Measure low DOM absorbance values in a microplate reader. Is there a suitable deep well microplate with flat bottoms available? I would like to use a microplate reader to measure the absorbance of dissolved organic matter (DOM), but the measured values are close to the lower detection limit. To overcome this, I thought of deep well plates as the pathlength is longer then. Nevertheless, I have not found any deep well plates with flat bottoms, which might be best for absorbance measurements. I want to avoid measuring in a photometer with 5 cm pathlength because the available amount of sample is limited and the microplate would need much less.

Does anybody have experience with similar problems?

Can anyone suggest a suitable microplate?

What is your opinion on the influence of the material?

I am not aware of suitable microplates for this application.  Here are some alternative approaches.

1. Low-volume quartz cuvettes.

There are quartz cuvettes in "micro" formats that may be suitable for you.  Mostly commonly they work with 1 cm pathlength.  The volume of the cells is controlled by the width of the non-optical dimension.  This can be made extremely small.

One example I found lists cuvettes that still have path lengths of 1 cm with total fill volumes as low as 15 μL.

2. Sample concentration

Details depend heavily on the details of your DOM and application, but you could concentrate your samples and then use existing microplates.

3. Longer measurement times.

Most plate readers (and cuvette-based spectrophotometers) let you control the number of "shots" or of the total amount of time a measurement of light intensity is integrated over.  You could crank this number up to get better sensitivity.


Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction? My book says:
  Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
  $$\ce{R-X + NaI ->[\ce{acetone,~reflux}] R-I + NaX}\\
(\ce{X}=\ce{Cl,Br};\ \ce{R}=\text{alkyl group})$$
    NOTE: Refluxing is a process of heating a liquid in a flask provided with a condenser. The vapours of the liquid which rise up get condensed back into the flask.
    Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.

MY QUERY

Can we prepare alkyl chloride or alkyl bromide by Finkelstein reaction?
As we can see, first line reads that Finkelstein reaction is particularly useful for preparing iodoalkanes. The lines followed by it signify that $\ce{NaCl}$ and $\ce{NaBr}$ are less soluble in acetone. If we want to prepare alkyl chloride or alkyl bromide, we will be needed to dissolve $\ce{NaCl}$ or $\ce{NaBr}$ respectively in acetone, which is not so possible. Is it true that, we can't prepare alkyl chloride or alkyl bromide by Finkelstein reaction because of the above reasons?

The thing is that you have to start with some compound $\ce{R-X}$ where $\ce{X}$ is a good leaving group. The quality of the leaving group is often thermodynamically defined, and in the Finkelstein reaction that you described, the energy motor for the whole thing is the insolubility of the salt that is produced.

I don't see what favourable starting material you could use for this in order to prepare other haloalkanes, especially since there are other ways for doing that.

That being said, it would in principle be possible to do
$$\ce{R-X + NaY -> R-Y + NaX v}$$
as long as the solubility of $\ce{NaY}$ is larger than the solubility of $\ce{NaX}$.


Is tungsten gas visible? I apologize if this isn't a sufficiently technical question to be posed in this forum.

I'm planning a short story, and unfortunately I've worked in a somewhat sience-y opening. Most of the information I need can be found online, but so far I haven't been able to find any comprehensive information about the properties of some of the less common gases, namely tungsten.

Would pure tungsten gas be visible to the human eye? Does it have any other unique properties that I should know before writing about it?

You can easily see it: "burn out" the filament of an incandescent lamp from a flashlight (torch) with over-voltage (e.g. put a 1.5 volt lamp across a 12 volt battery). The bluish flash you see is vaporizing tungsten (W), but to the unaided human eye, it looks little different from, say, the "spark" in xenon flash lamps.

Since the filament is so hot, it radiates heat rapidly and cools quickly, so is visible only as a flash. If you were to vaporize a few kg of W, it would be visible longer as a blindingly-bright object.

BTW, notice the film on the "dead" lamp: it is condensed W. If the lamp has a vacuum, the film is silvery, but collision with argon or other inert filler makes it rougher, or grayish.


When to include the mass of the dissolved substance in calorimetry problems? In some calorimetry problems, in which a substance is dissolved in water and we want the heat of the reaction, the mass of the dissolved substance is included in the mass plugged into mcat but other times it isn't. When should I include the mass of the substance?
For example, in this problem the mass of $\ce{NH4NO3}$ is not added to the mass of water to find the enthalpy:

$\pu{5.44 g}$ of ammonium nitrate was added to $\pu{150.0 mL}$ of water in a coffee-cup calorimeter. The temperature changed from $\pu{18.6 °C}$ to $\ce{16.2 °C}.$ Calculate the enthalpy change for dissolving $\ce{NH4NO3(s)}$ in water in $\pu{kJ mol^-1}.$ Assume the solution has a specific heat capacity of $\pu{4.18 J g^-1 K^-1}.$

But in this problem the mass of $\ce{CaCl2}$ is added to the mass of water:

In a coffee cup calorimeter, $\pu{2.6 g}$ $\ce{CaCl2(s)}$ was dissolved in $\pu{260 g}$ of water at a combined initial temperature of $\pu{23 °C}.$ The final temperature was $\pu{26.4 °C}.$ Calculate the enthalpy change of the reaction.

I would really appreciate if someone could explain to me when to include the mass of the dissolved substance.

There are multiple issues here, and the answer is: it depends.

How much of your substance is there relative to the solution? If the mass is, say, 1% of total and the specific heat is comparable, then your answer may be off by 1%, which is probably smaller than the systematic measurement error from heat loss in the calorimeter.

What is the specific heat of your substance? If it's very small, then it will hardly require any heat to raise the temperature of your substance.

Frequently, you have to make assumptions. Note that in your first example, it asked you to make assumptions about the specific heat of the solution. This is the specific heat for a mixture of your substance and water. That basically addresses both (1) and (2) above. Note that in this case, since it's the total solution, you should use the total mass.
In the second question, the total mass of calcium chloride is very small, so whether or not you include it is almost irrelevant.


Neighbouring group participation with and without retention of configuration in nucleophilic substitution of 2,6-dichloro-9-thiabicyclo[3.3.1]nonane Treatment of the dichloride A under the conditions indicated gives B with retention of configuration, while treatment of dichloride C gives D with inversion of configuration.

I am having difficulty drawing out the inversion ring product to find out if the s lone pair can attack from anti to the $\ce{C-Cl}$ σ* bond on the vertical ring. If it does, then, there is retention of product in the first reaction after the $\ce{N3-}$ attacks.
Can someone please help using diagrams?

The only difference between substrates A and C is disulfide vs sulfone nature at 9-position. Actually, that is good enough to display this major stereochemical outcome. As explained in Ref.1, the disulfide nature at 9-position of A has an advantage of having two lone pair on sulfur, one of which is proximate enough to attack backwardly on $\ce{C-Cl}$ bond ($\mathrm{S_N2}$ in nature) to give an episulfide ring (see following diagram and the extended comment of ron). The reaction center must have an inversion of configuration since the mechanism is $\mathrm{S_N2}$.

This episulfide intermediate undergoes ring opening backward attack by the intended nucleophile (e.g.,$\ce{N3-}$) to give the final product under 2nd $\mathrm{S_N2}$ mechanism. This stepwise replacement of chlorine by a nucleophile, the first substitution being considerably faster than the second, was first observed by E. J. Corey and E. Block (Ref.2). The nucleophile of their reaction was methoxy group of methanol (solvolysis). However, although they are the first group to suggest the 1,2-epi-sulfonium ion intermediates in these reactions, it was suggested for the initial dichloro compound A formation, not for the solvolysis by methanol. These types of neighboring group participation (NGP; or anchimeric assistance) have been observed previously for bridging carbon, nitrogen, and boron, all for the bicyclooctane series. In mechanistic point of view, this double inversion process preserves stereochemistry in substitution reactions on 2,6-Dihalosubstituted-9-thiabicyclo[3.3.1]nonanes (Ref.3).

Finally, C cannot follow the fast double inversion process due to the lack of a lone pair on $\ce{S}$ bridge, yet undergoes usual $\mathrm{S_N2}$ transformation (slow because the leaving group is $\ce{Cl}$) under the given conditions.

References:

D. D. Díaz, A. Converso, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold,” Molecules 2006, 11(4), 212–218 (doi: 10.3390/11040212).
E. J. Corey, E. Block, “New Synthetic Approaches to Symmetrical Sulfur-Bridged Carbocycles,” J. Org. Chem. 1966, 31(6), 1663–1668 (DOI: 10.1021/jo01344a001).
A. Converso, K. Burow, A. Marzinzik, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: A Privileged, Bivalent Scaffold for the Display of Nucleophilic Components,” J. Org. Chem. 2001, 66(12), 4386–4392 (DOI: 10.1021/jo015632y).


Why does ethene react explosively with fluorine? In this website it is claimed that "[e]thene reacts explosively with fluorine to give carbon and hydrogen fluoride gas", via this equation:

$$\ce{CH2=CH2 + 2F2 -> 2C + 4HF}$$

Why does the fluorine hog the hydrogen atom instead of attacking the $\pi$-bond?

Additionally, may I ask for the mechanism of this reaction?

The equation gives only the reactants and final products and tells us nothing about the mechanism; stoichiometric equations rarely do.This is an explosive reaction so it will proceed by many steps usually involving radicals. To get feedback, and so explosion, one step must be of the branched chain form $$\ce{R_1^. +  M -> R_2^. + R_3^.}$$ where $\ce{R^.}$ is a radical and $\ce{M}$ a reactant species (e.g. $\ce{CH2=CH2}$). As two radicals are produced for one reactant radical a chain reaction ensues, the reaction rate accelerates and so bang!

Any explosive reaction has initiation, propagation, branching and termination steps, some also have an inhibition step. The initiation step in your reaction probably forms the radical $\ce{FCH2-CH2^.}$ although I don't know the mechanism.


Why do surfactants lower the surface tension Everybody knows that detergents, or generally amphiphilic substances, lower the surface tension. Of water, usually.
I wonder if that's true for any polar solvent (most likely), i.e. will the same detergents work in, say, ethanol or liquid ammonia?
What happens in a nonpolar solvent? The detergents aggregate on the surface there too, but now with the polar end sticking out. Does the surface tension go up, or down?
And what exactly is the reason for the effect on water anyway, quantitatively? Imagine a surface of water which is covered by exactly a single layer of surfactant molecules, e.g. in a Langmuir trough. Now go to half or double that amount. Is there a discontinuity in the change of surface tension? If yes or no, why?
(sorry for the jumble of questions, I'm not quite sure from which angle to tackle this. The effects seems quite obvious, but I'm not sure it is. My real question is how to explain the effect on a molecular level.)

We usually say that at an interface, molecules will orient themselves at a "preferred" position, but we usually ignore that such a "preferred position" can actually be their "least unfavorable" position.

Energetically speaking, the interaction of a lipid chain with a gas is less unfavorable than interaction of a polar group with gas, because interactions from a lipid chain is mainly based on van der Waals interactions which are quite weak compared to polar (i.e. electromagnetic) interactions with a polar group. Also, they cannot make stabilizing bonds with the medium, i.e. water usually.

It must also be noted that the surface tension is linked to the interactions of molecules at the interface.

So "why do surfactants lower the surface tension"?

