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Does heavy water taste sweet?
In this YouTube video from Cody's Lab, Cody claims that heavy water tastes sweet.
He does some fairly convincing comparisons but still expresses a little doubt that the effect is real.
Has this been studied by others? Is the effect verified and if so, what possible explanations exist?

Just as a side note: You might find the vibrational theory of olfaction interesting, see for example https://en.wikipedia.org/wiki/Vibration_theory_of_olfaction

dude dont try anything of that sort! I just saw that guy drinking cyanide https://www.youtube.com/watch?v=bWNpO5vvhpk

@Tetrahydrocannabinol he also took a bite out of sodium metal...

I find cold water sweeter than warm water and he did say that heavy water feels colder.

@Tetrahydrocannabinol - I found Cody's youtube channel a while back, when he was embedding magnets in his fingers.  He's certainly ... interesting.  But I don't think most people would be advised to do many of the things he does.

From the abstract of the article by Abu et al. [1]:

Here we conclusively demonstrate that humans are, nevertheless, able to distinguish $\ce{D2O}$ from $\ce{H2O}$ by taste. Indeed, highly purified heavy water has a distinctly sweeter taste than same-purity normal water and adds to perceived sweetness of sweeteners. In contrast, mice do not prefer $\ce{D2O}$ over $\ce{H2O},$ indicating that they are not likely to perceive heavy water as sweet. For humans, the sweet taste of $\ce{D2O}$ is suppressed by lactisole, which is a known sweetness inhibitor acting via the TAS1R3 monomer of the TAS1R2/TAS1R3 sweet taste receptor. HEK 293T cells transfected with the TAS1R2/TAS1R3 heterodimer and the chimeric Gα16gust44 G-protein are activated by $\ce{D2O}$ but not by $\ce{H2O}$. The present study resolves a long-standing controversy about the taste of heavy water, confirms that its sweet taste is mediated by the human TAS1R2/TAS1R3 taste receptor, and opens way to future studies of potential sites and modes of action.

Reference

Abu, N. B.; Mason, P. E.; Klein, H.; Dubovski, N.; Shoshan-Galeczki, Y. B.; Malach, E.; Pražienková, V.; Maletínská, L.; Tempra, C.; Chamorro, V. C.; Cvačka, J.; Behrens, M.; Niv, M. Y.; Jungwirth, P. Sweet Taste of Heavy Water. bioRxiv 2020, 2020.05.22.110205. DOI: 10.1101/2020.05.22.110205

According to H.C. Urey and G. Failla, Science  15 Mar 1935,
Vol. 81, Issue 2098, pp. 273, there's no difference in the taste of ordinary and heavy water.

Very small sample size but an interesting reference indeed! (+1)

And now one definitively shown to be wrong by the Abu, et. al. paper.

OK, first of all I want to say I really dislike the videos that this guy puts out, since he is promoting unsafe handling of chemicals. He might know what he's doing, but it's a terrible example.
To the actual question: surprisingly there are nearly no sources on this. I can understand that this isn't something a lot of people would try nowadays, but I imagined that a lot of people tried 50 years ago or something like that.
There's one science article describing a blind taste test of three persons in comparison to distilled water, and they couldn't spot a difference.
Another paper describes a difference, and also cites older research that showed that rats avoid $\ce{D2O}$ if they have the option to choose. But that might also be smell. It is known that deuterated substances can have a different smell and taste.
Heavy water also has a higher density and viscosity, so that might also be a factor which makes it "taste" different. Fun fact: it also has a different color.
And well, he didn't do the testing in a scientific manner, which would minimally include blinding the observer.

Watch the video starting at 6:47. He let his girlfriend (or his sister, or someone, idk) taste taste taste it and she didn't know which tube contained normal water and which one contained heavy water.

@UTF-8 The odds of it being sheer luck is 50%. No conclusions can be drawn from that.

"he is promoting unsafe handling of chemicals" - ..and explosives. I hope he doesn't learn the hard way what static electricity can do to an unstable explosive.

@Fermiparadox Well, my sister started studying biology.They had a group visit a lecture.That group tested some gene manipulation on mice. They had 8 mice in total so I guess probably at most 4 which got a treatment. My sister has no idea what's meant by some number and then "sigma" but the group said that everything equal to or above 0.5 σ is relevant for confirming their hypotheses. So if you compare it with what's called "science" in biology ... :D ;-) Of course it's just 2 people on YT allegedly tasting it. But a study involving 3 people stating the opposite isn't exactly meaningful either.

@UTF-8 - I watched that video recently and IIRC the test tube that purported to be heavy water was marked with a sticker. Less than subtle choice forcing in my opinion.

@oPless Some way of reliably telling them apart is needed. He could've attached the sticker to either one.

Maybe his videos are promoting dangerous behaviour. Or, since he seems to take good precautions, they are bait for the Darwin awards.

@matt_black Maybe? Really? He is biting into sodium metal. There's also no good precautions you can take for that. It's just dangerous and extremly stupid.

It's educational, though. Extremely stupid is being reckless for no apparent gain.

@CeesTimmerman what's the educational value of biting into sodium? And I doubt it's for education, it's more likely just for money, and doing dangerous stuff for a little bit of money is stupid in my eyes. There are so many interesting and educational experiments you could do instead of eating cyanide or biting into sodium (but you will make less money).

He explains why biting sodium sounds like a terrible idea, and why his attempt is likely to succeed while warning people not to replicate it. He doesn't mess around doing the same stunt until it kills him and/or others. Was sodium metal generally known to taste tangy and turn tongue tips into soap before? I think not.

I tastet heavy water assuming it has not taste and was surprised about the sweet taste (Aldrich, Deuterium oxide, 99.9%). The bottle was old but still sealed. Three more people also defined the content of the same bottle as slightly sweet without knowing what my perception was.

Please expand on your answer - the OP asks, in addition to if this effect has been studied (not observed) and verified by others, what possible explanations exist.


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How to compare stability of following two alkenes

How to find which alkene is more stable ?. As same number of hydrogens are added across the double bond so I am not able to use the fact that more substituted alkene is more stable.
I was initially thinking in terms of steric effect (Repulsion Between the two methyl groups) which led me to believe the second alkene should be less stable but in answer given the second alkene is shown more stable .
Can somebody explain on what basis this result is deduced. Any help is appreciated.

related           http://chemistry.stackexchange.com/questions/27346/why-is-2-methylpropene-less-in-energy-than-its-alkene-counterparts

The second alkene is more stable since it is symmetric. Achiral compounds are always more stable than chiral compounds due to the Gauche Effect. The explanation for why the Gauche Effect takes place is here.

source: Stack Exchange

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sort: votes


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How to solvethis problem related to redox titration?

A $1.0\ \mathrm{g}$ sample containing $\ce{BaCl2*2H2O}$ was dissolved and an excess of $\ce{K2CrO4}$ solution added. After a suitable period, the $\ce{BaCrO4}$ was filtered, washed and redissolved in $\ce{HCl}$ to convert $\ce{CrO4^2-}$ to $\ce{Cr2O7^2-}$. An excess of $\ce{KI}$ was added, and the liberated iodine was titrated with $84.7\ \mathrm{mL}$ of $0.137\ \mathrm{M}$ sodium thiosulphate. Calculate the percent purity of $\ce{BaCl2*2H2O}$.

I first found the number of mili-equivalents of sodium thiosulphate  and equated that to the number of moles of pure $\ce{BaCl2*2H2O}$ times the n-factor i.e. $2$.
From this equation I get the number of moles of pure $\ce{BaCl2*2H2O}$ and then multiply this by its molecular mass to get the mass of pure $\ce{BaCl2*2H2O}$. But it comes greater than $1\ \mathrm{g}$. How to solve this?

Chromium(VI) is reduced to chromium(III) while oxidizing iodide to iodine. So one equivalent of chromate corresponds to 3 equivalents of iodine ($\ce{I}$, not $\ce{I2}$).
$$\ce{BaCl2 \cdot 2 H2O} \equiv \ce{BaCrO4} ~\mathrm{(chrome(VI))} \equiv \frac{3}{2}~\ce{I2} \equiv 3~\ce{S2O3^{2-}}$$
If you take this into account you should probably end up with 87.5% purity.

K 2 Cr 2 O 7 + 7H 2 SO 4 + 6KI → 4K 2 SO 4 + Cr 2 (SO 4 ) 3 + 7H 2 O +3I 2 , Is this not the reaction? Also the what would be the product when I reacts with  Na2S2O3? Can you please elaborate a little ?

Your reaction equation is correct. However, spectator ions are not very interesting for understanding what's going on. The essential thing is that 1 mole chromium(IV) oxidizes 3 moles iodide, and 1 mole $\ce{I2}$ will oxidize 2 moles thiosulfate, hence 1 mole iodine atoms oxidize 1 mole thiosulfate. Therefore 3 moles thiosulfate are equivalent to 1 mole chromium(VI) and hence to 1 mole barium chloride.

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Is there a physical explanation behind the derivation of spectroscopic term symbols?
In deriving the spectroscopic term symbols, one method seems common among several sources, the clearest presentation for me being located here, a derivation of the term symbols for a $d^2$ ion.
So, the execution of this method seems pretty straightforward. My confusion is in why it works. It currently seems rather arbitrary (a $L=4$ G term symbol subtracts a microstate from eg. $L=0$). Stumbling blocks for me are:

1. Some singlet microstates seem redundant
http://sli.mg/j4t7mg.png
If I look at, for example, the wiki diagram for LS coupling by vector addition, I would assume S = 0 and L is equal to 1 in both of the above microstates.

2. Some higher multiplicity states also seem redundant

From the same vector addition diagram above, these top row microstates should be equivalent to the bottom row microstate cross-diagonal from it. As far as I have understood, the spectral term takes the absolute value of L and does not distinguish between -L and +L (eg. L=1 and L=-1 both yield P term symbols), so $|-L-S|=|L+S|$.
Alternatively, I have also considered that maybe the absolute value is only taken on L and not S. In this case, $|-L|+S=|L|+S$. Then the microstates would not be equivalent cross-diagonally but instead horizontally (top row microstates equivalent to each other and bottom row microstates equivalent to each other).

