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Why do all atomic masses have decimals if there are some elements that don't have isotopes?
I understand that the way to calculate the atomic mass is to obtain a weighted average of the masses of all the isotopes of an element. I also understand that there are about 26 elements that are monoisotopic. If the reason the atomic masses have decimals is because of this weighted average, then it should follow that the monoisotopic elements should have a whole-number mass. Therefore, why do all the atomic masses have decimals?

The mass number of an atom’s nucleus is a unitless integer: it is the sum of the number of protons and number of neutrons in the nucleus. The atomic mass of an atom’s nucleus is the mass of the nucleus. It is a mass and has units. The answers discuss this in detail, and discuss binding energy and how the carbon 12 isotope gets defined as having an integer mass. But all other atomic masses are not integers, they have mass units and weighted atomic weight averages have nothing to do with mass numbers.

This is a simple question which has a complicated answer.
In simplest terms, there is one isotope, $\ce{^{12}C}$, which does have an integer atomic mass by definition.

unified atomic mass unit (u) - Non-SI unit of mass (equal to the atomic mass constant), defined as one twelfth of the mass of a carbon-12 atom in its ground state and used to express masses of atomic particles.

So for $\ce{^{12}C}$, there are exactly 12.000... grams of carbon per mole of the  $\ce{^{12}C}$ atoms.
Now using the $\mathrm u$ as a stake in the ground, the weight of the various atomic particles are:

neutron =  1.008 664 915 88(49) u
proton =   1.007 276 466 879(91) u
electron = 0.000 548 579 909070(16) u = $5.48579909070(16)\times10^{-4}$ u

Now a carbon atom has 6 neutrons, 6 protons and 6 electrons. A simple addition of all the individual particle masses yields about 12.09893977602 u. This is not exactly 12.000...
You can also see that the mass of a neutron isn't equal to the mass of an electron and a proton (1.007276466879 u + 0.000548579909 u = 1.007825046788 u).
You can also see that an electron and a proton have a different mass than the $\ce{^1H}$ isotope (1.00782503224 u)
The problem here is the binding energies of the nucleons and to a lesser extent the electrons. From Einstein's famous formula, $E=mc^2$, some of the mass of the particles is converted to the energies necessary to hold the particles together.
We could use a different scale other than $\ce{^{12}C}$, but the neutron and proton couldn't both have an integer mass. When you also consider the binding energies of the nucleus and the electrons, only a maximum of one isotope will be able to have an integer mass. (Why $\ce{^{12}C}$ was picked as the standard is another complicated answer.)

The answer to your question has to do with Einstein's most famous formula
$$
E=mc^2
$$
which relates energy to mass, but what does it mean? Let's consider a simple example first. The simplest atom is of course the hydrogen atom which consists of one electron and of one proton. If you add the mass of the proton $1.672621898\times 10^{-27}$kg, to the mass of the electron $9.10938356\times 10^{-31}$ kg, you find a value of $1.67353284\times 10^{-27}$ kg. If you compare this to the mass of the $^1$H atom of $1.6735326915926\times 10^{-27}$ kg, you see that the hydrogen atom is lighter than the combined mass of the proton and the electron by $2.44017148\times 10^{-35}$ kg. What happened to the mass when the electron and proton got bounded together? Well, according to Einstein the difference in mass is converted to the binding energy of the electron! If you use Einsteins formula you find a binding energy of $-2.19311675\times 10^{-18}$ J or $-13.7$ eV which is very close to the Rydberg constant!
Let's now get back to other atoms. In the same way that the formation of a hydrogen atom from an electron and a proton lowers the mass of the atom by the corresponding binding energy, the binding energy between the neutrons, protons and electrons that make up heavier atoms also results in a reduction of the mass of the atom. Because the binding of the neutrons and protons is much stronger than the binding energy of the electrons to the core, the effect of the mass is also much larger. See for instance here for further details.

So if that is true, Paul,  that some of the mass is converted to energy and the mass of the atom is less than the sum of the nucleons' weight, then the atomic mass should be LESS than the mass number ( which is the number of protons and electrons - or essentially the rounded up weight of the nucleons), and that is not is not always the case......????

It also has to do with the definition of the atomic mass. The atomic mass is defined as 1/12th of the mass of C12. In this definition part of the binding energy of the carbon atom is contained as well. As a consequence, sometimes you find that the mass difference is positive and the other time it's negative, depending on the relative binding energy with respect to carbon.

There is also the fact that neutrons are heavier than protons.  In fact an isolated neutron is heavier than a proton + electron + antineutrino, into which isolated neutrons decay with a half life of about 20 minutes.


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Meso and achiral does both term means the same?
When a molecule has plane of symmetry,  or centre of symmetry it is termed as an  achiral compound (though it contain some chiral centres).  Same goes for the meso compounds due to the internal compensation.  So both terms mean exactly same.  But I am eager to to know the applications of these terms  in different cases.  All meso compounds can be told achiral and as well as all chiral compounds as meso?

Methane is achiral, but nobody would call it meso.

From the Gold book:

meso-compound:
A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members

Achiral compound:
A compound that can be superimposed on its mirror image.
The difference lies in the presence of the chiral centre. All meso compounds are achiral (due to internal compensation as you have mentioned) but all achiral compunds aren't meso.

Look carefully and you'll be able to spot chiral centres in all the above compounds but you'll also notice that they have a plane of symmetry (internal compensation) which makes them achiral.

Bromocylohexane has a  plane of symmetry which makes it achiral. But it has no chiral centre, that's why it doesn't qualify to be a meso compound.

A meso compound is one which satisfies both of the following criteria: (1) it must have chiral centers and (2) it must have an internal plane of symmetry. On the other hand, an achiral compound is one which has an internal plane of symmetry. Thus you can say because a meso compound has an internal plane of symmetry, it is in fact achiral and optically inactive. You could not however say that all achiral compounds are meso... the terms are not synonymous. For an example, benzene is achiral but not meso. It has many internal planes of symmetry but does not have any chiral centers. It does not meet the first criterion for being meso. All chiral compounds cannot be considered meso as you suggested. The presence of chiral centers does not necessarily mean a compound is chiral.

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Intramolecular or intermolecular?
I am extremely confused about these types of forces.
Some websites state that the forces in the covalent bonds of carbon dioxide are inter. Other websites say they are intra?
Also, some websites say that the electrostatic bond between ionic compounds is intra, others say they are inter.
So, which is inter and which is intra?

Hey Daniel, welcome to Chem.SE! Can you please cite those websites, which state these things, so we too can have a look?

Inter- means in between molecules (like the interstate highway - in between states)and intra- means within one molecule ( like an intrastate highway- within one state). So a covalent bond of CO2 OR an electrostatic ionic bond would be an intramolecular bond. The type of bond that keeps water molecules close to each other ( hydrogen bonding) is an example of an intermolecular attraction.

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Why is the number of protons and neutrons equal to the atomic mass rounded?
The mass number is defined as the NUMBER of protons and neutrons. Electrons aside, why does the number of protons and neutrons equal the atomic mass rounded to the nearest whole number? That would mean, to me, that the number of protons and neutrons equals almost exactly to their relative mass! Why is that true?

Because proton weighs (well, almost exactly) 1 and neutron also weighs 1; what's unclear about that?

How about the isotopes?

This is incorrect. The atomic mass you read off the periodic table is the weighted average of all isotopic masses. For the most part, the masses of the isotopes are just the sum of the protons and neutrons, but nuclear binding energy causes some deviation. Also, protons and neutrons don't have the same mass, so for larger atoms, that will also cause deviation from a simple sum.
The biggest issue like I said is that atomic mass is the weighted average. For example, the two most prevalent isotopes of bromine are Br-79 and Br-81, present in about the same relative abundance. However, the atomic mass is 79.9 because it is the average. That rounds to neither 79 or 81.

What is incorrect specifically? The atomic mass rounded to the nearest whole number is defined as the mass number. And 79,9 does NOT round to 81. I'm unclear as to how you are explaining this.

That's my point. The mass number is the sum of protons and neutrons. The atomic mass is the weighted average of the masses of individual elements weighted by abundance. The atomic mass of bromine is 79.9. Bromine atoms do not have a number of protons and neutrons totaling 80.


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Reactivity of meta- and para-bromo-nitrobenzene towards nucleophilic substitution

Among 1-bromo-3-nitrobenzene and 1-bromo-4-nitrobenzene, which would be more reactive towards nucleophilic substitution?

From what I can infer, the answer should have been 1-bromo-3-nitrobenzene. Nitro is a −M/−R group, so it would create regions of high positive charge in the ortho and para positions with respect to the nitro group.
Now in 1-bromo-4-nitrobenzene, Br is in the para position with respect to the nitro group. Hence the positive charge would be created only in the ortho position, as the para position is blocked. On the other hand, in 1-bromo-3-nitrobenzene, there would be regions of positive charge in the both the ortho and para positions wrt to the nitro group.
But the answer given in my book is 1-bromo-4-nitrobenzene. Am I missing something here?

Have you been taught about SNAr mechanism? This is its direct application.

p- Bromonitrobenzene is certainly more reactive towards Nucleophilic Substitution than m-Bromonitrobenzene. The reasons are as follows,  Firstly, due to $-R$ effect of $\ce{NO2}$ group, the positive charge is created on the carbon attached to bromine in para position, which helps to increase thre electrophilicity of that carbon and that makes easier for the electrophiles to attack the benzene ring. If $\ce{NO_2}$ group would have been in  meta position, the positive charge would have been created on the adjacent carbon, so, the electrophilicity of the bromine attached carbon is not increased so much which in turn will not help it to undergo Nucleophilic attack easily. The image below demonstrates this effect on para position.
Secondly, the Nucleophilic substitution in these systems undergo by addition-elimination mechanism. So, when the nucleophile attacks and forms the intermidiate complex, that complex is much more stabilised due to charge delocalisation in the $\ce{NO2 }$ group also if it is in para position.
But if $\ce{NO2}$ is in meta position, the charge can only delocalise in the ring, but it can not delocalise in the $\ce{NO2}$ group (you can verify by looking at the  image also), so it is not so stabilised as the para isomer intermidiate.  These effects are mainly dominating which makes the p-Bromonitrobenzene more reactive than m-Bromonitrobenzene in Nucleophilic substitution.

The mechanism generally followed in nucleophilic aromatic substitution when an activating group (like $\ce{-NO2}$) is present on the ring is SNAr.
The first step, the formation of the carbanion, is usually the rate determining step.
You are aware that the nitro group has a strong -I effect as well as -M effect. Out of these, the -I effect can operate in both the isomers of nitrobromobenzene. However, the -M effect can only operate in the para isomer. (Draw the resonance structures and verify)
Thus, the carbanion formed in the case of p-isomer can be stabilised by two effects, taking full advantage of the presence of the nitro group which makes it more reactive towards nucleophilic substitution.


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How to correctly mix solutions?
Imagine you have two ingredients - 45 grams of 60% ethanol, and as much water as you wish. The only accurate tool you have is a scale, which measuring error is up to $\pm \pu{1gram}$. How would you create five different solutions of 1, 2, 3, 4, and 5% ethanol?
It can be a solution of % ethanol by volume or by weight, depending on which can be created more accurately with the tools at hand. It might be super easy to you, but as an electrical engineer the logic and math eludes me.