If you add some "surfactant", aka "tensioactive" agents, in water, its "least unfavorable" position would be at the interface between water and the other medium (usually gas). It will bring down the cohesive energy at the surface, compared to water alone, because it relies on van der Walls interaction. And that is why it lowers surface tension.


Which is discharged? Why Na+ ion is discharged at cathode rather than H+ ion when Mercury is used as cathode in electrolysis of aqueous NaCl solution ?

Mercury has a high overvoltage for hydrogen evolution; that means a barrier exists so that molecular hydrogen is not formed easily. Evolving hydrogen requires two protons to be neutralized, get together to form a molecule and clump up with a few more hydrogen molecules to form a gas bubble. If a sodium cation is neutralized right next to mercury liquid, it can quickly amalgamate/dissolve into the mercury, where its activity is much reduced. Sodium amalgam reacts with water, but much slower than metallic sodium reacts with water.
Similarly, ammonia amalgam can be made by electrolysis, and the mercury will foam up - it's not very stable.


Colour in muriate of Potash? Alkali earth metals produce white coloured salts with halogens like chlorine. But Muriate of potassium available in India generally used as a fertilizer is pink coloured(reddish white actually), why is it so?

If you have seen pink salt crystals from Khewra mines in Pakistan (the so-called Himalayan pink salt) sold at exorbitant rates in grocery stores, you would find out it is nothing but NaCl with a small amount of iron oxides as stated above. If you dissolve large tablespoons of such salt crystals in water, you will see a brownish precipitate sitting at the bottom on the container.

The word muriate reminded me of ancient chemistry books of the 1940s. It is a very very old name for a salt of hydrochloric acid.


What is the difference between Cyclic Process and Reversible Process? I want to ask the difference between Cyclic Process and Reversible Process.
Consider this graph:

If we have a process that goes from A to B then go back to A through the same path, then will it be called a Cyclic Process or Reversible Process or Both?

The key concept in a reversible process is that the system is in equilibrium with it's surroundings at every instant. A cyclic process is different: in a cyclic process, the initial and final state are the same.
The graph that you have drawn consists of a reversible isothermal expansion(AB) and a reversible isothermal contraction(BA). Overall, the process A->B->A is a cyclic process made up of two reversible processes.


Using water model in a computer program I am writing a small simulator for situations where water is the medium of all reactions.

A quick googling confirmed that there are water models available in computer readable formats. Being a beginner in molecular chemistry, I was hoping someone can direct me to already mature libraries which I can use to read water models.

Following seems to be good candidates:

http://msl-libraries.org/index.php/Main_Page
http://ball-docs.bioinf.uni-sb.de

Libraries which support Water model Simulation:

Assisted Model Building with Energy Refinement(AMBER)
Chemistry at HARvard Macromolecular Mechanics(CHARMM)
The Molecular Modelling Toolkit

Cool Site for Water Model Generation:

Water Models

Image Source: http://biomodel.uah.es/en/water/index.htm


Why can curcumin cross the blood-brain barrier, but not congo red? I've read that both congo red and curcumin are able to inhibit the clumping together of amyloid $\mathrm{β_{42}}$ in the brain, which would otherwise result in Alzheimer's disease.

Apparently, congo red cannot cross the blood-brain barrier (BBB) while curcumin can.

From what I learned at my biology classes, the BBB is lined by squamous, endothelial cells. Those cells are, as far as I know, the most permeable kind of cells you could come across in the human body, and if congo red doesn't pass through that, then it must be one stubborn molecule.

Now the cell membrane, itself being composed of a phospholipid bilayer, normally shouldn't have a problem transporting lipophilic molecules across the membrane. From an earlier conversation with Jan, I believe congo red is a lipohilic molecule. Yet inspite of it being lipophilic, it still isn't able to cross the BBB.

Comparing the structures wasn't particularly enlightening.

But the possibility congo red could act as a bidentate ligand, does seem to make it stand out from  curcumin. Am I onto something here? Apart from that, and a somewhat larger size, I don't see why congo red shouldn't be able to cross the BBB.

Now, I'd like to know two things,

1) What enables curcumin to cross the BBB?

2) Why isn't congo red able to cross the BBB, even though it has a structure similar to that of curcumin?

Preface: both molecules could act as bidentate ligands but the β-diketo structure found in curcumin is typically a much better bidentate ligands — as the 2-oxophenol moieties would be. So I would expect curcumin to be a much better ligand than congo red.

The key difference between the two molecules is comparing the extent of their lipophilicity and their hydrophilicity. Curcumin has a rather low water solubility of $3.2~\mathrm{mg/l}$, in line with the fact that it has rather little features that can enhance solubility. Basically, it boils down to the two keto-groups and the phenol moiety.

Congo red has two sulphonic acid groups and two aromatic amino groups. This means that there is a much greater relative charge present, and indeed its solubility in water is reported to be $1.16 \times 10^{2}~\mathrm{g/l}$ — notice the different unit. So congo red is already much more hydrophilic than curcuminis.

In terms of lipophilicity, the sulphonic acid groups present in congo red reduce its lipophilicity while the amino groups themselves don’t have too strong an effect. On curcumin’s side, the phenol groups can comfortably hydrogen bond to the neighbouring methoxy residues and the β-diketo fragment can tautomerise to a keto-enol form and form another intramolecular hydrogen bond. Thus, curcumin is a lot more capable of coping with low-hydrogen bond solvents and is thus expected to have a high lipophilicity.

This is the edge that curcumin has over congo red to cross the blood-brain barrier.


What is the point of introducing virtual orbitals in Hartree-Fock calculations? Let us for simplicity discuss RHF formalism. For $2n$-electron system we have $n$ Hartree-Fock equations written for $n$ spatial orbitals $\{ \phi_{k} \}_{k=1}^{n}$
$$
\newcommand{\mat}[1]{\boldsymbol{\mathbf{#1}}}
$$
\begin{equation}
    \hat{F}(1) \phi_{k}(1) = \varepsilon_{k} \phi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .
\end{equation}
Once we introduce finite basis $\{ \chi_{q} \}_{q=1}^{m}$ and express spatial orbitals as a linear combination of basis functions $\chi_{q}$
\begin{equation}
    \phi_{k}(1) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(1) \, , \quad k = 1, 2, \dotsc, n \, .
\end{equation}
we end up with $n$ Roothaan–Hall equations
\begin{equation}
    \sum\limits_{q=1}^{m} F_{pq} c_{qk} =
    \varepsilon_{k} \sum\limits_{q=1}^{m} S_{pq} c_{qk} \, ,
    \quad k = 1, 2, \dotsc, n \, ,
\end{equation}
which can be rewritten in the following matrix form
\begin{equation}
    \mat{F} \mat{c}_{k} = \varepsilon_{k} \mat{S} \mat{c}_{k} \quad k = 1, 2, \dotsc, n \, .
\end{equation}
The Fock matrix $\mat{F}$ and the overlap matrix $\mat{S}$ are both $m \times m$ square matrices, $\mat{c}_{k}$ is a column $m \times 1$ matrix, $\varepsilon_{k}$ is just a scalar value.

We can then collecl all $n$ $\mat{c}_{k}$ column $m \times 1$ matrices into one $m \times n$ matrix $\mat{C}$ and all $n$ values $\varepsilon_{k}$ into $n \times n$ square matrix $\mat{\varepsilon}$
\begin{equation}
    \mat{F} \mat{C} = \mat{S} \mat{C} \mat{\varepsilon} \, .
\end{equation}

In practice, however, we extend both $\mat{C}$ and $\mat{\varepsilon}$ to $m \times m$ matrices from $m \times n$ and $n \times n$ respectively, which results in having $m-n$ virtual (unoccupied) orbitals.

Taking into account that virtual orbitals are even more unphysical than their occupied counterparts the question is what is the point of such extension of $\mat{C}$ and $\mat{\varepsilon}$? Why do not we just leave them of $m \times n$ and $n \times n$ sizes respectively?

Virtual orbitals are automatically 'given' by the solved eigenvalue equations and are necessary for solving said equations. These orbitals are important for some post-HF methods such as CI and perturbation theory and coupled-cluster theory.


Reason for the generation of London Dispersion Forces While talking about gaseous state of matter we came to the topic of London Dispersion Forces while are caused by the generation of dipole in one atom which induces a dipole in another. While talking about the cause of such I stated that since Schrodinger's equation for multi-electron atoms is time dependent and hence at some moment a dipole may get generated (due to the increase in the probability of electrons in that region). Though as for now I am thinking that what I said was quite vague and may not be the reason for it.

So what is the reason for generation of dipole moment in one atom to cause the London forces?

You are correct that simply stating that the Schrodinger equation is time
dependent does not really answer the question.
In fact, time-dependence is not a component of the derivation of the dispersion forces. At a surface level, they just represent how much the ground state energy of a system of polarizable molecules is lowered by their interactions.
It's the same idea as any of the other intermolecular forces (coulomb, dipole-dipole, etc.) where some interaction between particles leads to a lower energy state than the free particles.

Generally, the London Dispersion interaction potential is not derived in a general chemistry class because it is purely quantum mechanical and rather involved. These slides provide a clear and thorough (albeit advanced) derivation, but I will explain the main points below.

Derivation Explanation

Let's imagine for this model that a polarizable atoms can be thought of as a simple harmonic oscillator with a positive charge ($+q$) on one end of a spring and a negative charge ($-q$) on the other end. The spring has an equilibrium length of 0, so that the atom has zero average dipole.
This model of an polarizable atom captures the physics we want because applying a force of some kind to the spring will induce a dipole in this atom  ($\vec{p} = q \vec{d}$).

So far we have only considered one of these atoms. What if we move to consider multiple of them together?
Remember we're really interested in the energy difference between two non-interacting atoms and two interacting atoms, so let's start by finding the non-interacting ground state energy of two atoms. We'll let the two atoms have two different spring frequencies $\omega_1$ and $\omega_2$ respectively.

Setting $z_1$ and $z_2$
as the displacements of the springs from equilibrium for atoms 1 and 2 respectively and $m$ as the mass of an atom,
we can write the total potential energy of the system as:
$$
U_{\text{non-interact}} = \frac{1}{2}\left(m \omega_1^2z_1^2 + m \omega_2^2z_2^2\right)
$$

Solving the Schrodinger equation for this potential yields the ground state energy for the two non-interacting harmonic oscillators:
$$
E = \frac{\hbar}{2} (\omega_1 + \omega_2)
$$

Now, we turn on the interaction and the problem becomes harder more interesting because the positions of the springs affect the magnitude of the interaction. Physically, because the dipole-dipole interaction can lower the energy, we expect that the interacting ground state of the two atoms will be a balance between minimizing the stretching of the springs (which increases the energy) and maximizing the amount of dipole-dipole interaction (which lowers the energy).

Setting $z$ as the distance between the two atoms, with some algebra one can show that the interaction energy between these two dipoles is given by
$$
U_{\text{interact}}(z,z_1,z_2) = \frac{q^2}{2\pi\epsilon_0}\frac{z_1 z_2}{z^3}
$$

Now, our total potential energy for the system is given by:
$$
U_{\text{total}}(z,z_1,z_2) = U_{\text{non-interact}}(z_1,z_2) + U_{\text{interact}}(z,z_1,z_2)
$$
While intimidating looking, this potential can also be solved exactly by changing variables to $z_+ = (z_1 + z_2)$ and $z_- = (z_1 - z_2$) and rearranging terms to make it look like the non-interacting potential energy. Once that is done the ground state energy can be found just as easily as in the non-interacting case.

