text:
Drawing bicarbonate ion in different way
I don't know why can't I draw $\ce{[HCO3]-}$ like that:

I tried to draw it like that and checked everything.
I checked formal charge and all was good.
Why they draw it OH I know its also alright but whats wrong with my draw? 24e checked and formal charge checked.
The upper left oxygen has 6 non bonding electrons the next one has 4 non bonding electrons. It's coordinative bond.

I suggest to review grammar, bring formatting in order and append the relevant comments into the question. Is this structure is what you have in mind?

there is no picture its black screen . i mean its simple question everyone can understand with the picture i drew why making life hard ?  if i knew the answer and someone asked i'll just tell him simple as that

It's a PNG with alpha-channel enabled. Change the background from black to another color or just enable checkerboard preview for images in your browser.

assuming this is the picture i want is it right to draw like that ?

I don't know, you tell us. I just assumed this is the Lewis structure you tried to display.

i will add a photo of my drawing through iphone but i am sure no one will answer and i will be wasting my time but i'll try maybe one good person shows up

I have not enough points for a comment, but all the downratings are for the insufficient formatting of the question, not enough information on what you want, no effort in your work. What you drew and i believe you want to draw is the conjugate base (deprotonated form of the acid) of performic acid. If not, then you may have carbonic acid in mind, if so then you drew the wrong structure. Your question is not clear to me, but you may want to try ACD ChemSketch (free) software for drawing molecules.

Technical note: please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. For drawing chemical strictures, see Software section of our Wiki.

Do you really suppose the strongly oxidizing hydroperoxide anion group directly bound to reducing aldehyde group would survive their marriage?? It is rather a recipe for an unstable explosive.

my course is general chemistry I am majoring as civil enginner so i dont know what you are talking about but in highschool i do remember OH group  the thing is i dont know whats wrong with my model ?

As mentioned in the Andrew's post,your structure does not belong to bicarbonate anion, but to performate ( peroxyformate ) anion, what are different compounds.

Who said you couldn’t? You absolutely can. What you have shown is performic acid $\ce{CH2O3}$ (or at least its conjugate base). Just to clarify, it is obviously different to carbonic acid $\ce{H2CO3}$.

i have answers that only show resonance drawings which are 2 models and didn't have my drawing

Your drawing is a valid molecular structure, it just isn't the structure of bicarbonate. As pointed out above, you have drawn the structure of performate (the deprotonated form of performic acid).

@RaziAwad what you have drawn is a peroxyacid which has vastly different properties to carbonic acid (which you were trying to draw), which is not, due to them having different functional groups.


text:
Ways of bringing a small peptide to gas-phase
I have a tripeptide molecule with an unnatural amino acid unit. I want to do the gas phase study of the molecule, hoping to understand the intrinsic interactions in detail. I need to bring the peptide to the gas phase for that. I don't think heating the sample will be a good idea, as I am worried about the denaturation of the peptide. Is there any other way to bring my sample to the gas phase. I want to know if there are any methods by which I can measure the melting point of the peptide without denaturing it.

What kind of measurements are you going to perform? What kind of samples does the instrument usually handle?

Let me ask: are you sure that you are talking about denaturation properly? You are worried about changes in conformation or about chemical transformations?

Do you know about the coupled matrix-assisted laser desorption-ionization MALDI-(e.g.https://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization) which is considered as soft technique?  Often, it is coupled with a time-of-flight mass spectroscopy (MALDI-TOF-MS). The residence time for a sample in a TOF (https://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry) is in the range of microseconds, but since you consume very little material per analysis, you may repeat the analyses again, and again.

I will be getting UV and IR spectrum of the molecule in the gas phase, from which I will be able to understand the conformation of the tripeptide in the absence of any external interaction (interaction between molecules and with the solvent). REMPI_TOF_MS is a common procedure for the study. A knowledgable person asked me about the melting point of my peptide. I can't just heat my peptide to find its melting point, as the peptide won't remain as such upon heating (due to denaturation, the structure, the bonds might break). I am looking for other methods.

source: Stack Exchange

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

sort: votes


text:
What exactly causes equilibrium to shift when pressure is decreased?
Take the reaction:
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
If pressure is decreased, equilibrium shifts to the left because it has more moles. I understand the basics of why this is, i.e. due to Le Chatelier's principle, but is there an actual explanation as to why the system does this (e.g. using collision theory)? What's actually happening to cause the shift?
My chemistry teacher says she doesn't know of an explanation.

I refer you to this question: https://chemistry.stackexchange.com/questions/4130/chemical-equilibrium-why-do-changes-in-pressure-cause-a-shift-in-the-ratio-of?rq=1

This happens because the equilibrium constant $K_p$ is independent of the pressure, although it does depend on temperature, which we will assume to be constant. The equilibrium constant can be written as
$$\displaystyle K_p =\frac{p_A^2}{p_Np_H^3}$$
where for simplicity I have abbreviated the chemical names. The partial pressure $p$ can be written in terms of the degree of dissociation and total pressure $P_T$, i.e. $p=aP_T$ where $a$ is a mole fraction term in the degree of dissociation. As partial pressure is proportional to the degree of dissociation and total pressure, for example, $p_A =a_A^2  P_T^2$ then the equilibrium constant
$$K_p=\frac{a_A^2}{a_Na_H^3} \frac{1}{P_T^2}$$
where the terms in $a$ we have not yet evaluated and contain the degree of dissociation $\alpha$. However, you can see that as $K_p$ is constant if $P_T$ decreases then $1/P_T^2$ increases and so  $\displaystyle \frac{a_A^2}{a_Na_H^3} $ must decrease. This means that the degree of dissociation decreases and so the equilibrium 'moves to the left'.
Working out the partial pressure for this reaction is a bit messy.
Let $\alpha $ be the degree of dissociation then $(1-\alpha)$ of nitrogen reacts , $3(1-\alpha)$ hydrogen and $2\alpha$ ammonia is produced. The total of these is $6-2\alpha$ and so the mole fraction of ammonia is, for example, $a_A=\displaystyle \frac{2\alpha}{6-2\alpha}$ and the partial pressure $p_A=\displaystyle \frac{2\alpha}{6-2\alpha}P_T$. You can make a similar expression for the other species and substitute to get the mole fraction term $\displaystyle \frac{a_A^2}{a_Na_H^3}=\frac{16\alpha^2(\alpha-3)^2}{27(\alpha-1)^4}$. This expression increases as $\alpha$ increases. As you see the expression is messy but the main point is that as the pressure decreases the mole fraction term must decrease which it does by reducing $\alpha$.

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


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Polystyrene properties after dissolved by acetone
Polystyrene plastic can be dissolved in acetone.
I would like to understand:

If the dissolved polystyrene will take-out of the acetone and becomes solid again (after all the acetone evaporates), will this solid polystyrene keep the same material properties as it was before the dissolving process?
How can I speed-up the solidification of the dissolved plastic?

What is happening is not actually the polystyrene melting but rather it dissolving in the acetone solvent. This is a physical change and not chemically altering the polymer. It should still retain its chemical properties as before.
However, other properties have changed. It is now a poorer insulator since the air pockets have been removed. It will also be more rigid for similar reasons.
The acetone ought to be able to be driven off by heating it in a ventilated space as acetone is quite volatile.

Thanks.So,if i dissolve in acetone a food-safe item that made of PS(like spoon/plate/cup etc),after the acetone will evaporate and the dissolved item will solidified again,the item will still be food-safe.Am i right?

I'm not certain as I have not done any analysis on this firsthand. If you use pure acetone (Not some sort of solution with other chemicals introduced) and be sure it's all driven out of the PS, then I suppose it should be safe. I would caution you though. You will likely need to perform tests to determine this material's food safety.

I understand.The problem is the traces of the solvent(acetone)in the PS,not the PS plastic itself that keeps its chemical properties after solidification.Thanks.


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Energy level of molecular orbitals in heteronuclear diatomic compounds
I have been studying the molecular orbital theorem. I read that the energy levels for the $\ce{p}$ orbitals in the first series of $\ce{p}$ block molecules are something to the tune of $$\pi_x=\pi_y<\sigma_z<\pi_x^*=\pi_y^*<\sigma_z^*$$
except for $\ce{O2 and F2}$, which have
$$\sigma_z<\pi_x=\pi_y<\pi_x^*=\pi_y^*<\sigma_z^*$$
But it got me thinking: what order would the orbitals take in a compound such as $\ce{CO, NO}$?
(Also, out of curiosity, what about $\ce{p-p}$ bonds in higher series of the $\ce{p}$ block? Does $\ce{Cl2}$ have the same order as $\ce{O2}$?)
My guess was that since in heteronuclear orbitals more electronegative elements have lower energy orbitals, compounded with the fact that the $\sigma$ orbitals it forms tend to be lower energy, it should follow the latter’s order.

This question requires idea of the Linear Combination of Atomic Orbitals (LCAO). It involves the following equation:
$\psi_n = \sum\limits_{i} c_{ni}\phi_{i} = c_{n1}\phi_{1} + c_{n2}\phi_{2} + ... c_{ni}\phi_{i}$
Here $\psi_n$ represents the wavefunction for the resulting molecular orbital, $\phi_{i}$ represents the atomic orbitals that contribute to the bond, and $c_{ni}$ represents the weight coefficient that can basically tell us how much of each atomic orbital is involved in the molecular orbital. If we look at the molecular orbital diagram for carbon monoxide, you are right, the bonding pi orbitals are lower in energy than the bonding sigma orbital:

The reason for this is due to the contributions from carbon. This comes from the relative energies of the atomic orbitals. From photoelectron spectroscopy, the energies of the carbon 2p orbitals and 2s can interact with the oxygen 2p orbitals because they are close enough in energy while the 2s oxygen orbital is a little too far to interact with the 2p orbitals of carbon. See image below

The important thing to note here is that the 2s orbital can interact as well. This means when we solve for the wave function using LCAO and the equation above, carbon can give a large contribution because two of its atomic orbitals interact, and it turned out it did. The same thing can be done for NO (which in fact is also similar to CO where the pi bonding orders are lower energy than the sigma bonding ones) and other molecules. However, it gets extremely complicated with larger and larger molecules.
I got this information from this public PowerPoint  https://www.chem.uci.edu/~lawm/10-9.pdf which come from lectures from UCIrvine which were released: https://www.youtube.com/watch?v=1CH6CeORL_g&list=PLaLOVNqqD-2GDVq2xFUuAltjIXPgQNNTL&index=7&ab_channel=UCIOpen

I found this exact point discussed indirectly in JD LEE's inorganic chemistry 5th ed in page 110. In it, the author discusses the CO molecule system where for C it's on order and O it's another, the following point is noted in the last sentence to assume the order was of the lighter element (in this case, C).
Then it is noted that if CO is is ionized to $CO^+$ by removal of an electron from $\sigma2p_x$ then bond order reduces and hence bond length increases (by the above approximation) however this is not what happens that is the bond length of 1.128 angstrom turns to a length of 1.115 angstrom.
It is then noted that same problem remains if we assume the order follows for heavier elements. Finally it is concluded that the order of MO energy levels applicable for simple homonuclear diatomic molecules is not automatically applicable when two different types of atoms are bonded together.

