Reaction of citric acid and calcium chloride I was planning to extract citric acid from lemon juice by reacting the lemon juice with calcium chloride hoping it would react to form calcium citrate and precipitate according to the reaction
$$\ce{2 C6H8O7(aq) + 3 CaCl2(aq) -> Ca3(C6H5O7)2(s) + 6 HCl(aq)}$$
However I was not sure that the reaction would occur.  I know that the calcium citrate should precipitate out and drive the reaction forward, however it seems that the $\ce{HCl}$ generated reaction would then react with any calcium citrate bringing back to calcium chloride and citric acid.  Which reaction would move to completion or would the two reactions reach an equilibrium.  If this will not work I plan to just use calcium hydroxide instead by reacting calcium chloride with sodium hydroxide.

This procedure is done in our laboratory class, but first you have to add sodium hydroxide until the lemon juice is somewhat alkaline. Then add the calcium chloride (use $1~\mathrm{ml}$ for every $2~\mathrm{ml}$ of the lemon juice + $\ce{NaOH}$ solution).
I think the calcium chloride would first react with the hydroxide, producing calcium hydroxide, which would then react with citric acid to form calcium citrate.

$$\ce{2 C6H8O7 + 3 Ca(OH)2 -> Ca3(C6H5O7)2.4H2O + 2H2O}$$

Filter out the salt (precipitate) formed, weigh it, then you would have to add sulfuric acid afterwards to react with the calcium citrate. This would then form citric acid (soluble) and a precipitate which is calcium sulfate.

$$\ce{Ca3(C6H5O7)2.4H2O + 3H2SO4 -> 2C6H8O7 + 3CaSO4.2H2O + 2H2O}$$

You would then have to concentrate the filtrate to obtain the citric acid crystals.


How to define equivalents here $3g$ of activated charcial was added to $50 ml$ of acetic acid solution ($0.06N$) and filtred after an hour and it was found to be $0.042 N$ .The amount of acetic acid adsorbed (per gram of charcoal) is

The solution for this question is exactly given as I have written below

No of equivalents before adsorption $=50\times0.06 =3$

No of equivalents after adsorption $=50 \times 0.042=2.1$

No of equivalents left $=0.9/1000 \times 60=54mg$
Acetic acid per gm $= 54mg/3=18mg$

Now I have some doubts regarding the question and it's answer

  Why is normality being decreased here?
  Why are they taking volume to be same when it generally increases when we add solutes?
  What does equivalent mean here? According to Wikipedia, it is said to be "the mass of one equivalent, that is the mass of a given substance which will 1)  combine or displace directly or indirectly with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine – these values correspond to the atomic weight divided by the usual valence; or  2) supply or react with one mole of hydrogen cations (H+) in an acid–base reaction or 3) supply or react with one mole of electrons (e−) in a redox reaction." The equivalent talked in the question doesn't meet any of this criteria so what is its role here. Can somebody please explain me this?

Number of equivalents and normality are out-of-date concepts in chemistry. For acids, they are just like number of mols and molar concentration (or molarity), with an extra factor indicating how many hydrogens can be effectively ionized in solution. Because acetic acid ($\ce{CH3COOH}$) is monoprotic, you can just exchange the terms by number of mols and molarity and you are good.

  Why is normality being decreased here?

Some of the acetic acid adsorbs - meaning it bonds - in the charcoal. So the concentration in solution is decreased. Pay attention to the fact that the charcoal remains solid and does not dissolve.

  Why are they taking volume to be same when it generally increases when we add solutes?

As I wrote above, the charcoal stays solid, so it has no effect on the solution volume. The loss of acetid acid in solution might have a slight effect on volume, but that is mostly negligible.


How tetraamminecopper(II) is formed? I'm trying to get a sense of how $\ce{Cu^2+}$ and $\ce{NH3}$ are bonded together because it doesn't make any!
I understand that $\ce{NH3}$ has a lone pair of electrons that can be donated (right?), but since there are four molecules of $\ce{NH3}$, then there are 4 pairs, and $\ce{Cu^2+}$ only needs two!
the reaction in my textbook was written like this:
$$\ce{CuSO4.4NH3 <=> [Cu(NH3)4]^2+ + SO4^2-}$$
I would appreciate any clarification.

