text:
How do I balance this redox reaction?
The reaction given was $\ce{FNO3 -> O2 + F^- + NO3^-}$
(in basic medium)
I've noticed that the atoms which have their oxidation numbers changed are the oxygen and the nitrogen atom (as the oxidation number in fluorine in $\ce{FNO3}$ should be $-1$, right?)
I'm having troubles at dissecting it into half reactions:
$\ce{FNO3 -> O2 + F-}$ (This one doesn't have a nitrogen atom on the right hand side)
$\ce{FNO3 -> NO3- + F-}$

Well, you lack something in there, that is in your medium (hint).

Is it the base in the form of $\ce{NH3}$? @Mithoron

No... there's just one thing that's in all basic water solutions...

Your equations are all wrong ! They should respect the first basic law valid in all chemical reactions : conservation of mass and of charges. For example, your first equation may produce F and NO3 ions, but one of these ions has to be positive, and O2 cannot be produced, whatever the nature of the ions. Your second equation is wrong for lack of nitrogen, and loss of oxygen atom. Your third equation is wrong for lack of positive charge somewhere. So please re-read your equations first !

@Maurice The question is exactly as written in the textbook, and old edition of Brady's Chemistry, translated at my language. So, I assume, it's just an invalid question?

You say in basic medium! @Mithoron has twice lead you to this! So put hydroxide ion on the left side of your reaction at the top and proceed from there!

@ Marchotridyo. I don't know your book. Was the translation miserable ? I don't know. Anyway, the three equations are wrong, and severely wrong. And the suggestion by Ed V to add OH- ions does not help. This would simply add a further difficulty, namely the H atoms.

From fluorine nitrate, fluorine nitrate decomposes in water to produce oxygen gas, oxygen difluoride, nitric acid and hydrofluoric acid. So the starting reaction is simply wrong for whatever reason.

Please consider giving the green checkmark (or upvote or both) to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

Unfortunately, the OP's starting reaction expression is simply wrong: the left hand side has only a neutral compound while the right hand side has anions among the product species. From the wikipedia article on fluorine nitrate, the unbalanced reaction of fluorine nitrate and water is
$$\ce{a FNO3 + b H2O  -> c O2 + d OF2  + e HF  + f HNO3  \tag 1}$$
For the same reason that this problem had infinitely many correctly balanced solutions, this one does as well. So here is the minimal coefficient solution.
Atom balance yields:
$$a=2d + e     \tag 2$$
$$a=f     \tag 3$$
$$3a + b = 2c + d + 3f    \tag 4$$
$$2b = e + f     \tag 5$$
The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:

From equation $(2)$, $a \ge 3 $, since d and e cannot be zero. From equation $(3)$, $a = f$. From equations $(3)$ and $(4)$, $b = 2c + d$, so $b \ge 3 $, because c and d cannot be zero.

Assume $a = 3 = f$. If $e = 1$, then $d = 1$ and $b = 2$, which is impossible. If $e = 2$, then $d = 0.5$, which is impossible.

Assume $a = 4 = f$. If $e = 1$, then $d = 1.5$, which is impossible. If $e = 2$, then $d = 1$ and $2b = 2 + 4 = 6$, so $b = 3$. Then $c = 1$.

Almost final result: $a = 4, b = 3, c = 1, d = 1, e = 2,$ and $f = 4$. Hence
$$\ce{4FNO3 + 3H2O  -> O2 + OF2  + 2HF  + 4HNO3  \tag 6}$$
There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is $5a + 3b$. In the present case, $5a + 3b = 29$.
But what about the basic medium mentioned by the OP? Add 6 hydroxides to both sides and cancel 3 waters from both sides.
$$\ce{4FNO3 + 6OH-  -> O2 + OF2  + 2F-  + 4NO3- + 3H2O  \tag 7}$$
Two other answers assume no oxygen difluoride, hence $d = 0$ in equation $(1)$. Then the atom balance equations reduce to $a = b = 2c = e = f$, so let $c = 1$ and the others are 2. Hence
$$\ce{2FNO3 + 2H2O  -> O2  + 2HF  + 2HNO3  \tag 8}$$
Then add 4 hydroxides to each side and cancel 2 waters on each side:
$$\ce{2FNO3 + 4OH-  -> O2  + 2F-  + 2NO3- + 2H2O \tag 9}$$

You got my vote as well. I like your way because it applies for every chemical reaction.

Thanks!. I was confused too by the textbook's question, it only asks exactly what I asked. I guess as it's an old version of the book it's some mistake by the translator. Thank you!

@Marchotridyo I think the problem was that the author made the standard assumption that, in a basic aqueous solution, water and hydroxide ions are freely available for use, as and if needed. In acidic aqueous solutions, water and hydrogen ions are freely available for use, as and if needed. Glad to have helped on this!

+1 I'm up voting a number of different answers  because I enjoy it when people get to the right answer in various ways, maybe enough hints and different approaches  will help the OP understand better. Liked this approach best to though.

@Ed V I actually did. I frequently disagree with AJKOER on some of the details, but I think this time it adds value, heck for someone battling with figuring out that the question mentioned the basic medium for a reason, every different approach that is not completely wrong adds value and may be the final repeat that suddenly leads to that "a-ha" moment, and hey, I can be nice ...

As noted elsewhere, the reaction as given is not balanced.  Assuming that other products such as $\ce{OF2}$ are ignored, apply the following rules for basic solution:

Use hydroxide ion to allow for balanced charges.

Use water to allow balancing hydrogen and oxygen atoms.

Here, hydroxide ions should he added on the left by Rule 1 and then water is needed on the right to balance atoms by Rule 2.  Thus, properly (prior to balancing):
$\ce{FNO3 + OH^- -> F^- + NO3^- + O2 + H2O}$.
The reduction half-reaction is difficult to render via atomic oxidation states because if we try to work out oxidation states in $\ce{FNO3}$, we find different oxygen atoms in different oxidation states.  Use the entire $\ce{FNO3}$ molecule as the oxidizing agent:
$\ce{FNO3 + 2 e^- -> F^- + NO3^-}$.
The oxidation half-reaction involves just the water and hydroxide ions giving the familiar result
$\ce{4 OH ^- -> 2 H2O +O2 + 4 e^-}$.
The combination of these half-reactions to give the balanced full reaction is as it would be for any redox reaction.

Besides having upvoted your answer, would you consider adding the first paragraph from your answer here? I think it is important and might help save some time later on: I can skip these questions! ;)

Hi @Ed, I tried your suggestion but it does not really fit because I refer to basic solutions here, that us not the case there.  Stay safe!

Ah, OK! I was just thinking that when there are N species in the unbalanced equation, but only N-2 elements, then the balancing act needs a bit more thought. By the way, I especially like the way you did the reduction half-reaction, nicely avoiding the oxygen issue and, indeed, staying true to the chemistry.

I agreed with EdV about OP giving an errotic equation. Since $\ce{NO3-}$ remained unchanged in RHS of the equation, we need a compound to reduce $\ce{F+}$ to $\ce{F-}$ in this redox reaction. Since $\ce{O2}$ is being a product, we can choose one of the following half reactions to do the trick based on the medium (acidic or basic):
$$\ce{O2 + 4H+ + 4e- <=> 2H2O } \tag1$$
$$\ce{O2 + 2H2O + 4e- <=> 4OH- } \tag2$$
Thus, if the reaction is in acidic medium (equation $(1)$), we can correct the OP's equation as:
$$\ce{ FNO3 + H2O -> F- + O2 + H+ + NO3- } \tag3$$
Or, if it is in basic medium (equation $(2)$), we can correct the OP's equation as:
$$\ce{ FNO3 + OH- -> F- + O2 + H2O + NO3- } \tag4$$
If it's according to the equation $(3)$, we can write two corresponding half reactions:
$$\ce{ F+ <=> F- } \tag5$$
$$\ce{ H2O <=> O2 + H+ } \tag6$$
You can balance these two equations as follows:

Balance center atom first.
Then, balance any oxygen by $\ce{H2O}$ molecules (we have plenty of water in the medium).
Adding water gives extra $\ce{H}$ atoms, which can be balanced by $\ce{H+}$ ions, because you are doing the reaction in acidic medium.
Now, balance the all chargers by $\ce{e-}$s, which is relatively massless.
If the reaction is in basic medium, neutralize the $\ce{H+}$ ions by available $\ce{OH-}$ ions and add same amount to other side of the equation to balance the mass because you already have balanced the equation.
Finally, add oxidation and reduction half reactions in order to cancel the electrons.

Lets follow this order in equations $(5)$ and $(6)$:
$$\ce{ F+ + 2e+ <=> F- } \tag7$$
$$\ce{ 2H2O <=> O2 + 4H+ + 4e-} \tag8$$
Add the equations $(8)$ and $(5) \times 2$ to cancel electrons:
$$\ce{2F+ + 2H2O -> 2F- + O2 + 4H+} \tag9$$
Since $\ce{NO3-}$ is a spectator ion in this reaction, you can add that to both side as follows:
$$\ce{2F+ + 2NO3- + 2H2O -> 2F- + O2 + 4H+ + 2NO3-} $$
Or:
$$\ce{2FNO3 + 2H2O -> 2HF + O2 + 2HNO3} \tag{10}$$
The equation $(10)$ is the complete balanced redox reaction in acidic medium. EdV has given it in basic medium.

(+1) My only hope of getting an upvote is to hope I get to one of these problems while you are busy! Well done! ;)

I have more succinctly arrived at Eq(9) as proposed by Ed V above, by noting observed reactions as reported in the literature.
For the record, I do not recommend working with FNO3, a shock-sensitive explosive. My expected action of FNO3, with say dry KOH, is better described simply by the term energetic. Working with dilute aqueous alkaline solutions is likely more feasible in practice.
Some supporting background, starting with Wikipedia on Flourine nitrate, to quote:

Fluorine nitrate is an unstable derivative of nitric acid with the formula $\ce{FNO3}$. It is shock-sensitive.[1] Due to its instability, it is often produced from chlorine nitrate as needed...It decomposes in water to form oxygen gas, oxygen difluoride, hydrofluoric acid, and nitric acid.[1]

So, an example of a possible water reaction could be given by:
$\ce{6 FNO3 + 4 H2O ->  O2 + 2 OF2 + 2 HF + 6 HNO3}$
Upon further adding 8 OH- to both sides of the above suggests an alkaline reaction (but with no cancelling of water as of yet) :
$\ce{6 FNO3 + 4 H2O + 8 OH- ->  O2 + 2 OF2 + 2 F- + 6 NO3- + 8 H2O}$
Further, any formed $\ce{OF2}$ only slowly dissolves in water per Wikipedia on oxygen difluoride. Also, per another source on the alkaline hydrolysis of OF2, it apparently proceeds, again not rapidly, as follows:
$\ce{OF2 (aq) + 2 OH- (aq) -> O2 (g) + 2 F- (aq) + H2O}$
So, upon adding the above intermediate reaction times 2 to my second cited equation above implies a net reaction (assuming all the oxygen difluoride is eventually removed from the system) of:
$\ce{6 FNO3 + 4 H2O + 12 OH- ->  3 O2 + 6 F- + 6 NO3- + 10 H2O}$
Or, upon canceling out like terms and dividing by 6:
$\ce{FNO3 + 2 OH- ->  1/2 O2 + F- +  NO3- + H2O}$
Those anticipating only a friendly oxygen gas, should be warned, again per the Wikipedia source, to quote:

Oxygen difluoride is considered an unsafe gas due to its oxidizing properties.

