text:
Using Mathematica to Plot an Isotherm at its Critical Temperature
Hello I'm new to mathematica and I have been trying to plot an isotherm for a substance at its critical temperature, using Van der Waals equation of state. I have done all the work except I do not know why my code is not outputting the graph; it simply outputs the graph with no line in it. Any help is appreciated!
Code:
P = R*T/(v - b) - a/v^2;

Tc = (8 a)/(27 b R);
R = 0.08205;
Tc /. {a -> 5.489, b -> 0.063}
Plot[P[Tc, v, {a -> 5.489, b -> 0.063}], {v, 0, 1}, AxesLabel -> {v(*L/mol*), T(*K*)}]

P is VDW equation, Tc is the critical temperature equation with parameters a and b and I am using the Plot function to graph the isotherm.

The way you have your code set up, your equation for pressure never calls for the value of Tc, instead it tries to use an unknown variable you have called T.
Additionally, if you're looking to plot an isotherm, your y-axis is incorrectly labelled. It should be pressure, not temperature. When plotting, make sure you have your equation's variables defined beforehand, so as to make the code used easier to follow. Here you go:
P = R*Tc/(v - b) - a/v^2;

Tc = (8 a)/(27 b R);
R = 0.08205;
Tc /. {a -> 5.489, b -> 0.063}
Plot[P, {v, 0, 1}, AxesLabel -> {v(*L/mol*), p(*atm*)}]

Hello thank you for your response! Should this make a plot for sure? I tried it right now but nothing was outputted as well.

Sorry my mistake it worked!!! Thank you so much!

No problem, glad to see it all worked out for you!


text:
pH of salt composed of the conjugate of strong acid and base
salt like [HA]B dissloves in water (here we don't consider its solubility) :
$$\ce{HA <=>[$K_\mathrm a$] A- + H+}$$
and
$$\ce{B- + H2O <=>[$K_\mathrm b$] HB + OH-}$$

usually we combine two and get
  $$\ce{HA + B- <=> A- + HB}$$
(Ex: $\ce{NH4CN}$)
and suggest this to be the dominant reaction so as
  to calculate pH
that is , the  amount of HA and $\ce{B-}$ being consumed are assumed to be the same
let x be the amount being consumed and solve  $\frac{x^2}{(C-x)^2}=\frac{K_aK_b}{K_w}$  where C is the initial concentration of the salt

My question is: under what circumstance the suggestion their amount (being consumed) are the same would be invalid ?
( Maybe both of $K_\mathrm a$ and $K_\mathrm b$ is really closed to $K_\mathrm w$ and three of their dissociation must be carefully inspected to get the right answer ? )
and if this is the case, how to solve it ?

It would be helpful if you organize this a bit. Describe the type of problems you want to solve, with a blockquote, then demonstrate your approach and ask where the strategy won't work. This was flagged as unclear, and I have to agree with the close reason unless you [edit] this.

For strong acid and base, the reaction from H+ and A- to HA simply 'happens negligibly' because of the common ion phenomenon. So, the conjugate base of acid (or vice versa) is considered neutral.
(that's why we don't taste any sour or bitterness from NaCl)

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sort: votes


text:
Galvanic Cell: Temperature vs Voltage

$$\ce{Cu^2+ + Ni -> Cu + Ni^2+}$$

The standard potential of this redox reaction is 0.57 volts (which was obtained by 0.23 + 0.34)
$$E = E_0 - \frac{RT}{nF}\ln Q$$
I used 0.1 M nickel sulphate and 0.5 M copper sulphate in my galvanic cell.
$$E_{25} = 0.57 - \frac{8.31 \cdot 298}{2 \cdot 96485}\ln(0.2)=0.59065\ldots$$
$$E_{45} = 0.57 - \frac{8.31 \cdot 318}{2 \cdot 96485}\ln(0.2)=0.59204\ldots$$
Shouldn't the two be inversely proportional? Is the effect of temperature really this minuscule?

Indeed, it should be inversely proportional. In case you divide the copper concentration by the nickel concentration, the sign of the RT/nF part should be plus and not minus. If you divide the product of the reduction by the precursor concentration, you have to use minus.
The temperature has a great effect on the current, especially in the "Cottrell-determined" area. The Nernst equation, which you use in this case, is true for a near-equilibrium environment with fast electrode kinetics. If you drive the potential too high, mass transport can become so slow (in comparison to electron transfer) that an increase in potential will not yield a higher current flow as predicted by Butler-Volmer kinetics.

Is there any practical significance of higher current and mass transport becoming slow?

It depends what you are looking for. You can either work at low potentials, have slow rate constants but mass transport will always 'bring enough reactants to the electrode'. At high potentials, the rate constants are fast enough, but mass transport is too slow and what is near the electrode will be quickly deposited. In electrodeposition/electroplating, working at low potentials (in the Nernst-controlled region) usually leads to more homogeneous electrodeposited films.

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text:
f-electrons in chemistry of heavy transition metals
In one lecture at a recent MCR-X conference I was puzzled by a side-note "f-hole is critical to describe DoS of $\ce{IrO2}$ correctly." The context was DFT periodic plane-wave calculations with pseudization of the core electrons.
This implies that f-electrons are somewhat involved into chemistry of Ir. More precisely, it seems that polarizability of f-shell affects subsequent d-shell in somewhat dramatic way.
Could someone point at relevant articles, preferably discussing the matter from both ab initio and experimental perspective? Bonus points for generalization of the matter: d-electrons affect chemistry of heavy p-elements in no less way.

I'm not sure enough to post as an answer, but I think these two articles are relevant: "Conduction-electron screening in metallic oxides: Ir" http://journals.aps.org/prb/abstract/10.1103/PhysRevB.22.4680 and "Many-body effects in transition metals: role of the density of states" http://iopscience.iop.org/article/10.1088/0305-4608/6/12/015/meta

@DavePhD At the very least, it gives some food for thoughts.

"f-hole is critical to describe DoS of IrO2 correctly."

[...]

This implies, that f-electrons are somewhat involved into chemistry of Ir.

Perhaps naïvely, but I would dispute that the second statement follows from the first. While f-electrons are generally not critical for the chemical reactivity of molecules or materials, their inclusion indeed is critical for a proper description of the Density of States. The magnetic and optical properties, notably, can be derived from the DoS and depend directly from f electrons.

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text:
Getting a cleaner GC Peak Resolution
I have a GC with a peak pair I'm interested in separating but are not getting resolved cleanly at the moment. Any tips on what temperature programming tweaks might help me?
The problem pair elutes at about 3.3 mins. The temp program was 120 C for 2 min followed by a 8 C/min ramp to 180 C & a final hold of 5 min.
Changing the column, detector or carrier gas is not an option. I want to just get the best I can by temp. programming optimization.
I've tried isothermal runs at 120 & 100 C but not much help. The peaks still come unseparated.
Since it is just this two peaks that interest me mostly, theoretically a ramp is superfluous isn't it? Just figuring out the right isothermal run ought to give me the separation of interest?
Edit:
The column is an OV-351 packed col. 2 m long and 1/8" in diameter. N2 carrier gas with FID. The molecules are:

Can you give any more details about the column/setup you're using, and the molecules you're trying to separate?

Try diluting your compounds.

You could also try derivatisation

If time is not an issue. Go for an isotherm at low temperature.

@QuantumCAPUCCINO Yep, time's not an issue. I'll try isothermals at 70 / 80 / 90 / 100 C.  What Temp. would you try? Any others? Even lower?

I would look at literature to be honest. If not trial and error going from high to low. You could also try adjusting the carrier flow.

Is it possible heat during the ramp or even in the injector is causing a rearrangement of the epoxide to the aldehyde?

2m seems short for a column. Is it 20m?

@Beerhunter Maybe. I can try. That would help by making the BPs further apart? Or.....?

@Beerhunter It was a packed column. Not  a capillary column.

Are you measuring relative amounts via GC, i.e., checking for reaction completion?

I just thought that the rearrangement might be occurring during the analysis I.e. The thermal ramp, implying a lower temperature would be better for you. The bp of each compound is too close to separate purely in that way.


text:
What's with the weird definition for the Calorie?
I came across this definition for the calorie, which I found... strange:
"The (15°) calorie is the heat required to raise the temperature of one gram of water from 14.5°C to 15.5°C."
Subsequent searches on-line revealed that this was (one of) the standard definitions for the Calorie.
Now two issues came to my mind after I read that definition:
1) Why is the term 'heat' and not 'energy' used?
From what I recall from my Thermodynamics classes at school last year, heat is energy in transit. I'm under the impression 'heat' is more of a 'surface phenomenon' if you will; there's 'heat' at the junction between two surfaces, but no matter how much you 'heat' up a body it can never contain heat (strictly speaking).
2) What's so special about the 14.5°C and the 15.5°C range?
Seriously, why that particular range? Why not 0°C to 1°C, or even 1°C to 2°C? Heck even 14°C to 15°C would go down (mildly) well with me, but 14 .5 °C to 15 .5 °C? What's with the inordinate fondness for the 0.5°C? A 1°C difference is a 1°C difference in temperature, so why specify any range at all? Why not define the calorie as 'The energy required to raise the temperature of one gram of water by 1°C' ?
I didn't have this question before because I, for all practical purposes, only used to see the calorie as being roughly equal to 4.19 Joules.
Would anyone happen to have the answer to those two queries?
EDIT-
I've already seen this particular question, and I believe it is necessary to state that this does NOT answer my question. My question is distinct from it in the following respects:
1) I'm asking for a clarification regarding the use of the term 'heat' over energy.
2) I want to know why a generalization of the definition, replacing the arbitrary temperature range with a '1° difference'  wouldn't suffice.
3) I want to know why the range is demarcated by two fractions (14.5° and 15.5°) instead of integers.

Related: Why a specific temperature for value of 1 calorie?

hope i get the satisfactory answer here for my question(http://chemistry.stackexchange.com/q/60321/7951)

I also use related, when it is related. You're asking about the same thing here.

A number of the Chemistry texts I've used or examined for adoption have indeed confused heat and thermal energy. They sometime refer to a body possessing heat. But Physics books are pretty unaminous that heat is the transfer of thermal energy and not a state property of matter.
I've never seen, in texts I've used, heat restricted strictly to the interface between the participants in the exchange. It refers to the total amount of thermal energy transferred between two bodies. It can also be used to describe how much energy was lost by one of the bodies in the exchange and how much was gained by the other. One way to reconcile this with your interface statement is that once the surface of a body warms due to contact with a hotter body, the higher temperature at the surface relative to the rest of the body will continue to cause heat to flow within the body.
As to the choice of temperature, it shouldn't near zero since this is where a phase change occurs.
Edit. In response to your edit, #2. Search for a table of "specific heat of water". Notice the specific heat is not constant as temperature varies.
I'll hazard a guess on the 14.5-15.5 part. Think 15 degrees. But, since an interval is required, let's begin a half degree before and end an equal amount after. As to why 15 degrees, hmmm. One answer to a recent question on the Physics forum says this is roughly room temperature. Pretty cold for me but maybe true when and where this was defined. (Sorry for no link, I'm not that fascile on iPad.)

So basically heat is a macroscopic phenomenon while energy is a microscopic phenomenon.