I finally found a practical answer on Quora, the reason of difference in energy levels is due to something known as s-p mixing and it occurs for diatomic molecules having less than or equal to 14 electrons.  Meaning the following order:
$$\pi_x=\pi_y<\sigma_z<\pi_x^*=\pi_y^*<\sigma_z^*$$
Is for molecules with $ \leq 14$ electrons.
For CO vs NO, CO has 14 electrons hence takes the order with $\pi_x = \pi_y < \sigma_z$ and NO takes the order $\sigma_z < \pi_x = \pi_y$ since it has 15 electrons

For homonuclear diatomics at and below period 3 the diagram is always the latter one, regardless of the group. See Chem. Rev. 2019, 119, 8781−8845.


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Confusion on ion product of water
The ion product of water is defined as $$ K_w = [\ce{H+}][\ce{OH-}] $$ and in pure water $ [\ce{H+}] = [\ce{OH-}]= 1.0 \times 10^{-7}$ and $ K_w = 1.0 \times 10^{-14}$. It is said that if we dissolve some acidic substance in water then $[\ce{H+}]$ will increase and $[\ce{OH-}]$ will decrease and the value of $K_w$ remains constant. I understand that $[\ce{H+}]$ increases because acids donate $\ce{H+}$ (according to Brønsted definition of acids) but how $[\ce{OH-}]$ decreases. I mean a water molecule i.e. $\ce{H2O}$ will certainly break into $\ce{H+}$ and $\ce{OH-}$ and to this $\ce{H+}$ our acid has added it's part and hence it's concentration has increased but why the value of $[\ce{OH-}]$ gets lowered from $1.0 \times 10^{-14}$ , if water molecule were to ionize then whenever $\ce{H+}$ gets formed simultaneously we would get $\ce{OH-}$.
I can cite a common example problem

The concentration of $\ce{OH-}$ ions in a certain household ammonia cleansing solution is $0.0025$. Calculate the concentration of $\ce{H+}$ ions.

We can solve this problem by using the equation $$ K_w = 1.0 \times 10^{-14}$$
$$ [\ce{H+}] [\ce{OH-}] = 1.0 \times 10^{-14}$$
and if put the value of $[\ce{OH-}]$ in the above equation and solve $[\ce{H+}]$, then we would get $ 4.0 \times 10^{-12}$. Here we observe that the value of $[\ce{H+}]$ has decreased from the it's original value in pure water, well this is understandable because ammonia being a base would consume $\ce{H+}$ but how the value of $\ce{OH-}$ has increased.
I want to know that how the chemical reaction can cause the increase or decrease of $\ce{OH-}$ when each molecule of $\ce{H2O}$ always going to yield one $\ce{H+}$ and one $\ce{OH-}$ always.
Thank you. Any help will be much appreciated.

The acid provides additional $\ce{H+}$. Hydroxyls present in water react with this to form back more water.

Two words: 'Detailed balance' would be a good place to start...

@BuckThorn Yes that's very agreeable, but how does the concentration of hydroxyl ions increases when we add base to water?

I just want to ask cordially that how my question is a homework problem, I mean it may be a trivial question for well educated and researchers over here but I can't see how it is homework problem.

Another way to look at the problem:
If we add $\ce{H+}$ to the water through adding acid, then some of the  $\ce{H+}$ would just remain in the solution as is, and some of them would react with  $\ce{OH-}$ in the reaction $\ce{H+ + OH- -> H_2O}$. How much of the added $\ce{H+}$  reacts? To calculate this, you need to know that $\ce{[H+][OH-] = K_w}$ remains constant. So while $\ce{H+}$  is increased, $\ce{OH-}$ is decreased through $\ce{H+ + OH- -> H_2O}$.
If we were to add $\ce{OH-}$ to the water through bases, the same thing would happen: a portion would react with $\ce{H+}$  and a portion would remain in the solution, and it will happen in such a way that is dictated by the constant value of $K_w$.

Thank you, your answer is very clear and acceptable. I just want to correct you (or you can correct me if I'm wrong) that base doesn't add $OH^-$ if we use the definition of BrØnsted base, it just consumes $H^+$. So, I think when we add a base to water $H^+$ gets consumed and hence equilibrium shifts to the right side and consequently we get more $OH^-$ . But I expect a more clear explanation from you, just the same way you have explained the acid problem.

You can look at the problem from any side you want. In my comment I argued that adding let's say $\ce{NaOH}$ to water increases the $\ce{OH-}$ concentration, since it dissociates into $\ce{Na+}$ and $\ce{OH-}$ . On the other hand this means that $\ce{H+}$ is consumed - exactly because of the constant value of $K_w$. Bronsted's theory always looks at the concentration of $\ce{H+}$ (hence the definition of $pH$), but it is equally reasonable to think about $\ce{OH-}$. It is the two sides of the same coin, and the connection between the two is exactly the $K_w$ equation in your question.

By the way when I used to teach chemistry calculations I liked the idea of introducing the quantity $pOH$, which is an analogue of $pH$, defined by $pOH = - lg[\ce{OH-}]$. Here the $K_w$ value means that $pH + pOH = 14$.

Is it true that all acids will always gonna release $\ce{H+}$ and all bases always release $\ce{OH-}$. Because the example problem which I have cited, $\ce{NH3}$ can't release $\ce{OH-}$ on its own, it need water for that. Please explain .

Release is not true. What is true is that acids increase the $\ce{H+}$ concentration and bases decrease the $\ce{H+}$ concentration (or, equally speaking, increase the $\ce{OH-}$  concentration). In the case of $\ce{NH_3}$, as seen from Poutnik's answer, we have $\ce{NH3 + H+ -> NH_4+}$. This reaction decreases the  $\ce{H+}$ concentration, but since  $[\ce{H+}][\ce{OH-]}$ is constant, tthis means that $\ce{OH-}$ is increased. But not from releasing from $\ce{NH3}$.

How $\ce{OH-}$ got increased? Is it due to the forward reaction?

Poutnik's answer explains this well. There are two equilibria present: one involving ammonia and one describing the dissociation of water. Both of the reactions involve $\ce{H+}$, so they are not separate, they influence each other. So by adding ammonia to water, you affect the dissociation of the water.

It is not a chemical problem, but a trivial mathematical problem.
If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.
The key part is to understand that  chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same rate. Like
$$\ce{H2O <=> H+ + OH-}$$
where the rate of ion creation equals the rate of ion recombination.
If there is an excess or deficit  of either of  $\ce{H+}$   or $\ce{OH-}$ ions, the rate of their recombination changes, while the rate of their creation remains the same. As consequence, the product of their concentrations converges quickly towards $K_\mathrm{w}$ to be in the equilibrium again.
The ammonia reacts :
$$\ce{NH3 + H+ <=>> NH4+}$$
That creates deficit of $\ce{H+}$. As consequence, dissociation
$$\ce{H2O -> H+ + OH-}$$
Is faster then recombination
$$\ce{ H+ + OH- -> H2O}$$
An alternative reaction mechanism is
ammonia reacting with water:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
what directly produces the excess of $\ce{OH-}$ that recombines with the most of the present $\ce{H+}$.
The production of $\ce{OH-}$ and elimination of $\ce{H+}$ continues, until their concentrations satisfy both the basicity constant of $\ce{NH3}$ ( or equivalently acidity constant of $\ce{NH4+}$ ) and the ion product of water  $K_\mathrm{w}$
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$
$$K_\mathrm{b, \ce{NH3}}=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}=\frac{[\ce{NH4+}]K_\mathrm{w}}{[\ce{NH3}] [\ce{H+}] }=\frac{K_\mathrm{w}}{K_\mathrm{a, \ce{NH4+}}}$$

Yes. If some acid is added to pure water, $[\ce{H+}]$ increases but $[\ce{OH-}]$ decreases. It means that:

in pure water, enough $\ce{H2O}$ will break into $\ce{H+}$ and $\ce{OH-}$.
in the presence of an acid, a smaller amount of water will break into $\ce{H+}$ and $\ce{OH-}$. The presence of an acid prevents $\ce{H2O}$ from being dissociated.

For example, if you add $10^{-7}\mathrm{mol}$ of acid in 1 liter pure water, the concentration $[\ce{H+}]$ is the sum of the added $\ce{H+}$ ($10^{-7}\,\mathrm{M}$) and of the smaller amount of water being broken. The concentration of $\ce{OH-}$ is decreased in the same way. Calculation shows that only $0.618\,\mathrm{mol}$ water are broken into $\ce{H+}$ and $\ce{OH-}$, if $10^{-7}\,\mathrm{mol}$ acid is added.
With this value, the final concentrations are: $[\ce{H+}] = 1.618 10^{-7}\,\mathrm{M}$, and $[\ce{OH-}] = 0.618 10^{-7}\,\mathrm{M}$. You may check that the product $[\ce{H+}][\ce{OH-}] = 10^{-14}\,\mathrm{M}$.

Okay. I have understood that the dissociation of $H_2O$ gets hindered due to the presence of an acid but speeds up in presence of a base, am I correct?

No ! The dissociation of H2O is hindered by both the presence of an acid and of a base.


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Plastic bags broken down by soap fumes
I've searched online, but have not found anyone that has experienced the same.
I have a 20 gallon plastic tub with a lid.
In it keep plastic shopping bags.
Also had a box of detergent in there.
I witnessed over time, the fumes (I am guessing) from the detergent affected the plastic bags.
The big 20 gallon bin is fine tough.
But a few of the plastic bags turned brittle to the touch and broke down into tiny bits like old dry leaves when I crumple it in my hand.
Just wondering what is happening to the plastic; chemistry wise. Has the breakdown of the plastic been accelerated by the detergent fumes?
Just found it very odd, and if the process can be used to help recycle plastic by breaking it down faster
I still have the bags on hand and can make a video as proof.

Share pictures please of plastic bags (brand?) and also the name of the detergent. Was it exposed to light or sunlight?


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If heavy metals bind to sulfur why isn't sulfur used as a safer way to chelate them?
I was wondering about sulfur as an alternative life-threatening chelators. Why isn't sulfur used as a chelator if heavy metals have an affinity for it?

'life-threatening", "safe" - what are you talking about?

Sulfur-containing molecules are indeed used in chelation therapy: prime examples are dimercaprol and penicillamine, both of which contain thiol functional groups.


text:
Angels Share seems bogus
What is the explanation of barreled spirits (at 62.5% ethanol) that undergo changes in concentration of ethanol. Whereas the volume after aging is typically reduced, the proof varies between low 100 to low 140 (50-70%). Does the presence or formation of azeoptopes play a role?
I contend that the barrel interface is important and not so much the differences in evaporation, that the industry states: Angel's Share and Devil's Cut.

