text:
Is the %V/V composition defined for homogeneous mixtures?
Why do major websites such as NASA and Wikipedia discuss the composition of the earth's atmosphere in terms of percentage by volume instead of percentage by amount of substance? Isn't the percentage composition by volume undefined since all gases in the atmosphere occupy 100% volume of it?

It's easier for me to understand there is 78% by volume of $\ce{N2}$ in the atmosphere rather than in molar fraction because every constituants have different molar mass. And also everybody (even those who never studied chemistry) is able to understand what is a volume fraction in an easy way. And I would add but not sure and no possible for me to do the calculation right now but the volume fraction of gases will be much less affected by a variation of temperature rather than the molar fractions. (but not sure) just idea.

Percentage composition by volume will assume ideal gas behavior. Then for 78% by volume of $\ce{N2}$ 78 out of 100 molecules of gas will be $\ce{N2}$.

Let's be clear on what percent by volume means.  If I say component $X$ is 25% by volume of a mixture of $X$ and $Y$, with a total volume of 1 L, it does not mean there is some 250 mL portion of the mixture that is solely occupied by $X$.  That wouldn't be a mixture.  It doesn't even mean that if I mentally subdivide that 1 L mixture into tiny identical cubes, one for each atom or molecule, then 1/4 of those cubes are occupied by atoms or molecules of $X$.
Instead, all it means is that I can prepare 1 L of the mixture by mixing some volume $V_X$ of $X$ and 3 times that volume ($V_Y = 3 V_X$) of Y.  There's not even any guarantee that $V_X = \pu{250 mL}$, since sometimes volumes are non-additive, that is, $V_X + V_Y \neq \pu{1 L}$.
Now as it happens, for ideal gases, which the atmosphere is close to being, you can mentally subdivide the volume into tiny little equal-sized cubes, one for each atom or molecule in the mixture, and on average there would be one atom/molecule of $X$ in each fourth cube if $X$ is 25% by volume of the mixture.  (I emphasize "on average" because of course since the gas atoms or molecules are continuously moving around randomly, they will bunch up and spread out momentarily all the time, so at any given instant there may be zero or many more than one $X$ atom/molecule per tiny cube.)  Real mixtures in the liquid state often have at least some nonideality, e.g. even in a dilute solution of $\ce{NaCl}$ in water, you will find "structure" in the solution, with the $\ce{Na+}$ cations and $\ce{Cl-}$ anions surrounded by a fairly fixed arrangement of $\ce{H2O}$ molecules, almost like a tiny piece of clathrate, so a mental arrangement of tiny boxes is an even less accurate atomic-scale description of the solution.
In short, the percent by volume description of a mixture is only a way to characterize the amount of material that goes into it, it is not intended as any kind of implication of what the  mixture looks like at the atomic scale.
As for why NASA reports the composition of the atmosphere by percent by volume: probably because it's closest to the actual experiments done to measure it.  You would typically measure the composition of a gas mixture by physically separating it (e.g. by lowering the temperature until each gas liquefied) and then measuring the volume of each component.  You certainly could from those measurements easily calculate a percent by moles, but – why?  You introduce a calculational step between the measurement and what you report, and scientists tend to prefer getting original data, right what comes from the instrument, if possible, as a way of avoiding even the smallest risk of some error introduced in calculation.

Yes, the composition by volume is defined for mixtures.
According to ISO 80000 Quantities and units — Part 9: Physical chemistry and molecular physics, the volume fraction of substance $\mathrm{B}$ is defined as
$$\varphi_\mathrm B=\frac{x_\mathrm BV_\mathrm{m,B}^*}{\sum\limits_i{{x_i}V_{\mathrm m,i}^*}}$$
where
$V_\mathrm{m,B}^*$ is the molar volume of the pure substances at the same temperature and pressure and
$x_i$ denotes the amount-of-substance fraction of substance $i$.
The IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book)  uses the simple but similar definition
$$\phi_\mathrm B=\frac{V_\mathrm B}{\sum\limits_i{V_i}}$$
where
$V_\mathrm B$ and $V_i$ are the volumes of appropriate components prior to mixing.
IUPAC also notes that other definitions are possible and that the term should not be used in accurate work without spelling out the definition.


text:
Why do most carboxylic acids have high pKa (~5) in spite of having a conjugate base ion that is stabilized by resonance?
This is from my textbook:

Carboxylic acids owe their acidity ($\mathrm pK_\mathrm a$ of about $5$) to the resonance-stabilized carboxylate anions formed by deprotonation.

Why are they such weak acids? If the anions are resonance-stabilized, I would expect them to be much stronger acids.
Stated another way: this screenshot of the textbook page shows a representative protonation/deprotonation equilibrium for a carboxylic acid:

Why is the reaction favoured in the reverse direction if the carboxylate ion is more stable than the reactant?

Resonance is not all that important in the grand scheme of things. If you really want to focus on just one factor, it should be the H-X bond strength.

Note also that you're asking about the aqueous stability of the anion. Now, solvation spheres and dipole effects and dielectric effects are all in play and make this a very complicated problem.

About 4-5 is typical pKa for acid with charge of 1 delocalised over 2 oxygen atoms strong acids need need delocalisation over 3 O atoms.

Everything is relative in acid-base reactions; the equilibrium will favor the side with the weaker acid. A carboxylic acid with $\ce{pKa = 5}$ will not protonate $\ce{H2O}$ significantly because $\ce{pKa(H3O+) = -0.7}$, but it will protonate something more basic like $\ce{NH3}$ because $\ce{pKa(NH4+) = 9.24}$.
Note: the pKa of $\ce{H3O+}$ is debated, see What is the pKa of the hydronium, or oxonium, ion (H3O+)? for more discussion.

Carboxylate anions are certainly not more stable than the parent acid.  That's exactly what the pKa tells you: the anion is about 10^5 times less stable than the parent acid.  The point of resonance stabilization is just that the factor is only 10^5 and not 10^15 which is what you'd expect for generically ripping an H+ off of an organic OH group.

reason for more stability of parent acids?plz explain ....also the bond OH in acids is polar so it can easily diassociate H in H2O ...

In the parent acid the presence of the H nucleus allows the electrons in the O-H bond to spread out, reducing their kinetic energy, without sacrificing any potential energy, because when they draw away from the O nucleus they come close to the H nucleus.    In the anion, these electrons have to stay closer to the O nucleus to enjoy the same low potential energy, and the greater degree of confinement means their kinetic energy is higher.  You can think of it as the intramolecular version of resonance stabilization.

I mean, you're just asking in effect why H2O is more stable than OH- and H+.  Because chemical bonds work!  Electrons have lower energy when they're in a bond than when they're confined to the atoms that form the bond.

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text:
is dental acrylic dangerous when heated?
I work in a research lab where we use small screws to attach onto hard objects and secure it with dental acrylic, to maintain durability. The screws are expensive, so when we are done with the particular project, we use a soldering iron to melt through the acrylic to get to the screws. I am concerned that when dental acrylic is heated, does it release toxins that can be harmful to my health?

VTCers, how is this a personal medical question?

It should be done in a hood, a device to extract fumes, whether the fumes are dangerous or just an annoyance.

What do you mean by acrylic, if it is PMMA. Then be aware that when it is heated that it cracks with unusual ease into the monomer (MMA) which is a strong irritant.
I have (in a fume hood) taken lumps of PMMA and heated them in a test tube with a bunsen, if you distill the product out and then redistill it you can get very pure methyl methacrylate. I have found that if you heat some of this MMA with some AIBN then it will then reform the PMMA.

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text:
Penny turns green with vinegar
What is the chemical explanation for pennies in vinegar turning green? What is the explanation at a macroscopic and atom-molecular level?
The procedure is:

Fold a paper towel so that it fits inside a dish.
Place a penny on the paper towel.
Pour vinegar over it so that the paper towel is fully saturated.

Experiment

Related: 1, 2, 3

This question may explain about what happens with the pennies at a microscopic level: What is the product of a copper and vinegar reaction?
Also, in his paper, Stephen DeMeo presents the electrochemical perspective:

The mechanism that has been proposed involves the formation of a free radical and an electrochemical corrosion process (1). [...], suffice it to say that the mechanism does involve formation of a strong oxidizer, peroxoacetic acid ($\ce{CH3COOOH}$), from dissolved oxygen, as well as the production of a copper oxide and a copper ion.

He also explains why the reaction just happens in the presence of oxygen:

Since copper is below hydrogen in the electrochemical series, copper is reluctant to displace hydrogen ions from electrolytes.  In terms of standard electrode potential, copper has a potential of 0.34 volts ($\ce{Cu -> Cu^{2+} + 2e-}$), while hydrogen has a greater potential of 0.00 volts ($\ce{2 H+ + 2e- -> H2}$). The electromotive force of the cell would be negative and therefore the reaction would be nonspontaneous under standard conditions. When oxygen is involved in the reaction between copper and acetic acid, the electrochemical series and the standard electrode potentials cannot be used as a guide to determine if the reaction is spontaneous.  A generalized relationship between copper metal and acids is nicely described by Massey: "Although copper metal will dissolve readily in an oxidizing acid such as nitric acid it will not dissolve in a nonoxidizing acid unless either an oxidizing agent or a suitable complexing agent is present also" (6).

Reference: DeMeo, S. Does Copper Metal React with Acetic Acid? J. Chem. Educ. 1997, 74, 844-846. DOI: 10.1021/ed074p844

You are probably making some combination of copper(II) acetate and basic copper carbonate (Cu2(OH)2CO3), both of which have a nice green color and are present in the classical painter's pigment verdigris (the name comes from "vert de Grece" in old French, "green of Greece"). The reaction is probably initially between the CuO "tarnish" on the surface of the penny and the acid, but may also involve Cu itself once the CuO gets eaten up in some locations.  If you start googling "verdigris" and "copper carbonate" you'll find lots of descriptions of the reactions, some of which have interesting historical roots.

This is a quote from your link:

A chemical reaction has occurred! (A chemical reaction is the combination of two reactants to form something entirely new.) A penny is made of copper. The vinegar on the paper towel helps the copper in the penny easily react with the oxygen in the air to form a blue-green colored compound called malachite.

If you want to know more, I would search Google for "copper tarnish", "malachite", "statue of liberty color", "green penny" or things like that.
Your link shows up first in the "green penny" search, but don't just look at the first one! The statue of liberty has enough copper to make 30 million pennies??? Really??? That is so cool.
Just look around. You will find amazing and interesting things.