Heat relies on the exchange of energy  via particle collisions. Internal energy, of which one aspect is thermal, relies on the motion and relative positions of particles with respect to one another within a body. I think it would be fair so say both exhibit facets of microscopic and macroscopic properties. Also, although it's particle motion and position involved in internal energy, there is also energy associated with the bulk motion or position of the body itself.


text:
Avogadro's number and percent error help?

Calculate the percent error in Avogadro’s number calculated by using the following experimental data.
  Pipette calibrated as 30 drops/mL
  Diameter of the watch glass: $15.5\ \mathrm{cm}$
  Number of drops to complete monolayer: 23
  Concentration of stearic acid solution: $1.2\times10^{-4}\ \mathrm{g/mL}$

Updated: My thought on this: I would take $23/30=0.77\ \mathrm{ml}$. Then take $0.77\times1.2\times10^{-4}=9.24\times10^{-5}\ \mathrm g$ of stearic acid. Then $(9.24\times10^{-5})/284.48\ \mathrm g=3.25\times10^{-7}\ \mathrm{mol}$ of stearic acid. Then how do I find the Avogadro's number (molecules per mole)?

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

The question is not entirely clear. However, the biggest error is in the solution concentration as 30 and 23 drops are defined exactly. The mass of stearic acid has a small error also although its error may be smaller than you quote. So propagate errors and then round to correct number of significant figures. The exact  mol wt contains exactly Avogadro's number of molecules.

When a problem requires calculations using values, always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.
Nevertheless, your approach is correct. You calculated the volume of the solution as
$$\begin{align}
V_\text{solution}&=\frac{23}{30\ \mathrm{ml^{-1}}}\tag{1}\\[6pt]
&=0.77\ \mathrm{ml}\tag{2}
\end{align}$$
the concentration as
$$\begin{align}
c&=\frac n{V_\text{solution}}=\frac m{M\cdot V_\text{solution}}\tag{3}\\[6pt]
&=\frac \gamma M\tag{4}\\[6pt]
&=\frac{1.2\times10^{-4}\ \mathrm{g\ ml^{-1}}}{284.48\ \mathrm{g\ mol^{-1}}}\tag{5}\\[6pt]
&=4.2\times10^{-7}\ \mathrm{mol\ ml^{-1}}\tag{6}
\end{align}$$
and the amount of stearic acid as
$$\begin{align}
n&=c\cdot V_\text{solution}\tag{7}\\[6pt]
&=\frac{1.2\times10^{-4}\ \mathrm{g\ ml^{-1}}}{284.48\ \mathrm{g\ mol^{-1}}}\cdot\frac{23}{30\ \mathrm{ml^{-1}}}\tag{8}\\[6pt]
&=3.2\times10^{-7}\ \mathrm{mol}\tag{9}
\end{align}$$
Furthermore, you can calculate the area of the surface as
$$\begin{align}
A&=\pi r^2\tag{10}\\[6pt]
&=\frac{\pi d^2}4\tag{11}\\[6pt]
&=\frac{\pi \left(15.5\ \mathrm{cm}\right)^2}4\tag{12}\\[6pt]
&=189\ \mathrm{cm^2}\tag{13}
\end{align}$$
For simplicity’s sake, you may assume that a single stearic acid molecule is a small block with a length $l$ and a width $w$.

Depending on the used model, you may find that the ratio is roughly
$$l=6w\tag{14}$$
Assuming that the surface is densely packed with stearic acid molecules, the number $N$ of molecules can be estimated as
$$N=\frac A{w^2}\tag{15}$$
The corresponding volume of stearic acid is
$$\begin{align}
V&=A\cdot l\tag{16}\\[6pt]
&=A\cdot 6w\tag{17}
\end{align}$$
You also know that
$$\begin{align}
V&=\frac m\rho\tag{18}\\[6pt]
&=\frac {n\cdot M}\rho\tag{19}
\end{align}$$
where $\rho$ is the density of stearic acid. Instead of using the density of solid stearic acid at room temperature, you may want to assume the density of liquid stearic acid close to its melting point, which is
$$\rho=0.847\ \mathrm{g\ cm^{-3}}\tag{20}$$
Solving $\text{(17)}$ for $w$ and inserting the result into $\text{(15)}$ yields
$$N=A\cdot\left(\frac{6A}{V}\right)^2\tag{21}$$
And inserting $\text{(19)}$ for $V$ yields
$$N=A\cdot\left(\frac{6A\cdot\rho}{n\cdot M}\right)^2\tag{22}$$
The Avogadro constant is defined as
$$N_\mathrm A=\frac Nn\tag{23}$$
Inserting $\text(22)$ for $N$ yields
$$\begin{align}
N_\mathrm A&=\frac{A\cdot\left(\frac{6A\cdot\rho}{n\cdot M}\right)^2}{n}\tag{24}\\[6pt]
&=\left(\frac{6\rho}{M}\right)^2\left(\frac An\right)^3\tag{25}\\[6pt]
&=\left(\frac{6\times0.847\ \mathrm{g\ cm^{-3}}}{284.48\ \mathrm{g\ mol^{-1}}}\right)^2\left(\frac{189\ \mathrm{cm^2}}{3.2\times10^{-7}\ \mathrm{mol}}\right)^3\tag{26}\\[6pt]
&=6.6\times10^{22}\ \mathrm{mol^{-1}}\tag{27}
\end{align}$$
Even when taking the uncertainty of the used values into account, this result is significantly smaller than the literature value of the Avogadro constant. Therefore, you may conclude that a large part of the difference may be attributed to experimental error. I guess, you did not use the floating device method to push the stearic acid molecules together, which makes sure that the molecules are densely packed and lined up on the surface of the water with polar heads on the surface and nonpolar tails sticking up away from the surface.


text:
Cyanide inhibitor of Catalase
Why is it that cyanide is an inhibitor of catalase but carbon monoxide isn't?
What makes cyanide bind well to $\ce{Fe^3+}$?

Jan has already covered this topic in a related question: Why is the cyanide ion toxic?
$\ce{CN-}$ and $\ce{CO}$ are isoelectronic, and one might expect them to behave exactly similarly. However, there is a key difference: the orbital energies in $\ce{CN-}$ are higher than in $\ce{CO}$, due to the lower electronegativity of $\ce{N}$ compared to $\ce{O}$.
This means that $\ce{CN-}$ is a much better σ donor, and a much poorer π acceptor than $\ce{CO}$. Consequently $\ce{CN-}$ is much happier binding to a metal in a higher oxidation state, whereas $\ce{CO}$ is much happier binding to a metal in a lower oxidation state.


text:
What is the difference between copper sulfate and copper sulfate pentahydrate?
I want to do copper plating and I have seen people using copper sulfate, but also there is copper sulfate pentahydrate.
What is the difference and which one should I use for electroplating?

Copper (II) sulphate pentahydrate, $\ce{CuSO4.5H2O}$, indicates copper (II) sulphate in crystalline form (for ease you can say copper (II) sulphate whose crystals contain  water).
Copper (II) sulphate, $\ce{CuSO4}$, indicates anhydrous copper (II) sulphate (copper sulphate which has lost its water molecules).
The difference is that $\ce{CuSO4.5H2O}$ has a blue colour due to the presence of water molecules in its structure (for same reason ferrous sulphate has a green colour), while anhydrous copper (II) sulphate, $\ce{CuSO4}$, is white.

It really doesn't matter for your purposes: most commercial "copper sulfate" is actually copper sulfate pentahydrate (copper sulfate plus five molecules of water). Pure (anhydrous, i.e. without water) copper sulfate absorbs water from air to become the pentahydrate, so it's easier to ship the stable pentahydrate.
You're paying for the extra mass of the unneeded water... c'est la vie.

Copper Sulfate is $\ce{CuSO4}$, 159.609 g/mol, and Copper Sulfate Pentahydrate is $\ce{CuSO4*5H2O}$, 249.685 g/mol. Either would work for copper plating.


text:
Sugden's concept of singlet linkage
I was reading about the octet rule and saw exceptions in it. Different theories I read further and one of those was Sugden's concept of singlet linkage which state that the octet rule is never violated.
My question is that why Sugden's concept is not taught us or is not as popular as covalent bond or dative bond? Is there any problem with this concept? Does it have any limitations?

Hint: singlet linkage is a more accurate description than octet expansion, but the octet rule is still violated by octet-deficient molecules such as BeH2.

What he said
He did not say that the octet rule is never violated.
On the contrary, he suggested that "the maximum number of electrons in the valency orbit can not [sic] exceed eight". (source)
This means, while he denies that the octet rule is violated by molecules like chlorine pentafluoride, he did not challenge whether molecules traditionally labeled as octet-deficient such as beryllium difluoride indeed violates the octet rule.
He argued for the existence of a two-center one-electron bond.
Why his theory was poorly accepted
Preconceptions?
The success of his theory
To quote Wikipedia:

In the 1940s and 1950s, Rundle and Pimentel popularized the idea of the three-center four-electron bond, which is essentially the same concept which Sugden attempted to advance decades earlier; the three-center four-electron bond can be alternatively viewed as consisting of two collinear two-center one-electron bonds, with the remaining two nonbonding electrons localized to the ligands.

Indeed, the three-center-four-electron explains the bonding in the triiodide anion well.
Also, another success is that this theory, unlike the expanded-octet theory, does not rule out the existence of hypervalent second-period element, such as carbon ("Frozen" Transition States: Pentavalent Carbon et al., J. C. Martin, Science, 05 Aug 1983, Vol. 221, Issue 4610, pp. 509-514, DOI: 10.1126/science.221.4610.509).
The problems with this theory
He was reluctant to accept that ionic characters do indeed play a larger in some molecules traditionally labelled as hypervalent, such as the sulfate anion.
The failures of this theory
In the sulfate anion, there is no two-center one-electron bond (Wikipedia):

For sulfuric acid, computational analysis (with natural bond orbitals) confirms a clear positive charge on sulfur (theoretically +2.45) and a low 3d occupancy.

Also, some of the traditionally-labeled hypervalent molecules involve very electronegative elements such as fluorine for sulfur hexafluoride, oxygen for sulfate, and chlorine for phosphorus pentachloride. A better explanation for these molecules would need to invoke ionic characters instead of the covalent singlet linkage.
For sulfur hexafluoride (source):

The sulfur d orbitals are found to contribute very strongly to the binding energy ($250$ kcal/mol) and to have a  total occupancy of around $0.25\mathrm{e}$, the valence electron distribution on sulfur being about $32\%$ in $3\mathrm{s}$, $59\%$ in $3\mathrm{p}$, $8\%$ in $3\mathrm{d}$, and $1\%$ in $4\mathrm{p}$. The occupancy of the two sulfur $3\mathrm{d}_{\sigma}$ orbitals ($0.16\mathrm{e}$), however, is only one-sixth of what would be required for $\mathrm{s}\mathrm{p}^3\mathrm{d}^2$ hybridization, and the energetic contribution of these orbitals is only two to three times larger than that of the sulfur $3\mathrm{d}_{\pi}$ orbitals. The sulfur $\mathrm{d}$ orbitals are important because they allow strong back transfer from the negatively charged fluorine ligands to the strongly positively charged ($+2.9\mathrm{e}$) sulfur, in turn allowing significant contraction of the $\ce{S-F}$ bonds and greatly increased molecular stability.


text:
What happens when you 'burn' ash?
This popped into my my mind while watching a match burn (don't ask me what I was doing with a burning match...).
Now as I know it, 'ash' is what you call the residual, grey-black powdery material left over following the combustion of wood.
I'm under the impression that ash is largely just carbon, with little or no organic constituents present along with it (since I'm pretty sure that pure powdered carbon is black, not grey-black).
Now I heated what remained of the matchstick with, well...another matchstick and found out, unsurprisingly, that the ash underwent practically no visible change. So I scooped up some ash into a crevice in a concrete block and then blow-torched it for about 2 minutes. Same result. However, while I was heating it this time, it glowed orange, it subsided as soon as I turned off the torch.
Thinking about it, if I did heat it strong enough it should decompose completely to black colored elemental carbon. If that is the case, to what temperature should you heat it?
I don't think I've considered everything there is to consider in this situation, which is why I've adopted a tentative tone while typing out this question.
Could there be other, side-reactions/effects as well? What would they be?
Additionally if anyone happens to know; what is it that contributes to the grey color that ash normally assumes?

