Why isn't organic chemistry taught in 3D? As the title says. What's the point of learning everything in 2D and trying to visualize and interpret stereochemistry, newman structures, isomers etc, with hundreds of hours wasted, when a 3d visualization suite could be used to teach everything instead, without any ambiguities as to what the structure looks like? The 2D format seems completely irrelevant besides the communication of chemical structures on paper.

Surely a better system would teach everything in 3D first to gain a true visual and theoretical understanding of the subject, and ONLY THEN taught in 2D on how to simplify things for proper communication?

Organic chemistry is often taught in 3D.  There have long been 3D molecular models used in classrooms, as well as Fischer projections.

Sometimes 2D is enough for a given situation.  Maybe curved 4-dimensional spacetime is needed for some situations, vicinity of neutron stars or black holes lets say.

Use the simplist description that includes all the relavent concepts.


Are primary, secondary, and tertiary amines considered functional group or chain isomers? Consider $\ce{C3H7NH2}$.  Can one make $3$ different isomers with this molecular formula? My question is that are they functional group isomers or chain isomers?

I have been told that they are functional group isomers but I intuitively think that they are chain isomers.

When we talk about chain isomers we mean isomers with different carbon skeleton.

If the carbon skeleton is changing as well as the functional group we report it as functional group isomerism.

For the given compound in your question, primary, secondary and tertiary amines will show functional group isomerism as they are forming different types of amines itself.


Question on assumption in Michaelis-Menten kinetics In Michaelis-Menten kinetics, it is assumed that the substrate concentration is much higher than the enzyme concentration. In which step of the derivation of the Michaelis-Menten rate law is this assumption used?

From the explanations given in different sources, I think this assumption is used in expressing the total enzyme concentration as the sum of free enzyme and enzyme-substrate complex. Is my understanding correct? What happens when the enzyme concentration is in the same range as that of substrate concentration?
Why is it important to make this assumption?

Could someone explain?

I think it is more about expressing[S]. If we assume the substrate concentration is much higher than the enzyme concentration, then we can ignore the substrate that is bound as ES in the derivation of the MM equation.  If the enzyme concentration is in the same range as the substrate (or if the enzyme affinity is very high), then the amount of substrate bound as ES cannot be ignored, and the derivation is not simplified. In this latter case, [S] in the equation can't be considered as the total, but must be corrected by the amount bound.


How to make this citric acid substance I made not sticky? So I made this solid material by boiling citric acid crystals and water then letting that cool to harden.  It hardened into this solid that is really easy to mold and is transparent but it has this really annoying sticky texture that I want to get rid of.  I don't know if its the water causing this texture or not but can someone tell me how I can get rid of this and whats happening?

Thanks

A vacuum oven with a solvent trap is a really good way to get rid of persistent water in your product. I'm not sure what your overall objective is so I'd read up on how aggressively (or gently) you can heat your product first though.


A possible reaction of heating Terephthalic acid There is no reaction of terephthalic acid heating effect on internet (atleast Wikipedia). I have written the most probable reaction possible as per my knowledge. Please tell if it is spontaneous and possible or not. Also give real conditions of doing the reaction if possible.

Following are the Statement from “On Terephthalic Acid and its derivatives” By Warren De la Rue and Hugo Muller February 7, 1861:

On heating, terephthalic acid sublimes without previously
fusing(melting). The sublimate, which is indistinctly crystalline, has
the same composition and properties as the original acid, and
therefore, unlike other bibasic acids, terephthalic acid cannot be
converted into an anhydrite by merely heating it

Terephthalic acid when pure is white opaque powder, insoluble in water and it sublimes at $\pu{300^{\circ}C}$ without melting.
When heated to decomposition it emits acrid smoke and irritating fumes. $\ce{FeOOH}$, nickel hydroxide and nickel oxide are suitable catalyst for thermal decomposition of terephthalic acid (at low temperature) but since nickel compounds are expensive so $\ce{FeOOH}$ is generally used. At higher temperature (in absence of catalyst) Terephthalic acid$\ce{_{gas}}$ is decomposed to carbon dioxide,benzene and benzoic acid.
For further reading refer to this more in depth paper
"Determination of the Thermal Decomposition Products of Terephthalic Acid by Using Curie-Point Pyrolyzer" by Mehmet uluturk


Why does the energy difference of a reversible process not equal that of an irreversible process in an adiabatic expansion? Suppose that one mole of an ideal gas at $P_1$ and $V_1$ adiabatically expands to $P_2$ and $V_2$, reversibly and irreversibly (two separate processes). Since $\Delta U$ is a state function, why is it wrong to say that $\Delta U_{rev} = \Delta U_{irrev}$? (Since in an adiabatic process, there is no heat transfer, we know that $\Delta U = w$)?

I think the premise of the question itself is false: if you start with a given state (P1, V1) and undergo a reversible, adiabatic (hence isentropic) expansion, you can get to state P2, V2. But if you don't have the requirement of reversibility, I don't think it is possible to gain the same second state - there will be either internal heating/cooling, causing the expected final volume (or pressure, depending on the mechanism of expansion) to not be P2 or V2.


Why isn't iron(III) oxide an effective catalyst for decomposing hydrogen peroxide? I've been working on a project involving the catalyst iron(III) oxide for decomposing hydrogen peroxide.
It barely seemed to produce any oxygen.

This YouTube video also seems to suggest that iron(III) oxide isn't a very effective catalyst.
But why?

Ah but iron(III) does act as a catalyst in peroxide decomposition (YouTube video).

It can be improved by the addition of some copper.

Some hydrogen peroxide is mixed with inhibitors (radical scavengers or chelating agents) to suppress the decomposition.


DIY spectrometers limits as a water quality sensor What pathogens or harmful stuff IN WATER can you detect with these homemade video spectrometers (webcam) + UV light?

https://www.youtube.com/watch?v=IA5BTD-aelo

I'm trying to do a cheap spectral analysis to analyze water quality, something electronic, so I can do continues monitoring remotely.

In general, with this setup, nothing. You would need a UV detector instead of a webcam. And, assuming you had a UV light source and the right detector, it would depend on the UV source spectrum. You could use a blacklight for mid 300 nm or a Steripen for 254. A broad spectrum lamp would be best. Water doesn't start to absorb UV strongly until you get below approx. 200 nm, so compounds that absorb above that will show up. This would be on a commercial unit like a Hach DR6000. On a DIY kit, I doubt you'll be able to detect much of anything unless it is either at high concentrations (very high) or your path length is long (ie Beers law). With all that being said, UV spec isn't a very good "fingerprint" analysis like ftir or mass spec. For example DNA and many organics all absorb at 254 nm, so you won't be able to differentiate them. Probably the easiest water quality parameter you could measure is called UV254, but that's really only useful for determining organic content and disinfection byproduct formation potential.

Okay, will all that being said, you might have a better option using visible light. Hach sells chemical kits for measuring certain constituents - they are coloromertric so they convert the constituent to a colored compounds that is detectable at a certain visible light wavelength. For example, you could test for nitrate or chloride. A used portable Hach meter will run you around 500, these are basically a fancy visible light spec. They work decent, but not great and even that would be worlds better than any of these DIY kits if that tells you anything.


Why is ice cubical in structure at very low temperature? My book says that at atmospheric pressure ice crystallises in the hexagonal form but at very low temperatures it condenses to cubic form.

Why and how does this happen?

First, a little clarification. The kind of ice you see at ambient pressure, ice I, exists as stacks of bilayers of water molecules. If they are stacked in an ABAB sequence, they form hexagonal ice, and cubic ice in an ABCABC sequence. The difference between the two allotropes is just the stacking of the molecules, so in fact they are structurally quite similar. The structure of cubic ice is analogous to the diamond structure of carbon, with hexagonal ice being analogous to lonsdalite. This picture is complicated by the fact that ice is generally proton-disordered, i.e. the molecules can have different orientations that break the crystal symmetry, subject to certain rules (the "ice rules").

There is evidence that ice crystallises as cubic ice in the upper atmosphere, namely the angle subtended by solar haloes and the angles formed in nascent snowflakes, for example. However, more recent evidence suggests that cubic ice doesn't not exist as a pure phase, but only in stacking-disordered hexagonal ice (http://pubs.rsc.org/en/content/articlelanding/2015/cp/c4cp02893g#!divAbstract). Curiously, there is also evidence that the lonsdalite form of carbon doesn't exist in reality, and is just stacking-faulted diamond (http://www.nature.com/articles/ncomms6447).

This is all very well, but does not tell us why hexagonal ice forms in preference under certain conditions, or cubic ice (?) under others. If you were to perform density functional theory calculation (essentially approximately solve the Schrodinger equation for a many atom system), you would find that hexagonal and cubic ices have almost identical energies, and are essentially degenerate at zero temperature, which makes it even more confusing why hexagonal ice is almost exclusively seen on Earth. I don't think there is a convincing answer to your question, since the best calculations do not show why the hexagonal structure of ice is lower in energy than cubic, but it could be due to a number of reasons: orientational disordering, which complicates modelling ices, zero point energy or free energy considerations, or a breakdown in the approximations used to perform such calculations. In any case, it looks like cubic ice doesn't exist under terrestrial conditions.


Shouldn't all of group 16 form double bonds? I notice that out of all of group 16 oxygen is the only element that can form a double bond. Why is that because shouldn't all of group 16 be able to form a double bond and be diatomic because they all have six valence electrons?

I notice that out of all of group 16 oxygen is the only element that can form a double bond.

This premise is not true. Sulfur in fact forms double bonds in many organic compounds.  Selenium and tellurium both form double bonds with triphenylphosphine as triphenylphosphineselenide and triphenylphosphinetelluride.

The larger chalcogenides do have difficulty in forming double bonds in organic molecules due to lower bond strength, higher reactivity, preference of molecules for oxygen, and steric bulk.  They can however indeed form double bonds; it's just not as common.


Which statement is correct about electron orbitals and energy levels? Which statement is correct about electron orbitals and energy levels?
A. Yttrium, $\ce{Y}$ ($Z = 39$) is the first element in the periodic table with an electron in a f sub-level.
B. The maximum number of electrons in one d orbital is 10.
C. The maximum number of electrons in the 4th main energy level is 18.
D. In a main energy level, the sub-level with the highest energy is labelled f.

I thought B and D are all possible solutions to this answer.
B--the maximum number of electrons in 1 d orbital is 10.
D--in main energy level, the sub-levels in order of increasing energy is s $<$ p $<$d $<$f
So why is the answer only D?
Please advise and sorry in advance for the wrong tags or title name, as I am new to the group and am trying to improve on it.

Although @orthocresol got this first, and it should be credited to him, I'm posting it as an answer.

In option (B), the statement is incorrect as it refers to $d$ orbitals, which can (according to Pauli's Exclusion Principle) hold 2 electrons each, rather than the $d$ subshell, which can hold 10 electrons, as it has a total of 5 $d$ orbitals which can hold 2 electrons each, and $5*2=10$.

Thus, only option (D) is correct.


Acid-base titration: Calculate pKa with only three values given There's an unkown acid, diluted with an unknown amount of water and titrated with $\ce{NaOH}$.