The barrel is sealed so simple evaporation can't explain the loss of volume. The liquid obviously permeates the wood and evaporates from the outside of the barrel. Water permeates wood better than ethanol, so the liquid inside the barrel becomes enriched in ethanol. // If the mixture were left in an open glass to evaporate, then the liquid would lose ethanol faster, and hence the proof would be reduced.

Thanks for the input. I think this is a big factor, but doesn't explain loss of proof. Most barrel proof bourbons are 55-59%; just slightly less than entry proof of 62.5.

Both the barrel/cask and the storage environment matter. For single malts matured in casks in Scotland, proof typically declines year after year, so really old whisky (30 YO and up) is rare. For straight bourbon matured in barrels in Kentucky, proof can go up, as in the famous G.T. Stagg whiskey. Too many variables ...


text:
Alkylation before ring closure?
So, another question from my side.
I have been thinking about the folowing reaction:
The reaction above exists and was already done on two occasions and reportedly works really well and it seems to be a no-brainer to me as well. Now, my question is related to the reaction below with the alkylated amine. First question: the alkylation of the compound above seems difficult, BECAUSE how can I make sure that I only alkylate the primary amine in meta position to the bromine? Can I use a protection group before for the other primary amine?
Second question: if the alkylation is possible as I imagine it here, can I still perform the reaction as depicted? Or is it unfavored because suddenly a positive charge appears?
Greetings

This shouldn't be much of a problem, but alkylated form would be probably better made using ethylamine in synthesis then alkylating diamine.

The ethylated material in the red box is accessible by SN2 on 4-Br-2-F-nitrobenzene with ethylamine followed by reduction of the nitro group. I think the condensation reaction to give the cationic species is viable


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Mechanism of Peptization
Can somebody please explain the mechanism of Peptization(or give a link to the website which explain the mechanism step by step). The Mechanism given in my book is somewhat different from the mechanism given on the Wikepedia. Also the Wikepedia one is baffling me.
The wikipedia Peptization states

The electrical charges present at the surface of the particles are so "neutralised" and disappear.

This process is same in the preparation of coagulated solution

The electrical repulsion no longer hinders the aggregation of particles and they can then coalesce to form a flocculent precipitate that is easy to filter.

Again my book has exactly same para that this step is used in preparation of coagulated solution
And isn't peptization reverse of it that is that making of colloidal solution from splitting aggregating precipitate molecules, but here we are forming precipitate again.

When colloidal particles bear a same sign electric charge.

Does wikipedia means precipitate particle here?
Also, how does coagulation differ from peptization. I mean one method to prepare a coagulated solution  is by adding electrolyte (containing ion of opposite charge ). In the process of peptization the only difference is that charge(should be opp charge) and one of the ion should be same as that in the precipitate. Does the mean in coagulation ion should  be different?

It would be nice to put both mechanisms (textbook vs. Wikipedia) together in your question, or at least provide links/references so that others could have a basis upon which the differences could be discussed.

@andselisk i just want a whole new perspective that will clear my confusion.I dont have mobile  phone with me therefore cant put snapshot of my book but i can copy and paste wikipedia part  and tell where my book and it got differ.Should i do that?

Go ahead and do that, making sure to reference with a link.

In coagulation you cancel the surface charge to allow particles to come close. In peptization you do the opposite, you want the particles to repel each other.

The wikipedia article might be confusing you because it starts out by describing coagulation or flocculation. It then explains how peptization is the reverse of that. See also for instance: https://www.mrwa.com/WaterWorksMnl/Chapter%2012%20Coagulation.pdf

@BuckThorn is it correct to say depending upon the type of electrolyte that is added   coagulation or peptization will happen . In nutshell coagulation will happen if the the electrolyte added doesn't contain any common ion with precipitate .


text:
Reaction with tin(IV) iodide and hydrogen peroxide
Today I reacted tin(IV) iodide $\ce{SnI4}$ (dissolved in water) with hydrogen peroxide. This produced a yellow/brown colour (I'm guessing due to formation of $\ce{I2}?).$
I then added some petroleum spirit which gave a bi-phasic system which was a clear/pink/ light purple colour on top and an orange-brown yellow underneath.
I'm struggling to find out what reactions have occurred here, can anyone please help?

I propose the following equation :
SnI4 + 2 H2O2 --> SnO2 + 2 H2O + 2 I2

$\ce{SnI4}$ quickly hydrolyses in water, so likely there will be hexahydroxostannic acid that hydrogen peroxide actually encounters.

Ahh ok, this would make sense, thank you for your quick reply! Would this explain the colours in each layer?

how would this account for the colours?

Clear/pink color is likely due to Mie scattering of white $\ce{SnO2}$ particles (sol formed from $\ce{H2[Sn(OH)6]}$), and orangey brown color is probably iodine from the reaction between HI (product of hydrolysis of $\ce{SnI4}$) and $\ce{H2O2}$. Do you need references from reputable literature sources to back this up? If so, feel free to add reference-request tag, I'll try to look them up and maybe post an answer with reactions.

Just in case, are you sure the brown yellow phase was underneath, and the clear pink one was on top and not vice versa? Also, are you sure it was an ordinary petroleum ether? If you have a photo, it would be nice to see it.

Note: this is a preliminary description, which I think answers the question, but disagrees in one detail (see asterisk).
Tin(IV) iodide is completely hydrolyzed by water [1, p. 120] to colorless hexahydroxostannic acid:
$$\ce{SnI4 + 6 H2O <=> H2[Sn(OH)6] + 4 HI}$$
Hydrogen peroxide can not oxidize $\ce{Sn^4+}$ further (in fact, hexahydroxostannate(IV) is used as a stabiliser for hydrogen peroxide), and is likely going to oxidize hydroiodic acid to iodine (which attributes to the brown color) instead:
$$\ce{H2O2 + 2 HI -> I2 + 2 H2O}$$
Hexahydroxostannic acid only exists in solution and tends to form cloudy/milky gel over time [2, p. 1021]; this can be simplified as formation of colloidal particles of tin(IV) oxide [hydrates]:
$$\ce{H2[Sn(OH)6] <=> SnO2 · n H2O + $(4-n)$ H2O}$$
When petroleum ether is added, iodine is extracted to organic phase (on top due to lower density of the ether). Tin(IV) oxide sol remains in aqueous phase and attributes to its slightly pink color due to Mie scattering of colloidal solution.*
* Phase order doesn't match OP's description which suggests opposite coloring of the phases (additional clarification on the type of petroleum spirit or a photo of the binary phase mixture would help).
References

Inorganic Syntheses; Bailar, J. C., Ed.; Inorganic Syntheses; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1953; Vol. 4. https://doi.org/10.1002/9780470132357.
Holleman, A. F.; Wiberg, E.; Wiberg, N. Lehrbuch der anorganischen Chemie, 102nd ed.; Fischer, G., Ed.; Walter de Gruyter: Berlin New York, 2007. ISBN 978-3-11-017770-1. (in German)

Thank you so much for this description, although yes the top layer was slightly pink but I suppose I should have described it as more clear? I’m not quite sure what’s going on @andselisk

@andelisk I should have taken a picture, I didn’t think at the time! I guess the top layer was just off a light purple, not a milky colour

@MarinaCalder Hmm, it's indeed hard to discuss colors in the absence of a photo. Purple color might be due to iodine remaining in aqueous phase. And by "milky" I didn't mean plain white, but a cloudy-white colloid solution, something like this (randomly picked from Google Images).

Which iodide is extracted to organic phase? Would it not be the iodine - I don't understand how the iodide would prefer a non-polar solvent.

@StianYttervik You are absolutely correct, this was a typo. Thank you!

Thank you so much for your help, I appreciate it so much!:) @andselisk


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Home test for the presence of sugar in erythritol
Is there a test to distinguish erythritol from sugar at home without expensive lab equipment or exotic reagents?  Sometimes I purchase erythritol online and wonder if it's been diluted with sugar.

I would consult with a doctor regarding symptoms. It should be relatively straightforward for a properly equipped lab to test composition but I can't think of a suitable and simple home test. There should be labels on the product - is this from a reputable vendor?

Clearly I didn't word this clearly.  I'm not asking for medical advice, just a simple chemistry question: how does one test for erythritol vs sugar.  Answering my own question: erythritol melts at 121 C, while sugar (sucrose) melts at 186 C.  I imagine there are other tests, such as solvents that dissolve one but not the other.

It's a good question, but it remains broad and unclear. It sounds like you want to run a home test. I would say a number of spectroscopic techniques will give you clear cut information, but you are unlikely to have access if you aren't working in a lab. On the other hand there are chemical tests for sugars that  distinguish them from polyols. I'm no expert on this aspect but I suspect this would be a good  start: https://en.wikipedia.org/wiki/Molisch%27s_test    You can't run just any test for sugars, because sucrose is a so-called nonreducing sugar.

See also: https://theskepticalchemist.com/demonstration-benedicts-test-reducing-sugars/

BTW remove any suggestion that this is for personal medical reasons (you can edit your post) and the question might be reopened. Hopefully my comments will provide useful pointers.


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Can Hartree-Fock be considered the quantum mechanical "Monopole" term in a quantum mechanical multipole expansion?
From what I understand, Hartree-Fock simplifies the exact electron-electron interaction with its two electron integrals. It reminds me of the classical energy of a charge distribution combined with the multipole expansion. Is it valid to think of it in this way?

One thing I'd like add is that Hartree-Fock does not simplify the "exact electron-electron interaction with its two electron integral" because electron-electron interaction is a two-body interaction term by definition. The simplification of Hartree-Fock is that the ansatz is a Slater-determinant. By the way, this is also a nice analogy since it is the same for multipole expansion: the interaction there is still a sum (integral) of two-body interactions, just the 'ansatz' the expansion is different.

And in this sense, you could say that Hartree-Fock theory is the 'mopole' term of the full configuration interaction expansion. I really like your analogy.


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Meaning of mass concentration "units" % m/v

In a saline solution that is 5.750% (m/v) potassium iodide $(\ce{KI}),$ there are:
a) 5.750 mL $\ce{KI}$ per 100.0 mL $\ce{H2O}$
b) 575.0 mg $\ce{KI}$ per 10.00 mL of $\ce{KI}$ solution
c) 5.750 g $\ce{KI}$ per 100.0 g $\ce{H2O}$
d) 5.750 g $\ce{KI}$ per 100.0 g of $\ce{KI}$ solution
e) 5.750 g $\ce{KI}$ per 100.0 mL of $\ce{H2O}$

I figured it should be e because for the equation of $(\text{m/v})\%$ is
$$(\text{m/v})\% = \frac{\text{mass solute (g)}}{\text{volume solution (ml)}}$$
and it's $\pu{g/100 mL},$ but it's not the correct answer.

5.75g of some salt isnt going to increase the volume of 100ml solution. if it does the change will be negligible.

But 100 mL of ........  ?

You wrote it yourself in the divider, "volume solution (ml)", and there is only one option for that (support your choice by calculating the concentration; consult Wikipedia if in doubt).