What were you doing with a burning match?

@DHMO you really don't want to know...plus, since I don't want to get arrested for arson, I'm not telling.

I have a feeling you sort of mistook burning to ash with a process like pyrolysis or making coke.

You know, everything kind of glows orange when you heat it for too long. https://en.wikipedia.org/wiki/Black-body_radiation

@DHMO Yeah, but not until (most of them) break down first....

The principal component of wood ash appears to be calcium carbonate. (Wiki entry, "wood ash".) Other components include compounds of potassium and phosphorus. If you heat the calcium carbonate strongly enough, it will decompose into CO2 and calcium oxide. Apparently, strong heating is likely to result in less carbon instead of a larger percentage. The remaining calcium oxide is pretty stable as to temperature although addition of water will readily convert it to calcium hydroxide.
Calcium carbonate is white. White mixed with the black of any carbon present would result in a grey color.

Calcium oxide - yes, but also (mainly) silcon dioxide and potassium oxide, along with some other oxides of metals. If calcium oxide was the main component of wood ash then it would glow brilliant white when heated, not orange (have you ever seen a limelight?).


text:
"4K-metal chelated" A chemical term used in Sagan's Cosmos
Maybe this question doesn't belong here. It's about a hypothetical xenobiology imagined by a prominent scientist in a seminal work of popular science. So I dunno. Tell me to go away and I'll see if some other SE can help me.
I'll never forget the races Sagan made up in his book Cosmos. He gave descriptions of them in the form of excerpts from his imagined Encyclopedia Galactica, a compendium of all the races in the galaxy.
There was one term that never made sense to me: "4K-metal chelated". What does the "4K" part mean? Here is the context:
Civilization Type: 2.3 R
Society Code: 1H1, "We Who Became One".
Interstellar civilization, no planetary communities,
  utilizes 1504 supergiants, OV, BV, AV stars and
  pulsars.
Civilization Age: 6.09 * 10^15 s.
First local initiated contact: 6.09 * 10^15 s ago.
Receipt first galactic nested code: 6.09 * 10^15 s ago.
Source civilization, neutrino channel.
Local Group polylogue.
Biology: C, H, O, Be, Fe, Ge, He. 4K metal-chelated
  organic semiconductors, types various. Cryogenic
  superconducting electrovores with neutron crystal
  dense packing and modular starminers; polytaxic.

(By the way, if anyone knows what "Local Group polylogue" is supposed to mean, I care about that too.)

If you can't find an answer here, maybe the people from https://worldbuilding.stackexchange.com will be able to help

Never actually heard of it, sorry.

If you didn't get it already there's some of fictional terminology. Fortunately you don't ask about 2.3 R type ;) Still it's quite possible that "4K" was made up.

Sounds suspiciously like 4 kelvin, i.e. the boiling point of liquid helium under atmospheric pressure, since it mentions cryogenic and superconducting in the next sentence.

@NicolauSakerNeto -- I think that's our answer. Helium is in that list of elements.

These guys are "cryogenic superconducting electrovores" who apparently have "no planetary communities" so my guess is that "4K" should probably be taken to mean "4 Kelvin" and "metal-chelated organic semiconductors" as that they are composed of various types of semiconductors whose molecules are organics chelated with a metallic ion. About the "Local Group polylogue" thing, it's probably fictional stuff like a sort of interstellar Commonwealth or something.


text:
What are overlap integrals?
I came across something called overlap integrals. I could understand that it is something about the overlapping of orbitals. What is it actually?

It basically gives you a quantitative measure for the overlap between orbitals. It should appear in any physical chemistry textbook. I don't feel confident giving a detailed explanation so I'll leave that to someone else.

Reduce the problem to one dimension first. The overlap integral between two functions, $f = f(x)$ and $g = g(x)$, is defined by
$$S_{fg} = \int_{-\infty}^\infty [f(x)]^*g(x)\,\mathrm{d}x$$
where $^*$ denotes complex conjugation. $S$ is the common symbol for the overlap integral; the subscript $fg$ indicates that it is the overlap between $f$ and $g$.
So, for example, let's try a simple case: let
$$f(x) = g(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$$
(the ground state of the quantum harmonic oscillator). This function is real, so the complex conjugation doesn't do anything. We have
$$\begin{align}
S_{fg} &= \int_{-\infty}^\infty \left(\frac{m\omega}{\pi \hbar}\right)^{1/2} \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\
&= \sqrt{\frac{m\omega}{\pi\hbar}} \int_{-\infty}^\infty  \exp\left(-\frac{m\omega x^2}{\hbar}\right)\,\mathrm{d}x \\
&= \sqrt{\frac{m\omega}{\pi\hbar}} \sqrt{\frac{\pi\hbar}{m\omega}} \\
&= 1
\end{align}$$
(I used a standard integral, $\int_{-\infty}^\infty \exp{(-ax^2)}\,\mathrm{d}x = \sqrt{\pi/a}$). So this tells you that the overlap of this function with itself is equal to 1. That's not very interesting, because if you substitute $g(x) = f(x)$ into the original definition, you'll find that that is simply the normalisation condition for a wavefunction. I just wanted to use it as an illustration.
In general, the overlap between two normalised states ranges from 0 (when they are orthogonal) to 1 (when they are identical). This applies to three dimensions as well.

Of course, the notion of overlap is usually talked about in a three-dimensional context. So, all you have to do is to generalise the definition to three dimensions. Let's say we have two functions $\psi_1(x,y,z)$ and $\psi_2(x,y,z)$. Then we simply have
$$S_{12} = \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \psi_1^*\psi_2 \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$
or in more concise notation
$$S_{12} = \int \psi_1^*\psi_2 \,\mathrm{d}\tau$$
where $\mathrm{d}\tau$ is the volume element (note that Wikipedia denotes it as $\mathrm{d}V$) and the limits of integration are assumed to be over all space.
This won't get you anywhere with atomic orbitals, though. Orbitals are frequently expressed in terms of spherical coordinates. So, you need to know what the volume element in spherical coordinates is, and that is:
$$\mathrm{d}\tau = r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$
(note again that Wikipedia can be confusing sometimes, because the symbols used for spherical coordinates are different: $\theta$ and $\phi$ in chemistry have exactly the opposite meaning from $\theta$ and $\phi$ in mathematics.)
The limits of the integral are still "over all space", but in spherical coordinates this is expressed differently. $r$ takes values from zero to infinity, $\theta$ from $0$ to $\pi$, and $\phi$ from $0$ to $2\pi$. So:
$$S_{12} = \int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^\infty \psi_1^*\psi_2 r^2 \sin\theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$
All you need to do now, is to find the form of the function, plug it into the integral, and evaluate the integral. Warning: That's best done by a computer.

I have a question - if you have $f(x) = g(x) = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$ and then you're doing $f(x) \times g(x)$ surely the integral would evaluate $\exp\left(-\frac{m\omega x^2}{2\hbar}\right)^2$ right since the exponential part of the function is squared? I can see you doubled the power that you raised the normalisation constant, but surely the same applies to the exponential part too?

@vik1245, $f(x) = g(x)$ has a factor of 2 in the denominator of the exponential. This was removed inside the integral as $\exp(a/2) \cdot \exp(a/2) = \exp(a)$. Tough to spot, I'll admit!

@orthocresol Can we have an overlap integral between a real and a complex function?

@adosar Of course, you can define any overlap integral you like using any functions $f(x)$ and $g(x)$, whether they are real- or complex-valued. Whether these functions $f$ and $g$ (and the resulting integral) are physically meaningful is a different question altogether.

I guess you are talking about the overlap integral in as used chemical bonding. If we consider the simplest case of $\ce{H_2^+}$. The $1s$ orbitals are $\psi_1(r)$ on one atom and $\psi_2(r)$ on the other, and are functions of the electron's position r in the molecule. The overlap integral (conventionally called S) is defined as $$ S= \int \psi_1(r)\psi_2(r) d^3r$$ (The $d^3r$ is shorthand for the fact that this is a triple integral as the orbitals are 3 dimensional and we have to work out the distance of the electron from each nuclei using special set of coordinates, which is important, but is a maths detail). As the atomic orbitals are normalised the maximum overlap is 1 and minimum zero. The overlap only contains contributions from regions of space where where the atomic orbitals are not zero. Thus as the nuclei become closer to one another the overlap increases and vice versa, thus the electrons are only ever partially 'overlap' in a molecule. Note, however, if the atomic orbitals are orthogonal to one another the overlap integral is zero at all distances.
On the same theme you will probably have have met the the Coulomb and exchange integrals. The Coulomb integral is $$ J= \int \psi_1(r)\frac{e^2}{r_2}\psi_1(r) d^3r$$
which is the electrostatic interaction between the charge distribution associated with the electron when in orbital 1 (associated with nucleus 1), with the proton in the nucleus of atom 2 (and similarly for orbital 2 and nucleus 1).  The Exchange or Resonance integral is defined as $$ K= \int \psi_1(r)\frac{e^2}{r_1}\psi_2(r) d^3r$$ and is difficult to explain in classical terms because it origin is quantum in nature.

Minor comment: You forgot to complex conjugate $\psi_1$ in all your integrals. The overlap integral vanishes as the distance between the atoms $R\to \infty$, and there is no bonding, I guess. @porphyrin

yes technically that is true, but the $\ce{H2} orbitals are real so it would make no difference in this case, and yes the overlap decreases rapidly as r increases because the wave function decreases exponentially with increase in r.


text:
Extraction of glucose and fructose from golden syrup
Not so long ago, I asked how to make golden syrup.
Now I am wondering how to extract glucose and fructose from it. I would then like to separate the fructose from the glucose (while keeping both of the materials.)
When I made the syrup I used citric acid as a catalyst.
How would you start the extraction if I only need glucose and fructose (separated)? I have no idea.