Have you seen the equation C1V1=C2V2 before?

Nope - What is it called and what does it refer to?

It's used to calculate the concentration of your solution after dilution. C1 is your initial concentration, C2 is your final concentration (e.g 5%) and V2 is your final volume (E.g 100 ml). V1 is what you calculate. http://www.mathcentre.ac.uk/resources/uploaded/c1v1c2v2.pdf

Would it work even though the 60% ethanol water mixture's density is about 0.9? https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/

Yep. So if you have 60% ethanol and need 5% solution,  v1=(5% x 100 m l) / 60%  Which means you need 8.33 ml of 60% ethanol in 100 ml solution. (8.33 ml ethanol, 91.77ml of water)

Well, in which case would you like to write it as an answer so I could mark it a solution? :-)

You should use the concentration dilution equation
$$C_1V_1 = C_2\cdot V_2$$
If you re-arrange it so the volume of 60% ethanol needed is:
$$V_1 = \frac{C_2\cdot V_2}{C_1}$$
You can then insert all the values you do know and there you go.
For example, if you need a $\pu{100 m}$ of 5% solution: $$V_1=\frac{5\% × \pu{ 100 ml}}{60\%}$$
Which means you need $\pu{8.33 ml}$ of 60% ethanol in $\pu{100 ml}$ solution.
As max pointed out, since you only have a scale, use d=m/v. D = density, m = mass and v= volume You re-arrange so mass = d.v

That doesn't work if all you have is a scale that weighs +/- 1 gram as the OP posed.


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Calculate Heat of Atomisation
What is heat of atomisation of P4O6(s) Given heat of sublimation of P4O6 is x kJ/mol & P–O bond energy is y kJ/mol.
What is heat of atomisation of P4O6(s) Given heat of sublimation of P4O6 is x kJ/mol & P–O bond energy is y kJ/mol.

Is it $x + 12y$ ?

yes the ans is x+12y ... @SoumikDas can u please let me know the solution i am stuck up...please asap     thanks in advance!

Do u know the structures of $\ce{P4O6}$  and  $\ce{P4O10}$ where two six membered heterocyclic rings are fused together ?

@SoumikDas i just know the individual structures of p4o6 and p4o10 not the fused one....please help!!

See this https://qph.fs.quoracdn.net/main-qimg-4a50afc1084fa22bd3a7f0ae8756dee7  and  https://ars.els-cdn.com/content/image/1-s2.0-S0079670015000088-sc2.gif

From the two pictures clearly you can easily count how many $\ce{P-O}$ bonds will be broken ?

@SoumikDas but what is the need of counting P-O-P bonds in P4O10??

Just saying. You don't need them in this question.

Without just giving you the answer:
Enthalpy of atomisation is "the enthalpy change that accompanies the total separation of all atoms in a chemical substance". You can read up more about it on Wikipedia1.

Look up the chemical compound shape and connection
Add up the number of connections
Include the heat of sublimation
You should have a reliable answer and know how to do future problems similar to this,

Also, see the comment from @gaurang-tandon


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What exactly is formed when K2Cr2O7 is reduced?
I read on Wikipedia and in many books that when $\ce{K2Cr2O7}$ is reduced, it forms $\ce{Cr^3+}$ and a change is noticed from orange color to green. I want to know in what exact complex/salt/compound $\ce{Cr^3+}$ exists that it shows green color.
I know $\ce{[CrCl3].3H2O}$ shows green color but in many of these reactions, there is no $\ce{Cl-}$ ion, so I guess that is not formed. Also, $\ce{Cr^3+}$ can exist as $\ce{[Cr(H2O)6]^3+}$ but since it is violet in color, that possibility is also ruled out. So basically, how does the solution becomes green in color?

"in what exact complex Cr3+ complex" The reduction of the dichromate anion does not lead to a complex. Did you mean to refer to a chromium salt instead?

I thought since it is colored, it is most probably a complex. But I edited it if that's not the case.

@GaurangTandon This is pretty bad nitpick. If you "salt" is hydrated then you do have aquocomplex in it. The concept of "salt" is one of worst in chemistry.

@Mithoron Oh! I was assuming the OP is looking non-hydrated salts though, like of the form $\ce{A_m^+ B_n^-}$. But, correct me if I'm wrong, the reduction of dichromate anion does not still lead to a complex, does it?

$\ce{[Cr(H2O)6]^{3+}}$ is rather acidic, in most solutions it will partially dissociate, forming $\ce{[Cr(H2O)5OH]^{2+}}$ and other hydroxocomplexes.

@GaurangTandon and to OP. What reduction? It can be reduced in so many ways that all this doesn't make sense.

The dichromate ($\ce{Cr2O7^2-}$) ions are strong oxidizing agents at low pH. During the redox process, each chromium atom in the dichromate ions (oxidation state = +6) gains three electrons and get its oxidation state reduced to +3. In redox reactions in aqueous acid solution, the aquated $\ce{Cr^3+}$ ion is produced according to following half-reaction (Electrochemical Series):
$$\ce{Cr2O7^2- + 14H3O+ + 6e- <=> 2Cr^3+ + 21H2O} \space \space \space \pu{E^0 = 1.36 V}$$
Assume that the $\ce{Cr^3+}$ ion in aqueous solutions is in its simplest ion form: the hexaaquachromium(III) or $\ce{[Cr(H2O)6]^3+}$. This ion is violet-blue-grey in color. However, when $\ce{Cr^3+}$ ion is formed during a reaction in aquous acid solution, it is often appeared in green color. Thus, we always describe the green color due to $\ce{Cr^3+ (aq)}$, implying it as the hexaaquachromium(III) ion ($\ce{[Cr(H2O)6]^3+}$), but that's actually not the case.
What's really happening in the solution is that one or more of the water molecules in each of $\ce{[Cr(H2O)6]^3+}$ ions get replaced by other negative ions in the solution, typically by $\ce{SO4^2-}$ (if the acid used is $\ce{H2SO4}$) or $\ce{Cl-}$ (if the acid is $\ce{HCl}$). For example, if one of the water molecules is replaced by a $\ce{SO4^2-}$ ion, $\ce{[Cr(H2O)6]^3+}$  becomes $\ce{[Cr(H2O)5SO4]+}$, which is green (notice also the change in the charge on the ion). You can check this by yourself: warm freshly prepared aqueous  chromium(III) sulphate hexahydrate solution, which would change its color of purple-blue or violet ($\ce{[Cr(H2O)6]^3+}$) to green ($\ce{[Cr(H2O)5SO4]+}$) due to the ligand exchange discussed above [Chem Guide].
In $\ce{CrCl3.6H2O}$, on the other hands, the green color of the solid itself or its aqueous solution is due to the $\ce{[Cr(H2O)4Cl2]+}$ ion. Hence, actually more proper way of writing the formula is $\ce{CrCl2(H2O)4Cl.2H2O}$. In contrast to previous case, the green $\ce{CrCl2(H2O)4Cl.2H2O}$ solution would slowly becomes violet in color by standing at room temperature overnight, by slow ligand exchange with solvent (water)[Chem Guide].
Interesting fact: Chromium was discovered by Louis Nicholas Vauquelin in 1797 who named his new element after the Greek word chroma — which means color.
Late Additions:
Although $\ce{Cr^3+}$ ions complexes with most ligands to give green color, it gives vivid line of colors with other ligands (note it names after chroma!). Some interesting examples are illustrated in the picture below:

Also see:
Chromium(III) nitrate: Blue-violet crystals (anhydrous), and purple crystals (nonahydrate), according to Wikipedia (see the picture below).

Nice answer, only problem is if you use hydrochloric acid in the oxidation process (as you mentioned, or I misinterpreted) it will be oxidised to chlorine. This may be a little slip of yours but it is not obvious to someone less experienced and can lead to disasters. This is why we use sulfuric acid mostly with dichromate or permanganate.

Potassium dichromate is a strong oxidizing agent and it helps any other compound to oxidize by itself getting reduced to $\ce{Cr^3+}$. This is a general redox reaction. Normally, the counterion i.e the anion comes from the acid used. If $\ce{H2SO4}$ is used then $\ce{Cr2(SO4)3}$ will form; if $\ce{HCl}$ is used, then $\ce{CrCl3}$ will form. Take for example:

But if you consider a pure chromium(III) compound devoid of any counterions, decomposing potassium dicromate leads to formation of chromium(III) oxide which is green in color. This is a decomposition reaction or can also be considered $\ce{ K2Cr2O7}$ reduction as oxidation state is reduced from +VI to +III.
$$\ce{4K2Cr2O7 ->[\Delta] 4K2CrO4 + 2Cr2O3 + 3O2}$$
Literally any chromium(III) compound is green in color. But in a reaction pot, the chromium(III) is present in aqueous acid solution of composition $\ce{[Cr(H2O)_{a}X]+}$  where X is the counterion from the acid used($\ce{SO4^2-, Cl-}$).  $\ce{a}$ depends on the charge of the counterion (credit @Mathew_Mahindaratne).

But shouldn't the two ions of the salt exist independently in the solution since its not a precipitate? So, the salt should exist as Cr3+ and SO42- in the solution and therefore the color is due to maybe a complex that Cr3+ forms but I don't know.

$$\ce{4K2Cr2O7 ->[\Delta] 4K2CrO4 + 2Cr2O3 + 3O2}$$
Chromium(III)oxide is formed in a normal decomposition reaction and the colour changes from orange to green.


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The Conductivity of Electrolytes based on structure of Ionic Crystal
Background
I want to run an  experiment to test the conductivity of various ionic compounds dissolved in water. I was hoping to see some sort of trend in the increase/decrease of conductivity based on my choice of salts. Except, conceptually, I don't quite get how my choice in the different compounds would affect conductivity.

Problems

I am somewhat aware of Kohlrausch's Law, but for my experiment I was not seeking to change dilution, temperature or electrode distances. What I wanted to do was to study the effect of the structure of ionic compound or charge of the ions on Conductivity.

I also wanted to compare my conductivities to some literature values.

Question
Just so this isn't a waste of time, I would like to know:

Does the structure of an ionic compound or the charge of its ions affect its conductivity, given that dilution and temperature will remain constant?
If so, how?
Is there a reliable Databse I can compare my results with?

I'm not too fussed about which ionic compounds they are, I would just like to understand how the compound itself plays a role in conductivity so I can confirm it with my experiment?
Thank you so much,

TL;DR
In a solution containing a dissolved ionic compound, to what extent is the ease of mobility of electrons or conductivity affected by either the electrophoric drag (presumably of polar water) and/or by the original structure (lattice shape) & charge of both the metallic and non-metallic ions?

This is like textbook example of too broad question.

Sorry, let me re-phrease: When I say, "Does the structure of an ionic compound or the charge of its ions affect its conductivity in a solution, given that dilution and temperature will remain constant?" Effectively, I am asking to what extent Electrophoresis or Electron mobility is slowed by the drag of the polar liquid (water) and to what extent the mobility is affected by the charge of the separated ions within the solution. Does the charge of the ions or the original structure of the Ionic crystal have any affect on the conductance or ease of electron movement in a solution?

