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Why doesn't Delta S total = 0 for this reversible process?
My textbook (Engel Physical Chemistry 3rd Edition) says:

For any irreversible process in an isolated system, there is a unique direction of spontaneous change: $\Delta S > 0$ for the spontaneous process, $\Delta S < 0$ for the opposite or nonspontaneous direction of change, and $\Delta S = 0$ only for a reversible process. In a quasi-static reversible process, there is no direction of spontaneous change because the system is proceeding along a path, each step of which corresponds to an equilibrium state.

It is my understanding that when a system is broken up into a system and the surroundings, $\Delta S + \Delta S_\text{surroundings} = \Delta S_\text{total} = 0$ for a reversible process.
I was doing the following question when I got stuck:

An ideal gas sample containing $\pu{2.50 mol}$ for which $C_{V,m} =3/2R$ undergoes the following reversible cyclical process from an initial state characterized by $T = \pu{450 K}$ and $P = \pu{1.00 bar}$:
a. It is expanded reversibly and adiabatically until the volume doubles.
b. It is reversibly heated at constant volume until $T$ increases to $\pu{450 K}$.
c. The pressure is increased in an isothermal reversible compression until $P = \pu{1.00 bar}$.
Calculate $q, w, \Delta U, \Delta H, \Delta S, \Delta S_\text{surroundings}$, and $\Delta S_\text{total}$ for each step in the cycle, and for the total cycle. The temperature of the surroundings is $\pu{300 K}$.

I was attempting the process denoted in question b, and my answers matched the books answers:

However, when I thought about the values, I was confused as to why $\Delta S_\text{total}$ is negative. Isn't it true that in the reversible process, $\Delta S_\text{total} = 0$?

How can you heat to 450 K if the temperature of the surrounding is 300 K?

Perhaps there is an electrical heating coil inside the system. Are you saying the question is incorrect?

Even if there is an electrical heater coil inside the system, topologically it is part of the surroundings (i.w., outaide the system)  and you need to consider the heat transfer at the coil temperature.

Furthermore, there is nothing reversible about how an electrical heater operates.

@DrPepper  In your solution to b) you wrote $w = 0$. That would exclude any non-expansion work such as the electrical work done on the system if you want the heating coil inside the system. "Are you saying the question is incorrect?". No, I am saying I do not understand the question. A p-V diagram would help, with each step on the diagram labeled to clarify the interaction with the surrounding.


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Why is Na+ reduced instead of H+ in the electrolysis of dilute NaCl(aq) with Mercury cathode?
In a class of electrolysis, my instructor told me that Hg forms Na-Hg in the electrolysis of dilute NaCl aqueous solution. For this reason, sodium cations are reduced in the cathode instead of hydronium ions. However, he did not give me any reason for how creating an amalgam affects the reduction of the metal or, why Hg creates amalgam in the first place.
So, I would like to know why Hg creates amalgam whereas other metals like Pt don't form them ? And also how this affects the reduction of Na.

It has to something with the hydrogen overpotential of Hg electrode.

Well,what do you mean by "hydrogen overpotential" ?And what has it got to do with Hg?

Umm, proper question would be rather why Na is reduced with Hg electrode and is this even really the case. While Na might perhaps be reduced also with other electrodes, it obviously won't  become part of amalgamate, as only solutions in Hg are called like that.

Please see “Castner–Kellner process” in wikipedia. Definitely sodium can be electrolyzed from brine. The reason it works is the overpotential for hydrogen reduction at a mercury cathode. In other words, there is a kinetic hindrance to hydrogen reduction and sodium forms an alloy with mercury, so back-reaction with the aqueous solution is impeded. Traditionally, mercury alloys with other metals are called amalgams. Liquid gallium, for example, also readily forms alloys with many metals, but gallium has not been known since ancient times.

@Ed V, is there an intuition behind this phenomenon (creating alloys by Ga,Hg but not others) at the atomic level?

In the case of sodium ion reduction at the liquid mercury cathode, the sodium atoms get “dissolved” in the liquid mercury. So they move around like in a ordinary solution. But this keeps them from contacting the brine, so they do not get to react with the brine. Those at the interface of brine and mercury can so react, producing hydrogen gas and NaOH. For platinum, for example, the sodium would, at best, form a plating: the atoms could not get sheltered, as it were, by getting into the interior of the platinum. So it is a matter of getting the sodium away from the brine and mercury does it.

The difficulty of two hydrogen atoms moving about on the surface of mercury to combine into a hydrogen molecule also favors the formation of ammonium amalgam (from electrolysis of solutions of NH4Cl), where NH4 dissolves into mercury rather than breaking into NH3 + H, then into H2.

The mental conflict occurs because two processes could theoretically happen (electrodeposition of sodium, or electrodeposition of hydrogen), and the theoretically unfavorable one occurs - and not only occurs, but is so favorable that it was used commercially to produce NaOH as EdV noted.
So why is the theoretically favored process not favored in this instance? Hydrogen evolution is a multi-step process, and if you wave your hands a lot, you can call it overpotential and be off to other projects.
When (or if) sodium is deposited (on mercury), it can dissolve and be removed from contact with water. So this process is a slam-dunk if you push with a high enough voltage.
But the hydrogen evolution should take place at a lower voltage. Why not? If/when hydrogen is deposited, its first appearance is atomic: one H atom sticks somewhere on the mercury surface; the second step is deposition of another H atom somewhere else. The next step, for the two atoms to form a molecule, requires some mobility on the part of the hydrogen and the mercury surface. At this point, conventional wisdom suggests that the molecule could dissolve into the solution, but eventually the aqueous solution becomes saturated. Then the first molecule must meet another molecule, and another, to begin the process of forming a bubble which will eventually rise to the surface of the aqueous solution. Somewhere in there is an impediment to the formation of bubbles, and it may involve mobility of atoms or molecules, or attachment of hydrogen to mercury, but the process is difficult enough to allow the voltage to be raised high enough to allow deposition of sodium to proceed.
And then we say that mercury has a high overpotential for deposition of hydrogen.
Now suppose we do the same, but with gallium. Sodium should deposit and become dissolved as with mercury, if the voltage is high enough. (Sodium alloys with gallium; amalgamation is a chemical term that should be reserved for mercury alloys.) But the hydrogen overpotential on gallium is less than on mercury (Ref 1), and gallium and hydrogen are deposited together from a solution of $Ga^{+3}$ (Ref 2). So, in the case of liquid gallium, it seems that hydrogen would be more easily deposited than sodium (unless, perhaps you used a very high voltage, which would override the thermodynamic preference because diffusion couldn't keep up with the current demands).
Ref 1. https://pubs.rsc.org/-/content/articlelanding/1966/tf/tf9666203524/unauth#!divAbstract
Ref 2. https://link.springer.com/article/10.1007/BF01016217

If you heat metallic sodium with let’s say gold / silver / mercury / platinum etc.. it will form an amalgam.
I have tried this with a tiny bit of gold some years ago.
I did heat Na in an inert argon atmosphere to several hundred degrees in an quartz tube and drop the gold inside.
It did take only fractions of a second till the amalgam was formed.
The result is a green looking amalgam when it is hot - when it cools down it is a brown solid substance.
Given the fact that gold is a pretty inert substance, I guess that sodium will do the same with Pt, Rh, Ir, Ru, Os, Ag, Hg ...
I guess during electrolysis you have metallic Na formed. I don’t mean junks of Na but Na atoms and some of them may live long enough to amalgamate with the Hg.

Not every metal amalgamates. See https://chemistry.stackexchange.com/q/117633/79678.

The question was not about Na amalgam nor amalgams in general, but specifically why it or metallic Na is formed at all during NaCl electrolysis with Hg as the cathode. And yes, other metal than mercury obviously cannot form amalgams as they are not mercury.

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Where does the RT term come from in the derivation for the activation enthalpy from the Eyring equation?
It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that $E_\mathrm{a} = \Delta H^\ddagger + RT$. Why is this?

Using vibrational partition functions to define the reaction rate constant produces an equation of the form $\displaystyle k=aT^be^{-\Delta U_0^\mathrm{O}/(RT)}$ where $a,b$ are constants independent of $T$ and $\Delta U_0^\mathrm{O}$ is the difference in zero point energies at the transition state compared to reactant, this is related to what is commonly called the activation energy. This approach allows molecules to have discrete vibrational levels and so a thermal population of these. The constants $a,b$ are found from the partition functions.
The (classical) Arrhenius equation is often written as $k_A=Ae^{-E_A/(RT)}$. Taking the log of each equation and differentiating wrt. $T$ gives
$$\frac{d\ln(k)}{dT} =\frac{bRT+\Delta U_0^\mathrm{O}}{RT^2}\equiv \frac{E_A}{RT^2}$$
thus the Arrhenius activation energy is $E_A\equiv bRT+\Delta U_0^\mathrm{O}$.
(Texts on statistical thermodynamics gives details of partition functions, some book on kinetics also give this detail.)

how would you have got here from the simpler delta G of activation, which is more commonly seen than the difference in the zero point energies?

The $\Delta U$  includes the difference in the minima of the potentials and also accounts for the difference in energy of the lowest vibrational energy level. The $\Delta G$ includes entropy changes, which can be incorporated into the pre-exponential term leaving just the $-\Delta H/RT$ you ask about in the exponential. The thermodynamic approach cannot account for any quantum effects such as zero point energy.

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How much glucose oxidase is usually deposited on a glucometer strip?
The concentration of serum glucose is on average roughly around 5 mmol/L and let's say a typical blood droplet applied to a glucometer is 1μL. We are therefore depositing 5 nmol of glucose on the test strip.
  These 5 nmol of glucose interact with the enzyme glucose oxidase and through a redox mediator a current is measured. This current is proportional to the concentration of glucose present and it is in the order of few μA (see Figures below taken from Ref.1 and 2). Each glucose molecule is oxidased to glucono-lactone by the enzyme glucose oxidase releasing 2 electrons to the amperometric circuit; that means that if we have 5 nmol of glucose, then 10 nmol of electrons are released or equivalently around 1 mC:
$$Equation 1: \frac{10*10^{-9}mol*6.022*10^{23}electron/mol}{6.24*10^{18}  electron/C} ≃ 1 mC $$
Once the oxidation of glucose starts, the current measured increases reaching its maximum for few seconds (time-frame of maximum stability) and then decays. The current used to calibrate and measure the concentration of glucose is taken during this stable time-frame of maximum current.  If the current measured is of few μA (i.e. let's say 10 μA), that means that in 4 seconds (I have read 2-4 s is typical), only 40 μC (or 0.04 mC) are released and this accounts for 4% of the total number of electrons that would be released if all the glucose contained in the blood droplet is oxidised (see equation 1).
 1.Does it mean that usually less than 10% of all glucose is effectively oxidised? if yes, why? What are the main factors that limit the yield so much?
It is also very difficult to find data on how much glucose oxidase is deposited on the strip. If the max current is roughly around 10 μA that means that 10 μC (or $6.24*10^{13}$electrons or ~ 0.1 nmol of electrons) are released every second; this means that 0.05 nmol of glucose are oxidised every seconds.
 Glucose oxidase can be considered to follow a Michaelis-Menten law:
$ Equation 2:  v = Kcat*[E_0]\frac{[S]}{Km+S}$, with $[E_0]$ the concentration of the glucose oxidase and $[S]$ of glucose (substrate)
$Kcat$ can be estimated to be around 740 $s^{-1}$, and Km 18 mM (see ref.3); in this example $[S]$ was 5 mmol/L and the enzyme was oxidising at a rate of 0.05 nmol/s.
 If we consider that the initial volume of the blood droplet (1 μL) is largely preserved during its travel to the reaction area of the strip, then $v = 5*10^{-5}M/s$.
 From this approximate data and equation 2, $[E_0]$ can be estimated to be around 0.3μM; considering the volume of 1 μL, then there are around $3*10^{-13}$ mol of glucose oxidase deposited on the strip or 50 ng (multiplying by the molecular mass of glucose oxidase, 160 kDa).
 2. Is 50 ng a reasonable estimate? Does anyone know (approximately) how much glucose oxidase is deposited on a strip?