The funny thing is that the correct answer is not available (unless posing Delta V zero and rounding 105 to 100). Totally unsuitable for a multi choice test. Acceptable for an oral exercise at the board/desk.

@Alchimista b looks fine to me without any approximations, why do you think there is no correct answer? An exercise looks decent to me and trains attentiveness well.

The fact is that writing "there are" means there are. Of course I understand what you mean, but we never write down = instead of ~ . Moreover the use of four digits is disturbing in this context. The solution isn't even in the ppm range, a case that I would have found less or no disturbing. ...

@Alchimista I suspect "there are $x$ grams", that's what they mean. I also don't see the problem with four digits, there is no mention how the mass has been determined (and it's really irrelevant for answering).

The task and the choice from given asnwers is really trivial. I do not understand the big deal @Alchemista makes about it.  The author has made a trivial trick with potentially misleading formulation of the correct answer.

@Andselisk sorry but the problem is that the solution is given at 5.750 %, you are using a circular thinking. The point is that either you know the KI mass at the four digit, or the percent concentration. Writing down both with the same precision exclude the standard approximation. Which by the way are correct in the ppm range. I can discern hundred grams from 105 g almost in my kitchen. That we use this conc. just for quenching or provide an environment is another story. In fact, we prepare 5% percent stuff, not 5.000%. The quiz in my opinion is really wrong, though answerable for the delight

.... of everyone.

YES B IS CORRECT. @Andselisk I miss the first line!  I think I must delete my comments not because ashamed but not to confuse op or future reader. Indeed I was attentive :))

@Alchimista It's up to you, really, but I see no harm in leaving your comments (the whole section is likely to be moved to the chat, so …:D).

Yes @Poutnik. Obviously the deal would have existed without the correct answer. I have taken the rest as minimizing that potential but not factual situation. Didn't see b) refers to ml of slt.

Since the question raised a discussion in the comments, I guess it won't hurt putting up a brief summarizing answer.
Notation % m/v, as you might've guessed, refers to the ratio between the mass and the volume, or, more precisely, mass concentration $ρ_i$ (sometimes denoted as $γ_i$ [1, p. 48] to distinguish from density, see How to distinguish mass concentration and density?):
$$ρ_i = \frac{m_i}{V}$$
where $m_i$ is the mass of solute and $V$ is the volume of solution, suggesting the SI unit $\pu{kg m-3}.$
Where does percent in % m/v come from?
Quoting Wikipedia:

In biology, the "%" symbol is sometimes incorrectly used to denote mass concentration, also called "mass/volume percentage." A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). The notation is mathematically flawed because the unit "%" can only be used for dimensionless quantities.

So, in order to account for this percentage “unit”, one should normalize the mass concentration to the mass of solute in grams in 100 mL of solution.
This can be done in one's head in most cases, but you must watch out for the units.
In this particular example, only option b satisfy the definition of the % m/v quantity we are expressing since only b lists volume of KI solution.
Let's check the math.
Since $\pu{575.0 mg} = \pu{0.5750 g},$ there is $\pu{0.5750 g}$ $\ce{KI}$ per $\pu{10.00 mL}$ of $\ce{KI}$ solution, and normalizing to $\pu{100 mL}$ we indeed obtain the aforementioned concentration:
$$ρ_\ce{KI} = \frac{\pu{0.5750 g}}{\pu{10.00 mL}} = \frac{\pu{5.750 g}}{\pu{100.0 mL}} = 5.750\%~(\text{m/v})$$
The exercise really just requires to pay more attention and follow the analytical chemistry definitions.
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


text:
Cyanogenic and Fluorine Compounds in Pharmaceutical Science
I was wonder what it was specifically that make things such as Lexapro safe when it came to its Fluoride and nitrile bonds. Why doesn't Lexapro break down into cyanide? Is it just minimal because the chemical is so large it has more mass? Is it like just priming genetic code so that it will chemically metabolize into a safe compound once pharmacokinetics steps in? I know that flourine in particular binds to calcium and sodium and so I'm assuming that's the pathway it bonds to, which is good because they have a wide range of uses.

Fluorine attached to a phenyl ring is not equivalent to a free fluoride ion hence it does not help in binding to calcium or sodium. The same goes for the cyano group. Fluoro group is usually added to the molecules in order to increase their "stay" in the body. Such compounds are more soluble in the fat and hence are excreted/metabolized slowly.

This is helpful, and I have also looked into Seligiline metabolism. Do they break down into a safe compound to be excreted, or do they just avoid breaking off of the chemical and goes out the body without breaking off? I'm assuming for the same reason the chemicals might increase their stay, they also would be harder to break apart.

Nitriles in latex gloves for example are a situation it seems where some people can have an allergic reaction, but it is a compound containing cyanide in which it does not break off and go into the body. However, there's examples of people's bodies reacting as though it may be possible as autoimmune.

Pharmaceutical companies, when they release medicines for public use, they have spent billions of dollars in research followed by very very rigorous testing following the regulatory procedure of the land. This testing does include excretion studies and studying the metabolic products. You have to consult a doctor who knows more it. –

Both are bonded to phenyl rings. There is no mechanism that would cleave either the $\ce{C-F}$ or the $\ce{C-CN}$ bond at any meaningful rate.


text:
How to calculate Revelle factor using only pH, DIC and TAlk for seawater?
I have the temperature, salinity, $\mathrm{pH},$ dissolved inorganic carbon (DIC) and total alkalinity (TAlk) values for an estuarine water sample. I want to calculate the Revelle factor.
The Revelle factor can be calculated from the partial pressure of $\ce{CO2}$ and DIC. But I don't have the value for partial pressure of $\ce{CO2}.$ I want to know whether it is possible to calculate Revelle factor using $\mathrm{pH}$, DIC and Talk values. And if it is possible then I want to know how to do it.

Use CO2SYS, it's a multi platform which links together the relevant equilibrium equations of the carbonate system.
Available at: https://cdiac.ess-dive.lbl.gov/ftp/co2sys/


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Why does ice get colder when you add salt?
I know that when you put salt on ice it may turn to a slush and it does feel cold. What is the reason for this?

State of matter changes from solid to liquid and from liquid to gas are endothermic reactions. When you sweat, water evaporates and that cools your body. You don't have to "put the energy into it" (intuitively as in heating up your skin over 100 degrees Celsius) for the state change to occur if it occurs spontaneously and the molecules in changed state are captured in that state, in the case of sweating by leaving the proximity of the liquid droplets on your skin (which is also why fans work to cool you even if the ambient temperature is higher than your body temperature).
Now back to your case of adding salt to slush it's the same thing: you have the salt holding the spontaneously molten ice in liquid state, and taking up heat in the process.
Conversely, when you have a cold beer can in a humid weather, the condensation drops forming on the can will heat your beer can quicker than if you were in super dry weather in the desert.
Oh, there is also solution enthalpy, which in case of NaCl is negative, i.e., dissolving NaOH is also endothermic. If you threw in NaOH instead, your slush might actually warm up.


text:
Crystal structures of heaviest nuclides' compounds
What is the heaviest element of the periodic table for which there is a compound whose crystal structure has been completely resolved by X-ray diffraction?

Since there were no further suggestions, I decided to use the references from my comment in an answer, which, I believe, is element 99, einsteinium.
Both single crystal x-ray diffraction (SCXRD) and powder x-ray diffraction (PXRD) have certain sample requirements/limitations:

Mass/amount

PXRD: lower limit is in micrograms region, ideally at least a couple of milligrams of polycrystalline sample;
SCXRD: a single crystal of a size of fractions of a millimeter.

Time

PXRD: solid polycrystalline sample must be homogeneous which sometimes requires prolonged synthesis time (minutes, hours) due to diffusion barrier (solid-solid, solid-liquid or solid-gas). A measurement itself on average only takes minutes;
SCXRD: in addition to the synthesis, growing a suitable single crystal for analysis normally takes days, maybe hours if you are lucky. A measurement itself on average takes hours, sometimes days.

Looking both at the end of periodic table of elements and these two constrains, I suspect it is unlikely that current techniques can be applied for transfermium elements and fermium itself as we neither have the necessary amounts nor time.
Among transuranium elements with $Z < 100$ plutonium $\ce{_{94}Pu}$ compounds appear to be of the greatest interest and probably the greatest number of published structures, too.
However, einsteinium $\ce{_{99}Es}$ exists as several relatively stable isotopes with half-lives of the magnitude of months and days and has a somewhat developed chemistry to speak of.
In 1968 Fujita et al. [1] having 3 μg of $\ce{^{253}Es}$ synthesized and reported crystal structures of einsteinium(III) chloride $\ce{EsCl3}$ and einsteinium(III) oxochloride $\ce{EsOCl},$ both determined via powder x-ray diffraction.
Reading the full text is encouraged, mainly to admire authors' ingenuity in overcoming technical difficulties (self-contamination of the isotope, film darkening by radiation, measurement of lattice parameters in temperature range from the room temperature to over 400 °C and so on) and overall concise writing style of the report.
$\ce{Es^3+}$ in both structures has C.N. 9 and coordinates in a single-capped square antiprism ligand environment:

Figure 1. Crystal structure of $\ce{EsCl3}$ (ICSD-109002). Color code: $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#B31FD4}{\Large\bullet}~\ce{Es}$.

Figure 2. Crystal structure of $\ce{EsOCl}$ (ICSD-109003). Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#1FF01F}{\Large\bullet}~\ce{Cl}$; $\color{#B31FD4}{\Large\bullet}~\ce{Es}$.
A few years later Haire and Baybarz [2] determined crystal structure of einsteinium(III) oxide $\ce{Es2O3}$ via electron diffraction, but this is another story.
References

Fujita, D. K.; Cunningham, B. B.; Parsons, T. C. Crystal Structures and Lattice Parameters of Einsteinium Trichloride and Einsteinium Oxychloride. Inorganic and Nuclear Chemistry Letters 1969, 5 (4), 307–313. https://doi.org/10/bxzkjh. (PDF)
Haire, R. G.; Baybarz, R. D. Identification and Analysis of Einsteinium Sesquioxide by Electron Diffraction. Journal of Inorganic and Nuclear Chemistry 1973, 35 (2), 489–496. https://doi.org/10/bbfgp4.


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Why is an alkyne substituent on benzene is electron withdrawing, but alkane and alkene are electron donating?
Hammett constants for alkyls/vinyls/etc. are all negative (i.e. toluene), and my understanding is that they donate electrons through inductive/resonance effects.
However, ethynylbenzene has a positive Hammett constant, and I am unsure how to rationalise this in terms of induction/resonance (or whatever other effects that might be responsible for this).

Bonds from alkynyl group to the ring have mor $s$ orbitsl character, lowercl electron energy level, therefore effectively more electronegative.

Basically electronegativy of a -radical is kind of directional, to be sensed along the bond direction.


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Copper sulphate use as mould deterrent in evaporative humidifier
Is it okay to use copper sulphate in an evaporative humidifier to prevent mould growth?

Maybe you can add more detail regarding how and where exactly you intend to use the copper sulfate.

