text:
Thermal vibration of bonds
So, the characteristic frequency of thermal motion is around 7E12 Hz at room temperature (300K), but from that information how can we conclude that the bonds are hard; they don't vibrate !!

Thermal energy does not affect the frequency of bond vibration;  all bonds vibrate with zero-point energy as well as at higher energy. Vibrational frequencies are particular to the type of molecule and depend on the exact nature of the interaction forming the chemical bonds; this is called the potential energy profile (or surface). Thermal energy just populates more energy levels and so more transitions are seen in infra-red or Raman spectroscopy which are the usual experimental techniques used to measure molecular bond vibration frequencies.

The bond energy is related to which kind of molecules are involved in the bond itself (their polarity,...). all the molecular bonds undergo some kind of vibration (they can bend, stretch, rock,...) which peculiar frequency varies according to the type of bond and molecules involved.


text:
Is the following procedure for antacid titration correct?
I'm trying to check which antacid is more effective by determining the amount of $\ce{HCl}$ they can neutralize.

Take $\pu{10 mL}$ antacid suspension and add excess $\pu{1 M}$ $\ce{HCl}$.
Back titrate the excess 1M $\ce{HCl}$ with 1M $\ce{NaOH/Na2CO3}$.
Do the required calculations.

Also, can I directly titrate the antacid with $\ce{HCl}$? Some solids remain.

Yes, you should titrate to pH 6 (the pH at the entrance to the small intestine, 7.4 (the end of the small intestine) and 4.5. Multiply the mL of HCl used by its normality to get the mEq of protons absorbed by the antacid. Multiply the pH 4.5 result by 84 to get the sodium bicarbonate equivalent (mg of NaHCO3 that neutralizes the same amount of acid as the test sample).
The antacid may contain some insoluble material (such as SiO2) but try to get everything dissolved. Also, when titrating wait a good long time before accepting the pH reading

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text:
Determine concentration of phosphate after copper(II) oxalate is precipitated
Solution with $\ce{Na3PO4}$ with $\pu{0.02 M}$ concentration and $\ce{Na2C2O4}$ with $\pu{0.03 M}$ concentration. I add excess $\ce{Cu(NO3)2}$ very carefully – tiny amounts.
What will be my $\ce{PO4^3-}$ concentration at the moment when precipitation of $\ce{CuC2O4}$ will occur?
Given: $K_\mathrm{sp}\left(\ce{Cu3(PO4)2}\right) = \pu{1.40e-37}$, $K_\mathrm{sp}\left(\ce{CuC2O4}\right) = \pu{4.43e-10}$.
I know that the reactions will be:
$$
\begin{align}
\ce{Cu3(PO4)2 &<=> 3Cu^2+ + 2PO4^3-}\\
\ce{CuC2O4 &<=> Cu^2+ + C2O4^2-}
\end{align}
$$
Now, solving some math I find that $\ce{Cu3(PO4)2}$ precipitates first. But then I start asking my self what is happening with the $\ce{Cu}$ amount. Never did lab lessons or thought about the process behind the scene.

The numeric value denoted by the cryptic letters $\rm Ksp$ is going to be useless to you unless you know what it is.

@Ivan I know that the very small value of Cu(PO4)2 tells me exactly that he is less prone to precipitate. As higher the value of our constant so the more likely he will 1st to precipitate. But in calculus i find the opposite, because the amount of Cu in Cu(PO4)2 is so small - he will be the first to precipitate. Little bit confused her.  Hence, I cant understand what remains with other amount of Cu that reacts with C2O4? And , how at all the precipitation will look like? the reaction will occur separetely different timing? or at the same time both mixed.

What you know is wrong, and calculus is right. Also, check the formula of that salt, otherwise it will turn out terribly wrong anyway. What is the charge of copper ion? What is the charge of phosphate ion? What is the formula, then?

The stoichiometry will be in 1 --> 1 : 2 ratio

dah... check your chemical formulas for the ppts. Copper has a +2 charge, phosphate a -3 charge, and oxalate has a -2 charge.

In the newly edited version of the question the equations are not even balanced, and the valence of Cu being +6 makes no sense.

You right. I updated the mistake.

If you don't mind I can make an edit for you and you can choose to accept it. It is still incorrect in the current form.

Yes, of course! @WeijunZhou

Clearly the phosphate will precipitate first. As you add the copper nitrate copper(II) phosphate will precipitate taking equivalent amounts of copper(II) and phosphate. Eventually there isn't enough phosphate left to take out the incremental additions of copper and its concentration  begins to rise. Eventually the saturation limit with respect to oxalate is reached. So find that limiting copper concentration (at which oxalate precipitation begins) from $\ce{[Cu^{+2}] =  KspCu2Ox/[Ox]}$. I get  4.92222e-07 M.  Then find the saturation level for phosphate at that copper concentration from
$\ce{[PO4^{-3}]}$  = $\sqrt{K_{sp\ce{Cu3(PO4)_2}}/\ce{[Cu]^3}}$
I get 1.08348e-09 M. A pKsp of nearly 37 suggests that copper phosphate isn't very soluble.

Without the correct formula for copper phosphate and copper oxalate (and the formula given in the question is clearly incorrect) correct calculation is impossible. Side note: use proper formatting. MathJax isn't that hard to learn.

It should be $\ce{Cu3(PO4)2}$. I don't expect redox reaction to take place here. And by the way, I am not the downvoter.

@A.J.deLange We try very hard to maintain a professional atmosphere here on the site. As such, please choose your language carefully when interacting with others (see more information at the [help/behavior]).

Agree, there is no place her for ego competition, i just need a little clarification. Yes, it can be called a "homework" question, but my point is not to do it and forget, but to understand deeply the process that stands behind all the theory and math of chemistry world.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

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text:
Heterogenous equilibrium nernst equation
In the nernst equation, when a gas - ion electrode is used eg when we write the equation for standard reduction potential for a particular electrode : we get a term of activity of h2 gas and h+ ion. We respectively put 1 atm and 1 molar. However how can we use partial pressure and molarity in the same expression for Q? Shouldn't it be one or the other?

You can mix activity with fugacity in the expression of equilibrium coefficient or reaction quotient. In fact that is quite common when the reaction involves both solution and gas phase reactants/products.
The definition of the equilibrium constant actually involves dividing the standard pressure $p^\unicode{x29b5}=1\,\rm atm$ or the standard concentration $c^\unicode{x29b5}=1\,\rm mol/L$. In your case, the equilibrium constant will be
$$K^\unicode{x29b5}=\frac{[a(\ce{H+})/c^\unicode{x29b5}]^2}{[f(\ce{H2})/p^\unicode{x29b5}]}$$
which is dimensionless, and is related to the change of Gibbs free energy by $\Delta G=\Delta G^\unicode{x29b5}-RT\ln K^\unicode{x29b5}$.


text:
How to dock aggregated structures comprised of "elementary" protein units like LEGO pieces?
There are proteins with intrinsic symmetries. For example:

I was wondering how to use transformations such as: rotations, translation, replications to construct possible structures using 2 types of proteins.
Assume I found a binding site of the 2 proteins using Molecular Dynamics simulators.
For example the above one and this one:

Can I use group theory way to find possible configurations of the 2 proteins?
How can I use the tensor product of the symmetry groups ($G_1 \times G_2$) to find possible structures?
What graphic/framework would you recommend do perform all the manipulations on the proteins?
Is it possible to use the transfer to create some algebraic structure that will help to construct a mechanical configurations comprised of proteins?
Is there any tool from Baker's Lab I can use for this task?
Amy recommendation for a tool I could use to replicate, rotate and shift structure would be appreciated us well.

TL;DR:
My goal is to use group theory or Monte Carlo simulation driven by group theory insight to create macro molecular structures using multiple (dozens) instances of the two proteins (the LEGO building blocks).
Any idea that relates symmetry, transformations, group theory to create structures is much welcome.

Clarification:
Given a binding point between 2 proteins, which software can I use to create the dimers into macro-structures like in LEGO namely only by replicating, translating and rotating the Lego pieces?

What "structures"? Proteins aren't a set of blocks. Their binding is very complicated and symmetry of crystal units doesn't need to have much to do with specific cases.

@Mithoron assume you find the binding of 2 proteins. Why couldn't you use them to build structures using multiple such building blocks. Like in Lego.

I think this question could be really interesting, but as is it's hard to understand what exactly you're looking for. Could you update the question with more details and examples?

@a-cyclohexane-molecule is it better know? The idea is to use group theory to get insight on macromolecular structures. Also possibly by symmetry of the protein to understand it's possible functions. I think that group theory can give a lot of insight into macro molecular structures and functions of proteins.

I think I understand the question. I think it is a really good one. I also have absolutely no idea how to answer it. Where is Eugene Wigner when you need him?

I don't see how group theory could help. Why not start by looking at known x-ray structures of dimeric/trimeric proteins such as the huge PSI trimer of photosynthesis to see what determines the shape/energetics of the interface. Then look at mis-folded aggregates to learn how they aggregate. Should keep you busy for a while :)

Maybe I should look into baker's work https://www.bakerlab.org/

@porphyrin, you are welcome to write it as an answer, although I don't agree.

In the link you give  the article the 'Computational design of Flu Glue' seems to describe what you are looking for.

@porphyrin, not sure. As far as I can tell in this work they simply bind 2 proteins. There aren't LEGO like structures.

Try DNA origami instead
While protein assemblies are well known, rationally using proteins to assemble larger supramolecular structures is still a difficult problem:

Predicting protein folding is an unsolved problem. Sometimes even small modifications to a sequence can change the folding propensity.
Predicting protein-protein interactions is often difficult, due to challenges in understanding protein dynamics, intermolecular interactions, and accurate predictions of protein electrostatics.

There has been some progress in terms of forcing particular protein-protein assembly through intentional side-chain modification and cross-linking, and reactive end-groups (e.g protein-based metal-organic frameworks (MOFs).
That said, use of proteins for supramolecular assemblies is far harder and less prevalent than using DNA assemblies. At the moment, DNA origami is far superior for these purposes.
Edit: If you're really looking to just copy and displace atoms, I don't think you need any particular software. I'd write a Python (or other script) that reads in the PDB files, and writes out duplicate atoms displaced in the XYZ directions as needed. Group theory can help predict the angles, but you'll still need to know the protein size to work out the geometry.

Assume I know the binding point between the proteins. How can use this to build as in Lego structures.

IMHO, that doesn't help you much. You're thinking that the proteins are rigid. In general, they can change geometry on binding - even binding small molecules into an active site.

For my project this what's needed: multiple rigid docking and manipulations based on known binding site between the 2 proteins.

@0x90 I think you might be too attached to the Lego analogy. If we were to keep that analogy, proteins are like Legos which change shape when they bind to another Lego. You can know what piece will fit on either Lego because it varies after the first binding.

@Tyberius this is my assumption, I just need a SW to take the proteins as rigid bodies and to dock them. I am surprised that there isn't such option in VMD auto dock or any other software. My assumption does make sense, because this is the result I can see in the experimental data. They construct Lego like structures.

