text:
Why does a photon have linear momentum, although it doesn't possess mass?
We know that photon doesn't possess mass. But it has a tendency of colliding with surfaces, as we see in the photoelectric effect. As it doesn't possess mass, then how does it have linear momentum?
$E=mc^2$ and $E=hc/\lambda$
$mc = h/\lambda =p$
Here $m=0 \Rightarrow p=0$.
But photons can collide.

https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation

Photon does have mass. It doesn't have rest mass, but that's another story.

Moving mass= rest mass/√1-(v/c)^2 now as rest mass =0 then moving mass must be zero

@jay it does not work like that for light.  The formula you give is undefined when $v=c $.

I'm voting to close this question as off-topic because it is really a physics question, with answers over on physics SE.

In general, Physics is good at answering "what" and "how" questions, it doesn't do very well at answering "why" questions except reductively or tautologically. There's several ways to answer your question, but they really don't satisfy (imho) anyone who seriously wants to know why photons have momentum - at least not in any fundamental sense. I think the better question is:"What led you to believe that having mass is a prerequisite for having momentum?" Chances are, you were told something that you interpreted to mean that. No. Photons have momentum because that is the way our Universe works. Or another way to answer is: Anything that has energy has momentum. The energy of anything can be divided into potential and kinetic energy. There is a intimate relationship between momentum and energy - for example what is the derivative of ½mv² with respect to velocity? (and what is the integral of mv dv?) It turns out that momentum and energy are two aspects of the same Laws of Physics. I'm not sure if these various answers can help or not. It may be possible to imagine a Universe in which only things with rest mass also carried momentum. If so, I am unaware of how you could go about such imagining.

The formula for the momentum, $p = mv$ is only valid in the context of classical mechanics. For photons, energy and momentum are related by $E = pc$, which, because $E = hv = \frac{hc}{\lambda}$ leads to the following formula: $p = \frac{h}{\lambda}$. So the momentum is always non-zero and depends only on the wavelength.
See this Wikipedia page for more details.

But mass of photon is zero how can it collide?

Mass isn't important; momentum is.

But this are contradicting!

No, it's not. Just like you can't use $E=mc^2$ to determine the energy of a photon, you can't use $p=mc$ to determine the momentum.

Means planks theory and Einstein's theory can not be co-relate?

Not in this simple way. It's something physicists have been researching for decades now.

Our teacher told us the experiment of plank was explained by Einstein theory of relatively. Its difficult to relate particle and wave nature of light!

Actually momentum is mass times velocity.  But the mass includes the equivalent mass due to the energy of motion.  What doesn't work from Newtonian mechanics is the assumption that mass is independent of motion.  Thus a photon moving with energy $E$ and speed $c$ has mass  $E/{c^2}$ and momentum $(E/{c^2})(c)=E/c$.


text:
How to estimate the enrichment of U-234 from the enrichment of U-235?
Fresh nuclear fuel made from naturally occurring uranium (i.e. neither MOX nor ERU fuel) is mainly characterised by the enrichment of $\ce{^235U}$. The mass and activity concentrations of $\ce{^235U}$ and $\ce{^238U}$ can be directly calculated from the given level of enrichment. If necessary, the mass and activity concentrations of relevant decay products can also be calculated easily.
However, the gas diffusion or gas centrifuge processes for $\ce{^235U}$ enrichment do not only lead to an increase in the $\ce{^235U}$ concentration in the fuel, but also to a more than proportional increase in the $\ce{^234U}$ concentration. This can be seen from the following table where the mass concentrations of uranium isotopes in natural uranium and typical fresh nuclear fuel (with $3.5\ \%\ \ce{^235U}$) are compared.
$$
\textbf{Mass concentration of uranium isotopes (in g/kg HM) in fresh nuclear fuel} \\
\begin{array}{lll}
\hline
 & \text{Natural U} & 3.5\ \%\ \ce{^235U} \\
\hline
\ce{^234U} & \hphantom{00}0.06 & \hphantom{00}0.4 \\
\ce{^235U} & \hphantom{00}7.2 & \hphantom{0}35.0 \\
\ce{^238U} & 992.8 & 965.0 \\
\hline
\end{array}$$
The corresponding activity concentrations are shown in the second table.
$$
\textbf{Activity concentration of uranium isotopes (in MBq/kg HM) in fresh nuclear fuel} \\
\begin{array}{lll}
\hline
 & \text{Natural U} & 3.5\ \%\ \ce{^235U} \\
\hline
\ce{^234U} & 13 & 93 \\
\ce{^235U} & \hphantom{0}0.58 & \hphantom{0}2.8 \\
\ce{^238U} & 13 & 12 \\
\hline
\end{array}$$
Is there a simple  and reliable calculation method or model that can be used to estimate the enrichment of $\ce{^234U}$ from a given enrichment level of $\ce{^235U}$?

It would depend on the enrichment method used, and how much somebody cared about the $^{234}$U.

Could a sequential off-line or continuous on-line monitoring by mass spectroscopy, discerning the isotopes (like $\ce{^{79}Br}$ / $\ce{^{81}Br}$  or $\ce{^{35}Cl}$ / $\ce{^{37}Cl}$ in small molecular organic chemistry) assist to establish a descriptive model about the enrichment of the uranium isotopes?

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Jahn-Teller distortion of [Fe(H2O)6]2+
I am wondering whether Fe(II) hydrated to an octahedral complex would undergo a Jahn-Teller distortion.
As far as I am aware, this distortion comes about because of asymmetry in filling orbitals, in this case the t2g and/or eg orbitals.
I am also aware that the distortion is most obvious for asymmetry in the eg  orbitals because these interact most with the ligands.
However, I deduced the configuration t2g4, eg2 for [Fe(H2O)6]2+, which has no asymmetry in eg but does in t2g.
My question is can it distort because of this asymmetry? Would it be possible that promotion to an excited state takes place for the distortion to then be possible, with an overall stabilising effect?

(1) It can and will distort. In general, any non-linear molecule with an orbitally degenerate ground state is susceptible to a first-order JT distortion. $\ce{[Fe(H2O)6]^2+}$ falls under this category.
However, the Jahn-Teller theorem has its roots in symmetry and group theory. These can only tell you whether an integral is zero or whether it is nonzero - i.e. whether there is no distortion, or whether there is a distortion. They cannot make any prediction as to the extent of the distortion, or the exact form of the distortion.
The ground-state distortion in $\ce{[Fe(H2O)6]^2+}$ is small, which may be rationalised as being due to an unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals, which point between ligands.
(2) No, the energy required to promote an electron is far larger than any stabilisation that can be derived from a JT distortion.

I guess the fact that this is often not discussed in inorganic chemistry textbooks is because it is probably negligible.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Gaussian Error: The Name of the Center is Too Long
I have been conducting research on $\ce{C10}$ - $\ce{C16}$ alkanes and encountered an unusual error when I attempted to perform an optimization and frequency calculation on Tridecane at the G3 level of theory at 298 Kelvin. Additionally, Tridecane was the only molecule within the set to experience an error when run. I believe I have isolated my error to the Z-Matrix.
Input File:
%chk=TDG3
#N G3 Temperature=298 opt freq

opt freq calc Tridecane

0 1
C
C     1    B1
C     2    B2   1    A1
C     3    B3   2    A2   1    D1
C     4    B4   3    A3   2    D2
C     5    B5   4    A4   3    D3
C     6    B6   5    A5   4    D4
C     7    B7   6    A6   5    D5
C     8    B8   7    A7   6    D6
C     9    B9   8    A8   7    D7
C    10   B10   9    A9   8    D8
C    11   B11  10   A10   9    D9
C    12   B12  11   A11  10   D10
H    13   B13  12   A12  11   D11
H    13   B14  12   A13  11   D12
H    13   B15  12   A14  11   D13
H    12   B16  11   A15  10   D14
H    12   B17  11   A16  10   D15
H    11   B18  10   A17   9   D16
H    11   B19  10   A18   9   D17
H    10   B20   9   A19   8   D18
H    10   B21   9   A20   8   D19
H     9   B22   8   A21   7   D20
H     9   B23   8   A22   7   D21
H     8   B24   7   A23   6   D22
H     8   B25   7   A24   6   D23
H     7   B26   6   A25   5   D24
H     7   B27   6   A26   5   D25
H     6   B28   5   A27   4   D26
H     6   B29   5   A28   4   D27
H     5   B30   4   A29   3   D28
H     5   B31   4   A30   3   D29
H     4   B32   3   A31   2   D30
H     4   B33   3   A32   2   D31
H     3   B34   2   A33   1   D32
H     3   B35   2   A34   1   D33
H     2   B36   1   A35   3   D34
H     2   B37   1   A36   3   D35
H     1   B38   2   A37   3   D36
H     1   B39   2   A38   3   D37
H     1   B40   2   A39   3   D38

B1  1.529941022
B2  1.529941022
B3  1.530022673
B4  1.529941022
B5  1.530022673
B6  1.529941022
B7  1.529941022
B8  1.530022673
B9  1.529941022
B10  1.530022673
B11  1.529941022
B12  1.529941022
B13  1.090066218
B14  1.090066218
B15  1.090008482
B16  1.090008482
B17  1.090008482
B18  1.090066218
B19  1.090066218
B20  1.090008482
B21  1.090008482
B22  1.090066218
B23  1.090066218
B24  1.090008482
B25  1.090008482
B26  1.090066218
B27  1.090066218
B28  1.090008482
B29  1.090008482
B30  1.090066218
B31  1.090066218
B32  1.090008482
B33  1.090008482
B34  1.090066218
B35  1.090066218
B36  1.090008482
B37  1.090008482
B38  1.090008482
B39  1.090066218
B40  1.090066218

A1  109.4723086
A2  109.4744707
A3  109.4744707
A4  109.4744707
A5  109.4744707
A6  109.4723086
A7  109.4744707
A8  109.4744707
A9  109.4744707
A10  109.4744707
A11  109.4723086
A12  109.4725572
A13  109.4725572
A14  109.4728391
A15  109.4704114
A16  109.4704114
A17  109.4714761
A18  109.4714761
A19  109.4704114
A20  109.4704114
A21  109.4714761
A22  109.4714761
A23  109.4704114
A24  109.4704114
A25  109.4725572
A26  109.4725572
A27  109.4693304
A28  109.4693304
A29  109.4725572
A30  109.4725572
A31  109.4693304
A32  109.4693304
A33  109.4725572
A34  109.4725572
A35  109.4704114
A36  109.4704114
A37  109.4728391
A38  109.4725572
A39  109.4725572

D1  180.0000000
D2  180.0000000
D3  180.0000000
D4  180.0000000
D5  180.0000000
D6  180.0000000
D7  180.0000000
D8  180.0000000
D9  180.0000000
D10  180.0000000
D11 -59.99687819
D12  59.99687819
D13  180.0000000
D14 -60.00082041
D15  60.00082041
D16 -59.99621620
D17  59.99621620
D18 -60.00082041
D19  60.00082041
D20 -59.99621620
D21  59.99621620
D22 -60.00082041
D23  60.00082041
D24 -59.99687819
D25  59.99687819
D26 -60.00015848
D27  60.00015848
D28 -59.99687819
D29  59.99687819
D30 -60.00015848
D31  60.00015848
D32 -59.99687819
D33  59.99687819
D34  119.9991796
D35 -119.9991796
D36  180.0000000
D37 -59.99687819
D38  59.99687819

Error:
Charge =  0 Multiplicity = 1
The name of the center is too long.
Authenticating with public key "imported-openssh-key"
?
Error termination via Lnk1e in /cluster/software/gaussian/gaussian-09-
D.01/g09/l101.exe at Mon May  1 12:25:02 2017.
Job cpu time:       0 days  0 hours  0 minutes  0.1 seconds.
File lengths (MBytes):  RWF=      5 Int=      0 D2E=      0 Chk=      2 Scr=
1

Not Solved; Work Around Established
I have not determined the source of the error; however, I have found a way to work around the cause of the error. By substituting a Cartesian Coordinate system for the Z-Matrix the error no longer occurs. This indicates the source of the error can be narrowed down to within the confines of the Z-Matrix.

