CO2 + High humidity = transperent substance? First I want to point out that I am not a chemist, sorry for bad description.
I have CO2 incubator. CO2 concentration is 10-20% and the humidity level is around 90%. I was using a humidity sensor to measure humidity and some kind of transparent substance have formed around the sensor. Sensor becomes unprecise and had a slow response time. Substance almost looks like plexiglass.
Is it possible that carbonic acid which is forming from CO2+H20 -> H2CO3 is somehow reacting with integrated circuit case or sensor and forming some kind of glass-like substance?
New sensor:

Used sensor:

Does it stay like thay when you remove the pressure and take it to the normal air? if yes, then it is likely co2 is being absorbed into the plastic part, and that plastic part is swelling. You can avoid it by using a glass window for optical sensor. Or use H2O permeable but not CO2 permeable membrane if this sensor needs contact with the gas. Or use another type of sensor that does not contain this plastic. Or reduce CO2 pressure. I dont know any CO2+H2O compounds that would stay stable and solid at normal pressure. Methanol could be forming from those materials if energy is present, but it needs catalyst and is very unlikely to happen on its own.


Why is sodium chloride edible whereas both sodium and chlorine are highly reactive? Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference?

Also, is there any difference in eating sodium chloride in powder form and after dissolving in water? It gets ionized, correct?

Sodium chloride is edible. But elemental sodium and chlorine are highly reactive and poisonous. What is the difference?

Solid sodium chloride is an ionic compound. This means it consists of positively charged sodium ions ($\ce{Na+}$) and negatively charged chloride ions ($\ce{Cl-}$) arranged in a cubic lattice.

On the other hand, sodium metal and chlorine gas are elements. $\ce{Na}$ has an electron configuration of $\mathrm{[Ne]~3s^1}$ and so it really wants to lose that single electron to form a full outer shell as a $\ce{Na+}$ ion. This makes it highly reactive towards species which can easily accept an electron such as $\ce{Cl2}$, $\ce{H2O}$ etc. $\ce{Na+}$ has a full outer shell, being isoelectronic with neon, and so it is quite happy to stay as an ion and not react. Chlorine, with an electron configuration of $\mathrm{[Ne]~3s^2~3p^5}$, really wants to gain a single electron to get a full outer shell, isoelectronic with argon, as the $\ce{Cl-}$ ion. This makes it highly reactive towards species which can easily donate an electron, such as $\ce{Na}$.

  is there any difference in eating sodium chloride in powder form and after dissolving in water?

Not really. Sodium chloride dissolved in water consists of hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions. If you eat solid sodium chloride then it rapidly dissolves in the water in your body, so the effect is the same. Interestingly though, solid salt tastes quite nice, but salty water tends to taste horrible.


Why does hydrogen bonding require a lone pair? Why does hydrogen bonding require a lone pair of electrons on the partially negative atom? They would still be partially negative even without the lone pair right? Or does a bond exist, but is much more negligible compared to when a lone pair is there?

You don't need a lone pair. Water can hydrogen bond to methane, attacking the back side of one of the carbon-hydrogen bonds in the methane. In a molecular orbital the sigma bonding orbitals in the methane are sufficiently delocalized to serve in place of lone pairs, overlapping the antibonding orbitals in the water [1]:
Abstract (from [1])

Quantum chemical calculations were performed at different levels of theory (SCF, DFT, MP2, and CCSD(T)) to determine the geometry and electronic structure of the HOH···CH4 complex formed by water and methane molecules, in which water is a proton donor and methane carbon (sp 3) is an acceptor. The charge distribution on the atoms of the complex was analyzed by the CHelpG method and Hirshfeld population analysis; both methods revealed the transfer of electron charge from methane to water. According to the natural bond orbital (NBO) analysis data, the charge transfer upon complexation is caused by the interaction between the σ orbital of the axial С–H bond of methane directed along the line of the O–H···C hydrogen bridge and the antibonding σ* orbital of the О–H bond of the water molecule. Topological analysis of electron density in the HOH···CH4 complex by the AIM method showed that the parameters of the critical point of the bond between hydrogen and acceptor (carbon atom) for the O–H···C interaction are typical for Н-bonded systems (the magnitude of electron density at the critical point of the bond, the sign and value of the Laplacian). It was concluded that the intermolecular interaction in the complex can be defined as an Н bond of O–H···σ(С–H) type, whose energy was found to be 0.9 kcal/mol in MP2/aug-cc-pVQZ calculations including the basis set superposition error (BSSE).

Reference
1. A. N. Isaev,"O–H···C hydrogen bond in the methane–water complex", Russian Journal of Physical Chemistry A 90(2016), 1978–1985.


Definition of "soluble" How to define "soluble" in chemistry?

For example, in 1:1 v/v, benzene does not mix with water, i.e. insoluble. Common sense.

However, in an analytical procedure, it called for making a "saturated" aqueous solution of benzene, which is about 1.5 g/L. So, benzene actually  has some solubility in water.

So, my question is, is there an official definition of "soluble"?

IUPAC defines "Solubility", but not "soluble". There is no single definition because there is no single meaning. The reason there isn't a single meaning is because it is useful to have different meanings in different contexts. (This is true of almost all words.) According to Wikipedia, 0.1g per deciliter is a common ceiling for insolubility. I've seen various values for the lower value for labeling a compound "soluble". I've seen 1g, 2g, 3g, 5g and possibly more claimed to be the (arbitrary) floor. I don't see any reason to call for standardization, because the use for such a property is varied and diverse.


Extracting sulphur from hot springs with medieval technology? For the sake of a dungeons and dragons game. Could sulphur be deposited on/extracted from hot springs water through evaporation or other proess?

Sulfur does naturally occur around hot springs and volcanos. It would be very reasonable to assume that a hot spring could have natural deposits of sulfur near it that could be collected without any special effort.


Generating electron density from protein coordinate (PDB) Normally protein coordinates are generated from electron density, such as that from X-ray scattering experiment. In my case, I want the reverse - PDB to electron density mesh. I thought it was more trivial and a program should be available. But Google did not give me any luck.

Is there any existing python library/program which converts protein coordinate (pdb) into electron density mesh (custom resolution)? If not, how should one proceed? The orbital and atomistic electron density varies among atom types.

Your suggestion appreciated.

Edited:

The precision of electron density I am interested in is ~ 1-2A so hopefully DFT level might not be necessary in my case. The aim is to visually resolve sidechain atoms. Since there could be manipulation of the PDB upstream, some crystal structure information could be lost.

Indeed the raw electron density might provide such information but a unified treatment of PDB coordinate homogenizes structures, making a more fair comparison between them, specifically in my case.

For the more recent structures, you can view the density (based on measured diffraction data and the model, so-called 2Fo-Fc density) directly in the protein data bank, e.g. http://www.rcsb.org/3d-view/6QU9?preset=electronDensityMaps:

For a theoretical model or a model without deposited diffraction data, you would first have to generate structure factors, and then calculate the electron density. In the simplest case, the electron density would just reflect the density of isolated atoms (i.e. no deformation density). Program suites such as CCP4 (originally in Fortran) or Phenix (in python) are available to do these steps.

In these calculations, the electron density of individual atoms is encoded in so-called atomic form factors (see e.g. http://lampx.tugraz.at/~hadley/ss1/crystaldiffraction/atomicformfactors/formfactors.php), which are fourier transform coefficients of the model electron density for each atom.


Can the light reaction of photosynthesis happen spontaneous in pools of water? I've just learnt that a part of the photosynthesis reaction is the light reaction, where sunlight is absorbed by chloroplasts, which then excites electrons to higher energy levels, allowing for water molecules to be broken down.

My question is whether this can also happen in pools of water? I mean, if all that is really needed is sunlight, then I don't see why water molecules in pools won't get excited and thus break down too.
If not, then how do chloroplasts assist in this?
And on top of that, why don't chloroplasts get damaged since this is similar to how UV rays damage our skin as it can excite electrons in cells and damage genes?

Photosynthesis is a lot more complex than just breaking down water

Yes, photosynthesis involves a reaction which partly involves absorbing photons and breaking down water. But overall it is far more complex than that as it also involves absorbing carbon dioxide and turning that into sugars and other molecules that plant can use a fuel. Breaking down water is a byproduct of that larger process as is the oxygen released by it.

The detail of the reaction involved is complex and consists of dozens of individual steps. It isn't as simple as absorbing a photon and splitting water. In fact, there are two separate and individually complex systems of proteins involved in the water-splitting part of the process: photosystem I absorbs light and generates higher energy electrons; photosystem II also absorbs light, breaks water molecules part and generates higher energy electrons. But these processes are so complex that we don't fully understand how they work (and our understanding of the structure of photosystem II was updated as recently as 2014). And that machinery is very complex involving dozens of individual proteins and components.

Given the complexity of the processes behind water-splitting in photosynthesis (which involves multiple photons and electrons) it should be clear that there is no easy way to perform the reaction.

You don't just need photons and water molecules. You need some of the most complex molecular machinery we know of and multiple photons. Sunlight + water isn't going to achieve it.


Is it possible to derive integrated rate law algebraically? Here is the calculus way:

I have no prior knowledge of calculus. But my teacher asks me to derive the integrated rate law for any order between one to ten.

It would be fine for me to memorize the derivations for zero to third orders. But I can't do all ten for sure.

Calculus is pretty much just a shorthand for algebra (plus the whole limits concept), so the answer is technically yes.

Your teacher might be intending for you to use the pattern @Klaus Warzecha mentioned:

  For a function $f(x) = x^n$, the primitive integral is $F(x) = \vert\ln(x)\vert$ if $n=−1$, otherwise $F(x) = \frac{1}{n+1}x^{n+1}$

So in that sense, you can use algebra if you let

$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]^n} \, \mathrm{d}[A] = \frac{1}{n - 1} \left[ \frac{1}{[A]^{n-1}_{\tau}} - \frac{1}{[A]^{n-1}_{0}} \right]$$

for $n \neq 1$

and

$$\int_{A_0}^{A_\tau} \! \frac{1}{[A]} \, \mathrm{d}[A] = \ln \frac{[A]_\tau}{[A]_0}$$

for $n = 1$

Note that this only works for single-species rate equations - if you have multiple species the integration result changes. Note that this also isn't really using algebra - it's memorizing and applying one of many general integration results. Which is what you do in integral calculus.

In any case, it doesn't seem fair to me to expect you to know how to derive integrated rate laws if integral calculus isn't a prerequisite for your course. If it is a prereq, then now you are seeing why. If it isn't, then what your teacher is asking you to do is, in my opinion, unrealistic.


Is the end of a cigarette on fire when lit? Is the end of a cigarette on fire when it is lit? Or is it smoldering? Is there a chemical difference between the two?

A lighted cigarette is burning, technically smoldering (no flame).  The stuff inside the paper bears only a passing resemblance to tobacco,

http://en.wikipedia.org/wiki/Cigarette#Construction


Stereoselectivity of Roush crotylboration One can use a tartrate derived boronate to enact enantioselective crotyl transfer. Eg in scheme below:

I do not understand how the stereoselectivity arises. We have a chiral boronate, true, and we have facial selectivity on addition to boron centre. But I don't understand why one face is selected over the other - With one ester group on the ring, I could see that would sterically block one face of the boron centre, but with 2 ester groups it seems that each side of the boron is equally blocked.

This figure from this reference (p. 170). might help you visualize things.