Extraction in terms of selectively extracting either glucose or fructose from the syrup is going to be difficult to impossible, at least as far as achieving any reasonable degree of purity is concerned.  The glucose-fructose separation method preferred by Dow Chemical is called ligand exchange chromatography.  According to this publication:

Unlike the macroporous structure of DOWEX™ resins used in deashing,
  DOWEX MONOSPHERE™ 99 separation resins are gel beads which have a
  smooth, uniform surface.  While macroporous resins are opaque, gel
  resins are translucent.  Chromatographic separation resins are
  functionalized and the beads contain a significant amount of water.
  While going through the column sections, the sugars to be separated
  dissolve in the water contained within the beads.  Inside the  bead,
  the dissolved sugars interact with the calcium ions held by the resin.
  Fructose, glucose and water form weak ligand complexes with the
  calcium ion.  A stronger interaction in the fructose/calcium ion
  complex than in the glucose/calcium ion complex is the basis of the
  mechanism of separation of fructose from glucose.

As this is a readily available commercial product designed for use on massive scales, constructing a simple separation column based on these separation resins as described in the publication above should be a very doable, cost-effective approach to your problem.


text:
Why did my iodine clock reaction not work?
For the iodine clock reaction, I'm using Potassium iodide, sodium thiosulfate ($\text{0.05 M}$) and sodium peroxodisulphate ($\text{0.04 M}$) and starch as an indicator. I am varying the concentration of $\ce{KI}$ ($\text{0.01, 0.02, 0.04, 0.08 M}$). Should the concentration of the starch be $\text{0.1 M}$?
I ran the experiment but it did not work:
In my procedure, I first mix 10 mL of $\ce{KI}$ with 1 mL sodium thoisulfate, then add 5 drops of starch and then add $\text{5 ml}$ sodium peroxodisulpahte, however no black solution was formed.

How about a copy of the procedural text? Are you using distilled water? Chlorine in tap water will interfere with the redox process.

Maybe try adding few drops of sulfuric acid. Peroxydisulfate can be sluggish in neutral pH.

Try repeating without the thiosulfate - you should get an instant blue/black colour.
If you do, then there's probably too much thiosulfate meaning the concentration is not 0.05 but higher.
If you don't get the instant blue/black without the thiosulfate the it's one or both of the other two solutions.


text:
Why is molar mass the same as atomic mass?
I am completely confused about why molar mass is the same as atomic mass. Could someone help explain why, thoroughly, but in simplified terms?

Do you know what a mole is, to begin with? Also, welcome to Chem.SE.

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass

Because chemists defined it that way, following the KISS-principle.

yeah i understand that a mole the number of atoms in 12 grams of C-12 and thanks.

Ok so I know that the atomic mass of a C-12 atom is 12 u but how is the molar mass 12 g/mole then?

Ar - relative atomic mass is the ratio of the mass of one atom of an element to 1/12 of the mass of one atom Carbon-12 which has Ar of exactly 12. It is a dimensionless quantity. The Ar values are what you see in the periodic table, reflecting the isotopic compositions of the elements (that's why Ar(C)=12.011 and not 12).
Mr - relative molecular mass - ratio of the mass of a molecule to 1/12 of the mass of one atom Carbon-12. As with Ar, Mr is also a dimensionless quantity. It can be calculated as the sum of the relative atomic masses of the constituent elements.
M - molar mass - this is the mass of a substance (m) divided by the amount of substance (n) and therefore has units of g/mol. I can be calculated as Ar (for atoms) or Mr (for molecules) multiplied by 1 g/mol (also called the molar mass constant or standard molar mass) to ensure units of g/mol.


text:
Why is cell potential defined as E0[Reduction]-E0[Oxidation] and not the reverse?
Why is the cell potential defined as:
$$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $$
or
$$ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} $$
as opposed to
$$ E^\circ_{cell} = E^\circ_{anode}  - E^\circ_{cathode} $$
or
$$ E^\circ_{cell} = E^\circ_{oxidation} - E^\circ_{reduction} $$
A few definitions which relate to the problem:

Cell potential is the measure of the potential difference between two half cells in an electrochemical cell
The potential difference is caused by the ability of electrons to flow from one half cell to the other
The difference between the potential for the reducing agent to become oxidized and the oxidizing agent to become reduced will determine the cell potential.   (Chem.Libretexts)

But why is this the case?
Also, why does it then say here that
$ E^\circ_{cell} = E^\circ_{reduction} +  E^\circ_{oxidation} $ rather than $ E^\circ_{cell} = E^\circ_{reduction} -  E^\circ_{oxidation} $

It is universally accepted to express potentials as reduction potentials. However, one of the reactions is always going to be oxidation so you need the reverse. Normally cell potential is the SUM of reduction and oxidation potentials but the value you get for the oxidation reaction from tables is actually the reduction potential so you would have to change the sign. So you actually get:
Cell potential = (Reduction potential for the reduction at the cathode) - (Reduction potential for the oxidation at the anode)
or
Cell potential = (Reduction potential for the reduction at the cathode) + (Oxidation potential for the oxidation at the anode)
because Reduction potential + Oxidation potential = 0 when the same species are involved

Sorry could you please explain what "Reduction at anode" means? I thought reduction occurs at the cathode and oxidation occurs at the anode?  Or are you just saying we treat one of the E0 values as if it were reduction (i.e. flipped) at the anode rather than oxidation?

Yes, by reduction at anode I am referring to the reduction potential for the oxidation reaction which occurs at the anode. For example, if you have the following oxidation half reaction Cu(0) - 2e- -> Cu(2+) you would actually find the value for the reduction Cu(2+) + 2e- -> Cu(0) which can be used in the first equation that I've written. I have edited my answer.


text:
Triiodide ion gravimetric analysis Reaction Equation
Last week I did a gravimetric analysis from the following reaction.
$\ce{AgNO3 + (CH3)4N+I3- -> AgI(s) + (CH3)4N+NO3- + I2}$
The objective was to identify if my reactant was triiodide or pentaiodide. In order to find out we reacted $\ce{AgNO3}$ in excess so all iodide ions precipitated and no $\ce{I2}$ was formed. Then, $\ce{AgI}$ could be weighted and calculations would tell you the valence of the ion.
Question is, if no $\ce{I2}$ is formed, you expect to have $\ce{3 AgI}$ precipitating. But how do you balance the equation?? This is the unbalanced equation:
$\ce{AgNO3 + (CH3)4N+I3- -> 3AgI(s) + (CH3)4N+NO3-}$
I don't need the answer, I am just curious because my professor could not solve the problem.
Thank you and good luck!

What is the oxidation state of iodine in $\ce{I3-}$? Without any redox reaction occurring, there's no way you can precipitate all three iodine atoms as $\ce{I-}$.


text:
What substance(s) will prevent superglue from hardening in its tube?
Superglue often sets in the tube, so I was wondering if ethanol or even water could stop it from hardening and going to waste. Ideas, anyone?

I doubt water would help. Water catalyzes the reaction.

Maybe acetone might help, according to this...

The right answers are below. I'd like to enlist you to use single serving packets, it is the best super glue related life hack I found. https://www.amazon.com/Scotch-Single-Super-Glue-AD119/dp/B00342VCGM

Based on what I gathered from this Wikipedia article on super glue :
Super glue is typically composed of a class of compounds called
 cyanoacrylates.
Cyanoacrylates are capable of taking part in this process we call "Polymerization", where chains of cyanoacrylate molecules, called "polymers" ( Poly- Many, mere- parts) are formed. Each constituent cyanoacrylate molecule in the polymer is called a "monomer" ( Mono- One, mere - part).
The chemical structure of methyl-2-cyanoacrylate:

Polymerization is known to occur for a number of reasons, and can follow a number of pathways. (So to keep this brief, I'll just skim over how methyl-2-cyanoacrylate polymerizes). Wikipedia provides the following mechanism:

Over there, the Nu bit is a common abbreviation we use for a nucleophile in chemistry. It is the presence of a nucleophile, that catalyzes the polymerization of methyl-2-cyanoacrylate.
So when you squeeze out a little superglue, how does it harden? Air around you, as well as the surface you're applying the superglue on will, inevitably, have trace quantities of water. To a small extent, water splits to provide positively charged hydrogen ions (essentially, it's just a proton) and negatively charged hydroxide ions (OH). It's the hydroxide ions that play the role of the nucleophile in the above-depicted mechanism.
So ironically, when super glue "dries out", it just transiently combined with (a part of) a water molecule. That is, as @MaxW points out in the comments, water catalyzes the "drying" of superglue.
Since it's water that causes superglue to harden/set, manufacturers go all out to make sure the packaging is water/air-tight. Ever noticed, that when you buy a tube of superglue, it has a seal at the mouth which you've gotta pierce before you can use it? That seal is what prevents the glue from drying out while it's still in the (unopened) tube. Once you pierce the seal, it's only a matter of days before a sufficient amount of water (courtesy: Atmospheric moisture) manages to seep in and render render the remaining superglue useless.
Note that in case of superglue, the manufacturers don't (typically) add anything to keep the superglue from drying...they just put on a seal and cross their fingers. Whereas your typical liquid/PVA glue is a water based emulsion. When PVA glue is squeezed out of the tube and onto a surface it, literally, "dries out" by losing water.
Once superglue is set, it is set... there's no turning back O:)
You can't revert it to usable superglue through any (simple) means.
However, since you now know that it's water that's responsible for superglue "hardening", if you want to make the superglue left in the (opened) tube last longer, store it an an air tight jar that's been packed with a desiccant, like silica gel ;)

Your use of the term "drying out" is dangerous. Superglue polymerises, it doesn't dry out. You would be clearer if you clarified that at the start and didn't keep using the incorrect description.

This MSDS is available from Super Glue Corporation / Pacer Industial.  It states that their superglue ranges from 85% to 100% ethyl-2-cyanoacrylate, which is soluble in acetone.
While it is also soluble in several other organic solvents, 99% acetone is available at you local pharmacy and clearly the most readily available and affordable to try.
Just to be clear, the solution here is to get past any of the glue that's already hardened so that you can still use the remaining glue.
Once it's hard, it will not work as glue again, and pre-mixing with acetone or anything else to prevent hardening will prevent it from working as glue in the first place.


text:
How can a molecular orbital with a nodal plane still be considered "bonding"?

Explain how a molecular orbital can have a nodal plane and still be a bonding molecular orbital.

I understand the concept of how nodal planes are formed and how they have a probability of 0 when trying to find an electron in the nodal plane, but I don't understand how it can be considered a bond especially since these nodal planes are usually called "anti-bonding". This question itself seems contradictory and I'm having a hard time trying to understand how this is possible.

The number of nodal planes isn't the determining factor. Yes, having more nodal planes increase the energy of the orbital, but you need to consider the fact that some MOs (specifically the anti bonding ones) are raised in energy. Overall, the total energy of the orbitals is the same as before, so you stabilize some and destabilize others. Take a look at the MOs and you can see that the there's two bonding MOs with one nodal plane perpendicular to the plane of the benzene with $E_{1}$ symmetry. Likewise, there's a pair of MOs with two nodal planes that with $E_{1}$ symmetry that are anti-bonding. These pairs happened to be correlated in that the increase in energy of the anti-bonding orbitals is essentially matched by the stabilization of the bonding orbitals.

