Can we tell if a compound is explosive by just looking at its chemical structure? If I showed you a compound's chemical structure will you be able to tell if it can be used as an explosive or we can only know that by conducting chemical experiments?

A compound, a mixture may explode under different conditions.  As an example, for some of them it is heat, other require electric discharge or mecanical impact to decompose.  There are classes of compounds with known examples of explosives (azides, picrates, fulminates) so one may infer new compounds within these classes equally might be explosives, too.

Among modern explosives, increasing the number of nitrogroups or / and increase of nitrogen content over other elements present is a theme, like octanitrocubanes.  See the Klapötke group in Munich / Germany working on 5,5’-azotetrazolate salts, too.

(dihydrazinium 5,5’-azotetrazolate dihydrazinate, source)


What would happen if we had a different kind of metal in our bloodstream? It is a common fact that we have iron in our blood. But what if we had something different in its place? Like instead of iron what if it was copper, or silver, or what if we had gold in our blood stream instead? Theoretically, what would that do?

To start, transition metal complexes with large porphyrin-like molecules are so electronically complex that standard chemical intuition doesn't apply and you have to either do the experiment or run a (nearly impossible) calculation to determine what effect a non-physiological metal would have on the protein activities. In the case of hemoglobin, your best bet would be to think about metals which have two accessible oxidation states because the interconversion of Fe(II) and Fe(III) is critical to the function of hemoglobin.

To give a broader answer, let's think about the general question of substituting non-native metals into a protein. This is, in fact, an very interesting and open question. On smaller metalloproteins there have been studies probing whether nature 'picked' the best metal for a task or just a convenient/abundant one [1,2]. The focus of those studies is on catalysis, and there are cases where many non-physiological metals can function in the protein and even catalyze the reaction as good or even better than the wild type.

[1] https://pubs.acs.org/doi/10.1021/ar500227u

[2] https://pubs.acs.org/doi/abs/10.1021/acs.jpcb.6b00997


Use of Δ (delta) to describe alkenes In a lot of (mainly older) papers discussing organic synthesis,  Δ is used to describe alkenes. As far as I know, this is no longer (if ever) IUPAC recommended nomenclature, so I haven't been able to find any reference to its intended use.

The main area I'd seen it used was in steroid synthesis, where I assumed it had some specific use relating to that class of natural product (i.e. related to the biosynthesis etc.). However it does appear in other places, such as in the synthesis of Rifamycin:

 Scheme from Nicolau Rapamycin synthesis. Taken from Classics in Total Synthesis I

"… the Δ35,36 double bond …" is used in the text to describe the alkene that gets reduced. Given that this could easily be referred to as the (E)-alkene or even just 206, I suspect that there is a special reason why Nicolau chose Δ35,36 however that reason isn't apparent to me.

@Loong's encyclopaedic knowledge of IUPAC nomenclature prompted me to have another look through the (irritatingly unsearchable) 2013 recommendations and the 2004 draft upon which they were built.
The only use of the uppercase Δ listed is when describing a double bond that links two cyclic systems:

P.28.2.2: Ring assemblies with a double bond junction
When two cyclic systems are linked by a double bond [...] the presence of a double bond is indicated by the Greek letter Δ, and the point of attachment of the ring is given by superscript locant numbers. This method is not continued in these recommendations

The example given is 1'1-bi(cyclopentylidene) (PIN) which should be used in preference of Δ1,1'-bicyclopentylidene.

Based on this, the example given in the question, using Δ to describe an unsaturated carbonyl would appear invalid, and used for convenience rather than to convey anything specific.


Converting a PDB file to XYZ file I am trying to convert a pdb file of gold nanoparticle to the xyz file format to get the atomic coordinates. But when I try to use openbabel to do this, I get the following error.

*** Open Babel Error  in ReadMolecule
ERROR: not a valid PDB file

My command is obabel shelxpro.pdb -O shelxpro.xyz

(shelxpro.pdb is at http://www.sciencemag.org/content/318/5849/430/suppl/DC1
inside the .zip folder 1148624_data.zip)

Also when I try to visualize this structure with Jmol or VMD, I get only part of the expected structure.

Is there a way to fix either one of these issues?

The conversion issue you're seeing is because the PDB you're using is not technically well-formatted. If you take a look at the output, you see a bunch of the following warnings.

==============================
*** Open Babel Warning  in parseAtomRecord
  WARNING: Problems reading a PDB file
  Problems reading a HETATM or ATOM record.
  According to the PDB specification,
  columns 77-78 should contain the element symbol of an atom.
  but OpenBabel found '  ' (atom 276)
==============================

This thread from the OpenBabel mailing list goes into more details. Basically, the official PDB specification requires that the atom names be present off to the right-hand side of each ATOM/HETATM line. Older PDBs (and lazy PDB-outputting software) don't do that.

For the most part that works, as the atom name entry normally contains the element designation. Parsing this is tricky, though, as it doesn't always work.

That's what's causing Open Babel to choke. Your PDB contains gold atoms, and Open Babel doesn't know how to handle the "AU" atoms. If you manually edit the ATOM lines to contain the letters "Au" in columns 77-78, then Open Babel is able to make the conversion, but still warns about the C/N/O atoms which don't have their elements correctly designated. - But for C/N/O it's okay, as Open Babel is able to figure out the element from the atom name.

If you don't want to manually edit the lines, you just need to pass the PDB file through a program that can make sense of the AU atoms, and will output the element name in the appropriate colum. VMD doesn't seem to help (it reads things fine, but doesn't add the element annotation), but PyMol is able to read in the file and when re-saving the molecule will put the correct values in columns 77-78.

Regarding the "missing atoms", the PDB file contains a crystallographic symmetry record in the first four lines. (CRYST1, SCALE1, SCALE2 and SCALE3 records). I'm guessing the atoms you're missing are related by crystallographic symmetry to the ones that are present. To generate those atoms, use a program which can use the PDB symmetry records to generate the symmetric partners. (PyMol is one such program.)


Henry's constant and solubility According to Henry's law

  The solubility of gas in a liquid is directly
  proportional to the partial pressure i.e S=Kp

My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid.
I don't understand how. I know I am overseeing something really simple over here as all the books that I have and online websites seem to conclude the same from the given equation. A little help maybe. Thanks.

Actually solubility $x$ is given as a function of pressure $p$ and henry's constant $K_H$:
$$p=K_Hx$$

  My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid. I don't understand how.

Since at a given pressure, $p$ is constant, if you increase $K_H$, logically solubility $x$ would decrease.


What chemical will dissolve the plastic coating on circuit boards I'm trying to reclaim the metals on circuit boards and the vast majority is underneath the plastic coating. It is a brittle plastic that chips off in small flakes when etched but that process will lose to much of the precious metal.

This seems to be a popular question judging from some Google queries.

Printed circuit boards are made of a variety of materials, and from what I understand, the specifics may vary somewhat based on the specific needs of the product and the manufacturer. For example, in some cases, I know the PCBs are coated with special anti-static coatings to prevent delicate RAM and CPU chips from being damaged (e.g., this article found via Google among others).

So I doubt there's a definite solution - more likely you'll need to try several different things.

That said, there's a recent article that looks very interesting about separating parts by dissolving and swelling the plastic resins with DMSO: "Dissolution of Brominated Epoxy Resins by Dimethyl Sulfoxide To Separate Waste Printed Circuit Boards" Environ. Sci. Technol., 2013, 47 (6), pp 2654–2660.

Certainly DMSO is considered a near-universal solvent and will swell many polymers. So I'd probably try that first.


why do atoms need to be ionised in a mass spectrometer? why do atoms need to be ionized in a mass spectrometer? why can't you determine the mass of the atoms when they are not ionised? Also what is the mass-to-charge ratio

why do atoms need to be ionized in a mass spectrometer? why can't you determine the mass of the atoms when they are not ionised?

There are various types of mass spectrometer
https://en.wikipedia.org/wiki/Mass_spectrometry#Mass_selection
but all of them require neutral particles to be charged in order to be accelerated and then detected. The spectrometers use electromagnetic fields to accelerate the ions, this effect only occurs if the species are charged as the field will not accelerate a neutral particle.

For example, in the magnetic sector machine, the ions are formed in an ionization chamber, accelerated in a uniform electric field and shot into the magnetic sector where they are sorted by their path through the field. The higher the charge on the particle, the more it is defected, the higher the mass, the less it is deflected:

http://orgchemguide.blogspot.com/2011/04/magnetic-deflection-or-sector-mass.html

  Also what is the mass-to-charge ratio

So the ions are sorted not by their charge or mass alone, but by the ratio of the two, the mass to charge ratio. The machine cannot tell the difference between an ion with mass 4 amu and charge 1+ and another with mass 8 amu and charge 2+, both have the same mass to charge ratio and will behave identically in the field.

Most detectors also work only if they are struck by a charged particle.

Here is a great guide to the simple principles of how a MS works:
https://www.chemguide.co.uk/analysis/masspec/howitworks.html


How it is possible to make Aqua Regia in the lab? How to make "Aqua Regia" in the lab conditions? Which lab equipment must be used?

Typically, 1 part of concentrated nitric acid is added to 3 parts of concentrated hydrochloric acid.

Personal protection (goggles, gloves, lab coat) is vital.
Prepare and use it under a well-vented fumehood only - hands inside the fumehood, head outside!

UPDATE

The comment of Crystal Lettuce is very important and points to a crucial aspect:

Aqua regia isn't a mixture to keep bottled on the shelf: it has to be prepared and used immediately!


Prove that a 10-Degree Temperature Increase Doubles the Rate Constant (k), when the Activation Energy is Approximately 50 kJ/mol I read that increasing the temperature by $10~^\circ\mathrm C$ will double the rate constant ($k$), when the activation energy for the reaction is relatively close to $50~\mathrm{kJ/mol}$.

However, no matter how hard I try to wrap my head around it (through graphical examples that require plotting the natural logarithm of $k$ vs the inverse of temperature), I just can't grasp this so called "rule of thumb".

As a result, I tried proving this statement through mathematical expressions:

(Note that the activation energy was evaluated for $50\,000$ because it needs to be expressed in $\mathrm{J/mol}$ to be used in Arrhenius' equation)

Prove that...

  A rise of temperature of $10\ \mathrm{^\circ C}$ will double the rate constant, when the activation energy for the reaction is approximately $50\ \mathrm{kJ/mol}$.