Hi Skeptycal, welcome to Chemistry! Take the tour to get familiar with this site. This doesn't really answer the question.

Thanks, I will. I suppose I took the context to be a question based on an article for "kids" so I gave a "kid level" response.


text:
How must elemental symbols be ordered in the formula for a compound?
An online problem asked me to find the molecular formula of a certain substance given its mass composition. I found that it was $\ce{N_5P_5Cl_{10}}$. It was incorrect and told me the ordering of my symbols was wrong. I switched it to $\ce{N_5Cl_{10}P_5}$ because that's the order that they appear on the periodic table. It marked me wrong again and told me that the answer was $\ce{P_5N_5Cl_{10}}$, but didn't tell me why it must be ordered this way. Is there a convention that must be followed here?

I would say that if there were to be a convention, it should be C, then H, then followed by everything else in alphabetical order, with single-letter elements coming before double-letter elements (i.e. N before Na). Now, your supposed answer does not quite follow that either, so this seems to simply be a case of the inflexible marking schemes typically associated with online worksheets.

Related: Why do we write NH3?

Ooh I totally forgot about the red book...

Just listing all symbols alphabetically works too.  See the JANAF tables.

Yes there is.  The most general rule is that elements are listed in increasing order of electronegativity, and in your case the Pauling ENs are 2.19 (P), 3.04 (N) and 3.16 (Cl), so that's why the application wanted it that way.
Most of the time we approximate that rule by saying elements are listed in increasing order of column of the Periodic Table, so e.g. OF2, CO2, and BF3, and when elements appear in the same column, in decreasing order of period, so e.g. SiC, BrCl and SO3.  The column rule takes precedence over the row rule, because EN varies more strongly left-to-right than up-and-down, generally.
The place where this approximation fails most obviously is with hydrides, since H has an EN close to C but appears at the far left of the Table.  Normally we skip the approximation and mostly use the underlying rule for hydrides, so e.g. BeH2 and GeH4 even though H is above and to the left of Be and Ge  (Pauling EN is 1.57 for Be, 2.01 for Ge, 2.2 for H), but H2O and HF. However CH4 (Pauling En 2.55 for C) doesn't quite follow that rule and NH3 (Pauling EN 3.04 for N) is wildly contradictory.  These may just be historical accidents.


text:
How many molecules are there in 17.53 grams of NH4ClO4?
How many molecules are there in 17.53 grams of $\ce{NH4ClO4}$?
I get to 1 mol = 118 then I get stuck.

LOL - Technically none. There is no such thing as a  "molecule" in a salt. There is not a unique to assign a particular $\ce{NH4^+}$ group to a particular $\ce{ClO4^-}$ group in the 3D crystal matrix

How about not technically, trying to help teach my daughter and it's been a long time

The site rules state that we're not supposed to do "homework" problems, but I'm a sucker for someone trying to help a daughter...

The question you need to ask with problems like this is molecules of what? If ammonium perchlorate were a molecular compound this would be simple: the answer would be the moles of the compound. But it is an ionic salt with two components. Unless the question is specific about what it wants you to count the question is ambiguous. The answer could be the umber of moles of ammonium perchlorate or it could be twice that (counting the ammonium and perchlorate ions separately).

Ammonium perchlorate has a molecular weight of 117.485 grams/mole. Since the mass was given to 4 significant figures you should carry at least that number in the molecular weight. I'd use all 6 and round the final result to 4 significant figures.
Moles Ammonium perchlorate = $\dfrac{17.53}{117.485}$
To six significant figures Avogadro's constant is $6.02214\times10^{23}$
$$\begin{align}
\text{Number of molecules} &= \dfrac{17.53}{117.485}\times 6.02214\times10^{23}\\ &= 8.9856674639315657317955483678768\times10^{22}\\ &= 8.986\times10^{22}\text{(rounding to 4 sign figures)}\\
\end{align}$$
Note that $\dfrac{17.53}{117.5}\times 6.022\times10^{23} = 8.984\times10^{22}$ hence the reason to carry 2 extra significant figures in the intermediate calculations. Significant figures are a crude way to doing error propagation. Not exact, but such use keeps you from stupidly assuming that all the figures in $8.9856674639315657317955483678768\times10^{22}$ are significant.


text:
Why does C=O have a larger dipole moment than C-O?
Why does $\ce{C=O}$ have a larger dipole moment than $\ce{C-O}$?
According to me, dipole moment directly depends upon bond length and electronegativity difference. In $\ce{C=O}$ and $\ce{C-O}$, (I guess, please clear if I am wrong) the EN difference is same, and bond length of $\ce{C=O}$ is less than $\ce{C-O}$. So, $\ce{C=O}$ must have less dipole moment than $\ce{C-O}$ but it is not so. Why?

Intuitively, you can also see that in the $\ce{C=O}$ bond, you "share" two electrons between $C$ and $O$, and they both contribute, while in a $\ce{C-O}$ bon, you only share one. The increase of distance is not enough to compensate. Of course the correct interpretation is more akin to resonance, so be careful with the electron sharing images

Always remember that EN is a pretty heuristic concept. It is good for getting a rough idea how some element will behave chemically, but one should never expect a phenomenon to depend only on bond lengths and EN values.

According to Wikipedia, bond dipole moment depends on:

Distance between atoms and
Overall charge difference, not just electronegativity difference.

Resonance tells us that there is some amount of charge separation in $\ce{C=O}$ bonds because of the $\ce{C+-O-}$ contributor. This difference in charge, in addition to the electronegativity difference, is more significant than the decrease in distance between atoms, hence the larger dipole moment for $\ce{C=O}$.


text:
How would 2-(chloromethyl)-1-methylpyrrolidine recyclize into 3-chloro-1-methylpiperidine with only water?

My Attempt

I always get thrown off by mechanisms with just water as a solvent. I've attached my work as far as I could get but I'm not sure where to go from there. Thank you!

Not sure but after you got the acyclic amine. The nitrogen  could attack the $\ce{C-Cl}$ via $\ce{S_N2}$. The liberated $\ce{Cl-}$ would in turn attack the $\ce{H2O+}$.

Forget about the water present - its just a solvent.
The best nucleophile present is the N-lone pair; if you think about it in 3-D you will see that it can do a nucleophile attack on the CH2-Cl to displace Cl- and form the aziridinium cation. Cl- can then attack this at either carbon of the [3] ring, one attack merely reforms starting material, the other gives the 3-Cl-piperidine product shown.


text:
Will there be change in electronegativity difference in C-O and C=O?
Is there any change in EN difference in C-O and C=O. If yes, why?
Does EN difference change if its bond is changed to single, double, or triple?

" its bond" that of the C-O or C=O?

s - character is proportional to electronegativity. In single, double and triple bonds, there is difference in s - character (triple bond > double > single) due to the participating hybridised orbitals of C, and hence difference in electronegativity.

I think that you didn't understand the answer to this question.
A dipole is created when two equal (only in magnitude) and opposite charges are separated by some distance. The dipole moment depends on the magnitude of the charges and the distance between them. Electronegativity is just a concept that helps us to get an idea of the magnitude of the charges. So, in order to find the dipole moment, you should focus on the magnitude of the two charges rather than the electronegativity difference.
Look at the hybridization of the atoms in the single and double bond. That will tell you that both the atoms should be more electronegative after forming a double bond, but won't help you comment on the electronegativity difference.

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


text:
Two reactions that catalyze each other
Is there an example of two reactions that catalyze one another? More than two is also helpful, but I guess it will be hard to find one.
By "catalyze one another", I mean that one of the products of the first reaction acts as a catalyst for the second reaction.

Reactions do not catalyze reactions.

Could we please reopen this question? It seems pretty clear what the OP is asking.

I agree.  There has to be more than the magnesium nitride/boron nitride thing I stumbled upon a while back.

An early synthesis of borazon from hexagonal boron nitride would be one possible example.  When other approaches did not work, Robert H. Wentorf Jr. added magnesium and voila!  The magnesium had reacted with some of the hexagonal nitride to form $\ce{Mg3N2}$, and the latter then catalyzed the hexagonal to cubic conversion.  (Today other methods have taken over.)


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Why does the C-C bond break earlier than the C-F bond in the pyrolysis of perfluroethane?
According to me $\ce{C-F}$ bond is highly polar and thus is a weaker bond as compared to $\ce{C-C}$. Hence, $\ce{C-F}$ should break first in the pyrolysis of perfluroethane. Instead, in J. D. Lee, it is stated that $\ce{C-C}$ bond will break first. Why?

I think that the main reason is that the CF3 radical would be delocalised (hyperconjugation?). Another reason could be stearic repulsion.

What makes you think the C-F bond is weaker?  In the tables of bond enthalpies I've seen, the C-F bond is considerably stronger than the C-C bond.

@Christopher Grayce i had misconception that high polarity always make a bond weak but now i got cleared that it can even make a bond strong by making bond ionic......

I would be cautious about the relationship between bond polarity and bond strength.   A bond can be strong and nonpolar (N2), strong and polar (C-F), weak and nonpolar (I2), or weak and polar (HI).  There's definitely correlation, but it isn't perfect, and should  be used with caution.  After all, the two phenomena arise from related but not identical causes.

I got this explanation from one of my seniors.
It is true that $\ce{C-F}$ bond is highly polar as fluorine is more electronegative than carbon. In fact it is this very difference in electronegativity that makes C-F bond very strong. This is because, in addition to the covalent bond strength between carbon and fluorine, there is an additional electrostatic attraction between positive carbon and negative fluorine.
We can think of it has the $\ce{C-F}$ bond has both covalent  and ionic character. This makes C-F bond extremely stable and hard to break.
In fact in organic chemistry we usually say that the $\ce{C-F}$ bond is not broken in reactions.

We can also verify this from their bond energies:
$\ce{C-F}$: $\pu{544 kJ mol-1}$
$\ce{C-C}$: $\pu{346 kJ mol-1}$

Another way to think about the difference in bond energies is to think about the stability of radical species formed after bond homolysis, which is likely the process occurring during pyrolysis. The general trend is that free radical stability decreases with increasing electronegativity. See: https://www.masterorganicchemistry.com/2013/08/14/bond-strengths-radical-stability/electronegativity.