Volume 2019, Article ID 5948182, The Influence of Blood Glucose Meter Resistance Variation on the Performance of a Biosensor with a Gold-Coated Circuit Board, https://doi.org/10.1155/2019/5948182,

https://slideplayer.com/slide/5728209/
Isolation, purification and characterization of a novel glucose oxidase from Penicillium sp. CBS 120262 optimally active at neutral pH, https://doi.org/10.1016/j.pep.2006.09.013


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What does it actually mean by stability of an atom
There are several rules related to the stability of an atom, like the octet rule, half-filled or fully filled d orbital, etc... But none of them (to the best of my knowledge) explain why an atom is stable when it obeys those rules. So, is there any reason at all or are they just based on experimental results?

It is an experimental fact that isolated atoms are stable. These rules are used to try to explain in simple terms, sort of 'rules of thumb' ,  how electrons are arranged. The proper understanding of these rules flows naturally from Quantum Mechanics.

The term "stability" is a big problem here. stability under what conditions? stability to what possible reaction? Defining the term would resolve many of the possible questions.

@porphyrin I don't think that is a strongly established empirical fact. I doubt that isolated fluorine atoms are stable to many reactions, for example. Fluorine molecules aren't exactly "stable" by most definitions either.

@matt_black I don't follow your argument: I'm sure an isolated fluorine atom remains as a fluorine atom for as long as you wish to observe it, or remain as a fluorine atom when it reacts, say to form HF.

@porphyrin If the question is related to isolated atoms, then fine: everything is stable, charged or neutral as absolutely no change can happen (mostly). If it reacts with something else, it isn't an isolated fluorine, is it? So I strongly doubt the question is about the esoteric stability of isolated atoms, rather it is about the general rules relating to chemical bonding.

@matt_black well its becoming a semantic argument, but (being pedantic for a moment) the OP did ask about atoms not atoms in molecules, its just not very clear what is being asked, but I agree they probably meant something like 'atoms bonding in molecules' which would have been better.

@porphyrin I may be misinterpreting what you are saying, but by suggesting all isolated atoms are stable, are we not including atoms that undergo radioactive decay?

@dval98 I was excluding that possibility as its not usually a consideration in chemistry.

First you need to understand what stability is. Normally, octet rule is based on noble gas configuration. Since, noble gases hardly react with other atoms, and they have 8 electrons as their shell configuration, thus we consider them "stable".
So, taking noble gas 8 electrons shell configuration, we tend to generalise all other atoms/elements to have 8 electron filled orbit to gain stability.
You must understand that scientist create models to explain natural phenomena. Rules are  designed to explain already existing phenomena.  Stability truly means "less energy" or inert behaviour where the atom neither gains nor lose electrons to other atoms/molecules. The lesser the energy of the atom/system, more the stability.
Thus your doubt on "does not explain why an atom is stable when it obeys those rules" is very vague. Think of inert gases, their configuration, their inert nature and you will understand.


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Is carbon–sodium bond mainly ionic or covalent?
We all know that carbon mostly forms covalent bonds (almost always) and sodium mostly forms ionic bonds (almost always). But if $\ce{NaOH}$ reacts with ethyne, then $\ce{HC#CNa}$ and $\ce{H2O}$ are formed. Then what type of bond will $\ce{C-Na}$ be: mostly ionic or covalent?

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

Well, I get the point that this bond is highly polar due to the high electronegativity difference between them . But if you were to classify it as either ionic or covalent, which one would you prefer to go for?

Generally, the difference between an ionic and a covalent bond is quite vague. Usually, it just comes down to an arbitrary difference in electronegativity on the Pauling scale of electronegativity. Most chemists agree that an electronegativity difference greater than 1.7 renders a bond ionic. Because the C-Na bond difference in electronegativity is 1.62, we would call it a polar covalent bond. Again, this definition is VERY arbitrary, and often times chemists will think of polar covalent bonds like the C-Na bond as ionic or ionic bonds as covalent (in some forms of organometallic chemistry).

@EliJones It can be done better then that https://chemistry.stackexchange.com/questions/17064/is-kf-the-most-ionic-compound

Electronegativity of Carbon is $2.55$ and that of Sodium is $0.93$ (from Wikipedia) so by the Haney-Smith equation (from J.D. Lee's Inorganic chemistry, third edition)
$$\%\text{Ionic Character} = 16 |\Delta\chi| =3.5 (\Delta\chi)^2,$$
where $\Delta \chi$ is the difference in electronegativity of the two atoms. Carrying out the calculation gives us
$$\% \textrm{Ionic Character} = 35.1054$$
So, it shows that the bond $\ce{C-Na}$ has a less ionic character.
In Organic Chemistry by Morrison and Boyd we find

The carbon-magnesium bond is covalent but highly polar, with carbon pulling the electrons from electropositive magnesium [...].
  But, whatever the metal, it is less electronegative than carbon, and the carbon-metal bond - like the one in the Grignard reagent - is highly polar. Although, the organic group is not full-fledged carbanion [...] it nevertheless has considerable carbanion character.

Note: No bond is completely ionic or covalent, what matters is which character is dominant and hence the bond will behave like that.


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Are raw unrefined silver ore rocks dangerous to touch?
A Farmer has come across an ancient, unrefined silver ore rock on his land. It cuts through like solid silver.
It was probably left over from medieval coinage times. He did get a small amount of dust.
Is this dangerous?
I hear it can contain arsenic, etc when heated, but I mean is it just generally dangerous to handle without gloves, does ashing after make it safe, or just generally have around the house on display?
The bare silver face is exposed to the air.

I am not an expert in geochemistry; however, I have worked with arsenic in the past, so I might be able to answer a few of your questions. To start, having the raw ore on display should be perfectly safe. Historically, the only problem with arsenic-based compounds was with Paris Green paint because the mold Scopulariopsis brevicaulis was converting the arsenic paint into toxic arsine gas. In your case, I highly doubt there would be enough arsenic (in the right form) or enough mold to cause any problem. I think it should be find to handle without gloves. Generally, silver ore contains only ...

... extremely small amounts of arsenic (and generally in its less form). There are really only specific types of ore that contain significant amounts of arsenic (arsenopyrite, realgar, orpiment, etc.); however, he probably does not have these.  It is never a good idea to breathe in any dust from metallic ore; however, I don't think this would necessarily be of serious concern -- as long as he wears a dust mask in the future. Again, I do not know everything about his situation and cannot be sure about any of this without testing his sample, but I would be fairly certain he would be fine!

Argyria

I've collected rocks and minerals. I wouldn't worry about it. The only precaution that I'd take would be to wash my hands after handling heavy metal mineral specimens.


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What is the difference in dehydration of cis- and trans-1,2-cyclohexanedicarboxylic acid in anhydride formation?
Is there some required stereochemistry for anhydride formation? Do both of them form the same product, or does one not even undergo dehydration?

I looked up cyclohexane-1,2-dicarboxylic anhydride's structure and found two images:

Does that mean both of them are formed?

Consider how close the two carboxyl groups are in the different strutures. Maybe try drawing them in 3D.

Think about the stereochemistry of the starting material (cyclohexane-1,2-dicarboxylic anhydride)

um so i think that the reaction wont happen in case of trans and in case of cis ill just eliminate a H2o.does that sound right?

The mechanism of formation of anhydride from a dicarboxylic acid is as follows: [1]

So as you can see, it involves a nucleophilic attack at the $\ce{C=O}$ of one of the carboxylic acids by the other one, following which internal rearrangement occurs and a water molecule is kicked out.
Stereo chemically, we are interested in finding out in which configuration can the lone pairs of oxygen from one of the $\ce{-COOH}$ attack the $\ce{π^* C=O}$ type antibonding orbital of the other $\ce{-COOH}$ group.
To achieve this, the nucleophile( oxygen in this case) has to align properly with the $\ce{π^*}$ orbital at the Bürgi–Dunitz angle. This configuration is necessary whenever a nucleophile has to successfully add to a carbonyl group. The interacting diagram will look like this for formaldehyde: [2]

(Green ball= nucleophile| Blue and Red balls= carbonyl)
We can also roughly estimate that the situation will be similar for the $\ce{C=O}$ part of the carboxylic acid group.
To see which of the two reactants in question will form the required anhydride, one can simply look at the 3-D conformers of both in the appropriate orientation, and compare to see which of the cases resembles the standard Burgi-Dunitz interaction more closely as illustrated above
Here is the cis configuration: [3]

And this is the trans configuration: [4]

1= oxygen| 2= carbonyl of acid group
As you can see, the oxygen seems much more aligned with the carbonyl in the trans configuration as compared to the cis. So, 2 might form the anhydride more easily than 1 (that is, with a lower activation energy barrier due to reduced torsional strain)
But again, it's not a strictly inaccesible barrier, as a few bond rotations in the cis configuration can also lead to a somewhat favorable configuration for the reaction. So it may ultimately lead to two different kinds of anydrides by each of the isomers like you said, 2 can lead to formation of 4, while 1 can lead to formation of 3

In your mechanism the carbonyl oxygen is the more nucleophilic of the two. If this were a Fischer esterification, I'm sure you would protonate the carbonyl oxygen.

