text:
Optical activity of substituted biphenyl
How come the given compound in the diagram is optically inactive? The two rings should be in different planes due to steric hindrance between $\ce{COOH}$ and bromine. How am I wrong?

There's a plane of symmetry

Can you plz explain further.. What I understand is that if the two rings stay in one plane then sure one can imagine a mirror plane. But aren't the two rings in different planes? How does a plane of symmetry come in the picture then?

As @Zenix states, planar or not, it's symmetrical. Make a paper cutout, or tinker-toy model, and flip it over.

Related: https://chemistry.stackexchange.com/questions/65064/chirality-of-biphenyls

As users Zenix and DrMoishe Pippik indicate there is symmetry in the compound.
If the two phenyl groups are planar, then there is one plane of symmetry which would be the plane of the image as shown. The bond between the two phenyl groups would be a $C_2$ axis of rotation.
If the two phenyl groups are perpendicular, then there are two planes of symmetry, one with each of the phenyl groups.  The two planes would intersect along the bond between the two phenyl groups, and that intersection line would again be a $C_2$ axis of rotation.


text:
Commercial chalk for kids composition
I have no background in chemistry whatsoever. However, I'm very interested in sidewalk chalk composition. I read that chalk can be both either made out of gypsum or chalk (calcium carbonate). However, I'm confused, since a person that it is connected to a chalk (final product) manufacturer told me that the main ingredient in their product (chalk for kids) is sulfuric acid carbon.
As I said, my knowledge in chemistry is very reduced. I was just hoping that someone that knows well the subject can say if there can be a connection between the sulfuric acid carbon and gypsum.

It seems that most "chalk" today is really based on gypsum, $\ce{CaSO4\cdot 2H2O}$, rather than the mineral chalk which is sedimentary carbonate rock, $\ce{CaCO3}$, derived from marine organisms.

I couldn't find any reference to Sulfuric acid carbon as some sort of industrial product. I'd guess that Sulfuric acid carbon is their name for some sort of calcium sulfate product which they buy and is a byproduct of some other process.

There are probably at least 2 translation steps in the terminology, with the second being into English and the first from the chemical composition to everyday language.  I can just about see how you could start from "mixed sulphate/carbonate" and get to "sulphuric acid, carbon"

@ChrisH: I suspect that there is a lot in what you say. In my native language, German, "Kohlensaurer Kalk" [literally translated "carbonic-acidic lime"] is an old-fashioned way of refering to CaCO3, similarly "Schwefelsaurer Kalk" [sulphuric-acidic lime] is CaSO4.

In addition, we have a chinese whispers-like joke of how such terms can turn into meaningless jumble of perfectly normal words for NaHCO3: Doppelsohlenkauendes Nashorn [double sole chewing rhinoceros] instead of doppelt kohlensaures Natron. It is a bit of a tautology, since German "Natron" is already NaHCO3, doppelt kohlensauer: 2 (double) equivialents of carbonic acid (compare bicarbonate) [per equivalent of sodium]. But Natron in this naming system can stand for  Na⁺ in general (or if you want, NaOH). So the tautology helps against this ambiguity.

Thanks a lot for all contributions!! I really appreciate.

I am afraid whoever told you "sulfuric acid carbon" does not know the basic terminology (must be from non-chemistry background). There is no such thing as "sulfuric acid carbon" to begin with. Any carbon obtained by charring with sulfuric acid would be jet black. Nobody would like to have a black chalk.
It is very easy to parse $\ce{CaSO4.2H2O}$ incorrectly.
I think he/she misread the calcium $\ce{Ca}$ symbol for carbon which is $\ce{C}$.
$\ce{SO_4^2-}$ can be easily misrepresented to be sulfuric acid.

+1 A weird but quite possible explanation!

It may also be that they used (in whatever language) the colloquial terms which were misunderstood and/or mistranslated by OP into English.

Could you also add what the correct name would be? (Calcium sulfate?)

Not all chalk is calcium sulfate I guess. But yes the you would call it (hydrated) calcium sulfate.

Rather pure chalk ($\ce{CaCO_3}$) can be found in the subsoil of huge regions of central France. This chalk can be used for writing on blackboard, but it is too soft. You quickly get whitened fingers and hands when using it in a classroom.
You can replace it with gypsum ($\ce{CaSO_4·2H_2O}$) which is made by mixing plaster with some water, to give a sort of paste, that could be moulded to form a stick useful for writing on a blackboard. But the stuff is too hard, and it will scratch the board.
The best piece of chalk to be used in classrooms is a appropriate mixture of these two white stuffs : $\ce{CaCO_3}$ and $\ce{CaSO_4·2H_2O}$
But in any case, sulfuric acid will never be used for making pieces of chalk or any other substance that may be handled by the hand. Sulfuric acid is a rather corrosive substance. It will punch skin and clothes. it is one of most corrosive substance known in the world.

I don't think you answered the question. The big question here is what is Sulfuric acid carbon?

Sulfuric acid carbon does not exist. It may be considered as a mixture of carbon and sulfuric acid, which are two independent substances which can be mixed of course. But I don't see the interest in doing this mixture. This mixture will be corrosive like sulfuric acid. It will never be used for any game for children. Too dangerous. So what for ?

Maurice - Sorry, I obviously didn't make my point clearly. The OP asked about Sulfuric acid carbon because someone who worked at a plant where "chalk" was made said that was the main ingredient. I agree that it doesn't seem to be a recognized industrial name. I'm guessing that is just what they call the raw feedstock within that particular plant. i'm guessing also that sulfuric acid carbon is a byproduct of making sulfuric acid or using sulfuric acid in some other industrial process. In other words, the chalk company is buying some other company's industrial waste.

@MaxW: still, I think this answer may help OP to dig down to what was actually said and meant in whatever language it was said.

source: Stack Exchange

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

sort: votes


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Order of adding reactants on growth of Cu
I am reading the following paper which details the steps to control growth of Copper. They fabricate Copper powders from $\ce{CuNO3, NaOH, and C6H8O6}$. Specifically, in this paper they add the sodium hydroxide first to the copper nitrate solution and then they add the ascorbic acid. The chemical equations that occur are $$\ce{Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3}\tag{1}$$
$$\ce{Cu(OH)2 + C6H8O6 -> Cu + C6H6O6 + 2H2O}\tag{2}$$
My first question is if the order of adding the acid and base was switched: first ascorbic acid was added to copper nitrate and then sodium hydroxide was added, would it yield the same products as (2)?
My second question is how could I calculate the theoretical pH of the overall reaction if I add 0.1M $\ce{CuNO3}$, 0.5M $\ce{C6H8O6}$, and 0.1 M $\ce{NaOH}$ following the procedure of the paper?
I am thinking that since NaOH is a strong base while ascorbic acid is a weak acid, the pH will be dominated by NaOH. Therefore, I think the pOH would be -log(0.1M $\ce{OH- from NaOH}$) and so the pH would be 14-1 = 13. However, I know this is too high a pH so my method is wrong. What am I missing?

Now is the time to finally learn, how is pH calculated in trivial and non trivial cases. Google and Wikipedia are good guides.

Uhh I know how pH is calculated

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Vapour pressure of a liquid subjected to pressure
I am following the derivation by Atkins in section 4B(c) (Physical transformations of pure substances; thermodynamic aspects of phase transitions), in which the authors derive the equation ${p = p_* * exp[V_m(liquid)* ΔP/RT]}$. The derivation starts by noting that chemical potentials of vapour and liquid are equal at equilibrium, $μ_{liquid} = μ_{gas}$, and follows by integration over the pressure to calculate the change in chemical potentials due to the added external pressure.
What is unclear to me is why the integration for the liquid phase is taken over the ambient pressure while the integration for the gas phase is taken over the vapour (NOT ambient) pressure. It seems counterintuitive to me because I would expect that $μ_{gas}$ is a function of the ambient pressure and therefore should be integrated over the ambient pressure. Am I missing something?
EDIT: I guess the question could be rephrased: in the definition of molar volume for gas, $V_m=(\frac{\partial \mu} {\partial P})_{T,n}$, is $P$ the partial or ambient pressure? More generally, does $\mu_{gas}(T,P)$ depend on the partial or ambient pressure?

It assumes that, in the gas phase, the mixture behaves as an ideal gas mixture.

Thanks @Chet Miller, it helps! So, $V_m(gas) = RT/p$, and we then change the integration variable from the total pressure $P$ to the vapor pressure $p$ by taking advantage of the fact that $dP = dp$ and replacing the gas-phase integration range accordingly. This, together with some approximation, will lead to the equation in the textbook. I see the logic now.

On second thought, it is not obvious why $dP = dp$. All we know is that the vapour pressure is some function of the ambient pressure but it is not necessarily a linear function.

Here's how it works.  The fugacity of the liquid at temperature T and total pressure P is equal to the equilibrium vapor pressure at T times the Poynting correction (assuming that the vapor phase behaves as an ideal gas mixture).  The fugacity of the same species in the vapor is equal to the partial pressure of the substance in the gas phase.  The vapor and liquid are at equilibrium at temperature T and total pressure P if the fugacity of the substance in the liquid is equal to the fugacity of the substance in the vapor phase.


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Chemical simulation software for a person who is not in the chemistry field but has a basic knowledge about chemistry
In my research I need to see, if I mix chemical A with chemical B, then what is the outcome? if it's another chemical, then I need to see the resultant chemical's 1. Density 2. Melting point 3.reaction time 4. if it will have any risks while performing the mixing in real life (explosions, fire etc etc)
What I'm actually trying to do is: I have a specific chemical, I want to know which chemicals or solutions will produce a rock-hard precipitate or crystals by reacting with my specific chemical.
The software should be accurate too, because I will use this for my PhD. I'm not in the chemistry field but my PhD involves inorganic chemistry.
I have only basic knowledge about chemistry, please don't ask me to go and read a book of chemistry or complete a chemistry degree as there's no time for that, finding the right chemical is a fraction of my research and I'm exhausted with the PhD already :)

IBM's Watson (Artificial Intelligence) hasn't reached that level of sophistication yet. Yes an exhaustive literature search is a pain in the neck.

I think this question is too broad as is. To be able to do what you describe for any two possible chemicals A and B is incredibly complicated. However if you know what A and B are, it may be a lot easier to predict or may already be known. You would probably be better served if you asked about your specific chemicals.