Lemal mentions that the charge on C is +0.76 while that on F is -0.19. This shows how large the separation of charge actually is and how strong the ionic contribution to the bond is. Lemal, D. M. (2004). Perspective on Fluorocarbon Chemistry [Abstract]. The Journal of Organic Chemistry,69(1), 1-11. doi:10.1021/jo0302556

One more reason is that-----Even CF being so polar bond but due to very small bond length and very small F size  IT IS MORE STRONG THAN CC BOND


text:
Stoichiometry yielding negative volume

$15~ \textrm{mL}$ of $4 ~\textrm{N}$ hydrochloric acid, $ 25~ \textrm{mL}$ of $2~ \textrm{M}$ $\textrm{H}_3\textrm{PO}_3$ and a certain volume of $2.5~\textrm{M}$ sulphuric acid are mixed together and made upto $2 ~\textrm{L}$.
$30~ \textrm{mL}$ of this acidic mixture exactly neutralises $42.9~\textrm{mL}$ of a sodium carbonate solution which contains $0.1~ \textrm{g}$ of sodium carbonate decahydrate in $10 ~\textrm{mL}$ of solution.
The volume of $\textrm{H}_2\textrm{SO}_4$ added is?

I calculated the number of gram equivalents of sodium carbonate and thus calculated the normality of the acidic solution required.
I am having trouble with the statement "is made up to", so I'm assuming that this means that the solution is diluted with water to make a volume of 2 litres.
This seems to be yielding a negative volume. Can you tell me what I'm doing wrong?

The answer will be 8 ml. Deal carefully with molarity & normality ; moles and equivalents. Also observe which are dibasic acid and the sodium carbonate is diacidic base. You will get the result.


text:
Why was the unit micrometers spelled backwards in old research papers?
I have recently been doing a literature research project on polymers and have been reading a lot of papers from the 1950s and 1960s (mostly in American journals from American authors). I saw this intriguing text:
 "the spectra was taken using light of $\pu{300 m\mu}$ through $\pu{600 m\mu}$. I think they mean micro meters. I would write that as $\pu{\mu m}$, not $\pu{m \mu}$.
I've seen this in more than one paper, but they're all at least forty years old. Was this some standard back then? Most of them are in the same journal (Journal of Polymer Science A), maybe that has something to do with it.

Oh my young padawan. Millimicrons they are. // millimicron. (mĭl′ə-mī′krŏn) n. A unit of length that is equal to one thousandth ($10^{-3}$) of a micrometer or one billionth ($10^{-9}$) of a meter

Oh, I guess that's a derp on my part. And it also makes sense, because nm light (which would be milimicron in that parlance, I guess) is much more relevant than $\mu$m light.
It still seems a bit oudated to use terms like that, but I guess that's what happens when you read a lot of older papers

Unfortunately when terminology changes there isn't any magical way to update old papers with the new terminology. So old papers use the terminology that was in vogue at the time.

Of course, I'm not expecting a revision or anything based on that. Merely commenting.
(I especially am not expecting it to be revised because I had to find this article on actual real paper instead of online, the horror)

Remarkably, the unit micron (symbol μ) was a valid SI unit from 1948 until 1967/68.

Similarly, electronics is littered with μFd, μμFd (and mFd and mmFd, when typesetters couldn't be bothered finding a Greek font), for microfarad and micromicrofarad. This is particularly confusing where the m- abbreviation means milli-. sigh


text:
Chemistry: A Chemical Dilemma
We accept that, via observations that an electron possesses a negative charge and a proton a positive charge, both of which attracting each another. My dilemma lies here, what is the reason for the force of attraction or repulsion? If it may be due to the desire to become more stable and is logically comprehensible via mathematics of potential energy – but why does a body seek the desire to become more stable at all in the first place?
Though this question may be silly, and some might say this goes into metaphysics, but does anyone know why, or do we assume that everything – or the universe to be more precise, acts as it is, accepting it as a fact, and construct our calculations and laws above it? Are all chemical and physical knowledge based on taking the mechanics of the universe for granted?
The question was bothering me for sometime and finally decided to ask in search of an answer.

I'm voting to close this question as off-topic because it is closer to metaphysics, and is better suited for Philosophy.SE.

Electromagnetism just is. Its a fundamental interaction: https://en.wikipedia.org/wiki/Fundamental_interaction You've reached the bottom of why in terms of chemistry. Why is metaphysics, not chemistry/physics.

@pentavalentcarbon Possibly Physics.SE instead, but yeah, definitely not on our site.

You have it a little backwards.  We don't observe charge, and then infer the existence of a force.  It's the other way around.  We observe forces between objects -- they repel or attract each other -- and then build models which explain which objects are attracted or repelled by which other objects, and by how much.  That leads us to invent the idea of charge and assign it to objects, and later still to connect charge to other properties of the objects.
If your question is why is there an electromagnetic force at all, the answer is probably that it is easier to imagine a universe in which there are forces between degrees of freedom than one in which there are not.  Our current understanding of the universe is that can best be described as a set of fields, like the surface of the sea, and everything we see is some kind of ripple or wave in that field.  It's easy to see why ripples on the surface of the ocean should interact with each other -- exert "forces" of some kind on each other -- and hard to see how they could just totally ignore each other.  That's how it is with nature.
Granted, it looks like there exist wholly separate bodies (particles, objects, things) and we wonder why they should interact with similar things far away.  But that may be because our perception is severely limited, and the underlying fields are much more complex than the surface of an ocean.  If you want to explore the topic further, start by googling "wave-particle duality in quantum mechanics."

First of all, I encourage you too see this short video, where Richard Feynmann answers, partially, your first question.
Secondly, as other answer said, charge is not observed. I would add forces aren't observed also; we believe we see forces because we took that framework as true.
It is even better to forget truth. Science works or it doesn't. So, some theories work, some doesn't. Some explanations work, some doesn't. What we might call true, I suppose, are the facts, but not its explanations. So, even scientific laws aren't true, but they work.
Interestingly, scientific explanations and theories can correct itself, that's called falsfiability, and it's possibly the best characteristic of science.
Why we use so much scientific method? It may have to do with evolution and fear of dead... But, how would you investigate it? Oops, we consider this method as the only possible. Are there more methods? Are there more ways to make sense of the world? I don't know.
We born in some world partially made and accept lot of frameworks, one of those is this point of view of science, where we create even ways to measure things.
In your example, if you accept the explanation that when something accelerates there is a force, and you measure it, you'll find something like Coulomb's Law. Then if you ask why again, we can talk about molecules, but you can continue asking why forever. Finally, if the answerer is honest, will say: 'I don't know'. That's in some sense the field.
Why things want to diminish its energy? Take your facts and look for the cause, if there is a way, or read, or learn to convive with that uncertainty of knowledge.
That's all I can say at least. Forget truth, look how it works.


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Are SN2 reactions in equilibrium?
I'm learning about SN2-reactions, and my text book gives the following examples of these reactions:

$$\ce{HO^- +CH_3Br -> CH_3OH + Br^-}\tag{1}$$
  $$\ce{I^- +CH_3Br -> CH_3I + Br^-}\tag{2}$$

Later I learned that $\ce{OH^-}$ is a worse leaving group than $\ce{Br^-}$, and that $\ce{I^-}$ is better than $\ce{Br^-}$. I can understand the reaction (1) is favored, since the leaving group gets worse (and therefore $\ce{CH_3OH}$ is more stable than $\ce{CH_3Br}$), but I cannot understand the reaction (2). Wouldn't the reverse reaction of (2) be favored instead? Or is (2) in fact an equilibrium?
I'm confused because $\ce{Br^-}$ seems like a better nucleophile than $\ce{I^-}$, which would favor the reverse reaction of (2)?

Equilibrium or not depends on the rate of interconversion. In the latter case, it's probably reasonably fast.

@Zhe So the reverse of $(2)$ is not more favored than $(2)$?

It also depends on the solvent. In the case of iodide the reaction is often done in a solvent that the bromide salt is insoluble in.

2nd Reaction is favoured in forward direction by using $\ce {NaI } $ in Acetone ( Finkelstein reaction )

When we talk about Sn1 or Sn2 reactions (and any other reaction to be honest), not only the reactants are crucial to describe what's going on but also the solvent you used. It is perfectly possible that the second reaction will be favored in the direction shown, as long as you use the right solvent for this.
There are two types of solvents one can use in an Sn2 reaction: polar protic (which has a reasonably acidic hydrogen in its structure) and polar aprotic (has a net dipole moment but no acidic hydrogens).
Examples of polar protic solvents: Water, methanol, ethanol, ammonia etc

Examples of polar aprotic solvents: DMSO, acetonitrile, acetone etc

So what does the solvent has to do with the reaction?

Nucleophiles are negatively charged species or dipoles, they will attack a sufficiently positive center (usually a carbon atom connected to an electron withdrawing group)
The solvent is commonly present in significant excess compared to the
reactants.

When you use a polar protic solvent, such as methanol or water, the nucleophile species (ions such as Cl- or molecules such as ethanol) will be surrounded by an overwhelming number of solvent molecules that will interact strongly with them by dipole-dipole interactions or hydrogen bonding. This creates a barrier around the nucleophile, making it considerably more difficult to approach the positive center and consequently slows the reaction.

Smaller nucleophiles (such as F-) will be surrounded by a bigger number of solvent molecules than bigger nucleophiles (such as I-), so this surrounding effect is stronger in smaller species.
If you consider only the nucleophile, Br- is stronger than I- since it has a smaller ionic radius and, therefore, has a more 'concentrated' electron cloud (less polarizable). If no solvent effect was considered, the reaction as shown in your textbook would be incorrect since the equilibrium would be towards the formation of methyl bromide.

HOWEVER, if you use a polar protic solvent, like methanol, the order of nucleophilicity is inverted (because of that surrounding barrier of solvent molecules) and I- becomes a better nucleophile than Br-.

So, to be consistent and correct, your textbook should also provide the solvent in which this reaction is done. If it is a polar protic one then it is correct.
(I took these pictures from master organic chemistry, they have a good text talking about it here: https://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/ )


text:
How to determine if an unknown is potassium chloride or potassium bromate?
In class, we were given unknown substances. I've narrowed the possible substances down to either potassium chloride or potassium bromate.
We were given several acids, indicators, and other chemicals (silver nitrate, calcium hydroxide, ammonium hydroxide) to test with. How could one tell them apart? Maybe with a precipitate reaction?