3. Not all term symbols for the potential microstates are a solution.
 $\Rightarrow$ $^3D_3$, $^3D_2$, $^3D_1$ microstates vanish in the derivation
After the method was so thorough in including all the potential microstates, some of these microstates turn out to be nonexistent or physically not relevant.

I am interested to know roughly why this method for the derivation of term symbols is physically logical (why it works), more specifically why the above states are not redundant, and most especially why certain spectroscopic terms / electron microstates are not relevant. I'm afraid the question may be too open ended, so if I can clarify any further, or there's a concise reference, please let me know.
Thank you for your time!

There isn't a one-to-one correspondence between those orbital diagrams and terms which makes it very hard to discuss the matter in this fashion. For example for the very first two diagrams you posted, the actual terms arise from linear combinations of those two orbital microstates (perhaps plus more, I'm not sure).

The terms that don't appear are not "physically unimportant" but rather "physically forbidden" - they violate the Pauli exclusion principle. Again those orbital diagrams aren't very useful to find out why. You need to construct the wavefunctions as linear combinations of those orbital microstates. See for example (a simpler $\mathrm{p^2}$ case - $\mathrm{d^2}$ is more complicated but same idea) Pauli-forbidden term symbols for atomic carbon (btw, I don't think this is too open-ended, so don't worry about that.)

Thank you for this link. I don't fully understand the spatial construction of microstates at a glance, but intend to clarify this later when I have had more time to look at it, as I am currently been torn away from the computer for tonight. Is a term symbol only allowed if every conceivable microstate corresponding to it is not Pauli forbidden (why $^3D$ is not allowed because one microstate for that term would involve a doubly occupied $m_l=1$, from your post)

also have a look at how this calculation is done http://chemistry.stackexchange.com/questions/63174/how-does-one-find-the-ground-state-term-symbol-for-a-configuration-that-is-exact/63261#63261 and some explanation also at http://chemistry.stackexchange.com/questions/54721/why-is-3p-1-an-atomic-term-symbol-for-carbon/54723#54723 which are two examples of how to do these calculations.


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Why is the answer specifically OH-?

When solid A is added to water, some dissolves to form a colourless solution.
When phenolphthalein is added to this mixture, it turns pink.
When dilute hydrochloric acid is added to the mixture, the temperature increases and a colourless solution forms, but no gas is given off.
(i) Identify, by name or formula, the anion present in A.

I believe the answer could be either $\ce{OH-}$ or $\ce{O^2-}$. What am missing?

Sounds about right.

There are many possible answers, not just $\ce{OH-}$ or $\ce{O^{2-}}$.

If you attempt to answer this question in the most literal way possible then it can only be a hydroxide. Note that the question says dissolved and not reacted. Anything that dissolves in water to create an alkaline solution and does not contain hydroxide ions must have reacted in some way. That specifically applies to oxide ions which are not stable in aquaeous solution.
Of course, if you remove the must not react constraint I just added by reading the question extremely literally, then you can have a lot more compounds, e.g. acetates.

Yeah thats what i was thinking too..the suttle difference of "dissolved" and "reacted"..Thanks alot for making it clearer..


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Visual explanation between Molecule vs Compound vs Element vs Atom vs Substance?
Im a visual learner and am trying to find a visual explanation between the meaning of those words.

Is this Venn diagram accurate?
Can you draw a better Venn diagram, perhaps one that would also include other words like homogeneous, heterogeneous, and mixture?
Even if you draw it in Paint, I can re-do it and see if you like it.
Or is this one better?

I don't think that Venn diagrams are the best way to express the relationship between these concepts - the circles in these diagrams are usually sets, and overlap indicates set intersection. Is the intersection of the set of atoms and the set of molecules the elements?

yes, perhaps the "Atoms" should be inside another circle that exists inside the elements circle? I uploaded a new version.

Not every compound consists of molecules.

Table salt ($\ce{NaCl}$) does not consist of molecules. It's a ionic compound.

Better. There are also intermetallic bonds, and probably coordinate bonds. https://en.wikipedia.org/wiki/Chemical_compound

does the updated one make sense?

The overlap of "Molecules" and "Atoms" does not make any sense. You either have an atom or a molecule. From Wikipedia: "A molecule is an electrically neutral group of two or more atoms ...".
Furthermore, "Elements" can either refer to the species of an atom, e.g. $\ce{O}$. In this case it is not a molecule! Or  it refers to the compound an element forms, e.g. $\ce{O2}$. In this case it is not an atom!

Here's a cone view.
Everything from large trees to atoms are viewed by our eyes in the cone view (just my perceptions...). This is a very interesting concept. It's like shooting an electron (into or out) of an element, that changes an physical & chemical properties of the atom, compound and substance etc..
Think about a pencil, and what it's made of.

This may be quite far from a deep chemistry expertise. So take this as a simple representation to accommodate the diagrams that you began with and looking from a different perspective. We aren't trying to fit a facts/theory into our understanding and visualisation but other way around.

An important distinction is the clarity of molecule, compound and mixture.
Molecules are constructed of atoms of same species. A compound is formed by atoms of different species. Compounds are of 2 different types mainly,
Covalent
Ionic
Intermetallic
Coordinate

Each of this bonds, origin goes back to a single atom and involvement of its orbitals energy and electrons.

For an example,
The water that we drink is a molecule and also a compound, made of two H atoms and oxygen atom covalently bonding. Isn't it odd to read in your paper or an article, "the water compound..." or wouldn't it feel fit to read as "a water molecule...".
The salt we add to balance flavour is a compound, made of one sodium atom (+charged ion) and chloride atom (-charged ion) ionically bonding.
The air we breath is a mixture of various molecules : $H_2O, H_2, O_2, N_2, CO_2$ etc (not atoms).

Put it this way, all compounds could be molecules, but (not) vice
  versa.

Here's the top view (transverse plane from the very top) of the cone.

In physical sciences the matter is classified in pyramids like the following. Reference is in the image itself.

There are more advanced and slightly different version of discussions on the same topic here on pubchemblog : "difference between xyz..."..
The starting cone isn't pretty as your images. But I will make them more user-friendly and eye-candy if you can comment. This almost feels like a drawing session now, apart from the conceptual outlook. I am happy to read more about intermetallic and coordinate bonding as well to address that in the answer. It is not something I come across on daily basis as I am not in electronics or soldering/alloy analytical chemistry. Apart from the cone, other picture references are taken from here.

Thank you for your answer. But could you please cite sources on your answer?

@iamcreasy [you have a funny username]. Well the cone view idea is my own perspective and it seems to fit well to the model. I have written information using my chemistry basic knowledge as well. I cited the links that I used. Let me re-check them for you.

@iamcreasy I rechecked the answer. There's no particular source per se. The picture sources are already embedded or the picture shows the reference. Cones are my own images. Do you want me to include wiki or references quoted in the comments by other users?

Yes, some other sources would be very helpful.


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Why do we use phenolphthalein as an acid-base indicator?
It seems that no resource I come across addresses the fact that the pH range for phenolphthalein's color change is around 8.2-10. In other words, it's still colorless at 8.2. Why then, is it claimed that the indicator changes colors when the solution changes from an acid to a base? Wouldn't you want an indicator that changes colors at about pH 7?

If you analyse a titration curve you will immediately notice the pH shooting up near equivalence point. For this reason, there is very little difference to the amount of base added for pH 7 and pH 8 (due to the logarithmic nature of pH). So, it's not entirely accurate but certainly good enough (for strong acid - strong base and weak acid - strong base), if you want something better use an electronic pH tester

Looking at a titration curve clears up a bit of my confusion, but now I'm wondering what the point is of having an indicator at all. If the "volume added" in this graph--http://www.chemguide.co.uk/physical/acidbaseeqia/sawb2.gif--matches up with all those different pH values, what's the point of trying to get it close to 7?

I believe the link was distorted: http://www.chemguide.co.uk/physical/acidbaseeqia/sawb2.gif

pH 7 is the equivalence point where the moles of H+ in the original solution equals the moles of OH- added, so we can construct an equation and solve for unknown

[volume of solution][conc(H+)] = [volume of added solution][conc(OH-)]

If you're wondering about that specific curve, the volume added is only specific to specific concentration of base, specific concentration of the acid, specific volume of acid. The volume added values all change based on the other 3 "variables"

Right, but if it didn't matter that we were at pH 8 with phenolphthalein, my question is why bother with accuracy? Why is such a big deal made when the solution becomes too pink (passed equivalence point)?

Due to the way the pH system works, the overshoot of base increases exponentially with an increase in pH. In this case, pH 10 contains around 100x more overshoot base than pH 8, which is quite often a significant amount (for precise measurements). (You can work this out by converting pH to pOH then finding concentration of hydroxide ions)

Say we have 1L of solution, with pH 8 it would be around 0.00001 moles NaOH (a very small amount), with pH 10 around 0.001 moles, at lot more. Say we overshoot to 11, that would be 0.01, pH 12 around 0.1 moles). Although an error of 0.01 right now may seem small, the concentration and volume of the initial acid is not likely to be high, which makes the error worse after we try to solve for what we want

Looking at the title, the other half of the answer should be that there are few indicators that go from colorless to a color. Phenolphthalein's color change is very easily detectable visually.

I think what OP was asking was since phenolphthalein only tested 8.2-10, why would it be used, as that directly relates to the post

In order to determine what indicator to use, you have to identify the reactants and strong or weak. For a strong base and strong acid reaction the pH shift will likely occur around 7. If its a strong acid vs a weak base, then the pH will probably lower than 7.


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How do self-relighting (trick) birthday candles work?
You may have seen those birthday candles, where, upon blowing them out they smoulder for a bit and then (magically) relight.

How do they work? Why can't regular candles do this (is it possible for a regular candle to relight itself, anyway?)