Proof starts here -----------------------------

For:  $\Delta T = T_2 - T_1$ ($T_2$ is the final temperature, and $T_1$ the initial temperature)

When: $\Delta T = 10 \rightarrow T_2 - T_1 = 10$

It follows that: The ratio between $k_2$ (at $T_2$) and $k_1$ (at $T_1$) will equal $2$, owing to the doubling of the rate constant ($k$):

$$
{k_2\over k_1} = 2
$$

Employing Arrhenius' equation:

$$
{k_2\over k_1} = {A\mathrm e^{\left({-E_\mathrm a\over RT_2}\right)}\over A\mathrm e^{\left({-E_\mathrm a\over RT_1}\right)}}
       = \mathrm e^{\left[{-E_\mathrm a\over R}\left({1\over T_2} – {1\over T_1}\right)\right]}
       = 2
$$

Or simply:

$$
{k_2\over k_1} = \exp{\!\left({-E_\mathrm a\over R} \cdot \Delta{1\over T} \right)} = 2
$$

So:

$$
\exp{\!\left({-E_\mathrm a\over R} \cdot \Delta{1\over T} \right)} = 2
$$

This might be the part where I messed up -------------------------

$\Delta{1\over T}$ can be set in terms of $\Delta T$:

$$
\Delta{1\over T} = \left({1\over T_2}-{1\over T_1}\right)
       = \left({T_1\over T_1T_2}-{T_2\over T_1T_2}\right)
       = \left({T_1-T_2\over T_1T_2}\right)
       = \left({-\Delta T\over T_1T_2}\right)
$$

Since Arrhenius' equation above holds true for $\Delta T = 10$ ...

And since $\Delta{1\over T}$ can be expressed as $-\Delta T \over T_1T_2$ ...

Then, the rate constant ($k$) will double when:

$$
\Delta{1\over T} = {-10\over T_1T_2}
$$

Substituting into Arrhenius' equation:

$$
\exp{\!\left({-E_\mathrm a\over R}\cdot{-10\over T_1T_2}\right)} = 2
$$

Simplifying:

$$
\begin{align}
\exp{\left({10E_\mathrm a\over RT_1T_2}\right)} &= 2 \\
{10E_\mathrm a\over RT_1T_2} &= \ln{2} \\
{1\over T_1 T_2} &= {R\cdot\ln{2}\over 10E_\mathrm a} \\
T_1 T_2 &= {10E_\mathrm a\over R\cdot\ln{2}}
\end{align}
$$

From $T_2 - T_1 = 10$ we can solve for $T_2$ as $T_2 = T_1 + 10$. Substituting above:

$$
\begin{align}
T_1 (T_1 + 10) &= {10E_\mathrm a\over R\cdot\ln{2}} \\
T_1^2 + 10T_1 - {10E_\mathrm a\over R\cdot\ln{2}} &= 0
\end{align}
$$

For $E_\mathrm a = 50\,000$:

$$
T_1^2 + 10T_1 - {500\,000\over R\cdot\ln{2}} = 0
$$

Solving the quadratic equation:

$$
\begin{align}
T_1 &= 290~\mathrm K \\
T_2 &= 300~\mathrm K \\
\Delta T &= 10~\mathrm K
\end{align}
$$

I don't think this proof is correct because substituting any value for activation energy ($E_\mathrm a$) onto the quadratic equation will produce different values for $T_1$, but that doesn't tell me anything.

What did I do wrong?
Can this be proved any other way?

Your method and your mathematics seem perfectly fine, and your calculated result is correct. You are also quite correct that the result will change depending on the value of $E_\mathrm a$ that you choose.

The problem you're running into is the assumption that the stated rule of thumb holds exactly, regardless of temperature. It doesn't.

I would set up the problem as follows. Consider the ratio $\rho$ between the reaction occurring at some Kelvin temperature $T$, and at $T+10$:

$$
\rho = {e^{-E_\mathrm a\over R\,\left(T+10\right)}\over
        e^{-E_\mathrm a\over RT}}
     = e^{{E_\mathrm a\over R}\left({1\over T} - {1\over T+10}\right)}
     = e^{{E_\mathrm a\over R}\cdot {10\over{T^2+10T}}}
$$

Plotting $\rho$ as a function of $T$ for $E_\mathrm a = 50~\mathrm{kJ\over mol}$, it becomes immediately clear that the rule of thumb does not hold over a very wide temperature range:

In fact, it only holds strictly for one specific temperature, which is why your math led you to a single temperature as your answer. By my calculations, using a value of $8.3144598~\mathrm{J\over mol\, K}$ for $R$, that temperature is $289.56~K$, which is equal to your $T_1=290~K$ to within three significant figures.

The gray box in the figure marks where the rule holds to within $20\%$ – that is, where $\rho$ falls between $1.6$ and $2.4$. This corresponds to a temperature range of $257~\mathrm K$ to $353~\mathrm K$.


Photosynthesis, science fair So I was thinking about something to do for a science fair and photosynthesis sounded like a good idea.  I was wondering what the exact process of photosynthesis was on a chemical level.  I know the formula, but how exactly is water and carbon dioxide converted into glucose and oxygen?  And where does sunlight come in, not in the equation but the chemical process.  Finally, any ideas for artificial photosynthesis?

I am assuming you are in grade school.
The problem is, as was mentioned before, that photosynthesis is very complicated.
The formula you know of is extremely simplistic and skips over 100 reactions on the way. you can find plenty of information in Wikipedia, or if that is too much you can use the Simple Wikipedia.

In high school we took some kind of water plant or algae and put it in an upside down test tube filled with water, as oxygen was formed it filled the end of the test tube. This is a fairly simple way to monitor photosynthesis and you can use it to study photosynthesis. for example: how does the temperature effect photosynthesis, how does the color of light affect it etc.


How can temperature be constant during a reaction? So I was reading my textbook about the topic thermodynamics.

Here the second paragraph States that the above equation describes the combustion of liquid ethanol at constant temperature and pressure.

But how can the reaction occur at constant temperature. The enthalpy change stated is accompanying the reaction at standard state. So it should be heat released at constant pressure and from temperature 25°C. So temperature of reaction mixture should be decreased.

But how can the reaction occur at constant temperature

Despite the dynamic name, thermodynamics is not concerned about time. Or more precisely, the theoretical model/world-view of thermodynamics allows the reaction to take place over an infinite amount of time, so you have all the time you need to dissipate the heat and thus keep the same temperature constant (within arbitrarily small limits).

To get to practical reality, we have a whole arsenal of possibilities that cover a range from very close to the definition here to experimental conditions that are quite different from those stated here:

The closest one is: The reaction takes place in a thermostat, i.e. in a device that keeps the temperature constant. You may measure the amount of heat that needs to be transported into or out of the reaction vessel to keep the temperature there constant (within practical limits).

As @MaxW says in their comment, you can also approximate this by using a reaction vessel that keeps the temperature approximately constant.

@Maurices comment goes yet a step further: as long as the outcome doesn't change (e.g., no irreversible reations at the intermediate temperature), you may deliberately allow intermediate temperature changes as long as you can keep track of the heat exchange until the reaction products reach the starting temperature again.

And lastly, you could allow temperature change and do a series of experiments that then allows you separate in the data analysis temperature-dependent from temperature-independent parts.


Why is oxygen so special? Why do reactions with oxygen produce flame (i.e. light), heat and hence a lot of energy? I mean, why only oxygen, why not any other element like nitrogen? What is special about the electronic configuration/arrangement or molecular geometry that confers it with this property?

Maybe you
will cite examples of other reactions involving a lot of energy and producing flame, but I am sure there is nothing as familiar to mankind (in terms of the frequency in their occurance) as the reactions involving this compound. Reactions with oxygen are so ubiquitous and vital to life and thus they were classified under a separate category, i.e. combustion.

Dioxygen, $\ce{O2}$ is a very special molecule. A good majority of organisms on earth use dioxygen to breathe and survive. Oxygen is also the second most abundant element in the sky, as well as the most abundant elements in the earth's crust. What makes it so special? Its because of its abundance, which is obvious, and also because of it's strong oxidizing power.

If you are a student below 11th grade, you might be taught that oxidation is addition of oxygen or the removal of hydrogen. I will now, introduce a new definition of oxidation.

Oxidation: The loss of electrons from a substance is called oxidation.

The reagents that cause substances to become oxidized are called oxidizing agents. $\ce{O2}$ is an example of an oxidizing agent. A popular oxidizing agent in analytical chemistry is Potassium Permanganate ($\ce{KMnO4}$ ).

Many times, an actual loss of electron doesn't take place, rather a shift of electrons from one element to another within the same molecule occurs, which is also considered as oxidation. To explain such stuff, we introduce the concept of oxidation numbers. An easy way to find oxidation number of an element in a compound is:

Ox.num=(Num. of bonds formed with electronegative atoms)-(Num. of bonds with electropositive atoms)

For example, take carbon dioxide. It's structure is:

Note that bonds made by an atom with a similar atom are not considered (like $\ce{O-O}$ or $\ce{S-S}$).

Carbon makes four bonds with more electronegative atoms (oxygen). So it's oxidation number is +4.

If oxidation number of any atom of a molecule is increased through a reaction, we say that oxidation has taken place. Note that as this happens, the oxidation number of the oxidizing agent will decrease.

Now, finally (with a deep breath), lets come to your question.

Speciality of Oxygen:

Oxygen in nature, as dioxygen has a 0 oxidation number. Since it's so electronegative, it almost always forms bonds with electropositive atoms, and so its oxidation number in compounds is always negative. Oxygen itself has such a high tendency to rip off electrons from other molecules. This is what makes oxygen very special.

There are so many examples of oxidation reactions with oxygen:

Combustion: Burning stuff itself is an oxidation reaction. If you burn coal(carbon), its oxidation number  goes from 0 (in pure carbon) to +4 (in carbon dioxide).

Respiration: Even our own bodies have to do an oxidation reaction to keep us with energy to do our daily work, like asking and answering questions on StackExchange. Glucose, which has a oxidation number of 0 (equal number of bonds with oxygen and carbon in its formula of $\ce{C6H12O6}$ ) is oxidized by the oxygen we breathe to carbon dioxide.

Much of the interesting properties of oxygen is due  to its oxidizing power. While many oxidizers are known, like fluorine for example; oxygen is, and will always, be the most well known one!


derivation of partial differential equations for changes in U Changes in internal energy at

  At constant temperature, $$d U = \left({\partial U \over \partial V}\right)_TdV$$
  At constant volume,
  $$d U = \left({\partial U \over \partial T}\right)_VdT$$
  When both temperature and volume is changing,
  $$d U = \left({\partial U \over \partial V}\right)_TdV +\left({\partial U \over \partial T}\right)_VdT$$

    I want to know how do we get these equations.

I think I can speculate a bit about how do we get the first and second equation.

For the first equation, since the temperature is kept constant, we can plot a U-V graph.

Tell me if I am wrong here but for that graph the slope at a given point is $\displaystyle \left({\partial U \over \partial V}\right)_T$.