Do you actually mean bromate and not bromide? Check out the solubilities of those compounds themselves and see if that gets you somewhere.

Noting the solubilities of potassium chloride and potassium bromate gives a nice point in the identification but apart from it, there are certain chemical test specific for certain anion that gives pinpoint information about the salt.
First is the chromyl chloride test for chloride ions. Perform the test and you will identify potassium chloride (potassium bromate will give -ve result to the test). Second is conc. $\ce{H2SO4}$ test. Potassium chloride will gives vapors of hydrogen chloride. Potassium bromate will give bromine gas.
$$\ce{KCl + H2SO4 -> HCl ^ + KHSO4}$$
$$\ce{4KBrO3 + 2H2SO4 -> 2Br2 ^ + 2O2 ^ + 4K+ + 2SO4^{2-} + 2H2O}$$
Third test is the heat test. Potassium chloride is stable to heat while potassium bromate is prone to decomposition $\ce{\approx 370°C}$.
$$\ce{2KBrO3 ->[\Delta] 2KBr + 3O2 ^}$$

About the silver nitrate test pointed by @Dallon Penney, precipitation will occur in both cases albeit depending on the concentration of respective ions, it seems that the precipitate forming has different properties. Silver chloride exist as a white, curdy precipitate, insoluble in water and in dilute nitric acid while silver bromate exist as a white crystalline precipitate soluble in hot water and sparingly soluble in nitric acid. Both precipitates however are soluble in ammonia solution.

Solubility of Silver Bromate in water: 0.167 g/100 mL
Solubility of Silver Chloride in water: 0.00052 g/100 g at 50 °C.
Assuming room temperature conditions, if you have potassium chloride in your solution it will form a white precipitate when silver nitrate is added due to the formation of silver chloride:
Ag+(aq) + Cl-(aq) -> AgCl(s)
Unless the concentration of potassium bromate in your solution is over 0.167g/100 mL, adding silver nitrate will not produce any precipitate.

It was quite nice until the last paragraph: the mass concentration of KBrO3 does not actually matter all that much, because it is AgBrO3 precipitating out, not KBrO3. Also you have a mixture of ions. Solubility products work much better here than mass concentrations

Put the unknown on filter paper and heat it. Sodium bromate, $\ce{NaBrO3}$, is a strong oxidizer and it's effect in speeding combustion should be obvious. $\ce{KCl}$, on the other hand, helps keep air off the surface and retards burning.


text:
Why does the rate equation between hydrogen peroxide and iodide ions not follow the stoichiometry of the rate-determining step?
The ionic equation of the reaction between hydrogen peroxide and iodide ions can be written as $\ce{H2O2 +2I- + 2H+  -> 2H2O +I2}$ which is the rate-determining step as well.
From what I have learnt, the order of reaction with respect to each reactant in the rate equation should be the coefficients of the respective reactants in the rate determining step.
So that makes the rate equation $= K\ce{[H2O2][I-]^2[H+]^2}$
However, from an experiment conducted, the rate equation I have obtained is rate $= k\ce{[H2O2][I-]}$ instead. Why is it so?

It is extremely unlikely that the rate determining step involves 5 separate species...

Could you provide a source for the reaction rate $\ce{R = k[H2O2][I-]}$?

I would guess because your equation does not represent an elementary step in the reaction mechanism.  That's hardly surprising, because it represents the simultaneous collision of five objects -- the H2O2 molecule and the four ions.  That would be an extraordinarily rare occurrence, even in a very concentrated solution.
Your experiment is suggesting the rate-limiting elementary step involves a collision between the H2O2 molecule and the I- anion.  Those will happen all the time.


text:
Differences between LIBS over LA-ICP-MS?
Which differences should I consider between the techniques Laser-Induced Breakdown Spectroscopy and Laser Ablation Inductively Coupled Plasma Mass Spectrometry for a geochemistry fingerprint focused research?
Both use Laser Ablation for particle generation and analysis, so the sample preparation step is minimal in both instruments, but I would like to know about the instrument costs, the cuantification limits if apply and the advantages that offer each one.

I think this question might be a bit too broad. One thing that might narrow it down is knowing a bit more about the sort of geochemical materials you will be using these machines on. @CristhianParedes


text:
How does cyclic voltammetry in a electrolytic cell measure the precise oxidation potential?
From what I understand, the potential necessary to induce a current to flow in an electrolytic cell depends on both reduction potentials of the anode and cathode reactions, where E(cell) is comprised of these two potentials and determines the spontaneity of the reaction, and hence the current flow. Thus, in cyclic voltammetry in an electrolytic setup, wouldn't the anodic redox potential thus depend on the cathodic redox potential?
As in, the position of a peak potential at a certain point in a cyclic voltammogram would change depending on what is reduced at the cathode in an electrolytic cell, because a change in the redox potential of the cathode changes the required voltage to induce spontaneous current flow from the anode to the cathode.


text:
Measure of partial pressure on blood
Here is the issue of blood gases partial pressure.
On medical reports is common to see partial pressure of gas on blood, usually oxygen, but there is no gas in equilibrium with blood, so it is confusing. What they mean with partial pressure of a gas on blood is, according to this answer:

the partial pressure which would prevail in a gas phase in equilibrium
with the liquid [blood] at the same temperature.

I suppose they use a partially filled vessel with blood and a manometer or something like that, but I would like to know the way they measure it.
Question
How do they measure $p(\mathrm{O}_2)$?

In principle you could measure it by getting a sample of blood and putting it in contact with gaseous O2 at various pressures.   Whatever external pressure of O2 results in the amount of O2 dissolved in the blood neither increasing nor decreasing would equal p(O2) in the blood.
As for how it's actually measured, I believe the usual answer these days is spectroscopically.  That is, you shine light through a sample of blood and look for the absorption at certain frequencies which are typical of hemoglobin with O2 bound to it.  The amount of absorption is proportional to the concentration of oxygenated hemoglobin, and your software has some internal chart that converts that to partial pressure.
As for why it's measured that way, in part it's probably just tradition, and medicine changes very slowly and conservatively, for obvious reasons (you're experimenting with health and life).  In part, though, it may be that it directly relates to something important: the partial pressure of O2 in the atmosphere, which is about 160 mmHg.  That's the concentration of O2 in the lungs.  What fraction of that reaches the blood at various points in the circulation is a natural indicator of how well oxygen is being taken up, transported, and delivered.  That is, you can directly relate this particular measure of O2 concentration to the form of concentration usually used to describe a mixture of gases, in this case the air you breathe, which is the origin and destination of the blood gases.

I believe you and nice reasoning, but any resources about what you say on second paragraph?

BTW because of a higher concentratiom of water vapor on lungs oxygen partial pressure is 150 approx...

For the casual setting, like the ER, I believe you use one of those pulse oximeters.  Wikipedia has an article on those.  I looked around for what's done for serious blood gas measurement, and this article: https://www.ncbi.nlm.nih.gov/books/NBK371/ suggests it's actually done with electrochemistry, by reducing the dissolved O2 and measuring the current required.

I'm not sure about your idea of measuring O2 spectroscopically as it does not have an ir spectrum (no dipole) . Oxygen uptake is measured (in your finger) by using  visible (red)  light and the absorption change O2 binding makes to the electronic absorption spectrum of haemoglobin vs that without O2.

Oh excellent point, I made an edit to make that clear, thanks.


text:
What is SO2Ar as a functional group and how is it relevant in this reaction?

This reaction came up in one of my problem sessions and I had never seen an SO2Ar function group before. What are it's general properties and how is it relevant to this reaction? Any help is greatly appreciated.

$\ce{-SO2Ar}$ means a sulfonyl group with an aryl ring attached to the sulfur.
It is totally irrelevant to this reaction. Your first step is to add the methoxide into the carbonyl group. Then collapse the tetrahedral intermediate.

Sulfonamide (NHSO2R) is an acid isostere, the proton on the N is pretty acidic

As @Zhe has indicated the first step is attack of methoxide on the carbonyl group. The tetrahedral intermediate collapses and sulfonamide anion (ArSO2N-) leaves giving the methyl ester. The sulfonamide anion then attacks the C-Br intramolecularly forming the product shown. This works because sulfonamide NH is acidic (pKa ~ 16) and the anion is a pretty stable species.


text:
Search by CAS Registry name to obtain trackable accession: ie InChI?
I'm updating a database and trying to move from CAS Registry names to InChIKeys. I have about ~800 compounds to track. Is there a way to do this mappings programatically? I'm looking for a web service or API that already has this implemented, does something like it exist?

It might help.

Chemical Identifier Resolver can convert several identifiers and has an API so you can convert with a script.
A Python example for CAS to InChIKey:
from urllib.request import urlopen

def CIRconvert(cas):
    try:
        url = 'http://cactus.nci.nih.gov/chemical/structure/' + cas + '/stdinchikey'
        ans = urlopen(url).read().decode('utf8')
        return ans
    except:
        return cas

casno = ['50-78-2', '69-72-7', '95-1E-6', '108-24-7', '31710-30-2']

for casid in casno:
    print(CIRconvert(casid))

Output:
InChIKey=BSYNRYMUTXBXSQ-UHFFFAOYSA-N
InChIKey=YGSDEFSMJLZEOE-UHFFFAOYSA-N
95-1E-6
InChIKey=WFDIJRYMOXRFFG-UHFFFAOYSA-N
InChIKey=CIPFDHFTBYJKQB-UHFFFAOYSA-N

note that your code is syntactically wrong

@mykhal Just tested - code and service works.

It really works, sorry for the inconvenience. But I have edited the "buggy code".

@mykhal For future reference: code prettify

Chemical Identifier Resolver has a python API also: https://cirpy.readthedocs.io/en/latest/index.html
import cirpy
print(cirpy.resolve('107-13-1', 'stdinchikey'))

Output
InChIKey=NLHHRLWOUZZQLW-UHFFFAOYSA-N


text:
Lewis Acid/Base Chemistry
I'm looking for some clarification on Lewis Acid/Base Chemistry.  For the Acids do they have to accept a full pair of electrons to be considered a Lewis acid, or do they just have to accept an electron, and vice versa for Lewis Bases?