Predicting chemical reactions is very hard.  See, for instance:
Why is predicting products of chemical reactions difficult?
and
Software for predicting chemical reactions
The only software that might have the capability to give useful predictions is software that has been highly optimized for a very specific subset of chemical reactions.  For instance, a phamaceutical company studying a certain class of drugs might have developed proprietary internal software that helps filter out what would and wouldn't be good drug development candidates.  But even that software would have difficultly attaining the degree of specificity you are looking for.
Your best bet would thus be to find an expert (ideally someone in the chemistry/chemical engineering/materials science dept. at the school where you're getting a Ph.D.), and describe exactly the type of reaction product you're looking for.  That person might then be able to tell you whether it's already known, from experimental work, what combination of reactants would give you that.  If so, you wouldn't have to rely on software, you'd just (with that person's guidance) need to do a literature search.
Alternately, if you could be specific about the reactants and products you had in mind, someone here (not me!) might have the expertise to tell you if it's known experimentally whether they (or a different combination) might give you what you want.


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A gram of salt added to pure (distilled) water introduces around $2 \times 10^{22}$ ions?
I am currently studying Practical Electronics For Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5.2 Resistivity and Conductivity, claims the following:

Adding an ionic compound in the form of common salt ($\ce{NaCl}$) to water increases the ion concentration within solution -- $\ce{NaCl}$ ionizes into $\ce{Na^+}$ and $\ce{Cl^-}$. A gram of salt added introduces around $2 \times 10^{22}$ ions. These ions act as charge carriers, which in turn effectively lowers the solution's resistance to below an ohm per meter. If we use the solution as a conductor between a battery and a lamp -- via electrodes placed in solution -- there is ample current to light the lamp.

How did the authors calculate that a gram of salt added introduces around $2 \times 10^{22}$ ions?
I would greatly appreciate it if someone would please take the time to clarify this.

Have you heard of Avogadro's number and the concept of molar mass?

They rounded down slightly: exactly 1 g of NaCl would have added $\mathrm{2.06088 \times 10^{22}}$ ions.

Look up  Avogadro's number, as @Tyberius statews, and the definition of a mole. Not the critter.

Using dimensional analysis, we get
$$\frac{1 \ \text{g}}{58.44 \ \text{g mol$^{-1}$}} = 0.017 \ \text{mol}$$
$$(0.017 \ \text{mol})(6.022 \times 10^{23} \ \text{mol$^{-1}$}) = 1.02 \times 10^{22}$$

You basically have it. The last little detail is that, when dissolved, NaCl breaks down into 2 ions, $\ce{Na+}$ and $\ce{Cl-}$. Also, in your first equation, the 1 is 1 gram.


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Reason behind markovnikov's rule in hydration of alkenes via mercury catalysis
Recently I came across an interesting reaction regarding the hydration of alkynes using mercury catalysis (HgSO4) in the presence of sulphuric acid. I've read that after the mercury complex acts like a protecting group by preventing any carbocation rearrangement, water attacks the more substituted site to open up the 3-membered ring formed as an intermediate. But this is a nucleophilic attack (SN2 type) and the more substituted site is more sterically hindered and carries less partial positive charge on it( due to influence of other groups). So why does this occur via Markovnikov's rule itself?

If I could postulate, the reaction is like an epoxide ring opening during acidic conditions, where the oxygen is protonated, making it electron-deficient and have a +1 formal charge. This creates a good leaving group (an electron-deficient O) and in your case, an electron-deficient Hg.

Credits: Clayden Organic Chemistry Vol 2
Since it is a good leaving group, it wants to leave and as predicted by the Sn1 mechanism, the C-Hg bond bonded to the more substituted carbon prefers to break in order to take advantage of the more stable carbocation intermediate. This makes that particular bond weaker. The nucleophile thus prefers to attack the more substituted end in order to take advantage of the weaker bond. As the bond is weaker, the net enthalpy change would be lower, making the attack at the more substituted end more thermodynamically favourable. Additionally, notice that in the transition state, the developing positive charge is more resonance stabilised if the nucleophile attacks the more substituted end, making the reaction more kinetically favourable. From both perspectives, kinetics and thermodynamics, an attack at the more substituted end is favoured.


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How to modify viscosity of ZnO ink?
I have stumbled upon these instructions on how to create $\ce{ZnO}$ nanoparticles.
I am interested in making a DIY ink, for use with a parallel metal plate pen and a piezo-actuated microdrop nozzle.
My question is:
What is a reasonable way of controlling the viscosity of such an ink? Is it possible to replace the water with a mixture of water and isopropanol, for example?

Sure ink usually includes the following ingredients: petroleum distillate solvent, linseed oil, some form of organic pigments, and soybean oil.  If an ink has a linseed oil base, it will dry through air oxidation. When alcohol is present, the ink will dry through evaporation. Considering that I wonder if glycerol would be a good candidate for your dye?

Idea: As you have photoactive ZnO, mix it with clear aqueous (NH4)2SO4 (from MgSO4  + ammonia, remove white Mg(OH)2 precipitate). Then, use it as a finger paint (ink) to write on colored paper (best dark colors). Let dry. Place in strong sunlight after very litely spraying with a B12/water mix. Should reveal your written (with your finger) message as the moist chemicals bleach the dark paper.

How useful would white ink be ?

@Tartrate Printing electronics is of increasing interest.  Did you consider a publication like 2020Crystals449 (open access) as an entry for further literature survey?

Regular black ink is pretty viscous it has a dynamic viscosity ($\eta$) of 4.177.
In comparison a 50% mixture of methanol and glycerol has a viscosity of 5.8
Some other solvents to consider are

Hexanol which has a viscosity of about 5.4
Heptanol which has as viscosity of
about 7.3

These long-chain alcohols, however, will not dissolve significantly in water.
Some more viscosity data can be found here (it's not an ideal place for viscosities but it is free)
Linseed oil is also not water soluble. In any case here is another link
Ultimately it seems like you'd need to try a few different mixtures to get desired viscosity, solubility and evaporation speed.


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why does galvanic cell stop working when charge neutrality is not maintained?
In galvanic cell, salt bridge maintains charge balance between solutions. If there's no salt bridge, charge neutrality will be broken as solution(anode) becomes positive and that of cathode becomes negative. Many textbooks just say galvanic cells thus stop working due to broken charge neutrality.
But I want to know why. Why broken charge neutrality stops electrons flow? does it have to do with any potential energy?

Haw you ever seen a thunderstorm lighning ? It is just a small charge disbalance that would be fixed by the current 1 A ongoing for few tens on seconds. If there is no half cell diaphragm or ion bridge connection, the being created charge disbalance very quickly creates a counter potential difference that stops such a process.

Related: Would a salt bridge work through which the ions could not diffuse into the half cells?.

Does this answer your question? Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?

Let's see what happens in a Daniel cell (Zn/Cu). In the beginning, a Zinc atom looses 2 electrons and becomes a zinc ion $\ce{Zn^{2+}}$. The electrons are attracted by the copper ions $\ce{Cu^{2+}}$ in the other compartment. And the $\ce{Zn^{2+}}$ enters the solution. But look ! After some time there will be 2, 3 or more $\ce{Zn^{2+}}$ ions in this solution. This creates a electrostatic positive charge which is diluted in the whole solution around the Zinc electrode. This positive charge repells new positive charges that could be produced on the Zinc plate. So the creation of electrons at this electrode stops after some microsecond.
The same thing happens on the Copper electrode. On this electrode the electrons coming from the zinc electrode through the outer connection react with copper ions $\ce{Cu^{2+}}$ and this produces a metallic copper deposit on this electrode. But this will not last long, because if the electrons destroy positive ions, they are progressively disappearing. As a consequence negative ions (probably $\ce{SO_4^{2-}}$) become more and more numerous. After some time, the solution gets negatively charged and will prevent new electrons from arriving to the copper electrode.
So after a short while, the cell stops working. If you want it to continue working, you must find a way of getting rid of the sulfate ion around the copper plate, and to send them to the zinc compartment. That is the goal of the salt bridge. The excess of $\ce{SO_4^{2-}}$ ions around the copper plate are attracted by the excess $\ce{Zn^{2+}}$ ions from the anode. They move to them through the salt bridge. And the cell can continue working.

I'll expand this just a bit. (1) As Maurice indicated there is no continuous current flow, only a brief pulse which polarizes the electrodes.  (2) An electron can travel through a foot of wire in about 1 nanosecond. So the current pulse that will polarize the electrodes will be very small and the pulse with travel between the electrodes very very quickly.

@MaxW You may confuse speed of the field propagation and speed of electron flow. The latter is very slow, some mm/s at the best.

@Poutnik - Thanks for keeping me honest. A particular electron doesn't travel through a wire in a nanosecond. So  the situation is sort of people on a moving sidewalk. Suppose that there is is a constant flow of a person getting off the moving sidewalk every 5 seconds. Well then there must be a constant flow of  people getting on the moving sidewalk every 5 seconds on the other end. But the sidewalk travels at 1.5 feet per second so the hundred yard ride takes a particular person 200 seconds.


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Find the number of geometrical and optical isomers of 1,2,3,6-tetramethyl cyclohex-1,4-diene
The compound given to me is 1,2,3,6-tetramethyl cyclohex-1,4-diene.
The geometrical isomers for the following compounds are:

The isomer on the right can also be an optical isomer with its enantiomer. So does this mean that there are 3 stereoisomers in total with one common geometrical and optical isomer?

Your analysis is in right track. Yet, I would like to point some few points on deciding stereoisomers. The step by step analysis is depicted in following diagram:

The two geometrical isomers are $3,6$-$syn$- and $3,6$-$anti$-isomers. The compound has 2 chiral centers, hence $2^2 = 4$ maximum stereoisomers possible. However, as shown in the diagram, $3,6$-$syn$-isomer has a plane of symmetry. Therefore, its mirror immage is the same as indicated by $(3S,6R)$- or $(3R,6S)$-. In addition, $3,6$-$anti$-isomer does not has a plane of symmetry, and hence, has two enantiomers (see the diagram).

I think syn and anti in the diagram should be switched. Differs from text.

Yes, and thank you. Just a mental mistake. :-)

Thanks a lot for your better explanation and clearing my doubt.

Given compound has 1 geometrical isomer having methyl groups at trans  , given is cis.Given compound has 2 chiral centers, C3,C6. Given structure is meso.C6 has S, C3 has R configuration. There exist its optical isomers having both chiral centers either R or S configuration leading to 2 optical isomers. Both having S configuration is also the trans isomer of given structure. Total 2 optical, 2 geometrical isomers(cis +trans).

your answer is wrong. There are 2 geometrical isomers not just one.

1 meant 1 structure other than you had given,cis. For language clarity I had edited other 1 to 2(cis+trans).