As you can see, the stereochemistry is given by electronic repulsion between the carbonyl group and the incoming aldehyde.


Why is the formation free energy of carbon dioxide almost constant in Ellingham diagrams? While reading about Ellingham diagram in my textbook, I noticed that for:
$\ce{C_(_s_) + O2_(_g_) -> CO2_(_g_)}$,
the standard Gibbs free energy ($\ce{\Delta_fG^\circ}$) doesn't seem to vary with temperature (I even compared the line's inclination with a horizontal line and they coincided completely!).
Going through the Wikipedia page for Ellingham diagram, I was able to find the following information:

The formation free energy of carbon dioxide ($\ce{CO2}$) is almost
independent of temperature, while that of carbon monoxide (CO) has
negative slope and crosses the $\ce{CO2}$ line near $\mathrm{700 ^\circ C}$.

I agree with this more as I believe that $\ce{\Delta_fG^\circ}$ should vary with the temperature. Upon comparing the $\ce{CO_2}$ line (in the Ellingham diagram in Wikipedia) with a horizontal line, this time the $\ce{CO_2}$ line is inclined slightly upwards.
I still have a doubt as to why the inclination of the $\ce{CO2}$ line is so less. One reason that I am able to think of is:
As there are equal number of gases on each side of the reaction, the entropy of the reaction is going to be less (as there is no net increase of decrease in the number of gases). As $\ce{\Delta G^\circ = \Delta H^\circ -T\Delta S^\circ}$, if $\ce{\Delta S^\circ}$ (whose negative value is also the slope of the lines) is really very less as compared to $\ce{\Delta H^\circ}$, then we can conclude that $\ce{\Delta G^\circ}$ is going to be almost constant. But I seem to have a doubt (more specific to thermodynamics). We know that the entropy of reaction is "less", but "less" as compared to what?, i.e., how are we able to tell that this magnitude of entropy is lesser than the magnitude of entropies of all the metal oxidation reactions?
I think there must be a better way to explain the less magnitude of entropy of formation of  $\ce{CO2}$.
Could anyone please provide a concrete method / reason to prove this?
Also, are there any other factors (other than the entropy factor) governing this?

Let's subdivide the CO2 standard entropy change reversible process into 3 parts:

Isothermal expansion of one mole of oxygen from 1 bar to the equilibrium partial pressure of oxygen in a reactor operating at equilibrium at a total pressure of 1 bar (a Van't Hoff equilibrium box)

Injection of one mole of oxygen at its equilibrium partial pressure into the equilibrium box reactor through a semipermeable membrane  while removal of one mole of CO2 through a semipermeable membrane at its equilibrium partial pressure in the reactor

Isothermal compression of one mole of CO2 from its equilibrium partial pressure in the equilibrium box reactor to a pressure of 1 bar.

The overall changes (standard changes) in G, H, and S for this 3-step process are -394359 J/mole, -393509 J/mole, and 2.8508 J/mole-K.  Now, let's break this down to the changes in the three steps.  Based on the standard change in G for the overall process, the equilibrium partial pressures of oxygen and CO2 in the equilibrium box are 8.08E-70 bars and essentially 1 bar.
In step 2, since the reaction is operating at equilibrium, the change in G is zero, while the changes in H and S are $\Delta H_2=\Delta H^0=-393509$ J/mole and $\Delta S_2=\Delta H_2/298.15 = -1319.8$ J/mole-K.
In steps 1 and 3, the expansion of oxygen and the compression of the CO2 are isothermal, so $\Delta H_1=\Delta H_3=0$.  In step 1, $\Delta G_1=R(298.15)\ln{8.08\times 10^{-70}}=-394359$ J/mole and $\Delta S_1=-R\ln{8.08\times 10^{-70}}=1322.7$ J/mole-K.  In step 3, the initial and final CO2 pressures are essentially 1 bar, so $\Delta G_3=\Delta S_3=0$
So, it seems that, merely by coincidence, the changes in entropy in step 2 compared to the sum in steps 1 and 3 are of nearly equal magnitude and opposite in sign (-1319.8 vs +1322.7).  This means that their difference is only 2.85, which represents the overall entropy change.  It is coincidental because the mechanistic changes in step 2 compared to steps 1 and 3 are very different.


What do different forms of ice look like? Everyone is familiar with the very common, hexagonal ice. Bit there are many other forms of ice - what do they look like? If you had big blocks of ice in all the different known phases next to a block of hexagonal ice, what obvious differences would you notice?

As you know, ice can exist in at least 15 or so solid phases. Of these, the first phase, $Ice$ $I$ (ice-one) is the only phase found naturally on earth, mostly in the form of $Ice$ $I_h$ hexagonal crystal as seen in glaciers and snowflakes.

The visible appearance of ice (eg: transparency and clarity) depend also upon the level of purity of the crystals formed. For example, air can get trapped within inclusions of the ice, making it translucent or opaque white, as shown in the image below.

The cubic form $Ice$ $I_c$ is metastable so is rarely found, generally in high clouds in the upper atmosphere, where it is formed at ambient pressure by the condensation of water vapour at very low temperatures (-80°C).

Most other forms of crystalline ice (designated as $Ice$ $II$ to $XV$ are metastable, being produced in the laboratory by the application of high pressures (above 200MPa) and/or low temperatures (below 72K), resulting of denser packing of the water molecules.

For example, $Ice$ $II$ (see image below) is formed by compressing 'normal' $Ice$ $Ih$ to 200MPa at temperature of 198K.

The SEM image above is of $Ice$ $I$ decorated grain boundaries in fine-grained $Ice$ $II$. The $Ice$ $I$ stands in raised relief relative to the $Ice$ $II$ due to volumetric expansion to the lower-density phase.

Scientists believe this form of ice can be found on the interior of some large icy moons orbiting Jupiter.

The macroscopic appearance of the 'exotic' forms of ice depends upon many variables such as crystal grain size. The laboratory production of these 'novel' forms of ice to ensure proper formation of fine crytallite grains (less than 10 microns) is time consuming, requiring careful control of temperature and cycling of very high pressures.

For example, $Ice$ $IV$ is formed by heating high-density amorphous ice at a slow (0.4K/min) rate from 145K and at a constant pressure of 0.81GPa. It forms a rhombohedral crystal (see image (a) below).

Similarly, $Ice$ $V$ is formed from liquid water at 500MPa by lowering its temperature to 253K. It has a more complex lattice structure and forms monoclinic crystals (see image (b) above).

"Square snow" or $Ice$ $VI$ is formed from liquid water by compressing it at 1.1GPa (10,850atm) and lowering its temperature to 270K. Its unit cell forms tetragonal crystals with all water molecules hydrogen bonded to four others, resulting in the formation of dendritic crystals (approx 45 microns across, see image below).

For more images from the laboratory, see: http://web.mit.edu/wbdurham/www1/short%20htmls/more%20ice%20photos.html

References:

Water structure and science, Ice phases, known ices, Martin Chaplin.

Grain size-sensitive creep in Ice II, Kubo, Durham, Stern & Kirby, Science, Vol 311 (2006).

Visual observations of crystal morphologies and melting of Ice IV at pressures up to 1GPa, Chou and Haselton, Review of High Pressure Science Technology Vol 7 (1998).

Dynamic pressure-induced dendritic and shock growth of ice IV, Lee, Evans & Yoo, Proceedings of the National Academy of Sciences, Vol 104, No. 22.


Hill function and stoichiometric matrix: confusion about reaction rates When deriving the Hill function (protein $X$ and $n$ molecules of the signal $S$), my book introduces the following rates:

$$
\text{collision rate} = k_{ON} \cdot X_n \cdot S^n\\
\text{dissociation rate} = k_{OFF} \cdot [nSX],
$$

where $[nSX]$ is the complex. This translates into the following ODE:

$$
\frac{d}{dt} [nSX] = k_{ON} \cdot X_n \cdot S^n - k_{OFF} \cdot [nSX].
$$

This seems fine if I follow the $\textit{rules}$ of, for instance, this table.

On the other hand, during the class, my professor introduced the stoichiometric matrix. If we have the reaction $2A + B \leftrightarrow C$, then the corresponding system of ODEs would be:

$$
A' = -2r_1A^2B + 2r_2C\\
B' = -r_1A^2B + r_2C\\
C' = r_1A^2B - r_2C.
$$

So, my question is: why in this case we put the $2$ in front of the reaction rates while for the Hill function we do not (put the $n$ in front of it)? Is it just because the $n$ molecules of signal $S$ are not consumed nor produced?

If the reaction is $\ce{aA + bB = pP + qQ} $ then we define the rates as

$$-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{p}\frac{d[P]}{dt}=+\frac{1}{q}\frac{d[Q]}{dt}$$

As to the Hill function it seems that the complex nSX is just one species.


Is the given stereoisomer of 1,2-dimethylcyclobutane chiral? I was going through some problems of chirality when I came across the following compound

  Is the given compound chiral?

My attempt

If we look at its mirror image about a mirror which is perpendicular to plane of the molecule, the two $\ce{-CH3}$ groups which is pointing upwards and downwards (implied by the wedge and the dash) will interchange their positions but then also the mirror image would be perfectly superimposable on the given compound. Hence it should be achiral but the answer key gives it to be chiral. I couldn't get it. Is this compound an exception?

One thing you should always remember is that you should always consider the actual three-dimensional structure when determining chirality. Sometimes there are elements of chirality that are not visible on paper but clear if the full structure is considered. For cyclobutane, this means you should consider the puckered structure of the core structure. However, this often rapidly equilibrates between different practically identical configurations and does not add a layer of complexity to this molecule. Thus, we are able to analyse with the paper plane only.

If you put a plane of symmetry through the centre of the $\ce{CMe-CMe}$ bond, at first glance it may look like it’s a perfect fit: all four carbon atoms of the ring are put on top of another and the wedge and hash simply switch positions.

However, note that wedge and hash mean something entirely different. In one case, the methyl group is above the paper plane (wedge), in the other, it is below it (hash). The corresponding opposite position is occupied by a hydrogen atom that completes the four valences of the cyclobutane’s carbon. Thus, this plane of symmetry you propose transforms hydrogen onto methyl and vice-versa. It is not a symmetric transformation.

You can maybe visualise this by taking a sheet of paper, marking the bottom left corner with a filled dot (the wedge-methyl group) and the bottom right with a circle (the hash-methyl group). Turn it around and put a circle behind the dot and a dot behind the circle, so that you may identify it from both sides. You may already notice that in both cases the dot is bottom-left while the circle is bottom-right — this already indicates that we have a $C_2$ axis, i.e. we can rotate around an axis that is perpendicular to the $\ce{CMe-CMe}$ bond and in the ring plane.

Prepare a second piece of paper but with the reverse orientation: this is the molecule you created by mirroring along the plane. Now attempt to orient the two pieces in such a way that dot aligns with dot and circle aligns with circle. Spoilers: you will fail. anti-1,2-dimethylcyclobutane is chiral.

You can retry the experiment with the syn-1,2-dimethylcyclobutane, i.e. two dots at the bottom of the sheet of paper (and two circles on the back side). But you should immediately see that another piece of paper with identical dots/circles can immediately be aligned perfectly with the original. Hence, syn-1,2-dimethylcyclobutane is achiral.

This can be explained well with point groups. The anti isomer has the point group $C_2$ which does not feature a plane or centre of symmery. The syn isomer has the point group $C_\mathrm{s}$ whose only element of symmetry is a plane, hence it is achiral. You can read more about point groups in this question.