In normal candles, as you blow them out, you will see burning embers in the candles, which vaporises the wax and thus causes a ribbon of paraffin wax to rise up (this looks like a wisp of smoke). These embers are often not hot enough to light the ribbon of wax, so the candle goes out.
However, in trick candles there is magnesium put in the wick. When normally burning, the liquid wax shields it from oxygen so it cannot ignite. However, when the candle is blown out, this magnesium is able to react with oxygen (ignite) and light up the paraffin wax ribbon rising, reigniting the candle (or rather, the volatile materials above the candle). The magnesium is often powdered or in thin strips for low ignition temperature and easy ignition.
Sometimes regular candles do reignite, in these cases, the embers are hot enough to light up the paraffin wax vapour


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How can either reactants or products be "favored" at equilibrium?
From Atkins' Elements of Physical Chemistry (6th ed.):

$$\Delta G_\text{r}^\circ = -RT \ln K \tag{8}$$
  This fundamentally important equation links thermodynamic quantities – which are widely available from tables of thermodynamic data – and the composition of a system at equilibrium. Note that:

If $\Delta G_\text{r}^\circ$ is negative, then $\ln K$ must be positive and therefore $K > 1$; products are favored at equilibrium.
If $\Delta G_\text{r}^\circ$ is positive, then $\ln K$ must be negative and therefore $K < 1$; reactants are favored at equilibrium.

It says products/reactants will be favored at equilibrium, but I thought at equilibrium neither are favored?

At equilibrium the rates of the forward and reverse reactions are equal. This means that you can have more product than reactant, or more reactant than product. It just means that the rates of formation of each are equal and so there is no net change in the amount of each. The amount of products and reactants at equilibrium can be favored, their amounts just won't change once they reach equilibrium.
Depending on the temperature and pressure, the composition of the system at equilibrium can change.
Recall that in the reaction $$\ce{aA + bB<=>cC + dD}$$
The equilibrium constant is: $$K=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$
These concentrations refer to the concentrations of $\ce{A}$ and $\ce{B}$ at equilibrium.
Therefore in a situation where $\ce{A<=>B}$ and $\Delta G_\mathrm{r}^\circ\lt0$, then $K>1$, meaning that there is more of $\ce{B}$ than $\ce{A}$ at equilibrium (and vice versa).

If you are interested to know why $\Delta G_\mathrm{r}^\circ = -RT \ln K$ gives the standard change in Gibbs energy for a reaction, then you should note that the general form of the equation at any point in the reaction is given by
$$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + RT \ln Q$$
Where $Q=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$ at any point in the reaction. At equilibrium, there is no more Gibbs free energy, as the system no longer tends to the formation of products or reactants. Therefore at this point $\Delta G_\mathrm{r} = 0$, and the original expression given in Atkins is what you get. Also we define this special case of $Q$, saying that at equlibrium $Q=K$, the equilibrium constant.

Source
You might find the above image helpful. The y-axis shows the amount of Gibbs free energy, $G$. The reaction will want to tend towards the state where it has the lowest amount of Gibbs free energy (the local minimum). The slope of the curve can be therefore seen as $\Delta G_\mathrm{r}$. Therefore the point where the system has the least Gibbs free energy will be when $\Delta G_\mathrm{r} = 0$ could occur anywhere in the extent of reaction. This shows that the equilibrium concentrations need not be equal, but shows that the reaction will not progress any further when $\Delta G_\mathrm{r} = 0$.

It's not at equillibrium; it occurs when we are going to achieve equillibrium. More the value of $K$, the reaction will get driven towards the product side and more amount of product is formed easily (as more negative value of $\Delta G$ suggests). If $K\lt1$ then $\Delta G$ becomes negative and the reaction will drive towards reactants side, opposite to what happens for $K\gt1$ case.

That doesn't make any sense and I'm not sure if it's a language problem or a problem in your understanding in thermodynamics.

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What does bond polarity have to do with acid strength?
Why does increasing bond polarity affect acid strength? For instance, why is $\ce{HF}$ a stronger acid than $\ce{CH4}$?

This can be extrapolated to why polar substances ionize more often in water than non-polar substances. This is because water too is polar. Always remember "like dissolves like." Polar dissolves polar. This is because the positive end of the water molecule pulls at the negative end of the solute while the negative end of the water molecules pull at the positive end of the solute. In a non-polar molecule this would obviously not happen. Hence, in the case of acids, polar acids will dissociate more and hence more hydrogen atoms will be released into the solution.


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Retinol — an oligomer of isoprene?
The following image is taken from Wikibooks

How does isoprene polymerise to form double bonds between two monomers and in each monomer? As far as I have learned is that upon polymerisation two monomer join by a single bond and form a double bond within each monomer which can be explained by the following image:

If you scroll to the bottom of the Wikipedia page you linked to, you've got the answer to your own question.  It's not actually isoprene reacting with itself, but rather a more complicated pathway. The isoprene idea is just a convenient way of recognising terpenes.

Biomolecules that consist of ‘isoprene oligomers’ — more typically labelled terpenes do not actually polymerise in a way that you may be used to from plastic synthesis. When generating plastics, a double bond of a monomer unit (e.g. propene) is somehow made more reactive (e.g. by the addition of a nucleophile to generate an anion or by the addition of an electrophile to give a cation) and this reactive species then continues to attack monomer units giving a saturated chain of polypropylene.
Nature chooses a different path. It is not actually isoprene ((3E)-2-methylbut-1,3-diene) that is oligomerised — the resulting reaction would be far too hard for nature to control efficiently. Instead, the actually employed monomers are isopentenyl pyrophosphate (IPP) and dimethylallyl pyrophosphate (DMAPP).

Figure 1: The terpene biomonomers isopentenyl pyrophosphate (IPP; left or top) and dimethylallyl pyrophosphate (DMAPP; right or bottom).
As you can see, these only have one double bond but a diphosphate group attached to C4 that can be eliminated to give another. The biosynthetic pathway to couple these into oligomers begins by the removal of diphosphate from DMAPP resulting in a delocalised 3,3-dimethylallyl cation. This is captured by IPP’s terminal double bond to give a tertiary carbocation. In a final step, one of the two α-hydrogens in direction of former IPP’s diphosphate is eliminated to give geranyl pyrophosphate (figure 2).

Figure 2: Geranyl pyrophosphate; the dimerisation product of IPP and DMAPP.
Another addition of IPP gives farnesyl pyrophosphate which is then fed into multiple different biosynthesis pathways.
Thus, terpenes can be considered formal oligomers of isoprene but one should never think that they are generated by actual condensation of isoprene units.


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How 'heavy' should an element be, to be a "Heavy Metal"?
I've come across the term "Heavy metals" innumerable times in articles, mostly pertaining to environmental issues.
Is there a weight range (of sorts) against which an element (metal) is classified as a "Heavy metal"?
Often I see "Heavy metals" refer to Cadmium (Atomic number 48), Mercury (Atomic number 80) and Lead (Atomic number 82), but does this term also apply to metals of intermediate atomic masses [such as Tungsten (Atomic number 50), Gold (Atomic number 79) and Tin (Atomic number 81)]?
I can't seem to find any chemical literature that deals with this, so I suspect that the term "Heavy metal" has no real serious scientific definition.

Beryllium, the second lightest metal, is pretty toxic. It should therefore (ironically) be included in many definitions of "heavy" metals based on toxicity. Note that even lithium is bioactive, and is used (normally as the carbonate) in appropriate doses to treat mental illness.

For some people all elements with an atomic number greater than helium are metals, and all elements with an atomic number greater than iron are heavy metals.

That's the astronomer's definition of metals.

According to Wikipedia, there is no universally accepted definition.

@Russell Borogove: Yes. The funny thing is that even hydrogen and helium exist as metallic phases.

The real question: "Does it djent?"

Thallium is 81. Tin is 50. Tungsten is 74.

You may want to define it as a metal whose nucleus is heavy enough such that relativistic theory applies.
Atomic radii contract due to relativity when their nuclei are big enough, which may account for more dense metals as well.

There is no true, accepted definition of heavy metal. I was taught to apply the option a metal that has density equal to or over $5.0\ \mathrm{g/cm^3}$.
Other variants include a different density range, specific gravity over density, environmental impact, atomic number, toxicity, or atomic mass, even chemical properties. See here$^{[1]}$ for further information and references.

$[1]$ John H. Duffus. '"Heavy metals"—a meaningless term?'. IUPAC technical report. Pure and Applied Chemistry, $(2002)$, 74(5), pp 793$-$807.

There is no widely agreed criterion-based definition of a heavy metal. Different meanings may be attached to the term, depending on the context. In metallurgy, for example, a heavy metal may be defined on the basis of density, whereas in physics the distinguishing criterion might be atomic number, and a chemist would likely be more concerned with chemical behaviour.

There is no set criteria, as far as I am aware.  I tend to think of iron as the median.
Thus, I would happily refer to metals above iron as being 'heavy', i.e. density in excess of $\pu{7.8 g cm-3}$.

I don't know of a single agreed way.  The way I use relates to the way elements are created.
The way elements are created in the universe is through fusion in stars.  Fusion releases energy when elements are light and heavy elements require energy ( soak up energy ) in order to fuse.
Iron is the element at which marks the boundary between releasing energy and soaking up energy.  Elements above Iron are consider heavy, so I would describe metals with an atomic number above Iron as heavy metals.
When a star begins to create Iron ( and heavier elements ) it starts to die ( rapidly, because energy generated is less than that required to resist gravitational collapse ), so it's a pretty definite marker for how to separate light and heavy elements.

I think it should be a bio-chemical term, because they act as toxic in biological forms and they block biological processes in a cell or an organ.
The term "heavy metal" should be an interactivity of a metal classified element on bioorganic reactions.


text:
Naming CHCl(NH2)OH
I'm a technical writer who is writing an instruction guide for a molecular modeling kit. I chose a compound based on my ability to construct two enantiomers from the kit in question, but I don't know what to call the compound or even which order to write the atoms in the formula. I tried ChemDoodle but by the time I figured out how to use the software to draw the molecule, I'd already exceeded my free structure-to-name limit.
The compound has a central chiral carbon with bonds to hydrogen, a hydroxide group, an amino group, and a chlorine. How do I write its formula? Is it CHClNH2OH? What order do I list the groups in, in other words? And what would I call it?  I'm not sure what order to name the groups. Something like methanolchloramine? Any help is appreciated. I'll take the IUPAC name or any accepted name.

Do you also need to express the configuration of the enantiomers?

I think you should choose compound which wouldn't decompose immediately...

DHMO already gave the answer on how the compound as you drew it should be called. However, it has been commented that your compound is not stable and it also won’t be chiral for long. It will rapidly eliminate $\ce{HCl}$ to give a compound typically by retained name called formamide $\ce{HCONH2}$. Since formamide is planar, it is no longer chiral.
For the benifit of education, it would be better if you chose a compound that is sufficiently stable; whose isomers can be separated and ideally are sold separately.