There are two geometrical isomers, cia and trans, but three stereoisomers, RS, RR and SS


text:
Does -M Effect increases rate of hydration of aldehyde and ketone?
I saw somewhere written, the rate of hydration of carbonyl compounds depends on:

-I group
Steric hindrance
Intramolecular H bonding

Can -M effect also increases rate of hydration?

https://chemistry.stackexchange.com/questions/55090/rate-of-o-exchange-with-h2o

https://chemistry.stackexchange.com/questions/90828/rate-of-formation-of-hydrate-in-a-carbonyl-compound


text:
Converting Fischer formulas to Newman projections
How are line diagrams of molecules converted to their Newman projections? All I got from looking it up is that one may convert a Fischer formula to an eclipsed conformation directly, and that this may further be converted to other conformations.
Doesn't that stop the molecule from having a unique stereochemical configuration?

The compound is not optically active (it's meso)... So making conformers won't have any change in "configuration"

Stereochemical configurations are independent of conformational changes since conformational changes just involve bond rotation and not inversion. Caveat: let's assume we're not dealing with atropisomers.

Oh, the picture's wrong, yeah. I took it down, but my question is general.

That doesn't change my comment. Stereochemical configurations are unique and not affected by conformations except in very limited cases, which you most likely need not worry about right now.

Think this as a comment, since I have low points I CANNOT actually comment
step 1: take the bottom carbon (in the line, not the substituent) as the front carbon of the newman
step 2 : in this method of converting line into newmans you will get the groups attached to the mainline in the form of Y (in Newmans) step 3: the left group will go on the left hand of the y and the right side group will go on the right hand of the y

This video gives a good explanation.

That's for converting Newman to Fischer projections, which is the opposite.


text:
How to calculate the half time of a unimolecular reaction given the Arrhenius coefficient and the activation energy?

Consider the following unimolecular reaction:
  $$\ce{H2O2 + M -> OH + OH + M}$$
The high pressure limit unimolecular rate coefficient for this reaction is $k_\mathrm{uni} = A\cdot \mathrm{e}^{\frac{-E_\mathrm{a}}{R \cdot T}}$, where $A = \pu{3E14 1/s}$, and $E_\mathrm{a} = \pu{305 kJ/mol}$. Calculate the half-life of this reaction at $\pu{1 atm}$ and $\pu{1000 K}$.

Source: Question 14.1 from John W. Daily: Statistical Thermodynamics: An Engineering Approach. Cambridge University Press, 2018. ISBN: 1108244645, 9781108244640. DOI: 10.1017/9781108233194.
I started by getting that $k_\mathrm{uni}$ through the formula, got $\pu{0.03515 1/s}$ and then, since it's unimolecular:
 $$t_{\frac{1}{2}} = \frac{\ln(2)}{k_\mathrm{uni}}.$$
But I got it wrong, I don't have the solutions manual to this book to check what is wrong. I also tried using the Troe form, using the $k_\mathrm{uni}$ from the starting formula as the $k_{\inf}$ but still got it wrong.

Okay, I finally found the solutions manual. And saw that the answer on it is completely wrong, so there was no way that introducing what that math gave would work. The halftime is actually 19.7 s, but in the book it is 9 $\mu$s, which doesn't make sense.

Do you know what the answer is supposed to be?

No, the book doesn't bring answers either, that's been making it more difficult. I've been looking for a solutions manual on the web to see if there's something wrong.

Did you remember to convert 305 kJ to 305000 Joules ? It is an easy slip to make :)

Are you sure about your numerical value for the activation energy ? The given data (305 kJ/mol) looks rather high, as it is more than twice higher than the reported O-O bond energy (146 kJ/mol)

That's what the problem tells, I thought it was too high too.

@AllanJosuéGonzlez-Villalobos - OK, I'm totally confused. If the book doesn't provide an answer, how do you know that you got it wrong?

You calculated the high-pressure-limit half life, where the rate of collisions with M is so high that it has no influence on the rate constant. Now all you have to do is to consider that the pressure is 1 atm instead of at the high-pressure limit. That slows down the reaction such that the half-time is in the second instead of the micro second range.
Something similar happens with ozone. It is not very stable at sea level, but has a substantial half-time in the upper atmosphere (unless there are certain man-made substance up there).
To solve numerically, you could assume a Lindemann mechanism (see https://en.wikipedia.org/wiki/Lindemann_mechanism) and that $k_1$ is diffusion-limited (i.e. no activation energy, every collision leads to activated complex).
The textbook cited by the OP has:

Strangely, in an earlier table they show $k_0$ and $k_\infty$ as (along with pre-exponential temperature dependence $T^n$ and ?activation energy? E - units unknown):

See https://pubs.acs.org/doi/pdf/10.1021/ed081p832

Yep, the problem is on the exercise itself, since it says you gotta use that $E_{a}$ of 305 kJ/mol, but it actually's right witht the one on the table. Thanks, man.

@AllanJosuéGonzlez-Villalobos Could you write up your calculation as an answer?

For $[A]_{0}$ the initial concentration of $H_{2}O_{2}$, we are looking for the in time in which the half of that concentration is left:

$\frac{[A]_{1}}{[A]_{0}} = \frac{1}{2}$

For unitary reactions like the one on the problem:

$[A]_{1} = [A]_{0}\cdot e^{-k\cdot t}$

To me, there's two ways of proceeding, either you take the $k$ you're given an substitute it on the equation, or you consider it as the $k_{\inf}$ and use the 1 atm pressure to find the $k$. Either way, you'll get that your halftime $\tau_{\frac{1}{2}} = \frac{ln(2)}{k}$ for the $k$ you chose or got.

Are you answering your own question or is this an edit/comment of the original post?

I wasn't able to comment till now, sorry.


text:
Why does aluminium chloride react with water in 2 different ways?
Aluminium chloride has 2 different reactions with water, depending on whether a small amount or an excess of water is added. One produces a solution of hexaaquaaluminium ions whereas the other produces a ppt. of aluminium hydroxide And HCl gas. Does the 2nd reaction occur due to the reaction 1 first occurring followed by vaporisation of the solution due to the heat given out?

Are you sure with the gas ? Is not it just common hydrolysis  of many salts   ?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumblestone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Im sorry. I’m relatively new here so I don’t know how to insert the relative equations with the correct syntax here. How do you want me to show my research? I’ve been through many sites on the internet, and  several of my text books and past exam marking schemes.

See Wikipedia  - Aluminium_chloride - Reactions_with_water

Yes, adding a small amount of water to anhydrous AlCl3 is different from an aqueous reaction. Here is an excellent educational institution's answer in the first case, depicting the action of water on anhydrous aluminium chloride, to quote:

If a small amount of water - a few drops - is added to anhydrous aluminium chloride there is a hissing sound and hydrogen chloride gas is given off as steamy acidic fumes. The reaction that occurs will depend on the amount of water added, but is probably
$\ce{AlCl3(s) + H2O(l) → AlCl2OH(s) + HCl(g)}$

In the second case, aluminium chloride in aqueous solution appears to undergo a two-stage reaction, starting with the formation of the hexaaquaaluminium ions, to quote again:

If aluminium chloride is dissolved in a large amount of water the solution is acidic, but this has nothing to do with formation of hydrochloric acid. The solution contains hydrated aluminium ions and chloride ions:
$\ce{AlCl3(s) + aq → [Al(H2O)6]^{3+}(aq) + 3 Cl -(aq) }$

This is followed by an acidic hydrolysis paralleling transition metals:

The hexaqua complex ion behaves exactly like ions of similar type formed from transition metals; the small, highly charged metal ion polarises (withdraws electron density from) the water molecules that are attached to the aluminium ion through dative covalent bonds. This makes the hydrogen atoms d+ and susceptible to attack from solvent water, which is acting as a base. The complex ion is deprotonated, causing the solution to be acidic from the formation of hydroxonium ions H3O+:
$\ce{[Al(H2O)6]^{3+}(aq)  + H2O(l) →  [Al(H2O)5OH]^{2+}(aq) + H3O+(aq)}$

This may be a stupid question but, Is there any specific reason why the reactions are different?


text:
How do you calculate the norm of a function describing a molecular orbital?
I'm a bit confused by a past paper question I came across for one of my courses, shown below.

The wavefunctions corresponding to the lowest, $\psi_{1}$, and second-lowest, $\psi_{2}$, energy molecular orbitals in 1,3-butadiene are:
  $$\psi_{1}=0.372\phi_{1} + 0.602\phi_{2}+0.602\phi_{3} +0.372\phi_{4}$$
$$\psi_{2}=0.602\phi_{1} + 0.372\phi_{2}-0.372\phi_{3} -0.602\phi_{4}$$
  where $\phi_{i}$ is the $p_{z}$ orbital on atom $i$. Calculate the norm of $\psi_{1}$, and the overlap between $\psi_{1}$ and $\psi_{2}$. Is the result what you would expect? Explain.

Neither calculating norm nor the overlap between 2 orbitals in the way the question asks came up.
I assume by norm it means the normalisationg factor, but that makes no sense to me since it's already normalised. I assume the overlap should be 0 since they should be orthonormal, but don't know how to show that. I'd really appreciate some help!
Sadly the lecturer for the course has contracted Covid-19 and so of course is not available to answer questions.

A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

Please add a citation to where this exercise comes from. Also note that the use $\psi_x$ in the text and $\Psi_x$ in the equation. They should be the same, and should probably be all lower-case.

The "norm" refers to https://en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm.

As I stated in my question, it comes from a past paper at my university. I take your point about the notation; that's simply how the question is in the exam paper.

if the $\phi$ are orthonormal then $\langle \phi_i\phi_j\rangle = \delta_{i,j} $ meaning that if $i=j$ the result is 1 else it is zero. So multiply out term by term to get overlap $\langle \psi_1\psi_2\rangle$

I'm unsure if you're meant to solve this via the Hückel molecular orbital theory (see Chemistry LibreTexts), however, if you need to calculate the norm of a wave function $\Psi$, this is $|\Psi |$. Given what we know in your post about both $\Psi_{1, 2}$ then
\begin{align}
  |\Psi_1|^2
    &= \langle\Psi_1 | \Psi_1\rangle \\
    &= 2\cdot(0.602^2)+2 \cdot (0.372^2) \\
    &= 1.001576 \\
  \to |\Psi_1| &= 1.000788 > 1\\
\end{align}
I write that this is greater than one, as, on closer inspection,  is down to rounding from the coefficients of each $p_z$ orbital.
The repeat process can be found for the overlap part, which I am interpreting as $\langle\Psi_1 | \Psi_2\rangle$.