Is mg/m^3 same as ppmw (CO2)? I'm trying to convert $\ce{CO2}$ ppm (volume) in the air into ppm (weight) in the water.

I found this formula

  $\mathrm{mg \ m^{-3}}$ = (ppmv)(MW) / [($0.08205$) (¦$\mathrm{K}$)]

where,

ppmv = air pollutant concentration, in parts per million by volume

$\mathrm{mg \ m^{-3}}$ = milligrams of pollutant per cubic meter of air

$\mathrm{K}$ = atmospheric temperature in degrees Kelvin = $273.15$ + $^\circ C$

$0.08205$ = universal gas law constant in $\mathrm{L\ atm \ mol^{-1} K^{-1}}$

MW = molecular weight of the air pollutant (dimensionless)

$\mathrm{atm}$ = absolute atmosperic pressure in atmospheres

$\mathrm{g \ mol}$ = gram mole

I've read the $\mathrm{mg \ m^{-3}}$ is a subunit of ppmw, so is it correct to say $\mathrm{mg \ m^{-3}}$ is ppmw?

No. ppmw means parts per million by weight (mass). This is not the same as mg/m3, which is milligrams (mass) per cubic meter (volume).

For example, in a gas of very low density (e.g. $\ce{CO2}$ at 10-6 Pa), 1 mg/m3 could exceed the weight of $\ce{CO2}$, i.e. would be greater than 900,000 ppmw.

One measure has to do with mass in a given volume, the other measure is mass in a given mass of another substance.


Where do we report yields in a research report? I'm in an integrated laboratory course at university that combines all the disciplines of chemistry that we have here. I was instructed to place the yield in the methods section of the report by my professor, but found that to be odd.
Is reporting the yield in the methods section a standard protocol in some paper formats?

Your professor is right! Giving the yield in the method section, that is, together with detailed information on synthesis, workup and characterization of the products obtained, is standard in scientific journal I know or I have published in.


How does precipitate affect cell potential? A cell is set up with copper and lead electrodes in contact with $\ce{CuSO4(aq)}$ and $\ce{Pb(NO3)2(aq)},$ respectively, at $\pu{25 °C}.$ The standard reduction potentials are:
$$
\begin{align}
\ce{Pb^2+ + 2 e- &-> Pb} &\quad E^\circ &= \pu{-0.13 V} \\
\ce{Cu^2+ + 2 e- &-> Cu} &\quad E^\circ &= \pu{+0.34 V}
\end{align}
$$
If sulfuric acid is added to the $\ce{Pb(NO3)2}$ solution, forming a precipitate of $\ce{PbSO4},$ what will happen to the cell potential?

I know the lead is the anode while the copper is the cathode:
$$\ce{Pb(s) + Cu^2+ -> Pb^2+ + Cu(s)}$$
The addition of $\ce{H2SO4}$ will cause less $\ce{Pb^2+}$ to be in the solution, thus causing the equilibrium to push to the right creating more $\ce{Pb^2+}.$ With more $\ce{Pb^2+},$ if you were to put it in the Nernst equation, would the greater number in the log expression cause the cell potential to decrease?

The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to
$$E_\ce{Pb} = \pu{-0.13 V} + \pu{0.0296 V}\times\log{10^{-8}} = \pu{-0.37 V}\tag{1}$$
As a consequence, the $\ce{Pb}$ potential decreases. But the cell potential increases and goes from the initial value (without sulfate)
$$E^\circ_\mathrm{cell, i} = \pu{0.34 V} - (\pu{-0.13 V}) = \pu{0.47 V}\tag{2}$$
to the final value (after precipitation of lead sulfate)
$$E_\mathrm{cell, f} = \pu{0.34 V} - (\pu{-0.37 V}) = \pu{0.71 V}\tag{3}$$


Comparing first, second, and third ionization energies between atoms Among the elements boron, carbon, nitrogen, and magnesium, what are the first ionization energies from least to greatest? What about the second and third ionization energies?

I understand the first ionization energy is given by the following formula: $\ce{X -> X+ + e-}$, the second ionization energy is given by: $\ce{X+ -> X^2+ + e-}$, and the third ionization energy is given by: $\ce{X^2+ -> X^3+ + e-}$.

The electron configurations of the elements listed are:

$B: 1s^2 2s^2 2p_x^1$

$C: 1s^2 2s^2 2p_x^1 2p_y^1$

$N: 1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$

$Mg: 1s^2 2s^2 2p^6 3s^2$

I know that magnesium has the lowest first ionization energy because it undergoes the most electron shielding; it is shielded from the positive nuclear charge by the ($1s^2 2s^2 2p^6$ electrons).

I can deduce that nitrogen has a higher I.E. than carbon because they are both shielded by the same electrons, the $1s^2 2s^2$ electrons, but nitrogen has a higher nuclear charge. The same reasoning can be used to conclude that carbon has a higher first ionization energy than boron.

I did the above reasoning with respect to the first ionization energies. How do I apply a similar logic with the second and third ionization energies? Do the second and third ionization energies follow periodicity with respect to periods and groups? What is that trend, if any?

If you know the trends for $1^\text{st}$ ionization energies, a good first approximation for the trends of $n^\text{th}$ ionization energies at any given point is the $1^\text{st}$ ionization trend $n$ elements earlier.

As an example, look at the first and second ionization energies for the second row elements (Table).
From $\ce{Li}$ to $\ce{Be}$, the $1^\text{st}$ ionization energy increases due to the greater effective charge on the outermost electron. From $\ce{Be}$ to $\ce{B}$, the $1^\text{st}$ ionization energy slightly decreases, as the electron we removing is in the p-subshell and is shielded by the inner s-subshell electrons. From $\ce{B}$ to $\ce{C}$, it increases again.

What about the $2^\text{nd}$ ionizations? Well from $\ce{Be}$ to $\ce{B}$, it increase, as the $2^\text{nd}$ electron from each is pulled from the s-subshell, but $\ce{B^+}$ has a greater effective charge than $\ce{Be^+}$. From $\ce{B}$ to $\ce{C}$, it decreases, as the $2^\text{nd}$ electron pulled from $\ce{C}$ is still in the less shielded p-subshell.

The $\ce{Be}$ to $\ce{B}$ trend in $2^\text{nd}$ ionization matches up with the $\ce{Li}$ to $\ce{Be}$ trend in $1^\text{st}$ ionization. The same is true for $\ce{B}$ to $\ce{C}$ (2nd) and $\ce{Be}$ to $\ce{B}$ (1st).


Distinguishing cis and trans positions What I fail to understand is why does trans isomer contain $2$ $\ce{Cl}$ at axial positions but $2$ $\ce{NH3}$ are not taken to be so. Isn't the two $\ce{NH3}$ in the trans isomer in the picture cis to each other?

The way I justify it relates to the main plane of the complex:
referring to your compounds, the plane I am talking about it the one which includes Co, the two NH3 and ethylenediamine.
If you want to "join together" with a line the two chlorides, you have to "cross that plane", therefore your compound is trans.

In the other case (the compound on the right), the plane includes Co, ethylenediamine, chloride and NH3. If you want to do the same thing, i.e. linking the two chlorides or the two ammonia together, you don't cross that plane. In the end, your compound is cis.

Also, take a look at this website as well.
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Coordination_Chemistry/Properties_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_Transition_Metal_Complexes_II

Feel free to add more questions about this!


Safety/health hazards: working with 2-aminobenzimidazole, iminoesters, and 1,3,5-triazine How dangerous is it working with 2-aminobenzimidazole, iminoesters and 1,3,5-triazine? Has there been any cases of health problems recorded concerning those three compounds?

Looking up the chemical you're working with is always a good idea, but you need to look closer and evaluate the specific danger those chemicals present to you. What you want to look at are the danger symbols and the hazard and precautionary statements. Those will tell you the specific dangers a chemical presents.

If you look up your first substance, 2-aminobenzimidazole, it looks pretty tame for organic chemistry standards. It has an exclamation mark as danger symbol, which doesn't tell you much, but the absence of a skull and crossbones and a health hazard symbol tell you that this probably won't kill you immediately. This doesn't mean the stuff is harmless, but it is also not extremely toxic. If you look at the hazard statements of this chemical you'll find that you shouldn't get it into your eyes, on your skin or breathe it. All of that is a good idea when working with any chemical, and this is exactly what you should learn when studying chemistry. This is a really tame chemical compared to a lot of stuff organic chemists work regularly with, but you still should avoid exposing yourself to it.

Your second item is an entire category of chemicals, I'll skip it for this reason. The third one is a bit more dangerous than the first one, it has a corrosive and a health hazard danger symbol. You want to avoid contact with it, but that's what all your protective equipment like safety goggles, your lab coat and your fume hood are for.

I'm not saying those chemicals are harmless, they're not. But they are less dangerous than many other common chemicals in an organic chemistry lab.


Is the kinetic energy change of a system (q) equal to the enthalpy change of a system (∆H)? When silver nitrate and hydrochloric acid are put together in a container, they form silver chloride. If you're given information about the reactants (including temperature change, heat capacity, and mass of reactant aqueous molecules), then you can find the kinetic energy change of the system.

What I want to know is: does the $q$ value you find for the system equal to the $\Delta H$ value for the system? Worded differently, does the kinetic energy change of the system equal to the enthalpy change of the system? Is this one of those obvious questions that most people understand instantly but people like me who lack common sense don't?

Under the restrictions of constant $p$, and no non-$pV$ work, we have:

$$q_p=\Delta H$$

q is the heat flow into the system (postive for flow in, negative for flow out).  However, q is not necessarily the change in the system's kinetic energy.  For instance, suppose you have an endothermic chemical reaction at constant temperature.  Heat will flow into the system.  However, some or all of that heat flow could go into increasing the potential energy of the chemical bonds.  And if the endothermic chemical reaction is accompanied by an increase in volume, some of that heat flow will go into the work needed to expand the system against the external pressure.

Indeed, if the temperature is constant, and the heat capacity of the reactants and products are the same, then there will be no change in the kinetic energy of the system.  I.e., in that case, none of the heat flow will go into changing the kinetic energy of the system.


What's the source of electrical energy in galvanic cells? Let's take a simple Daniell cell for example:

Where does the energy required to drive electrons through the external circuit and power up devices (like a light bulb) come from? Does the oxidation of zinc release that energy? And, does the reduction of copper result in more energy being released? (since we're calculating the cell potential by adding up the oxidation and reduction potentials of the two half cells). I've always wondered about this question, and I'd really appreciate a clear and concise explanation regarding this particular topic.

Judging by your above comment to beginner's answer, you seem to be confused about how reactions proceed in general.

Take a look at this graph : http://en.wikipedia.org/wiki/Activation_energy#/media/File:Activation_energy.svg

The oxidation in the beginning certainly requires energy, but at the end of the reaction, the system ends up at a lower energy state overall. This causes the reaction to be "spontaneous".


What happens if water is treated to AC power supply of 12 kV and 14 kHz? Generally water splits to oxygen and hydrogen at $\pu{1.23 V}$. What happens if we treat water at several $\pu{kV}$ with frequency range of $\pu{10 kHz}$ to $\pu{15 kHz}$ AC supply.

Using up to 30 kilovolts is not uncommon with aqueous systems in a technique called capillary electrophoresis. Similarly, applying kilohertz current is also common in a technique called conductometry to measure the resistance of aqueous solutions.
In general, when you pass high frequency current, electrolysis does not occur.