As a plus point, the PIN is also 'formamide' (P-66.1.1.1.2.2).

The compound should be named aminochloromethanol.
You have correctly identified the parent hydrocarbon to be methane.
The substituents are $\ce{*Cl}$, $\ce{*NH2}$ (amine), and $\ce{*OH}$ (alcohol).
The order of precedence is:

Cations
Carboxylic acids
Carboxylic acid derivatives
Nitriles
Aldehydes
Ketones
Alcohols
Hydroperoxides
Amines

Therefore, $\ce{*OH}$ (alcohol) goes to the suffix, while $\ce{*Cl}$ and $\ce{*NH2}$ (amine) go to the prefix.
The suffix for $\ce{*OH}$ is -ol.
The prefix for $\ce{*Cl}$ is chloro-.
The prefix for $\ce{*NH2}$ is amino-.
Since "amino-" comes before "chloro-" alphabetically, the combined prefix is aminochloro-.
Therefore, the name would be aminochloromethanol.

According to current IUPAC recommendations, a minimum of parentheses must be used around prefixes when no locants are necessary. Therefore, the correct name is amino(chloro)methanol.

@Loong 'Correct' as in preferred IUPAC name?


text:
How to find out the IUPAC names of Alkenes?
While finding out the IUPAC names of organic compounds, should I count the number of the carbon atoms which is attached to the functional group and having the bond in the very same direction in which I counted the former (i.e., the attachment of the functional group)?
I am facing this confusion while solving questions based on the IUPAC nomenclature.

As it’s currently written, it’s hard to tell exactly what you’re asking. Please clarify your specific problem or add additional details to highlight exactly what you need. It would be best if you would include an example structure that illustrates your problem. You might also want to consider reading our guidelines on how to ask nomenclature questions.

To name Organic compound the functional group priority must be followed. For which below is the Priority Table :

Follow the function group priority. Count from the carbon of highest priority such that lowest locant number are given to every carbon. Denote the low priority groups as their corresponding prefixes and the greatest priority group is included in the name of the parent chain.

Please give proper credit to the author and site where you found the image, including a direct link to it. Also note that this image could be misleading since it mixes suffixes, multiple bonds (which are not included in the seniority order of suffixes), and prefixes (which do not have any seniority order).

@Loong I could not post the link of the original author because I do not have required reputation.

First of all, alkenes are hydrocarbons. So they do not contain any functional group except the double bond, '='(though i prefer to regard and call it as bond and not functional group). By reading the details of your question, i presume that you are facing difficulties in writing the IUPAC names of organic compounds in general and not only hydrocarbons since you use the word "bond" as well as the phrase "functional group".
To write the IUPAC name for any organic compound in general proceed as below:
1.First, inspect the molecule of the compound for functional groups, multiple bonds and substituent groups.
If more than one type of functional group is present, then, from the seniority table of functional groups find out which of the functional group(s) present in the compound is the located highest in the table. That functional group is known as the primary functional group and the rest are known as secondary functional group(s).
If only one type of functional group is present then that functional group is the primary functional group and there is no secondary functional group.
2.Then, find out the longest carbon atom chain (in the molecule of the organic compound) which contains (if they are present) the maximum number of functional groups (including the primary functional group), multiple bonds and substituent groups.
3.If functional group(s) are present then the carbon atoms are numbered from such an end that the sum of the position numbers of the primary functional group is the least.
If functional group(s) are not present but multiple bonds are present then between the double bond and triple bond whichever multiple bond is nearest to an end, numbering is done from that end. If, different multiple bonds occupy equal position numbers from both ends, then, the double bond gets greater priority over the triple bond and the numbering is done from the end to which the double bond is the nearest.
If both functional group(s) and multiple bond(s) are absent but substituent groups are present, then the numbering is done from an end corresponding to which the sum of the position numbers of the substituent groups is the lowest.
(In this rule, i have used the sum of locants rule which is actually a wrong generalization of the first point of difference rule but it works in most of the cases)
Note that priority(functional group)>priority(bond)>priority(substituent group)
4.Find out the number of carbon atoms in this chain.
5.A typical IUPAC name has the following structure-
$\ce{secondary prefix + primary prefix + word root + primary suffix + secondary suffix}$
SECONDARY PREFIX
The complete secondary prefix consists of the name(s) of the substituent group(s) along with its(their) position numbers and the secondary functional group(s) along with its (their) position numbers arranged in alphabetical order.
PRIMARY PREFIX
The primary prefix indicates whether the compound is cyclic or not. If it is cyclic, the secondary prefix is cyclo otherwise it is left blank.
WORD ROOT
The word root indicates the number of carbon atoms in the longest carbon atom chain.
If it is 1 use meth, if 2 use eth, if 3 use prop, if 4 use but, for 5 or more use greek names for numbers, that is, pent, hex, etc.
PRIMARY SUFFIX
The complete primary suffix consists of ane, if the compound has no multiple bonds.
If multiple bonds are present then it consists of
position number(s) + (di/tri and so on)ene for double bonds and position number(s) + (di/tri and so on) yne for triple bonds.(If the number of double bonds or triple bonds is more than 1 then correspondingly di, tri and so on is added before ene and yne) ( Note that the double bond part is written before the triple bond part)
SECONDARY SUFFIX
The secondary suffix consists of the following :
position number(s) + (di/tri and so on)name used for primary functional group
6.Use '-' to separate a letter and a number and ',' to separate two numbers. If two vowels occur in succession drop the the second vowel.
Here is the seniority table for functional groups.


text:
Why do diacidic and triacidic alkalis dissociate in one step?
My book states that dibasic and tribasic acids dissociate in water in two and three steps respectively but for diacidic and triacidic bases it states that they dissociate in one step. I do not know why.
For example:
$\ce{H2SO4}$ dissociates in water in two steps as follows:
$$\ce{H2SO4(aq) <=> HSO4^{-} + H+}$$
$$\ce{HSO4^{-}(aq) <=> SO4^{2-} + H+}$$
but
$\ce{Cu(OH)2}$ dissociates in water in one step as follows:
$$\ce{Cu(OH)2(aq) <=> Cu^{2+} + 2OH-}$$
Why is this so. Why does not $\ce{Cu(OH)2}$ dissociate in two steps to form $\ce{CuOH+}$ in the first step and $\ce{Cu^{2+}}$ in the second step?

Your sulphuric acid example is two steps, not one. I am also not certain your book states what you are implying that it does, or at least, that your book is correct to say that. Can you clarify?

@Nij sorry that was a discrepancy. I have edited my answer.

You are comparing very different reactions: copper hydroxide dissociation is not an acid-base reaction at all but a solubilisation.

@matt_black iam talking about dissociation in water.

We do have separated dissociation constants for the heavier alkaline earth hydroxides whose solubility allows a high enough pH to access both equilibria.
Calcium hydroxide

Acidity ($pK_a$): 12.63 (first OH−), 11.57 (second OH−)[1][2]

(The $pK_b$ values are obtained by subtracting the above from $pK_w=14$.)
In contrast with hydroxy acids such as sulfuric or phosphoric acid, the bonding between the central atom and the hydroxyl groups is predominantly ionic, with little covalent overlap; so the induction effect we would expect to see with increasing charge buildup is diminished. This shows up as the two dissociation constants being only about one order of magnitude apart in calcium hydroxide, versus about five orders of magnitude for successive deprotonations of phosphoric acid. The closeness of the dissociation constants for hydroxide bases combined with the limited solubility in most cases makes separate dissociation constants usually hard to measure.
Cited References

"Sortierte Liste: pKb-Werte, nach Ordnungszahl sortiert. – Das Periodensystem online".
ChemBuddy dissociation constants pKa and pKb

Polyprotic acids

In most acids, even the strong acids, the bond to the acidic hydrogen atom is covalent (though frequently very polar). Thus, for acids, the more correct description of what happens in solution is ionization - a covalent bond is breaking heterolytically to form ions. The solvent is involved in solvating the proton, which is why we usually write $\ce{H3O+}$ for protons in water and not $\ce{H+}$.
Polyprotic acids ionize in separate steps because after the first ionization, the new species has a different strength covalent bond to break. Consider your example of sulfuric acid. Sulfuric acid is a strong acid that ionizes readily and (nearly) completely in water:
$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$
The resulting hydrogen sulfate anion is a different chemical species. Its bonds have different strengths. The second proton is not as easily ionized. In fact, the hydrogen sulfate anion is considered a weak acid. It does not completely ionize in water. An equilibrium is established with an equilibrium constant of $K_a = 1.02 \times 10^{-2}$. This means that at $\mathrm{pH = 2}$, the hydrogen sulfate anion is 50% ionized: half is hydrogen sulfate and half has ben converted to sulfate.
$$\ce{HSO4- + H2O <=> SO4^2- + H3O+}$$
What evidence do we have for this step-wise ionization? We have titrations! If we add small volumes of a solution of a stong base (say $\ce{NaOH}$) of known concentration to a solution of sulfuric acid and monitor the pH, we can see two distinct equivalence points (the inflection points on the graph where the pH change is very large over a small volume of base added) corresponding to the two ionization events. However, because the first ionization of sulfuric acid is complete before the base is added, a titration  curve of sulfuric acid will only show one equivalence point when two equivalents of base are added (though there is a blip at 1 equivalent).

If we consider maleic acid $(\ce{H2C4H2O4})$, for which both ionizations are "weak" $(K_1=1.26\times 10^{-2},\ K_2=8.51\times 10^{-7})$, we get a different looking titration curve with inflections at 1 and 2 equivalents of base. If the ionization of both protons occurred simultaneously, we would not see a curve like this.
$$\ce{H2C4H2O4 +H2O <=> HC4H2O4- + H3O+}\\
\ce{HC4H2O4- + H2O <=> C4H2O4^2- + H3O+}$$

Bases

Bases that are the hydroxide salts of metals are ionic compounds. Though many of them have poor solubility in water, what is dissolved is completely ionized. Thus, for copper (II) hydroxide in water at equilibrium, there is only solid copper (II) hydroxide ad solubilized copper(II) and hydroxide ions:
$$\ce{Cu(OH)2 (S) <=> Cu^2+ (aq) + 2OH- (aq)}$$
A titration curve for copper (II) hydroxide with a strong acid like hydrochloric acid would show only one equivalence point at two equivalents of acid:

All ionic bases will behave similarly.
However, some bases are covalent and ionize by reacting with water:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
If there was to be a base of this type that had two ionizable groups, we would see two equivalence points.

