text:
What elements in Tomato and Spinach make them cook quickly when cooked together?
I have started experimenting on cooking these days and I have noticed that when I add spinach leaves and tomato, they cook very quickly. The spinach melts very fast and the tomato melts fast also - after adding a spoon of table salt. My question is what is the reason behind it. Is it something to do with oxidation?

The table salt is probably irrelevant to why the veges cook down quickly, although it does cause veges to sweat, it may leach the moisture out of the tomato and spinach faster than if you didn't add it but I think heat would be the main variable. Also do you stir the food more when you add the salt than if you didn't?
Most organisms have a ratio of about 《80% water, spinach and tomatoe may have a high water content and also contain oils and other volatile components that are released when heated so when you heat / cook them the water and volatile components evaporate and the solid mass(other 20% of initial weight) is left behind.
They are soft veges too so the material can be sweated easier than say a carrot and will lose its structure and water content at a faster rate.
Some oxidation could be occurring along with other chemical reactions between components such as decarboxylation, hydrolysis, hydration, dehydration and many more depending on what compounds are present in the process, the kind of environment (eg. aqueous) and the techniques used (eg. frying, baking).

Food has a solid consistency due large macromolecules (protein, cellulose, collagene etc), which degrade and/ or dissolve during cooking. I would guess  vegetables like tomato and spinach has most its firmness from pectins and polysaccharids. Many of these polysaccharids have acidic groups that makes them interconnect and firm in presence of divalent ions like magnesium or calcium, and soft in presence of lo of sodium. Maybe this is what happening with your food


text:
Boys function for Gaussian integrals in ab initio calculations
A few days ago I mentioned a problem with an Hartree-Fock program I am writing (HF using cartesian Pople's STO-3G basis set). I can reproduce overlap and kinetic integrals of some references for ($\ce{H2}$, $\ce{HeH+}$ and $\ce{H_2O}$). Things start to change (at the third/fourth decimal digits) in the nucleus-electron attraction integral. At the beginning I thought the error I get on the total energy was mainly caused by electron-electron repulsion integrals (as errors on nucleus-electron attraction seem minimal). But now I am quite convinced the problem comes from my calculation of the Boys function
$$
    F_n(x) = \int_0^1 t^{2n}e^{-xt^2}\, dt.
$$
A clue of this comes from the fact that for $\ce{H2}$ and $\ce{HeH+}$ everything works fine: in this case Boys function is only computed for $n = 0$.
In order to compute this function I used numerical integration routines of SciPy (Python library) or directly QUADPACK (I traduced my Python program into Fortran90). However, looking at the literature it seems that nobody compute Boys function via numerical integration. I tried to implement different versions (recursive, asymptotic values, ...) of Boys function but all of them eventually had problems (depending on the range of $x$ or on the value of $n$).
Since I am loosing too much time on this problem, which seems to be numerical, I was wondering if there is some standard implementation of Boys function that I can simply use in my code.
Do you know any Python or Fortran library containing a good (safe and efficient) implementation of Boys function? If not, there is a simple algorithm that works well in any range of $x$ and is not sensible to numerical errors?

I recommend asking questions about code and/or libraries at the computational science SE. I'm not saying is off topic, I just think it's better suited there.

I just found a library called "LibInt" for computing molecular integrals: http://sourceforge.net/projects/libint/ It contains a file 'Boys.h', so what you want seems to be implemented. I havn't tested it yet (I'm currently working on writing a Full CI code, so I keep running into problems similar to yours ^^), but it's used in several quantum chemistry packages, so I'm sure it'll give good results.

Thank you for the hint. I managed to implement this computation using the incomplete gamma function (as suggested in the answer) contained in SLATEC, but I will certainly give a look at the library!

As an insider to ORCA, I can state that the Boys function is computed by various methods depending on the argument and that higher orders are computed recursively. I can also recommend looking at open-source implementations such as NWChem for inspiration on this.

I am not aware of any existing Fortran code for direct numerical quadrature of this problem, but it is worth pointing out that Mathematica can perform this integral symbolically:
Integrate[t^(2 n) Exp[-x t^2], {t, 0, 1}]

(* 1/2 x^(-(1/2) - n) (Gamma[1/2 + n] - Gamma[1/2 + n, x]) *)

where the $\Gamma$ function can be computed by exponentiating easy-to-find $\ln \Gamma$ functions, often called lngamma().  (One of them is the Euler function $\Gamma(z)$ and the other incomplete $\Gamma(a,z)$ function).
As a check:
With[
  {x = 1, n = 2},
  1/2 x^(-(1/2) - n) (Gamma[1/2 + n] - Gamma[1/2 + n, x])
  ] // N

(* 0.100269  *)

and numerical quadrature yields:
With[
 {x = 1, n = 2},
 NIntegrate[t^(2 n) Exp[-x t^2], {t, 0, 1}]
 ]

(* 0.100269  *)

One has to be a little bit careful, as the $\Gamma$ function is closely related to the factorial, and it can explode as the arguments get too large.  I would experiment and see whether you can get by with this analytic solution, which essentially takes you from numerical quadrature into evaluation of special functions.
The $\Gamma$ function approach is faster by over two orders of magnitude  than the quadrature approach.  This is critical when evaluating molecular integrals.  I have seen implementations where the Boys function is actually interpolated rather than evaluated, and there is discussion about the error in interpolation being almost insignificant if suitable interpolation basis functions are used.

I already found this analytical formula in one publication, but the incomplete gamma function is not part of Fortran standard library. I will look for Python, though. The values of n I have at the moment range from 0 to 3, therefore I don't think an overflow will be a problem.

See: http://people.sc.fsu.edu/~jburkardt/f_src/asa032/asa032.html

I don't know how it is possible I did not saw it! Thank you very much!

Unfortunately this does not solve the mentioned total energy problem... =(. I will have to search elsewhere.

I know this is an old question, but I would like to give a small comparison regarding efficiency when evaluating the Boys function $F_n(x)$. Below are some implementations (in Julia) with simple benchmarks.
In the end, I also give some useful approximations to the Boys function, both for small and large values of $x$.

For benchmarking, I'll use BenchmarkTools to measure the average time it takes for a particular implementation to calculate $F_1(x)$ $\forall x \in \hat{x}$, where $\tilde{x}$ is the 10000-vector below:
using BenchmarkTools

n = 1
x̃ = range(1e-15, 15, length=10000)

Implementations are in order, from fastest to slowest. (Of course, benchmarks should be compared with each other; different computers will produce different results.)
Evaluations
Using the incomplete gamma function
As mentioned by Eric above, $F_n(x)$ can be written in terms of the incomplete gamma function as
$$\frac{1}{2 x^{\frac{1}{2} + n}} \Gamma \left( \frac{1}{2} + n \right) \gamma \left( \frac{1}{2} + n, x \right)$$
using SpecialFunctions

# Incomplete gamma function implementation.
boys2(n, x) = gamma(0.5 + n) * gamma_inc(0.5 + n, x, 0)[1] / (2x^(0.5 + n))
@benchmark boys2.(n, x̃)

BenchmarkTools.Trial:
  memory estimate:  406.52 KiB
  allocs estimate:  1005
  --------------
  minimum time:     1.317 ms (0.00% GC)
  median time:      1.345 ms (0.00% GC)
  mean time:        1.380 ms (0.33% GC)
  maximum time:     2.880 ms (0.00% GC)
  --------------
  samples:          3614
  evals/sample:     1

Using quadratures
The simplest option is a quadrature of
$$\int_0^1 t^{2n} e^{-x t^2} dt$$
using QuadGK

# Adaptive Gauss-Kronrod quadrature implementation.
boys1(n, x) = quadgk(t -> t^(2n) * exp(-x * t^2), 0.0, 1.0)[1]
@benchmark boys1.(n, x̃)

BenchmarkTools.Trial:
  memory estimate:  8.32 MiB
  allocs estimate:  326405
  --------------
  minimum time:     23.747 ms (0.00% GC)
  median time:      24.629 ms (0.00% GC)
  mean time:        25.839 ms (3.78% GC)
  maximum time:     45.680 ms (0.00% GC)
  --------------
  samples:          194
  evals/sample:     1

Using the confluent hypergeometric function
Yet another relation, mentioned by Joshua above, uses the Kummer confluent hypergeometric function in
$$F_n(x) = \frac{{_1}F_1 \left( \frac{1}{2} + n, \frac{3}{2} + n, -x \right)}{2n + 1}$$
using using HypergeometricFunctions

# Confluent hypergeometric function implementation.
boys3(n, x) = pFq([0.5 + n], [1.5 + n], -x) / (2n + 1)
@benchmark boys3.(n, x̃)

BenchmarkTools.Trial:
  memory estimate:  18.03 MiB
  allocs estimate:  162806
  --------------
  minimum time:     58.511 ms (0.00% GC)
  median time:      59.524 ms (0.00% GC)
  mean time:        60.007 ms (1.19% GC)
  maximum time:     65.223 ms (0.00% GC)
  --------------
  samples:          84
  evals/sample:     1

Approximations
Small values of $x$
We can expand the integrand $t^{2n} e^{-x t^2}$ as a series around $x = 0$ and obtain successive polynomial approximations to the Boys function. Below is a systematic way of doing that (using SymPy in Julia):
using SymPy

I = t^(2n) * exp(-x * t^2)

order = 2  # The order of the final polynomial.
I_taylor = series(I, x, 0, order)
B = SymPy.integrate(I_taylor, (t, 0, 1)) |> simplify

$$B^2_n(x) = \frac{3 + 2n - (2n + 1) x}{(2n + 1) (2n + 3)} + O(x^2)$$
Such approximations can be used for small values of $x$ (they are safer for $x \approx 0$ than some implementations above). An approximation of order 10 is compared below with a quadrature of the Boys function.

Large values $x$
Since the Boys function is decreasing, we could, for large $x$, approximate
$$\int_0^1 t^{2n} e^{-x t^2} \approx \int_0^\infty t^{2n} e^{-x t^2}
= \frac{1}{2x^{\frac{1}{2} + n}} \Gamma \left( \frac{1}{2} + n \right)
= \frac{(2n - 1)!!}{2^{n + 1}} \sqrt{\frac{\pi}{x^{2n + 1}}}$$
(I learned this trick here.) Compare this to the Boys function below.

Further reading
The above are obviously toy examples. Here are some nice reads:

Journal of Mathematical Chemistry 54, 2022-2047 (2016)
Journal of Computational Chemistry 36, 1390-1398 (2015)
Journal of Mathematical Chemistry 40, 179-183 (2006)

What a great response. Thank you!

@Rick glad it helped!

The Boys function is $F_n(x)$ a special case of the Kummer confluent hypergeometric function $M(a,b,x) = {_1}F_1(a,b,x)$, which can be found in many special function libraries, such as scipy.special. According to equation (9) of this paper, the relationship is
$F_n(x) = \frac{{_1}F_1(n+\frac{1}{2},n+\frac{3}{2},-x)}{2n +1}$
If you use scipy you can get ${_1}F_1$ here.
Otherwise, you can grab a FORTRAN implementation here. I believe the routine you want is CHGM. There are probably other implementations.

the leading subscript 1 in front of F appears to be a typo in the WIREs paper.  StackExchange does not allow me to correct this because the edit is less than 6 characters.