An intuitive way of looking at this is that when you have many atoms to consider, there will be multiple bonding/antibonding interactions between atoms. Whether the MO is overall bonding or antibonding will depend on the sum of these interactions. This is rather simplified, but using the example of benzene again, we can examine the three bonding π MOs:

MO 1 (the lowest-energy π MO) has one nodal plane in the plane of the ring. However, because this isn't a plane separating two constituent atomic orbitals (the p orbitals), this nodal plane doesn't tell us anything about bonding/antibonding properties of the MO. Indeed, this nodal plane only arises because the building blocks – the p-orbitals – have a node in their centre. The interactions between different p-orbitals are all bonding in nature, because the shaded (positive) lobes of each p-orbital are overlapping with the shaded lobes of their neighbours (and vice versa for the unshaded lobes).
MO 2 has a nodal plane between p-orbitals, which suggests the presence of antibonding character. That is true: there are two major antibonding interactions, one between the top-left and top-right carbons, and one between the bottom-left and bottom-right carbons. (The middle-left and middle-right carbons are too far apart). However, there are also multiple bonding interactions: the three carbons on the left, for example, are all interacting with each other in a bonding fashion. Overall, this MO is bonding, although not quite "as bonding" as MO 1.
Likewise, MO 3 has antibonding interactions between the top-left and bottom-left carbons (for example). It also has bonding interactions between the top-left and top-right carbons. However, the antibonding interactions are occurring over a greater distance than the bonding interactions. Consequently, they are weaker than the bonding interactions, and again we have a case of net bonding.

I'm not sure if Hückel theory is rigorous enough to invoke, but in that theory the coefficients are also different, which is why even though MO3 has "fewer bonding" orbitals, it can still be degenerate with MO2. This can be intuitively depicted by varying the sizes of the orbitals. Page 5 in the following link shows this: https://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2007/lecture-notes/lecture31.pdf

I wish one would separate the concept of bonding/ anti-bonding interactions from energy considerations. There are probably examples where an 'overall bonding' MO is anti-bonding with respect to one bond, which will be elongated due to that fact. After all these considerations led to the believe that the HOMO of CO is anti-bonding, which cannot be supported by any kind of theory. There is one universal truth: All molecular orbitals have nodal surfaces. Within every MO there are bonding and/or anti-bonding interactions.

@Blaise Symmetry alone dictates that MO2 and 3 depicted above are degenerate (and any linear combinations thereof). Hückel's calculations show that in a very simple and accurate way.

@Martin-マーチン in Huckel theory, any interaction between atom $i$ and $j\neq i$ leads to a contribution of $2\beta_{ij}c_ic_j$ to the energy $E = \mathbf{c^THc}$; so AFAIK I think it is fair enough to link bonding ($c_ic_j > 0$) interactions to a decrease in energy and antibonding ($c_ic_j < 0$) interactions to an increase (since $\beta_{ij} < 0$). The way I'm seeing it (correct me if I'm wrong), the problem is linking this to the overall bonding/antibonding character of an MO. For benzene it is simple, because every p-orbital has the same value of $\alpha$; so for the MO if $E < \alpha$ [...]

[...] we can say it is bonding, and if $E > \alpha$ it is antibonding. In the absence of any interactions between pairs of AOs (which corresponds to turning every off-diagonal element $\beta_{ij}$ to zero), every MO simply has the energy $\alpha$, so any form of bonding interactions will bring the energy below $\alpha$. Thus we regain the qualitative and rather simplified picture in my answer. For CO we don't have the privilege of all $\alpha_i$ being the same, and consequently I don't know if this kind of approach remains valid.

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


text:
Difference between Lindlar and Rosenmund catalysts
Is there any difference between the Lindlar and Rosenmund catalysts? I've checked around, and it seems the same compounds are used to make both. Is there a difference in their reactivities or are they essentially the same?

The additive.$%stuff in comments =D$

This is a rather interesting question because these names actually refer to classes of reactions (specific to certain reagents and products), and aren't constrained by specific proportions of substances or even the identity of these substances.
$\hspace{4cm}$
A Rosenmund catalyst is used to reduce acyl chlorides to their corresponding aldehydes, and is typically composed of palladium supported on $\ce{BaSO4}$, and poisoned by sodium acetate, N,N-dimethylaniline, thiourea, thiophene, dibenzothiophene, ethyldiisopropyl amine, or, most commonly,
quinoline.
A Lindlar catalyst is used to reduce alkynes to cis alkenes, and is typically composed of palladium supported on $\ce{BaCO3}$, $\ce{CaCO3}$, or carbon, and poisoned by lead acetate and quinoline.
The broad list of poisons which cause nearly identical catalytic activity of the supported palladium in both Lindlar and Rosenmund catalysts tells that the poisons are not involved in the catalytic process itself, but rather are affecting the surface of the catalyst. Interestingly, Lindlar-active palladium shows no activity for the conversion of benzoyl chloride to benzaldehyde under various conditions. This demonstrates that despite similar catalyst preparation procedures, the two reactions are catalyzed by different active sites.$^{[1]}$

$[1]$ Mcewen, A. B.; Guttieri, M. J.; Maier, W. F.; Laine, R. M.; Shvo, Y. The Journal of Organic Chemistry 1983, 48 (23), 4436–4438.

Lindlar catalyst: we see it was used in the preparation of alkenes from alkynes, where an alkyne is partially reduced with a calculated amount of dihydrogen in the presence of palladised charcoal $(\ce{Pd/C})$, and it is deactivated with poisons like sulfur compounds or quinoline give alkenes.
$$\ce{RC#CR' +H2  ->[Pd/C]     RHC=CHR' (\text{cis-Alkene})}$$
Palladium on barium sulfate $(\ce{Pd-BaSO4})$ is sometimes called Rosenmund catalyst. This is used in Rosenmund reduction, where acyl chlorides leads to the preparation of an aldehyde.
These both catalysts have palladium in them but the compound used with it is different in both above cases. Thus we can conclude that these catalysts look similar, but they are not the same.

$\ce{R-H-C=C-H-R}$ does not make sense chemically. Also, why did you put the reaction arrow where it is?

$LaTeX \ce{CaCO}3$ formatting


text:
Can isopropyl alcohol vapor damage polymers?
Since isopropyl alcohol is found in many cleaning products, can the vapor emanating from the isopropyl alcohol corrode, stiffen or react with polymers?
For example, if I use isopropyl alcohol to clean my desk, will its vapor cause any damage to the mousepad, cables and synthetic rubber?
I used 70% isopropyl alcohol to clean my mousepad before, and it ruined the surface by hardening the polymer fibers. I wonder if the vapor can cause the same thing.
Thanks

I will use your example of the mouse pad to answer your more general question.
As you implied in your question, the mouse pad will be exposed to some isopropyl alcohol vapor if it is used to clean in the vicinity of the pad. As the vapor will quickly disperse to negligible levels, a simple solution could be to move the mouse pad (or other rubber-based or otherwise isopropyl alcohol sensitive item) to the other side of the room while cleaning the area around the computer. Once the area is dry and the smell is gone, there should be negligible isopropyl alcohol vapor remaining and you can put the mouse pad back. For the reasons discussed below, even this much effort is likely overkill and you can probably safely just clean around the pad and other items with isopropyl alcohol.
The key reason that isopropyl alcohol vapors are much less damaging to the rubber component of the mouse pad is simply how much more of the compound comes into contact with the rubber when in direct contact with the liquid phase vs secondary exposure via the vapor phase. Depending on a number of factors (contact time with the liquid, temperature of the surfaces that are cleaned with the isopropyl alcohol, proximity and quantity of liquid used in vicinity of mouse pad, etc.) the mouse pad could be exposed to a few orders of magnitude more isopropyl alcohol when in contact with the solution than when merely exposed to vapors from cleaning some area in the vicinity of the mouse pad. This means that the exposure of the sensitive items to the isopropyl alcohol vapor is very small and not likely to have any noticeable effects on the exposed items like the mouse pad.

Technically yes, since IPA is found both in the liquid and vapor. The rate at which the vapor will damage your mousepad would be determined by how much IPA goes into the vapor and how long it stays in contact with the mousepad.


text:
On Wavefunctions of Hydrogen atom
In the context of the wave functions of electron in a hydrogen atom, I just want to get clear regarding the terms probability and probability density of an finding an electron. To be precise, we know that the value of $|\Psi|^2$ at a point gives the probability density of the electron at the point. Now, does that imply that if the value of $|\Psi|^2$ is maximum at a point, the electron is most likely to be found at that point?
Reason for my doubt: The question given by my instructor goes:

Given that the normalized wave fuction of the electron in $\mathrm{3d}_{z^2}$ orbital of an Hydrogen atom is:
  $$
\Psi = N \sigma^2 \mathrm{e}^{-\frac{\sigma}{3}}(3\cos^2\theta-1) \, ,
$$
  where $\sigma={\frac{r}{a_o}}$
Now where is the electron most likely to be found?

The solution given goes like:
If $P$ represents probability of finding the electron and $\mathrm{d}V$ the volume element then,
$$
\frac{\mathrm{d}P}{\mathrm{d}V} = |\Psi|^2
\implies
\mathrm{d}P=|\Psi|^2 \mathrm{d}V$$
Also since,
$$\mathrm{d}V = r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi$$
We get,
\begin{align}
\mathrm{d}P &= |\Psi|^2 r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi \\
&= N_1 \sigma^4 \mathrm{e}^{-\frac{2\sigma}{3}}{(3\cos^2\theta-1)^2} \sigma^2 \sin\theta \mathrm{d}\sigma \mathrm{d}\theta \mathrm{d}\phi \\
& = N_1 [ \sigma^6 \mathrm{e}^{-\frac{2\sigma}{3}} \mathrm{d}\sigma ][ {(3\cos^2\theta-1)^2} \sin\theta \mathrm{d}\theta ][ \mathrm{d}\phi]
\end{align}
He later maximized the $[\sigma^6 \mathrm{e}^{-\frac{2\sigma}{3}}]$ and $[{(3\cos^2\theta-1)^2} \sin{\theta}]$ terms to get the values of ${\sigma}$ i.e. hence $r$ and ${\theta}$ at which the electron is most likely to be found. Infact, he stressed on the maximization of ${\theta}$ part which came out to be $0$. And hence, a conclusion made that for an electron in $\mathrm{3d}_{z^2}$ orbital, the electron is most likely to be found in the XY plane.
I find all this stuff confused and fishy. Can someone explain me this solution or prove it wrong via a right way of solving this question.

I thought the difference between probability and probability density is that the former is a value while the latter is a function....

@DHMO So, how does the solution to the question go. Are the conclusions right?

I made an attempt to improve math expressions a bit. Note to OP: the probability density is $|\Psi|^2$ and not $|\Psi^2|$. I'm not sure if the later is even defined for a complex-valued function. :)

I can assure you, $|\Psi|^2$ and $|\Psi^2|$ are exactly the same thing.

@IvanNeretin, I don't think so. Show me the definition of the square or a complex number. I've never seen it.

@Wildcat The square of a complex number is that number multiplied by itself. You can multiply arbitrary complex numbers, can't you?

@IvanNeretin, please, show me any credible source where $c^2$ is defined.

@IvanNeretin, cause if you define the square of a complex number $c=a+b\mathrm{i}$ as $c^2:=c \cdot c$, then $|c|^2 \neq |c^2|$ since $|c|^2:=c^* \cdot c=a^2+b^2$ while $|c^2| :=c \cdot c = |a^2+2ab\mathrm{i}-b^2|$.