  After adding $\pu{10.00 mL}$ $\ce{NaOH}$, a $\mathrm{pH}$ value of $4.65$ is measured.
  After adding another $\pu{12.22 mL}$ ($\pu{22.22 mL}$ $\ce{NaOH}$ in total), the equivalence point is reached.

  Calculate the $\mathrm{p}K_\mathrm{a}$ of the acid!

I'm currently facing this uncommon task regarding acid-base titration. There's quite few information in this text, only three values are given:

\begin{align}
V_1(\ce{NaOH}) &= \pu{10.00 mL}: &\quad &\mathrm{pH = 4.65} \\
V_2(\ce{NaOH}) &= \pu{22.22 mL}: &\quad &\text{equivalence point reached}
\end{align}

What I've got so far:

Before adding any $\ce{NaOH}$, the reaction should look like this:

$$\ce{HA + H2O <=> A- + H3O+}$$

I expect the neutralisation reaction to be:

$$\ce{A- + H3O+ + Na+ + OH- <=> NaA + 2 H2O}$$

In general $n = c V$. At the equivalence point $n(\ce{NaOH}) = n(\ce{HA})$, so:

$$C(\ce{NaOH}) \cdot V_2(\ce{NaOH}) = n(\ce{HA})$$

$$C(\ce{NaOH}) = \frac{n(\ce{HA})}{\pu{22.22 mL}}$$

Henderson-Hasselbalch:

\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K_\mathrm{a} &= \mathrm{pH} - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K_\mathrm{a} &= 4.65 - \log{\frac{C(\ce{A-})}{C(\ce{HA})}}
\end{align}

I'm unsure how to express this equation as a function of $V_1(\ce{NaOH})$. I may be on the wrong track as well. Any help is appreciated!

First, we note that this is not a strong acid, since when it is only partially neutralized, the $\mathrm{pH}$ has already risen to almost $5$.

At that point, we know that $\frac{10}{22.2}$ of the total acid has been neutralized. Therefore, the ratio

$$\frac{\ce{[A-]}}{\ce{[HA]}} = \frac{10}{22.2} \cdot \left(1 - \frac{10}{22.2}\right)^{-1} = \frac{1}{1.22}$$

Substitution into Henderson-Hasselbalch gives that the $\mathrm{p}K_\mathrm{a} = 4.73$.

Note that I assumed that the solution wasn't super dilute. If that were the case, I couldn't plug directly into the Henderson-Hasselbalch equation because the acid-base equilibrium would need to shift, but the change in the concentration of hydronium would be dominated by auto-ionization of water.

And in general, the total volume of solution does matter. You could consider a solution of hydrochloric acid that were dilute enough that even the first assumption of a weak acid were wrong. It's pretty easy to construct the solution such that the partially neutralized solution had a $\mathrm{pH}$ of $4.65$. I'll leave that as a follow-up exercise for you to explore.


Can citric acid protonation produce benzene and water Is this reaction possible: $\ce{C6H8O7 + 12 H3O -> C6H6 + 19 H2O}$?
Considering that the $\ce{H+}$ can be induced by ionization of the citric acid solution the benzene can be produced in this way?

Here's a drawing of citric acid and benzene.  You would have to spend a lot of energy in order to break a large number of bonds (carbon-carbon bonds, carbon-oxygen bonds and carbon-hydrogen bonds) to rearrange citric acid into benzene.  This would be unlikely to occur in a natural process and would require many steps in the laboratory.


Constrain bond angle In Gaussian molecular structure optimization I am computing the optimized molecular structure of an organic molecule in the Gaussian 09 computational chemistry package.
As I have it now, I do not think that the optimized structure that I am getting from Gaussian is correct, in part due to the symmetry of some of the bonds.  To try and correct this, I want to fix two of the bond angles in the molecule in order to help/force it to produce a more believable structure.  Even though I want to fix these bond angles, I want to leave everything else variable and open to optimization, including those bond lengths.

How can I specify this in the input file?

Fixing the bond angle to a set value seems like an extreme solution to this problem.  Perhaps I can just constrain the angle within a certain range, is this possible?

After a lot of work and playing with settings, I have figured out how to fix bond angles, but now my difficulty is fixing them to 180 degrees (my desired value), for linear bonds. It seems that this is very hard to do.

Fixing the bond angle to 179 degrees (or 179.9 degrees), results in an error saying "Error in internal coordinate system." A lot of people online say this is because of linear/nearly-linear bonds in the molecule being optimized. Some people also say this might be avoided if I try to optimize in Cartesian coordinates, instead of internal coordinates, using the cartesian parameter after opt. However, I do not think I can use ModRedundant and cartesian parameters at the same time.

How can I fix a bond angle to be linear, but still optimize the rest of the structure in Cartesian coordinates? Is it even possible to fix angles to 180 degrees?

Fixing the bond angle to a set value seems like an extreme solution to this problem.

It is. To be honest, fixing parameters in a calculation is almost always no solution at all. The resulting geometry has little to no meaning and all derived properties are for a purely hypothetical case. If the located geometry is no stationary point, what does that even mean? A wrong solution is worse than no solution at all. You have to be very careful with what you fix and when you fix, and you should always confirm what you have found with a full optimisation and a frequency calculation. (On various levels of theory.)

I think Greg's advice is also very important. Symmetry can be a very powerful tool if utilised correctly, but you should make sure that the implied symmetry is correct. You also need to check whether your methodology matches the system you are computing.

All that being said, there are certainly a few very productive uses for partial optimisations:

finding higher or unusual local minima
scanning the potential energy surface
modifying converged geometries
fixing transitional modes
etc.

There are a few possibilities how to make partial optimisations in gaussian. All have some caveats and all may lead to undesired results. There are certainly restrictions to all of them. I'll give a short overview the most common ones with a few practical examples.

You have already noticed, that fixing a bond angle to 180 degrees is a challenge. This is due to the fact, that it is a requirement to have linearly independent variables, which is simply not possible with threes point on a line. However, there are certain ways to work around the problem. Spoiler alert: It is only possible with z-matrices or very rigid symmetry (where you fix the symmetry, not the bond angle).

Partial optimisations with z-matrices

Let's start with something very simple. A full optimisation of methanol without any symmetry restrictions.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt
int(ultrafinegrid)
sym(none)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc4         2 hch4          3 dih4
 o    1 oc5         2 och5          3 dih5
 h    5 ho6         1 hoc6          2 dih6

hc2           1.1
hc3           1.1
hch3        109.5
hc4           1.1
hch4        109.5
dih4        120.0
oc5           1.4
och5        109.5
dih5       -120.0
ho6           0.9
hoc6        109.5
dih6        180.0

This will result in the following optimised geometry:

 c
 o    1 oc2
 h    1 hc3         2 hco3
 h    1 hc4         2 hco4          3 dih4
 h    1 hc5         2 hco5          4 dih5
 h    2 ho6         1 hoc6          3 dih6

oc2           1.414851
hc3           1.109697
hco3        107.476
hc4           1.118853
hco4        113.586
dih4       -118.329
hc5           1.118853
hco5        113.586
dih5       -123.342
ho6           0.974680
hoc6        107.408
dih6        179.999

Fixing an angle in a z-matrix is straight forward. You need to specify that you are doing a partial optimisation (popt) and move the fixed parameter to the constants section. As an example, I am fixing the COH angle, and also invoke symmetry.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc3         2 hch3          3 dih4
 o    1 oc5         2 och5          3 dih5
 h    5 ho6         1 coh           2 dih6
  Variables:
hc2           1.100000
hc3           1.100000
hch3        109.5
dih4        120.000
oc5           1.400000
och5        109.5
dih5       -120.000
ho6           0.900000
dih6        180.000
  Constants:
coh          90.0

This will give you the following list of optimised parameters:

       hc2         1.1104
       hc3         1.1173
      hch3       107.0973
      dih4       116.0195
       oc5         1.4471
      och5       108.92
      dih5      -121.9694
       ho6         0.9796
      dih6       180.0092
       coh        90.0

Now we will try the same, but we will make the bond angle 180 degrees. Here we need to introduce dummy atoms:

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)

title card required

0 1
 c
 h    1 hc2
 h    1 hc3         2 hch3
 h    1 hc3         2 hch3          3 dih4
 o    1 oc5         2 och5          3 dih5
xx    5 xxo6        1 xxoc6         2 dih6
 h    5 ho7         6 xxoc6         1 dih7
  Variables
hc2           1.100000
hc3           1.100000
hch3        109.500
dih4        120.000
oc5           1.400000
och5        109.500
dih5       -120.000
dih6        180.000
ho7           0.950000
  Constants
xxo6          1.000000
xxoc6        90.000
dih7        180.000

Which will give you once again the following set of parameters:

       hc2         1.1209
       hc3         1.1211
      hch3       106.3981
      dih4       113.09
       oc5         1.3644
      och5       112.3772
      dih5      -123.4351
      dih6       180.0
       ho7         0.9465
      xxo6         1.0
      xxoc6       90.0
      dih7       180.0

And the following Cartesian coordinates:

    6
scf done:  -115.581115
 C     0.000000     0.000000     0.000000
 H     0.000000     0.000000     1.120932
 H     1.075500     0.000000    -0.316498
 H    -0.421786    -0.989342    -0.316498
 O    -0.695163     1.052866    -0.519431
 H    -1.177401     1.783245    -0.879763

Quick and Dirty fixing with Cartesian coordinates

Appending -1 to an element in a row in the Cartesian coordinates block will fix it. This is a very quick and dirty approach, which may lead to a brute force computation. You are basically restricting the algorithm to rotate the molecule, which leads to less effective calculations. Fixing Cartesian coordinates also fixes the bond length, but it is the only way to force a 180 degree angle with this choice of input.
Here is an example input:

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
popt
int(ultrafinegrid)
symm(loose)

title card required

0 1
 C  -1    0.000000     0.000000     0.000000
 H        0.000000     0.000000     1.300000
 H        1.221600     0.000000    -0.444626
 H       -0.479083    -1.123738    -0.444626
 O  -1   -0.695162     1.052869    -0.519426
 H  -1   -1.177400     1.783250    -0.879755

This will result in the following optimised geometry:

    6
scf done:  -115.581115
 C    -0.663620    -0.000005     0.000000
 H    -1.090591     0.003267    -1.036538
 H    -1.090556     0.896080     0.521076
 H    -1.090648    -0.899281     0.515456
 O     0.700780    -0.000004     0.000001
 H     1.647272    -0.000004     0.000001

Note that this calculation takes about five times longer.

Fixing parameters with modredundant

This option is described quite straight forward in the Gaussian online manual. But let's make an example where we try to restrict the COH angle to 90 degrees.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt(modredundant)
int(ultrafinegrid)

title card required

0 1
 C     0.000000     0.000000     0.000000
 O    -0.659966     1.143095    -0.466667
 H    -0.000000     0.000000     1.089000
 H     1.026720     0.000002    -0.362997
 H    -0.513360    -0.889165    -0.363000
 H    -0.502133     0.869719    -1.359507

A 5 6 1 F

With this option it is impossible to fix the angle to be 180 degrees, since this is not an allowed value. You also cannot use dummy atoms, as these do not exist in the redundant coordinate system. It might be a good idea, to kill this variable completely, to ensure that the  program will not fail trying to construct it, or when an optimisation is getting close to that value.