Now since the change is a infinitesimal change, the slope of tangent at that point is given by $\displaystyle \left({\partial U \over \partial V}\right)_T$.

So $\displaystyle {\Delta U \over \Delta V} = \left({\partial U \over \partial V}\right)_T$, since the change is infinitesimal, $\Delta U = dU, \Delta V = dV$ and substituting this we will get out equation 1. Same can be said for equation 2.

I have no idea how do we get the equation 3, I think it is a clever use of  product rule or something like that.

Is my understanding for the equations 1 and 2 correct ?
How do we get equation 3 from 1 and 2 ? because simply adding 1 and 2 does only give the RHS of 3 not the LHS.

One definition of a state function such as internal energy $U$ is that its infinitesimal change is equal to an exact differential. Thus in general

$$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$

There are various variables with regard to which one can apply the partial derivative. Entropy, volume, temperature to name a few. Usually we assume that our thermodynamic state functions are two-dimensional but this is only if the number of moles is constant.

$$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i \overset{\mathrm{d}n_j\ =\ 0}{\implies} \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V.$$

A good way to think of this assumption is that other functions are also simply functions of $T$ and $V$, e.g., $S = S(T,V)$. From the equation

$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V$$

you can reach the other two equations by imposing $\mathrm{d}T = 0$ and $\mathrm{d}V = 0$, respectively.


How can nitrogen be both inert and bond with everything in periodic table? Aren't below statements from this link contradicting each other?

Nitrogen bonds to almost all the elements in the periodic table    except the first three noble gases, helium, neon, and argon

Nitrogen gas is mostly used as an inert atmosphere whenever the    oxygen in the air would pose a fire, explosion, or oxidising problems

How can one element be both inert and bond with everything at the same time? Aren't inertness and bonding mutually exclusive?

Nitrogen as an atom can be bound to any other atom of the periodic table. But nitrogen as a molecule is something else. It is $\ce{N2}$ and the two atoms are strongly bound in the molecule. It is even extremely difficult to separate the two atoms from the molecule $\ce{N2}$, because $\ce{N2}$ contains a triple bond : there are three covalences between the two atoms. That is why $\ce{N2}$ is nearly inert. A huge amount of energy is needed to break these three bonds. And it is usually impossible to do it in a lab at ordinary temperatures. Some plants (beans, peas) are able to do it and to metabolize $\ce{N2}$ by a somewhat mysterious process.


Search for the last step of an altered Raschig process reaction mechanism The Raschig process (according to the german wikipedia) is divided in the following steps:

$\ce{2NaOH + Cl2 -> NaOCl + NaCl + H2O}$
$\ce{NaOCl + H2O -> HOCl + NaOH}$
$\ce{HClO + NH3 -> NH2Cl + H2O}$
$\ce{NH2Cl + NH3 -> N2H4 + HCl}$
$\ce{N2H4 + 2NH2Cl -> N2 + 2NH4Cl}$

But what happens now if one changes $\ce{NH3}$ into $\ce{R-NH2}$, with $\ce{R}=$?

Starting from step 3 of the Raschig process I came up to the reaction of the produced hydrazine derivate to give an azo compound.

$\ce{HClO + R-NH2 -> R-NHCl + H2O}$
$\ce{R-NHCl + R-NH2 -> R-NH-NH-R + HCl}$
$\ce{R-NH-NH-R + x~R-NHCl -> R-N=N-R~+~?}$

The last reaction step bothers me a little ... is it simply
$$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$
but without the cleavage?

It can't be

$$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$

The reason is that while the atoms are balanced, the electrons aren't. Going from $\ce{R-NH-NH-R}$ to $\ce{R-N=N-R}$ involves the loss of two protons and two electrons, but the $\ce{R-NHCl}$ are really only picking up two protons. (Nitrogens with four substituents are positively charged.)

When thinking of mechanisms, it may help to explicitly split the final listed reaction into two steps.

5a. $\ce{H-NH-NH-H + NH2Cl -> H-N=N-H + NH4Cl}$

5b. $\ce{H-N=N-H + NH2Cl -> N2 + NH4Cl}$

That is, each chloramine is involved in pulling one dihydrogen (two protons and two electrons) off of the diazo. - The addition of a "hydride" (a proton and two electrons) to the nitrogen kicks the chlorine off as a negatively charged ion, and the remaining proton protonates the lone pair on the nitrogen.

The analogous last step for the $\ce{R-NH2}$ case might be:

$$\ce{R-NH-NH-R + R-NHCl -> R-N=N-R + R-NH3Cl}$$

This is equivalent to reducing the $\ce{N-Cl}$ covalent bond to $\ce{R-NH2 + HCl}$ followed by an acid/base reaction.


Urea + water yields ... urea + water? I added urea to water and, predictably, observed a temperature drop (10 C).  I wanted to place the beaker under (reasonably) high vacuum (for a classroom) and recover the urea as a dry substance.  Well, I didn't.  It's been a couple of months now, and there's white scale all the way up the inside, and crystaline filaments on the bottom, with an unchanging fraction of the original H2O.  Is the substance still urea?  There's been no odor.  Is it ammonium carbamate? In either case, can I use vacuum to dessicate it?

As water evaporates, the vapor pressure above the more concentrated urea solution drops. Either a better vacuum pump is needed, or perhaps gentle warming, to extract more water.

You might also try chilling to (partially) separate the solution. At ~260 K, achieved in many home refrigerators, much of the urea will have crystallized from the remaining saturated solution... just don't mix this up with food in the freezer.


Why is lead sulfide found in nature, whereas lead oxide is less common? This question probably applies to other heavier metals as well.  The only rationalization I can figure out is that lead in the +2 oxidation state (which is most common) is a borderline soft acid and prefers to take electrons from a soft sulfur instead of trying to react with a hard oxygen.
Is this the right direction?  Thank you in advance.

Lead(IV) and sulfide ions are large and polarizable, making them soft in the context of hard/soft acid/base theory (HSAB), which states that ions with similar hardness will form stronger bonds.  The oxide ion is hard and therefore has a tendency to form weaker interactions with Pb(IV).

HSAB theory has been discussed on Chem.SE in this question and this answer.


Equilibrium Constant of a Salt knowing the solubility Calculate the equilibrium constant in $\ce\\M ^ 3$ of the heterogeneous system in aqueous solution:
$\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$
Knowing that the solubility of the salt is $15.9 \ce{mg/L}$.

They are three and gave the same result.

The molar solubility $s$ equals:
$$s= \frac{15.9\times 10^{-3}}{78}= 0.2038\times10^{-3} \ce{mol/L}$$

According to the equilibrium: $\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$, the  equilibrium constant is:

$$K_{sp}=4s^3$$

By substituting the value of $s$ in the above equation, we find:
$$K_{sp}=0.034\times10^{-9}$$


Computational methods to obtain relative energies of electronic configurations of atoms When one learns the Aufbau principle to “predict” electronic configurations, and the $(n + \ell)$ ordering rule (or Madelung rule), one also learns of the exceptions to the rules… some of which (Cr, Cu, Mo, Ag, Au) can be half-explained by adding another “rule”, some of which (Nb, Ru, Rh, …) cannot. We have on this site our fair share of electronic-configuration questions dealing with these: Nb, Pt, Pd, Cu, Th, etc.

My question here is not about rules or theories to rationalize and understand these complicated electronic configurations, but rather about computational tools to predict them. What computational methods (and software) would one use to compute the different energies of several electronic configurations of an atom?

For example, take niobium. How would I answer, with computational quantum chemistry methods, the following question:

  what is the energy difference between the 5s1 4d4 and 5s2 4d3 electronic configurations of niobium?

And would that method also give some interpretation in terms of what these energy differences arise from?

The different occupations you suggest correspond to different spin states. Therefore, they are quite simple to distinguish.
On the other hand, the ground states are often degenerate (in the terms of 5 equivalent d orbitals available). Therefore the lowest (and sufficient) level of theory is the CAS-SCF with inclusion of the s- and d- orbitals.
I tried several calculations using ORCA in def2-svp basis and the results correspond to what one would expect, the time to perform the calculations is in order of seconds.

Zr (should be $4d^2$ $5s^2$, ie. MULT=3)

    CAS-SCF STATES FOR BLOCK  1 MULT= 5 NROOTS= 1
    ROOT   0:  E=     -46.4247189198 Eh
    0.62280 [     0]: 211110
    0.37690 [     2]: 211011

    CAS-SCF STATES FOR BLOCK  2 MULT= 3 NROOTS= 1
    ROOT   0:  E=     -46.4308999710 Eh
    0.49082 [     0]: 221100
    0.47988 [     2]: 221001
    0.01125 [    18]: 210111
    0.00276 [    27]: 201120

Cr (should be $3d^5$ $4s^1$, ie. MULT=7)

    CAS-SCF STATES FOR BLOCK  1 MULT= 7 NROOTS= 1
    ROOT   0:  E=   -1043.1949498785 Eh
    1.00000 [     0]: 111111

    CAS-SCF STATES FOR BLOCK  2 MULT= 5 NROOTS= 1
    ROOT   0:  E=   -1043.1559672632 Eh
    0.40000 [    15]: 111111
    0.30000 [     0]: 211110
    0.30000 [    30]: 011112

Nb (should be $4d^4$ $5s^1$, ie. MULT=6)

    CAS-SCF STATES FOR BLOCK  1 MULT= 6 NROOTS= 1
    ROOT   0:  E=     -56.3066169626 Eh
    0.96220 [     0]: 111110
    0.03771 [     2]: 111011

    CAS-SCF STATES FOR BLOCK  2 MULT= 4 NROOTS= 1
    ROOT   0:  E=     -56.2843641219 Eh
    0.89474 [     0]: 211100
    0.04255 [     2]: 211001

The lower the energy the better.

From the numbers I'd conclude that it is possible to obtain the ground state configuration by reasonably simple quantum chemical calculations. Nevertheless, I'd expect also less elaborate theories to predict such a fundamental properties.

Edit: I added the multiplicity to the expected ground state configuration


Superheating metal alloy above solidus Is it possible to heat a material above its solidus without melting occurring?

More specifically, at a heating rate of approximately 200K/s, will a metal alloy exhibit superheating and if so by what order of magnitude i.e. fractions of a degree versus tens of degrees. What is the mechanism behind this?

There is a lot of information about supercooling but I am not certain whether this applies!

Yes, superheating can happen in some alloys. Whether it occurs at all, and its extent if it occurs, is very specific to each alloy (and heating rate dependent, obviously). It also depends on things like sample geometry (particles, solid, etc.) and size. This paper reports delays in melting by up to 100 K for some Al-Pb and Cu-Pb alloys.


Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation? Take for instance the reaction

$$\ce{H2(g) + I2(s) <=> 2HI(g)}$$

The equilibrium constant would not include the solid $\ce{I2}$, but why is this? I have read that its concentration is a constant, but what does that mean? If you increase the amount of $\ce{I2}$ the reaction should shift towards the right so logically it does appear to play a part in the equilibrium.

For liquids, I can understand why you don't include it when you have something like water which (usually) does not play a part in the reaction, but why do you still not include when the liquid takes part in the reaction?

Also, do you include solids/liquids for reaction rates? Increasing the amount of a solid should increase the surface area and thus make for a faster reaction, so it seems like you should, but I am not too sure.

It very much depends on what definition of the equilibrium constant you are looking. The most common usage of the same has quite a variety of possible setups, see goldbook:

  Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

  Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{RT}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $R$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

It is worthwhile to note, that the first definition is always an approximation to the standard definition and in the standard definition all compounds regardless of their state are included in the equilibrium.

If you are looking at a reaction that takes place in gas phase, solid materials will play a constant role, since their partial pressure will solely depend on their vapour pressure and can therefore be regarded as constant. Therefore it can be seen as a part of the equilibrium constant. (The same applies to liquids while deriving for gas phase.) If you would increase the amount of solid in the system you would still not change it's concentration in gas phase.

For the reaction $$\ce{H2 (g) + I2 (s) <=> 2HI (g)}$$ you can form the standard equilibrium constant with activities/ fugacities
$$K^\circ = \frac{a^2(\ce{HI})}{a(\ce{H2})\cdot{}a(\ce{I2})},$$
with $$a=\frac{f}{p^\circ}.$$
The activity for a pure solid is normally defined as one ($a(\ce{I2})=1$) and therefore
$$K^\circ\approx K = \frac{a^2(\ce{HI})}{a(\ce{H2})}$$

For concentration dependent equilibrium constants, the following assumption is to be considered: $c(\ce{H2O})\approx55.6~\mathrm{mol/L}$ and is usually always much larger than that of any other component of the system in the range where the equilibrium approximation can be used. Also in most solution based reactions, the solute itself does not directly influence the reaction. Its concentration will therefore not change (significantly) and can be included in the equilibrium constant.

You are correct in assuming that the kinetics of a surface reaction will only depend of the actual area of the surface (and of course the reactants forming the products). But here again this area is most likely to be considered as constant and will result in a scalar for the reaction rate (in the area where an accurate description of the reaction is possible).

As of Greg's comment, He is of course right. In an equilibrium process the surface area will affect both directions of the reaction, hence it will not show up in the reaction rate at all. (The only effect it might have is, that the equilibrium state will be reached faster.)


What is the reaction between oxalic acid and potassium permanganate? I've already searched online some information about this equation, but everywhere I read, there's also sulfuric acid, which we did not use in the laboratory. We only used potassium permanganate and oxalic acid and water, there was no sulfuric acid. The purpose of the laboratory was to calculate experimentally the rate of the reaction at different concentrations, but in order to do so, I need the balanced equation. Can anyone help me out?

By the way, the oxalic acid was dihydrate.

If anyone is wondering, one of the experiments was as follows:

First we took $1.5~\pu{mL}$ of $\ce{H2C2O4}$ and added $1.5~\pu{mL}$ of $\ce{H2O}$, and finally, we added $1~\pu{mL}$ of $\ce{KMnO4}$ and then we shook the test tube until the colour changed from purple to ochre. Because from what I read, they said the $\ce{H2SO4}$ was necessary so the reaction takes place, but in the experiment, the reaction did occur and we didn't use it.

We had a titration of oxalic acid vs. potassium permanganate for an experiment in which we used concentrated $\ce{H2SO4}$.

Actually, the reaction requires an acidic medium i.e. $\ce{H+}$ is involved as a reactant... I suppose the $\ce{H+}$ released by the oxalic acid would be enough since you are studying the rate of the reaction though I am not sure.

The balanced ionic equation is:
$\ce{2MnO4- + 5H2C2O4 + 6H+ -> 2 Mn^2+ + 10CO2 + 8H2O}$

As you can see, $\ce{H+}$ is involved as a reactant and that is why you might have read that conc.$\ce{H2SO4}$ is necessary.


Is there an equilibruim in an unsaturated calcium hydroxide solution? Imagine an unsaturated(No Ca(OH)2 (s) just Ca2+ (aq) and OH- (aq))  aqueous solution of Calcium Hydroxide.

Is the following equilibrium present in that solution and does the transparency of an unsaturated Calcium Hydroxide vanish after a while?

$$\ce{ Ca^2+_{(aq)} + 2OH- _{(aq)} <=>  Ca(OH)2_{(s)}}$$

Equilibria in solutions are given as solubility products, $K_{\text{sp}}$. Your reaction is:

$$\ce{Ca(OH)_2 <=> Ca^2+ + 2OH-}$$
for this, $K_{\text{sp}}=5.5\times 10^{-6}$ (from Wikipedia). What this means, is that in a saturated solution, at equilibrium, you will have:
$$K_{\text{sp}}=[\ce{Ca^2+}]  [\ce{OH-}]^2=5.5\times10^{-6}$$

However, for an unsaturated solution, you will have:

$$[\ce{Ca^2+}]  [\ce{OH-}]^2 <K_{\text{sp}}$$

As mentioned the solution does get cloudy for different reasons, namely the formation of $\ce{CaCO3}$ by reaction with $\ce{CO2}$.


Esterification and dehydration synthesis I know there are many more terms out there like polymerization, condensation, etc. But the important thing is whether they can be differentiated or not.

In Esterification, which I just learned, needs to have a dehydrating agent when involving bad leaving group, such as $\ce{-OH}$.They are typically Brønsted-Lowry acids. Correct me if I'm wrong.

How does heterolysis work in the case of $\ce{CH3COOC2H5}$. [the reverse of the esterification of ethanoic acid and ethanol]

The reverse of esterification is called "hydrolysis."  The hydrolysis reaction can be performed under acidic or basic conditions.  This link shows the mechanisms involved for both the acidic and basic hydrolysis of an ester.  Each step in the reaction is explained in detail and accompanied by a picture.


Silver in defect chemistry When talking about conductivity in combination with lattice defects you can often see that AgX is taken as an example for a Frenkel defect where the $\ce{Ag^+}$ moves to intersticial voids. I remember that my professor once called this a property of silver, that silver likes itself so much that it can be close to other silver cations. And I think we see trends like this in for example $\ce{Hg2}$-dimers. Is this some sort of relativistic effect or can I explain it with certain orbitals overlapping that I can get a quite close contact for two $\ce{Ag^+}$ ions?

(partial answer)

Interesting question.  Silver also forms an unusual "subfluoride", Ag2F.  My understanding is the Ag is NOT in dimers.  But the structure has sorts of "sheets" of silver in that all of the octagonal holes are filled but only in alternating layers.  This allows silver atoms to be very close to each other, similar to the metal.  Presumably there is a lot of silver-silver bonding going on.  But more like sheets of metal than dimers.

The explanation of Frenkel defects you usually hear is that the cation is small enough versus the structure to allow dislodging by radiation (i.e. not a thermodynamics distribution of defects but a provoked one). See the Wikipedia page for instance.  However defect models are idealizations.

It would further be interesting to know how often the Frenkel defect is associated with some higher conductivity.  A purely billiard ball dislodged model would be less prone to conductivity than one where silver-silver bonding is the driver.


How does the composition of ice differ from that of water in a partly frozen solution? Take any solution, then start freezing it at sufficiently low temperature. Soon a small amount of ice forms. I'd like to know how the composition of dissolved salts and gases in the bit of ice differs from that in the water that's still liquid.

Assume that the solution is far from saturation and whatever other extreme conditions there can be. I don't have any particular solution in mind - it's the general principles that I'm interested in.

It sounds like you're talking about freezing-point depression.  If the solid component of a system is pure (oft assumed so; crystals are good at excluding impurities), the equilibrium is shifted because the chemical potential of the solution drops.

For a system with two pure species, the chemical change from say, water ($\star$ denoting pure) to ice is in equilibrium when their chemical potentials (denoted $\mu$) are equal:

$$\mu_{\ce{H_2O}(\text{s})} = \mu^\star_{\ce{H_2O}(\text{l})}$$

Note: Chemical potentials are always dependent, to varying degrees, on both temperature and pressure, so I'm not going to bother writing $\mu(T, P)$ everywhere

On the following graph that's at $T_0$; the usual freezing point of water where the "pure water" and "ice" curves intersect.  If water exists at a temperature lower than $T_0$, its higher chemical potential translates to a higher free energy, so it's thermodynamically driven towards freezing.

(image source: a website that is currently down; retrieved via the Web Archive)

If something non-volatile (e.g. salt) is dissolved in the water, its chemical potential changes to $\mu_i^\star+RT\ln \chi_i$, where $\mu_i^\star$ is the chemical potential of the pure species $i$, $R$ is the gas constant, and $\chi_i$ is the molar fraction of species $i$.

Originally, the fraction was $1.0$ (it was pure), and because $\ln1 = 0$, that term was ignored.  But now, because some of the solution species is something else, $\chi_i$ drops, thus $\ln\chi_i$ becomes negative, so the term lowers the overall chemical potential.  Because the chemical potential is lower, its free energy is lower, so it is thermodynamically more stable at lower temperatures.  At some $T$, we will find that:

$$\mu_{\ce{H_2O}(\text{s})} = \mu_{\text{solution}} = \mu^\star_{\ce{H_2O}(\text{l})} + RT\ln\chi_i$$

And there, the two species will be in thermodynamic equilibrium.

For more, check this site, or Wikipedia articles on chemical potential and freezing point depression (a basic Physical Chemistry text book, or chapter in a Gen. Chem. one might be better; the WP articles are not that great to get started with).


How could uranium be extracted from ore to be melt? I would like to go hunting for uranium ore, and because it has a relatively low melting point (1132.2 °C), I thought it would be cool to melt it just like any other metals. According to this video extracting yellowcake (mainly U3O8, UO2 and UO3) is relatively easy, however, how could I remove the oxigen from the oxides? Or is it easier to obtain the uranium directly from the ore in another way?