@user55119 Well, in hybridization terms, I would expect the carbonyl oxygen to have more s character than the hydroxyl oxygen, thereby the former being more electronegative and less nucleophilic. From what I understand, we protonate the carbonyl oxygen in Fischer esterifcation as a last resort. The initally preferred protonation of the hydroxyl oxygen would lead to no useful result,and so the carbonyl oxygen gets protonated after that to make progress in the reaction

To each his own.

but the answer turns out to be the other way round, its happening in cis but not in trans:( and the possible reason proffessor suggested is that after dehdration(as you mentioned the chair form) its harder to form a ring in trans(axial,axial or equitorial,equitorial)

Also I guess the more interesting point here is that both the cis and trans anhydride forms can be formed and can even be bought online for almost 95% purity here and here respectively. So your professor will have to specifically clarify what exactly does he means when he claims that "it happens in cis but not in trans"

Your professor has evoked a form of reasoning where the reactants are closer to each other and so it's perhaps easier for them to react, while I tried to evoke a stereo chemical perspective and analyse through better approach for the nucleophile. Both the reasonings are okay on a qualitative basis, but what can really settle matters is a published finding which shows the yield, and also the probable reason behind it found experimentally. Also, your prof's statement is pretty vague as of now because in my prev. comment, you can see that both the products are formed and even commercially sold


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Covalent bond nature in electronegativity gap

Does a large difference in electronegativity mean the covalent bond is weaker?

In a covalent bond between two atoms of different electronegativities, the bonding electrons are pulled towards the more electronegative electron. This makes the bond polar (eg. $HCl$).
In a diatomic molecule, the electronegativity is the same so the electrons are equally attracted to either atom. Does this make diatomic covalent bonds stronger than the former? And if so, as the difference in electronegativity increases, does the strength of the covalent bond decrease?

No. If anything, it makes the bond more ionic.

Strength of covalent bond depends on many factors, one of the key factors is "extent of overlapping of orbitals". For e.g., if you consider $\ce{H2}$ molecule, there is a head on overlapping of $\mathrm{1s-1s}$ orbitals involving two hydrogen atoms, but in the case of, say, $\ce{HCl}$, there is a head on overlapping of $\mathrm{1s-3p_z}$ orbital ($\mathrm{1s}$ of hydrogen, and $\mathrm{3p_z}$ of chlorine), and we all know that $\mathrm{p-s>s-s}$, that is, overlapping between a $\mathrm{p}$-orbital and a $\mathrm{s}$-orbital is stronger than $\mathrm{s}$ and  $\mathrm{s}$ orbital, and stronger the overlapping more is the covalent bond energy.
So, it largely depends on the type of orbitals involved and the nature of overlapping.
Check this to learn more about overlapping.

Well, I would say that the strength of the covalent bond is related to the level of polarisation abilities of the molecule in addition to the point stated by peaceHoper . In an ionic compound ,  polarisation is defined by the distortion of the anion by the cation . The more the polarisation , the more the covalent characteristics of the molecule . And thus , it results in higher strength of the covalent bond . It also depends on the type of orbitals involved but it's already been covered.
According to Fajan's rule , the smaller the anion and the larger the cation ,the greater the polarisation of the molecule . In H-H , both the atoms are the same . But in H-Cl , the cation H+ is really small compared to the relatively a lot bigger Chloride anion (let's hypothesise the molecule to be ionic ; not a bad one as well since there are some ionic characteristics present in it) . Hence, the polarisation of HCl is much more than H-H and therefore it has a stronger covalent bond.


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How do you explain partial charges in molecules that have a dipole?
For example if i have hydrogen and fluorine covalently bonded with each other, it would result in a polar molecule with the valence electrons being pulled closer to the fluorine than it is to the hydrogen. At some time the electrons will be at the fluorine end more often as it has higher electronegativity and thus making the hydrogen end partly positive. How does this work? If one electron is pulled to the fluorine end (the one that is shared), doesn't this just revert the hydrogen to its neutral state (as an element) and not be a partly positive charge?
Or does the electrons that are shared also including the lone pairs are going to orbit the shell of the other covalently bonded element, so that in some time there won't be any electrons in one end?

It is very hard to understand what you are describing. The positive nuclear charges attract electrons. More charged nuclei can be expected to attract electrons more strongly, all else being equal (but all else isn't equal because neutral atoms with more highly charged nuclei also have more electrons). Quantum mechanics tells you how those electrons spread about the nuclei. It is the outermost electrons in the valence shell that are shared. The electrons are not localized, their position is described by something called a wavefunction (often represented as orbitals), ...

... you can think of the electron probability being smeared about the nuclei.

I should have written "to a first approximation it is the outermost valence electrons that are shared".

RE: If one electron is pulled to the fluorine end (the one that is shared), doesn't this just revert the hydrogen to its neutral state (as an element) and not be a partly positive charge? Hydrogen only has one electron. So if the fluorine atom pulls that electron away then you're left with $\ce{H+}$.

At some time the electrons will be at the fluorine end more often as it has higher electronegativity and thus making the hydrogen end partly positive. How does this work?

This is the charge distribution on the "surface" of the HF molecule, estimated by the program molcalc.org:

In a thought experiment, if you placed a negative sample charge (ideally with less charge than the elementary charge of the electron to not mess with the electron distribution) on the surface of the molecule, it would feel a force that is color coded in the picture (blue and cyan - attractive, red and yellow - repulsive). If you go sufficiently far away, the sample charge would not have any interaction because the molecule is net neutral. On the other hand, if you approach either nucleus, the sample charge would experience strong attraction (so the fluorine nucleus is not negative, even though we correctly assign the fluorine atom a negative partial charge).
To get a quantitative measure of the partial charge, you have to divide the space into two parts - that "belonging" to fluorine and that "belonging" to hydrogen. Then, you can sum up the charges due to the electrons and compare it to the nuclear charge.

If one electron is pulled to the fluorine end (the one that is shared), doesn't this just revert the hydrogen to its neutral state (as an element) and not be a partly positive charge?

Two electrons are shared. If you start with atoms, electron density from both atoms moves toward the center of the bond (very little, surprisingly, and drawing a stick for a bond is not very realistic but is great for counting electrons).
Your idea is that one electron is pulled to the fluorine. This is a very simple model, but let's explore it a bit. In this picture, the electron "originating from the hydrogen" would still be involved in binding, so we are not back to atoms.
What really happens is that instead of being spherically distributed around atoms, the electron distribution changes a bit when the bond is formed, with the bonding orbital (in the simplest valence bond or molecular orbital model) having electrons occupy the space near either nucleus, but not in a symmetric way. That leads to the charge separation.

Or does the electrons that are shared also including the lone pairs are going to orbit the shell of the other covalently bonded element, so that in some time there won't be any electrons in one end?

I did not understand this scenario. Maybe edit the question to clarify before it gets closed.

Do the blue, green and red regions represent charge distribution $δ^{+}$ or $δ^{-}$ or the electrostatic potential which is related to the partial charges?

Red is negative, blue is positive and green neutral potential. Based on that you would assign negative partial charge to fluorine and positive to hydrogen. For a quantitative treatment, you would have to integrate over areas belonging to either atom.

Apologised if I was misunderstood. What troubles me is the interpretation of the coloring. Does it denote partial charge or electric potentital?

@Anton Electostatic potential

In a covalent bond, it is true that if one atom is more electronegative than the other atom, there will be creation of charges, (negative on the more electronegative atom). But, it doesn't mean that the shared electron will only stick towards Fluorine (in your case) which is more electronegative. The shared electron will also revolve around the other less electronegative atom, i.e., Hydrogen (which is less electronegative).
So, it will not revert back Hydrogen into neutral state, moreover due to covalent bond energy between Fluorine and Hydrogen this is highly unlinkely (without any external stimuli).
Check this to understand the concept clearly!


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Can I make UV-Vis calibrant in the lab?
I have a Thermo Nanodrop 1000 spectrometer that I have reason to believe is out of calibration. After consulting the manual, it appears as though there is a way to check the calibration of the spectrometer using a premade CF-1 calibration solution. This solution itself appears to be aqueous potassium dichromate, but I cannot find the concentration and exact specifications of the solution online. I was curious if anyone has attempted to make the calibrant from reagent they had in lab, and if so, what exactly does one use to make the solution?

The composition of CF-1 solution can be seen on its packaging label (Source): potassium dichromate: 0.07–0.10%; perchloric acid: 0.02–0.04%; water: 99.8%. Preparing such solution on your own doesn't look like rocket science to me granted you have pure enough chemicals at your disposal.

Of course it's right in front of my face, thank you very much!

The last time I calibrated an absorbance spectrometer, it was one I made from parts (for an undergraduate Instrumental Analysis teaching laboratory) and I used a 0.0509 M holmium chloride solution I prepared. The visible region spectrum I obtained is here:

Holmium solutions are well known as UV-Vis calibration standards and you can purchase from, e.g., www.starnacells.com. Below is a screenshot from the Starna Cells web site, indicating that their ready-made sealed holmium reference standard cuvette cells have NIST traceability. Evidently, they are good down to 241.0 nm. I would have purchased one, if I had had the money at the time, but the solution I prepared worked well and was not particularly expensive. Erbium and praseodymium solutions also work well.
Sharp peaks in an aqueous solution at room temperature are always nice. Maybe this will help you!


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Water evaporation from calorimeter giving less exothermic enthalpy value?
To find the enthalpy change of combustion for an alcohol, you use a calorimeter filled with water and heat it by burning an alcohol.
I was told that usually experimental values for the enthalpy change of combustion are less exothermic than data book values, because of many reasons (eg- heat loss, incomplete combustion etc.)
One of the reasons why the experimental value may be less exothermic is because water has evaporated from the calorimeter.
I don't understand how water evaporating from the calorimeter would make the enthalpy change less exothermic. If the water evaporates, then the mass of water left in the calorimeter is lower, so it would heat up faster, so wouldn't this give a more exothermic result?

Heat of vaporization is quite high, so if in the local area of the combustion, you vaporize water, that heat won't be distributed to the rest of water to warm it up.

When water evaporates the heat to vaporize the water comes from the water itself, so the water cools. // Apologies to the "young" chemists, but I still think in calories. It takes 540 cal/gram to evaporate water, but 1 cal/g to change water temperature by $\pu{1 ^\circ C}$.

First, we have some things to consider and they are:

the specific heat of water;
the latent heat of vaporization of water;
how the terms mentioned above are compared to each other.

The specific heat of water is $\pu{4200 J kg-1 K-1}$ and it means that it takes $\pu{4200 kJ}$ amount of heat to increase the temperature of $\pu{1 kg}$ water by $\pu{1 K}.$ On the other hand, the amount of heat needed to turn $\pu{1 kg}$ water at $\pu{373 K}$ into $\pu{1 kg}$ water vapor is $\pu{2268000 J kg-1},$ i.e. the latent heat of vaporization of water is $\pu{2268000 J kg-1}.$
Now, if we consider two pieces of data above, it's clearly visible that it takes much more heat to vaporize water than it takes to increase the temperature of the water (given the same mass). As it's this heat which absorbs heat from the reaction materials (mainly products), the high latent heat of vaporization makes the reaction much less exothermic than increasing the temperature of the water (since it requires more heat).
So, reducing the exothermic characteristic of the reaction depends more on the rate of vaporization than it does on the increase in temperature of the water. As a result of this, despite the fact that the amount of water available to absorb heat via increasing its temperature decreases as more and more water is vaporized, the amount of heat absorbed by vaporization is more than enough to compensate for this. Thus, the reaction becomes less exothermic than expected with the vaporization of water.


