and kinetic factors, it predicts whether a chemical compound will undergo a chemical reaction (either by itself or with other compounds) or not and if it does how fast is the reaction going to completion i.e., rate of reaction.
Reactivity is also used to compare or study rates of two or more chemical reaction. For example when we say reactivity of HBr is more than reactivity of HCl - here we refer to how fast the compund will give acid-base reaction.
Reactivity is sometimes referred to bulk and sometimes to a single molecule but mostly it is referred as bulk, but it is not predictable.
In pure compounds, reactivity is regulated by the physical properties of the sample. For instance, grinding a sample to a higher specific surface area increases its reactivity. In impure compounds, the reactivity is also affected by the inclusion of contaminants. In crystalline compounds, the crystalline form can also affect reactivity. However, in all cases, reactivity is primarily due to the sub-atomic properties of the compound.

Hi there, so I am currently learning about redox reactions. So is the “oxidation state of a potassium atom” the same as the “oxidation state of potassium”?

The only oxidation state of potassium is +1. It cannot go to higher values, like +6 !!!

@Maurice Thank you for pointing out, I forgot why I did that there.

When we say that potassium is reactive, it is understood that it is more reactive than sodium. This can be easily demonstrated in classrooms by dropping a small piece (max 0.1 g) of those metals into a tub of cold water.
Sodium reacts in a relatively calm way by melting and moving quickly at the water surface, yielding Hydrogen gas H2 which can be set on fire with a match. The H2 flame is one or two millimeters high. It stays and move erratically around the sodium globule. The whole is not dangerous. It is worth the trouble showing it in a classroom. No smoke is produced.
$$\ce{2 Na + 2 H2O ->  2 NaOH + H2}$$
$$\ce{2 H2 + O2 ->  2 H2O}$$
Potassium reacts so violently with water that both metal and gas take fire in such a way that it is a dangerous reaction. Intense smoke is produced, made of potassium oxide, hydroxyde and superoxide KO2, which are corrosive and toxic. It should not be shown in an ordinary classroom.
$$\ce{2 K + 2 H2O ->  2 KOH + H2}$$
$$\ce{K + O2 ->  2 KO2}$$
Don't try to use more than 0.1 g of these metals. The reaction becomes explosive with higher amounts.


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Multiplication symbol in SMARTS
There is a SMARTS pattern that was published which looks like the following[1]
[#6 × 4]

and there are actually spaces around the × symbols between the 6 and 4.
Question is, should I change this multiplication notation to a lower case "x", or really use some sort of ASCII symbol?  I didn't see a multiplication symbol anywhere in the Daylight SMARTS description.
References:

Hou, T. J.; Xia, K.; Zhang, W.; Xu, X. J. ADME Evaluation in Drug Discovery. 4. Prediction of Aqueous Solubility Based on Atom Contribution Approach. J. Chem. Inf. Comput. Sci. 2004, 44 (1), 266–275. PDF

The question was answered on the Open Babel mailing list
Suffice to say, there's no multiplication symbol in SMILES or SMARTS - only standard ASCII characters. It's likely a lower-case 'x' that's been mangled by a journal.


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Why do we call O2 oxygen?
I have been taught that oxygen is a chemical element, in other words a certain type of atom that has 8 protons in its nucleus. So why is O2 called oxygen? It is not a type of atom but rather a molecule.

You can call it molecular oxygen all the time, or at least until someone tells you it's a waste of time.

Or maybe gaseous diatomic molecular oxygen, just to be really sure no one thinks you're somehow talking about ozone or red oxygen.

Note that the same thing applies not just to oxygen, but also to hydrogen, nitrogen, fluorine, bromine, chlorine, and iodine. All of these elements prefer diatomic molecular form.

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

What's in 02 that is not oxygen?

:/ I think the original misunderstanding here was not getting that oxygen as an element is not a type of atom - an element is a collection of all atoms with same number of protons.

I think what you may find most helpful is to know a bit of the history of element discovery and atomic theory.
The first pure substance containing only the element oxygen to be isolated was dioxygen ($\ce{O2}$), in 1774, though it was called "dephlogisticated air" until 1777 when Lavoisier used the term "oxygen" for the first time. This was some 30 years before John Dalton even proposed the first empirical atomic theory. Back then, we only barely had an understanding of stoichiometry, such that Dalton famously claimed the molecular formula for water was $\ce{HO}$. The fact that dioxygen is a substance made of molecules containing two atoms of oxygen probably wasn't widespread knowledge until at least 1811, with the gas stoichiometry experiments of Amadeo Avogadro.
Basically, for a point in time, we knew that there was a substance composed of a single type of atom, which could not be broken down into anything simpler. This fit the then-prevalent definition of an element; "a pure substance that could not be decomposed into any simpler substance". We knew that Lavoisier's "oxygen" had to be $\ce{O_n}$, for some n, but we had no reason to assume $n \neq 1$ for decades. By the time we figured out $n = 2$, the name "oxygen" was already widely used to refer to dioxygen. The fact that $n = 3$ also forms a stable compound in ambient conditions (ozone) would also not be known until 1867. A similar story happened with (di)nitrogen (octa)sulfur, (tetra)phosphorus, and so on. The only elements which form stable monoatomic substances in reasonable conditions are the noble gasses.
There is an interesting aspect to consider behind all this. There are some (such as Eric Scerri) who claim we are doing Chemistry a disservice in muddling together the properties of elements and the pure substances which they make. Nowadays our definition of an element is solely dependent on the number of protons inside an atomic nucleus, with no reference to reactivity or in what form the pure substance can be found. In this sense, the elements do not have "reactivities", "melting points", etc.; these are all properties of the pure substances. The only true properties of the elements are things such as electronic distribution, ionisation energies, and so on. However, it is common to see periodic tables stating the melting and boiling points the pure substances of each chemical element, and even Wikipedia bundles the physical properties of dioxygen with the atomic properties of elemental oxygen. For better or for worse, we're stuck with this subtle ambiguity in nomenclature.