You have to check the dielectric breakdown voltage of water, perhaps with high voltage and high frequency current, severe Joule heating would be observed, if there is no dielectric breakdown (which means internal sparking).


Stability of cyclic radicals How do we arrange the following in increasing order of stability?

The answer given is: $5<3<1<2<4$ (numbering according to their order in the image).

My Approach:

I can figure that in the second and the fourth structure, resonance stabilization will be applicable and the fourth structure will be more stable than the second.

But I am not able to figure out how do I arrange the remaining structures.

Radicals are electron-deficient compounds. Therefore, according to Bent’s rule a radical should always populate an atomic orbital with a maximised $\mathrm p$ contribution. Ignoring 2 and 4 for a second, the radical occupies an $\mathrm{sp^2}$ orbital in 3 and 5 while it occupies an $\mathrm{sp^3}$ orbital in 1.[1] This already gives us $3\approx5<1$.

In a next step, let’s consider 2 and 4. These, as you noticed, are adjacent to a π bond, so the radical can occupy a $\mathrm p$ type orbital and participate in π resonance. This is the main way to differentiate 1, 2 and 4, since all three could essentially adopt a planar configuration with the radical confined to a $\mathrm p$ orbital — yet in 2 it can also resonance with the allyl system and in 4 it can resonate across two double bonds. Thus, we now intermediately have $3\approx5<1<2<4$.

All that remains is to determine the difference between 3 and 5, which is not trivial. In 3, you have a single double bond. This structure can distort a little bit, increasing the bond angle on the radical-bearing carbon towards something closer to $\mathrm{sp}$. The freely rotatable single bonds making up most of the ring allow this distortion. The phenyl radical can distort much less, since its geometry is basically already defined from the six-membered aromatic ring. Thus, we may assume 3 to be more stable than 5. Therefore: $5<3<1<2<4$.

Note:

[1]: In fact, rehybridisation is also possible in 1, giving the radical a pure $\mathrm p$ orbital and making the arrangement of substituents planar around the radical centre.


Molecular dynamics (Mac OSX user): what software can I use to plot RDFs for NPT simulations? VMD only works for NVT (constant volume) I would be very grateful if someone could recommend a way to plot RDFs for NPT ensemble molecular dynamics simulation results.

I use CP2K to run NVT simulations and usually use VMD to process results, but I don't see a way in VMD to deal with NPT results, where the volume is different at each time step.  I'm hoping that someone can recommend a piece of software that can deal with my CP2K output (i.e., a position file in terms of atomic xyz coordinates and a file that contains volume output) on a Mac or Linux system.

I have spent several hours online looking for a solution, but haven't even come across discussions of this problem.  This seems odd so maybe I'm using the wrong search terms.  If anyone can point me to online discussions of this issue, that would be great.

I'm also not having much luck tagging this question!  If anyone knows of any molecular dynamics-related tags I could use to reach people or perhaps of a more appropriate StackExchange site, I would welcome any suggestions.  Thank you.

Thank you, Buck, for all your suggestions.  You're right that a good way to check the influence of the volume is to try varying it.  I did that by using the maximum and minimum volumes of the fluctuations in one simulation.  There was some variation, so the volume does matter!

I was lucky enough to hear about "travis," an amazingly simple-to-compile code (even for Macs) available online for free.  I thought I'd share that information here because other people are bound to run into this same problem.  This code, travis, takes my trajectory file and my volume file and then asks a number of questions interactively.  It outputs a nice CSV file that can be plotted.

Thanks again for your help, Buck.  Really helped me think this through.  I'd like to ask the moderators or whoever is in charge of tagging to please add some molecular dynamics-related tags if possible!  Thank you.


Is there an experiment that visibly shows the effects of metastability? I am attempting to create an experiment to show the effects of metastability to a public audience and get people excited about science. More specifically, I am wondering if there is any sort of liquid that can mimic the effects that water had in the book Cat's Cradle.

In the book, the idea was that a shard of a more stable polymorphic crystal of ice with a high melting point could "infect" water. This crystal would quickly spread, solidifying oceans and destroying most life on earth.

The basic question I'm asking is as follows. Is there any public display I could make to show a liquid quickly solidifying from a seed crystal? It is preferable (but not necessary) for the materials to be reasonably inexpensive and require no special equipment, so as to allow for the experiment to be done on a grander scale. Additionally, the experiment should have a low failure rate (unlike using supercooled water, which is difficult to control in large amounts).

The standard demo for this is crystallization of sodium acetate trihydrate from its melt.  The trihydrate melts at 58 °C.  The molten liquid is easily supercooled, after which point dumping in a few crystals of unmelted stuff results in a rapid solidification of the mixture.

Several videos on YouTube demonstrate the reaction; Nurdrage's is my personal favorite.


What are the effects of salt additives and impurities on the CMC? I understand what a critical micelle concentration (CMC) is, however I do not know why or how impurities or salt additives affect it when measured in specific ways. Any help is much appreciated.

So what determines whether they are stabilizing or destabilizing? If it interacts with the head group, how does this change the CMC and why? The head group already has a counter ion present?

Every chemical present can interact with the free lipid and/or with the micelles. Organic chemicals will tend to interact with the lipophilic portion of the lipids, while ionic ones will interact with the polar head.

Depending on their structure, all those species can have a stabilizing or destabilizing effect on the formation of micelles and so the CMC will change accordingly.

Of particular importance is the "ionic strength" of the solution (you should research that term).


Dirac notation when bra and ket aren't vectors So, I know that
$$
\langle \psi_j(x)|\psi_k(x) \rangle =\int_{-\infty}^{\infty} \psi_j^*(x) \psi_k(x) \,\mathrm{d}x
$$
and I was wondering that if this is because of the inner product rule as with vectors? Like $\langle a|b \rangle$, but the $\psi$ isn't a vector, so how would this work in this case?

Wave function for a quantum system is a vector in a Hilbert space, which is a special kind of an inner product space. Specifically, for a spin-0 particle constrained to move in 1 spatial dimension the wave function $\psi(x)$ is an element of $L^2(\mathbf{R})$ space, i.e. the space of functions that are square-integrable on the whole real line $\mathbf{R} = (-\infty;+\infty)$. And the inner product between any two elements $\vec{\psi}_{i} = \psi_i(x)$ and $\vec{\psi}_{j} = \psi_j(x)$ of $L^2(\mathbf{R})$ space is indeed defined as follows,
$$
\newcommand{\inner}[2]{\langle{#1},{#2}\rangle}
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}} := \int\limits_{-\infty}^{+\infty} \psi_i^*(x) \psi_j(x)\, \mathrm{d} x \, .
$$

Just note here that, first, it is the matter of convention in which argument the inner product is linear: in physical literature it is often defined to be linear in the second argument, while in mathematically oriented literature it is usually defined to be linear in the first argument. Consequently, mathematicians usually define the inner product between any two elements $\psi_i$ and $\psi_j$ of $L^2(\mathbf{R})$ space in the following way,
$$
\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}} := \int\limits_{-\infty}^{+\infty} \psi_i(x) \psi_j^*(x)\, \mathrm{d} x \, ,
$$
as it is in fact done in some Wikipedia articles referenced above.

And secondly, physicists usually prefer to use even a different notation, the so-called Dirac bra-ket notation, in which elements of a Hilbert space are denoted as $\ket{\psi_i}$ and $\ket{\psi_j}$ (rather than $\vec{\psi}_{i}$ and $\vec{\psi}_{j}$ as it is usually done in linear algebra), while their inner product is designated as $\braket{\psi_i}{\psi_j}$ (rather than $\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}}$).


State solid synthesis of LiMn2O4 mechanism I have difficulties finding the mechanism of the state solid synthesis of $\ce{ LiMn2O4}$ in literacy.

Starting from lithium acetate and manganese acetate, I've found the general equation easily:

$$\ce{Li(OAc) + 2 Mn(OAc)2 + 43/4 O2 -> LiMn2O4 + 10 CO2 + 15/2 H2O}$$

What I can't find is a mechanism of this reaction. I know that $\ce{Mn}$ is oxidized and $\ce{Li}$ keeps its oxidation state. Is $\ce{Mn3O4}$ a precursor in this reaction? Is $\ce{Mn2O3}$ a by-product?

$\ce{LiMn2O4}$ has been made by rotary kilning of $\ce{Li2CO3}$ (or $\ce{LiOH}$ or $\ce{LiOH.H2O}$) + $\ce{MnO2}$. I made hundreds of grams on a small scale and 1000 pounds on a larger scale about 25 years ago.

The fine powders are added to a rotary kiln (similar to making Portland cement) and heated to about 800 degrees C. The kiln rotates, stirring the mix; the lithium compound melts and reacts with the $\ce{MnO2}$ as the powdery mass gets mushed around and around. The product was used as the cathode material in a lithium battery.

The oxidation state of the manganese drops from +4 to +3.5. The significant process is that oxygen gets lost, so to balance the equation, take one lithium, two manganese and four oxygens on the product side to make $\ce{LiMn2O4}$; everything else from the raw material side goes away.

Perhaps $\ce{LiMn2O4}$ can be made in water from dissolved salts, but gaseous $\ce{O2}$ would not be a convenient reagent.


What is a 'corona phase'? What is a 'corona phase'? I have come across this term in several papers, including this one

but all I have found concerning its definition is:

  [...] synthetic heteropolymers, once constrained onto a single-walled carbon nanotube by chemical adsorption, also form a new corona phase that exhibits highly selective recognition for specific molecules...

What is the formal definition of 'corona phase'?

One of the authors has recently established a research group at my university, so I spoke to a member of their lab:

The corona phase is not exactly a phase in the thermodynamic sense, but rather the portion of the polymer in contact with the nanotube. In the below image (taken from the Nature Nanotechnology paper in question), the blue portion and the parts of the red portion in contact with the tube would be the corona phase.


Elements other than carbon that can form many different compounds by bonding to themselves? My textbook says the following:

  Unique among the elements, carbon can bond to itself to form extremely strong two-dimensional sheets, as it does in graphite, as well as buckyballs and nanotubes.

Is carbon the only element that can do this?

If not, then what are the other elements can also do this? Is there a term to describe such elements?

What is the chemical characteristic that allows this to occur?

I would greatly appreciate it if people could please take the time to clarify this.

Is carbon the only element that can do this?

No, carbon is not the only element with such characteristics.

  If not, then what are the other elements can also do this?

There is a whole number of elements such as silicon, arsenic, germanium.

  Is there a term to describe such elements?

At least I'm unaware of such a term, which might be furnished by our far wiser community.

  What is the chemical characteristic that allows this to occur?

Catenation.

More information:

According to the Molecular Orbital Theory, the condition for a compound to exist is that it should have more electrons in the bonding orbitals than in the anti-bonding orbitals. So, as long as you have the bonding orbitals filled more, you can have pretty anything, more than just chains of atoms.

Thus, the existence of a compound also depends on the precise conditions in which the compound is kept, for example sodium forms different types of chlorides under different conditions and that as pointed out by Poutnik in the comments, $\ce{He2^1+}$ and a ton of others are discovered and still more awaiting discovery.