So, with sufficient time elapse, the cited REDOX per Ed V is likely observed.

(+1) Another interesting route to an answer! Funny how we all do these problems differently, but end up at the same place.


text:
The Solubility of Tyrosine
I find very little data on the solubility of tyrosine, other than that it has a very low solubility at room temperature for an AA. (1)
Does anyone know different ways to dissolve fair amounts of it in water (preferred) or an organic solvent?
I suspect I could increase its water solubility by decreasing the $\mathrm{pH}$ and increasing temperature, but I'm not sure what is the best way to do it, which would be the most effective, and by how much it would increase the solubility.

(1) https://pubchem.ncbi.nlm.nih.gov/compound/Tyrosine#section=Solubility

Can either go low in pH or high, depending on what you plan to do with it. Sigma-Aldrich reports 100 mg/mL solubility in 1 M HCl (with heating) and ~4 mg/mL at pH 10. I used to make stocks in 1 N NaOH but I don't remember the max solubility. https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/t3754pis.pdf

According to Product  Information of Sigma-Aldrich for L-Tyrosine, the solubility is $\mathrm{pH}$-dependent:

Preparation Instructions: This product is soluble in $\pu{1 M} \ \ce{ HCl}$ ($\pu{100 mg/ml}$), with heating. The solubility in water ($\pu{25 ^\circ C}$) is $\pu{0.45 mg/ml}$ in the pH range $3.2-7.5$ (Ref.1). At $\mathrm{pH}$ $1.8$, the solubility is $\pu{2.0 mg/ml}$; at $\mathrm{pH}$ $9.5$, the solubility is $\pu{1.4 mg/ml}$; and at $\mathrm{pH}$ $10$, the solubility is $\pu{3.8 mg/ml}$.

The above claims can be justified by the study of the solubility of tyrosine in the acidic and basic conditions (Ref.2), the summery of which states that:

Measurements have been made of the solubility at $\pu{25 ^\circ C}$ of tyrosine in hydrochloric acid and in sodium hydroxide solutions varying from $0.001$ to $\pu{0.05 M}$, and also in distilled water. The $\mathrm{pH}$ of the saturated solutions was measured with the hydrogen electrode. The following values for the ionization constants of tyrosine have been obtained from the measurements: $k_\mathrm{b} = 1.57 \times 10^{–12}$, $k_\mathrm{a1} = 7.8 \times 10^{–10}$, $k_\mathrm{a2} = 8.5 \times 10^{–11}$. The changes in solubility with $\mathrm{pH}$ can be satisfactorily explained by the use of these ionization constants.

According to plot of solubility versus $\mathrm{pH}$ of their results, it was concluded that the highest solubility of tyrosine can be achieved in the $\mathrm{pH}$ range of $1-2$ or in the $\mathrm{pH}$ range of $9-10$. For example, the concentration of tyrosine in $\pu{0.05 M}$ $\ce{HCl}$ is found to be $\pu{0.0165 M}$ with the $\mathrm{pH}$ of solution being $1.450$. And also, the concentration of tyrosine in $\pu{0.0498 M}$ $\ce{NaOH}$ is found to be $\pu{0.0358 M}$ with the $\mathrm{pH}$ of solution being $9.953$. The least solubility of tyrosine is shown to be with in the $\mathrm{pH}$ range of $3-8.5$ (Ref.2).
Note: The extrapolation of the acidic side of above graph has shown the solubility increases up to about $\pu{0.04 M}$ at $\mathrm{pH} = 1$ (Ref.2).
References:

T. A. Brown, In Molecular Biology: LabFax; 1st Edition; Bios Scientific Publishers: Oxford, England, 1991, p. 29 (ISBN: 1‐872748‐00‐7).
David I. Hitchcock, "The solubility of tyrosine in the acid and in alkali," Journal of General Physiology 1924, 6(6), 747–757 (https://doi.org/10.1085/jgp.6.6.747)(PDF).

What about temperature? Do you know anything about the max solubility in HCl when increasing the temperature (to say 80C gor instance)?

@Hans: I did not concentrate on solubility with temperature since it wasn't a concern according to the question. Nevertheless, solubility decreases with the decreasing temperature as rule of thumb. Since solubility of tyrosine is highly depend on its ionization (total positive or negative charge by acid or base, respectively), difference in solubility might not be too big of a deal (c.f., solubility of $\ce{NaCl}$).

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Electrolysis of aqueous solutions
Water competes with any ions in solution for both oxidation and reduction. There are $1000\ \mathrm{g\ l^{-1}}/18\ \mathrm{g\ mol^{-1}}\approx55.56\ \mathrm{mol\  l^{-1}}$. That is truly huge in comparison to any amount of salt we can dissolve. Why doesn't water dominate any electrolysis reaction since even if a small percentage of the collisions of water can accomplish the reaction? The sheer number of collisions on its side will ensure it wins. Also, if hydrogen ions and hydroxide ions alone participated, a concentration of $10^{-7}$ means that it's several orders of magnitude away from the dissolved salt's ion concentration, which means that water can never dominate. Yet, we use a comparison of the thermodynamic feasibility in the form of reduction potentials to decide which competing ion gives the product. How is it that the concentration effects become comparable enough to allow this in most cases?

Do you know Nernst equation ?

Two down votes on a very reasonable valid question and surely enough the down voters have no clue of the answer.

@M.farooq just a theoretical question.

It is good question given the fact that this idea is not commonly discussed. The main source of your confusion is the following idea

Why doesn't water dominate any electrolysis reaction since even if a small percentage of the collisions of water can accomplish the reaction? The sheer number of collisions on its side will ensure it wins.

Unlike chemical reactions, electrode reaction do not solely require collisions. It requires an additional process: electron transfer. No matter how much collisions are occurring, the electrode is unable to transfer or withdraw electron from water in high salt concentration. Thermodynamics tells us (from the Nernst equation) how easy it would be to remove or add an electron to a water molecule at a given salt concentration. It is silent as to why this happens and what is the mechanism.
Electrode processes are very complicated. Marcus got a Nobel Prize on electron transfer.

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text:
Process of a substitution reaction

In this question I am not able to decide by what process this reaction is happening.
I first thought that one of the $\ce{H}$ from $\ce{CH3}$ will depart $\ce{Br}$ from $\ce{CBrCl3}$ but I am wrong. Would someone tell me what is the right process for this reaction?

This reaction is already in the literature (peer-reviewed paper), so I don't want to change any outcome already explained by Waylander. But, I like to explore the reaction  a bit more for the benefit of novice students of organic chemistry. The reaction seemingly follows the same mechanism as that of a general free radical reaction: Initiation; propagation; and termination. The starting material is active enough to get initiated by light:
$$\ce{Br-CCl3 ->[$h\nu$] Br^. + ^.CCl3} \tag1$$
$$\ce{Ph-CH3 + Br^. -> Ph-CH2^. + HBr} \tag2$$
Now propagation begins:
$$\ce{Ph-CH2^. + Br-CCl3 -> Ph-CH2-Br + ^.CCl3} \tag3$$
$$\ce{Ph-CH3 + ^.CCl3 -> Ph-CH2^. + HCCl3} \tag4$$
After all limiting reagent is used up, the remaining radicals react with each other to terminate the reaction:
$$\ce{Cl3C^. + ^.CCl3 -> Cl3C-CCl3} \tag5$$
According to the reactions shown in the original paper (Ref.1), the most probable limiting reagent should be toluene. If that's the case, the propagation ends at equation $(3)$ after going through many propagation cycles. At the end, what remains are
two $\ce{^.CCl3}$ radicals from the equation $(1)$ and last circle of the equation $(3)$. The reaction ends with dimerization of these two radicals.
As pointed out by Waylander, the ratio of $\ce{HBr:Ph-CH2Br}$ is ~$1:20$. This result supports the mechanism since formation of  $\ce{HBr}$ is only by the initiation reaction $(2)$. The formation of $\ce{Ph-CH2Br}$ is by the propagation (chain) reactions $(23)$ and $(4)$, hence the larger portion. For the same reason, amounts of $\ce{Ph-CH2Br}$ and $\ce{CHCl3}$ are equimolar.
Note: When the same reaction applied to toluene derivatives with substituted aromatic nucleus, a relatively large polar effect was found in the reaction rates. The effect is depend on $\sigma^+$-values of the substituents (Ref.2).
References:

Earl S. Huyser, "The Photochemically Induced Reactions of Bromotrichloromethane with Alkyl Aromatics," J. Am. Chem. Soc. 1960, 82(2), 391–393 (https://doi.org/10.1021/ja01487a034).
Earl S. Huyser, "Relative Reactivities of Substituted Toluenes Toward Trichloromethyl Radicals," J. Am. Chem. Soc. 1960, 82(2), 394–396 (https://doi.org/10.1021/ja01487a035).

This chemistry was first described in this 1960 JACS paper here.

The first step is cleavage of the $\ce{Br-CCl3}$ bond to give $\ce{Br^.}$ and $\ce{^.CCl3}$.
The second step is $\ce{Br^.}$ abstracting a proton from toluene to give $\ce{HBr}$ and a benzyl radical.
The third step is the $\ce{^.CCl3}$ radical abstracting a proton from toluene to give chloroform and a benzyl radical.
The fourth step is a benzyl radical abstracting $\ce{Br}$ from $\ce{Br-CCl3}$ to give benzyl bromide and $\ce{^.CCl3}$.
The fifth step is two $\ce{^.CCl3}$ radicals quenching to give hexachloroethane $\ce{Cl3C-CCl3}$.

So overall the products are $\ce{HBr}$, Benzyl bromide, Chloroform, and Hexachloroethane.
Reactions 3 and 4 are the major portion of the reaction so equimolar amounts of benzyl bromide and chloroform are formed along with small amounts of $\ce{HBr}$ and hexachloroethane (approx 20:1).


text:
Which has a higher enol content between 4-Pyridone and Cyclohexa-2,5-dien-1-one?

I am new to organic chemistry but here is what I feel. 4-Pyridone will have a lower enol content as in its keto form it has higher resonance stabilisation due to the involvement of lone pairs on nitrogen in conjugation while in the enol form the lone pair on nitrogen being in the molecular plane will not participate in resonance and thus will have an impact on resonance stabilisation. Thus the second compound(cyclohexadienone) will have higher enol content. Is my justification right?

Hint: how much relative stability does enol form provides in both the cases?

I think B>A is correct

What's more acidic? Ammonia or methane?


text:
Electron tunneling between orbitals
Electrons can move through potential barriers by tunneling. Atomic/molecular orbitals are separated by energy differences. Therefore I was wondering if an electron can tunnel from one orbital to another with higher energy, say from a 2s to a 2p, or if it is limited to spatial barriers.
Clarification:
It is possible to tunnel to a lower potential (as in scanning tunneling microscopy, that provides a voltage (potential difference)).
It is also possible to tunnel to inside the potential barrier (see this question). If at the other side of the barrier there is a state of medium energy (higher than the previous but lower than the barrier), the wavefunction (and therefore the probability of finding the particle) cannot drop immediately to zero, because the wavefunction must be continuous. So, even if by a small probability, the particle must be able to tunnel to another state of higher energy. The question is if this applies only to spatial barriers (as the space between the tip and the sample in STM) or to different states in general, as in atomic orbitals.