With this done, we can find that the (one dimensional) energy stabilization is given by:
$$
U(z) = \frac{-\alpha^2 I}{2}\frac{q^2}{(4\pi\epsilon_0)^2}\frac{1}{z^6}
$$
Where the final step was to replace some of our less measurable parameters with their atomic counterparts: polarizability ($\alpha$) and ionization potential ($I$)


Predicting the pH of a weak acid and weak base solution I know that it is possible to predict whether combinations of acids and bases will be acidic, basic, or neutral:

weak acid (WA) and strong base (SB) reacts basic
strong acid (SA) and and weak base (WB) reacts acidic
SA and SB reacts neutral

How does a WA react with a WB? I've been told that it isn't neutral. How do you predict what it is and why is it not neutral?

Note: The expected $\text{pH}$ values refer to 1:1 mixtures as suggested in the question

One major feature of weak acids and bases is that they do not dissociate completely in water, producing less free oxonium or hydroxide ions than strong acids and bases do (Brønsted). For example in Sodiumacetate, a weak base, there is an equilibrium where the reactions left to right and right to left occur at the same speed, hence cancelling each other.
\begin{aligned}\ce{
   2NaOAc_{(aq)} + H2O &<=> 2Na+_{(aq)} + {}^{-}OAc_{(aq)} + HOAc_{(aq)} + {}^{-}OH_{(aq)}} &(1)
\end{aligned}

The same also applies to weak bases like ammonium hydroxide $\ce{NH4OH}$. Note, that ammonium itself is not a Brønsted base as it lacks to donate hydroxide ions.
\begin{aligned}\ce{
   NH4OH_{(aq)} &<=> NH4+_{(aq)} + {}^{-}OH_{(aq)}} &(2)
\end{aligned}

Weak acids can be expanded in the same way, example benzoic acid:
\begin{aligned}\ce{
   Ph-COOH_{(aq)} + 2H2O&<=> Ph-COO^{-}_{(aq)} + H+3O_{(aq)}} &(3)
\end{aligned}

In the Lewis definition, weak acids are not able to complex the hydroxide ions of water's autoprotolysis.

\begin{aligned}\ce{
   2H2O &<=> H+3O_{(aq)} + {}^{-}OH_{(aq)} &\\
   BF3_{(aq)} + 3H2O &<=> [HOBF3]^{-}_{(aq)} + 2H+3O_{(aq)} + {}^{-}OH_{(aq)} &(4)
}\end{aligned}

These equations are also in equilibrium, meaning all species are present in aqueous solution.

In the Lewis definition ammonia $\ce{NH3}$ is a base and it reacts analogous to the above.

\begin{aligned}\ce{
   2NH3_{(aq)} + 2 H2O &<=> NH4OH_{(aq)} + NH4+_{(aq)} + {}^{-}OH_{(aq)}} &(5)
\end{aligned}

Now the rate of dissociation is strongly dependent on the weakest bond. In $(1)$ that is $\ce{Na-O}$ and successively $\ce{H-O}$. Overall the reaction will produce excess hydroxide ions, hence the solution will have a $\text{pH}>7~(@25~^\circ{}\!C)$.
The same applies analogously to $(2)$ and $(5)$

In $(3)$ the opposite is the case. The reaction will produce excess oxonium ions, hence having a $\text{pH}<7~(@25~^\circ{}\!C)$. The same applies to $(4)$.

So when adding a weak base to a weak acid, one has to evaluate the whole equilibrium:
\begin{aligned}\ce{
   NH4OH_{(aq)} + HOAc_{(aq)} + H2O
   <=>&\\ [(NH4)(OAC)]_{(aq)} + NH4+_{(aq)} &+ {}^{-}OH_{(aq)} +  {}^{-}OAc_{(aq)} + H+3O_{(aq)}
} &(6)
\end{aligned}

It is therefore very much dependent on the chemical properties of all compounds involved in the solution. These systems are also very well known as buffer solutions.

So a good start would be comparing $\text{p}K_a$ and $\text{p}K_b$ values of the involved components, giving you a hint whether it will be more acidic or more basic. But you also have to pay attention to the amount of substance in particular in these cases. Dilution also plays an important role in this kind of chemistry.


What is the difference between "reaction pathway" and "reaction mechanism"? What is the difference between "reaction pathway" and "reaction mechanism"?

Are these expressions synonyms?

Some textbooks authors use both without making clear to the learner if they are synonyms or not.

Usually with the reaction mechanism the sequence of events (electron transfers) is indicated, while the term reaction pathway usually refers to the reaction coordinate diagram associated with the reaction, i.e. the change in energy with an extra notion on the transition states.


Why are Alcohol Dehydration Reactions Favored by High Temperatures? If I go by the Gibbs' free energy formula, it makes sense that high temperatures make the dehydration of alcohols more feasible, as elimination reactions usually have a positive value for the change in entropy ($\Delta S$) and the hydration of the sulfuric acid used is highly exothermic ($\Delta H$). This decreases the value for the change in free energy ($\Delta G$).

However, if I go by the modified Vant Hoff's formula (relationship between an initial equilibrium constant at an initial temperature and a final equilibrium constant at a final temperature), the $K_2/K_1$ ratio would be higher at lower temperatures, owing to the product $\Delta H \Delta T$ being positive (rather than negative if the reaction mixture is heated).

These are the formulas I'm referring to:

\begin{align}
\frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\
\Delta G &= \Delta H -T \Delta S \tag{2}
\end{align}

Let's take: $\,\,\,\,\ce{ethene(g) + H2O(g) <=> ethanol(g)} $

The standard enthalpy for hydration of ethene is $\Delta H^\circ_r=-45\frac{KJ}{mol}$. So we have $\Delta H^\circ_r=45\frac{KJ}{mol}$ for dehydration, which means that it is endothermic.

The expression:

\begin{align}
\frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\
\end{align}

(This equation is valid when Ts are similar, because $\Delta H$ varies with $T$ and you are considering it constant.)

with $\Delta T= T_2-T_1$ and $\Delta H_r^\circ>0$ (dehydration) equation (1) gives $K_2>K_1$.

As $$K = \frac{[\ce{H2O}]\,[\ce{CH2CH2}]}{[\ce{CH3CH2OH}]}$$

and $K_2>K_1$. In consequence dehydration enthalpy is favoured by temperature-increase.


Gamma Spectroscopy: 511 keV peak? Given a source with no gamma rays at 511 keV, what are two reasons for a peak at this location in a gamma spectrum (on HPGe detector)?

I know the annihilation peak from pair production appears at 511 keV and if a gamma ray had an energy of 1533 keV then the double escape peak (1533- 1022 = 511) would also appear at this energy.

Is there another reason that a peak could appear at this location, or are these the only reasons?

A true annihilation peak at $E = 511\ \mathrm{keV}$ is caused by the detection of individual annihilation photons. The source of such annihilation photons can be:

1) Pair production in the surrounding material caused by high-energy gamma radiation
If the gamma energy is large enough to make pair production in the surrounding material (typically in the high-$Z$ shielding material surrounding the detector) relevant, the gamma quantum (a high-energy photon) may disappear and be replaced by an electron and a positron. The electron and positron may travel a few millimetres before losing their kinetic energy to the absorbing medium. When its kinetic energy becomes low, the positron may combine with an electron in the absorbing medium. Then both disappear and are replaced by two annihilation photons of $E = 511\ \mathrm{keV}$ each.

2) Positron emission of the source
If the source consists of a radionuclide that decays by positron emission (β+ decay), and the positron is stopped and undergoes annihilation before it reaches the detector (i.e. the positron is stopped within the source volume or in the material that is encapsulating the source), this region acts as a source of annihilation photons of $E = 511\ \mathrm{keV}$ each.

If one annihilation photon reaches the detector and is absorbed within the detector, an annihilation peak appears in the spectrum at an energy of $E = 511\ \mathrm{keV}$.

Furthermore, a peak at $E = 511\ \mathrm{keV}$ could be caused by mere coincidence due to

1) Coincidence summing
If two gamma rays are detected approximately at the same time, the spectrum will show only one count at the sum of their gamma energies. If the sum amounts to approximately $E = 511\ \mathrm{keV}$, the resulting peak would look like an annihilation peak.

2) Double escape
If the gamma energy is large enough to make pair production in the detector material relevant, the photon may disappear and be replaced by an electron and a positron. The electron and positron may travel a few millimetres within the detector before losing their kinetic energy and finally combining with an electron. Then both disappear and are replaced by two annihilation photons. If one annihilation photon escapes from the detector without interaction within the detector, a single escape peak appears in the spectrum at an energy of $511\ \mathrm{keV}$ below the full-energy peak. If both annihilation photons escape, a double escape peak appears at an energy of $1.02\ \mathrm{MeV}$ below the full-energy peak.
If the remaining energy is approximately $E = 511\ \mathrm{keV}$, the resulting escape peak would look like an annihilation peak. This effect is not possible for single escape peaks since their energies are always well above $E = 511\ \mathrm{keV}$, but it is theoretically possible for double escape peaks.


Freeware that converts structure in a conformational model? In order to draw chemical structures I use ChemBioDraw Ultra.

As an exercise I've to draw the most stable conformation.

I expect a twist-boat like structure (instead of usual chair conformation for substituted dioxanes)

Is there a freeware or an easy computer online tool to check if my propose (right side picture) is correct?

When I go to >Edit > Get 3D Model the program gives me a small picture with a 3D-Model:

By double-clicking this small image it opens ChemBio3DUltra and I can rotate this molecule:

@Martin: Is this correct (I've expected a twisted cyclic ring)?

@Martin I draw the structure at right with i-Bu and H exchanged in Chemdraw and do >Edit > Get 3D Model

The reason why I first thought it should be twist boat like was that one C-O is pointing up and the other pointing down. May be someone knows why people gives this stereoinformation inside the cyclic ring?

As I've drawn it as a twist boat I get exactly the same with ChemBioDraw. Is there a way to calculate by how many kcal/mol the first one is more stable then the second one with ChemBioDraw?

What's interesting: I searched for a structure in which the twist-boat conformation is the most favorable one and found one:

  However, when there are two large substituents in the cis-1,4 arrangement on
  a cyclohexane ring, neither of which will go easily into an axial position, then
  it appears that the twist-boat conformation (Section 12-3A) is most favorable
  [http://authors.library.caltech.edu/25034/13/BPOCchapter12.pdf, page 458]

(The A-value of a tert-Butyl is in the range of 5 kcal/mol which is around the value of the energy difference between cyclohexane in twist-boat and chair conformation.
I've drawn cis-1,4-di-tert-butyl-cyclohexane with ChemDraw and used >Edit>Get 3D Model and ChemBioDraw gave me the structure which is NOT the energetical most suitable one! Why?

Short answer: yes, the dioxane has a minimum when it's in a twist-boat form.

I use Avogadro which offers a "auto-optimize" tool. You can grab atoms and throw them around to force rings into different conformations while the force field is minimizing.

So here's what I did.