What do you define as "stable monoatomic substances" or "reasonable conditions"?  Mercury in a puddle will evaporate at a significant rate in open air at 25C, producing a vapor whose molecular weight matches mercury's atomic weight, but mercury is hardly a noble gas.

@supercat That's an excellent point. I considered metal vapours briefly, but evidently I forgot about mercury.

Technically, $\ce{O2}$ is molecular oxygen. $\ce{O}$, atomic oxygen, only exists under rare conditions; e.g. for a very short time after the molecule is dissociated by radiation, or where air pressure is so low that dissociated oxygen does not find a partner to recombine quickly.
Whenever one refers to an element, it is likely in the form of molecules, or metallically bound, except for the (usually) inert gases. The common yellow form of sulfur is as $\ce{S8}$ molecules. However, it is still considered elemental sulfur, not a compound, because all the atoms are of the same type.
BTW, later comments indicate some confusion with the number of protons, found in the nucleus, and the electron cloud, found around it. Think of an atom as a fluffy ball of cotton candy (electron cloud), with a few sugar crystals (neutrons and protons) in the center. When atoms combine to form a molecule, only the cotton candy sticks together; the nuclei are separate, with the same number of protons and neutrons (no croutons, though) as in the atomic element. Nuclei can combine, with enough pressure, and that process is nuclear fusion -- which could be discussed better in the physics section of StackExchange.
In short, an element can be found in molecules; it's still an element.

But why do we call O2 oxygen is my question? I thought only an atom with 8 protons can be called oxygen?

Each atom does have 8 protons. A molecule of two atoms with 8 protons each is not one atom with 16 protons. BTW, it's called oxygen because the discoverer thought all acids contain it (oxy means biting, sharp or acid in Greek, ὀξύς). So there is a misnomer.

Not much of a misnomer, though, because oxygen is a good oxidiser; it completely destroys heavier alkali metals, for instance, and eats very quickly through warm hydrocarbons.

@wizzwizz4 The term "oxidiser" was created later, and it's named that way because oxygen is the most common oxidiser.

@NieDzejkob But oxidisation does bite through things.

To be really specific, one should call the $\ce{O2}$ molecule dioxygen, to distinguish it from ozone ($\ce{O3}$) and other, less common allotropes of oxygen. And in some contexts one might want to further specify that one is talking about the usual triplet state, rather than the metastable singlet oxygen ($\ce{^1O2}$).

To elaborate on @DavidRicherby's point, words don't generally have unique meaning.  Soy milk doesn't come from a mammal; the town hall is neither a corridor nor a large room; and iron can denote a substance or an object, as well as an element.

@J.Smith "I thought only an atom with 8 protons can be called oxygen?" I'm not sure why you would think that. As you know, the word "oxygen" is used all the time, by everyone, including professional chemists, to refer to two of these atoms joined together into a molecule. The meaning of English words is determined entirely by usage -- there's no organization with the authority to declare "This word means X" -- so the fact that people (especially including experts) use the word "oxygen" to refer to the molecule implies that the molecule is one of the meanings.

It's simply a colloquialism for ease of reference. It is sort of like the way we call a heap of corn, corn. Instead of corn off the cob, like the comedian Mitch Hedberg jokes about. He says "corn" should mean "corn on an ear of corn", and a pile of corn kernels should be called "corn off the cob", but since we encounter "corn off the cob" much more often than a big ol ear of corn, we simply call it corn.
The more common something is, the less likely it is to be accurately named, and more likely it be colloquialized.

12.5% keyword density of the word 'corn' in an answer to a question about oxygen, that's got to be a world record

I am not sure the rest of us get these "corn" (AKA maize) references.

Words can mean more than one thing. There is nothing whatsoever inaccurate about referring to individual kernels of corn as "corn".

@DavidRicherby Duh; after all, corn originally meant "a grass with big seeds we use to make food", and still does in Traditional English. But the point mark makes is a correct one anyway - the uses that are more common get preferential treatment, because humans like to optimize their communication. Therefore, "molecular dioxygen" is called "oxygen", while "monooxygen" is called "elemental oxygen", if the distinction even needs to be made; just like you're more likely to use "car" rather than "a horse-less horse-drawn carriage". Accurate terminology helps communication, but so does efficiency.

@Luaan but is it a colloquialism?  It's just one sense of the word oxygen.  How is it a colloquialism?