Check out for instance: http://npic.orst.edu/factsheets/cuso4gen.html


text:
How do you know is a solution is acid or basic when you dissolve a weak acid/base?
If I dissolve the weak acid HCN into water, I get a solution of oxonium and CN ions and since there is an equilibrium there is more HCN left.
So I thought CN ions can consequently react with water to form hydroxide-ions, which is a weak base-reaction, by which we also get the original HCN molecules.
I am confused because if its true what I am saying, there are two equilibriums.
Could anyone explain this, and tell me in general whether the bases that are formed when you dissolve an acid can react reversibly to an acid?

Related https://chemistry.stackexchange.com/q/123365/35806

Why not put some on litmus paper and confirm?

There are 3 equilibriums, described by 3 equilibrium constants:
The first one is the equilibrium of water dissociation:
$$\ce{H2O <<=>H+ + OH-}$$
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$

The second one is for the ( very weak )  hydrogencyanide acid dissociation:
$$\ce{HCN <<=>H+ + CN-}$$
$$K_\mathrm{a}=\frac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}$$

The third one is for the  cyanide anion hydrolysis:
$$\ce{CN- + H2O <=> HCN + OH-}$$
$$K_\mathrm{b}=\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]}=\frac{[\ce{HCN}] \cdot K_\mathrm{w}}{[\ce{CN-}][\ce{H+}]}=\frac{ K_\mathrm{w}}{K_\mathrm{a}}$$

All 3 equilibriums are therefore related by equation
$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$
Remember all 3 equilibriums are dynamic, with chemical reactions  ongoing simultaneously in both directions at equal rate.
In the case the rates were not equal, the dominant reaction shifts the system toward the equilibrium for the rates to be equal again.

If you're dissolving into neutral water, then you are overthinking the problem.
If you dissolve a weak acid into neutral water, the result is an acidic solution.
If you dissolve a weak base into neutral water, the result is a basic solution.
It is easy to confuse yourself if you only choose to look at one component of the equilibrium and try to understand the system. You want to look at the total equilibrium. In your example, yes, the cyanide will react with water but it does that to a smaller extent than hydrogen cyanide will dissociate. That's why hydrogen cyanide is an acid.


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Hydroxy Protecting Groups that are not Cleaved Under Acidic Conditions
I am currently designing a synthesis requiring the use of hydroxy protecting groups. The problem I am having is that the protecting group must be able to survive highly acidic conditions. I would have loved to use a TMBS protecting group since it can withstand LAH, lewis acids, and strong base; however, not acidic conditions. I really do not want to have to use a t-butyl ether protecting group, as its removal requires very harsh conditions. I will provide a scheme indicating all of the conditions I have proposed so far (this scheme is just a rough draft, so there are a few reagents that I have changed to be compatible with the reaction, but the general conditions should remain the same). Does anyone know any protecting groups that would match this criteria? I am only referring to replacing the TMBS protecting group on the second intermediate.

A substituted benzyl ether such as 4-nitrobenzyl, or 2,2,2-trichloroethyl ether. I wouldn't do the sulfonation step to block that site

Do you know of any other good blocking groups? If I could use a blocking group that didn't involve such harsh conditions, I might be able to use a silyl ether protecting group (if I altered some of the other steps of course)

Why use a blocking group? If you form methoxymethyl ethers of the hydroxyls of the diphenol then you can put in both your carbon chains by sequential directed lithiations and cut your number of steps drastically.

How about Br? Then you can put your second carbon chain in by a Pd catalysed coupling

If I first converted the hydroxy groups from the second step into -MOM groups, would that not direct the metalation to occur where the -SO3H is now since both of the -MOM groups are strong ortho-directors?

It would, just put your carbon chains in different order. Or put in Br

Monoprotection of phloroglucinol will be difficult. You plan to protect it under basic conditions with a tertiary bromide? What about elimination? It is helpful to number structures to facilitate discussion.

I've included an updated scheme with numbered intermediates. The first step is an esterification via an Sn1 mechanism. I will achieve this by heating a dilute solution of my tertiary iodide (dilute to prevent poly esterification) in phloroglucinol as my solvent and nucleophile.

I'm concerned about the selectivity of the bromination 2 to 3. There may be enough steric hindrance that 3 is the major product, won't know until you try

On step 6, the diels-alder reaction, there are two possible ways in which the dienophile can attack the diene; however, in my mechanism, the less sterically hindered route is shown as the product. Will this be the case, or will there be a large percentage of the minor product?


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Racemic mixture vs enantiomers in drugs
a racemic mixture has 50-50% of the 2 different enantiomers. There are racemate drugs out there in the world. The misunderstanding I have is that usually 1 out of the 2 enantiomers molecules have a different reaction with the body, so if you perform racemisation to form a racemic mixture, ain’t that drug going to be still harmful in your body ? Or does FDA not accept racemic mixture where 1 of the enantiomers in it are harmful to the body? Thus they only accept racemic mixture where both enantiomers have to perform the same therapeutic effect on the body ?

https://chemistry.stackexchange.com/questions/105/what-are-known-examples-of-drugs-that-racemize-stereoconvert-in-vivo-and-how-ar

The 'wrong' enantiomer is far more often inactive than actually harmful. FDA  does not usually approve racemic mixtures unless the compound racemises under physiological conditions.

Thalidomide is one classic example which is an approved medicine. It racemizes in the body and one of the enantiomers damaged a lot of babies in pregnant mothers. It is called thalidomide tragedy.


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What does bulk mean when talking about polymers?
I recently read a paper by Lyulin et al. [1] where they talk about thin films and bulk polymers, I just don't quite understand what they mean by "bulk". Can someone please explain?
References

Lyulin, A. V.; Balabaev, N. K.; Baljon, A. R. C.; Mendoza, G.; Frank, C. W.; Yoon, D. Y. Interfacial and Topological Effects on the Glass Transition in Free-Standing Polystyrene Films. The Journal of Chemical Physics 2017, 146 (20), 203314. https://doi.org/10/ggcghf.

I think the meaning of the word "bulk" — as it is used in English language — is preserved here despite chemical context.
I suspect that it's been used to stress some aspects or features of the surface chemistry: thin film is a 2D object, whereas bulk polymer is a 3D object (e.g. continuous in three directions) possessing a significant volume.
From IUPAC's Compendium of polymer terminology and nomenclature [1, p. 199]

3.12 continuous phase domain
matrix phase domain
Phase domain consisting of a single phase in a heterogeneous mixture through which a continuous path to all phase domain boundaries may be drawn without crossing a phase domain boundary.
  Note: In a polymer blend, the continuous phase domain is sometimes referred to as the host polymer, bulk substance, or matrix.

More general from McGraw-Hill dictionary of chemistry [2, p. 171]:

gross sample [ANALY CHEM] One or more increments taken from a larger quantity of a material that is to be analyzed. Also known as bulk sample […]"

Gooch's Encyclopedic Dictionary of Polymers [3, p. 98] keeps it even simpler:

Bulk Material n A material or product in large quantity such as a drum or sack.

References

IUPAC “Purple Book” Compendium of Polymer Terminology and Nomenclature; International Union of Pure and Applied Chemistry, Jones, R. G., Eds.; RSC Pub.: Cambridge, 2009. ISBN 978-0-85404-491-7.
McGraw-Hill Dictionary of Chemistry, 2nd ed.; McGraw-Hill: New York, 2003. ISBN 978-0-07-141046-5.
Encyclopedic Dictionary of Polymers; Gooch, J. W., Ed.; Springer New York: New York, NY, 2011. https://doi.org/10.1007/978-1-4419-6247-8.

Yes it means a piece of polymer having almost comparable xyz dimensions. It is important as for a lot of properties of thin and very thin films differs from those within a big piece of solid. Sometimes it is said that thin films are a further state of the matter. *This is dictated - as the answer suggests - by characteristics of the surface weighing more or, sorry but it even helps, of the "bulk" within the film. The polymer/material arrangement might be dictated by the interaction with the substrate if any. Also the process used for casting might lead to out of equilibrium frozen states. +1

In this case. Otherwise there is bulk polymerisation, bulk as in bulk chemicals, etc.

I agree with the interpretation given by andselisk. The authors wish to differentiate thin films vs. bulk of the same polymeric material. One of meanings of bulk given in the unabridged Oxford English Dictionary [by subscription] is

Magnitude in three dimensions; volume.
1825   J. R. McCulloch Princ. Polit. Econ. ii. ii. 141   They [gold and silver] possess great value in small bulk.
  1878   T. H. Huxley Physiogr. (ed. 2) 57   Sea water is denser or heavier, bulk for bulk, than fresh water.

For example, when the authors say:

Our studies reveal that the fraction of the chain-end groups is larger
  in the interfacial layer with its outermost region approximately 1 nm
  below the surface than it is in the bulk.

You can interpret that as "Our studies reveal that the fraction of the chain-end groups is larger in the interfacial layer with its outermost region approximately 1 nm below the surface than it is in the three dimensional body of the [polymer]."


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Estimation of Molar Entropies
There are plenty of compounds out there with unknown standard molar entropies. I know that molar entropy has a strong dependance on the physical phase, molecular weight, and number of atoms. So is it feasible to use these parameters to roughly estimate the molar entropy of, for example, ammonium acetate?

I'm not sure what the question is about; there are several dozens handbooks and databases of thermodynamic data, some are readily available, some are paid only, which list temperature-dependent molar entropies for thousands of compounds. Sure, there wasn't always so much data, but there was a huge gain in accumulating these parameters over the cold war period by both USA and USSR for the needs of rocket scientists. Is there a certain system or compound you are having difficulties finding molar entropy for?

Yes. Currently I'm looking for the std molar entropy of ammonium acetate. This is a reoccurring problem for me, as I keep coming across compounds for which I cannot find the molar entropy. I don't mind if the database is paid-only, could you point me to a fairly large database. Or in this case, at least one which has the molar entropy for ammonium acetate?


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Chemical Anti-bonding and Pauli repulsion
We can explain chemical anti-bonding just using the Pauli repulsion correct?Let's take He2.
2 atoms of He share 4 1s electrons and since the magnetic spin for electrons has 2 values there would be 2 electrons with the same wave function which would violate the Pauli exclusion principle.This makes He2 unstable and it disassociates in 2 He atoms
Am I correct?

The anti-bonding orbitals still obey the Pauli principle. You mixed up two different unrelated things.

No I say that the antibonding orbitals exist due to Pauli exclusion principle.

Let's put it this way: all orbitals except the lowest one are occupied due to Pauli exclusion principle. This applies to atoms as well. He2 is the only system where this is important for the molecule, but not for the atoms.

Orbitals do not exist in a physical sense. You can argue, as @IvanNeretin that occupation of higher energy levels occurs because of the Pauli principle (without that all electrons would go to a single orbital), but those different orbitals (single-electron energy level) exist because they are solutions for the Schrodinger equation.

I know all of this ok?Just the existence of antibonding orbitals is due to the Pauli exclusion principle.In He2 there are 4 electrons in the same orbital ->2 of them must have the same wavefunction and this makes He2 unstable.