@Jeff Do you have a citation? I don't see any typo here. The relationship to Kummer's function 1F1 and the Boys function Fn is derived in Sec (9.8.6) of Helgaker, Jorgensen, and Olsen; Molecular electronic-structure theory; (2000).

I see now.  ${}{1}F{1}$ is a completely different $F$ from $F_{1}$.  How nice.

@Jeff Correct, $_1F_1$ is the Kummer confluent hypergeometric function, not the Boys function. The Boys function $F_n$ is a weighted version of $_1F_1$. I agree, the notation can be very confusing here.

I think that the CHGM have troubles calculating the Boys function for what is needed for Molecular Integrals. It performs well enough for H2O/6-31++G**. But for bigger molecules it seem to fail p-nitrophenolate/4-31G. Do you know any other implementations of Boys function in Fortran?

I published an implementation in C/C++ in http://vixra.org/abs/1709.0304 which uses Gauss-Jacobi quadratures. It is limited to quantum numbers of 130 and to real arguments; a relative accuracy near 1.e-15 is achieved. The source code is available in https://zenodo.org/record/3603249 .

While the asker may find this useful, note that the OP requests implementations of python and/or fortran specifically.

It is worth stating that it is roughly as easy to call C from Python as it is to call Fortran from Python, so this is useful in that regard. Less useful is source code embedded in a PDF.

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text:
Solubility product when you add an acid to a solid?
I ran into some trouble when I was resolving some exercises about solubility product principle.The exercise says:

A concentrated, strong acid is added to a solid mixture of $0.015\ \mathrm{mol}$ samples of $\ce{Fe(OH)2}$ and $\ce{Cu(OH)2}$ placed in $1.0\ \mathrm{L}$ of water. At what values of pH will the dissolution of each hydroxide be complete?

So, I don't know how start when you have an acid and an insoluble salt when using the Ksp equation.
I know I can't to do this.
$\ce{Fe(OH)2 + HCl -> FeCl2 + H2O}$
$\ce{Cu(OH)2 + HCl -> CuCl2 + H2O}$
Not even do I have a common ion in order to establish a reaction. Do you know where I should begin? I tried using $K_\mathrm{sp}$, but I already have molar solubility, is $0.015\ \mathrm{M}$ for both. I can't ignore that. I don't find useful $K_\mathrm{sp}$ in this case. I know there's something that I'm not understanding. Hope you can help me to figure it out.

Think of the problem this way. At what pH would the Fe(II) and Cu(II) solutions just start to ppt? At that balancing point if the pH is any higher the ppt will form. If the pH is lower then no ppt will form.

As @MaxW has stated, look up the solubility data for the different salts in water at different pH values.

Instead of using the data from the equation of formation of the different salts, you need to write the equation of both salt products dissolving in water then use the Ksp equation with that data. Also 1M of each of the hydroxides reacts with 2M Hcl to form the dichloride.

Are not you complicating the problem?
The solution is simple. You have $\pu{0.015M}$ of $\ce{Me(OH)2}$ and according to the reaction you will need twice as much acid (based on stoichiometry): $\pu{0.030M}$. Knowing that $\pu{pH = - \log [H^+]}$, the dissolution will be complete at $\pu{pH = 1.52}$.

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text:
Polarity and Solubility of Tetracyclone
In one of my organic chemistry experiments, I synthesized tetracyclone (Tetraphenylcyclopentadienone), and we were supposed to dissolve it in DCM to do our TLC.
DCM is polar and looking at the structure of tetracyclone I think it is non polar, isn't there a saying that "like dissolves like"? Then how could tetracyclone dissolve in DCM?
Below is the structure of DCM and tetracyclone:

Edit: After searching online, I found that due to the lack in ability to form hydrogen bonds, we treat DCM as a non polar solvent, even though it is weakly polar.
Website: https://www.quora.com/Is-CH2Cl2-both-polar-and-non-polar-Why

Tetraphenylcyclopentadienone is a polar moleule.  Look at the resonance structure with a positive charge on oxygen and a negative charge delocalized within the 5-membered ring.  This would make the ring aromatic with 6 pi electrons.  Therefore such dipolar structures are significant descriptors of the molecule, making it polar.

My comment is: surely it dissolved as the experiments are pre-checked? The molecule has a dipole moment, same as DCM.@Ron - I've seen the charge split you're talking about for cyclopropenone where the positive charge is on the carbon to give a ring with 2-pi electrons, with a negative charge on the oxygen. I've not seen anything with a positive charge on the oxygen and negative on the carbon.

@Beerhunter Yes, tetracyclone dissolved in DCM no problem. I asked my professor and she said that DCM is in between polar and non polar, so it can dissolve both polar and non polar substances.

@ron, but the tendency for oxygen to have a positive charge, plus octet deficiency isnt quite stabilized by aromaticity. Molecule might be polar, but not so polar.


text:
Simultaneous Equilibrium Reactions, One Goes to Completion
Say I have the following reactions:
$\ce{CH4 + H2O <=> CO + 3H2}$
$\ce{C2H6 + 2H2O -> 2CO + 5H2}$
Which undergo these processes in the same container. The first reaction goes to equilibrium and the second to completion. Clearly the reactions are entangled as the products are the same. The equilibrium concentration of methane will intuitively depend on the initial amount of ethane introduced. I'm not sure how to go about calculating the equilibrium concentration of methane however. My first thought was to find the amount of carbon monoxide and hydrogen gas generated by the second reaction and use this as initial product in the first reaction and to calculate equilibrium concentration from there but I'm having trouble doing this.

Initially
Suppose that initially, before any reaction, these are the species and amounts present:
$$\ce{CH4}: m \textrm{ moles} $$
$$\ce{H2O}: w \textrm{ moles} $$
$$\ce{C2H6}: e \textrm{ moles} $$
$$\ce{H2}: h \textrm{ moles} $$
$$\ce{CO}: c \textrm{ moles} $$
After the second reaction completes
Let's assume that the second reaction is limited by the amount of ethane present, not the amount of water.  (This must be true, otherwise the first reaction could not be at "equilibrium" because one of the equilibrating species would  head to a concentration of zero.)  If the reaction is limited by ethane, then when it completes, $2e$ moles of carbon monoxide and $5e$ moles of hydrogen are formed.
$$\ce{CH4}: m \textrm{ moles} $$
$$\ce{H2O}: w - 2 e \textrm{ moles} $$
$$\ce{C2H6}: 0 \textrm{ moles} $$
$$\ce{H2}: h + 5e\textrm{ moles} $$
$$\ce{CO}: c + 2e\textrm{ moles} $$
After the first reaction equilibrates
The first reaction doesn't go to completion.  Instead, let's assume the extent of the reaction from our current state to the equilibrated state is $\xi$ moles.  That means that the species inventory becomes:
$$\ce{CH4}: m - \xi \textrm{ moles} $$
$$\ce{H2O}: w - 2 e - \xi \textrm{ moles} $$
$$\ce{C2H6}: 0 \textrm{ moles} $$
$$\ce{H2}: h + 5e + 3 \xi \textrm{ moles} $$
$$\ce{CO}: c + 2e + \xi \textrm{ moles} $$
These values can now be plugged into an equilibrium equation:
$$ K = \frac{[\ce{H2}]^3 [\ce{CO}]}{[\ce{CH4}][\ce{H2O}]}$$
$$ K = \frac{(h + 5e + 3 \xi)^3(c + 2e + \xi)}{(m-\xi)(w - 2 e - \xi)} $$
Now "all" you have to do is solve that equation for $\xi$.


text:
Can grignard reagent contain secondary amino group?
Is there any reason why this grignard reagent wouldn't work? My best guess is that perhaps the nitrogen could act as a second nucleophilic "head", thereby not giving a good yield on the intended product. Or maybe this isn't stable enough?

The variables regarding why the reagent may or may not be effective for certain processes depends a lot on what your using it for?

I'm really just wondering if there's a way it could deactivate itself, like how you shouldn't make a grignard reagent that can reach a carboxyl group on itself. Originally I intended to use it to attack benzoyl chloride. Can the grignard really deprotonate itself? It seems to me like the nitrogen's proton is too close?

I'm almost sure it can and will deprotonate and thus deactivate itself. Whether or not the nitrogen's proton is close to the Grignard site, does not matter at all.

@Ivan and the OP—Please (self-)answer this question so it wouldn't show up in the 'unanswered' list. i.e. close the case.

I guess it wouldn't even exist. Grignard's reagent is basic enough to easily take off an imine proton, so as soon as you try to make such a thing, it will turn upon itself.
Probably they have a suitable protecting group for this, but here I'm stepping outside my expertise.

Why are you calling the amine an imine?

Obviously because I mistook the secondary amine's $\ce{-NH-}$ for imine's $\ce{=NH}$. Who cares. Grignard would instantly destroy both all the same.

Just wondering, thought maybe I was missing something.


text:
Troubles with cinnemaldehyde extraction - A few general questions
I'm trying to do a cinnemaldehyde extraction.  I've tried both a simple and a steam distillation.  I'm using the video: [https://www.youtube.com/watch?v=bDoaPIKT_Xg] - as a guide.  One of the main problems is that, at the end, the receiving flask is full of water and has very few drops of oil whereas the video shows I should have a foggy mixture.
First off, what's up with that?  Second, third and fourth:
What is a good solvent to use other than DCM/ether?
I have read a lot but haven't figured out: What information does one use/how does one know what makes a good solvent given a situation?  Through some research, I decided to use methanol as an alternate solvent - I ended up with a purely clear substance in the separation flask - No layers to separate.  I have isopropyl, methanol, acetone and naphtha to choose from as solvents.
Finally, in the video, I'm having difficulty understanding what the guy is doing after separation.  I've watched the video several times but it still makes little sense to me.  If this last paragraph doesn't make sense, please just focus on the first two - Theoretically, I could figure out the last part through trial and error - that's why I'm doing such a benign experiment as my first.

One of the main problems is that, at the end, the receiving flask is full of water and has very few drops of oil whereas the video shows I should have a foggy mixture. [...] what's up with that?

It's tough to say without seeing your exact setup, but here are some possibilities for you to think about:

Too much water and not enough cinnamaldehyde is coming over in your distillation.  Try heating the still less rapidly.  You might also try using more cinnamon bark and/or less water.
You are using a degraded or very impure cinnamon bark that does not contain very much cinnamaldehyde.  Is there a way you can test a different batch?  What does the distillation smell like?  What does the condensate smell like?
Other possibilities I haven't thought of...

Second, third and fourth: What is a good solvent to use other than DCM/ether?