It would be very odd to try to deny all allotropes of an element that element's name, or to require an additional description as in the phrases "molecular oxygen" or (to exclude non-modal allotropes) "diatomic oxygen". Why would it be odd? Because it may require some very specific descriptions that quickly become distracting, difficult to get right, and unimportant to the point you want to make. This is especially true if an element exists in multiple allotropes, whose chemical reactions of present interest might only care about the overall atom count present.
It's easy to get used to monatomic and diatomic forms of elements (metals usually treated as honorary examples of the former), and forget about such creatures as:

$\ce{As4}$ and multiple allotropes that don't admit as neat a formula (which will be a theme),
$\ce{B12}$ and other allotropes of boron
$\ce{O3}$, $\ce{O4}$ and $\ce{O8}$
$\ce{P2}$  vs $\ce{P4}$ and multiple polymeric forms of phosphorus
$\ce{Se8}$ and its crystalline and polymeric counterparts
$\ce{S_n}$ where $n$ can be any of $2$ to $9$, $15$, $18$ or $20$
and some elements having pretty much no natural $X_{\mathrm{small}\,n}$ allotropes such as antimony, carbon, germanium and tellurium.

What would you like to call them all?

When we say that certain elements “exist as molecules” what does that mean? From my understanding, an element is a species of atoms with the same number of protons, so when we say that oxygen, a species of atoms “exists” as a molecule, what does that mean precisely?

@J.Smith Well, from my understanding, an element is a collection of atoms that all have the same proton number, said atoms possibly not separate. Obviously we can define terms however we want, but given what atoms usually do it would be strange to define a lot of matter as neither an element nor a compound nor mixture.

So like a physical collection of atoms? If that’s the case, then when we talk about the reactivity of sodium why do we say that “sodium has one electron in its valence shell” as to imply that sodium is a type of atom?

@J.Smith Well, there are contexts where you can assume a particular form is intended. Never mind sodium or oxygen; you see the Swiss-Army knife nature of language when we discuss something as simple as hydrogen. It means very different things to stellar astronomers vs Jupiter experts vs spectroscopers vs quantum mechanists vs organic chemists vs fuel storage researchers vs those discussing acids.

@J.Smith Why wouldn't we? Sodium is an atom. It's also a metallic substance composed predominantly of sodium atoms. We only need to distinguish the two when it's important to what we're talking about, and the intended meaning isn't obvious. Whenever you're discussing anything, you need to be extremely mindful of context. Valence electrons are important because they affect how chemical bonds form - it should be pretty obvious why you want to discuss them when you're considering how sodium reacts with itself and other elements.

So when we say sodium is reactive, how do you know that we are talking about the atom "sodium" and not the metallic substance?

@J.Smith for the purposes of having the property of being reactive, they are the same thing.

@J.Smith if you're writing a technical paper, you define your terms more precisely.  If the distinction between atomic sodium and metallic sodium is significant, you mention it explicitly before or after making the statement that sodium is reactive.

"Oxygen" is a set which has "atomic oxygen" and "molecular oxygen" as elements(of a set).
It's like asking why is a "Ford automobile" and a "Chevy automobile", each called an "automobile". So, calling "atomic oxygen", "oxygen" is not inconsistent. Neither is calling "molecular oxygen" oxygen inconsistent.
We can add ozone to this as we can say, "ozone is the triatomic form of oxygen".
There are no inconsistencies here.
     Your question is not of science but of grammar. The singular and plural forms of the chemical elements are the same. We don't say, "oxygens", or "aluminums" unless we are talking about a property as "oxygens melting point is...". This question doesn't really belong here, but belongs in the category of English grammar.

But I have been taught that “elements” are species of atoms. If oxygen is a species of atoms with 8 protons then how can we call O2 oxygen?

Diamond and graphite are both arrangements of carbon atoms, each with very different properties and yet they are both pure carbon, as is a single carbon atom, of course. Likewise $\ce{O2}$ is as much oxygen as atomic oxygen is. The only complication is that what we habitually think of as oxygen is oxygen as a gas comprised of $\ce{O2}$ molecules. Like Humpty Dumpty in Alice in Wonderland, "a word means what [we] choose it to mean" and often we have to add modifiers or alternate terms to avoid ambiguity. Such is language...

If O2 is a type of molecule, can we refer to a collection of O2 molecules simply as O2? Because it doesn’t matter how much we have, it is still O2 right?


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Substances - A Macroscopic Concept
Does the term “substance” refer to any sample of matter that is tangible? In other words is “substance” a macroscopic concept or can we also refer to individual particles such as atoms, molecules, etc. as substances?

It depends on the context
"Substance" is not a tightly defined term so the argument is largely linguistic not about chemistry. Unless, that is, you are talking about observable chemical properties and use the word. Then, the context of use matters and you might need to be careful in how the word is used.
With gases it doesn't much matter how much stuff you have as the individual molecules don't interact much. But for solids and liquids when you talk about "substance" the apparent chemical properties may depend on how many molecules or atoms you have. So you might need to be careful how many of the components are present. The reason why I say this is that some collections of small numbers of components have different properties to larger collections.
Nanoparticles of gold, for example, have different properties to bulk metallic gold. Some show much stronger colours that metallic gold (eg small colloidal gold clusters are used to impart a red colour to some specialty glass products). So the small clusters have different properties to bulk gold and you don't see the bulk properties until the amount of gold is large enough.
So although the substance of both bulk gold and colloidal clusters of gold atoms might be said to be the same, it is probably better to be more specific and talk about colloidal clusters as being a different "substance" to bulk gold. But this isn't universally agreed as usage of substance is not precisely defined.
In short, if you are talking purely about what something is made from then the amount doesn't matter and you can say substance for any amount but if you are talkingg about the properties of the material it can be better to be specific as the properties do change with the amount of the components involved.

So if I were to say that “sodium” is a reactive substance would I be referring to sodium atoms or sodium metal?