EDIT: because of all of the comments I've received I'm going to explain better the question:

I'm well aware of the dangers of handling uranium, a weak radioactive materials, the fact that it's dangerous as an heavy metal and that the ore is more radioactive than the pure metal because there are products of the decaying chain that are more active tha the uranium itself.
I know all the regulations and laws in my country, so that I know how much I can keep it (about 0.1 curie) and what I can and what I cannot do with my uranium source. I also know where I can get it withouth getting in trouble.
what I need to know is what chemical peocess I could use to extract pure uranium metal (238U + 235U) that I can melt in whatever shape I please. And no, I don't want to make a nuke because that would be impossible

There is another process for extraction of Uranium from the uranium ore in the form of UO2(NO3)2 by dissolving the Uranium ore in nitric acid and extract of uranium  using 20% solution of Tributyl phosphate/dodecane. TBP form strong complex with uranium nitrate and thus you can easily remove uranium from other impurities. Obtained UO2(NO3)2 can easily be converted to (NH4)2U2O7 by reacting with ammonia and finally reduction with hydrogen gave you UO2.
You can't obtained pure Uranium because it is susceptible to oxidation easily.


Recipe for paper glue I need to produce paper bags. Is there any simple receipt for making paper glue so that brown (packaging) paper will stick to itself?

There are a few different ways of making your own glue, here are four different versions you can try :

Simple Glue Ingredients:
1/2 cup flour
1/3 cup water

Paper Mache Glue Ingredients:
1 cup flour
1/3 cup sugar
1-1/2 cup water with 1 tsp white vinegar mixed in

Cornstarch Paste Ingredients:
1-1/2 cups cold water
2 tablespoons cornstarch
2 tablespoons corn syrup
1 teaspoon white vinegar

Simple No-Cook Flour Paste:
1 cup flour
1/2 cup water
1/8 teaspoon salt

Taken from http://www.wikihow.com/Make-Glue


About nitrogen acidity in function groups I have a compound and it has several nitrogen atoms. I wish to know how do we differentiate between the basic and acidic nitrogens?

This is an example compound:

First of all, a terminology correction. While a nitrogen atom may be basic, a nitrogen atom itself won't be acidic, rather, the hydrogen atom attached to that nitrogen atom would be acidic, if at all.

Now, whenever given an arbitrary compound, and this also holds true for atoms like carbon, oxygen, etc., here's what you need to do to determine the acidic and the basic atoms.

What does acidity mean? Acidity means the ability to lose a proton from the compound. The higher the ability to lose a proton, the higher the acidity. If a compound, on losing a proton, forms a more stable conjugate base, it is considered more able to lose a proton. Hence, it is also more acidic (than other similar protons in that compound).

What does basicity mean? Basicity means the ability to donate a lone pair of electrons to a proton. The higher the ability to donate a lone pair to a proton, the higher the basicity. If a compound, on donating a lone pair, forms a more stable conjugate acid, it is considered more able to donate a lone pair. Hence, it is also more basic (than other similar lone pairs in that compound).

They both sound similar, don't they? That's because they're on the same underlying principle: a chemical species would want to accept a proton or donate a lone pair if the resulting species is more stable.

Now, you can simply check for this in any given compound by deprotonating every H-atom and protonating every lone pair, while applying the simple rules of aromaticity, resonance, hyperconjugation, and inductive effects individually to each.


Can fully-hydrogenated oils be mixed with unsaturated oils for use in food (margarines, etc)? Would it be feasible to mix fully hydrogenated oils with unsaturated oils (eg. soybean oil) to create an edible fat with the consistency of margarine?

If so, what is the motivation for interesterified oils? If not, then what is the constraint that prevents this? Do the fats not form a solution?

I'm trying to understand how palm oil came to replace partially hydrogenated oils in margarines in response to trans-fat bans, and what possible alternatives might be.

The purpose of hydrogenation is to convert liquid oils into semi-solids at room temperature.  Palm "oil" is unusual among natural vegetable oils/fats in that it is naturally fairly unsaturated and is thus a semi-solid at room temperature.

Economics is probably the driving factor preventing significant research into the solution (pun intended) you propose with mixing saturated and unsaturated fats to get the desired consistency.  Hydrogenation techniques are cheap, well established, agency approved, etc.  Same idea with palm oil.

I hope I hit the points of your question, comment if I missed anything.


Why Atomic size decreases left to right in a period in MPT? In modern periodic table,  atomic size decreases as we move from left to right in a period. As no. of electron increase left  to right but why atomic size decreases?

The Modern Periodic table is made according to increasing number of their atomic numbers . So if we move from left to right in a period the no of electrons as well as the no of protons goes on increasing . The electrons are added to the same orbit or energylevel but the nuclear charge increases by one in each step due to increase in the number of proton by one in each step. So the inward pull increases and the outer orbital shrinks. As a result the atomic size decreases moving towards right in the modern periodic table.


Does sublimation happen only on the surface or throughout the body? Like evaporation happens only on the surface of a liquid, and boiling/vapourisation takes place throughout the body, how does sublimation occur? Why?

The situation is akin to boiling.  If a molecule at the surface frees itself of its attractions to its neighbors, it can move into the vapor phase easily.  Same thing with evaporation at the surface of a solid.

Can a molecule evaporate from the bulk liquid?  The answer is no.  Imagine such a thing happening.  The escaping molecule(s) form a bubble inside the liquid.  The pressure inside the bubble is the vapor pressure of the liquid at that temperature.  But that temperature is less than the boiling temperature so the vapor pressure is less than the pressure external to the liquid.  The result is that the bubble collapses.  The boiling temperature is the lowest temperature at which the bubble pressure is equal to the external pressure.

Now think of sublimation in the bulk of a solid.  Here the "bubble" has to have enough pressure to break apart the solid.  It generally won't develop that much pressure.  Indeed, one can watch "dry ice" sublime well above its freezing point and nothing happens inside the bulk of the solid.

So the general answer is no, sublimation does not take place in the bulk of a solid.


What does bond polarity have to do with acid strength? Why does increasing bond polarity affect acid strength? For instance, why is $\ce{HF}$ a stronger acid than $\ce{CH4}$?

This can be extrapolated to why polar substances ionize more often in water than non-polar substances. This is because water too is polar. Always remember "like dissolves like." Polar dissolves polar. This is because the positive end of the water molecule pulls at the negative end of the solute while the negative end of the water molecules pull at the positive end of the solute. In a non-polar molecule this would obviously not happen. Hence, in the case of acids, polar acids will dissociate more and hence more hydrogen atoms will be released into the solution.


Why does spontaneity of a reaction depend on temperature? As we know, $$\Delta G=\Delta H - T\,\Delta S.$$
Both in my textbook and on the internet, it was given that temperature was a determining factor in the spontaneity of a reaction. For example, when $\Delta S$ is negative (unfavorable) even then the reaction can still be favorable at lower temperatures.
If $\Delta S = q/T,$ then wouldn't the entropy also decrease by the same factor on decreasing the temperature? In other words, why don't the temperature terms cancel leaving $\Delta G$ independent of temperature?

The issue here is in your definition of entropy. According to your definition:
$$\mathrm dS = \frac{\text{đ}q}{T}.\tag{1}$$
However, there is a small difference.
As you may know, the entropy of a system is a state function and so it only depends on the initial and final states. Therefore, if the entropy of the system were to be calculated for a specific path for which the parameters are simple to find, then the value would not change irrespective of the path.
The actual formula of change in entropy is:
$$\mathrm dS = \frac{\text{đ}q_\mathrm{rev}}{T}.\tag{2}$$
This value is calculated for a reversible isothermal process.
Now, let us assume that we have taken such an isothermal reversible process and discovered the value of $\Delta S$ and now used an isobaric process to discover the value of $\Delta H.$ Now, we solve for $\Delta G$ in our isothermal process (this is done so that we can assume constant temperature in our formula).
Therefore, for such a process
$$q_\mathrm{rev} = nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{3}$$
Thus, we get change in entropy as
$$\Delta S = nR\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{4}$$
Now, enthalpy would be a constant value that we discerned from the isobaric process.
Therefore, since $\Delta G = \Delta H - T\Delta S,$ we get
$$\Delta G = \Delta H - nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{5}$$
As you can see, there is still a temperature dependence.
This is because $\Delta S$  is temperature independent and so you would still have a temperature dependence for $\Delta G.$
Another thing that you may have forgotten to take into consideration is the fact that $q$ is dependent on the temperature.
The above example is that for an ideal gas expansion.
As Poutnik stated in the comments in case of a reaction, the calculation of $\Delta H$ becomes more complicated as we have to take into consideration the different molar capacities of the reagents and products which would $\Delta H$ would also be a function of temperature.
For standard value of entropy, we use the formula
$$\Delta S^\circ_\mathrm{rxn} = \sum nS^\circ_\mathrm{products} - \sum nS^\circ_\mathrm{reactants}.\tag{6}$$


Hydrazine as a reduction agent When using hydrazine as a reduction agent to reduce graphene oxide, which is the product hydrazine oxidises to?

I'm sorry for the wrong answer initially given; you might consider to unaccept this one.

I'm not sure whether this has ever been examined in detail. It is likely that nobody bothered to collect the (volatile) reaction products of hydrazine.

According to a procedure provided by the group of Rodney Ruoff (DOI), the reduction was performed in an open flask equipped with a reflux condenser. An aqueous suspension GO was first sonicated (to support exfoliation) until it became clear, then hydrazine hydrate was added and the mixture refluxed at 100 °C for 24 hours. During this time, the reduction product precipitated as a black solid.

No remarks on what actually happened to the hydrazine were given in this article.


Regarding units of water potential and chemical potential If water potential is nothing but the chemical potential of a given solution, then why do their units differ?

If the unit for the latter is taken as 'Energy per molecule' (which makes sense to me mathematically), then why is the former measured is 'Pressure'(Pascal)?

According to German Wikipedia water potential is defined as
$$\psi := \frac{\mu - \mu_0}{\bar{V}},\tag{1}$$
where $\mu_0$ is the standard chemical potential (usually pure water at atmospheric pressure at a specified reference height), and $\bar{V} \approx \pu{18 cm3 mol-1}$ is the molar volume of pure liquid water.
This means that water potential $\psi$ is defined as energy $E$ per volume $V$, which is equivalent to pressure $p$ if you perform dimensional analysis:
$$[\psi] = \frac{[E]}{[V]} = \frac{\mathsf{L}^2·\mathsf{M}·\mathsf{T}^{−2}}{\mathsf{L}^3} = \mathsf{M}·\mathsf{L}^{−1}·\mathsf{T}^{−2} = [p]\tag{2}$$
Megapascal $\pu{MPa}$ is usually used as the unit of measurement; tensiometers are usually graded in hectopascals $\pu{hPa}.$


What should you couple with the reference material in testing galvanic corrosion? I am currently studying the effect of metal coating to steel. I want to test the galvanic current between steel and aluminum by getting the current that passes through a wire that connects them in an electrolyte solution, to know if the specific aluminum will fit as a good coating for the steel. What should I use as the couple material for the steel? Should it be the coated steel or the metal to be coated? I'm using ASTM G71 as the reference method.