@0x90 - I think your best bet is to write a script to do this - read in the two PDB files and write out duplicate atoms after translating in the XYZ directions as needed. There's no software because what you're describing is your unique research project.

@0x90 - Avogadro can copy and paste and translate, although with multiple protein assemblies, you'll get much better performance by writing a Python script.

@GeoffHutchison I guess this is my question how to do this? I prefer python or Matlab. Also to be the least hackie way.

@0x90 - I'm not intimately familiar with the capabilities in Pymol or Biopython, although I'm sure it's possible to write a script for that. I'll add an example using OpenBabel / pybel tomorrow.

I wrote my answer hoping to get some clarification and encourage other responses - not for the bounty. So I feel bad that you haven't gotten other answers.

@GeoffHutchison I know. It's ok. I appreciate your effort. :)

I don't see how to rotate and translate a PDB using pybel. Also I am not sure how to load the same PDB multiple times in different locations.

@GeoffHutchison, I am wondering if you had the chance to craft the simplest demo using pybel and openbabel. I tried to do it by myself, but I have to many technical struggles.


text:
Aromaticity of annulenes
Why are some annulenes such as [14] and [18]annulenes are aromatic even though they are not fully planar as cited by Wikipedia? What does it mean that they can achieve planar form?

Related: https://chemistry.stackexchange.com/questions/51484/does-aromaticity-require-a-co-planar-pi-system/51485#51485

It means very little, at least in terms of stabilization.  Perfectly planar or not, larger aromatic rings bring less stability than smaller ones, so they react more like mere conjugated polyenes than what we think of as "aromatic rings".  Wikipedia:

When the annulene is large enough, [18]annulene for example, there is enough room internally to accommodate hydrogen atoms without significant distortion of bond angles. [18]Annulene possesses several properties that qualify it as aromatic.[1] However, none of the larger annulenes are as stable as benzene, as their reactivity more closely resembles a conjugated polyene than an aromatic hydrocarbon.

Cited reference:
1.
Oth, Jean F. M.; Bünzli, Jean-Claude; De Julien De Zélicourt, Yves (1974-11-06). "The Stabilization Energy of [18] Annulene. A thermochemical determination". Helvetica Chimica Acta. 57 (7): 2276–2288.

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text:
Determining Acidity in organic compounds
Among the following compounds,the strongest acid is:
(a) $\ce{C2H2}$
(b) $\ce{C6H6}$
(c) $\ce{C2H6}$
(d) $\ce{CH3OH}$
My doubt is how do we see acidity in such cases.....arhenius or lewis definition or we just see acidic hydrogen in such cases

Arrhenius definition.

Arrhenius definition is about water, so it is no good here. Use Brønsted instead.

@IvanNeretin but who will they be donting hydrogen to?

Who cares? Their relative acidic strength does not depend on the nature of the base  you use.

@IvanNeretin can I just say that conjugate base of ethanol is more stable because of negetive charge on oxygen...will that be enough

Yeah, that's about right, except that you don't have ethanol here.

@IvanNeretin Oops ...that would be methanol

True. I guess you must be able to order the remaining three by their acidity as well, but the question doesn't ask for that, so we're done.

They can only donate protons to super bases because they are such weak acids.

The conjugate base of ethanol would be very unstable in aqueous solution because it would deprotonate water: $\ce{EtO-  + H2O \bond{->} EtOH + OH^-}$

To judge the acidity in any case, just check out the stability of the conjugate base of the acids given. More the stability of the conjugate base, more is the acidic character in the acid.
So according to what I said, the $\ce{CH3OH}$ is most acidic compound among all of them. This can be easily seen if we compare the stabilities of their conjugate bases.
For (a): If ethyne loses a proton, its conjugate base will be $\ce{HC#C-}$. The negative charge resides on a $\mathrm{sp}$ hybridized carbon.
For (b): If benzene loses a proton, in its conjugate base, the negative charge resides on a $\mathrm{sp^2}$ hybridized carbon.
For (c): If ethane loses a proton, its conjugate base will be $\ce{H3C-CH2-}$. The negative charge resides on a $\mathrm{sp^3}$ hybridized carbon.
For (d): If methanol loses a proton, its conjugate base will be $\ce{H3C-O-}$. The negative charge resides on an oxygen atom(no need to take care of the hybridization).
We can see that the point of difference among all the conj bases above is the atom on which the negative charge resides.
As we know greater the electronegativity of the element holding the negative charge, more is its stability. Also more the $\mathrm{s}$-character of the atom, more is its electronegativity.
So the order of electronegativities of the atoms holding the -ve charge is: $\ce{O}>\mathrm{sp}>\mathrm{sp^2}>\mathrm{sp^3}$.
Now the order of the stability of conjugate base is: $\mathrm{(d)>(c)>(b)>(a)}$
And hence the order of their acidic character is: $\mathrm{(d)>(c)>(b)>(a)}$

The acidity of a substance is quantified by the energy it takes to remove a proton from it, $RT \ln(K_\mathrm{a}) = RT \mathrm{p}K_\mathrm{a}$. Thus to find which of these substances is the most acidic you would look up the individual $\mathrm{p}K_\mathrm{a}$ values and list the compounds in ascending order of $\mathrm{p}K_\mathrm{a}$. The compound with the smallest $\mathrm{p}K_\mathrm{a}$ is the 'strongest' acid and I put that in quotes because these are all very weak acids indeed - all weaker than water. But compare the structure of water, $\ce{HOH}$, to $\ce{CH3OH}$. Pretty similar, no? If this suggests to you that the acidity of $\ce{CH3OH}$ may be close to that of water then your surmise would be correct. The other compounds listed are much, much weaker acids.


text:
What is the modern hardware/supermarket term for 'laundry bluing'?
I want to make this amazing paper Crystal Tree with my kids. One of the ingredients is laundry bluing. It is described thus:

The magic crystals that are left behind are a combination of the Mrs. Stewart’s Bluing and the table salt. The solution that you created is supersaturated by the bluing and salt that you add to the water. The bluing is a colloid, with many tiny particles suspending themselves within the water. It’s just like when you shake up a snow globe, except the particles of bluing are much smaller than the snow. As the bluing and salt water make their way up the tree, the water begins evaporating. These means there is less water able to support the bluing particles and dissolved salt. This evaporation allows the salt and bluing particles to crystallize, resulting in your beautiful tree.

No one in my supermarket has heard of laundry bluing. Is it possible it is hazardous and has been banned in some states/countries? Perhaps it has another name? Perhaps this has a chemical name?
My question is: What is the modern hardware store/supermarket term for 'laundry bluing'?

I think it's just this: https://www.amazon.com/Mrs-Stewarts-Bluing-H-PC-81402/dp/B001NEMV3Q

Agree but the person asking might not be in the US so might need a chemical name

@hawkeye but the person asking says they are from Texas.

@hawkeye either get it at Ace Hardware or read this: https://sciencing.com/can-use-instead-liquid-bluing-crystal-experiments-17750.html

The active ingredient is prussian blue which you can make yourself from potassium ferrocyanide and an Fe(III) salt.


text:
Name of corresponding anion of a hemiacetal
If a hemiacetal is deprotonated at its hydroxyl group, what is the correct name for the resulting anion (or salt)?

I could not find anything, and assume that these compounds are probably not stable (just like most hemiacetals themselves), but I'd still be interested what's the right nomenclature here - something with -ate, -alkoxide, or the like?

Regenerates the carbonyl compound rather quickly

Hemiacetals can actually be named as ‘hemiacetal’ using functional class nomenclature, e.g. pentanal ethyl hemiacetal. However, the preferred IUPAC names are generated using the usual systematic substitutive nomenclature, i.e. as substituted hydroxy compounds, e.g. 1-ethoxypentan-1-ol. The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-66.6.5.2 Hemiacetals and hemiketals
Compounds with the general structure $\ce{RR'C(OH)(O-R'')}$ have the class name ‘hemiacetals’. They are named substitutively as ‘alkoxy’, ‘alkyloxy’, ‘aryloxy’, etc. derivatives of an appropriate hydroxy parent compound, such as an alcohol; these names are the preferred IUPAC names. Other names are formed by functional class nomenclature using the class name ‘hemiacetal’; similarly derivatives of ketones are denoted by the class name ‘hemiketal’.

According to P-72.2.2.2.2, an anion formed by subtracting $\ce{H+}$ from a hydroxy compound that can be expressed by a suffix ‘ol’ is systematically named by using compound suffix ‘olate’, formed by addition of the ending ‘ate’ to the suffix ‘ol’.

P-72.2.2.2.2 Anions derived from hydroxy compounds
An anion formed by subtracting a hydron from the chalcogen atom of a hydroxy characteristic group, or a chalcogen analogue, that can be expressed by a suffix such as ‘ol’, ‘thiol’, ‘-peroxol’, etc., is preferably named by using suffixes ‘alate’, ‘thiolate’, ‘peroxolate’, etc., formed by addition of the ending ‘ate’ to the suffixes ‘ol’, ‘thiol’, ‘peroxol’, etc. The multiplying prefixes ‘bis’, ‘tris’, etc. are used before these suffixes, to avoid any ambiguity.
(…)

Disappointingly, the current IUPAC recommendations still retain traditional names as preferred IUPAC names for anions of several simple alcohols (such as ‘ethoxide’ instead of the systematic name ‘ethanolate’).
The anion of my example 1-ethoxypentan-1-ol, however, is not affected by such exceptions; the systematic name 1-ethoxypentan-1-olate is also the preferred IUPAC name.


text:
pH value change with dilution

How does pH value change with dilution for the question shown in the picture? I expect the pH value to increase with dilution. Is there any case where the value remains same with dilution?

You were correct in stating that the pH value would increase with dilution, though this explanation is incomplete. To understand why you must understand what pH actually is. In simple terms, pH is a scale based on the concentration of H+ ions in a volume of a solution. As you likely know, the higher the number, the higher the basicity; the lower the number, the higher the acidity; with pure water being exactly 7 in the middle of the scale. This is where the issue arises. Yes, diluting the acid with pure water would decrease the concentration of H+ in the solution and therefore increase the pH, but only until it approaches 7, the pH of water. It is possible to go higher than that by adding bases, but pure water is specifically mentioned in the question, so that does not apply here. As for your last question, I do not believe that any acids are "immune" to dilution, as one can always add more water to decrease concentration, and therefore increase pH (to 7).


text:
What is the difference between n-heptan (CAS 93924-37-9) and branched, linear and cyclic heptane (CAS 426260-76-6)?
I am looking at the substitution of hazardous substances in my company. I would like to explain what makes the branched, linear and cyclic heptane (CAS 426260-76-6) safer than n-heptane (CAS 93924-37-9).
I do not know what is the influence of molecular structure of heptane on the properties of this substances. I mean here two different primers (chemical adhesion promotor) on heptane basis. One of them includes branched, linear and cyclic heptane and methylacetate mixture (this primer is not carcinogenic) and the second one n-heptane only (is carcinogenic).
Is there someone who can explain it?