It turns out, according to this paper, stated:

The solvent-berg model which put forward the suggestion that smaller ions are strongly interacting with the nearest neighbour shell of solvent molecules. This was considered to be particularly true of cations since these are generally smaller in size than the corresponding anions and have a higher charge density. The ion essentially carries this shell of solvent molecules long enough that this leads to a larger effective diameter which lowers its conductivity to a value smaller than the conductivity of larger ions which have no strongly attached shells of solvent.

And in terms of drag from the polarity of the solvent, namely water, they showed a formula for dielectric friction:
ζDF = 3qi2(ε0 − ε∞)τD/(cri3(2ε0 + 1)ε0)
Where:
ζDF is the dielectric friction,
ri and qi are the radius and charge of the ion,
ε0 and ε∞ are the static and high frequency dielectric constants of the solvent,
c is the product of the limiting ionic conductivity and viscosity of the solvent,
So, I assume that the ionic radius will play a role in its drag by the surrounding polar solvents which will affect its mobility, and therefore the conductance.


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Does the probability function for a molecule to have energy in a certain mode depend on the minimum required quantum energy level of that mode?
I understand that the probability density for a particular velocity v+dv in one dimension for a molecule in a system is:
$$f(v) = \sqrt{\frac{m}{2\pi k_B T}} \cdot e^{\frac{-mv^2}{2k_B T}}$$
I'm noticing that this formula does not distinguish the type of freedom in which the concerning velocity is used. So basically this formula would say that a certain velocity in a translational degree of freedom has an equal probability as the same velocity in a rotational or vibrational degree of freedom, which is incorrect from what I know due to the different required minimum quantum energy for each type of mode (depending on the molecular structure) with respect to the energy that the molecule possesses.
Is there a formula that takes the type of mode into account for the probability?
I found a source that takes the frequency of a degree of freedom into account and calculates its minimum required energy according to hf = E but I don't know how to incorporate that formula into the probability function.

This function considers molecules as hard spheres and is used to study the macroscopic behavior of matter. Treatment of movement of molecules taking into consideration their quantum states is something else and as far as I know very complicated.

The formula you give is the Maxwell-Boltzmann distribution for the velocity of a gaseous molecule in one dimension.  It is derived assuming translational kinetic energy only of the gas and that there is thermal equilibrium. It is also assumed that components in the y and z direction do not affect motion in the x direction. Translational motion is quantised but quanta are so small that we treat the system classically. This is not true of rotations and vibrations of small molecules. Proteins, &colloids could be treated classically for rotational motion. Similar equations may result.

There's a couple of conceptual problems at the base of your question that need to be addressed.
1) Free translation of particles in space is not discrete - the distribution of quantum states is continuous, and therefore so is the distribution of energies or of velocities - but rotation or vibration are quantised, so they have discrete distributions, so the formula definitely can't be applied to them. You could apply it to molecules as classical objects, ignoring quantum behaviour, but what you are describing then is not a molecule; you won't get realistic results that you could compare to experimental results.
2) The equipartition theorem states that rotation or vibration degrees of freedom will have the same energy as translation degrees of freedom, but the same energy will not translate into the same velocity.
For instance, in a classical treatment, the energy distribution function per degree of freedom in equilibrium that we obtain from Maxwell-Boltzmann is
$$ f_\epsilon = \sqrt{ \cfrac{1}{\pi \epsilon_i k_B T} } \exp \left[ \cfrac{-\epsilon_i}{k_B T} \right] $$
Taking into account that classical rotational energy is
$$ E_{rot} = \frac{1}{2} I \omega^2 $$
we can derive the classical angular velocity distribution:
$$ f_\omega = \sqrt{\cfrac{I}{2 \pi k_B T}} \exp \left[ \cfrac{-I \omega_i^2}{k_B T} \right] $$
with the moment of inertia $I$ being dependent on the geometry of the molecule. For instance, in the simplest case of a diatomic molecule approximated as a rigid rotor, it is
$$ I = \mu d^2 $$
where $\mu$ is the reduced mass and $d$ is the interatomic distance.
However, as we said, this is not a correct description of an actual molecule, in which vibration and rotation movements are quantised. In a quantum treatment, the allowed values for rotational energy in a linear molecule are
$$ E_{rot} = J \left( J + 1 \right) \cfrac{\hbar^2}{2I} ~ \mathrm{with} ~ J=0,1,2... $$
note that rotational levels are degenerate (with unequal degeneracy $g_j = 2J+1$) and discrete, so we need to use a discrete partition function $Z_{rot}$:
$$ Z_{rot} = \sum_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] $$
which in principle we can't calculate analytically. For molecules with a very large moment of inertia $I$ and/or at high enough temperatures $T$, the gap in energy between levels becomes very small and we can approximate this sum as a continuous integral over $J$ (the so-called high temperature limit):
$$ Z_{rot} \approx \int\limits_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] dJ = \cfrac{ 2 I k_B T}{\hbar^2} $$
Under that assumption, the (discrete) angular momentum distribution of a rotational degree of freedom can be derived as
$$ F_J = \left( 2 J + 1 \right) \cfrac{ \pi \hbar^3}{ I k_B T} \exp \left[ \cfrac{-2 \pi \hbar^3 \left( J + 1 \right) }{ I k_B T} \right] $$
Note that we're using angular momentum and not angular velocity here because, while the former is a well-defined magnitude in quantum mechanics, it's not clear the later makes physical sense in a quantum treatment of molecules. We could naively calculate a semi-classical "angular velocity" distribution from that angular momentum distribution, but, again, we'd not be actually describing a molecule.
And a final conceptual problem:
3) Rotation and vibration (and electronic) motion is coupled; to give a simple example, in a diatomic molecule, as rotation angular momentum increases, interatomic distance (and therefore moment of inertia) changes; a simple approximation is to introduce a rotational term representing centrifugal distortion, but for higher precision both movements have to be considered at the same time, what makes for much more complex equations.
Keep also in mind that we've restricted ourselves to the simplest cases in rigid diatomic molecules; more complex systems have much more complex mathematics involved and often have to be treated empirically.

Thank you so much for your detailed and clear explanation. You have answered my question.
One extra thing that I was wondering about. Suppose, for simplicity’s sake, a system has a total of 3 possible rotational energy states, the gaps between the energy levels being too large to continuously intergate over J, and you would like to calculate the probability of energy state J = 2. Should one then calculate this probability by E(rot) / Z(rot), using J = 2 and the upper summing limit of Z(rot) being 3 instead of infinity? (the total number of possible rotational states)

If there were a limited number of states, yes, that's how you'd do it.
In fact, even if you have infinite states and can't do the high temperature limit approximation, you can still approximate the partition function by doing a discrete sum of as many terms as you can afford, as higher states will most likely be very sparsely populated, so while the solution won't be exact, it will be a good approximation.

Thanks for clarifying. Looking back at my previous statement, I realised that Z(rot) is actually the total number of possible states, not the total summed energy of all the possible states. This means that the probability calculation for a state J should be (J(J+1)) / Z(rot) instead of Ej(rot) / Z(rot). Please correct me if I'm wrong on this.


text:
Why don't all atoms stick together due to dispersion forces if these forces are ubiquitous?
If dispersion forces are always there wherever there are electrons, why don't two pieces of copper for instance meld together when put next to each other? Or why don't other substances stick to each other? For instance, I saw on a video that geckos have "sticky feet" due to dispersion forces.

They do. But the dispersion forces are weak. More importantly, real-life objects are very rough on atomic level.

In principle you can stick objects together by polishing their surfaces to a roughness of about 1 nm. See optical contact bonding.

In particular liquids of non-polar molecules such as benzene for example but also any liquid.


text:
Equilibrium constant Kp for vaporization of water?
For the reaction H2O(l) -->H2O(g) it is given that Kp = p(H₂O(g))/p° where p° is the atmospheric pressure. Why is it not just equal to p(H₂O(g))?

It is (numerically, I mean).

@IvanNeretin But that is only for the case of unit atmospheric pressure right? And this is the first time I'm seeing any mention of atmospheric pressure in equilibrium constant. Is this the case with all equilibrium constants?

Let's put it this way: you want your constant to be greater than 1 when and only when the reaction is shifted to the right, don't you? Well, here you go.

@IvanNeretin would you please explain a bit more, I didn't really understand.. if we didn't put the atmospheric pressure would it be greater than one? And why do we use the atmospheric pressure in this particular reaction when we don't use it in many others?

Recall the origin of the equilibrium constant.  It comes from the fact that the chemical potential of an ideal gas can be written in the form u(T,p0) + RTln(p/p0), with p0 some reference pressure.  What this lets you do is measure the equilibrium composition at a reference pressure, then calculate the equilibrium composition at a slightly different pressure with a simple formula, without having to know the general dependence of chemical potential on pressure and temperature.  Very handy!  But the existence of the reference pressure is essential.  It goes into the numerical value of the equilibrium constant.  For example,  for just two reactants with stoichiometric coefficients c1 and -c2, i.e. the equilibrium is c1 R1 = c2 R2, you have:
dG = 0 = c1 (u1 + RTln(p1/p0)) - c2 (u2 + RTln(p2/p0))
-(c1 u1 - c2 u2)/RT = ln((p1/p0)^c1 / (p2/p0)^c2)
exp(-(c1 u1 - c2 u2)/RT) = (p1/p0)^c1 / (p2/p0)^c2
The thing on the left is just a function of T and the reference pressure p0, and we set it equal to the measured equilibrium constant K(T,p0), which of course we get by measuring the equilibrium composition at T and p0.  The thing on the right you'll then recognize as the usual expression for Kp.
K(T,p0) = (p1/p0)^c1 / (p2/p0)^c2
With this equation in hand (and the value of K) we can calculate the equilibrium composition {p1,p2} at some temperature and pressure not too
far away from the reference T and p0, using the usual freshman chemistry tricks.
Now you can certainly choose a different reference pressure if you like, it doesn't have to be 1atm, that's just convention.  If you chose a different pressure, K would be different, but the expression for Kp would be the same, only you'd be dividing by the new reference pressure.  But you have to have some reference pressure for this method to work.


text:
Trichloroacetyl carbamate cleavage to unsubstituted carbamate
I studied that when alcohol is treated with isocyanate and then undergone hydrolysis (or methanolosis), carbamate is formed. But when we use carbamate as trichloroacetyl isocyanate, why hydrolysis doesn't give amine containing trichloroacetyl group rather than the one containing ester group?

http://reag.paperplane.io/00002775.htm#5
In any material, there is no explanation about it, and I'm really curious.

