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Would I use PBr3 or HBr in this synthesis?

In the above reaction, I know that I have to use either $\ce{HBr}$ or $\ce{PBr3}$ and then $\ce{NaCN}$, DMF, $\ce{H+}$ and $\ce{H2O}$. However, I'm just confused on when to use $\ce{HBr}$ or $\ce{PBr3}$. How is the stereochemistry controlled?

Try to think in terms of desirable configuration. It seems to me that neither would work as desired.

Who says you have to use PBr3?

As @permeakra has said neither reagent will get you the stereochemistry you want in the product. The HBr route goes through a carbocation which scrambles the stereochemistry completely.
PBr3 will invert the stereochemistry, then SN2 -CN addition will invert it again, giving you the wrong product.
You need something that turns the -OH group into a leaving group while preserving the stereochemistry, so that the CN can displace it by a SN2 process. This will invert the stereochemistry giving you the product, after hydrolysis, you want.
I suggest forming a sulfonate, either tosylate, triflate or (my choice) mesylate. This should be displaced smoothly by cyanide under the DMF conditions you refer to.

What if we use a Lewis acid like AlCl3 or FeBr3 at the beginning? Will that work?

No. That is still going to go by a carbocation


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How to make acetic acid from 2-methylpropan-2-ol?
How to make acetic acid from 2-methylpropan-2-ol? I tried oxidation but it didn't work.

Hi neel gohil! Take a [tour] to know about this community!

Welcome to Chem.SE! We require you to show your efforts on this problem. What formulae/steps did you try? Please add this to your question. Thanks!

I made alken from the reactant and then I am confused.

First dehydrate the given tert-butanol to form 2-methyl prop-1-ene. Then do reductive ozonolysis and you will have acetone and formaldehyde. These two products can be easily separated if you treat the solution with Tollen's reagent. Thus formaldehyde will form formic acid and acetone will remain as it is.
Now if you pour this solution into a solution of polar basic solvent, the acetone will form a layer, which is easily separated, but formic acid will be easily soluble. Taking the acetone, you will get the acetic acid via the haloform reaction and then hydrolysing under acidic conditions.

If you like it , just upvote..

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Radial probability for orbitals

Questions

Question F-2:
I think question f1 requires one to relate $\mathrm{\psi^2}$ or. Probability density with radius but I am just in high school and have not learnt about Hamilton operator
Question F-1:
In second I think A should be 1s and B should be 3p

I think question f1 requires one to relate ¥^2 or. Probability density with radius but I am just in high school and have not learnt about Hamilton operator ..

In second I think A should be 1s and B should be 3p

I have suggested an edit if you feel it's an improvement you can accept it here. As for F-3 if I remember correctly there is a formula for radial nodes $\mathrm{n - l - 1}$ .As for the first question i don't know either( i'm in highschool).

Is this what they do in school? Where you are from? Randomly memorizing such things without learning about Hamiltonians and so on? "the maximum probability of finding the electron in the dxy orbital is" - I hope such a nonsensical approach to learning about nature will not discourage you.

Actually I am studying above the curriculum

Then start at the beginning. There is zero value in knowing this forumla

For the 1st one, the maximum probablity of finding the electron occurs at the Bohr's radius,i.e. at $\ce{r=r_0}$. This can be guessed by thinking this result classically. But if you want to know the exact method, the wave function for $1s $ orbital is, $$\ce{$\psi$_{1s} = 1/$\sqrt{\pi a_0^3}$ e^{-r/a_0}}$$ So, the radial probablity distribution for the $1s$ orbital will be,$$\ce{P(r) = |$\psi$|^2 4$\pi$r^2dr}$$ which will be eqal to , $$\ce{P(r) =4/a_0^3 r^2e^{-2r/a_0}}$$ To find the value of $r$ at which $\ce{P(r)}$ is maximum , you just have to find the solution of $\ce{dP/dr =0}$. You can verify very easily that the solution is $\ce{r=r_0}$.For the next one, the graph of B) is definitely for $2s$. Now, according to judgement of no. of radial nodes of $2s$ and $3p$, they  both have 1 radial node each. So, that is inconclusive. But if you would have known the wave functions, it would be easier.The wave-function of $2s$ is $$\ce{$\psi$_{2s} = c. (2- r/a_0)e^{-r/2a_0}, where c is a constant.}$$But, wave function of $3p$ orbitals are, $$\ce{$\psi$_{3p_z} = c_1(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}cos$\theta$}$$ $$\ce{$\psi$_{3p_x} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$cos$\phi$}$$ $$\ce{ ($\psi$_{3p_x} is actually an additive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$ and, $$\ce{$\psi$_{3p_y} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$sin$\phi$}$$$$\ce{ ($\psi$_{3p_y} is actually an subtractive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$So, you can see at $r= 0$, all the $3p$ orbitals have value of wavefunction (i.e. |$\psi$ |) = $0$ (simply check by putting $r$ as $0$ in all given $3p$ wave functions). But at $r =0$,$\ce{ $\psi$_{2s}}$ has non-zero finite value so, $\ce{|$\psi$|^2}$ does not vanish at origin for$ 2s$ orbitals which the given graph is showing. So, it can't be $3p$ as the value of the wave function doesn't match at origin. If you consider higher orbitals like $4d $ also, their wavefunction will also vanish at origin. So, the figure B) corresponds to $2s$.

This answer could be improved by distinguishing between the radial probability density discussed in the first question, and the probability density discussed in the second question.

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Number of atomic and molecular orbitals in different Hartree-Fock references
For a given molecule and electronic configuration,

Are the number of contracted atomic orbitals (AOs) different in a restricted open-shell Hartree-Fock (ROHF) calculation from an RHF calculation? What about unrestricted Hartree-Fock (UHF)?
Are the number of molecular orbitals (MOs) different in a ROHF calculation from an RHF calculation? What about UHF?

For example, if we are using the STO-3G basis set for a water molecule, we get 1 + 1 + 5 = 7 contracted functions for AOs in an RHF calculation, leading to 7 MOs. Are any of these numbers different in an ROHF calculation? What about UHF?
Essentially, what are the core differences between AOs and MOs within each approach?
For water molecule using sto-3g basis I obtained the orbitals in the following order For RHF: Occupied (A1) (A1) (B2) (A1) (B1) Virtual (A1) (B2) For ROHF: Occupied (A1) (A1) (B2) (A1) (B1) Virtual (A1) (B2) But if we take alcohols for an e.g., the number of such orbitals is not the same. I want to know how they are calculated, so that atleast I could do them by hand. For the RHF one, it is nicely explained in "Modern Quantum chemistry" by Szabo-Ostlund. I want some similar information on ROHF and UHF (if possible).

Straight answer to your question: whatever method you use, RHF, ROHF or UHF, your calculations will use the same number of basis functions (7 contracted functions for your water example). How will they be used and combined, however, is different. In your comment on the question you talk about molecular orbitals - it's important to keep these conceptually separate from the basis set you are using to construct the orbitals.
How many functions do you use for an orbital in ROHF?
In general, UHF/ROHF approaches use a separate molecular orbital for each electron, so you would use basis functions twice to describe an unrestricted pair compared to a restricted pair.
In other words, basis functions aren't "split" between the electrons when treated as an unrestricted pair; rather, two orbitals are calculated using the same basis functions, one corresponding to the alpha electron and another for the beta electron.
So, if you compare UHF with RHF, you'll be using the same basis functions to construct twice the number of molecular orbitals.
Let's take an example to illustrate it: $\mathrm{O_2}$.
For singlet $\mathrm{O_2}$ ($\mathrm{S=0}$), there are 8 alpha and 8 beta electrons. Using STO-3G, we'll have 10 contracted functions, those corresponding to $\mathrm{1s}$, $\mathrm{2s}$ and 3 x $\mathrm{2p}$ for each oxygen atom, each of them composed of 3 primitive gaussians.

In RHF, we can combine these 10 basis functions in a single Slater determinant, producing 10 molecular orbitals. The first 8 of these will be doubly occupied, the last 2 will be vacant.
In ROHF, we have exactly the same as for RHF, because ROHF is exactly the same as RHF for molecules with $\mathrm{S=0}$.
In UHF, we use the same 10 basis functions, but we use them to build two sets of 10 molecular orbitals - 10 for the alpha electrons and 10 for the beta electrons - in a single Slater determinant. We occupy them with 8 alpha electrons and 8 beta electrons, 1 electron per orbital, with 2 alpha and 2 beta orbitals left vacant.

For triplet $\mathrm{O_2}$ ($\mathrm{S=1}$), there are 9 alpha and 7 beta electrons. We have the same basis functions as before.

In UHF, we do the same as before, and we also get 10 alpha and 10 beta molecular orbitals in a single Slater determinant. Occupation changes, though, so now we have 9 occupied and 1 vacant alpha orbitals and 7 occupied and 3 vacant beta orbitals.
In ROHF, and since now we have an unequal number of alpha and beta electrons, we have a more complex construction of the Fock operator - one that is not unique. Essentially, we'll calculate alpha and beta electrons separately under the hood, and then combine them; and we'll be treating full, partially full and empty orbitals as three coupled sets. We'll produce 10 orbitals, of which we'll doubly fill the first 7 and partially fill the following 2, leaving 1 unoccupied.

In conclusion: the number of basis functions in all cases is the same: 10 contracted functions comprising 30 primitive gaussians. However, these basis functions are used a different number of times for RHF, UHF and ROHF; RHF uses them once in a simple Slater determinant, UHF uses them twice to form a Slater determinant with a two sets of orbitals, one for alpha and the other for beta electrons, and ROHF uses them twice but ends up producing a single set of full, partially or non-occupied orbitals that share their spatial part.

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Does melanostatin peptide whiten the skin in a similar manner as to how melanotan darkens it?
I was reading about peptides and came across an interesting peptide that bodybuilders use to darken their skin.
How can peptides change the genetic makeup of someone's skin and darken it?
I also found another peptide called melanostatin that inhibits alpha-MSH. So my question is, can it whiten the skin?
Because bodybuilders aren't very much interested in whitening their skin, there were no trials and experiences with that peptide.
So does melanostatin work similarly to how Melanotan darkens the skin?

Disclaimer: I do not intend on using this peptide on animal experiment, nor humans for that matter. My question is purely theoretical and for educational purposes.