Yes, we did cover Huckel in this course. I suspect that the fact that it's greater than 1 is a rounding error rather than intentional.
Thank you - that's really useful

Great I'm glad to have helped, in terms of thr rounding, you may have a point actually, looking at various other Wave functions for the Energy bounds of 1,3-Butadiene I see close enough numbers that do indeed sum-square to unity.

@Martin mod - thanks for the improvment in my Bar-Ket notation.

Happy to help. MathJax (i.e. Latex syntax) is not the most intuitive. There are some guides on our meta, if you want to learn more.


text:
Would the 1st excited state of Li2 be vibrationally excited?
I'm quite torn about an interesting past paper question that's come up during my revision, shown below.

On the one hand, you're promoting an electron in a bonding orbital. In general, since this means you would substantially change the electronic wavefunction, I would expect overlap from the initial vibrational state with a range of excited vibrational states and so due to the Franck-Condon principle the molecule would be vibrationally excited in the electronically excited state.
But the bond order of the excited molecule is 0, and so it will presumably dissociate as soon as it vibrates.
Which of these considerations wins out?

You are correct in assuming that several vibrational levels would be excited and in fact this excited state is known in Li2 (and there are two others at least). This excited state has a lower vibrational frequency than the ground state and a longer bond length as might be expected. This means that it does not dissociate immediately on excitation although the barrier to dissociation should be small so possibly only a few vibrational levels exist. (btw the transition is allowed so there is no trick in the question, i.e. if it were forbidden then one might not expect an excited state anyway).

Thank you! I foolishly didn't think about s-p mixing which would stabilise both the bonding and antibonding sigma orbitals formed by the 2s orbitals, which is presumably why it doesn't dissociate.

The only way of knowing whether $Li_2$ is vibrationally excited is measuring its spectrum, by heating lithium metal at high temperature in a vacuum, because lithium vapor contains  small proportion of Li2. If vibrational lines are visible, they will be easy to recognize


text:
Production rate of parallel reactions

The degradation of hydrogen peroxide in water $\ce{(2 H2O2 (aq) → 2 H2O +O2 (g))}$ can be catalyzed by $\ce{Br-}$ and $\ce{H3O+}$. Derive an expression for the production rate for the following reactions:
$\ce{H2O2 + H3O^+ <-> H3O2+ + H2O}$ , where the reaction constants are $k_1$ and $k_{-1}$
$\ce{H3O2+ +Br- -> HOBr + H2O}$ , where the reaction constant is $k_2$
$\ce{HOBr + H2O2-> O2 + H3O+ + Br-}$, where the reaction constant is $k_3$

I known that $\ce{H3O2+}$ and $\ce{HOBr}$ are intermediates and so I can use the steady-state approximation and set these rates to zero. From this I get that:
$$\frac{d[O_2]}{dt} = \frac{k_1k_2k_3 [\ce{Br-}][\ce{H3O2+}][\ce{H3O+}]}{k_1k_3[\ce{H2O}] + k_2k_3[\ce{Br-}]}$$
Which I hope is correct.
Then they ask:

Under what conditions does: $$\frac{d[O_2]}{dt} = k [\ce{Br-}][\ce{H3O2+}][\ce{H3O+}]$$

Here I am unsure if the conditions are that we could suppose that $\ce{H2O}$ is in excess and thus not include it in our rate expression, and that $\ce{Br-}$ also can be excluded since it is a spectator ion.
I would appreciate if someone could confirm that my expression for the rate of production is correct and that my conditions are valid. Thank you!

"are intermediates and so I can use the steady-state approximation and set these rates to zero" Which rates? For intermediate $\ce{I}$, the steady state approximation says that because $\ce{[I]}$ is small, then change in its concentration is also small: $\frac{d}{dt}\ce{[I}] = 0$. I hope that's what you are doing when you are try to "set these rates to zero."

The following must be true  $$k_1k_3[\ce{H2O}] \gg k_2k_3[\ce{Br-}]$$, then you can neglect the influence of  $\ce{[Br-]}$ on the value of the denominator. Another option is excluding $\ce{H2O}$ from the $k_{-1}$ reaction

You still have an intermediate in you equation for oxygen production so this cannot be correct yet.

As both $\ce{[H_2O}$ and $\ce{[Br^-]}$ are not consumed, their concentration do not change during the whole process. So there is no conditions for the last equation to be applied, namely : $$\ce{d[O_2]/dt = k [Br^-][H_3O_2^+][H_3O^+]}$$ This expression is valid at all times.


text:
Enthalpy change of dissolving air in water?
I'm looking for a value of the enthalpy change of dissolving air in water but can't find it. In other words, if I pull a vacuum over some water to remove all the air, per Henry's Law, how much energy change would I expect in the water as the normally-dissolved air evolved?
I tried searching for enthalpy of solvation of air in water and looking up N2(aq), etc., in standard thermodynamic  tables, but didn't get results for either one.

Given that you can look up the solubility of air in water as a function of temperature, you could calculate the enthalpy change using the van't Hoff equation (which shows the relationship between the enthalpy change and the temperature-dependence of the equilibrium constant):  https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation


text:
Notation in atomic structure
In the book, The theory of atomic structure and spectra, Robert D. Cowan says that the allowed terms of $p^2$ are $^1S$, $^1D$, $^3P$ and so the possible terms of $p^2 + p\quad$ (obtained by using the vector model) are:
$$^2P^\circ,\quad ^2P^\circ D^\circ F^\circ,\quad ^{2,4}S^\circ P^\circ D^\circ
 \tag1$$
I am having trouble in understand the notation in $(1)$. For example  the term $^2P^\circ$ means that $s=1/2,l=1$ and $^\circ$ means odd parity, but for the term $^{2,4}S^\circ P^\circ D^\circ$, I do not know what it means. Can anyone explain me it please?

My guess is that $\quad ^{2,4}S^\circ P^\circ D^\circ $ means $^2S^\circ, ^2P^\circ,^2D^\circ,^4S^\circ, ^4P^\circ,^4D^\circ$

Copied table below from "Quantum chemistry", Donald A. McQuarrie, University Science Books,  2008, page 472
This illustrates the notion that $p^2$ and $p^4$ are equivalent.

I think the 2,4 superscript in Cowan's book for $^{2,4}S^\circ P^\circ D^\circ$ is a peculiar notation of his. Most other books seem to presume the reader knows the equivalency.


text:
What is the recipe for the Belousov-Zhabotinsky reaction for correct simulation with the Oregonator model?
For a school project I'm trying to model the Oregonator and perform some experiments with the Belousov-Zhabotinsky reaction. Like predicting the reaction and the swings in concentrations. For that it is of course crucial that I perform the correct BZ-reaction, the reaction where the Oregonator is based on.
In the paper 'An Analysis of the Belousov-Zhabotinskii Reaction' by Casey Gray I found a mathematical analysis of the Oregonator. For the experiment at school I want to perform the experiment exactly as described in the images below (the images are from the same paper).
Now I have a problem, because there are many different recipes on the internet, all a bit different. I am no experienced scientist, so I have difficulty determining which recipe I should use. So my question is: what is the exact recipe I should use for the (below) BZ-reaction? Which concentrations, how much of them and in what order I should add them (and any additional things that I should do).
Second, are the ingredients used in the BZ-reaction common in most laboratories? I need to perform the experiment at school; what are the chances they have those ingredients? And if school does not have some of chemicals, are they expensive?
Third, is ferroin the best (visual) indicator for the BZ-reaction? Is ferroin expensive (to make)? Are there alternatives that visually indicate the changes in concentrations?
Description of BZ:

FKN mechanism:

An the Oregonator model (based on the FKN):

CORRECTION: HBrO2 is not hypobromous acid, but bromous acid.

Link to paper links to image.

Label error: HBrO2 is Bromous acid and HBrO is Hypobromous acid. By the way, a simple oscillating system is H2O2 + HCl. More colorful perhaps, use HBr.

An experimental procedure for one variant of the B-Z reaction is here: https://chemistry.stackexchange.com/q/34442/79678. You really should check if your school, during the present global situation, has the chemicals you need or would give any priority at all to ordering ones they do not have. And you need to fix the link, as @RodrigodeAzevedo commented.

To emphasize Ed's point - (1) All chemicals are expensive. This is not the same as buying table salt at the grocery store. (2) It is not just chemical cost but shipping too. Solids are generally easy to ship, but sulfuric acid has to be a freight truck shipment. (3) For such a complicated series of reactions you'll probably need "good" quality reagents to not have some contaminate that poisons the cycle.

I second what @MaxW commented and have one last comment. Once you commit to one of the variant published protocols, you will know what reagents and apparatus you will require. Then you need enough reagent to do the experiment multiple times because complicated experiments often do not work correctly the first or second time. As well, your school’s laboratory facility has to be open and a supervisor needs to be present. So I wish you success with this ambitious project, but please do not underestimate the time and effort required to achieve that success!

@RodrigodeAzevedo I fixed the link, thanks!

@AJKOER thank you for pointing that out. I can't change the image, because they're from the mentioned paper, but I've corrected the question.

@EdV thanks for the link to that experimental procedure. In your expert opinion, do you think that a simulation with the Oregonator is realistic if I perform the procedure like described in that post? Second, I checked with my school if they have all the chemicals. They do not have sodiumbromate (I now asked if they have potassium bromate) and ferroin, but the person I e-mailed assumed that school could order some. I'm waiting for the final decision.

@MaxW thank you for letting me realise these things. I indeed did not think of all those measures that need to be taken for transporting chemicals. Luckily my school has sulphuric acid (I thought that was a common chemical, but probably not regarding the risks). As far as I know, most of the chemicals come from Sigma Aldrich. What are your experiences with Sigma Aldrich?

@EdV I'll make sure that has enough reagent, provided that they can order the missing ones. I did not know that most complex experiments fail the first or second time, so I'll certainly keep that in mind.

I am not an expert on the B-Z experiment or computationally simulating it, though I have seen the experiment demonstrated by an actual expert (Prof. Irving Epstein of Brandeis U.). The reaction is very visually compelling, so if it was me, I would focus on getting that to work. Anyway, the key to success is often dogged perseverance and paying meticulous attention to details. Easier said than done! Good luck!


text:
Reason for nomenclature of azeotropic mixtures
The book states the solution which shows large positive deviation from Raoult's Law form minimum boiling azeotrope
What is meant by " minimum boiling azeotrope" or I should say what does minimum signify here

I think the meaning is very clear from the attached figure. A minimum boiling azeotrope boils at lower temperature than any other mixing composition of the two compounds.

I wanted the reason as well and Max w has done it lots of thanks

What is meant by minimum boiling azeotrope?

Two red lines have been drawn in the phase diagram below. Both are parallel to the horizontal axis. The first is at the boiling point of ethanol, and the second is at the boiling point of water.