The intended application of the solvent is extraction of cinnamaldehyde from an aqueous emulsion of the water-insoluble cinnamaldehyde.  Thus, the main criterion for the solvent is that it be insoluble in water.  If it isn't insoluble, adding it to the aqueous emulsion will just dilute the emulsion.  Nothing will be transferred from one phase to the other, because there will only be one phase.
A secondary feature of DCM in the video is that it is heavier than water.  This means that in the separatory funnel, the DCM/cinnamaldehyde layer will be lower than the aqueous layer.  In this particular extraction, the aqueous layer is repeatedly washed with fresh DCM, and the DCM layer is removed each time.  Doing this is a bit easier if the solvent layer is the bottom layer.  If it were the aqueous layer at the bottom of the funnel, each wash with the 60 mL of solvent would require draining the aqueous layer into one flask, then draining the solvent/cinnamaldehyde layer into another flask, then re-adding the water to the separatory funnel and washing it with fresh DCM, etc.  This requires more flasks and more manipulations than just being able to leave the aqueous phase in the funnel the whole time.

I have isopropyl, methanol, acetone and naphtha [available].

Now that you know why DCM makes a good solvent, you can probably see why methanol wasn't a great choice.  It is miscible with water, so that explains why you didn't get multiple layers.  Of the solvents you have, only one, naptha, is immiscible with water.  The other two, isopropanol and acetone, are miscible with pure water, but are not miscible with very salty water.  So you might be able to extract the cinnamaldehyde by adding a lot of salt such as calcium chloride, magnesium sulfate, or sodium chloride to the distillate, and then trying to wash the now-salty emulsion of cinnamaldehyde to the solvent for washing and phase separation.  However this is likely to be more difficult than if you could access DCM.  (Getting DCM may be less difficult than you think, by the way.)

I'm having difficulty understanding what the guy is doing after separation. I've watched the video several times but it still makes little sense to me.

I don't know what part is confusing or exactly what "separation" you mean when you say "after the separation".  But if you mean after the separatory funnel phase separations, then roughly the steps are:

"Drying" the DCM solvent.  Even though DCM and water don't completely mix (they form two phases, which, as explained above, is the reason why it works to extract cinnamalehyde from an aqueous emulsion), some water does leach into the DCM phase during the separation.  "Washing" the DCM with brine helps rid the DCM  of the small amounts of water dissolved in it after the phase separation.  Adding calcium chloride powder to the DCM is an even more effective drying agent that removes the last bit of water from the DCM.
DCM distillation.  After drying the DCM, all the cinnamaldehyde is still dissolved in a (now "dry" or water-free) DCM phase.  To get rid of the DCM and obtain the pure cinnamaldehyde oil, another simple distillation is performed.  Because DCM boils at around 40 °C and cinnamaldehyde doesn't boil until 248 °C, it is fairly easy to remove nearly all the DCM by simple distillation, leaving the oil.
NMR.  The very end shows a proton NMR of the oil.  This is something that requires big fancy equipment and won't be possible for home chemists.  Don't worry about it and enjoy your cinnamaldehyde!

Wow thanks for all the info.  You answered pretty much every question I've had.  I'll be looking back at this for a while.


text:
Similarity between all gases
How are gases all so similar unlike any other state of matter?
For ex:
1) 1 mole of any gas will occupy 22.4 litres at atmospheric pressure. (Cannot be said of all solids/ liquids)
2) partial pressure of any gas is directly proportional only to mole fraction. ( even weight of atoms is irrelevant.)
I guessed it would be due to negligible intermolecular attractions but I'm not too clear. Could someone explain.

Those would only be true if you assume that they are ideal gases, which by definition have no intermolecular forces. For real gases the molar volumes etc. would be a little different. These results can be derived from kinetic theory of gases or statistical mechanics - if you go through the working, you will find that the molar mass of the gas does not come into play.

@orthocresol how would you explain pressure being independent of weight of gas? Even in ideal condition gases have different weights, right?

Well, it's not easy to explain qualitatively. If you really wanted to know, then you should look for a derivation of the ideal gas law. A handwavy argument would go, if the molar mass increases, the force exerted by the gas molecules upon collision with the walls would be larger; but the gas molecules are also on average moving more slowly since they are heavier, which reduces the force. These factors balance exactly as long as the temperature (which is a measure of average kinetic energy of a molecule) is kept the same.

Related: Gas Pressure and Molecular Mass

The way to get a handle on why gases are all similar is to think about the relationships between how the particles of the gas behave and the macro properties temperature and pressure.
The temperature is a measure of the average kinetic energy of the particles. The pressure is a consequence of the particles of the gas bouncing off the walls of the vessel the gas is contained in or the surfaces in contact with the gas. In a gas made of heavy particles the average speed of the particles will be much lower than for a gas made of light particles. When you do the calculations about the effect at a surface (where the pressure is a result of the particles bouncing elastically off the surface) the effects of the mass cancel out because the heavier particles travel more slowly at the same temperature. Overall this means that, for an ideal gas, the mass of the particles doesn't matter for the pressure and all gases behave the same regardless of the particles they are made from.


text:
Transition of electron from HOMO to LUMO?
Molecules usually have certain number of (doubly) occupied molecular orbitals and an infinite number of unoccupied molecular orbitals but the probability of electron makes the orbitals shapes so, how LUMO (empty of electrons) orbitals exist despite no electron occupying it?
Also, when electron absorb sufficient energy will go excited from HOMO to LUMO, but is it allowed that the other electron still in HOMO will also go excited to LUMO leave HOMO without any electrons?

the probability of electron makes the P orbitals so, how LUMO orbitals exist despite no electron making  I don't quite understand what you mean here and also in the second question. Please clarify your question.

sorry, I donot know if my question clear know or not?

An empty LUMO does not exist in the same sense that atoms and things and buildings and Earth do. Until it receives an electron, it is more like an abstract idea.

Thanks, it seems clear know but what about the transition of electron , why always one electron not both travel from HOMO to LUMO?

Well, it is possible (though with low probability) for the two electrons to go up to some vacant orbitals; that would be called a doubly excited state.

Thanks for your short but valuable answers for me.

@IvanNeretin You should write an answer based on that.

@AhmedMohammed Lots of molecules have normally orbitals with 1 electron.


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Why are negatively charged ligands behind the neutral ligands in spectrochemical series?
I had doubt that shouldn't negatively charged ligands be better electron pair donors than neutral ligands like $\ce{CO}$ or $\ce{PPh3}$. But its not so. Why?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your posts by using MathJax; for more information on how to do so yourself, visit the [help], this meta-post or this one.

I think you are confused by what the spectrochemical series actually shows. It is not a rating of stronger or less strong ligands to mean they can bind more or less strongly to the metal. It is only a rating of which ligands induce a high crystal field split in the resulting complex.
The splitting derives from the metal $\mathrm{e_g}$ orbitals — $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$ — actually being antibonding to the metal-ligand interaction and thus higher in energy than the (unmodified in a complex with only σ type interactions) $\mathrm{t_{2g}}$ orbitals.
For systems more complex, i.e. including π interactions, the $\mathrm{t_{2g}}$ orbitals (the other three d-orbitals) interact in a π manner with ligand p-orbitals. If the ligand has populated p-orbitals such as halides have, the $\mathrm{t_{2g}}$ energy is raised as it is turned into a slightly antibonding interaction (the bonding one being on the ligands’ side). Since this brings $\mathrm{t_{2g}}$ closer to $\mathrm{e_g}$, the field split is reduced and the ligands are termed weak-field ligands. In simplified Lewis schemes, one could represent this by a double forwards bond from the ligands to the metal.
In the case of empty p-type orbitals such as in $\ce{CO}$, metal to ligand back-bonding can occur by populated metal orbitals interacting with unfilled ligand orbitals; now it is the metal’s side of things that is stabilised. Since this π backbonding stabilises the $\mathrm{t_{2g}}$ orbitals, their energy is lowered and the field split becomes greater. Hence they are termed strong-field ligands. In simplified Lewis structures one could represent that by drawing a donor-acceptor arrow from the ligand to the metal (σ forwards bond) and one the other way (π back bond).
This all has nothing to do with the overall strength of metal-ligand interaction, as I mentioned earlier. In fact, strong complexes are typically obtained for certain metal-ligand combinations and the same ligand can be very bad in interacting with other metals. Classic example: $\ce{CO}$ ligands are extremely good at stabilising low metal oxidation states such as in $\ce{[Fe(CO)4]^2-}$, while they are rather lousy in stabilising high ones: $\ce{[Fe(CO)6]^2+}$ can only be synthesised in the extreme absence of all other ligands, i.e. in $\ce{SbF5}$ from $\ce{FeF2}$ where the fluorides will be drawn towards antimony to generate the even more stable $\ce{[SbF6]-}$ complex.


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I know PV=nRT but that's total what about partial?
What is the formula for finding the partial pressure ?
I know PV=nRT but that's total what about partial ?

From Wikipedia: In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature. Thus, for a mixture of (ideal) gases you get an equivalent equation for gas $i$ with partial pressure $p_{i}$ and amount $n_{i}$: $p_{i} V = n_{i} R T$.

I'm voting to close this question as off-topic because it is too simple (but I'm not sure whether it is covered by homework).

@Jan I don't think we should close something for being 'too simple'. That sounds very condescending.

@bon You’re right and I probably should have let someone else do the judgement. I apologise.

To find partial pressures of a mixture of non-reactive gases, you can use Dalton's Law of partial pressures.
If you know the total pressure of the mixture and the number of moles of each gas, you can find the pressure of each gas by multiplying its mole fraction with the total pressure.

By Dalton's law of partial pressure (any high school chem textbook will have this law), the partial pressure of a single gas in a mixture of non-reactive (or relatively inert gases) is directly proportional to its mole fraction in the mixture, with the constant of proportionality being the P in PV=nRT (i.e., the total pressure of the system)

authors: خالد ابن احمد


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Why are noble gases 'gases'?
Why do group 18 elements exist as gases, why are they not found in any other physical state ?

Why, they surely become liquids and eventually solids if you cool them down enough.

Although Helium needs a little added pressure. What is the electron configuration of the noble gases? How will they bond together at room temperature?

Liquids and solids form because intermolecular forces hold together different molecules/atoms/ions/biscuits in a strong enough fashion to prevent them from drifting away from each other. In a very basic sense you could say that gases are non-ordered, liquids have near-range ordering but not far-range, and solids are well ordered into crystal lattices.
Noble gases have very little intermolecular forces acting between them since they are monoatomic and unpolarised. Thus, only London dispersion forces act and these directly depend on the number of electrons in a compound. If you compare each noble gas with the element to its left in the periodic table (i.e. to the neighbouring halogen; hydrogen for helium) and take $n$ to be the number of electrons the noble gas has, then the next-door neighbour will always form a diatomic molecule with $2(n-1)$ electrons. In every case except for hydrogen, this is substantially more than in the noble gas, thus there is a substantial increase in melting/boiling point when going from the noble gas to the corresponding halogen.
Note that radon’s boiling point is already at $-61~\mathrm{^\circ C}$, higher than that of the highly polar molecule $\ce{HCl}$ ($- 85~\mathrm{^\circ C}$). And note that ununoctium has been predicted to have a boiling point of $(+80 \pm 30)~\mathrm{^\circ C}$ and would thus better be termed a noble liquid. Unfortunately, ununoctium has a half-life of about $0.89~\mathrm{ms}$ so actually observing the noble liquid could prove tricky.
(All values taken from the respective gases’ Wikipedia articles.)