Melanotan
The "Melanotan" you're referring to is probably Melanotan I, also known as afamelanotide, which is an analogue of α-melanocyte stimulating hormone (α-MSH).  In other words, it activates (some of) the same receptors as α-MSH.
α-MSH has several functions, one of which is to stimulate the cells that darken the skin (melanocytes), so they produce more melatonin.  It activates these cells via the MC1 receptor.  Afamelanotide is more specific to that receptor, so it darkens the skin without some of the other effects.
It acts at the hormone level, so it won't be changing anyone's genetic makeup - stop using it and you'll go back to your normal colour.
Melanostatin
Confusingly, the name Melanostatin has been applied to more than one compound.  I've found two short peptides that might fit the bill.
The most likely candidate is described in this leaflet, which says it is used for skin lightening.  It's a receptor antagonist - it prevents α-MSH from binding to the MC1 receptor, so reducing melanin production.  So yes, it's working at the same site as Melanotan I but having the opposite effect.
The other possibility is also known as Melanocyte-inhibiting factor, which inhibits production of (natural) α-MSH, so it would be expected to lighten the skin.  However, since α-MSH has several other functions, reducing its production would also affect other systems - there would be side-effects, and skin lightening doesn't seem to be the main thing it's known for.  More here: Wikepedia

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Type of floor for a chemistry lab
I'm currently building a small chemistry lab and I'm wondering about which type of floor would be the best to use. Currently I'm mainly doing electrochemistry experiments with lithium and other metals. Although my interests might change in the future.
I was thinking about ceramic tiles. Since it's not flameable and unreactive. Are there better types of floor for my application ?

Here is link that might be useful for you : Lab Flooring

Thank you for your response. This helped me to progress in my decision making process.  Still not 100 % sure for what i will decide but i now have a better overview

Good luck with that! Always willing to help:)


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Ipso substitution and its rules
I was taught that the following reaction happens:

It is called the "ipso-substitution". My professor had told me that it occurs in the case of $\ce{-SO3H}$ and $\ce{-COOH}$. But he had ended the topic there. My textbook doesn't detail any more either.
My searches landed me up at the fact that ipso-substitution is also employed in the preparation of picric acid from nitrobenzene. But I couldn't find answers all my other questions, which are:

Does the ipso-substitution occur only for these two groups, or are there more? I guess there is a "threshold" value for some parameter, above (or below) which ipso substitution cannot occur, but I am unable to come up with what that parameter is. I guess it's the high stability of the leaving group $\ce{SO3-}$, but I am not sure.
Is ipso-substitution affected by the choice of the electrophile and the solvent? What if, instead of bromine water, I chose bromine in acetic acid? Or something even simpler as $\ce{FeCl3/Cl2}$. Would ipso substitution occur then?
I know that no organic reaction produces 100% of only one compound, as its other isomers are always also formed. Ipso substitution should be no exception. But how do we decide whether the ipso product is a major product or a minor product? I know how to decide major/minor products between ortho/meta/para substitution (electron density/steric hindrance/H-bonding), but I haven't been taught any logic to decide the same for ipso substitution.

I can't say that there is a set of rules for ipso substitution. For the loss of a more complex group by an ipso  protonation, see https://chemistry.stackexchange.com/questions/50866/mechanism-of-a-dephenylation-reaction/93946#93946

All "ipso" substitution means is that it occurs where the leaving group was. That's all. Essentially all of the $\mathrm{S}_{N}\mathrm{Ar}$ reactions you learn are ipso substitutions.

Trick to remember. All o/p positions wrt strongest activating group becomes br on addition on bromine water(irrespective of the group's there)

The shorthand way that we draw structures in organic chemistry, with implicit hydrogens, leads us to often forget that the hydrogens are there, or to neglect considering them in our analyses. In electrophilic aromatic substitution (EAS) reactions, the generally accepted mechanism involves attack of the aromatic ring on the electrophile to give a Wheland intermediate, which then loses a proton from the $\mathrm{sp^3}$-hybridized ring carbon to restore aromaticity. Ipso substitution, then, is really nothing special - we simply substitute a position on the aromatic ring that bears a non-hydrogen (e.g., $\ce{X}$) group, presumably via an analogous mechanism. For this to be viable, both steps of the mechanism need to be favorable:

$\ce{X}$ needs to enhance, or at least not greatly reduce, the nucleophilicity of the carbon it is attached to. For example, inductive effects would great disfavor ipso attack at groups like $\ce{F}$ or $\ce{CF3}$.
$\ce{X+}$ should be a reasonable species to lose in the second step, to regain aromaticity, or there should be some mechanism by which $\ce{X}$ can be lost, with electrons flowing back towards the aromatic ring.

Sulfonic acids and carboxylic acids are rather electron-withdrawing, but if you drill down into the details of these ipso-substitution reactions, many are actually conducted on the conjugate base (i.e. sulfonate or carboxylate), which also provides a pathway for loss of a neutral group in the second step, i.e. $\ce{SO3}$ or $\ce{CO2}$. Alternatively, harsh conditions are used, e.g., sulfonation/desulfonation.
It turns out that ipso attack is more common than it first appears, because the product of the initial ipso attack can re-arrange. So, for example, nitration of p-cresol involves 40% attack at the ipso carbon bearing the methyl group, followed by acid-catalyzed rearrangement to the expected product.[1] There is more discussion here.
Closer to the original thrust of your question - there are some groups that are even better at being substituted in an EAS reaction than a proton. Trialkylsilyl and trialkylstannyl groups are often used. Both $\ce{Si}$ and $\ce{Sn}$ are less electronegative than carbon and highly polarizable, making the ipso carbon a good nucleophile. These groups also help to stabilize the Wheland intermediate through hyperconjugation. Finally, they both have energetically-favorable pathways to leave with electron density flowing back towards the aromatic ring. For these reasons, ipso substitution at $\ce{SiR3}$ or $\ce{SnR3}$ substituted carbons can often be achieved rapidly, under mild conditions, with high regioselectivity. This is often helpful in specialist applications, like radiolabelling.
I haven't specifically addressed each of the original three numbered questions, but they can be tackled using the principles above. Ipso substitution is possible for any group - it's a competition, $\ce{H}$ vs $\ce{X}$. Is ipso substitution affected by reaction conditions? I'm sure it could be - factors like protonation states, hard vs soft electrophiles, reversibility, etc. would have an influence.


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Editions of IUPAC Green Book
Why does the URL https://www.iupac.org/cms/wp-content/uploads/2015/07/Green-Book-PDF-Version-2011.pdf (ending in 2011) contain the Green Book labelled 2007, and reprinted in 2008? Does a 2011 edition exist and they just uploaded the wrong file?
Same goes for the URL http://media.iupac.org/publications/books/gbook/IUPAC-GB3-2ndPrinting-Online-22apr2011.pdf (containing the exact same file).
And there is the version https://www.iupac.org/fileadmin/user_upload/publications/e-resources/ONLINE-IUPAC-GB3-2ndPrinting-Online-Sep2012.pdf with different water mark, one page more, and the front matter in a different order.

Wikipedia reads:

The third edition of the Green Book (ISBN 978-0-85404-433-7) was first published by IUPAC in 2007. A second printing of the third edition was released in 2008; this printing made several minor revisions to the 2007 text. A third printing of the third edition was released in 2011. The text of the third printing is identical to that of the second printing.

So yes, maybe they uploaded the wrong file. But the content of 2011 and 2008 should be 100% identical.


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Numerical on acetic acid - sodium acetate buffer
Starting with 1 L of 2.0 M $\ce{CH3COOH}$, we wish to make a buffer solution of pH=4.00.
Consider two ways to make the buffer:
a) One way would be to add sodium acetate. How many moles must be added?(assume no volume change)
b) Another way would be to add NaOH and have it react with acetic acid to produce the correct acetate ion concentration. How many moles of NaOH do we need?
Part a) - My work:
$$ \ce{CH3COOH <=> H+ + CH3COO-} $$
From Henderson’s equation,
$$ \ce{pH = pK_a+ log \frac{[base]}{[acid]}}$$
$\ce{pK_a}$ of acetic acid = 4.74
Plugging in, I get: 4= 4.74 + log$\ce{\frac{[CH3COO-]}{[2]}}$
For my final answer, I got $\ce{[CH3COO-]}$ = 0.364 M. So 0.364 moles of $\ce{CH3COO-}$ need to be added to 1L of solution.
Part b)-My work:
$$ \ce{CH3COOH + OH- <=> H2O + CH3COO-}$$
$\ce{K_a = \frac{[CH3COOH]}{[OH-]}}$
I have no idea where to go from here. Any help would be greatly appreciated!

Suppose you add $x$ moles of NaOH, So, you acetate ion will form in $x$ moles. NaOH will be totally consumed and acetic acid will be left $2-x$ moles. Now you solve the Henderson-Haselbach equation as, $$\ce{4=4.74 + log(x/ (2 - x))}$$  You will get $x$ as $0.308 $ moles, which is your solution.


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Ionic Character, on which factors do it depend
I was watching a lecture of chemistry teacher. He gave the following question:
Q Arrange in terms of increasing ionic character: LiF,NaF,KF,RbF,CsF?
My try: I tried two approaches:
Approach 1: As F is common to all compounds, therefore its metal that affects the ionic character. As we move down the group as in this case, the shielding effect increases and hence it becomes more easier to abstract an electron from the outermost shell or in other words, Ionization energy decreases as we move down the group, therefore CsF>RbF>KF>NaF>LiF should be the order.
Approach 2: Li, Na, K, Rb, Cs have same electrovalency. As Cs has more larger size, the positive charge intensity on the atom decreases and thus by formula of electrostatic force F=-kq1q2/r^2, the electrostatic force of attraction decreases and thus lattice energy decreases and therefore order is:
LiF>NaF>KF>RbF>CsF.
According to me lattice energy should determine ionic character as lattice energy determines ionic bond strength I think and therefore approach 2 should be more legitimate but when I checked the answer, the answer by Approach 1 was correct. Why is it so? Which of the approach is right and which is wrong and why? Is there any other approach?

When you are considering lattice energy as the parameter for judging the ionic character of compounds, you are assuming two things.

The constituting elements of this alkali metal halides are purely ionic,and thus the Lattice energy (U) is the released energy when those two isolated gaseous ions forms the crystal lattice. But, in practice the constituting ions are never purely ionic (especially smaller alkali metal halide such as $\ce{LiF} $ or $\ce{NaF}$.)
The ions constituting the crystal lattice are hard non-elastic spheres. They hardly (in general) obey this postulate. This postulate is again not totally valid for smaller alkali metal halides due to distortion of their electron cloud due to large polarization.

So, it is seen that if you just theoretically say that the electrostatic force as simply Coulomb's law and thus lattice energy decreases, it doesn't conclude anything as it is seen that the Lattice energy is not a proper parameter for judging ionic character.
The correct approach in judging the covalent character present in an ionic compound is to follow Fajans' rules. According to this rule, $\phi _{cation}$ is inversely proportional to $r_{cation}$ and directly proportional to $Z_{cation}$ Here, $Z_+$ for all the cation is same. But radius of $\ce{Li^+}$ is smallest and ionic radius gradually increases from $\ce{Li^+}%$ to $\ce{Cs^+}$. So, $$\ce{$\phi$_{Li^+} > $\phi$_{Na^+}  > $\phi$_{K^+}  > $\phi$_{Rb^+}  > $\phi$_{Cs^+} }$$ Here  $\phi$ is the measure of polarising power of cation. As, the value of $\phi$ increases, the polarising power increase and thus the anion's electron cloud gets attracted towards cation nucleus which tends to form a kind of overlap between the orbitals of cation and anion, which increase covalency and decreases ionic character.
So, according to Fajans' rules, increasing ionic character (i.e. more covalency to less covalency) is, $$\ce{LiF < NaF < KF < RbF < CsF}$$

thank you. I got it. I have to work on understanding Fajan's rule.