This is rather unclear - does not make sense. In tunnelling particle doesn't get energy.

Remember that all electrons in a molecule are indistinguishable from each other, so there's no practical meaning to the concept of an electron being confined to a specific orbital. What you're really considering is the transition between two electronic states of a whole molecule, for example between a ground and excited state. The barriers to different types of transitions are different for different cases (eg "spin-forbidden" or "symmetry-disallowed"). Since spectroscopy makes use of these transitions, a book on that subject is a good place to start.

There are no spatial barriers, much like there are no sky-high blue walls along all meridians and parallels, even though the map of the world might have convinced you otherwise.
There is, however, the energy barrier. You can't skip that by tunneling. The tunneling is about getting to the state of equal energy when getting there seems to be topologically impossible. Getting to a state of other energy requires energy. There is no going around that.
So it goes.

Hey, what about tunneling to a lower energy state? Isn't that what happens in a tunneling microscopy (because of the applied voltage)?

Sure, that's possible all right.

By looking at a "tunneling diagram", I see that by the continuity of wavefunction requisite, the probability of finding the particle inside the barrier is non zero (also these answers). If at the other side of the barrier there is a higher energy region (significantly less than the barrier but still higher than the other side), the probability cannot vanish to zero immediately as well. Could this then provide a really small but possible way to tunnel to a higher energy state?

Let's say there is a really small but possible way for a particle to act as if it where in a higher-energy state for a short while. That's pretty much what tunneling is about. To reside in that state permanently, though, would violate the conservation of energy and hence is impossible.

Yes, it is possible to for an electron to move between orbitals, provided it can gain or radiate away the energy involved. The energy involved is often in the range of visible light or ultraviolet radiation, which means that UV/vis spectroscopy typically deals in electron transitions. More precisely, the transitions are between electronic states, which may involve "movement" of more than one electron.


text:
How to compare size of subshells?
For the same principal quantum number, on increasing the value of the azimuthal quantum number does the average radius of the subshell increase or decrease?
In other words, which out of, say, 3s, 3p, 3d would have the largest average radius and why?

This looks like a homework question.  To get answers on this site for homework questions, you need to show the work you've done first.

@theorist no, it's not. It is just something I thought about while reading my text and found nowhere on the internet. It may feel so due to the way I have worded it, but I did that only to be crystal clear. And it's a theoretical question, what work should I show?

Well, you'd want to clarify if you're thinking about a single-electron atom (typically this would be hydrogen), or a multi-electron atom.  Probably the former would be what you'd want to consider, since then you'd be looking at just the inherent nature of the orbitals themselves.   And you could  then do some searching on the internet for effect of azimuthal quantum number on average electron distance in hydrogen and, based on that, say something to the effect of: this is what I've found/understand thus far, but I'm still confused about this, etc etc..

For a hydrogen atom, according to Hertel and Schulz [1]:
$$\langle r\rangle_{n,\,l}=\frac{a_0 n^2 \left(\frac{1}{2} \left(1-\frac{l (l+1)}{n^2}\right)+1\right)}{Z}$$
and
$$\langle r^2\rangle_{n,\,l}=\frac{a_0^2 n^4 \left(\frac{3}{2} \left(1-\frac{l
   (l+1)-\frac{1}{3}}{n^2}\right)+1\right)}{Z^2},$$
where $\langle r\rangle_{n,\,l}$ is the average distance of the electron from the nucleus as a function of n and l, and  $\langle r^2\rangle_{n,\,l}$ is the average distance squared of the electron from the nucleus as a function of n and l, $a_0$ is the Bohr radius, and Z is the nuclear charge (Z = 1 for hydrogen).  These are known as expectation values, because these are the expected values based on the probabilistic distribution of electron density.
Using these, we can compare $\langle r\rangle_{n,\,l}$ and $\langle r^2\rangle_{n,\,l}$ for the $3s \,(n = 3, l = 0), 3p \,(n = 3, l = 1) \text{ and } 3d\, (n = 3, l = 2)$ subshells:
$$\langle r\rangle_{3,\,0}= \frac{27 a_0}{2 Z}$$
$$\langle r\rangle_{3,\,1}= \frac{25 a_0}{2 Z}$$
$$\langle r\rangle_{3,\,2}= \frac{21 a_0}{2 Z}$$
$$\langle r^2\rangle_{3,\,0}= \frac{207 a_0^2}{Z^2}$$
$$\langle r^2\rangle_{3,\,1}= \frac{180 a_0^2}{Z^2}$$
$$\langle r^2\rangle_{3,\,2}= \frac{126 a_0^2}{Z^2}$$
You can see that, as the azimuthal quantum number (l) increases, the electron's average distance from the nucleus decreases.  Note, however, that in hydrogen (or any single-electron atom, such as $\ce{He^+}$ or $\ce{Li^{2+}}$, etc.), all orbitals in the same shell have the same energy. Thus the energies of the 3s, 3p, and 3d orbitals are identical.  As a consequence, a hydrogen electron excited to the third energy level is actually in a weighted linear combination of all of these orbitals, and its actual average distance would likewise be a weighted average of these.
But still, why does the average distance decrease with increasing azimuthal quantum number?  Well, quantum is not my field, but I believe this is what's going on (perhaps a quantum expert can let us know if this is correct):
As I noted, in any single-electron atom, the 3s, 3p, and 3d orbitals have the same energy. The electron's energy is a combination of potential energy due to its distance from the nucleus, and the energy due to its angular momentum.  As the azimuthal quantum number increases, the electron's angular momentum increases (it has no angular momentum in 3s, some in 3p, and more in 3d), and thus the energy associated with this angular momentum increases.  But since the total energy is constant, as the energy due to angular momentum increases, the potential energy associated with distance from the nucleus must decrease, and hence the average distance from the nucleus gets lower.
Finally, you can see from the formula that $\langle r\rangle$ is much more sensitive to changes in n than l.  For instance, if we compare the sizes of the s, p, and d subshells in the 3rd shell (n = 1, l = 0 to 2), $\langle r\rangle$ ranges from $\frac{27 a_0}{2 Z}$ to  $\frac{25 a_0}{2 Z}$ to $\frac{21 a_0}{2 Z}$.  By contrast, if we compare the sizes of the s-subshells in the first three shells (l = 0, n = 1 to 3), $\langle r\rangle$ ranges from $\frac{3 a_0}{2 Z}$ to $\frac{12 a_0}{2 Z}$ to $\frac{27 a_0}{2 Z}$.

Hertel, Ingolf V., and Schulz, Claus-Peter. Atoms, molecules and optical physics. Berlin: Springer, 2015

Sorry if this was too much of a bother, I tried searching for what you suggested but wasn't able to find something useful to me. Well, I think I would, but the thing is, I am not yet familiar with the maths in quantum. Actually, I had thought of posting the question again after I had gotten better at it and done my research. But thanks, your answer was very comprehensive and helped me understand.

Just one question- if energy is a function of distance from nucleus and the distance is a function of both the principal and azimuthal quantum number, then how is energy of 3s,3p,3d equal?  https://chemistry.stackexchange.com/questions/76111/energy-of-electrons-in-the-same-subshell#:~:text=In%20a%20single%2Delectron%20atom,with%20distance%20to%20the%20nucleus) I previously saw this answer and in it they said that the radius was only a function of the principal quantum number (in hydrogen-like atoms) so I understood then but now I am confused because the formula you used is a function of both

@l1mbo For your first question, that's what I explained in my answer: If one contributor to the energy (e.g., the angular momentum) goes up, the other (poential energy due to distance from nucleus) has to go down, to keep the total energy the same.  I.e., you start with the fact that the total energy depends only on n, and then understand the effect of changing l from that.

And as to the second, clearly the average radius is not a function of n alone, so whoever said that either didn't know this, or meant to say it depends principally on n, with only a smaller added effect from l (which is correct) (i.e., changing n has much more effect on  than changing l).  When I get a chance I'll add some examples illustrating the latter to my answer.

Before we start answering the question, we should first understand what the radius of electron is. According to quantum mechanics, the electron density is smeared across the space, and strictly speaking, the density becomes radially only if you have radial nodes, and at $r=0$ (at the nucleus) and at $r=\infty$. This means that you cannot specifically say that this is the radius of my shell. In quantum mechanics, the average radial distance of the wave function is defined as an "average radius", and the formula is given as
$$\langle r\rangle=\frac{\langle\psi|r|\psi\rangle}{\langle\psi|\psi\rangle}$$
For hydrogen atom, the average radius of the $1s$ orbital is known as Bohr radius ($0.529 Å$).

Considering this idea, you can take the wave function of the system you like and calculate the average radius. However, you have to keep the following in mind:

The notion of orbitals ($1s,2s$ etc.) exists for hydrogen or hydrogenic atoms (one electron systems) only. Beyond that, the analytical form of the wave function (for multi-electron systems) is impossible to obtain. In fact, it is one of the major challenges in the field of quantum mechanics. In many cases, people use approximate wave function to estimate the orbitals and their energies, and in such cases, it does not make sense to speak about the hydrogenic orbitals at all.
Does it really make sense to talk about average radius in an non-radially symmetric system? This is because, if you change the value of $\theta$ and $\phi$ for a non-radially symmetric system, you might get different values for average radius.

I know I have not answered your question directly, but in case you are interested in orbitals of hydrogen (or hydrogenic systems) only, you can get the expression on internet. Combine that with a bit of knowledge of mathematics and/or coding, you will get the average radius of any (or any property) of any orbital you wish to know.


text:
How is methane removed from the atmosphere?
I am not a chemist, but I've studied anthropology, geology, and history for decades. Obviously, my reason for asking this community is climate change. Specifically the concept of the methane ice on the sea floor melting and exponentially increasing warming.
What is the mechanism that brought all of that methane down to the sea floor in the first place? As I understand, methane is quite light and doesn't react with $\ce{H2O},$ or $\ce{O2},$ or $\ce{CO2}.$ Does it react with the photons over time and become reactive and sink?

Think of it under the water column pressure and bottom temperature. . It should be methane that formed as natural gas, there - is not really methane sinking from the atmosphere to the ocean bottom.

Methane does react with H2O to form methane hydrates, given the right conditions.

Oh, okay, so the seafloor methane wasn't sunk down from the atmosphere sometime before the pleistocene, when things were a lot hotter. Thank you for that, that's really cool information. That still doesn't answer the question of how atmospheric methane can be bound into the ground. I know t

I think methane forms through the (bacterial?) degradation of organic matter. In some stinking still waters you can poke holes in underwater bubbles with a stick and light the gas as it reaches the surface.

Methane does oxidate, and this is probably how it is removed from the atmosphere. Perhaps it is catalized somehow (by light or by specific compounds). It can oxidate to methanol, or entirely degraded to H2O & CO2. It can also be biologically anaerobically oxidized with nitrite.

Methane reacts with the hydroxyl radical (OH$\cdot$) in the atmosphere, and the new radical produced then reacts with oxygen; see the Wikipedia page .

I'm confused by "That still doesn't answer the question of how atmospheric methane can be bound into the ground." Do you have an example of methane being bound into the ground from the atmosphere? Generally, methane that is now below ground was never in the atmosphere. Methane in the atmosphere is removed by chemical or biological oxidation, so it does not end up as below-ground methane.

I'd take a look at the Wikipedia article on the topic and links from there. There is no direct connection between methane in the atmosphere and that in the deep sea, other then oil exploration and maybe natural emissions resulting in atmospheric methane originating from "oceanic deposits".