@Interestinggg For sodium it would make no difference. Pure metals contain atoms not molecules of their element so, in reactions, act like they react one atom at a time in simple reactions.

Alright so elements are substances “made up of one type of atom”. And sodium is an element. So why can we say that sodium “has one valence electron” when sodium is a substance not a type of atom?

@Interestinggg Context. And note that just because sodium is a substance doesn't mean we can't talk about sodium atoms. Sodium is both a type of atom and a substance. If we talk valence electrons the context tests us we are talking about atoms; if we talk "soft, reactive metal" we are talking about the bulk substance.

Could you just explain what the “made up of one type of atom” thing means. This would clear a lot of confusion. Regards.

@Interestinggg It isn't complicated. A lump of pure sodium metal only contains sodium atoms and no other type of atom. An oxygen molecule only contains oxygen atoms; oxygen gas contains oxygen molecules which consist only of two oxygen atoms.

An element is more like a form of matter isn’t it? As opposed to a physical aggregation of particles?

@Interestinggg "element" is a much broader term than that. Single atoms are by definition elements. small aggregates of atoms can be mede up from a single element. Large aggregates can be mad from a single element but don't have to be. Many of the properties of an element in large aggregates are often called the properties of the element but some elements form more than one type of aggregate. Ozone and dioxygen are both molecular forms of oxygen; diamond and graphite are both aggregates of carbon.


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Sugar: A molecule or crystal
Is sugar a molecule or a large crystalline structure composed of many molecules? What do we mean when we say sugar?

Sugar is white powder used for sweetening our tea; the moment you realize it consists of molecules, it ceases to be called "sugar". The related molecule, and also the crystalline structure composed of these molecules are both called "sucrose".

You're asking another time about the same misunderstanding! Sugar is compound which can be in multiple states, just like water, oxygen etc. Compound or element doesn't equal atom/molecule/crystal /gas/liquid but can be in one of these forms. Is it clear now?

So basically, “element” is a way of referring to a certain type of material made of one type of atom regardless of how much you have? So one atom of sodium is “sodium” and a collection of sodium atoms can still be called “sodium”?

yes, but generally we use extra words to help remove ambiguity, for example sodium metal, or sodium vapour or sodium ions, or sugar solution or sugar crystals.

Terminology can be confusing especially when some terms are only vaguely defined.
"Sugar" is one of those vague terms. There is a whole family of sugars recognised by chemists. In casual use what people usually mean is sucrose, a specific sugar molecule made from two other sugars, fructose and glucose, joined together. this is produced from sugarcane or sugar beet and is the dominant form of sugar seen by non-chemists.
But is sugar the molecule or the crystal? Both. Many pure chemicals form crystals, indeed crystallisation is a very common way of producing purer chemicals. Those crystals consist of a regular 3D array of a single molecule (for a pure chemical). So "sugar" describes both the molecule and the crystals of that molecule which consist of a large number of them arranged in a regular structure.

To answer your question I will separate it into two parts, crystal and sugar.
I wills start with the concept of a crystal, for that I will cite wikipedia since I couldn't find the IUPAC definition of a crystal:

A crystal or crystalline solid is a solid material whose constituents (such as atoms, molecules, or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

So basically stating a crystal is a structure that has a unit-cell that repeats in all direction. Every unit-cell is exactly the same as the other ones.
Sugar is a carbonhydrate. It may constitute of one (example glucose), two(example sucrose), or even a chain of sub-units (example starch).
The white sugar that one buys at a store is crystalline sucrose. It is not a big crystal as a slab of Aluminum is. But it is still a crystalline structure made of a repeating unit-cell which is sucrose molecules.

A slab of aluminum is not one big crystal either.

"carbonhydrate" is not a correct spelling


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Valence Electrons of Molecules
Why do we say that hydrogen has 1 valence electron? Hydrogen (H2) is a molecule so how does this make sense?

Don't misunderstand it. When you say Hydrogen has 1 valence electron, it means you are referring to Hydrogen atom (H) by default. Although, it's ambiguous saying Hydrogen has 1 valence electron. It should be said as Hydrogen atom has 1 valence electron.

Come to think of it, there is no ambiguity. Molecules don't have valence electrons; atoms do.


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Apple Cider vs. Cleaning Vinegar
My grandparents stumbled on some cheap  Heinz Apple Cider Vinegar while cleaning their house.

Ingredients: Made from the juice of apples diluted with water to a uniform pickling and table strength of 5% (50 grains) acidity.

Barring the origin and 1% difference in acidity, how does it differ from Heinz® All Natural Cleaning Vinegar? Is Apple Cider Vinegar equally as potent for cleaning clothes and the house?

Ingredients:
  Made from select sun-ripened grain, diluted with water to a special cleaning strength of 6% acidity.

Beverly Leestma purports that the brand beneath at Walmart didn't indicate "the acidity but did state it was “Not For Food Consumption." Some vinegars (like this Industrial Strength Cleaning Vinegar) are still 5% but aren’t refined or purified to be food quality."

Assuming that the industrial strength cleaning vinegar's acidity is also 5%, then why can't it be consumed?
Why must raw vinegar be refined or purified for human consumption?
Is Beverly Leestma correct in her wariness beneath?

Part of me thinks Heinz is just looking to get in on the popularity of homemakers who want natural cleaning.  I’m not convinced that the 1% of extra acidity warrants the higher price tag and that maybe they’re just trying to cash in on a trend at our expense.