File -> Import -> Fetch by chemical name: (2R,4R)-2-isobutyl-4-methyl-1,5-dioxaspiro[5.5]undecane. This grabs the 3D geometry from the NIH Resolver (i.e., PubChem)
Auto-Optimize Tool: Force Field MMFF94

Then I started minimizing and looking at the energies. I pulled various atoms around to try to minimize steric clashes and flip rings from chair to boat, etc.

For the figures that follow, I've removed the hydrogens for clarity, but they exist as far as the energies and minimization. I just removed them to save the graphic.

Chair/Chair: 24 kJ/mol

Twist-boat/Chair: 14 kJ/mol

So yes, there's a decent stabilization in the twisted form of the twist-boat. You have to get the hydrogens right, or you'll end up with a bunch of eclipsed interactions between the dioxane oxygens and hydrogens on the cyclohexane.


Will ice be cold or not when formed using pressure? We know that ice can be formed either by lowering the temperature of water or by applying more pressure. If we lower the temperature of water the ice so formed will be cool which is obvious.

If ice is created by applying pressure, will it be cold as well?

Water can be formed by applied pressure, but only at certain temperatures. Looking at the phase diagram below, you can see that water must be at 0 C until extreme pressures are reached. This can create ice, but it would only remain solid under these pressure conditions at that temperature. The ice would have a cooling effect on surroundings that have a higher temperature, in fact the rapid change back to liquid would cool the surroundings, but the ice would not be cold. This ice would also quickly revert to another phase if not held under these pressure conditions.

So, if at the bottom of a deep, high gravity sea, the ice will form at temp above 0C. If it starts floating upward, the lower pressure will allow it to melt.

If you create warm ice in a hydraulic ram, it will melt as soon as pressure is relieved.


Does more branching lead to lower boiling point in alkanes? List the following alkanes in order of increasing boiling point.

  A. $\ce{CH3(CH2)4CH3}$
  B. $\ce{(CH3)2CHCH(CH3)2}$
  C. $\ce{CH3CH2CH(CH3)CH2CH3}$

  Answer: $C<B<A$, low to high.

Why is it that B has not a lower boiling point than C? Is it not more branched? I believe branching decreases surface area, leading to less intermolecular interactions, and to a higher boiling point.

$A$ n-hexane, boiling point 69 degrees C.

$B$ 2,3 dimethyl butane, boiling point 58 degrees C.

$C$ 3-methyl pentane, boiling point 63 degrees C.

$B<C<A$


Chemical structure of Turquoise I know nothing about chemistry, I managed not to take it in correspondence high school. I don’t know how to read or make sense of this: $\ce{CuAl6(PO4)4(OH)8.4H2O}$ the chemical formula of turquoise. From looking at a children’s periodic table I figured out some of the letters like Copper and aluminum, but I don’t know what the parenthesis or numbers mean. Also, is the PO polonium?

I don’t know how to read or make sense of this: $\ce{CuAl6(PO4)4(OH)8.4H2O}$ the chemical formula of turquoise.

It means for every atom of copper (Cu), there are 6 atoms of aluminum, 4 phosphate groups (which each contain a phosphate atom and four oxygen atoms), 4 hydroxide groups (which each contain an oxygen and a hydrogen atom) and 4 waters (which each contain two hydrogen atoms and one oxygen atom). So the parentheses are for grouping, and the numbers are to count how many atoms or groups there are.

The structure is a crystal, so it goes on and on in all directions. That is why I was careful to say "for every atom of copper...". For other substance like carbon dioxide that are found as molecules, it is less complicated (there, the formula actually tells you the number and kind of atoms in a single molecule).

  Chemical structure of Turquoise

Some of the components in turquoise are positively charged (copper and aluminum) and some are negatively charged (hydroxide and phosphate). The building blocks are arranged that positive and negative come close together, and like charges are further apart. The entire structure is very complicated.

The picture below does not show all the atoms, but instead shows copper, aluminum and phosphate as 3D shapes (the metals and the phosphorus would be in the center of each of the shapes). The shapes have common corners occupied by oxygen atoms (in the formula, these appear as part of phosphate, hydroxide and water). The hydrogen atoms are shown at tiny spheres.

Source: http://rruff.info/uploads/MM64_905.pdf


Why is CO ligand higher than CN- in the spectrochemical series? I've drawn MO diagrams for both molecules and they are isoelectronic. The main difference is that $\ce{C}$ is closer in electronegativity to $\ce{N}$ than it is to $\ce{O}$ so there will be less stabilisation of the bonding orbitals for $\ce{CO}$ due to the increased energy difference in the AO's.

Does this difference lead to $\ce{CO}$ having its $\pi^*$ orbital (LUMO) lower than that of $\ce{CN^{-1}}$ so it more readily accepts $\pi$ backdonation from a metal centre? Does this alone account for why $\ce{CO}$ is higher in the spectrochemical series? What about the difference between the  $\sigma$ orbitals (HOMO) in $\ce{CO}$ and $\ce{CN^{-1}}$?

You're on the right track. It has to do with the energies of the frontier orbitals. As you rightly said, both species are isoelectronic, and the orbital energies in CO are lower than those in CN−.

The lower HOMO energy means that CO is a poorer σ donor orbital towards the metal than CN−. Likewise the lower LUMO makes it a better π acceptor.

These two factors are conflicting: stronger π acceptors are stronger-field ligands, but poorer σ donors are weaker-field ligands. Empirically, since CO is higher in the spectrochemical series, it stands to reason that the π acceptor effect is the most important factor here. This isn't really a surprise, as π-effects tend to outweigh σ-effects when it comes to the spectrochemical series.

I don't pretend that the above is a rigorous proof, but short of crunching the numbers, I don't think there's a much better explanation.


How to calculate the formal charge of sulfate ion? I found a site with this formula if you will,

  Formal Charge = [Number of valence electrons on atom] – [non-bonded electrons + number of bonds]

I don't think I am using it correctly in finding the formal charges of each atom in $\ce{SO_4^{2-}}$. $\ce{S}=2$, $\ce{O}=-1$ and the other $\ce{O}=0$ and if you add them together I don't get the overall charge of -2.
 How do you work out formal charges?

Your formula is correct. If $V$ is the number of valence electrons originally assigned to an atom, $N$ is the number of nonbonding electrons, and $B$ is the number of bonds ($\frac{1}{2}$ of the number of bonding electrons to be more precise), then the formal charge $FC$ is:

$$FC = V - (B+N)$$

Sulfate ion has two valid structures that you could draw, one with sulfur having a formal charge of zero and one with sulfur having a formal charge of +2. The structures below (from the Wikipedia page, released into the public domain) follow the annoying but allowable convention of replacing lone pairs with bars.

If the structure on the left (#1) is correct, then the sulfur atom ($V=6$) has six bonds ($B=6$) and no lone pairs ($N=0$). Two oxygen atoms (V=6) have two bonds ($B=2$) and two lone pairs ($N=4$), while the other two oxygen atoms have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom and two oxygen atoms have $FC=0$ and the remaining two oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.

$$FC_{\ce{S}}=6-(6+0)=0$$
$$FC_{\ce{O_{1,2}}}=6-(2+4)=0$$
$$FC_{\ce{O_{3,4}}}=6-(1+6)=-1$$

If the structure on the right (#2) is correct, then the sulfur atom ($V=6$) has four bonds ($B=4$) and no lone pairs ($N=0$). All four oxygen atoms (V=6) have one bond ($B=1$) and three lone pairs ($N=6$). The sulfur atom has $FC=+2$ and the oxygen atoms have $FC=-1$ for an overall charge on the ion of $-2$.
$$FC_{\ce{S}}=6-(4+0)=+2$$
$$FC_{\ce{O}}=6-(1+6)=-1$$

But, wait! A more fundamental question might be "How can the sulfate ion have two very different structures that puts different formal charges on the sulfur and oxygen atoms?"

Formal charge is a nice bookkeeping mechanism, but has no experimentally valid relationship to the actual charge on any atom in most molecules or ions. Formal charge only equals actual charge on monoatomic species. Formal charge falls into the category of models that we use in chemistry that are 1) helpful, 2) produce the correct answer when used correctly, and 3) completely bogus.  Other models in this category include oxidation number, VSEPR, resonance, and electronegativity. The experimental evidence suggests the real structure of sulfate combines features of both structure #1 and structure #2, but would be challenging to draw using the formalisms we have adopted:

All four $\ce{S-O}$ bonds are equal in length (#2).
The length of the $\ce{S-O}$ bonds is shorter than a normal $\ce{S-O}$ single bond and longer than a normal $\ce{S-O}$ double (#1).
The sulfur atom has a partial positive charge (#2) (note that partial charges, unlike formal charges, have some experimental basis).
The four oxygen atoms have equivalent partial negative charges (#2).
The partial negative charges on the oxygen atoms add to more than $-2$ but not anywhere near $-4$ (#1).


When, how and why does sodium metal react with cycloalkenes, cycloalkadienes, cycloalkatrienes? I've been searching for the conditions necessary for the reaction of sodium (Na) metal with cycloalkenes, cyclo-di-alkenes, cyclo-tri-alkenes all over the internet but I cannot find a proper answer. Can someone explain to me the necessary conditions?

For example why does sodium metal react with cyclopentadiene but not with cycloheptatriene?

The negatively charged particles are conjugate bases of respective compounds, which (reactants) acts due to presence of active methylene group. Depends on the stabitily of conjugate base the reaction goes forward.


Do Na or Li need heat in order to react with Cl₂? I've read how putting a burning piece of sodium in chlorine would produce white $\ce{NaCl}$ fumes. However, I can't find out whether Na or Li can react with $\ce{Cl2}$ without heating or burning.

From personal experience, Na(s) and Cl2(g) react at room temperature without added energy. Li(s) and Cl2(g) should do so also.

Consider the following heat of formation data harvested from the NIST Chemistry Webbook:

For NaCl:
$$\ce{Na(s) + } \frac 1 2\ce{Cl2(g)->NaCl(s)}\quad \Delta H_f^o=-411.12 \; kJ/mol $$

For LiCl:
$$\ce{Li(s) + } \frac 1 2\ce{Cl2(g)->LiCl(s)}\quad \Delta H_f^o=-408.27 \; kJ/mol $$

These data suggest that both reactions should occur without added heat. However, the rate of reaction is not determinable from these data. The reaction between Li(s) and Cl2(g) will likely be slower.


Is there a relationship between loss of electrons and EM wave reflection? Is there a relationship between the loss of electrons in a cation and the amount of electromagnetic spectrum, a compound of this cation would subsequently reflect or transmit?

For example, any compound with $\ce{Fe^3+}$ ions is dark orange. This must mean compounds with $\ce{Fe^3+}$ reflect the longer wavelengths of the visible light spectrum more strongly, so the range of $\approx\pu{635–590 nm}$.

While a compound with a $\ce{Fe^2+}$ cations is light green. This, I think means that compounds that house $\ce{Fe^2+}$ ions reflect wavelengths of the range $\approx\pu{560–520 nm}$ more than others.

Now,  I am not aware at all of any link between atomic electron loss  influencing the transmission/reflection of wavelengths at all, so it could very well be something else entirely that may instead be influenced by the gain/loss of electron. I do not know.

So is there a relationship between the two? If so, what is it?

Actually, I think you have it backwards. The color you see is due to the colors of light which are removed from white light when traveling past the iron. That is, if you go and find an absorption spectrum of iron cations in different oxidation states, you will find they primarily absorb the complementary colors of the color you actually see.