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Stability of Peroxynitric Acid in cold temperature
I have a question about the stability of Peroxynitric Acid i.e. $\ce{HO2NO2}$
at cold temperature. I know that it becomes stable when temperature decreases, but what is the specific reason for it? I guess it has to do with the temperature changing the potential hypersurface, but I'm not sure. I also seem to not be able to find a good explanation about it in the internet, since most is about the kinetics I'm not interested in. Possibly, this question relates to all PAN-like species i.e. $\ce{R(O)OONO2}$ such as the acetylperoxy nitrate. Any good physical explanation? Thanks

High entropy of decomposition, I guess.

Can you elaborate a bit more?


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Why are lead and zinc commonly found together?
Copper, silver, and gold are often found in combination. That makes sense to me, because they are members of the same chemical family. Likewise, I would expect nickel palladium, and platinum to be found together.
But lead and zinc are often found together, even though they are not of the same chemical family. At least, these metals seem to appear "jointly" more often than lead and tin, even though these are members of the same family.
Why would that be, or am I missing something?

Lead and Zinc have in common two electrons "s" in their outer shell. That is why they may sometimes have a similar behavior. You would object that Lead has also two electrons "p" in its outer shell, which are missing in Zinc. It is true. But these p electrons are "modified" and can be "forgotten" by an unusual phenomena in chemistry, namely Einstein's relativity.
When the number of protons in the nucleus is as high as about $80$, and it is $82$ for $\ce{Pb}$,  the nucleus charge is so high that the electron has to move at a relativistic speed to stay rotating around the nucleus, according to Bohr's model. Of course Bohr's model is wrong and the electron is not rotating and has no speed. But the calculations made with it are still valid, as was shown by Pekka Pyykko, Relativity and the Periodic Table, Accounts of Chemical Research, Vol. 12, No. 8 (1979) p. 276 - 281.
At relativistic speeds, the dimensions of all moving objects decrease. And Pekka Pyykko has shown that this effect is proportional to the sum of the two quantum numbers n and l. Lead has the following electronic configuration : $\ce{[Xe] 4f^{14} 5d^{10} 6s^2 6p^2}$. The sum n+l is equal to $8$ for $\ce{4f, 5d, 6p}$. It is $7$ for $\ce{6s}$. So the electrons $\ce{4f, 5d, 6p}$ are so contracted that they are mixed with inner electrons and "disappear" from the periphery. Only the two electrons $\ce{6s}$ , with n+l = 7, remain in the outer shell of the atom. That is why Lead $\ce{Pb}$ often behaves like an atom of the $2$nd column : The most usual oxidation number of Pb is $2$ (like in $\ce{Pb^{2+}}$), and not $4$ (like the $4$th column), which exists but is not frequent. Also $\ce{PbSO_4}$ is insoluble in water, like $\ce{BaSO_4}$. So $\ce{Ba}$ is made of $\ce{[Xe] 6s^2}$, and Pb may be considered as a sort of [pseudo-$\ce{Xe] 6s^2}$...
To go back to $\ce{Zn}$ and $\ce{Pb}$ configuration, relativity explains why $\ce{Pb}$ looks like a noble gas (Xenon) plus $2$ electrons "s" in its outer shell. By comparaison, Zinc is $\ce{[Ar] 3d^{10} 4s^2}$. Like $\ce{Pb}$, it also looks like a noble gas (Argon) plus $2$ electrons "s" in its outer shell, if the filled shell $\ce{4d^{10}}$ is "forgotten". Mother Earth is probably not able to recognize the existence of this $\ce{4d}$ shell and has often mixed $\ce{Zn}$ and $\ce{Pb}$ minerals.

I think this answer is focused on the wrong details. The most common minerals of  lead and zinc are galena, PbS and sphalerite, ZnS which are often found together. Trying to understand why those minerals are found together would require a complex discussion of geologic processes.

@MaxW: I think that the above is a very good answer within the realm of Chemistry. You're right that it overlooks non-chemistry issues such as geology. But that is basically an issue for another site, say Earth Sciences.

@MaxW I was under the impression that lead and silver ores are often found together, or, at least, where there was lead, there was probably also silver. Any truth to that?

@EdV - Yes sulfide minerals are often found together. The mineral $\ce{Ag2S}$ is acanthite. Galena itself can also contain significant quantities of silver.


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Surface Tension of Water and Breaking through that surface tension
I know that the high surface tension of water makes it relatively more difficult for light objects to break through the surface (i.e. it's why insects can "walk" on water).  So that seems to work when an object moves from the air into the water (submerged).  But what about from the other direction?  By that I mean, an object trying to break through the surface of the water but moving from a submerged location to one above the water (in air).  In this scenario, is it also relatively more difficult to break through water than if an object was trying to break through ethanol, for example, from a submerged location?

Yes.$\mathstrut$

It all depends on the nature of the sample.

Notice small bubbles do not immediately break through the surface.


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Calculating the pH given number of moles of HI and 500. ml of water
Shouldn't we also add the volume of the HI to the volume of the water (500mL) when we calculate the molarity? My book simply divided the number of mol by 500 mL.

Hydrogen iodide is a gas that is highly soluble in water, so any volume change caused by its dissolution was likely considered too small to bother with. The problem is just an exercise in computing pH.

oh ok thanks! but when calculating the pH, shouldnt we also consider the pH of water by itself?

Only unless you have a molar concentration down around 100 nM or so! An aqueous solution of HI is like an aqueous solution of HCl, i.e., they are strong acids.

Simple questions are often interesting.
(1) Most importantly, volumes are rarely additive. So the notion of adding the volume of the HI and the volume of water just doesn't work.
(2) In an answer to a question about hydroiodic acid,  Curt F. showed a chart of concentration and density. For the $\pu{20 ^\circ C}$ values the liters of water in a liter of the acid solution can easily be calculated. So for the OP to get an error of 1% or less, the molarity would have to be about 0.2 molar or less.
\begin{array}{|c|c|c|c|}
\hline
\text{Concentration, %}(w/w) & \text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) &  \text{Molarity} & \pu{L(water)/L(acid)}  \\ \hline
5.2 & 1.0342 & 0.420 & 0.980 \\ \hline
10.8 & 1.0812 & 0.922 & 0.964\\ \hline
16.4 & 1.1226 & 1.44 & 0.938 \\ \hline
22.4 & 1.1765 & 2.06 & 0.913 \\ \hline
27.2 & 1.2333 & 2.62 & 0.898 \\ \hline
33.1 & 1.2918 & 3.34 & 0.864 \\ \hline
38.7 & 1.3605 & 4.12  & 0.834\\ \hline
42.9 & 1.4208 & 4.77 & 0.811 \\ \hline
48.7 & 1.5072 & 5.74 & 0.773 \\ \hline
53.0 & 1.5913 & 6.59 & 0.748 \\ \hline
57.0 & 1.6933 & 7.55 & 0.728 \\ \hline
\hline
\end{array}
You can see from the red line that the data is not linear.

hi, thanks for the detailed reply! do you think i should just do the simplified version and disregard that the water actually has some H3O by itself, too, in the test?

oh wow this is getting confusing since the H3O isnt additive...

@12345bird - Yes, just assume that the volume doesn't change. I did the calculation as a matter of curiosity.

thanks! sad to know that what we learn is just simplification of reality :/

@12345bird - Chemistry uses good enough rather than perfect extensively. Mathematicians can calculate $\pi$ to million of places. Chemistry just doesn't work like that.

oh ok! just knew that now!


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Why isn't PBr3 used to halogenate tertiary alcohols?
I've been looking into activation of alcohols for nucleophilic substitution, and I've come across a series of reactions that differentiate between primary, secondary and tertiary alcohols. I've seen it written in multiple places that reagents like thionyl chloride and phosphorus trihalides aren't used for tertiary alcohols, but that strong acid is used instead. I was wondering why this was the case, because surely an activated alcohol has a good enough leaving group to progress through an SN1 mechanism.
The only thing I can think of is that the alcohol is too sterically hindered to effectively react with the activating agents. But I have seen a reaction scheme that substitutes a tosyl group from a tertiary alcohol via SN1, which confused me even more.
Any help is greatly appreciated.
Edit: I forgot to mention that I've also seen specified that the Mitsunobu reaction works only on primary and secondary alcohols, and I really hope all of these are linked in some way because that would make this a lot easier.
Edit: from https://www.masterorganicchemistry.com/2015/03/10/tosylates-and-mesylates/#four


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Octahedral Crystal Field Splitting Orbital Degeneracy

How are the $\mathrm{e_g}$ orbitals degenerate with each other?
Note: This isn't a homework question. After the semester ended (I don't go to MIT), I ended up on MIT open course-ware to watch some videos about areas of chemistry I haven't covered yet or haven't covered well. I am asking the question here because I have no other avenue in which to ask questions to other people. I'm just trying to use my time during the pandemic to build my knowledge of chemistry. I don't have a lot of understanding about this, I'm just looking for some help

It is because perfect octahedral symmetry is normally assumed; the two $\mathrm{e_g}$ levels and three $\mathrm{t_{2g}}$  are degenerate.  If there was a distortion, say by lengthening both $z$-axis ligand positions then the $\mathrm{e_g}$ degeneracy would be removed as the $\mathrm d_{z^2}$ becomes more stable than $\mathrm d_{x^2-y^2}$. This happens because the $z$-axis ligand has more effect on $\mathrm d_{z^2}$ than on $\mathrm d_{x^2-y^2}$ orbitals simply due to its position. The $\mathrm d_{xy}$ orbital also increases in energy removing the degeneracy of the $\mathrm{t_{2g}}$ and the  $\mathrm d_{yz},$ $\mathrm d_{zx}$ lowered.

The crystal field splitting is based on where the ligands (modelled as point charges) are in relation to the orbitals. In an octahedral complex, the ligands are all at 90° from each other and are placed on each of the $x,$ $y,$ $z$ axes. The orbitals that lie on these axes will experience the most repulsion and will rise in energy, while the orbitals between the axes $(\mathrm{t_{2g}})$ will lower in energy as they experience less repulsion from the ligands, and the average overall energy is maintained.
I can only assume that the degree by which the $\mathrm{e_g}$ orbitals are raised is the same due to the $\mathrm d_{z^2}$ orbital technically being a linear combination of what would have been the $\mathrm d_{z^2-x^2}$ and $\mathrm d_{z^2-y^2}$ orbitals. This means that the $\mathrm{e_g}$  orbitals lie on the axes to the same extent as each other, so experience the same overall repulsion from the ligands.


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Can a solvent be a gas?
A solvent is an element which dissolves something, basically the solute. Normally, all the solvents we encounter are in liquid phase. So, I was wondering if there could be any solvent in a gaseous phase. Can there be any gaseous solvent, after all?

Yes, it can be also be solid, or a supercritical fluid.