Errgh, Ben, you can't call Cu(OH)2 here as an example since it is most definitely not dissolved completely under normal circumstances and in case you are talking about Cu(OH)2 solubilized by addition of complexone, you are not talking about Cu(OH)2 dissociation. The only potencially relevant base I can come immidiately is Ba(OH)2, all other bases have complications involved.

As complex as the actual species generated by dissociation may be (remember that metal cations typically form complexes and all the associated shenanigans with those) I think it is worth pointing out in a simplified way that there is a difference between ionic dissociation and compound ionisation as this answer does. Granted, it may be fitted with a caveat section but one lacking doesn’t make it bad imho.

They most definitely do. The problem with bases is that most of the common ones that are soluble are monobasic, polybasic ones are usually insoluble and thus, the intermediate states are hard to detect.
Various polyamines that are polybases most definitely react step-by-step. However inorganic polybases often form insoluble basic salts and usually are insoluble themselves and thus partially dissociated forms in solution exist only in miniscle amounts in fast equilibrium with precipitate. And no, basic salts such as various $\ce{Cu(OH)_xCl_y}$ can't surve as a proof, since stability of solids is ruled by very different set of rules.
Furthermore, when an inorganic polybase dissociates incompletely it usually forms various polynuclear particles with potentially very complex structures. This buries any attempts to create a simple and neat picture. This chemistry is extremely complex, studied only a little and completely unsuited for introductory courses. So most such courses keep such chemistry in mind but pretend it does not exist.


text:
Is it Cu²⁺ or Cu⁺²?
Is $\ce{Cu^{2+}}$ or $\ce{Cu^{+2}}$ correct? In my book the author uses both but at different places.

The standard notation for ions is 2+, not +2.

The net charge is written with the magnitude before the sign; that is, a doubly charged cation is indicated as 2+ instead of +2.


text:
Why does Graphene have a higher melting point than Diamond?
I'm not 100% sure about this, so I just wanted to double-check.
I think it has something to do with each carbon atom in Graphene having a delocalized electron and this creates carbon ions that can exert an electrostatic attraction. This would reduce the inter-particle distances between carbon atoms in Graphene relative to Diamond, hence Graphene's higher melting point. But again, I'm not quite sure.

Neither graphene nor diamond have a "melting point". https://de.wikipedia.org/wiki/Kohlenstoff#/media/File:Carbon-phase-diagramp_german.svg

This is a poorly defined question. Melting is a change in stare, and it's based on thermodynamics. But at room pressure, neither diamond nor graphene are stable allotropes: graphite is. They are metastable.
So you start to mix in kinetics with the thermodynamics.
It's a bit like asking why does ice melt at a lower temperature than water.
If you look at the effects of pressure, then you can construct a phase diagram, and then the "melting" point of any given allotrope is defined by the boundary between solid and liquid.

Graphite(collection of graphene layers) has higher melting point than diamond, because of the partial double bond character of C-C bonds. So, the bond in each layer(graphene) is strong. Also, there are attractive forces between successive layers of graphene, increasing the melting point of Graphite

So does Graphene form a resonance hybrid?


text:
[3+1]-Cycloadditions of Sulfur Ylides and 1,3-Dipoles(?)
My question is basically regarding the stereoselectivity of the situation.
I am not being able to decide on the positioning of the sulfur ylide with proper justification.
In the attached hand-drawn image, I have considered the (1'-oxoalkyl) group away from the alkyl group $\alpha$ to the carbonyl. The other possible orientation is to direct in on the same side as the alkyl group,  but coming out of the page.
Each positioning leads to a different outcome, which is why I drew the final product with a "wiggly" bond.
Furthermore, the transition state that I have drawn really seems sketchy, and I feel I need some guidance on that part.
[3 + 1]- and [3 + 2]-Cycloadditions of Azaoxyallyl Cations and Sulfur Ylides

I can't read this paper. Do the authors think that this proceeds via a carbene? That carbon in the transition state has too many bonds to seem reasonable...


text:
Equilibrium Constant in terms of Molar Concentration of Gases versus Equilibrium Constant
Suppose the gases involved in a reaction behave as real gases with
significant intermolecular attractions.
I just learned that $$K_c(RT)^{\Delta n_r} = K$$
Would the value of Kc calculated
from the equation above be too high, too low, or the same as the actual value?
I thought that it would be the same, but I don't think that's right because deviation from ideal behavior should deviate from equilibrium.

I believe that the real $K_c$ would be higher. This is because in ideal gases london dispersion forces, hydrogen bonding, etc. is ignored. These would increase the number of collisions between particles since they would be attracted to each other more.
$$\ce k=zp*e^{-E_a/RT}$$
where k= rate constant, z= frequency of collisions, p= fractions of collisions with proper orientation to cause a reaction, $E_a$= activation energy, T= temperature. So Since z would increase, the K would increase. Hence the $K_c$ would increase.

Wow, that was actually a really helpful answer. I do want some confirmation from other users as well but your answer actually really helped me out. Thank you!

The first time I read this I was confused.  Are you saying that $K_c$ would be higher for all ideal gases, or simply saying that $K_c$ would be higher for gases that have significant intermolecular attractions?  (I.e., would $K_c$ be lower for gases that have intermolecular repulsions?)

Well it would be higher for all gases. Ideal gases don't technically exist. All gases have some intermolecular forces thought it may be small in certain cases. The difference between the true $K_c$ and "fake ideal" $K_c$ of a gas would increase with stronger intermolecular interactions. Since the question specifically asked for gasses with significant intermolecular attractions I aimed it towards attraction not repulsion. Other than heat, the intermolecular attractions are much larger than the repulsion in all gases. And heat is summed up in T in the equation which is assumed constant.


text:
What does the term "simple" in the phrase "simple whole number ratio" mean in Gay Lussac's law?
What does the term "simple" mean in the phrase "simple whole number ratio" in Gay Lussac's law?
I have searched this site as well as different sites on the internet. The answer to one question on this site states that it indicates that the numbers are small whole numbers. Another site stated that the terms of the ratio are not decimal numbers while another site states that the terms of the ratio are in simplest terms that is reduced form.

Related: Law of Multiple Proportions: What is the significance of small whole numbers?

3/2 is simple. 45498334/9232493 is not.
The reason it works is because reactions of gases producing gases involve relatively simple compounds. If they were complicated, the molecular weight would likely be high enough that the substance would not be a gas.
This is akin to Loong's comment about the Law of Multiple Proportions, which also mentions "small whole number ratios." The difference is that for the Law of Multiple Proportions, "small" ratios doesn't apply in the case of something like palytoxin ($\ce{C129H223N3O54}$).
But palytoxin isn't a gas. For the reason mentioned above, you're almost certain to get simple coefficients for gas to gas reactions.

So "small" here only means that it is not large and not anything else like decimal, in simplest terms. Correct?

Yes. It's just from the fact that the stoichiometric ratios for gas reactants producing gas products are small and simple.

One more question: Don't you mean that the volumes are simple after reducing them to the lowest terms? Otherwise they may be large(since we can take any volume of gases as reactants) but still, they have a common factor and on dividing them by this common factor they become small (simple), i.e., to say that, in the ratio of the volumes in their lowest terms, the volumes are simple.

This doesn't have anything to do with volumes. It's just ratios of atoms. You just happen to be using gases, so assuming their ideal (they're not) you could argue that the number of moles are related to the volumes, but that hides the information encoded in the stoichiometry of each substance.

What do you mean by saying " but that hides the information encoded in the stoichiometry of each substance"?

And by the way, since ratio of volumes and the ratio of amount of the substance are equivalent, isn't what i said still valid?

What you said is only valid if (1) you assume the gases are ideal and (2) if you assume that the reactants are gases. But the law of multiple proportions applies to the formulas for all simple compounds. You've just selected a very specific example to look at, but doing so hides the full extend of where the law can be applied.

It means that both terms are reduced to the smallest whole number possible so that the numerator and denominator are co prime.


text:
Number of alkanes which are chiral
How many alkanes of molecular weight $M$ are chiral?
As a particular case let us consider M=100.The general formula for such a alkane is $C_nH_{2n+2}$ where $n$ is the number of carbon atoms.
Putting in values of molecular weight we get $12n+2n+2=100$ which gives $n=7$. So there are 7 carbon atoms. So far I could think of the following two structures.

Since their enantiomers are also chiral giving us a total of 4 chiral alkanes. However I am interested in the general case when the molecular weight is given M. Is there some generalised result to the situation?

No, there isn't. You could only try to get max number, not accounting for steric factor.

check out http://chemistry.stackexchange.com/questions/16135/how-to-determine-number-of-structural-isomers

The first problem to solve would be the one in the question Mithoron linked. If you got that solved, you can start figuring out which one of the actually synthesisable compounds can or cannot be chiral. It’s very non-trivial.


text:
Do perfect gases react when placed in a container whose volume can vary?
Suppose that $n$ moles of four generic perfect gases are made to react in a container, whose volume can vary, at a constant temperature and pressure, according  to the formula
$$\ce{A(g) + B(g)<=>C(g) + D(g)}$$
and suppose also the the equilibrium constant $K_c$ is known.
Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, regardless of the number of particles present in the container (this is more or less the same argument that my chemistry textbook makes to justify the exclusion of concentration of solids and liquids from the expression of the equilibrium constant).
But if the concentrations are constant, then at the outset of the reaction, when there are $n$ moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.
My question is the following: is there something wrong about this argument? If not, how will the number of moles of each gas vary with time?

You have yourself all wrapped around the axle with odd conditions.

Suppose that $n$ moles of four generic perfect gases...

This is a really odd way to specify the initial concentrations of four gases. Say I have 5 moles total. There are four gases, so I have four unknowns and only two equations (total moles and equilibrium equation) which is unsolvable.
This is also ambiguous. It is $n$ moles total, or $n$ moles for each gas?

Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, ...