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Does a live electrical wire change the surrounding atmospheric chemistry?
Background: I'm world building an Earth-like world for a writing project. One of the goals of this project is to talk about changes in atmospheric chemistry across multiple scales ('nano'-scale through the entire atmospheric air column). My project has a scientist-type character to describe how atmospheric chemistry is affected by different processes. Example: Oxygen is released from leaves during the day, and carbon dioxide is released from leaves at night. Inversion layers occur regularly through the day-night cycle. Altitude impacts the density of air.
Here, I'm curious about the intersection of electricity and atmospheric chemical makeup.
Question: What chemical changes to atmospheric nitrogen and oxygen would/wouldn't occur in the presence of a wire carrying a current (not insulated)? (if any) Put another way, is the chemistry of the air around a live wire different in any way, shape, or form to the chemistry of the air around a wire with no current?
My internet research indicates a few possibilities but I'm not getting a firm sense of the reliability of the results. Example: Maybe ozone is produced by an electrical charge, or maybe it isn't - there seems to be conflicting ideas out there.
Here are some ideas I've seen:

charged particles in the air might naturally gravitate differently towards/away from an electrical current? This would be a change in the movement of existing molecules, perhaps physics rather than chemistry.
dinitrogen and dioxygen might become charged species (will these be positive or negative charges? Is the charge lost easily?
ozone might form?
Other species might form? nitrogen oxide species?

My best estimates for the current are:
~10000 volts and 120 milliamps in one portion of the writing project (electric fence); 120 volts and 20 amps in another portion (electrical cord.) I have some latitude here but those are ballpark estimates.

How much current are you talking about?


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Is is possible to Generate ~1ppm Copper in Solution from Elemental Copper?
I am designing a hydroponic system and need to control algae.  Copper is a known algaecide and lettuce actually needs some copper to grow well.  I had read stories about people using copper pennies in things like bird baths to reduce algae.  Pennies in the nutrient solution reservoir did not seem to affect the algae much.
I had a thought that by increasing the surface area, I might get a better result.  Fine copper wool is available for finish sanding, and I was thinking about running the nutrient solution through that to try to get some copper into solution.  The nutrient solution is ~pH 6 and 25C and has the (somewhat generic fertilizer) contents shown below at about a 1g/L (~2000uS EC) concentration.  The flow rate is ~4L/min, but only for about a 10 seconds every 300 seconds.  Another option is just to leave the copper in the water reservoir so that it is always in contact with the nutrient solution.
Is it possible to get ~1ppm of copper into solution from an elemental copper source in the nutrient solution?
It doesn't matter if it takes a week or so, as long as there is some chance of the copper going into solution.
EDIT:
I found this paper on Cu/Zn soluability vs pH.  It contains the graph below.
Is it true that assuming that Cu2+ ions have 63.546 g/mol weight, and 1ppm = 1mg/L, that 0.001g/L * 1/63.546 mol/g => 15uM solution, and thus the graph indicates that at pH 6.5 there will definitely be more than 1ppm of copper in solution at 25C?

Just buy better fertiliser.

@Mithoron Is my question edit sufficient for removing the hold?

This graph is for equilibrium with hydroxide not metal, also it suggests the opposite of what you state. And my earlier comment still holds...


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Does specific heat capacity of the container have an effect on the exactitude of a calorimetre?
I'm doing a lab on what is the best material for a calorimetre. I chose glass, aluminum, styrofoam cup, paper/cardboard cup and a plastic cup. I've found that the cardboard cup was the best followed by the styrofoam cup. I thought at first it was related to the specific heat capacity of the calorimetre itself, as those values lined up with the results I got from my experiment.
However recently I've been wondering if the specific heat capacity would really have an effect on the exactitude of the results? Should I base myself more on the thermal conductivity values of the materials even though there doesn't seem to be a correlation between those values and those I got from my experiment?

Of course it does, which is why you characterize it by running calibrations using the empty containers.

Ok thanks. For my experiment I repeated the same exothermic reaction so as to be able to calculate the error percentage between my experimental values and the theoretical values. I tried running calibrations with hot and cold water but my data didn't really make any sense and so wasn't of any use as the data was all over the place. As for the specific heat capacity, would a lower heat capacity, for example with the cardboard mean that more heat would be dissipated to the calorimetre instead of remaining in the solution?


text:
Is a low specific heat capacity for the material of the calorimeter good or bad?
I wish to ask if a low specific heat capacity of the material of the calorimeter means that the calorimeter would be more exact as it takes less heat for the material to reach equilibrium. I thought it would make more sense if it was the opposite and that you would want a substance with a higher specific heat capacity.

I would assume that if your absolute error in a measurement is fixed across all samples, higher heat capacities are better, since then the relative error in a measurement is lower.


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PTFE-to-PTFE or PTFE-to-Copper pressure seal
I am currently working on an engineering project with a friend. Without giving too much away, I have two parallel strips of Zirconium-Copper alloy sandwiched between an upper and lower piece of smooth PTFE. Between those two strips of metal runs a channel, and I wish to run liquid nitrogen (or another cryogenic coolant) in there.
My question is will a pressure seal alone (light pressure being applied from top and bottom uniformly) be enough to keep the liquid nitrogen in the channel? Will PTFE seal when pressed against itself, or when pressed against a metal?
To be clear, this is oil-free PTFE without any filler.

Some data on the materials:
PTFE data sheet
Zirconium Copper Alloy
An image of the setup. PTFE is in white, copper in orange... This is just a crude representation.

PTFE forms an air tight seal against glass (J. Youngs valves), so i would assume it should be fine with smooth metal as well as long as your design is correct (the one in the pic doesnt seem great for forming a seal). However the issue here is that due to the liquid nitrogen PTFE will shrink and you will lose the seal (this even happens in the freezer) so i dont think the above setup is suitable for liquid nitrogen. If you share more info on what you are trying to do i might be able to help.

Hi, thanks for the comment, this mostly answers my question. The image is really just a crude visual, in reality there are PTFE fasteners (nuts and bolts) going through both PTFE and copper, and tightening them together. My concern was that some materials really don't seal well, as well as some fluids really tend to go through the smallest gaps. But from what you commented, PTFE should have no issue forming a seal once the fasteners are in and tightened. I mostly wanted to avoid an additional gasket or sealant.

Actually, i should also have mentioned that there are PTFE swagelok ferrules and fasteners used to seal gases under pressure which is the best example of such use of PTFE as well. My only concern is that i dont think any of that will work well at very low temperatures.


text:
Can a magnesium atom combine with just one chlorine atom?
If I have a magnesium atom and a chlorine atom why can't the magnesium atom combine with just one chlorine atom? Wouldn't the magnesium atom become more stable, than it was before combining, as it gets closer to octet after combining with one chlorine atom?

I don't know how are you going to stop the second chlorine atom from attaching.

There isn't a second chlorine atom as i have mentioned in the question. I just want to know why wouldn't the Mg atom combine with one Cl atom.

If you had bare Mg and Cl atoms it gas phase then they would bind (if their energy wasn't too high), just like you say. You have a misunderstanding, probably based on  ordinary behavior  in more common situation.

So we can have MgCl atom, right? Is it true for all other elements?

MgCl molecule certainly exist but in gas phase and ordinarily also high temperature, so it's a bit exotic and reactive. What is true for other elements? Molecules without full octet are found for most of elements, but are often very reactive or  very high or low temperature is needed to make/detect them

Very little chemistry is done with single atoms. In almost all practical chemistry you have to worry about what all the other atoms in your system will do.

Be aware than many unusual and bizarre molecules and ions are know in context of astrochemistry of interplanetary/interstellar space, where atoms "get very bored" between collisions.

In solid state $\ce{MgCl2}$, can be approximately described assuming fully ionic bond, in which $\ce{Mg}$ atoms lose two electrons to form $\ce{Mg^2+}$, while a $\ce{Cl}$ atoms gain one electron to form $\ce{Cl-}$. To balance the two electrons released by $\ce{Mg}$, you need two chlorine atoms. The electrostatic force of attraction (aka the "ionic bond") binds them, and it has the formula $\ce{MgCl2}$.
Now, you wish to form $\ce{MgCl}$. To maintain charge neutrality, you'll want the $\ce{Mg}$ to lose only one electron. This means the valence shell of $\ce{Mg}$ would still have a single electron left in it. This is a very unstable configuration.
However, it is incorrect to say that $\ce{MgCl}$ cannot form. A research paper1 has explored the very $\ce{MgCl}$ molecule. Note that the conditions were very drastic, $\pu{500^\circ C}$ for pyrolysis and $\pu{2500^\circ C}$ for vaporization respectively. They ensured "the complete formation of $\ce{MgCl}$ molecular species, since $\ce{Cl}$ was the limiting reactant" as you had correctly identified yourself.
Another paper3 describes the formation of the $\ce{MgCl}$ molecule:

Excited $\ce{MgCl}$ molecules were produced by two different techniques: (i) at Orsay, they were created by mixing Mg vapor with a flow of $\ce{He/Cl2}$ and excited in a “heated” Schuller-type discharge tube, and (ii) at Waterloo, they were generated by using a copper hollow cathode lamp loaded with $\ce{MgCl2}$ powder.

Note that these techniques are indeed quite "unnatural", in the sense that you might not find $\ce{MgCl}$ naturally, but you can certainly create it through artificial conditions. In fact, $\ce{MgCl}$ is fairly popular otherwise. Several papers2 have already discussed the emission spectrum of the $\ce{MgCl}$ molecule. Some4 also explore its laser cooling. Hence, by no means it is an unknown compound and in fact, can definitely be formed.