Cider vinegar is made from apple juice so it has stuff from apples left in it. Thus it will not dry "clean". If you leave some out to evaporate until dry you'll get a gummy spot. //  "White" vinegar can be made from cider vinegar by distilling off  the acetic acid (vinegar) and leaving the apple gunk in the pot. So white vinegar is just acetic acid and water. It will dry "clean" without leaving a residue.

The acetic acid in "Cleaning vinegar" may come from an industrial process. One way to produce acetic acid industrially, is carbonylation of methanol. And methanol is toxic!
So unless the product has been analyzed for residues of methanol and approved for food consumption, it is probably unwise to use it for consumption.


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Bond length in NO3-
I am doing a multiple choice question for which one of the possible answers is, for the Lewis structure of NO3- (one of the oxygens is double bonded to the nitrogen and the other two are single bonded),
"one of the N-O bonds is shorter than the other two".
Apparently this is not the right answer, but my understanding was that a double bond is shorter than a single bond. I just googled it and I'm reading that the more bonds there are the shorter the bond length. So shouldn't that answer be correct?
Thanks!

Resonance, resonance... all three bonds are equal.

Could you explain that? Is it just that the "double bond" moves around equally around the N?

It's not really a duplicate (IMO), but it contains the answer to this question nevertheless.

The main reason for your answer being wrong is the phenomenon of resonance in nitrate ion dur to presence of a conjugated system. The resonance is shown in the figure below. Dur to resonance none of the cannonical structures explain the bond length in nitrate except the resonance hybrid ( which happens to be the last drawn structure). Due to resonance all bond lengths would be the same


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Long linear molecule with alternating elements?
In the classic freshman physics text Electricity and Magnetism by Purcell, one of his first examples is the computation of the electrical binding energy of a $\ce{NaCl}$ crystal. He models the crystal in terms of perfect ionic bonding and sums up the energy $q_1q_2/r$ pairwise, computing the $r$ values in three dimensions. (Actually he just gives the first few terms explicitly and then says he carried out the computation numerically on a computer.) The result is $U=-0.8738Ne^2/a$, where $N$ is the number of ions, $a$ is the edge length of the cubic lattice, and the units are cgs, so that the Coulomb constant equals 1.
I thought this was a nice example, but it seemed unfortunate that the actual result could only be found through a fairly complicated numerical computation. I thought it would be nice if I could come up with a similar but simpler example for use with my freshman E&M classes, using a linear molecule with charges $\pm e$ arranged like $\cdots +-+-+-\cdots$. A relatively simple calculation gives an energy of
$$-\frac{Ne^2}{4a}\sum_{k=1}^\infty \frac{1}{k(k-1/2)}.$$
I imagine that the sum can be done in closed form, but anyway it's clearly finite and positive, so that the result is a finite binding energy that has the right sign to keep the system bound.
Is there any molecule that actually has this structure, or something close to it? Googling and asking a chemist in the hall resulted in a variety of examples that are not exactly what I had in mind:

There are various 3-ion molecules such as $\ce{BeF2}$.
You can have long carbon molecules, with hydrogens hanging off of them, but they're not exactly linear, and in any case the carbons are all the same element, so there is no ionic bonding.
Acetylene is linear.

Does nature provide anything like the $+-+-$ chain that I'm describing? Do they not occur because they're too unstable with respect to crumpling up? Could they exist on a substrate (the kind of thing people build using scanning electron microscope probes) or as linear impurities in a bulk solid?
For my purposes, I don't think it really matters whether the bonding is perfect ionic bonding, as long as there is some segregation of charge.
Related:

Why does sulfur, but not oxygen, catenate?
Are long oxygen molecules possible?

I believe I read many years ago in a solid state physics textbook that the calculated speed of sound in a 1D crystal is imaginary. If there is no context I am forgetting, wouldn't that suggest they cannot exist?

While bonding in molecule may have more ionic then covalent component, they are rather treated as covalent - any simple ionic model shouldn't be useful.

My main criticism is on line with the comment of @Mithoron. Treating a long chain with alternating (partially) oppositely charged fragments as kept in place by electrostatic force isn't the proper treatment. Perhaps one can estimate a coulombic term but then we enter in the computational and quantum chemistry field, truly.

@Alchimista: By Earnshaw's theorem, no system can be in a stable, static equilibrium due to electrical forces. That isn't what Purcell is claiming. He goes into a lengthy discussion of this.

Therefore I say. It is somewhat the Peierl instability in a  charged version. It is not clear what you want calculate.

related: https://math.stackexchange.com/questions/3026370/evaluating-the-series-sum-k-1-infty-1-kk-1-2

It is even visually clear that a linear configuration of + and - ....can be at best merastable. Your question boiled down to " is there any covalent structure stabilizing such above mentioned linear configuration?". Answers have provided what you were looking for. But for clarity: remind that a system bound is not a system in its more stable configuration. If I orbit earth I'm bound to it. Still I can only decay to a lower orbit. Shortly: the molecules you can use as example are kept as they are not (only) by the electrostatic term you want to calculate.

..... that is why Purcell treat NaCl as it is, a three dimensional array, in spite this requires a tedious computation.

@Alchimista: No, the three-dimensional NaCl is also classically unstable. Earnshaw's theorem doesn't have anything to do with whether the structure has one geometry or another.

I don't get what are you looking for anymore. Do you agree that NaCl does exist? And that is a three dimensional array?