Linear and n are the same.  The rest is  about flash point boiling point flammability.

After your edit I just note that both primers have n-heptane (name for linear heptane) in their formulation. So it is not clear why or  what something makes one carcinogenic and the other not.also is not clear what are you really asking. Their structure?  Their mechanism of carcinogenic potential? Safety in general

Hi, thank you for your response. I am asking about their mechanism of carcinogenic potential and safty in general. Once again. I have two primers, one of them includes only n-heptane isoalkane, cyclic alkane. The second one is a mixture of branched, cyclic and linear heptane (40-60%), methylacetat, naphtha (dearomatized and hydrodesulfonated)  and tributhyl-O-acetylcitrat. The first one is cancirogenic, the second one is not. I can not explain what causes the second mixture more safety. Both include heptan as a main substance.

Go through all the components and look for the carcinogenic ones. For safety in general you have safety datas . For the formulations trust the companies. At least now is more clear what you are asking and perhaps someone can answer


text:
Determining the theoretical mass of gas produced through the process of electrolysis
I have a question that has been stumping me for quite some time. I wish to calculate the mass of gas that is produced from varying concentrations of salt (NaCl) through the process of electrolysis.
I know the calculation for finding the moles of gas produced.
n = Q/(Fz)
The problem is that I do not have the number of coulombs. Therefore using the equation:
Q = It
I am able to find the coulombs but now I do not know how many amps. I know that amps can be found by dividing volts by resistance:
I = V/R
Now I need to know my resistance. I have found on the Internet that the conductivity can be determined by multiplying the molar conductivity by the concentration but I'm not 100% sure.
Can anyone please help me solve this soon?
My mass of salt used was table salt (NaCl)
20g
40g
60g
80g
100g
My volume of water was
255mL
The battery I used was
9v energizer(alkaline)
I placed the batteries directly into the solutions

But I need to find the resistance

Without knowing conductivity

https://sciencing.com/calculate-conductivity-due-concentration-2691.html

This is the site where I found the calculation of conductivity from varying concentrations

Finding the resistance or the conductivity of an electrolysis cell is not that easy when doing it with pure math.
Why don't you just apply a voltage to your cell and measure the current?
R = U / I (resistance = voltage / current)
The resistance of your cell depends on many factors:
a) Cell design

bottlenacks
distance of electrodes
number of electrodes
material of the electrodes
surface area of the electrodes

b) Electrolyth

What sort of salt you are using
The ion concentration of the dissolved salt

c) Temperature

the temperature changes the resistance of the electrolyth a lot.

You see, that things are not that easy and there is no simple formula, at least not that I know of, to calculate the resistance of an electrolysis cell.


text:
Why does the size of hybrid orbitals vary as sp³>sp²>sp?
In my text book it said that

Size of hybrid orbitals vary as sp³>sp²>sp

Does this size variation in hybid orbitals means that S orbitals are smaller than P orbitals?
But we know that larger the quantum number "n" is, the larger the orbital. I guess it has nothing to do with type of orbital.
What could be the reason for this variation in sizes?

S orbitals are smaller than p orbitals.

$n$ is identical for the valence s and p orbitals of the same element.

Another thing to point out is the character of each of these hybridized orbitals. We know that the s orbital is smaller than the p orbital. Thus, sp3 is 25% s character, sp2 is 33.3% s character and sp is 50% s character. Since hybrid orbitals are a type of average, this would explain the decreasing size of the hybridized orbital.

Looking at the shapes of atomic orbitals shown in many textbooks can be deceptive. The probability density of the radial part of the wavefunction ($r^2|R_{n,l}(r)|^2$)  shows that the maximum probability for 2s orbitals is noticeably greater than for 2p, and for d orbitals the same applies with p greater than d.  (As the principle quantum number increases so does the position of the maximum probability density of an orbital, i.e. increasing $n$ increases the size of the orbital.)
The hybrid orbitals form from  the atomic orbitals and are added together in different proportions. So the position of maximum probability is a type of average value of the s and p. The sp hybrid has $2s \pm 2p$, the $sp^2$ in the z direction $2s+\sqrt{2}.2p$ and the $sp^3$ also in the z direction just as for the sp orbital. It seems to be that the maximum position of probability taken along, say, the z direction is the same in each case.

Re: "maximum probability is a type of average value of the s and p." in your next step, have you done vector addition of the $s$ and $p$ orbitals?

@Guarang Tandon no just simple addition of radial functions but just along one axis as by symmetry it is the same along others. To see formula look in D. McQuarrie & J. Simon 'Physical Chemistry' chapter 10.

Can you please illustrate with a diagram? Or anything you deem more useful? For someone who hasn't seen those diagrams before, this addition of radial functions is not obvious at all, and could use a bit of detailing. [Besides that, another thing, that $sp^2$ orbitals are (trigonal) planar, so won't they have a zero addition value along an axis perpendicular to their plane?]

Simply as p character increases length of the hybridised orbital increases. As value of wavefunction of p orbitals is root 3 times of the wavefunction  of s orbital. What else you expect??


text:
Sulphonic acid group in benzene ring
Is the sulphonic acid group a benzene ring activating [ortho-para directing]  or deactivating group [meta directing] ?
My thought: As highly electronegative O atoms are attached, the sulphonic acid group will take electron from benzene ring by negative inductive effect...So it would be meta directing.

Your thoughts  are missing in the question .

Take a look at the homework policy https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange

@OscarLanzi Don't talk like typical homework OP.

It is a meta-directing group and a deactivating group because the $\rm -I$ effect of the group is quite strong and significantly outweighs that stabilization caused by conjugation (that you can describe with resonant structures).
Your question has some confusion in the concepts. A ortho-para-directing group is not necessary an activating group. Halogen is a typical counter example.
Your reasoning that highly electronegative O is attached and hence the group is meta-directing is incomplete because you fail to consider the conjugation. $-\rm OH$ is an example. It has $\rm -I$ effect, but is an ortho-para-directing and activating group.


text:
Which is more acidic BF3 or AlCl3?
Which of the species $\ce{BF3}$ and $\ce{AlCl3}$ is more acidic, both are electron deficient.
In case of $\ce{BF3}$ one can argue that there is extensive p-π-p-π back bonding and as a result the electron deficiency is less.
That means that the positive charge on the overall molecule is less.
In case of $\ce{AlCl3}$ one can argue there is p-π-d-π back bonding and as a result the electron deficiency is more in case of aluminiuim.
Therefore the overall positive charge on the molecule is more, meaning it is more acidic.
So $\ce{AlCl3}$ is more acidic that $\ce{BF3}$ is conclusion of the above factors. Is this plausible?
I also faced the counter logic that $\ce{BF3}$ being a small compound tends to react more readily and therefore is a strong acid.

Back bonding would dominate

D-orbitals in aluminium or chlorine? No.

@Martin-マーチン why can you explain?

[AlCl6]3-  exists what about that

@starunique2016 The molecular ion $\ce{[AlCl6]^3-}$ is predominantly bound by ionic interactions. The covalent part is best described with multi-centre bonds. In valence bond theory you have to invoke resonance. In none of these have the d-orbitals a role beyond polarisation effects.

Actually, there are three reasons why BF3 is more acidic than AlCl3.

BF3 is smaller in size and can easily attract the incoming pair of electrons.
Fluorine is more electronegative than chlorine. So boron has less ionisation potential as compare to aluminium thus forms anion easily.
Boron is more electronegative than aluminium, therefore boron has more tendency to attract electrons. It means that BF3 behaves more like Lewis acid than AlCl3.

This photo has been taken from this article.

It would be preferable to include a human readable citation instead of just a link to a paper.

I have inserted a photo for better understanding the matter.

The image certainly is a good addition. I kindly ask you to include a citation for it though. One that you would use in other articles. Thank you.

really well explained, even i think that this is the correct answer...i was trying to convey the same in my answer

@Exeplone According to the attached photo, Lewis acid strength order is: $\ce{BF3 < AlCl3 < BCl3}$

Applying Pearson's HSAB concept-
Toward hard Lewis bases such as ethyl acetate, the Lewis acidities of the halides of group 13 elements would decrease as the softness of the acceptor element increases:
$\ce{BX3 > AlX3 > GaX3 > InX3}$
Toward soft Lewis bases such as dimethyl sulphide, the Lewis acidities of the halides of group 13 elements would increase as the softness of the acceptor element increases:
$\ce{BX3 < AlX3 < GaX3 < InX3}$
But for comparison of $\ce{BF3}$ and $\ce{AlCl3}$, experimental data is required as provided by @Exeplone

Lewis acids are those which can accept electrons easily. Now, in BF3 molecule, B is small in size and thus can easily attract the incoming pair of electrons. Also, the presence of three fluorine atoms makes the boron acquire a partial positive charge, due to which boron easily accept electrons acting as Lewis acid. But, in the case of AlCl3, the size of Al is large so incoming electrons are less attracted.
(Since Boron is more electronegative than Aluminium, it can attract electrons more easily and since Lewis acids are electron acceptors, BF3 is a stronger Lewis acid than AlCl3).
Thus BF3 is better Lewis acid than AlCl3.

Unfortunately that is not correct, and not thorough enough for such an involved problem.

$\ce{BF3}$ is less acidic than $\ce{AlCl3}$ as there much stronger pi-backbonding between the Boron and Fluorine atoms while there is no backbonding in AlCl3 as well. The tendency of boron to accept electrons from the base, is way lesser than that of Al.
And, There is no backbonding in AlCl3, check out this link:
https://byjus.com/questions/why-alcl3-does-not-show-back-bonding/

No.  There can be backbonding in Al-Cl systems.

@KartikAnand That is one of the problems of this answer. You should have said that there is backbonding in $\ce{AlCl3}$, because there is. Also, nobody ever said that there is noc backbonding in $\ce{BF3}$. The whole reasoning with "octet completion" as a drop in for acidity is hand-wavy at best.

isn't BF3 a stronger Lewis acid in comparison to AlCl3, considering that it is more electronegative and thus attracts electrons more easily?

@Martin-マーチン I edited my answer. Let me know if that still doesn't sense.

@RaghavendraSingh No. Electronegativity is not the deciding factor here, as effect of Backbonding is way stronger than electronegativity. It is the least to effect factor in any comparison.

That source page you are referring to probably has so much tracking on it that they decided it's not going to be available in the EU. In any case, judging just from the title, the content will be completely wrong.


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General organic chemistry
What is the correct order of boiling points of the following compounds ?
1) butanol
2)butanal
3)butanoic acid

It takes like a minute to google that?

Boiling points depends on strength of intermolecular interaction. In case of butanol and butanoic acid there is H bonding which is a stronger intermolecular attractive force. Further butanoic acid remain in dimeric form so there is more extent of H bonding. So order of boiling point is butanoic acid > butanol > butanal.