Wikipedia states:"Galvanic corrosion (also called bimetallic corrosion) is an electrochemical process in which one metal corrodes preferentially when it is in electrical contact with another, in the presence of an electrolyte." What are the two metals? How would we know??!! The general answer is:"They are the two metals which form the galvanic cell (in contact with the electrolyte)." Depending on your exact circumstances, this might be the metallic coating and the base metal, or it could be the metallic coating and the metallic coating,or even some other metal component in electrical contact. Of course, if the metallic coating is perfectly homogeneous and completely encloses the base metal, then there will be no galvanic corrosion (at least, not until other corrosion mechanisms have changed that situation).


Gibbs Free Energy : What is it trying to say actually? "Gibbs Energy is the useful work that can be extracted from the heat of a reaction or a thermodynamic process." I understand how it predicts the feasibility of a chemical reaction , considering the Entropy statement of the Second Law. But how does it relate with the useful work that can be extracted? (This definition is implicitly used in my textbook in deriving from Nernst Equation, the useful work that can be extracted from an electrochemical cell and I am missing the point why this actually works.) Is there an intuitive explanation? Also if this much of energy is "free", what happens to the energy that is "not free"? Why can't we extract the whole energy that is released?

Also I am interested to know if there is any direct relation between Gibbs Function and Kelvin-Planck Statement, which apparently expresses the inherent inability to convert the whole heat to useful work, or am I mixing up two very different aspects?
Thanks.

You can prove that the Gibbs free energy is equal to the maximum amount of work that a certain reaction (or any physical process in general) can perform, i.e. $\Delta G = w_{max}$. Such a proof can be found in any decent Physical Chemistry textbook (such as Atkins'). The remaining energy must be "lost" as heat to basically account for the observation that is essentially stated with the Clausius inequality $\Delta S \geq 0$ (i.e. the entropy of the universe is always increasing) . Of course the definition of entropy $\Delta S = \frac{q_{rev}}{T}$ and the Clausius inequality that is derived from it (and hence the Gibbs free energy) are inherently connected to Kelvins (and Clausius') statement of the second law. You can see from the definition (don't forget the change of entropy of the surroundings!) that (since $\Delta U = q + w$) the entropy is decreasing when heat is completely converted to work.

Ask yourself the following: if heat could be completely converted to work, then the collision of an inelastic ball with the earth (or any other object) would be reversible, have you ever seen a tennis ball suddenly jumping back from the ground?


Characterisation of gold-silver alloy nanoparticle I have various compositions of gold-silver alloy nanoparticles.
I would like to characterize my nanoparticles and confirm that they are alloys.
Aside from the qualitative analysis of the colour and using UV/vis to identify the number of plasmon bands, what other methods can I use to confirm the presence of alloys?

I Think you are in right track using UV-vis spectroscopy as a tool to identify your alloys. Other than UV-vis spectroscopy, researchers around the world have used many other analytical techniques to evaluate the synthesized nanomaterials, including X-ray diffractometry (XRD), Fourier transform infrared spectroscopy (FTIR), X-ray photoelectron spectroscopy (XPS), dynamic light scattering (DLS), scanning electron microscopy (SEM), transmission electron microscopy (TEM), atomic force microscopy (AFM), and etc. (For review of these techniques: Read Ref.1).

According to your notes, it seems like your main concern is how to confirm that your samples are alloys. UV-vis spectroscopy is a very useful and powerful tool in that regards, and reliable technique for the primary characterization of synthesized nanoparticles as well. Silver absorbs around $\pu{400 nm}$ and gold around $\pu{550 nm}$, yet these values vary depends on particle sizes of the dispersion (See Figure 1A & 1B, respectively). For example, absorption spectra calculated using the quasi-static approximation for silver and gold nanospheres with diameters $\mathrm{D} = 26$ and $\pu{20 nm}$, respectively, have shown absorption maxima (surface plasmon resonance wavelength $\lambda_R$) at ~$\pu{375 nm}$  for $\ce{Ag}$ and ~$\pu{525 nm}$  for $\ce{Au}$ in water (Ref.2). This reference also discussed the behavior of $\lambda_R$ as a function of the dielectric constant of the surrounding matrix ($\epsilon_m$), which is 1.77 for water.

This phenomenon was used to the best in identifying $\ce{Ag-Au}$ alloys in various compositions in an article (Ref.3), abstract of which states:

  Gold−silver alloy nanoparticles with varying mole fractions are prepared in aqueous solution by the co-reduction of chlorauric acid $\ce{HAuCl4}$ and silver nitrate $\ce{AgNO3}$ with sodium citrate. As the optical absorption spectra of their solutions show only one plasmon absorption it is concluded that mixing of gold and silver leads to a homogeneous formation of alloy nanoparticles. The maximum of the plasmon band blue-shifts linearly with increasing silver content. This fact cannot be explained by a simple linear combination of the dielectric constants of gold and silver within the Mie theory. On the other hand, the extinction coefficient is found to decrease exponentially rather than linearly with increasing gold mole fraction $x_{\ce{Au}}$. Furthermore, the size distribution of the alloy nanoparticles is examined using transmission electron microscopy (TEM). High-resolution TEM (HRTEM) also confirms the formation of homogeneous gold−silver alloy nanocrystals.

According to this reference, only one plasmon band for particular alloy (Two bands would be expected for the case of a mixture of gold and silver nanoparticles). The band shows a blue shift with increasing $\ce{Ag}$ amount in the alloy (See Figure 2) and a linear relationship with increasing gold mole fraction, $x_{\ce{Au}}$ (See the insert in Figure 2):

References:

X.-F. Zhang, Z.-G. Liu, W. Shen, S. Gurunathan, “Silver Nanoparticles: Synthesis, Characterization, Properties, Applications, and Therapeutic Approaches,” Intl. J. Molecular Sci. 2016, 17, 1534 (34 pages) (doi:10.3390/ijms17091534).
F. Vallée, “Chapter 7: Optical Properties of Metallic Nanoparticles,” In Nanomaterials and Nanochemistry; C. Bréchignac, P. Houdy, M. Lahmani, Eds.; Springer-Verlag: Berlin, Germany, pp. 197-228, 2007.
S. Link, Z. L. Wang, M. A. El-Sayed, “Alloy Formation of Gold−Silver Nanoparticles and the Dependence of the Plasmon Absorption on Their Composition,” J. Phys. Chem. B 1999, 103(18), 3529–3533 (DOI: 10.1021/jp990387w).


Boiling of water does not involve temperature change, so why is there a change in internal energy during the process? in an isothermal process, the value ΔT = 0 and therefore ΔU = 0 but during boiling if water at constant atmospheric pressure temperature doesn't change so why internal energy changes

The change in internal energy $U$ is dependent on the heat transferred $q$ and the work done $w$.

$$\Delta U = q + w$$

$\Delta U = 0$ in two cases: when $q=-w$ and when $q=w=0$.

The heat transferred is not necessarily zero if there is no temperature change. The most common expression of heat transfer is $q=C\Delta T$, where $C$ is the heat capacity of the system and $\Delta T$ is a temperature change. We can also write $q=ms\Delta T$, where $m$ is the mass of the system and $s$ is its specific heat capacity (per unit mass). However, heat is being transferred during boiling (else how does the liquid boil?). We need a new equation for heat that does not require temperature.

In a constant pressure system (most boiling happens in a constant pressure,  open to the atmosphere, system), change in enthalpy is equal to $q$ since all of the work and other pressure-volume terms cancel. The enthalpy change of vaporization is usually given in $\mathrm{kJ}$ per mole or unit mass. Now we have a new equation for heat:

$$q_{vap}=m\Delta_{vap}H$$

Let's say we are boiling 100.0 g of water. Water's enthalpy of vaoprization is $2257\ \mathrm{kJ/kg}$. Thus the heat transfer is:

$$q=(0.1000\ \mathrm{kg})(2257\ \mathrm{kJ/kg})=225.7\ \mathrm{kJ}$$
Note that in this case, the work done is also not zero. There is a volume change of the system. Let's take our 100 grams of water.

$$V=0.1000\ \mathrm{kg}\times \dfrac{\mathrm{m^3}}{958.4\ \mathrm{kg}}=1.040\times 10^{-4}\ \mathrm{m^3}$$

At the boiling point, the density is $958.4\ \mathrm{kg/m^3}=0.9584\ \mathrm{kg/L}$. Using the ideal gas law (water vapor is reasonably ideal at atmospheric pressure and its boiling point), we can determine the volume of 100.0 g of water vapor, given that

$$100.0\ \mathrm{g}\ \ce{H2O}\times \dfrac{1\ \mathrm{mol}\ \ce{H2O}}{18.02\ \mathrm{g}\ \ce{H2O}}=5.549\ \mathrm{mol}\ \ce{H2O}$$

$$V=\dfrac{nRT}{P}=\dfrac{(5.549\ \mathrm{mol})(0.080257\ \mathrm{\frac{L\cdot atm}{K\cdot mol}})(373.15\ \mathrm{K})}{1\ \mathrm{atm}}=169.9\ \mathrm{L}$$

We can now calculate work. So that we get joules out, I am going to convert pressure to pascals and volume to $\mathrm{m^3}$.
$$P=1\ \mathrm{atm}=101,325\ \mathrm{Pa}\\
V_{gas} = 169.9\mathrm{\ L} =0.1699\ \mathrm{m^3}\\
\Delta V= 0.1699\ \mathrm{m^3} - 1.040\times 10^{-4}\ \mathrm{m^3} =0.1698\ \mathrm{m^3}\\
w=-P\Delta V=-101325\ \mathrm{Pa}\times 0.1698\ \mathrm{m^3}=-17.20\ \mathrm{kJ}$$

Now we calculated $\Delta U$:

$$\Delta U = q +w = 225.7\ \mathrm{kJ} - 17.20\ \mathrm{kJ} = 208.5\ \mathrm{kJ}$$


Safety in handling of concentrated ammonium fluoride I recently saw a protocol for extracting organic materials that are bound in the silica matrix formed by diatoms.  The extraction protocol used 10 molar ammonium fluoride ($\ce{NH4F}$).

I briefly considered using the protocol and then endeavoring to remove the ammonium fluoride (and water) by lyophilization or with a rotary evaporator.

However, I am concerned about the safety of doing this.  My first worry is that any glass equipment or vessels in the lyophilizer or rotary evaporator systems would be damaged by the fluoride.  My second worry is that even if the equipment used did not contain any glass, dangerous solutions of hydrogen fluoride (i.e. hydrofluoric acid) could be formed in the cold trap of the equipment.

My first question is, am I right that it is a bad idea to evaporate ammonium fluoride in glass equipment?