Alternatively the angle can be restricted via a linear bend. This is the best option, if you know that the angle is 180.0 degrees.

%chk=bp86svp.chk
#p BP86/def2SVP/W06
DenFit
opt(Modredundant)
int(ultrafinegrid)
sym(none)

title card required

0 1
C        0.000000000      0.000000000      0.736983516
H        1.037151577      0.000000000      1.103476471
H       -0.518575789     -0.898199613      1.103476471
H       -0.518575789      0.898199613      1.103476471
O        0.000000000      0.000000000     -0.663016475
H        0.000000000      0.000000000     -1.563016262

L 1 5 6 2 F

The last line freezes the linear angle between atoms 1, 5, and 6. Atom 2 needs to be specified to determine the plane in which the linear bend lies, or is orthogonal to.

Fixing linear angles with general internal coordinates (GIC)

A new way of handling the coordinate system was introduced with Gaussian 16: generalised internal coordinates. It is in many ways superior to earlier versions, however, also a little bit harder to learn. Nevertheless, it is absolutely worthwhile trying.

Here is a sample input for oxoketene $\ce{O=C=C=O}$. In this case I specified the complete set of coordinates necessary.

#p PM6
opt geom(AddGIC)
scf(xqc,MaxCycle=250)
int(ultrafinegrid)
symmetry(none)

Oxoketene (constrained)

0 1
O        0.000000000      0.000000000     -2.210000000
C        0.000000000      0.000000000     -0.770000000
C        0.000000000      0.000000000      0.770000000
O        0.000000000      0.000000000      2.210000000

BondC1O1(Active)=R(2,1)
BondC2O2(Active)=R(3,4)
BondC1C2(Active)=R(2,3)
LinBend1O1C1C2(Freeze)=L(1,2,3,0,-1)
LinBend2O1C1C2(Freeze)=L(1,2,3,0,-2)
LinBend1C1C2O2(Freeze)=L(2,3,4,0,-1)
LinBend2C1C2O2(Freeze)=L(2,3,4,0,-2)

The format is

L(     i,      j,      k,      l,                                 M)
L(Atom 1, Atom 2, Atom 3, <Atom 4 | Orthogonal to Plane>, component)

The linear bend is therefore defined between i, j, k as Atom 1, Atom 2, Atom 3 and must not be dummy atoms (and with GIC these are hardly necessary anymore). The fourth value l, can be another atom to determine the orthogonality/ plane to, or it can be defined as a plane where l = 0, -1, -2, -3 standing for automatic, $yz$, $xz$, $xy$, respectively. The last value M gives the component, of which two are necessary, and must be M = -1, -2.

The methanol example is as follows:

#p PM6
DenFit
opt(AddGIC)
int(ultrafinegrid)
sym(none)

title card required

0 1
C        0.000000000      0.000000000      0.736983516
H        1.037151577      0.000000000      1.103476471
H       -0.518575789     -0.898199613      1.103476471
H       -0.518575789      0.898199613      1.103476471
O        0.000000000      0.000000000     -0.663016475
H        0.000000000      0.000000000     -1.563016262

L(1, 5, 6, 2, -1) Freeze
L(1, 5, 6, 2, -2) Freeze


Calculations for making a buffer from a weak base and strong acid How many moles of $\ce{HCl}$ must be added to $\pu{100 ml}$ of a $\pu{0.100 M}$ solution of methylamine ($\mathrm pK_\mathrm b = 3.36$) to give a buffer having a $\mathrm{pH}$ of $10.00$?

(The answer is supposed to be $\pu{8.1 mmol}$.)

My thought process:

First, I found the $\mathrm pK_\mathrm a$ of methylammonium ion:

$$\begin{align}
\mathrm pK_\mathrm a(\ce{CH3NH3+}) &= 14 - \mathrm pK_\mathrm b(\ce{CH3NH2}) \\
&= 10.64
\end{align}$$

Substituting this into the Henderson–Hasselbalch equation

$$\begin{align}
\mathrm{pH} &= \mathrm pK_\mathrm a + \log\left(\frac{[\ce{CH3NH2}]}{[\ce{CH3NH3+}]}\right) \\
10.00 &= 10.64 + \log\left(\frac{\pu{10 mmol}}{x}\right)
\end{align}$$

where $x$ is the amount of $\ce{HCl}$ that must be added. However, when I solve for $x$ I find $x = \pu{43.7 mmol}$, nowhere near the correct answer. What am I missing?

Since $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$, you get $\mathrm{p}K_\mathrm{a} = 10.64$ for the methylammonium cation.

$\ce{HCl}$ protonates methylamine. The amount of methylammonium increases by the same amount methylamine decreases. So

$$\mathrm{p}K_\mathrm{a} = \mathrm{pH} + \log\left(\frac{[\ce{HA}]}{[\ce{A-}]}\right) = \mathrm{pH} + \log\left(\frac{x}{10~\pu{mmol} - x}\right)$$

lets you calculate the value $x$ of $\ce{HCl}$ needed to obtain the desired $\mathrm{pH}$.


How to find enthalpy of solvation for gas phase cation using Hess's law? Find enthalpy of solvation $\Delta_\mathrm{solv}H$ using Hess's law, and show the expression that allows to calculate the enthalpy of solvation.
$$\ce{Ca^2+(g) -> Ca2+(aq)}$$

I am not told which enthalpies of reaction to use. I don't need the exact value, but I do need to show the sum of each enthalpy. I managed to find the enthalpy of the following reactions:
$$
\begin{align}
\ce{Ca(g) &-> Ca+(g) + e-} &\quad \Delta H &= \pu{590 kJ mol-1} \\
\ce{Ca+(g) &-> Ca^2+(g) + e-} &\quad \Delta H &= \pu{1145 kJ mol-1} \\
\ce{Ca(s) + 2 H+(aq) &-> Ca^2+(aq) + H2(g)} &\quad \Delta H &= \pu{1925.9 kJ mol-1} \\
\ce{Ca(s) &-> Ca(g)} &\quad \Delta H &= \pu{178 kJ mol-1}
\end{align}
$$
I nearly have the reactions I need, but I am still needing
$$\ce{H2 -> 2 H+ + 2 e^-}$$
I haven't been able to find the answer.
P.S. I am aware of the enthalpy of formation of $\ce{H2}$, but don't know how/if I can use it here.

One approach to compute the enthalpy change for the reaction
$$\ce{H2(g)->2H+(aq) + 2e-}$$
is to combine the bond dissociation enthalpy and hydrogen ionization energy (both refer to gas phase species) and the hydration enthalpy of the hydrogen ion:
$$\ce{H2(g)->2H(g)}~~~~~~~~{\Delta H = \pu{439 kJ/mol}  }\\
\ce{2H(g)->2H+(g) +2e-(g)}~~~~~~~~{\Delta H = \pu{262 kJ/mol}  }\\
\ce{2H+(g)->2H+(aq)}~~~~~~~~{\Delta H = \pu{-2260 kJ/mol}  }$$
If you add up the enthalpies you obtain as the solvation enthalpy of $\ce{Ca^{2+}}$ the value $\Delta_{\text{solv}} H = -1546 \pu{kJ/mol}$ which is quite reasonable (see eg a table here)
The result is clearly very different from my earlier answer which I am still pondering:
The catch is that $\Delta H^\circ=0$ for this reaction. The reason is because $\Delta G^\circ=0$ for the reaction (by definition, see standard reduction potential) at all temperatures, and then, since
$$\Delta H = -T^2\left( \frac{\partial (\Delta G/T)}{\partial T}\right)_p$$
it follows that  $\Delta H^\circ=0$. I suspect the problem with this approach is due to a missing term involving enthalpy of solvation of the electrons (or a similar term involving the electrons) but I haven't put my finger on it.


What is the energy change of ethene reacting with hydrogen gas? $\ce{CH2CH2(g) + H2(g) -> CH3CH3(g)}$

Using this table, source: http://www.kentchemistry.com/images/links/Kinetics/TB08_003.GIF

C-C enthalpy = 348 kJ/mol
C-H enthalpy = 413 kJ/mol
H-H enthalpy = 436 kJ/mol

The way I see it, H-H has to break, C-C has to break, and then two C-H bonds have to form.

(Units omitted) Therefore, should the answer be: $413 \cdot 2 - 348 - 436 = 42 \, \text{kJ/mol}$

So then it has to be an endothermic reaction.  Is this correct, or am I wrong?

Your reasoning is correct, but your numbers are wrong.

You are not breaking a C-C bond, but changing a C=C bond to C-C. So you will also need the energy for C=C which is 614 kJ/mol.


Which is the best rocket fuel? I searched on internet about rocket fuel and the most common fuel is $\ce{KNO3}$ + sugar (rocket candy), but I think that I could use another fuel methods, like using can with gas or using $\ce{HCl + Al}$.

My question is: Which one ($\ce{KNO3}$ + sugar, can with gas or $\ce{HCl + Al}$) gets more performance? Can I increase the performance by increasing the pressure and the durability?

I used Zn + S as a teenager. It is a lot safer than Al, and you can mix it cold.

Stick to solid propelants because it's simpler, safer and smaller. You don't want an explosion but something very quick burning.

Mix the two fine powders well.

You can mix them in a thick plastic bag or wide mouth bottle and shake until you see an even color in all the mix with no lumps in sight.

(You can make a paste with alcohol 100% but I never tried this.)

Then tap it in with wood or plastic. You don't want a spark! Nozzle reduction is 1/3 and not smaller. The mix is 3:1. Light with a filament, it is safer than a spark.

In those days all I had was a book but you can find a lot of info out there.

We switched to this after a (not too serious) accident with rocket candy.

Safety first allways, the rocket can go anywhere!


Why is M1*V1 = M2*V2 only used for diluting solutions? Why is $M_1V_1 = M_2V_2$ only used for diluting solutions? Why can it not be used to figure out how many moles to add to make a solution more dilute?

As Satwik has stated in the comments, the equation relies on the conservation of moles principle.

  $$\dfrac{mol_1}{V_1} \times {V_1} = \dfrac{mol_2}{V_2} \times {V_2}$$

Based on the above, given a constant volume of a solution, adding more moles of solute will only make the concentration increase, not decrease.  Adding in more solvent is the only answer in that case.


Chemistry-bond energy Considering the hyperconjucation effects with C-T,C-D and C-H bonds, the maximum efficiency is given by C-H bond.This should mean that C-H must be the weakest bond and must hve the least bond energy compared to others.While all are isotopes of the same element with completely identical outer electronic configuration,then with is the difference in the bond energy?

Heavier isotopes make stronger bonds, therefore C-D and C-T bonds are stronger than C-H bonds. The difference between the bond strengths is however not very significant and can be ignored for most practical purposes.

Therefore, the extent of hyperconjugation can be given as: C-H > C-D > C-T (the difference however is not very significant).

The exact reason for this little difference involves a quantum mechanical explanation, which you should not bother about if you are studying high school chemsitry. If you want to read that, you can refer the links in the comment section.


How can I measure the pH of high-viscosity fluids like dough? I would like to measure the pH value of dough with a pH-Meter, to ensure a pH value of 4.1 (max). Because of the high viscosity I would thin down the dough with demineralized water for two reasons:

I expect more accurate measurements
This reduces pollution of the measurement device.