Finding heat of formation The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,\text{and } c$ cal respectively. The heat of formation of $\ce{C_xH_y}$ will be:
The answer is
$-\left(x b+\frac{y c}{2}-a\right)$ cal
But I am getting negative of it by adding equations:
$$\ce{C_xH_y} + (x+\frac{y}{4})\ce{O2->}x\ce{CO2}+\frac{y}{2}\ce{H2O} \text{ , }\Delta H=a \tag{1}\label{1}$$
$$\ce{C} + \ce{O2->}\ce{CO2}\text{ , }\Delta H=b \tag{2}\label{2}$$
$$\ce{H_2} + \frac{1}{2}\ce{O2->}\ce{H2O} \text{ , }\Delta H=c \tag{3}\label{3}$$
$\Delta H_f=-1\times\eqref{1}+x\times\eqref{2}+\frac{y}{2}\times\eqref{3}=-a+bx+\frac{cy}{2}$

The question: The heats of combustion of $\ce{C_xH_y}$, $\ce{C}$, and $\ce{H2}$ are $a, b,$ and $c$ $\pu{cal mol-1}$, respectively. What will be the heat of formation of $\ce{C_xH_y}$?

We have three combustion reactions to consider:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = b \tag{1}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = c \tag{2}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = a \tag{3}$$
Assuming the given answer is correct, I have to tell this straight: The wrong doing in this question is giving all three heat of combustion values as their numerical values and not mentioning it in the question (thus, solver has to assume it). That, I called being tricky at this level of education. For example, if the question is a multiple choice, then it may include both $-\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ and $+\left(x b+\frac{y c}{2}-a\right) \ \pu{cal mol-1}$ as chices to pick as an answer. Too bad for students.
Let's assume $\Delta H_{\mathrm{comp}(\ce{CO2})}$ is $-b$ in the equation $(1)$, and so on. Now we can rewrite above equations as:
$$\ce{C + O2 -> CO2} \quad \Delta H_{\mathrm{comp}(\ce{CO2})} = -b \tag{4}$$
$$\ce{H2 + 1/2 O2 -> H2O}  \quad \Delta H_{\mathrm{comp}(\ce{H2})} = -c \tag{5}$$
$$\ce{C_xH_y + ($x$ +y/4) O2 -> x CO2 + y/2 H2O}  \quad \Delta H_{\mathrm{comp}(\ce{C_xH_y})} = -a \tag{6}$$
From the equation $(4)$:
$$\Delta H_{\mathrm{comp}(\ce{CO2})} = \Delta H_{f(\ce{CO2})} - \left(\Delta H_{f(\ce{C})}  + \Delta H_{f(\ce{O2})}  \right) = \Delta H_{f(\ce{CO2})} - \left(0  + 0  \right) = -b\\ \therefore \ \Delta H_{f(\ce{CO2})}  = -b \tag7$$
Simillarly, from the equation $(5)$:
$$\ \Delta H_{f(\ce{H2O})}  = -c, \tag8$$
and from the equation $(6)$ :
$$\Delta H_{\mathrm{comp}(\ce{C_xH_y})} = \left(x\Delta H_{f(\ce{CO2})}  + \frac{y}{2}\Delta H_{f(\ce{H2O})}  \right) - \Delta H_{f(\ce{C_xH_y})} = \left(-xb - \frac{yc}{2}\right) - \Delta H_{f(\ce{C_xH_y})} = -a$$
$$\ \Delta H_{f(\ce{C_xH_y})}  = +a - \left(xb + \frac{yc}{2}\right) \tag9$$
The equation $(9)$ is the correct answer.


Why are heterogeneous catalysts calcined? Most of the literature I read says to calcine heterogeneous catalysts when preparing them. When synthesizing heterogeneous catalysts, what is the reason for calcining them?

For heterogeneous catalysis calcining of the catalyst allows for the following:

mechanically and thermodynamically stable macrostructure (matrix) with defined geometry, surface and pore size;
predictable activity based on calcining regime (generally, there is a temperature peak for maximum activity, and further extreme temperature treatment often lowers catalyst's activity);
pure composition without traces of organics and majority of other undesired impurities;
production convenience and scalability (for example, metal oxide catalyst manufacturing process can be easily implemented within mining and processing plants);
when distributed over support, more uniform distribution of metal oxide is achieved, and upon slow treatment crystallinity of the catalyst can also be tweaked.


Equilibrium expression for dissolution of carbon dioxide I was asked to give the equilibrium expression for:

$$\ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}$$

As water is a liquid and a solvent here, it shouldn't be included in the expression, hence I wrote:

$$K_c = \frac{[\ce{H2CO3}]}{[\ce{CO2}]}$$

However, the answer of the question included water in the denominator.

What is the correct answer in this situation?

In the case of heterogenous equilibria when molar concentration of pure solid or liquid involved, water in this case is constant or is independent on the amount present its concentration is omitted as per the general norm. The equilibrium constant is just modified to

$$K_c' = K_c[\ce{H2O}]$$

So the book and you are referring to different constants.


Removing HCl from water I have water contaminated with dissolved chlorine:

$\ce{H2O + Cl2 <=>HCl + HOCl}$

During electrolysis, this is an issue. These are my products:

$\ce{NaCl + H2O ->T[electrolysis] NaOH + Cl2 + O2 + H2 }$

However, $\ce{NaOH}$ will be neutralised by $\ce{HCl}$:

$\ce{NaOH + HCl-> NaCl + H2O}$

I guess this can be considered hydrochloric acid. I want to remove the chlorine ions from water and prevent future dissolving.

Does increasing temperature help to remove the chlorine? Does boiling make the chlorine leave the mixture?

$\ce{H2O + HCl + HOCl ->T[???] H2O + Cl2(gas)}$

PS.: I understand that there are means of separating the electrolysis columns. But this question is really about removing $\ce{Cl2}$ or $\ce{HCl}$ from aqueous solution.

1) Do NOT heat the solution, if you want chlorine out. Electrolysis of hot $\ce{NaCl}$ solution produces $\ce{NaClO3}$, then $\ce{NaClO4}$, not chlorine. You apparantly don't want this.

2) traditionally, in such setups some barrier between cathode and anode spaces is used. The easiest one would be unglazed ceramic pot (with no holes and no direct contact of anode and cathode solutions), though I'm not sure if it will work with chlorine ions. It is recommended to perform electrolisys of $\ce{MgSO4}$ solution first. In this case chlorine ions would migrate from cathode space. With barrier introduced you may expect chlorine to evolve from anode space and sodium hydroxide to accumulate in cathode space.

3) do NOT blow air atop sodium hydroxide solution. It reacts with $\ce{CO2}$ from air.


Boiling points of Butan-2-ol and Butan-1-ol Butan-1-ol has a boiling point of approx. 117.7 celsius, and Butan-2-ol has a boiling point of approx. 99.5 celsius. What causes this difference in boiling points?

My initial idea would be that Butan-2-ol essentially has a side-branch (the OH-group), whereas Butan-1-ol does not. This should disrupt the London dispersion forces and thereby reduce the strength. Another idea would be that the position of the OH-group being in the middle for Butan-2-ol creates fewer Hydrogen bonds than Butan-1-ol does. My third idea is that the position reduces the strength of the regular dipole-dipole bonds (when I mean regular I mean not hydrogen bonds), because the polarisation is weakened.

Am I correct in any of my assumptions? Is there something else to it?

You have certainly understood the main idea.

In general the boiling point of a substance is determined by the way the molecules come together. The closer they are, the bigger are the intermolecular forces making it harder for one molecule to leave the entire group.

In this particular case the molecules are similar, since both substances have a hydroxyl group, thus being to able to create H-bonds. But in the case of 1-butanol, the molecule is linear while 2-butanol is not. This stereochemical difference affects the way the molecules attract each other. The linear molecules are able to come closer, thus creating stronger bonds or in other words raising the boiling point, because a higher amount of energy has to be applied in order for one molecule to obtain enough, to leave the liquid phase.


If nothing sticks to Teflon, how does Teflon stick to frying pans? The most notable characteristic of polytetrafluoroethylene (PTFE, DuPont's Teflon) is that nothing sticks to it. This complete inertness is attributed to the fluorine atoms completely shielding the carbon backbone of the polymer.

If nothing indeed sticks to Teflon, how might one coat an object (say, a frying pan) with PTFE?

It has to be so common a question that the answer is actually given in various places on Dupont's own website (Dupont are the makers of Teflon):

  “If nothing sticks to Teflon®, then how does Teflon® stick to a pan?"
  Nonstick coatings are applied in layers, just like paint. The first layer is the primer—and it's the special chemistry in the primer that makes it adhere to the metal surface of a pan.

And from this other webpage of theirs:

The primer (or primers, if you include the “mid coat” in the picture above) adheres to the roughened surface, often obtained by sandblasting, very strongly: it's chemisorption, and the primer chemical nature is chosen as to obtain strong bonding to both the metal surface. Then, the PTFE chain extremities create bonds with the primer. And thus, it stays put.


Confusion Regarding Nitrogen Inversion In nitrogen inversion, the lone pair changes its direction continuously.

Does this in any way delocalize the charge density on the nitrogen making it less nucleophilic and less basic than another nitrogen attached to, say, a bridgehead carbon?

Consider the specific case of
Quinuclidine and tri-methyl amine. Which will be more nucleophilic among them and why?

Yes, amine inversion does effect(reduce) basicity and nucleophilic character of amines

In amine inversion a nitrogen which is sp3(tetrahedral) converts into sp2(planar) and then again coverts into sp3(tetrahedral).

This happens because nitrogen has a lone pair and this lone pair moves through nucleus towards the opposite side. This is known as quantum tunneling. While the electron pair is moving through the nucleus nitrogen has sp2 hybridisation. So, if nitrogen inversion is effective availability of lone pair decreases.

Now let's come to the case of quanidine. In quanidine, nitrogen inversion doesn't happen occur because of the impossible geometry it would have to form to do so.
However, in tri-methyl amine there is no such hindrance and hence achieve planar state relatively easily.


When writing the net ionic equation, if one of the products ionizes, what is the most appropriate way to account for this in the answer? Take these two practice problems and their solutions from Ebbing (8th ed) that involve completing the molecular equation, then writing the net ionic equation:
4.42a: $\ce{Ca(OH)2(aq) + 2H2SO4(aq) -> }$
soln:
$$\ce{Ca(OH)2(aq) + 2H2SO4(aq) -> 2H2O(l) + Ca(HSO4)2(aq)}$$
$$\ce{OH-(aq) +2H+(aq) +SO4^{2-}(aq) -> H2O(l) + HSO4-(aq)\\ \text{ (ionizes to } H+ + SO4^{2-})}$$
4.42c: $\ce{NaOH(aq) + H2SO4(aq) -> }$
soln:
$$\ce{NaOH(aq) + H2SO4(aq) -> NaHSO4(aq) + H2O(l)}$$
$$\ce{OH-(aq) + H+(aq) -> H2O(l)}$$
In 4.42a the authors do not ionize the product, only note to the side that it does ionize. In 4.42c the authors do ionize the product and account for this in the net ionic equation.
In general, is one practice preferable or more correct than the other?

Please note that I did not intend to answer my own question. After reading some of the answers, I feel it is necessary to provide clarification for future readers. Hopefully, my answer is enlightening.
Let us begin with a point of interest, reading again from Ebbing:

Phosphoric acid is an example of a triprotic acid. By reacting this
acid with different amounts of a base, you can obtain a series of
salts:
$$\ce{H_3PO4(aq) + NaOH(aq) \rightarrow NaH_2PO_4(aq) + H_2O(l)}$$
$$\ce{H_3PO4(aq) + 2NaOH(aq) \rightarrow Na_2HPO_4(aq) + 2H_2O(l)}$$
$$\ce{H_3PO4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)}$$

And now, let's consider the problem.