As far as the relationship between charge and the color, the ions with a smaller charge will likely have a larger HOMO-LUMO gap which means you will see colors of shorter wavelength, which is what you report. This property will depend very sensitively in the presence of ligands and other environmental changes, so it is not always possible to guess the color of a compound based only on charge.

This is typically how colors work for systems that absorb light. It is, however, possible for the we see to be due to reflection. Pure reflection, though, is independent of wavelength, so the phenomenon you describe is actually scattering, which does depend on wavelength. This type of phenomenon is responsible for the color of the sky.

Finally, you sometimes see the color which is emitted after absorption, as is the case in fluorescence and phosphorescence.

Hopefully that clears up your confusion


Reaction between NaOH and Carbonic acid I was presented with the following reaction when we were just learning about acids and bases:

$$\ce{2NaOH + H2CO3 -> H2O + Na2CO3}$$
however I had thought it might've been something like

$$\ce{NaOH + H2CO3 -> H2O + NaHCO3}$$

They both seem to fit the notion of a neutralization reaction creating a salt so how do I know that the first one is correct? The second reaction looks fine to me (is it that bicarbonate is too reactive?) and more intuitive as the carbonic acid just gives a proton away ending up as an ion to bond with $Na$, so how do I distinguish between these kind of cases?

You're right, both of these reactions take place.

From Wikipedia, the Free Encyclopedia

  In aqueous solution, carbonate, bicarbonate, carbon dioxide, and carbonic acid exist together in a dynamic equilibrium.

Now, the answer to the question depends on context. If you're asked to do this reaction in a mildly acidic/weakly basic medium, then the answer would be the bicarbonate, while in a strongly basic medium, the answer would be the carbonate.

Interesting for further reading:

This $\ce{pH}$ dependent equilibrium between carbonate and bicarbonate is one of the ways in which the blood $\ce{pH}$, and the $\ce{pH}$ of the mouth is maintained. It also helps aquatic life to survive in ocean waters. (Also see urea cycle and carbon cycle)


Determining Enthalpy Change of the Combustion of Ethanol A worked example went something like this:

Use the following experimental data to determine the enthalpy change when 1 mole of ethanol (C2H5OH) is burnt:
Mass of water = 150.00 g
Initial temperature of water = 19.5 °C
Maximum temperature of water = 45.7 °C
Initial mass of spirit burner = 121.67 g
Final mass of spirit burner = 120.62 g

Being a worked example, I saw the worked solution and understood the process completely. What's bothering me is a fundamental flaw in the procedure that has seemed to be neglected completely.
We are relying on the idea that the combustion of ethanol will produce some energy, the amount of which can be calculated with the increase in temperature of the beaker of water on top. However, if there was no ethanol, no combustion, the flame giving rise to the combustion would still manage to heat the water up; in this sense, we do not know whether it is the flame supplying the energy to the rise in water, or the exothermic reaction that takes place with ethanol. How does the existence of such a flaw make the procedure useful to any extent? Is there anything I'm missing?

I see your concern, and it's absolutely valid. Any time you perform a calorimetry experiment, you need to be certain that you're not introducing extra energy into the system. Typically, the spark used to ignite the sample has a negligible energy input, but it is still there. However, any heat released by the flame (other than the initial energy put in to generate the spark) is generated by the combustion process. Without the combustion, the flame would cease to exist. If you were running this experiment, you could use an electric sparker and measure the power input for the sparker to know how much energy was inputted. Another option would be to run a control experiment with just an ignition (but no ignitable material) and see how much, if any, the water heats up.

Of course, if you used a lighter to start the flame, and then held the lighter open throughout the experiment, you would be introducing extra energy from the combustion of lighter fluid, and this would give you false results as you feared. That's why we don't run calorimetry experiments in that way.


Forms of Henry's law versus ideal gas law This might be a slightly daft question as I'm a physicist rather than a chemist, but I have a slight problem in using Henry's law which I think I can circumvent using the ideal gas law – I'd be grateful if anyone can confirm or deny my logic!

I have an equation that yields the volume of oxygen gas per unit mass of tissue. We'll call this quantity $C$ with S.I units of $\mathrm{m^3\: kg^{-1}}$. We also have the partial pressure $p$ measured in $\mathrm{mmHg}$ and we wish to relate the two; several texts have suggested using Henry's law of the form $C = kp$ where $k$ is Henry's constant for oxygen. However, this causes a problem, as I can find no form of Henry's law with the correct units to allow this conversions (units of $\text{vol} / ({\text{pressure} \times \text{mass}})$).

My first question is does such a series of constants exist and does anyone know where?

In the interim, I decided to try to circumvent this problem by using the ideal gas law, which I believe holds at the relatively low pressures in very shallow water. Knowing one mole of oxygen gas has a mass of $32\:\mathrm{g}$, and occupies a volume of $22.4\:\mathrm{l}$ at standard temperature and pressure of $760\:\mathrm{mmHg}$, I re-wrote the expression for the mass of $\ce{O_2}$ in a given volume;

$m_{\ce{O_2}} = \frac{0.024\,p}{(0.032)760RT} = \frac{0.7\,p}{760RT}$

Using the ideal gas constant, and a temperature of $310.15\:\mathrm{K}$, I get

$m_{\ce{O_2}} = 3.572 \times 10^{-7} \cdot p$

If I define $M_{\ce{O_2}}$ as the unitless mass fraction of $m_{\ce{O_2}}$ per unit mass, then knowing the density of oxygen gas is around $1.331\:\mathrm{kg/m^3}$, I can divide this into $M_{\ce{O_2}}$ to get the expression

$C = 2.6835 \times 10^{-7} \cdot p$

which has the correct units. My second question is then is this valid, and if not, why not?

I wrote out an answer to this but realized I was wrong. But I've noticed a number of things:

First off, your form of the ideal gas law has the Molar Mass inverted. If you make the substitution that $n=m/M$, then the rearrangement reads $m = MVp/RT$, you have the $M$ on the bottom.

Secondly, You are selecting a Volume of 1 Liter, so your calculation will solve for a relation of mass and pressure @ 1L, which I don't think you want.

Thirdly (?), I'm guessing that the 760 in the denominator is there so you can use a gas constant in terms of atm? Or did I miss something?

So those are my observations about your attempt with the ideal gas law. I also am concerned over the bit with the mass fraction, as I think you converted a tissue mass into an $\ce {O2}$ volume using $\ce {O2}$'s density, which would be wrong if that was the case.

Looking again at Henry's law in the $p=kC$, you can see that in order to use the standard k units available, you would need to convert the $C$ you have from $L/kg$ to $mol/L$. You can convert $L$ to $mol$ using the ratio you used earlier, and you end up with $C$ in terms of $mol/kg_{tissue}$. The trick here is that you need the density of the tissue (not of $\ce {O2}$) to convert this into the volume you're looking for in order to use the equation. I think haha.


Why is N₂ stable but HCN and C₂H₂ unstable? Compounds with triple bonds generally seem to be unstable. $\ce{HCN}$ and $\ce{C2H2}$ are high-energy, relatively short-lived molecules that will readily polymerise or react with other organic molecules. My naïve mental picture is that they don't really want to have a triple bond because the geometry makes it awkward, so they'll do anything to transfer an electron somewhere else and turn it into a double or a single bond instead.

But $\ce{N2}$ seems to be an exception. By my naïve reasoning, when an $\ce{N2}$ meets an organic molecule it should be eager to shrug off its triple bond and join the party. But in fact this doesn't happen, and nitrogen fixing organisms have to do quite a bit of work to get nitrogen to participate in organic molecules.

So my question is, what is it about $\ce{N2}$ that makes its triple bond so energetically favourable while $\ce{C#C}$ and $\ce{C#N}$ bonds are so unfavourable?

Note that this question isn't about reactivity with $\ce{O2}$ but rather with organic molecules. I'm interested in why it's so difficult for $\ce{N2}$ to react with organic molecules in general, given that other molecules with triple bonds seem to react with them very easily.

(Information to guide the scope of answers: I'm coming at this from a physics background, so thermodynamic concepts can be taken as understood, but I quickly get lost when it comes to electrons and orbitals and so on, hence the rather basic question.)

I think this question can be answered reasonably well with thermodynamic considerations.  When we say something is stable, it is (or should be) assumed that the compound does not engage in a chemical reaction.  It is important when we discuss the stability of compounds to focus not only on the compound itself but what the products of the reaction are.

The three triple bonds you mention ($\ce{N#N}$, $\ce{C#N}$ and $\ce{C#C}$) all have very high bond dissociation energies (from various sources such as this one they are $\approx950, 940\text{ and }960\:\mathrm{kJ/mol}$) so all of these triple bonds require a large amount of energy to break.

Now, when we think about the stability of acetylene, I suspect most people are thinking about its stability towards ignition (which has been studied, by the way).  That gives us a reaction we can explore, namely the combustion of acetylene in the presence of oxygen to produce carbon dioxide and water.  WolframAlpha can calculate the thermodynamics of this reaction, giving a result of $-2600\:\mathrm{kJ/mol}$; clearly a very exothermic reaction.

Now let's look at the combustion of nitrogen in the presence of oxygen, which could form nitrogen dioxide.  Again, WolframAlpha (Note, if you happen to play around with WA, make sure you put your reaction in quotation marks or it won't be interpreted properly) gives us a result of +66 kJ/mol or an endothermic reaction.

What do we make of these results?  Focusing on the products we see that for the acetylene combustion, there is a large amount of energy released when creating the products.  The same is not true for the products of nitrogen combustion.

In summary, when considering stability, we must think about we mean by the word, since that meaning will lead to a chemical reaction, and the tools we use to describe chemical reactions help us quantify that stability.


Calculating the electrode potential from thermochemical data How can I calculate the electrode potential from thermochemical data?

My attempt: I know that the electrode potential is dependent on the enthalpies of: atomisation, first ionisation, second ionisation and hydration. So I summed up all to get $\Delta G$ and used the formula $\Delta G=-nFE$,but I am not getting the same value as given in the table.

Does the electrode potential depends on factors other than those given in the table, or there is something more to it than just addition?

As stated above by @orthocresol, these values of $\Delta H$ alone are not sufficient to find the electrode potential. To do so, you would need $\Delta G$ for each reaction, so you would either need to find a table of $\Delta G$ for these reactions, or you would need a table of $\Delta S$ along with the table given.

But your question is why do we care about those particular values to begin with. The reason for this is that those reactions summed together give the reduction reaction for that metal. For example, I will look at the reactions of iron.

The reduction reaction for iron: $$\ce{Fe(s) \to Fe^2+(aq) + 2e^-}$$

All the other reactions in the order they are listed in the original post:
\begin{align}
\ce{Fe(s) &\to Fe(g)} \\
\ce{Fe(g) &\to Fe^+(g) + e^-} \\
\ce{Fe^+(g) &\to Fe^2+(g) + e^-} \\
\ce{Fe^2+(g) &\to Fe^2+(aq)}
\end{align}

If you add together these reactions, you should obtain the reduction reaction. Consequently, by Hess's law for Gibb's free energy, the sum of the $\Delta G$'s for these reactions will give you the $ΔG$ for the reduction reaction.