Stop thinking antibonding orbitals are a different thing from bonding orbitals , just their existence makes the energy of the system very big.

The reason we speak of orbitals as being exclusively inhabited by 2 (max) electrons (with opposing electron spin quantum number) is because electrons are fermions and therefore observe the exclusion principle. The exclusion principle constrains the allowed electron configurations, disallowing occupation of lower E orbitals by more than 2 electrons. In the case of $\ce{He2}$ it requires occupation of orbitals that raise the total E above the energy of the atoms at greater separation. So, in a word, the answer is yes.
The wikipedia entry on the Pauli exclusion principle alludes to its role:

It has been shown that the Pauli exclusion principle is responsible for the fact that ordinary bulk matter is stable and occupies volume. This suggestion was first made in 1931 by Paul Ehrenfest, who pointed out that the electrons of each atom cannot all fall into the lowest-energy orbital and must occupy successively larger shells. Atoms, therefore, occupy a volume and cannot be squeezed too closely together.[12]

I should add that dispersion interactions mean that there actually is a very weak attractive interaction between He atoms before the repulsive term "kicks in". See e.g. this wikipedia description of the London dispersion force, which has this illustration of the potential between two Ar atoms:

See also the table at  the bottom of the wikipedia article, which explains that for the Ne dimer, dispersion contributes 100% of the total intermolecular interaction energy (I would modify that perhaps to attractive interaction).

Not just in He2 , but in Be2 as well and many more.


text:
Mechanism for the reaction between trisilylamine and hydrogen chloride
In Concise Inorganic Chemistry by J. D. Lee (adapted by Sudarsan Guha), in the chapter "Chemical Bonding", under the topic "Back Bonding with Nitrogen as a donor atom" the following reaction is given:

$\ce{(SiH3)3N +4HCl \to NH4Cl + 3SiH3Cl}$ (bond cleavage through $\mathrm{S_N2}$ mechanism)

I have learnt about $\mathrm{S_N2}$ in organic chemistry. I know its mechanism there and tried to do the same here (there is no reaction mechanism given for this in my book).
As suggested by this comment, $\ce{(SiH3)3N}$ is still basic so the following set of reactions takes place:
$$\ce{\color{orange}{(SiH3)3N} + \color{blue}{HCl} \to (SiH3)3NH+ +Cl-}\tag{Protonation}$$
$$\ce{(SiH3)3NH+ +Cl- \to (SiH3)2NH + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$
$$\ce{(SiH3)2NH + \color{blue}{HCl} \to (SiH3)2NH2+ + Cl-}\tag{Protonation}$$
$$\ce{(SiH3)2NH2+ +Cl- \to (SiH3)NH2 + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$
$$\ce{(SiH3)NH2 + \color{blue}{HCl} \to (SiH3)NH3+ + Cl-}\tag{Protonation}$$
$$\ce{(SiH3)NH3+ + Cl- \to NH3 + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$
$$\ce{NH3 + \color{blue}{HCl} \to \color{green}{NH4+Cl-}}\tag{Protonation}$$
Summing up all the above reactions we get the overall reaction as follows:
$$\ce{\color{orange}{(SiH3)3N} + 4\color{blue}{HCl} \to \color{green}{NH4+Cl-} + 3\color{red}{SiH3Cl} }$$
I got the mechanism I was looking for! But the main problem is the first step - protonation. In my book, it's given that in $\ce{(SiH3)3N}$, due to the presence of back bonding, the lone pair on nitrogen atom is not available for donation. Or in other words, $\ce{(SiH3)3N}$ cannot act as a lewis base (electron-pair donor). Thus the first step of protonation (acid-base reaction) is ruled out.
So, how does the above reaction take place? Do we have any other reaction mechanism for the above reaction? If yes, could you please specify the mechanism? Or is the above one is itself correct?
I am unable to find relevant information regarding this on the internet.

https://chemistry.stackexchange.com/questions/69568/hybridization-of-nitrogen-in-trisilylamine-sih33n

Hmm, I have a feeling your textbook may be wrong after all, and while what you wrote is just what I thought about, it still may be not exactly SN2, but more like addition-elimination, because Si has no problem with being pentacoordinated.

I am not sure what back bonding you are referring to. It may be related to the outdated concept of silicon using its vacant, high-energy d orbitals in any meaningful way which is not actually the case.
There is a second effect which consists of overlap of nitrogen’s p orbitals with the σ* orbitals of the $\ce{Si-H}$ bonds (which are silicon-centred due to the low electronegativity of silicon). This indeed reduces nitrogen’s basicity slightly compared to other nitrogen compounds but the effect is not that important overall. If you check page 2 of the Evans $\mathrm pK_\mathrm a$ table, you can see that the difference between $\ce{(Me3Si)2NH}$ and $\ce{(iPr)2NH}$ is about 10 logarithmic units which is noticeable (i.e. the HMDS salts are milder bases than LDA) but it does nothing to counteract their overall behaviour. (I am aware that I am arguing with the wrong $\mathrm pK_\mathrm a$ values but sadly the table does not contain data for $\ce{(Me3Si)2NH2+}$ so this is the closest equivalent.)
It is also important to point out that you are using a very strong acid $(\mathrm pK_\mathrm a \approx 8)$ which will protonate practically any nitrogen lone pair that it can come across.


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Why is potassium ferrocyanide considered safe for consumption, when it is just one reaction away from the highly toxic potassium cyanide?
According to this answer, heating potassium ferrocyanide will decompose it into potassium cyanide.
Now, potassium ferrocyanide is an approved anti-caking agent, and hence used in table salt, for example.
Table salt is often heated as part of cooking. So, isn't there a chance for trace quantities of potassium cyanide entering the body?
NB: I understand that the dosage here, if any, might not be lethal; am more concerned about trace amounts causing inefficiencies in bodily functions.

That reaction needs very high temperature (you need melt it) and high concentration. So in short,no.

The same question: why are peptides considered safe for consumption when they are just one reaction away from highly carcinogenic nitrosoamines? I hope you see the parallels.

Consider that when heated to 600 Celsius or more a lot of things might decompose and release toxic compounds. This is a major problem in caser of fire in closed environment, when gases can be as harmful as the fire itself. This is to put the example you are giving in place.

I remember my chemistry teacher in the high school in early 80s, who was approaching then his retirement. He told us, he had once been a crazy young teacher, who had demonstrated this safety by eating potassium ferrocyanide in front of his class.

Technically, sodium chloride (table salt) is one typo/reaction away from sodium chlorate (weed killer, oxidizer)....

The reaction described in the linked answer involves producing molten KCN in a crucible. KCN melts at 634 °C. No food could survive at that temperature.

https://chemistry.stackexchange.com/questions/85352/will-squeezing-lemon-juice-on-top-of-foods-with-ferrocyanide-additive-release-hy

@GoodbyeMsChipps No eater will survive that temperature. Assuming the person ate the KCN before the reaction. Boiling is a greater concern.

When I tried to melt (tiny amounts of) household salt in the flame of our gas stove my chemistry teacher told me to better stop my experiments, because of the ferro cyanide in it. Still better than trying to strip a teflon coated pan with fire.

@Mindwin: if I knew I was going to be boiled or heated to 600C, the reaction of potassium ferrocyanide to potassium cyanide would not be foremost on my list of concerns.

Under biological conditions it is almost impossible to release HCN
Free cyanide can be released from potassium ferrocyanide by heating or by strongly acidic conditions (and some heat). Neither of these conditions is possible in a living organism.
The reason why this is so is because the cyanide ligands are very strongly bound to the iron (in slightly more technical terms the dissociation constant is very, very small (ca 10-24)). Most of the mechanisms that generate cyanide require much more violent chemical attack: waiting for the cyanide to dissociate and then protonating it to give deadly HCN will never result in a notable amount of NCH. This means that, in biological systems, ferrocyanide is effectively inert.
This explains why the LD50 in mice, for example, is about 6g/kg which is a lot ([see the safety info here]2 for both ferri- and ferrocyanides).
It is also worth noting that food safety authorities have studied the compounds and agree that:

The Panel concluded that ferrocyanides (E 535–538) are of no safety concern in these current authorised use and use levels.

So, while it might look like you are "one reaction away from cyanide" in practical terms this is not remotely true.

You said ferricyanide.  The question is about ferrocyanide.  They are different.  Is there a typo?

@OscarLanzi yes, typo. But the central issue that Fe(CN)6 complexes are very strongly bound is the same for both oxidation states.

Worth pointing out that cyanide is toxic because it binds so tightly to iron (it prevents hemoglobin from carrying oxygen). If it's already bound to iron, it won't affect your hemoglobin.

I must disagree that ferricyanide is harmless.  It reacts with mineral acids to release HCN and the stomach containsxsuch an acid.  Ferrocyanide, in which the additional electron contributes more metal-ligand bonding, is more inert.

I was asking about the decomposition that might happen during cooking, not inside a living organism. Are cooking acids like vinegar and typical oven temperatures (260 centigrade) strong enough to trigger the decomposition, partially or otherwise?

@OscarLanzi It won't release measurable amounts of HCN in biological conditions which explains why the LD50 is probably a dose of hundreds of grammes for an adult.

@HRJ This is going to be irrelevant in cooking as the amounts in common salt is far too low to be significant even if it were converted to HCN. The estimated LD50 would require hundreds of grams of the pure potassium ferrocyanide to kill you and your food would be inedible with that much table salt alone never mind the small proportion of ferrocyanide in it.

@HRJ And decomposition releasing cyanide requires strong acids and heat. Acetic acid isn't strong enough.

@HRJ it is also worth noting that you need about 0.25g of pure potassium cyanide to kill an adult. This is a lot. Typical table salt contain maybe 10 micro grams of the ferrocyanide per gram. You would die far faster from excess table salt even if all the cyanide were released.

@matt_black "Acetic acid isn't strong enough." What about something like concentrated lemon juice?

@nick012000 Citric acid (principal acid in lemon juice) isn't notably stronger than acetic acid. Besides, how are you going to get enough ferrocyanide to notice anything given how little there is in table salt?

@HRJ Potassium ferrocyanate (referring to $\ce{K4[Fe(CN)6]}$, -- I wonder why neither question, nor answer adds the formula as clarification -- indeed is used as stoichimetric regent to introduce the cyanogroup to (electronrich) arenes via Pd catalysis. Under basic conditions at 125 C (doi 10.1021/op400204g, about 75kg used in DMAc/toluene), or at 70 C in MeCN/water (doi 10.1021/ol1028892) to mention just two examples showing that it provides cyanide, but slowly enough to react with the arene without accumulation.

@matt_black, where would these HCN go in human body, chronically accumulate in some form somewhere or metabolized in some other way, Is there any long term toxicity research available?

@Ben HCN does not accumulate. It is rapidly destroyed or metabolised in biological conditions unless there is so much (say 0.25g of pure KCN taken in a single dose) it overwhelms the body's systems. Given the level in table salt (micrograms per gram) this isn't possible even if there were mechanisms to release CN from ferricyanide, which there are not under biological conditions. This has been studied by food safety authorities.