Per a source discussing the Methane Cycle, apparently it is slowly removed on average of approximately 9 years as a consequent of sunlight-induced radical activity. To quote:

The main mechanism for removal of methane from the earth’s atmosphere is oxidation within the troposphere by the hydroxyl radical (OH).

that reaction is given by:
$\ce{CH4 + .OH -> .CH3 + H2O}$
with the formation of the transient methyl radical which subsequently can interact with, for example, O2 leading to formaldehyde, water vapor, and eventually possibly CO2. The reaction chains are long and provided in this Wikipedia commentary.
The second sink for methane is the chlorine radical:
$\ce{CH4 + .Cl -> .CH3 + HCl}$
The associated source of radicals is strong UV light found in the upper atmosphere acting on water vapor, for example.
Per the cited Methane Cycle article, ascribed sources of methane are attributed to:

It is estimated that up to 60% of the current methane flux from land to the atmosphere is the result of human activities. Some of these activities include emissions from fermentation processes associated with livestock, from cultivated rice paddies, from fossil fuel use and biomass burning, and from landfills.


text:
Why do the peaks in a radial distribution function graph have a probability >1
Why do RDF plots have the probability at a particular distance >1. How can probability be more than one?
I am asking this because I want to understand what do the peaks of different length but at the same position on the x axis mean
The below is an example of a RDF plot:

This is not probability.

The radial distribution function isn't a probability or a probability density. The usual definition is $$g(r)=\frac{\rho(r)}{\rho_0}$$ where $\rho(r)$ is the number density at a particular distance $r$ away from some specified atom and $\rho_0$ is the bulk density of the material.
It can be larger than one if the local density is greater than the bulk (which you see in the first peak of your graph), so it can't be a probability. It also can't be a probability density, as integrating the RDF from $0$ to $r$ should give you the number of particles (divided by the bulk density) in that spherical volume, which should continue to increase even up to macroscopic distances (in contrast to a probability density which should integrate to $1$).
So what does the RDF tell you? Well if you integrate it over a spherical shell and multiply by the bulk density, you obtain the number of atoms in that shell. One particular shell we are often interested in is the first coordination sphere. Using the RDF, we can define the number of particles in the first coordination sphere as $$n(r')=4\pi\rho_0\int_0^{r'}g(r)r^2dr$$ where $r'$ is the location of the first minimum of the RDF. We can similarly determine the number of atoms in the $n^{\text{th}}$ coordination sphere by integrating from the$(n-1)^{\text{th}}$ minimum to the $n^{\text{th}}$ minimum.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/08%3A_The_Hydrogen_Atom/8.02%3A_The_Wavefunctions defines RDF for bound electrons.

This is not about electrons

Why at longer distances converges to 1, that is it behaves like bulk density? Is it because when we have larger spherical shells we approximate better the bulk?

@Anton considering very long (i.e. macroscopic) distance, we know it has to converge to the bulk density because eventually we are including the whole solution. For moderate microscopic distances, it should already start to approach 1, as you start to include more of the bulk structure rather than just individual local clusters.

@Tyberius When you say we are including the "whole solution" you mean that the shell dimensions are so big that it mimics the bulk right? Because $ρ(r)$ refers to density between $r$ and $r+dr$ and not $0$ and $r$ (the whole solution can be thought as the limit of $r$ going to infinity).

@Anton Looking over this again, I realize my statements about $g(r)$ converging are incorrect. While the integrated RDF should eventually converge to the bulk density, the local density doesn't in general have to approach the bulk and consequently $g(r)$ doesn't have to converge to 1. A simple example case would be a sphere of constant density $\rho_0$, surrounded by a large shell of constant density $\rho_1$. The bulk density is some average of these two, but the bulk density is always one or the other.


text:
Mechanism of conversion of Cr from FeCr2O4 (chromite)
The basic steps involved in the conversion of $\ce{FeCr2O4 -> Cr}$ involves,

$$\ce{FeCr2O4 ->[NaOH / air, \Delta] Na_2CrO4 ->[C, \Delta] CrO3 ->[Al, \Delta] Cr}$$

Basically, I'm having problems in learning so large amount of inorganic reactions, so I thought it would be easy to learn the mechanism behind these reactions, just like we do in organic chemistry.
Therefore, I'm keen to know the mechanism behind first conversion (i.e., from $\ce{FeCr2O4 -> Na_2CrO4}$). I've already tried to search for this, but found nothing related.

My attempt
I know some basic sets of inorganic mechanisms, as follows

Ion-exchange
Decomposition
Water-addition
Redox
Complex reaction

From the above listed, I don't think it would be ion-exchange (bcoz, $\ce{FeCr2O4 -> FeCr+ + CrO4-}$ is very unlikely to happen), water-addition or complex reaction. I guess it could be decomposition or redox, but I'm not sure on the "bond-movement" mechanism part.

This paper may help

@SafdarFaisal: From the paper you've linked, it seems that inorganic mechanisms don't have bond movement visualizations / e- or nu- attacks / addition or subs, unlike organic ones. Is it true?

unfortunately feels so.. There are mechanisms for some reactions like hydrolysis..

@RahulVerma The majority of the reactions you quoted occur either completely in solid state, or have at least one solid reactant, which suggests the main limiting factor is likely a diffusion. For the same reason one doesn't use arrow pushing schemes in IC as there are usually no molecules at all, the change occurs to a network solid. For molecular inorganic species, however, one can use similar schematics from OC, and several examples are covered, for instance, in the textbook Arrow pushing in inorganic chemistry by Ghosh and Berg.

This answer by @Maurice may be of interest. If so, please consider upvoting that answer.

The first reaction is a double oxidation. Both Fe and Cr are oxidized by O2 from the air. The reaction is made of two half-reactions. The first half-reaction is not easy to establish, because two elements (Fe, Cr) are oxidized simultaneously, Fe from +II to +III, and Cr from +III to +VI. Sorry to say it : It is one of the most difficult half-equations I ever had to state. It is :    $$\ce{FeCr_2O_4 + 11 OH^- -> Fe(OH)_3 + 2 CrO_4^{2-} + 4 H_2O + 7 e-}$$ The second half-equation is easier to write.
$$\ce{O_2 + 2H_2O + 4 e-  -> 4 OH^-}$$ The overall equation is obtained by multiplying the first equation by $4$ and the second by $7$ and adding the whole. After simplification, it yields :
$$\ce{4 FeCr_2O_4 + 16 OH^- + 7 O_2 -> 4 Fe(OH)_3 + 8 CrO_4^{2-} + 2 H_2O}$$
or, without ions :
$$\ce{4 FeCr_2O_4 + 16 NaOH + 7 O_2 -> 4 Fe(OH)_3 + 8 Na_2CrO_4 + 2 H_2O}$$
This was the first step, defining how to pass from the mineral $\ce{FeCr_2O_4}$ to $\ce{Na_2CrO_4}$, at high temperature, and in the presence of air and $\ce{NaOH}$. The final mixture can be washed with water, which dissolves easily $\ce{Na_2CrO_4}$, as $\ce{Fe(OH)_3}$ is insoluble in water.
Then $\ce{Na_2CrO_4}$ is transformed in $\ce{CrO_3}$ by adding moderately concentrated sulfuric acid : $$\ce{Na_2CrO_4 + H_2SO_4 -> CrO_3 + Na_2SO_4 + H_2O}$$ $\ce{CrO_3}$ is not soluble and can be separated by filtration. Then it is mixed with aluminum powder to get a exothermic reaction when engaged by a match : $$\ce{2 Al + CrO_3 -> Al2O3 + Cr}$$

This is an example of industrial recovery process of $\ce{Cr}$ from toxic chromium(VI) waste, which usually use high temperature reactions. However, the other answer is acceptable since OP's main focus is on the first reaction of converting $\ce{FeCr2O4}$ (one of the products from initial reaction of chromium(VI) waste) to $\ce{Na2CrO4}$. Because these reactions are done in high temperatures (usually, $\pu{1000-1200 ^\circ C}$; Ref.1), I'd like to make few comments:

$\ce{CrO3}$ would decompose to $\ce{Cr2O3}$ upon heating above $\pu{197 ^\circ C}$ (Wikipedia). The relavent reaction is:
$$\ce{4CrO3 ->[$\Delta \gt 197 \ \mathrm{^\circ C}$] 2Cr2O3 + 3O2}$$ Nonetheless, $\ce{Cr2O3}$ can also be easily reduced to elemental $\ce{Cr}$.
A large amount of chromic residues is discharged during the production of chromates by calcination of chromic ores. Statistics show that to produce a ton of sodium dichromate, the calcination process will discharge 2.5-3 tons of poisonous chromic residues. That  chromic residues contain 3-7% of residual $\ce{Cr2O3}$, 8-11% of $\ce{Fe2O3}$, and 0.5-1.5% of water-soluble $\ce{Cr(VI)}$ (Ref.1). This reference has claimed a re-calcination of these poisonous chromic residues:
Chronic residues are dried and crushed to 80-100 mesh, and mixed with chromic ore powder and sodium carbonate. Water is added to the the mixture to make it homogeneous (coke powder is also added as necessary), and then is heated in furnace at $\pu{1000-1200 ^\circ C}$. The main chemical reaction for this mixture is: $$\ce{4(FeO.Cr2O3) + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}.$$ Keep in mind that $\ce{(FeO.Cr2O3)}$ is representing $\ce{FeCr2O4}$ in this equation and $\ce{Na2CO3}$ replaces $\ce{NaOH}$ in OP's equation. Yet, the end results are the same.
The OP's second reaction, $\ce{Na2CrO4 + C ->[C, \Delta] CrO3}$ does not make any sense since oxidation of chromium does not change. I'd say it may be $\ce{Na2CrO4 + C ->[C, \Delta] Cr2O3}$. For example, the following conversion has been achied directly from $\ce{FeCr2O4}$: $$\ce{FeO.Cr2O3 + C -> Fe + Cr2O3 + CO}$$ Regardless, I also found following reaction to support my suggestion: $$\ce{2Na2CrO4 + 3C -> Cr2O3 + 2Na2O + 3CO}$$
Nevertheless, the reference 1 gives a different set of reactions for the conversion of $\ce{Na2CrO4}$: $$\ce{4Na2CrO4 + 6S + 7H2O -> 4Cr(OH)3 + 3Na2S2O3 + 2NaOH}$$ $$\ce{8Na2CrO4 + 6Na2S + 23H2O -> 8Cr(OH)3 + 3Na2S2O3 + 22NaOH}$$ $$\ce{2Cr(OH)3 ->[\pu{1200 ^\circ C}] Cr2O3 + 3H2O}.$$
For the 3rd reaction, $\ce{Al}$ is not the only reducing agent able to reduce $\ce{Cr2O3}$ to $\ce{Cr^\circ}$. The following reactions are listed in Ref.1:
$$\ce{Cr2O3 + 3CO -> 2Cr + 3H2} \\ \ce{Cr2O3 + 3CO -> 2Cr + 3H2O} \\ \ce{Cr2O3 + 3C -> 2Cr + 3CO}$$ All of these reactions have been done at high temperatures.