Now that's totally right, and these two expressions are in fact equal! See, $|c^2| = |a^2+2ab\mathrm{i}-b^2|=\sqrt{(a^2-b^2)^2+(2ab)^2}=\sqrt{a^4+b^4-2a^2b^2+4a^2b^2}=\sqrt{a^4+b^4+2a^2b^2}=a^2+b^2$.

Let us continue this discussion in chat.

Now, does that imply that if the value of $|\Psi|^2$ is maximum at a
  point, the electron is most likely to be found at that point?

It certainly does imply something similar to what OP is saying, but not quite. As always in quantum theory one has to be very careful with his or her wordings. Namely, we are talking about a continuous random variable here, so the only probability which makes sense is that of it falling within a particular range of values, i.e. probability of finding an electron in some infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ and not on this single point. Still, if you divide the whole space into volume elements of an identical (infinitesimally small) size, the bigger the value of $|\Psi|^2$ is, the bigger is the above mentioned probability.

Now to the actual findings. So, for an electron occupying the $\mathrm{3d}_{z^2}$ orbital, the probability of finding it in an infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ is given as $|\Psi|^2 \mathrm{d}V$, where $|\Psi|^2$ is the probability density. We analyze the probability density
$$
|\Psi|^2
=
N^2 \sigma^4 \mathrm{e}^{-2\sigma/3} (3\cos^2\theta-1)^2 \, ,
$$
or, more precisely only its angular $(3\cos^2\theta-1)^2$ part. So, when does this expression have its maximum value? The global maxima are at $\theta = 0$ and $\theta = \pi$, i.e. alongside $z$-axis, plus there are local maxima at $\theta = \pi/2$ and $\theta = 3\pi/2$, i.e. in a plane perpendicular to $z$-axis. 1

From the discussion with OP, it seems it must be noted that terms "probability density" and "probability" are defined with respect to a particular random event. Above we analyzed the probability density of finding an electron in an infinitesimal element of volume $\mathrm{d}V$ centered at a point $r$. The radial distribution function arises when considering a probability of a different event, namely, the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$. We start from the probability of finding an electron in infinitesimal volume element $\mathrm{d}V$ at a point $r$
$$
  |\Psi|^2 \mathrm{d}V
  =
  \underbrace{R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2}_{|\Psi|^2}
  \underbrace{r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi}_{\mathrm{d}V} \, ,
$$
and integrate out $\theta$ and $\phi$ dependences
$$
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  |\Psi|^2 \mathrm{d}V
  =
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2
  r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi \, .
$$
And since the spherical harmonics $Y_l^m$ are normalized (to unity)
$$
  \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi}
  |Y_l^m(\theta, \phi)|^2 \sin\theta \mathrm{d}\theta \mathrm{d}\phi
  =
  1 \, ,
$$
for the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$ we simply get
$$
  R_{nl}^2(r) r^2 \mathrm{d}r \, .
$$
The corresponding probability density is, of course, $R_{nl}^2(r) r^2$. Such analysis is different from what we did above for the angular $\theta$ dependence: we simply ignored the radial and the other angular dependence instead of integrating them out. This ignorance makes our analysis a bit rough, but not worthless: at least we were able to identify some general trends in the angular $\theta$ dependence.

1) See Wolfram Alpha for details.

I'm going fine through the mathematics and I agree to your idea. But I doubt the way the question was solved. If the same idea is applied for an electron in 1s orbital, wouldn't we get that sin{\theta} should be maximum for the probability to be maximum in an infinitesimal dV volume element around that point. Now which would imply {\theta}=pi/2. Don't these conclusions seems counter-intuitive or awkward in the sense that in a 1s orbital, the probability of finding an electron in a dV volume element should be spherically symmetric.

Sorry, there was a mistake: $\sin\theta$ should not have been counted in the probability calculations, since it is a part of $\mathrm{d}V$.

This particular mistake was in the question itself in the first place. @user104014, your intuition serves you right: the whole thing was clear as mud.

BTW, I think that the conclusion that an electron in $\mathrm{3d}_{z^2}$ orbital is most likely to be found in the $xy$-plane is wrong. It is certainly likely that it will be there, but even more likely it will be somewhere along the $z$-axis.

@Ivan Neretin Ok! Let's leave this solution for a while. Now how you would have approached to this question. Like, when you get the wavefunction of an electron in an orbital (say 1s or 3dz2), then what conclusions can you make (in regard to electron's probability to find).

@user104014 I'd write along the same lines, except I'd point out in the answer itself from the very beginning that including $r^2\sin\theta$ was plain wrong, precisely for the reason illustrated in your first comment.

@Wildcat Even I have the same intuition as in your last comment. But the analysis gave us that {\theta}=pi/2 is the global max. Whats happening. If the methodology followed in the solution mentioned by me in the question is wrong, what is the right one?

I agree with @Ivan: OP went wrong on including parts of the $\mathrm{d}V$ in the probability calculations. $r^2 \sin\theta$ play exact same role as $\mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi$: they are parts of the differential $\mathrm{d}V$ that indicates the variable of integration, and not parts of the integrand.

Yes, @user104014, you wrongly included $r^2 \sin\theta$ into the integrand.

@Ivan OK! If the inclusion of $r^2\sin\theta$ is wrong, then I have two arguments. One that if we don't take that term, then we will end up maximizing probability density which  doesn't seem to be correct (or atleast I think so). Second, if you don't include that term, then in case of 1s orbital, you won't get that for $r=a_o$ i.e. the bohr radius, the probability of finding the electron is max. Instead we'll get that $r=0$ is where electron is more likely to be found.

Maximizing probability density, though it has no immediate physical meaning, at least might give you an insight at what the whole 3D density plot looks like. Maximizing probability density times $r^2\sin\theta$ has no meaning whatsoever. As for 1s, that's right, the probability density has a maximum at the origin. What's wrong with that?

@user104014, you need to analyze the probability density. There is absolutely no need at all to consider what the element of volume $\mathrm{d}V$ is.

Why do you all say that $dV$ should not be included?

Because it is not a part of the probability density.

@Ivan Then for 1s, why do many books find the radial probability distrubution function i.e. $4pir^2dr{{\psi}^2}$ and then conclude that this function is max at bohr radius i.e. 0.529 A° nearly. What all this stuff then?

I don't quite understand what are you talking about, but the so-called radial distribution function $P(r)$ does not contain $\mathrm{d}r$. $P(r) \mathrm{d}r$ gives you the probability of finding electron at a distance between $r$ and $r+\mathrm{d}r$.

@user104014 Because, I suppose, they are interested in that radial probability distribution. To obtain it, you take the overall probability density and integrate over all possible values of $\varphi$ and $\theta$, thus making these two go away. Mind you, during integration you will introduce $r^2\sin\theta$.

@Wildcat As Ivan mentioned just now, why are we introducing ${r^2}\sin{\theta}$ in case of radial distribution functions?

@user104014, it seems like you're completely lost in the topic. :| To get the radial distribution function you need to integrate out the angular dependencies, so you need to "resolve" $\mathrm{d}V$ as $r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi$ for the purpose of the integration. Here we're doing a different analysis: we just look at the angular part of the probability distribution $|\Psi|^2$.

Thanks @Wildcat, I did not realise that we need to leave out the dependence in the volume.

Please try to avoid extended discussions in the comments section. You can use [chat] or open a chatroom yourself. Since the discussion seems to have ended, I refrain from transferring it to a chatroom of its own. If that is not the case, flag this comment and ask to transfer comments to a chatroom. (cc @user104014)

Although you ask about hydrogen atoms the same principle is true for any wavefunction.  In quantum mechanics we often need to find the probability at some position x. However, x is a continuous variable with an infinite number of values so it does not make sense to ask what is the probability of being at some exact position, say $x=2/3$, as this will be vanishingly small.
Instead we imagine that the probability is that of being at position x to $x+dx$ where dx has some very small value. Suppose there is some general function $f(x)$ from we wish to find the probability density then $f(x)dx $ is probability of being between x and $x+dx$ and $f(x)$ is the probability density which must be a real,  non-negative number. It turns out that quantum mechanics postulates that the probability density is given in terms of the wavefunction is $|\psi|^2$ and the $|~|$ means take the absolute value. If the wavefunction is a complex quantity, and it often is, then we write $\psi^*\psi$, where the * indicates the complex conjugate of the wavefunction $\psi$, i.e. replace i with $-i$ . Finding the maximum of the probability density is done in the same way as you would find the maximum/minimum  of any function, differentiate, set equal to zero and solve. Check if maximum or minimum.

I agree to everything you said. But if why are we maximizing probability density? If thats the case, for an electron in 1s orbital, shouldn't it be that the electron is more likely to be found at the center, i.e. the nucleus instead of thr bohr radius? [because for 1s orbital, the wavefunction is an exponentially decreasing function like $ke^(-r)$]

Notice that for your d-orbital, the wave function is actually something like $r^{2}e^{-r^{2}}$. The maximum value of the norm will occur at some non zero value of $r$.

The atomic wavefunctions are often defined in  spherical polar coordinates and so the volume element over which to integrate is in three dimensions and is $r^2\sin (\theta )dr d\theta d\phi$. The radial part can be separated out and the volume element is therefore  $r^2dr$ which means that the probability is not at zero. For the 1s orbital the probability of being at r to $r+dr$ is $(4/a_0^3) r^2\exp(-2r/a_0)dr$  where $a_o$ is the Bohr radius.

@porphyrin Exactly, thats where my doubt is! If we know seperate the angular part, we'll be left with $\sin{\theta}d{\theta}d{\phi}$. Now maximizing this gives us ${\theta}=\frac{\pi}{2}$. Whats this?

You asked specifically about the radial part. I should have mentioned that to calculate the total probability of finding an electron at r to $r+dr$, say in 1s orbital for simplicity, we need to account for the angular parts also, and do this by integrating over them as we seek radial probability, $r^2dr\int_0^{\pi}\sin(\theta)d\theta\int _0^{2\pi}\psi ^*(r,\theta ,\phi)\psi(r,\theta ,\phi) d\phi = (4/a_0^3)r^2\exp(-2r/a_0)dr $

as to the angular probability this is $\sin (\theta) d\theta d\phi \int _o^\infty r^2\exp(-2r/a_o)dr =(a_0^3/4) \sin (\theta) d\theta d\phi $ (probability of being in that small volume element) and the angular probability density (the 'shape' of the orbital in a plot) is therefore $a_0^3/4$ a constant as expected for a sphere.


text:
How do I arrive at the mass of solvent from a molar fraction concentration?
So the question says:

If the mole fraction for $\ce{H2SO4}$ in an aqueous solution is $0.325$, what is the mass of water (in grams) in $100\ \mathrm{mL}$ of solution?

Me, and my other classmates tried to solve this one but we didn't reach a fully correct answer and our teacher arrives at a different answer.
My answer is:
Since X(H2SO4) + X(H2O) = 1
So: 1-0.325=0.675
X(H2O)=0.675
The mole percent is (0.675*100)=67.5%
1 mole percent = 0.01 mol
H2O moles = 0.675mol
H2O mass (g) = (0.675) * (18) = 12.15 g
Teacher is not convinced with this answer and say its wrong because the total must be 100g, and if the H2O mass is 12.15g then H2SO4 mass will be 31.85g, and 12.25+31.85 dont equal 100.
His answer is simple as follows:
1-0.325=0.675
0.675=67.5% of H2O
H2O mass= 67.5g

Since you worked and found an answer, could you post it and your work? Your teacher's answer would also be helpful. That way we can give you better answer.