I once found this error when moving input files across computers. There break marks were present ^M but only visible in vi. I eliminated them with the usual procedure
:1,$s/^M//

and it worked. I didn't find the exact source of error either but that was the last time I found the error.

The tool dos2unix does the same job, with a smaller overhead and possibly more reliable. Another possibility is to use a sed (on unix-like) command (see one-liners): sed 's/.$//'. In vim the abbreviation for 1,$ (first to last) is % (whole file).

I'm not able to reproduce that error, but your input file also has another problem, which might be causing your problems but you should get a different error. Something along the line of:

Undefined symbol, bond length, card  14.
Undefined symbol, angle alpha, card  14.
Undefined symbol, bond length, card  15.
Undefined symbol, angle alpha, card  15.
Undefined symbol, angle beta, card  15
Undefined symbol, bond length, card  16.

Gaussian is very sensitive to wrong formating. Blank lines usually mean end of "block" so it doesn't find the variables/constants Ax and Dx. You need to remove those blank lines. After that I can open it with Gaussview and Gaussian09 has no problems starting it (did only some seconds to look if there's an input error).
That Z-Matrix also isn't very "clean", since there's no reason to define different variables/constants with the same value, one would be sufficient. For example B1 and B2, B4, B6, B7 and B9 are all the same, you can just use B1 for all of those which cleans up the whole Z-Matrix.
For example here 6 variables/constants are sufficient:
%chk=TDG3
#N G3 Temperature=298 opt freq

opt freq calc Tridecane

0 1
C
C   1   R1
C   2   R1           1   B1
C   3   R1           2   B1          1   D1
C   4   R1           3   B1          2   D1
C   5   R1           4   B1          3   D1
C   6   R1           5   B1          4   D1
C   7   R1           6   B1          5   D1
C   8   R1           7   B1          6   D1
C   9   R1           8   B1          7   D1
C   10  R1           9   B1          8   D1
C   11  R1           10  B1          9   D1
C   12  R1           11  B1          10  D1
H   1   R2         2   A2        3   D1
H   1   R2         2   A2        3   D2
H   1   R2         2   A2        3   -D2
H   2   R2         3   A2        4   -D2
H   2   R2         3   A2        4   D2
H   3   R2         4   A2        5   D2
H   3   R2         4   A2        5   -D2
H   4   R2         5   A2        6   -D2
H   4   R2         5   A2        6   D2
H   5   R2         6   A2        7   D2
H   5   R2         6   A2        7   -D2
H   6   R2         7   A2        8   -D2
H   6   R2         7   A2        8   D2
H   7   R2         8   A2        9   D2
H   7   R2         8   A2        9   -D2
H   8   R2         9   A2        10  -D2
H   8   R2         9   A2        10  D2
H   9   R2         10  A2        11  D2
H   9   R2         10  A2        11  -D2
H   10  R2         11  A2        12  -D2
H   10  R2         11  A2        12  D2
H   11  R2         12  A2        13  D2
H   11  R2         12  A2        13  -D2
H   12  R2         11  A2        10  D2
H   12  R2         11  A2        10  -D2
H   13  R2         12  A2        11  D1
H   13  R2         12  A2        11  -D2
H   13  R2         12  A2        11  D2

R1 1.52996824
B1 109.47388108
D1 180.00000000
R2 1.09003735
A2 109.47134891
D2 59.99856213

But both your version(without the blank lines) and the one posted by me are running without problems for me.

Do same bond lengths by one variable B1 for B2,4,6,7,9 mean that you force it to have symmetry?

@pH13-YetanotherPhilipp you mean that I force the length of all those bonds to be the same in the optimized structure? That's not the case.

This is interesting! I always thought that If you are using the same variable, then only that variable gets optimised, but I guess you'll have to force g09 to not use redundant coordinates to achieve that. Just a side note: the supplied values in the z-matrix would be treated as constants (as they are read from the constants block).

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


text:
Can any compound be a Lewis acid but not be a Brønsted acid?
Is it possible for a compound to act as a Lewis acid but NOT act as a Brønsted acid? When would this happen? Could you please give a few examples where/why this happens ?

Certainly.  There are many compounds that can accept a pair of electrons from a Lewis base, yet don't have any Brønsted-Lowry acid (proton) moiety. The two are really rather different things, although there are of course many compounds that are both Lewis and Brønsted-Lowry acids ($\ce{Al(OH)3}$ for example).
Some examples of Lewis acids that are not also Brønsted-Lowry include:
$\ce{BF3}$
$\ce{(CH3)3B}$
$\ce{AlCl3}$
Metal cations like $\ce{Li+}$, $\ce{Na+}$, $\ce{Mg^2+}$, etc.

Water isn't Lewis acid - it would need to make an adduct where it would accept electron pair. Normal action as proton donor is an exchange of base in adduct. It seems it's sometimes considered as you are doing it but I think it shouldn't be.

Although I've always seen water to be considered as a Lewis acid and considered it as such, technically, I think you're probably right. Even though water is in equilibrium with the strong lewis acid $\ce{H+}$, that still isn't the same as $\ce{H2O}$. I figured my answer would be better without a controversial (or possibly wrong) example and changed it to a more certain example. Thx for the input.

Yeah, if there's only one other guy connected with post it pings him. If someone isn't connected even @ doesn't work.

@ airhuff What about HCl?

@Nightshade , HCl is a Brønsted–Lowry acid, as it readily donates it's proton.

@airhuff Thanks. Sorry for these silly questions. I actually got confused because of some questions I saw somewhere. Thanks for clarifying the concept.

@Nightshade , Absolutely no problem :)

In addition to above nice answers, there is a trick that can be used to confuse students: A compound for example  $\ce{Al2O3}$;
It can acts as a Lewis acid and Brønsted base but not as Brønsted acid
Consider these reactions:
Recall the Lewis Theory of acids and bases which states that an acid is an electron pair acceptor and a base is an electron pair donor.
Applying the theory to  $\ce{Al2O3}$:
$$\ce{Al3+(aq) + 6 H2O(l) <=> Al(H2O)6^3+(aq)~~~~~~~(1)}$$
Thus, in (1) the  $\ce{Al(H2O)6^3+}$  ion is formed when an  $\ce{Al^3+}$ ion acting as a Lewis acid picks up six pairs of electrons from neighbouring water molecules acting as Lewis bases to give an acid-base complex, or complex ion
the Brønsted-Lowry theory is also observed:
$$\ce{Al2O3 + 6H+ -> 2Al^3+ + H2O~~~~~~~(2)    a typical Brønsted-Lowry base    }$$
The compound accepts the proton
This compound somewhat behaves as both a Lewis acid and Bonsted-Lowry base but not a Brønsted-Lowry acid.
Clearly, the compound cannot be considered to be Brønsted-Lowry acid since the compound in question has no protons to donate.

There is also boric acid which is electron deficient but instead of donating a proton it accepts hydroxyl ions.


text:
What would the name of this organic compound be?

I am a little lost in naming this compound.
I think it would be something like 2 bromo 4 methylene pentene??
Is that close?
It is a five carbon chain, so pentane, with a Br on carbon two, a methylene substituent on carbon 4, and a alkene on carbon 1.

2-Bromo-4-methylpenta-1,4-diene

Perhaps you could [edit] this to show how it was arrived at? Whilst it may be the correct answer, its not really helping the OP learn anything, and certain not of help to anyone else looking to follow and apply this to similar molecules.

It is reasonably self-explanatory: keep the double bonds in the chain, alphabetical order in the substituents [switch Br and Me is still understandable] Both double bonds located.


text:
Does solubility of acetic acid depend on the pH?
Does the solubility of acetic acid increase or decrease in water if I adjust the pH?
Conversely, a water-insoluble molecule (like retinol), can its solubility be increased by adjustment of the pH?
I would say no (in both cases) because the protonating or deprotonating these molecules does not change their overall hydrophilic or hydrophobic character.

Acetic acid is miscible with water, so it's solubility definitely can't be increased ;) It's solubility simply won't be affected by pH. If retinol could be protinated, that would increase it's solubility, but it can't be protinated or deprotinated to any significant degree so it's going to be insoluble in water regardless of pH.

If you increase the pH some acetate will precipitate eventually. ;)

You are correct in both cases, though your reasoning isn't quite on target.
Regardless of the pH, retinol will not be protonated or deprotonated to any significant degree (which would greatly affect it's solubility if it were). It's just a nonpolar molecule that will stay that way regardless of pH.
Acetic acid also would not be protonated to any significant degree at low pH, but that's a mute point as it's fully miscible in water anyway.
The one situation in which you would see deprotonation of the compound is that of acetic acid at high pH. For the most part, both acetic acid and the acetate ion produced by the basic conditions should remain highly soluble. It is possible however, that if you added enough strong base, say NaOH, to a concentrated solution of acetic acid, you could eventually precipitate sodium acetate.
Summary, TL;DR:
You are correct that the answer to both questions is no. Acetic acid is miscible with water (so it's solubility definitely can't be increased) and it's solubility simply won't be affected by pH. If retinol could be protonated or deprotonated, that would increase it's solubility, but it can't be protonated or deprotonated to any significant degree so it is insoluble in water, regardless of pH.

How about an emulsifying agent such as ascorbylpalmitate?

@TMOTTM , sorry to have neglected your comment. I'm not sure I understand what you are asking with respect to using an emulsifying agent with two fully miscible liquids. If you are referring to affecting the solubility of retinol, I'm don't think that would have much effect in this case, but that's really just a gut feeling.

I ment, does the pH affect the solubility of emulsifying agents?

For the two specific substances, read the other answers and comments, they answer that part.

I would say no (in both cases) because the protonating or deprotonating these molecules does not change their overall hydrophilic or hydrophobic character.

When you protonate or deprotonate a neutral molecule, you get an ion. This will allow ion-dipole interactions with water and change the solute:solvent interactions. In liquid-liquid extraction, you get very different results at different pH values of the aqueous solution. So in general, solubility in water does change a lot with protonation state.
As an aside, retinol could be protonated or deprotonated in solvents other than water, and then you would expect a change in solubility. According to this source, you would have to go outside of the -2.2 to 16.44 pH range to achieve that (and these numbers are theoretical predictions).


text:
What is behind of the attraction between a burned matchstick with a magnet?
I do not know how to understand the reasons of why a burned matchstick is attracted by a magnet (most probably) of Neodymium.
Here are some screenshots of a video in which this natural phenomenon is presented:

(In the video, see at 1 min 25 s):

Source
Source

Don't think anything will be reduced on the match tip except the potassium chlorate. The heat and access to air will probably promote oxidation, not reduction of a species as noble as iron. If stackexchange allowed guesses, I would answer with something involving the antimon oxide residue and thermal magnetisation. +1 to the question, totally derailed me from work.