As far as I understand this will affect the measured value, so I would thin down the dough in a defined ratio of 1:10.

Questions:

Is it correct that the thinning with demineralized water changes the pH value?
The ratio 1:10 should cause an offset of 1 in ph value, right? In which direction? Do I have to add 1.0  to or subtract 1 from the measurement to estimate the real value?

Edit #1

As it seems to be important: We are talking about weak acids produced by fermentation of rye flour (or wheat flour) using sourdough (lactose bacteria + yeast). Some samplings are taken during the fermentation process to check that the pH value is low enough (value of 4.1 or lower).
I'm less interested in the -details- of the theoretical background, more in a practical solution for the issue. The actual real value isn't that important (I need to check, if the pH limit is reached), but it would be interesting how to measure it as there might be future experiments where the value matters...

The use of additional chemicals except (demineralized) water should be avoided for simplicity. Of course the use of tab water would be preferable, but I guess this becomes more complicated because of unknown water hardness and varying ion concentrations.

In addition to buckminst’s answer, for weak acids, which are not fully ionized, the pH of the solution depends on the ratio of the concentrations of the conjugate base $\ce{A-}$ to the neutral acid $\ce{HA}$, especially in the pH regime close to the $\mathrm{p}K_\text{a}$ of the acid (the buffer region). Diluting the solution will change the concentrations, but in the same way. The ratio will not change. The Henderson-Hasselbalch equation can be derived from the mass action expression for $K_\text{a}$ of a weak acid:

$$\ce{HA <=>H+ + A-}$$
$$K_\text{a}=\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}$$
$$\text{p}K_\text{a}=-\log{K_\text{a}}=-\log \left(\dfrac{\ce{[H+][A^{-}]}}{\ce{[HA]}}\right)=-\log[\ce{H+}]-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
$$\text{p}K_\text{a}=\text{pH}-\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$
$$\text{pH}=\text{p}K_\text{a}+\log\left(\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}\right)$$

If the volume increases by a factor of 10, then $[\ce{A-}]$ and $[\ce{HA}]$ both decrease by a factor of 10, and $\dfrac{\ce{[A^{-}]}}{\ce{[HA]}}$ remains constant.

Since the acids in dough are likely acetic or lactic from fermentation; carbonic or tartaric from leavening; or citric or ascorbic from preservatives – all of which are weak acids – diluting the dough might not change the pH.


Why is it that less useful energy can be extracted in explosive reactions? If you extract energy from a substrate (e.g. glucose) step-wise, a lot more free energy is released that can be used. However, if you were to do this in one single step, very little useful energy could be harnessed. Why is that?

I think there are two relevant points here.

1) explosive reactions can give significant useful energy - e.g. internal combustion engine

2) step-wise processes may go slower and release less heat / light other energy that is not so useful. This is a bit like in thermodynamics you get maximum useful energy from slow reversible processes.


working with ratios and RMS to find variables question
consider separate 1.0L samples of helium and $UF_6$ both at 1.00atm and containing the same number of moles. what will the ratio of the separate temperature need to be in order to have the same root-mean-square speed?

my steps
So i set up an equation
$\sqrt{\frac{3(8.314T)}{0.004}}$=$\sqrt{\frac{3(8.314T)}{0.352008}}$
but i soon realized that that would be impossible
and that is where i got stuck

As I said solve using variables first and simplify as much as you can, then plug in the numbers. From the Maxwell-Boltzmann Distribution:

$v_p = \sqrt{\frac{2RT}{M}}$

$v_{rms} = \sqrt{\frac{3}{2}}v_p$

where:

$v_p$ most probably velocity
$v_{rms}$ rms velocity
R universal gas constant
T temperature
M molecular mass

so let:

$v_{p,He}$ - most probably velocity for He
$v_{p,U}$ - most probably velocity for $\ce{UF6}$
$v_{rms,He}$ - rms velocity for He
$v_{rms,U}$ - rms probably velocity for $\ce{UF6}$

we're given:
 $v_{rms,He} = v_{rms,U}$

but $v_{rms,He} = \sqrt{\frac{3}{2}}v_{p,He}$
and $v_{rms,U} = \sqrt{\frac{3}{2}}v_{p,U}$
so:
$\sqrt{\frac{3}{2}}v_{p,He} = \sqrt{\frac{3}{2}}v_{p,U}$
which can be simplified to:
$v_{p,He} = v_{p,U}$

Now substituting from the Maxwell-Boltzmann equation for $v_p$ for both He and $\ce{UF6}$ we have:

$\sqrt{\frac{2RT_{He}}{M_{He}} } =  \sqrt{\frac{2RT_U}{M_U}}$

simplying
$\frac{T_{He}}{M_{He}} = \frac{T_{U}}{M_{U}}$

rearranging:
$\frac{T_{U}}{T_{He}} = \frac{M_{U}}{M_{He}} = \frac{352.02}{4.003} = 87.94$

Superfluous information is that $\ce{UF6}$ decomposes about $1800^o \text{ K}$ and is a solid below about $330^o \text{ K}$. Helium liquefies about $4.2^o \text{ K}$. So this only works if Helium is in range $4.2^o \text{ K}$ to $18.4^o \text{ K}$ which would mean that the uranium hexfluoride is between $369^o \text{ K}$ to $1800^o \text{ K}$.


How many carbons does an alkane need to have an enantiomer? How many carbon atoms does an alkane need before it is capable of existing in enantiomeric forms?

If we allow our substituents to be isotopes, then 2 is all that is really necessary:

$\hspace{6cm}$

If we exclude such compounds from consideration, then cyclopropanes will give you smallest number of carbons at 5:

$\hspace{5.5cm}$

As a side note, for a non-alkane, only 1 carbon is necessary:

$\hspace{5.7cm}$


Vapors of Aqueous solutions I'm dealing with a common photoresist developer TMAH (tetramethylammonium hydroxide) and attempting to design a vapor sensor for a sealed chamber. Now I havent been able to confirm what exactly is evaporating from the solution causing it to smell.
I assumed that tetramethylammonium hydroxide would equilibrate with trimethylamine and methanol, and the liberated non-ionized species would evaporate before shifting back into their respective ions.
My question is whether those species as individual molecules would evaporate, or do droplets of the ionized solution evaporate together? And if droplets of water/TMAH solution evaporate, would water absorb energy, throwing off the reading made by a photoionization detector?

You may want to read the Wikipedia article on TMAH.  While it does decompose, it may not produce the methanol you are expecting.  Also, heat is also involved in this process, so you may need to heat and, in the process, vaporize much of the water to get this decomposition to go.

TMAH, in common with many other TMA salts containing simple anions, decomposes on heating into trimethylamine.[1] Dimethyl ether is a major decomposition product rather than methanol.[2] The idealized equation is:

$\ce{2 NMe4+OH− → 2 NMe3 + MeOMe + H2O}$

Such a rearrangement suggests an irreversible reaction.  The Wikipedia article also indicates that TMAH is not obtained by combining trimethylamine with methanol anyway.  Instead, a metathesis between a tetramethylammonium salt such as the chloride and an alkali hydroxide is used.
Cited References:
1.
Lawson, A. T.; Collie, N., "The action of heat on the salts of tetramethylammonium". J. Chem. Soc., Trans., 1888,53, 624-636. doi:10.1039/CT8885300624.
2.
Musker, W. Kenneth. (1964). "A Reinvestigation of the Pyrolysis of Tetramethylammonium Hydroxide". Journal of the American Chemical Society. 86 (5): 960–961. doi:10.1021/ja01059a070.


Are refractive or density based Brix meters better for use with alcoholic drinks? Assume that somebody wants to measure the sugar content of a liqueur. It seems that there are two types of Brix meters for it, ones which use the solution density and ones which use refraction of light.

If we have a meter intended for measuring the content of a sugar-in-water solution (and assume that we are talking pure sucrose here, not other sugars), but are using it for measuring a solution which has, beside sugar and water, also 15 to 35 percent alcohol and also unknown organic (colored or transparent) substances dissolved:

which type will give less error, refractional or density-based?
how large will the error be? For example, if the true value is 20 g sugar per 100 ml solution, how far are the values delivered by the meter likely to stray from those 20 grams (I am looking for a back of an envelope estimation, not for an exact interval)?

Once I got going, this turned out to be a lengthy answer. So here's the summary: the alcohol will introduce a significant error in both methods. The density measurement will probably be closer to the actual sugar content. Th density measurement will underestimate the sugar, and the refractive index will overestimate the sugar.

Both the refractive index and the density-based methods work for measuring sugar content because adding sugar to water raises the refractive index and the density in amounts proportional to the concentration of the added sugar. As you and others noted in that thread, ethanol interferes with that measurement because it also affects the density and refractive index of the resulting solution. So how large is that effect?

To estimate this effect, I am going to restrict myself to the 15-35% alcohol (by volume) range you suggested. The charts I found for the density and refractive index of water/ethanol mixtures gives the ratio by weight; your range converts to 12-28% by weight. Pure water has a density of 0.99823 g/mL, and a refractive index of 1.3330. 12% ethanol has a density of 0.97910 g/mL, and a refractive index of 1.3410. 28% ethanol has a density of 0.95710 and a refractive index of 1.3524. Note that adding ethanol increases the refractive index, but decreases the density. I'll come back to that at the end.

This page gives a table of the index of refraction and density of sucrose solutions in water, as a function of the sucrose weight fraction. The 20 g sucrose per 100 mL solution is approximately 20% by weight, so that will have a refractive index of 1.3639, and a density of 1.0810 g/mL.

The density and refractive index changes for both sucrose and ethanol are pretty linear over the range you're interested in, if you take them independently. Assuming that the sucrose effects and the ethanol effects are independent (NB: this may not be a good assumption), the refractive index measurement will be higher by 0.01-0.02, which corresponds to an overestimate of 7-14% sucrose. The density measurement will be lower by an amount 0.02-0.04 g/mL, which corresponds to an underestimate of 5-10% sucrose. So it looks like the hydrometer measurement will be a bit better, but the two methods will be off in different directions. It's a noticeable effect: depending on the method and the alcohol content, you might be off by somewhere between a quarter and half of your intended sugar content.

There is one neat thing you could do to take advantage of the fact that the alcohol moves the density in the opposite direction to the sugar. If you had a bottle of liquor with unknown alcohol content and unknown sugar content, you could take both the density and the refractive index measurement. Again, assuming the four effects are linear, and the sugar effects are independent of the alcohol, you could write two equations for the refractive index $RI$ and density $\rho$ as functions of the alcohol and sugar: $RI = Ax + By, \rho = Cx + Dy$. $x$ and $y$ are your alcohol and sugar content in whatever units you prefer, and $A, B, C,$ and $D$ are the empirical parameters describing each of the four effects. If you take a measurement for $RI$ and for $\rho$, you can solve for $x$ and $y$.