Since you're holding temperature and pressure constant, and given the reaction:
$$\ce{A(g) + B(g)<=>C(g) + D(g)}$$
then the density will always be the same. Even if you have all A and B which are totally converted to C and D, the density of the gas mixture will be the same (assuming ideal gas behavior which you stipulated).
If at a constant volume and the reaction had been:
$$\ce{2A(g) + B(g)<=>C(g) + D(g)}$$
then the density would change as the equilibrium shifted.
But since you stipulated constant temperature and pressure then the volume would also have to change as $\ce{2A(g) + B(g)}$ was converted to $\ce{C(g) + D(g)}$ to maintain constant pressure, so the overall density of the gas remains constant.

But if the concentrations are constant, then at the outset of the reaction, when there are n moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.

Let's break this down. If the concentrations are constant, then the reaction is already at equilibrium.
It does not necessarily follow that when there are $n$ moles of each gas then the reaction is at equilibrium. That would only be true if $K_c = 1$, but $K_c$ has not been specified.

If not, how will the number of moles of each gas vary with time?

Ok, let's assume $n$ moles of each of the four gases. You'd also need to provide the rate constants for the forward and reverse reactions to calculate concentrations over time. Without the rate constants it isn't clear if the equilibrium will be established in seconds or years.

Density is mass over volume. We do not know that the molar mass of A+B=C+D so hence the mass would change. Now since we do not know what kind of container it is being held in, we cannot assume that the volume would change in a way as to keep the density of the entire container the same.

@ClangorousChimera - you have the formula for density right. Since we don't have real numbers to work with then the actual value of the density is unknown. But given the reaction, the density of the gas mixture won't change. Let's say I have 1 mole of A and 1 mole of B which is wholly converted to 1 mole of C and 1 mole of D. For both the starting case and the products I have two molar volumes. Chemical reactions don't change masses, so the total volume and the total mass are both some unknown constants.

Ah! You're right I forgot about that. Thanks! I deleted my answer

Sorry, but either you are misunderstanding me or I am misunderstanding you. What I meant to say was that at the outset of the reaction you have n moles of A, n moles of B, n moles of C, and n moles of D. Now, we agree that in the conditions specified density is constant, but since density is directly proportional to concentration, then the concentrations will not change, hence Q will be a constant. But then either 1)Q=K, which entails that the reaction is at equilibrium right in the beginning or 2)Q=/=K, in which case it never reaches equilibrium

does not seems plausible. Therefore we're left with 1), which is still kind of odd

edit: when I say "density is constant", I mean that the density of each of the gases is constant

If you always have n moles of A, n moles of B, n moles of C, and n moles of D then the system is at equilibrium and $K_c$ would have to be exactly 1. But you didn't specify the either $K_c = 1$ nor that the system was at equilibrium. If you start with only A and B you may essentially convert all the starting reactants to C and D, you may convert none, or some fraction of the starting reactants.

Re: edit: when I say "density is constant", I mean that the density of each of the gases is constant -- Well that is a rather weird notion since you can't really measure the density of individual gases in a mixture of gases. You can only measure the density of the mixture.

Let's change the question a little: suppse you have a homogeneous reaction involving liquids (or solids) $\ce{A(l) + B(l)<=>C(l) + D(l)}$. What happens to  Kc and Q, since concentrations of liquids are not supposed to be appear in their expressions?

@Nicol = re: since concentrations of liquids are not supposed to be appear in their expressions // Again this is wrong. at equilibrium $K_c$ is a function of the concentrations.  $$\ce{K_c = } \dfrac{[C][D]}{[A][B]}$$

But the concentrations of liquids and solids don't appear in heterogeneous equilibria (because they are constant). But why do they in homegeneous ones? Because the concentrations of liquids and solids in homogeneous reactions are not constant (in this case, why is there a difference?) or for some other reason?

@Nicol - Back to basics. The activity https://en.wikipedia.org/wiki/Thermodynamic_activity should really be used instead of concentration for the equilibrium constant. In most cases we just assume that the activity is equal to the concentration. If we look at a reaction like the solubility product of barium sulfate, then the activity of the solid barium sulfate is by definition 1 since it is in a separate phase (solid) than the barium cations and the sulfate anions which are in the liquid phase. So reactants and products only "drop out" if they are in another phase. (bit over simplified...)


text:
Number of picoseconds in a week is exact?
My chemistry textbook states that the number of picoseconds in a week is not exact because a week is defined by the Earth's rotation.
Does that also mean that 3600 seconds in an hour also isn't exact because an hour is also defined by the Earth's rotation?

The SI base unit of time is the second ($\mathrm s$). The only non-SI units of time that are generally accepted for use with the SI are the minute ($1\ \mathrm{min}=60\ \mathrm s$), the hour ($1\ \mathrm h=60\ \mathrm{min}=3600\ \mathrm s$), and the day ($1\ \mathrm d=24\ \mathrm h=86,400\ \mathrm s$); however, that not necessarily mean that other units of time are not exact.

So in Chemistry: the hour is defined by seconds and the week is defined by the earth rotation? And the second must be defined by the rotations of an atom maybe?

A week is defined in terms of seconds. The second (nowadays) is not defined through the earth's rotation. So your book either is wrong or you might have misinterpreted it. See https://en.wikipedia.org/wiki/Second

@eromod The question is a good one. But it isn't a chemistry question which is why readers here are saying it is off topic (even if it appears in a chemistry textbook). And the real issue is that the definition of time in science is exact but that we have to adjust the length of a calendar year (sometimes by days and sometimes by seconds) to keep astronomical/calendar time in sync with precise atomic clocks because the earths rotation and orbit are not exactly in line with the clocks.

@matt Are you sure that the definition of time isnt exact in science? Thats like saying the definition of the kilogram isnt exact in science.  We start with a base and derive everything else from it right?

@eromod The definition of time is exact, its the calendar that isn't because it has to match to uneven rotation of the earth.

The calendar tries to please both the sun and the digital measurements. Thats why we have leap days and leap seconds. So even if you base your measurements on the official calendar, you would still get a specific amount of picoseconds, though the amount of picoseconds per day may be different so its not as useful to use the calendar for such a precise measurement. Better to use something digital, like a satellites do when they need very precise measurements of time.

This is a case where if you need high levels of precision then you need to be clear about your definitions.

A week is 7 days

but what is a day?
The SI talks about the units of "minute", "hour" and "day" in "Non-SI units accepted for use with the International System of Units". It defines them as 60s, 3600s and 86400s. This is a definition and so it is by it's nature an exact number.
But this is only one definition. A day can also be defined as the time taken for the earth to rotate so the same place is facing the sun (a solar day) or the time taken for the "time of day" on clocks to come back to the same value.
The length of a solar day varies cyclicly by nearly 30 seconds over the course of a year due to the shape of the Earth's orbit. The average over longer periods also varies slighly.
A day of UTC is normally 86400 seconds as measured by averaging a load of atomic clocks but occasionally a UTC day may by 86401 or 86399 seconds due to leap seconds.
A day of local civil time is also normally 86400 seconds but daylight savings time shifts may lead to a day as short as 82800 seconds or as long as 93600 seconds.

So if this question appears on a multiple choice test (like mine did) what would you answer? Number of pico seconds in a week is exact or not?

Im guessing you would answer no because a week isnt an accepted SI standard unit? But  the amount of people in a classroom is an exact amount, even though "number of people" isnt an accepted SI unit either.

Unfortunately this is one of those crappy questions where the only way to ensure a "correct" answer is to know what the test-setter is expecting.

I think they are expecting a correct answer, no?

@eromod "Correct" depends on the assumptions that have been made in the course of construction the question. Different premises can lead to different "correct" answers.

@hBy, that makes sense, but as Peter pointed out, the International System of Units website shows that time isnt based on the sun anymore. Its based on specific and exact ratios like 1 min = 60s. So the default assumption should be that the amount of pico seconds in a week is indeed exact. Just like how there is 1,000 mm in a meter.

@eromod I reject your implicit premise that all time units in use today are SI units, or directly derivable therefrom; I further reject your explicit premise that all measures of time are no longer based on the sun.

Where would you draw the line? At pico seconds, at seconds at hours, at days, at weeks, at months, at seasons? Wouldint it be more consistent to choose just one? This is sort of like saying: I accept that a meter has 1000mm but I dont accept that a light year has x amount of meters in it.

You have a problem in the statement of the question because you are asking for a simple answer. There is no simple answer:
Time is defined so that the laws of science look simple. You can define it any way you like, but it's most useful when computation and expressing quantitative relationships are easy.
A week is not a unit that a chemist generally needs to deal with, unless they are meeting a grant deadline that is due in a week. Frankly, the same is true for physicists, but it is the time keepers who decide on minute adjustments of leap seconds who need to account for the discrepancies.
No measurement is perfectly precise, so time measurements are not perfectly precise. Even the hyperfine transitions of a cesium atom that we use to define a second have some variability.
Here's the nail in the coffin. Is time even measurable? The Einsteinian Equivalent Principle says that gravitation is like an acceleration so the very act of living on Earth means that our time measurements are dilated with respect to an observer who is far away from massive bodies (whatever that means). Technically, a week is a different number of seconds for someone on the top of Everest versus someone at the bottom of the Marina Trench.
My key point here is that measurement and all of its complications is part of the crux of science, and it makes science what it is. You're asking the right kinds of questions, but don't expect simple answers.

Archived discussion in chat..


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What is the chemical process which causes paints to dry?

How does paint dry?
What is the chemical process?

Related: What is the difference between wet paint, dry paint, and in between those stages?