Are you sure the video is not a fake? Camera trick? Modified matches containing iron?

The only thing that I can identify is that the magnet is of Neodymium

@StianYttervik Generally increased temperature should lead to less magnetism...

Its a mini thermite reaction iron III oxide used for pigment is converted to iron

Some matchstick heads contain iron(III) oxide as a colorant.

The yellowish color of the burning match indicates that it has low oxygen, i.e. a reducing flame. It reduces the iron oxide to iron which is attracted by the magnet.
The reduction reactions that occur are probably quite complex. Below are two simplified possible reaction equations. $\ce{(CH2)_n}$ in equation (2) stands for paraffin from the splint's impregnation.
$$\ce{Fe2O3 + 3 CO -> 2 Fe + 3 CO2} \tag{1}$$
$$\ce{Fe2O3 + \frac{1}{n} (CH2)_n -> 2 Fe + CO2 + H2O} \tag{2}$$

When you burn things you are oxidizing them not reducing them, right?  Of course it's more complex than that.  What we need to see in a full answer is how the flame has reducing and oxidizing regions, and where athend how the iron oxide gets reduced to get that iron which is attracted to the magnet.

Initially I was somewhat uncomfortable, the reason is that the links are hiding additional information.

Can we get a summary of the key information in the links?  For instance, "as seen here [link], a typical flame has both a fuel-rich reducing flame and an air-rich oxidizing flame, therefore the reducing flame will [insert its reaction with the iron (III) oxide]."  That shows the logic up front without making the user chase down links.  Common courtesy, IMHO.

The reduction of iron oxide to iron involves many reactions, especially in the case with matchsticks where wood, paraffin, and different chemicals are involved. Added to simplified equations and hope they give hints what's going on.


text:
Is cyanidin aromatic?
The structure of cyanidin is as follows:

There are eight double bonds in the entire compound, and hence $8 \times 2 = 16$ π-electrons, which does not satisfy the $4n + 2$ rule. However, the molecule is usually depicted as aromatic in literature. MarvinSketch (a software) also reports the entire molecule as aromatic.

Is this molecule aromatic? Why?
Are all anthocyaninidins aromatic?

I guess you have to count the electrons for the upper benzene ring and lower fused rings seperately. Then the lower one has 10 electrons (aromatic) and upper one has 6 (again aromatic)

Probably that's true. Upper ring is aromatic and lower ring is also aromatic if counted separately.

No doubt it is aromatic.
The $4n+2$ rule is intended for a single cycle of π-electrons, which may possibly include multiple rings but can't have any interior atoms or pendant π- bonds (which would not fit in a cycle). Here, as Kartik indicates in a comment, there are evidently two such cycles and you have to consider them separately. One cycle has six π-electrons and the other has ten. You have a molecule (or more accurately, a molecular ion) with two aromatic cycles.

Thanks for the reply. What do u mean by interior atoms?

Look at pyrene.  The presence of two conjugated atoms surrounded by the others means you can't fit them all into a single cycle.  Thus the $4n+2$ rule is not reliable.  And true to form it doesn't work.

Hückel's rules can only be applied to the compounds they were derived from: monocyclic, planar hydrocarbons. For most aromatic compounds they are a lot less reliable and only a very rough approximation. https://goldbook.iupac.org/html/H/H02867.html


text:
When conducting metal/metal ion displacement reactions in a high school laboratory, why does the precipitate metal commonly look black in colour?
For example, take the redox displacement reaction of a strip of magnesium metal (silver color) and a solution of copper nitrate (blue color). The reaction is obviously able to progress due to the strength of copper as an oxidising agent; it will be reduced out of solution to form a layer of copper metal on the magnesium strip.
However, the Chemistry Data Booklet which is currently used in our state (and, I hope, is up to date) states that the precipitate, in this case, copper, should be salmon pink in colour (come on, who knows what salmon pink looks like?). This was certainly not the case; my observations for this reaction state: "Silver metal dissolves into light blue solution. Black precipitate formed."
I apologise for the lack of pictures of the experimental setup, but I was wondering if there is an explanation for the mismatch of colours. I have noticed this black precipitate in other reactions, notably zinc metal in a copper nitrate solution, and zinc metal in a lead nitrate solution. This is the data booklet currently authorised for use in our assessments.


text:
Does adding water to a reaction mixture shift equilibrium?
Just like decreasing volume in a gaseous reversible reaction shifts the position of equilibrium towards the side that produces more moles (increasing pressure), shouldn't adding water shift equilibrium to side that produces more moles in a liquid equilibrium mixture (increasing concentration)?
On research, I've seen an explanation that it does not because the concentration of both reactants and products decrease evenly, but I don't understand this. The side with more moles should have a greater decrease in concentration because the concentration is proportional to the number of moles. Also, by the same argument, decreasing volume in a gaseous reaction should not shift equilibrium.
My guess is that adding water does affect equilibrium for a reaction with different number of moles on each side. Because the equilibrium constant is affected. Is this right?

Apparently it is right (I asked at work, to be sure). If you consider a reversible reaction $\ce{A + B = C}$ as the effect of two opposite reactions with rates $\ce{r_f = k_f \cdot [A] \cdot [B]}$ and  $\ce{r_b = k_b \cdot [C]}$, equilibrium is reached when the forward and backward rates are the same: $\ce{r_f = r_b}$, which also gives you the definition of the equilibrium constant  $\ce{K=\frac{k_f}{ k_b}}$. If you suddenly double the volume, the two rates change, so all concentrations have to readapt, for the equilibrium to be re-established (as both rate constants stay the same).

The equilibrium constant is not affected. But the number of moles of products and educts may change to hold the equilibrium when the amount of solvent changes.

Yes, the equilibrium does indeed readjust after a change in concentration. This can be seen by careful examination of the formula of the equilibrium constant $K$. Consider an arbitrary reaction that produces two products from a single reactant and is in equilibrium:
$$\begin{align}\ce{A &<=> B + C}\tag{1}\\[0.5em]
K &= \frac{[\ce{B}][\ce{C}]}{[\ce{A}]}\tag{2}\end{align}$$
Let’s assume initial concentrations of $\pu{2M}$ for each species; that gives us $K = 2$ in dimensionless units (you can also use M as a unit here). Remember that $K$ is a constant so we can examine whether equilibrium has been reached by examining $K$.
Now let’s double the volume of solvent. The new concentration of each reactant is $\pu{1M}$. Therefore:
$$K' = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]} = \frac{1\times 1}{1} = 1\tag{2'}$$
Since $1\ne2$, we are no longer at equilibrium. Thus the concentrations must readjust. How? Obviously, the value of the fraction is too low. To increase its value, we can increase the numerator or decrease the denominator — luckily for us, both of these mean that we must perform the reaction in a forward direction to get think right again. Since $[\ce{B}]=[\ce{C}]$, we can actually calculate the ratio of concentrations and from $2[\ce{A}]+[\ce{B}]+[\ce{C}]=\text{const.}=4$, we can calculate exact values.
$$\begin{align}&&K &= \frac{[\ce{B}]^2}{[\ce{A}]}\\[0.5em]
\Longrightarrow&&[\ce{A}] &= \frac{[\ce{B}]^2}2\\[1.3em]
&&2\times\frac{[\ce{B}]^2}{2} + [\ce{B}] + [\ce{B}]&=4\\
&&[\ce{B}]^2 + 2[\ce{B}]-4 &= 0\\
\Longrightarrow&&[\ce{B}]_{1/2} &= \frac{-2\pm\sqrt{2^2+4\times4}}{2}\\[0.5em]
&&[\ce{B}]_{1/2} &= -1\pm\sqrt{5}\\[0.5em]
&&[\ce{B}]_{1/2} &\approx 1.24\end{align}$$
Thus, the product side has increased.

A more real-world example is maybe the dimerisation of hydrogenchromate $\ce{HCrO4-}$ according to $(3)$:
$$\ce{2 HCrO4- <=> Cr2O7^2- + H2O}\tag{3}$$
This is also pH-dependent, since hydrogenchromate is a weak acid and can deprotonate to give chromate $\ce{CrO4^2-}$. However, it is fairly easy to try this experiment yourself: take an orange coloured dichromate solution and add water. A colour change to yellow should be observed indicating that the reaction has equilibrated to the reactants’ side.

So just to confirm, for the final example, this occurs because the reactant side produces a greater amount of moles, not because the amount of water has increased thus increasing the reaction rate? I take that it is the former since you're not actually increasing the concentration of water however I just wanted to make sure.

@bio The real underlying reason for the hydrogenchromate/dichromate example is a change in pH which corresponds to more or less water being present in the system. In fact, it is often written with $\ce{CrO4^2- + H+}$ instead of $\ce{HCrO4-}$.

Dilution of an aqueous ionic equilibrium favors the reaction that produces more ions. This changes the position of equilibrium. When in doubt, consider the expression of K and know that the equilibrium shifts to satisfy the constant K at a given temperature.


text:
Enthalpy Change
In calculating the change of enthalpy in converting Fulminic to Isofulminic acid, would one simply take the sum of the negative enthalpies of the bonds of Isofulminic acid minus the sum of the negative enthalpies of the bonds in the Fulminic acid, following the definition of enthalpy change as heat of product - heat of reactants?

I am not too sure what your negative enthalpies refer to. In general, when defining enthalpy change in terms of bond enthalpies, enthalpy of reaction is equal to the bond energies of the bonds broken, after substrating the bond energies of the bonds made. This is defined as such as bond making is an exothermic process and bond breaking is an endothermic process. Note that bond energy is always a positive vale (i.e. it is endothermic) as it is defined to be the energy needed as input to break a bond.
So in your case, to obtain the enthalpy change of reaction: Substract the sum of the bond energies of the bonds in isofulmunic acid from the sum of bond energies of the bonds in fulmunic acid. This would be the standard way of going about it.
Hope this helps.


text:
What are the products in this reaction?
I came across this question

I thought that the answer should be option (a)but the answer is given option (d)
I thought that other products might be a result of the following mechanism
But this cannot be the case as a positive charge still remains on the carbon atom. Can you please help me figure out where I am going wrong?
UPDATE:
Or is this a possible mechanism
But I am still unclear about how the double bond will reappear between 3 position and the carbon that carried Chlorine

FWIW, in my browser both links say "The image cannot be displayed because it contains errors". That might just be me and my somewhat odd browser, but it's good if you can insert images into your question rather than links. Anyway, you got an answer, so I guess someone can see them ;)

https://en.wikipedia.org/wiki/Arene_substitution_pattern

@airhuff I am rather new to this site and so I don't know how to add an image in my question. I admit adding links of images is a cumbersome process.

List item

All the reactions eliminate chloride (a) by the obvious direct attack, the other products b and c arise from methoxide attack on the 3 and 5 positions of the furan and chloride loss via an exomethylene intermediate. The system is somewhat similar to a polyallyl choride where attack can be direct or at the 3 or 5 position.