An example of a linear structure is given by cycopentadienyl lithium, whose structure is explored in this question.  There is some covalent interaction in this polymeric structure, as in most organolithium compounds, but the bonding between lithium and cyclopentadienyl groups is highly polar.

Check for polyoxymethylene (POM).

It has partial positive charge on carbons and partial negative on oxygens, and it's "long" enough to serve as plausible example for sum with large maximum k. But I should mention that there are no perfectly linear polymers which are like string of beads.

Not true how about indefinitely long polyynes & cumulenes? https://bit.ly/2Sp81yb I'd imagine any deviations from linearity would be pretty minor! ; )

@ManRow: I'm having trouble parsing your comment. What are you saying is not true? By the way, it's not necessary to give shortened bit.ly links; you can just give a link straight to Wikipedia, which makes it easier for people to understand what the link is going to be about. The link re linear acetylenic carbon is interesting, but it isn't ionic bonding, since every atom in the chain is a carbon...?

Hi! I was commenting on what Kelly Shepphard said about linear polymers "like a string of beads" not existing, and just provided some counterexamples : )

ManRow, they are still not perfectly linear (and I've said about perfect linearity), and  definitely have impurities (as any other polymer). Besides, usually they exist only as rigid part of molecule whis is linear copolymer, dendric structure or, at least, crosslinked polymer. I even bet that such "strings of beads" would crosslink spontaneously.

@KellyShepphard Haha well those polymers I gave are definitely quite unlikely to be stable in any sort of "real world" conditions (aside from some extremely supercooled or weird other  setups). And, while probably quite more "linear" than the other more common polymers that we do see around, they aren't really going to be "perfect" by any stretch, like most things in the real world!

Furthermore, I'm guessing they may get even less "linear" the longer they get too (im guessing), as deviations with the extra length (& having more subunits) increase https://en.wikipedia.org/wiki/Cumulene So propadiene may be the "most linear' of them so far, while butatriene and longer cumulenes even less!

Then it's not a conterexample for now. But maybe one day humanity will master some new reactions and ways to support perfectly linear long structures - this day will be the day when your example becomes counterexample :) For now, it's like example that polymers can't be perfectly linear - even if parts of them are.

Invite both to consider that you are about right and wrong at the same time :)


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What is the resulting shape all orbitals are added up?
In all scientific books and Wikipedia (e.g. this video), the shapes of s, p, d, and f orbitals are always drawn individually, but not the sum of them.
If we add up all the orbitals of an atom, for example for the atom of krypton, what will be the resulting shape, or what space will be occupied?

If you take a look at all of the electron density of an isolated atom, it must be spherically symmetric over a time average. Why? Because space is isotropic, that is, it looks identical in any direction. Specifically, our choice for $x$-, $y$-, $z$- axes is arbitrary.
If you say that the electron distribution is not symmetric, there must be some preferred direction, call it $z$. But our choice of $z$ is arbitrary in space, so some other $z$ will do just as well, and we expect that the atom will have the same preferred direction, which is a contradiction. The original assumption of non-spherical distribution is wrong.


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Does OH- attack carbonyl carbon in aldehyde during aldol reaction?
Why does $\ce {OH^-}$ not attack the carbonyl carbon in an aldehyde during the aldol reaction and instead, deprotonate the $\alpha$ hydrogen to form an enolate intermediate?
Also, why does the enolate intermediate attack the carbonyl carbon of another aldehyde once it is formed?

It does, but the addition is reversible! It is an unproductive step. Welcome to ChemSE!

Thank you.Many reactions are reversible i.e in SN1 reaction carbocation formation is reversible but the reaction still proceeds. could you please provide an alternate explination

Related: https://chemistry.stackexchange.com/questions/64425/why-do-some-nucleophiles-attack-at-the-carbonyl-group-while-others-at-an-alpha-c?rq=1

Could you clarify your question? It isn't clear at the moment. The phrase "But instead enolate intermediate attacks the carbonyl carbon" seems to be rather out of place.

As noted by Jan here, both pathways do proceed, i.e. the hydroxide attacks the carbonyl carbon and also deprotonates the $\ce {\alpha-H}$. To give you a rough gauge of the extent of each reaction, I will now provide some data from Carey & Sundberg (2007).

In general, the hydration of carbonyl compounds (aldehydes, ketones) is unfavourable, i.e. the equilibrium constant has a value that is less than 1. Only in cases where there are inductive (e.g. presence of halogen substituents) or mesomeric effects (e.g. presence of phenyl substituents) operating will the value be much greater than 1. Also, it is noted that the hydration proceeds very quickly, i.e. the rate of reaction is pretty fast. Note that this is reversible.
Having a greater number of steps, amongst other factors, results in the aldol reaction being slower in general. The equilibriums of each individual step in the aldol reaction mechanism are likely to not establish themselves as quickly as that for the hydration of the carbonyl.
From Carey & Sundberg (2007), p. 684:

In general, the reactions in the addition phase of both the base- and acid-catalyzed mechanisms are reversible. The equilibrium constant for addition is usually unfavorable for ketones. The equilibrium constant for the dehydration phase is usually favorable because of the conjugated $\alpha$, $\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall reaction can go to completion, even if the equilibrium constant for the addition step is unfavorable.