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Ozone is above, nitrogen is below
Ozone is heavier than nitrogen. Then why is ozone in stratosphere and nitrogen in troposphere?

Weight does not matter at all, otherwise all atmosphere would separate and we would drown in $\ce{Ar}$ and $\ce{CO2}$. Ozone is formed where UV is the strongest; being not very stable, it does not reach us down below.

Why doesn't the weight matter?

Because, in an ideal gas mixture, the gases behave as if they are independent of one another.  There will be a slight density stratification of all the gases.  But, if you had ozone and air in a room and calculated the ratio of the mole fractions at the ceiling compared to the ratio at the floor, you would find that the difference would only be a tiny fraction of a percent.  As Ivan points out, more important than this, ozone in the atmosphere is controlled by photochemistry, while air is not. https://www.physicsforums.com/threads/if-air-is-a-mixture-why-dont-the-gases-separate.843424/

In addition to this, the stratosphere is not entirely composed of ozone, as you seem to be suggesting. Most of the stratosphere is still the regular nitrogen oxygen mix (albeit in slightly different proportions).

LOL - molecular weight does matter. The earth's gravity isn't enough to hold helium.

@MaxW Good point! Well, it does matter in a way, but keep in mind the time scale.

I have to point out that the Physics Forums thread linked by Chester has an amazing reference inside. Definitely worth a look.

@MaxW Weight does not matter at all. Inside the homosphere (nomen est omen), close to 100km, the concentrations of all gases is perfectly constant, unless they have a local source, like ozone, or water. Stratification only takes place below pressures that are, for a chemist, nearly at perfect vaccum. Like what your well serviced oil pump can do.

@NicolauSakerNeto The link does not work anymore, I have probably found it here citeseerx.ist.psu.edu:  Schlatter: Atmospheric Composition and Vertical Structure

The entirety of the Earth's atmosphere can be divided into two regions: the homosphere and the heterosphere. The former extends from the earth's surface up to about $\pu{100km}$.  Within this region, turbulence and mixing, even within the relatively calm, persistent temperature inversion known as the stratosphere, prevents any segregation of gases based on molecular/atomic weight.  With the exception of water vapor, the composition of the major constituent gases only varies by a few hundred ppm from any given region of the homosphere to another.
The heterosphere extends from about $\pu{100km}$ above the earth's surface upward until merging with the interplanetary medium at around $\pu{100,000km}$. Within this region, the mean free path of any gas is great enough that gravitational forces have a chance to segregate the different species based on their mass, rather than by bulk processes like turbulence.
The Encyclopaedia Britannica describes the heterosphere this way:

In the heterosphere, the concentrations of the heavier constituents,
  such as nitrogen or oxygen, decrease more rapidly with increasing
  altitude than do the concentrations of lighter gases, such as hydrogen
  or helium; and eventually the atmosphere is dominated by the lighter
  gases. Under mean daytime conditions, helium and its ion become
  dominant around $\pu{1,000 km}$ ($\pu{620 miles}$) and hydrogen and protons above
  $\pu{2,500 km}$ ($\pu{1,555 miles}$).

At sufficient altitude, $\ce{H+}$ becomes the dominant species and hydrogen is the only element of significance that escapes the atmosphere based on its low mass.
So, to reiterate the answer to the question; the troposphere and stratosphere both lie well within the homosphere, and thus segregation based on mass plays no roll in their distribution throughout this region of the atmosphere.

What you say is correct but it doesn't answer the actual question.

The primary reason is because ozone is unstable and the major reactions that form it only occur in the upper atmosphere.
The lower part of the atmosphere (the troposphere) is well-mixed and there is no separation of gases by density. This is less true in some of the higher layers of the atmosphere, but this is irrelevant to why we have an ozone layer.
Ozone is unstable and reactive. The reactions that create it in the upper atmosphere are photochemical. The concentration is a balance between the rate of formation and the rate of decay and it is this balance that leads to an ozone layer. Lower in the atmosphere there is less UV and the ozone would decay faster so the concentration is usually very low. That is why we find it up top: nothing to do with density.
Having said that, we sometimes do get ozone in the lower atmosphere. This happens when certain other chemicals promote its formation (as many pollutants like nitrogen oxides do and also hydrocarbons from trees). But, being reactive, it doesn't hang around long.
It is the reactivity that matters not the density.


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Why anions of weak acids are weak bases?
This is what I saw from Chemistry: The Central Science 13th ed. by Theodore brown, page 698. It stated that there are two types of weak bases: one is those with lone pairs of electron and the other is anions of weak acids, which means conjugate bases of weak acids. My question is, aren't conjugate bases of weak acids strong bases? Because weak acids are less readily to give off proton but it's conjugate base can accept proton easily.

Your understanding is correct - on a relative scale, the conjugate base of a weak acid is a strong base, and the conjugate base of a strong acid is a weak base. For example, hydroxide ion (conjugate base of water) is a stronger base than acetate ion (conjugate base of acetic acid) which is a stronger base than chloride ion (conjugate base of hydrochloric acid).

If mine is correct, why does the book says that weak bases are anions of weak acids since conjugate bases of weak acids are strong bases.

Maybe it is a typo. These kinds of errors sometimes occur. Unfortunately I do not have a copy of the book to look at.  Or maybe the author was thinking non-rigorously about, say, acetate ion as an example - it is indeed a "weak" base (compared to some bases) and acetic acid is a "weak" acid (compared to some acids). Just not the same ones.

It's not a typo but problem with definition - if any base stronger than water would be called strong this primitive rule would be ok.

Theresa, could you please quote exactly what Brown's book states, or post a picture of the offending paragraph?


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Why are intensive properties not additive?
The extensive properties are additive. An intensive property is a ratio of two extensive properties.  So I think we can add intensive properties.

Its because intensive properties like temperature, refractive index, density are same for all the portions of the container of gas. You can't say that if the temperature of one portion is x and and other portion is also x, then the total temperature is $2x$.
But extensive properties, they are different for different portions of the container like mass, volume, length. So they can be added.
It doesn't matter if the intensive property is a ratio of two extensive properties. If two extensive properties are in ratio then if first extensive property changes then the second one should change.

An intensive property is a ratio of two extensive properties. So I think we can add intensive properties.

This is a tempting thought but is wrong for several reasons.
Mathematical inconsistency
This doesn't make mathematical sense.  Consider two extensive properties for system 1, called $A$ and $B$, and let intensive property $X = \frac{A}{B}$.  Consider another system, system 2, for which the two extensive properties are $a$ and $b$. The corresponding intensive property is $x=\frac{a}{b}$.
The question is, what are the properties of the combined systems 1 and 2 when considered as one single larger system?  You seem to be saying that you can approximate $X + x \approx \frac{A + a}{B + b}$ but this approximation is wrong.
Instead, $X + x = \frac{A}{B} + \frac{a}{b} = \frac{bA + Ba}{Bb}$.  So adding the intensive properties is not the same as adding each extensive property separately and then taking the ratio again.
Conceptual inconsistencies
It is true that all ratios of two extensive properties are intensive properties.  But I don't think it's true that all intensive properties are easily defined as the ratio of two extensive properties.  For example, temperature is an intensive property, but it isn't at all obvious to me what two extensive properties temperature is a "ratio" of.

I remember Richard Feynman's book Surely you are joking, Mr. Feynman!, where he  quotes from some Brasilian textbook of physics brought to him to review it. It was listing surface temperatures of stars of various colours, asking what is their total temperature.  :-D

$E_\mathrm{therm}=\frac 32 N k_\mathrm{B} T \implies T = \frac{2}{3k_\mathrm{B}}\left(\frac{E_\mathrm{therm}}{N}\right)$, where $N$ is the number of degrees of freedom. Both quantities are extensive.

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


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Expansion of ideal gas in isothermic conditions
I'm doing this problem and I get a different value than the answer in a book. I'm not sure if the book is wrong or me.

At a constant temperature of $298~\mathrm{K}$, calculate $\Delta S_{\mathrm{sys}}$ and $\Delta S_{\mathrm{univ}}$ for the free expansion of $3.0~\mathrm{L}$ of an ideal gas at $1.0~\mathrm{atm}$ to $11.0~\mathrm{L}$.

My solution
Firstly, I calculated the moles of gas $n$:
$$\begin{align}
n &= \frac{PV}{RT} \\
&= \frac{(1~\mathrm{atm}) \cdot (3.0~\mathrm{L})}{(0.0821~\mathrm{L~atm~mol^{-1}~K^{-1}})\cdot(298~\mathrm{K})} \\
&= 0.1226~\mathrm{mol}
\end{align}$$
Secondly, I used this to calculate $\Delta S$:
$$\begin{align}
\Delta S &= nR \ln{\frac{V_2}{V_1}} \\
&= (0.1226~\mathrm{mol}) \cdot (0.0821~\mathrm{L~atm~mol^{-1}~K^{-1}}) \cdot \ln{\left(\frac{11.0~\mathrm{L}}{3.0~\mathrm{L}}\right)} \\
&= 0.0130~\mathrm{L~atm~K^{-1}}
\end{align}$$
However, the answer in the book is $0.027~\mathrm{L~atm~K^{-1}}$. I'm not sure if that's an error.

I don't see any obvious errors in your arguments. Is the question you state all that was given? No other minor conditions?

I agree.  I get the same answer as you guys.

@SatwikPasani, I copied it exactly as given in the book. I forgot to mention that the book is "Study Guide for Chemistry 5th Edition by Zumdahl" The author of the study guide is Paul B Kelter. Copyright 2000. ISBN: 0-395-98586-2  Might be a typo.


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What are the physical properties of the cation tris(ethylenediamine)cobalt(III)?

I am aware of the (pictured) structure of $\ce{[Co(en)3]^3+}$, but what are the physical properties of this ion? Namely, what is its colour?

@我叫叶莎 You can safely assume that the additional counter-ions have no effect on the solution’s overall colour. In general, for all transition metal complexes, it is always the central metal that determines the colour. $\ce{[Co(en3)]Cl3}$ is a salt formed of a complex cation with ‘standard’ counter-ions, hence why they are outside of the square brackets.

The Wikipedia page actually lists a few sources, one of which is the Acta Cryst. B paper by Witiak, Clardy and Martin Jr who solved the structure of the nitrate. Who would have guessed that they obtained yellow-ish crystals? (CC @Joel)

[Co(en)3]Cl3 was important in coordination chemistry for its stability and stereochemistry and  has been isolated as yellow, golden, needle like crystals. It is also described as a yellow, orange solid, soluble in water. The racemic complex is commonly obtained as the di or tri-hydrate although other hydrated forms are known.(Ref:Wikipedia)
The computed physical properties and other chemical data are well documented on @pubchem.