I'm sorry I don't know what an exomethylene intermediate is.

Double band from carbon at 2 position (next to oxygen) to carbon that carried chlorine

But how will chlorine leave? Will you please add the mechanism of its loss in your answer

I have no access to a drawing package. Chlorine leaves as Cl- just as in the direct attack.  Consider product c to be result of Michael attack at the 3 position and move the bonds around to kick out Cl-

from what i understood,I've added a mechanism in my question, I am very sure that it's not correct. will you please go through it once ?

@DaenysTargaryen  The first two steps are probably more concerted than that, also, $\ce{CH_3O^-}$ will act as base, formally yielding NaCl.

The mechanism you added is fundamentally correct. As TAR86 says methoxide will remove the proton leading to re-aromatisation. The same mechanism applies to attack at the 5 position.

Ok,but when the double bond between carbon at position 2 and the CH2 break,there will be a positive charge left on it. How do I get a -CH3 group there

The exo carbon will pick up a proton from the MeOH. You could visualise it as a 1,3 proton shift


text:
Is there a way to tell which one is the racemic or meso compound with 1H-NMR if all you had was the structure and spectra?
I know that diastereomers will have different chemical shifts but can you predict which one is which via up/down field shifts in the spectra?
For instance, is it like alkenes vicinal H’s when cis and trans are up and down field respectively?
I ran a 1H-NMR of hydrobenzoin and when comparing it to the literature I noticed that the racemic mixture was further upfield than the meso.
I made the Newman projections below and assumed the most energetically favorable positions to show the cis and trans H’s in the diastereomers.
Is this the right way of thinking about this?

You might wish to take a look at the Caplus (not sure if I spelled that correctly) relationship

@tillyboy it's Karplus

Hydrobenzoin has 3 stereoisomers: meso, R,R and S,S. You have not included the S,S. Also, your Newman projections might make sense based on steric grounds, but largely ignore electronic factors such as pi-stacking and hydrogen bonding. A very quick run through the Avogadro Bable geometry optimisation has these molecules optimising with the phenyl rings somewhat gauche.
While there are many methods in NMR that can assist in determining relative and absolute stereochemistry, there is no method that offers universal application. Generally it is not possible to identify stereoisomers from first principles in non-rigid skeletons. This is because conformers are rarely locked into a single position. Properties such as spin coupling, and nOe are useful, and the method by Murata is considered groundbreaking in this area. It relies on measurable  3J(HH) and 2,3J(CH).
Derivatisation, database comparison, measurement of residual dipolar couplings in aligned media (liquid crystals), and computational methods. Determination of absolute configuration usually involves complexation with a enantiomerically pure reagent.
And I'm not confident that your method for assigning cis/trans geometry based on chemical shifts is reliable.


text:
In which process $\Delta G$ is zero
We have to find in which of the process $\Delta G$ is zero .

I know it zero when there is equilibrium . and according to me every reaction is in equilibrium except the last reaction .
As in them physical conditions are same .
But in my book it is written that on reaction b,c,d,e will have $\Delta G=0$

You can't just look at whether the values of temperature and pressure are the same on either side. You also must consider the whether those states would coexist under those conditions. For example, water liquid and water vapor are not in equilibrium at $0 ^{\circ}\text{C}$ and $1\text{atm}$.

@Tyberius then how c and d options are in equilibrium

The info at the bottom says the vapor pressure of water at $27^{\circ} \text{C}$ is $0.04$. From the wikipedia page for vapor pressure, it is defined as "the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system"

Firstly, your statement "I know it zero when there is equilibrium" is not true. $\Delta \text{G}$ can take any value for a reaction, and the system can be at equilibrium if the concentrations of the reactants and products are at appropriate concentrations for that value. However, $\Delta \text{G} =0$ implies that the equilibrium constant for the reactions or phase change is 1.

2  The equilibrium constant

See the link it shows $\Delta G =0$ at equilibrium


text:
Is the 'central Tin atom' in crystalline tin oxide really offset like this?
Is the position of the central Tin atom as shown in the first figure ($\ce{SnO2}$) drawn offset towards the oxygen to the right for a reason? The oxygen also seems to reciprocate and move towards the tin. Usually the rutile structure is shown as symmetric with the metal the center. I'm showing a GIF alternating images of $\ce{SnO2}$ and $\ce{RuO2}$, with complete image source information following.
note: I'm interested to know about the locations of the atoms within a unit cell as determined by crystallographic information - not one's impression of the drawing. I'd like to know if the offset shown in the drawing is there to represent an actual asymmetry, or if it's just an error.

Figure 1. Tin Dioxide (Tin IV oxide), also found as Cassiterite, image from Rutile unit cell 3D balls.
Figure 2. Ruthenium(IV) oxide, image from Ruthenium(IV) oxide unit cell 3D vdW.
Figure 3. Rutile structure, image from Rutile structure
 Fig. 1
 Fig. 2
 Fig. 3

I can't find any information which suggests that the Sn atom at the centre is not at (0.5, 0.5, 0.5). The unit cell parameters are slightly different, and the oxygens are shifted by a very tiny bit (the oxygens are at (x,x,0), (1-x,1-x,0), (0.5+x, 0.5-x, 0.5), (0.5-x,0.5+x,0.5), and x = 0.305 for TiO2 and 0.307 for SnO2) but nothing on the tin atoms.

@orthocresol "nothing" means (0,0,0),  (0.5, 0.5, 0.5), or nothing as in no mention of the location of the tins? I'm asking this to personally learn more on the subject, and it seems that there should be some sort of symmetry argument against (0,0,0), (0.5, 0.6, 0.6) for a crude example - the unit cell would have a net direction and the material might show particular optical or dielectric behavior. Answer that included a non-paywalled way to see the numbers you are seeing would be greatly appreciated! Also there's still this question... https://chemistry.stackexchange.com/q/73254/16035

@orthocresol Also I realized that that the unique motion of the one oxygen atom on the right could not be explained with the simple expression in your comment, so I have a hunch you've got the answer.

No mention of the location of the tins, so I expect it to be the usual (0,0,0) + (0.5,0.5,0.5). I don't know much about it, I just flipped through the books that I had. It's in A.R. West "Solid State Chemistry and Applications"

The tin is not offset. Likely the images have been distorted.

@JonCuster thanks. I have a hunch I'm not going to get an answer to accept on this one, but I can move now at least. :)

From http://www.crystallography.net/cod/1000062.html one can see that Sn sits in the middle, without any displacement, and mimics the classic Rutile-type structure:

I checked original Wikimedia image, and according to metadata

CIF retrieved from The American Mineralogist Crystal Structure Database

is no longer available. I also failed to discover this source:

X-ray crystallographic data from:
  R. W. G. Wyckoff (1963) Second edition. Interscience Publishers, New York, New York. Crystal Structures 1, 239-444

Ideally, if you have *.cif file, it might be worth checking occupancies for the atoms you are interested in. Probably there is a 50-50 disordered Sn atom about center of unit cell, and you only observe a single site. As long as this is not the case, I would rely on the available COD data from 1971 instead.

Thanks very much for your post! Each figure in my question is numbered, and the captions contain the word "from" followed by the link to the original image. I did not draw any figures myself. Can you help me understand your comment about "50-50 disordered Sn atom"? Does this mean that the atom will in fact be more likely to be found on one side or the other rather than the center? Would this show up crystallographically as an elongated charge distribution for that atom?

I slightly edited my answer to clarify my point of view. By "50-50 disordered Sn atom" I suggested that there might be statistically disordered Sn atoms, most likely with 50% probability, either due to poor quality of crystals, or experimental technique, or just because of different phase. Anyway, this recent publication reveals that this is most likely not the case. I'm also not sure what you mean by "elongated charge distribution" (dipole?), in crystallography this entity is usually called "thermal ellipsoid" (also see ORTEP).

Oh, my ellipsoid was part of my trying to understand the 50-50. I thought that you were suggesting that in 50% of the unit cells within a physical crystal the center Sn would be on one side, and in 50% of them it would be on the other side of the center, and so this might show up as a non-spherical electron distribution when modeling diffraction data. But it looks like that's not what you meant. Since I didn't make these drawings, can you adjust the sentence "I'm not sure what data you used to draw this structure." to perhaps "...what data was originally used..."? Thanks!

I edited my answer a bit more, please let me know if you are not happy with the new one. Regarding disorder: you are right, "when modeling diffraction data" it really is sometimes looking as an elongated ellipsoid, and we embrace it as a mathematical model for representation of experimental data; whereas in reality it might be two (or more) separate atoms with a set of own coordinates each statistically distributed among all the unit cells, so that we see their average electron density.

Ha, I actually found your entry in COD: http://www.crystallography.net/cod/9009082.html, this is the exact data used by that person who made image for Wikipedia, but still there is a single Sn in the middle of unit cell. Maybe it has something to do with visualization software. Wiki says Accelrys DS Visualizer has been used, but I could never afford that one, so I have no idea what causes optical shift on that picture.

Your answer and follow-up are both fantastic, thank you for your help!

No prob at all, I'm very glad I could help.

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


text:
How to name this compound
2 bromo 3 methyl-butane or 3 bromo 2 methyl-butane. In general, if we have a symmetrical situation I know the functional groups should be ordered alphabetically and I also know that the smallest possible number on the functional group should be used. But what if each f-group is equidistant from the ends of the chain. Which one gets the smallest number?

https://chemistry.stackexchange.com/questions/68624/alphabetization-rule-in-case-of-consecutively-placed-substituents

ok, does this take priority over alphabetical order then?

Sorry, I'm not really sure what you mean. Alphanumeric order dictates that the name is of the form x-bromo-y-methylbutane. That's a given, and you didn't seem to have an issue with that. Your question is whether (x,y) should be (2,3) or (3,2). As described in the linked question, rule P-45.2.3 would lead to (2,3) being the preferred locant set. Quoting myself: "The correct way to do it is [...] first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants."

Also related: Order of citation of alkyl and halide prefixes in IUPAC names


text:
How to derive the Gibbs-Helmholtz equation?
Can someone please explain to me how to derive the Gibbs-Helmholtz relationship from $G = H - TS$?

https://en.wikipedia.org/wiki/Gibbs%E2%80%93Helmholtz_equation#Derivation

@orthocresol can you please outline the chain rule step in the derivation? This is the part I don't understand:

It's more of the product rule rather than chain rule. It may be more familiar to you to look at it this way: you should know that $\mathrm{d}(fg)/\mathrm{d}x = f(\mathrm{d}g/\mathrm{d}x) + g(\mathrm{d}f/\mathrm{d}x)$. Now let $x = T$, $f = G$, $g = 1/T$.