PS: to answer one question suggested by the comments of the [cooking.se] question you linked, density-based brix meters (I'm used to seeing them called hydrometers) are typically used for home brewing beer and wine because the only source of alcohol is the fermenting sugar. You take a density measurement before fermentation begins to get a baseline. Then, at any point after fermentation begins, the difference between the current density and the baseline density is directly proportional to the alcohol content. I'm sure the same reasoning applies to the refractive index method, but I've never used that for beer-brewing (I've only used it for my real research).  See Cutaia, A.J., Reid, A-J. and R.A. Speers,  2009. Examination of the relationships between Original, Real and Apparent Extracts, and Alcohol in Pilot Plant and Commercially Produced Beers. J. Inst. Brew. 115:318-327.


Why did permanent marker come off of glass after rubbing it with my fingers? No, my fingers don't look stained but I accidentally made a stroke of permanent marker on glass and I was looking for nail polish remover to use, couldn't find it. So I started rubbing the stroke of permanent marker with my fingers and it started coming off. I was able to get it off with 3 fingers, with the thumb getting most of it off.

I wonder, is it because permanent marker dissolves in skin oil that I was able to get it off? Or was it the small amount of sweat? Or both? Or was it just that I was rubbing my fingers on the permanent marker and the shear force got it off?

I mean yes, this was just a few minutes after the accidental stroke that I got it off with my fingers but that would have been plenty of time for it to dry, I think.

Glass is a very smooth surface and therefore things adhere to it less effectively such as the ink in your permanent marker. Maybe your marker is not of a very good quality or maybe you didnt leave it long enough for the ink to stabilize on the surface.

Why dont you conduct an experiment to try your various hypotheses by varying one factor while keeping the rest constant? For exaple:

1) Try different brands of markers on the same glass surface leaving them to dry for the same time and then rubbing them in the same way.
2) Use the same marker but leave it to dry for different amount of time and then rub it in the same way each time.
3) Use the same marker on different surfaces and rub in the same way.
4) Try with different amounts of dilute saline solution to simulate sweat or cooking oil to simulate skin oil etc.
etc.

Or even better: Try at glass surfaces that have been etched with different sandpaper grits.

Your experiment is reproducible though, i tried as well and small amounts come off easily.


ᴅ-amphetamine, ʟ-amphetamine The following structural formula (source: Wikimedia) is labelled ᴅ-amphetamine:

A Fischer projection of this molecule would contain, I think, the following line:
$$\ce{H2N-C-H},$$
and I would have labelled the molecule therefore as ʟ-amphetamine.
Why am I wrong? Why is it called ᴅ-amphetamine?

The L and D nomenclature is directly transform from its origin, D- and L-phenylalanine. The following scheme summarize the transformation:

The blue box in right-hand corner explains how L- and D-$\alpha$-amino acids get their names using Fischer projection (this is similar to D- and L-monosaccharides, replacing $\ce{OH}$ group by $\ce{NH2}$ group). If you reduced the carboxylate group of amino acid to a methyl group, the optical rotation of $\alpha$-carbon change from $(S)$ to $(R)$ because second highest priority of $\ce{COOH}$ group to third highest ($\ce{CH3}$ group) based on the size of the $\ce{R}$ group (see general formula of amino acid in top right corner). For example, when $\ce{R=CH2Ph}$ group, the amino acid is phenylalanine, which has $(S)$-rotation, but reduction of caboxylate group to methyl make the rotation $(R)$. However, the product $(R)$-amphetamine still carries the original L-nomenclature (see examples in bottom red box).
Note: The conversion of phenylalanine to amphetamine in several steps were achieved in beginning of 21st century (Ref.1).This is the ultimate enanthimerically pure version.
Reference:

Dominick A. Quagliato, Patrick M. Andrae, Edward M. Matelan, “Efficient Procedure for the Reduction of α-Amino Acids to Enantiomerically Pure α-Methylamines,” J. Org. Chem. 2000, 65(16), 5037–5042 (https://doi.org/10.1021/jo000242h).


How to calculate the solubility product for PbS from galvanic potential? Knowing that at 25 °C the following galvanic cell:
  $$\ce{Pb~|~Pb(NO_3)_2~1M~||~PbS~saturated~|~Pb}$$ shows an $\mathrm{EMF} =0.413~\mathrm{V}$, find the $K_\mathrm{sp}$ of $\ce{PbS}$.

My Approach
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the cathode and the right one the anode.
So we have the following semireactions:
\begin{align}
 \ce{Pb^2+ + 2e- &-> Pb}  && \text{(cathode)} \\
 \ce{Pb  &-> Pb^2+ +2e- } && \text{(anode)}
\end{align}

And for the anode we also have $$\ce{PbS <=> Pb^2+ + S^2- }$$ where $\ce{[Pb^2+]} = \ce{[S^2- ]}= \sqrt{K_\mathrm{sp}}$.
So the semicell potentials are:
\begin{align}
E_\text{cathode} &= E^\circ\\
E_\text{anode}   &= E^\circ - \frac{0.059}{2} \log_{10}{[\ce{Pb^2+}]}\\
\end{align}

Thus:
$$0.413~\mathrm{V} = E_\text{cathode} - E_\text{anode} = \frac{0.059}{2} \log_{10}\ce{[Pb^2+]} \Rightarrow \ce{[Pb^2+]} = 10^{14} $$
And: $$K_\mathrm{sp} = \ce{[Pb^2+]}^2 = 10^{28}$$

I'm sure that I'm wrong for a sign but I don't understand where is the error.

The galvanic cell must be written like this:

$$\ce{Pb | PbS saturated || Pb(NO3)2 \pu{1 M} | Pb}$$

in order to have a positive  $\mathrm{EMF} = \pu{0.413 V}$.

Your approach is right with a slight error.
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the anode and the right one the cathode.
So, we have the following semireactions:

$$
\begin{align}
 \ce{Pb^2+ + 2 e^- &-> Pb} &\quad &\text{(cathode)} \\
 \ce{Pb &-> Pb^2+ + 2 e-} &\quad &\text{(anode)}
\end{align}
$$

And for the anode we also have

$$\ce{PbS <=> Pb^2+ + S^2-}$$

where $[\ce{Pb^2+}] = [\ce{S^2-}] = \sqrt{K_\mathrm{sp}}$.
So the semicell potentials are:

$$
\begin{align}
E_\text{cathode} &= E^\circ\\
E_\text{anode}   &= E^\circ + \frac{0.059}{2}\log_{10}[\ce{Pb^2+}]\\
\end{align}
$$

Thus:

$$\pu{0.413 V} = E_\text{cathode} - E_\text{anode} = -\frac{0.059}{2}\log_{10}[\ce{Pb^2+}] \implies [\ce{Pb^2+}] = 10^{-14}$$

And

$$K_\mathrm{sp} = [\ce{Pb^2+}]^2 = 10^{-28}$$


How does boiling water affect its hardness? Does boiling water reduce its hardness, or on the contrary, increase it?

It seems there are ambivalent phenomena at play here:

boiling will evaporate pure water, thus increasing the concentration of minerals including calcium, thereby increasing hardness
boiling will remove $\ce{CO2}$, making apparently $\ce{CaCO3}$ precipitate, thereby reducing hardness

This is based on rather uncertain sources, as I didn't find any better, and maybe I missed other aspects.

I would appreciate any clarification, but I understand that the answer might depend on the context and type of water considered.

You may need to think outside chemistry a little bit and pay a little bit attention to your everyday life. This is what happened when you boil water in a container:

Most hardness in water, calcium/magnesium, will precipitate onto the hot surface and is removed from water. Now if you can get more something by removing it, is it against the very fundamentals of science?


What can I use as a sacrificial electrode to stop the oxidation of aluminium? For a solution of chromimum phosphate when putting in to solution aluminum there is removal of the aluminum.

It is possible to prevent a galvanic cell, stop the etching of the aluminum, by putting in a sacrificial anode.

In this situation I have access to metals like tantalum, titanium, tungsten and copper.

Which one of these would stop the oxidation of the aluminum.

I have slightly familiar with electronegativity and I understand that aluminum is 1.6.  I just do not know if you want some thing with a higher or lower electronegativity.

Thanks

I think that you are interested in the corrosion protection of aluminum. You have basically two choices depending on your experiment setup.

You can use an impressed current protection or a sacrificial anode method. The sacrificial anode method uses a more reactive metal (less noble) in contact with the metal requiring protecting. The less noble metal reacts first within corrosive environment.

The impressed current protection uses electric current to charge the metal requiring protecting with excess electrons. It changes the electrode potential in the negative direction until the metal is not reacting.

If you want to use the sacrificial anode method look in the table where you are looking for metal that is positioned higher than your aluminium (is more electronegative - so non of the mentioned metals).

More useful information can be found here CORROSION PROTECTION OF METALS
 or here metals.about.


Poly(3-hexyltiophene-2,5-diyl) (P3HT) reactions with inorganic salts in aqueous solution I would need for my research at least as a starting point to know what kind of reactions occur at the interface of P3HT and an electrolyte. The electrolyte is simple $\ce{KCl}$ solution, maybe with $\ce{HCl}$ or $\ce{KOH}$ added. The reactions of polytiophene with the same type of electrolyte will do, if there is no information on P3HT. The two molecules don't seem to be that much different to me from the point of view of electrolytic reactions.

What I really need is information about reaction kinetics, at least in the form of some reaction constants, but simple stochiometric equations are still better than nothing at all, so I'm interested in those, too. I have been searching for some time now, but I'm not a chemist, and I don't really know where to look to find something like this.

I'll point to a very recent paper on oligothiophenes for solution-gated FET sensing that suggests there are no chemical reactions. (This article popped up in my feed - there are many others on solution-gating organic electronics.)

$\alpha,\omega$-dihexyl-sexithiophene thin films for solution-gated organic field-effect transistors. Appl. Phys. Lett. 108, 073301 (2016)

The authors look at many factors influencing the device performance:

  Finally, excellent transistor stability is confirmed by continuously operating the device over a period of several days, which is a consequence of the low threshold voltage of DH6T-based SGOFETs. Altogether, our results demonstrate the feasibility of high performance and highly stable organic semiconductor devices for chemical or biochemical applications.

The devices in this article use KCl and phosphate buffer. They vary the pH and ionic strength.

I haven't seen anyone suggest chemical reactions, and I'm not sure how it would occur. Certainly the oligo- or polythiophene could degrade over time, but most articles argue these are stable for long-term use.

The authors do note a small drift on the time scale of days:

  both changes may possibly be attributed to the passivation of trap states at
  the semiconductor-electrolyte interface by water molecules penetrating into the thin film

I can't speak to the drift you see, but I'd make sure that:

You aren't seeing evaporation of solvent over time (e.g., from device heating) that's changing your electrolyte concentration.
You use a proper reference electrode
You try removing water via vacuum and repeat the experiment (i.e., you drive off any water penetrating the semiconductor film)

Certainly organic semiconductors are porous, and while mostly hydrophobic, it's not unreasonable to expect that after days of use, some water may migrate into the film.


Why do only chiral molecules rotate the plane of polarized light and how do they rotate it? I have learnt that only chiral or unsymmetrical molecules can rotate the plane of linearly polarized light. But, why is it so?

And how can molecules rotate the plane or what does it actually mean by 'rotating'?
Does it mean that the electric field vectors are tilted or turned left or right depending on the type of isomer?

See also: How do Optically Active Compounds Rotate Plane Polarized Light?

This is because optical rotation is a chiral phenomenon.