Reference:

Qi-Jiang Situ, Ke-Ming Xu, Pei-Nian Huang, Xing-Qin Li, De-Han Zeng, Zhi-Fa Hu, Zhi-Quan Wen, "Re-calcination and extraction process for the detoxification and comprehensive utilization of chromium residues," United States Patent 1995, 5,395,601 (PDF).


text:
Can work be done on a body when temperature and pressure are changed without changing its volume?
Can work be done on a gaseous system when temperature and pressure are changed without changing it's volume?
In chemistry the formula of work is always $p\,\mathrm dV,$ whereas we can also do work on on a gas by changing its pressure without changing the volume. Work can also be done by changing the number of moles and changing the pressure. Is the above hypothesis correct?

You can't do mechanical work on a gas by changing its pressure without changing its volume, unless the work is some sort of stirring.

"Work", loosely speaking, is an useful quantity because of it's conservative nature, and that's what first law of thermodynamics says. Work is actually defined as
$$W_\mathrm{ext} = \int\vec{F}\,\mathrm d\vec{x}.$$
Hence you see that it's not only force that matters the force should cause a displacement of the particle which it acts on.

we can also do work on on a gas by changing its pressure without changing the volume

You see that its not possible because there is some external force acting on the particles, but that doesn't cause any net displacement. That is the volume doesn't change. And you should self realize that changing the amount of substance won't do work on a system.
In your case, can you now prove that only $p\,\mathrm dV$ causes a net displacement on the system? Remember, even though pressure is a scalar quantity, it is not force acting per unit area but thrust acting per unit area, hence the dot product in original expression is not violated. For more queries you can visit Physics.SE.
Edit: Wherever I say "work" it should be understood as "mechanical work". Other forms of work are also possible as suggested by the other answer and they also obey the first law of thermodynamics. This mechanical work by stirring is a nice special case where it's hard to define pressure, but we have other easier methods in physics, have you ever heard of "Joule's paddle wheel"?

The root of your confusion, as I understand, is that you are not much familiar with other kinds of work done. Although, $p\,\mathrm{d}V$ is everyone's favourite but it's not the only one in the league.
Always remember, the $p\,\mathrm{d}V$ is work done during an expansion/compression. You can change the pressure or concentration to get electric work.

You could mechanically rotate a generator (i.e., do work) to make an electrical current which flows into a resistor within the gaseous volume and heats it. So you could say you did work on the gaseous system (indirectly), and some of it is recoverable as work by releasing the pressure.

This is not work done on the system,  it is heat transfer Q to the system( unless you consider the resistor as part of the system).


text:
How does the polarizability of an apolar compound influence properties such as its boiling temperature?
I have understood that the more polar is a compound, the higher is its boiling temperature. But I would like to know what is the justification for this answer.
I am new to chemistry so I would appreciate a simple answer.

Do you mean perhaps "polarity" rather than polarizability? They are related but not the same.

In non-polar liquids, the intermolecular forces are known as London dispersion forces (or dipole induced dipole interaction). This is the weakest interaction that exists between molecules.
The strength of this interaction depends on polarisability of non-polar molecule. As an example, let's take the following examples:
$$
\begin{array}{lr|lr|lr|lr}
\hline
\text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} \\
\hline
\ce{CO2}  & -78.5 & \ce{CCl4} & 76.7  & \ce{CBr4}  & 189.5 & \ce{CI4}   & 329.2 \\
\ce{CS2}  & 46.3  & \ce{CBr4} & 189.5 & \ce{SiCl4} & 57.7  & \ce{SiBr4} & 153.0\\
\ce{CSe2} & 125.5 & \ce{CI4}  & 329.2 &            &       &                 \\
\hline
\end{array}
$$
In these molecules, polarisability depends on polarisability of anions $(\alpha),$ which depends as
$$\alpha \propto q \times r,$$
where $q$ is the charge and $r$ is the ionic radius.
As charge on each anion is same, hence we can conclude that, greater the radius, greater is the polarisability, and thus greater is London dispersion forces.
Therefore, greater boiling point.


text:
Help with determining concentration of magnesium hydroxide for a specific pH level
This seems like such an obvious and elementary-level question for someone who has taken high school chemistry, but yet I'm having a difficult time solving it. I need to find the right amount of $\ce{Mg(OH)2}$ to have a $\pu{250 mL}$ solution with a $\mathrm{pH}$ of $8$ and $9.$
I first started by trying to determine the concentration of $\ce{Mg(OH)2}$ needed to obtain a solution with a certain pH. Since $\ce{Mg(OH)2}$ is a weak base, we must use a RICE table to determine its concentration at equilibrium.
$$
\begin{array}{lcccc}
  & \ce{Mg(OH)2 &<=> & Mg^2+ &+ & 2OH-} \\
\text{I} & x & & 0 && 0 & \\
\text{C} & x & & +x && +2x \\
\text{E} & x & & x && 2x
\end{array}
$$
However, the equation involving $K_\mathrm{sp}$ only involves the $x$ values in the RICE table above, not the actual concentration. How would I calculate the $\mathrm{pH}$ level for $\ce{Mg(OH)2}$ with varying molarity? I thought it was a weak base, so I'd have to use a RICE table.

Mg(OH)2 is a strong electrolyte/base. It is an ionic compound. Do you mean to say that it is sparingly soluble as the reason for using the RICE table

There is an easy way to do what OP wants, assuming OP wants to prepare solutions at $\pu{25 ^\circ C}$. So, OP can prepare saturated solution of $\ce{Mg(OH)2}$ solution:
$$\ce{Mg(OH)2_{(s)} <=> Mg^2+_{(aq)} + 2OH-_{(aq)}}$$
Since $K_\mathrm{sp}$ of $\ce{Mg(OH)2}$ is $\pu{5.61 \times 10^{-12} M3}$, you can find the solubility of $\ce{Mg(OH)2}$ at $\pu{25 ^\circ C}$ ($s$):
$$K_\mathrm{sp} = s \times (2s)^2 = 4s^3 \ \Rightarrow \ s = \left(\frac{K_\mathrm{sp}}{4}\right)^{\frac13} = \left(\frac{\pu{5.61 \times 10^{-12} M3}}{4}\right)^{\frac13} = \pu{\pu{1.12 \times 10^{-4} M}}$$
Thus, $[\ce{Mg^2+}] = \pu{1.12 \times 10^{-4} M}$ and $[\ce{OH-}] = 2 \times \pu{1.12 \times 10^{-4} M} = \pu{2.24 \times 10^{-4} M}$.
$$\therefore \ \mathrm{pOH} = -\log {[\ce{OH-}]} = -\log (\pu{2.24 \times 10^{-4} M}) = 3.65$$
Thus, $\mathrm{pH} = 14.00 - 3.65 = 10.35$. This means the $\mathrm{pH}$ of saturated $\ce{Mg(OH)2}$ solution is a little higher than what OP anticipated. The dilution of the saturated solution with deionized water do the trick as demonstrated in following example:
Suppose you want to make $\pu{250 mL}$ of $\ce{Mg(OH)2}$ solution with $\mathrm{pH} = 8.00$. Thus, $\mathrm{pOH} = 14.00 - 8.00 = 6.00$. Thus, $[\ce{OH-}] = \pu{1.00 \times 10^{-6} M}$. For the calculation for the dilution, you can use $c_1V_1 = c_2V_2$ equation.
In OP's case, $c_1 = \pu{2.24 \times 10^{-4} M}$, $c_2 = \pu{1.00 \times 10^{-6} M}$, and $V_2 = \pu{250 mL}$, the volume of anticipated solution with $\mathrm{pH} = 8.00$. The unknown $V_1$ is the volume of saturated $\ce{Mg(OH)2}$ solution ($\mathrm{pH} = 10.35$) needed to be diluted:
$$c_1V_1 = c_2V_2 \ \Rightarrow \ V_1 = \frac{c_2V_2}{c_1} = \frac{\pu{1.00 \times 10^{-6} M} \times \pu{250 mL}}{\pu{2.24 \times 10^{-4} M}} = \pu{1.12 mL}$$
Thus, you can measure $\pu{1.12 mL}$ of saturated $\ce{Mg(OH)2}$ solution into $\pu{250 mL}$ volumetric flask and diluted it with DI water to the $\pu{250 mL}$ line mark. After shaking well to get homogeneous solution, its $\mathrm{pH}$ should be anticipated $8$ (or closer to $8$ based on the accuracy of the measurements).
Note: It would be better if you can measure the $\mathrm{pH}$ of saturated solution before do the calculations. That's because, the factors such as temperature influence the realtime $\mathrm{pH}$.

Dear Mathew, It would be better if you don't attach units to the equilibrium constants.

@M. Farooq : Thanks for your kind comment. I'll keep it in mind for future perspectives. I just want to show how we get concentration term at the end for $s$. Thanks again. ;-)

$\ce{Mg(OH)2}$ is a strong base since it is ionic in nature; it usually dissociates completely and so it's degree of dissociation is one.
For weaker salts, the concentration values you assigned for the ions, $x$ and $2x$ respectively, do depend on the molarity of the $\ce{Mg(OH)2}.$ You have to define $x$ in the terms of its degree of dissociation $(\alpha),$ and the concentration $(c)$ as $c\alpha.$ The RICE table would come out as
$$
\begin{array}{lcccc}
  & \ce{Mg(OH)2 &<=> & Mg^2+ &+ &2 OH-} \\
\text{Initial} & c & & 0 && 0 \\
\text{Change} & -c\alpha & & +c\alpha && +2c\alpha \\
\text{Equilibrium} & c - c\alpha & & c\alpha && 2c\alpha
\end{array}
$$

If $c$ is the concentration of $\ce{Mg^2+}$ in the $\ce{Mg(OH)2}$ solution, the concentration $[\ce{OH-}] = 2c.$
At $\mathrm{pH}~9,$ $[\ce{OH-}] = \pu{1E-5 M},$ then $c = \pu{5E-6 M}.$ So you have to dissolve $\pu{1.25 μmol}$ $\ce{Mg(OH)2}$ in $\pu{250 mL}$ water. This is $\pu{71.3 μg}$ of $\ce{Mg(OH)2}.$ This is difficult to do in practice, as such a small amount is difficult to weigh. It may be done in two steps: first prepare a moderately concentrated solution, then dilute it to $\mathrm{pH}~9.$

Now much base of any kind you need to get $\mathrm{pH} = 7$? None, the water itself is enough.
For $\mathrm{pH}$ levels near 7 (8 is pretty much eligible) you want to account for the $\ce{OH-}$ ions that come from water.
$[\ce{OH-}] = 1 \times 10^{-6} (\mathrm{pH} = 8)$
$[\ce{OH-}][\ce{H+}] = 1 \times 10^{-14} $ (water at room temp)
$[\ce{OH-}] = [\ce{H+}] + 2[\ce{Mg^2+}]$ (electric neutrality)
(everything in molar concentrations)
... solve it for $[\ce{Mg^2+}]$.
For $\mathrm{pH} = 9$ you can skip the water $[\ce{OH-}]$ and the error will be less than 1%. This error may or may not be acceptable in your context.
You may next want to check if $\ce{Mg(OH)2}$ is at all soluble that much.


text:
How do orbitals overlap in difluorosilane?
In $\ce{SiH2F2}$ four orbitals of silicon $(\mathrm{3s},$ $\mathrm{3p}_x,$ $\mathrm{3p}_y,$ $\mathrm{3p}_z)$ overlap with two hydrogen $(\mathrm{1s})$ and two fluorine $(\mathrm{2p}).$
I want to know how these orbitals are paired for overlapping.
Is it based on strength of overlapping?