You need to know that
$$\require{begingroup} \begingroup \newcommand{\d}[0]{\mathrm{d}} \d G = V\,\d p - S\,\d T$$
from which you can determine that
$$\newcommand{\pdiff}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} \pdiff{G}{p}{T} = V \qquad \pdiff{G}{T}{p} = -S$$
Therefore (by the quotient rule)
$$\begin{align}\pdiff{}{T}{p}\left(\frac{G}{T}\right) &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p}
{T^2} \\[8pt]
&= \frac{T(-S) - G(1)}{T^2} \\[8pt]
&= \frac{-TS-G}{T^2} \\[8pt]
&= -\frac{H}{T^2}
\end{align}$$
as desired (since $G = H - TS$).

$$dG=dH-TdS-SdT=dU+VdP+PdV-TdS-SdT$$
But, $$dU=TdS-PdV$$
Adding the above two equations together, we get $$dG=VdP-SdT$$
At constant pressure, we have $$dG=-SdT$$But, from the definition of G, $$-S=\frac{G-H}{T}$$Substituting for -S yields:
$$\frac{dG}{dT}=\frac{G-H}{T}\tag{constant P}$$
The rest is strightforward math.


text:
Why does adding salt to water reduce the pH?
I recently heard someone claim that adding table salt to vinegar caused HCl to form which helped them clean coins - clearly not the case - however I thought I'd disprove it by doing an experiment.
So I added NaCl to ethanoic acid and measured a 0.5 reduction (3 to 2.5) in pH using a pH probe. This seemed a bit weird so I repeated using tap water and the same happened (7 to 6.5).
The pH of water when NaCl is added should not change from 7!
The probe measures by detecting a change in voltage through the solution so it could be that, but you'd think that salts shouldn't have such an effect or be adjusted for. Can anyone repeat please?
What's going on?

If you add NaCl to any neutral or acidic solution, the pH will decrease by a factor of at least 0.5. This is due to the fact that the pH is not the log of the concentration calculated with respect to the total volume, but the concentration should be calculated with respect to the volume of free water. In concentrated NaCl solutions, nearly all water molecules are adsorbed around Na+ and Cl- ions. The apparent concentration of H3O+ increases a lot for lack of water.

I know salt lowers the measured pH because when I sprinkle a tiny bit of salt into distilled water I can watch my pH meter drop to a reading of 6.4 so it does affect the pH meter reading whether or not it actually changes the true pH.

First: the experiment is lacking in details, Tapwater exactly 7!! The data presented lacks precision. If both readings changed 1/2 a unit that is 30mv, That is a large difference in a[H+]in the two solutions. That makes the reference electrode junction potential suspect. Another possibility is the actual change in Kw and Keq of acetic acid. First ensure the equipment is measuring appropriately.

I’m voting to close this question because it lacks information about the type of electrode, the calibration procedure and the definition of pH at high ionic strength, see e.g. https://core.ac.uk/download/pdf/38093394.pdf

This is probably an effect in which sodium ions from the salt are being weakly detected by the glass electrode, giving the appearance of a lower pH. Try add the same concentration of salt to samples of your reference buffers and recalibrate your pH meter to those before measuring the pH of the salt water. I would expect you would see pH 7. Also, keep in mind that the pH of unbuffered water is very sensitive to low concentrations of certain impurities -- from the air or from the added salt. For example, carbon dioxide in the air will lower the pH over time.

If this were true, a basic solution would also shift towards lower pH after adding salt. Arguments involving ionic strength would predict the opposite. These predictions are easy to test experimentally - add some sodium chloride to a solution of a weak base and compare to adding water, making sure the final volume is the same. Then, measure pH values and compare.

I think that the response of the glass electrode to sodium ions is more important than the activity coefficient changes.

Modern electrodes have a minimal Na+ or other monovalent ion sensitivity.

Please be aware that adding salt changes the ionic strength of your solution. Therefore, the activity $a_\mathrm H$ of $\ce{H+}$ is changed. This changes $\mathrm{pH}=-\log_{10}(a_\mathrm H)$. In a 1 molar $\ce{NaCl}$ solution ($I\approx 1\ \mathrm{mol/l}$) the Debye-Hueckel (DH) approximation gives a shift of the activity by around 0.5 pH units (compared to salt free water):
The activity coefficient $f$ is given by DH
$\log_{10}(f_i) \approx - A \sqrt{I/(1\ \mathrm{mol/l})}$, with ionic strength $I$ and $A=0.509$
This gives $\mathrm{pH}=-\log_{10}(c_\mathrm H/(1\ \mathrm{mol/l})) -\log_{10}(f_\mathrm H)=-\log_{10}(c_\mathrm H/(1\ \mathrm{mol/l}))+A \sqrt{I/(1\ \mathrm{mol/l})} $.
This equation would predict an increase of pH by 0.509 pH units for a 1 molar salt solution.
You might want to use a better theory than Debye-Hueckel. In the above form it is only applicable for ionic strengths below $\sim 0.001\ \mathrm{mol/l}$, so the calculation above is very crude and might explain the failure.

Alternative explainations are given by https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3638298/

+1 for mentioning activity coefficients and ionic strength. However, if the activity coefficients of hydrogen ions and acetate ions are different from one, the position of equilibrium has to be recalculated (more acetic acid dissociates). This reference gives such a calculation for a different weak acid/base system at ionic strength 0.1 M. In that example, the pH of the solution changes marginally towards lower pH.


text:
What is "mineral spirits"?
I'm working in a project, and for a specific thing, I need to work on a possible recovery of $\ce{HCl}$ in aqueous phase. I found a patent on the net in which they use an amine to make a complex with $\ce{HCl}$.
They say in the summary:

where a complex between the
tertiary amine and the methyl chloride is formed, a hydrocarbon
solvent, such as mineral spirits, is added to the phase containing the
complex and the complex is then thermally dissociated.

I have never heard about that until yet.
I did some research on Sigma-Aldrich and I found that but I have not too much information, if you have ever heard about that can you please tell me a bit more about? Is that a molecule or a mixture?

It is a solvent mixture, commonly used as a solvent/thinner of oil-based paints. Since those disappeared from use decades ago, it is not surprising that you did not recognize the name. But, once upon a time, one could buy gallons of mineral spirits at the hardware store.

Mineral spirits goes by many names, including "white spirit" and "paint thinner", and is basically just a hydrocarbon solvent mixture. According to this Wikipedia article:

White spirit is a mixture of aliphatic and alicyclic C7 to C12 hydrocarbons with a maximum content of 25% of C7 to C12 aromatic hydrocarbons. A typical composition for mineral spirits is > 65% C10 or higher hydrocarbons,[6] aliphatic solvent hexane, and a maximum benzene content of 0.1% by volume, a kauri-butanol value of 29, an initial boiling point of 145 °C (293 °F) to 174 °C (345 °F), and a density of 0.79 g/ml.

There are also three different types based on whether or not it's undergone processing to remove sulfur, solvent extraction or hydrogenation. Then each type has a separate grade: low flash, regular and high flash grades. So with three different types each having three different grades, there are nine different sorts of mixtures that "officially" fall under the term white spirits, or mineral spirits.

minerals spirits are also called ligroin, naphta, or petroleum ether. Its a matter of favorites


text:
Lone pairs in MO theory?
I have learned that non-bonding MOs are lone pairs. However when considering the MO diagram for the $O_2$ molecule, I do not see any non bonding orbitals yet there are two  lone pairs on each oxygen atom. Where do these come from?

Short answer:
Non-bonding MOs are usually lone pairs, but a lone pair does not necessarily mean there is a non-bonding MO.
Long answer:
You are trying to match what you learned from Lewis structures to the orbital picture. In general, that does not work (there may be exceptions).
Let's look at the MO diagram:

It is tempting to assign the bonding $\sigma_{2p}$ to the first (sigma) bond in $\rm O_2$. The two $\pi_{2p}$ bonding orbitals would then results in 2 $\pi$ bonds, hence we get a O-O triple bond. The further two electrons in the anti-bonding $\pi_{2p}^*$ orbitals could be assigned to one oxygen each, thus we have a biradical. The remaining two electron lone pairs might be assigned to the two $\sigma_{2s}$ orbitals. This is actually one of the two valid Lewis structures for $\rm O_2$.
(The other possible Lewis structure has only a double bond and two lone pairs at each oxygen atom.)
The problem is that the lone pairs are ascribed to only one atom, but the MOs are delocalized over the molecule. Thus there is no direct relationship between them. A non-bonding orbital is usually just an atomic orbital, which you may thus ascribe to a lone pair. But that does not mean each lone pair leads to a non-bonding orbital.
It is however possible, to localize the molecular orbitals (by making appropriate linear combinations of MOs) in such a way you get atomic orbitals, which you might match again with the lone pairs. The thing here is, that there are many different sets of orbitals which are all valid descriptions of the bonding situations.
Also note, that MO diagrams are just a simplification as well.
Source of the MO Diagram

As you started to mention, linear-combination of atomic orbitals (LCAO) with each molecular orbital being required to obey the rules of the full group ($\text{D}_\infty$) for $\text{O}_2$ using only the atomic orbitals designated in the subscripts of the MO. A better choice of orbitals would be using the natural bonding orbitals (NBO) derived from an accurate wavefunction. In either case, some should appear to be more bonding, while others are less so.

Actually, I was not talking about LCAO, but about linear combinations of molecular orbitals, which in turn are usually (but not necessarily) linear combination of atomic orbitals. I am not so familiar with NBOs, but I think, simply put, they are a variant of the localized orbitals I was referring to.


text:
XPS analysis intrepretation
A lab presented me with an elemental composition (mainly metals in char) derived from an XPS analyzer and mentioned that binding energies suggested a list of compounds (various oxides, sulfides and silicates etc). Is there a way to show the composition in terms of these compounds instead of just the elements? Only familiar in bucket chemistry (aka chem eng), these analyzers are a bit of a black box to me even after reading up on them.
Also, is analyzing the surface of a batch of tiny particles <<1 micron somewhat tricky to get reliable results?

What do you mean by saying "in terms of these elements"? The xps analysis suggests these substances, there are other ways of determining the elemental composition of a particle mainly AAS (atomic absorption spectroscopy).

Having done work similar, the surface is often not representative of the bulk.

The elements list shows for example: 3%Ni, 1%Fe, 2%Zn and so on and the accompanying annotation mentions compounds identified by binding energies are Fe2O3, FeS2, ZnO, ZnSO4 etc without showing the mass composition of these compounds. Is there any way of getting to compound percentages rather than element percentage?

If the lab didn't break down the elemental fractions into compounds, then you probably can't readily estimate that either. If they also supplied processed, well-labelled spectra, then you may be able to get the proportion of FeS2 vs Fe2O3, for example, and work from there. But note that chemical state assessment in XPS is not 100% reliable - not all states are clearly resolved and uniquely identifiable. So I'd treat that compound list as a guide, not an assay.

This is a bit late, but I'll add an answer anyway. XPS is already a method that can be used for the speciation, so you should be able to differentiate between the different the different oxides of the same metal or metalloid. This is because the work function to eject an electron from a shell of the metal centre is dependent on its chemical environment.


text:
How can I measure amount of ozone byproduct coming out of hydroxyl generator
A Hydroxyl generator releases hydroxyl radicals into the air to neutralize odors and kill mold.
Most hydroxyl generators also release some amount of ozone as a byproduct. How much ozone is released varies from one hydroxyl generator to the next.
Hydroxyl is supposed to be safe to use in occupied premises. Large amounts of ozone are not.
How can I measure the amount of ozone released by my hydroxyl generator?

Ozone is very reactive, so unless you have an industrial scale hydroxyl generator, the ozone will readily react with organic matter in the air and surrounding surfaces before it builds up to dangerous levels. There are commercial portable instruments that are sold to monitor ozone concentration, they will be in the \$500 to $1500 price range for the instrument. You can also look at this simple experimental apparatus described in the Journal of Chemical Education for a cheaper alternative.