Because like so much more, in small doses it's fine. Unlike the previous answer though, I think you may be on to something when you say

I understand that the dosage here, if any, might not be lethal; am
  more concerned about trace amounts causing inefficiencies in bodily
  functions.

I think the only concern being about the conversion to cyanide is a little misled though.
See here for the first case of poisoning by potassium ferrocyanide. The guy drank two glasses of it of course, so what's in your salt will never kill you.
Please don't let me speak to any health implications though, I'm not a chemist or biologist. Just providing a source I found and adding my own two cents.
For why it's allowed, there's no evidence it hurts people, and certainly isn't lethal when used as allowed. whether or not it causes slight inefficiencies in bodily functions... I think it's likely it does. But I have no basis for saying that, and it certainly won't stop me from eating table salt with potassium ferrocyanide. Again, fully an assumption, but I assume the 'harm' is negligible. Also, peppers can be poisonous in large doses and have a host of health benefits when eaten regularly and in small amounts. I think we need some serious funding and studies on it's effect on the human body over a real sample of the population to give a definitive answer.
TL;DR; I don't really know, but it's clearly poisonous in large doses, so as far as I can see small doses have to have some, maybe negative, effect.

This is an attempt to answer what I think might be more the spirit of the question than literal translation, because I think @Jan's comment (2nd on OP) answers the exact question very well

Thanks for reading the question fully and taking the time to answer! Since it is not conclusive I won't mark it as an accepted answer. In my naive opinion, anti-caking property seems like a small benefit compared to the potential downside.
BTW, I can't access the linked article (some problem with cookies / javascript).

authors: Bob Jarvis - Слава Україні


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Electrical and thermal conductivity of graphite
I was reading about carbon family where it was mentioned that electrical conductivity of graphite increases with temperature while thermal conductivity decreases.
However, I could not find a suitable explanation for that. According to me, electrical conductivity should decrease with rise in temperature as it happens in case of conductors.
Any help will be appreciated.

Graphite is not a typical conductor. In a way, it is related to semiconductors.

It is a very non-linear effect. The conductivity increases in a certain range but above 100 °C it starts to decrease. I don't think there is any rock solid explanation. See https://link.springer.com/article/10.1007/BF01135029

The mechanisms of heat and electrical conduction are different. Heat conductivity is encouraged by collective motions (phonons) that may become increasingly incoherent at high T. Electrical conductivity is harder to explain. Presumably pi electron delocalization is important, but why this should show a significant T dependence is not clear to me. A quick search suggests conductivity goes through a maximum with increasing T. Can you produce some references or specific temperature intervals?

My book does not mention any intervals.However ,from the above answers I understood that it depends upon the temperature intervals we are talking about.

I don't understand your heat conductivity explanation(I don't know quantum mechanics ).Can you simplify ?

Heat transmission in solids may be the result of collective vibrational motions which convey kinetic energy from one part of a lattice to another (analogous to waves in a body of water). Phonons are collective motions in the lattice: https://en.wikipedia.org/wiki/Phonon


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How can methane absorb infrared light within wide range of frequency but CO2 can't?
Methane is one of the most potent greenhouse gases in Earth's atmosphere.I saw a lot of answers related to it but lacks de facto exegesis to it.

CO2 can absorb too. It is just that their ranges are different.

Comparing the vibrational spectra of two different molecules is just like wondering why two people do not have the same fingerprints. The simplest answer is that these two molecules are different molecules with different atoms hence their vibration modes are supposed to be different.
IR Spectra of Methane and Carbon dioxide are available in NIST. This is how real spectroscopy looked like before the focus shifted from true science to fancier presentations and instruments became black-boxes. Anyway, all you have to note both methane and carbon dioxide absorb quite strongly in the typical range of infrared light 400 to 3800 $cm^{-1}$. The question of wide range does not arise as you can see from the spectra, their absorption band positions are different. Methane's spectra is more complex just because it has more active vibrations (dipole moment change is non-zero) and one can see the fine rotational structure of methane but carbon's dioxide's rotational lines are not resolved.

Oh, we are no longer doing true science? /s (Seriously, that sentence could do with a more neutral tone ;) )

Sometime ago, someone asked here if I did not know LaTeX how could I publish in science? My answer was that most of good fundamental science was done when we had type-writers. Sorry to say these days, presentation matters more, I mean the quality of papers and journal covers look more like comic magazines. Just a fews ago, a Wiley journal suggested that we publish a cover, and the price $ 1000 dollars. Angew. Chem. charges $2000 for a cover.

@M.Farooq Ah, that someone was me:) I couldn't agree more about the rest. Although thousand of dollars is a reasonable price for a chance to end up on TOC ROFL…

@andselisk, Now I remember :-)...Excellent examples for those comic cover articles. I dislike the graphical abstract too...what is the point of those cartoons. Wastes so much time in making them. What about $2000 for a cover option in Angew Chem. That is sad.


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Isomerism and namegiving
Is this the right chemical formula of the $\ce{Rb[Pt(NH3)Br2Cl]}$ complex: rubidium amminedibromidochloridoplatinate(II)?
And is it correct that it has geometric isomerism, being square planar with a cis- and trans-isomers, leaving the rubidium out?

Your formula, $\ce{Rb[Pt(NH3)Br2Cl]}$ is correct for the named complex, rubidium amminedibromochloroplatinate(II) (For clear understanding of naming, you may read Naming Coordination Compounds-1 and Naming Coordination Compounds-2).

Many coordination compounds have distinct geometric structures. Two common forms are the square planar, in which four ligands are arranged at the corners of a hypothetical square around the central metal atom, and the octahedral, in which six ligands are arranged, four in a plane and one each above and below the plane. Altering the position of the ligands relative to one another can produce different compounds with the same chemical formula.
Coordination compounds often exist as isomers—i.e., as compounds with the same chemical composition but different structural formulas. Many different kinds of isomerism occur among coordination compounds (Source).

Cis-trans-isomerism  is one of the more common types. Cis-trans (geometric) isomers of coordination compounds differ from one another only in the manner, in which the ligands are distributed spatially. Most of platinum(II) complexes are square planner. Most common example is Cisplatin ($\ce{Pt(NH3)2Cl2]}$) where the each of two ammonia molecules and the two chlorine atoms are situated next to one another (see the diagram in the blue box below), and hence isomer is called the cis (Latin for “on this side”). Thus, coordination name for Cisplatin is  cis-diamminedichloroplatinum(II) (Wikipedia).

When the identical groups are in opposite sides, it is called the trans-isomer (Latin for “on the other side”). To emphasize this nomenclature, I sketched cis- and trans-isomers of diamminedibromoroplatinum(II) (above) and your compound (below) in the maroon box below for better understanding (cis-isomers are in LHS and trans-isomers in RHS).


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What compressible rubber like gasket material is the least permeable to CO2?
I am trying to can a carbonated solution at 120 PSI.  I am trying to figure out the best material to use for the gasket on the valve cap. I am carbonating my fluid with $\ce{CO2}$ and it is loosing pressure over time due to the permeability of $\ce{CO2}$ through my gasket. I have tried about 8 materials (aluminum, copper, graphite,  buna, FMK A, Viton, ECH, PVDF) I was not able to compress the metals enough with my equipment we also dismissed a number of other materials that are known to be highly permeable to $\ce{CO2}$ or are not feasible and have not found a suitable gasket yet. I also tried using $\ce{N2O}$ as my gas but it seems to have the same permeability as $\ce{CO2}$. my gas has to be soluble in water and non toxic. I know that Viton and FMK A are the same but one supplier said theirs was more pure than anyone else.

Viton is a trade name for a fluoroelastomer type. There are different grades which have different properties. I've had a Viton seal be deformed in a vessel when using tetrahydrofuran as a solvent, whereas Kalrez does not. I've seen different Viton grades again not have great compatibility with acetone. So do check out the compatibility of each supplier's product.
We have two seals that we use for $\ce{CO2}$ applications. EPDM and PTFE.  These both cope with 100 bar pressure.


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Methylammonium lead(II) iodide perovskites in bulk form
I am trying to obtain $\ce{(MA)PbI3}$ as a bulk powder using lead(II) iodide and methylammonium iodide as precursors.
I have not yet tried simply mixing the two in methanol/ethanol and removing the solvent in a rotavap, but this will simply result in a mixed salt with very little actual perovskite crystals if I am not mistaken.
So, how could I obtain $\ce{(MA)PbI3}$ in the highest purity possible using lead iodide and methylammonium iodide?

Procedure from the paper by Xie et al. [1] (Powder XRD didn't reveal any other impurity phases in the bulk samples):

Synthesis of $\ce{CH3NH3PbI3}$. The as-prepared $\ce{CH3NH3I}$ and $\ce{PbI2}$ (Sigma-Aldrich, 5N) were mixed in stoichiometric ratio and dissolved in γ-butyrolactone (Sigma-Aldrich), and then a 40wt% solution was formed. The solvent of as-prepared solution was evaporated in an oven at 60 °C for 2 days and $\ce{CH3NH3PbI3}$ precipitate was collected.

Similar approach has been utilized by Yamada et al. [2] for the synthesis of monocrystalline samples. From the SI:

Sample Preparation
$\ce{CH3NH3PbI3}$ $(\ce{MAPbI3})$ single crystals were prepared from γ-butyrolactone solution at 100 °C. Purified lead iodide (2.73 g, 5.92 mmol) and methylammonium iodide (0.93 g, 5.84 mmol) were dissolved in dry
γ-butyrolactone (4.8 mL) at 60 °C by stirring. The resulting solution was poured into a screw-cap vial which was placed on a hot plate at 100 °C for three days to produce a single crystal with dimensions of around 5 mm x 5 mm x 5 mm. We obtained a clean surface by cleaving the single crystal with thin knife under an Ar atmosphere before being used for the measurements. The experiments were conducted in an Ar atmosphere. We confirmed that the same results could be obtained with multiple samples that were cleaved and kept in Ar atmosphere.

Alternative method by Hoke et al. [3]; from the SI:

Crystals of $\ce{(MA)PbBr3}$ (orange, rhombic dodecahedra) were first
grown by combining stoichiometric ratios of $\ce{PbBr2}$ and $\ce{(MA)Br}$ in concentrated (9 M) hydrobromic acid. Diffusion of acetone into this solution afforded crystals within 2 days, which were filtered and dried under reduced pressure.