Now look at the liquid curve.
(1) As the % Water increases from 0% the curve starts down, outside of the zone defined between the two red lines. The curve can't go down forever since at 100% water the curve must reach 373 K. So between 0% Water and 100% Water there must be a minimum in the curve.
(2) Since the curve falls below the zone defined between the two red lines an azetrope must form at the minimum. Remember that if the liquid is changing composition so must the vapor.

The converse is true too. If the vapor line goes above the zone defined between the two red lines, then a maximum boiling azeotrope must form.


text:
Why does meta directing group also pull electrons from ortho and para positions?
I have been taught that electron withdrawing groups are meta directing. If it cannot modulate electron density over ortho and para positions during resonance, why does it help in nucleophilic substitution in a aromatic ring in which the leaving group is situated at ortho/para positions?

Draw canonical structures...

@Zenix sir are you asking me to give you a diagram or Are you asking me to draw and check.If you are asking me for a example, 1 chloro 4 nitro benzene undergoes nucleophilic substitution more easily than chloro benzene.

So? Draw the reaction with transition state and paste it into your question.

I was asking you to draw and conclude, also we use the terminology of "directing" groups in EAS, ArSN (your example) are a whole different reactions, you are mixing two different things...

@Karl Sir it does not allows me to paste the photo.

@Zenix sir it is not allowing me to upload picture to this question.

Don't use mobile app, rather use cellphone's browser (in case you are using  cellphone)....

@Zenix Ok sir.Thank you for your response.


text:
What is the efficiency of UV lamp in ozone generation (Ozone g vs. energy consumed )
re.:
What is the efficiency of UV lamp in ozone generation (Ozone g vs. energy consumed )
I read a lot of papers from the Internet on using UV lamp for generating ozone but none discussed the efficiency of such process.
How much energy (kWh) gets consumed by UV lamp to generate 1g of ozone ?
I read  Ozone Formation from UV Light
https://www.oxidationtech.com/ozone/ozone-production.html#answer2
and related web links
and all I get as pros and cons is:
"
Ozone production from UV light for commercial applications
Advantages of UV Ozone Generator
Simple Construction – only a UV lamp with ballast and a fan

Low cost – due to simple construction

Lower nitric oxide production

Disadvantages of UV Ozone Generator
Low ozone output – g/hr

Low ozone concentration – less than 1% by weight

Bulbs and ballasts will need to be replaced

"
Let me know your opinion

"How much energy (kWh) gets consumed by UV lamp to generate 1g/h of ozone ?" I think you can spot the problem with this question yourself. Do you?

I have asked alike question in Physics and was asked it to ask in Chemistry to get the correct answer

Either g per kWh, either g/h  per kW. Not g/h per kWh.

OK, two problems. ;-) The other is that UV light sources have their own efficency. Electricity -> UV photons (of what wavelength?) -> ozone

I read about UV light source efficiency from one boomarked site  but just visit UV Light Ozone Generators https://www.boie.us/uv-ultraviolet.php claiming ozone output: 10g/h and still no data on power consumed in the specification
"Lamps:
Long-life VH** 9,000-hour high-output ozone lamps
not rated
"Electrical Rating:
120V~60Hz 2.6 Amps
So in theory, power consumed is less than 120V x 2.6 Amps = 312W.
Hard to believe, low efficiency UV lamps can generate 10g/h ozone comsuming less than 1,000W x hr

great image of the opened box
BoiE UltraViolet 10K Shock Model
https://www.boie.us/images/image-slider-uvozone.jpg
not sure 10K stays for what
" 10,000 mg/hr
Eureka!!!
10K stays for 10,000 mg/hr
to be verified

and one more 12V model of 12-volt corona discharge ozone generator
https://www.boie.us/dc-sportsman-pro.php
since ozone output is selectable, matching electrical rating, so
"Electrical Rating:
12V DC 1.55A (Low) 4.45A (High)
matches
selectable ozone output (1,000 or 4,500 mg/hr)
so 1,000 mg/hr consumes 18.6W x h
4,500 mg/hr .. 53.4W x h
4.5 x  18.6W = 83.7W
so a single generator consumes less energy than 4.5 stacked in parallel, generating the same amount of ozone.

ok, ..20W rated power, 12-volt corona discharge ozone
generates 1,000 mg/hr
vs.
312W UV light ozone generator generates 10g/hr,
so 30W UV ... generates 1g/hr
20W (corona) matches 30W (UV) rated 1g/h
since corona generator is much more complicated HV
I am going to build UV ozone generator

I read about UV light source efficiency from one boomarked site but just visit UV Light Ozone Generators boie.us/uv-ultraviolet.php claiming ozone output: 10g/h and still no data on power consumed in the specification "Lamps: Long-life VH** 9,000-hour high-output ozone lamps not rated "Electrical Rating: 120V~60Hz 2.6 Amps So in theory, power consumed is less than 120V x 2.6 Amps = 312W. Hard to believe, low efficiency UV lamps can generate 10g/h ozone comsuming less than 1,000W x hr – a a Apr 28 at 20:11
great image of the opened box BoiE UltraViolet 10K Shock Model boie.us/images/image-slider-uvozone.jpg not sure 10K stays for what " 10,000 mg/hr Eureka!!! 10K stays for 10,000 mg/hr to be verified – a a Apr 28 at 23:07
and one more 12V model of 12-volt corona discharge ozone generator boie.us/dc-sportsman-pro.php since ozone output is selectable, matching electrical rating, so "Electrical Rating: 12V DC 1.55A (Low) 4.45A (High) matches selectable ozone output (1,000 or 4,500 mg/hr) so 1,000 mg/hr consumes 18.6W x h 4,500 mg/hr .. 53.4W x h 4.5 x 18.6W = 83.7W so a single generator consumes less energy than 4.5 stacked in parallel, generating the same amount of ozone. – a a Apr 28 at 23:17
ok, ..20W rated power, 12-volt corona discharge ozone generates 1,000 mg/hr vs. 312W UV light ozone generator generates 10g/hr, so 30W UV ... generates 1g/hr 20W (corona) matches 30W (UV) rated 1g/h since corona generator is much more complicated HV I am going to build UV ozone generator – a a Apr 28 at 23:24


text:
Comparing between cleavage of ethers
We have to find out which ether would undergo cleavage faster.
My Attempt
IV. I couldn't figure out, so i carried out simple SN1 cleavage. Since +M effect of OH > OCH3  1st reactant seemed  right answer. But I am wrong. Why and what would be process?
V. I believe since there are no 3° Carbon, so process of cleavage should be SN2. The answer comes out right, but is my analysis right?
Please help.

(Nitpick: Tell you teacher we´re now in the 21st century, not the first half of the 20th. It´s called a tertiary carbon atom.)

The first example has an ether, but that's misleading. You're trying to cleave a hemiacetal/acetal. But given that it's cyclic, I'm not even sure what the question is asking, since it's likely that no matter what the conditions are, then equilibrium will favor the closed form. In (V), you need to look how the ether will be hydrolyzed. That requires understanding that the second one is a vinyl ether and has other modes of reactivity with acid.

Zhe i think they are not much concerned about the reaction of double bond. In solution they have simply written acetals are more reactive and less stable. But can u tell why?

eyeroll This is for understanding how mesomeric stabilisation affects reactivity of acetals and vinyl ethers. More precisely, that oxocarbenium ions are very stable in comparison with carbocations.

The NCERT Book of class 12 demonstrates this well. Read it.

No, there are precisely concerned with the location double bond because that's the difference in reactivity.

I found this answer for IV
And in V-B is formed because it is more conjugated


text:
What effect does lone pair have on bond angle?
I have read that the presence of lone pair decreases the bond angle.Shouldn't it increase the bond angle since there is repulsion between the electrons?

Lone pairs on central atom or on "non-central" atoms? In case of latter, your statement is true..

Does this answer your question? Application of the electronegativity effect on bond angles

Lone pairs on central atoms are like bulgy water filled balloons(diffused electron cloud of unshared electrons). They push down the neighbouring bond pairs causing a decrease in bond angle. In case of bond pair you may imagine the balloon being pulled (electron density attracted) by two persons from opp side (nuclear attraction of participating atoms) which reduces the bulgyness and hence bond pairs repel less than lone pairs.This is just my understanding. Hope my analogy is correct. If someone has a better answer I'll be happy to learn.

It varies tremendously on specific case. With metals, there's like no effect at all.

You are considering the angle between the lone pair and the  bonds but bond angle is always between two bonds which will be reduced due to repulsion only as the bonds will get closer to each other.

@Loveforphysics bond angle is reduced due to the presence of lone pairs on non central atoms but not on central atoms.


text:
Difference between calcination, roasting and pyrolysis
Hie everyone. I have been going through pyrometallurgical processes for recycling lithium-ion batteries and l have been coming across these terms; calcination, pyrolysis and roasting which l need assistance to be able to differentiate them as there is no notable difference when it comes to recycling lithium-ion batteries.

Related: https://chemistry.stackexchange.com/q/39281/30836

Thanks but l was hoping to get a clear differentiation, maybe the chemical mechanisms at play because when using these techniques sometimes the process conditions are all the same but one journal would choose to call it pyrolysis, roasting or calcination. The material in the battery is Lithium cobalt oxide (LiCoO2). Like l had said the distinction when you are doing extractive metallurgy of ores is clear but for recycling its not.....


text:
Reduction of (4-methylcyclohex-3-en-1-yl)methoxymethylbenzene to 4-methylcyclohex-3-en-1-ol

Find the suitable reagent to carry out the following transformation:

The answer provided says that $\ce{H2/Pd-C}$ can be used to bring about the given transformation, but I am failing to understand why would the double bond remain unaffected. Moreover, the answer does not specify the number of equivalents of $\ce{H2}$ used.

You are right, they are wrong. There is no way you can carry out this hydrogenation and leave the alkene untouched.


text:
What will be the product of following Wolf-kishner reduction?
Question

Attempt
1.Wolf kishner would reduce double bond oxygen to alkane.
2.Also since water is present acetal should break.
3.SnAr doesn't seems possible due to adjacent bulky groups.
So according to me answer should be C
 But why am I wrong?
Please help.

Acetals are stable to aq base. They need acidic conditions for removal. SNAr seems a very real possibility, the Cl is doubly activated.

But wouldn't the bulkynadjacent group reduce SnAr

Not that much. The potential nucleophiles are fairly small and potentially intramolecular. Is this a reaction drawn from a literature example? It would not surprise me if this gave a benzopyrazole.