A Lewis acid as an electronic gap somewhere. A Lewis base as a lone pair somewhere. $\ce{AlCl3}$ is a Lewis acid with an electronic gap on the aluminum. $\ce{NH3}$ is a Lewis base with a  lone pair on the nitrogen. You'll find more details on wikipedia


text:
Liquid that burns in a vacuum
This may seem like a silly question, but is there a flammable liquid that could oxidise itself, so it could effectively burn in a vacuum?

Rocket propellants, which are generally mixtures of fuel and oxidiser

Nitroglycerine?

Indeed yes, and it's been used in all kinds of airless situations, including torpedos and rockets and I think even submarines.  It's hydrogen peroxide, which when exposed to a catalyst will oxidize and reduce itself:
2 H2O2(l) -> 2 H2O(g) + O2(g)
You'll notice that the oxygens on the left are in oxidation state -1.  Half of them become reduced to the O in water (oxidation state -2) and half become oxidized to the O in O2 (oxidation state 0).

thanks, could the concentration be 12% or less, as UK laws prohibit me from owning hydrogen peroxide  in a concentration greater than 12%?

Nope.  And those laws were probably passed specifically for the purpose of keeping rocket propellant out of your hands!  If you'll settle for vicarious living, I've heard this book https://www.amazon.com/Ignition-Informal-History-Liquid-Propellants-ebook/dp/B076838QS2/ is a whole lot of fun.


text:
Which white powder first "boils off" and then catches fire at high temperatures?
I'm a high school student doing a project where we need to identify chemicals and one of the substances had a very unique change when heated.
Before heating it was an extremely fine, flour-like white powder. I placed it on a hotplate at 350 degrees Celsius in its solid form and it "boiled off", I presume a hydrate evaporating. Before the hydrate appeared to evaporate, the powder had small pink dots appear all over it. This also happened when exposed to $\ce{NaClO}$ and $\ce{NaOH}$.
Once the temperature was raised to 400 degrees Celsius, the remaining solid turned yellow and then spontaneously caught fire, turning it all black before the fire ran out.
I had ran other tests on this substance, being insoluble in water, ethanol, and hexane. It also produced no apparent flame color.
My initial thought was that this compound was starch, however now that I think about it, it may be an insoluble magnesium compound, due to the lack of flame color? Just curious as to if anyone knows what substances display these properties when heated.
Here is a list of potential substances based off of a predetermined list.

Aluminum nitrate
Ammonium chloride
Ammonium nitrate
Ammonium sulfate
Barium acetate
Barium nitrate
Barium oxide
Barium sulfate
Calcium carbonate
Calcium chloride
Calcium oxide
Calcium nitrate
Calcium sulfate
Iron(II) sulfate
Iron(III) sulfate
Lauric acid
Lithium carbonate
Lithium nitrate
Magnesium carbonate
Magnesium chloride
Magnesium hydroxide
Magnesium oxide
Magnesium sulfate
Potassium bromate
Potassium carbonate
Potassium chloride
Potassium iodide
Potassium metabisulfite
Potassium nitrate
Potassium persulfate
Potassium phosphate
Potassium sodium tartrate
Potassium thiocyanate
Sodium acetate
Sodium bisulfate
Sodium borate
Sodium carbonate
Sodium citrate
Sodium chloride
Sodium fluoride
Sodium nitrate
Sodium nitrite
Sodium oxalate
Sodium sulfate
Sodium sulfite
Sodium thiosulfate
Starch
Strontium carbonate
Strontium hydroxide
Strontium nitrate
Sucrose
Tin(II) chloride
Urea
Zinc nitrate
Zinc sulfate

Of these, as stated earlier, an Mg compound (due to insolubility) or starch is most likely.

The fact that it turned black when heating tells you that:

It was an organic substance and did not burn completely leaving some carbon behind, or
It was a transition metal that had a lower oxidation state and was oxidised during burning.

Let's remove the obvious ones:

Not a nitrate. These will release toxic brown gas, which is not something you give to high school students.
Not an alkali (Na, K, etc) or alkali earth (Ca, Mg, etc) element because they're not pink nor they turn black.
I doubt it's a sulfur compound of any kind. You should have detected a sulfur smell when burning.

Maybe this can help you a bit.

Let's look at magnesium carbonate.  It's white.  When heated it remains white, and upon decomposition it gives off carbon dioxide leaving behind magnesium oxide -- which is white.  So can your color changing substance really be magnesium carbonate?
Only one of your remaining candidates really matches up with the color criteria you present.


text:
Why is ascorbic acid (vitamin c) readily oxidised in the presence of oxygen?
I know that vitamin c is notoriously unstable in aerobic conditions, but I'm not sure how to explain why it is - is it due to redox potential, acid strength etc?

Welcome to Chemistry! Take the tour to get familiar with this site. This may be helpful: https://chemistry.stackexchange.com/q/46577/194

According to Juan Du, Joseph J. Cullen and Garry R. Buettnera[1]:

Ascorbic acid ($\ce{AscH2}$, vitamin C) is a water-soluble ketolactone with two ionizable hydroxyl groups. It has two $\mathrm{p}K_\mathrm{a}$’s, $\mathrm{p}K_1$ is 4.2 and $\mathrm{p}K_2$ is 11.6; thus, the ascorbate monoanion, $\ce{AscH−}$, is the dominant form at physiological pH. Ascorbate is an excellent reducing agent and readily undergoes two consecutive, one-electron oxidations to form ascorbate radical ($\ce{Asc•−}$) and dehydroascorbic acid (DHA). The ascorbate radical is relatively unreactive due to resonance stabilization of the unpaired electron; it readily dismutes to ascorbate and DHA ($K_\mathrm{obs} =\pu{2×10^5 M−1 s−1}$ at $\mathrm{pH} = 7.0$).

Also,

Ascorbate readily undergoes pH-dependent autoxidation producing hydrogen peroxide ($\ce{H2O2}$).

So, in presence of oxygen, this process is accelerated due to formation of hydrogen peroxide. It is also known that this reaction may happen even faster with a metal as a catalyst[2]

References:

Du, J.; Cullen, J. J.; Buettner, G. R. Ascorbic acid: Chemistry, biology and the treatment of cancer. Biochim Biophys Acta. 2012, 1826 (2), 443–457. DOI: 10.1016/j.bbcan.2012.06.003.

Jiang, D.; Li, X.; Liu, L.; Yagnik, G. B.; Zhou, F. Reaction Rates and Mechanism of the Ascorbic Acid Oxidation by Molecular Oxygen Facilitated by Cu(II)-Containing Amyloid-β Complexes and Aggregates. J. Phys. Chem. B 2010, 114 (14), 4896–4903. DOI: 10.1021/jp9095375.


text:
Does the van der Waals force not affect the alkali metal 's trend of decreasing melting points down the table?
Melting and boiling points increase further down the halogen group, but they decrease further down the alkali metal group. I know that the former's trend has to do with the van der Waals force, but I don't understand why it is not applied to the alkali metal group. Please explain why the halogens are affected by the van der Waals force and the alkali metals are not.

Because, in the condensed phases, you don't have molecules of metallic elements.  You have, as a first approximation, a "rice pudding" structure where cations are embedded (like raisins) in a "pudding" of delocalized valence electrons (metallic bonding).  The main cohesive force in the condensed metals is then electrostatic attraction between the opposing charges of the cations and electrons.  In alkali metals where each atom breaks up into the same charge cation (+1 charge) and produces the same number of delocalized electrons (one per atom), this electrostatic attraction goes down with larger atoms.


text:
Can we manipulate the rate of water's autoionization?
I just got introduced to this concept and these thoughts came into my mind. Is there a way to increase the equilibrium constant of the autoionization of water?

Are you talking about the rate (kinetics) or the equilibrium position (thermodynamics)?

Heat it.  Autodissociation is an endohermic reaction.

As a follow on from @Oscar Lanzi,  supercritical water can start to pit/corrode steel due to the increased ionisation causing increased acidity.

First of all, I'm assuming you mean the equilibrium position. This is not the same as the rate of the reaction.
Sure there is. The easiest way to increase the equilibrium position water's ionization is by heating it.
To find out how much increasing the temperature of water will affect $K_w$, the Arrhenius equation will work for values of $T$ close to $0 °C$. $K_w=Ae^{\frac{-E_a}{RT}}$.
Of course, you'll have to find $A$ from experimental data.
Here is a graph of how changing the temperature affects $pK_w$. For the most part, as temperature increases, you will see an increase in K.

The curve turns around because of the effect of decreasing density of the (supercritical) water at higher temperatures.  Is there a similar curve at constant density available?

@OscarLanzi you are right. I'm not sure if there is a graph at constant density, but I will look for one.


text:
What is the minimum mass of iron(III) chloride required to be added until the dark blue precipitate first appear?
How do I solve this question?

The presence of iron(III) ion is confirmed in a qualitative analysis by adding potassium hexacyanoferrate(II) ($\ce{K4[Fe(CN)6]}$) solution.
A dark blue precipitate, iron(III) hexacyanoferrate(II) ($\ce{Fe4[Fe(CN)6]3}$), is formed.
What is the minimum mass of iron(III) chloride (in grams) required to add to a $\pu{200 cm3}$ of $\pu{0.0010 mol dm-3}$ potassium hexacyanoferrate(II) solution until the dark blue precipitate first appears?

solubility product of $\ce{Fe4[Fe(CN)6]3} = \pu{3.30 x 10^-41 mol^7dm^-21}$
relative molecular mass of $\ce{FeCl3} = \pu{162.3 g/mol}$

Choices:

$\pu{1.35 x 10^-8 g}$
$\pu{4.38 x 10^-7 g}$
$\pu{5.48 x 10^-7 g}$
$\pu{2.19 x 10^-6 g}$

This is what I came up with just now :

$K_\mathrm{sp} = \ce{[Fe^3+]^4 [Fe[CN]6^4-]^3}$
$\pu{3.3x10^-41}$ = $\ce{[Fe^3+]^4[0.001]^3}$
$\ce{[Fe^3+]^4}$ = $\pu{3.3 x 10^-32 mol dm^-3}$
$\ce{[Fe^3+]}$ = $\pu{1.35 x 10^-8 mol dm^-3}$
Mass of $\ce{FeCl3} = \pu{1.35 x 10^-8 mol dm^-3 \times 0.2 dm^3 \times 162.3 g/mol}$
= $\pu{4.38 x 10^-7 g}$
I am still relatively new to this topic so I am not sure if I am doing this correctly. Any help will be appreciated. Thank you !