Well, I have simply no idea as to how a solid can be a solvent. Same applies to a super critical fluid.

https://chemistry.stackexchange.com/questions/39815/what-is-the-solvent-and-what-is-the-solute https://chemistry.stackexchange.com/questions/102884/what-is-to-be-called-as-solvent

@Mithoron, these links don't explain how a solvent can be a solid. I have posted a new question on it.

It's a bit of a looser definition, but any uniform mixture that has a major and minor component can be thought of as solvent and solute, respectively. Uniform is a not quite well-defined, so that leads to some boundary cases.

@ Habib. Glass is a solid that becomes progressively liquid by heating, without having a real melting point. When liquid, it can dissolve small amounts of metallic compounds, that gives it a color. Adding 1% of some cobalt oxide, for example, produces a nice blue color. The cobalt compound is dissolved in the liquid glass. When cooled down to ordinary temperature, it becomes a solid and remains blue. The obtained solid glass is a solution of cobalt oxide in solid glass.

In order for a substance to act as a solvent, there has to be a solution. According to the IUPAC Gold Book, a solution is:

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.

Wikipedia's article on Solutions has a section on gaseous solutions, but calls them "trivial" and admits that part of the literature does not consider solutions in the gas phase (without citation, unfortunately):

If the solvent is a gas, only gases are dissolved under a given set of conditions. An example of a gaseous solution is air (oxygen and other gases dissolved in nitrogen). Since interactions between molecules play almost no role, dilute gases form rather trivial solutions. In part of the literature, they are not even classified as solutions, but addressed as mixtures.

Of course, solutions are mixtures (specifically, homogeneous ones), as implied by the IUPAC definition talking about a phase (singular) rather than phases.
So if you go with IUPAC, gasses are not solvents because solutions are either in the liquid or the solid phase. They don't mention supercritical fluids, but neither does the OP, so those would have to be considered elsewhere.
Reference: IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version:  (2006–) "solution". doi:10.1351/goldbook.S05746

Solvents imply dissolving process as intermolecular interactions between solvent and solute, leading to disperging the solute from the condensed form.
"Permanent" gases near normal conditions do not provide such interactions. Volatile compounds evaporate with or without presence of gas. Gas is rather a bystander/spectator of such evaporation.
Fluids near critical conditions like supercritical CO2 share many properties of liquids and gases, so such interactions are present, so in some sense and with both eyes closed, one can consider it as solvent and gas.

Yes, solvents can be in gaseous form. Air itself is an example of a gas-gas solution, where other gases are "dissolved" in nitrogen. But liquids and solids can not be dissolved in gas to form a true solution, instead they form colloids.
Reference

Physical Chemistry Textbook for Competitions by Dr. O.P. Tandon and
Dr. A. S. Singh,  Ninth Edition, 2008.

The trouble with defining gases as "solvents" is that they don't behave like other solvents chemists talk about all the time. All gases mix. Most of the other things we might worry about in a typical solvent are irrelevant (like limited solubility). So it it seems better to talk about gases as a separate category. Unless we are talking supercritical gases where the situation is different.

No, unless you use a very loose definition of solvation
The essential feature of solvation is some relatively strong molecular interaction between the solvent and the substance being solvated. Ethanol is soluble in water (and vice versa) in any proportion because their hydrogen bonds have strong intermolecular interactions. Sodium chloride is partly soluble in water because the dipolar nature of water molecules can interact sufficiently strongly with the ions of the salt to keep some of them in solution. But only up to a certain point when the water's capacity to do this is saturated and no more salt will dissolve.
These concepts are irrelevant for gases. If two compounds are a gas at a given set of conditions, they will mix. No interactions between the components are relevant to this. It is inherent to the nature of things being gases. The normal things happening in solvation are irrelevant. The mixing of gaseous substances in any ratio is a property of them both being gases and does not relate to the normal processes of solvation. So there is little benefit to describing the process as "solvation" of one gas being a solvent for another.
There is one exception to this: supercritical fluids. Above the critical point for a substance (usually at higher than normal pressures or temperatures for typical gases) there is no distinction between the gas and liquid phase. For example, Carbon dioxide is a supercritical fluid beyond about 30 °C and 7 atm pressure and is used in some industrial processes. Supercritical fluids have many of the properties of liquids (but not all eg a lack of surface tension) but, importantly, things can be solvated in them in the same way that liquids solvate solids.
Apart from the (extreme) example of supercritical fluids, it makes no sense to talk about gases being solvents.

Hot carbon dioxide gas can act very much like a solvent, and for an otherwise unlikely solute at that -- elemental carbon. See here for the not so exotic chemistry involved.
The concept of carbon "solubility" in a hot carbon-dioxide bearing gas is useful in optimizing temperature and gas composition for cleaning carbonaceous contaminants off a steel strip during some continuous annealing/hot dip coating processes.

In the case of spraying perfume ,can we say that the air is the solvent ?

Not quite the same thing.  Perfume is a mixture of different substances whose molecules would have to separate if they truly dissolved. You either have a multiple component solution in which air is indeed a solvent, or microscopic droplets of the perfume in suspension, or maybe a combination of both.

I think it is pushing the definition of solvation to include reacting with the compound being dissolved.

My point us, it can be fuzzy.  Since solvation occurs and it involves the same forces as a chemical reaction, where to draw the line?


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What's the most effective way to reliably trigger and observe hard water scale buildup at a micro level?
We live in an area that's got some really hard water. It's so hard that it kills most of the fish we try to care for in our aquariums and shrimp?... forget it. While this is sad and challenging for aquarium lovers, this constant quest my spouse is on got one of my kids really interested the concept of "hard water." One particular point that I've been struggling with was to figure out how to "see" hard water. They asked me if we could build an apparatus that would show them the build up of the scales more quickly so they could look at it under a microscope and do various experiments.
I'd really like to encourage their curiosity and so I wanted to ask, what options exist to:

trigger buildup (some kind of a "seed" on a glass pipette or...?)
capture or isolate buildup so that it can be observed with a
microscope
what kind of microscopic resolution would / should be
used to observe this (preferably semi-realtime, perhaps via
stop-motion capture, with a photo/video attachment as it
accumulates)?

You have to find out what kind of hardness is present in your water. Is it permanent hardness or temporary hardness? The latter goes away with boiling.
The easiest and visual demo for children is to show the behavior of hard water with a soap bar. Soap when added to hard water does not lather very well. Watch the video here:
https://www.youtube.com/watch?v=jDgSohKVLio
https://www.youtube.com/watch?v=bBYsfEHKYCk
As a control, you would need to buy distilled water.
Does your electric kettle develop crusts with time?

On the final question: yes, everything from pots and sinks to faucets and bidets have build up and are frequently ruined if we aren't very cautious and use filters.

I think a more advanced and interesting experiment would be to corrode Zn metal in a solution containing your hard water, H2O2 (or pumped in O2 from an air-pump) in the presence of a noble electrode (perhaps activated carbon (AC) or some graphite rods). This results in the formation of white Zn(OH)2.
Interestingly, adding amphoteric Zn(OH)2 to hard water containing either Mg(HCO3)2 or Ca(HCO3)2, could also result in the formation of insoluble carbonate salts. This can perhaps be easily verified by leaving in sunlight a mix of your hard water, basic H2O2 (add a base to acidic store purchased 3% H2O2), some NaCl (a good electrolyte) and a piece of Zinc from scrapping a penny on concrete exposing the Zn core leaving Cu to serve as the cathode.
In other words, pumping air into hard water (rich in electrolytes) in the presence of Zinc (best employed as a low surface area small piece of metal) and a noble cathode (like water purifying AC), may actually reduce the water hardness, increase dissolved oxygen and remove toxins (via the activated carbon).
Disclaimer: This is an original experiment and should be tested. As to whether any soluble zinc compounds are present in a problematic concentration for other purposes, or the water treatment process is entirely successful in mitigating the water hardness, should be determined/evaluated, but an interesting experiment nevertheless. I recommend proceeding with caution if contemplating introducing fish.
[EDIT] Performed the following quick verification experiment: aqueous H2O2 (acidic store purchased 3% H2O2), added MgSO4 and a penny with exposed Zinc, heated for 90 seconds in a microwave. Repeated experiment replacing MgSO4 with NaCl. The first experiment has to date, produced a more evident concentration of a white salt (most likely an insoluble magnesium salt together with ZnO). Clearly, so far, not all of the MgSO4 has been transformed, likely more time and added H2O2 required. Note, this electrochemical experiment was based on MgSO4 (and not Mg(HCO3)2 as found in hard water) since magnesium bicarbonate breakdowns to MgCO3 (s) on warming.
Also, just found a corresponding supportive electrolysis commentary, in contrast to my battery (electrochemical) cell approach above. To quote from a Science Direct source:

It is sometimes possible to reduce both cathode and anode reactions by ‘artificial’ polarization (for example, by adding inhibitors which stifle the electrode reaction). Calcium bicarbonate, naturally present in hard water, deposits calcium carbonate on metal cathodes and stifles the reaction. Soluble salts of magnesium and zinc act similarly by precipitating hydroxide in neutral solutions.


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Determining solvent and solute
On which conditions does determining solvent depend on?
1. the higher boiling indicates the solvent

the material with the higher concentration is the solvent
the solvent is kept first in a pot. Then the solute is added.

Here, pls identify the solvent and solute. I hope that it will clear all my doubts about them.

Note that solvent and solute are terms of convention and many cases are questionable, especially in case of unlimited miscibility. In some cases both can be considered solvent and solute at the same time, forming 2 separate phases.

@Poutnik, then what is the point of defining the terms- solvent and solute? If they are just mere conventions , what is the absolute basis of their definition?

For solid + liqud, it is easy. For 2 liquids, not so much. Lower melting point and higher abundance is just a guidance. Is ethanol in 70 % mixture with water solvent or solute ? It has lower melting point and higher mass ( but lower molar fraction ). Phenol with water forms 2 liquid phases, 1 with phenol majority, one with water majority. Which one is solvent ? What is considered solvent or solute is rather matter of their status then properties. Solute - substance of interest. Solvent - substance forming the environment for solutes. One may also consider solute=analyte, solvent= medium.

1/ Are traces of crude oil in water solvent and water solute ?
2/ Is a salt with extreme solubility in its saturated solution solvent ?
3/ If solid is put first, does it become a solvent ?

Chemists need terminologies. NO terminology is perfect and you will always find exceptions. In general, in an n-component mixture, the component which has the highest concentration is the solvent.*
Therefore your point no. 3 is invalid because pouring does not determine the solute-solvent label. Only point 2 is relevant. Point 1 is irrelevant.
*Corollaries & Exceptions:
If we mix 50 mL of hexane and 50 mL of heptane, which one is the solvent and which one is your solute? It is up to you to label one as solvent and solute. Both components were liquids.
If I take 100 gram of water, I can dissolve about 140 grams of potassium iodide! The mixture remains a liquid. Now which one is the solute and which one is the solvent? Traditionally, you may call the liquid component as the solvent and the solid phase as the solute, because here water molecules are helping to solvate potassium and iodide ions.
The same goes with sugar in water.
So how to overcome these paradoxical labeling issues?
Physical chemists and analytical chemists prefer to use the term mole fraction x when they want to extra precise to what they are saying about mixtures. It is defined as "Amount of a constituent divided by the total amount of all constituents in the mixture. It is also called mole fraction. Amount fraction is equal to the number fraction: the number of entities of one constituent divided by the total number of entities in the mixture."
Just as nobody cares about "physical" vs. "chemical" changes taught in school, no one will chastise you for identifying one component as solvent and the other component as solute in complicated cases.