Take a molecule, and draw arrows depicting the polarization of incoming and outgoing light on it. Even if the molecule is achiral, the molecule with the arrows is chiral.

Chirality can't spring out of nowhere, it can only arise from a chiral process.

Also, see my answer here. Since molecules will exist in all rotations in a given fluid, for an achiral molecule the mirror image molecules cancel each other out.


Hydrogenolysis of benzylic C-N bond Why does N,N-dimethyl-1-phenylmethanamine (the left molecule in the picture) undergo hydrogenolysis of the C–N bond readily while 2-methyl-1,2,3,4-tetrahydroisoquinoline (right) doesn't react at all?

Hydrogenolysis typically involves a metal surface ($\ce{Pd, Pt}$, etc.) and hydrogen gas. In the reaction a weak bond, that is, a bond that is either strained or a bond that can generate a stabilized radical is broken. In your examples there is no strain, but a resonance stabilized benzylic radical can be generated in one case.

In the benzylic amine pictured on the left, the benzylic carbon can easily adopt a geometry such that when the p-orbital on the benzylic carbon is generated it is already aligned with the p-orbitals in the aromatic ring and is therefore stabilized while it develops.

The molecule on the right is different. I built a model and as the C-N bond starts to break and the p-orbital starts to form, the developing p-orbital is twisted 45-60° away from alignment with the p-orbitals in the aromatic pi system - it is not effectively stabilized due to conformational constraints imposed by the ring system.

Therefore this latter molecule will not undergo hydrogenolysis of the C-N bond as readily as the molecule on the left.

Here are some photos of my model:

the molecule - note the blue $\ce{N-CH3}$ nitrogen as a point of reference

conformation A - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is axial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted about (90-30) 60° away from being aligned with the p-orbitals in the aromatic ring - poor overlap potential

conformation B - looking down the benzylic carbon - aromatic carbon bond on the left side of the photo, the N-methyl group is equatorial and in the foreground, the aromatic ring is planar and in the background, the $\ce{C-N}$ bond is tilted about 30-45° from the plane of the benzene ring, therefore the (soon to be) p-orbital is tilted at least (90-45) 45° away from being aligned with the p-orbitals in the aromatic ring - poor overlap potential


How does sulfuric acid dehydrate sugars? I know sugars have $\ce{-OH}$ groups, and that sulfuric acid is a strong acid. What I'm failing to see is how exactly sulfuric acid dehydrates sugars. Does it protonate the hydroxyl groups, making them good leaving groups? What about the case of sucrose?

For example, how are the oxygens "taken care of" by the sulfuric acid? Are they doubly protonated? Is there some established mechanism for the dehydration of sugars by sulfuric acid, because I can't find any - I just find general equations and I've looked on Google and through textbooks.

After some further research I learned that the gas, sulfur dioxide, is evolved. This suggests to me that this reaction is also redox in nature. So is this reaction both an acid-base reaction and a redox reaction? Sulfur in sulfuric acid has an oxidation state of +6; sulfur in sulfur dioxide has an oxidation state of +4. Plus the elemental carbon formed has an oxidation state of 0 while in sucrose, carbon has various oxidation states.

$\ce{H2SO4}$ protonates the alcohol to make $\ce{R-OH2+}$ and $\ce{HSO4-}$, the $\ce{HSO4-}$ then takes a hydrogen from the carbon atom adjacent to the alcohol, yielding an alkene and water.  Alcohol dehydration reactions using acid are just elimination reactions.

Additionally, at temperatures above 340 degrees Celsius, sulfur trioxide exists in equilibrium with sulfuric acid in significant concentration.  The presence of sulfur trioxide explains the presence of sulfur dioxide  (via dissociation), or some redox chemistry occurring with the sugar.


What are the products of the Grignard reaction with methyl-magnesium iodide? I want to react my ketone with methylmagnesium iodide (the Grignard reagent). I then need to use a dilute acid solution to protonate the $\ce{MgI-O-R'}$ product to produce my alcohol: $\ce{H-O-R'}$

Now I was trying to figure out what acid to use. The first thing that came to mind was sulfuric acid. What would the products be?

In general the reaction should look like: $$\ce{R'=O + XMgI + H-A -> X-R'-OH + MgI+ + A-}$$

Now if I use sulfuric acid the conjugate base reacted with the Grignard products is something like $\ce{MgIHSO4}$. Is this correct?

This molecule doesn't look very stable so my thinking is that the $\ce{Mg}$ donates an electron to $\ce{I}$ in $\ce{MgI+}$ to form elemental iodine and $\ce{Mg^2+}$, then this reacts with $\ce{HSO4}$ to produce $\ce{Mg(HSO4)2}$ and $\ce{MgSO4}$.

Your reaction scheme doesn't make much sense. $\ce {R'=O}$ is not a ketone. Is $\ce X$ supposed to be $\ce Me$? This is confusing since $\ce X$ is most commonly used as the symbol of a halogen.
I would rewrite as:
$$\ce {R'RC=O + MeMgI +H_2O-> R'RC(Me)OH + Mg(OH)I}$$
$\ce {R'(R)C=O}$ is implied.

Make sure you use very dilute aqueous sulphuric acid since your product is a tertiary alcohol. I would rather use saturated aq. $\ce {NH_4Cl}$. Even water will do but I suspect that the work-up might not be as smooth (a less water soluble magnesium salt will be formed). There will be no iodine formed and in any case I don't find the idea of recycling iodine cost effective. It will cost you more to isolate iodine that can be pure enough to work in the notoriously capricious Grignard reagent formation. In fact I would suggest you to buy a commercially available solution of $\ce {MeMgX \ (X=Cl, Br, I)}$ since preparing a Grignard can be very challenging and you will also have to handle toxic and volatile $\ce {MeI}$, have access to dry ether/THF etc. I assume you follow a detailed experimental procedure from the literature.


Dipolar moment between HI and HF Which between $\ce{HI}$ and $\ce{HF}$ has a greater dipolar moment?
I think it is $\ce{HI}$ because the atomic radius of $\ce{I}$ is greater.

The dipole moment of $\ce{HF}$ is greater than $\ce{HI}$:

$\ce{HF} = \pu{1.91 D}$
$\ce{HI} = \pu{0.42 D}$

This is due to high electronegativity of fluorine. Hence from $\mu = \vec{q} \cdot \vec{d}$, the charge of fluorine is larger than iodine, but the bond length changes only a small amount.


What is the yellow precipitate formed from reacting diamminsilver fluoride and potassium iodide? These two compounds, silver diamine fluoride and potassium iodide (actually a solution called Lugol’s solution), are being used together in dentistry.

When combined they form a yellow precipitate. I cannot find any articles that state specifically what is formed. I would like to know what the yellow precipitate is and the steps involved in the reaction of the two chemicals. The equation is:
$$\ce{Ag(NH3)2F + ssKI -> ?}$$

The yellow precipitate that forms is silver(I) iodide, $\ce{AgI}$. It is very, very insoluble and hence will precipitate immediately if silver ions and iodide ions meet in solution.

In fact, all three common halides other than fluoride form insoluble salts with silver: silver(I) chloride is a white precipitate, silver(I) bromide is somewhere between white and yellow and silver(I) iodide is yellow. These precipitates are so characteristic that they were historically (and still in qualitative inorganic analysis lab courses) used to identify the different halide ions in solution. To further improve the distinction, silver(I) chloride can be dissolved in dilute solutions of ammonia while silver(I) bromide needs concentrated ones. Silver(I) iodide does not dissolve in ammonia.

$$\begin{align}\ce{Ag+ (aq) + Cl- (aq) -> AgCl v &->[NH3 (dil)] [Ag(NH3)2]+ (aq) + Cl- (aq)}\tag{1}\\
\ce{Ag+ (aq) + Br- (aq) -> AgBr v &->[NH3 (conc)] [Ag(NH3)2]+ (aq) + Br- (aq)}\tag{2}\\
\ce{Ag+ (aq) + I- (aq) -> AgI v &->[NH3] \text{no reaction}}\tag{3}\end{align}$$

I assume that the original silver solution was the diammine complex to prevent silver chloride from precipitating as soon as chloride ions are added (due to not perfectly deionised water being used).


How do I type a simple chemical equation in Microsoft Word? How do I type a simple chemical equation in Microsoft Word? I can do subscripts, but long arrows are more difficult. I can't get them to align with the text.

Also, I can't figure out how to put a delta above the arrow for heat. I have tried the Chemistry add-on from Microsoft, but that does not seem to help with equations.

If you are using MS Word 2007 or newer, use the equation feature. It is designed for math but works okay for chemistry.

Go to the insert tab. (For shortcut you can press Alt+= sight together)

Click on the equation button on the far right.

Type in your equation. Use the buttons in the ribbon to do superscripts and subscripts. Alternatively you can use _ for subscript and ^ for superscript.  The default is to have letters italicized (as variables), so you will want to fix that.

There are also shortcut commands to render most the common things you want. For example, underscore _ creates a subscript and a caret ^ creates a superscript Shortcut for typing subscript and superscript in MS Word 2007|2010|2013|2016 and office 365  . You have access to a wide range of arrows from a pull-down menu, but -> will give you a simple right arrow (although it is not very long). This feature on Word will also accept some (but not all) tex commands for formatting equations.

To get a long arrow, click on the operator but and choose the arrow with the word "yields" written over it under common operator structures. For up arrow and down arrow showing gas liberation and precipitation use \uparrow or \downarrow followed by space Shortcut for typing arrows of chemical equation in Word 2007 and above.

Click on the word "yields" and replace it with as many spaces as you need to create an arrow of whatever length you want. Shortcut for other types of arrows is.

Finally, finish your equation.

If you need to type above or below arrow just type "\above(text above arrow goes here)[space]".Similarly tying below arrow just type \below(test below goes here)[space]". How to type chemical equation and arrows in Word 2007 and above.

For older versions of MS Word, go to the insert menu and click on the equation, which launches the Equation Editor Program (you can also find this program on your computer by searching for eqnedt.exe), which gives you the same ability to create equations.


How does physical adsorption of mixed gas occur on solid? (eg I mean how the gas molecules of compete each other for a slot on solid surface) I am working on sorption phenomenon of gases (including inert gas) on coal seam surface.

For a mixture of 2 gases at a certain pressure, carbon dioxide and methane, for instance, $\ce{CO2}$ is definitely adsorbed quicker and more comparing to methane. How come? I try to explain by Langmuir Isotherm and IAS model but it doesn't help much. For Langmuir isotherm, the affinity of gases to solid is determined by "b=1/PL" which comes from experiment. Langmuir isotherm assumes that all gas have equal access to the solid surface; however, this is untrue in reality.

Can you help me explain it in the thermodynamics point of view or molecular forces please? ($\ce{CH4}$, $\ce{CO2}$ are both non-polar gases)

Your supposition that gases do not have equal access to teh surface does not seem to me to be correct.

At the surface  physisorption and chemisorption can occur. The former is due to inter-molecular interaction with the gas molecule and the surface molecules so clearly will depend on the nature of both of them. In chemisorption a chemical bond is formed. The physisorption occurs at longer range and has a smaller potential well (as measured by potential energy vs distance from surface) than chemisorption. (See Atkins & DePaula 'Physical Chemistry' for a figure.)