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Understanding an Electrochemical Cell

Up there is a Electrochemical cell. I have marked the anode (-) and cathode (+) (by taking help from google) but I still need the explanation because such assignment of charges on given electrodes is pretty unacceptable to me.
Here is why.
If the copper electrode is (+) and since we have considered electrons negatively charged, the CU++ ions will not be attracted by copper electrode. Then how come CU++ ions go towards copper electrode and reduction takes place at all? (Like charges repel.)
Actually the book I am reading has not marked the electrodes (+) and (-) by itself however it asks 'will the zinc rod be positive or negative" as I have circled it in the image.
Help taken from following image

Well, it is clear from your reasoning that the copper electrode can't be  (+). Then maybe it can be (-)?

Do you really think that people should bend their necks to see your image?

@IvanNeretin I have added an image you can check out dude.

I seriously don't see why it got downvoted, it is a brilliant question although a very low level question but the problem is genuine and I am sure many many beginners find and are going to find the same problem and this post is going to help them out.

Don't think of the copper rod as positively charged but as less negative.
On both electrodes there's an electrochemical equilibrium:
$\ce{M<->M2+ + 2e-}$ (charge may vary depending on the metal)
For more precious metals this equilibrium is further to the left than for non-precious metals. Therefore there are more electrons on the zinc electrode than on the copper electrode which will travel to the less negative electrode in order to compensate for the potential differences.
As ions of more precious metals are more likely to form neutral atoms and there's a constant flow of electrons coming to the copper electrode, the copper gets reduced.
Please note that this answer is strongly simplified.

Anode is the place where the oxidation reaction occur; Cathode is the place where the reduction reaction occur. The reaction is $\ce{Zn +Cu^2+->Cu + Zn^2+}$ so you can tell that Cu is undergoing a reduction reaction, hence it is the cathode.
To elaborate, copper rod's standard potential is more positive than zinc rod's ($E_{\ce{Cu}} = + 0.35\;V, E_{\ce{Zn}} = -0.76\;V$). So, copper will withdraw the electrons from zinc if electrical connection is established. Why? You pointed that out - because of different electrochemical potential.
So a) why is cathode positive and b) why does a reduction of positively charged Cu ions take place at the "positive" cathode?
a) Why is Cu cathode positive?  $\underline{First\; case:\; electrodes\; are\; not\; connected.}$ Immersion of Cu and Zn in the dilute solution of their salts leads to the dissolution of metals and to the formation of an electrolytic double layer (see Galvani potential). Both metal electrodes are negatively charged and the double layer prevents further dissolution of metals because of electrostatic repulsion. In other words - an electrochemical (not chemical!) equilibrium is established. Note that the metal electrodes are both negatively charged and are neither anode nor cathode! However, the chemical potential of the zinc electrode is greater than that of the copper electrode:$$ \mu^*_{Zn}>\mu^*_{Cu}.$$
One can easily see it in the picture where Zn electrode has an excess of electrons. You can think of that as a kind of electron pressure.
${\underline{Second\;case: electrodes\; are\; electrically\; connected.}}$

The two electrodes now have to seek for a new state of equilibrium, with the electrons running from zinc to copper. This happens because the relationship $$ \mu^*_{\ce{Zn}}=\mu^*_{\ce{Cu}}$$ has to be valid. Just imagine that the electrons flow from an area of "high electron pressure" (Zn) into a "low electron pressure" area (Cu). According to the convention, electrons flow from the negative pole to the positive pole. So, copper electrode has a plus sign and the Zn electrode has a minus sign.
b) Why does a reduction of positively charged Cu ions take place at the "positive" cathode?
Once connected, the electrochemical equilibrium (see above) is disturbed at the electrodes. The cells or systems (electrode + solution) try to restore equilibrium. Reaction and equilibrium constant for Cu are:
$$ Cu\rightleftharpoons Cu^{2+} + 2e^- $$
$$ K_{Cu}=\frac{\left [\ce{Cu^2+} \right]}{\left[\ce{Cu} \right]} $$
As the concentration of electrons increases, so increases the equlibrium constant. The equilibrium point thus shifts to the left or to the educt side (similar to the Le Chatelier principle). This is the reduction reaction and therefore Cu electrode is the cathode.
Same for Zn:
$$ Zn\rightleftharpoons Zn^{2+} + 2e^- $$
$$ K_{Zn}=\frac{\left [\ce{Zn^2+} \right]}{\left[\ce{Zn} \right]} $$
 If electron concentration decreases, the equilibrium point shifts to the right or to product side. This is oxidation, and Zn electrode is the anode.

Maybe let us talk about the potential values of copper and zinc after we get why is copper rod negatively charged?

@Mostintelligent If you don't like numbers, focus on the reaction, you do know that the reaction should be $\ce{Zn +Cu^2+=Cu +Zn^2+}$, don't you?

what about reaction ? @weijunZhou

@ Most intelligent Sorry, it was wrong! Of course, as you said, the cathode has to be positively charged.

Don't put 2e in the expression of $K$! It makes no sense. $K=\ce{[M^2+]}/c^\circ$ is already fine. Putting $[\ce M]$ in the expression of $K$ is not that bad, but it's a constant anyway so it is usually omitted.

@Mostintelligent If you know the direction of the reaction (at standard state), you can figure out which side is going through a reduction half reaction and which side is going through an oxidation half reaction, hence figuring out which electrode is a cathode, as explained in this answer.

@WeijunZhou put that comment as an answer.


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When there's a photoelectric effect, do the electrons leave altogether or leave one by one?
When a strong light, or even a laser, hits a metal surface or some molecules, and causes lots of electrons to fly away, do the electrons leave together, or do some of them leave first? If it's the latter, then do the valence electrons leave first and then inner layers? If so, how long are the intervals between their leavings?

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

In 1905, Einstein demonstrated that the photoelectric effect was a quantum phenomenon. That is, for each photon of light of a minimal specific wavenumber, one electron in a specific orbital is released.
For example, for cesium in its base electron configuration, $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 6s^1}$, about $\pu{3.43E-19 J}$ is needed to raise that electron to, effectively, an "infinite" orbital (where it is unlikely to be found near the atom). This corresponds to a wavelength of $\pu{\approx 580 nm}$ or wavenumber of $\approx \pu{17200 cm-1}$.
That said, very energetic photons may knock out more than one electron apiece, with some of the photon's surplus energy applied to the kinetic energy of the first electron, and some applied to ionizing the atom further. However, at the minimum photon level, for every photon absorbed, one electron is freed. A brighter light has more photons, so a brighter light releases more electrons (with limitations due to the build-up of free electrons near the $\ce{Cs}$ cathode, the space charge).

In the photoelectric effect the energy of the photon is completely transferred to the electron. If the energy is higher than the minimum required to eject the electron, it will leave with higher kinetic energy. Then, the electrons only have to be elevated to the conduction band, not to an infinite orbital.

Actually, photons, particularly high energy photons, can cause multi-electron emission: https://www.sciencedirect.com/science/article/pii/S0368204816301074  &  http://pubs.acs.org/doi/abs/10.1021/jz500489s?journalCode=jpclcd

Only electrons that are hit by a photon with the appropriate wavelength will be moved. So that means that they are leaving one by one. This can however ocurr at several places at a time so that means that they can leave simultaneously. In fact, if a current of 1 mA is maintained during 1 second, $\frac{10^{-3}\cdot 1}{1.6\cdot 10^{-19}}=6.25\cdot 10^{15}$ electrons will have left the metal.
Here $I=Q/t$ ($I=$ intensity, $Q=$ charge, $t=$ time) and the charge of one electron is $1.6\cdot 10^{-19} $ Coulomb.
The thing is that the metal remains electroneutral. It is not that a huge number of electrons leaves the metal, stripped off by the incident light. The whole setup works because there is an external electric circuit. Every electron that leaves the metal (or the semiconductor, for that matter) is immediately replaced by another electron returning from the circuit.
The deeper electrons in the atoms are never moved. Once and again the external electrons are pushed to the conductive band and replaced by fresh electrons from the circuit.

What if the atoms or molecules are not charged by anything? And the energy of the light is strong enough to make multiple electrons to escape from one single atom or molecule? Then are these electrons flying away simultaneously or one by one or pair by pair?

Only the electrons from the outer shell are involved in the photoelectric effect. So if there is only one electron, only one can escape. If there were two, after the first one has gone, the atom is charged $+1$ so the second one will need more energy to leave. To leave at the same time means that two photons hit both electrons simultaneously, which is an unlikely event, but there should not be a problem for them to leave both if they also could do it sequentially. "Strong enough" here means that you have a lot of photons, not that their energy is very high.

Do you mean that only after the outer shell electrons are gone can the inner ones be gone? If the photons are very dense, then can the inner electrons and outer electrons absorb photons simultaneously and thus fly away simultaneously? Or only after outer ones leave can inner ones absorb photons and leave?

As far as I know, the inner electrons cannot be moved. Even in the Compton scattering, in which much higher energy photons are used, only free electrons are ejected. See for instance https://physics.stackexchange.com/questions/31581/compton-scattering-vs-photoelectric-effect?noredirect=1


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Why don't particles of a colloid settle?
Why don't particles of a colloid settle but those of a suspension do?
- Is it because they are really small and so remain in constant motion which overcomes the force of gravity and prevents them from settling?
- Is it because they are charged and so they repel each other and stay suspended in the solution as they try to go away from each other?

Particles in colloid absorb either positive or negative ion preferentially from solution depending on there surface. This develops a potential on colloid surface called Zeta potential. Due to this potential colloid particles repel each other and can't coagulate.
Particle size also play a major role. With decreasing size surface to volume and hence surface to weigh ratio increases. So, same volume  of colloid  particles  have higher surface area than suspension. So, colloid surface have enough charge to keep them apart whereas suspension don't have.
Conclusion Both the point you've mentioned is responsible for prevention of colloid coagulation.

Settle down of colloids is thermodynamically favored. So the charge on the colloid particles only alters the kinetic barrier?


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Ethanol Water mixture viscosity
It is found that increasing composition of $Ethanol$ in $Ethanol$-$water$ mixture increases the viscosity of the mixture upto certain composition, but after that the viscosity begins to decrease again. I could not figure out any theoretical explanation to this phenomenon. Any help with theoretical reasons will be appreciated.

Maybe because Ethanol-Water interactions are most effective at the point of highest viscosity. I guess this is related to partial molar volume in a mixture of ethanol and water which depends on the mole fraction of the components.

Alcohol water mixtures have v. unusual behaviour, probably not fully understood.  At different compositions  but with the same viscosity molecular properties are different as measured by  probe molecules having different rotational diffusion times. The reason was rationalised as their being large clusters of ethanol or large ones of water depending on composition. Not what one would expect.

That's a homework question, isn't it? Please write what behaviour you would expect (e.g. for a simpler mixture), and why.


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is there any evidence that there are ~75 extra elements on the periodic table?
there are 118 currently existing elements on the periodic table, and recently I've found like 77 elements that quote on quote "may exist" on our periodic table. Is this true?

Search the internet and you  will get the answer

Questions are downvoted, not users - but I know that they can be off-putting either way. Do be sure to use the help center, and the search. As pointed out, there are resources out there that could give you some idea, eg : https://en.wikipedia.org/wiki/Extended_periodic_table

oh okay, I thought people were just downvoting because they could.