Thank you, J. I am not looking for a lot of precision, so the cheap, high-school experiment set up should do it for me.


text:
With which 2 of these 5 chemicals would Chlorine not react? N, NaOH, O2, NaBr, Fe
A question in a Russian EGE test asks to choose 2 chemicals out of 5 with which Chlorine does not react:

Nitrogen - I checked and it reacts
$\ce{KOH}$ -- reacts
$\ce{O2}$ -- reacts
$\ce{NaBr}$ -- reacts
$\ce{Fe}$ -- reacts

I'm at a loss. Which two should I pick then? Maybe Fe and Nitrogen, because they would be the least likely to react?
Here's the question:

Checking unknown reactions on obscure sites might lead you into the dark. Nitrogen and oxygen do not react with chlorine.

I would suggest nitrogen and oxygen.

@IvanNeretin  - yes it can, but I could not think of a way to investigate this by other means. There is no authoritative source on "non-reactivity of elements". I thought of oxygen, but then I thought that maybe it does react a bit.

I now can see why Oxygen does not react (they are both oxidants). But why not Nitrogen..

Chlorine and oxygen react as $\ce{Cl + O2 =ClO + O; 2ClO = ClO2 + Cl}$ as well as $\ce{2Cl + N2 = Cl2 + N2}$ and are described in a famous paper Porter & Wright,  Faraday Society Discussions, 1953, No.14, p23 'Studies of the Free Radical Reactivity by the methods of Flash Photolysis: the photochemical reaction between chlorine and oxygen'.

@porphyrin You can react everything with everything if you use appropriate conditions. This is dumb test question and if you simply mix Cl2 with air there is no significant reaction. Cowper I have a bad feeling... Tests aren't meant to be solved via digging through internet, but by smart thinking and clever guess

@Mithoron - I only started refreshing my chem knowledge on April 21, so I ran out of smart thinking and clever guess pretty fast with this question.

Minor point here.  When I read the Russian words for (2) they look like "gidroxid kaliya". How is this sodium hydroxide not potassium (kalium) hydroxide?

@OscarLanzi -ooops my mistake! Thanks for spotting it.

Chlorine reacts with all of the chemicals except nitrogen and oxygen. We expect elemental nitrogen/oxygen and chlorine to directly combine to form nitrogen trichloride/chlorine oxides but chlorine doesn't have enough energy to break the triple bond of nitrogen and overcome the activation energy of the reaction (also see collision theory). But there has been attempt to get singlet/atomic oxygen react with chlorine. See here and here.

All nitrogen chlorides and chlorine oxides are endothermic. That doesn't preclude a reaction but, especially given there's no huge entropic driver in any case I can think of,  it does make it unlikely. The reference given for nitrogen is very strange, and the one for oxygen while correct is poorly worded, or at least titled.

Its most likely $\ce{Fe}$ and $\ce{N}$ like you typed. $\ce{N}$ is most accurate because nitrogen is nonreactive so a spark will be needed to react it with chlorine. $\ce{Fe}$ will react with Chlorine but its reactivity is rather low compared to other transition metals such as zinc.

but iron does react with chlorine. It's not about low reactivity but whether it can react at all or not. The answer should be binary(Y/N).


text:
Can molten Strontium and Barium Hydroxides be electrolysed to give the metals?
Can strontium and barium hydroxides be electrolysed to give the metals? I've noticed they have relatively low melting points. What challenges would this present?

They contain hydrogen and the hydrogen is reduced, maybe?

Expanding on @Oscar Lanzi comment: first, all the hydrogen (and half the oxygen) would be electrolyzed before then alkaline earth metals.  At that point, if kept liquid, then the $\ce{BaO}$ and $\ce{SrO}$ remaining would be split. That would not be easy, as the melting point of $\ce{BaO}$ is 2,196 K and that of $\ce{SrO}$ is 2,804K (though a eutectic mix of the oxides would be a bit lower).
[It's more efficient to dissolve the oxides in a halide salt. The electrolysis of $\ce{CaO}$ in $\ce{CaCl2}$ has been explored here and here.


text:
How can I define alkenes using graph theory?
I want to represent alkenes in a mathematical fashion but I want to make sure of what I'm writing. Here is what I tried by defining "alkenic" graphs:
Alkenic graphs are connected non-looped multigraphs where:

nodes are of degree less than 5 (octet rule);
there is at least a double edge and no triple or quadruple ones (edges like in bonds);
cyclic submultigraphs with conjugated single-double edges given in any conjugation order are to be considered isomorphic (anti aromatic or aromatic isomorphism).
For example, the graphs representing "1,4-dimethylcyclobuta-1,3-diene" and "1,2-dimethylcyclobuta-1,3-diene" are isomorphic.

In this case I'm not taking stereoisomers into account.

Assistance requested in Math.SE Chat.

Perhaps you could just represent them as simple weighted graphs? Edges of weight 1 represent single bonds, while edges of weight 2 represent double-bonds.
*Edit: For condition #3 simply pick a canonical ordering for the vertices/atoms. See
https://en.m.wikipedia.org/wiki/Graph_canonization
http://depth-first.com/articles/2006/08/12/inchi-canonicalization-algorithm/
for examples

Ok, but what about condition 3?

Sure, I've updated my answer


text:
Oxidation of tertiary alcohols
Why is it that tertiary alcohols cannot be oxidized? I was told that it's because there is no hydrogen to remove, but I don't understand why another OH could not be formed where there is a C-H bond.

What are you oxidizing to? You can just burn the alcohol. That will attempt to replace all of the C-H bonds with C-O bonds...

The meaning of the answer you've gotten, "it is because there is no hydrogen to remove", is with respect to oxidizing the carbon of the tertiary alcohol itself. You could oxidize the rest of the molecule into any number of things, but there is no way for the tertiary carbon to be further oxidized than the alcohol.

I read in one of the books that even tertiary alcohols can be oxidised under drastic conditions. Can you please explain if it can be done and with what kind of oxidants.


text:
Bohr Model and Exciting Electrons
So I've got a relatively simple question; I'm not currently taking any courses in chemistry but while I was helping someone review a quiz I came across a question we couldn't figure out. According to the Bohr Model, is the energy required to excite an electron from the n=3 to n=4 orbit generally greater or less than the energy required to go from n=4 to n=5? Also, why is this so? I can't seem to find a clear answer anywhere online, as just about every result involves directly computing energy required to bump the electron up a few levels in very specific cases.

Asking why is the wrong question. That's the way the universe is. Once you accept quantization of energy levels, that's the result you get back.

The energy difference between the orbits gets smaller the higher "up" you go.
For the classical bohr model the energy of a given orbit $n$ can be approximated by:
$E_n=-\frac{Z^2R_E}{n^2}$
where $Z$ is the charge of the nucleus and $R_E$ is the Rydberg energy. Since there is a $n^2$ in the divider the difference $E_{n+1}-E_n$ gets smaller and smaller.
This is explained in more detail here.
As for why this is the case: here we have to look into quantum mechanics and I guess this is outside the scope here. But if you are interested look up explanations of a particle in a 1D box. The spacing of the energy levels there follows the same trend and you can find explanations there on why this is the case.


text:
The number of geometrical isomers of complex of type [Ma₃b₂c]

Find out the number of geometrical isomers of the complex $\ce{[Ma_3b_2c]}.$

I applied simple combinations to get the number of isomers as
$$\frac{6!}{3!\,2!} = 60.$$
However, my book states that only three geometrical isomers are possible. I could not see why simple combination is not giving the correct answer.

You have a permutation, not a combination.

First, let's establish that we're all talking about octahedral complexes to eliminate any possible confusion.
Fix $\ce{c}$. You can have either $\ce{b}$ trans or $\ce{a}$ trans.
If $\ce{b}$ is trans to $\ce{c}$, there's only one distinct way to arrange the remaining $\ce{b}$ and three $\ce{a}$'s. So that's one isomer.
If $\ce{a}$ is trans to $\ce{c}$, then the remaining two $\ce{a}$'s can be cis or trans to each other, i.e., all three $\ce{a}$'s can be facial or meridional, respectively. That's two more isomers. That's it.
I think your confusion comes about from symmetry because some of the combinations you calculated can be rotated into others.

Your problem comes from the fact that multiple permutations may be rotated into each other and thus count as only one isomers.
Let's illustrate this with a simpler problem. Suppose you seek the number if isometric tetrahedral complexes having the type $\ce{Ma3b}$. There are four permutations for placing the $b$ ligand, but they may all be rotated into each other and thus count as only one isomer.
Bookkeeping
To properly count isomers in your case (octahedral $\ce{Ma3b2c}$) you have to count out the number of permutations each one can be rotated into. You identified 60 permutations, so you will have all your isomers once you account for all 60 of these permutations. You also need to know that a regular octahedron can be left unchanged by 24 different rotations, for a regular tetrahedron it's 12, for a planar square there are 8 (including flipping the square), a triangular bipyramid or triangular prism has 6 and a square pyramid has 4. (These are all the shapes you're likely to see with four-, five- and six-coordinate complexes.)
Start with an isomer that has all three $\ce{a}$ ligands adjacent to each other. There are 24 ways to rotate a regular octahedron that gives you back the same octahedron with the same set of vertex locations, but none of those rotations (except the zero rotation) keeps the same permutations of $\ce{a}$, $\ce{b}$, and $\ce{c}$ ligands. You can keep the $\ce{a}$ ligands in place with a 120° turn around the appropriate axis, but then the $\ce{b}$ and $\ce{c}$ ligands are differently placed. Therefore this isomer rotates into 24 out of the 60 permutations.
No other isomer is possible with the $\ce{a}$ ligands all adjacent, but you may consider the case where two of these ligands are opposite each other. You may then arrange the other ligands so that the $\ce{b}$ ligands are either adjacent or opposite, these representing two clearly different cases. In the former case you again have no rotational symmetry with the ligands different, so rotating the octahedron must again take up 24 permutations.
When you place the $\ce{b}$ ligands opposite each other, however, there is a different outcome. This isomer still has a twofold rotational symmetry even with the different ligands; you get the same permutations back if you rotate 180° around the axis containing the $\ce{c}$ ligand and the opposite $\ce{a}$ ligand. So the number of permutations taken up by this isomer is cut in half by the retained rotational symmetry: 24/2=12. That together with the 24 permutations taken up by each of the other two isomers equals 60, so you have accounted for all the permutations with three isomers.
Look in the mirror
One final note: Since I considered only rotations I do not address whether two mirror images can be superimposed on each other. If they cannot, that is considered two separate isomers (called enantiomers); here that situation does not arise because all of your isomers still have mirror-reflection symmetry. You will learn about cases where that does not apply when you study optically active or chiral complexes.

it is not obvious to me how all these isomers have mirror reflection symmetry, is it possible to draw a plane of symmetry for these molecules?

If two of the three A ligands sre on opposite vertices, you have a mirror through the plane of the other tgree ligands. If the A vertices are mutually adjacent, the mirror passes through the C ligand and the midpoint between the two B's.

There are three possible cases in this combination $\ce{MA3B2C}$.

3 $\ce{A}$'s are in both cis and trans positions.