A simple model for bonding in this molecule is taking $\ce {Si}$ to adopt $\ce {sp^3}$ hybridisation, as suggested from the tetrahedral geometry of the molecule. Subsequently, we can refine our model by applying Bent's rule (see more about it here), which tells us that $\ce {p}$ orbital character concentrates towards more electronegative substituents. Since the electronegativity of the elements has the following order: $\ce {F}$ > $\ce {H}$ > $\ce {Si}$ with Pauling electronegativity values of $\ce {3.98}$, $\ce {2.20}$ and $\ce {1.90}$ respectively, the $\ce {3p}$ character of $\ce {Si}$ would concentrate towards $\ce {F}$ while $\ce {3s}$ character concentrates towards $\ce {H}$. Thus, we would have $\ce {Si}$ using $\ce {sp^{3+x}}$ orbitals to form the bonds with $\ce {F}$ and $\ce {sp^{3-x}}$ to form the bonds with $\ce {H}$, where $x$ is some small value between $\ce {0}$ and $\ce {1}$.

It means one hydrogen approach to S orbital to form 1s-3p overlapping and other hydrogen will approach to p orbital of Si to form 1s-3p overlapping and the rest orbitals for fluorine to for 2p-3p.Is it correct?

@Private5661 Are you familiar with the theory of hybridisation? Have a look here: https://en.wikipedia.org/wiki/Orbital_hybridisation


text:
Defining general acidic strength
The $K_\mathrm{a}$ or $K_\mathrm{b}$ values provide an idea of how likely a specific step is to happen.
However, how do you specify one acid just being stronger than another, as in the case of $\ce{H2CO3}$ $(\mathrm{p}K_\mathrm{a1} = 3.6,$ but $\mathrm{p}K_\mathrm{a2} = 10.32)$ and phenol $(\mathrm{p}K_\mathrm{a} = 9.95)?$ How do you compare them?

Compare pKa1 for both compounds

Isn't it used to measure the likelihood of just the first deprotonation?

Isn't that what acidity is a measure of?

Acidity is, by definition, the tendency of a compound to act as an H+ donor for Bronsted acids. For Lewis acids, acidity relates to the compound's ability to accept an electron pair. That doesn't necessarily mean the first deprotonation every time, does it?

It generally is.. atleast to the best of my knowledge

Look up polyprotic acids; https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Polyprotic_Acids_And_Bases#:~:text=Polyprotic%20acids%20are%20specific%20acids,H%20%2B%20atom%20per%20molecule).

Let us continue this discussion in chat.

According to the Brønsted-Lowry acid base theory, an acid is a compound that releases a $\ce{H+}$ ion to give a conjugate base. For most acids, this reaction exists in a dynamic equilibrium. The equilibrium constant of this reaction is what defines $K_\mathrm{a}$. Generally more stable the product, the more of the forward reaction takes place.
Using this last statement, when comparing two compounds, we can say that if the conjugate base of one compound is more stable, it would mean that the value of its $K_\mathrm{a}$ would be greater. This implies its $\mathrm{p}K_\mathrm{a}$ would be less, meaning that the compound would be more acidic than the other.
For example, taking the case of two compounds, $\ce{HCOOH}$ and $\ce{C6H5-OH}$: Here formate ion would have resonance via oxygen whereas phenoxide ion would have resonance via carbon. This makes phenol less acidic than formic acid.
$\mathrm{p}K_\mathrm{a} \ {\text{ of phenol} = 10.0}$
$\mathrm{p}K_\mathrm{a} \ {\text{ of formic acid} = 3.75}$
Now, if we consider polyprotic acids, we can only consider one hydrogen at a time. Successive $K_\mathrm{a}$s decrease in magnitudes of very large orders.
For example, $\ce{H3PO4}$ has $\mathrm{p}K_\mathrm{a1} = 2.12$ whereas $\mathrm{p}K_\mathrm{a2} = 7.21$ and $\mathrm{p}K_\mathrm{a3} = 12.68$.
Hope this answered your question.

@RahulVerma just trying to follow andelisk's formatting..

Ah, I didn't noticed it. But, in general, I use $\ce{}$ in most of my equations. I was also suggested in the beginning by some user to use $\ce{} instead of ${}$. Anyways, if @andselisk is using, then it'll be fine. btw, I edited just to make numerals beaultiful. Feel free to make further edits :)

What is the difference between using "dollar\ce{...}dollar", with \ce, and using "dollar{...}dollar" without \ce ?

@Maurice, see this answer for mhchem extension, i.e., $\ce{}$

$K_\mathrm{a}$ is the measurement of equilibrium constant when an acid dissociates as:
$$\ce{HA <=> H+ + A-}$$
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
The term $\mathrm{p}K_\mathrm{a}$ is defined as $\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$. So lower the $\mathrm{p}K_\mathrm{a}$, the better is the acidity.
In the example given, if you wish to compare the acidity, you have to mention which proton of $\ce{H2CO3}$ you are interested to abstract. In other words, $\ce{H2CO3}$ is more acidic then phenol but $\ce{HCO3-}$ is less acidic then phenol.


text:
Stability of dialkyl substituted alkenes
The stability order of alkyl substituted alkenes with respect to heat of hydrogenation was given as
$$\ce{R_2C=CH2} > \ce{RHC=CHR}~(\textit{trans}) > \ce{RHC=CHR}~(\textit{cis}).$$
While I understand that trans alkene is generally more stable than cis, I don't understand why $\ce{R_2C=CH2}$ is the most stable. Shouldn't there be repulsion between the bulky alkyl groups which leads to decrease in stability?

See "Why is 2-methylpropene less in energy than its alkene counterparts?".

Following is an image of hyperconjugative effect. We can see a $\ce{\sigma}$-orbital delocalising electron pair with an empty $\ce{\pi^\ast}$-orbital.

The explanation to your problem can be given by deciding whether the effect is "centralized" or "decentralized". Let me explain it by an example.

A quick scene:
Suppose, you and your friend is sitting across a boat ($\ce{C_{you}=C_{friend}}$) in the middle of a river, and both of you have to reach on either bank soon (a process similar to delocalization, probably). Assume the methyl groups to be oars you need to move across the river.

In $\ce{R2C_{you}=C_{friend}H2}$, both the oars are with you. So, only you can row the boat and hence both of you can reach to the bank soon. Technically, the $\ce{\beta-H}$'s are at one carbon, so the effect can happen only on it, hence "centralized" and super-fast.
But, in $\ce{RHC_{you}=C_{friend}HR}$, both of you have the oars. So, both of you can row the boat and therefore both of you will reach to either bank a bit later. Technically, the $\ce{\beta-H}$'s are distributed equally among the carbons, so the effect can happen on either side, hence "decentralized" and not as fast as in the previous case.
As per your concern on steric factors, you may know that "alkyl" is referred to a straight hydrocarbon chain, and it isn't as bulky as to create "steric hindrance".
I guess that makes some sense :)

EDIT: The above stuffs explains how is first isomer the most stable. I'd previously said that "steric effects" doesn't matter, but in case of cis and trans isomers, it matters and it's the only thing that differentiate them (along with polarity). Due to this, cis isomer is least stable among the three.
Reference table of $\ce{\Delta H^°_f}$: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/thermo/heats-formation.html#Alkenes (in #C = 4)

Interesting analogy. I'm wondering whether it was a reflection of the wolf, goat and cabbage problem:)

Is this a way of saying that in the trans case, the C-H bonds are effectively competing over hyperconjugation, but not in the case with 2 alkyl groups on the same side?

@Jabbamanga: Not exactly, but it's very similar. Actually, it is a random phenomenon. The orientation of sigma orbital at time of effect decides which C drives the effect.

@andselisk: ...or the fox, chicken and grain. ;)

Could you please add thermodynamic considerations of stability. While the analogy is great I would appreciate a more rigorous answer.

@trinitrotoluene: added some more points + reference


text:
At what composition is Maximum deviation from Raoult's law observed?

Raoult's law states that in an ideal binary solution of two volatile components, the partial vapour pressure of each component is proportional to its mole fraction in the solution.

A plot of total pressure to mole fraction of one of the components gives:

If A and B form a non-ideal solution with positive deviation, then the following graph is observed:

At what composition of the two components is the deviation (difference in total pressure) maximum?
My guess would be that maximum deviation is observed when mole fraction of both the components are equal.
My reasoning is that the deviation is observed because of the fact that interactions between A-A and B-B particles are not equal to the A-B particle interactions. When the quantity of both the components are equal, there is a maximum amount of A-B interactions. Hence, the deviation must be maximum.
Is this reasoning correct?

Are you talking for the postive deviation then for postive deviation vapour pressure should be greater than the value of individual vapour pressure mathematically you can differentiate the value it would be in the range

There's no definitive answer to your question. You might want to look up azeotropic mixtures.
Essentially, to make an azeotrope, you need a set of liquids with really high deviation from "ideal" behaviour, and at the composition where the deviation is maximum, the slope is zero. At that composition, there isn't much variation in vapor pressure as you vary the mole fractions, and so both components of the liquid boil equally, as stated here:

An azeotrope or a constant boiling point mixture is a mixture of two or more liquids whose proportions cannot be altered or changed by simple distillation.Because their composition is unchanged by distillation, azeotropes are also called (especially in older texts) constant boiling point mixtures.

Now, if you look at a few examples of azeotropes, it becomes clear that the maximum deviation is not achieved at the same composition ratio:

Sulfuric acid (98.3%) / water, boils at 338 °C
||Nitric acid (68%) / water, boils at 120.2 °C at 1 atm||
Perchloric acid (71.6%) / water, boils at 203 °C||
Hydrofluoric acid (35.6%) / water, boils at 111.35 °C

References: Azeotropes, wikipedia

Be very cautious when you put Wikipedia as your [single] reference. As written in Wikipedia itself, it "is not a reliable source". Please also look at Citing Wikipedia.


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What is the correct electronic configuration of darmstadtium?
Some of the places, the electronic configuration of darmstadtium ($\ce{Ds}$) is given as $\mathrm{6d^9 7s^1}$ while at some other places, it is $\mathrm{6d^8 7s^2}$. Which among the two is correct?

Electron configurations are merely fictional or just a book keeping work. Don't worry too much about it. Find me a chemist who can tell you how to experimentally determine the electron configuration of sodium. Forget about Ds.

They should not be asking such questions whose answers they do not know themselves. As I said, ask them how did someone experimentally determine the electron configuration of a simple element such as Na (Z=11)? I bet nobody can answer because I have been searching for this answer for decades without any decent response.

Anyway, what is the use of such a question ? Up to now, the number of Ds atoms produced in the world is increasing with the time, but it is still much lower than 100. And they all live less than 10 seconds, often less than 1 millisecond...

Sir could you please elaborate we cannot determine the electron configuration?

According to Wikipedia, the electronic configuration of darmstadtium is predicted to be [Rn] 5f14 6d8 7s2.

From practical point of view: electronic configuration of an atom strictly speaking referring to the isolated, single atom. If there are more than one low energy state, chances you can observe the ones that are preferred by the given environment. Ie when the answer is difficult, the answer generally doesn't matter.