See step 4 --> 5 here: https://chemistry.stackexchange.com/questions/124901/why-does-phenolphthalein-form-in-this-reaction/124916#124916

Check the temperature required for the Wolff -Kishner reduction and then determine if you think nucleophilic substitution could occur given the other functional group on the ring.


text:
Solutions and Precipitates
If a solution of $\pu{20mL}$ of $\pu{ 0.050M}~\ce{K+ }$is added to $\pu{80mL}$ of
$\pu{0.50M}\ce{ClO4-}$ will a precipitate form and what is the value of $Q_\mathrm{sp}$? For $\ce{KClO4}, K_\mathrm{sp}= 1.07 \times{10^{-2}}$
*I know that we have to show what we tried originally, which is great and I think the best way to learn. I am honestly looking at this problem and have no idea what any of it means. I'm self-teaching chemistry from a textbook, and the sample problems were nothing like this! I looked online for videos but I don't know what I am looking for, and I am unsure of the difference between $Q_\mathrm{sp}$ and $K_{\mathrm sp}$?? Does anyone have some kind of video, or sample problem, or explanations and definitions that could help me solve this on my own? I don't know what I am doing in this problem! Any help is appreciated!

Let's start with the basics. If [K+] is the concentration of potassium cation and [ClO4-] is the concentration of perchlorate anion, what is the formula for the solubility product, Ksp?

Effort for a question can be as little as writing a reaction equation and looking up a few key words. Adding a citation to the exercise also often helps for context. Generally spending some time with your post to make it look nice will go a long way in attracting answers.

@Martin-マーチン Thank you for your insight, I will remember that!

If we take it generally, e.g. for a reaction $$\ce{A + C <=> C + D}$$
then $K$ in $$K = \frac{\ce{[C][D]}}{\ce{[A][B]}}$$ means the reaction equilibrium constant and $\ce{[X]}$ is a convention for the molar concentration of $\ce{X}$ in mol/L, in this case the equilibrium concentration.
OTOH. very similar quantity $Q$ in $Q = \frac{\ce{[C][D]}}{\ce{[A][B]}}$ is called the reaction quotient with plugged in the actual concentrations of the involved compounds.
If $Q \lt  K$, the reaction is ongoing forward, toward the products on the right reaction side.
If $Q \gt  K$, the reaction is ongoing backward, toward the reagents on the left reaction side.
If $Q =  K$, the reaction is in equilibrium, with the net reaction rate being zero. It means, on the molecular level, there reaction is ongoing in both directions at the equal rate.
Now, $K_\mathrm{sp}$ is a specific equilibrium constant, called a solubility product and $Q_\mathrm{sp}$ is a reaction quotient of eventual precipitation/dissolution process.
For the reaction $$\ce{AB(s) <=>[\ce{H2O}]A+(aq) + B-(aq)}$$,
$$K_\mathrm{sp}=\ce{[A+][B-]}$$ is for the equilibrium between solution and precipitate.
$$Q_\mathrm{sp}=\ce{[A+][B-]}$$ is for the current concetration of ions.
If $Q_\mathrm{sp} \lt K_\mathrm{sp}$, all precipitate is dissolved and the solution is not saturated yet.
If $Q_\mathrm{sp} = K_\mathrm{sp}$, the solution is saturated and the dynamic equilibrium is ongoing between the solution and precipitate.
If $Q_\mathrm{sp} \gt K_\mathrm{sp}$, the precipitate is being formed, or the solution is oversaturated, what is ametastable state, that is easily disturbed.

Thank you for your help, I appreciate your time! That makes sense. I calculated the moles of both K+ and ClO4-, which is .001 moles and .04 moles, respectively. And then, Qsp is simply those two concentrations multiplied together, for which I obtained .00004. Qsp, as a result, is less than the given Ksp so no precipitate forms. That is because Qsp is the value we have obtained from our solution and because it is less than equilibrium, a full chemical reaction will not be able to occur! This helped a lot, it was simpler than I thought it would be, assuming I am on the right track! Thank you!

Thank you for your help, I appreciate your time! That makes sense. I calculated the moles of both K+ and ClO4-, which is .001 moles and .04 moles, respectively. And then, Qsp is simply those two concentrations multiplied together, for which I obtained .00004. Qsp, as a result, is less than the given Ksp so no precipitate forms. That is because Qsp is the value we have obtained from our solution and because it is less than equilibrium, a full chemical reaction will not be able to occur! This helped a lot, it was simpler than I thought it would be, assuming I am on the right track! Thank you!


text:
How can temperature be constant during a reaction?
So I was reading my textbook about the topic thermodynamics.

Here the second paragraph States that the above equation describes the combustion of liquid ethanol at constant temperature and pressure.
But how can the reaction occur at constant temperature. The enthalpy change stated is accompanying the reaction at standard state. So it should be heat released at constant pressure and from temperature 25°C. So temperature of reaction mixture should be decreased.

The reaction obviously doesn't happen at a constant temperature. For such a reaction the heat generated is measured in a calorimeter. The calorimeter basically has a large reservoir of water. The amount of water is enough so that the calorimeter only heats up a few degrees. Knowing how much heat it takes to raise the temperature by 1 degree and how many degrees the temperature increased allows the heat of reaction to be calculated.

Another way of handling this problem is to say that the temperature may change during the reaction, but the only important data is the fact that the initial and the final temperatures must be the same.

As an addition to @Maurice: this is allowed as long as the intermediate temperature change does not lead to a different outcome (other than wrt.reaction time, which is not regarded in thermodynamics).

Since enthalpy is a state function, it doesn't matter if the temperature is constant during the reaction.  All that matters is that the final temperature is forced to equal to the initial temperature.  This can be done by holding the reaction mixture in contact with a constant temperature reservoir until the system equilibrates.  You just need to figure out how much heat you had to add (or remove) to make the final temperature the same as the initial temperature.

But how can the reaction occur at constant temperature

Despite the dynamic name, thermodynamics is not concerned about time. Or more precisely, the theoretical model/world-view of thermodynamics allows the reaction to take place over an infinite amount of time, so you have all the time you need to dissipate the heat and thus keep the same temperature constant (within arbitrarily small limits).
To get to practical reality, we have a whole arsenal of possibilities that cover a range from very close to the definition here to experimental conditions that are quite different from those stated here:
The closest one is: The reaction takes place in a thermostat, i.e. in a device that keeps the temperature constant. You may measure the amount of heat that needs to be transported into or out of the reaction vessel to keep the temperature there constant (within practical limits).
As @MaxW says in their comment, you can also approximate this by using a reaction vessel that keeps the temperature approximately constant.
@Maurices comment goes yet a step further: as long as the outcome doesn't change (e.g., no irreversible reations at the intermediate temperature), you may deliberately allow intermediate temperature changes as long as you can keep track of the heat exchange until the reaction products reach the starting temperature again.
And lastly, you could allow temperature change and do a series of experiments that then allows you separate in the data analysis temperature-dependent from temperature-independent parts.


text:
Reactions of Depressants
In the froth flotation process, the concentration of a sulphide ore takes place using $\ce{NaCN/KCN}$ as a depressant. It forms a soluble complex with the impurity sulphide.
Taking the particular case of lead extraction from Galena($\ce{PbS}$), $\ce{NaCN}$ reacts with the impurity $\ce{ZnS}$ present in it forming a soluble complex.
$$\ce{ZnS + 4NaCN → [Zn(CN)4]^{2-} + 4Na+ + S^{2-}}$$
We also know $\ce{Pb^2+}$ ions react with $\ce{NaCN/KCN}$ to form a white precipitate of $\ce{Pb(CN)2}$.
My question therefore is why when both $\ce{ZnS}$ and $\ce{PbS}$ are present in the solution only $\ce{ZnS}$ reacts whereas $\ce{PbS}$ doesn't? Or is it simply that it reacts but from an insoluble compound?

You have indicated that the lead sulfide does react and forms an insoluble precipitate, as opposed to zinc which forms a charged soluble complex. $$\ce{PbS(s) + 2CN- -> Pb(CN)2(s) + S^{2-}}$$

@MaxW Ikr, but that's what generally happens. That's basically a test for Pb2+ ions. But when you read through a metallurgy book you find 'No reaction' written.

HUH?!? you wrote: "We also know $\ce{Pb^{2+}}$ ions react with NaCN/KCN to form a white precipitate of $\ce{Pb(CN)2}$." So do you KNOW that the reaction happens, or are asking IF the reaction happens?

@MaxW Come on the test for Pb2+ takes place in an aqueous solution. But in the extraction of Pb the same reaction is taking place in a solvent made of water, oily components, collectors, activators, depressants etc. In which I'm asking about the particular reaction between the depressant and the principal ore. Anyone who has relevance to the topic will probably understand! I didn't mean you don't have, but please open your book it must be written.

Ok, after having poked around a bit I doubt that $\ce{Pb(CN)2}$ forms.
I'm guessing that you're collecting the froth to get a concentrate of the galena $\ce{PbS}$ out of some finely crushed ore.
An aspect of the operation seems to be chemically reducing the concentration of $\ce{ZnS}$ by dissolving via the following reaction:
$$\ce{ZnS(s) + 4NaCN(l) -> Zn(CN)4^{2-}(l) + 2Na2S }$$
Now there are two ways that $\ce{ZnS}$ could be mixed in with the $\ce{PbS}$.
(1) The individual $\ce{Zn^{2+}}$ cations replace individual $\ce{Pb^{2+}}$ cations in the $\ce{PbS}$. In this case the random distribution of the $\ce{Zn^{2+}}$ cations wouldn't allow much extraction of $\ce{ZnS}$ as $\ce{Zn(CN)4^{2-}}$. The $\ce{Zn^{2+}}$ cations just are not going to diffuse through the solid galena particles to any appreciable extent.
(2) There could be also be that the crushing creates a mixture of relatively pure $\ce{PbS}$ and $\ce{ZnS}$ particles which are mixed together. In this case an appreciable amount of the Zn impurity could be removed.

Now for what seems to be the question. $\ce{PbS}$ is very insoluble with a $K_\mathrm{sp} = 9.04\times10^{−29}$. I couldn't find a solubility product for $\ce{Pb(CN)2}$ but it is doubtful that the solubility would be enough to cause the lead sulfide to dissolve and then form lead cyanide.