Problem 4.42 Complete the right side of each of the following
molecular equations. Then write the net ionic equations. Assume all
salts formed are soluble. Acid salts are possible.
a. $\ce{2H_2SO_4(aq) + Ca(OH)_2(aq) \rightarrow} $

To give us insight into the nature of this reaction, let us first consider a complete neutralization reaction for calcium hydroxide and sulfuric acid and see what that looks like.
$$\ce{1H_2SO_4(aq) + 1Ca(OH)_2(aq) \rightarrow CaSO_4(s) +2H_2O(l)}$$
Now, let us again consider our original problem in terms of the complete neutralization. We divide the coefficients by two.
$$\ce{1H_2SO_4(aq) + \frac{1}{2}Ca(OH)_2(aq) \rightarrow \text{ incomplete neutralization}} $$
This helps give us an idea of how to predict the products for the molar ratio provided. We know the acid salt formed is soluble as we are told this in the problem statement.
$$\ce{2H_2SO_4(aq) + Ca(OH)_2(aq) \rightarrow Ca(HSO_4)_2(aq) + 2H_2O(l)}$$
For the net ionic equation, we consider that $H_2SO_4$ is a strong acid and will ionize completely in solution and $Ca(OH)_2$ is a strong base and will dissociate in solution to be present entirely as ions.
$$\ce{4H^+(aq) + 2SO_4^{2-}(aq) +} Ca^{2+}(aq) \ce{+ 2OH^-(aq)} \rightarrow  Ca^{2+}(aq) \ce{+2HSO_4^-(aq) + 2H_2O(l)}$$
Eliminate the spectator ions.
$$\ce{4H^+(aq) + 2SO_4^{2-}(aq) + 2OH^-(aq) \rightarrow 2HSO_4^-(aq) + 2H_2O(l)}$$
Reduce coefficients.
$$\ce{2H^+(aq) + SO_4^{2-}(aq) + OH^-(aq) \rightarrow HSO_4^-(aq) +  H_2O(l)}$$
Note that $\ce{HSO_4^-}$ ionizes to $\ce{H^+}$ and $\ce{SO_4^{2-}}$. Substitute and reduce.
$$\ce{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$$
We accept either form of the answer, though if to go with the former you must notate the species that ionizes, otherwise is incorrect.
$$\ce{2H^+(aq) + SO_4^{2-}(aq) + OH^-(aq) \rightarrow H_2O(l) +  HSO_4^-(aq) \text{ (ionizes to } H^+ \text{ and } SO_4^{2-}\text{)}}$$

We consider mixing of $\ce{H2SO4(aq)}$ and $\ce{Ca(OH)2}$ in molar ratio 2:1.
Molecular equation version:
$$\ce{2 H2SO4 + Ca(OH)2 -> Ca(HSO4)2(aq) + 2 H2O}$$
Eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp,}\ce{ CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca(HSO4)2(aq)  <=> CaSO4(s) + H2SO4(aq)}$$
Ionic equation version:
Separated solutions:
\begin{align}
  \ce{H2SO4(aq) + H2O &-> H3O+(aq) + HSO4-(aq)}\\
  \ce{HSO4-(aq) + H2O &<<=> H3O+(aq) + SO4^2-(aq)}\\
  \ce{Ca(OH)2(aq) &-> Ca^2+(aq) + 2 OH-(aq)}
\end{align}
More exactly:
\begin{align}
  \ce{Ca(OH)2(aq) &<=>> Ca(OH)+(aq) + OH-(aq)}\\
  \ce{Ca(OH)+(aq) <=>> Ca^2+(aq) + OH-(aq)}\\
\end{align}
Mixing solutions with 2:1 molar ratio:
$$\ce{2 H3O+(aq) + 2 HSO4-(aq) + Ca^2+(aq) + 2 OH-(aq)-> 4 H2O + 2 HSO4-(aq) + Ca^2+(aq) }$$
And after elimination of spectator ions:
$$\ce{2 H3O+(aq) +  2 OH-(aq)-> 4 H2O }$$
Then, further ionizations of the bisulphate, as $\ce{[H3O+]}$ decreased:
$$\ce{ HSO4-(aq) <=> H3O+(aq) + SO4^2-(aq)}$$
Again, eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp, }\ce{CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca^2+(aq) + SO4^2-(aq)  <=> CaSO4(s)}$$


Phase diagrams (weird region) On the left, there is a region for $\ce{MgO(ss)}$.

I've been told such regions exist because stuff are not miscible. However, does that mean the composition axis on the bottom is a lie? Namely, because it says ≠100% $\ce{MgO}$ and the region signals pure $\ce{MgO}$.

There is nothing weird in here. What you've been told is wrong, or maybe you have misunderstood it. Regions like this exist because stuff are miscible. (ss) stands for solid solution, meaning that the solid crystal structure can accommodate some amount of the other component (Al2O3 in this case).

The diagram is a bit misleading, because naming something as MgO(ss) is not entirely correct. Adding the (ss) implies that it is not pure MgO. A more correct way to name the fields in the diagram would be to rename MgO to "periclase" and MgAl2O4 to "spinel", which are the mineral names for these compounds. They will not imply pure compositions of the solids, and conceptually allow the solid solutions to occur.

Some applications of the lever rule: For example, take a look at 10% Al2O3 and 2000 degrees. You basically have periclase, which in pure form as the composition MgO. However, it has 10% of Al2O3 in it. It is still periclase though. Let's take a more complicated example: At 40% Al2O3 and 1800 degrees. You have two solid phases, periclase and spinel in roughly equal proportions. The periclase has 10% Al2O3 in it, and the spinel has 65% Al2O3 in it. Note that it's less than the ideal spinel should have!

Also, notice that Al2O3 in the diagram does not have (ss) in it. This means that it's always going to be Al2O3, with no MgO in it at all.

The curves that separate solid solution (ss) fields are appropriately called solvi (singular: solvus). They mark phase changes that are purely in the solid state, without any melting. For example, let's say you are at 70% Al2O3 and 2000 degrees. You have only one phase: a spinel. By cooling down the system, you will cross the solvus line and periclase will be added to the system, on the expense of spinel. You will be able to microscopically see it as patches or stripes or something like that of periclase in the spinel. Wikipedia has an image to demonstrate that (although it's built for eutectic compositions, the morphology is similar).


What are the subfields of chemistry that work very closely with biology? Background information:
I'm not sure if this question belongs here; but i'll risk the down-votes to ask it.

I'm currently doing an undergraduate program, and it's in chemistry. I am torn between doing chemistry and biology and ideally the nicest compromise would be biochemistry. However I have a horrible ineptitude for biochemistry.

My question is, what subfields of chemistry work very closely with biology?

In addition to jonsca's link of chemistry subfields, Wikipedia also has a list of biochem subfields. Find what research interests you, perhaps not based on a course you've taken.

You can look up what professors are doing in their respective fields of research by visiting faculty research group sites online. Check out their publications, current projects, probably would be good to join a lab too. (I'm blindly assuming you're interested in research and/or graduate studies)

The list on Wikipedia first lists (for whatever reason these are "main branches" but I'm surprised biotech is categorized under "other branches"...):

Animal Biochemistry
Plant Biochemistry
Molecular Biology
Cell Biology
Metabolism
Immunology
Genetics
Enzymology

Getting into more or less random scopes of research topics which may or may not be within the above fields:

Biotechnology,
Bioluminescence,
Molecular chemistry,
Enzymatic Chemistry,
Genetic engineering,
Pharmaceuticals,
Endocrinology,
Hematology,
Nutrition,
Photosynthesis,
Environmental Chem,
Toxicology
(Probably many, many others...)


What are good lubricants for Bunsen burners? We've been running into a lot of problems the last several years with our laboratory Bunsen burners becoming jammed and/or corroded.  I've just come to find out that our lab tech has been lubricating the burners with glycerol, which seems wholly inappropriate to me. I guess she figured that if it worked for inserting glass tubing into rubber holes, it would work for anything. To me, something like a WD-40 would seem more appropriate based on household experience, but I'd like to hear if anyone else has found something for this exact application that works well for them.

I'm currently experimenting with white lithium grease.  An engineer friend also recommended Permatex anti-seize lubricant.  These seem to work well on a very short term, but we'll have to see how it holds up throughout the next several weeks.

Anyone have a favorite lubricant for something like this?

We've been running into a lot of problems the last several years with our laboratory Bunsen burners becoming jammed and/or corroded. I've just come to find out that our lab tech has been lubricating the burners with glycerol.

Does the issue itself seem to have been caused by the glycerol?

Admittedly, we don't use many Bunsen burners nowadays as they're seen as being fairly unsafe in the kind of organic lab I work in where we handle large amounts of solvent, however we do have a well vented hood with several burners in.

The ones we do have are very old indeed (pre 50's) and still work fine without the need for any maintenance (i.e. greasing).

Bunsen burners themselves are incredibly simple pieces of kit, if you do have issues with corrosion (i.e. because of an overly humid environment or harsh chemical vapours), the most sensible thing would be to disassemble, remove the corrosion and re-assemble.  The use of any kind of grease should be un-neccesary, especially as very few of them (other than very specialised products) can withstand long term exposure to heat without degradation.


How to find volume of substance from molar concentration The concentration of a substance dissolved in water $i$ is known in (mol/m3), so is the molar mass in (g/mol) and the density in (g/m3).

I want to find the volume of this substance in m3.

I have two possible ideas for equations to use:
$$\mathrm{volume}_i= \frac{\mathrm{mass}_i}{c_i  \times M_{W,i}},$$
however, to use this equation I need to know the mass of the substance, which is not obvious.

Also I could use something like this:
$$\mathrm{volume}_i = \frac{c_i \times M_{W,i} \times \mathrm{volume}_{?}}{\rho_i },$$
however, I'm not sure what volume$_?$ would be, as I don't know the volume of the solution and I am trying to find the volume of the substance $i$.

Any ideas how to find the mass in the first equation or the volume in the second equation, or other equations to use?

Extra comments:
My real aim is to find the volume of the solution so I can use the following formula for volume concentration, where I already know the volume concentration $\phi_i $:
$$\phi_i = \frac{\mathrm{volume}_i} {\mathrm{volume}_\mathrm{solution}} $$
 or
$$\mathrm{volume}_\mathrm{solution} = \frac{\mathrm{volume}_i} {\phi_i} $$

Nothing has been said about the actual volume of the solution, only concentration of the solution has been given.

Assume that $c$ moles of solute is present in 1m$^3$ of solution.

Then mass of solute=c*molar mass

volume of solute * density of solute = c* molar mass

From this equation, volume of solute can be known.

Clearly, if the volume of the solution increases, the volume of the solute will also increase. Since the volume of the solution has not been given, it must give you a hint to take the above assumption.


How can I find whether there is oxygen present inside a well of diameter 1 m at 50 m's deep? How can one find whether or not there is enough oxygen for breathing present inside a (dry) well of diameter 1 m at a depth of 50 m?

Many people in Nepal are killed while cleaning wells because of a lack of oxygen.

The ideal solution would be a single gas personal detector, such as this one (it's only the first example I found, I don't know this brand, but there are many equivalent apparatuses from other vendors).