The triple chlorine substitution substantially increases the electrophilicity of the carbonyl group, making is more susceptible to nucleophilic addition. Given that this step is probably endothermic, we expect the transition state to be late and having significant build up of negative charge. The inductive effects from the substitution will stabilize the transition state and increase the rate of addition and hence, the overall rate of hydrolysis of carbonyl group over the other.


text:
Medicines in the fridge
Why are there some medicines that are put in the fridge?
Thanks

Perhaps you should ask yourself why we use fridges at all.

some medicines contain unstable compounds, which will react with their environment and could cause them to degrade before they have a chance to be taken. chemical reactions require energy to occur, so by reducing the thermal energy available to these compounds, they will react slower
so basically, making things colder means reactions happen slower and your medicine keeps longer


text:
Why do NON -Transition elements exhibit variable oxidation state?
I know transition elements exhibit variable oxidation states because valence electrons are in d block and s which are very close so they have similar energies ,thus with slightly different energies electrons can be removed and thus different oxidation states ,but why do non-transition elements exhibit variable oxidation state ..
Is it due to
1.Hybridisation
2.D-block
?

Why not?
Elements with only a single oxidation state are rare things.  In fact technically, only helium and neon are known in just a single oxidation state, zero for each.  Everything else is known to form at least some (conventional) compounds, and then you have the zero oxidation state of the pure element plus any oxidation states there are in the compounds.
Multiple nonzero oxidation states are the norm for most nonmetals, as well.  Just check out common compounds of, say, carbon, nitrogen, sulfur, phosphorous, ... .
I would not ask why, just revel in it.

why? is asking why a bad question?

and are there factors other then hybridisation and closer orbitals because of which non -transition elements have variable oxidation states.?


text:
What material most closely resembles/emulates dog's excrements?
Please get me right - this is no joke!
Our engineering team has a terrible contract - we need to find a way to get height from which some dog excrements were dropped judging by what was found on earth when it hit the ground. We need to find some invariant in excrements form, scattering of droplets/material, I don't know whatever else...
We did not start math modelling yet, because it makes no sence - we need to get somewhere physical propeties of "material", besides it is clear that the best way to start with is real-life experiments - but throwing such stuff from different heights over and over again is definitely a bad idea.
The best solution would be to find some material which is maximally close to what our target is - like ballistic gel instead of a human body.
Sorry to ask, but what material most closely resembles/emulates dog's excrements?
So far we found nothing in literature, maybe at list please a hint where to look for...
Sorry once again - this is absolutely no joke, this is for real. The height is within 70 meters limit (this is the max height), the surface is hard - it is pavement (just where people walk). Actually we need to create methodology allowing to calculate from what floor of a building it was dropped (or proove that required resolution of 2-3 meters cannot be reached) by teenage pranksters who do it regularly in different locations (they play it from their rooms and they are networked through some kind of viral social network). This is a real-life problem and we understand why it arose, but it is terrible from the technical point of view.

LOL - First you assume all excrement comes from the standard spherical dog...

Human bodies are all alike, hence the gel. Feces, on the other hand, can be extremely diverse. Ever heard of the Bristol stool scale?

Would have thought it depends on the diet of the dog

Have you ever interviewed the teenage pranksters about that subject? They probably have the best expertise in telling apart good from bad stuff.

You'll never be able to get to 2-3m accuracy with this. With a reproducible sample, you could make a few dozen tests and create a convincing statistics.
But dog excrements? Even if they always used the same dog, which they kept on a constant diet, and which you had impounded, it'd be impossible. The sample is much, much too complex, and, unfortunately, very much alive. (I don't mean the dog.)
I say stick to old-school policework: Keep interrogating your pranksters until one of them sings. ;-)


text:
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)

As you can see in the image diazonium can attack at 2 different positions to form different products then why only the first product is correct?
As there are two ortho positions available why does it attack only one?

I tried to draw the intermediates see if they are useful and correct
is it because there are 7 intermediates in the first case?
or steric factors play some role here?

Not sure if it's alright to ask another question in the comments, but it's closely related to this question here... What would happen if instead of $\ce{\beta-napthol}$, $\ce{\alpha-napthol}$ were to be used? Would the attack take place at the ortho or para position?

it takes at para but idk why

Perhaps it could be solved using quantum mechanics? Like a computational calculation of the LUMO coefficients for the activated naphthol? Not sure if that helps though.

i just finished high school so i don't know what you are talking about @Justanotherchemist

HOMO has the biggest coefficient at that site.

The most important thing to point out here is that your arrows are backwards. The diazonium is not the nucleophile. It is the electrophile. The diazonium ion has one lone pair. If it were a nucleophile, there's no way that you'd end up with the diazo compounds drawn, where each nitrogen has a lone pair.
This also means that drawing out all of your resonance structures doesn't mean anything because that is not the intermediate. The real question to ask here is which site on the naphthol ring is more nucleophilic.
Consider the naphthol as an enol for the purposes of the addition. Your two starting structures are exactly the right ones for focus one. The unsaturation in the enol in the top example is such that the nucleophilic site is benzylic while, in the other case, it is far from the ring.

The intermediates after attack by an electrophile are shown (technically, the carbonyls should be protonated still, but I'm too busy to fix that now). Only in the top case, where we substitute adjacent to the other ring, is aromaticity still preserved in the other ring. This is why the substitution is strongly preferred there as opposed to the other position, where we proceed via an intermediate that is not aromatic.

The formation of para-red is an example of electrophilic aromatic substitution.
 The electrophile reacts with the ring carbon that is ortho to
both the activating group ($\ce{-OH}$) and the other ring because even the ring is weakly activating through $\text{+M}$ effect.
The reaction is similar to the sulfonation of napthalene. If the electrophile attacks the first ortho position, then for the resulting carbocation two resonance structures can be drawn which contain aromatic sextets.  On the other hand if the electrophile attacks the second ortho position, only one resonance structure can be drawn which has an aromatic sextet.
Thus, by the Hammond postulate, the first product is major.

can you please elaborate the sextet part

@amishdua Look at the resonance structures you have drawn for both cases and spot the ones with an aromatic benzene ring.

I don't think the Hammond postulate is particularly useful in predicting the major product here. Perhaps we can say that the TS will resemble the reactants as it is an endothermic reaction,but beyond that, I don't think you can comment on the stability of final product using this(In fact, your product is mostly kinetically controlled)..


text:
Titrations in three equibilums
You have three solutions:

0.1 M solution of a weak monovalent acid
0.1 M solution of a strong monovalent acid
0.1 M solution of a weak divalent acid

Each solution is titrated with 0.15 M NaOH. What will be the same for the 3 titrations?
I know the answer is:

The volume of the NaOH solution will be the same when you reach the first equivalence point

But why?

Acid-base equilibria always exchange only one proton at a time - so if you have a diprotic acid, that's equivalent to having two monoprotic acids with different $\mathrm{p}K_\mathrm{a}$.
When you titrate either of the three solutions with $\ce{NaOH}$, the first equivalence point is reached when the strongest (lowest $\mathrm{p}K_\mathrm{a}$) acid reacts completely - i.e. when you have added the same amount of moles of base as moles of the strongest acid were there in the solution.
Remember that the volume of titrating solution added at the equivalence point can be calculated from the respective concentrations:
$$ V_\ce{NaOH} = V_\ce{HA} \cfrac{C_\ce{HA}}{C_\ce{NaOH}} $$
Since all three examples have the same concentration of acid, the first equivalence point will happen when the same amount of $\ce{NaOH}$ has been added to all three solutions. What will be different will be the $\mathrm{pH}$ corresponding to this equivalence point, but the amount of base added will be the same.
To illustrate this, I did a quick plot of $\mathrm{pH}$ vs. volume of $\ce{NaOH}$ solution added for three acids that fit your question: strong monoprotic hydrochloric acid ($\ce{HCl}$, $\mathrm{p}K_\mathrm{a} \approx -7$), weak monoprotic acetic acid ($\ce{HAc}$, $\mathrm{p}K_\mathrm{a} = 4.8$) and weak diprotic glycine ($\ce{H2Gly}$, $\mathrm{p}K_\mathrm{a1} = 2.34$,  $\mathrm{p}K_\mathrm{a2} = 9.6$).

As you can see, although starting $\mathrm{pH}$ values are different, and although glycine goes on to have a second equivalence point, the first equivalence point (solid vertical line) occurs at the same volume of base solution.


text:
A strange case of aromaticity
The compound in question is given below

This is one of several compounds that was listed in a JEE Advanced 2017 question about aromaticity. From the solutions it seems this compound is aromatic.
However this compound has 2 sp3 carbon atoms and there isnt conjugation either, so why is this compound aromatic?

Arguably, compounds can't be aromatic at all; individual rings can.

If thats the case how would compounds like Pyrene be aromatic? From what i understad if the rings are attached via bonds like in Biphenyl or in triptycene, it must be considered single cycle at a time. But in fused ring systems like anthracene, napthalene, we consider them together. Hence the question. Please correct me if Im wrong.

You're right. The 4n+2 rule was only made for monocyclic ring systems. for other types of compounds, it may not be necessarily correct. Also see Is pyrene aromatic?

I know how pyrene is aromatic, I just mentioned that to say that we need not always look single ring at a time.

@ULTIMATEGAMER07 Consider if there weren't the whole other ring, but just a benzene ring bound to an ethene. Would the benzene ring still be aromatic?

The compound is aromatic, it can be regarded as a modified styrene which in turn is a benzene derivative.
The compound contains a cyclic pi system which has 4n+2 pi electrons and is planar. Thus it is aromatic. It is normal not to count the pi electrons in side groups when counting the pi electrons in possible aromatic compounds.
For example benzoic acid has a carboxylic acid which has 2 pi electrons in it, these are not added to the total of the pi electrons in the benzene ring despite the fact that the benzene ring and the carboxylic acid can become coplanar thus allowing the pi systems of the two to overlap and join into a larger system.
If we consider naphthalene then this is a compound which has ring of ten carbons, half way around the ring there is a sigma bond which links one side of the ring to the other. Thus it can appear that naphthalene has two rings.
When we consider aromatic systems we should ignore sigma bonds, the sigma bonds may help to hold the molecule together but the sigma bonds which are within the nodal plane between the two sides of the pi system are unable to take part in the pi system.
We should be careful of the lone pair which I lightheartedly describe as "the bond to nowhere". In some aromatic compounds such as indole and thiophene one of the lone pairs on the heteroatom starts off in an atomic orbital which is a p orbital which then takes part in the pi system of the aromatic ring.
In other heterocycles such as pyridine the lone pair is in the nodal plane where the angular function of the wavefunctions of all the atomic orbitals which make up the pi system is zero. Thus there is no orbital overlap between the lone pair of a pyridine nitrogen and the pi system of the pyridine.
Be careful as in some cases such as pyrazole and oxazole there are both heteroatoms which contribute lone pairs to the pi system and other heteroatoms which only contribute a single electron (rather than a lone pair) to the pi system.

I am down-voting this, because it needlessly oversimplifies aromaticity. The compound itself is not aromatic, the phenyl ring is an aromatic subsystem. The application of Hückel's rules is also borderline correct here.


text:
How can a vacancy defect be a stoichiometry defect?
NCERT (India) explain stoichiometric defects as those defects which do not disturb the stoichiometry of the solid. Vacancy defects are stated to be under the category of stoichiometric defects. When some lattice sites are vacant, the crystal is said to have a vacancy defect.
My question is: if lattice points are missing in vacancy defects, shouldn't this disturb the stoichiometry of the compound? I think yes but the book states no. Why?