Yes that is right.
However I would not round intermediate results to avoid round off errors. In fact I'd carry two extra significant figures for intermediate calculations and only round off final answer. (Easy now with a calculator, not so easy to do when I was doing these problems with a slide rule...)  So I'd get Fe as $1.3478\times10^{-8}$ molar. Using that number I get 4.3749893530170255598967635197789e-7 as the final answer which rounds to 4.37e-7.
The problem's answer is really wrong since the concentration of the hexacyanoferrate(II) is only given to two significant figures. So the answer shouldn't have more than two either. If the problem wanted three significant figures then the hexacyanoferrate(II) should have been specified as 0.00100 molar.
You have to realize that significant figures are a crude way to do error propagation. So two significant figures for 0.0010 molar means +/- 0.00005 molar, so a +/-5% specification. However if the answer to two significant figures is 9.9 then that means +/- 0.05 which only has an error of +/-0.5%. So although crude using significant figures keeps you from stupidly assuming that all the figures in the calculator's answer, like 4.3749893530170255598967635197789e-7, are significant.

The only trouble is that by adding small amounts of a $\ce{FeCl3}$ solution in a $\ce{K4[Fe(CN)6]}$ solution, the rection does not produce a blue precipitate of $\ce{Fe3[Fe{CN}6]4}$ as given in the text. It produces a blue solution of "soluble Prussian blue" which is $\ce{KFe^{III}[Fe^{II}(CN)6]}$ by the reaction $$\ce{Fe^{III}Cl3 + K4[Fe^{II}(CN)6 -> KFe^{III}[Fe^{II}(CN)6 (aq) + 3 KCl}$$ and no precipitate does happen at all !!
If insoluble Prussian blue is wanted, the contrary should be done : small amount of $\ce{K4[Fe(CN)6]}$ should be added to a $\ce{FeCl3}$ solution.
Ref.: 1) G. Fornosier, A. Bleuzen, Bleu de Prusse, Act. chim. 444, Oct. 2019, p. 16-21 (in French)
2) N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon, Oxford, 1984, p. 1271


text:
2- or -2 as superscript?
When writing the formula for an ion, should the operator come before or after the number that indicates the ion's charge?
Here are some examples:
$\ce{SO_4^{-2}}$ or $\ce{SO_4^{2-}}$
$\ce{Fe^{+2}}$ or $\ce{Fe^{2+}}$
I'm not great at chemistry, so I'm sorry if it's a stupid question.
I just haven't been able to find many answers.

$\ce{SO4^{2-}}$ and $\ce{Fe^{2+}}$

@MaxW I have made your absolutely correct comment into an answer. I hope you do not mind.  Just trying to return friendly answers to new users. Can delete it if you feel uncomfortable

As @MaxW has said in comments, the correct form is:
$\ce{SO_4^{2-}}$ and $\ce{Fe^{2+}}$
It is just as in math.
If you have $a+a$ it results in $2a$, not $a2$, and thus:
$\ce{SO_4^{--}=SO_4^{2-}}$ and $\ce{Fe^{++}=Fe^{2+}}$
It is not a stupid question.
IMHO it is a perfectly legitimate question.
I apologize for those who are downvoting your question.

Last I checked it was -1 + -1 = -2 not 1- + 1- = 2-, so "just as in math" is a slightly flimsy argument.

@R.M. $a+a=2a$. Charges are entities, not numbers.

My point was that "just as in math" is somewhat dismissive, as you can come to the exact opposite conclusion from math conventions from the -1 + -1 = -2 argument (which is probably where most people who want to put the sign in front of the number get it from). -- "Charges are entities, not numbers" might be the rationale, but currently it appears nowhere in your answer.


text:
How can Arrhenius bases also be Brønsted-Lowry bases?
According to ChemGuide,

The Brønsted-Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.

According to Study.com:

All Arrhenius acids and bases are also Brønsted-Lowry acids and bases.

This makes sense for Arrhenius acids, such as hydrochloric acid. Hydrochloric acid produces hydrogen ions when dissolved in water. Additionally, it donates a proton to water.
However, sodium hydroxide is an Arrhenius base, since it produces hydroxide ions when dissolved in water. However, sodium hydroxide cannot accept a proton to form $\ce{HNaOH+}$. So, how can it be a Brønsted-Lowry base? Similar reasoning can be applied to other Arrhenius bases, such as potassium hydroxide.

Sure it can.  You're just incorrect about the products when $\ce{NaOH}$ accepts an $\ce{H+}$ in aqueous solution.  The result is $\ce{H2O(l) + Na+(aq)}$ instead of $\ce{HNaOH+(aq)}$.  That is, the $\ce{NaOH}$ doesn't stay together.
Now mind you, most people would point out to you that the $\ce{NaOH}$ doesn't stay together as soon as it's dissolved, meaning $\ce{NaOH(s) -> Na+(aq) + OH-(aq)}$ happens right away, and then as soon as you have some $\ce{H+(aq)}$ around it reacts only with the $\ce{OH-(aq)}$ and the $\ce{Na+(aq)}$ doesn't get involved, probably isn't even anywhere near.  But this doesn't change the main point, which is that if you add $\ce{NaOH}$ to water, you have a solution that can react with $\ce{H+(aq)}$, which means you added a Brønsted-Lowry base.

Quoting from Peter Atkins, NaOH is merely the provider of the Bronsted-Lowry base, the hydroxide ion.

Sure, that's a point of view @TanYongBoon.  The only complication is that you could really say the same thing about strong acids.  Since HCl(g) + H2O(l) -> H3O+(aq) + Cl-(aq) happens as soon as you put the HCl in, and any addition acid-base reaction is between the H3O+ and base, you could equally well say HCl is merely the provider of the Bronsted-Lowry acid instead of a BL acid itself.  But we don't.  There is definitely a mild logical lacuna here in our definitions, over which reasonable men could and probably will disagree.

@ChristopherGrayce Is H2O (l) + Na+(aq) considered to be the conjugate acid of NaOH (aq)? Also, does NaOH (aq) have a Kb value?

@user62238 yes, if we're asked we might say Na+(aq) is the conjugate acid of NaOH(aq), although the idea of a conjugate is a little strange for strong acids and bases, since the conjugates have no acid/base activity.  Na+(aq) for example will not react with bases.  Similarly the conjugate base of HCl is Cl-(aq) but  Cl-(aq) will not react with acids.  It's not of great use to attach Ka or Kb values to strong acids and bases (at least in aqueous solution) like HCl or NaOH, because of what's called the "leveling effect."

In essence, every acid stronger than H3O+(aq) and every base strong than OH-(aq) gets immediately converted to these species on dissolution, and then all further acid/base chemistry happens with those species.  So in aqueous solution, there's no point in ranking how strong an acid or base is, once it reacts completely on dissolution.  (This is not true in organic solvents, by the way.)


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The reaction mechanism for epoxides to react with LiAlH4
Does anyone know the mechanism for the reaction of epoxides with Lithium aluminium hydride (product of course being an alcohol)

Depends how precise.

I suspect that this paper will go a long way to answering your query, but as it is behind a paywall I cannot be sure  https://pubs.acs.org/doi/pdf/10.1021/ja01038a042

I'm experiencing paywall problems


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Is work done by a system negative delta G or positive delta G?
My textbook states that a process is spontaneous and can perform work only when it is moving toward equilibrium. Since the process is spontaneous, wouldn't that mean that the work done is attributed to negative free energy?
Another way of considering it is the question of what should the sign of the amount of work done on objects outside the system (or heat released from the system) should be when the reaction is exothermic or endothermic.

Depends what you mean by that.  I agree it can be a little bewildering to figure out exactly what all these words in thermodynamics mean -- energy and work and heat and free energy and God knows what else.
It may help to remember in all such cases we're just talking about energy.  That's it, that's all there is.  The various types of energy that we call enthalpy or free energy or whatever are just different ways to identify part of the energy with the system and part with the surroundings.  Most of the time, changes in enthalpy or free energy or energy are just statements that the energy got transferred under certain conditions, for example with the system at constant temperature and pressure, or maybe constant pressure and volume.
But the important thing to remember is it's all just energy, and you can use the same intuitions about it you'd use if you just used the word "energy" everywhere (up to a point, of course, because the distinctions aren't meaningless).
So for example in this case, I hope it makes intuitive sense to you that if a system does work in its surroundings (causes weights to be raised, springs to be compressed, batteries to be charged, whatever) that it has to lose energy for that to happen.  Energy has to flow out of the system and do the work.  So that would suggest the energy -- or enthalpy -- or free energy -- has to go down. And so they would, generally.  That's a good intuition to have, a good starting point, although there are certainly more complicated tricky situations where you'd want to refine your first guess.
Usually if you're talking about free energy, you're talking about a system at constant T and P.  For that system, to do work, it has to lose energy, which means G goes down, dG < 0.  On the other hand, if work is done on the system -- springs within the system are compressed, batteries charged, et cetera, then dG > 0 and the free energy goes up.  All just as intuition would suggest for what happens to the energy.


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Synthesis of P2NP
I am trying to synthesize P2NP for an experiment I'm documenting online, and I have a hard time.
17ml benzaldehyde were added to 35ml of (kind of anhydrous) ethanol, with 14ml nitroethane (the benzaldehyde was not distilled beforehand or washed with baking soda solution but was tested and had ph around 6, which meant no much benzoic acid contamination) and 4ml di-n-butylamine, and something like 2ml glacial acetic acid was added to reach neutral ph of 7.
No dean-stark trap was used, but it is said water would prevent good yield, which implies some yield at all.
This was boiled on reflux for about 3h and become dank orange. it was put off heat and cooled in the fridge. No crystals formed  >:(. The solution still smelled like benzaldehyde, which makes me think the reaction is not complete.
So, what am I doing wrong?

Distill your benzaldehyde then try using the ammonium acetate in acetic acid conditions. This has been detailed here: J. Org. Chem. 1953, 18 (1), 1–3. To quote:

The aldehyde (5g), 5 ml. of the nitroalkane, and 2g of ammonium acetate were added to 20 ml of glacial acetic acid. The resulting solution was refluxed for two hours and then poured into ice-water. If a solid product was obtained it was collected and recrystallized from methanol, ethanol, or acetic acid. If the product was an oil it was separated and crystallized, if possible, from one of the above solvents.

The authors record a 55% yield for the product you want.