I am well aware. But I wanted to know more about the cation [Co(en)3]3+ (tris(ethylenediamine)cobalt(III)), not the compound  [Co(en)3]Cl3 (tris(ethylenediamine)cobalt(III) chloride).

Oh OK, then why even mention the chloride anions in the structure?

I couldn't find any picture of the structure of Co[en]3 without the chloride anions mentioned. There is actually very little easily accessible information about Co[en]3 alone as a cation and not as a component of a compound. Thus, I asked this question.

Have you tried looking at the site I mentioned @pubchem? Or @chemspider.


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Sulfur from pyrite roasting process
I'm solving a quantitative problem involving the process of making oleum from pyrite.
The problem specifies that, at the end of the reaction, we find sulfur which does not come from the impurities present in pyrite.
Where is that sulfur coming from? Is it from the thermal decomposition of $\ce{FeS2}$?
$$\ce{FeS2 -> Fe + 2S}$$

What reaction? There is more than one, if I remember correctly.

@Ivan Well, I didn't include the reactions used for obtaining sulfur trioxide (which is used for obtaining oleum) because the products ($SO_2,\  SO_3, \ H_{2}SO_4$) are unlikely to decompose into sulfur.

Guilin Hu, Kim Dam-Johansen, Stig Wedel, Jens Peter Hansen. Decomposition and oxidation of pyrite, Progress in Energy and Combustion Science, Volume 32, Issue 3, 2006, Pages 295-314, ISSN 0360-1285, http://dx.doi.org/10.1016/j.pecs.2005.11.004.

"In inert atmospheres, the transformation of pyrite proceeds through a multi-step process in the sequence of pyrite→pyrrhotite→troilite→iron. The whole process is determined by the temperature and the total sulfur gas pressure in the system."

Since there isn't any oxygen in an inert atmosphere, the sulfur in the pyrite must be converted to elemental sulfur to maintain charge neutrality and mass balance as you proposed.


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What is the adverb for a "hydrogen bond"?
I am looking for some opinions concerning the terminology around hydrogen bonds. Other types of bonds can be expressed as an adjective: "The atoms are bonded covalently." or (though less common) "(...) are bonded ionically.". Sometimes the sentence can be made clearer, and its length reduced, by using the adverb form rather than the noun. In the case of hydrogen bonds, I often end up with awkward and longer-than-they-need-to-be formulations, because I feel I need to use the noun form.
However, I have come across the term "(...) are hydrogen-bonded" or "(...) are H-bonded". This is inventing a new verb, and some people do not like (I have gotten feedback on reports). "The two water molecules are bonded hydrogenly" seems odd, and no results are found in Google. So, what do you do?
My own opinion is to use "H-bonded", as I feel this gets the message across without ambiguity in a concise way.

Well, problems start because hydrogen bonds are not covalent or ionic bonds. Maybe that was what the report feedbackers were getting at?

No, the feedback was about how "H-bonded" was not a real verb, and that the sentence should be reworded to adhere to standard grammar. (Of course, it is hard to argue against that, but I feel that unnecessary complexity is added to the sentence by not accepting the "fake" verb)

"(...) are hydrogen-bonded" or "(...) are H-bonded" I have used this and I have seen lecturers use this (if it lends any credence to what I say, I am at Oxford). In my opinion, this phrasing is perfectly acceptable - it conveys exactly what you are trying to say, but I suppose there will always be purists out there who insist that you write "there is a hydrogen bond between (...) and (...)".

I agree. I will stick with the "H-bonded" formulation, and argue for it if need be! Thanks.


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The addition reaction between alkene and gaseous hydrogen bromide: why is it an addition reaction, rather than electrophilic addition?
I have come unstuck on why - despite the fact that hydrogen bromide is a polar molecule - why it does not react with ethene via electrophilic addition in its gaseous state? For example, bromine reacts with ethene by becoming an induced dipole and then reacting via electrophilic addition. Why does H-Br(g) not? Is is to do with boiling temperature?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to answer any remaining questions on how it works. Unfortunately, your question is rather unclear (and the big image isn’t helping), hence it may get closed. You can (and probably should) [edit] your post by clicking the grey edit-link below the tags to include additional context and maybe better descriptive images.

Did you actually read the textbook?? I quote ethene reacts readily at room temperature with a solution of hydrogen bromide in a polar solvent. It is another example of electrophilic addition.

Sorry, I meant to ask specifically why gaseous HBr doesn't react via electrophilic addition. See paragraph down from the one you quoted from. Many thanks,

Presumably because HBr won't readily dissociate in the gas phase and so there are no ions to do the electrophilic addition.


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Can a very high concentration of hydrogen peroxide, inhibit or denature the enzyme catalase?
My experiment, where I am investigating the effect of increasing substrate concentration of hydrogen peroxide on enzyme activity hasn't worked out as I hoped. The last result, highest concentration of hydrogen peroxide, shows a decrease in enzyme activity compared to the lower substrate concentration and I don't understand why?

We need more details to be able to help you.  What specific concentrations of hydrogen peroxide did you try?  Nanomolar?  Or molar?  It makes a big difference in possible explanations for what is happening.

It was 1, 3, 5, 10, 15 and 20% solutions of hydrogen peroxide with 2% catalase. I put 10cm^3 of hydrogen peroxide and 5cm^3 of catalase solution in a measuring cylinder with o.1cm^3 of washing liquid. The volume of bubbles produced corresponds to enzyme activity. Its a pretty basic high school experiment

20% hydrogen peroxide sounds HUGE. But something can happen that is called substrate inhibition. It can be detected if the substrate acts as a non-competitive reversible inhibitor (i.e. binds only to the active ES Michaelis complex). In the case I studied during my PhD, the inhibition constant ($K_\mathrm{i}$) was roughly 1000 times higher that the Michaelis constant ($K_\mathrm{M}$).
The steady-state rate was well fit with :
$$r = \frac{r_\infty}{1+\frac{K_\mathrm{M}}{s} + \frac{s}{K_\mathrm{i}}}$$
So the activity would go through a maximum ($s$ is substrate concentration).
Tip : if it is possible, do you assays with substrate concentrations growing exponentially, like $1\,\mathrm{\mu M}$, then 2, 5, 10, 20, etc.
A good way to decide between enzyme denaturation and substrate inhibition would be to know if it is reversible : denaturation must not be reversible. You could first make a baseline assay (at low conc.), expose the enzyme to a high concentration, then change the composition of the solution (eg dialysis, dilution) and run another assay at low concentration.
A "volume of bubbles" does not sound very accurate. Could you use a Clark electrode to measure the activity instead ?

the simplest answer for me is that;(at a higher concentration) there was not enough enzyme for all the hydrogen peroxide to react with which slowed down the process. I did the same experiment and the lab report I read gave me this explanation. And my biology teacher approved.

If there is not enough enzyme, this would imply (in first approximation) a constant, maximum value. The reaction slowing down requires further explanation.


text:
organic chemistry of alkyl halides and alcohols
What happens if a primary alcohol that also bears a halide substituent on the alcoholic carbon (the carbon to which the hydroxyl group is attached) when it is reacted with copper at 300 °C ?
   I know primary alcohols, on treatment with copper at 300 °C react to give  aldehydes.  But I am confused what happens if halides are present on the same alcoholic carbon.

You are correct that copper metal is a catalyst for alcohol dissociation to aldehyde + $\ce{H2}$.  The reaction of ethanol, for example, can be written as:

$$ \ce{CH3CH2OH  ->[\ce{Cu}~\textrm{metal}][300~^{\circ} \textrm{C}] CH3CHO + H2} $$

Your question about halo-substituted alcohols is very interesting, because those species are very unstable.  If I understand you correctly, you are not referring to halohydrins such as 2-chloroethanol, which are species that have a halo substituent beta to the alcoholic carbon, but rather to species such as 1-chloroethanol where the halo and hydroxyl substituents are on the same carbon atom.
alpha halo alcohols are very unstable, and tend to eliminate halo-acid and form the corresponding carbonyl compound.  For example, gaseous chloromethanol spontaneously eliminates hydrogen chloride to form formaldehyde at 296 K (~23 °C) with a half-life of about 300 seconds.
Thus I think the best answer to your question is that:

Halo alcohols like 1-chloroethanol or chloromethanol are already oxidized to the same degree as carbonyl compounds.
Even before reacting them with copper metal at 300 °C, they are likely to decompose to the carbonyl compound via elimination of hydrogen halide.
Unlike the reaction of simple alcohols on hot copper metal, this reaction is not a redox dissociation, but is a simple elimination reaction.$%edit$

I wouldn’t call that a disproportionation, because different atom kinds get reduced and oxidised. In my understanding, disproportionations only include reactions where atoms of the same element get simultaneously reduced and oxidised such as in $\ce{H2O2 -> H2O + 1/2 O2}$.

Jan, I can see your point.  What would be an appropriate term?  Dissociation?

I just edited to make it say dissociation instead of disproportionation.  Thanks for pointing out my non-standard use of the term!

authors: Βασίλης Πεσματζόγλου


text:
Gels that increase in volume
I would like to know if there are any gels that would increase in volume or swell due to electric field or current?!


text:
Number of resonance structures of nitrobenzene
I was studying about the resonance and here's my doubt:

How many resonating structures are there in nitrobenzene? Is it 5 (as drawn) or is it 4 because first and the last structures are same or is it 3 because second and fourth structures are also same?
In short, do we count fifth structure when we take the number of resonating structures?

We never really need to know the number of resonance structures. The whole concept is qualitative, so we can't draw any quantitative conclusions from it. But if you want to do it anyway, then of course 1 and 5 are the same, while 2 and 4 are not (they have positive charges on different atoms).
Also, both oxygens are in fact equivalent, so you might want to add the structures with their roles reversed.

The first and fifth are equal contributors but not the same.

The first and fifth are not the same, considering the position of the bonds has changed @IvanNeretin. so the number of resonating structures is 5

If you consider strict symmetry rules, you can only draw 3 really different resonance structures (the first three, from left to right), because all other structures you can draw are symmetrically equivalent to one of those 3 if you rotate it properly, those structures are only useful to show you that there is delocalization of electronic density within those atoms.


text:
Hydrogen's confusing purple glow
While reading about hydrogen on Wikipedia, I read that hydrogen has a purple glow in its plasma state. This confused me because although hydrogen is present in its plasma state on the sun, the sun doesn't appear purple to us (rather, it is yellow-orange). Why is this the case?

Weird as it is, Sun is actually a decent approximation of a black body.

And it is not yellow or orange (or even green).  It looks that way through our atmosphere, whose scattering properties cause color separation, but its actual spectrum corresponds closely to what we interpret as white light.

The radiation of hydrogen in plasma with low density comes from discrete energy levels, resulting in spectral lines. Those spectral lines can be observed in sunlight, too.
But the sun consists of a plasma with a much higher density. Most photons are the result of a long series of energy exchange inside the sun following the emission of high energy photons and kinetic energy of the nuclear processes. They are part of a continous spectrum as there is not a discrete number of energy levels at the interactions in this high density plasma.
In the photosphere, some of them are absorbed again and result in negative spectral lines (Fraunhofer lines), but the main portion of light remains a continuous spectrum.