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Can a solvent be a solid?
Can a solvent be a solid? I do not think so. But is it possible for a solvent to be a solid?

When most people think of the term solvent, a liquid medium comes to mind; however, in the technical sense of the definition, this does not have to be the case. Here is the definition of a solution present in the IUPAC gold book:

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. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

The definition of a solvent is incorporated into the definition of a solution. This definition states that the substance present in the larger quantity is the solvent while that present in the smaller quantity is the solute. Based on this definition, it is entirely possible to have a solvent that is in the solid phase. For example, brass is a mixture of 64% copper and 34% zinc. According to the IUPAC definition, a solution can be in the solid phase. Because copper is present in a larger amount, it is considered the solvent -- a perfect, everyday example of a solvent in the solid phase.

"This definition states that the substance present in the larger quantity is the solvent" - no it doesn't.

The term solvent is not restricted to liquids. You can very well use it for gases, it may not be wrong to say that in air, nitrogen is the solvent in which oxygen is dissolved. We regularly use supercritical $\ce{CO2}$ in chromatography. There is no harm in calling a solid solvent- think of alloys.
You can read the concept of solid solutions
here
.

copying from Poutnik's post on "whether a solvent can be a gas",to create a solution a solvent has to engage in inter molecular forces with the solute. But volatile materials can disperse without the presence of solvents . So it can be deduced that if the material is in gaseous surrounding, the gas acts like spectators. So, in this case, the gas is not a solvent. As a result , normal gas in general circumstances cannot be a solvent

There is no such thing as no interaction situation. London dispersion forces/ Van der Waals forces are always present among gas molecules.

@M. Farooq It was meant as no interactions principally involved in separation of molecules of solutes. Volatile compounds would evaporate without gas.  Soluble solute would not dissolve without solvent. With solvents comes solvatation with hydration as its special case.  There is no solvatation in air, pulling  molecules from a condensed phase.

@Poutnik, true. This is all semantics of solvent/solute. Don't forget the principles of GC where nitrogen is the carrier gas, it does pull the molecules from the condensed (actually pretty viscous) stationary phase.

@M. Farooq Well, it is not true pulling. It is dynamic disbalancing of the diffusion equilibrium. Analytes are "pulled" by stopping dissolution  in stationary phase, while the opposite - evaporation - continues. They would be "pulled" by vacuum as well, but we would lose the carrier functionality.

Both your question and the answers seem to rely on a common linguistic/mental shortcut which regards phase as some kind of inherent property of a compound, but I would argue that it becomes a confusing oversimplification when regarding this specific question.
I would say that the proper statement here would be "During the process of dissolving, the solvent has to be in a liquid state", or maybe "in a fluid state" depending on how you like to define the semantics of a solution. The reason for this is that the solvent has to get "between" the molecules of the solute to do its job, and that can only work when the solvent flows.
As for the answers which discuss solid solutions, they are correct, under the implicit assumption that the phase change happened after the dissolving process was finished. Sure, there are some kind of edge cases of diffusion between metals, but really, nobody makes brass by putting a block of copper and a block of zinc in a press at room temperature and waiting a century, to then harvest a fraction of a millimeter thick alloy. So, if we discount such unrealistic edge cases, the statement as I formulated it above is still true, and absolutely compatible with the statement "There are solid solutions" which could be rewritten more explicitely as "A liquid solution can change its phase to solid". (The opposite is of course also true - you can melt an alloy - but not especially relevant to the apparent contradiction).
The comments under M. Farooq's answer do a good job of describing the point about gases. In that case, you have a simple linguistic/categorization problem. The prototypical process of dissolving something involves a liquid, and the process of mixing two gases is in some ways similar and in some ways different. So it is up to the individual language user to "decide" (typically not a conscious decision, but a gut feeling that can be rationalized post-hoc) whether they consider this process to be similar enough to deserve to be called "dissolving" or not. Depending on that decision, you can have a gaseous solvent (at the time point of creating the solution) or not. The above point about phase changes still apply - if you would mix, say, water vapor and ethanol vapor and then cool it down to a liquid, I am pretty certain that everybody would term it "a solution of ethanol in water" (or water in ethanol, depending on ratios) even though there will be a dispute whether the process of dissolving happened when the mixture was still in a gas state.
A separate linguistic case which deserves a mention: there is a widespread usage of "dissolving" among laypeople that is actually incorrect by scientific standards. It is used to describe a situaion in which you have a chemical reaction by a corrosive gas or liquid with a solid. If you pour liquid acid over a basic salt, chances are that you will get a chemical reaction - but many people don't recognize that detail and, seeing the crystal disappear, will state that the acid dissolved the substance (they are also unlikely to call it "the salt", btw, reserving that name for sodium chloride only). Even if one of the two reactants happened to be in a gaseous phase, I would say that it is incorrect to consider this an example of a solution with a gaseous solvent, even though you might meet people (without advanced scientific education) who would classify it as such.

It is more a question of appearances. An acquaintance was playing with some mercury and got a drop or two on his gold wedding ring and watched it disappear into the gold the ring broke in two. What dissolved in what? Place a cup of rice in two cups of boiling water and simmer for about 20 minutes. Where did the water go? Do the same with equal volumes of pasta, then with sugar and water. Hydrogen that is generated on some metals such as Pt, Pd, and some high nickel stainless steels will pass thru thin sheets of the metal most likely as H atoms. This works best if the H is generated galvanically by corrosion processes [hydrogen embrittlement] but can happen with very high pressures of H2 gas [ask the petroleum people]. In real life real things happen that do have relatively simple answers based on chemistry but require some imagination beyond the textbook.


text:
What ratio of 73% calcium hypochlorite powder to water mimics the strength of sodium hypochlorite (bleach)?
Calcium hypochlorite is sold as pool shock in varying strengths.  I've read that it can be dissolved in water to create a 'bleach' (typically sodium hydrochloric or sodium hydroxide) and is often used as such to treat drinking water in developing countries, but every source I've read prescribes a different ratio. Strictly looking at chlorine levels for water sanitation, cleaning, etc., how much powder (at 73% calcium hydrochlorite) would you need to dissolve into 1 gallon of water to match common household bleach?

Warning, swimming pool shock often contains other ingredients (copper algaecide, cyanuric acid) when it isn't something else entirely (monopersulfate often).  Generally be cautious about any product not designed to be used for sterilization for the purpose of human consumption--quality control doesn't have to be quite as strict if you are not expected to drink the result of the chemical process!

To start, we must first understand that we can ignore the counterions to the $\ce{Ca(ClO)2}$ and the $\ce{NaClO}$. Typical household bleach claims to have a concentration of 6% (w/v). Now we can do some basic arithmetic to convert the percent weight by volume of sodium hypochlorite in water to a molar concentration:
$\mathrm{\frac{0.06\,g}{1\,mL}}$$\mathrm{\frac{1\,mol}{74.44\,g}}$$\mathrm{\frac{1000\,mL}{1\,L}}$$\mathrm{\,= 0.8060\,M \,OCl^-}$
The first fraction in the above equation is based on the definition of w/v, which is the number of grams of a substance dissolved in one milliliter of a solution times 100%. In the next fraction, we convert the grams $\ce{NaClO}$ into moles of $\ce{NaClO}$. Finally, in the last fraction, we convert milliliters into liters to give the number of moles per liter of $\ce{NaClO}$, or simply $\ce{ClO^-}$ without the counter ion.
Now that we know the molar concentration of $\ce{ClO^-}$ we are aiming for, we can convert this into the number of grams of pure $\ce{Ca(ClO)2}$ we would need to to achieve this concentration in one gallon of water
$\mathrm{\frac{0.8060\,mol}{1\,L}}$$\mathrm{\frac{1\,mol}{2\,mol}}$$\mathrm{\frac{142.98\,g}{1\,mol}}$$\mathrm{\frac{3.78541\,L}{1\,gal}}$$\mathrm{\,=\frac{218.1\,g\,OCl^-}{gal}}$
In the second fraction, we converted the moles of $\ce{ClO^-}$ into the number of moles of $\ce{Ca(ClO)2}$ (as there is only one unit of $\ce{Ca(ClO)2}$ per two units of $\ce{ClO^-}$). In the third fraction, we converted the number of moles of $\ce{Ca(ClO)2}$ into grams. In the last fraction, we converted the number of liters into the number of gallons.
As you stated, the powder only contains 73% $\ce{Ca(ClO)2}$ by mass. We know we need a total of 218.1 g of $\ce{Ca(ClO)2}$, so we can use the following conversion factor:
$\mathrm{\frac{218.1\,g\,OCl^-}{1\,gal}}$$\mathrm{\frac{1\,g}{0.73\,g}}$$\mathrm{\,=298.8\,g\,powder}$ (Rounded to 0.3 kg since we only have one significant figure)
In the second fraction, we know that per every gram of your powder, 73 percent (0.73 g) of the $\ce{Ca(ClO)2}$ will be present. Thus, the conversion factor is valid.
Final Answer: 0.3 kg of Powder

Actually, as I have elsewhere cited Bleaching powder is formed from the action of Cl2 on moist Ca(OH)2. This results in a mix of CaCl2,  Ca(OCl)2 and some residual Ca(OH)2.  The figure of 73% appears to be high, see, for example, http://www.wiredhealthresources.net/presentations/58-C/story_content/external_files/16.Chlorine_Solution_Formulas.pdf . It is close to correct upon adding H2SO4  as half of the Cl2 created comes from the CaCl2, as HCl + HOCl = Cl2 + H2O.

Note, adding aqueous CO2 to aqueous Ca(OCl)2, creates only HOCl. If light decomposes it, per the reaction 2 HOCl-> 2 HCl + O2, again Cl2  per HCl + HOCl = Cl2 + H2O, but only half as much as would be formed with a strong mineral acid.

Just to be clear, the conversion I did ignored the acid-base equilibria of OCl- + H2O = HClO + HO-, so the solution is by no means actually 0.8 M OCl-; however, it is the same concentration as household bleach (although simple conversions cannot determine the precise concentration of OCl-). To be clear, I also treated the filler as completely intert and as not interacting with any equilibria (this is, of course, not the case); however, it should be good enough for a rough approximation!


text:
Doubt concerning Andrews Curve

In Andrews Curve of $\ce{CO2}$, what do the horizontal lines signify?

I know that the gas changes to liquid in that region, but since the pressure and temperature is held constant, how does the gas liquefy more and more, even when the pressure and temperature are constant?