References:

Rina Lourena da S. Medeiros, Sidnei de Oliveira Souza, Rennan Geovanny Oliveira Araujo, Djalma Ribeiro da Silva and Tatiane de A. Maranhão, Chlorine determination via MgCl molecule in environmental samples
using high resolution continuum source graphite furnace molecular absorption
spectrometry, Talanta, http://dx.doi.org/10.1016/j.talanta.2017.08.026
Darji, A.B., Shah, N.R., Shah, P.M. et al. Pramana - J Phys (1985) 25: 571. https://doi.org/10.1007/BF02847235
Vervloet, M. Fourier Transform Spectroscopy 2001. https://doi.org/10.1006/jmsp.2002.8514
The Journal of Chemical Physics 143, 024302 (2015); https://doi.org/10.1063/1.4926389

Aside from the reactive gas phase monochloride, magnesium is well known in the +1 oxidation state.  Magnesium(I) compounds are generally dimeric, $\ce{Mg2^{2+}}$ (similar to mercury) complexes with complex organic, anionic ligands.  The magnesium atoms are  covalently bound to each other with one valence, and the other valence is used for ionic bonding to the ligand.
An example is taken from the Wikipedia article linked above*:

The structure of Magnesium Complex by Makecat
Calcium has been made in the +1 oxidation state, too.
*This structure shows aromatic rings, which is misleading because of the ionic bonding between magnesium and nitrogen. There are six pi electrons in each ligand, but the ionic bond prevents them from being well-delocalized all the way around the ring.

This is a nice answer, so I apologize for nitpicking technicalities. May I assume the WP is an abbreviation for Wikipedia? https://en.wikipedia.org/wiki/WP Apart from that, it is not necessary to point to the source of an image in the public domain, but when you do, and for any other cases where attribution is required, a link under the word "source" is not the proper way to do it. When in doubt, apply scientific quoting and citing rules on Stack Exchange, too.

How should I source the image, then? If you can point me to an example I can aim for it.

Well, the image is in the public domain, so just use it. I think structural diagrams are not copyrightable anyway, I remember Loong writing that somewhere. In other cases, like with a creative commons license, attribution is required and the conditions are explained in their licensing information. Which is also the license our content uses. Something like the following: Title by User at Link (Date or version).


text:
What does this chemical formula mean?
Wikipedia says the formula of aluminum oxynitride is this $\ce{(AlN)_x·(Al2O3)_{1−x}}$,
$0.30 \leq x \leq 0.37$ . What does this formula mean?

I don't even understand it.

Go through Non-stoichiometric defects of Crystal Lattice , such as metal excess defects, metal deficiency defects etc. You may find your answer

@Hark Have you been taught the simplest of on-stoichiometric defects? Forget the $x$, instead, do you understand what $\ce{Fe_{1.98}O}$ means?

No i don't know about that

@GaurangTandon what does $\ce{Fe_{1.98}O}$ mean?  Usually I see $\ce{Fe_{0.947}O}$ or something similar.

@OscarLanzi Oh, sorry. I was about to write $\ce{Fe_{0.98}O}$ to highlight the non-stoichiometric metal deficiency defect. I made a typo by mistake.

OK.  So much for iron suboxide :-).

@Hark That just means non-stoichiometry. The composition varies according to crystal defects.


text:
Color of potassium permanganate and potassium dichromate
In $\ce{KMnO4}$, the Oxidation state of $\ce{Mn}$ is +7, which results in an electronic configuration as $\ce{[Ar] 4s^0 3d^0}$
In $\ce{K2Cr2O7}$,the Oxidation state of $\ce{Cr}$ is +6, which results in an electronic configuration as $\ce{[Ar] 4s^0 3d^0}$
As it is clearly visible there are no $\ce{d}$ electrons in either of these compounds but still, they are coloured.
My Doubt:

If there's no $\ce{d-d}$ transition in these compounds what gives these compounds their colour?
Are there other transition element compounds whose colour is not because of $\ce{d-d}$ transition?

See here.


text:
Acetic acid in vinegar
How do I calculate the number of moles of acetic acid present in 25 mL of a 10% vinegar solution?

10% solution means 10 g of solute in 100 mL of the solution.  So we have 10 g $\ce{CH3COOH}$ in 100 mL. Therefore , we have 2.5 g in 25 mL.
No. of moles of acetic acid = $\frac{m}{M}$
No. of moles = $\frac{2.5}{60}$ = 0.0417 moles of $\ce{CH3COOH}$.


text:
π-ligands in trigonal prismatic complexes
We recently had a discussion here about the symmetry problems for the often postulated s- and p-mixing with the d-orbitals in complexes to somehow allow a d-d-transition. It was pointed out back then that the s- and p-orbitals of the ligands do not possess the $t_2$ symmetry, which is required to mix with the $t_2g$-orbitals in an octahedral geometry.
I looked for some literature but could not find much. The only answer I came up with was that this was only true for pure σ-ligands, because at least the π*-orbital has the $t_2$ symmetry. That is why only π-type ligands influence the ligand field splitting on the side of the $t_2g$.
I think it was Mingos who wrote in his new book that you can predict this by looking at the dual body of your coordination geometry and only if the dual contains triangular faces it will leave no non-boding orbitals. I will finish that sentence with 'in a pure σ-complex' (I guess this is what he wanted to say).
So the dual body of the triangular prism is the triangular bipyramid. So this should leave no non-boding orbitals. And I also checked the character table for the corresponding point-group, all d-ortbials find a symmetric partner in the s- and p-orbitals of the ligands (if I'm not mistaking). So this leaves me with the question, whether π-donor/acceptor ligands have any influence on the ligand field splitting?
This should not be the case here, so for all ligands it's basically up to the electronegativity in the end?
EDIT: I checked it again and just saw that I jumped one row I think, it seems like the $d_xy$ and $d_yz$ orbitals can only be represented by the π-ligands.


text:
How to calculate RESP charges in a system with addition of user-defined extra-point charge in AMBER?
I would like to add in extra point of charge in my molecule system (EP) and calculate the resulting RESP charges. Parameters for the EP has been defined in a separate frcmod file. However, I am not sure on how to incorporate the EP into the RESP calculation. I have generated the .esp file beforehand from a QM calculation (Gaussian 09).
Any help would be greatly appreciated.
Thanks!
FYI, I am using AMBER16.
P.S. Somehow cannot post in the AMBER mailing list, hence asking here.

http://upjv.q4md-forcefieldtools.org/RED/ take a look through the documentation section. Hope it will help You

@alex personally I couldn't run RED  programs properly. They are quiting error again and again. I run "perl Ante_RED-1.5.pl file.pdb" after that it gave you guassian *.com files, I  run these gaussian files and then tried to run RED-vIII.52.pl, first it couldn't recognize gaussian log file and then I opened codes and read them, and change "guass-out.log" (which red program generated itself) Mol_red1.log, then I run again and this time it says: The initial P2N file can not be found I corrected and run again, this time gave different error. Is there a basic way for this?


text:
What is happening to my chlorine dioxide when I add too much bleach or citric acid?
I have some sodium chlorite and Chinese mystery powder which are meant to be mixed with water, then mixed together to form chlorine dioxide. I wanted to try a more food safe (and less mysterious) route for making small amounts of chlorine dioxide, so I diluted the sodium chlorite, then mixed in some bleach and a tiny bit of citric acid. I don't have a firm understanding of the reactions, but know $\ce{Cl2}$ and hypochlorous acid will react with sodium chlorite, but plain bleach may not, so the acid is to acidify the bleach and shift its equilibrium away from being hypochlorite. It worked, as evidenced by a pale yellow color. (It was very dilute, which is not a bad thing, for this chemical).
What I don't understand is that when I add more citric acid or more bleach, the color turns clear again. Either the chlorine dioxide is reacting or coming out of solution. What are the reactions with excess citric acid or bleach that might eliminate $\ce{ClO2}$?

How will we know if it's 'your' chlorine dioxide haha

The "mystery powder" was proposed to be sodium bisulfate, in a recent answer to another question. Sulfuric acid is used in one process to turn sodium chlorite into $\ce{ClO2}$; bisulfite would be slower, but should do the same.
Citric acid will acidify the sodium chlorite and sodium hypochlorite, but will itself be oxidized, perhaps all the way to $\ce{CO2}$. $\ce{HCl}$ is used to convert $\ce{NaClO2}$  and $\ce{NaOCl}$ to $\ce{ClO2}$ and citric acid could well do the same (be oxidized at the same time).
$$\ce{2 HCl + NaClO + 2 NaClO2 -> 2 ClO2 + 3 NaCl + H2O}$$
$$\ce{4HCl + 5NaClO2 -> 4ClO2 + 5NaCl + 2H2O}$$
Chlorite ion and hypochlorite ion are known to react rapidly to form chloride and chlorate (Lister, Can. J. Chem., 30, 879-889, 1952).
The unknown amounts of oxidizing agents and citric acid (a reducing agent) and the complex reactions make the pale yellow color a poor indicator of what might be going on. The presence of $\ce{ClO2}$ does not seem to be positively identified and might not even be needed to explain the observations.

And regarding the redox reactions, an excess of citric acid would use up the ClO₂ as it is oxidized? That makes sense. About the excess bleach, the hypochlorite could be being oxidized to chlorate (and using up the ClO₂)? Is the chlorate ion more stable so the reaction isn't very bidirectional? Because I was unable to correct it by adding more of either of the other two substances.

I found a reference to an alkaline equilibrium: 2 OH- + 2 ClO2 = ClO2- + ClO3- + H2O, and another in acid: 4 HClO3 = 4 ClO2 + O2 + 2 H2O (but it went on: If concentrated sulfuric acid is added to solid chlorate, the above reaction takes place, and usually the ClO2 formed explodes with great violence.) I didn't mean to say that no ClO2 was present, but that there could have been lots of compounds, especially from oxidation of the citric acid.

James please edit and correct  your cited reaction as it is NOT correct:2HCl+5NaClO⟶2ClO2+5NaCl+H2O. The action of HCl on OCl- forms HOCl + Cl-. With excess HCl, HOCl + HCl = Cl2 + H2O (sources: See https://en.wikipedia.org/wiki/Hypochlorous_acid ), The action of HCl, or Cl2, or HCl + OCl-, acting on ClO2-  are known paths to ClO2 (see  https://en.wikipedia.org/wiki/Chlorine_dioxide).

James, if you do not correct, I will downgrade as it is an egregious error to those intimately acquainted with this chemistry.

@AJKOER: Thank you for the correction! I must have messed up something from Lister. Interestingly, my edit was accepted by one but rejected by two referees, so you and I made the deciding votes. (My edit was posted to the referees as from "anonymous". Perhaps I should have expanded my reason for editing to include your comment.)


text:
Acid/base separation in nanofiltration
How do semi-permiable membranes handle salts that can separate into an acid and a base based on the "rules" of the membrane?
For example, say I have a nanofiltration membrane that rejects multivalent ions (eg $SO_4^{-2}$) but passes monovalent ones (eg $Na^+$) (common in nanofiltration membranes). If I supply a pressurized feed of an aqueous mixed-valency salt solution, say $Na_2SO_4$, what happens?
I can't find any literature that describes what happens in a simple case like this without other ions around. I can imagine two options are most likely:

The salt gets rejected. This is implied by lots of literature, but I can't find it explicitly stated.
The monovalent ion passes, in the above example creating $NaOH$ solution permeate and $H_2SO_4$ solution retentate. Evaluating the osmotic pressure in this case would be very odd as I guess you would have to include the disassociation energy of water.