From "Introduction to Paint Chemistry and principles of paint technology" by John Bentley:

In the can of paint the mixture of substances must remain stable for long periods, so the ingredients must not react with one another chemically. Yet the paint must dry on the coated surface. Under some conditions some paints do not dry, which suggests that the drying process does not merely consist of evaporation of the liquid. In fact, there are three broad mechanisms of drying, two of which involve chemical reaction, while one does not.
Drying without chemical reaction
In this case the paint does dry solely by evaporation of liquids. The polymer is fully formed in the can and, when free of solvent, is relatively hard and not sticky. During the drying process there is no chemical change in the polymer...Nitrocellulose lacquers and decorative emulsion paints dry by this process.
Drying by chemical reaction
Drying by Chemical Reaction Between Paint and Air
...oxygen reacts with drying oils and other unsaturated compounds to produce free-radicals and bring about polymerization. Also, the reaction between water and isocyanates can cause condensation polymerization. Both reactions can give a cross-linked film. In both cases the principle is the same: the air is used as a chemical reactant and is kept apart from the reactive ingredients in the paint by the tightly fitting lid on the can. If it is not, it may cause the formation of a skin, which seals off the paint below from further reaction with the air.
Drying by Chemical Reaction Between Ingredients in the Paint
Obviously the paint must remain chemically stable. The reactants must not react until the paint has been applied, yet they must all be in the paint. This paradox is resolved, either by separating the reactive ingredients in two or more containers and mixing just before use, or by choosing ingredients which only react at higher temperatures or when exposed to radiation.

Tldr: paint may dry through evaporation, or chemical reactions (these chemical reactions may be driven with heat or radiation)


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Does the solubility of iron (iii) oxide increase when the pH of the water it is in decreases?
I did an experiment and I got these results:
pH 4: 3.46
pH 3: 3.77
pH 2: 2.18
pH 1: 3.69
The numbers on the right is the amount of iron oxide remaining that was not filtered.
Do these numbers make sense? (The measurements are in grams.) The original amount of iron oxide in each solution (4 different solutions as there are 4 different pH's) was 3.81 grams
I basically put iron (iii) oxide in different pH's, left them, and filtered out the water after. I measured how much iron (iii) oxide remained. I am not sure if my numbers make sense though because I thought that if the pH drops it increases in solubility. However, my pH 2 was the highest in solubility. Do these numbers make sense chemically?

How are you making different pH's?

Do not repost your question. [Edit] the original instead, please.

I used acetic acid to create the different pH's (90 mL water, 10 mL acetic acid)

Iron (III) oxide is insoluble in water. If you add an acid (e.g. $\ce{HCl}$) it might react with the oxide and form a soluble iron salt.
How much of the oxide gets dissolved depends, amongst others, on what acid is used, its concentration and amount, and the time of reaction.

I used acetic acid to create the different pH's (90 mL water, 10 mL acetic acid)

With acetic acid it is even more difficult to tell what's going on since iron(III) and acetate form complexes https://en.wikipedia.org/wiki/Iron(III)_acetate


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Reducing property of alkali metals
In my book it is given

Reducing property increases down the group in gaseous or molten state .
  The increasing order is  Li , Na , K , Rb ,Cs

In according to me reducing property is based on ionisation property . As Cs has highest ionisation property so it can easily loose elctron (oxidation potential)

But in aqueous solution incresing order is given as
  Na , Cs ,Rb , k ,Li

For this what could be the reason .

Reducing property in aqueous phase depends upon (1)  enthalpy of sublimation (2)enthalpy of ionisation (3)enthalpy of hydration.
First two processes are endothermic while last one is exothermic.
For lithium enthalpy of hydration is very high because of its small size hence Li acts as stronge reducing agent in aqueous phase. For Cs enthalpy of hydration is very small hence overall process is less exothermic .

Then why Na shows different trend in enthalpy of hydration

It is because of high ionisation enthalpy of Na compared to K,Rb and Cs.


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a structure of compound related question
Why the structure of XeF6 is capped octahedral but not pentagonal pyramidal although there is a possibility of being pentagonal pyramidal (with lone pair below the pyramid) ?

It is because to minimize lone pair and bond pair repulsion. In pentagonal bipyramidal arrangement lone pair electron will occupy eqatorial position which results in more lone pair -bond pair repulsions.


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how are colloids and suspensions solutions?
my definition of solution is a homogenous mixture of a solute and solvent.
and in my textbook they call colloids and suspensions solutions . how is this possible if both of them are heterogenous mixtures.

The definition of solution is a little bit cloudy. IUPAC define a solution in this way:

A liquid or solid phase containing more than one substance, when for
  convenience one (or more) substance, which is called the solvent, is
  treated differently from the other substances, which are called
  solutes.

And indeed this is a quite general approch that takes in account that we usaually call solutions systems (such as metallic solution) that are not compose of single molecules or ions dissolved in a solvent. Besides this approch chemists generally call solution a system with the folowing charachteristics:

It is stable.
It has only one phase.
It has to be an homogeneous mixture of two or more substances.
The solute can't be distinguished from the solvent (no scattering, can't be filtered)

And so I would not define colloids and suspension solution because they have a dispersed phase and often they are not stable.

Reducto ad absurdum, all matter is discrete, none is "homogeneous". So, there are no "homogeneous" solutions, since not only are there no solutions without particles present, but there are no solutions which contain exactly the same number of atoms in every cubic volume element. You may be thinking that my argument is silly, and you wouldn't be wrong. OTOH, your definition uses the words "solute" and "solvent" which is quite useless (circular definitions are really not appropriate nor necessary here). The only information you give us in it, is that the mixture must be homogeneous - which you don't define. Did you know that you can take a solution of some heavy metal salts, and centrifuge them and concentrate the salts at the bottom? So, the idea that there is a single criteria you can use to separate dispersions from solutions is incorrect - it changes with context. To be perhaps a bit clearer, some would consider colloidal mixtures, and even suspensions to be "homogeneous enough" to be "solutions". Generally, a colloid has a lot of interactions between its surface and the solvent, so that it makes sense in some contexts to call it a solution. I'm less enthused about calling a suspension a "solution", I'd call it a dispersion. Bottom line here is that "phases" are macroscopic things, and both suspensions and colloids are microscopically distributed, so a mixture of either may for some purposes be considered homogeneous. This is about terminology. I won't say its a bad question, but I will point out that this type of question ultimately depends on definitions, so may not be useful to worry about. (I decided not to "go there" as far as inquiring what qualifies you to say "my definition" - generally, you should be widely published in the appropriate area, and even experts may use terminology differently. This question isn't about things, its about words. It would be more interesting if YOUR definition was obtained from the text book, in which case it would seem to be inconsistent. I have no problem with the fact that the usage in the textbook is different than what you've seen elsewhere.


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Predicting the product of a hydrogenation reaction

The question is to predict the nature of the product(s) of the above reaction, that is if the product(s) form a racemic mixture, meso, diasteromers, or are structural isomers.
I am aware of the fact that addition of $\ce{H2}$ is a syn addition. So it should form a meso compound. However the answer key gives it as a mixture of diastereomers. I am not quite sure if the reaction is strictly selective and some anti products are also formed.

Ask yourself first if the methyl group is up or down....

To answer and understand this, let’s first consider the products of the hydrogenation ($\ce{H2; Pd/C}$) of 1-methylidene-4-methylcyclohexane (the same reactant except the methyl group is in para-position). This compound is symmetric and therefore achiral. We can draw the methyl group pointing upwards iwthout loss of generalisation. The hydrogenation is syn-selective and it can occur from the top or the bottom. We expect a $1:1$ mixture of the former methylidene group pointing upwards (the syn-isomer) and pointing downwards (the anti-isomer). Both are achiral (the plane of symmetry is retained in the reaction) but not identical; therefore, they must be diastereomers.
Now let’s move the methyl group back into the 2-position. By moving the methyl group, the plane of symmetry is lost; it suddenly makes a difference whether the methyl group points upwards (S) or downwards (R). For each of the two isomers, we can again hydrogenate from both above and below; we can expect two anti and two syn isomers at first approximation: we enter one stereocentre into the reaction and we form an additional one making two stereocentres in total and thus 4 theoretically possible isomers.
In explicit forms, starting from the (2S) isomer the syn-product could be called (1R,2S) and the anti-isomer would be (1S,2S). From the (2R) isomer, another syn (1S,2R) and anti (1R,2R) isomer is expected. Only when taking a closer look at the products, it becomes obvious that the two syn isomers are in fact identical because they can be transformed into each other by rotation; and that they are achiral because they generated a new plane of symmetry. Thus, the syn isomer is a meso isomer.
Putting this together, the hydrogenation products are two enantiomeric anti products and a meso (syn) product. These two classes are diastereomers of each other, thus ‘a mixture of diastereomers’ is the only correct answers.


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Influence of rotational degrees of freedom on gas phase stability of ions in CID
In mass spectrometry it is possible to determine the stability of an ion in the gas phase using e.g. collision induced dissociation (CID). In CID, we fragment an ion by colliding it with a neutral buffer gas. By gradually increasing the collision energy, it is thereby possible to get a relative measurement of the stability of the ion in the gas phase (so called survival yield analysis).
Now, I have the following question:
The picture below illustrates an effect that is commonly observed in CID namely that bigger ions tend to be more stable as they have more rotational degrees of freedom available to handle the collision energy transferred using CID. This can lead to counterintuitive effects, as for example in this case, one would expect acetic acid to be more acidic (and thus a more stable anion), than propanoic and decanoic acid, but it is the least stable.

In my work, I wanted to show that increasing chain lengths of the ligand increase the stability of a complex by weak interactions but I don't know how to factor in the effect that the increasing chain length has on the stabilization by the increased rotational freedom. Obviously I want to rule out that the increase in stability is due to the longer chains absorbing the collision energy better. How large could that effect be? Is there any systematic approach to take this effect into account? Is there any literature on the subject?

I have a number of concerns with your question, and maybe some suggestions. (a) In the definition of the stability of your molecule, you tend to focus on the acidity whereas in a survivor ion analysis, stability is defined relative to the fragmentation pathway (difference between starting product and fragments or starting product and transition state). Thus, unless the fragments or TS are very close in stability, you cannot focus your analysis based purely on the acidity carboxylic acids as the basis of your analysis.

(b) How do you consider the energy deposition in your molecule? Energy transfer in CID is dependent on the mass of the ion if you do not consider energies in the center of mass reference frame. In the laboratory frame, you will observe a decrease in energy deposition as ion mass increases. This would apply for a single collision conditions. If your conditions are multi-collisional conditions, things get even more complicated.

(c) The displacement in fragmentation energy as size increases is generally known as "kinetic shift", as it involves a reduction in the fragmentation kinetics when the degrees of freedom of a system increase. I guess that this is what you refer to as the number of rotational degrees of freedom. RRKM theory could probably be used to model this effect, but calculation times would probably scale up rapidly with the size of the systems under study.