But why is the plasma purple if the main emission line is the red Hα line at 656 nm?

Because apparently a second emission gets excited, too. The second line of the Balmer series is at 486 nm, which is blue or turqois. The resulting visual impression of both emissions should be purple.


text:
How does dimethyl sulfate react with water to produce methanol?
I came across the reaction of dimethyl sulfate with water to yield methanol. I was curious how that works as $\ce{H2O}$ isn't a strong nucleophile.
From my understanding the $\ce{H2O}$ should pop off a $\ce{-CH3}$ and you have $\ce{[CH3OH2]+}$ which has a $\mathrm{p}K_\mathrm{a}$ of -2.5 so a relatively strong base. Does the $\ce{[CH3OH2]+}$ protonate the $\ce{H2O}$ to form $\ce{H3O+}$ and $\ce{CH3OH}$?

@Joel There's no redox reaction, also I wouldn't be so sure with this mechanism, please be more careful with your posts.

@Mithoron , I'll be more careful thank you I should have checked. Apparently the DMS is hydrolyzed first to methyl sulfuric acid then to methanol and sulfuric acid in a two step process occurring slowly in cold water and rapidly in warm water or acidic solution. (Ref:Chemours/dimethyl sulfate/chemical properties)

The mechanism of methyl transfer of $\ce{Me2SO4}$ is generally thought to be an $\mathrm{S_N2}$ reaction. Thus, water would attack the methyl group, releasing $\ce{MeO-SO3-}$, a good leaving group generating methanolium $\ce{H3C-OH2+}$. The proton is later transferred back to a methyl sulphate to give methyl hydrogen sulphate $\ce{HSO4Me}$ and methanol.
One could imagine the proton from methanolium to initially be transferred to water to generate hydronium. That’s actually pretty probable and it will be hard to tell where it will end up. Water, methanol and hydrogen sulphate (and therefore likely methyl sulphate, too) all have similar $\mathrm{p}K_\mathrm{b}$ values and are equally eager to displace it/non-eager to accept it. adding some base, e.g. sodium carbonate should provide a good ‘resting point’ for such an unhappy proton.
The question is why water can do this even though it is a lousy nucleophile. Well, it turns out that for one, $\ce{MeSO4-}$ is a great leaving group enhancing the reaction rate and for two, there is nothing else around if you mix just $\ce{Me2SO4}$ and water that would be a better nucleophile. ($\ce{OH-}$ is present in too low concentrations to actually count.) Thus, the two reagents have the choice of doing nothing or slowly doing something thermodynamically favourable (= reacting). Kinetics dictate that it must happen slowly, but it can happen.


text:
What is the molecular interpretation of Raoult's law?
I was reading about Raoult's law which states:

It states that the partial vapour pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Regarding the molecular origin of this law, Peter Atkins in his book writes:

The molecular origin of Raoult's law is the effect of the solute on the entropy of the solution. In the pure solvent, the molecules have a certain disorder & a corresponding entropy; the vapour pressure then represents the tendency of the system & its surroundings to reach a higher entropy.
[...] Because the entropy of the solution is higher than the pure solvent, the solution has a lower tendency to acquire an even high entropy by the solvent vaporising. In other words, the vapour pressure of the solvent in the solution is lower than that of the pure solvent.

I've not understood his explanation especially the bold lines above.
$\bullet$ Firstly, how could vapour pressure represent the tendency of the system & its surroundings to reach a higher entropy?
$\bullet$ Why does the solution has a lower tendency to acquire an even high entropy by the solvent vaporising ? After-all every system has greater tendency to acquire high entropy, isn't it? Is not this statement contradictory to the Second law by saying lower tendency to acquire high entropy?
Could anyone please explain that to me?

I believe what the quote is saying is that once equilibrium is reached between the vapor-liquid mixture, then there will be no tendency to increase entropy. This is because entropy is not the only factor to consider. In any system, one must always consider the interplay between both entropy and energy. For instance, at equilibrium, $$\Delta G=\Delta H-T\Delta S=0$$ which means that at equilibrium, $$\Delta H=T\Delta S$$So, while you raise a good point that in most cases, a system will tend to maximize entropy, in a closed system, one must consider the interplay between entropy and energy which in this case are exactly equal (technically they differ by a scaling factor which is the absolute temperature).
Additionally, what is meant by "vapour pressure represents the tendency of the system and its surroundings to reach a higher entropy" is that one can make an inference about the energetic and entropic properties of a solution based on the vapor pressure of that solution. For instance, if a system has a very low vapor pressure (consider a solid as an extreme example) then the solution is very energetically stable and thus it has a very low tendency to reach higher entropy. But, if you think about acetone, which has a vapor pressure of $30.6 ~\text{kPa}$ at $25^{\circ}~\text{C}$ (ten times that of water at the same temperature), one could say that this then has a strong tendency to increase its own entropy and the entropy of the surroundings.

Peter Atkins in his book writes:

The molecular origin of Raoult's law is the effect of the solute on
  the entropy of the solution. In the pure solvent, the molecules have a
  certain disorder & a corresponding entropy; the vapour pressure then
  represents the tendency of the system & its surroundings to reach a
  higher entropy.

What is meant by "vapour pressure represents the tendency of the system and its surroundings to reach a higher entropy" is that one can make an inference about the energetic and entropic properties of a solution based on the vapor pressure of that solution
Raoult's law states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture
p=($\ce{pA^⋆}$) x ( $\ce{xA}$)

In this equation, $\ce{pA^⋆}$ is the vapor pressure of the pure
  solvent and $\ce{xA}$ is the mole fraction of the solvent. This law
  allows us to calculate the vapor pressure of a given solution, and
  shows the influence of all of the molecules in the solution.

Source: Boundless. “Vapor Pressure of Nonelectrolyte Solutions.” Boundless Chemistry. Boundless, 08 Aug. 2016. Retrieved 23 Oct. 2016 from https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/solutions-12/colligative-properties-of-nonelectrolyte-solutions-95/vapor-pressure-of-nonelectrolyte-solutions-406-7509/

I don't think the Atkins book's explanation is correct, but if I explain it locally, it's as follows.

The molecular origin of Raoult's law is the effect of the solute on the entropy of the solution. In the pure solvent, the molecules have a certain disorder & a corresponding entropy; the vapour pressure then represents the tendency of the system & its surroundings to reach a higher entropy.

Vapor pressure means the number of molecules in the vapor phase, and since gas phase generally has much higher entropy than liquid phase, large vapor pressure means that the entropy of the system is that high.

[...] Because the entropy of the solution is higher than the pure solvent, the solution has a lower tendency to acquire an even high entropy by the solvent vaporising. In other words, the vapour pressure of the solvent in the solution is lower than that of the pure solvent.

Assuming that the entropy $S$ is the same whether there is a solute or not, the existence of solute creates mixing entropy, which additionally contributes to $S$. In order for the $S$ value to be the same, the vapor pressure must be reduced, reducing what this contributes to $S$.
However, these arguments have one major problem and one minor problem. The state is different between the two systems with only pure solvent and when solute is added. Therefore, the second argument that the entropy $S$ should be constant for both cases is not valid.
In addition, according to Raoult's law (1), the more solute you add, the less mole fraction of the solvent and the less vapor pressure you should have.
$$ p_A = x_A p_A^* \tag{1}$$
This is qualitatively the same as the conclusion of the above argument, but quantitatively, it is not known why the changed vapor pressure should be multiplied by the mole fraction in the vapor pressure of the pure solvent.
The simple molecular interpretation of Raoult's law is as follows. Raoult's law is a law established in an ideal solution that is mixed with similar chemical species. In other words, both in liquid phase and in vapor phase, solvent A and solute B cannot distinguish each other.
Now suppose that system1 contains 100 solvent A and 10 vapor pressure (A) of these occurs. Let's also suppose that system 2 contains 50 solvent A and 50 solute B. At this time, A and B are not distinguished, so the total vapor pressure (A+B) of system2 will have to be generated by 10.
But how much of this should be the vapor pressure corresponding to "solvent A"? I hope the answer to this question will help you understand why (1) holds.


text:
Which are stronger intermolecular forces: ionic bonds or hydrogen bonds?
Do strong ionically attracted molecules have stronger intermolecular forces than hydrogen bonding forces?

Have a look at @chemistry department/Elmhurst College/hydrogen bonding. The hydrogen bond is usually considered stronger than normal dipole forces between molecules. The normal hydrogen bond is about 1/10 as strong as a normal covelant bond.

Please clarify what you mean by ionic bonding! If it is Coulomb interaction between monopoles, it is far stronger than H bonding

Question is about ionic interactions between charged molecules, like tetramethylamonium acetate imo. @anna Could you confirm it?

a molecule is "An electrically neutral entity..." http://goldbook.iupac.org/M04002.html

@DavePhD Then betaine may be better example.

@Mithoron I think the question is unclear as to whether it is asking about a force between separate ions (ionic bonding) or between charged portions of molecules, as in intermolecular salt bridges.  https://en.wikipedia.org/wiki/Salt_bridge_(protein_and_supramolecular)

Yes, looks unclear.

Hydrogen bonds are the strongest intermolecular forces. Stronger than van der Waals' forces, namely (ordained by strength):

Permanent dipole - permanent dipole (Keesom's Force);
Permanent dipole - induced dipole (Debye force);
induced dipole - induced dipole (London's dispersion force);

Strong hydrogen bonds are created by the FONs (fluorine, Oxygen and Nitrogen), which are very electronegatives, what produces a strong attraction between one them and hydrogen (which is less electronegative). As example of this interaction we can cite the water, that's a polar molecule in a "V" shape, take a look at the figure below.
Then, the both hydrogens yields a $\delta^{+}$ that's attracted by the Oxygen (very strong electronegative, the $\delta^{-}$ side) of the another water molecule, yielding the hydrogen bonds, take a look at the figure below.

Hydrogen bonding , an interaction involving a hydrogen atom located between a pair of other atoms having a high affinity for electrons. Such a bond is weaker than an ionic bond or covelant bond but stronger than van der Waals forces.(Ref:@Brittanica/hydrogen bonding). Ionic bond and van der Waals forces are both similar regarding electrostatic attraction but are describing different things specifically.

Won't lie... I thought your oversized water diagram was going to be a picture of a Voltorb. In all seriousness though, you might consider removing it, since the more reasonably sized picture below also seems to demonstrate what you were talking about, or at least try resizing it.