Then, it is just a matter of converting the $\Delta G$ to a reduction potential using the equation given in the original question:$$\Delta G^{\circ}=-nFE^\circ$$ where $n$ is the moles of electrons, $F$ is Faraday's constant, and $E^\circ$ is the standard reduction potential.


Wine tasting like soot I'm not sure if this question is for chemistry.

I had a bottle of (cheap) wine. It had a strong taste of soot, to an extend that I suspected that in the filling process some machine oil might have gotten into it.

I discarded the bottle, but some time later, I've found the same taste, to a lesser extend, in some other wine. Now it looks like my taste buds have become more sensitive to that, as I can taste it now (depending on what, I don't know) quite frequently, mostly on a low scale.

Any idea to explain that?

The odour and aroma of wine consists of about 600–800 components. Many of these compounds have been described with a range of odour/flavour sensations, even in more than one category, and often have differing flavour impressions, dependent upon their concentrations in aqueous solution or in the air-space above.

The odours and flavours in wine tasting are typically divided into groups, for example: Floral, Woody, Rustic/Vegetal, Balsamic, Fruity, Animal, Empyreumatic, Chemical, Spicy, and Etherish. [Bakker, J.; Clarke, R. J. Wine flavour chemistry; 2nd ed.; Wiley-Blackwell 2012]

What you describe as “strong taste of soot” probably belongs to the Chemical group, which is associated with synthetic chemical manufacture.

Typical compounds in this group that might be responsible for the soot aroma are volatile phenols:

3-methylphenol (tarry/leathery odour)
4-methylphenol (tarry/smoky odour)
2-ethylphenol (smoky odour, phenolic flavour)
2-methoxyphenol (smoky/woody odour)

A special substance that is believed to be responsible for a strong, petrolly kerosene-like aroma developed during aging is 1,1,6-trimethyl-1,2-dihydronaphthalene, which is formed from carotene breakdown.


Density of an object To complete homework I am working on, I need help finding the density of a penny. I'm not given the mass or the volume. All I know about this penny is it's zinc coated in copper, and I am given the percentage of copper in the penny: 3.0% (by mass). We are to use these values for the density of copper and zinc:

Copper = 8.93, Zinc = 7.14

My friends did this:

$$\mathrm{8.98 \times 0.03 + 7.14 \times 0.97 = average~density~of~a~penny}$$

I told them "I think you're wrong, you can only calculate the average density of something if you use the volume", but I couldn't explain it any better.

Here's what I did:

Let $M$ be the mass of this penny in grams.

$$\frac{0.03M~\mathrm{g~mL^{-1}}}{8.98~\mathrm{g}} = \frac{0.03}{8.98}M~\mathrm{ml}$$

$$\frac{0.97M~\mathrm{g~mL^{-1}}}{7.14~\mathrm{g}} = \frac{0.973}{7.14}M~\mathrm{ml}$$

Adding them together we get $7.1952M$.

I'm not sure what this number represents but we can use it to find the ratios of volume.

$$\frac{\frac{0.03}{8.98 M}}{7.1952 M} = 0.0374$$ $$\frac{\frac{0.97}{7.14 M}}{7.1952 M} = 0.963$$

Using these numbers, we do what my friends did:

$$8.98 \times 0.0374 + 7.14 \times 0.963 = \mathrm{average~density~of~a~penny}$$

Am I correct?

If I am, is there a formula or any other simpler way to do this.

This is my first question on this site, so let me know if I did anything wrong.

You have the right idea, but somewhere your math is incorrect.
Let the mass of the penny be $M$.
The mass of the copper is $\pu{0.03M}$ and the mass of the zinc is $\pu{0.97M}$.
The volume of copper is $\pu{0.03M}/(\pu{8.93g/cm^3})$ and the volume of the zinc is $\pu{0.97M}/(\pu{7.14g/cm^3})$.
The total volume is
$$\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})$$
The density of the penny is
$$\frac {M}{\pu{0.03M}/(\pu{8.93g/cm^3}) + \pu{0.97M}/(\pu{7.14g/cm^3})}$$
Factoring out M gives
$$\frac {1}{0.03/(\pu{8.93g/cm^3}) + 0.97/(\pu{7.14g/cm^3})}$$
So the density of a mixture works out to be:
$$D=\frac {1}{(\%_1/D_1)+(\%_2/D_2)}$$


Does it matter what electrolyte we use for a Galvanic Cell? A typical Galvanic Cell, say a Daniell cell, consitsts of two separate beaker, one containing Zn rod dipped inside aq $\ce{ZnSO4}$ and the other beaker containing Cu rod dipped inside $\ce{CuSO4}$. Does it matter what electrolyte we dip the metal in?

If we change the electrolyte of the beaker containing Zn rod to say $\ce{AgNO3}$. I would expect a displacement reaction to take place where a redox reaction occur between the Zinc rod and the Ag+ ions. No electrons will flow through the external wire. What if I replace the electrolyte with a solution of a metal that is less reactive then the metal dipped in? eg. $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{KCl}$(aq). Would my cell still work?

Also, I don't understand the need of a porous pot OR salt bridge in a galvanic cell. Why would it matter if the solutions are mixed together? Why not just have one single beaker, containing an electrolyte $\ce{KCl}$(aq), and have the two different metals dipped into it. In fact, if the solutions are mixed together, there will not even be unbalance charge accumulation, and the porous pot/salt bridge would not be necessary?

The electrolyte is very important for the functioning of the galvanic cell. The emf that the cell generates depends on the type and the concentration of the electrolyte used. In the Daniell cell, the reaction would be:

$$\ce{Zn + CuSO4 -> ZnSO4 + Cu}$$

As per the Nernst's equation, you get the emf of the cell to be:

$$\text{E}=\text{E}^\text{o}-\frac{2.303\text{RT}}{\text{nF}}\log_{10}{\left(\frac{\ce{[Zn^{2+}]}}{\ce{[Cu^{2+}]}}\right)}$$

Clearly, the emf of the cell depends on the concentrations of the cell. Dipping the rods in $\ce{NaCl}$ would not make the cell work.

And for why the salt bridge is needed, think about why you would build a cell in the first place. To generate power in a usable way right? If you let the solutions mix, they would react directly, and the energy would be released as heat instead, and not as electrical power. You want the ions to move towards the electrodes and get discharged there.


Color changes of a solution of acetone and sodium hydroxide I have put acetone and sodium hydroxide solution (approximately $\pu{1M}~ \ce{NaOH}$) in a beaker and didn't see any colour change for a day (actually, it was transparent, no colour).

After almost 2 days, I saw a dark brown colour (no precipitate, just diffused colour). Now (after 25 days) its colour seems black, but when I take a small amount and dilute it with water, it is brown. What chemical compound has formed that has this colour? What reaction has taken place? (I doubt it is an aldol condensation)

Under basic condtions, acetone does condense with itself, first to form mesityl oxide, which is colorless, and then to form isophorone, which has a yellow color.

Both mesityl oxide and isophorone are still reactive carbonyl compounds likely to be attacked by deprotonated acetone carbanions.  That leads to bigger and more conjugated condensation products, which leads to darker and darker colors.

I found a paper which looks in some detail at the base-catalyzed polymerization of acetone, albeit with a different base and at high-temperature conditions in the gas phase, but I suspect a similar pathway happens, more slowly, at ambient temperatures in the liquid.  In the liquid, even larger molecules can likely form because they will stay solvated and can keep reacting (as opposed to in the gas phase where big molecules "crash" out of the phase as soot/gunk/etc.)

The color is likely in part from isophorone but much moreso from large polyconjugated molecules derived from isophorone containing many consecutive sp2 hybridized carbons. Such compounds are very often colored due to intense absorption of visible light.


What are the parent compounds of the following? I just need these two answered correctly so I have examples to work with, also if someone could label the longest continuous chain, that would be great too!

Thanks!

Find the longest chain by starting at one end of the molecule and counting the number of carbon atoms along each possible path.  I've attached a figure with numbering of the longest chain.  Hopefully it will answer your question and you will understand how to find and number the longest chain.  In the future, post your thoughts, so we can better understand how to help.


Shape of C₂H₂ molecule What is the shape of the $\ce{C2H2}$ molecule?

The fact that is has a triple covalent bond between the 2 carbon atoms and 2 other bonding pairs between the hydrogen and carbon atoms leads me to believe that it could be linear shape.

However, I'm not sure if the triple bond counts as a single electron bonding pair or 3. If it counts as 3 bonding pairs then the molecule would have 5 overall, and therefore be a trigonal bipyramidal shape.

So is it octohedral or linear? Pretty sure it's linear

According to VSEPR theory, each carbon has 2 electron "pairs".  These electron pairs are not what you would typically think of as electron pairs because VSEPR theory considers double bonds and triple bonds to still be a single "pair".

So considering each carbon atom, the shape is $AX_2E_0$ ($A$ is the carbon atom, $X$ is a bonding pair of electrons, and $E$ is a non-bonding pair (lone pair) of electrons).

$AX_2E_0$ corresponds directly to a linear molecule.

To directly answer your question, when trying to determine the shape of a molecule, consider double and triple bonds as only a single electron bonding pair.


Why is osmotic pressure only for solvent I believe that the definition of osmotic pressure means that it is only about solvent passing through a semipermeable structure.  But is there a such thing as a solute passing through a semipermeable structure because it is very small too?

This has been bothering me for a couple days.

Osmosis depends on the separation of a solution from its pure solvent by means of a  membrane permeable to solvent but not to solute. The semipermeable membrane can take several forms, cellophane for example is impermeable to high molecular mass compounds, such as polymers, but allows water to pass though. Palladium foil is permeable to hydrogen gas but not nitrogen or oxygen so osmosis can be performed in the gas phase.

The experimental arrangement is for the pure solvent to be separated from the solution by the semipermeable membrane. The natural tendency is for the solvent to flow from the pure solvent chamber into the solution. The tendency is counteracted by applying pressure only to the solution side. The excess pressure is called the osmotic pressure.

Now if the solute as well as solvent could diffuse through the semi-permeable membrane there would soon be a solution of different concentrations on either side of the membrane. Osmosis would now work both ways across the membrane and result in equilibrium when the same solution concentration is on both sides.


Rocket propellants: accuracy of vapor pressure estimation techniques I would like to determine the vapore pressures of a couple of substances (rocket propellants) over a significant temperature
range within a ~10% error margin.

For a first test, I tried to approximate the vapor pressure curve of Dinitrogen Tetroxide (N2O4) via the Clausius-Clapeyron (CC) relation, the Lee-Kesler (LK) method and the Ambrose-Walton (AW) method and compare the results to each other and to literature values that are available for N2O4.
Unfortunately my results deviate alot (up to 60%) from the published values.

Attached you find a plot of my attempts and a small table that compares the different methods and their deviation from the literature values.
Don't let the first picture mislead you! As the table and the detailview show, even at low pressures/temperatures the deviation is quite significant!

Is this large error margin really within usual boundaries? What are your experiences with vapor pressure approximation? Could you provide some tips on how to improve my results?

Some Details on my approach

Clausius-Clapeyron:

I chose the boiling temperature at atmospheric pressure as input, but I have no idea which value to
choose for the enthalpy of vaporisation. Many source state the assumption of a constant enthalpy,
but that puzzles me, since for N2O4 it varies by some 16kJ/mol over the temperature range I'm interested in.