The reason is that anions and cations will be vacant in pairs.

Yes, you are not completely wrong but try to answer these questions to check your understanding (mouse over for the answers):
First of all, what is a Non-stoichiometric compound?

 Non-stoichiometric compounds are chemical compounds, almost always
 solid inorganic compounds, having an elemental composition whose
 proportions cannot be represented by integers.

In which way a vacancy affects the ratio of the elemental composition?

 If the compound it's a "pure" metal the vacancy will not affect at all the ratio of the elemental composition. However, it could affect the ratio if, for example, we have $\ce{NaCl}$ crystal where a lot of $\ce{Na+}$ ions are missing.

How is likely to have an ionic compound where a lot of anion or cation are missing?

 Not at all, the ionic compound, usually, maintain an overall neutral charge see Schottky defect.


text:
What happens to the gas inside cold drinks as the cold drink is frozen?
According to Henry's law the amount of gas dissolved is directly proportional to the pressure exerted by a gas on the surface of the liquid.
And we know if we increase the temperature, the amount of dissolved gas decreases as it is an exothermic process and due to increase in entropy of system too , and if we decrease the temperature the amount of gas dissolved increase.
So what will happen to the gas inside the bottle if we freeze the cold drink bottle

Does all the carbon-dioxide dissolve??

Assuming the bottle doesn't burst

Well, you are certainly right in that the solubility of the gas will grow if you decrease the temperature, and so it will continue all the way down to the freezing point. That's when things get hairy.
Ice is quite different from liquid water. It does not dissolve any $\ce{CO2}$ to speak of.(*) So pretty much all $\ce{CO2}$ will leave the solution and go to the gas phase, adding its pressure to that of ice in an effort to tear the bottle open. (Some plastic bottles manage to withstand it, some don't.) The dissolved gas and whatever little salts are there will lower the freezing point of water and in fact make it an interval rather than a point, but that's another story.
After thawing, much of $\ce{CO2}$ will happily go back to the solution (though not instantly), and the drink will be more or less restored to its original state.

(*) For any physical quantity, there are layers upon layers of scientists from different fields, each confined to a certain range of the said quantity and calling everything above it "infinity" and everything below it "zero". The levels of dissolved $\ce{CO2}$ in the ice core samples from lake Vostok are quite a big deal for paleoclimatology. Not my field, though.


text:
Why are bicarbonates of alkaline earth metals stable only in aqueous solution?
Numerous internet sources (1, 2, 3, etc.) state this fact, including a  Wiley publication "Alkaline Earth Metals: Inorganic Chemistry - William S. Rees Jr" (DOI: 10.1002/0470862106.ia005) on page 9:

The bicarbonates of group 2 elements are stable only in solution.

Why is it so? Why do solid alkaline earth metal bicarbonates decompose/are unstable?
In other words, (thanks Zhe!), group 1 bicarbonates are stable. Group 2 bicarbonates are not. They decompose into the group 2 carbonate, water, and carbon dioxide. What is the principle difference between group 1 and group 2 bicarbonates that leads to a thermodynamic preference for the carbonate in one case and the bicarbonate in the other? Also, why is the preference only shown in the solid state, but not in the aqueous state?

I am not asking about thermal stability. I am just asking that what's the problem if we take these bicarbonates in solid state at room temperature and leave them there. Why would they decompose?

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

Not all Group 1 bicarbonates are stable as solids.  Lithium misses the bus.

The unstability of alkaline earth bicarbonates can be traced back to their hypothetical crystal structure. Let us start with calcite CaCO3. There, each Ca2+ cation coordinate with 6 oxygen atoms in an octahedral shape. Besides, each oxygen coordinate with a C and two metal atoms. Thus, each carbonate group (containing three oxygens) coordinate with six metal atoms (valence 2+) without much distortion, creating a structure that is very stable.

Now, the bicarbonate structure is a distortion of  CaCO3. There, on oxygen atom coordinates with three metals, another with two metals, and another with one metal plus a hydrogen atom. The latter is stabilized via hydrogen bonding with an oxygen atom which is coordinated with two metals. On average, you still have 1 metal atom (valence 1+) for each HCO3- group. Besides, both anion and cation are still hexacoordinated.

Li atoms are too small and can hardly be stabilized by coordinating with 6 oxygens. The anionic nature of such oxygens would make them "feel" and repel each other. This is the reason behind the instability of LiHCO3, previously mentioned by Oscar Lanzi.
Conversely, to create a solid bicarbonate of an alkaline earth metal, eg Ca(HCO3)2, each Ca atom would need to coordinate with 12 different oxygen atoms, which would be awfully unstable. Alternatively, each oxygen atom would need to "uncoordinate" with a metal atom, making them a lot less stabilized by electrostatic forces. In any case, the crystalline CaCO3 and their decomposition products (CO2, H2O) would be a more stable arrangement.

If there is a number with a plus or minus sign, it cannot be valence. Considering the context, it's either charge, or oxidation number.


text:
Which is a better oxidising agent: concentrated nitric acid or dilute nitric acid?
I was reading the Wikipedia page on nitric acid and saw the reaction of concentrated $\ce{HNO_3}$ and dilute $\ce{HNO_3}$ with $\ce{Cu}$.
It is mentioned that:

Copper reacts with dilute nitric acid at ambient temperatures with a 3:8 stoichiometry:
$$\quad\qquad\qquad\qquad\ce{3 Cu + 8 HNO_3 -> 3 Cu^{2+} + 2 NO + 4 H_2O + 6 NO_3^−}$$
With more concentrated nitric acid, nitrogen dioxide is produced directly in a reaction with 1:4 stoichiometry:
$$\ce{Cu + 4 H^+ + 2 NO_3^− -> Cu^{2+} + 2 NO_2 + 2 H_2O}$$

My question is: If concentrated nitric acid is a better oxidising agent, then why does it oxidises only 2 moles of $\ce{Cu}$ per 8 moles of nitric acid against the 3:8 ratio in dilute one. Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?
Does this indicate dilute $\ce{HNO3}$ is better oxidising agent than concentrated one?

https://chemistry.stackexchange.com/questions/17067/how-does-copper-reduce-dilute-nitric-acid-to-nitric-oxide-and-concentrated-nitri?rq=1

I think your assignments for the equations need to be inspected closely. Let's look at those equation with their redox potentials:
In dilute conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 8:3):
$$
\begin{align}
\ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)} &E^\circ = \pu{0.957 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\
\ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.615 V}
\end{align}$$
In concentration conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 4:1):
$$
\begin{align}
\ce{2NO3-(aq) + 4H3O+(aq) + 2e- &<=> N2O4(g) + 6H2O(l)} &&E^\circ = \pu{0.803 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &&E^\circ = \pu{-0.342 V}\\
\ce{2NO3-(aq) + 4H3O+(aq) + Cu (s) &-> Cu^2+(aq) + N2O4(g) + 6H2O(l)}&&E^\circ_{\text{rxn}} = \pu{0.461 V}
\end{align}$$
Because  this concentrated reaction is exotermic, the colorless $\ce{N2O4}$, gas decomposes to reddish brown $\ce{NO2}$ gas according to $\ce{N2O4 (g) <=> 2NO2 (g)},~\Delta H^\circ = \pu{+57.2 kJ mol^{-1}}$
Based on $E^\circ_{\text{rxn}}$ values, one can assume that the concentrated reaction is less spontaneous. However, some dissolved $\ce{N2O4}$ can still undergo following redox reaction which is the most spontaneous reaction among three:
$$
\begin{align}
\ce{N2O4(aq) + 4H3O+(aq) + 4e- &<=> 2NO(g) + 6H2O(l)} &E^\circ = \pu{1.035 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\
\ce{N2O4(aq) + 4H3O+(aq) + 2Cu(s) &-> 2Cu^2+(aq) + 2NO(g) + 6H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.693 V}
\end{align}$$
If we can consider the loss of $\ce{N2O4}$ to $\ce{NO2}$ is negligible, then total oxidation of $\ce{Cu}$ with both conc. $\ce{HNO3}$ and $\ce{N2O4}$ can be written as:
$$
\begin{align}
\ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{1.154 V}
\end{align}$$
Although it is identical reaction to that in dilute conditions, the reaction proceeds much more favorably (compare $E^\circ_{\text{rxn}}$ values in both conditions).
Late edition:
I have tried to avoid criticizing literature values on this subject, but my answer so far arose questions from concerned people. Thus, I'd try to do my best to handle it as delicate as possible.
To my knowledge, there shouldn't be a question about whether conc. $\ce{HNO3}$ is better oxidizing agent than dilute $\ce{HNO3}$. The oxidizing reagent here is the $\ce{NO3-}$ ion, as my equations described above. For example, any $\ce{NO3-}$ ion (say $\ce{KNO3}$) will oxidize $\ce{Cu}$ in an acid medium (say, $\ce{HCl}$). In using dilute or concentrated $\ce{HNO3}$, we conveniently provide both $\ce{NO3-}$ ions and $\ce{H3O+}$ for our redox reaction. This reaction is highly based on reaction conditions. I think nobody knows exact process, but everybody speculate by the results. That's the reason you are seeing too many different equations in different sources. I'm using the comprehensive table in: http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrochemical%20series.pdf, which I believe more realistic. Now, consider following 2 questions:

The question 1: If concentrated nitric acid is a better oxidizing agent, then why does it oxidizes only 2 moles of Cu per 8 moles of nitric acid against the 3:8 ratio in dilute one?

Answer: Conc. $\ce{HNO3}$ has more $\ce{NO3-}$ available to react. When available, two moles of it react with 1 mole of $\ce{Cu}$ to give more stable product, $\ce{N2O4}$ ($\Delta H^\circ_f = \pu{+9.16 kJ/mol}$) compared to $\ce{NO}$ ($\Delta H^\circ_f = \pu{+90.25 kJ/mol}$).

The question 2: Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?

Answer: Look at the last series of reactions involving $\ce{N2O4}$ given in my original answer. Eventual change of oxidation state of $\ce{N}$ is same in both cases. This reaction is not just a matter of power. I don't think anybody studied this process in details yet. I'd like to make my point by giving some known redox processes of of $\ce{NO3-}$.
$$
\begin{align}
\ce{NO3-(aq) + 3H3O+(aq) + 2e- &<=> HNO2(aq) + 4H2O(l)} &E^\circ = \pu{0.934 V}\\
\ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)}&E^\circ = \pu{0.957 V}
\end{align}$$
Both processes are under relatively similar conditions (say, $\pu{2M}$ vs $\pu{3M}~\ce{HNO3}$). In process (1), the oxidation state of $\ce{N}$ changes from +5 to +3 while in (2), it was +5 to +2. Which process dominate when they were subjected to oxidize appropriate amount of $\ce{Cu}$?

Sorry, but I am unable to grasp what is your final answer to the two questions: "(1) If concentrated nitric acid is a better oxidising agent, then why does it oxidises only 2 moles of Cu per 8 moles of nitric acid against the 3:8 ratio in dilute one. (2) Why is there change of -1 in oxidation state of N in concentrated solution as opposed to -3 in dilute solution?"