I have ammonium acetate but it is very wet. Is that going to be a problem?

Probably not, try it.

Ok, I will, thank you. Accepted your answer. Check my other question if you like: https://chemistry.stackexchange.com/questions/99120/p2np-synthesis-ammonium-acetate-catalyst Thank you very much.


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Reaction direction change based on phase of reactants and products
I am curious specifically how the entropy of a reaction will change if the phases of the reactants and products at different conditions change.
For instance, a reaction
$$\ce{O2(g) + C(s) -> CO2(g)}$$
has a positive entropy because there is no net change in the moles of gas in the system and the product is more complex in its degrees of freedom (secondary). If one were to change the conditions (temperature and pressure) such that oxygen is a gas but $\ce{CO2}$ is a liquid or solid, would the change in entropy for the system become negative?
$$\ce{O2(g) + C(s) -> CO2(l)}$$
Or
$$\ce{O2(g) + C(s) -> CO2(s)}$$
If the $\Delta H$ is still negative but the $\Delta S$ is now positive, isn't the Gibbs energy now able to be positive or negative, making this a valid method of manipulating the directionality of a given reaction?

Quite correct.  And manipulating which direction a reaction tends to go by changing the phase of reactants or products is a common trick in chemical engineering.


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Why is Rf called Retardation Factor?
I understand that $$R_\mathrm{f} = \frac{\text{distance traveled by center of analyte spot } (b)}{\text{ distance travelled by solvent front } (a)}$$ What I do not understand is why this is called retardation factor; an analyte that travels further in the mobile phase seems to me to be less retarded by the stationary phase, while one that doesn't travel as far seems more retarded. If we're using this terminology why doesn't $R_\mathrm{f} = 1- \frac ba$?

I always learned Rf as being Retention Factor, not Retardation Factor. Your question still stands though...

They're often used interchangeably in TLC, but IUPAC uses Retardation factor (Rf) for planar chromatography , and Retention factor ( k ) for column chromatography.

You have the Rf inverted. It should be a/b or distance traveled by the spot divided by distance traveled by the solvent front. Nonetheless, a low value is more retarded or retained, so your question is still a valid one. It seems to be a matter of  definition.

See also retention factor $k$ and retardation factor $R_\mathrm F$.

Ah yes,  it is inverted, thanks Dr.J. I'll fix that.

The clearest answer is that an analyte is retained by both the mobile and the stationary phase and the retention factor is a measure of the ratio between retention in the mobile phase and in the stationary phase.
So you could think of it as a higher Rf value indicates a higher retention in the mobile phase.
(Also, in answer to some of the comments, the retardation factor is the inverse of the retention factor)

That makes sense to me, but I thought that "retention factor" was reserved for discussing column chromatography?

I'm not sure, I haven't come across that as a limitation, it may be a convention but I don't think it's a hard rule; I have definitely seen the retardation factor used in reference to TLC, for example, but very rarely.


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Does sheer amount affect reaction rates?
For example, in the iodine and starch reaction, if I use 20 mL of a set concentration of each, will it be slower to completion if I use 40 mL of each at the same concentration?

Depends, how you use them.

In a simple bimolecular reaction between A and B with rate constant k the rate = k[A][B], where [A] is the concentration of A,  and [B] for B, so if the concentration is the same the volume does not matter if other things remain constant, temp , pressure etc.

If you define "completion" by a certain percentage of conversion then no, they will behave exactly the same, provided that things like cooling and stirring are the same in both systems.

Problem is if you scale up process, those things can't be exactly the same.

Here some elements to add to your reflection:

Some chemicals (among them, many bio-polymers or proteins) can highly change fluid properties (mainly viscosity and surface tension) when dissolved, like you can see when solubizing Starch;
Sheer stress affects viscosity of non-Newtonian fluids (moderate concentrations of starch are not Newtonian fluids, looks like jelly), and then chemical advection and diffusion. Thus it can affect reaction rate by modifying mass transfer rate;
Sheer stress may change protein or polymer conformations (think about Starch which is a polymer) and then can affect their physical or chemical properties (eg.: it can produce foam a complex polyphasic system, or denature protein).
Dark-purple color you see when mixing Iodine with Starch comes from a very specific complex of Iodine-Starch chains;
Practically, you will apply sheer stress differently to a tank about 1L than 1000L.

Regarding to that aspects, it explains why sheer stress may impact reaction rates, in your case: Iodine Starch complex coordination reaction rate.


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Why can we still breathe in valleys?
Air is 1% argon.  Argon is heavier than air.
Why doesn't the argon concentrate in low-lying areas, choking out life there?

Related: Why don't heavy and light gases separate in the atmosphere?

Also related: Why does the composition of the air does not change with altitude? (Physics.se)

A part from the mixing aspect, note that almost all valleys have a "drain". They are not surround by walls on all sides. Otherwise, they wouldn't only fill up with gases, but also with rain water - at least to the point, where the water level is high enough to find a drain, and at that point the gases would flow out, too, and distribute over the lowest surfaces on earth, which are almost exclusively seas. So there would be a huge space for the gases to distribute.

While the question already has an answer elsewhere, the answer given here is truly remarkable (esp. the last two paragraphs, which help show why the question is indeed a reasonable one despite appearing silly at first), and shows me (at least) that sometimes letting another answer appear isn't a bad thing at all.

It does.  You would find the average percentage of the atmosphere that is argon is very slightly higher at the floor of valleys.  However, bear in mind first of all it wouldn't be anywhere near a complete stratification -- a layer of pure argon, then another of pure N2, and so on.  A mixture of nearly ideal gases doesn't do that, at least at equilibrium, because it would eliminate the considerable entropy of mixing.  (It can happen in liquids because liquids have strong intermolecular forces that normally favor separation and oppose the entropy of mixing.)  Another way to think about it is that since the atoms and molecules in gases don't (much) interact, there's nothing stopping an individual argon atom going slightly faster than nearby nitrogen and oxygen molecules from bouncing up higher than they do.
What you would get in a theoretical ideal (uniform gravitational field, complete stillness -- no wind -- and uniform temperature) would be an exponential fall of pressure with altitude, and the exponential for heavier gases would be steeper than for lighter gases.  That would result in enrichment of the heavier gases at lower altitudes.  A little work starting from the Boltzmann distribution of gravitational potential energies of each type of atom and molecule would get you an ideal estimate of the argon excess as a function of altitude.
In practice the lower atmosphere has so much mixing due to wind and big thermal gradients that I doubt you could even measure the mild excess of argon and other heavy gases.
There is one fascinating short-term exception, which bears directly on your question.  Sometimes volcanoes will belch out a considerable quantity of CO2, which is significantly denser than air, and this CO2 can accumulate briefly in a thick layer at the bottom of a valley or over a lake, if there isn't much wind.  It can persist for some hours, perhaps days, before it diffuses away and is mixed with the rest of the atmosphere.
Then indeed the valley bottom becomes an invisible death trap for humans and animals: walk into the valley, or be unable to exit fast enough when it happens, and you will suffocate for no reason you can see.  The most famous example of this is the Lake Nyos disaster in 1986 which killed thousands of humans and animals.  I think the government now has mixing devices installed in that lake to prevent any future sudden release of CO2.

Hey Christopher, great answer! Though do you mind adding in some citations/references to back it up? Thanks!

@GaurangTandon: I can't find the citation due to it being in a book I don't have but I can pretty much confirm this. So if you don't get any ...

$\text{CO}_2$: https://en.wikipedia.org/wiki/Mazuku

Not just carbon dioxide.  There can be sulfur dioxide that would be even heavier and slower to disperse.  And more toxic.

@OscarLanzi True, but sulfur dioxide is much easier to notice - it smells awfully, and as it increases to (acutely) dangerous levels, it's painful to breathe. In contrast, carbon dioxide pretty much just kills you before you have any chance of noticing anything bad. Large carbon dioxide releases are also a lot more common in nature. The "sudden release" bit is pretty important, since a more gradual release will be readily dispersed throughout the atmosphere.


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Process of transferring kinetic energy in regard to photons.
Now I know that temperature is the average kinetic energy of atoms, molecules and compounds. Photons increase the kinetic energy/temperature of molecules since photons are pure energy. Hence, wouldn't the transfer of kinetic energy mean that photons need to be transferred from a higher temperature molecule to a lower temperature molecule?

No. Photons are pure energy all right, but the converse is not true.

Photons increase the kinetic energy of molecules. All I was asking is if photon transfer is a mechanism of heat transfer in molecules. Christopher Grayce directly answered this question, so I don't know why it is on hold.

You may be experiencing a very common misconception, which I blame on Star Trek and Star Wars and all their talk of pure energy that always comes as bright lights, with the implicit assumption that energy is some kind of fiery bright substance, like a glowing fluid or something.  It isn't.  Energy is a property of matter, and of the rather peculiar substance we call electromagnetic radiation, chunks of which are called photons.  Think of energy as kind of like an adjective that describes a noun (some kind of matter, or radiation).  An atom can be energetic or not energetic, and so can a photon, just as a cat can be furry or not furry.  But you would  never mistake "furriness" as some kind of object itself.  In the same sense, you should not mistake energy as some kind of substance itself.  It's a property of substances, just like furriness is a property of cats.
So energy has to be transferred from a hot substance to a cold one, and indeed it could go by way of photons.  One particle could emit a photon, transferring part of its energy to the photon, and maybe slow done, and then this photon could be absorbed by another particle, which would receive the transferred energy, and maybe accelerate (go faster).  That's definitely one way it could happen.  But it doesn't have to happen that way, because energy can be transferred other ways.  For example, the first particle could simply run into the second, bang, and the impact itself could transfer the energy directly, no photons involved.  Or the first particle could run into a third particle, or the wall of a container, and the second particle could pick up the energy from the third particle, or the wall of a container.  And so on.

To add a little to the answer by @Christopher Grayce, molecules have kinetic energy due to the fact that they move about and also internal kinetic energy because they have bonds that vibrate and some bonds can rotate as can whole molecules. For a photon to give up all of its energy to a molecule it has to be absorbed and in doing so the photon is destroyed and the molecule is left with this extra energy which, depending on just how much energy the photon had, can excite rotations, vibrations or form electronic excited states or all three. Because molecules can collide with one another this extra energy is eventually spread out among all molecules. (Some energy may also be lost as fluorescence from an excited state).
We see things as being coloured due to visible photons producing electronically excited states and feel heat from a fire or radiator because the molecules in our skin absorb infra-red photons which excite vibrations in these molecules.