@PLD Right, rotational degrees of freedom is wrong here as the energy will be stored in vibrational modes I should fix that. So there is no generic rule of thumb like a methylene group increases the stability in CID by around this value, right?

@PLD I guess that besides CID, other fragmentation methods (e.g. IRMPD) will suffer from the same problem, as they are also ergodic processes that will deploy the energy in the vibrational modes. Right? It would be really nice to find a way to determine the gas phase stability for differently sized substituents though. Hmm.

I see some thermodynamic rules of thumb that could fit, but in this case it is a kinetic effect, so it would also depend on the exact experimental setup (time frame of the fragmentation event). And yes, other activation methods (IRMPD, UVPD, ...) will suffer from similar experimental artefacts. One option that would work would be to allow long fragmentation times such as in an FT-ICR cell, or in the same vein, to use BIRD experiments.

I see, that makes sense : )

By the way, for the acidity of carboxylic acids in the gas phase, you should consider having a look at this J. Phys. Chem A article by McMahon et al. http://pubs.acs.org/doi/abs/10.1021/jp9908202 which shows that gas phase acidities of carboxylic acids increase with chain lengths due to intramolecular solvation of the charge.

@PLD "One option that would work would be to allow long fragmentation times such as in an FT-ICR". I see that then, e.g. in a BIRD experiment, the fragmentation would be controlled by the thermodynamics due to the long time frame. However, wouldn't complexes with larger substituents still be more stable as they are able to uptake a larger amount of energy?

@deux ex machina From a thermodynamic point of view (infinite time), the only requirement for a fragmentation to take place is that the internal energy is above the activation energy. In the RRKM theory framework, the presence of larger substituents will only translate in a reduced probability of the internal energy to populate the vibration mode leading to the transition state. The difference is really not that larger substituents are able to store more energy (smaller ones can also store large amounts of energy), but a reduced probability for the system to follow the fragmentation pathway.

@PLD Thanks for all the helpful hints. I guess I will take a look into RRKM theory a bit... The CID results I have gotten really just depend on the size of the substituent and I guess I will try BIRD now. Thanks! : )


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The reaction which forms a pair of diastereomers

The reactions among the following which results in the formation of a pair of diastereomers are:

I am aware of the fact that $\ce{HBr}$ gives anti addition product and $\ce{BH3}$ forms syn addition product. In part (a) since the back side is less sterically hindered ( due to presence of only $\ce{H}$ ) $\ce{BR-}$   preferentially attacks from the back side.Similar is the case with (b).c) proceeds with a free radical mechanism.d) however occurs via anti-markinicoff addition so that $\ce{OH}$ adds to the primary carbon.Since there are two possible ways in which the $\ce{OH}$ group may add( either from the back side or the front side)my idea is that it may result in a pair of diastereomers. However I am not quite sure about it. Thanks.

Consider that, in order to form diastereoisomers here, you need to form a new chiral center. So which of these reactions will generate a new chiral center in the product?

@MarcoB that implies that the correct answer should be b).

@aniline To quote the Lion King: ‘Look harder!’


text:
Wearing a tooth guard
Since I have been advised by my Dentist to wear a 'Tooth Guard' at night because apparently I grind my teeth, my top front teeth under the guard are disappearing, some now only about 3mms from the gum. Are they being dissolved by acid in my saliva?
Regards,
David Bowley (Scotland)

Hi David, please do ask your dentist what is happening - if you trust random Internet people more than you trust your dentist, perhaps it's time to find a new one ;)


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pH Curve in Acid/Base Titrations

So I came across a question on the equation of the pH curve, and I wanted to know how the user here arrived at the matter balance equation. Any help would be appreciated!
Here is the question:
The reason behind the steep rise in pH in the acid base titration curve

Nicolau already outlined it quite clearly in his answer.

Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ($\ce{HA}$ or $\ce{A−}$) is simply $n_A = C_A V_A$

so you have $n_{\ce{HA}} + n_{\ce{A-}} = C_A V_A$

We will assume that volumes are additive (total volume $V_T = V_A + V_B$)

and given that $[\ce{HA}] = n_{\ce{HA}}/V_T$ and $[\ce{A-}] = n_{\ce{A-}}/V_T$ it should be fairly clear that
$$[\ce{HA}] + [\ce{A-}] = \frac{C_A V_A}{V_A + V_B}$$
as desired.


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Racemic mixture or not?
Will the following reaction give a racemic mixture or not?

According to me, it should give. But my teacher disagrees, but I cannot understand the reasons he gave. Shouldn't there be equal approaches from above and below the molecules?
I am confused when he says that the adjacent 3° Carbon can change its configuration too.
Update
Original question:-

Solution provided:-

The answer hinges on whether the starting material is enantiomerically pure or not. As drawn, I would assume it is a racemate. Can you confirm?

@jerepierre I assumed a racemic mixture. The answer given in the book is 'No racemic mixture will be formed'. I don't know what to assume as I am confused with my assumptions.

Is the methyl group bolded (or dashed)? If not, then this is a poorly designed question, given the book's answer.

@jerepierre I have added the original question as well as the solution provided.

IUPAC advises against using the term racemic mixture because of its ambiguity.

racemic mixture (obsolete) The term is confusing since it has been used as a synonym for both racemate and racemic conglomerate. Usage strongly discouraged.

The term that applies in this case is racemate.

racemate An equimolar mixture of a pair of enantiomers. It does not exhibit optical activity.

There are three qualitatively different initial conditions to consider.
1. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is not $\ce{OH-}$
Both enantiomers behave similarly, each yielding a stereocenter at the carboxylic carbon.

Verdict: not optically active but contains two pairs of equimolar enantiomers: $\mathrm{a), d)}$ and $\mathrm{b), c)}$. Even three if you count the reagents. Is a racemate if one relaxes the single pair criterium.
2. A racemate of (3S)-methylpentan-2-one and (3R)-methylpentan-2-one, nucleophile is $\ce{OH-}$
Verdict: not optically active, equimolarity is ensured, a pair of enantiomers. Is a racemate.
3. Either (3S)-methylpentan-2-one or (3R)-methylpentan-2-one, nucleophile is / is not $\ce{OH-}$
Verdict: optically active, does not contain enantiomers. Not a racemate.

In 1, shouldn't that be two pairs of diastereomers instead? And what has that to do with optical activity? In 3, this is the step where my teacher says that since a pair of diastereomers are formed, it is not optically active and we should not talk of racemisation, and I am confused.

@ReeshabhRanjan 1. Enantiomers are a type of diastereomer. In this case, if you compare $\mathrm{a), d)}$, you see that both stereocenters are in fact reverses of each other. Same applies for $\mathrm{b), c)}.$ Thus, these in effect "cancel" polarisation of light. 3. If the nucleophile is $\ce{OH-}$, then indeed you get a mixture of diasteromers that are not enantiomers. It should still be more likely to be optically active than not. But since they are already not enantiomers, optical activity is extraneous.

how are enantiomers a type of diastereomers? We say diastereomers are compounds which are neither the mirror image, nor superimposable on each other, and enantiomers are the compounds which ARE mirror image but NOT superimposable.

@ReeshabhRanjan Depends on your definition of diastereomer. If you define it as stereoisomers which are not enantiomers, then you are correct. You can also define a diastereomer as a stereoisomer that differs from its isomer in at least one stereocenter. And so if all sterocenters are different, we have enantiomer as a special case. But still: $\mathrm{b), c);\ \ \ a), d)}$ are enantiomers; similarly, $\mathrm{a), b);\ \ \ c), d)}$ are diasteromers in either definition.

In this reactant one stereogenic carbon is already present and its configuration will not change during reaction. When Nu- attack on carbonyl carbon two products are formed due to top side and bottom attack but these two products are Diastereomers.

And what if a racemic mixture of the substrate had been given?

Two pairs of diastereomers will be formed

Why in the the other case too will a disteriomeric pair will be formed?

When you add to the planar carbonyl group, attack from both sides of the plane is equally probable. If there are no other carbons bearing four different groups in the molecule (chiral centres), a pair of enantiomers will be formed in equal amounts (racemate). Keep in mind that diastereomers and enantiomers apply to relationships between pairs of  molecules. If you create two pairs of enantiomers, then the molecules in each pair will be diastereomers of the other pair. So you also create two pairs of diastereomers.
Now, there is already a chiral centre in your molecule - but its configuration is not given. Without knowing that, the question cannot be answered properly. The molecules could be either all R or all S; assuming it is R addition of the nucleophile will give (R,S) and (R,R) in equal amounts. These are diastereomers because the mirror image of (R,S) is (S,R) and the mirror image of (R,R) is (S,S). The same applies if all your molecules would have an S chiral centre before addition. The molecules that are formed are not mirror images.
If your molecule would be present as 50% R and 50% S beforehand (racemate), addition will create equal amounts of (R,R), (R,S), (S,R) and (S,S). Now, you have two pairs of enantiomers: (R,R)/(S,S) and (R,S)/(S,R). Every stereoisomer in the mixture has one mirror image (together they are enantiomers) and 2 others that are diastereomers. I'd also like to add that enantiomers are definitely not diastereomers. In fact, that's how IUPAC defines diastereomers: they are stereoisomers that are not enantiomers. This mixture is not a racemate because there are four stereoisomers. According to IUPAC, a racemate is a pair of enantiomers present in equal amounts.
If the nucleophile is hydroxide, then addition will not create an additional chiral centre. So then it would depend on whether you had a single stereoisomer to start with or a mixture, whatever stereochemistry you started with will be retained since no bonds to the chiral centre are formed or broken.

Oh, the racemic mixture is only defined for a pair of enantiomers only? Even if n (>1) pairs of enantiomers present making the net optical activity zero?

Out the two sites, one site is adjacent to the Me group So attack to that site is hindered because of Steric Reason. Only one site is available for attack. Therefore, enantiomers are not formed in equal ratio.

Your answer seems much practical, but what if we had been given an equimolar mixture of all the 4 possibilities? Will it still be NOT a racemic mixture?

You are saying that due to steric hindrance, there will not be an equimolar mixture of any pair of enantiomers. I agree. But what if we were already given equimolar mixture of the enantiomeric pairs?

Pair of enantiomers.

I am saying that I myself add same amount of each enantiomer into a common container.