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effect of overpressure on lowering viscosity of polysaccharide solution
it has been reported that agarose dissolved at 110C under slight over-pressure has a viscosity at least 10X smaller than dissolved at atmospheric pressure at 90C (ref:https://doi.org/10.1016/j.ceramint.2004.06.007).
I have not independently verified this result (but I plan to).
The context of this result is ceramic but the solution viscosity result is independent of this context.
I see this result as quite interesting with many other applications BUT
I cannot find ANY other reference of similar results with agarose gel or any other polysaccharide for that matter, which surprise me.
Do I miss something?
I am looking for references to better understand this result and the physico-chemistry behind it.
If somebody could provide some pointers it will be greatly appreciated
Thanks
JF

The pressure is irrelevant, it is only applied so the solution can reach 110°C. Water boils at 100°C at ambient pressure.
Agarose is a polysugar, and at 110°C in water, I expect it breaks down into smaller molecules, disugars, monosugars etc.
Solutions of smaller molecules have lower viscosity. https://en.wikipedia.org/wiki/Mark%E2%80%93Houwink_equation
For typical polymers, the scaling exponent $a$ of Mark and Houwink is 0.5-0.8, but that's for infinitely low concentrations. It goes up drastically for higher concentrations, reaching a typical 3.4 in the melt.
(That's a bit sloppy, Mark-Houwink predicts the intrinsic viscosity, while in an entangled melt the actual viscosity scales with $M^{\approx3.4}$.)
So, if you have a saturated solution, and the average agarose polymer molecule breaks just twice, a factor ten in viscosity decrease is very reasonable.

Thanks this is useful as I was ignorant of the Mark-Houwink equation. On the other hand and on a qualitative level, breaking down polymer chains should impact gel properties which was not observed in the mentioned reference.Is it possible that other effect could be at play such that the quality of the unfolding of the chains.. which is a pure speculation on my side...

I don't have access to that paper. What gel, and what properties?

agarose  used to gel alumina green body and the property is gel strength after cooling. The increased solubility and greatly reduced viscosity ( measured at 60C after dissolving agar at 110C ( vs 90C  for the reference case) allow for a much better gel (defined as  increased solid loading and greater mechanical strength after gelling). I find this result impressive  (in my field) and I dont understand its root cause(s), but I am not a chemist...

OK I think I got iy now and your answer was correct.Temperatures*time hydrolysis the agar molecule reducing the chain length and viscosity (ref:https://arxiv.org/pdf/1603.00778.pdf) . Thanks for pointing me in the right direction. I cannot vote for the answer because of a low reputaiton but it answered my question.


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Preparing 1M acetic acid buffered with 1 M Ca-acetate to specific pH
I am prepping a buffer solution of 1M acetic acid and 1M calcium acetate with a pH of 5, but it has been a long time since chem 101. I have some notes from an old lab mate that suggested that 600mL of acetic acid and 150mL Ca-acetate (with water added to make 1L) makes about a pH of 4.5. However when try using Henderson-Hasselbalch to get amount for a pH of 5 I get vastly different results - 640mL Ca-acetate and 360 mL of acetic acid.
Where am I going wrong?
Any tips on how to calculate the amounts needed correctly?
Thank you

This is the way I would approach the problem, for what it's worth.
Henderson Hasselbalch equation is pH = pKa + log [base]/[acid]
If we take the pKa of acetic acid (HAc) to be 4.74, then we have
5 = 4.74 + log [B]/[A] where B is the base and A is the acid (B = acetate)
Solving for B/A we get
log [B]/[A] = 5 - 4.74 = 0.26
[B]/[A] = 1.82 and ...
[B] = 1.82[A]
But since you are using calcium acetate as the salt, there are two acetates for each calcium acetate, i.e. CaAc_2 ==>2Ac^- + Ca^2+
Thus, [B] = 0.91[A] and
A + B = 1000 ml (to make 1 liter of buffer)
Simultaneous equations and solve for A
0.91A + A = 1000 ml
1.91A = 1000 ml
A = 524 ml
B = 476 ml

Sorry for poor formatting.  I'm new here.

Ok, thanks. I was definitely missing the two acetate ions part.


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Energy at constant Entropy and Volume

The system of constant entropy and constant volume will attain the equilibrium in a state of minimum energy. Why?

I know that:
$\Delta G= \Delta H-T\Delta S$
And at constant volume and entropy: $\Delta G =\Delta U$
But from above equations how can we justify the give statement?

The fundamental relationship is dU=TdS-PdV

We can't. $\Delta G$ is of no use here.

A function such that the equilibrium state of a system is determined by its maximum (or minimum) is called a characteristic function. There is a whole bunch of those. All they are ultimately derived from the Second law after some fiddling around with differentials.
The said law basically postulates that entropy is one such function. But it is only good for closed systems, that is, those with constant $U$ and $V$. Other kinds of systems use other kinds of characteristic functions. To derive them, we typically go along the lines of Legendre transformations. Say, you have $f(x,y,z,\dots)$, which is a characteristic function in the variables $x,y,z,\dots$, and you want to switch from $x$ to $f'_x$ to use it as your independent variable. Then $f-xf'_x$ is a characteristic function in your new variables.
So let's begin with switching from $U$ to another variable. The new characteristic function is
$$S-U\cdot\left({\partial S\over\partial U}\right)_V=S-{U\over\left({\partial U\over\partial S}\right)_V} = S-{U\over T}$$
(Cryptic as it might seem, the identity $\left({\partial U\over\partial S}\right)_V=T$ is true, and moreover, is the definition of temperature.)
To bring this characteristic function to a more familiar appearance, we switch the sign (so we'll be looking after a minimum, and not maximum, as we did with entropy), then multiply it by $T$, since it is only going to be used in systems with $T=const$ anyway, and get the Helmholtz energy $F=U-TS$. That's what determines the state of equilibrium for fixed $(T,V)$.
Now let's make the final step. Luckily for us, $-S$ is the thermal derivative of $F$, hence the characteristic function in variables $(S,V)$ is
$$F-T\cdot\left({\partial F\over\partial T}\right)_V=F-T\cdot(-S)=F+TS=U$$
just as we were seeking to prove.
So it goes.
One more variable change from $V$ to $P$ will give us Gibbs energy, but that's another story.


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Can methanol be oxidised by PCC?
Pyridinium chlorochromate (PCC) in $\ce{CH2Cl2}$ can be used to oxidise primary alcohols only to aldehydes. I can't however, find any reference of it oxidising methanol (which is strictly speaking not a primary alcohol). My question is, does PCC oxidise methanol to formaldehyde and if not, what can I use to achieve this reaction?

My textbooks don't spend much time on reactions with methyl derivatives and I'm struggling to find brief discussions of 'methyl chemistry' that isn't on a higher level.

www.chem.libretexts.org/Textbook_Maps/Organic_Chemistry_Textbook_Maps/Map%3A_Organic_Chemistry_(Smith)/Chapter_12%3A_Oxidation_and_Reduction/12.07_Oxidizing_Agents

Yes it can. Why would you want to do this?

Look at the examples given here:

(Ignore the stuff written with pen; I have a habit of taking notes when reading)
PCC is taking two hydrogen atoms – one from the carbon atom attached to the -OH group and the other one from the -OH group itself, and then making a C=O double bond. Methanol has both of these. So, PCC should be able to oxidise it to formaldehyde.


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How to find the combining valence of an anion when it exhibits variable valency?
I'm doing an online course and can't for the life of me figure out how to write chemical formulas from the chemical name.
Take the example of calcium chloride.
I know the the valence for calcium is +2 because my periodic table shows me this, but I cannot figure out the valence of chlorine, (or any other element) with multiple valences. My periodic table lists chlorine as having a possible valence of -1, +1, +3, +5, or +7. How am I to know which one is to be used?

See, you have to combine plus (Ca) with minus (Cl).

Per Ivan the compound has to have a neutral charge. So if calcium is +2, then there must be -2 somehow. The rub here is that there isn't a -2 charge on a single chloride anion, but two chloride anions each with a -1 charge. The name "calcium chloride" doesn't give you a clue as to how many chloride atoms there are.  I can't imagine any chemist saying calcium dichloride. Truthfully if you study chemistry or a while you just learn how inorganic salts are named and the appropriate valences. The naming convention for inorganic "molecules" just isn't as formulaic as the naming of organic molecules.

https://chemistry.stackexchange.com/questions/4547/how-to-get-the-valencies-of-elements

Generally, $\ce{Cl}$ shows a valency of -1. As $\ce{CaCl2}$ is an ionic bond, also neutral therefore, valency of $\ce{Ca}$ * 1 + valency of $\ce{Cl}$ * 2 = 0   which gives 2 + valency of $\ce{Cl}$ * 2 = 0. Therefore, valency of $\ce{Cl}$ = -2 / 2 = (-1).

First of all, sorry for the (very!) long post!
Determining the Number of Valence Electrons
The number of valence electrons of atom can be determined by the group in which the atom is in. For example, Na is in group 1A, and thus has one valence electron (1+ charge). Chlorine is in group 7A and thus has 7 valence electrons (1- charge). A quick (but very simplified) way of determining if an element has a positive or negative charge is to look again at its group. Groups 1A, 2A, and 3A are positive; while groups 5A, 6A, and 7A are negative. An element in group 4A can be both.
Ionic Compounds:
If the charges are the same, no "balancing" is necessary:

if Na (1+) were to bond with Cl (1-) it would form NaCl.

If the Charges are different, a quick way to determine the formula unit is to assign each element the charge of the other:

if B (3+) were to bind with Cl (1-) it would form BCl3: Boron get the 1 from Chlorine; Chlorine gets the 3 from Boron.

Covalent Bonding
Covalent Bonds are a little trickier and has more contributing factors. It seems like you're having trouble with Lewis-Dot diagrams so that's what I'll address. In order to correctly draw the Lewis-Dot diagram of a compound, you have to determine the formal charge of each atom. For demonstration purposes I'm going to use CH4.

The formal charge of an atom can be determined by:

Number of Valence Electrons - non bonding electrons - number of bonds

So in our example, the formal charge of Carbon is :

4 Valence electrons - 0 non bonding electrons - 4 bonds = formal charge of 0.

Then Hydrogen:

1 Valence electron - 0 non bonding electrons - 1 bond = formal charge of 0.

So if we add the formal charged together we get 0, which makes sense because the charge of the compound is 0.
In general, the more compounds that can have a formal charge of 0, the more likely it is that that particular structure will occur, as it will be more stable. If the compound is a poly atomic ion, the more electronegative atom should get the negative formal charge (electronegativity increases as you move up the group and to the right in a period).
One more significant detail is the ability of some atoms to expand the octet, which can occur if the element is in the third period or beyond. In this case, the charge of the atom can be made 0 (or whatever appropriate) if more bonds are made than typically seen.

i.e. in ICl4-, Iodine, though typically only making one bond, bonds to four Chlorine atoms. In this case formal charge = 7 valence electrons - 4 bonds - 4 non bonding electrons = 1-, the charge of the polyatomic ion. While Cl is more electronegative, only I can expand its octet, thus I has the negative formal charge. The 4 non bonding electrons on Iodine were determined by counting the total number of valence electrons (7 from I + 4*7 for Cl + 1 from the charge = 36) and adjusting accordingly.