The answer is nobody knows.  See the referenced question in the "duplicate question" block for some ideas.

okie dokie, I'll go check it out :)

Also you should probably cool down a bit, and check [help].


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Why are copper salts less soluble than zinc salts?
Compounds with cations having pseudo-inert gas electron configurations have more covalent character than those that don't, therefore they should be less soluble.
Here are some solubility products:
$$
\begin{array}{lcc}
\hline
\text{Anions}    & \ce{Cu^{2+}}                 & \ce{Zn^{2+}} \\
\hline
\text{Arsenate}  & \color{red}{ \pu{7.95e-36} } & \color{red}{ \pu{2.8e-28} } \\
\text{Hydroxide} & \color{red}{ \pu{4.8e-20} }  & \color{red}{ \pu{3e-17} } \\
\text{Sulfide}   & \color{red}{ \pu{8e-37} }    & \color{red}{ \pu{2e-25} } \\
\text{Phosphate} & \color{red}{ \pu{1.4e-37} }  & \color{red}{ \pu{9e-33} } \\
\hline
\end{array}
$$
Zinc ($\ce{Zn^{2+}}$) has a pseudo-inert gas configuration ($\ce{[Ar] 3d^{10}}$), and its salts are more soluble than those of copper ($\ce{Cu^{2+}}$), which does not ($\ce{[Ar] 3d^{9}}$). Why?

I don't know. I've never seen a comparison between Cu and Zn covalency simply based on electronic configuration. Now if we were talking Ca vs Zn, then yeah, I can agree that $\ce{Zn^2+}$ is more polarising than $\ce{Ca^2+}$, hence more covalency. Incidentally, the example given on Wikipedia is $\ce{Ca^2+}$ vs $\ce{Hg^2+}$. If there's a reason for the difference in solubilities listed in the question, I doubt it has much to do with pseudo inert gas configurations - but I may be proven wrong.

Why would complexes of cations with pseudo-inert gas electron configuration be more covalent? I am lost here. If anything, I would expect the opposite.

C.F.S.E., maybe?

It is because the difference in electronegativity.
The higher the difference in electronegativity between the cation and the anion, the stronger the ionic bond.
Since we are comparing species with the same anions and different cations, we only have to look at the values of the latter ones.
cation    electronegativity   configuration
  Cu             1.9         [Ar] 3d^10 4s^1
  Zn             1.65        [Ar] 3d^10 4s^2

$\ce{Cu}$ is more electronegative so the bonds have less ionic character which makes for stronger bonds.
The same happens in the same period between $\ce{K}$ and $\ce{Ca}$, but here the situation with regard to the electronic configuration is reverse.
Potassium salts are much more soluble than calcium salts, and again it is explained by the electronegativity:
cation    electronegativity   configuration
  K              0.82           [Ar] 4s^1
  Ca             1.0            [Ar] 4s^2

Never thought of interpreting that $\ce{K+}$ salts are more soluble than $\ce{Ca^2+}$ from this perspective. Interesring point, just wondering whether there are any counterexamples (fluorides don't count).

I wonder if it makes any sense to compare cations with different charge this way. They form different kind of lattices, and with eg organic anions there is a clear difference based on the charge instead of electronegativity.


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Calculating the density of a saturated salt solution
Is there a way to calculate the density of a saturated salt solution from the solubility limits or does it have to be experimentally determined?
First instinct, is to add the mass of the salt to mass of the water e.g. a solubility of $\pu{80 g}$ of salt in $\pu{100 mL}$ would have a solution density of $\pu{180 g/100 mL = 1.8 g/mL}$. However, it seems the salt should affect the volume of the solution. What additional information is needed or does this quantity just have to be empirically determined?

You can't calculate that.

This is the sort of thing where just measuring it is far easier than trying to calculate it.

Often, solubility is given as mass per volume of solution (not solute). Nevertheless, unless you have the mass percentage (w) too, you are not able to calculate the density of the solution.

This is something you can calculate, provided you know the partial molar volumes of all the substances in your solution. But someone must have measured those at some point, and tabulated it for your conditions (or close enough that you can interpolate). You can probably estimate it closely enough, depending on your needs.
The experiment would be as simple as taking a known volume aliquot of your saturated solution and weighing it.

Your proposed method (adding the masses, but using the volume of the solvent) can work under certain circumstances; it is often a good first approximation for the times when it is not a good assumption.
For low solubility salts, you can safely assume that the volume of the solution is the same as the volume of the solvent.  "Low" here is somewhat open for debate.  As a general rule, I feel safer with this assumption for each order of magnitude below 1 molar / molal.  So, I consider this method valid to one (or fewer) sigfigs for solutions near 1.0 molar; 0.1 M makes me feel "2 sig figs safe", etc.  (Again, these are just guidelines, don't get hung up on specific numbers.)  With a salt that can only dissolve fewer than 10 grams into 100g of water, this method is usually fine to 2 or 3 sig figs.  A salt that dissolves less than 1.0g into 100g of water is now getting into the territory of measurement errors of accurately measuring the volume of the solution (in most high school or even some college labs).
Seawater runs about 3% salt to water by mass.  Estimating it's density at 1.03 g/mL is pretty valid.  In the example in the OP of 80g of salt in 100mL of water (giving a value of 1.8 g/mL) is a good first order approximation, but I would take that value with... a grain of salt.


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Why an acid form h3o+, not h3o
When a lone proton bonds onto an H2O molecule in an acid, why doesn't the proton bond with a pair of the lone electrons in H2O and satisfy its electrons needs by having 3 single bonds? Why is it still charged?

Okay let’s look at the H2O molecule first. The oxygen atom has 6 protons and each of the hydrogen atoms have a proton. Hence the H2O molecule has a total of 8 positively charged protons. Oxygen has 6 valence electrons and each of the hydrogen’s have 1 valence electron, directly involved in covalent bonding with the oxygen atom. Hence the H2O molecule has a total of 8 negatively charged electrons. As the number of protons = number of electrons in the molecule, the molecule has no overall charge. When a lone proton forms a dative covalent bond with a lone pair on the oxygen atom of the H2O molecule, there are now overall 9 protons and still the same 8 valence electrons in the molecule. Hence due to the imbalance between protons and valence electrons in H3O, there is an overall positive charge on the molecule, hence why it exists as H3O+ when a lone proton (no electrons) bonds with H2O.

Why doesnt the OH- molecule kick out a valence electron so it will acheive electrical neutrality. It looks like the ions are driven by a tendency to satisfy the octet rule, over the tendency to balance electrical imbalances.

There aren't any free protons floating around it water. They are attached to water molecules or acid molecules. If a positively charged proton is transferred to a neutral water molecule (from another water molecule or from an acid molecule) then clearly that molecule becomes positively charged.
Now how is the new proton bonded to the water molecule? If you consider it to be ionically bonded then you are assuming the oxygen is hanging onto its electrons (at which it is pretty good) in which case the proton retains its positive charge. But if you assume, as you seem to be suggesting, that the bond is purely covalent, then the bond would be formed by sharing the electrons equally between the proton and the oxygen. In this case you would indeed write $\ce{H_3O^\oplus}$  i.e. the positive charge becomes a formal charge on the oxygen.


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Molecule where an sp3 chalcogen is connected to an sp2 atom
I am looking for a small molecule where an $\mathrm{sp^3}$ atom from Group 16 (O, S, Se,...) is connected to an $\mathrm{sp^2}$ or resonant atom of another column?
Background: The universal force field defines the above case in its torsion
rules and I would like to find an example/application of such a structure
to understand it better.
Thanks!

You mean something like phenol?

@GeoffHutchison That was actually my first thought, too. But oxygen within phenol is considered to be $\mathrm{sp^2}$-hybridized, right?
The original UFF paper (Rappé et al, 1992) states: "For a single bond involving a sp3 atom of the oxygen column and an sp2 or resonant atom of another column, eq 17 is used directly. For this case the periodicity (n) is 2 and the equilibrium angle do is $90^\circ$."
I think that the equilibrium angle $90^\circ$ is wrong and that is why I was looking for an example.

I actually used the implementation in Open Babel to check when this rule is triggered which is not the case for phenol. I could only find the single bond in benzenesulfonic acid which is technically d2sp3?!
In the meantime I found that towhee and Gulp are using $180^\circ$. At last, I discovered the deMonNano documentation which explicitly mentions this error.

The examples below by permeakra make sense to me (no chemist), but they are too complex for the SMARTS matching/atom typing in Open Babel and hence don't trigger the rule as well. Moreover, my experiments showed that using $90^\circ$ or $180^\circ$ does not really make a difference, since the torsional barrier height is rather low. So I conclude that it is probably a niche case that was included for completeness.
[PS: I plan to submit patches to Open Babel, will take some time to assemble my findings.]

@Andi there is no simple molecules matching your requirements since, indeed, oxygen enters conjugation readily. And honestly, exhaustive enumeration of all possible cases for general use is futile, it would be best to print inferred atom types and let the user to override misdetected types by hand from the command line.

Out of the blue, consider triphenyloxonium cation. It is non-planar thanks to bulkiness of the phenyl groups.
Another possible example is mono-epoxide of allene. Considering oxygen atom as $sp^2$ here is unconvincing since it would involve huge angular strain, much higher than for a structure with $sp^3$ oxygen.


text:
Will the solubility product constant differ in this case?
I am considering the solubility product of sodium citrate. I need this value, I do not have it. However, I do know that the solubility product of calcium citrate is (7±2)×10^−14.
Is it safe to assume that sodium citrate will have a very similar Ksp?
I am concerned about whether or not the addition of sodium citrate will force borate ions out of the water as sodium tetraborate.
Furthermore, could I hypothesise on the solubility based on the likely strength of the hydrogen bonds? I am able to view the charges of each individual atom within a molecule, and the citrate's oxygen anions are far more charged than those in the borate ions. The sodium citrate's hydrogen bonds should therefore be stronger, and result in a higher solubility?
Thank you.

No, because sodium is singly charged and calcium is doubly charged, so there will be a different number of terms in your solubility product formula, and you should expect quite a large difference in the $K_\text{sp}$ between the two salts.

No. The reason you can't find the $K_{\rm sp}$ of sodium citrate is most likely because it is highly soluble. Except for a few exotic cases, sodium salts are very soluble in water.
Adding sodium citrate to the solution does reduce the solubility of borax, however this is not going to be a problem unless you are using solution with really high concentration. As a reference, the solubility of sodium citrate and borax are $\pu{770 g/L}$ at $\pu{25^\circ\!C}$ and $\pu{51 g/L}$ at $\pu{20^\circ\!C}$, respectively.
The solubility of a solid in water depends on many factors. In this case where the two solids share the same cation, we can say that thermodynamically it depends on the extent that water stabilizes the anion compared to the crystal lattice. Your reasoning that the citrate anion's hydrogen bonding with water is stronger than borate anion's is valid and I think it's fine to say that sodium citrate is more soluble than borax based on this. Just that you need to keep in mind that the crystal energy also plays a role here as well, and need to be examined with care if the oxygens are not "far more" charged in one species than in the other.