So it's the sulphide part that prevents such a reaction. The test is probably shown by soluble salts of Pb. But ZnS is too an insoluble salt in water (to show such reaction). That's exactly why I emphasized on the role of the sovent!


text:
Entropy and Equilibrium Constant: Chemical Thermodynamics

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant $K$ in terms of change in entropy is described by:
(A) With the increase in temperature, the value of $K$ for exothermic reaction decreases because the entropy change of the system is positive
(B) With the increase in temperature, the value of $K$ for endothermic reaction increases because of an unfavorable change in entropy of the surroundings decreases
(C) With the increase in temperature, the value of $K$ for endothermic reaction increases because the entropy change of the system is negative
(D) With the increase in temperature, the value of $K$ for exothermic reaction decreases because the favorable change in entropy of the surroundings decreases

Now, I am having trouble understanding option (B) (which in fact, is one of the correct options).
Since the process is endothermic and the temperature is rising, the value of $K$ increases, so heat must flow from surroundings to the system to favor the reaction which implies increase in an unfavorable change in entropy of surroundings rather than decrease as given in the option.
Kindly clarify my doubt. Also, any insights (other than the ones already given in the links attached below) into other options are very appreciated.
This question might look Duplicate of the questions already posted here:

relation between equilibrium constant and entropy change

Effect of temperature on equilibrium constant in terms of entropy change

But since this question is not focusing on the same problem, so I hope, it will not be marked as duplicate.

It's hard to understand exactly why this is not a duplicate. I tried making my answer a little clearer in case that helps.

@BuckThorn I believe it's not a duplicate b/c, at least based on my read of the question, the OP got himself confused about a side-issue that wasn't addressed in your answer -- he though $\Delta H$ is determined by the enthalpy change in going from reactants to the equilibrium mixture (as opposed to what it actually is: reactants to products), and thus was thinking that, if you increase T for an endothermic rxn, as it shifts to the right it's actually absorbing more heat (b/c you need to go further to the right to get to equilibrium).

@theorist That doesn't seem at all obvious from re-reading the post, but you are welcome to post an answer. Looks like a dupe to me.

@theorist Exactly, I was having that misconception.
Thanks, I understand my mistake as well as the concept now.

I agree it's not a duplicate because, while you are asking about the same homework problem discussed in the other question you linked, the issue you are having is different.  Hence, because the last answer didn't address your specific confusion, you are asking a different question about that same homework problem.
Here's where you've got things mixed up:  Whether a reaction is exothermic or endothermic depends on the inherent nature of the reactants and products, not the equilibrium constant.  I.e., $\Delta H^{o}_{rxn}$ is the enthalpy change for a complete conversion of reactants to products at standard state, irrespective of the equilibrium constant.  It's not, as I believe you were thinking, the enthalpy change to go from pure reactants to the equilibrium mixture. Hence if you could, say, change $K_{eq}$ purely by tweaking $\Delta S^{o}_{rxn}$, then $\Delta H^{o}_{rxn}$ wouldn't change just because you've changed $K_{eq}$.
So B is correct.   If you raise T (and here we're assuming $\Delta H^{o}_{rxn}$ is independent of T), and $\Delta H^{o}_{rxn} > 0$, the reason the reaction shifts to the right is this: In calculating $\Delta S^{o}_{surr}$, we have to divide $q_{rev, surr} = q_{surr}$ by T, and  T is now higher.  Hence, for the same heat flow from the surroundings (which is given by $-\Delta H^{o}_{rxn}$), the resulting unfavorable entropy change in the surroundings is less.
[I've used standard state here, which means it's carried out at a constant external pressure of 1 bar, but what I've written applies at any constant external pressure.]

Since the process is endothermic and the temperature is rising, the value of K increases, so heat must flow from surroundings to the system to favor the reaction which implies increase in an unfavorable change in entropy of surroundings rather than decrease as given in the option.

This describes what happens if you let a reaction go to equilibrium, and then increase the temperature. K increases, the reaction goes forward to reach equilibrium, and heat is transferred from surrounding to the system (because it is endothermic), decreasing the entropy of the surrounding.
This is no surprise. Whenever an endothermic reaction approaches equilibrium, the entropy of the environment decreases and the entropy of the system increases (by a larger amount). At equilibrium, further forward reaction would go against the 2nd law of thermodynamics (as the reaction proceeds, the change in entropy of the system gets smaller and smaller: $\Delta_r S = \Delta_r S^\circ - R \ln(Q)$).

Now, I am having trouble understanding option (B) (which in fact, is one of the correct options)

The exercise does not specify any specific scenario. It is just asking about the change in equilibrium constant. You can solve this by looking at the relationship between standard Gibbs energy of reaction and K, as well as the definition of Gibbs energy via enthalpy and entropy:
$$ \ln(K) = - \frac{\Delta_r G^\circ }{R T} = - \frac{\Delta_r H^\circ }{R T} + \frac{\Delta_r S^\circ }{R}$$
If reaction enthalpy and entropy are independent of temperature in the range of interest, raising the temperature will increase ln(K) if the reaction enthalpy is positive. In a more advanced treatment, you will learn that $\frac{d \ln(K)}{dt} =  - \frac{\Delta_r H^\circ }{R T^2}$ exactly.

"At equilibrium, further forward reaction would go against the 2nd law of thermodynamics" beautifully sums it up since it incorporates the very definition of Equilibrium. And shouldn't it be $\frac{d \ln(k)}{dt}=-\frac{\Delta_r H^\circ }{R T^2}$?


text:
What makes metals malleable and ductile?
Two well-known Physical Properties of metals is that they are malleable and dutile
I was just wondering about what causes metals to be malleable and/or ductile and non-metals to be brittle? What factor determines this?
Thanks!

The metals are malleable because they don't make molecules : they exist as individual atoms. In a metallic piece, the atoms are identical, and they are piled up like layers of beads in the bottom of a flat box, each bead touching its neighbors. The first layer of beads covers the entire surface of the surface. You may imagine a second layer identical to the first one covering it, then a third layer, and so on. Hopefully you see that is not very difficult to push a whole layer, or a whole pile of such layers from the side. A whole layer can slide easily on a lower layer. This would not break the whole piece, which gets distorted but not broken. It would be much more difficult to do it if suddenly one or some of these beads are replaced in a given layer by bigger beads or by non-spherical objects. ${}{}{}{}{}$

Well, fcc metals are pretty much this way, and things like Au are certainly both malleable and ductile. However, bcc metals do not have good slip systems, and are not particularly malleable or ductile. Its all about dislocations...


text:
What is the correct electron dot notation of carbon?
My book says that the dot notation for Lewis structure of carbon has 2 possible ways:

two dots on the right, one on the top, one on the left
one dot on all four sides

Why are both of these correct? Isn't only the first one correct by Hund's rule? By the way, we are now learning Lewis structures and the book randomly used the 2nd dot notation of carbon.

why is it allowed to move around the valence electrons for bonding?

Both are correct. Electrons are free to move around the atom. They change position all the time. Don't forget they may move and repel each other.

oh ok, so it is allowed for a valence electron to move from s sublevel to p sublevel and vice versa freely?

Umm, you're not really asking about dumb dot drawings there... It's  not "freely" at all in reality. Four unpaired electrons are high excited state, but I kinda doubt people using "dot structures" usually care about it.

@Maurice There's this dumb rule with drawing dots in pairs in "dot structures".  It's no wonder you didn't get OP's question though, as like no-one serious about chemistry cares about them IMO.

This is how it works, but you don't really need to know it.


text:
Normal Mode Analysis of proteins
I'm trying to do Normal Mode Analysis for my MD trajectory. I'm following Amber tutorial (http://ambermd.org/tutorials/advanced/tutorial3/py_script/section5.htm) which uses an example of estrogen receptor protein and raloxifene ligand. Since I'm only getting errors from this, I'm having a couple of technical questions.

As for solvent, my system is explicitly solvated. Skjaerven et al. in J Mol Str: THEOCHEM 898 (2009) 42–48 claims as follows:

"normal modes of proteins are most often calculated in the absence of solvent or with simple solvent models, as opposed to explicit solvent molecules in molecular dynamics simulations. (…) a common approximation is the absence of (explicit) solvent. Studies have shown that explicitly present water yields an increase in the density of local minima compared to vacuum simulations. In addition, water is found to increase the friction and amping on the atomic fluctuations."

Could NMA be used for proteins? I've heard an opinion that rigid rotor model used in NMA is not appropriate for proteins which are flexible.
The molecule in Amber tutorial example is a protein and it's solvated in explicit solvent (I think).

So, to conclude, how does it all make sense?

This would be an interesting question to post on the new materials (and molecular) modeling page. https://materials.stackexchange.com/


text:
What question could be answered by "because the metabolites in the cell are in ionized form" in the topic of anaerobic metabolism?
I know it might sound a bit absurd, but my teacher gave us some "answers to question", that is, given one answer, write at least one question the answer to which is the given one.
I've solved most of the answers, but there's one I just can't figure out: "because the metabolites in the cell are in ionized form", and the topic is anaerobic respiration and fermentation. No additional details from her.
I'm in high school (a "scientific" school in Italy), if that can help for the knowledge level.
Any idea?

Hmmm, scientific Jeopardy, interesting idea...

"Why does ethanol escape from cells but metabolites do not when anaerobic respiration and fermentation are happening?". BTW: Not all metabolites are in ionized form (e.g. glucose), but many are, including neutral molecules attached to charged carriers (phosphorylated glucose, fatty acids with CoA etc).

That's a pretty valid question, thanks! Any other idea?

Interesting reverse question. I could figure out three questions:

Why milk changes from a liquid to a gel when it turns into yoghurt?
Why venous blood has lower pH than arterial blood?
Why do we get tired on intense physical exertion?

The answer to the first one would be because the metabolite of lactose fermentation in the lactobacilli cells is lactic acid, that in ionized form releases $\ce{H+}$, lowering the medium pH until the isoelectric point of the milk proteines is reached, causing their agglutination. The answer to the second one is that because the venous blood carries away carbonic acid produced as a metabolite in cellular respiration, that has two ionized forms, $\ce{HCO3-}$(bicarbonate ion) and $\ce{CO3^2-}$(carbonate ion). The answer to the third one would be that, as in lactobacilli case, our body resorts to anaerobic lactic fermentation in activities that require a large power output, beyond the oxygen delivery capacity of the circulatory system. So lactic acid builds up in the muscles, lowering the pH and causing the fatigue sensation. This last explanation was broadly accepted, but seems to be controversial in light of recent studies, as we can see in the abstract for this article:

In addition, many textbooks report that muscle fatigue is mainly the
  result of a decrease in pH within the muscle cell due to a rise in
  hydrogen ion concentration ([H+]) resulting from anaerobic metabolism
  and the accumulation of lactic acid.6–8 Recent literature, however,
  contradicts this assertion.

All good questions, but none of these are related to "anaerobic respiration and fermentation". The third one comes the closest, but we are not anaerobic organisms!