Hi and welcome to chemistry.stackexchange.com. Note that this is a [tag:homework] question. We cannot tell you whether and where you went wrong unless you post your work. It would also be helpful to include the answer your teacher has. Please [edit] your post accordingly or this question will get closed. Learn more about closing, editing and everything else in the [help]. By the way: You can use MathJax for mathematical (and chemical) equations. See also this post.

Both your teacher and your answer are wrong!
There are two steps to this problem.  First, we need to convert between mass fractions and mole fractions.  Second, we need to convert from a per-amount basis to a per-volume basis.  "Per-amount" is my made up word for all units such as mass fraction, mole fraction, molality, etc that are expressed per amount (whether moles or mass) of substance (whether total or solvent).  "Per-volume" units include things like molarity, grams per liter, etc., where the basis is the volume of solution.  Converting from an amount basis to a volume basis requires knowing the density of the solution!
Step by step:

Converting to mass fraction

$$ 0.325 \frac{\mathrm{mol\;\ce{H2SO4}}}{\mathrm{mol\;total}} \Rightarrow  \frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}}$$
$$\frac{0.325\;\mathrm{mol\;\ce{H2SO4}}}{(1-0.325)\;\mathrm{mol\;\ce{H2O}}} \times \frac{98.1\mathrm{\frac{g\;\ce{H2SO4}}{mol\;\ce{H2SO4}}}}{18\mathrm{\frac{g\;\ce{H2O}}{mol\;\ce{H2O}}}}=2.62\mathrm{\frac{g\;\ce{H2SO4}}{g\;\ce{H2O}}}\Rightarrow \frac{2.62}{2.62+1}\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}=0.724\mathrm{\frac{g\;\ce{H2SO4}}{g\;total}}\Rightarrow (1-0.724)\mathrm{\frac{g\;\ce{H2O}}{g\;total}}=0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}$$
This step is pretty easy to do using the information in the problem.  There are 72.4 grams of sulfuric acid present in 100 g of the solution.  So far, so good.

Converting to a per volume basis

But at this point the problem gets very tricky:

...what is the mass of water (in grams) in 100 mL of solution?

It says 100 mL, not 100 g.  This makes the problem much harder.  This also makes your teacher's answer wrong:

Teacher is not convinced with this answer and says its wrong because the total must be 100g

The total doesn't need to be 100 g because we are apparently dealing with 100 mL of solution.  Depending on the density of the solution, the total mass will be more or less than 100 g.
According to Wikipedia, sulfuric acid at a mass fraction of 0.7 has a density of 1.60 kg/L and at a mass fraction of 0.78 has a density of 1.70 kg/L.  Let's suppose a mass fraction of 0.724 has a density of ~1.65 kg/L.
$$0.276\mathrm{\frac{g\;\ce{H2O}}{g\;total}}\times\frac{1650\mathrm{\;g\;total}}{\mathrm{L}}= 46\mathrm{\frac{g\;\ce{H2O}}{L}}$$
With the final number, now it is easy to see that if there are 430 grams of water per liter, then in 100 mL there are 46 grams of water.

Im sorry, but 0.325 is a mole fraction and it doesn't have a unit, why did you said (0.325 mol of H2SO4) how did that happen? Also can you give me a brief explanation of how to convert from mole to mass fraction? because I didn't understand a lot of the first step. another thing i tried to do the first step in my calculator but it gave me 2.62 g not 2.95 as you mentioned above. Sorry for the many questions

Mole fractions "don't have units" in the sense that "mol" cancels out "mol".  You could say it has units of "mol per mol".  The problem in all these kinds of problems is keeping track of "mol of what?"?  I tried to make the what as explicit as possible in my answer.  I hope that helps.

You were right about the 2.62.  Updating my answer.

Converting from mole fraction to mass fraction first requires converting to mole (or mass) ratios.  If in a two-component mixture one of the components has a mole fraction of $x_i$, it has a mole ratio of $x_i/(1-x_i)$.  (If you have two blue marbles and one red one, then the one third of all the marbles are red, but the ratio of red marbles to blue ones is one-to-two.)  One things are in ratio form you can just divide or multiply by formula weights to convert.

Continuing the marbles analogy: if you had two blue marbles and one red marble, and blue marbles weighed 1 oz while red marbles weighed 5 oz, what would be the (a) mole fraction, (b) mole ratio, (c) mass fraction, and (d) mass ratio of your marbles?

So mole fraction can also present as (mol)?secondly: after figuring out the mole ratio, why did you multiple it with (mole mass of H2SO4/ mole mass of H2O)? What does the 2.62 represent? And why you added 1 to it in the next step?

Total is 7 oz
Blue marbles mole fraction would be (1/total marbles oz)?
mole ratio would be (1/total — 1)?
as for the mass fraction, still confused with it because so far we didn't get to this lesson in our school, please excuse my unawareness, but thank you i appreciate your patience with me.


text:
Why does the BET equation take into consideration a multimolecular-layer, even though it deals with chemisorption?
I came across the Brunauer-Emmet-Teller (BET) equation recently, and one thing that I found strange was an implicit reference to the formation of molecular multi-layer.
In the BET equation
$$
\begin{align}
\frac{P}{V(P^0 - P)} &= \frac{1}{CV_m} + \frac{C-1}{CV_m}\cdot \frac{P}{P^0}\, ; \tag{1}\\
C &= e^{(Q_1 - Q_2)/RT}\quad (C:\text{constant})\, ,\tag{2}
\end{align}
$$
where $V_m$ – monolayer adsorption amount, V – adsorption amount at the equilibrium pressure $P$, $Q_\mathrm{L}$ refers to the heat of adsorption of the second and higher layers of adsorbate onto the adsorbent.
Now according to my textbook, chemisorption only results in the formation of a molecular monolayer. Multilayers are formed by physisorption.
The thing is, I'm told the BET equation is an extension of Langmuir's adsorption isotherm for chemisorption at low pressures. If the BET equation applies only to chemisorption, then what is the point of including a term dependent on the formation of multilayers, when chemisorption doesn't even form a multilayer all by itself.
Is there something I'm overlooking here? The internet has not yielded any decent result in that regard.

BET equation does not deal with chemisorption.

Brunauer–Emmett–Teller (BET) explains the physical adsorption of gas molecules on a solid surface, and doesn't address chemisorption. Even the wikipedia link that you provided, makes that quite explicit. However, you are correct to think that it is in a extension of the Langmuir adsorption model, which only accounts for monolayers.
The assumption that underscore this model are as follows:

Adsorptions occur only on well-defined sites of the sample surface (one per molecule)
The only molecular interaction considered is the following one: a molecule can act as a single adsorption site for a molecule of the upper layer.
The uppermost molecule layer is in equilibrium with the gas phase, i.e. similar molecule adsorption and desorption rates.
The desorption is a kinetically-limited process, i.e. a heat of adsorption must be provided:

these phenomenon are homogeneous, i.e. same heat of adsorption for a given molecule layer.
it is E1 for the first layer, i.e. the heat of adsorption at the solid sample surface
the other layers are assumed similar and can be represented as condensed species, i.e. liquid state. Hence, the heat of adsorption is EL is equal to the heat of liquefaction.

At the saturation pressure, the molecule layer number tends to infinity (i.e. equivalent to the sample being surrounded by a liquid phase)

References: The original article
P.S: Good reading material on BET theory can be found here

My book (G. Wedler, H.-J. Freund, Lehrbuch der Physikalischen Chemie, 6. Edition ... obviously a German book on physical chemistry) begins with the following, when introducing the BET equation.

Besonders bei der Physisorption beobachtet man, dass häufig Adsorption in mehreren Schichten übereinander, eine sog. mehrmolekulare Adsorption eintritt.
Esp. during physisorption one can observe that often adsorption happens in multiple layers on top of each other, a so called multilayer adsorption occurs.

Further on it is mentioned that:

Als ersten gelang es Brunauer, Emmett und Teller, eine für die praktische Anwendung geeignete Adsorptionsisotherme für die Mehrscheichtenadsorption abzuleiten.
Brunauer, Emmett and Teller were the first who developed a useful adsorption isotherm for practical purposes regarding the multilayer adsorption.

From that point of view, I would assume, that the BET equation is used for physisorption.

Wait, just confirming; the Langmuir adsorption isotherm ist für niedriger pressure, right? And it's used in the analysis of Chemisorption?

a) Please write english and not a mix of different languages. While I understand what you want from me and I really appreciate your wish to use "my" words, that is not helpful for a lot of people. b) Why do you think that?

Ah, well my German sucks, and I'm trying my best to practice using it. Trouble is most of the time I'm not able to form complete sentences in German every time (in this case it's because I don't know what German for 'Pressure' is). Also, my bad, I ended up putting more effort trying to transliterate what I was trying to convey, than in the actual content of what I was trying to convey. So correcting that: Langmuir's is for chemisorption at high (I know I said niedriger, my bad) pressure right?


text:
Why can't alcohols form hydrogen-bonded dimers like carboxylic acids?
Carboxylic acids such as acetic acid are capable of forming dimers:

I'm wondering why alcohols like ethanol don't generally form dimers. In the diagram below, the oxygen atom on the left ethanol molecule should be capable of accepting a hydrogen bond from the partially positive $\ce{H}$ atom on the right ethanol molecule. Does it have something to do with the $\ce{-OH}$ bond angle?

Because the ring would be too strained.

That I don't believe in, hydroxyl groups are just not acidic/basic enough to engage in that type of hydrogen bonding, no matter the ring size. If you would add in water molecules the rings would get extended and still you wouldn't get dimers with incorporated water.

The strength of a hydrogen bond somewhat depends on the $\ce{X-H\bond{...}X}$ angle that the hydrogen-bonding hydrogen forms with the two electronegative elements $\ce{X}$. In our case, carboxylic acids or alcohols, $\ce{X} = \ce{O}$ so the angle is $\ce{O-H\bond{...}O}$. The ideal angle for this fragment is $180^\circ$.
As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire $\ce{(R-COOH)2}$ feature as a benzene-like ring streched in a single direction. Most importantly, the acceptors and donors line up nicely, the carboxyl function features an angle of $120^\circ$ and the $\ce{C-O-H}$ angle (and the $\ce{C=O\bond{...}H}$ one!) is also close to $120^\circ$. These arae the theoretically predicted, unstrained angles.
For two alcohol (e.g. ethanol) molecules to attempt a similar arrangement, we would end up with only four atoms that have to form a rectangle with $\ce{O, H, O, H}$ at the four corners. Thus, the $\ce{O-H\bond{...}O}$ hydrogen bond angle would be much closer to $90^\circ$ — a very bad arrangement. Furthermore, this would put the two electronegative oxygens (which thus feature a negative partial charge) closer together with nothing in-between which would cause destabilisation due to like charges approaching. The hydrogen-bonding hydrogens cannot alleviate this unfavourable interaction since they are at the corners of the square.
The situation for alcohols is much better if they create a network of a number of molecules that allow for much more favourable angles. With four molecules, you could already create a cyclic structure of approximately $100^\circ$ $\ce{H\bond{...}O-H}$ angles and $170^\circ$ $\ce{O-H\bond{...}O}$ angles — much more favourable. Of course, in the actual solution this number will fluctuate strongly as hydrogen bonds are broken and reformed constantly and different ring sizes happen all the time.