@GunaPrashant Updated


text:
Why do gas molecules travel in a straight line?
As we know, a single particle moves in a straight line until it collides with another particle. The question is how scientist figure out that gas particles move in a straight line?

This is an immediate consequence of the Newton's first law: when not acted upon by a force, a body either remains at rest or continues to move in a straight line.

Also check out How do gas molecules break newton's laws of motion and discussion therein.

I'm voting to close this question as off-topic because it's not exactly about chemistry.

I'm voting to close this question as off-topic because it duplicates an existing question on Physics Stack Exchange.

Newton's First Law of Motion states that an object at rest stays at rest and an object in motion stays in motion at a constant velocity, unless acted upon by a force.
Put more simply, stuff stays still until something else (like a force or another object) makes it move, and moving stuff moves at a constant speed unless something else changes its speed. (I substituted speed for velocity in this sentence even though they don't really mean the same thing to make my explanation simpler.)
For example, if one were to throw a tennis ball into the air, it would stay moving at the same speed and direction if it weren't affected by the "forces" of gravity, air resistance, or wind. (I put forces in parentheses because wind isn't actually a force.)
Newton's First Law of Motion doesn't just apply to macroscopic objects like tennis balls; it also applies to microscopic objects like particles. (Except perhaps at the quantum level; I'm not an expert in quantum mechanics so I wouldn't know too much about that.) Therefore one assumes that a gas particle will move in a straight line until it collides with another particle, and this assumption will generally work when experimenting with small quantities of gas in labs.
Of course, in nature gas particles don't ALWAYS move in straight lines. For example, planets form when gas and dust particles orbiting around a star clump together due to gravity, and those gas particles will be moving in a curved line relative to the star. You could also argue that no gasses on earth are really moving in a straight line since their paths curve around the sun along with everything else on our planet. It really depends on our reference frame, but Newton's First Law of Motion allows us to know a gas particle will move in a straight line until it collides with another particle within our reference frame.

I respectfully disagree with those that mark this question as off-topic. There are many areas where different fields of science overlap. For example, physical chemistry is "a branch of science applying physical methods and theory to the study of chemical systems" according to Merriam-Webster. Physics is often useful in answering questions about chemistry and vice-versa, and this question serves as an example of that overlap. So please stop marking this question as off-topic just because it makes use of physics.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Identifying amphiprotic species
I'm a bit confused about amphiprotic species – specifically why some species can be amphiprotic and others can't.
For instance, $\ce{SH-}$ is an amphiprotic species (in aqueous solution).
$$
\begin{align}
&\text{Acid:} &\ce{SH- + H2O &<=> S^2- + H3O+}\\
&\text{Base:} &\ce{SH- + H2O &<=> H2S + OH-}
\end{align}
$$
On the other hand, isn't $\ce{HNO3}$ an amphiprotic species too? I don't understand why $\ce{HNO3}$ isn't amphiprotic whilst $\ce{HS-}$ is; they seem to both be able to have an acid and a base reaction:
$$
\begin{align}
&\text{Acid:} &\ce{HNO3 + H2O &<=> NO3- + H3O+}\\
&\text{Base:} &\ce{HNO3 + H2O &<=> H2NO3+ +OH-}
\end{align}
$$
I hope this makes sense.

I guess it ultimately comes down to circumstance, consider superacids, some of which can protonate methane. Usually, one thinks about diluted aqueous solutions when talking about amphiprotic species. $\ce{HNO3}$ undergoes a base reaction like you listed in the course of nitration of aromatics such as benzene.

Amphiprotic are species that have both acidic and basic properties. A classic example of an amphiprotic ion is dihydrogen phosphate $\ce{H2PO4^-}$, which reacts in the presence of a $\ce{H3O+}$ as
$$\ce{H2PO4^- + H_3O^+ <=> H_3PO4},$$
where $\ce{H3PO4}$ is the conjugate acid of the original base.
In the presence of an $\ce {OH-} $ the $\ce{H2PO4-}$ reacts in the following way
$$\ce {H2PO4^- + OH- <=> HPO4^2- + H2O}.$$
Aminoacids are amphiprotic, as they contain both, a weak acid and a weak base functional group. Also water behaves amphiprotic.
The second reaction you wrote, even if in theory possible does not happen since $\ce{HNO3}$ is such a strong acid, that in water is totally dissociated and has no tendency to behave as a base.
You can check this by writing down the general equilibria and calculate the equilibrium constant $K$ of each reaction you wrote. The $K_\mathrm{a}$ of nitric acid is $2.4 \times 10^1$. This means that, since
$$K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w} = \pu{1.0*10^{-14}}$$
the $K_\mathrm{b}$ of nitric acid is at least in the order of $10^{-15},$ which gives you an idea why this reaction won't happen.


text:
Which easy-to-acquire clear liquid would be best for testing whether or not a twenty-sided die is balanced?
In another question on another stack, I asked if the Saltwater float present in a specific youtube video was capable of determining any differences in mass between the faces of a die in determining the balance of a group of dice, and was met with very good results and a positive affirmation that yes, it was very possible to do so.
Which led me to believe that there was a possibility that denser liquids might have the possibility of testing the balance of the dice even better, with much less mess. This led me to go through and search the densities of many liquids to determine which would be best for the job.

Water came in at ~1g/mL at room temperature, which was the medium for the test initially.
Saltwater was at about ~1.028g/mL. Dice were able to float in this.
H2O2 (Hydrogen peroxide) clocks in at ~1.44g/ML, but I tested this one myself with a commercially available H2O2 and was unable to float dice in it just on its own. It isn't dense enough to float a die on its own.
Originally I thought Galium might be a good choice since it has a high density and is very easy to melt, but it isn't transparent.

So the question is this:
Is there a liquid, that is easy to acquire, that has a higher density than H2O + MgSO4 (Epsom Salt), that it would be easier to test a dies balance in and at what density does a d20 float?
Generally when referring to dice, the most well known manufacturer is probably Chessex. The dice are made out of some kind of plastic polymer of some sort. I don't really know any specifics about their chemical composition, but they could just be injection molded plastic dice.
As for the approximate mass of said dice.. This site states that the dice themselves are ~5mg (That doesn't sound right. I learned in school that a Paper clip is a gram, Dice weigh much more than a paper clip..)

You don't want a denser liquid. You want a liquid with just the right density so the dice would barely float, almost fully submerged. To this end, you mix the solutions of different concentration and so fine-tune the density, as is beautifully described in the accepted answer to your rpg.SE question.
A denser liquid is worse for your purpose. Ultimately, with a very dense liquid (like mercury) the dice would just sit on one face, much like they do on a tabletop.
In the relatively rare case your dice might be made of different kind of plastic and wouldn't float at all. That's when you genuinely do need a denser solution. Chemistry knows a lot of salts; which of these are available to you is another question. Saturated solution of $\ce{MgSO4}$ has the density of 1.3 or so; $\ce{CaCl2}$ will get you to ~1.4, and other salts farther yet, but they are also more difficult to come by.
As for the reference books with some density values, you have to read those really, really careful to see what they actually refer to, and then read again, just in case. 1.028 looks like the density of seawater; it is salty all right, but that's pretty far from saturation. The solution you use is a lot more dense. 1.44 for hydrogen peroxide looks like the density of the pure compound, which you surely can't acquire (which in turn is probably good for your survival); what you had is probably the 30% solution which is a lot less dense.
As for the density at which your d20 will float... well, how are we supposed to guess that? For all we know, your die may be carved out of wood; also, chances are it is cast in bronze. The resulting figures may vary pretty widely. Upd. So it is plastic, after all. Well, that still leaves a pretty wide room for wiggling. You can't tell the density until you measure it. Chemistry is an experimental science, they say.
So it goes.

I've amended the question to include the type of dice and as much information about them as I have. As is currently however your answer doesn't quite answer either portion of my question.
You have neither listed an easy-to-acquire clear compound that has an approximate density of a sturated MgSO4 Solution, nor an approximate density.
The information about denser liquids is graciously accepted, I had thought about galium when I was considering easy to acquire liquids, but I can already assume it'd likely roll on the surface of the liquid instead of being suspended in it.

Do you actually have any dice that would not float in the saturated solution of MgSO4? If no, you don't need a denser liquid. If yes, try CaCl2.

Near impossible to say at what density of solution some unknown die will float. I've seen hollow plastic dice and metal dice.
I'm also not sure how reliable some float test would be. I'd guess that the various gaming commissions have standard tests for dice. Using such a standard test would be preferable to trying to invent some test of your own.


text:
Understanding the increase in pH of a buffer solution upon incremental additions of NaOH analytically
So let's say we have $200\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{CH3COOH}$ solution. In this solution we have the equilibrium $\ce{CH3COOH <=> CH3COO- + H+}$. To that we add $100\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{NaOH}$ solution. Then after reacting we get a buffer solution where $[\ce{CH3COOH}] = [\ce{CH3COO-}] = \frac{1}{3}\ \mathrm{mol\ dm^{-3}}$. In moles, we have $0.1\ \mathrm{mol}$ of both $\ce{CH3COOH}$ and $\ce{CH3COO-}$.
Then suppose we add $x\ \mathrm{mL}$ of $0.1\ \mathrm M$ $\ce{NaOH}$. This is, in effect, reacting $0.1\ \mathrm{mol}$ of $\ce{CH3COOH}$ with $0.0001 x\ \mathrm{mol}$ of $\ce{NaOH}$, the result of which is that now $n(\ce{CH3COOH}) = (0.1 - 0.0001 x)\ \mathrm{mol}$ and $n(\ce{CH3COO-}) = (0.1 + 0.0001 x)\ \mathrm{mol}$.
Could we not then model, by the Henderson-Hasselbalch equation, $\mathrm{pH} = \mathrm pK_\mathrm a + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right) = 4.76 + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right)$?
But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82\ \mathrm{cm^3}$ of $0.1\ \mathrm M$ $\ce{NaOH}$ solution, which sounds absurd. Moreover, for different acids, like propanoic acid and butanoic acid, the same line of logic could be used to deduce that the volume required to increase the $\mathrm{pH}$ by $1$ would be the same for all, but they have different buffering capacities. So what's wrong with the logic?