There are other exceptions such as free radicals and atoms that can have less than a full octet, but the "rules" above should be sufficent guides for basic Lewis Dot structures.
Hope that covered everything :)


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Is it possible to define the coordinates of an atom using other coordinates while optimizing in gaussian 16?
I know it is possible to fix the value of composite expressions using generalized internal coordinates (gic) in gaussian 16. But is it possible to assign that value to the coordinate of another atom.
For example if I have the following coordinate file
H x1 y1 z1
H x2 y2 z2
O x3 y3 z3
Is it possible to assign x3 = (x1 + x2)/2 . When optimizing this structure I would like that the x coordinate of oxygen always be the average of the x coordinates of the two hydrogen atoms.

I just figured out that there is a way to do this. Though its not that clean I think. I can define the initial structure where x3 = (x1+x2)/2 and then freeze the following expression using gic.
x3 - (x1+x2)/2


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Zinc and copper salt bridge replacement
If I were to replace $\ce{Zn}$ with $\ce{Al}$ in  $\ce{Zn|Zn^2+ || Cu^2+|Cu}$. Would it produce the same results?

No, the results would be quite different. You might still be able to get some voltage, though.

No, it would not.
$\ce{Al}$ is normally covered by oxide layer which effectively renders it inert to most things. Your cell would stop dead, as if the electrodes were made out of wood.
Then again, you might want to know whether it is possible to remove the said layer. Yes, it can be done, but then $\ce{Al}$ turns out to be just too active. It reacts with water quite readily (albeit not as spectacularly as $\ce{Na}$). Merry little bubbles of $\ce{H2}$ will pop up, flakes of $\ce{Al(OH)3}$ will drop to the bottom, and in a matter of an hour or so your electrode will be gone. While it lasts, the cell will produce some voltage; as Dr.J. correctly noted, it will be even greater than that of the normal $\ce{Cu/Zn}$ cell.
So it goes.

The reduction potential for Al is different from that for Zn, so you will definitely obtain different results.  The voltage produced with Al will be almost twice the voltage obtained with Zn.

There are more important differences than that.

@Ivan For example? I would think one of the most important factors in a galvanic cell would be the voltage that is generated.


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Products formed on cleavage of epoxides using HI
There seems to be multiple forms of $\ce{HI}$ used to cleave epoxides:
conc. $\ce{HI}$, anhydrous $\ce{HI}$, cold $\ce{HI}$, hot $\ce{HI}$, cold conc. $\ce{HI}$, hot conc. $\ce{HI}$, etc.
So, does concentration of $\ce{HI}$ and temperature have any significance in determining the product?

Cleavage of epoxides can happen in 2 ways, resulting from the cleavage with the 2 carbons to which the oxygen atom is bonded.
If the medium is acidic, it will follow the SN1 pathway, proceeding via the most stable carbocation. This makes sense, since in acidic medium, the alcohol can get protonated and break away from one of the carbon atoms.
If the medium is basic, it will proceed via the SN2 pathway.
In other words, the type of cleavage doesn’t depend on the concentration of $\ce{HI}$, the only thing that matters is the acidity or basicity of the medium.

This seems to be a repeat of the previous answer.. what about temperature? See Martin's comment below..

Temperature shouldnt matter. This is a competition between sn1 and sn2, and temperature should be considered only when elimination is involved, like in the case of beta unsaturated epoxides, where elimination will lead to a conjugated product.

for unsymmetrical epoxides or oxiranes reaction takes place in this way:
in acidic medium- ring on reaction with HI ring opens from more hindered side i.e., the more hindered C-O bond. So the carbon in which more number of groups are attached which is bonded to the oxygen in ring gets -I.
in basic medium- the ring opens from less hindered side.
temperature is not a much important factor in this reaction.

Welcome to chemistry.se! While there is certainly some truth to what you have written, I think it would be better to beck it up with some references; especially the last sentence. I also think this does not address the question completely.


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Stat Mech: Maxwell-Boltzmann Distribution of Momenta
I'm trying to show that the MB distribution of scalar momentum that I've found is properly normalized but I'm stuck. My process: I started with the MB distribution of momentum in 2D and found the normalization factor. Then I'm asked to find the distribution for scalar momentum (where only the magnitude of momentum matters) and prove that it's properly normalized. I transformed from Cartesian to polar coordinates and used a 2D Jacobian matrix. The thing is, once I try to prove it's normalized, I get stuck with an expression that goes to zero and I can't find where I went wrong. I'd appreciate any input. Thanks!
MB dust of momentum in 2D, normalized:
$$ f(\textbf{p}) = \frac{\beta}{2\pi m} e^{\frac{-\beta \textbf{p}^2}{2 m}} $$
To find $F(p)$, we integrate over the polar angle and integrate out the angle-dependence.
From Cartesian to polar coordinates:
$$ dp_x dp_y  = det(J)* dp d\theta = p*dp d\theta$$
The probability of finding p between p and p+dp:
$$prob(p,p+dp) = \int_{0}^{2\pi}f(\textbf{p}) p*dp d\theta = 2\pi p \frac{\beta}{2\pi m} e^{\frac{-\beta p^2}{2 m}} dp = p \frac{\beta}{m} e^{\frac{-\beta p^2}{2 m}} dp$$
Therefore,
$$ F(p) = p \frac{\beta}{m} e^{\frac{-\beta p^2}{2 m}}$$
Then to prove F(p) is properly normalized,
$$ 1 = \int_{-\infty}^{\infty} p \frac{\beta}{m} e^{\frac{-\beta p^2}{2 m}} dp $$
$$ = \frac{\beta}{m} \int_{-\infty}^{\infty} p e^{\frac{-\beta p^2}{2 m}} dp $$
but by symmetry,
$$\int_{-\infty}^{\infty} p e^{\frac{-\beta \textbf{p}^2}{2 m}} dp = 0$$

I can't see where the error is exactly, but it has to be in your conversion to polar coordinates. In Cartesian, the integral is clear if you note that $p^2=p_x^2+p_y^2$ and then evaluate each part of the double integral using the formula for the standard Gaussian integral.

I've actually spotted it now. In your conversion to polar coordinates, you did not account for the change in the limits of integration. Magnitude of p can only go from zero to infinity, and if you make this change you will get the same result as doing it with Cartesian.

Yes, I see that the bounds of integrations was where I went wrong. Thanks for the help!


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Reactions of Grignard Reagent: 1,2 vs 1,4 addition
So I was reading through Grignard reagent and I came across this answer by jerepierre, which mentions:

In general, Grignard reagents and organolithium reagents add directly to the carbonyl carbon, while organocuprates (organocopper reagents) add to the beta-position of an unsaturated ketone.

Now, I want to know, what is the meaning of 'in general'? Can anyone please provide me some cases where the general rules are not followed.
Also, I came to know that:

give 1,4 addition with $\ce{PhMgBr}$ to give:

(that reference is difficult to provide at the moment).
Can someone please explain why this happened?

Can you work on getting that reference?  I am getting feedback on my answer asking for it, but you made the claim.  Thanks!

@OscarLanzi Sorry... but at my current level its impossible to procure citations. All claims made here are according to what my instructor taught me...

This is not a 1,4 addition.  It is a 1,2 addition across a carbon-carbon double bond.  This mode of reaction is unusual for Grignard reagents, but here a highly stabilized "double benzylic" anion is formed and the competitive reaction of adding to the carbonyl group is sterically hindered.

Do you have a reference for that?

I'm going by what the asker posted.  The 1,2 attack forming the carbanion is the only way to get the claimed product.  I think the asker needs a reference for the claim.

@OscarLanzi Can you explain formation of 'carbanion'?


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Can the alkyl-oxygen bond in lactones be cleaved in an SN2 reaction?
I understand that nucleophiles typically attack the electrophilic carbonyl carbon of the lactone and open the ring through a nucleophilic acyl substitution. Has there ever been the case where the nucleophiles open the lactone ring by attacking the carbon adjacent to the oxygen (not the carbonyl oxygen) via a SN2 reaction instead? Is there a preferred pathway for certain nucleophiles to take and why so?

Yes, it is possible. If you consider the chemistry of a four-membered lactone (a β-lactone), it is able to act as an alkylating agent, as nucleophiles can attack the C–O bond in an SN2 fashion.
One example is in the synthesis of a series of β-substituted amino acids (J. Am. Chem. Soc. 1985, 107 (24), 7105–7109). An intramolecular Mitsunobu reaction of Cbz-protected serine 1 forms the β-lactone 2 which is then opened by a variety of nucleophiles. Here, benzylthiolate is used as an example, giving N-Cbz-S-benzylcysteine 3 as the product:

The authors note that:

Only relatively “hard” nucleophiles like ammonia and methoxide attack the carbonyl to give acyl-oxygen cleavage. In earlier reports serine β-lactones were used mainly for synthesis of seryl peptides by attack of amines at the carbonyl. However, with ammonia as the nucleophile, simply altering the solvent [from MeCN to THF] inverts the 3:1 amide/amine ratio.

The following article has a wide selection of more examples (see references 6–8): J. Org. Chem. 2002, 67 (14), 4680–4683.

The reactivity you describe also occurs in α-lactones. The most common example is the diazotization of amino acids, followed by nucleophilic ring opening (e.g. Org. Synth. 1988, 66, 151):

The reaction proceeds with overall retention of configuration, via an initial SN2 displacement of N2 by the carboxylic acid, followed by a second SN2 ring-opening at the α-carbon by the nucleophile (here chloride).
See also Organic Chemistry (2nd Ed), Clayden et al., p 875.


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Rutherford's alpha particle scattering experiment for sheets made of different metals
The correct order of increasing number of alpha particles passing undeflected through the foils of Au, Ag, Cu and Al of 1000 atoms thickness each in an alpha scattering experiment of Rutherford would be?

Are you talking about the Rutherford's gold foil experiment? I read that the gold foil was 400 atoms thick. Why did you title this "size of atom question?"   Size of which atom?

@suse The thickness of the foil is same in all four options here, so I think that's irrelevant. I think we have to look at the number of protons in the nucleus here, or maybe the size of the nucleus.


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Comparing basic strength of nitrogen heterocycles

Compare the basic strength of the following compounds:

d > b > c > a
c > b > a > d
c > a > b > d
a > d > c > b

I know that in compound (c), the π-bond between nitrogen and cyclopropenyl will polarize (due to aromatic stability) such that it increases electron density of the nitrogenous ring. Whereas, in compound (d), the π-bond will polarize reducing the electron density of nitrogenous ring. Hence, comparing compounds with similar nitrogenous rings, I get that the basic strength of (c)>(b)and that of (d)<(a). But I thought the ring containing more number of nitrogen atoms will be more basic as there will be more lone pair donors, so I thought the answer to be option (4), but the correct answer happens to be option (2). I would like to know that how is basic strength of the nitrogenous rings being compared?