So what happens? In the aldol condensation reaction, the dehydration step is what essentially moves the whole equilibrium forward. As the carbonyls react in this aldol pathway and are depleted, the hydrates would convert back to this carbonyl form to react in this aldol pathway. Eventually, if the reaction were to go to completion, the amount of hydrates present would go to zero and all the carbonyls would have been converted to the $\alpha$, $\beta$-unsaturated carbonyl.
Reference
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.


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Why does diamond have no molecular formula?
Why does diamond have no molecular formula, while buckminsterfullerene does have it, despite both being allotropes of the same element?

somewhat related to https://chemistry.stackexchange.com/questions/75170/why-is-the-buckminsterfullerene-the-purest-form-of-carbon/75177#75177

To quote my answer to Is a diamond a single molecule?:

Diamond is a covalent network solid, like a number of other common materials (quartz, graphite, glass, and a whole bunch of stuff).
Because they are not discrete molecules - there is no 'diamond' molecule the same way there are molecules of caffeine, benzoic acid, citric acid, N,N-dimethylaminopyridine, etc.

Another way to think about it, since diamond is a covalent network solid, each diamond is a single giant molecule. Since each diamond is one molecule, there is no universal diamond molecule with the exact same number of carbon atoms every time it occurs.
Fullerenes form distinct molecules with the same number of carbon atoms in each species every time. Thus buckminsterfullerene is always comprised of molecules of $\ce{C_{60}}$.


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Why does ligand substitution occur when OH- is a better ligand than NH3?
$$\ce{[Cr(H2O)3(OH)3](s) + 6NH3 -> [Cr(NH3)6]^3+ + 3H2O + 3OH-}$$
When excess $\ce{NH3}$ is added to triaquatrihydroxychromium(III), why does ligand substitution occur when $\ce{OH-}$ is a better ligand than $\ce{NH3}$?
Wouldn't the 3 $\ce{H2O}$ molecules be deprotonated to $\ce{OH-}$ and therefore form $\ce{[Cr(OH)6]^3-}$?
This also seems to occur when excess $\ce{NH3}$ is added to tetraaquadihydroxocobalt(II):
$$\ce{Co(H2O)4(OH)2(s) + 6NH3 -> [Co(NH3)6]^2+ + 4H2O + 2OH-}$$

"Excess" may mean that this occurs simply because the equilibrium is pushed to the right to form the hexaammine complex due to overwhelming conc. of NH3

Have a look at Jan's answer here: https://chemistry.stackexchange.com/q/68547/44877

This is an equilibrium reaction. Using e.g. LeChatelier's principle, you can figure you that if you increase the concentration of $\ce{NH3}$ sufficiently, you shift the equilibrium so far to the right that you get the hexamine as the main product. At neutral pH in water, the concentration of hydroxide is quite low (100 nanomolar). You would expect this reaction to be pH-dependent, and of course depend on the concentration of ammonia in solution.


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Isotopes of Elements
When we talk about the substance $\ce{H2O}$ with a molar mass of $\pu{18.0 g/mol}$, what hydrogen and oxygen isotopes are present in the $\ce{H2O}$ molecules? In other words, what isotopes do the symbols represent?

Most periodic tables quote the molar mass of an element as the average mass of the element given the natural abundance of the isotopes. Chlorine, for example, has a standard atomic weight of 35.45 which is a weighted average of ~75% 35Cl and ~25%  37Cl, the two common isotopes.
But isotopes of hydrogen and oxygen are rare. So the vast majority of water molecules will not contain them. So, at least to one decimal place, most water will have a molar mass of 18.0.
At it's natural abundance of about 0.25% 18O won't noticeably affect the molecular weight of water until you can measure it to at least 3 decimal places of precision. Slightly simplifying (assuming pure isotopic masses are exact integers, which they are not), water containing that isotope at its natural abundance will have an an average mass of 18.005 rather than 18.000.
When working with normal (unenriched) substances the best assumption is that they will contain the normal isotopic abundance of the component elements (which are well known and, for most elements, don't very much by source). But the effect on molecular mass will often be small and you will need a fair amount of precision to notice it. If you are not working to very precise masses, it is fair to assume natural isotopic abundance.

It's impossible to find isotopic composition of a compound with a molecular weight containing only one decimal point ($\pu{18.\underline{0} g mol-1}$). Typically, 6th decimal place is required in order to refine percentage of each isotope.
Symbols themselves (such as $\ce{H}$ and $\ce{O}$) don't carry any additional information except atomic number $Z$. For isotopes you need to use so-called AZE notation: $^A_Z\ce{E}$, for instance $\ce{^1_1H}$ and $\ce{^{17}_8O}$. Since $Z$ is already known from $\ce{E}$, simplified notation is often used instead: $\ce{^1H}$ and $\ce{^{17}O}$.

I'm just confused as to where the molar mass of a substance comes from. As in how do we know how much a mole of a substance weighs exactly? I could have 2 samples of 1 mole of water, and one could weigh more than the other because one may contain more heavy water (D2O) than the other.

Unless you took one water sample from the nuclear reactor and the other one from the river, isotopic distribution is going to be the same as elements have uniform isotopic distribution in nature. You probably want to read these topics: https://chemistry.stackexchange.com/questions/71942/what-is-the-standard-molar-weight-of-carbon-and-why and https://chemistry.stackexchange.com/questions/38082/what-is-the-difference-between-molecular-mass-average-atomic-mass-and-mola and any introductory chemistry textbook.