I just answered what I thought was this question but realized that it was essentially the same question asked 5 years ago. So here's that answer modified for the slight difference in the way it is asked here:
We need to know the the pH of the buffer i.e. what does it measure before any NaOH is added. Later we will assume that it is 4.76, the pK of acetic acid.
$[\ce{HAc}] = [\ce{Ac^-}]$ ([·] symbolizes the molar concentration of ·). In order to solve the general problem we need to be able to calculate the fraction of the total Ac that is dissociated and the fraction that isn't.
The fraction that is dissociated comes right out of the Henderson - Hasselbalch equation and is $$f_1 = 1/(1 + r_1)$$ where $$r_1 = 10^{(pH - pK_1)}= [\ce{Ac^-}]/[\ce{HAc}]$$ as is clear from inspection of the Henderson - Hasselbalch equation. The subscript 1 indicates that $r_1$ is the ratio of the number of acid ions that have lost 1 proton to the number that have lost $ 1 - 1 = 0$. In $f_1$ the subscript is indicative that $f_1$ is the fraction of the total Ac molecules that has become singly charged by loss of a single proton. When dealing with monoprotic acetic acid the subscripts aren't that important as there is only one proton to loose. But if the acid is polyprotic we have unionized, once ionized, twice ionized etc. ions to consider. The fractions of those ions are $f_0$, $f_1$, $f_2$... and we have other ratios as well
$$r_j = 10^{(pH - pK_j)} = [H_{n-j-1}Ac^{−j}]/[H_{n-j}Ac^{-(j-1}]$$ Here "Ac" stands for the acid anion and n is the number of protons it can yield when fully dissociated. For acetic acid. $\ce{CH_3COOH}$, and Ac is $\ce{CH_3COO}$, n = 1 and j only has values of 0 or 1. With phosphoric acid, $\ce{H_3(PO_4)}$, Ac is $\ce{(PO_4)}$, n = 3 and j = 0,1,2 or 3.
So suppose now we have a polyprotic acid with values for $r_1$, $r_2$, $r_3$... and that there are x moles of the acid in a solution. Then there would be $xr_1$ moles of the singly deprotonated species, $xr_1r_2$ moles of the doubly deprotonated, $xr_1r_2r_3$ moles of the triply deprotonated and so on.Then the total number of moles of Ac would be the sum of the number of moles of each$$C_{Ac} = x + xr_1 +xr_1r_2 +xr_1r_2r_3...$$ The fraction of the total that is undissociated is $$f_0 = x/(x + xr_1 +xr_1r_2 +xr_1r_2r_3...) = 1/(1 + r_1 +r_1r_2 + r_1r_2r_3...)$$ The fraction that is singly dissociated is $r_1$ times this $$f_1 = r_1f_0$$ and the fraction that is doubly dissociated is $r_2$ times that $$f_2 = r_2f_1$$ and, in general $$f_j = r_jf_{(j-1)}$$
At this point let's remember that $f_j$ is a function of the solution pH and all the pK's of the acid in question and that it is the fraction of the anions of that acid that carry charge -j. Thus we can write an expression for the total charge on all species of Ac at a given pH. This is $$Q_{Ac} = -C_{Ac}(0f_0 + 1f_1 + 2f_2 + ...)$$
Before going on to show you how Q solves buffering problems let me stop to suggest that the simplest way to work with it is to make an Excel (or other) spread sheet. Designate a column for pKs and a cell into which the pH goes. For example, put pH into cell A1 and start the pKs list in A2, Then in B2 put =10^($A$1 - A2).  Using $A$1 lets you copy and paste B2 into as many cells as you have pKs. The B column now contains the r corresponding to the pK in the cell to the left of it. Now enter the formula for $f_0$ in a cell and make another column with the $f_j = r_jf_{(j-1)}$ formula in it. So how many pK's. I say make the spread sheet for 5 or 6. Why? Well lets go back to $\ce{CH_3COOH}$ for a minute. It doesn't have 1 proton to give, as we have been assuming. It actually has 4. Are the other three ever coming off? Not with any base in my lab but we can model those other protons simply by assigning pKs that are so high (say 50) that the f values for anything other than $f_0$ or $f_1$ are 0. The point being that if you are going to go to the trouble to make the spreadsheet you might as well make it big enough to handle any acid you may ever encounter as it easily handles anything up to its maximum size using this trick.
Now how to use $Q(pH)$. If there are a total of $C_{Ac}$ moles of Ac in a solution the negative charge on them at $pH_0$ is $C_{Ac}Q(pH_0)$. At $pH_1$ it is $C_{Ac}Q(pH_1)$. Thus to move the solution from $pH_0$ to $pH_1$ you must supply or remove charge of $$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = C_{Ac}Q(pH_1) - C_{Ac}Q(pH_0)$$ by adding or absorbing protons. If $ pH_0  > pH_1$ then $Q(pH_1) > Q(pH_0)$ (less negative) and so the difference will be positive indicating that protons (acid) will need to be added to effect this pH shift.
The acid species in the solution are not the only thing that emits or absorbs protons when pH changes. The solvent does too.
$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-pH_0} + (10^{(pH_0 - pK_w)} - 10^{(pH_1 - pK_w)})$$ represents the number of protons that must be supplied (or absorbed) to change the pH of water from $pH_0$ to $pH_1$.
Now let's use this to solve the original question. We have 300 mL of 4.76 buffer with $C_{Ac}$ = 0.2 mol ( 0.1 mol of Ac and 0.1 mol of $Ac^{-1}$. This implies that $f_0 = f_1 = 0.5$ Also
$$\Delta Q_{Ac}(pH_0\ce{->}pH_1) = 0.02(Q(pH_1) - Q(4.76))$$
$$\Delta Q_{W}(pH_0\ce{->}pH_1) = 10^{-pH_1} - 10^{-4.76} + (10^{(4.76 - pK_w)} - 10^{(pH_1 - pK_w)})$$
Aparently we want to know how much NaOH we would need to raise the pH to 5.76.  With our formulas in a simple Excel spread sheet and using 5.76 for $pH_1$ we quickly find that $\Delta Q_{Ac}(4.76\ce{->}5.76) = -0.0818 mole. IOW protons must be absorbed. At the same time we find $\Delta Q_{W}(4.76\ce{->}5.76) = -1.56E-5 mole per liter. We only have 0.3 L so the protons to be removed to adjust the water are insignificant. We apparently need 81.8 mL of 1 M NaOH. Note that your digits are pretty similar so one of us is off by a power of 10. As 0.8 mol OH- to shift an 0.2 mol buffer is a lot, I think it's you. As we both used Henderson Hasselbalch, we had better get the same answer.
So why would you do what I'm suggesting instead of what you are doing (assuming you find the factor of 10). If you do a lot of problems like this it makes cranking them out a snap.
Now this post has gone on long enough so I won't have space to tell you about extension of this technique to much more complicated systems. You can use it to find the pH of complicated mixes of weak and strong acids and bases or any materials which have buffering capacity. I ginned it up (this is just using the 'proton condition' to solve a complex system - I didn't invent it) to predict the pH of brewer's mash which is a mixture of weak acid (bicarbonate) and several malts (each of which has its own buffering properties).


text:
What is the mathematical relation for the change in water density at different temperatures?
I am curious to the relation of the density of water at different temperature and as to also how one would come to an equation that took the temperature of the water and would output it's density? I have graphed density vs temp from tabulated values and it seems to look logarithmic, but wanted to make sure. I am interested in this as I am currently taking analytic chemistry.

No real analysis, just curve fitting. See http://nvlpubs.nist.gov/nistpubs/jres/097/jresv97n3p335_A1b.pdf

To do this one must measure density accurately at many temperatures and then fit a curve to the measurements. This has been done by several investigators in several industries. For example in the sugar industry we have
$$\rho =(((((-281.03006e-12*t +105.84601e-9)*t-46.241757e-6)*t-7.9905127e-3)*t+16.952577)*t +999.83952)/(1+16.887236e-3*t)$$
where $t$ is the temperature in °C and the density $g·cc^-1$. The coefficients were taken from the ICUMSA formula with the sugar concentrations set to 0.
And
$$\rho = 0.99984+t*(6.7715e-05-t*(+9.0735e-06-t*(1.015e-07-t*(+1.3356e-09-t*(1.4421e-11-t*(+1.0896e-13-t*(4.9038e-16-9.7531e-19*t)))))))$$
returns the density of water, also in $g·cc^-1$,  as a function of centigrade temperature (ITS 1990). It is based on a fit to data from
Bettin, H.; Spieweck,F.: "Die Dichte des Wassers als Funktion der Temperatur nach Einführung der Internationalen
Temperaturskala von 1990. PTB-Mitt. 100 (1990) pg 195-196
The referenced data set lists densities for each 0.1°C over the range (0,100). The data used in determining this
polynomial are the subset on integer degree values. Sporadic checking shows that the fit appears to be
accurate to 1 count in the 6th decimal place. The rms residual wrt the fit points is 3.1E-7 and the peak
residual 5.5E-7. The residuals appear noiselike.
You ought to be able to paste those formulas into a visualization program and plot values. I don't see any logarithmic behaviour. I've left the * in the formulas so you can do that. Note, when reading them, that 16.887236e−3∗t means (16.887236e−3)∗t.


text:
Balancing other atoms in a redox reaction using half reaction method
I have this balanced redox reaction:
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$
I think its unbalanced form should be (I guess water added to balanced it)
$$\ce{Al + NaNO3 + NaOH -> NH3 + NaAlO2}$$
I wrote the half reactions as
$$
\begin{align}
\ce {6H2O + 8e- + NO3- &-> NH3 + 9OH-}\label{rxn:1}\tag{1}\\
\ce {Al &-> Al^3+ + 3e-}\label{rxn:2}\tag{2}
\end{align}
$$
Then by $3\cdot\eqref{rxn:1} + 8\cdot\eqref{rxn:2}$, final half reaction is:
$$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$
Now I am really confused how to get above balanced redox reaction by the final half reaction I have obtained. What is the missing idea?
$$\ce{UPDATE:- Anybody please show me how to proceed from here,}$$
$$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$
$$\ce{to here}$$
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$

What you call "final half reaction" is not a half; in fact it is a complete reaction, balanced and consistent. It is just that you don't want ions on either side, so you add counterions to them. Also, note that $\ce{Al^3+}$  reacts with $\ce{OH-}$.

@IvanNeretin yes. Thanks for correction. How to get NaAlO2 as aproduct by doing so? Could you please show the procedure?

What procedure? Add as much $\ce{OH-}$ as it takes. Remove the water.

@IvanNeretin I think you didn't get what I said. So from the final reaction that I have, I don't see any obvious way to get the firstly I mentioned FULLY BALANCED reaction with all compounds (Not as IONS) . NO3- become NaNO3 and NH3 remain as same. no problem there. But how Al3+ become NaAlO2? What procedure to get that final complete reaction after having the reaction (with ions)

You are supposed to "add" the reaction between Al3+ and OH-, which is also known in its own right (say, as AlCl3 + NaOH), not necessarily related to any redox. Can you balance it?

@IvanNeretin Yes. I can understand balancing of $$\ce{Al^3+ + 3OH- -> Al(OH)3}$$ Then is this right? $$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 3OH- + 8Al(OH)3}$$ How does that  18H2O  become 2H2O as in the first reaction??

I guess you've heard about the amphoteric species, didn't you? What will happen with that precipitate of Al(OH)3 if you add more NaOH?

@IvanNeretin YES! Thank you sir for showing me the way to tackle the question rather than giving the right out answer. I got it. Al(OH)3 + NaOH → Na[Al(OH)4] So it means my balancing was right but I needed to understand what happens to the products if they ALSO mutually react. Am I right? If so I understood. Thanks.

Yes, that's about right.