While digital personal detectors are expensive ($200+ range), there are simpler systems which also provide some basic safety (better than nothing). The Davy lamp is perhaps the best known example of these, and it was a huge breakthrough in mine safety when it was popularized after 1815.

(Source: http://lindal-in-furness.co.uk/MinersLamps/MinersLamp.jpg)


Epoxidation for dienes I have been asked to determine the major product for the reaction:

From the reagent given, one can tell that this is an epoxidation reaction. But, as only one mole of the peroxyacid is given, epoxidation can only take place in one of the two bonds.
How do you determine on which double bond the epoxidation will take place?

MCPBA preferentially reacts at the more electron-rich alkene, in this case the more substituted alkene according to this entry on chemistryportal.net here and this review RSC Advances


Number of mercury atoms in a drop of mercury Question:

  A) Calculate the mass of a mercury atom in kg.
  B) How many atoms of mercury does a drop of mercury contain, knowing that the volume of a drop of mercury is $\pu{0.05mL}$ and that the density of mercury is $13.6 \times 10^3 \mathrm{kg/m^3}$

I know that $\mathrm{1 mL = 1 cm^3}=1.0\times10^{-6} \mathrm{m^3}$

And the mass of a drop of mercury is $\mathrm{m=v}\times \mathrm{density}
= 5.0\times10^{-18} \times 13.6\times{10^3}
=6.8\times10^{-14}$

What I want to know is how to proceed to find how many atoms mercury are present in its one drop? Do I simply divide the mass of a drop of mercury by the mass of a mercury atom, or do I use the Avogadro's number?  I don't want a straight-off answer, I only wish to know how to calculate it.

You can use dimensional analysis. It is a very useful tool to solve equations such as these.

What you have is the volume of one drop, $V(ml)$, the density of mercury, $\rho(kg\cdot{}m^{-3})$, the molar mass of mercury, $Mr(g\cdot{}mol^{-1})$, and the Avogadro's constant, $N_A(mol^{-1}$).

To find out the number of atoms in a drop of mercury, all you have to do is to manipulate those variables via dimensional analysis.

You have: $\frac{Mr(g)}{(mol)},\frac{N_A(atoms)}{(mol^{})},\frac{\rho{(kg)}}{(m^3)},\frac{V(ml)}{drop}$ and you know that $\frac{1(kg)}{1000(g)},\frac{10^6(ml)}{(m^3)}$

Now, try arranging them to form an equation such that the units cancel off,in order to get the answer you need.

A: Mass of mercury:

$$\frac{kg}{atom}=\frac{1(kg)}{1000(g)}\cdot{}
\frac{Mr(g)}{(mol)}\cdot{}\frac{(mol^{})}{N_A(atom)}
$$

Notice that this is dimensionally equivalent. Therefore,

$$\frac{kg}{atom}=\frac{Mr}{1000\cdot{}N_A}
$$

B: Atoms of mercury per drop:

$$\frac{atoms}{drop}=
\frac{N_A(atoms)}{(mol^{})}\cdot{}
\frac{(mol)}{Mr(g)}\cdot{}
\frac{1000(g)}{1(kg)}\cdot{}
\frac{\rho{(kg)}}{(m^3)}\cdot{}
\frac{(m^3)}{10^6(ml)}\cdot{}
\frac{V(ml)}{drop}$$

Notice that this is dimensionally equivalent. Therefore,
$$\frac{atoms}{drop}=
\frac{N_A\cdot{\rho}\cdot{V}}{Mr\cdot{10^3}}$$


Exact meaning of constant volume heat capacity From Wikipedia:

$$
\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{\partial Q}{\partial T}\right)_V = C_V,
$$

$C_V$ is what known to be constant volume heat capacity. I don't really get the exact meaning of 'constant volume'. Does it mean that (a) the whole process takes place under a constant volume, i.e, it can be any volume as long as the volume does not change during the process, the volume is maintained at its initial value throughout the process, or (b) the $C$ is measured under a fixed constant volume $V$?

If (a) is correct, are the rates of change of $Q$ with respect to $T$ same at, for example, $V = \pu{1 dm3}$ and $V = \pu{2 dm3}$? We are not considering only perfect gases.

If (b) is correct, what is the value of $V$ for standard $C_V$?

(a) is correct. (b) is wrong. $U$ depends on $V$ only in the combination $v=V/N$. $v$ is the specific volume, an intensive quantity that does not depend on the size of the system.

We can derive this from first principles. Take $U = U(T,V,N)$. Because $U$ is extensive, $$U(T,\lambda V,\lambda N) = \lambda U(T,V,N) \quad \Longleftrightarrow \quad U(T,V,N) = \lambda^{-1}U(T,\lambda V,\lambda N),$$ by taking $\lambda = N^{-1}$, we can write this expression as $$U(T,V,N) = NU(T,V/N,1) =: Nu(T,v),$$ which produces the desired functional relationship.


Where do the bubbles in a glass of water come from? This weekend I accidentally left a glass of water on my kitchen table for a few days while I was out of town. Unsurprisingly, bubbles formed on the walls.

As I understand from the last time I attended a chemistry class quite a while ago, this happens due to water self-ionizing (at about a ratio of $10^{-14}$ if I recall correctly).

$$\ce{H2O + H2O \rightleftharpoons H3O+ + OH-}$$

However, for some reason, the bubbles only formed up to a certain depth:

What is the reason for this? Does self-ionization happen uniformly in the glass? Do bubbles start forming from the top down and would they have reached the bottom if left untouched for longer?

Gases are less soluble in warm water than cold water.

The bubbles are likely from dissolved gas coming out of solution as the water warmed.

The bubbles are unrelated to self-ionization of water.  The self-ionization does vary somewhat with pressure, so at an extreme depth in the ocean you might need to consider what Kw is, but not in a glass of water.

The solubility of gases increases with pressure, so on this basis it is more likely that you would have bubble appear near the top.  However, the pressure is only 1-2% greater at the bottom of the glass compare to the top.  There may be fewer nucleation sites (such as tiny scratches) near the bottom of that particular glass.  Also, if the bubbles move up along the wall sometimes, it would be logical that there are less at the bottom.


How does the oxidation of NaI by sodium hypochlorite to give I+ work? I'm in trouble with the oxidation of NaI by $\ce{NaClO}$, the reaction should give $\ce{I+}$ which then reacts by electrophilic substitution with an aromatic complex. As of now I've come to:

$$\ce{ ClO- (aq) + 2I- (aq) + 2H+(aq) -> I2 (s) + Cl- (aq) + H2O (l)}$$

But I can't then find a way to obtain then the electrophilic $\ce{I+}$ that should react.

I think the reaction should be done in basic solution, not acidic solution.  "One equivalent each of sodium iodide (2.77 g) and sodium hydroxide (0.74 g) was added, and the solution was cooled to 0 C. Aqueous sodium hypochlorite (34.50 g, 4.0% NaOCl) was
added dropwise over 75 min at 0-3 C" J. Org. Chem. 1990,55, 5287-5291.

So perhaps:

$$\ce{ ClO- (aq) + I- (aq) + H2O (l)  -> I+ (aq) + Cl- (aq) + 2OH- (aq)}$$

Edit: after looking at the cited article more closely, the hydroxide was only being added because of the acidity of the iodophenol product, so you could write

$$\ce{ ClO- (aq) + I- (aq) + 2H+(aq)  -> I+ (aq) + Cl- (aq) + H2O (l)}$$

also.


Predicting isotope patterns in mass spectrometry Imagine you have a mixture of $\ce{^1H}$, $\ce{^2H}$ and $\ce{^3H}$ (hydrogen, deuterium and tritium) present as diatomic molecules and that the numbers of the atoms of the three species are the same. Sketch the mass spectrum.

The book's answer is

For peaks of mass= 2, 3, 4, 5, ratio of heights = 1:2:3:2:1.

I’m not getting why the answer is this?
Why there is no peaks of mass 1, is it because $\ce{H2}$ consisting of $\ce{^1H}$ can break in mass spectrometer?

The first step is to write all the possible combinations of H atoms: $$\mathrm{H^1H^1~~ H^1H^2~~ H^1H^3 ~~H^2H^1 ~~H^2H^2~~ H^2H^3~~ H^3H^1~~ H^3H^2 ~~H^3H^3}$$ Now it is important to note that each of these parings have an equal probability of forming and exist in equal quantities in the mixture. Their molar mass are respectively: $$2, 3, 4, 3, 4, 5, 4, 5, 6$$ So there is $1$ way to make $2$, $2$ ways to make $3$, $3$ ways to make $4$, $2$ ways to make $5$ and only $1$ way to make $6$. Since each way has an equal probability of occurring, then the ratio of the molar masses should be in the ratio -  $~1:2:3:2:1$.

Hence for peaks of mass = $2, 3, 4, 5, 6$ ratio of heights = $1:2:3:2:1$

The reason why there is no peak for mass $1$, I am not too sure. However I think it is because if the hydrogen molecule was to break up, it would form a proton and a H radical. Both of these are extremely reactive and unstable and will probably just get lost in the machine and never actually get recorded. This could make sense as the mass spectra for $\ce{HCl}$ or $\ce{HBr}$ doesn't include a peak for mass $1$. Also in mass spectra for hydrocarbons, such as heptane, the smallest mass peak recorded is usually 29 which is the ethyl group. Therefore any thing lower than that (such as the methyl group) is probably too unstable.


Why is the units of rate constant dependent on the order of the reaction? In many textbooks, it is written that

For order (m + n), the rate constant, k has units of mol1−(m+n)·L(m+n)−1·s−1

However, by the Arrhenius equation, k can be expressed as

k = PZexp(-Ea/RT)

and its unit doesn't seem to change with the order.
I think if reaction velocity, v is defined as

v = k([A]/cref)m([B]/cref)n

like the definition of equilibrium, the unit of k becomes mol·L−1·s−1 and it looks more reasonable. Am I wrong?

The rate has units $\displaystyle \frac{concentration}{time}$ and is $\displaystyle \frac{d[C]}{dt}$ and the right hand side of the equation has always to have the same units as the left hand side, so for first order reaction consuming C;
$\displaystyle \frac{d[C]}{dt} = -k_1[C]$
and the rate constant $k_1$ therefore has units $\displaystyle \frac{1}{time} $ usually $\mathrm{s^{-1}}$ and for second order
$\displaystyle \frac{d[C]}{dt} = -k_2[C][C]$
the units of rate constant $ k_2$ are $\displaystyle\frac{1}{time \times\,concentration}$ which is $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$.
The units for other examples follows in the same way.
In the Arrhenius equation as the exponential is dimensionless the other terms must have the units of the rate constant, which means units of PZ in your notation. These units must always depend on the order of reaction as described above and if they don't then there is an error somewhere.


Do various soaps work better when dissolved/used in different water temperatures? My wife and I have had an ongoing argument about various cleaning agents, and I wanted to get a Chemists' view on it.

When we are doing laundry, she will generally clean using the cold water setting and a standard amount of detergent.

If I do laundry, I will use the warm water setting with a standard amount of detergent.

Either way, the clothes come out (seemingly) clean. Using cold water saves energy from the heating process, which has been her excuse for doing such a dastardly thing. Her argument against using warm water is that it will damage/shrink the cotton. While I don't disagree on that point, I am curious if the use of warm water causes the detergent to work better or not.

In addition, there are various soaps that work best in cold water.

What is the actual mechanism at work here? Why do some soaps work better in cold (or hot) temperatures?