Both $\ce{A}$ and $\ce{B}$ are in trans positions simultaneously.

Both $\ce{A}$ and $\ce{B}$ are in cis positions simultaneously.


text:
Changes in UV-Vis spectra in host-guest complexes
I'm doing some background reading before conducting an experiment, but am getting a little confused.  What serves as evidence for the formation of a host-guest complex when conducting analysis through UV-visible absorption?
From the sources that I have read and have personally observed in the lab, formation of host-guest complexes can push equilibrium to favour a specific species, like an acidic form or a basic form of the guest molecule due to stabilizing interactions within the host (think cyclodextrins + phenolphthalein).  This is reflected in absorption spectra as an increase/decrease in absorbance of the peaks to reflect the increasing/decreasing proportion of acidic/basic species of the guest molecule.
Another way of telling if the host-guest complex is formed is if the there is an increase in the molar absorption coefficient or a shift in the absorption peaks of the guest molecule.  Another conflicting source tells me that when a guest molecule is complexed, absorption of the guest molecule should decrease.  Can anybody clarify which of these changes serves as evidence for formation of a host-guest complex?

Can anybody clarify which serves as evidence for formation of a host-guest complex?

There's no single, general answer to this.
As you aptly described in the body of your post, there is a wide variety of changes to the UV-Vis absorption of a host-guest system on binding of the host and guest.  The changes to the recorded spectrum could fall into any of these categories, depending on the particular structural features of host and guest, and the exact manner that they bind to one another.
AFAIK, the only way to know how the spectrum changes when host and guest bind, is to perform careful experimentation and observe the changes that occur.

Would you happen to know how molar absorptivity of a guest molecule can change upon binding?  Going out on a limb here but is this to do with a change in oscillator strength due to changes in the transition dipole moments when the guest molecule interacts with the host?

@logicallyabsurd As I said above, the molar absorptivity could increase, decrease, or remain unchanged upon binding. It's all heavily system-dependent. AFAIK your second sentence is essentially a textbook definition of what's involved, so: yes.


text:
Will the concentration of 3% w/v H2O2 be lowered after 3 weeks?
I have read online that $\ce{H2O2}$ decomposes if exposed to sunlight and air. If this is the case then will the concentration of 3% w/v of $\ce{H2O2}$ be lowered if it is left out in the open at around 25-degree celsius for a few hours?
Background:
The question arises from a titration experiment that I conducted where I had to verify the concentration of $\ce{H2O2}$ to be 3% w/v. But the titration found the concentration to be lower and other groups in the class found the same.

I would not think a solution as dilute as 3% would be changed much over an "hours" timescale. By all means, it will decompose, I just cannot see how the rate of that happening is anything but negligible. NB: I assume well out of reach from sunlight.

Hydrogen peroxide is somewhat unstable even when not exposed to air and sunlight. It decomposes to water and oxygen gas over time.
According to this Wikipedia article:

Hydrogen peroxide is thermodynamically unstable and decomposes to form water and oxygen with a $\mathrm{\Delta H}$ of $\mathrm{-98.2}~\pu{kJ/mol}$ and a $\mathrm{\Delta S}$ of $\mathrm{70.5~\pu{kJ/mol}\ce.\pu K}$:
$$\ce{2H2O2 -> 2H2O + O2}$$
The rate of decomposition increases with rising temperature, concentration and pH, with cool, dilute, acidic solutions showing the best stability.

So, not knowing the exact conditions during the 3-week period or other factors, it's not possible to say for certain whether the $\ce{H2O2}$ degraded or whether it just was a little low to begin with. But the potential for degradation over that period of time is certainly there.


text:
Where the equibrium will shift when we add one mole to both reactants and products in PCl5 giving PCl3 and Cl2
I know that if we add 1 mole to reactant side then the reaction will shift forward and if on product side the it will shift backward according to Le Chatelier's principle.Then what will  happen if I add 1 mole on both sides?Please explain.

Specify exactly what are you adding to the reactants and products side. Do you mean you are adding 1 mole of PCl3 and Cl2 both on the products side

I mean that I know when we increase the concentration   of reactants the reaction will do forward and if I increase concentration of products the it will go backward.Now where will it go if I add 1 mole of PCl5,1 mole of Cl2 and 1 mole  of PCL3 in the equilibrium mixture

Write Kc expression for the equilibrium, now write Qc expression when you add 1 mole of each of your reactants and products in the mixture. If Qc<Kc the equilibrium will move forward and vice versa

Can we tell it directly by Le Chatliers principle without knowing the concentrations of the mixture

Le Chatelier's principle would have helped if you were changing the equilibrium concentration of either the reactants or products(it's very qualitative). To be sure you could always approach the problem In a standard way,that is what I just described.

In a reaction of the form $\ce{A <=> B + C }$ if $a_0$ is the initial amount of A (and B and C are initially zero) then if an amount x reacts the equilibrium constant $K_e = x_e^2/(a_0-x_e) $ where $x_e$ is the equilibrium amount reacted so equilibrium A is  $a_e = a_0-x_e$. Solving gives $x_e = (-K_e+\sqrt{K_e^2+4K_ea_0})/2$.
Now adding 1 mol to each side gives $K_e= (x_e'+1)^2/(a_0+1-x_e')$. Solving again gives $x_e' = (-K_e-2+\sqrt{K_e^2+4K_ea_0+8K_e})/2$. You can work out the ratio  $x_e'/x_e$ for different values of $k_e$ for simplicity, say, with $a_0=1$.
You can repeat the calculation for different initial amounts of B and C and the equations are far more messy but the final result similar.


text:
Formula for finding how much base to add to buffer to obtain a desired pH
Given: molarity and volume of the weak acid, molarity of the strong base, desired $\ce{pH}$, $K_{a}$ of the acid. I tried coming up with a formula to solve for the amount of base to add in order to reach the $\ce{pH}$.
$$\ce{HA + OH- -> H2O + A-}$$
After addition of $\ce{OH-}$:
$n_{\ce{HA}}=M_{\ce{HA}}V_{\ce{HA}} - M_{\ce{OH-}}V_{\ce{OH-}}$
$n_{\ce{OH-}}=M_{\ce{OH-}}V_{\ce{OH-}} - M_{\ce{OH-}}V_{\ce{OH-}} = 0$ mol
$n_{\ce{A-}}=0 + M_{\ce{OH-}}V_{\ce{OH}} = M_{\ce{OH-}}V_{\ce{OH-}}$
$$\ce{HA + H2O <=> H3O+ + A-}$$
Using an ICE table, the concentrations at equilibrium become:
$[\ce{H3O+}]=10^{-\ce{pH}}$ (given the desired $\ce{pH}$)
$\ce{[HA]}=\frac{n_{\ce{HA}}}{V_{soln}}=\frac{M_{\ce{HA}}V_{\ce{HA}} - M_{\ce{OH-}}V_{\ce{OH-}}}{V_{\ce{HA}} + V_{\ce{OH-}}}-[\ce{H3O+}]=\frac{M_{\ce{HA}}V_{\ce{HA}} - M_{\ce{OH-}}V_{\ce{OH-}}}{V_{\ce{HA}} + V_{\ce{OH-}}}-10^{-\ce{pH}}$
$[\ce{A-}]=\frac{n_{\ce{A-}}}{V_{soln}}=\frac{M_{\ce{OH-}}V_{\ce{OH-}}}{V_{\ce{HA}} + V_{\ce{OH-}}}+[\ce{H3O+}] = \frac{M_{\ce{OH-}}V_{\ce{OH-}}}{V_{\ce{HA}} + V_{\ce{OH-}}} + 10^{-\ce{pH}}$
Writing a $K_{a}$ expression gives:
$$K_{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}=\frac{10^{-\ce{pH}}\left(\frac{M_{\ce{OH-}}V_{\ce{OH-}}}{V_{\ce{HA}} + V_{\ce{OH-}}} + 10^{-\ce{pH}}\right)}{\frac{M_{\ce{HA}}V_{\ce{HA}} - M_{\ce{OH-}}V_{\ce{OH-}}}{V_{\ce{HA}} + V_{\ce{OH-}}}-10^{-\ce{pH}}}$$
I tried rewriting this to solve for $V_{\ce{OH-}}$ and got:
$$V_{\ce{OH-}} = \frac{ V_{\ce{HA}}(10^{-2\ce{pH}} - K_{a}M_{\ce{HA}}) }{ -10^{-\ce{pH}}(-10^{-{\ce{pH}}}+M_{\ce{OH-}}) -K_{\mathrm{a}}M_{\ce{OH-}} }$$
Is this a valid formula? Could this be used to determine how much strong base to add to a buffer in order to reach a desired $\ce{pH}$? Is there an easier way to write this? Is it useful at all? Thanks.


text:
Calculating Boiling Point of solution
I wish to calculate the boiling point of the following solution.
(w/w)
78.8% Glycerol,  11.6% Propylene Glycol,  8.3% Water,  1.2% Nicotine
(molar fraction)
0.5790 Glycerol,  0.1037 Propylene Glycol,  0.3121 Water,  0.0052 Nicotine
I am trying to use Raoults law, after ruling out Boiling point elevation (due to the number of solutes), but I'm stuck with getting the vapor pressure values for each solute since I don't yet know the temperature.
So as I understand it, at some temperature, the sum of each component's vapor pressure will equal 1atm.
This temp is what I wish to calculate. Am I on the right track? Any advice appreciated.
============================================
EDIT: - Additional;
I've found Antoine's equation.
Also found Antoine values for all ingredients from webbook.nist.gov
Note: Antoine values used for Glycerol (57% mol.frac) are only valid over 456°K, a full 48° more than my calculated BP.
Have made a spreadsheet and got a result of 135°C, which is considerably lower than what I had expected.
View/Edit spreadsheet here
Have I used a sensible method to calculate BP for this solution?
Obviously, this is for my DIY e-cig juice, not commercial, just read something that piqued my interest and needed to investigate.
============================================
EDIT 2: Additional.
Thank you Ivan, porphyrin and airhuff for your comments, they were helpful to my understanding. I wish I had some equipment to actually experiment instead of calculate. But in saying that I would have been happy with a +-10°C result.
Also thank you user44840, I had not heard of bubble point or known the difference from boiling point.
So now as i understand it (please forgive me for being persistent, and correct wherever I'm wrong)
Dew point is used (as a upper limit) due to the changing molar fraction of the mixtures components as it evaporates, which in this case increases the boiling point as glycerine concentration increases.
So (looking at user44840's diagram, and over simplifying) if I put the solution into a beaker and apply heat I would see the first bubbles at approx 150°C and the last of the solution would remain until ~290°C? (Ignoring PG, Nic and Glycerine decomposition at 280°C)
I am having a problem reconciling this information with my actual experience using an e-cig. I have never noticed a viscosity difference between the start and end of a tank, which would be easily noticeable even if only 50% of the water was removed. (Water is used as a thinner since glycerine is far too viscous).
However I quickly found on google people claiming to experience this exact thing. I guess also glycerine being hygroscopic may also account for my experience above.
It is now obvious my question is of limited theoretical use, and reality has far too many variables for such a simple analysis.
As Ivan suggested "Chemistry is an experimental science, after all."
Anyone want to let me in their lab for a day?
Thanks once again folks.