It appears that methods described in [1] and [2] are based on the preparation of the solution of $\ce{MAPbI3}$ for the thin film deposition on $\ce{TiO2}$ layer, which was published earlier by Noh et al. [4] (SI).
References

Xie, J.; Liu, Y.; Liu, J.; Lei, L.; Gao, Q.; Li, J.; Yang, S. Study on the Correlations between the Structure and Photoelectric Properties of $\ce{CH3NH3PbI3}$ Perovskite Light-Harvesting Material. Journal of Power Sources 2015, 285, 349–353. https://doi.org/10/f7c87m.
Yamada, Y.; Yamada, T.; Phuong, L. Q.; Maruyama, N.; Nishimura, H.; Wakamiya, A.; Murata, Y.; Kanemitsu, Y. Dynamic Optical Properties of $\ce{CH3NH3PbI3}$ Single Crystals As Revealed by One- and Two-Photon Excited Photoluminescence Measurements. J. Am. Chem. Soc. 2015, 137 (33), 10456–10459. https://doi.org/10/f7p5s9.
Hoke, E. T.; Slotcavage, D. J.; Dohner, E. R.; Bowring, A. R.; Karunadasa, H. I.; McGehee, M. D. Reversible Photo-Induced Trap Formation in Mixed-Halide Hybrid Perovskites for Photovoltaics. Chem. Sci. 2015, 6 (1), 613–617. https://doi.org/10/gc3g49.
Noh, J. H.; Im, S. H.; Heo, J. H.; Mandal, T. N.; Seok, S. I. Chemical Management for Colorful, Efficient, and Stable Inorganic–Organic Hybrid Nanostructured Solar Cells. Nano Lett. 2013, 13 (4), 1764–1769. https://doi.org/10/gc3g8w.


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Computational tool for identification of activity cliff
Are there computational tools for identification of activity cliff between two analog compounds? This paper  list chemotypes that with a high propensity to induce activity cliff.

This may fit well on the new Materials Modeling SE: https://materials.stackexchange.com/


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Order of reaction for hydrogen ions being ignored
I have been reading through a case study about calculating the order of reaction for the following reaction:
$$\ce{H2O2 + 3 I- + 2 H+ -> I3- + 2 H2O}$$
The paper uses an iodine clock reaction with vitamin C to calculate its order of reactions, experimentally. However, when writing out the rate law, the author writes it as such:
$$\mathrm{rate} = k[\ce{H2O2}]^x[\ce{I-}]^y$$
How come it is not written as follows:
$$\mathrm{rate} = k[\ce{H2O2}]^x[\ce{I-}]^y[\ce{H+}]^z\;?$$
For what reason is the order of reaction of hydrogen omitted? My thinking was that it has an order of reaction of $z = 0$ and therefore it does not affect the rate at all. However, I was unable to find a source to verify this.

H+ isn't in the rate determining step so can be ignored.

That would explain it, but would you perhaps have a source for this statement?

Another possibility is that if you have vitamin C (the acid) in a large excess, then $\ce{[H+]}$ might not change significantly. If this is the case, then you can actually use an effective rate coefficient defined by $k = k' \ce{[H+]}^z$ . I am not sure if this is the case here, but if it is, then this reaction is a so-called pseudo-first order reaction.

But according to my experiment, HCl concentration can influence the rate of the iodine clock reaction...

I assume [H+] is kept constant – being in excess – and can be present implicitly in the value of k=f([H+]). Similarly as a reaction with general kinetics of the 2nd order can be considered of the pseudo-first order when one reactant is in excess.

Interesting that without H+, the reaction would yield I3- + 2 OH-.  Peroxide is less reactive in base and I3- complex might be interfered with by OH-.  So, yes, run it with excess H+ (H+ does not participate in the redox, but reacts with a product of the redox).  Lack of it could be considered rate limiting.  (Runs with various conc. of H+ could be tested).

Although the comment by Ezze is a valid and helpful answer, according to chegg.com, the order of reaction with respect to the $\ce{H+}$ ion is $\ce{0}$ and therefore it can be omitted from the rate expression as $\ce{[H+]^0= 1}$.

Imagine you are doing a process which takes 'n' steps. Label the steps : 1,2,3..n, now consider the time it takes to do step 1 , step -2 and so on. Now, wouldn't you agree that the time taken to complete process would be entirely dependent on the step which takes the most time(slowest step) to complete?
So, in your example the hydrogen ions are in one of the faster steps. So, if the concentration of hydrogen is 'enough' such that once you come to the step that their ample hydrogens to fuel the reaction, then it would seem that rate is independent of hydrogen concentration ( I say seem because there may be minute effects of it on the rate)


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What does it mean "polyester doesn't breathe"?
Often people say buy a cotton shirt because cotton "breathes" but polyester doesn't. (i.e. it makes you hot sweaty).
What is the scientific meaning behind this? How does polyester trap the heat more than cotton?
I would have thought it would be the other way round. Is there some chemical explanation?

That is not a scientific term. It is about permeability and thermal insulation combined. It might depend on several things. One could even be how tight is the textile but the main should be hydrophobic/hydrophilic behaviour. Advanced textiles might be also structured. Sweating does not necessarily mean polyester is "warmer", but that water cannot escape/evaporate through. Or less than with cotton. Cotton also comes with different wettability, by the way. Not sure if on chemical base or for how the fibers are worked out.

Fabric "breathing" is as Alchimista states "permeability and thermal insulation combined". Polyester is worse than cotton at allowing sweat to evaporate and pass through it, so it makes you feel hotter because the sweat is still on your skin.

I'm wondering why water can't escape, if the clothing is full of microscopic holes? Does cotton have more holes microscopically? I guess polyester fabric is more "solid"

As stated earlier this is not a scientific term, but I'd say the key is the rate at which water is transferred from one side of the fabric to the other. Because water is absorbed better by cotton, it can soak through to the outside and evaporate more quickly. Permeation of water via "microscopic holes" is likely slow compared to this process. Also the heat of evaporation of water is large and likely dominates heat transfer via conduction through the fabric. I'll admit these are kind of hand waving arguments based on observation and experience, and thus are given as a comment, not an answer.


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Standard Reduction Potentials and Electropositivity
I was looking through a table of standard reduction potentials and the trend seemed like it correlated with electropositivity. Is it reasonable to qualitatively compare two metals to see which has a higher standard reduction potential based on electropositivity? If not, why is this unreasonable?

Your reasoning is reasonable. However, although we can make the association between electropositivity and reduction potential qualitatively, relying on the former one to imply about the reduction potential is not always safe. For instance, following the trend of electropositivity on the periodic table, we would imply based on the reasoning that you proposed, that Zn which has a greater electropositivity than Cu (check the table bellow) has also a greater reduction potential, which is not the case. When it comes to transition metals, there are actualy some other effects that play a role on the reduction potential of the atom, ex.: shielding effects, final electron configuration, etc. The bottom line is, although electropositivity could help to justify a certain trends in reduction potential, it should not be used to make rigorous predictions of these trends since it could easly lead to mistakes mainly when considering transition metals.
P.S. The table shows the trend according to electronegativity. You can visualize the electropositivity trend by inverting the color scale at the bottom.


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Do secondary alcohols react faster than primary alcohols or is it the other around?
My questions is that do secondary alcohols react faster than primary alcohols or is it the other around? This for the oxidation of the alcohol to a ketone or an aldehyde.
This reaction is with KMnO4 with a catalyst of H2SO4
Example of 2 alcohols:
butan-1-ol (primary)
propan-2-ol (secondary)
I thought that secondary has a faster oxidation reaction because its carbon chain is shorter?
I was wondering if someone could provide the answer with multiple reasons?
Thanks in advance!

In general, primary alcohols react faster due to sterics. Exceptions may occur.

To reason about the kinect reaction you must consider the mechanism of the reaction. Honestly I can't say much about this reaction specificaly since the oxidation mechanism by KMnO4 is considerably complex and not much studied probably until graduate courses. I can say, however, that you can't reason about the reaction kinectics under the consideration of chain sizes. How fast a reaction will occur will depend, as said, on the mechanism and the intermediates formed (speed limiting step) during the reaction. These intermediates, in turn, will depend on the stereochemistry of the molecule and the sites where possible charges are placed. These last factors do not depend on chain size.
Hope it helps.


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Photosynthesis-Equation
Everyone knows the simplified photosynthesis-equation from school:
$\ce{6CO2 + 6H2O -> C6H12O6 + 6O2}$
Most people assume that the carbondioxide is being split into carbon and oxygen during photosynthesis, and the equation seems to confirm this. But we know that is not true, and that the oxygen derived from photosynthesis actually comes from the watermolecule being split into hydrogen and oxygen (see e.g. this course note and wikipedia).
Clearly there is much more going on as the formula most of us learned in school suggests, so my question is: what is the correct and complete equation for photosynthesis?

https://en.wikipedia.org/wiki/Light-dependent_reactions and https://en.wikipedia.org/wiki/Calvin_cycle.

Thanks for the links. Found the answer as to why i was confused by the simplified equation:
"Hexose (six-carbon) sugars are not a product of the Calvin cycle. Although many texts list a product of photosynthesis as C6H12O6, this is mainly a convenience to counter the equation of respiration, where six-carbon sugars are oxidized in mitochondria. The carbohydrate products of the Calvin cycle are three-carbon sugar phosphate molecules, or "triose phosphates", namely, glyceraldehyde-3-phosphate (G3P)." https://en.wikipedia.org/wiki/Calvin_cycle#Calvin_cycle

If you labeled the oxygen atoms in carbon dioxide in green and the oxgyen atoms in dioxygen in red (because they derived from photosynthesis, while $\ce{CO2}$ is used up in the Calvin cycle), you would have to add some water molecules to be able to write an equation like this:
$$\ce{6C\color{limegreen}{\ce{O2}} + 12H2\color{red}{\ce{O}} -> C6H12\color{limegreen}{\ce{O6}} + 6\color{red}{\ce{O2}} + 6 H2\color{limegreen}{\ce{O}}}$$
This is problematic because oxygens in carbonyls can interchange with oxygens in water fairly rapidly (e.g. hydration of carbonyl to diol, and reverse reaction losing the "original" oxygen on the carbonyl). It would be fine to write the net reaction of the Calvin cycle separately from the net reaction of photosynthesis. This would also nicely show that hydrolysis of ATP is required for some of the steps, and that NADPH is involved.
On the other hand, if you are talking about the relationship between photosynthetic organisms vs other organisms, it is nice to have a simple net reaction for the entire organism (which also ignores that plant cells do break down glucose as well).

Clearly there is much more going on as the formula most of us learned in school suggests, so my question is: what is the correct and complete equation for photosynthesis?

There is no official complete equation. You can add as much or as little detail you want. It really depends on where you want to go with the chemical equation. The links in the comments are great starting points.

[OP, from comments:] Found the answer as to why i was confused by the simplified equation: "Hexose (six-carbon) sugars are not a product of the Calvin cycle. Although many texts list a product of photosynthesis as C6H12O6, this is mainly a convenience to counter the equation of respiration, where six-carbon sugars are oxidized in mitochondria. The carbohydrate products of the Calvin cycle are three-carbon sugar phosphate molecules, or "triose phosphates", namely, glyceraldehyde-3-phosphate (G3P)."

Great, it feels good to figure things out! In biochemical pathways, it can be difficult pinpointing products. The Calvin cycle and pentose phosphate pathway can be tweaked to make a variety of products, and those tweak affect what the "real" net reaction is.