To make things a bit better, let's remove some more variables. We use this detergent:

Tide Pods

Unfortunately I can't find information on the normal temperatures those machines operate at.

Also who is right? (I bet it's her)

To start, soaps and detergents have two different chemical makeups, meaning that they act different under different conditions.

For general soaps, here is an explanation of how it works:
http://chemistry.about.com/od/cleanerchemistry/a/how-soap-cleans.htm

As a general chemistry note, as the temperature of a solution increases, the ease of a chemical to change (whether it is to react or change shape) increases.  This means that colder temperatures will allow a chemical to remain in the same state.  If the soap depends on a specific conformation/shape to act then colder temperatures may be preferable.  However, if a reaction needs to occur in order for the soap to act some heat may be required.

For Tide Pods, it states that it "dissolves quickly in hot & cold."
http://www.tide.com/en-US/product/tide-laundry-detergent-pacs.jspx

The chemical makeup of Tide Pods is composed primarily of salt compounds and enzymes.  Both salts and enzymes tend to operate better at higher temperatures.  So as long as you don't pass the denaturing temperature of the enzyme, higher temperatures would be preferred.


Aluminum deoxidation gas I am new to chemistry. I am looking for a gas that would de-oxidise aluminium.
Which gases would work for that and at what temperature?

Details: I need to sinter aluminium powder. But because the top layer of aluminium is oxidized when exposed to the atmosphere, I need to remove the oxide film before the sintering would work.

Thanks so much,

Sergey

Since Al2O3 is very stable, powerful acids such as HF and chromic acid are used to remove the oxide surface. Since you're working with fine particles and wish to use a gas, rather than liquid, you'll need to consider what will become of the removed oxides. Gaseous HF would leave fluorides behind. Perhaps atomic hydrogen might work, but you'd have to remove the resultant water before the Al reoxidizes.

It is easier to prepare powdered aluminum in an inert atmosphere from bulk aluminum. There are a number of sites showing how to grind aluminum using anything from a file to a coffee grinder; you'd need to adapt this to run in argon, perhaps with a glove-box.

Sintering might be difficult, since Al is so soft that pore spaces would close.


What is the chemical structure of the cocoate ion vs. the olivate ion? I am wondering how the cocoate ion (in natural soaps, in the form of sodium or potassium cocoate and from the saponification of coconut oil) differs structurally from the olivate ion (from the saponification of olive oil)?

My family is attempting to implement the methods in solveeczema.org and believe that our children have detergent-reactive eczema and our baby is most sensitive.  In moving away from detergents and to natural soaps, we have started using soaps that result from the saponification of vegetable oils - mostly coconut oil, palm kernel oil, and olive oil.  We discovered that our other child (a toddler) seems to react to these soaps in a similar way that he would react to a cleaner containing much stronger/harsher surfactants (such as sodium lauryl sulfate).

Our toddler seems to get bad eczema flares using soaps that contain cocoate, but he has no reaction to soaps that contain 100% olivate.  I have no idea why this would be, since I would assume that both cocoate and olivate have the R-COO- structure.  Is cocoate much more hydrophilic than olivate?  Is the functional group different?  Is the cocoate ion a longer molecule that looks much more like a harsh detergent than a salt resulting from saponification of olive oil or tallow?

Please note that this question is related to two questions already asked:

1)  Is soap the natural form of Sodium Lauryl Sulfate?

2)  https://parenting.stackexchange.com/questions/5047/how-to-test-whether-my-toddler-is-allergic-to-detergent/5049#comment7644_5049

Note 1. Both coconut oil and olive oil are natural oils. The soap, produced from them, is mixture of several compounds.

The difference of olive and coconut oil may be easily found in wikipedia: http://en.wikipedia.org/wiki/Coconut_oil#Composition_and_comparison

In short, coconut oil contains mostly saturated acids esters, while olive contains mostly unsaturated ones. The unsaturated acid esters are responsible for olive oil ability to solidify on air. But here is

Note 2: In my knowledge, allergy is usually caused by carbohydrates or proteins. Fatty acids and their ions have too few binding sites to be recognized by immune system as something distinct from our own. The real source of reaction may be

Minor components of initial oil. There are quite many of them and removing them completely is not easy.
Other components of soap, such as perfume or pigment.
Soap solution pH, that may differs from soap kind to kind, but this is unlikely.


What makes certain zeolites better at ion exchange? I would have thought that the bigger the pores in the zeolite, the more ions it can contain, and therefore the better the use of the zeolite for ion exchange.

However, zeolite A has small pores but its main use is as an ion exchanger, as opposed to many other zeolites which have larger pores but are not used as ion exchangers.

Therefore, why are some zeolites better than others at ion exchange? Or, what influences the ion exchange capabilities of zeolites beyond pore size?

Zeolites are selective ion exchangers.

What is usually required for a good zeolite is that it selectively absorbs the ion you care about rather than any old ions in the mixture.

That selectivity is a product of the shape of the holes in the 3D structure. Bigger holes will be less selective than smaller holes. If all the ions in the mix fit in the hole the zeolite won't reject the ones you don't want. So, ideally, the holes will be big enough for one ion but too small for another one. This allows selective exchange.

That is about it. It isn't about general equilibration of ions; it is about selective exchange of ions.


Are there magnetic rocks, and not hematite? I dug up some sand and small rocks in a stream bed near me and moved a powerful magnet through the wash picking up numerous small rocks and black sand.

Now, the black sand is definitely small crystalls of magnetite, but what about the small rocks which tend to be gray in color, some being brownish?

I thought pretty much anything you would be likely to encounter commonly (like in a stream bed) that is magnetic is magnetite. In other words, my understanding is that the only COMMON mineral that is magnetic is magnetite. But these small rocks do not appear to be magnetite crystals, and the brown ones definitely are not.

Does this mean there are magnetite crystals INSIDE the rocks? Like if I break them up will I find little tiny magnetite crystals within?

Magnetite is indeed very magnetic. But other minerals are magnetic too, and given a strong enough magnet you will attract more than magnetite. Chromite is one that comes to mind that's also common in some black sands.

You might not always see the magnetite. Magnetite is mixed valence of iron 2 and 3 and as such it isn't stable in our oxygen rich atmosphere. Given enough time it will alter to a variety minerals such as hematite and iron hydroxides. So you might have magnetite in the core of a grain mantles by non magnetic hematite. It will look brown but it will still be attracted to a magnet.

Same thing with your grey rocks: they most likely have magnetite inside, hidden from view.


Why there is no bichloride or binitrate ion like bicarbonate or bisulphate? I have been thinking lately that if there is bicarbonate why there is no bichloride or binitrate, please someone  give an easy explanation.

Okay, you want simple, here's a real simple way to put it.

How many limbs does a dog have? 4 right?
Now if it loses one of its limbs, a prosthetic one can be placed,(We'll get to this: with its whole body, being able to cope with the leg). Now consider the case of a human being, same scenario, the same thing can be done, since we can place the limb in the "slot"(somebody put a better word here), where the leg previously was(although, the limb cant do everything a leg couldve done). Now consider the case of a creature with just one limb, if it loses the sole limb it had, life's gonna be hard, simply because of the amount of discomfort its going to face while adapting to the limb. Similarly, sulphate ($\ce{SO4^2-}$) and carbonate ($\ce{CO3^2-}$) have a valency greater than 1. Hence can accept an H+ ion and still exist as a stable specie, and further accept a metal ion;

In the case of $\ce{HCl}$, it has a valency of just 1, hence upon accepting an $\ce{H+}$ to get the "bi" tag to its name, it has lost its charge, and now is incapable of accepting a metal ion to form a salt(while keeping the $\ce{H+}$ to itself), as it has simply become hydrochloric acid.


What stops intermediates of alcohol hydration and dehydration reacting in the same way? I am aware they both occur under different conditions and that water attaches in a secondary position in elimination but primary position in hydration.

(Stage 2)

(Stage 3)

But both appear to be the same intermediate, why exactly do different reactions follow this stage?

As you have spotted, the reaction is reversible and can proceed both ways. Which way it takes depends on the conditions and by implication on Le Chatelier’s principle.

In concentrated $\ce{H2SO4}$, any water will be immediately ‘captured’ by the strongly hygroscopic sulphuric acid. Thus, water is actively drawn out of the equilibrium and the alkene is produced by elimination.

If the reaction is run in aquaeous solution (or at least with water as a cosolvent) and only catalytic quantities of acid, then the nucleophilic attack of water is more likely and the alcohol hydration product is produced.

(In modern synthetic lab-scale chemistry, neither of the reactions are used due to their rather poor selectivity (especially hydration) and harsh conditions (especially elimination). Instead, hydration is typically performed with boranes (anti-Markovnikov) or Markovnikov-selective reagents while elimination is typically achieved by transformation of the alcohol functionality into better leaving groups such as mesylate.)


At what point does Avogadro's law not apply? A mole of hydrogen gas takes up 22.4 L, but so does a larger molecule like $\ce{CO2}$. Methane ($\ce{CH4}$) is an even larger sized molecule, but a mole of it will still take up 22.4 L. I would assume that an even bigger molecule like decane ($\ce{C10H22}$) would still act the same as the others. A mole of sand however does not take up 22.4 L. This begs the question: is there a certain size particle at which Avogadro's law is not valid anymore? Perhaps at a molecular size larger than buckyballs ($\ce{C60}$)? (This also happens to be the point that wave-particle duality does not apply.)

My feeling is that the size of the molecule matters much less than you would think. For instance, from a purely ideal gas perspective (as Avogadro's law assumes), the expected value for the speed of a particle is $$v=\sqrt{\frac{8kT}{\pi m}}$$
where $k$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the particle.

So, that means that as the mass of your particle doubles, the average speed of the particle is decreased by a factor of $1/\sqrt{2}$. But, since pressure, from a kinetic molecular theory perspective treats pressure as being the average force which the particles apply to a wall (over a certain area), and force is the time derivative of momentum $p=mv$, we see that $p$ increases by a factor of $\sqrt{2}$ when the mass is doubled (because $v$ decreased by $1/\sqrt{2}$.) That means, the average force on the wall will be greater per collision, but not very much greater.

For instance, I went ahead and calculated the pressure which a mole of hexane exerts on $22.42\ \mathrm{L}$ according to the van der Waals (VDW) equation and found it was $1.05\ \mathrm{atm}$. So, as expected, the value is larger, but not much larger. That's because VDW actually accounts for a parameter (b) which is the volume excluded per mole (meaning the volume that particle actually takes up, sort of).

So, hexane, which is significantly larger than $\ce{CO2}$ or anything else mentioned, only increases pressure by 5% when confined to the same volume. That also means the volume should be 5% larger.

Be wary of scaling that to larger molecules though! As hydrocarbons get larger (and being linear helps) they have a much greater polarizable volume which results in (how about this for a name) greater van der Waals attractions. This extra attraction counteracts the increase in size of the molecule a pretty significant amount.

All that to say, the deviation from Avogadro's Law at SATP is actually quite small.

You're only going to get noticeable deviations at SATP when you have very polar molecules, even then, however, it's not too bad. Water for instance takes up $22.45\ \mathrm{L}$ at SATP which isn't even deviating in the direction we would expect, but if you increase the pressure to $1.1\ \mathrm{bar}$, then the volume occupied for water is $22.33\ \mathrm{L}$.

Intermolecular forces are more significant than the particle's size, particularly at lower temperatures and higher pressures.