What happens in this type of situation?


text:
Calculate the formula, and moles

If $\ce{1.84\times10^{-4}}$ moles of element A was completely reacted with element B to create $\ce{1.231\times 10^19}$ molecules of $\ce{(A3B2)_x}$ a hypothetical compound, what is the value of x?

I tried this:

$\pu{1 mol} = 6.02 \times 10^{23} ~\pu{molecules}$
$y~\pu{mol} = 1.231 \times 10^{19} ~\pu{molecules}$
$y = \pu{2.0449 \times 10^-5 mol}$

$\ce{(A_3B_2)_x} : 1.84\times10^{-4}+ {2\over 3}(1.84\times10^{-4})=3.067\times10^{-4}$
So it produces $3.067\times10^{-4}$ mol of the compound

and, $3.067 \times 10^{-4}~\pu{mol} \div 2.0449 \times 10^{-5}~\pu{mol} = 15$
And since 1 compound has 5 atoms, $15 \div 5 =3$
Therefore, $x$ is 3

I am not sure if I did this question correctly. Even if I did, is there a more efficient and organized way of solving this problem?

Can you describe what exactly you did below the first horizontal line? ("1 mol = 6.02×1023 molecules;x mol =1.231×1019 molecules;x = 2.0449×10−5 mol") what are you trying to do by that?

well, 1.84×10−4 is in moles, while  1.231×1019  is the actual number of molecules. So I'm trying to convert number molecules into moles

@didgocks use different variables. “x mol = $\mathrm{1.231 x 10^{19}}$ and we also have $\ce{(A3B2)_x}$ “.

At first you said that $x = \pu{2.0449\times10^−5 mol}$ and then later said $x=3$. It is clearly wrong even without going into the maths!
That said, I believe you are confusing the stoichiometry of the compound with the number of moles of the compound produced. The $x$ in $\ce{(A3B2)_x}$ is a part of the chemical formula for that compound, and not the moles of $\ce{(A3B2)}$ produced.
For example, $\ce{AlCl3}$ usually exists as a dimer in gas phase, so we write it as $\ce{Al2Cl6}$ or simply $\ce{(AlCl3)2}$ for clarity. You should hence realize that the $x$ in $\ce{(A3B2)_x}$ represents the number of $\ce{A3B2}$ molecules that have polymerized together to form a single molecule. And the question states that $\pu{1.231\times10^19}$ of such molecules was produced.
With the conceptual clarity attained, we'll now solve the numerical. Note that the question says that all the moles of elemental A were obtained in the compound. This implies:
$$
\begin{align}
\text{moles of A in the compound} &= \text{original moles of elemental A}\\
\implies 3x\times(\text{moles of compound }\ce{A3B2})&=1.84\times10^{-4}
\end{align}
$$
Solving this, you'll get approximately $x=3$, which is the expected answer.


text:
Which melting point would indicate a metallic structure?

Question
  If you have three substances with three different melting points;

445 °C
1907 °C
-7 °C

Which of these would most likely have a metallic structure?

I understand that the third option would most likely be covalent. But between the first and second option, there are many bonds which have a similar melting point to both answers. Which of these would most likely indicate a metallic strucure to be involved.

Now, before we get to the "expected theoretical" answer, I should make one point clear: predicting the solid state just based on the melting point data can not be perfect! Browsing the data on melting points on Wikipedia, you'll find that vanadium has a melting point of $\pu{1910 ^\circ C}$, while zinc has a melting point of $\pu{420^\circ C}$. Both these values are very close to the options (1) and (2), hence, one can very well argue that both the options represent a metallic structure.
But this is definitely not the answer your teacher expects. A theoretical (and again, practically wrong) approach instead is to realize that the values of intermolecular forces increase in the order of molecular solid < metallic solid < ionic solid < covalent networked solid. Thus, the option (3) is probably a molecular solid, option (1) can be either an ionic solid or a covalent networked solid, while option (2) must be the metallic solid as expected.

I understand that the third option would most likely be covalent.

Just get rid that misconception already. Covalently bonded networked solids have higher melting points than several metals and even ionic solids! Diamond melts at $\ce{3800 K}$, while boron melts at $\ce{2349K}$. Our beloved ionic solid sodium chloride melts at a poor $\ce{1073K}$, while all the 3d series metals melt under $\ce{2200K}$! Hence, the assumption, that low melting point indicates a covalent solid, is false.

And lets not forget gallium or mercury (+27C, -37C), both of which are decidedly metallic. The bottom line seems to me to be that the question given by the teacher is, well, horrible.


text:
Increasing concentration of products in a reversible reaction
My teacher says it is not possible to add more products to increase its concentration. This is because products are complex ions. Maybe I just misunderstood what she said. Can somebody verify and explain this to me?

More details would help in order to answer this question. As it stands, it's unclear what's being asked.


text:
Is silicon dioxide more acidic than the oxides of nitrogen?
Is silicon dioxide more acidic than the oxides of nitrogen? How to find acidic strength of compounds which don't have hydrogen atoms?

https://chemistry.stackexchange.com/questions/22189/how-to-decide-acidic-or-basic-nature-of-oxides

Possible duplicate of How to decide acidic or basic nature of oxides

I think your question boils down to:
What is the meaning of acidic strength of oxides? How do we determine the acidic strength of oxides?
In general, the electropositive character of the oxide's central atom will determine whether the oxide is acidic or basic. The more electropositive the central atom, the more basic its oxide. The more electronegative the central atom, the more acidic its oxide.
Electropositive character decreases from left to right across the periodic table and increases down a group.
The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (for example, aluminium oxide) in the middle.
This trend applies only to the oxides of the individual elements  in the highest oxidation states for those elements. The pattern is somewhat less clear for other oxides.
A non-metal oxide's acidity is defined in terms of the acidic solutions formed as a result of reaction(s) with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.
In general, acidic oxides are oxides of nonmetals and basic oxides are oxides of nonmetals.
Note: There are three nonmetal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have low (not the highest possible) oxidation numbers for the central atom that they give neutral aqueous solutions.
Since the acidity of a cation rises rapidly with charge, d-block elements which exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties.
In general, the higher the oxidation number, the more acidic the corresponding oxide (similarly, reverse trend for basic character of oxides). Chromium makes a decent example here, as it exhibits a variety of oxidation states. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and $\ce{CrO3}$ is acidic.

P.S.
Let me introduce you to a theory developed by Hermann Lux and Håkon Flood, named Lux-Flood theory to explain the acidic or basic character of oxides. Here's what they had to say:

An acid of Lux-Flood is an acceptor of $\ce{O^2-}$
  A base of Lux-Flood is a donor of $\ce{O^2-}$

For example:
$\ce{CaO}$ is a base of Lux-Flood because $\ce{CaO}=\ce{Ca^2+}+\ce{O^2-}$
$\ce{SiO2}$ is an acid of Lux Flood because $\ce{Si}$ has vacant d orbitals, allowing it to accept more $\ce{O^2-}$ ions.
The two compounds may react to give you $\ce{CaSiO3}$.


text:
Writing disproportionation half reactions

Question
  Write the half equations for the following reaction:
  $$\ce{3OCl- -> 2Cl- + ClO3-}$$

I know that $\ce{Cl}$ is reduced from an oxidation state of 1+ to 1- (in $\ce{Cl-}$) and is oxidized from an oxidation state of 1+ to 5+ (in $\ce{ ClO3-}$)
Below is as far as I have gotten:
$$\ce{3OCl- -> ClO3- + 4e}$$
$$\ce{3OCl- + 6e -> 2Cl- }$$
I am sure this isn't correct, as I can't balance it and together they do not add up to the full equation

How did you go about balancing the reactions?

@Avnish I tried to multiply each half reactions so that they both had 24 electrons.

The atoms of an element need to be balanced in the half reactions as well.

@EllaLewis Avnish is right. Also, we need to know the medium of the reaction (acidic or basic) to be able to completely balance the half reaction. Was it given in your book?

Without supplying the details of the reaction medium, this question seems unanswerable.

Hint:
$$\ce{2OCl- + 4e -> 2Cl- + 2[O] }$$
$$\ce{2[O] + OCl- -> ClO3- + 4e}$$
P.S.
$$\ce{2[O]  -> O2}$$ is likely to happen, if [O] (nascent oxygen) is not used up in some reaction.
Here, as soon as it is released in the first reaction (reduction), it is used up in the second (oxidation).
Hope that helps.

How can you balance the reaction without knowing the medium of the reaction (acidic or basic)?

If it's not mentioned, this is the best we can do.

Yeah, but still it is a point worth noting. The answer will differ for both mediums.

Yeah, you're right!


text:
How to find the bond order in delocalized pi bonding of more than two atoms?
J.D Lee has written down some rules for calculating the bond order.

I have no difficulty following step 1 and 2. It's easier to work with an example so I'll highlight my issues with step 3 , 4 by using $\ce{CO_3^{2-}}$ .
It's been used by the book.

Here's my thought process

The structure is trigonal planar
Number of electrons; 24

Now, the parts I have difficulty handling

3$\mathrm{\sigma}$ bonds .They've written that oxygen has four non-bonding electrons.
Is it because oxygen's valency is 6 and after assigning the $\mathrm{\sigma}$ bonds.Each oxygen gets 2 of bonding electrons and a leftover of four which are satisfied as the non bonding electrons?
I follow that 4 molecular orbitals will form but how is it decided how many bonding , non-bonding and anti-bonding ?