The gases approach the surface at random and the number of collisions of each will be in proportion to their partial pressure. What happens on collision depends, as mentioned above, on the strength of the intermolecular interaction of gas and surface and so depends on what they are. At higher temperature the collision energy is greater than at lower ones and so one would expect less physisorption as the collision energy will be greater than that of most available physisorption sites.

It is also possible for multiple layers to be formed (as in BET isotherm) and for a physisorbed molecule to diffuse around on the surface until it leaves again (by random thermal activation) of finds a more stable site or becomes chemisorbed.

As to your particular case we can only surmise but $\ce{CO2}$ is far more polarisable than methane so should have the larger dispersion forces so larger induced dipole - induced dipole intermolecular interaction and so preferentially occupy the surface. Both gases have transient dipoles due to molecular vibrations of which the $\ce{CO2}$ should be the larger. This may also be important.


Safely storing hydrogen I have built a solar panel which is used to add electrodes to $\ce{H2O}$ and $\ce{NaCl}$, evolving $\ce{H2}$ and $\ce{O2}$.

How can I safely store the hydrogen in the long-term?

Hydrogen can't and won't explode as long as it is kept separate from oxygen, unless you're storing it under high pressure.  So as long as your hydrogen container can't break or leak under all sorts of environmental hazards (heat, fire, dropping, collision, degradation, etc.), you should be fine.  Essentially, this question is more about risk management to consider all the possible scenarios of a potential leak.

Ultimately it depends on how and where you will store your hydrogen container, and the potential dangers of that environment.  So storing hydrogen under low pressure within an argon-filled safe would be a lot safer than storing it under high pressure in a mobile oxygen tent.


Do hydrogen atoms not bound to carbon also appear in 1H NMR spectra? I would like to know if hydrogens that are not bound to a carbon (such as in groups $\ce{-OH}$ and $\ce{-NH2}$) would lead to the presence of a singlet peaks in the NMR spectra or not, and if we have to count or omit them in the integration of the spectrum. I thought I was sure about it but lately I noticed that it depends on the spectra.

Yes, with a but.

First the yes: they do appear, albeit at slightly different chemical shifts as, of course, the environment around a proton in $\ce{-CH2\!\;\!-}$ is slightly different to the electronic environment around the proton in $\ce{-NH2}$. The $\ce{N-H}$, $\ce{O-H}$, and $\ce{C-H}$ bonds all have slightly different electronegativies.

Now for the but: the thing that more significantly distinguishes $\ce{N-H}$ and $\ce{O-H}$ from $\ce{C-H}$ is their lability in protic solvents. Exchange of labile protons in the sample molecule with deuterons (i.e. deuterium ions, $\ce{^2H+}$) in the solvent can can remove the proton peak from the spectrum (this exchange is an equilibrium process, but as the solvent is in large excess the equilibrium lies over towards full deuteration of the sample's labile groups. This can actually be useful in identifying peaks: first run a sample in, for example, $\ce{CDCl3}$, then add a couple of drops of $\ce{D2O}$ to your NMR tube, give it a shake, and re-run the spectrum -- your labile protons will have vanished.

Edit 1: Furthermore, as @user1573870 points out in the comments, exchange of protons (exhanging protons for protons, rather than exchanging protons for deuterons),  on "the NMR timescale" can lead to broadening, and the resulting peak can be, while technically present, very difficult to spot. I'd suggest anyone doing NMR do background reading on that, because it's both useful and interesting (and I need to do some, because I'm out of practice with NMR).

Edit 2:To directly answer @Chewie's question whether "we have to count or omit them in the integration of the spectrum", you only count each of the peaks that are present. Using the example of ethanol, as asked in the comments, $\ce{CH3CH2OH}$ should give three peaks:

The $\ce{CH3}-$ peak should integrate to show three protons,
The $\ce{-CH2}-$ peak should integrate to show two protons, and
The $\ce{-OH}$ peak should integrate to show one proton.

If due to exchange with the deuterated solvent the $\ce{-OH}$ peak is not present, or if the $\ce{-OH}$ peak is 'there' but so broad as to be invisible to you (or so broad that it is difficult to integrate accurately), then you would only give the integrals for the other two peaks, and those two peaks should give integrals with a ratio equivalent to their respective number of protons, i.e. 3:2. If the $\ce{-OH}$ peak is not there, its proton does not just get thrown in with one of the other peaks/integrals (unless, sneaky proton, it just happened to have the same integral as another peak, and was hidden behind it, but for something as simple as $\ce{CH3CH2OH}$ that is unlikely).

The punchiest way to say all of thise would be that one does not "integrate spectra"; one integrates peaks.


Is the equation dG = Vdp − SdT valid only for a reversible process? Can it be applied for an irreversible one too? In the derivation, we write:

$$
\begin{align}
G &= H - TS &\quad&\to&\quad \mathrm dG &= \mathrm dH - T\,\mathrm dS - S\,\mathrm dT \\
H &= U + pV &\quad&\to&\quad \mathrm dH &= \mathrm dU + p\,\mathrm dV + V\,\mathrm dp \\
U &= Q + W &\quad&\to&\quad \mathrm dU &= \mathrm dQ - p\,\mathrm dV \\
  & & & & &= T\,\mathrm dS - p\,\mathrm dV
\end{align}
$$

So solving these gives

$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$$

But what we actually write is

$$\mathrm dH = \mathrm dU + p_\mathrm{int}\,\mathrm dV + V\,\mathrm dp$$

And in

$$\mathrm dU = \mathrm dQ - p\,\mathrm dV = T\,\mathrm dS - p_\mathrm{ext}\,\mathrm dV$$

so on solving, it should be

$$\mathrm dH = T\,\mathrm dS - (p_\mathrm{ext} - p_\mathrm{int})\mathrm dV + V\,\mathrm dp$$

And only in reversible process, should it be valid to write

$$p_\mathrm{ext} - p_\mathrm{int} = 0$$

But my teacher argued that because this is a state function, it should be valid for all types of processes. I'm a bit confused about how his claim is valid. Can someone please explain what he means or even if he is correct or not?

The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:

$$\mathrm dw_\mathrm{rev} = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV$$

where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_\mathrm{ext} = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $\mathrm dH$ are not correct.

The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(\mathrm dU = -p\,\mathrm dV + T\,\mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $\mathrm dH = V\,\mathrm dp + T\,\mathrm dS.$

  But my teacher argued that because this is a state function, it should be valid for all types of processes.

The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.

In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.


How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂? When a solution contains $\ce{Hg^2+}$ and $\ce{I-}$, depend on the concentration ratio of $\ce{Hg^2+}$ and $\ce{I-}$, the following reactions can happen:

$$\ce{Hg^2+ + 2 I- ⟶ \underset{insoluble red color solid}{HgI2}}$$

If there is excess $\ce{I-}$, then this red solid will continue to react with excess $\ce{I-}$ to form an colorless soluble complex:

$$\ce{HgI2 + I- ⟶ [HgI3]^-}$$

$$\ce{[HgI3]^- +  I- ⟶ [HgI4]^2-}$$

Is there a way to revert $\ce{[HgI4]^2-}$ back to $\ce{HgI2}$?

As noted in the comments, to attain mercuric iodide, you need to maintain a balance between the concentration of $\ce{Hg^2+}$ and iodide ion. If there is slight excess of iodide ion, it will form a complex (much like $\ce{KI3}$ from $\ce{KI}$). The complex is popularly known as Nessler's reagent and is prepared pretty much in this way.

Another solution is to decompose the complex to revert back into original products. But I won't recommend this solution. Though, it is expected to give the original reactants, occasionally, decomposition reactions give rise to undesired side products and may release toxic fumes of mercury, iodine, iodide, potassium oxide($\ce{K2O)}$ etc. Do not try to heat the complex. Instead opt for the 1st solution.


Reference request - Nomenclature of organophosphorus compounds? Can someone please provide me a good reference for the nomenclature of organophosphorus compounds such as pesticides and chemical warfare agents? Something from the IUPAC would be great.

Thanks.

The usual substitutive nomenclature for organic compounds was extended to various other elements including phosphorus. The current rules for name construction of organic compounds containing phosphorus are included in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). These recommendations cover typical organophosphorus compounds that are used as pesticides and chemical warfare agents. However, they do not include the nomenclature for organometallic compounds, coordination compounds, polymers, and some natural products.


Electrochemistry - Understanding Diffusion Potential I have a big problem trying to wrap my head around this for several hours and I cannot seem to make sense of it.
I have to link this scheme with the 3 colored potential profiles on the bottom of the picture. Considering the electrolyte AK, which splits into A- and K+, that it is 10x more concentrated in compartment I (left) and that the 3 different cases are the assumed:

1) Mobility of A- >>> than mobility of K+
2) Mobility of K+ >>> than mobility of A-
3) Both ions have an equal mobility

Now, I know that in the 1) case, anions will diffuse through the membrane and work toward balancing their concentration. However, this will generate a potential difference because K+ cannot move and compartiment I will be left with a higher potential. Now, I can imagine that if we set our reference (left electrode) at an arbitrary number, let's say 0, then we can assume compartiment II being negative potential in respect to I. This leads to the constant value of the Nernst eq. for same metal electrodes and electrolyte with 10x concentration difference of 59 mV. Since II is considered lower potential VS I in case n°1, should I assume that the potential difference at the membrane's level is -59 mV (green profile)? And another +59 mV between the electrodes because of Nernst equation, which would bring to a final measured potential of 0V. Is this a legit reasoning?

Thank you in advance.

Let's discuss the problem in a qualitative way. In such a cell, the metal will get dissolved in the right-hand side of the cell, where the concentration is low, so that $[M^{z+}]$ increases in this compartment, which becomes the anode. In the left-hand side, the ions $[M^{z+}]$ are discharged  and deposited as a metal layer on the electrode, which is the cathode.

The only problem is what is going on in the solutions. If the anion $A^-$ has a high mobility, it will flow from left to right through the membrane to compensate the chang of concentration of the metal ion. And after some time, the two solutions will have the same intermediate concentrations, and the cell will not deliver any current.

But if the ion $A^-$ has a low mobility, the cation $[M^{z+}]$ has a higher mobility and will easily cross the membrane from right to left. It will take a much longer time for the concentrations to be equalized in the two compartments. The reason is that a much smaller amount of ions $A^-$ will cross the membrane from left to right. But at the end, the two solutions will have the same concentrations.


n-factor for ammonium nitrite? The given reaction is $$\ce{NH4NO2 \rightarrow N2  +  2H2O}$$

The way I did it was
$$\ce{NH4+ \rightarrow \frac{1}{2} N2 + 4H+ + 3e-}$$
and
$$\ce{NO2- +3e- +4H+ \rightarrow \frac{1}{2} N2 + 2H2O}$$
Therefore n factor is 3

But the book I am referring to says 6. Can someone please help?

"n factor" of oxidizing or reducing agent = total number of electrons transferred per a mole of the reactant.

When more than one element in a substance is oxidized/reduced, the n-factor of the substance is taken as the sum of the n-factors of the individual elements of the substance.