Wikipedia suggests the electronic configuration of Ds as $\ce{[Rn] 5f^{14} 6d^8 7s^2}$ while some others suggests $\ce{[Rn] 5f^{14} 6d^9 7s^1}$.
Relativistic stabilization of 7s orbital in darmstadtium explains its predicted configuration to be $\ce{[Rn] 5f^{14} 6d^8 7s^2}$. It contracts the $l = 0$ subshell i.e. the s subshell which in this case contains two electrons. According to this reference:

The relativistic stabilization of the 7s orbital is 2.6 eV

The effect of the ns orbital contraction reaches its maximum in the 6th period with Au (17.3%) and in the 7th period with element 112
(31%)

The shift of the maximum to element 112 in the 7th period in contrast to gold in the 6th period is due to the fact that in both elements 111
and 112 the ground state electronic configuration is
$\ce{d^\ce{q}s^2}$

which means the relativistic effects on 7s subshell are quite significant in the element adjacent to element 111 which is darmstadtium (element 110).
These factors makes darmstadtium to have a configuration of $\ce{[Rn] 5f^{14} 6d^8 7s^2}$ instead of $\ce{[Rn] 5f^{14} 6d^9 7s^1}$.
Reference: The Chemistry of the Actinide and
Transactinide Elements, $\ce3^{rd}$ edition, by Lester R. Morss, Norman M. Edenstein, Jean Fuger & Joseph J. Katz

Are there other sources, especially computational articles, apart from just Wikipedia? The energy levels between s/f/d are probably not very large (the gaps get smaller with larger $n$) and I wonder if there are other factors (particularly relativistic effects) that are important in determining the electronic configuration.

I dont think we can trust  wikipedia for things books or university lectures dont have much about.Though ofcourse it is better than nothing....

@Proscionexium what I am saying is that we havent studied much elements on the 7th row of the periodic table and predictions rely mostly on the trends of lighter elements which we dont know if they apply to these elements.

@Proscionexium yes but because we havent been able to isolate Darmstadtium we dont know its electronic configuration we can only make educational guesses based on trends of lighter elements of the same block.So I dont think there is a correct answer.

@Proscionexium I really believe its a GOOD idea not to continue the discussion in chat and leave the comments in plain sight because the problem here has its roots at the scientific procedure.

@Volpina For sure, experimental evidence would be the best. But in the absence of that, you can still do theoretical calculations (and that's what I was hoping might exist out there). It's not right to say that we can only make an educated guess. Wikipedia is perfectly fine as a place to look up infromation but it is not a source in and of itself — you must always look for the sources which it cites.


text:
Is there an opposite to shielding effect?
I recently read about shielding effect and lowering of effective nuclear charge  due to penetration of other electrons. I wonder while doing calculations involving Slater's rules the electrons from outer shells are neglected.
I think they can reverse the shielding produced by inner shell electrons.

Hmmm... Why would you think that the outer shell electrons can "reverse the shielding" provided by the inner shell electrons? By what mechanism does this occur?

I answer your question here: https://chemistry.stackexchange.com/questions/136873/while-calculating-the-effective-nuclear-charge-why-dont-we-consider-effect-of

Slater's rules are an attempt to lump the effect of all other electrons on the wavefunction, and thereby other properties such as energy, of an electron (described by a hydrogen-like wavefunction). The effect of electron-electron repulsion is modeled indirectly by saying that inner shell electrons effectively screen the attractive nuclear charge sensed by electrons further away from the nucleus. For simplicity screening is modeled as a one-sided effect: electrons further from the nucleus are not expected to significantly shield those closer to it, since an outer electron is close to the nucleus less frequently than an inner one (or, using more accurate language, the density near the nucleus of the outer electron is lower).
Slater's rules amount to a method of estimating appropriate exponents in a hydrogenic approximation of the electron wavefunction. They are semi-empirical, useful as a guide to explain why certain trends are observed, and fit data because they contain "fudge-factors" (parameters obtained from fits to data, not from fundamental theory). Still, that they work at all indicates what amazing insight Slater had in selecting the shape of the function and using the method to predict a number of atomic properties across the periodic table. If you haven't you should look at his original paper (Ref. 1), it's impressive and accessible. This is also fairly well explained in a Wikipedia article.
Reference

Slater, J.C. Phys. Rev. 36 (1): 57–64. doi:10.1103/PhysRev.36.57.


text:
Misconception of gas laws
I face a misconception with gas laws: Boyle’s law, Charles’ law, and Gay-Lussac’s law.
This is because each law of these assumes the constancy of one of the three following variables (the pressure, the volume, the absolute temperature) and then studies the relation between the other two.
Isn’t this a contradiction?
I mean for instance: Charles’ law states: "the volume of an ideal gas at constant pressure is directly proportional to the absolute temperature.”
But how the pressure is constant, didn’t Boyle just say that pressure is inversely proportional to volume?

Suppose the overall law is PV-nRT.  Mathematically, doesn't this mean that, at constant pressure, the volume is proportional to temperature?  And doesn't it also require that, at constant temperature, pressure is inversely proportional to volume?

@ Chet Miller. The ideal gas law should be clearly written, without a minus sign. It should be : PV = nRT. And it should be mentioned that P is the pressure, V the gas volume, n the amount of gas in moles, R is a constant (8.314 J K-1 mol-1), and T is the absolute temperature in Kelvin

See also: https://chemistry.stackexchange.com/a/127772/79678 for a bit more detail.

Possible duplicate of https://chemistry.stackexchange.com/q/58330/79678.

Sorry @Maurice I accidentally typed a minus sign instead of an equal sign, and my vision is not what it used to be (so I didn't notice that I had done that).  Please excuse an old man.

Does this answer your question? How can we combine all the three gas laws into a single ideal gas equation?

There is no contradiction. Each law is valid under a certain conditions.

This is because each law of these assumes the constancy of one of the three following variables (the pressure , the volume , the absolute temperature) and then studies the relation between the other two. Isn’t this a contradiction?

This is how scientist design experiments. Suppose a teacher wants to assess your chemistry knowledge in an examination room in writing. There are so many variables that would affect your performance. For example, difficulty level of questions, room temperature, humidity, room light, chair comfort and noise level.
What if the teacher, the starts changing all these variables simultaneously in the exam room while you are writing the exam? Certainly, anyone would become crazy under those conditions and it would be hard to assess your true exam results.
Would anyone be able to assess the outcome? So it is better to keep all the variables constant and then vary the difficulty level of questions.
Now let us say, we wish to see the affect of room temperature on your exam performance. What would you like to do? Keep the remaining variables constant, just change the room temperature.
So experimentally, it is easy to alter one variable at a time.
Boyle law studies the pressure behaviour of a gas with constant temperature.
Charles law studies the volume behaviour of a gas with constant pressure.
Do you still see a contradiction?

I am trying to imagine what it'd be like to take an exam under those conditions. You might try a subway train!

@BuckThorn, Not a bad idea ;-)

Boyle, Gay-Lussac, Charles laws can all be deduced from the more general ideal gas law, namely : $\ce{PV = nRT}$, provided you know that P is the pressure of the gas, V is the volume of the container, n is the amount of gas, in moles, R is a constant ( $\ce{R = 8.314 J K^{-1} mol^{-1}}$), and T is the absolute temperature in Kelvin.
Let's show that Boyle's law is a particular case of PV = nRT. In Boyle's experiments, P and V may change, but n and T are constant. So the product of the three constants nRT is also constant: it is a new constant, in all experiments by Boyle. And Boyle's law is PV = constant = nRT. Or : $\ce{P_1V_1 = P_2 V_2}$
Let's show that Charles's law is also a particular case of PV = nRT. Charles's law states that a given volume V of gas is proportional to the temperature T, at constant pressure P. It has nothing to do with Boyle's law, which was valid at a given constant temperature. As the pressure P (and the amount of gas n) is a constant in all Charles' experiments, the constants n, R and P can be grouped into a new constant nR/P = k,  and the volume V changes with T. The initial law can be rewritten : V = (nR/P)·T = k·T. The volume V is proportional to the temperature T.
Similarly, Gay-Lussac's law is also a particular case of PV = nRT. Gay-Lussac's law states that the pressure P of a given amount of gas is proportional to the temperature T if the volume V does not change during the experiment. As the volume V (and the amount of gas n) is constant, the three constants n, R and V, are such that their product nR/V is a constant in all Gay-Lussac experiments: These constants n, R and V can be regrouped into a new constant nR/V = k', and the law can be rewritten : P  = (nR/V)·T = k'·T
This shows that the three gas laws, by Boyle, Gay-Lussac and Charles, are all particular cases of the more general law PV = nRT.


text:
Iron and copper oxide for thermite
I recently successfully made thermite from iron oxide powder and aluminum powder. However, I have a couple of questions about copper and iron oxides.
The kind of iron oxide that is made by electrolyzing iron in a salt water bath. You rinse the solution well to get some of the salt out then boil off/dry the remaining sludge. What is left is a red powder. What kind of iron oxide is this? (II,III)
Lastly, if I repeated the same process with copper, would I get copper(I) oxide or copper(II) oxide?
Apologies for my lack of knowledge in chemistry.

Iron(I) and iron(IV) oxides do not exist. The only iron oxide with a reddish color is iron(III) oxide $\ce{Fe_2O_3}$. Iron(II) oxide FeO is not stable in air. It gets quickly transformed into iron(III) oxide. The thermite reaction works well with aluminum powder and iron(III) oxide.

Copper oxides are nowhere near as stable as aluminum oxides.

@Maurice  $\ce{FeO}$ oxidizes to black $\ce{Fe3O4}$, not to red/brown $\ce{Fe2O3}$, as the former is the most stable at low/medium temperatures. So does $\ce{Fe2O3}$ converts to $\ce{Fe3O4}$ when heated enough. Note that these oxides as many oxides of transition metals are not fully stoichiometric.

@ Poutnik. Thank you for the complementary information.

Per a source : Time-Resolved Mass Spectrometry of Nano-Al and Nano-Al/CuO Thermite under Rapid Heating: A Mechanistic Study, to quote:

The ignition of aluminum particles has been found to be very sensitive to particle size and closely related to the properties of the oxide shell. It has been found that Al-NPs can be ignited at as low as ∼930 K which is close to the melting point of the Al (933 K). The microsized aluminum particles ignite at a much higher temperature of ∼2300 K that is close to the melting point of the oxide shell.7

Here is a comment on the nature of the Al/CuO thermite:

3.2. Nano-Aluminum/CuO Thermite Reaction in TJump/TOMFS. A more vigorous reaction of nano-Al was probed with a thermite mixture using CuO as the oxidizer at a heating rate of ∼5 × 10^5 K/s.

where the use of the adjective 'vigorous' does not reflect, in my opinion, the intensity of the reaction.
However, as noted above, metal particle size is very material, and I would expect primarily molten copper with varying aluminum oxides presence (as cited in the study).


text:
Thermionic Emission Conservation of Energy
I am learning about thermionic emission and how it led to the vacuum tube. It is said that thermionic emission was a way to increase the flow of electrons from a signal (e.g. in a radio receiving a weak signal, boosted by thermionic emission). What I don't understand is how it "generates" electricity; even if the coil is heated to it's work threshold, wouldn't a coil only be able to emit as many electrons as are supplied?
Thank you for your time

I was mistaken on the principle that governs vacuum tubes. It is not that the signal is being amplified by thermionic emission; rather, the radio wave is being used as a gate to allow or block electrons from being received. Thermions are being emitted from a coil using electricity, emitting at a higher current, modulated by the radio wave as a control grid (preventing or allowing the thermions from being collected at the anode).