You'd rather not put too much trust in the calculations. Chemistry is an experimental science, after all. I would expect a mixture like this to be markedly non-ideal and hence deviate from Raoult's law in an unpredictable way.

I agree with @Ivan Neretin, calculation will not be that helpful as alcohol water mixtures show considerable deviations from Raoult's law. A good place to start might be to look for glycerol water boiling temperatures at different compositions, some data is here http://www.engineeringtoolbox.com/glycerine-boiling-freezing-points-d_1590.html

I agree with porphyrin that the link he gave is a good place to start. The only comment I would make is to not waste your time and efforts accounting for the effect of nicotine. Whatever BP estimate you come up with will be in the +/- 10's of degrees arena anyway, and compensating for the effect of that small amount of nicotine will not improve this at all. Your spreadsheet kind of validates this too.  Anyway, good luck!

You may want to contact the team to merge your accounts.

Raouls law is a good method to estimate about what the "boiling point" will be. The bubble point is the lowest temperature that a mixture will be in vapor liquid equilibrium. What you actually want is the dew point. Mixtures don't have a boiling point like pure water. Instead, mixtures have a range at which they will be in a state of VLE or vapor liquid equilibrium. So you need to find the temperature that no liquid exists i.e. The dew point Phase diagram glycerol and water


text:
Obtaining number of functional groups in a compound from SMILES, SDF or MOL2 format
I'm looking for a tool that can give how many methyl groups, hydroxyl groups etc are in a compound's structure presented as SMILES, SDF or MOL2 format.

Possible duplicate of https://chemistry.stackexchange.com/questions/47537/is-there-an-algorithm-that-counts-joback-groups-in-a-structure

There are numerous programs, but it's pretty easy to use the obgrep tool in Open Babel to do this using SMARTS for the various functional groups:

obgrep -c '[CX3]' file.sdf # count the number of methyl groups
obgrep -c '[#6][OX2H]' file.sdf # count the number of hydroxyl groups

etc. Examples of functional groups using SMARTS can be found on the Daylight SMARTS examples page.
I've also written scripts in Python for these purposes - depends on how you want to run the SMARTS searches.

Thanks a lot. For testing I've used this file, it contains two methyl groups but the command:
obgrep -c '[CX3]' HMDB37025.sdf  returning 0!


text:
Primes in symmetry elements/operators
When classifying the molecules in symmetry point groups some non-principal rotation axes $C_2$ and some reflection planes $\sigma_v$ and $\sigma_d$ are primed or double primed, like $\sigma_v'$, $\sigma_v''$, $C_2'$ or $C_2''$. What determines which is the unprimed, primed or double primed $C_2$-axis/$\sigma_v$-plane, or is it arbitrary?

Search for Mulliken Symbols, there is a table here http://mathworld.wolfram.com/MullikenSymbols.html

Sure the Mulliken symbols of irreducible representations also have primes given in the table you linked, but this is not what I am asking about (i think). I ask bout the primes in the symmetry elements themselves, like the $\sigma_v'$ plane, or the $C_2''$ axis.

I misunderstood; they are just to identify between similar operations, say,  the three different $C_2$ axes in $D_6h$. Some tables use $C_2(x), C_2(y)$ etc. rather than $C_2'$ etc. and $\sigma (xy)$ instead of $\sigma '$.

OK, now that makes sense. So whether the plane of the water molecule is called $\sigma_v'$ or $\sigma_v''$ may differ in different books, for example?

yes or even $\sigma (xz)$ and $\sigma(yz), which I prefer, as different authors use different axes orientations, for example, all the  atoms in water may be chosen to lie on the zx plane or on the zy plane, z being the principal axis. So its often necessary to check what is being used and convert to your preferred orientation if needed.


text:
How and why does this Cu change into C (graphite)?

Source: gutefrage.net
At the right side of the graphic, the Cu changes while discharging into C as graphite. How and why does this happen?
I thought it happens because of the dissolving of the Cu when the electrons stream from the Cu to the Al.

The electrode in here has two parts – the Cu and the graphite (C). The Cu won't dissolve. Only the graphite (C) will react. So there is no changing from Cu to C.


text:
How can you identify the anode and cathode of a car battery of indistinct shape by using a potassium iodide solution?
Now I know that the anode in a car battery is lead, the cathode is lead dioxide and the electrolyte is sulfuric acid, but still: how could a potassium iodide solution be used to differentiate between the anode and the cathode?
I mean, lead can't replace potassium in its salt because it lies below it in E.C.S., so how are we supposed to identify them by using potassium iodide?

Surely you can't replace potassium, but maybe iodine would respond to electric current in some way? Also, welcome to Chem.SE.

Thanks for your help....and it's a pleasue to be a part of this site :)

By using the battery as a source of electricity in an electrolytic cell that contains potassium iodide as an electrolyte.
Electrolysis takes place where iodide ion is oxidized at anode and brown colour of iodine solution appears at anode.
At cathode hydrogen gas evolves which is indicated by formation of bubbles at cathode.

Given the conditions, iodine is by no means violet, nor is it a gas. Also, potassium would never precipitate from water solution. Other than that, you are right.

But won't iodide ion lose an electron at the anode and iodine will evolve as a gas at the anode and the opposite happens for potassium?

You ask too many questions at once. In short, the answers are: yes, no, no. "Yes" means that iodide would indeed lose an electron.

What about the"no,no" ?

@marawanhassan No, iodine would not evolve as a gas. And no, potassium would not evolve at all.

Apparently what will actually happen is that brown colour appears at anode indicating formation of iodine solution

And hydrogen gas evolves at cathode which is indicated by formation of bubbles potassium will not precipitate since it has lower reduction potential than water(hydrogen) and potassium ions and hydroxyl ions remain in solution making the solution at cathode basic

Obviously i don't study that much information at school so i neglected the presence of water.....deeply sorry about that but on the bright side i got to learn something new ☺

Good. Now please edit the new data into your answer, so as not to spread inaccurate information.


text:
filling of a ballon and gas law
When we fill a balloon with gas the pressure and volume inside it increases. Will this not go against the gas law as  pressure and temperature are inversely proportional to each other.

The said law is applicable if we have constant amount of gas.

@Ivan You mean the OP's interpretation of the law? Glorfindel's answer seems plausible, yet he doesn't "pin-down" one of the parameters in the Ideal gas law. [Side note: Why on earth do you keep away from The Periodic Table? It'll be interesting to have you there ;) ]

@paracetamol I mean Boyle's law, not the full PV=nRT thing. What is The Periodic Table?

No. The Ideal gas law states that
$$PV = nRT$$
You're right that the left hand side (pressure times volume) increases, but so does $n$, the amount of substance of gas. So the ideal gas law is still satisfied.


text:
Expected accuracy of experimental IR spectra compared to literature reference
I have a table from a journal which indicated signals of an alpha chitin that I used as theoretical values. I did my own analysis to identify a chitin sample and it had signals that slightly varied from the theoretical (ex. $3106\ \mathrm{cm}^{-1}$ symmetrical N-H vibration and I got $3103\ \mathrm{cm}^{-1}$). Should I consider this? And what caused the variation?

seems pretty close to me, only 3 in 3100;  what is the resolution with which the spectra was taken, are the lines overlapping , was the sample environment the same etc? These may explain the very small difference as may slight changes in calibration.

Hello @porphyrin, the lines weren't overlapping and the sample environment were the same.

It is a very good agreement, most likely within experimental error.

In addition to @porphyrin's comment, the interaction of your sample with the surrounding may overall cause some changes in shape and place of the absorption bands; namely if your analysis

follows the classical sample preparation, grinding your compound altogether with excess of KCl to press a pellet (ionic environment) / mixed in nujol (apolar environment) that is then characterised in transmission

.or.

is an analysis by ATR with single or multiple reflection where your sample basically is a neat film on top of your optical window; either probe head or between crystal and anvil. (The ATR correction offered by the spectrometer's software, not every time active, basically will account for the penetration depth varying by the wavelength used, hence influencing rather the displayed %transmittance, then the position of the barycentre of the absorption band in question.)

If you aim for a qualitative IR analysis of a small organic molecule, I would not worry too much if the reported IR band was reported  at 3106 cm$^{-1}$ and is now spot at 3103 cm$^{-1}$.  The iS10 ATR-FT-IR spectrometer by Nicolet/ThermoScientific, for example, provides a data spacing of around 0.7 cm$^{-1}$.  Performing a library search in an electronic database to identify / narrow which compound the sample in question is hopefully would not be affected, either.
(A quantification, like determining the composition of gasoline by PCA, would of course need some calibration.)
The original question however may be read as if you aim to monitor the (partial) digestion of larger proteins and polysaccharides, to say something about their secondary structures.  A shift of 10...20 cm$^{-1}$ of an amide absorption band may be relevant; yet then more often the ones around 1600...1700 cm$^{-1}$ are taken into consideration.

Hello @buttonwood, I used a ThermoScientific FT-IR spectrometer with an old OMNIC software and proceeded with the KBr pellet method. I was aiming for a qualitative analysis and got confused with the slight difference. Thanks for clarifying it though!


text:
How do you know if SN1 orSN2 reaction occurs when it is not so obvious?
For instance, when you have a primary halide with a weak nucleophile, do you think that the reaction mechanism would be SN1 or SN2, neither or would it even be both, like sometime leaving group can leave by itself creating carbocation and nucleophile because it is late later attack in second step. Same when you have a tertiary halide and really strong nucleophile and polar aprotic solvent (which we know facilitates SN2 rxn). I am not sure how to think of the process. Like when it comes to primary and tertiary halides,  does the substrate nauture determine mode of mechanism from the get go, with no input of the strength of nucleophiles, or type of solvent. Or do nucleophiles and type of solvent have a role in determining whether sn1 or sn2 would occur also?

You should never invoke $S_{\mathrm{N}}1$ on a primary halide unless it's heavily stabilized. And in cases where it's truly not clear, the answer is that they're both operating.


text:
Alternative to mercury bubbler for increasing boiling temp?
A common reaction for forming the octachlorodirhenate(III) anion is to heat tert-n-butylammonium perrhenate in benzoyl chloride at 209 ºC. If you look up the info for benzoyl chloride, one sees that the boiling point is only 197 ºC. To increase the boiling point, publications suggest closing off the system with a mercury bubbler, which increases the pressure enough to bump the boiling point of the liquid to 209 ºC.
In the teaching lab, use of mercury is highly frowned upon… are there any reasonable alternatives to a mercury bubbler to increase the boiling point (and thus, the reaction temperature) for this reaction? Publications report a 30-40% decrease in yield if the reaction temperature is lowered to 196 ºC, so there's a lot of motivation to keep the temperature up.

Silicone oil bath it is then I guess

Can you just use a sealed tube?

According to Wikipedia: "A gas bubbler is a piece of laboratory glassware which consists of a glass bulb filled with a small amount of fluid — usually mineral or silicone oil, less commonly mercury." I am not particularly familiar with this procedure and may be missing something basic, but it seems like a mineral or silicone oil would be the more common way to go (as Mithron also said).

Silicone oil and mineral oil don't provide the weight (literally, weight of liquid providing the over-pressure) needed to raise the boiling point high enough. I suppose that a bubbler of sufficient length could be made to make the weight of oil equal the weight of mercury in a standard-size bubbler.