This is the case here.You have  ammonium ion that undergoes oxidation reaction (n-factor= 3) and  nitrite ion that undergoes reduction reaction (n-factor= 3).
So, the n-factor of ammonium nitrite is 6.


Why did roasting of copper(II) sulfate pentahydrate hurt my nose? I want to use copper(II) sulfate as rocket propellant additive in hope that it will alter the flame colour. Because water is not welcome in rocket fuel, I decided to dehydrate the crystals.

First I tried stainless steel mess tin, but that produced some smoke warning me that it's reacting. So I switched to aluminium mess tin, that one didn't smoke.

But still some fumes, presumably steam was produced. I continued roasting. Almost at the end, I noticed that my nose hurts in the inside a little.

I suppose copper sulfate has got in there. Do you think I risked badly, or is it just weak irritation? How can I avoid it next time, apart of using fumehood?

Roasting of copper(II) sulfate at elevated temperatures generates sulfur trioxide, which would hurt your nose.

$$\begin{align}
\ce{$\underset{\text{(blue)}}{\ce{CuSO4.5H2O}}$ &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][100\ \mathrm{^\circ C}] CuSO4.H2O + 4H2O}\\[6pt]
\ce{CuSO4.H2O &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][\gt\ 200\ \mathrm{^\circ C}] $\underset{\text{(white)}}{\ce{CuSO4}}$ + H2O}\\[6pt]
\ce{CuSO4 &->[\hphantom{\gt\ 200\ \mathrm{^\circ C}}][] CuO + SO3}
\end{align}$$


How are turbostratic graphite phases formed? Turbostratic graphite is graphite in which there is quenched rotational disalignment between adjacent graphene sheets, i.e. one sheet is rotated with respect to its neighbor. I suppose this could be considered a crystallographic defect of sorts. How do these turbostratic layers form, and can they possess long range order, such as glide or helical symmetry (e.g. similar to a twisted nematic phase)?

"High-temperature phase transformation and low friction behaviour in highly disordered turbostratic graphite," doi:10.1088/0022-3727/46/39/395305

"Phase Transformation Mechanism of Graphite-Turbostratic Graphite in the Course of Mechanical Grinding," CHEM. RES. CHINESE U. 19(2), 216 (2003)

http://www.onxlti.com/product-divisions/contract-manufacturing-products/on-x-pyrolytic-carbon/


Half cell reactions for oxidation of water by acidified solution of potassium dichromate From Chemguide:

I don't understand the first equation. The chromium atoms in the dichromate ion receive 6 electrons, and their oxidation state declines from 6+ to 3+. But where do the 14 hydrogen cations get their 14 electrons in order to form water with the seven oxygens?

Look closely at the oxidation state of all atoms in the equation.
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) + 6e- <=> 2Cr^3+ (aq) + 7 H2O (l)}$$

The oxidation state in chromium changes from +6 to +3. This is what the electrons in the equation are necessary for. Oxygen retains its oxidation state of -2 in both sides of the equation, as well as hydrogen which remains its +1 oxidation state.

In the other equation
$$\ce{O2 (g) + 4H+ (aq) +4e- <=> 2H2O (l)}$$

the oxygen gets reduced by changing its oxidation state form 0 to -2. Again for hydrogen, the oxidation state of +1 is retained.

The total equation for this process would therefore be
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) <=> 4Cr^3+ (aq) + 8H2O (l) + 3O2}.$$

From the standard potentials you can see, that this reaction may happen, it does not tell you anything about the conditions under which this reaction will happen. There are a lot more factors to consider.


Maximum (Z) to (E) ratio in the photoisomerization of azobenzene? Considering the photoswitching of azobenzene, I was told the following argument: Considering that the photoexcited state of azobenzene has a about 50% to 50% chance of yielding either the (E)- or (Z)-isomer, the maximum (Z)/(E) ratio achievable should be 1:1.

However, I have seen UV-VIS spectra where the (Z)-isomer seems to be in excess. So I figured that the above argument must be wrong. I would analyze the situation in the following way:

In this picture, the fact that the photoexcited state has about equal probability to yield (E) or (Z), i.e. $k_{-1} = k_2$.

So wouldn't the overall (Z) to (E) ratio be given by the expression
$\frac{k_1\cdot k_2}{k_{-1}\cdot k_{-2}}$?

I would then argue that in many cases, the cis isomer exhibits lower absorption at the wavelength of excitation, meaning that $k_{-2} << k_{1}$, which would explain that in many cases, an excess of the cis isomer is obtained.
By this reasoning, the initial argument would be completely wrong, as in principle an infinite excess of the cis isomer could be obtained if $k_{-2}$ is sufficiently smaller than $k_{1}$.

Is this analysis correct or am I overlooking something?

Your question is not completely clear to me, so I am answering based on thinking that you are asking about why, after photoisomerization, the system can relax with roughly equal probability to either the E or Z minimum.

Strangely I know something about the photochemistry of azobenzene. One overarching theme is that azobenzene molecules undergo very complicated isomerization processes which are sensitive both to how you substitute them and to the solvent in which the experiment is done. Additionally, the photoisomerization and thermal isomerization processes don't necessarily share anything in common. There is a pretty in-depth interesting literature on different types of azobenzene molecules discussing the relative roles of inversion and rotation relaxation pathways, solvent effects, and so on... I just bring this up in case you are interested to read more.

The case you describe, I think has a relatively straightforward explanation actually. In ref. [1], a computational study is done on azobenzene to find which pathway, rotation or inversion, is dominanant upon photoisomerization of azobenzene. To clarify, rotation means rotation about the CNNC dihedral, and inversion means moving one of the rings along the CNC angle.

The paper shows that there is a conical intersection along the rotation pathway which is only 2 kcal/mol above the $\ce{S1}$ excited state minimum. Thus, the molecule will follow this conical intersection upon relaxation. Because the molecule is already above both states, and the molecule can essentially end up going toward either the E or Z minimum from this conical intersection, to first order, the two should end up equally populated on relaxation. Of course, from there, one would expect to move back to a Boltzmann distribution of the two states thermally.

[1]: Cembran, A., Bernardi, F., Garavelli, M., Gagliardi, L., & Orlandi, G. (2004). On the mechanism of the cis− trans isomerization in the lowest electronic states of azobenzene: S0, S1, and T1. Journal of the American Chemical Society, 126(10), 3234-3243.


Referring to numbered atom (e.g. C-1 vs C1) Is there a recommended style how to designate an atom by its number according to the structure numbering (given by nomenclature, or any other one in the text context)?

The are (at least) two different styles used by IUPAC itself. E.g.[1]

  Diastereoisomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.

or[2]

  Note: The configuration at C-4 (method 2) or C-1 (method 1) cannot be given because the attached substituent is unknown.

or[3]

  Small substituents on the C-8 and C-9 atoms of steroids should be drawn within the ring system in this fashion even if there are no substituents on the adjacent fusion atoms C-10 and C-14.

vs[4]

  The oxygen atom of the $\ce{CF3CO}$ portion of the ligand is attached
  to C2, while that of $\ce{MeCO}$ is attached to C4.

I can see some disadvantages for both of the styles.

With hyphen (dash):

Might be mistakenly thought, in favorable cases, that it is an isotope abbreviation. E.g. C-14 for carbon-14 ($\ce{^{14}C}$).
Not very likely, but eventually might be thought as a mathematical expression (e.g. when talking about 1st halogen atom X-1).

No hyphen:

When the text formatting was lost, it might be interconfused with molecular formula. Maybe not for carbon / organic chemistry, more likely for other elements in inorganic compounds, e.g. P4. Especially at chem.SE with Georgia font ☺.

References:

Basic terminology of stereochemistry (IUPAC Recommendations 1996), p. 2199 (anomers, via IUPAC Gold Book)
Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, p. 551
Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), p. 356
Nomenclature of Inorganic Chemistry (IUPAC Recommendations 2005), p. 156

How about using the label plus the number encircled? Would be hard to confuse and misinterpret. I actually use this myself when writing on paper or typesetting on computers.


Is the amount of cubic inches in a cubic foot exact up to infinite significant figures? I know we are supposed to use SI units and I know they are defined very well.
But does that mean that Imperial units of length/volume aren't exact? Don't they  have infinite significant figures in the equations?
I couldn't find it in the SI brochure.
I'm asking because I need to convert from cubic feet to liters and trying to decide how many significant figures to use. My first thought is that the conversion factors are exact and I should use all the numbers my calculator comes up with. But the conversion factor wasn't in the back of my book so maybe cubic inches in a cubic foot aren't exact?

Dealing with imperial measures in science is problematic. The main reason for that is that you have (at least) two different definitions (fixed values) for a lot of them, one based on the laws of the United States and one based on the laws of the United Kingdom. While the differences are quite insignificant for everyday use (as in baking a cake), in the science world these small deviations can cause further inaccuracies. On Wikipedia you can find more, but for the argument, let's look at the values for the yard:
United States    0.914401829 m
United Kingdom   0.9143993   m
International    0.9144      m

In principle the definition of cubic inch to cubic foot is exact, while the conversion to the SI system may carry an error.
Unit        Abbreviation    Definition  SI equivalent (rounded 4sf)
Cubic inch  cu in or in³                16.38 cm³
Cubic foot  cu ft or ft³    1728 cu in  0.02831 m³
Cubic yard  cu yd or yd³    27 cu ft    0.7646 m³

Most of the time this shouldn't affect your measurements more than the use of significant figures, which is used as stating an error implicitly. That means it is in itself inaccurate. In any scientific capacity you would state the error explicitly.
When a value is stated as $1.00$ it has three significant figures it actually means that the measurement falls within the interval $0.99$ to $1.01$, or $1.00\pm0.01$.
Let's have a quick example. You measure the volume of water in a beaker to be $0.35~\pu{in^3}$. The beaker doesn't specify how accurate the measurement is, so we simply assume $\pm0.01~\pu{in^3}$. You would state that value as $0.35\pm0.01~\pu{in^3}$, or $0.34-0.36~\pu{in^3}$, or simply $0.35~\pu{in^3}$ with implied error. On the other hand, when we know significant figures are used, we would interpret a value of $0.35~\pu{in^3}$ as $0.35\pm0.01~\pu{in^3}$.
When you convert this to the SI system, your value doesn't become more or less accurate. You cannot use significant figures to express the error in one system in the same way you would do it in the other, because the implicit error is different.
The exact conversion (of the international yard) is
$$\frac{0.9144~\pu{m}}{1~\pu{yd}}=\frac{0.9144~\pu{m}}{3~\pu{ft}}=\frac{0.9144~\pu{m}}{36~\pu{in}}\implies1~\pu{in^3}=\left(\frac{0.9144}{36}\right)^3~\pu{m^3}.$$
The actual trust of the value is $\pm0.01~\pu{in^3}=\pm0.16387064~\pu{cm^3}$. When you convert your value $0.35~\pu{in^3}=5.74~\pu{cm^3}$ (keeping the same number of significant figures since the conversion is exact), you would imply an error of only $0.01~\pu{cm^3}$, which is far less than the actual trust. Instead your measured value falls somewhere between $5.58-5.90~\pu{cm^3}$. So the correct way of implicitly stating the error would be $6~\pu{cm^3}$.
TL;DR
The conversions are exact, the system of significant figures is not. In doubt, state error margins explicitly.