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Why is dipositive dilithium more stable than neutral dilithium?
According to J.D Lee, compounds with fraction bond number are unstable. I calculated that the bond order of $\ce{Li2^2+}$ is 0.5 while that of $\ce{Li2}$ is 1. Hence, $\ce{Li2^2+}$ must be less stable than $\ce{Li2}$ due to half bond character.
But, in reality, $\ce{Li2}$ is more stable than $\ce{Li2^2+}$. Why is it so?

Your question makes no sense: "Hence Li2+ must be unstable than Li2 but then why Li2 is more stable than Li2+" says the same thing twice. What are you trying to ask?

i m trying to ask why li2+ is more stable than li2?

I suggest you edit your question to say that then. Further stable with respect to what? The dissociation products of the two molecules are different. Do you mean bond strength? I also suggest you think more carefully about the tags you use for your question, organic-chemistry, periodic-table and electronegativity are all totally irrelevant.

Rafael, please don't rollback edits that attempt to improve your post. If you have some reasons to completely revert the changes, please also let the particular editor know that you're reverting their edits via the @ notification. Thanks.

@GaurangTandon He's asking about bond order 1/2 so definitely not +2 charge.

@Mithoron I agree but I cannot introduce changes on my own. The latest revision by the author was this.

The comment asks "Why is Li2+ more stable than Li2". Li2 has a relatively low bond energy (gas phase) of 27 kcal/mol (Cotton and Wilkinson, Inorganic Chemistry). The energy cost to make one mole of Li gas is 37 kcal (CRC Handbook). The ionization energy of Li is 5.39 eV = 124 kcal/mol.
So, breaking a Li2 gas molecule into 2 Li (gas) costs 27 kcal/mol.
Breaking a Li2+ gas molecule into Li + Li+ involves not the separation of two uncharged atom, but the separation of a very small Li+ ion from a Li atom which provides some charge accomodation. The heat of formation of Li+ in water is 66 kcal/mol (presumably from aquation with 4 waters); the hydration bonding is worth about (124 + 37 + 66)/4 = 54 kcal/mole each. So I would estimate the bond between Li+ and Li to be perhaps 40 kcal/mol, at least a significant fraction of 54 kcal/mol, and probably more than 27 kcal/mol.
The important point is that the unusually small size of the Li+ ion makes it able to polarize a neutral atom to form a strong bond (and come in closer!), whereas the neutral atoms have only uncharged molecular orbitals to spread their electrons over. It is highly unlikely that larger atoms (Na, K) would show similar stability for an ion-molecule compared to a neutral molecule. Magnesium might be a similar exception, however, since both Li and Mg have small radii; it might be interesting to compare stabilities of Mg2 and Mg2+.

Here you're saying that Li2+ is more unstable than Li2

Hence Li2+ must be unstable than Li2

And here you're confirming that Li2 is more stable than Li2+, corroborating with your previous statement

but then why Li2 is more stable than Li2+

But in your question you're saying the opposite, that Li2+ is more stable than Li2. So i'm confused with what you're trying to ask.
In my knowledge, based on inorganic text books, you can explain the stability of a molecule using molecular orbital theory (MOT). But MOT is a bond theory and Li2+ has no bond unless you're talking about (Li-Li)2+. If this is the case, Li2 is more stable than (Li-Li)2+ -(Li-Li)2+ is not even going to exist according with MOT - since the bond order for Li2 is 1 and (Li-Li)2+ is 0 as shown in the diagram below.

If you're trying to compare Li2+ and Li2 I don't know if it's possible, i've never seen something like that.

He's probably asking about $\ce{Li2+}$. If you don't understand question then why to answer? (Rhetoric question)

Someone commented a question about my answer and deleted after a while. I didn't see that the person deleted it.


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Why does hydrolysis of OF2 give a neutral solution?
I found this equation on the internet for the hydrolysis of $\ce{OF2}$: $\ce{OF2 + H2O -> HF + O2}$. So, according to me, $\ce{HF}$ must make generate an acidic instead of neutral solution.
But in JD Lee, it is stated that it is a neutral solution. Why so?

Who says it is not so?

It is neutral -i found in many books

Beyond the ocean there is an island inhabited by two-headed people. I saw this in many books. Should I believe them?

@RafaelNadal we can help you better if you cite the specific books and page numbers. Thanks.

@Ivan Neretin sir i had uploaded book snapshot from (j.d lee concise inorganic chemistry )

OK, I got it. The reaction does indeed produce acid, but it takes some time. Before that, you will have just a solution of OF2, which is neutral.

@IvanNeretin this reaction is itself a slow reaction and requires temperature of about 200 degree C ,so  it gets proper time to attain its nature till the reaction is fully completed  ....So, it is not the reasons that you have mentioned

To me, this looks just like what I said: you will only have acid after the reaction, and the reaction is slow.

@IvanNeretin sorry sir, but the overall reaction is a neutral solution

This is nonsense. The overall reaction gives HF, which is an acid (and there is nothing to neutralize it), and O2 that simply flies away. The solution after the hydrolysis just can't be neutral, nor does your book say so.

Hydrolysis and dissolution are different.  The dissolution of OF2 in water gives a neutral solution. Subsequent hydrolysis, which is very slow, eventually produces HF and hence acidic solutions.

@IanBush But not a strongly acidic one as HF is a weak acid.

Everyone is comments it is acidic but in my Question Paper answer was neutral

Hydrolysis doesn't give a neutral solution, but the hydrolysis reaction is slow
Don't confuse dissolving something in water with hydrolysis. Some chemicals hydrolyse rapidly in water but not OF2. Oxygen difluoride will dissolve in water forming a neutral solution but will only slowly hydrolyse giving a solution containing HF which will be acidic.
So your book is correct, at least in the short term, the solution will be neutral.


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Which carbon electrode gets plated with Zinc on a beaker with zinc sulphate solution?
In a beaker containing zinc sulpate and two carbon electrodes connected to a battery, which electrode gets electro plated with Zinc? Why?

This is an interesting question to solve this question one needs an understanding of exchange current densities for zinc cation reduction and proton reduction on the electrode.


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Why does physisorption change to chemisorption at high temperatures?
I read in my textbook:

In certain cases, physisorption changes to chemisorption as temperature is increased.

Could anyone explain why it is so? I know the basics of both physisorption and chemisorption, but I am unable to relate them to this statement.

Presumably your book assumes that there is a larger activation barrier for chemisorption than physisorption (normally the case) and that that cannot be crossed at a low temperature, so here physisorption is favoured.


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Can hydrogen peroxide be oxygenated?
As per the heading, would peroxide be able to dissolve oxygen by whatever method, to have free oxygen and hydrogen peroxide mixed (under pressure?) similarly to $\ce{CO2}$ or $\ce{O2}$ in water?  I understand peroxide is trying to break down  & would produce similar over time anyway, though with a loss of peroxide. Question is just for the sake of knowing, no pressing need. Thanks...

In general, all gases have some non-zero solubility in any solvent. It is generally a small number. The amount of gas dissolved also depends on the partial pressure of that gas. So, sure, why not?

If we assume hydrogen peroxide solution (often 3%) is quite similar to water in its ability to dissolve oxygen and carbon dioxide, then yes some oxygen is always dissolved in the solution at STP.
However, the amount of carbon dioxide that can dissolve in water is much higher as compared to oxygen. Carbon dioxide reacts with water to form carbonic acid and that drives water to dissolve something like 25 times more (according to a cursory internet search) carbon dioxide than water under similar conditions.
Secondly, the addition of hydrogen peroxide to the water, which is constantly decomposing to produce free oxygen, would lead the solution to dissolve even less atmospheric oxygen than normal water.
So, yes, it dissolves some. Nowhere near similar to carbon dioxide. I'm not sure it would be useful in any context where it might compare to soda water =)
This is really what hydrogen peroxide already is if it wasn't for the brown bottle slowing down the decomposition. You would see slow bubbles of "fizz" as the peroxide gradually goes "flat."
Now higher concentrations of peroxide would release more oxygen and can be quite dangerous. Lab quality 30% solutions are corrosive, strong oxidizers, and can be an explosion risk. But the oxygen still would not remain dissolved in the water for long, thus the danger. They have special handling precautions and even special expandable "accordion" containers for storage.
Interesting to think about...


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How can chemical energy be stored in chemical things like candle wax, gasoline,...?
I have learned about a type of storing energy is chemical energy. But energy isn't sth we can see, can touch. It just simply gives the ability to work. So how can energy be contain in such chemical things like above?

Well, just like that: you lit a match, and you see the energy coming out of it.

Yes, energy isn't a "substance" it is the ability to do work. But chemicals can still store it.
The key to understanding how is the relative amount of energy in different chemicals. Some chemical reactions release heat because the products of the reaction are lower energy than the chemicals that react (reactions are a bit more complicated than this in reality and some are driven by changes in entropy but that isn't important for understanding simple reactions). A candle, for example, consists of a mixture of long chain hydrocarbons with a formula approximating CH2. A candle will burn in air (reacting with molecules of oxygen) to give a mixture of CO2 and H2O. This reactions releases heat (simplifying slightly) because the bonds making up the products have lower energies than the bonds making up the candle wax.
This is why we can say that chemicals "store" energy. But we should always be careful to specify the reaction involved as chemical energy is stored relative to the possible products from a reaction not in some absolute sense.

I'm still not 100% understand but this has satisfied me. Thank you very much!


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Finding pKa from equivalence point on titration curve
Is there a mathematical proof/explanation of why $\mathrm{p}K_\mathrm{a}$  corresponds to the $\mathrm{pH}$ at $\text{Volume of titrant}/2$  at the equivalence point?
A concise proof will suffice.

You've got a weak acid, since you're contemplating a positive pKa, which means when you're halfway to the end point you're in the buffer region and you can use the Henderson-Hasselbalch equation:
pH = pKa + log [A-]/[HA]
You've titrated half your initial HA, so half of it is still around and half got turned into A-, which means [A-] = [HA].  For example, if you started with 0.1 mol/L of HA, you now have 0.05 mol/L HA and 0.05 mol/L of A-.
That means [A-]/[HA] = 1, and the log of 1 is zero.  The HH equation then tells you pH = pKa.