We are not obligate anaerobes, but in fact most living things can rely on some degree of fermentation, even us, as a backup of sorts.


text:
How do you optimize an organo-metalic compound containing cerium in in Gaussian?
I am trying to perform optimisation of the molecule attached. I need to get partial charges and frequencies. The issue is that it doesn't converge with following input:
%nprocshared=8
%mem=16GB
%chk=CTB.chk
#n opt b3lyp/gen pop=(ReadRadii, savechelpg) SCF=QC
CTB
0 2
Ce -6.41954 5.21611 0.61820
...
N -4.38490 3.87720 0.35349

H -5.52145 4.13584 -2.66372
Ce 0
SDD
****
H Si C N 0
6-31G
****
Ce 2.88

As well I get error message:
Restarting incremental Fock formation.
Gradient too large for Newton-Raphson or scaled steepest descent --
use steepest descent instead.

Hello, it is necessary to add the keywords "pseudo=read" after "b3lyp/gen" because you use pseudopotential SDD. Moreover, I would suggest to use "scf=xqc".

Baffled by the negative vote. Seems to me to be a perfectly valid question.

It's hard to know your problem with convergence without seeing more of the output (e.g., how the SCF progressed). I agree that XQC is a good idea, but I'd actually suggest starting by minimizing your compound with xtb an approximate method from the Grimme group that generally provides good starting geometries and usually converges more easily.

Other options to try and find a better starting geometry and save time: simplify the ligands to $\ce{N(SiH3)2}$ or even $\ce{NH2}$, use a GGA such as PBE for a pre-optimization. As a side-note, use a modern dispersion correction.

With a little more detail, this question would likely be well text on the new Materials Modeling SE


text:
Dry paint becoming lubricant with heat
I am undertaking a physics experiment and have come up with a chemistry problem. Could paint perhaps become a lubricant under mild temperature. (It would take a while to explain my experiment, but an object sliding down a ramp is increasing velocity with increasing mass, this should not happen if if coefficient of friction stays constant). I am wondering if anyone has information about paints lubrication vs temperature, or anything similar. The coefficient of friction is the same before and after the experiment, so I am wondering if paint changes its coefficient of friction or lubrication when temperature increases.
I'm sorry if this questions doesn't make sense, if anyone can point me towards anything paint and friction related, thankyou.

If you take "paint" in the very broad meaning of substances used to treat surfaces:
wood can be waxed, and waxes melt at not too elevated temperature and can then behave like oils.
I'd expect the behaviour of waxes here as very prone to surprises, since they can also become sticky when some compounds are liquid while others are solid.

Certainly skiers have opinions about skis and waxes! ;)

@EdV :-D never met anyone who really liked klister... But I think the grip wax, glide wax, liquid water (lubricant), snow/ice crystals system is quite a bit more complicated than what OP has here (giving room also for lots of opinions). At least I hope their system is easier...


text:
Defintion of Solubility in Different cases
How do we define solubility (in moles per liter) in cases when concentration of cation is not equal to the concentration of anion?
e.g. When we put AgCN in an acidic solution (Leading to formation of HCN) or when AgCl is put in CH3COONa (Leading to precipitation of CH3COOAg)
Do we take concentration of Ag dissolved as solubility or concentration of CN or Cl as solubility?
Please answer for both cases.

Are you familiar with the quantity called solubility product?

Yes. That's why I used examples.

There are two ways of dealing with this.
Solubility product
The more general way is to state the solubility product. Taking your first example, this would have the same value no matter how much acetate is present in solution before adding the silver ions. The disadvantage is that for highly concentrated solution (in fact, in the presence of ions, even for moderate concentrations) this gets inaccurate unless you switch from concentrations to activities.
Solubility (of a solid)
The other way is to specify the solubility of $\ce{AgCl}$ in pure water, in 0.1 M sodium acetate, in 0.5 M sodium acetate and so on. In this case, because silver chloride is much less soluble than silver acetate, no silver acetate would precipitate (the silver ion concentration will be sufficiently low in solution so that silver acetate will not be exceeding the solubility product).
Similarly for AgCN, you could specify its solubility in pure water, in aqueous solution buffered at pH 7, or buffered at any other pH. For the latter example, you should probably also specify ionic strength or the entire makeup of the solution prior to adding AgCN. In this case, there will be a strong dependence of the solubility on the pH because HCN is soluble and its formation will decrease the concentration of cyanide ions while the concentration of silver ions will remain the same up to the point where it precipitates.
In both of these cases, there is no ambiguity in terms of concentration because there are no sources of the ions forming the precipitate other than what was added as a (stoichiometric) solid.
On the other hand, if you add NaCl to a solution already containing LiCl, the apparent solubility of NaCl will be lower than in pure water because of the common ion effect. You could still talk about the solubility of NaCl in 2 M LiCl, and the quantity would refer to the concentration of sodium and chloride ions due to the addition of NaCl, not due to ions already present before addition.
Concentration of ions vs. concentration of added solid
With any salt that does not have 1:1 stoichiometry, we give the concentration of the salt, not that of the ions. For example, we talk of a 1 M $\ce{MgCl2}$ solution even though it is a strong electrolyte containing 1 M magnesium ions and 2 M chloride ions. The concentration given is based on the formula of the solid added, not on the concentration of ions after dissolution. In a similar way, we would say we added acetic acid to a concentration of 1 M even though we know full well that some of it will be ionized.

I think you misunderstood my question. I asked wether we consider concentration of Ag+ as solubility or the concentration of Cl- the solubility

@ParthKamra The concentration of AgCl. I added some material to my answer to that effect.


text:
Why there is no need for reference standards in High Resolution Mass Spectrometry (HRMS)
I have a question about HRMS (High  Resolution Mass spectrometry). HRMS is a powerful technique that can discriminate between possible target compounds without the need of reference standards, unlike QqQ (triple quad systems). My question is simple why do HRMS don't need reference standards for analysis?
A short explanation would really help I cannot find the answer online or anything related to it.
Thank you, hopefully I can be helped with this part. I put a section of a book as an attachment maybe that will help.
Greets,
Student

I smell a rat. I've never work with an instrument that didn't use some sort of standard for setting it up. I can believe that you don't need to run standards with each particular sample, but never run a standard? No way.

I assume that the quoted text is about a GC-MS or LC-MS setup. With a low res MS you'd need retention time and MS to identify a compound. With a high res MS you just need LC or GC to give a clean separation then the MS alone could identify the compound. However the MS itself would definitely need standards to tune it. Say injecting a pure compound of some sort.

The statement is not 100% correct. Ordinary mass spectrometry will not give you exact masses to 5 or 6 decimal places, say, m/z 415.14 for a singly charged molecular ion. Now there will be hundreds of compounds with this mass. On the other hand HRMS gives you a accurate and precise mass say 415.14509. This detailed number drastically reduces the number of matches in a spectral library. Not only that, the isotopic distribution pattern and their masses reduces the choice to one compound. Hence the claim that HRMS will not need a reference.
Like any claim, there are always exceptions. Structural isomers can be a problem. If we have isotopologues (isotopic isomers), even HRMS will fail.  Say you have two analytes but both are acetonitriles. Say one is
$^{13}$CH$_3$CN, and the other analyte is CH$_3$$^{13}$CN, HRMS will give the identical molecular mass to five decimal places. One might look up their fragmentation pattern but MS is not that great at low masses (high noise). Here you would need another spectroscopic technique such as microwave rotational spectroscopy. These techniques are used an in emerging field called position specific isotopic analysis.

Thanks for the answer. I understand your part. But still I then want to know how do they use HRMS without reference standards? Short procedure, I can't still imagine that part.
Is is due the fact they created a database with accurate masses and can then conclude the unknown compound or am I not right.
Thank you very much

Yes, you are right. HRMS is used in analytical problems where the analytes are  unknown. Therefore you have no clue which standard to buy? So HRMS would allow the users to match exactly with the library or database with very high confidence.

Even if the compound is not listed in the library, one can deduce the exact chemical formula (not the structure) with HRMS.


text:
Why is the boiling point of homologous primary > secondary > tertiary alcohols whilst their solubility in water is the reverse?
From my understanding, for homologous alcohols, primary alcohols have the highest boiling points, because the OH group is the most accessible for other molecules to hydrogen bond to, therefore stronger intermolecular forces result in higher mp/bp.
One the other hand tertiary alcohols have the highest solubility in water because the packed structure of branches means the non-polar part of the molecule is condensed and therefore water near any part of the molecule can receive these hydrogen bonds. Doesn't this mean tertiary alcohols have the most accessible hydrogen bonds and therefore should have the highest melting point too?
Alternatively does this have to do with water being small and alcohols being larger therefore 'access' requirements to hydrogen bonding are different as larger alcohols need to be even closer than water to make a difference therefore the packing is less relevant as even being slightly away from the molecule means failure to access hydrogen bonds but water will be ok with being slightly away.

The key concept here is that for solubility, you need to compare alcohols with the same number of carbon atoms. For example, the solubility order of isomers of butanol is as follows:
$$\ce{CH3CH2CH2CH2OH < (CH3)2CHCH2OH < (CH3)3COH}$$
This difference in solubility follows the order 3°>2°>1° and can be explained by the branching concept that you have used in your question. However, If you compare alcohols with different number of carbon atoms, say butanol and methanol, then the alcohol with a smaller hydrocarbon part will be more soluble in water.
The boiling point of alcohols increases with the increase in the number of carbon atoms, as stronger van der waals forces start acting in between the molecules. For example, Phenol boils at around 180°C whereas methanol boils at around 60°C. For alcohols with the same number of carbon atoms, a more exposed hydroxyl group will result in stronger intermolecular H-bonding.
TL;DR: your comparision can only be applied for alcohols with the same number of carbon atoms


text:
Why is [Ni(en)3]2+ paramagnetic?
I could not find a satisfactory answer for the above question anywhere on the internet/on SE. Why does $\ce{en}$, which is a strong field ligand, not cause the electrons to pair up in $\ce{[Ni(en)3]^2+}$ the same way it does in $\ce{[Fe(en)3]^3+}$ or $\ce{[Co(en)3]^3+}$?

See this one https://www.toppr.com/ask/question/nicl2pch332-is-a-paramagnetic-complex-of-niii-analgous-pdii-complex-is-diamagnetic-how-many-geometrical/. The only problem is they mix up CFT and hybridization.

@M.Farooq that answer is for $\ce{[Ni(Cl)2(PPh3)2]}$ which is tetrahedral and not for the complex in question.

https://chemistry.stackexchange.com/questions/35081/can-a-high-spin-complex-be-diamagnetic?r=SearchResults&s=1|61.1726