It is likely that the electron pair (from the HOMO) being donated to the acid is from a π molecular orbital. Since the π bond is rather weak, the π electrons should be rather high in energy and is thus able to interact well with the LUMO of the acid. When nitrogen atoms are incorporated into the π system, the energy of the HOMO is lowered due to the high electronegativity of N. Thus, incorporation of N atoms into the system would decrease the basicity of the molecular ion. As such, A and D would be inferior bases, compared to B and C.
Choosing between A and D, and between B and C, we would have to base our criteria for the basicity on the extent of delocalisation. In D, the π system is larger, thus greater delocalisation was achieved, with a lowering of the energy of the HOMO. Thus, D is likely to be less basic than A. Similarly, the extent of delocalisation is also larger in C, thus it would have a lower-energy HOMO, compared to B. Thus, B is likely more basic than C.
Therefore, option (2) is likely the correct order of basicity.

Can it be explained without the use of Molecular Orbital Theory?

@AumkaarPranav Well... You could ignore the MO theory and simply just base your considerations on delocalisation of the $\pi$ electrons and electronegavity effects. The use of MO theory is just to further concretise the analysis.

Read the answers to this question to see why more no. of lone pairs does not always mean more basicity.
Here, the common feature of all four compounds here is the middle nitrogen atom, so let's focus on that.
For it to be a good base, we need to split the double bond heterolytically to give the nitrogen a negative charge.
Done with step one? Now notice that a and b will have an aromatic ring each, c will have two aromatic rings and d will have one aromatic and one anti aromatic. Based on this, c is the most basic and d the least. Now the confusion is between compounds a and b.
To decide the strength of a and b, notice that in b, there is just one nitrogen pulling electron density away (inductive effect), while in a there are two of them. So, b is more basic.
One thing to keep in mind when deciding b/w compounds a and b –one might say that since there are two nitrogens which can donate lone pairs in a, there will be more  resonance stabilisation of the positive ion formed after acid attacks the middle nitrogen which we are focusing on. But remember that the lone pairs of these two nitrogens are not participating in resonance. So, decide based only on inductive effects.

Why will the middle nitrogen atom act as the basic one? Its lone pairs will be involved in resonance with the nitrogenous ring which will also be responsible for its aromaticity. I think the remaining nitrogen atoms will be donating their lone pairs like in imidazole . That is why I considered the basicity to be dependent on no. of nitrogen atoms present.

@keenlearner $NH_2$- has two lone pairs.

@keenlearner Also, the middle nitrogen is common in all four options and also bonded to the other ring, which gives me reason enough to focus on it. If you can provide an alternate explanation which does not focus on the middle nitrogen and is satisfactory enough, I'll be happy to accept it. :)

@keenlearner When solving a question, you will need to look for patterns like this. You will have to ask yourself -- "Why would this question have been created? What is the concept behind it?" Which in this case would be aromaticity. If you see something which can create aromaticity, it is a huge hint for you that the question setter has provided.


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Ammonia complex
I am desperately trying to find an aqueous chemical that will selectively form a complex with ammonia gas. Any suggestions please?
I am building a gas sensor for ammonia

Why use a complex in the first place? Is litmus paper not good enough?

Five seconds on google give this: https://www.sciencedirect.com/science/article/pii/S0925400504008196

also: Cu2+, but how do you make a sensor out of this info?

Why not use a Drager tube for ammonia, this is a porous pad with the acid form of a pH indicator and a known amount of acid in each unit length of the pad. You could base your sensor on a pH sensor combined with an electrochemical cell which generates acid.
If you can devise electronics which can maintain a constant voltage from a glass electrode (combined with a reference electrode) by adjusting the flow of current through a cell which generates acid at one electrode. The other electrode would need to be behind a membrane.
Another apporuch would be to use cobalt or another transition metal. The formation of ammonia complexes will alter the redox properties of an electrode. If you had copper electode exposed to the air with a thin layer of liquid on it, which is one electode and the other electode is a reference electode isolated by a salt bridge or membrane from the copper electrode then you could see a change in cell emf as a function of the amount of ammonia which the cell is being (or has been exposed) to.


text:
Why is 5,6-dimethylidenecyclohexa-1,3-diene non-aromatic?
The compound 5,6-dimethylidenecyclohexa-1,3-diene is supposedly non-aromatic:

But I have no idea why this is so. The compound is probably planar. All the atoms on the ring are $\ce{sp^2}$ hybridized. So, the p-orbitals are in conjugation. There are 4-$\ce{\pi}$ electrons, as I suppose the external pi bonds do not participate in resonance. So, it should be antiaromatic. Why is it non-aromatic?
I am a school student, so an detailed description would be helpful.

Not a conjugated ring system.  With the pendant pi electrons we can't apply the 4n+2 rule to it.

It is conjugated though, in effect it is electronically similar to a linear polyene. To make a 3rd double bond in the ring would mean making a radical species of unreasonably high energy.

Or you would need a zwitterionic structure where one methylene carbon is negative and the other, seemingly identical carbon would be positive and short of an octet.  This is also not pretty.  We are just not getting this thing to be aromatic, period.

@OscarLanzi, what is 'pendant pi electrons'? I am guessing it means the pi electrons outside the ring. But if we only consider the pi electrons in the ring, then it should be anti-aromatic (conjugated ring system, 4 pi electron.

Like the "pendant" decoration hanging off a necklace or chain.  Pendant pi electrons, or even vacant pendant pi orbitals, modify the ring orbitals and bust the 4n+2 rule no matter how you try to count.  DO NOT apply the 4n+2 rule, or any rule, beyond its range of validity!

@OscarLanzi, But what about tropolone? We apply the 4n+2 rule there, and tropolone is indeed aromatic. Tropolone has a hanging pi bond with oxygen.

@OscarLanzi, I meant tropone, not tropolone.

@SRMaiti, tropone is formally aromatic since the carbonyl is polarised towards the more electronegative oxygen atom which leads to a charge-separated aromatic resonance form. For more detailed explanation see https://chemistry.stackexchange.com/questions/15272/is-tropone-aromatic.

In this question, conjugation of p-orbitals in the system is misrepresented. It should be noted that, a π-system to be an aromatic or anti-aromatic, there should be a closed loop(s) of p-orbitals (conjugated), which is planner. What makes it an aromatic or anti-aromatic whether it follows Hückel's
rule of (4n+2) π-electrons or not.
Aromatic (most stable): If a planer, closed loop(s) of p-orbital system has (4n+2) π-electrons (in the closed system), it is aromatic. Example:, benzene (a closed p-orbital system with six π-electrons) and naphthalene (a closed p-orbital system with ten π-electrons).
anti-Aromatic (least stable): If a planer, closed loop(s) of p-orbital system has (4n) π-electrons (in the closed system), it is anti-aromatic. Example:, cyclobutadiene (a closed p-orbital system with four π-electrons) and cyclopentadiene carbocation (a closed p-orbital system with four π-electrons).
non-Aromatic (stability depends on the system): Every other systems are non-aromatic. Examples: 5,6-dimethylidenecyclohexa-1,3-diene and heptafulvene (In both cases, each p-orbital system is not closed).
See following diagram:

What about tropolone? The pi electron system is not closed, but the molecule is aromatic.

Sorry, not tropolone, it should be tropone.


text:
A hint for multiple choice question

Give me a hint to solve this question,
what should my first step ?
should convert the grams into moles ?
aluminum= (54g/150g)= 36% ( percent composition of aluminum)
sulphur= 64%
0.36x=20 (36 percentage of what is 20)
x=20/0.36
x=55.5 grams
is there a more easy way ? to do in mind

Stoichiometry problems should generally be solved by converting to moles, using the appropriate stoichiometric rations, and converting back to a desired quantity.

You might consider writing a correctly balanced equation, and then convert grams to moles.

i have edited the question now,

That is certainly one way to do it, and the answer appears to be correct.  There are other ways to approach this problem, but I'm not sure they are any easier than your approach. The answer would be given to 1 significant figure, so the correct choice would be 60 g.

i really appreciate your answer ,

but this website is very dumb, physics forums is best, any one can ask any type of question, ,, ,,,,, they are saying you have to wait one day for your next question ,,,,,,,, very dumb thinking,

please block my acount, I don't want to be her e, thankyou for your dumb service , meanwhile physics forums has given me 32+ replies .


text:
Synthesis of Polypropylene glycol

Polypropylene glycol is produced by ring-opening polymerization of
propylene oxide. The initiator is an alcohol and the catalyst a base,
usually potassium hydroxide. When the initiator is ethylene glycol or
water the polymer is linear. With a multifunctional initiator like
glycerine, pentaerythritol or sorbitol the polymer branches out.
Source: https://en.wikipedia.org/wiki/Polypropylene_glycol
Polypropylene glycols ... are formed by action of base on propylene oxide in the presence of 1,2-propanediol as an initiator.

I don't understand the meaning of an initiator here. My mechanism is the first the OH- adds to the propylene oxide giving $\ce{HOCH2CH2O-}$ which again adds to another molecule of the oxide and the chain continue and the polymer is formed.
My mechanism requires no role of the "initiator". I want to know what is the requirement of the diol in initiating the polymerisation, whether my mechanism is correct and if wrong, what is the actual mechanism?

If you just use hydroxide, then the initiator is water.

One of the best one-liners I've seen so far.

There are two broad classes of polymerisation, there are the condensation polymerisations and the addition polymerisations. The classic addition polymerisation is the reaction of a free radical with an alkene (olefin) such as styrene to form a macromolecular radical which is a little larger. The growing polymer chain has a reactive site at the end which reacts with more of the monomer thus making a longer polymer chain.
In the case of a hydroxide or alkoxide anion, it can react with ethylene oxide or propylene oxide to ring open the epoxide and form a new alkoxide which has a higher formula mass than the original anion.
This process can be repeated again and again to form either polyethylene oxide (polyethylene glycol) or polypropylene oxide.
The hydroxide anion (or an alkoxide) can start the polymerisation reaction, while many polymerisation initiators are compounds such as tert-butyl peroxide, benzoyl peroxide and AIBN which are able to create radicals which in turn start (initiate) the polymerisation these radical forming initiators are a subset of polymersation initiators.
Another example of a non radical initiator would be a bis-cyclopentadienyl methyl zirconium compound with a vaccent site on the metal. This will be able to start the formation of high density polyethylene (HDPE) if ethylene is added to this compound. This is an example of a Zieger Natter polymersation.
I think that you need to accept that a polymerisation initator is any substance which can start the formation of a polymer chain from monomer molecules by means of an addition polymerisation.


text:
Can a stable vinylic carbocation be formed?
Is it possible to prepare/form a vinylic primary carbocation? Will it be stable (low energy) or not?

Related: Why are vinylic and arylic carbocations highly unstable?

Would you like to know the reactions which consist of vinylic carbocation intermediates? Please clarify. Thanks!

@Abcd Yes i would love to, it will help a lot. Thank you for support.

What does this question mean? Low energy? Compared to what?

https://www.sciencedirect.com/science/article/pii/S1010603016311649 Yes, you can make such things.

@Zhe Oh thats a good one article. Is there any evidence for a non aromatic vinyl carbocation ?

In general yes, google vinyl iodonium salts, a primary one though is unlikely

See references at bottom. https://en.wikipedia.org/wiki/Vinyl_cation