My second (but related) question is, even if glass is not used, what are the safety risks of evaporating ammonium fluoride under vacuum?  Imagining that a glass-free apparatus could be found, is there a way to do the evaporation safely?

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.

The rate of etching at 25 degrees C is proportional to :

$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$

according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.

The relavent equilibrium constants at 25 degrees C and ionic strength of 1M are

$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$

$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$

From the second equilibrium, it can be seen that if the fluoride concentation is 10M as in the question, the more active $\ce{HF2-}$ will be present at a concentration of about 100 times $[\ce{HF}]$.

Multiplying the two equilibrium equations together, and substituting $[\ce{F-}] = 10M$:

$$\frac{[\ce{H+}]}{[\ce{HF2-}]} = 1.4 \times 10^{-6} $$

Therefore at $pH = log (1.4 \times 10^{-6}) = 5.9$, there will be about 1M $\ce{HF2-}$, diminishing by a factor of 10 for each pH unit increase.

Experimentally, concentrated Ammonium Fluoride has a pH of 7.2 according to http://www.solvaychemicals.us/SiteCollectionDocuments/sds/P20117-USA_CN_EN.pdf (which also says to avoid glass).

At pH 7.2, [$\ce{HF2-}$] is 0.05M.

In conclusion, there will be a significant concentration of the most active etching species, $\ce{HF2-}$, present in 10M ammonium fluoride and I would be concerned about glassware.


Practical safety of storing potassium Potassium ($\ce{K}$) seems to be the transitional alkali metal in terms of safety. Lighter than $\ce{K}$ and we can store under oil with no worries. Heavier than $\ce{K}$ and only ampules will work. $\ce{K}$ is such that even under oil it can absorb enough oxygen to form superoxides that are contact explosives. But this is the theory: does anyone have any experience with extended proper storage of potassium?

I ask for two reasons. First, there are lots of online videos showing $\ce{K}$ under oil that the scientist then proceeds to cut, smash, and otherwise manhandle with no fear of detonation. Second, my $\ce{K}$ arrived tarnished, which I have stored under oil, and then vacuumed sealed in a bag, which has shown zero change in visual oxidation. So while I get that leaving an open jar of $\ce{K}$ under oil is problematic, I'm less clear on well stored $\ce{K}$.

So I'd appreciate any comments from practitioners that use $\ce{K}$ regularly. What situations are genuine dangers? Thanks!

I have never worked with potassium personally, but have experience with other alkali metals.   You are spot on about potassium superoxide.  I have direct knowledge of an explosion that occurred when potassium metal was stored for an extended period of time in mineral oil.  I would therefore be extremely cautious if you intend to store it in this manner.  Factors such as length of storage time, permeability of your container seals and storage temperature can all affect the amount of superoxide that forms.  Based on the evidence I have seen, it is not safe to assume that there will be no transport of oxygen through a mineral oil barrier.  The only way I would work with this metal is in an inert glovebox, period.  That is my own preference, of course, and I always tend to err on the side of caution.


How does one calculate oxidation states? For example, how could I calculate the oxidation state of lead in $\ce{PbS}$? Or for that matter how would I calculate it for sulfur? Or I have two unknowns, for example sulfur in $\ce{PbSO4}$? Do I assume that lead has a 2+ charge and therefore a +2 oxidation state and since it has ? Or is there something that I am missing? Additionally, from all the reading that I've done I've only found a rule for bromine, chlorine, and iodine in binary compounds, what happens if they are in a compound with more than two elements? For example $\ce{COCl2}$. How would someone calculate the oxidation state of carbon? This stuff is really confusing for me and I apologize if this seems really basic or I have missed something.

I have read the introduction to oxidation states post, and it was a very well written answer, however it still did not help things click. I still feel like I am missing something

For compounds where you have a non-metal with a metal, there is likely an ionic bond. The oxidation states are just the charges on the ion. PbS has Pb2+ and S2-, so those charges are simply the oxidation states.

For PbSO4, you know sulfate is a 2- ion, so it must be bonded to a Pb2+. Therefore, again, Pb has a +2 oxidation number. When you look at the sulfate ion, you know that the sum of the oxidation states of the elements must equal the ion's overall charge. Oxygens are -2, so set up the equation: (X = oxidation number of sulfur)

X + 4(–2) = -2
X = +6

Calculating the oxidation state of carbon can be more complex. If you draw a dot and cross diagram showing the covalent bonds made by carbon, this will help. Look at the electronegativities of the elements in the bonds. Carbon takes the electron pair if it is the more electronegative, but this is rarely the case. For every electron pair carbon takes, it has a -1 added to the oxidation state, and +1 added if it loses the electron pair. For example, if you draw the methane diagram you find that the oxidation state of carbon is +4. Using this method, you can find the oxidation state of carbon in compounds.


Which white powder first "boils off" and then catches fire at high temperatures? I'm a high school student doing a project where we need to identify chemicals and one of the substances had a very unique change when heated.

Before heating it was an extremely fine, flour-like white powder. I placed it on a hotplate at 350 degrees Celsius in its solid form and it "boiled off", I presume a hydrate evaporating. Before the hydrate appeared to evaporate, the powder had small pink dots appear all over it. This also happened when exposed to $\ce{NaClO}$ and $\ce{NaOH}$.

Once the temperature was raised to 400 degrees Celsius, the remaining solid turned yellow and then spontaneously caught fire, turning it all black before the fire ran out.

I had ran other tests on this substance, being insoluble in water, ethanol, and hexane. It also produced no apparent flame color.

My initial thought was that this compound was starch, however now that I think about it, it may be an insoluble magnesium compound, due to the lack of flame color? Just curious as to if anyone knows what substances display these properties when heated.

Here is a list of potential substances based off of a predetermined list.

Aluminum nitrate
Ammonium chloride
Ammonium nitrate
Ammonium sulfate
Barium acetate
Barium nitrate
Barium oxide
Barium sulfate
Calcium carbonate
Calcium chloride
Calcium oxide
Calcium nitrate
Calcium sulfate
Iron(II) sulfate
Iron(III) sulfate
Lauric acid
Lithium carbonate
Lithium nitrate
Magnesium carbonate
Magnesium chloride
Magnesium hydroxide
Magnesium oxide
Magnesium sulfate
Potassium bromate
Potassium carbonate
Potassium chloride
Potassium iodide
Potassium metabisulfite
Potassium nitrate
Potassium persulfate
Potassium phosphate
Potassium sodium tartrate
Potassium thiocyanate
Sodium acetate
Sodium bisulfate
Sodium borate
Sodium carbonate
Sodium citrate
Sodium chloride
Sodium fluoride
Sodium nitrate
Sodium nitrite
Sodium oxalate
Sodium sulfate
Sodium sulfite
Sodium thiosulfate
Starch
Strontium carbonate
Strontium hydroxide
Strontium nitrate
Sucrose
Tin(II) chloride
Urea
Zinc nitrate
Zinc sulfate

Of these, as stated earlier, an Mg compound (due to insolubility) or starch is most likely.

The fact that it turned black when heating tells you that:

It was an organic substance and did not burn completely leaving some carbon behind, or
It was a transition metal that had a lower oxidation state and was oxidised during burning.

Let's remove the obvious ones:

Not a nitrate. These will release toxic brown gas, which is not something you give to high school students.
Not an alkali (Na, K, etc) or alkali earth (Ca, Mg, etc) element because they're not pink nor they turn black.
I doubt it's a sulfur compound of any kind. You should have detected a sulfur smell when burning.

Maybe this can help you a bit.


How long does a wine sauce need to simmer for to reduce the alcohol to less than one percent Perhaps you can shed some light on this topic for me.  I've recently researched the topic of boiling off alcohol while cooking--specifically how long one needs to simmer alcohol to boil it off.

I'd come across a couple of sites which showed tables of simmering times and percentage of alcohol remaining.  Since I am a math-head, I noticed that something didn't quite jive in the tables I was viewing.

I thought to ask a chemistry expert to see if there is something unusual going on chemically to explain these inconsistent values in the tables.

I don't want to make this a novel, so I will use only a few entries in the tables to illustrate my point.

Simmering time                    Alcohol remaining

15 minutes                                  40%

30 minutes                                  35%

2 hours                                     10%

So what the table suggests is that if you simmer a solution containing alcohol, after the first 15 minutes, you will have 40% of the original amount of alcohol in the solution.  If you simmer for an additional 15 minutes (total of 30 minutes) you will have 35% of the original amount of alcohol left in the solution.

So, (according to the table) if I have a solution which originally contains 100 ml of alcohol in the solution, and I simmer it for 15 minutes, I will have 40 ml of alcohol left in the solution.  Then if I simmer it another 15 minutes (30 minutes total), I will have 35 ml of alcohol remaining in the solution.

However, If I start with a solution containing 40 ml of alcohol in the solution (let's say it's the solution from above after the first 15 minutes), and I simmer it for 15 minutes, I will be left with 16 ml of alcohol in the solution (again, according to the table).  You can see my dilemma; 16 ml does NOT equal 35 ml.  The math doesn't work.

I am wondering about the nature in which they'd come up with these values or the methodology they used in their experiments.

Can you help me to understand what is going on here?

The original source of the data is Alcohol retention in food preparation Journal of the American Dietetic Association vol. 92, April 1992, pages 486-488.

After 2.5 hours of simmering at 85 degrees C, pot roast Milano still retained 4-6% of the original alcohol content.


How do I find the hybridisation of the orbitals of carbon in a molecule? What are hybridisation states of each carbon atom in the following compounds?

$\ce{CH2=C=O}$
$\ce{CH3CH=CH2}$
$\ce{(CH3)2CO}$
$\ce{CH2=CHCN}$
$\ce{C6H6}$

How do I find the hybridisation of the orbitals of the carbon atom in such compounds?

My method for finding the hybridization of the orbitals of an atom in a molecule:

Look at the molecular structure (geometry), apply the following:

Tetrahedral: sp3
Trigonal planar: sp2
Linear: sp


Material to change its color by temperature condition meets over time There are thermochromic materials which their color changes reversible by temperature. Simply they show their current temperature by color. We can say their color is a function of current temperature:
$$c_t = f(temp_t)$$
Where $c_t$ is color in time $t$ and $temp_t$ is temperature in time $t$

I wonder if there is similar material which their color changes irreversible by history of temperature they meet over time. In comparison to thermochromic material, we can say their color changes by integrations of temperature:
$$c_t = f(\int_0^t temp_x dx)$$
Where $c_t$ is color in time $t$ and $temp_t$ is temperature in time $t$.

This seems to be more general and simpler than thermochromic materials as they are like regular chemical reactions such as oxidation, ... happens every day.