@IvanNeretin I understood that the products mutually react so, finally I got the reaction, $$\ce{18H2O + 3NO3- + 8Al-> 3NH3 + 5Al(OH)3 + 3Al[(OH)4]-}$$ BUT still, I see no way to get the fully balanced reaction. Could you please kindly post a 'stepwise procedure' as answer to show,$$\ce{how 18 H2O become 2H2O on left side?}$$ $$\ce{how 5Al(OH)3 on right side disappear and 5NaOH appear on left side? }$$ I could balance the other reactions I tried but this one got me stuck really bad.

Add more OH- to the left so that all Al ends up in the anions.

Wait, something is wrong here. You half reaction has $\ce{Al^{3+}}$ but your total reaction has $\ce{AlO_{2}-}$. You can't go from one directly to the other. Ideally, you have find a half reactions that contain all of the species in the total reaction. Otherwise, you need to do some manipulation.

@Zhe I'm really confused with this reaction. Could you please tell me a stepwise procedure to balance the reaction(top of page) using method of half reactions?

If it's reasonably straightforward, I would start by removing all spectator ions. Then you'll notice that your oxidation half reaction has to include something of the form: $\ce{Al -> AlO_{2}-}$. If you don't see that half reaction, you might not have the correct half reactions to do this decomposition...

@Zhe Man seriously, how would I missed that! Yes it worked like charm. I was messing the whole time with $$\ce{Al  ->Al^3+ + 3e}$$ Please post as answer I'll definitely select it. In your answer, if possible please mention why Al oxidizes to AlO2- instead of Al3+ Thank you so much BTW.

Your total reaction is:
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH⟶3NH3 + 8NaAlO2}$$
But your choice of oxidation half reaction is:
$$\ce{Al -> Al^{3+} + 3e-}$$
Notice that the right side of the half reaction doesn't represent the right side of the total reaction (once you've accounted for the sodium spectator ion).
You must find the corresponding oxidation half reaction, and if it's not available, then you're out of luck.
Your half reaction will have $\ce{Al}$ on the left, and $\ce{AlO2-}$ and 3 electrons on the right; with water, hydrogen ions, and hydroxides to balance.
In comments, you also asked for why this is the oxidation product. That is the wrong question to ask. All you're doing is breaking down the total reaction into component half reactions. The simple half reaction ($\ce{Al|Al^{3+}}$) is totally valid. It's just gives a different total reaction from the one you started with. There's nothing wrong with that reaction. It's just different.

so you mean I cannot "guess" the reaction products by considering the medium and reactants etc? Does that mean i have to remember the products (by heart)? This is a question for me.

You can. You just might be wrong. The reaction will do what it wants to do, not what we would like it to do. With appropriate experience and insight, you might have a pretty good success rate, but again, you're just never guaranteed to be right.

The idea of matching the left & right of half reactions with that of the full reaction to be balanced was something I should've focused. Thanks for pointing that out. Great help and thank you very much sir!


text:
Is reactant concentration = product concentration at equilbrium?
When a reaction attains chemical equilibrium, it means that the forward reaction rate = backward reaction rate. Since rate is defined as the change in concentration per unit time,does it mean that at equilibrium,
1) concentration of reactants = concentration of products?
2) Also, is the time taken for the forward reaction = time taken for the backward reaction?

Suppose you lose 100$$$. Suppose that Bill Gates also loses 100$$$. The change is the same. Does that mean the amount of money is also the same?

1) Not necessarily. In chemical equilibriums Concentrations could be any of following variations after attaining equilibrium. $$\ce{[reactants] < [products]}$$ $$\ce{[reactants] = [products]}$$ $$\ce{[reactants] > [products]}$$
Although reaction rates of forward and backward are equal, something like this could also happen. Therefore concentrations are not ALWAYS equal at equilibrium of all reactions, but could be in some reactions.

image source and more reading:-http://faculty.chem.queensu.ca/people/faculty/mombourquette/FirstYrChem/equilibrium/index.htm
2) Since the general idea of "Rate" means some change (say in this case reactants consumed or products produced) divided by time, thus AT EQUILIBRIUM, as you said, $$\ce{forward reaction rate =  backward reaction rate}$$ But I don't see how it can strictly be said like their times are equal. I THINK it depends. Maybe someone can correct me.

I think that it depends on your definition of equilibrium. There is alwasy an equilibrium between reagents and products (if we are considering reversible reactions). It can be thermodinamically shifted in one of the two directions
Products
$$\ce{A + B <=>> C}$$
Reagents
$$\ce{A + B <<=> C}$$
Same
$$\ce{A + B <=> C}$$
it means that in the first case you will have a thermodinamical tendence to one of them or that in the latter case that there are not actually reagents and product but that for the two chemicals (it can be also one form of same chemical) is equal.
As an example C can be in the form of Graphite or Diamond. Only one of them is the thermodimically stable but it will take quite a lot of time to see them chance in to each other.. you can have a look here for a more detailed explaination here http://www.ch.ic.ac.uk/rzepa/mim/century/html/diamond.htm
If you reach the equilibrium in a reaction it means that even if the quantity of reagents and products are different the inteconversion (which is still present !) won't alterate their final ratio of concentration of each of them. It is important to notice that there is still an interconversion (that's why reversible equilibrium) . Also you have to keep in mind that changing the condition of your sistem (P,T,conc) can alterate your equilibrium so there is not ONE equilibrium. Just one caveat about concentration. As chemist we generally refer to molarity but it is important to define it before any reasoning ;) Hope this helps


text:
Is there a reliable way to determine if a chemical system is more suited to be calculated using a multireference method instead of DFT?
Some time ago I recall hearing a professor say that looking at a DFT result (I don't remember exactly what, a TD-DFT, spin contamination etc...), he could tell that the system was multireferenced.
My question is: do we have anything in a DFT calculation that is indicative of a multireference system?

I don't have the time to post an answer right now, but there is a paper by Peter Taylor that introduces the idea that if the norm of the $\hat{T}_1$ operator from a coupled cluster calculation (or roughly $a_1$ from CI) is > o.o2 (I think), the system may have multiconfigurational character.

I note that while the T1 measurement has been used before, many dispute its usefulness and have introduced other such measurements, such as T2 etc.

@TAR86 Do you have a reference handy? There's at least one Don Truhlar paper where he evaluates a handful of diagnostics, it could be in there too.

However, $\hat{T}_1, \hat{T}_2, \dots$ cannot come from an SCF-type calculation, they must come from CCSD or CISD, so they aren't useful. I wouldn't be surprised if there is a TD-DFT quantity related to something like number of $(\mathbf{X}+\mathbf{Y})^2$ elements above some threshold that's useful.

@pentavalentcarbon Unfortunately, I no longer have access to publications. My old group did some work on strong correlation recently, I assume that at least some references are in there. It may also be of interest to OP: http://dx.doi.org/10.1002/chem.201604682 .

You should have a look at the so called FOD (Fermi Occupied Density) analysis by Grimme et al. That is accessible by ORCA 4 very easy and straight forward by specifying "FOD" as a keyword. [1-3]

From ORCA Input Library [1]:

The FOD plot above (plotted with Chemcraft) indicates a significant and delocalized FOD. Multiconfigurational methods should be used to study this molecule and any single-reference result (HF, DFT, MP2, CC should be regarded with suspicion)

Via Multiwfn you can access the "local total/dynamic/nondynamic electron correlation function" and "total/dynamic/nondynamic electron correlation index" [4,5].

Visual abstract from [5]. You can see the dynamic and non-dynamic correlation from which you also can see if single reference methods are acceptable or not.

[1] ORCA Input Library, FOD analysis
[2] S. Grimme, A. Hansen, Angew. Chem. Int. Ed. 2015, 54 (42), 12308-12313
[3] C. A. Bauer, A. Hansen, S. Grimme, Chem. Eur. J. 2017, 23 (25), 6150-6164
[4] E. Ramos-Cordoba, P. Salvadorc, E. Matito, Phys. Chem. Chem. Phys. 2016, 18, 24015-24023
[5] E. Ramos-Cordoba, E. Matito, J. Chem. Theory Comput. 2017, 13 (6), 2705–2711

FOD = fractional occupied density acc. [2]


text:
Difference between tendency of benzene and thiophene to undergo sulfonation
Why does thiophene possess a greater tendency to undergo sulphonation than benzene? I came across this reaction which employs the use of sulphuric acid to separate thiophene from commercially prepared benzene.

The activation energy is less for the the thiophene reaction because thiophene is less aromatic than benzene.

Not because aromatic conjugation in thiophene imparts negative charge to the carbon?  Draw thiponene Kekule type structures and see ... .

As ron said thiophene is less aromatic than benzene, but perhaps more importantly thiophene is an electron-rich ring which leads to greater reactivity with electrophiles (which are electron-poor). Pyridine is as aromatic or slightly less aromatic than benzene, but pyridine is very unreactive towards electrophiles.

I came across a Quora answer saying that in thiophene, the 6 electrons are delocalised over 5 atoms but in benzene, the delocalisation is over 6 atoms. Therefore, electron density is higher in the former.

@Orthocresol as an aside -- pyridine can be made to "wake up" for electrophiles by N-oxidizing it.

In the chapter Aromatic five-membered ring heterocycles with one heteroatom in Organic Chemistry by  J. William Suggs (2002) the following explanation can be found (pp. 403-404) which summarizes the thoughts already pointed out in the comments:

The resonance stabilization energy of benzene is greater than that of these heteroaromatic compounds.
  The order of aromaticity is benzene > thiophene > pyrrole > furan.
[…]
All three of these ring systems undergo electrophilic aromatic substitution and are much more reactive than benzene.
  In part, this reactivity difference arises because the rate-determining step in electrophilic aromatic substitution is the first step, which breaks up the aromatic $\pi$ system.
  Since thiophene, pyrrole, and furan have less stabilization to lose than benzene, the intermediate is lower in energy and the overall reaction proceeds more rapidly.
[…]
All three of these heteroaromatic rings undergo electrophilic aromatic substitution, preferentially at C-2.
  The reactivity order is pyrrole > furan > thiophene because of several factors, including the electronegativity of the heteroatom and the resonance stabilization of the aromatic ring.


text:
How do you arrive at the second form of the 1st order integrated rate law?
As I understand it, the first-order integrated rate law is:
$$\ln[\ce{A}] = \ln[\ce{A}]_0 - k t$$
However, I'm also told that this can be expressed as a ratio of $[\ce{A}]_0$ and $[\ce{A}]$, as follows
$$\ln \left(\frac{[\ce{A}]_0}{[\ce{A}]\,\,}\right) = kt$$
How did they arrive at this expression of the integrated rate law?  I don't see a way to arrive at it algebraically, unless I'm greatly missing something.

This is just using logarithm rule $\log(a/b) = \log(a)-\log(b)$.

To get to ln([A0]/[A])=kt from ln[A]=−kt+ln[A]0 just requires some understandings of log properties and a few log rules.
First rearrange ln[A]=−kt+ln[A]0
to
kt = ln[A]0 - ln[A]
From there, the log rule
log(a/b)=log(a)−log(b) means that you can state this as
kt = ln([A]0/[A])