Correct... perhaps delete both question and answer. That said, thermionic emission can generate electricity, e.g. https://www.geographynotes.com/energy-management-2/generators/thermionic-generator-with-diagram-energy-conversion-energy-management/4195 , but radio tubes are called "valves" for good reason.


text:
How can magnesium hydroxide be a strong base and be sparingly soluble at the same time?
A strong base means it completely dissociates forming $\ce{OH^-}$ ions when added to aqueous solution.
Sparingly soluble means it dissolves very little when added to aqueous solution.
Yet magnesium hydroxide is said to be both. These two properties seem to contradict each other. How can this be?
A very similar question was asked (What does it mean to say that a strong base is only slightly soluble?), the most relevant answer states that

Some strong bases like calcium hydroxide aren't very soluble in water.
That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions.
Calcium hydroxide still counts as a strong base because of that 100% ionisation.

But I don't understand the difference — how can something dissolve and not be ionised? (wouldn't that imply a covalent structure like glucose?)

There is no contradiction: calcium hydroxide is an ionic compound, so it already consists of ions. That portion which dissolves in water consists of calcium and hydroxide ions. They just get to move around in the solution. The rest of the solid simply does not dissolve. Same for magnesium hydroxide.

How would one know it's a strong base? By reaction with an acid. Does it behave as you would a strong base to behave?

Your question: How can something dissolve and not be ionized? How about acetic acid? It completely dissolves in water, but only partially ionizes. Hence we call it weak acid. Both $\ce{Ca(OH)2}$ and $\ce{MgOH)2}$ are sparingly soluble in water, but completely ionize whatever amount dissolved in water giving $\ce{OH-}$ ions. Thus, they are considered strong bases.

@MathewMahindaratne I think this then becomes a very technical matter. From what you say, you seem to suggest that $\ce {Mg(OH)2 (aq)}$ (i.e. the dissolved form of magnesium hydroxide) is a strong base while $\ce {Mg(OH)2 (s)}$ (i.e. the undissolved form of magnesium hydroxide) is not a base. Is that right?

If you agree that you have implied the above, then you are essentially saying that the strong base are the solvated ions $\ce {Mg^{2+}}$ and $\ce {OH^-}$ since those aqueous ions are effectively what $\ce {Mg(OH)2 (aq)}$ refers to.

I think we should not conflate "solubility" with "dissociation". They are different phenomena. The former has to do with a molecule breaking away from the other solute molecules and forming interactions with the solvent. The latter is a process which occurs after dissolution, whereby the solute molecules break apart to form ions, separately solvated by the solvent molecules. For a simple covalent substance, the two phemomena are clearly different.
For an ionic solid, the two may seem to occur together since when a soluble ionic salt is placed in water, its ions are in fact pulled apart by water molecules. In this sense, it is dissolving and dissociating at the same time. Hence, by the Arrhenius definition of acids and bases, insoluble hydroxides such as $\ce {Mg(OH)_2}$ may not even be considered to be strong bases. Arrhenius theory would tell us that the strong base is the hydroxide ion, that is produced from the dissolultion and dissociation of the salt.
Actually, in particular for some metal oxides such as $\ce {MgO}$, the first reaction it undergoes upon contact with water is $\ce {MgO + H2O -> Mg(OH)2}$. Hence, we can rightfully say $\ce {MgO}$ is not an Arrhenius base, but it is the precursor to the weak Arrhenius base $\ce {Mg(OH)2}$.
I would like to disagree with the part of the answer you have referenced in your post because as aforementioned, dissolution and dissociation are the same phenomena for ionic salts. It is thus not correct to say that "what does dissolve is still $\ce {100}$% ionised into [ions]" and hence, say that insoluble (or sparingly soluble) ionic salts are strong bases.
To generalise, Arrhenius theory would consider soluble ionic hydroxides such as $\ce {NaOH (s)}$ to be strong bases while it would consider sparingly soluble ionic hydroxides such as $\ce {Ca(OH)2}$ to be weak bases. In other words, the strength of these metal hydroxides as bases is directly linked to their solubility and this makes perfect sense since dissolution and dissociation for an ionic salt are the same process.

@ Tan Yong Boon. You seem to forget that $\ce{Mg(OH)_2}$ and $\ce{Ca(OH)_2}$ are only partially dissociated in water. A significant fraction of these substances produce ions like $\ce{[Mg(OH)]^+}$ and $\ce{[Ca(OH)]^+}$ in solution. There also exists ion groups like  [$\ce{Mg(OH)_2}$] or [$\ce{Ca(OH)_2}$] in solution, as if they were not formed of ions. This can be shown by the measurement of the experimental solubility of these hydroxides, which is much higher than the result of the calculations based on the corresponding solubility products.

@Maurice Thanks for the additional info. I am aware that the metal hydroxides I have mentioned are only sparingly soluble in water.

Not sure I completely agree with @maurice.  The solubility of magnesium hydroxide may be low enough so the saturated solution is close to full dissociation because of dilution (cf. 0.001 M sodium bisulfate).  You're more likely to see partial dissociation with heavier alkaline earth hydroxides as the increase in solubility outruns the increase in dissociation constant.

Whether magnesium hydroxide is a strong or weak base is a matter of disagreement. Usually it is considered a weak base because of its low solubility.
Whether it's a strong or weak base in terms of dissociation is not completely known.  IUPAC reports a single $\mathrm{p}K_\mathrm{b}$ value, generally between 2 and 3, based on hydrolysis measurements. But because the hydrolysis is dominated by the reaction
$$\ce{Mg^{2+} + H2O <=> H^+(aq) + MgOH^+},$$
this says that only the second dissociation constant is (slightly) weak.
The first dissociation constant could be stronger, but magnesium hydroxide has very little solubility in solutions that would be sufficiently alkaline (probably $\mathrm{pH} > 12)$ to equilibrate the first dissociation of the hydroxide.  Both dissociation constants are known for calcium hydroxide and heavier alkaline earth hydroxides where their binary aqueous solutions provide sufficiently alkaline $\mathrm{pH}.$
Magnesium hydroxide is, however, much stronger and more soluble than most transition metal hydroxides. This answer describes how magnesium hydroxide with its superior solubility displaces iron almost quantitatively from aqueous solution, yet an excess does not dissolve into the water.
This self-regulating property makes magnesium hydroxide (or the oxide which becomes the hydroxide in situ) and the similarly behaving calcium hydroxide especially suitable for water and soil treatment applications. A fully strong base would risk making a caustic solution.

If you're wondering if it's a strong base or weak base, just titrate it and see what happens...

Let's look at it in reverse. Were magnesium hydroxide a weak base, solutions of magnesium ion in water would hydrolyze in water removing hydroxide ions and making solutions of magnesium salts such as Epsom salts [magnesium sulfate], or magnesium chloride acidic. Quick anecdotal search says these salts give neutral solutions in water; this indicates that magnesium hydroxide is a strong base. If someone is truly interested some actual research in the literature or in the lab with some salts, DI water, a GOOD pH meter and good technique can answer this without all the hand waving.


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Manganese oxide reacts with tin(IV) hydroxide?
I have a problem on my chemistry summer homework in which it asks me to predict the product of “solid Manganese-Oxide when it reacts with tin(IV)-hydroxide in aqueous solution”. My issue is that there are multiple manganese oxides (https://en.wikipedia.org/wiki/Manganese_oxide) And that, normally,  neither hydroxide nor tin are water soluble.  How do I solve this?  I am allowed to answer that there will be no reaction, but how do I prove this, and how am I sure?  This is for A.P. Chemistry summer homework.

You guessed it right, it is a vague question. You can mention it is neither can acid-base reaction nor a redox reaction.

You may have already constructed an answer, it does not react under ordinary conditions. Now, there may be an issue with the question where reagents required, reaction conditions,..., have been omitted, so qualify your answer.

@M.Farooq: Also I'm not sure $\ce{Sn(OH)4}$ exissts or not. I think OP should look into that too. Any suggestion?


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Is it safe to launder clothing that had a small amount of isopropyl alcohol on it?
I just have a safety question.  I work in a clerical position at an outpatient office located in a hospital.  Since mid-March when the pandemic began to hit our state, we’ve been wiping things down in our work area, but regular cleaning products are in short supply, so each morning until recently, I'd used a bottle of 70% isopropyl alcohol to wipe the desk, keyboard, phone, and I also sprayed my fabric chair with it.  Not thinking, I’d also used the small alcohol pads on occasion to wipe off small areas of my clothing or the bottom of my shoes, and have used it to wipe off my car seat, steering wheel and floor mat a handful of times, too.  Once I mistakenly washed a load of laundry that contained a sealed alcohol pad hidden in a pocket; the package did not seem to leak into the load as when I discovered it and opened the package, the outside packaging did not smell of alcohol but the intact inside smelled strongly as is typical.  At this point, do I need to discard clothes (which have been washed multiple times already), my desk chair (which I've wiped down instead with soap and water in recent weeks) and car seats/floor mats (again, I've wiped these with soap and water)?  Would any clothing which I'd slightly wiped with an alcohol pad (small area) affect my laundry machines or loads of clothing that were washed after that load?  Thanks so much for any thoughts!

First, this is not the forum for medical advice. Second, why would you be concerned about ordinary "rubbing" alcohol?

Isopropyl alcohol is volatile.   It would quickly evaporate from any clothing to which you've applied it.  By the time it goes in the washer, the alcohol should have long ago dissipated.

@DrMoishe Pippik It doesn't sound as if they want medical advice, just handling/disposal advice. The OP probably read up on the stuff, and all the dire warnings about spillage handling such as recommendations to evacuate everyone in an 800m radius of the spill in the standard MSDS scared them.  While rubbing alcohol isn't really an issue in small amounts, municipalities also really don't want it in the sewage works, which is why in many countries factory workers' overalls have to be laundered on site in case of contamination.

I would try to avoid in general mixing of organic compounds with chlorine bleach. In small amounts, no immediate issue, but possible long term exposure to chloro-organics is likely not healthy.

As noted isopropyl alcohol evaporates very quickly. It is also completely miscible with water so washing clothes would effectively remove it completely from the wet clothes making the clothes safe to put in the dryer. // The only thing that would make me very nervous would be using a spray bottle of the solution around any sort of ignition source (cigarette, space heater etc.) since isopropyl alcohol is flammable.

Always consult the Material Safety Data Sheet of any substance if you are unsure about handling.
In the case of isopropyl alcohol and the 70% aqueous solution, unless you can still smell it on the clothing or other fabric it should be fine. It is not dangerous on contact with skin although it tends to dry out your skin with frequent use (e.g. in hand sanitizers). With larger spills, you would have flammability worries and of course fish and other aquatic life don't like it and neither does the bio-processes at the sewage works, but a swab or a few drops from a spray bottle is not going to do much harm.
You should be more worried about using it inside your car before driving than about small spills on clothes.  The stuff will evaporate reasonably fast, so if you air out the vehicle well before driving you should be fine, however if you use it inside a closed up vehicle (or other confined space) the vapour may irritate your eyes and upper respiratory tract and could contribute to drowsiness.

In the US at least Material Safety Data Sheets (MSDS) have been replaced with  Safety Data Sheets (SDS).

I want to point out that SDS most of the time are so that a person without chemical or technical knowledge will dismiss the use of many things, perhaps even NaCl. This state, I also promote the use of the maximum precautions.