Even when pressure and temperature might be constant, the system might be releasing heat energy i.e. the enthalpy is changing. This, precisely, is what is called latent heat.
When CO2 liquifies, the heat from the system is taken away as latent heat, otherwise the enthalpy of condensation would raise the temperature instead of keeping it constant.

If pressure and temperature are held constant, it does not mean the system anthalpy is constant as well.
The heat is taken away to keep temperature constant, otherwise it would heat up due $\ce{CO2}$ enthalpy of condensation.
Similarly, for water boiling, heat is brought in to keep pressure and temperature constant, otherwise it would cool down because of the enthalpy of evaporation.


text:
How to find the oxidation numbers of individual ions of a transition metal with an average fractional oxidation state
In the classical example compound of $\ce{Fe3O4}$, we know that $\frac 23$ of Fe ions display +3 oxidation state and $\frac 13$ of Fe ions display +2 oxidation state fulfilling charge neutrality condition. However, this is the only solution considering the most common oxidation states of Fe are +2 and +3.
Now, consider the compound $\ce{La(Ni_{0.1}Mn_{0.9})O3} \rightarrow \ce{La10NiMn9O30}$,
Assume, the oxidation numbers of La, Ni and O are +3, +2 and -2. Therefore, under charge neutrality condition, the charges of 9 Mn ions should add up to 28. Knowing that the most common oxidation states of Mn are +2, +4, +7, we will have two solutions where charges add up to 28.

4 Mn ions showing oxidation number of +2 and 5 Mn ions showing oxidation number of +4
7 Mn ions showing oxidation number of +2 and 2 Mn ions showing oxidation number of +7

How could I decide what is actually happening?
If it helps, I need to know this to calculate the average ionic radius of a B-site cation of a perovskite structure.
EDIT
Edited the typo by changing ox. state of La to +3.
Furthermore, I am looking for an answer that makes sense theoretically (real world formation may be different). It would be great if somebody could guide me for a relevant publication, if there is any.

I don't think you can do this without a practical analysis of the compound in question.

For clarity to quote from Wikipedia on Fe3O4: "sometimes formulated as FeO ∙ Fe2O3". So, look for related mixed salt compounds.

@AJKOER Please don't suggest OP the outdated "mixed salt" concept. First, it's called a formal addition compound or a double oxide at best, not "mixed salt".

@AJKOER Second, writing "FeO ∙ Fe2O3" was acceptable in the past when there were no determined crystal structures available. These days the structure of $\ce{Fe3O4}$ is well-defined (cubic; monoclinic) and using "FeO ∙ Fe2O3" makes little to no sense as there are very little structural similarities and the stoichiometry as well as oxidation numbers can be well expressed with $\ce{Fe3O4}$ formula unit.

Mn(VII) is unlikely in this context...

Nickel could be +3. Then the manganese would be so, too.

To accurately determine oxidation states in compounds, you need to have a good idea of what the actual underlying bonding model is like. Only in rare cases you can resort to shortcuts.
One such shortcut which is often very helpful for binary compounds is to take one element’s ‘obvious’ oxidation state and use maths to derive the other one. That, for example, works well with ethane (hydrogen must have $\mathrm{+1}$, so carbon has $\mathrm{-III}$), iron(III) chloride ($\ce{Fe^{+III}/Cl^{-I}}$) or manganese dioxide ($\ce{Mn^{+IV}/O^{-II}}$).
In some cases, such a shortcut will lead to a fractional oxidation state such as in $\ce{Fe3O4}\ (\ce{Fe^{+\frac83}/O^{-II})},$ $\ce{HN3}\ (\ce{H^{+I}/N^{-\frac13})},$ or $\ce{S4O6^2-}\ (\ce{S^{+\frac52}/O^{-II})}.$ One might be tempted to take the closest common or known oxidation states of the element in question and use maths to derive a potential distribution. That works by accident in $\ce{Fe3O4}$ ($\frac83$ is $2\frac23$ which suggests $\ce{Fe1^{+II}Fe2^{+III}}$) and $\ce{HN3}$ ($-\frac13$ suggests $\ce{N2^{\pm0}N1^{-I}}$) but it already fails with $\ce{S4O6}$ because although $\ce{S3^{+II}S1^{+IV}}$ gives an overall oxidation state of $+\frac52$ so does the correct solution of $\ce{S2^{\pm0}S2^{+V}}$.
Since the simple model already broke down for binary compounds, there is little hope it will be of much use in ternary or even quarternary compounds—especially if there are less leads to go on. So while it might be possible to use the simple shortcut to derive the oxidation states in aluminium potassium sulphate or potassium alum ($\ce{KAl(SO4)2}$) and by extension that of chromium in chromium potassium sulphate or chrome alum ($\ce{KCr(SO4)2}$), it is a lot harder if you have two redox-promiscuous elements (nickel and manganese) to consider.
In short, there is no simple way to derive the oxidation states just by looking at the formula and I myself would have to research what publications say and how they arrived at this conclusion.

(+1) Nice answer. By S4O6 did you mean S4O5? I do understand the problem complexity increases with number of elements in the compound, especially when the compound includes transition metals. But please do come back here if you find a theoretical investigation on such compounds. I will too.

He probably means the tetrathionan anion $\ce{S4O6^2- = O^- - SO2 - S - S - SO2 - O^-}$, formed by iodine oxidation of thiosulphate $\ce{S2O3^2-}$

@AchinthaIhalage As Poutnik guessed, I meant tetrathionate ($\ce{^-O-SO2-S-S-SO2-O-}$). Sadly for you, I don’t really intend on researching that oxide and coming back to it; it’s really out of my field of interest.


text:
Which ethanol derivatives display the gauche effect?

How many of the following ethanol derivatives of the form $\ce{GCH2CH2OH}$ have a more stable gauche conformer compared to their anti conformer?
$G =  \ce{F}, \ce{Cl}, \ce{OH}, \ce{NH2}, \ce{OMe}, \ce{NHMe}, \ce{NMe2}, \ce{NO2}$

I was sure that $\ce{F}$, $\ce{OH}$, $\ce{NH2}$ and $\ce{NO2}$ display the gauche effect. I'm not so sure regarding all the other groups: all of them seem bulky and won't form hydrogen bonds with the $\ce{OH}$ group on $\ce{C}_1$. How many of the other substituents are capable of displaying this effect? Also, is there a quantitative way to determine if the gauche conformer would be more stable than the anti conformer?

Nice question, I am pretty sure that ethylene chlorohydrin also has gauche form more stable. Also I think for knowing most stable form we need to optimize the molecular geometry on different levels of quantum chemistry, starting from a conformer halfway between staggered and eclipsed. Levels are like HF/def2-TZVP, PBE-D3/def2-TZVP... Don't know direct way to find that. Further, I think G=NHMe would also form more stable gauche form (not sure). What's the answer given?

The answer is 7. I'm ignoring it for now until I have an explanation.

Then, I guess, all except NMe2.

@Zenix More probably NO2 - it would have H-bond but with 6-membered ring. BTW even N(CH3)2 isn't all that bulky.

See here

See the last page here

@YusufHasan Thanks a lot! could you post this as an answer so that I can accept it?

As per this source, there is evidence for the gauche effect for most of these groups on basis of intramolecular hydrogen bonding:

I tried looking for a similar evidence for 2-nitroethanol, and found this paper on the same:

  Figure 4. OCCN gauche structures with an intramolecular H-bond for 2-aminoethanol (7) and 2-nitroethanol (9); Conformations 8 and 10 indicate disrupted H-bonds after rotations by approximately 120º about the O-C axes

So it seems like all the groups shown here can show the gauche effect on the grounds of intramolecular hydrogen bonding,and so your answer should be 8


text:
Total number of stereoisomers of truxillic acid

Total number of stereoisomers of the compound will be:

This was a question asked in our mock test. I've tried by considering pseudo-chirality on the carbon atoms. But I don't know where to start.
I need help finding the different stereoisomers. It would be appreciated if you could draw the different isomers.

Wikipedia already has the answer. However, I strongly suggest you to try one more time before looking up the answer.

At first glance, not only does it seem like there are no stereogenic centres in truxillic acid, it is also highly symmetrical. However, if we consider different conformations of the chemically identical units (H, COOH or Ph), which give rise to different isomers, some carbons at the 4-membered ring can possibly become stereogenic.
Stereogenic centres are labelled with an asterisk: ε-truxillic acid and peri-truxillic acid do not have stereogenic carbon centres as their 4-membered ring carbons have chemically and conformationally identical neighbours.
It took me a while to get this.Thanks to @andselisk and wikipedia.


text:
Nitrogen dioxide fluorescence quenching and lifetime
Nitrogen dioxide fluorescence quenching:
$$
\begin{align}
\ce{NO2 + h\nu &->[$\varphi_\mathrm{Ia}$] NO2^{\ast}}\tag{I}\\
\ce{NO2^{\ast} &->[$k_2$] NO2 + h\nu'}\tag{II}\\
\ce{NO2^{\ast} + NO2 &->[$k_3$] 2 NO2}\tag{III}\\
\ce{NO2^{\ast} + Xe &->[$k_4$] NO2 + Xe}\tag{IV}\\
\ce{NO2^{\ast} + NO2 &->[$k_5$] 2 NO + O2}\tag{V}\\
\end{align}
$$
The fluorescence lifetime of $\ce{NO2^{\ast}}$ in the presence of all the reactions, $\tau$, measured at $\pu{298 K}$ at different concentrations (molecules per litre) of the reactants:
$$
\begin{array}{crrr}
\hline
\text{Experiment} & \tau/\pu{μs} & [\ce{Xe}]/\pu{L-1} & [\ce{NO2}]/\pu{L-1}
 \\
\hline
1 & 3.38 & \pu{1.6E19} & \pu{3.2E18} \\
2 & 1.89 & \pu{1.6E19} & \pu{6.4E18} \\
3 & 3.64 & \pu{0.8E19} & \pu{3.2E18} \\
\hline
\end{array}
$$
What is the real fluorescence lifetime of $\ce{NO2^{\ast}}?$
Is this question valid? Isn't the lifetime vs the concentration of the quencher supposed to be linear?

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. It might also help if add some more details as to where this data is taken from: is this a textbook exercise, or an experiment conducted by you?

Thanks! it's a HW assignment by my tutor. I'm not sure if this is solvable or not.

@andselisk why did you use \phi instead of \varphi?

@Zenix No reason, both are equally usable to me. Probably an old habit to avoid \var prefix from using $\mathrm\LaTeX.$ If you are interested in the whole var-thing, have a look at TeX.SE. But you are right, since the source image had φ and not ɸ, I changed my edit to \varphi. Thank you for pointing this out.

Given that the question was constructed by your tutor, I suspect the question is really about half-life (a clue is a halving of the concentration).