Point 3)
Why some electrons are $\mathrm{\sigma}$-bonding, $\mathrm{\pi}$-bonding or non-bonding is easier to understand if you first think about the orbitals involved. I'll discuss it in terms of hybridisation.
The central $\ce{C}$ atom has an $\mathrm{sp^2}$ hybridisation (corresponding to a trigonal planar structure) so it presents three $\mathrm{sp^2}$ orbitals, one towards each of the $\ce{O}$ atoms, plus one non-hybridised $\mathrm{2p_z}$ orbital, perpendicular to the plane of the molecule. $\ce{O}$ atoms, for their part, we can picture as aligned towards the $\ce{C}$ atom - so, for each $\ce{O}$ atom, their $\mathrm{2p_x}$ orbital will point to the central $\ce{C}$ atom, their $\mathrm{2p_z}$ orbital will be perpendicular to the plane of the molecule (and so parallel to the carbon $\mathrm{2p_z}$ orbital), and the remaining $\mathrm{2p_y}$ orbital will be in the plane of the molecule but perpendicular to the $\ce{C}$-$\ce{O}$ axis.
Each of the $\mathrm{sp^2}$ orbitals has symmetry properties suitable for $\mathrm{\sigma}$-bonding with the $\ce{O}$ atoms - they are axially symmetric around the line connecting the $\ce{C}$ and $\ce{O}$ nuclei and have no nodes. They are aligned and overlap with the $\mathrm{2p_x}$ orbitals of the oxygen atoms, allowing for effective bonding. So we can build three $\mathrm{\sigma}$ bonds that take up 6 electrons, and we have 18 electrons left.
Two of the orbitals in each $\ce{O}$ atom cannot overlap in a bonding way: the $\mathrm{2s}$ orbitals because they are too small and localised, and the $\mathrm{2p_y}$ orbitals because they have opposite sign lobes at each side of the $\ce{C}$-$\ce{O}$ axis - so the overlap with the carbon orbitals is zero. These two orbitals per $\ce{O}$ atom will hold two pairs of non-bonding electrons - for a total of 12. So, we have 6 electrons left.
Finally, we have four $\mathrm{2p_z}$ orbitals (one on the carbon atom, three on the oxygen atoms) which have a symmetry suitable for $\mathrm{\pi}$-bonding (they are parallel, with the nodal plane in the plane of the molecule). Combining them will require a molecular orbitals approach and take us to point 4.
Point 4)
So, we have 4 suitable $\mathrm{2p_z}$ orbitals and 6 electrons left. Combining these to form $\mathrm{\pi}$ molecular orbitals is a bit more complex, though, as symmetry plays a very important role, so outside from simple cases, the results may not be immediately obvious.
The three oxygen $\mathrm{2p_z}$ orbitals, under the trigonal planar symmetry of the molecule, form three symmetry adapted linear combinations. These three symmetry adapted versions all have the same energy as the separate $\mathrm{2p_z}$ orbitals, but they have symmetries that are compatible with the trigonal planar molecule, which is something we need if we want to see if bonds can be formed. Since they all have the same energy as the three equivalent $\mathrm{2p_z}$ orbitals, they will also be degenerate.

Bird, P. Chemistry 241 - Inorganic Chemistry I. http://faculty.concordia.ca/bird/c241/notes_ch3-shriver.html (accessed 19/03/2018)
So now we wonder: how do these symmetry adapted orbitals interact with the lone carbon $\mathrm{2p_z}$ orbital?
It turns out, one of those orbitals (identified in the image as $\mathrm{\Psi_1}$) has a non-zero overlap with the carbon $\mathrm{2p_z}$ orbital, while the other two ($\mathrm{\Psi_2}$ and $\mathrm{\Psi_3}$) have an overlap of zero (because of signs and symmetry). And since overlap is a requisite for bonding, these two linear combinations will be non-bonding orbitals (i.e. they will have the same energy in the $\ce{CO_{3}^{2-}}$ molecule as they would have in the isolated oxygen $\mathrm{2p_z}$ orbitals).
So, what about the symmetry adapted linear combination that did have the correct symmetry to interact with the carbon $\mathrm{2p_z}$ orbital? They do interact, forming two combinations: one which is more stable than the independent orbitals, and therefore a bonding molecular orbital, and another that is less stable and therefore anti-bonding.

Bird, P. Chemistry 241 - Inorganic Chemistry I. http://faculty.concordia.ca/bird/c241/notes_ch3-shriver.html (accessed 19/03/2018)
So, we have four molecular orbitals out of our four atomic $\mathrm{2p_z}$ orbitals: in ascending order of energy, those would be one bonding orbital, two degenerate non-bonding orbitals, and one anti-bonding orbital. We also had 6 electrons yet to assign - which will fill the bonding and two non-bonding orbitals. Those non-bonding orbitals are the molecular equivalent of "lone pairs" - so they contribute nothing to the bond order. The fully-occupied bonding orbital does; so we have a single $\mathrm{\pi}$ bond shared by the four centres, leading to a bond order of $\mathrm{1}$ (for the $\mathrm{\sigma}$ bond) + $\mathrm{\tfrac{1}{3}}$ (for the $\mathrm{\pi}$ bond) = $\mathrm{1 \tfrac{1}{3}}$.
It is important to stress that it's three bonds of order $\mathrm{1 \tfrac{1}{3}}$, and not one bond of order $\mathrm{2}$ and two of order $\mathrm{1}$: the interaction between the central $\mathrm{2p_z}$ orbital and those on the oxygen atoms required adapting them to the trigonal planar symmetry, and, as you can see in the images above, all three oxygen atoms contribute to the bonding and anti-bonding molecular orbitals.


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Chemical equilibrium and stoichiometry
In reactions that go until the end the stoichiometric number indicates the exact amounts of reactants and products you will get. However, this is not the case in equilibrium reactions and you will get a different amounts (concentrations) than what exhibits the stoichiometric numbers.
$$\ce{Na + Cl -> NaCl}$$
You will get in the end 1 mole of NaCl from 1 mole of Na and one mole of Cl.
$$\ce{2SO3<=>2SO2 + O2}$$
In equilibrium you will not get 1 molar of $\ce{O2}$ and 2 molar of $\ce{SO2}$ from 2 molar of $\ce{SO3}$.
Then what do the stoichiometric numbers indicate in equilibrium?

They indicate that for each mole of O2 you will get two moles of SO2.

How is it true? You will get totally different mollar concentrations based on Kc. I can't catch the issue exactly.

This is true because the stoichiometric coefficients tell you so. Yes, you might get totally different molar concentrations. But whatever they might be, the concentration of SO2 will be twice that of O2.

Let us take the example of your equation:
$$\ce{2SO3<=>2SO2 + O2}$$
The stoichiometric coefficients represent two facts:

If 2 mole $\ce{SO3}$ reacts, it will give 2 mole $\ce{SO2}$ and 1 mole $\ce{O2}$ but one needs to remember here that all of the 2 moles will not react. But the number of moles that do dissociate, give products in the above ratio i.e. if initially have 2 moles of $\ce{SO3}$ and degree of dissociation is 50%, then we will have 1 mole of $\ce{SO3}$ reacting to give 1 mole of $\ce{SO2}$ and half mole of $\ce{O2}$
The stoichiometric coefficients also provide the powers to which the concentrations of each compound will be raised to in the expression for equilibrium constant $$K_\mathrm c = \frac{[\ce{SO2}]^2[\ce{O2}]}{[\ce{SO3}]^2}$$

OK thanks. The soft spot is that not all the reactants dissociate however the stoichiometric coefficients do represent the rations between the substances.


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Methodology and basis set for study in UV-Vis
I am currently working with a system of 30 atoms of a organic molecule. I need to make a study related to determine the UV-Vis graphic as precisely as possible.
I am using DFT method. However, I am having trouble to find a methodology and basis set to obtain a proper result of the experiment.
Does anybody know what methodology and basis set I should use?
P.S: I have already tried B3LYP with 6-31G(d), 6-31G(d,p) and 6-311++G(2d, 2p) in sequence but the results weren't good.

Simple functionals and Pople-style basis sets are notoriously bad for UV/Vis.  DFT and TD-DFT in general aren't that great, unfortunately, AFAIK, but there may be some decent options. I'll ping an expert in chat to come take a look.

Ooof. Said expert is in the middle of an international move, and not reachable in chat at the moment.

Look for benchmark papers from Denis Jacquemin.

I think you need to be more precise about "weren't good." There are many benchmarks, but results are generally accurate to ~0.2 eV. Are you including a solvent model, for example?

The other question is whether you're just trying to get low-lying $\pi-\pi^*$ transitions or you want CT or Rydberg states.

If you can provide more details, it would be helpful.


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When can a Secondary standard solution be used for standardising something instead of a Primary?
I am writing an assignment on 'Preparing and Using Standard solutions for Use in Analysis in Titrations and Spectroscopy' and one part is asking about primary and secondary standard solutions.
I know the different between them and have stated what they are and some examples, but one question I have to answer is 'Explain why Primary standard substances are not always used in standardization, and why Secondary standard substances are used instead'.
I have looked at many sources online and they all just tell me the differences between primary and secondary, was wondering if anyone could shine some light on why a secondary standard solution would be used to standardize a solution instead of a primary?

If you are doing pH titrations, and your primary standard is an acid - say - potassium hydrogen phthalate, you can't use that to standardize an acid directly. Instead, you would make a secondary standard of - say - sodium hydroxide, titrating against your primary standard. Then, you are ready to standardize the acid you want to use in your titration.
Secondary standards are also used if the primary standard is expensive. Hydrochloric acid solutions are cheap and stable, but you can't make them by weighing in a solid. So instead, you make an expensive primary standard and a lot of cheap hydrochloric acid of unknown concentrations, and determine the titer of the hydrochloric acid. Then, you can use it as a secondary standard.


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How to experimentally measure number of protons or electrons in an element?
How can one measure the numbers of protons or electrons (the atomic number) in a substance?
Like if the element I examine is carbon I expect the result to be 6.
Preferable a simple experiment which can be done in the kitchen with kids attending (cheap and safe). Any atoms will do. A mass spectrometry is not available.

Moseley measured the atomic number in 1913 with relatively modest equipment (by modern standards) but I doubt that you could reproduce his experiments in a typical kitchen.
Atomic number at Wikipedia
Expansion as suggested in comments.
Here is a quote from that article of what he did.
"Moseley measured the wavelengths of the innermost photon transitions (K and L lines) produced by the elements from aluminum (Z = 13) to gold (Z = 79) used as a series of movable anodic targets inside an x-ray tube."
It might be just possible at home if you really knew what you were doing but I doubt that it would be cheap or safe.  What is regarded as safe has probably changed a lot in the last 100 years.


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Why does acetic acid (CH3COOH) conduct electricity the worst in comparison to HBr, HCOONa and NaNO3
Which of the given substances (in an aqueous solution) conducts electricity the worst: $\ce{CH3COOH}$, $\ce{HBr}$, $\ce{HCOONa}$ and $\ce{NaNO3}$?
I've found this assingment included in the 2017 national chemistry exam paper, and the published solution is that the correct answer is $\ce{CH3COOH}$.
I would be very grateful for a thorough explanation behind the science of this assignment. Thank you for the time taken.

What did you try?

HBr, NaNO3 and HCOONa are all fully ionised in aqueous solution so all of their material is available as ions to conduct electricity.
Acetic acid with a pKa of 4.75 is not fully ionised so the solution will be a poorer conductor

It's because acetic acid won't fully ionize in solution, it's a weak acid. Less ionization means less total amount of ions in solution, which implies lower electrical conductivity. The wording of the question is bad IMO however, because it already assumes you are comparing over equal molar concentrations of each solution. If the question was about mass concentration the analysis would have to be different.