Thank you for the suggestion @ringo! I'll try to resize this! :)

I edited your answer to remove some errors you can revet some of the changes if you disagree. vdW doesn't mean ionic - question is about ionic interactions between charged molecules, like tetramethylamonium acetate also @Greg


text:
Is there a way to liquefy a super absorbent polymer without using salt?
I understand that once a SAP absorbs a liquid it becomes a type of gel and that this gel can be liquefied by applying salt, are there other methods to liquefy this gel without using salt?

change the pH to less than 3.   absorbents do not like low pH.
since absorbents  are a polymer of sodium acrylate.  lowering the
pH covert  sodium acrylate to the acid form (acrylic acid) the acid form is not a super absorbent.


text:
'Chemical' First-order Phase Change
First order phase changes occur when one local minima of the Gibbs Free Energy becomes deeper than another. Thus at 1 atm and 99 °C, the Gibbs Free Energy of liquid water is less than the Gibbs Free Energy of steam at those conditions (though both states form local minima). As the temperature increases, the relative depths of the two minima change: at 100 °C they are the same depth, and above 100 °C the vapour-phase minimum is lower.
I'm wondering if similar first-order changes can occur in chemically reacting systems? It seems to me that equilibrium constants always change continuously with temperature: are there any situations in which this isn't the case, and there exist discrete jumps in chemical compositions? And if these jumps in the deepest minima of the Gibbs Free Energy never occur in chemically reacting systems, do we know why not? Of course, situations in which a chemical reaction is accompanied by a phase change don't count!

Well, let's see. As you know, the equilibrium constant is determined by the reaction's $\Delta_rG:\;K=e^{-\frac{\Delta G}{RT}}$, so a discrete jump in the former means a discrete jump in the latter. What is that $\Delta_rG$? That's the sum of $\Delta_fG$ of products minus that of starting compounds. Now what is the $\Delta G$ of formation of a compound? A continuous function, unless there is a phase transition, which by your problem definition is not the case. Well, a linear combination of a few continuous functions is still a continuous function.

What if there are two competing equilibria?  With different equilibrium constants, that vary with temperature in different ways?  Isn't that exactly analogous to the phase transition example and contrary to your answer?

Sounds mighty like there is a flaw in my reasoning, yet still I can't pinpoint one. Let's look at it this way. If $\Delta G$ of a phase equilibrium crosses 0, a phase transition happens. That's not our case. What happens if $\Delta G$ of an equilibrium within one phase crosses 0? Looks like nothing. Looks like the equilibrium concentrations would behave like continuous functions of $K$ and hence of $\Delta G$. Is that so?


text:
How to save the force on a molecule in GROMACS?
I am a new student of molecular dynamics. I have started using GROMACS a month ago.
I have a small box of 8192 molecules each with 3 atoms. I froze the last molecule and ran the simulation. I even converted the .trr file from binary to .gro and checked that the last molecule indeed stayed in the same place.
However, I want the .trr file to print the force on each molecule or each atom - in the x, y, and z-direction.
I even set the nstfout in the grompp.mdp file to a reasonable non zero number - 5000.
However, the .trr file - when converted to .gro - did not show the force. Can someone tell me what I am doing wrong and how to get GROMACS to save/reveal the force in the .trr file?

It may be considered off-topic as the question is about using program - more appropriate on computational science SE

I don't think this is off-topic here.  However, it may well be true that other sites may be able to answer better.

The GROMACS Users mailing list is a good place for such questions http://www.gromacs.org/Support/Mailing_Lists/GMX-Users_List

Gro output format does not support forces, so using it as an intermediate output format defeats your purpose.
Use
gmx traj -f your.trr -of

which runs the gmx wrapper binary, calling the traj module to do trajectory processing, names your input trajectory file after -f and triggers the output of forces with -of (optionally named after -of, defaulting to force.xvg), or similarly
gmx dump -f your.trr

to dump the entire contents of the file including forces) to the terminal.
See also
gmx help traj

or
gmx traj -h

or http://manual.gromacs.org/documentation/5.1/onlinehelp/gmx-traj.html
Note that I currently develop GROMACS.


text:
How can the thermal conductivity of a binary gas mixture be estimated?
Recently, I am seeking for a simple but relatively accurate method to estimate the thermal conductivity of a binary gas mixture, say ethanol vapor + nitrogen. I assume the mixture as ideal gas.
The molar fraction and thermal conductivities of the components are known. The parameters for a try are: mole fraction of ethanol vapor: $x_A=0.2$, thermal conductivity of ethanol vapor and nitrogen are $k_A=0.016\ \mathrm{W/(m\,K)}$, $k_B=0.026\ \mathrm{W/(m\,K)}$, temperature of the mixture gas $T=330\ \mathrm{K}$, and the bulk pressure is $101\ \mathrm{kPa}$.
I really hope someone could give me a hint or point me to a classical reference.
Thank you in advance!

I'm thinking you could use the Fourier equation for heat flux
$$
\phi =K \frac{S\theta}{e} \,\, ,
$$
where

$\phi$ = Heat flux
$K$ = Termal conductivity coeficient in units of  $kcal  s^{-1}\ m^{-2}\ K^{-1} /m^{-1}$
$S$ = Area of the section you're dealing $m^{2}$
$θ$ = Difference of temperature $K$
$e$ = Thickness in $m$.

And use for the resulting thermal conductivity the fraction
$$
K_{r}=\frac{\sum K_{i}J_{i}(M_{i})^{1/3}}{\sum J_{i}(M_{i})^{1/3}} \,\, ,
$$
where:

$K_{r}$ = Resulting termal conductivity of the moisture;
$K_{i}$ = Termal conductivity of each gas;
$j_{i}$ = Molar fraction of each gas;
$M_{i}$ = molar mass of each gas.

I think it'll work with some 2% of standard deviation. If you're able to read in portuguese, here is the link of my reference:
Reference
And here is the link of the Book cited by @Chester Miller.
I hope I hope I was Helpfull!

See also Section 9.3 Theory of Thermal Conductivity of Gases at Low Density in Transport Phenomena by Bird, Stewart, and Lightfoot.

Hi @ Chester Miller, thank for your answer, but I can not find this reference in the library database... Could you please share me your soft copy? Many thank!

I don't understand.  This is a very well known book.  It has been in use for over 50 years, and has stood the test of time.  A new edition came out in around 2000.  Here is the reference again:  Bird, R.B., Stewart, W.E., and Lightfoot, E. N., Transport Phenomena, John Wiley, New York, 2nd Edition, 2002.

Unfortunately I may have to disagree that this answers the question. Could you include some of the relevant parts of the book here?

Isn't mentioning Section 9.3 good enough?  I'm uncomfortable with the IP implications of copying some of the text.  If you want, I can try to paraphrase some of it.

@ChesterMiller, I think I have misunderstood your citation. The title of the book is  Transport Phenomena, which is indeed a wonderful book!

In my long career as a ChE in industry, I used Transport Phenomena much more than all my other books combined.

While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference.  Link-only answers can become invalid if the linked page changes. - From Review

@Jan Thank you for your "timely" response.  It's only a year late.

OP asked for a reference, so in my opinion, your answer is fine, albeit rather brief. It might be better if you could include some of the information from there, but that's up to you. (It's not NAA.)

@ChesterMiller Some things only turn up if they are flagged. I saw the flag in the Low Quality Posts review queue and responded accordingly. I wouldn’t have opened the post in itself since I don’t know much about the topic, but imho this is not an answer, just a reference point. Four other reviewers agree with me.


text:
What are the products in a reaction between a metal and an acid?
When magnesium metal and hydrochloric acid react, hydrogen gas is released (and not chlorine gas). However, when metals such as copper is placed in nitric acid, NO(g) is released and not hydrogen. So, how do we know that chlorine gas is not released in the first reaction? Does this have to do with reduction potential?

Nitric acid ($\ce{HNO3}$) is a strong oxidizing agent. The hydrogen gas ($\ce{H2}$) produced in the reaction with metal is oxidized to $\ce{H2O}$, in turn evolving nitric oxides like $\ce{NO2}$ and $\ce{NO}$. Even magnesium can be observed evolving nitric oxides when reacted with $\ce{HNO3}$.
Additionally, you know that chlorine gas isn't released by the reaction of $\ce{HCl}$ with magnesium because you would see it evolved from the solution as a pale, yellow-green gas. Instead, only clear hydrogen gas is evolved.

Thanks for the answer. Though, my main intent of asking this question was to be able to predict which gases would be produced in an acid-metal reaction, so is there any theoretical way to know that Cl2 isn't produced in the reaction? Sorry for being unclear!

It sounds like you are looking for a reduction potential table.

Thanks for the reply. I assume that Cl2 gas isn't produced because it has a high reduction potential to become Cl- ions - I think I get it now :)


text:
What is the role of the slash ("/") in chemical reactions?
I have always been confused about the role of slash in chemical reactions.
In some reactions, e.g. $\ce{NaBH4}/\ce{OH-}$ it means we use both compounds, right?
And in reactions where we write $\ce{Pt/Pd/Ni}$ do we mean platinum or palladium or nickel?

Usually in a reaction slash means both. H2, Pd/C is the prime example. If it's in a sentence, i.e. plain text, I'd say it probably means or, although context is important as well.

I always thought the everyday meaning is implied and it seems to have been a correct assumption every time.

I'm not aware of the standard, but the textbook usually specifies it above the reaction when it is first mentioned. It would be useful to have convention.

The ambiguity seems to come not so much from Chemistry, but from English punctuation. Slashes are often used in place of "and", "or", or the latin "cum" (meaning combined with). It seems that it unfortunately is just context dependent, though I would argue its use as a replacement for "or" is most consistent grammatically and defining this to be the meaning of the slash would eliminate ambiguity in these situations.

The "/" mainly means the medium in which the compound is present. NaBH4/OH- means NaBH4 is being used in basic medium. After all the medium in which a particular compound is present matters much in a chemical reaction. Similarly, Pt/Pd/Ni (which is commonly used when H2 is a reactant) implies H2 is being passed over any of the above 3 metals (which help to dissociate H2 into [H] atoms to help in reaction).

The two uses of '/' that you are describing seem to be completely different, not similar.

Agree with bon...in the first example of yours / plays role of and in 2nd it plays role of or..this is exactly my confusion


text:
Which is the central atom in Lewis Dot Structures?
What is the correct Lewis structure for hypochlorous acid, a compound containing chlorine, hydrogen and oxygen?
As shown in the following video, the central atom in Lewis Dot Structures is always the least electronegative atom of all the elements that make up the atom.
https://www.khanacademy.org/science/chemistry/chemical-bonds/copy-of-dot-structures/v/drawing-dot-structures
This way, you would expect chlorine to be the central atom. However, it is shown that oxygen is the central atom.
The ambiguity exists for molecules like $\ce{OCl_2}$ as well. What's the issue? Is this not a very consistent rule of thumb?

isn't Chlorine the most electronegative?

That's fluorine.

No in your question you said that the central atom is the least electronegative, but then you state that chlorine should be the central atom. Is that a typo? Because the least electronegative in hypochlorous acid would be, between oxygen and chlorine, oxygen.

Isn't oxygen more electronegative than chlorine?

Yes you are correct, sorry I just checked. I guess it kind of depends on the moles in play. For OCl2, there's one oxygen atom and two chlorine, so it would seem like oxygen would be in the center with 2 chlorines branching out. Even if you tried to put chlorine in the center for OCl2 while drawing a dot structure, it would inevitably come out to a structure similar to H2O.