Are there any industry suited solutions out there based on this behaviour which shown in the second equation?

Compounds that irreversibly record the thermal history of the surface they are attached do do exist and others are being developed.

One class, used in harsh industrial environments is described on the Sensor Coating Systems site.

There are chemical systems used to detect high temperatures in refrigerated food, but most don't seem to be single chemicals. There are certainly many devices in development, but how many involve single chemicals is unclear.


Is it important to use arrows as bonds in writing coordination compounds? Why are the coordination compounds representated by arrows? Is it necessary to write every coordination compounds with the arrow side of the bond facing the metal ion and the other part the ligand?

Historically, coordination compounds have been depicted in various ways.

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), bonds representing coordination from one atom to a single other atom should be represented as normal plain single bonds. The use of dative bonds (i.e. arrows pointing to the central atom) to represent coordination is not acceptable.


Must a desiccant be made anhydrous before it can be used for drying? I'm writing about the treatment of methane gas, and in one part of the process, the methane gas is dried in a tank called a scrubber. Here, a desiccant called triethylene glycol is used.
Thinking about another desiccant, magnesium sulfate heptahydrate (Epsom salt), the water must be taken away through evaporation before the salt can be used for drying. The chemical must be anhydrous, so that it'll steal away all the water from the object in need of drying.
However, is this the case with triethylene glycol? Must it be made anhydrous before it can be used for drying, or is its attraction to water molecules great enough to begin with?

In the oil and gas industry, glycols are used in a scrubber to remove water from natural gas [methane] for a number or reasons: water is very soluble in glycols, water is extracted into “dry” glycols due to equilibrium forces, and there is efficient contact between the gases and the liquid.  Another huge part of the stribber is the regeneration portion of the system, the stripper – getting to the point of your question.  Heat is applied to the water/glycol mixture after having been in contact with the “wet” methane, driving off the water and regenerating the sorbent, the glycol.  The glycol can then be recycled.  Although I have described this as a batch process, the actual process is continuous.
In an industrial setting, “throwing something away” is used as a last resort.
Below is a schematic of a glycol dehydrator and regeneration unit.

https://www.sciencedirect.com/science/article/pii/S2352484719301039#fig1


Single electron reduction of ketone by Na or Zn Whenever we add Na or Zn metal to a ketone, why does the electron prefer to attack the carbonyl bond and form a radical, instead of removing the acidic hydrogen and forming an enolate and $\ce{H2}$ gas?

I was thinking that usually when we add Na and there is a sufficiently acidic Hydrogen present, it reacts and releases $\ce{H2}$. I just wanted to know if there is an advantage in attacking the carbonyl bond instead of an acid-base reaction. Also, is this dependent on kinetic or thermodynamic control.

Let the carbonyl compound under consideration be acetone.

Case I: Hydrogen Abstraction

The species formed are NaH and an allyl-type radical. The resonating system includes an electronegative oxygen atom, which also has an incomplete octet in one of the canonical forms.

In essence, products are not really stable.

Case II: Attack on Carbonyl

The species formed is a radical anion, which are relatively stable as the filled orbital of the oxygen relieves some electron deficiency of the radical. This effect is stronger than the one offered the pi-bond in Case I.


N2O production vs air distillation After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)? I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air? When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other? These questions are in reference to an automotive application.

After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)?

It's not clear to me that $\ce{N2O}$ is produced from $\ce{N2}$: Rather, it appears that commercial quantities of $\ce{N2O}$ are produced by thermally decomposing ammonium nitrate:
$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$
as described (at least) by this manufacturer and supplier of nitrous oxide gas plants. See the comprehensive Wikipedia entry for more information.

  I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air?

I'm not sure what you mean here.

  When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other?

Nitrogen is liquid in the range 63 to 77 K; for oxygen, it's 54 to 90 K. Thus, the temperature range for which both are liquids is bounded by the range on nitrogen, or roughly between 63 and 77 K.

The temperature at which nitrous oxide solidifies (182 K per Wikipedia) is much higher than the temperature range at which oxygen and nitrogen are in the liquid phase, and thus it would have come out of solution as a solid long before the requisite temperature was reached. Thus, we would not expect to see liquid $\ce{N2O}$ co-existing with liquid oxygen and nitrogen.

Note bene: I am using the pressure range of 5 to 10 bars as cited here in assuming that liquid and solid phases of the above moieties do not coexist during industrial gas production at these relatively low pressures.


Temperature profile at steady state Consider a solid wall consisting of three layers where the layers are of equal
  thickness but made out of different materials: $\ce{Layer 1}$ is made out
  of copper (high thermal conductivity), $\ce{Layer 2}$ is made out of a
  highly porous ceramic (low thermal conductivity), and $\ce{Layer 3}$ is
  made out of concrete (medium thermal conductivity).

  How does
  the steady state temperature profile inside the wall look like?

I'm not sure if this question is asked in a way to mislead me into drawing the T-profile as (2) when at steady-state it actually should look like (1).

I apologize if this question is seen as off-topic or simple, but I would appreciate it if someone would explain if I am correct in saying that the T-profile at steady state should be linear.

Thank you!

(1)
 (2)

No, the temperature gradient is not uniform across all three layers.

To see why, start from the Fourier law for heat flow, shown in my solution to this problem.

Since at steady state the heat flux $q$ must be equal for all layers, we have that

$$q_{1}=q_{2}=q_{3}$$

And since the layers are equal in thickness,

$$ k_1\Delta{T_1}=k_2\Delta{T_2}=k_3\Delta{T_3} $$

Clearly, if the $k$ are not equal, the $\Delta{T}$ cannot be equal either, so the profile will not be that in (1).


Understanding colour changes of systems in dynamic equilbrium Consider an aqueous reversible reaction of two chemicals $\ce{A <-> B}$, where $\ce{A}$ is red and $\ce{B}$ is yellow. The reaction is in equilibrium when there is $\pu{1 dm^3}$ of A with concentration $\pu{1 M}$ and  $\pu{0.5 dm^3}$ of $\ce{B}$ with concentration $\pu{1 M}$, such that $K_C = 0.5$ and the solution is orange (due to the equal parts of red and yellow).

If I were to add $\pu{1 mol}$ of $\ce{A}$ to the system without increasing the total volume of the solution, Le Châtelier's principle predicts that the equilibrium position would move to the right, hence there would be more of $\ce{B}$ than $\ce{A}$ and the solution would become more yellow, however this does not seem to agree with my understanding:

$K_C = 0.5$, and is constant when the concentrations of chemicals are changed, so some of $\ce{A}$ must be converted to $\ce{B}$ by the forwards reaction to keep $K_C$ as $0.5$. This will make the concentration of $\ce{A}$ $\pu{\frac{5}{3} M}$ and the concentration of $\ce{B}$ $\pu{ \frac{5}{6} M}$ at equilibrium. The chemicals are still in the same ratio, so shouldn't the colour remain the same instead of becoming more yellow as Le Châtelier suggests?

This is a simplified problem, and I am wondering if the logic applies to more complicated cases (if it's correct in the first place of course!) Essentially, in my understanding, if you increase the concentration of a chemical in a dynamic equilibrium, when equilibrium is re-established, some but not all of the added chemical would be converted into the products on the opposite side of the reaction. This means the amounts of chemical on both sides of the reaction have increased, so it doesn't make any sense to say that the equilibrium position moves to one side of the other - both sides have increased in concentration, and in terms of coloured chemicals, the colour would remain the same.

Please use this more complicated reaction for demonstrative purposes:

$$\ce{Fe^{3+}_{(aq)} + 3\,SCN^-_{(aq)} <--> Fe(SCN)3_{(aq)} }$$

where $\ce{Fe^{3+}}$ is yellow, $\ce{SCN^-}$ is colourless and $\ce{Fe(SCN)3}$ is red. How would the reaction behave if the concentration of $\ce{Fe^{3+}}$ ions is increased (e.g. add iron III nitrate).

Apologies in advance if this question is a little convoluted, I'm not a professional chemist (I'm studying A-level chemistry), and I'm quite confused, so I'm struggling to form a coherent question.

Addition (edit):
In the case of the more complicated reaction, my understanding is as follows:

If I add $\ce{Fe^{3+}}$, the forwards reaction will convert some of the added $\ce{Fe^{3+}}$ and some $\ce{SCN^-}$ (already present) into $\ce{Fe(SCN)3}$.
Not all of the added $\ce{Fe^{3+}}$ will be converted (the equilibrium cannot completely counteract the change, as this would involve converting all the $\ce{Fe^{3+}}$ into $\ce{Fe(SCN)3}$, and this would be an increase in the amount of $\ce{Fe(SCN)3}$ which the reaction would have to counteract).
Therefore the concentration of reactants and products increases, because the added $\ce{Fe^{3+}}$ is split between some which is converted into $\ce{Fe(SCN)3}$, and some that stays as $\ce{Fe^{3+}}$.
Since this gives an overall increase in the concentrations of both chemicals, neither has increased relative to the other, and the equilibrium position hasn't moved, so the colour doesn't change.

In the general case of $\ce{A + B <=>C}$ if  there are initially $a_0, b_0,c_0$ amounts present , then if an amount x of A and B react to produce C then the equilibrium constant $K_e$ is given by
$$ K_e=\frac{k_1}{k_{2}}=\frac{c_0+x_e}{(a_0-x_e)(b_0-x_e)}$$
where $k_1, k_2$ are the rate constants for the forward and reverse reaction and $x_e$ is the amount reacted at equilibrium.
Solving the equation and  for (algebraic) simplicity let $c_0=0$, gives the amount at equilibrium
$$x_e=\frac{1}{2}\left(a_0+b_0+K_e^{-1}-\sqrt{(a_0-b_0)^2+2(a_0+b_0)/K_e +K_e^{-2}}\right)$$
Using this the amount of each species can be obtained as for example the equilibrium amount of A is $a_e=a_0-x_e$. However, in your question you ask about the colour not the concentrations and this alters things a little. The quantity that is measured optically as in a spectrophotometer is the optical density of for example species A is $D=\epsilon_\lambda[A]l$ where $\epsilon_\lambda$ is the extinction coefficient of the molecule at wavelength $\lambda $ and is effectively how strongly the molecule absorbs at that wavelength, $\ce{[A]}$ is the concentration and l the path-length of the sample through which the light passes.

The total absorption is the sum of that for all species so depending on the values of $\epsilon$ and how the concentrations change the absorption may go up, down or remain constant at any given wavelength.

Furthermore, if you do not use a spectrophotometer and measure optical density but you are looking through the sample you will observe the transmitted light which is by the Beer-Lambert law given by $I_{trans}=I_0e^{-D}$ for each species present at each wavelength, so what you may see becomes pretty complicated.