"As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire (R−COOH)2 feature as a benzene-like ring streched in a single direction." Well except that in benzene, no bond is linear as they are all tilted by 60° (so that the bond angle is 120°)." Also the two alcohols dont necessarily have to form a rectangle, as the h bonds might just as well be longer than the covalent O-H bonds. Im still not convinced...

@ketbra Re angle: The first sentence is about a general $\ce{X-H\bond{...}X}$ angle. The second sentence tells you which atoms we are dealing with. And since $\ce{X}$ is $\ce{O}$, it also specifies that the angle is $\ce{O-H\bond{...}O}$ in this case. An angle will always be made up by three points in space; in the context of molecules/chemistry, typically three atoms.

Re streched benzene ring: Yes, benzene consists of six atoms while a carboxylic acid dimer consists of eight. Hoever, imagine two bonds of benzene which are transformed onto one another by a mirror image parallel to both extended in a way such that an additional atom fits between the two carbons. These eight atoms are now a ‘benzene-like ring structure’; six $120^\circ$ angles and two $180^\circ$.

Re rectangle: A rectangle is a shape with four corners, two sets of parallel sides and four $90^\circ$ angles. I don’t see how that should be unable to accomodate H bonds ‘longer than the covalent $\ce{O-H}$ bonds’? I explicitly said rectangle, not square.

Four 90 ° angles would be the definition of a square though... But then I dont get the next statement that the "hydrogen bond angle would be much closer to 90° — a very bad arrangement". How do you come up with that? I would say that there is considerable stabilization between hydroxyl groups (i.e. alcohols) by hydrogen bonding e.g. in water and there also exists the water dimer (or methanol) so that more or less refutes this statement.

@ketbra No, the definition of a square requires four equal side lengths in addition to four $90^\circ$ angles. The next statement says that a hydrogen bond angle of $90^\circ$ is rather unfavourable (again talking about $\ce{O-H\bond{...}O}$ which is a general feature. Note how even the question draws a $180^\circ$ hydrogen bond. Generally, the larger the $\ce{O-H\bond{...}O}$ angle, the more stable the hydrogen bond.

I haven’t seen any ‘water dimer’ structure to date; the most common structure of solid water I know is the ‘extended diamond’ (See Wikipedia). Of course methanol undergoes hydrogen bonds, but the structures it forms are not dimeric.

The carboxylic acid dimer is more likely than the alcohol dimer because the carboxylic acid dimer has two attachment points, whereas the alcohol dimer only has one. While the hydrogen bonds may be of comparable strength in the two cases, the carboxylic acid case is also favored entropically.
Consider, by analogy, the chelate effect. Chelating (multidentate - more than one binding site) ligands bind more strongly to metals than monodentate ligands.
Here are some data for the comparable systems of copper/ethylenediamine $\ce{Cu(en)_2^2+}\ (\ce{en}=\ce{H2NCH2CH2NH2})$ and copper/methylamine $\ce{Cu(CH3NH2)_4^2+}$. Here, I am presenting the overall formation constants  $\beta$ (whcih are the products of the equilibrium/formation/binding constants of each individual association step). The formation constant for $\ce{Cu(en)_2^2+}$ is higher than for $\ce{Cu(CH3NH2)_4^2+}$ by a factor of  $\sim 10^4$, which corresponds to the chelated complex having a more negative free energy of binding by 23 kJ/mol. The major component in this lower energy state is entropy. Data from here.
$$\begin{array}{|c|c|c|c|c|}\hline
\mathrm{Equilibrium} & \beta & \Delta G^\circ& \Delta H^\circ& -T \Delta S^\circ\\ \hline
\ce{Cu^2+ + 2en <=> Cu(en)_2^2+} & 4.17 \times 10^{10} &-60.67 & -56.48 & -4.19 \\ \hline
\ce{Cu^2+ + 4CH3NH2 <=> Cu(CH3NH2)_4^2+} & 3.55 \times 10^6 & -37.4 & -57.3 & 19.9 \\ \hline
\end{array}$$
(data of last three columns in $\mathrm{kJ\ mol^{-1}}$).

It's all a matter of relative energies. Carboxylic acids form two aligned hydrogen bonds, so the $\Delta E$ associated with this process is fairly negative. There's also some extra electron density on the carbonyl O that likely contributes to a better hydrogen bond. Now, alcohols can form one hydrogen bond. You almost certainly make the hydrogen bond in solution, but any collision from a solvent molecule could break the hydrogen bond.
Edit: Apparently, my answer was not clear. No, the bond angle is not the problem.

The main reason for the higher extend of H bonding in carboxylic acids is due the higher acidity of the carboxyl group (e.g. ethanol: $~\mathrm pK_\mathrm a = 16$, acetic acid: $~\mathrm pK_\mathrm a=4.5$) making it a good hydrogen bond donor, and the ability of the carbonyl oxygen to act as a good hydrogen bond acceptor.
Thus the extend of H bonding can be explained solely based on acidity/basicity arguments. Compared to that a hydroxide is just too basic to be formed out of the hydroxyl group, so hydrogen bonded dimers of alcohols make no sense. What is also important to point out, is that the resulting dimer will be completely symmetric (due to resonance) making the situation energetically even more favourable.

I have a feeling you were rather confused here...

@Mithoron Thanks for getting personal : ) Maybe try finding arguments for your position next time, if you have any. But I really don't believe in the points mentioned in the other answers. It just has to do with the strength of the O-H bond, i.e. its lower aciditiy, compared to carboxylic acids and not with the alignment.

Well we could talk about it in chat (http://chat.stackexchange.com/rooms/3229/the-periodic-table) or maybe separate room. Or you could just have faith in Jan's answer :D


text:
How could you distinguish between an irreversible electron transfer couple and a chemically irreversible couple using cyclic voltammetry?
What is the difference between an irreversible electron transfer couple and a chemically irreversible couple?
How could you distinguish between these two processes using cyclic voltammetry?

Please do not significantly change a question after it was answered. Chameleon questions that invalidate any answers are very much frowned upon here and will usually be rolled back. If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. However, your new question “What is cyclic voltammetry and how can we use it?” seems to be too broad.

To this question there are black-and-white as well as grey answer parts.
Chemical reversibility refers to the generation of a species that cannot be reduced or oxidised furthermore. In case you scan at moderate speed (scan rate), you may reduce or oxidise a compound and this intermediate can further react to an electrochemically inert species. If the follow-up reaction occurs fast, this couple is chemically irreversible.
Electrochemical reversibility refers to the rate constant of electron transfer. High rate constants (fast electron transfer) are found in electrochemically reversible couples whereas low rate constants are found in electrochemically irreversible cases.
Okay, you can trick chemical reversibility by scanning so fast, that you oxidise/reduce the intermediate of a redox-reaction at the electrode so fast that it has no time to further react. In this case, you should have only a forward-peak using slow cyclic voltammetry without a backward-peak; but you should start to see a backward-peak as you increase the scan rate.
For electrochemical reversibility, you can use Butler-Volmer kinetics. The main message of Butler-Volmer is that you increase the rate of electron transfer by increasing the potential. Electrochemical reversibility is seen on the peak separation between oxidation and reduction peak. Reversible systems show a peak separation at standard conditions with one electron transferred of 59 mV.
So, how would I dinstinguish them using cyclic voltammetry? By the peak shape and the change of CV by varying the potentials and scan rate.


text:
Shielding effect- specifically defined?
In all the sites I have searched, shielding effect has been defined as the reduction of the nuclear attraction upon the outer electron/electrons, due to the core electrons that repulse the outer electron.
So, is the shielding effect specifically defined for outer shell electrons? Can't the shielding effect be defined for inner shell electrons?
If it can be defined, do the electrons in outer shells than that shell have any impact on the shielding effect?(They too must repulse the electron in the inner shell, possibly helping the nuclear attraction)
If not, why?
Can someone help me out? Tks

Generally, inner electrons shield outer electrons. In order to shield you have to be closer to the nucleus, but it's harder for an electron in a higher energy orbital to be closer to the nucleus than an electron in a lower energy orbital. Therefore the shielding effect from inner on outer is a lot more pronounced than outer on inner.
EDIT (after better understanding of OP's question):
One thing to remember is that the inner electron can be on the same side of the nucleus as the outer electron, or it can be on the other side of the nucleus as the outer electron. In one case, the inner electron is being pushed towards the nucleus but the other case results in the inner electron being pushed away from the nucleus. A hand-wavy application of Gauss' Theorem suggests that these effects might cancel out over time. So, the net effect is that the outer electrons have minimal impact on the inner electrons. This argument fails if the outer electrons have greater nuclear penetration, but that doesn't happen by definition.

The question poster probably meant that the outer shell electrons will repel the inner shell electrons towards the nucleus, thus strengthening the attraction.

Yes,he is is asking exactly same

@DHMO My sense is that that wouldn't be the case. We're dealing with an electron density (yes, it's going to be altered by the positions of the other electrons). Even though the distribution isn't spherical, I would posit that Gauss' theorem would say that the impact on the inner electrons is minimized on average over time.

@Zhe can you explain your comment further? I feel that it might clear up my confusion, Pls keep it as simple as possible

If you consider a spherical shell of charge, what is the net electric field anywhere inside the sphere? Zero. In other words, the time averaged effect of outer electrons on inner electrons probably cancels out. This is probably grossly oversimplified, but from a purely physical perspective, I think it makes sense that the outer electrons have minimal effect on the inner ones. One other way to think about it is, when the outer electron pushes the inner electron closer to the nucleus, what happens when the outer electron ends up on the other side of the nucleus?

Much clearer now, :) But, what do you mean by "what happens when the outer electron ends up on the other side of the nucleus?"

If that's the case, isn't it repelling the inner electron away from the nucleus?

The shielding effect describes the attraction between an electron and the nucleus in any atom with more than one electron shell. Shielding effect can be defined as a reduction in the nuclear charge on the electron cloud, due to a difference in the attraction forces of the electrons on the nucleus. You may also have heard of screening effect,these two are same.
Since the shielding effect is reducing the nuclear charge on the outer electrons,the net charge experienced by outer electron is called effective nuclear charge.
If you look at the formula of effective nuclear charge,you will see that Z*= Z-S
{Where Z is the number of protons in the nucleus (atomic number) and S is the average number of electrons between the nucleus and the electron in question }
So it is evident that not only valence electrons but inner electrons experience shielding effect too but for inner electrons, shielding effect is less(a quick look to formula will clear it to you).
Now your next question is  Do the electrons in outer shells than that shell have any impact on the shielding effect? the answer is NO
According to wikipedia(as shown in the image) electron in the higher energy levels do not contribute in anyway to sheilding effect.

Nicely explained, but do you have any argument as to why outer electrons don't have any impact on the shielding effect?