You are looking for the rate constant $1/\tau^0=k_2$. The rate equation for the $\ce{NO2^*}$ is (using $N^*$ for $\ce{[NO2^*]}$ and N for ground state NO2), $\displaystyle \frac{dN^*}{dt}=-k_2N^*$ and when quenched $\displaystyle \frac{dN^*}{dt}=-k_2N^* -N^*((k_3+k_5)N+k_4[Xe])$ which can both be integrated to give exponentials, i.e $N^*=N^*_0e^{-k_2t}$ and similarly for the other equation. Writing the rate constants as $1/\tau^0=k_2$ and $1/\tau_Q=k_2+(k_3+k_5)N+k_4[Xe]$ you can find the values after looking at the pattern of the quencher concentrations in the question.


text:
How to derive composite acidity constant for H2CO3*?
I see the equation (19) and (20) on Stumm and Morgan's Aquatic Chemistry book (Fig. 1 here). They said "a combination of equations 19 and 20" gives Equation 21. But I don't know how the Equation (21) is generated? Because if I make summation of (19) and (20), it will be:
$$\ce{2H2CO3 <=> H2O + CO2(aq) + H+ + HCO3-}$$

@Zenix Yes, but here K_H2CO3* seems not the K times K_H2CO3. So I am confused about the way to derive from Equation (19&20) to Equation (21)

[OP] Because if I make summation of (19) and (20), it will be: [...]

They meant they combine the mathematical expressions, not the chemical equations.
To express
$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K}$$
in terms of concentrations, you substitute equations (1) and (2) into the expression:
$$K = \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}\tag{1}$$
$$K_\ce{H2CO3} = \dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}\tag{2}$$
So
$$\frac{K_\ce{H2CO3}}{1 + K} = \frac{\dfrac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}{1 + \dfrac{\ce{[CO2(aq)]}}{\ce{[H2CO3]}}}$$
Multiplying top and bottom of the fraction by $\ce{[H2CO3]}$ gives:
$$\frac{\ce{[H2CO3]}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]} + \ce{[CO2(aq)]}}$$
If you want to make the intention clearer, you define a total concentration of carbon dioxide as
$$[\ce{H2CO3^\huge{*}}] = \ce{[H2CO3]} + \ce{[CO2(aq)]}$$
and then write
$$K_\ce{H2CO3^\Huge{*}} = \frac{K_\ce{H2CO3}}{1 + K} = \frac{\ce{[H+][HCO3-]}}{[\ce{H2CO3^\huge{*}}]}$$

+1 Ah... very nice. A much less convoluted derivation than mine.

@AdnanAL-Amleh - I deleted my answer since Karsten's is so very much better. $K_\ce{H2CO3^*}$ is the fictional $K_{\alpha 1}$ for $\ce{H2CO3}$ assuming that the dissolved $\ce{CO2}$ is $\ce{H2CO3}$. In reality $\ce{CO2(aq) > \text{actual }H2CO3}$

@MaxW Good point about the actual concentrations, so it should really be called $$\ce[\ce{CO2(aq)}^*]$$

I take perhaps a more intuitive view of the system, and in the case of, for example, Iron bicarbonate, it intuitively explains its formation in well water under pressure and its subsequent decomposition upon exiting a faucet at ground level.
I start with a modified cited equation (19) citing gaseous and not aqueous carbon dioxide in the equilibrium reaction:
$\ce{ H2CO3 (aq) <-> CO2 (g) + H2O (l) }$
I also cite Equation (20) and introduce dissociated CO2 as represented by carbonic acid:
$\ce{ CO2 (d) = H2CO3 (aq) = H+ (aq) + HCO3- (aq)}$
Now, at high pressure, examining the modified (19) above, per Le Chatelier's Principle, the position of equilibrium will move to produce more carbonic acid, and more bicarbonate per the second equation, resulting in the creation of Iron bicarbonate.
However, with the release of pressure, the bicarbonate equilibrium concentration shifts to create more carbonate acid, which can decompose releasing CO2 gas resulting in the ultimate removal of the bicarbonate ion and the decomposition of Iron bicarbonate, as reported.
Note, to preserve the total concentration of carbon dioxide in the system:
CO2 (d) + CO2 (g) = [CO2]


text:
What method do I use to synthesize propyl-2-(2-methylpropyl)-ether and why?
I need to know which method to use to synthesize propyl-2-(2-methylpropyl)-ether.
I considered electrophilic addition of alcohols to alkenes, but I'm not really sure how the method works, and anything I found on it was hard to understand. From my understanding I need to "add" electrophilically activated alkenes to alcohol.
I don't really know what that means. I was thinking of using t-butanol and propylene, but the whole part of electrophilically activated alkenes threw me off. Any advice or help is greatly appreciated!

Electrophilic addition simply means whatever you are adding across your alkene is electrophilic. For example, in the acid-catalyzed addition of water across an alkene, the hydronium acts as an electrophile and hence is being added to the alkene electrophillically in the rate-determining step.  The alkene is NOT the electrophile in an electrophilic addition, it is the nucleophile. This might be where you are confused.

What is "propyl-2-(2-methylpropyl)-ether" supposed to mean? It doesn't resemble any valid IUPAC name. Do maybe mean 2-methyl-1-propoxypropane?

I think what OP meant is t-Bu-O-Pr.

Thing is i think we might use different ways to write chemical names in sweden, which is where im from.

There are two basic choices here if you are trying to do this on a lab scale. The alcohol plus alkene with catalytic acid is not an approach I would consider unless I needed a lot of this. t-BuOH plus acid is a recipe for isobutylene. Sodium or potassium t-butoxide is commercially available and reacted with an excess of iPr-I should give a decent yield of the ether.
The other approach is to use t-Butyl 2,2,2-trichloroacetimidate compound details here - a general reagent for the formation of t-Butyl ethers originally reported by Armstrong et al, Tet. Letts 29 2483 (1988)


text:
Why And How are Unbranched chains of hydrocarbons easily biodegraded?
My chemistry textbook says

If detergents are highly branched then they can't be easily degraded by bacteria.So,nowadays the branching of hydrocarbon chain is controlled and kept to a minimum.

Upon searching this I got the answer that "side chains stop bacteria from attacking and breaking the chains".
I really don't understand how does this work.Why would having side chains prevent bacteria from attacking ? Won't they provide more sites for attack of bacteria ?

I don't know the appropriate Stack topics available for this question.If it should be posted on a different website please notify me via comment.

Hm, it´s not "bacteria", it´s enzymes. All modern standard detergents are completely linear, so is soap made out of natural oils. The enzymes (quite a zoo of them) afaik have trouble reaching a quarternary carbon atom.

Does this have anything to do with steric hindrance? If 'breaking chains' is classified as reducing the total length of a compound by making a scission in the detergent backbone, then having too many branched groups may provide a spacial limitation for the enzyme to get to where it needs to.

Here is a study1 showing this is the case, that generally speaking, branched hydrocarbons are more difficult to degrade, especially those exhibiting anteiso-terminal branching (branching  on the third carbon). Both the study and a book on oil spill forensics2 hypothesize that steric hindrance is the  mechanism responsible. They state, respectively:

If alkyl branches are located near the terminus, it seems likely that decreased biodegradability could be the result of steric inhibition of terminal oxidizing enzymes

Complex branching hinders both the initial oxidation and the subsequent lipid catabolism apparently because tertiary and quarternary carbon atoms interfere by steric hindrance with the oxidation enzymes.

However, this is not always the case as Boethling3 explains when discussing design objectives for molecule biodegradability:

Similarly there is a commonly held view that any branching, for example, even a single methyl group on an otherwise linear alkyl group, is to be avoided. This is a gross oversimplification. All that can be said with any confidence is that quaternary carbon is usually to be avoided, as is extensive methyl chain branching, which has no strict definition.

1 Schaeffer, T. L., Cantwell, S. G., Brown, J. L., Watt, D. S., & Fall, R. R. (1979). Microbial Growth on Hydrocarbons: Terminal Branching Inhibits Biodegradation. Applied and Environmental Microbiology, 38(4), 742-746. Retrieved from https://aem.asm.org/
2 Wang, Z., & Stout, S. (2006). Oil Spill Environmental Forensics: Fingerprinting and Source Identification (1st ed.). Burlington, MA: Academic Press.
3 Boethling, R. S., Sommer, E., & DiFiore, D. (2007). Designing Small Molecules for Biodegradability. Chemical Reviews, 107(6), 2207–2227. https://doi.org/10.1021/cr050952t


text:
Why is Finkelstein reaction not conducted using NaF?
I was studying the Finkelstein reaction and all sources through which I have studied this reaction indicate that the reaction uses $\ce{NaI},$ where $\ce{I-}$ acts as the nucleophile which will then replace $\ce{X}$ from the alkyl halide $\ce{R-X}.$
This is an $\mathrm{S_N2}$ reaction, hence we use a polar aprotic solvent (PAS, usually acetone) as PAS stabilizes the $\ce{Na+}$ cation. We know that while using a PAS the order of nucleophilicity is
$$\ce{F-} > \ce{Cl-} > \ce{Br-} > \ce{I-}.$$
The part I am struggling with is that even though $\ce{F-}$ is a better nucleophile than $\ce{I-}$ in a PAS, why do we still use $\ce{NaF}$ rather than $\ce{NaI}?$

Search for the solubility of NaF in acetone.

Further to my comment - the solubility of Sodium Fluoride in Acetone is 2.4 microgrammes per 100g of solvent i.e. essentially insoluble source. There will be no F- in solution so the reaction will not occur.


text:
Has anyone actually prepared Zn(HCO3)2 absent the presence of pressurized aqueous CO2?
Under normal conditions, I am not certain, as per Atomistry on $\ce{ZnCO3}$, the following quote:

The hemihydrate, $\ce{2ZnCO3.H2O}$, was obtained by precipitating the solution of a zinc salt with ammonium bicarbonate and digesting the gelatinous precipitate in excess of the precipitant till it became crystalline.

where I would have expected zinc bicarbonate, and further:

The basic zinc carbonate, $\ce{5ZnO.2CO2.4H2O}$, can be prepared by the direct hydrolysis of the normal carbonate, or by boiling a solution of zinc, zinc hydroxide, or zinc carbonate in an excess of aqueous carbonic acid.

And finally, a reference to a presumably formed zinc bicarbonate, and at times, citing the employment of pressure:

Basic zinc carbonates can be converted into the normal carbonate by contact with water charged with carbon dioxide under pressure. The normal carbonate is itself somewhat soluble in water containing carbon dioxide - presumably forming the bicarbonate.

The last quote is unclear, in my opinion, as to whether under normal experimental conditions, should one expect the creation of a stable $\ce{Zn(HCO3)2}$?

We have to distinguish between the bicarbonate in solution and the bicarbonate as a solid.  While many metal cations are compatible with bicarbonate anions in dilute aqueous solution, only in a few cases can a solid bicarbonate be expected as a precipitate.  In most cases, such as with lithium or with hard-water forming metals such as calcium and magnesium, the carbonate is formed instead.  The only solid metal bicarbonates I know of are those of the heavier alkali metals than lithium.