Which value is usually taken - the enthalpy corresponding to the values provided as input or the one for the target temperature?

Lee-Kesler / Ambrose-Walton:

As expected both methods produce very similar results and are convenient to use, since they only
rely on the critical pressure, critical temperature and the acentric factor. But MY results are really
disappointing, especially since Wikipedia promises a maximal deviation of 10%.

As a side note:

For different reasons, I don't want to use interpolation of substance specific vapor pressure tables,
but instead I would like to use one of the many vapor pressure equations. It is important that the
approximation results in a closed representation of the vapor pressure curve , therefore I would like
to avoid methods that have points of discontinuity (e.g. Antoine equation?). Since I don't have access
to public databases that cover the substances I'm interested in, approximations that rely on "exotic"
empirical parameters (Wagner, DIPPR, etc.) are also out of the question.

I would like to determine the vapor pressures of a couple of substances (rocket propellants) over a significant temperature range within a ~10% error margin.

The best way to determine vapor pressures is to measure them.  Is that not an option for you?

  It is important that the approximation results in a closed representation of the vapor pressure curve , therefore I would like to avoid methods that have points of discontinuity (e.g. Antoine equation?).

The Antoine equation isn't discontinuous by nature, it's only discontinuous if you use two different parameter sets for a single substance in order to widen the temperature range of applicability of the equation.

  What are your experiences with vapor pressure approximation? Could you provide some tips on how to improve my results?

I would recommend the Antoine equation as a first step beyond the Clausius Clapeyron equation.  Unlike the CC equation, the Antoine equation does not assume that the enthalpy of vaporization is independent of temperature.  It also provides a single, closed-form representation of the vapor pressure dependence on temperature.

Unfortunately, the Antoine equation requires compound-specific parameters.

However, many of the other methods you report using, such as the Lee-Kesler and Ambrose-Walton methods, also require compound specific parameters: specifically, the critical temperature and pressure of the compound of interest.  Even the Clausius-Clapeyron equation requires an (assumed constant) enthalpy of vaporization, which is also a compound-specific parameter.

Thus, the Antoine equation is probably the best place to start.  As Wikipedia shows, it can have accuracy better than 10% deviation for some compounds.  Ideally you would measure vapor pressures for your two compounds and then fit the data to the Antoine equation.  If measuring the vapor pressure of your compound is not possible, you could start by examining the Antoine equation predictions for a series of compounds that approximate your unknowns in chemical structure and molecular weight.  However, with this latter approach, I doubt error under 10% will be possible.

Antoine equation parameters for your compounds of interest

UDMH has Antoine parameters available from NIST.
The case of dinitrogen tetroxide is complicated by the equilibrium $\ce{N2O4 <=> 2 NO2}$.  It is unusually difficult to fit vapor pressure data to.
Hydrazine has Antoine parameters available from NIST.
In general NIST has Antoine parameters for a variety of substances.


What is the fate of sulfur in cysteine when it participates in gluconeogenesis? Amino acids can be broken down to yield pyruvate, which in turn can be used to construct glucose molecule in the process of gluconeogenesis.  What happens to the sulfur atom in the amino acid cysteine when it is processed into pyruvate?

Image source: NCBI Bookshelf

Cysteine is a dispensable amino acid required for synthesis of protein and non-protein compounds. These non-protein compounds include pyruate (hence acetyl coenzyme A), taurine, sulfate, and glutatione (GSH). The fate of the cycteine is hence sketched in following diagram (Ref.1):

As evidence from the diagram, the sulfur atom in cysteine has been converted to sulfate ion ($\ce{SO4^2-}$) through $\ce{HS-}$ ion. This fact has been previously proved experimentally (Ref.2). Accordingly, the first major step in cysteine catabolism in mammalian tissues is the oxidation of cysteine to cysteine sulfinic acid catalyzes by cysteine dioxygenase (CDO). The reaction mechanism of this non-heme iron dioxygenase has been recently reviewed (Ref.3).

References:

G. Courtney-Martin, P. B. Pencharz, “Chapter 19: Sulfur Amino Acids Metabolism From Protein Synthesis to Glutathione,“ In The Molecular Nutrition of Amino Acids and Proteins; Dominique Dardevet, Ed.; Academic Press, Elsevier Inc.: London, United Kingdom, 2016, pp. 265-286 (https://doi.org/10.1016/C2014-0-02227-7).
Catherine L. Weinstein, Rudy H. Haschemeyer, Owen W. Griffith, “In vivo studies of cysteine metabolism. Use of D-cysteinesulfinate, a novel cysteinesulfinate decarboxylase inhibitor, to probe taurine and pyruvate synthesis,” The Journal of Biological Chemistry 1988, 263(32), 16568-16579 (http://www.jbc.org/content/263/32/16568.full.pdf+html).
Crisjoe A. Josepha, Michael J. Maroney, “Cysteine dioxygenase: structure and mechanism,” Chem. Commun. 2007, (32), 3338-3349 (DOI: 10.1039/B702158E).


How can I know when I have added an excess of magnesium oxide? The following instructions were used to prepare magnesium sulfate crystals, $\ce{MgSO4 . 7H2O}$.

Measure $50~ \mathrm{cm^3}$ of dilute sulfuric acid into a beaker and warm the solution.
Using a spatula, add some magnesium oxide and stir the mixture. Continue
adding the magnesium oxide until excess is present.
Separate the excess magnesium oxide from the solution of magnesium
sulfate.
Heat the solution until crystals form. Obtain the crystals and dry them.

The question I have an issue with is this: How would you know when excess magnesium oxide is present in step 2?

The answer key says that I would know when no more solid (that is, no more magnesium oxide dissolves). However, since we are talking about dilute sulfuric acid, there should be water, and I read on the internet that $\ce{MgO}$ is soluble in water.

$\ce{MgO}$ is insoluble in water (solubility: 0.0086 g/100 mL). On the other hand, $\ce{MgSO4}$ is soluble in water (solubility: 25.5 g/100 mL).

The reaction that is taking place is an acid-base reaction:

$$\ce{MgO + H2SO4 -> MgSO4 + H2O}$$

Now for your question, how would you know when excess $\ce{MgO}$ is present? Here you are trying to make $\ce{H2SO4}$ the limiting reagent. When all the $\ce{H2SO4}$ has reacted, no more $\ce{MgO}$ can further react. Therefore, you will see solid in your reaction mixture that cannot be further dissolved.

At the end of your reaction,

$$\ce{MgO_{excess} + H2SO4 -> MgSO4 + H2O + MgO_{remaining}}$$


Why carbon nanotubes can act as a good electrical conductor? Bulk carbon is an insulator, but at nanoscale (e.g. carbon nanotubes) carbon may become a good electrical conductor. If I am correct this means  carbon nanotube band structure is metal-like. E.g. it has partly filled valence band (like in the rightmost case in the figure below).

At nanoscale materials are influenced by quantum confinement. Because of quantum confinement semiconductor nanoparticle has higher band gap and discrete energy levels compared to bulk semiconductor (see the figure about quantum confinement below).

I am assuming that quantum confinement phenomenon also occurs for carbon at nanoscale. Then, how certain carbon nanotubes can show good electrical conductivity?

I am sorry to say, but your question contains several false assumptions, most importantly:

Nanotubes are good conductors:

No, not all of them. Certain types have metallic/semi-metallic conductance, others are semiconductors. Single wall carbon nanotubes are generally indexed with a so-called chirality index (m,n) that tells you how the carbon sheet is wrapped up into a tube. See wikipage.
Nanotubes behave exactly like 3D nanoparticles in terms of band structure:

No, nanosheets (graphene) and nanotubes behave in different ways and your description of quantum confinements don't work in that way in 1D infinite (nanotubes) and 2D infinite (graphene) cases.

You can derive an approximate band structure of graphene/nanotubes using tight-binding models, which in most cases predicts correctly, whether the structure is metallic or semiconducting. See simulation of nanotube band structure.


Stronger ESI signal for deuterated substances I observe a constant phenomenon in my ESI measurements, which indicates that deuterated substances give a stronger signal than the same substance in non-deuterated form. Unfortunately I can't make any sense of it, because I can't see any reason why a deuterated substance should be protonated any better. Since I have not seen or read anything about this phenomenon so far, I would be very pleased if someone could clarify whether this is something well known, or whether my samples just happen to show the same artifact over and over again. If it is a known phenomenon, I would also be happy to hear an explanation.
Further bits of information: LC-MS/MS in ESI mode, same acquisition method for both forms of the compound, observed in EPI scans, increase in intensity c.a. 3-fold.
To make this very clear, the two compounds (deuterated and non-deuterated) are measured separately in separate vials with a blank in between, so no co-elution can occur. One measurement is done with a solution of 100 ng/mL estradiol in acetonitrile and one with 100 ng/mL estradiol-d5 in acetonitrile. They are not mixed!

There are a couple reasons this could be happening.
The most interesting possibility: isotope effects during ion fragmentation
You are performing LC-MS/MS, i.e., your mass spectrometer is chemically reacting the estradiol and estradiol-$d_5$ with gas molecules in the collision cell of the instrument.  This process, known as CID or collisionally induced dissociation, will break chemical bonds in the parent molecule and form fragment ions.
On triple-quad and qTOF instruments, the CID is of the "beam" or non-resonant type.  This means a fairly large amount of energy is imparted to parent ions.  In general, this means that a wide variety of fragment ions are formed.  This is true of estradiol; publicly available fragmentation spectra show many fragment ions for this compound (see several GNPS spectra for example).
Deuteration is likely to change the relative propensity of the different fragmentation pathways.  For example, these reactions could represent the fragmentation of a parent ion $\ce{P}$ into two different fragments $\ce{F1}$ and $\ce{F2}$.
$$\ce{P -> F1}$$
$$\ce{P -> F2}$$
Isotopically normal $\ce{P}$ might give an intensity ratio of say 1 to 1 for $\ce{F1}$ and $\ce{F2}$.  But due to deuterium isotope effects, the pathway to $\ce{F1}$ may be much less favored in deuterium-labeled $\ce{P}$.  This could lead to an enhanced signal for $\ce{F2}$ as a much higher fraction of the $\ce{P}$ ion remains available to proceed down that fragmentation pathway.
However, a three-fold increase is pretty large.  I was able to find good measurements of various deuterated tocopherols, and judging from Table 1 in that paper, the response ratio for $d_0$ (unlabeled) tocopherol vs. $d_9$ is 11.8/0.967/10, or 1.22, not 3.  For measurement of trimethylamine N-oxide, authors note "the intensity ratios of the two product ions are different between TMAO and $d_9$-TMAO due to the difference in bond energy among C and H and C and D (deuterium)" but did not quantify this difference.  Eyeballing Figure 1, the intensity ratio of the light and heavy fragment is about 2.5:1 for $d_0$ but 1.4:1 for $d_9$.  None of those values are close to 3-fold.  But maybe estradiol and its site of deuteration and its fragmentation pathway differ and that a value of 3-fold is possible.
Errors in sample preparation
Maybe the concentration of one of your samples is off.  How many times have you independtly prepared the standards and repeated the measurement?
Ways to tell
If you can see a signal with a single quadrupole (i.e. do MS1 instead of MS/MS), do you see the same three-fold difference?  If so, error in sample prep is likely to blame.  If not, then isotope effects during ion fragmentation are to blame.