The rules for drawing dot structures leave a lot to be desired, that's the main issue. For example, most organic compounds are outright impossible to draw using only a formula and the rules. You couldn't draw OCl2 in any other way, and that goes for most other violations of the dot structure rules. Molecular orbital theory gives more accurate results. That said, other than "you couldn't follow the rules", I don't have a proper answer to this. I'm not so sure there is one.

This question is a duplicate of several questions: http://chemistry.stackexchange.com/questions/25128/what-is-the-lewis-structure-for-hclo?rq=1 , http://chemistry.stackexchange.com/questions/13615/determining-central-atom-in-ncl3/13617#13617 , http://chemistry.stackexchange.com/questions/16624/which-atom-goes-in-the-middle-of-a-lewis-dot-structure?rq=1

The video is wrong if it says "the central atom ... is always the least electronegative atom".
There are chlorine oxides with Cl central such as perchlorate, chlorate and $\ce{ClO2}$, even though $\ce{Cl2O}$ has O central.
There are $\ce{SeCSe}$, $\ce{CAt4}$, $\ce{NI3}$ and $\ce{Li6C}$ each with the more electronegative element central.

How would you answer the question then? Is there always some level of doubt one would have when coming up with Lewis structures when just given the constituent atoms?

@Arjun I would have to say there is always some doubt.  For example HNCO, HOCN, and HCNO all exist, although HNCO is most stable.  https://en.wikipedia.org/wiki/Isocyanic_acid  Additional guidelines beyond "the more electronegative element central" would be "H" is almost always on the outside, usually no lone pairs on "C", almost never more than one lone pair on nitrogen, no more than two lone pairs on "O" in a neutral molecule.


text:
What determines whether a double displacement reaction will occur?
In normal displacement reactions, reactivity plays a large role and sometimes the reaction doesn't even happen. So is there anything limiting double displacement reactions? For example $\ce{2KI + Pb(NO3)2 -> 2KNO3 + PbI2}$. If reactivity plays a role in this reaction, why does it occur? Isn't the halogen ion (iodide) more reactive than $\ce{NO3-}$?
(I'm a high school chem student so this may come off as a newbie question, sorry)

That name (double displacement) is nothing more than a name. At university level it is not used to the best of my knowledge. It is always the overall balance of reactivity or in your case solubility that matters.

@Jan could you elaborate on why potassium ion prefers the nitrate ion, if that is the case?

There is no $\ce{NO3^2-}$, only $\ce{NO3^-}$. Also, there is no such thing as reactivity in general. As for potassium ion, it does not prefer nitrate. Potassium ion does not really "know" that a reaction takes place at all.

@S.Mo It is not potassium ‘preferring’ nitrate — in fact there are no ‘preferences’ (i.e. choices) involved at all. Rather, $\ce{PbI2}$ has a very low solubility, thus iodide and lead(II) in solution will meet each other and precipitate. Potassium and nitrate are merely spectators.

@Jan can we predict such equations with other elements using some kind of rule of thumb?

No.$%comments~$

@S.Mo generally speaking, the alkali metal ions (Li+, Na+, K+, etc.) and ammonium as well as nitrate and perchlorate will remain in solution. http://www.chem.sc.edu/faculty/morgan/resources/solubility/

The simple answer is whether the reaction leads to the formation of a solid (precipitate) or a gas otherwise there is really no reaction

Generally speaking, that's correct, but it would be great if you could elaborate a little bit or perhaps give some examples.

It's often helpful to use state symbols when dealing with solutions, especially in this case. Let's say you mix $\ce{KI}$ with $\ce{NaNO_3}$. Because the $\ce{KI}$ and $\ce{NaNO_3}$  are soluble and fully dissolved (or assumed to be) in the solutions, before mixing, we write them as $\ce{KI}$ $\ce{(aq)}$ and $\ce{NaNO_3}$ $\ce{(aq)}$. The aq stands for aqueous. After you pour them together, the ions in both solutions mix into the larger volume. However, they're still fully dissolved, so no reaction happens. Before, you had $\ce{K^+}$ $\ce{(aq)}$ and $\ce{I^-}$ $\ce{(aq)}$ with $\ce{Na^+}$ $\ce{(aq)}$ and $\ce{NO_3^-}$ $\ce{(aq)}$, and that's exactly what you have afterward. ($\ce{KI (aq)}$ means the same as $\ce{K^+}$ $\ce{(aq)}$ + $\ce{I^-}$ $\ce{(aq)}$, and the same with sodium nitrate).
However, with $\ce{KI}$ $\ce{(aq)}$ and $\ce{PbNO_3}$ $\ce{(aq)}$, they are both initially fully dissolved, but after pouring them together, the $\ce{Pb^{2+}}$$\ce{(aq)}$ ions react with the $\ce{I^-}$ $\ce{(aq)}$ ions and form a yellow solid ($\ce{PbI_2}$) that IS NOT SOLUBLE in water. It comes out of solution as a precipitate. Thus, we write it as $\ce{PbI_2}$ (s). Something actually happened: you had ions that mixed to form a solid.
The equation you have is better written this way:
$\ce{KI (aq) + Pb(NO_3)_2 (aq) -> KNO_3 (aq) + PbI_2 (s)}$.
Notice the solid. If a reaction produces only aqueous products, then nothing happens beyond mixing. If it produces anything that's non-aqeuous (like in this case), THEN there's a chemical reaction.

Even better is to cancel the spectator ions, i.e., potassium ion and nitrate ion.


text:
Why does the copper anode dissolve?

In such a setup, the copper anode is known to dissolve. My question is, why? Does it not receive electrons from the $ \ce{OH^-}$ ions. And if you say that those electrons just flow to the battery, let me provide the counterargument: the electrons flowing to the cathode do not prevent the copper cathode from dissolving because copper ions in solution accept the ions, so there is actually nothing stopping the cathode from dissolving as well. So, very simply, my question is this:  Why does the copper anode dissolve?

The anode is where oxidations happen (the cathode is where reduction takes place).
In the vicinity of your anode, you have the following compounds: $\ce{H2O}$, $\ce{Cu}$, $\ce{Cu^2+}$, $\ce{SO4^2-}$. If you check oxidation states, you will realise that oxygen is always $\mathrm{-II}$, hydrogen is $\mathrm{+I}$, sulphur is $\mathrm{+VI}$ and copper is present in both $\mathrm{\pm 0}$ and $\mathrm{+II}$. Thus, the only species available for oxidation are copper($0$) and oxygen($\mathrm{-II}$).
Checking standard potentials:
\begin{align}
\ce{O2 + 4 H+ + 4 e- &<=> 2 H2O} \qquad &E^\circ = \pu{+1.23 V}\tag{1}\\
\ce{Cu^2+ + 2 e- &<=> Cu} \qquad &E^\circ = \pu{+0.35 V}\tag{2}
\end{align}
It should be clear why copper is oxidised.
On the reduction side, hydrogen, copper and sulphate all could accept electrons. Copper is again the happiest of the three to accept electrons.

I don't fully understand the potential part because I haven't learnt it yet, but it's definitely a more solid answer than my textbook can give me. Thanks.


text:
Reaction of furan with chloroform in base

This reaction looks like the Riemer-Tiemann reaction. The only difference is the reactant. First step is the formation of carbene. Second step involves the attack of π-electrons of furan on the vacant orbital of carbene which results in two possible type of products:

The answer given is of type (B) (That is, dichlorocarbene is attached like product B). But both results in the same position of carbocations. Aren't both products produced in equal quantity?
Also, in the case of pyrrole, the opposite happens. That is, the major products are:

The key to this question lies in realising that the Reimer–Tiemann reaction is not an electrophilic aromatic substitution but it is in fact

addition of a carbanion to a strongly electron-deficient carbene.

This was discussed by @KlausWarzecha and @ron in answers to a previous question.
Therefore the major product will be the one where the attack comes from the carbon with the greatest electron density. Using MolCalc (partly developed by @JanJensen) I generated an electron density plot of furan which clearly shows that there is more electron density on the 3 and 4 carbons than on 2 and 5. This explains the given major product.

Is there a reason why there is a difference in electron density?

This answer by @Ron says otherwise...  http://chemistry.stackexchange.com/questions/26743/electrophilic-substitution-of-pyrrole-regioselectivity

@AdityaDev How does it say otherwise? The Reimer-Tiemann reaction is not an electrophilic aromatic substitution so the reasoning given in that answer does not apply.

I could not get why reaction of furan is a reimar teimar type rection and not pyrole.


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Can the concentration of a sodium carbonate solution be determined using volumetric analysis?
In the titration of sodium carbonate with hcl , isn't the first end point determined by phenol phthalene is equivalent to the volume of sodium carbonate
because my text book say thats the volume at the first end point is just half the sodium carbonate and the seconds end point is the whole sodium carbonate.
I know that at the first end point all the sodium carbonate all be converted to sodium bicarbonate but why shouldn't i stop here?!

The titration of hydrochloric acid with sodium carbonate is a common school practical, used to determine the concentration of the acid (standardisation).
In the first step, as you rightly point out, the sodium carbonate reacts with one equivalent of hydrochloric acid. The phenolphthalein shows the pH change associated with the formation of the sodium hydrogen carbonate (aka bicarb). You could use this value to calculate concentration, however, in order to ensure your answer was correct, you'd need to run this titration at least twice more, averaging your answers and discounting any anomalous titres.
$\ce{Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)}$
As the bicarbonate is also reactive with hydrochloric acid, you can do something known as a 'double titration'. After the first titration, you can immediately do a second titration, this time using a different indicator, such as methyl orange. By doing this, you get a second value. If your titration is good, the values for the first and second titrations should be equivalent.
$\ce{NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) }$


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Effect of nitrobenzene on solution UV/Vis spectra
I am wondering if there is a difference in the maximum absorbed wavelength of light if nitrobenzene is added to an acidic or basic solution.
I've figured that resonance plays a part in some structures like phenol, where the electron density in the benzene ring increases when in basic solution and therefore the energy required to go from a $\pi$ to $\pi^*$ orbital decreases and thus $\lambda_{\text{max}}$ increases. But could resonance play a role in nitrobenzene too? And if it does, how?

http://pubs.rsc.org/en/content/articlepdf/1980/c3/c39800000394?page=search

I am not able to access this article

Nitrobenzene does protonate at the nitrogroup in concentrated acid. Obviously, then the group becomes even more electron-withdrawing, and a bathochromic shift is observed. W. F. Forbes provided an excellent article for You in 1958, freely available now:
https://cdnsciencepub.com/doi/pdf/10.1139/v58-200
The "B-band" (CT band from the ring onto the nitrogroup) of nitrobenzene is at 252 nm in hexane, at ca. 265–266 nm in neutral water, at 266 nm both in 1M NaOH and 0.1M HCl but at 287.5 nm in concentrated H2SO4.
I am not sure about the intensities Forbes et al. report, as newer sources (but not that rich in experimental material) cite the transition at 252 nm (in non-polar solvents) to have intensity about 10 times lower. Maybe another scale was in common use back then.