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Which set of reactants is better for preparation of 1-methoxy-4-nitrobenzene?
I've got two set of reactants:

4-nitrobromobenzene + sodium methoxide
sodium 4-nitrophenoxide + bromomethane

I've got to tell which reaction is more preferable among these two.
The first reaction takes place through nucleophilic addition of methoxide on the ring followed by elimination of bromine. I've been taught that such reactions take place under drastic conditions, but the presence of an electron withdrawing group at ortho and para position increases the reactivity, so the presence of $\ce{NO2}$ group should increase the reactivity .
The second reaction takes place by SN2 mechanism, where bromomethane is attacked by the nitrophenoxide ion. This reaction "seems" to be better than the first. (Is it?)
Hence, my attempt is that the second reaction should be more favourable than the first since the first reaction should take place under more drastic conditions than the second.
Am I right? Or does the presence of $\ce{NO2}$ group increase the reactivity so much that it is comparable to the second reaction or maybe better?

Drastic conditions for reaction 1? Not really. Reflux the reactants in $\ce{MeOH}$ and the reaction should go smoothly and in high yield.
Reaction 2 - two difficulties.
The anion of 4-nitrophenol is very stable and, because of the electron withdrawing power of the nitro group, it is not very nucleophilic. Second difficulty is that bromomethane has a boiling point of 3C so the reaction mixture has to be kept cold to keep it in the reaction mixture. This does not help you overcome the poor nucleophilicity of your phenoxide.
Without running the two reactions it is guesswork to predict which will give the better outcome, but I favour the SNAr.

My book says that reaction 2 is more favourable. (There's no explanation given , just the answer) . I have a reference book and it says that both the reactions are favourable so I'm kind of confused .

So now you have two ways of making the product. That's synthetic chemistry for you. There is often more than one viable way of making a product.

If you have the reagents for one reaction and not the other, then the issue is settled.

Path1 is by SnAr mechanism where Br from benzene is substituted by an nucleophile , but with loss of aromaticity by intermediate , a carbanion.Path 2 is an SN2 reaction  of aliphatic halo compound.This does not involve loss of aromaticity and therefore could be a better path.


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Size consistency in computational chemisty methods
I am looking for a clear answer that would explain what exactly size consistency is, why configuration interaction (CI) is not size consistent and how this leads to errors when using CI for large systems.

What is your current understanding of size consistency? Is there some particular part of the definition that is confusing?

Size consistency is essentially the idea that if you were to determine the energy of two noninteracting systems A and B, you should obtain the same result if you sum the energy of A and B separately and if you calculate the energy of A and B as one supersystem (usually with A and B separated by a large enough distance where they don't interact). Expressed mathematically, we might write: $$E(A+B)=E(A)+E(B)$$ provided that A and B don't interact.
As to the second part of your question, I want to address a finer point. Full CI is size consistent; the issue comes when you have to truncate CI to make it computationally feasible. Truncated CI is only exactly size consistent up to the point of truncation. For example, CISD (CI with single and double excitations) would only be exactly size consistent for systems with $N=2$ electrons.
The issue becomes a problem for large systems because this lack of size consistency leads to the wrong limiting behavior of the correlation energy. Specifically, truncated CI leads to the false conclusion that $\lim_{N\rightarrow \infty}\frac{E_{corr}}{N}=0$, while the correlation energy should vary linearly with $N$.
For more information on this, I would read chapter 4, section 6 of Szabo and Ostlunds Modern Quantum Chemistry. The chapter as a whole gives a thorough primer on CI, with this particular section discussing the size inconsistency problem.


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the product form between phosphorus(V) and ethanol
How do I know the main product of the reaction between phosphorus(V) chloride and ethanol?

You could try googling "reaction of PCl5 with alcohols"
In which case you might find this https://www.chemguide.co.uk/organicprops/alcohols/halogen.html  or this  https://crab.rutgers.edu/~alroche/Ch11.pdf
or even this  Reaction of alcohols with PCl5 and PCl3
TL:DR  The product is ethyl chloride and POCl3

But I was thinking, can't the product formed be chloroethane? Or what else do you think?

Ethyl chloride and chloroethane are the same thing!


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Is an allene or a conjugated diene formed?
If we dehydrate the following molecule, which hydrogen ($\ce{H1}$ or $\ce{H2}$) gets removed?

The conclusion that I have drawn from the molecule: The carbon to which $\ce{H1}$ is attached is $\ce{sp2}$ hybridized which makes $\ce{H1}$ slightly more acidic than $\ce{H2}$.

I would go for the conjugated product .

As the intermediate passes through an allylic carbocation which is fairly stable, it does not matter which H is 'more acidic' - the whole system is in equilibrium. The product you get is the thermodynamically most stable which will be a conjugated diene.


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Is there a fixed corresponding orbital for every specific magnetic quantum number?
Does each value of $m$ have a fixed orbital (given the value of $l$ as well as $n$)?
For example, if I know that $n = 2$, $l = 1$, and $m = -1$, which orbital ($\ce{p_x}$, $\ce{p_y}$ or $\ce{p_z}$) does it represent?

They are assigned arbitarily, no significance whatsoever . For diagrams the the orbital along the x axis will be $p_x$ though similarly for y, z.

It is not arbitrary: $m = -1$ is a linear combination (a quantum superposition) of $p_x$ and $p_y$ orbitals. I agree this is a duplicate.

Actually, $n=2 ,l=1$, and $m=0$ represents $\ce{p_z}$ orbital. But if you look carefully into the wave functions, you will find that $\ce{p_x}$ and $\ce{p_y}$ orbitals are additive and subtractive linear combinations of $\ce{$\psi$_{2,1,1}}$ and $\ce{$\psi$_{2,1,-1}}$ respectively.
The wavefunction for  $n=2,l=1,m=0$ is, $$\ce{$\psi$_{2,1,0} = Are^{-r/2a_0}cos$\theta$}$$ You can see that this wave function is proportional to $\ce{rcos$\theta$}$ which is actually $z$ in polar coordinates. That's why due to this proportionality to $\ce{rcos$\theta$ or z}$, this $\ce{$\psi$_{2,1,0}}$ corresponds to $\ce{p_z}$.
But if you look into $\ce{n=2,l=1,m=+1}$ or $\ce{n=2,l=1, m=-1}$ wavefunctions, $$\ce{$\psi$_{2,1,1} = A're^{-r/2a_0}sin$\theta$e^{i$\phi$}}$$$$\ce{$\psi$_{2,1,-1} = A're^{-r/2a_0}sin$\theta$e^{-i$\phi$}}$$But if you take additive linear combination of the above two, you will get $\ce{e^{i$\phi$} + e^{-i$\phi$} = 2cos$\phi$}$, and you will get the resultant wave function proportional to $\ce{rsin$\theta$cos$\phi$ }$ which is actually wavefunction for $\ce{p_x}$ orbital( as $\ce{x=rsin$\theta$cos$\phi$ }$ in polar coordinates)If you take subtractive combination , $\ce{e^{i$\phi$} - e^{-i$\phi$} =2sin$\phi$}$, and you will get resulting wave-function proportional to $\ce{rsin$\theta$sin$\phi$}$ which is the desired wave function of $\ce{p_y}$ orbital (  as $\ce{y=rsin$\theta$sin$\phi$ }$ in polar coordinates. So, you can see $n=2,l=1,m=-1$ doesn't correspond to any of the three orbital . $m=0$ only corresponds $\ce{p_z}$ orbital. $\ce{p_x or p_y}$ orbital have wavefunctions which are linear combinations of the actual solution of Schrodinger's solutions $\ce{$\psi$_{2,1,1}}$ and $\ce{$\psi$_{2,1,-1}}$(i.e $m=1$ and $m=-1$ orbitals)


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Why do alkanes have higher boiling point than their ether counterparts?
Based on my understanding of inter-molecular forces, I expect dipole-dipole interactions to be significantly stronger than van der Waal's interactions. Hence, I expect ethers (which obviously have dipole-dipole interactions) to have higher boiling points (as boiling point $\propto $ inter-molecular forces) than alkanes (assuming approximately the same molecular mass). But, the experimental data suggests otherwise.
$$
\begin{array}{|c|c|c|c|}
\hline
\textbf{Compound}&\textbf{Boiling point / °C} &\textbf{Compound}&\textbf{Boiling point / °C} \\ \hline
\textit{n}\text{-pentane}  & 36.1   & \text{ethoxyethane}  & 34.6    \\ \hline
\textit{n}\text{-heptane}  & 98.42  & \text{1-propoxypropane} & 90 \\ \hline
\textit{n}\text{-nonane}   & 151    & \text{1-butoxybutane}  & 141   \\ \hline
\textit{n}\text{-undecane} & 196    & \text{1-pentoxypentane}& 184   \\ \hline
\end{array}
$$
As evident, experimental data suggests that alkanes are on par with ethers when it comes to inter-molecular forces, infact the former are slightly higher.
I tried to reason this discrepancy as follows. Ethers have weak dipole, so their dipole-dipole interactions will also be weak (though stronger than their van der Waal's interactions). Alkanes on, the other hand, have van der Waal's forces only but I suspect that size is playing a dominant role here. Ethers have a v-shaped bent structure, and therefore, have reduced surface area as compared to alkanes. Therefore, I believe that alkanes lead over ether in this case and this compensates the weak dipole of ethers. Though this is only a hypothesis from my part.
But there is another problem:
\begin{array}{|c|c|c|c|}
\hline
\textbf{Compound}&\textbf{Boiling point / °C} &\textbf{Compound}&\textbf{Boiling point / °C} \\ \hline
\text{propane} & -42 & \text{methoxymethane} & -24 \\ \hline
\end{array}
Now at this point I am out of ideas. From $\ce{Et-O-Et}$ to $\ce{Pe-O-Pe}$ we had ethers leading over alkanes which felt odd but when we compare di methyl ether and propane, everything just seems fine with ether having the dominating lead (as expected initially). What is real cause? Am I missing some important concept here? Or is there any other factor that I am not considering?

I did not intentionally skip decane and octane, it's just that there is no symmetrical ether counterpart of these (I am avoiding asymmetrical ethers).

Why did you avoid non-symmetrical ethers? That might give you important insight into why this trend is such. Also, polyethers. Someone actually studying this problem would probably have looked at different permutations.

I actually did look at non-symmetrical ethers, but the trend was way too irregular. I had to consider many possibilities like one carbon difference non-symmetrical ethers, 2 carbon difference, etc. It extensively messed up my data and no conclusion could be drawn. Their irregular dipole is also a problem both understanding and (i suppose) explaining. Under these consideration, I thought I would first keep my observations confined to the simplest case, once I get a strong theory I would expand it to other cases.

I suspect London forces dominate in the longer hydrocarbons, whereas they are disrupted by the ether bond. For shorter hydrocarbons, the dipole of the O-C bond is greater than the London forces, but this is reversed as the # of carbons increases.

Isn't this more of a observation rather than a concrete theoretical reason?