Hydrogen has 2 stable isotopes,H-1 and H-2, Oxygen has 3 stable isotopes H-16, H-17 and H-18.
There is very little of the minor isotopes so the molecular weight of water is 18.
However hydrogen and oxygen are light elements and differences in atomic weights have detectable differences in physical and chemical properties. For example H-2 compounds can have lower reaction rates than the H-1. Boiling points and freezing points of various isotopic forms of water differ slightly. Indeed it possible to distil water to separate the various isotopes.
This also happens in natural systems where minute variation in isotopic contents trapped in ancient  ice can be used to gauge the ocean temperature of that time


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What explains the stark white colour of my iodometric titration analyte solution?
During my Advanced Higher (secondary school) investigation, I reacted pure copper with concentrated nitric acid to acquire a copper nitrate solution. This was neutralised with excess sodium carbonate and brought back to a clear solution with ethanoic acid.
I then reacted the solution with potassium iodide in a conical flask before titrating it with sodium thiosulfate.:
$\ce{2Cu^2+(aq) + 4I^-(aq) → 2CuI(s) + I_2(aq)}$
$\ce{I_2(aq) + I^-(aq) ⇌ I_3^-(aq)}$
$\ce{I_3^-(aq) + 2S_2O_3^2-(aq) → S_4O_6^2-(aq) + 3I^-(aq)}$
This analyte solution was at first brown, but when titrated it turned a pale coffee colour. When starch was added just before the end point, it turned a dark grey-brown colour, as expected - the iodine forming a complex with starch. https://en.wikipedia.org/wiki/Iodine_test
But when all of the iodine had reacted with the sodium thiosulfate, I was left with a stark white solution. I tested to make sure no iodine was left over by adding in another drop of starch - no reaction. So, this seems to be the endpoint.
Copper iodide seems the likely explanation, but the wikipedia page shows that it is mostly insoluble in water. It is soluble in iodine solutions but all the iodine is gone, presumably.
Does it take time for the copper iodide to precipitate? I quickly rinsed out the conical flask. There was some white staining on the sides which I think is copper iodide.

You probably have the solution of potassium iodide in excess for the oxidation. $\ce{CuI}$ is somewhat soluble in in iodide solutions (not in iodine solution), due to common ion effect (see Wikipedia for solubility).

@MathewMahindaratne Yes, that makes sense! I can't believe I forgot about the excess potassium iodide. It's the small things that always end up being the key. Can you post your comment as an answer so I can accept it?

I agree with one of answers provided that what you obtained at the end point is a colloidal solution of copper(I) iodide. Since $\ce{CuI}$ is insoluble in water (0.0042% w/v), the stark white color (probably milky) solution is nano particles of the solid suspended in $\ce{KI}$ solution, assuming you have used significantly excess $\ce{KI}$ for the reaction. Thus your solution should be in following equilibrium:
$$\ce{CuI + I- <=> [CuI2]-}$$
Wikipedia states that:

$\ce{CuI}$ is poorly soluble in water (0.00042 g/L at 25 °C), but it dissolves in the presence of $\ce{NaI}$ or $\ce{KI}$ to give the linear anion [$\ce{[CuI2]-}$. Dilution of such solutions with water reprecipitates $\ce{CuI}$. This dissolution–precipitation process is employed to purify $\ce{CuI}$, affording colorless samples.

Thus, your intention is to isolate solid $\ce{CuI}$, you must add extra $\ce{KI}$ while moderate heating your solution until you receive a clear solution. Let the solution to cool to room temperature, and add adequate amount of pure water (about 4 times by volume) to cause the precipitation. Upon standing for about $\pu{1 h}$, $\ce{CuI}$ will precipitate.
A typical procedure is given below in literature (Ref. 1):

Commercial copper(I) iodide or a sample that has become discolored ($\pu{3.80g}$, $\pu{0.02 mole}$) placed in a 250-mL Erlenmeyer flask is dissolved with magnetic stirring and moderate heating by the addition of a minimum volume (about $\pu{125 mL}$) of $\pu{ 3.5 M}$ potassium iodide (about $\pu{3 hr}$ is required). Activated charcoal (e.g., Matheson, Coleman, and Bell "Darco" (3-60) (about $\pu{5 g}$) is added to the resulting solution, which varies in color from light tan to dark brown, depending upon the extent of contamination. The mixture is stirred magnetically with moderate heating until the supernatant solution becomes colorless, at which time stirring is stopped and the charcoal is allowed to settle, indicating that the contaminating iodine has been completely adsorbed (about $\pu{10 min}$). The mixture is suction-filtered through a 150-mL (60-mm i.d.) sintered-glass funnel (medium porosity), and the filtrate is diluted with sufficient water (about $\pu{500 mL}$) to cause reprecipitation of the copper(I) iodide upon standing for about one hr.

Important: Vigorous heating in air causes $\ce{CuI}$ to decomposes to $\ce{CuO}$.
Ref. 1: Inorganic Syntheses, Volume 22; Smith L Holt, Jr., Ed.; Inorganic Syntheses, Inc.: New York, NY, 1983, Chapter 2: Transition Metal Complexes and Compounds (sub Title: 20. Purification of Copper (I) Iodide; submitted by G. B. Kauffman & L. Y. Fang; Checked by N. Vishwanathan & G. Townsend), pp 102–103 (https://doi.org/10.1002/9780470132531.ch20).

You’re correct that this is just a suspension of copper(I) iodide. If CuI were soluble in water, the solution would be clear; the white colour is a sign of (very small) particles of the solid suspended in solution.