Why are alkyl halides used as solvents for relatively non polar compounds and not polar compounds? It is written in my textbook that :

  Alkyl halides are used as solvents for relatively non polar compounds.

Alkyl halides are themselves polar with carbon having delta positive charge and halide having delta negative charge. If polar compounds are dissolved in them, then there will be more interactions between the compound and alkyl halide. The polar compound should dissolve in it.

What do they mean "relatively non polar compounds are soluble in alkyl halides" ? How will non polar compounds become soluble and why the textbook does not mention solubility of polar compounds?

Polarity is relative. Dipole moment (macro scale) and dielectric constant of the solvent (molecular scale)  are probably the most decisive. Water has ε~80, ethanol 25 and DCM is "only" 9. By comparison toluene, benzene or tertachlormethane have ε~2. The last example also illustrates why differences in electronegativity are not enough. Being symmetric (Td) tetrachloromethane has dipole moment of zero, even if there is a big difference between electronegativity of carbon on chlorine.


Why do we measure the initial rate of a reaction? Why do scientists usually measure the initial rate of a reaction, instead of measuring the rate some time into the reaction?

First, note that one does not always measure initial rates. It is fairly common to fit an entire reaction curve (or nearly entire), but it requires a little bit more math. It is also common to measure an "initial" rate that is not quite initial. That is, it is the rate after a burst phase has occurred, and the system has settled into something close to a steady state.

That said, measuring initial rates is done for two practical reasons.

1) we can approximate the product concentration as 0. For a reaction that is at all reversible or product-inhibited, the math is much easier if [P]=0.

2) we can often approximate the concentration of one (or more) of the reactants as fixed. For example, if you measure the rate only for an amount of time required for a few percent of the reactant to be consumed, you can get a pretty good fit to the data by assuming that reactant is fixed at its initial concentration. That makes it easy to produce plots of the initial rate vs the concentration of that reactant.

Both of these factors make it much easier to determine basic rate laws and the approximate value of the associated rate constants.


Dissolution of lead chloride because of the formation of- Na2PbO2 or [Pb(OH)4]-2? I read that $\ce{NaOH}$ precipitates lead hydroxide which dissolves with excess hydroxide.

In some sources, I read that lead chloride dissolves due to the formation of complex $\ce{[Pb(OH)4]^2-}$ but my teacher taught that it is because of the formation of $\ce{Na2PbO2}$

I don't think that both are same. I cannot find much information about $\ce{Na2PbO2}$ either.

Which of these two is actually formed?

Actually there are two reactions:

$$\ce{Pb(OH)2 + NaOH → Na[Pb(OH)3]}$$

  Lead(II) hydroxide react with sodium hydroxide to produce sodium
  trihydroxoplumbate(II). Sodium hydroxide - concentrated solution. (chemiday.com)

$$\ce{2NaOH + Pb(OH)2 → Na2[Pb(OH)4]}$$

  Sodium hydroxide react with lead(II) hydroxide to produce
  tetrahydroxoplumbate(II) sodium. (chemiday.com)

In the Wikipedia article of lead hydroxide, it is written that:

  In solution, lead(II) hydroxide is a somewhat weak base, forming
  lead(II) ion, $\ce{Pb^2+}$ under weakly acidic conditions. This cation
  hydrolyzes and under progressively increasing alkaline conditions,
  forms $\ce{Pb(OH)+}$, $\ce{Pb(OH)2(aq)}$, $\ce{Pb(OH)3−}$, and other
  polynuclear species like $\ce{Pb4(OH)4^4+}$, $\ce{Pb3(OH)4^2+}$,
  $\ce{Pb6O(OH)6^4+}$

Sodium plumbite ($\ce{Na2PbO2}$) is formed by the reaction of lead(II) oxide and sodium hydroxide.

$$\ce{PbO + 2NaOH -> Na2PbO2 + H2O}$$

Some extra info.

The oxoions and hydroxoions of lead are called plumbates like  hydrated plumbate anion, $\ce{[Pb(OH)6]^2−}$ or the anhydrous anions $\ce{[PbO3]^2−}$
(meta-plumbate) or $\ce{[PbO4]^4-}$ (ortho-plumbate). The hydrated plumbates are formed by dissolving lead(IV) oxide in alkali and the anhydrous ones are formed by heating alkali or metal oxides with $\ce{PbO2}$. All plumbate(IV) salts are very strong oxidising agents and the hydrated plumbate(IV) salts decompose upon dehydration.


Na2CO3 + FeBr3 + H2O: why do we get Fe(OH)3? $$\ce{3Na2CO3 + 2FeBr3 + 3H2O -> 6NaBr + 2Fe(OH)3 + 3CO2}$$

Why $\ce{Fe(OH)3}$? Where do these $\ce{-OH}$ groups come from?

This is from a series of reactions in a Russian sample exam question.

P.S.

Following Ivan't hints, I've gone through the process of reaction, and I get some water in the right-hand side of the equation. I wonder why the answer has no water in the right-hand side.

You have certainly already recognized $\ce{Na2CO3}$ as the base in your equation.

Did you also consider the following:
When $\ce{FeBr3}$ is dissolved in water, the hexaaqua complex $\ce{[Fe(H2O)6]^3+}$ is formed. This complex is fairly acidic:
$$\ce{[Fe(H2O)6]^3+ + H2O<=> [Fe(H2O)5OH]^2+ + H3O+}$$

Ignoring ligand exchange processes, you might state that the $\ce{OH}$ groups come from the deprotonation of the water ligands of your initial complex.


Is it possible to make synthetic oil without fossil fuel? If human ran out of crude oil and all other fossil fuel, are we still able to create synthetic oil? I've searched online about the raw material of synthetic oil, but I'm still not sure if none of the raw materials is from crude oil.

Also is there any synthetic oil with viscosity less than 10cSt?

Yes, it is being theorized and it's being done:

Example A:
https://www.researchgate.net/publication/264810861_Engineering_cyanobacteria_for_production_of_a_cyclic_hydrocarbon_from_CO2_and_H2O

Example B:
https://www.sciencedirect.com/science/article/pii/S2211926416304982

Example B is more theoretical with economic considerations, and example A is more methodical. In these papers, we see cyanobacteria being used to make the aromatic hydrocarbon limonene with the limonene synthase gene derived from a plant.

As for can we make synthetic oil that is chemically proportional to fossil fuel, Probably not ... but by adjusting the ratios of aromatic hydrocarbons, we can measure some useful properties like the heat value of the fuel for which limonene has a heat value very close diesel fuel and the density is close as well as you can see if you read the aforementioned Life Cycle Analysis paper. If you want to explore the chemical composition of fossil fuel then you need to do some spectroscopy work.

But fossil fuel isn't perfect. There is residual incombustible material, and perhaps by adjusting the proportion of combustible aromatic hydrocarbons in fossil fuel maybe we can get more energy from fossil fuel. This is also a function of the funding that these research projects get and also industrial support which is also a function of money ... which is why you need high yield in these reactions, hence genetically engineered cyanobacteria.


Most stable conformational isomer of 3-methoxycyclohexan-1-ol Among the following, the most stable isomer is?

I am aware of the fact that equatorial substituents are more stable than axial substituents but couldn't proceed to apply it here. However the answer key gives the answer as d) in which both substituents are in axial position.

Sometimes, the intuitive rule of thumb that favours conformations with $\ce{OH}$ substituents in equatorial positions miserably fails.

For 1, a cyclization product of geranyl acetate, one might assume that the favoured conformation in solution and in the crystal is 1eq. However, this is not the case! Both NMR spectroscopy and X-ray crystallography indicate that 1ax is the actual conformation.

In the case of 1, the reason most likely isn't a (stabilizing) intramolecular hydrogen bond that would form a seven-membered ring, but the sterical interaction of the methyl groups. The unfavourable interaction of the 1,3-diaxial methyl groups in 1eq is avoided in the conformer with the axial $\ce{OH}$ group.


Stabilisation of anions by silicon In my lecture notes it was stated that in $\ce{Me3SiCH2SMe}$, deprotonation with tBuLi generates an anion in between Si and S. I've been not able to find the $\mathrm pK_\mathrm a$ values (not listed in Reich nor Evans).

Since Si is electropositive it should actively destabilise the anion, meaning that I'd expect to see terminal methyl group deprotonated.

No additional detail about conditions was given, but since it was tBuLi I presume low temperature (kinetic deprotonation).

Is there a reason why this deprotonation takes place? Are the $\mathrm pK_\mathrm a$ values relatively different?

3rd period heteroatoms can stabilize adjacent carbanions charges via hyperconjugation.

Consider ylides used for Corey,-Chaykovsky, Peterson, and Wittig olefinations. Granted those have formally positive sulfur, silicon, and phosphorus, but you will not see similar easy deprotonation with tetraalkylammonium for example.

The reactions of silicon-stabilized carbanions are described more fully in Clayden 1st ed., pp 1304–5 (this material seems to have been cut from the 2nd edition).


What is the link between electrode surface areas and EMF? I was wondering if someone would help me with my question.
I was completing a question which asked about the effect of increased surface area of a platinum electrode on the EMF of the overall cell.

This is the cell: Pt | H2O2(aq), O2(g) || IO (aq), I2(aq) |Pt

When I answered, I thought the EMF would increase due to the surface area of the electrode being able to now accept more electrons but I found out the answer is that the EMF is unaffected by any change in surface area, why is this?

Thanks for any help anyone can provide!

If by EMF you mean the ideal, thermodynamically determined, voltage then the answer is that increasing the surface area does not affect voltage/EMF. If by EMF/Voltage, the questions means operational voltage then it will increase because increasing the surface area of electrodes decreases the overpotential losses (Tafel equation) by essentially decreasing the resistance due to the kinetics of the reaction thus increasing operational/real voltage of a cell.


Relationship in equilibrium constants $\ce{2CH4(g) + O2(g) \rightleftharpoons 2CO(g) + 4H2(g)}$

  Equal numbers of moles of $\ce{CH4}$ and $\ce{O2}$ are drawn into a
  sealed container and the system is left until it reaches equilibrium.
  Which of the following relationships pertaining to this particular
  system are true at equilibrium?

  $\ce{[CH4]<[H2]}$
  $\ce{[CH4]>[O2]}$
  $\ce{[H2]>[CO]}$

I know that the equation for $\ce{K_{eq}}$ is

$$\ce{\dfrac{[H2]^4[CO]^2}{[CH4]^2[O2]}}$$

I think that 3. is wrong because if $\ce{CH4}$ and $\ce{O2}$ are equal, then $\ce{H2}$ and $\ce{CO}$ have to be the same. Now I'm confused about 1 and 2. I'm not sure how to distinguish between the two.

Let's say that the amount of $\ce{CH4}$ and $\ce{O2}$ at the start is $x$.

After the reaction is complete, let's say that $y$ moles of $\ce{O2}$ reacted.

The amount of moles of $\ce{CH4}$ left is $x-2y$;

The amount of moles of $\ce{O2}$ left is $x-y$;

The amount of moles of $\ce{CO}$ formed is $2y$;

The amount of moles of $\ce{H2}$ formed is $4y$;

That should make one of the choices clearly false, one clearly true and one might not be so obvious, unless you assume that the forward reaction path is less favourable ($\ce{CO}$ is not very stable)