I don't recognize the distinction between observation and theory in your (?) question. The bp data you've tabulated is observational/empirical. The theory is that cooperativity of Van der Wals forced exceeded the energy of the dipole interaction for values on n, where n = # of carbons, with n being

@AdamWorth I believe that your comment is a potential answer, can you please elaborate further, explaining why the Van der Waals forces exceeded the dipole interactions?

I second that the Van der Waals forces will dominate, because the dipole of the symmetrical ethers is perpendicular to the long axis of the molecule.  In this case the number of electrons over a large surface area is going to play a bigger role in holding the molecules together.
Oxygen has the same number of electrons as -CH₂-, but  because oxygen is more electronegative, it’s holding those electrons more tightly, making it less polarisable, and reducing the magnitude of the induced dipoles.
I suspect that ᵀ he effect is less apparent for dimethyl ether because the polar C-O bonds play a bigger role because the partial-positive charges on the carbons is not attentispread down the carbon chain, muting it, but remains a concentrated, bare, bright spot of charge, which can interact with the lone pairs on other ether oxygens. Additionally, there is less steric hindrance from the floppy alkyl groups. To prevent this kind of dipole-dipole attraction.


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Relative importance of electronegativity and bond length for dipole moment magnitude
In this answer to the question "Why is the dipole moment of chloromethane larger than the dipole moment of fluoromethane?", the order of the dipole moments was given as $\ce{CH3Cl > CH3F > CH3Br > CH3I}$, with an explanation that bond length is also a contributing factor; specifically, electronegativity dominates for $\ce{Br/I}$ and bond length/order dominates for $\ce{F/Cl}$.

Can more detail be given as to why there is this division between $\ce{F/Cl}$ and $\ce{Br/I}$?
In general, is there a method to determine which effect (electronegativity or bond length/order) is more important for comparing the magnitude of dipole moments?

These anomalies usually happen in compounds having period 2 elements.

Are you referring to introduction of d-orbitals? If yes, how does that help here?

Not d-orbitals in this case, but the small size of flourine atom causing interelectronic repulsions among its own electrons.

I need help understanding, in that trend specifically, as to when the length factor dominates, and when the electronegativity factor does?

There is nothing much to understand here. All you need to understand is that this is basically an experimental observation.
We are trying to use qualitative arguments (higher bond length/more electronegativity) to explain the actually calculated values. When we observed a contradiction in our answer for fluoromethane and chloromethane, we corrected our reasoning, by saying that the bond length dominated over the electronegativity difference in this case. And that is true, because this correction is based on the calculated values only.
So, we first try to predict the order by our own qualitative arguments (logically reasoned, of course). If they work, great! If they don't, we look into the factors that we originally deemed to be minor instead. We calculate their values, and see what effect they had. Then, we justify those minor factors by saying that they dominated over the other in the particular case.
Another such interesting case is the order of the basicity of ammonia derivatives in water. It is $\ce{NEt2>NEt3>NEt1}$ (for the ethyl derivatives) and $\ce{NMe2>NMe1>NMe3}$ (for the methyl derivatives). Here, three factors are operating simultaneously! The solvation effect, the steric hindrance of alkyl groups, and the positive inductive effect of alkyl groups - all either reinforcing or hindering each other. In such edge cases, it becomes extremely tough to predict any order solely based on these factors, and hence, computated values are much preferred.
If you prefer to read the theory of those experiments (be warned it's scary), here is one such paper discussing the detailed calculation of the dipole moments of various fluoromethane derivatives.

So is there no logical way to explain this?

@YUSUFHASAN The point of my answer was that there are logical qualitative arguments, but they cannot predict the order of the calculated values with 100% accuracy. You can only guess which factors would be minor and which major beforehand. If the calculated data doesn't agree with your major factor, you would have to say that the minor factor dominated here instead. And that is true, if you would consult the values, you would find that the minor factor did dominate.


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Which one is more basic: acetamide or aniline?

Which one is more basic: acetamide or aniline?

I had approached the question by protonating both of the compounds. Then I judged them on the basis of the stability of the corresponding conjugate acid.

In $\ce{CH3CONH3+}$, due to resonance, a positive charge is created on the oxygen atom. So, resonance is not very effective here. Hence $\ce{ CH3CONH2}$ is not a good base.
In $\ce{C6H5NH3+}$, there is far better resonance which takes place.

From the above observations, I concluded that aniline is more basic.
I had another view that $\ce{C6H5NH2}$ is a basic type of compound rather being salt-like as is $\ce{CH3CONH2}$.
Am I going in the right direction to solve the problem?

No. Acetamide, if it protonates, does so on oxygen. Consider the pKa's of these species or the pKb's of aniline and acetamide (which one can be extracted by aqueous acid from an ether solution of the two?)

Also you can consider availability of electron pair on N or O in both bases. Better resonance means lesser electron availability and weaker base.. So it seems O is most basic by that logic.

https://chemistry.stackexchange.com/questions/14017/why-are-amidines-so-much-more-basic-than-amides

You can consider this in the following way.
If amide undergoes protonation, it forms a positively charged nitrogen atom next to the strong electron-withdrawing group. This destabilises the molecule, so most probably it will not act as a base. Instead, it will act as a moderate acid with a strong base, since conjugate anion will be stabilized by the adjacent carbonyl group.
In the case of aniline, the phenyl group is not a strong withdrawing group. Still, the phenyl group withdraws electron density from the nitrogen atom through resonance. However, anilines can be protonated by strong acids. You can observe that during the formation of a diazonium salt, where aniline is protonated by strong acids like $\ce{HCl}$.

on resonance in CONH3+ group oxygen get a positive charge on it and gets destabled


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Why do we obtain racemic iodoalkane in nucleophilic displacement reactions?
In my textbook, the following statement is given :

Attempts to prepare optically active iodides by nucleophilic displacement on optically active bromides using the iodide ion normally produces a racemic mixture of iodoalkanes.

The reason given is that this is due to the following equilibrium:

My doubts:

$\ce{I-}$ is a stronger nucleophile than $\ce{Br-}$. Then, why does the backward reaction even occur?

Is the reaction SN2?

Does this happen for other types of nucleophilic displacements also?
Book: Arihant Organic Chemistry (DPP Problems)

This suggests the reaction has some SN1 character. Formation of a carbocation leads to loss of stereochemical integrity.

Yes the backward reaction is an sn2 reaction.and I'm sure that you have the knowledge of iodide being nucleophilic catalyst-it is a very good nucleophile as well as a good leaving group.that is why the backward reaction is not an illogical one.you can't generalize chemistry.so for this reaction it occurs but it does not mean that it will happen in others also.


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What is the order of the catalytic decomposition of hydrogen peroxide?
I recently did an experiment to determine the activation energy of the decomposition of hydrogen peroxide using iron (III) chloride as a catalyst. I repeated this experiment at different temperatures. I know how to figure out the rate constant and the rate law, but I am not sure if this is a second-order reaction. Is there any way I can figure out the order of this reaction? (because that would impact the units of the rate constant, k, and the rate law)

Check out this paper. There they studied the decomposition of hydrogen peroxide by iron hydroxide and found that the reaction is a second order reaction for low concentrations of hydrogen peroxide.
In your case, you are working with a soluble iron species which will change the reaction kinetics because the hydrogen peroxide does not need to adsorb to the surface of insoluble iron hydroxide particles. This is assuming you worked under neutral or better slightly acidic conditions, otherwise the iron will precipitate as iron hydroxide.
Therefore, in your case the kinetics should depend on the relative concentration of iron chloride and hydrogen peroxide. As long as the hydrogen peroxide is in excess, I would expect a zero order reaction because the iron will limit the reaction speed. If the iron is in excess, I would expect a second order reaction because you need two molecules of hydrogen peroxide to form molecular oxygen.
How you experimentally determine the reaction order is described on Wikipedia. Briefly, you need to perform the experiment at different concentrations of hydrogen peroxide and measure the initial reaction speed.  Then you plot the logarithm of the initial rates against the logarithm of the concentrations. The slope of the resulting line should be your reaction order.


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What is the meaning of 'mass action'?
I understand the statement of 'Law of mass action', but I don't understand clearly what 'mass action' means.
I investigated it and found the following descriptions.

The law was originally suggested as a law about 'active mass'. [from Wikipedia]
The term 'active mass' is a historical term that professional chemists do not use anymore. 'Active mass' meant 'amount per unit volume'. [from Chemistry Stack Exchange #58141]

From these explanations, I understood that 'active mass' corresponds to 'concentration' in modern words, which is usually expressed in the unit 'mole per liter' today.
However, a question still remains for me. What does the term 'mass' intend to originally? I am not a native English speaker, so I cannot understand the nuance well.
The word 'mass' has some different meanings.

a physical quantity expressed by the unit kilogram,
a large amount of a substance,
a lot of people, etc.

Of all these meanings, which do native English speakers understand the word 'mass' as?
Background of this question
In Japan, 'Law of mass action' has been translated as '質量作用の法則' (shitsuryou-sayou-no-housoku). '質量' corresponds to 'mass'. Nowadays, however, this is accused as a mistranslation because the law states the nature of concentration and it does not relate to mass. On the other hand, if native English speakers also understand the word 'mass' as 'a physical quantity expressed by the unit kilogram', the translation is not a 'mistranslation' but just a historical word originally. So I would like to consult those who are familiar with both chemistry and English.

"Mass" in this context refers to the large number of particles and consequent behavior in a stochastic manner, as opposed to a small set of particles, where Maxwell's demon might have been able to influence outcomes.Of course, Maxwell didn't know know of Heisenberg's uncertainty principle.


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Enthalpy of formaton
today I was doing a set of HW problems regarding enthalpy and stumble upon one problem that I found confusing.It was the standard enthalpy of formation for CaCO3 is -1207.6 kJ mol when 30.7 g of calcium is reacted with an excess of carbon dioxide, how much heat is transferred?
So I was wondering if the reaction was exothermic or endothermic. For my answer, I said it was exothermic because the standard enthalpy of formation had a negative value which indicates it released heat and it evolved 370.4 kJ of heat. So I was wondering if my answer was correct? Any help would be appreciated.I am a bit lost for this unit cause I was sick for most of the thermodynamics unit.

Come on everybody, downvotes with no comments for someone with 1 rep? Daniel, I suspect the downvotes are because SE isn't intended to be a replacement for your classmates, teacher, TAs, or textbook, all of whom would be a better reference if you missed class. If you're still unsure after consulting those sources, please edit the question to explain why you don't understand how the standard rules laid out there apply to this specific problem.

Also, please check out the tour if you haven't yet: https://chemistry.stackexchange.com/tour

@Erick people don't or forget to comment. That's why we have you, or anyone else who's willing to comment. :)

Yes, you are correct.
If it helps, you could visualise the internal energy of the system: If the enthalpy change is negative, then the internal energy of the system is decreasing and so energy is being released as heat; as we know this is an exothermic enthalpy change. Vice versa for an endothermic enthalpy change.