There are many ways to describe chemical bonds. Copper II ion binds to some anions with ionic bonds in solids, sometimes in covalent bonds (Ref 1).
In aqueous solution, copper II ions are surrounded (hydrated) with six water molecules, with the oxygen ends pointing in, toward the positive copper ion. Water is acting as a Lewis base. Ammonia is a stronger Lewis base - after all, it is basic in water. So an ammonia molecule will replace a water molecule in a hexa-aquated copper ion. Eventually, you get $Cu(NH_3)_4)^{2+}$.$(H_2O)_2$. Oh, I forgot, sometimes we leave out the water molecules, since we just assume that they are there - everybody knows that, since the teacher has explained that (unless the teacher hasn't explained that).
In the equation of the question, only four of the actual six water molecules are shown, because they are important for describing the interaction with $NH_3$, and the other two are just a complication, because I know you would ask, why don't they get replaced too, to give $Cu(NH_3)_6^{2+}$? After all, nickel, cobalt and chromium III do! Well, zinc doesn't; it forms a tetra-ammonia complex. And iron, manganese and aluminum don't even dissolve to form soluble ammonia complexes in alkaline solution. Reference 3 suggests that it is an equilibrium situation, but there are deeper explanations which you can explore later.
Ref 1. https://aip.scitation.org/doi/10.1063/1.1677079
Ref 2. https://www.chemguide.co.uk/inorganic/transition/copper.html
Ref 3. https://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top


Mechanism for decarboxylation Propose a mechanism for the following transformation.

I gather that this mechanism might be similar to decarboxylation of keto acids:

However, I am unsure of how to begin.

In acidic media, the ester carbonyl will be protonated and the carbonyl carbon attacked by a water molecule. The intermediate will have a proton abstracted by a water molecule and the ethoxy group will be protonated (it can happen either by intermolecular reaction with an H3O+ or by an intramolecular reaction with a proton migration from the protonated intermediate after the water attack). The carbonyl is restored and an ethanol molecule leaves the structure. A subsequent proton abstraction by water forms the carboxylic acid, which will then suffer decarboxylation following the mechanism you presented above. (I'm sorry about my ChemDraw, I suck at drawing even on a computer).


Electron orbital diagram of vanadium I am learning about electron configurations and I came across the "exceptions" of $\ce{Cr}$ and $\ce{Cu}$ where electrons actually half fill the $\mathrm d$ orbitals instead of the expected $\mathrm s$ orbitals. I thought maybe vanadium would do the same, but my textbook gives the answer of an electron configuration exactly following the Aufbau principle.

I've attached a picture of my drawing for the textbook answer and what I had thought might happen — could someone explain why vanadium is not an "exception" like chromium or copper?

Electrons always fill in the lowest energy configuration possible.

Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order.  In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion.  In their case, the energy to promote an s electron to the d orbitals is compensated for by the effects of exchange energy and no repulsion.  This effect is called "the stability of half filled subshells", or something to that effect, in most textbooks.

The energy cost to promote 2 electrons from the s subshell, as would be the case for Vanadium, is too much to be compensated for by the exchange energy gained and the repulsion lost, so the lower energy configuration is to have two electrons in the s subshell and 3 in the d.

Here's a nice website that shows the electron configuration for all elements: with http://www.ptable.com/#Orbital


Is cis-1,2-dimethylcyclohexane a meso compound? I was watching a video lecture today on hydrocarbons and came across this. The instructor says that, as there is a plane of symmetry, cis-1,2-dimethylcyclohexane is a meso-compound:

However, the cyclohexane ring is never planar as it adopts a chair conformation. I cannot find any plane of symmetry in chair form. Like this:

Is the compound meso or not?

cis-1,2-Dimethylcyclohexane is achiral, not because there is a plane of symmetry, but because it consists of two enantiomeric conformations which interconvert rapidly via ring flipping at normal temperatures.

This is exactly the same case as amine inversion. "Chiral nitrogens" such as that in $\ce{NHMeEt}$ do not lead to chirality or optical activity because of rapid inversion of configuration at the nitrogen atom, leading to interconversion of the two enantiomeric forms 1a and 1b.

Likewise, the ring flip in cis-1,2-dimethylcyclohexane leads to two different conformers. I have deliberately chosen to depict the ring flip in the following fashion, to make the mirror image relationship more obvious. The green methyl group, equatorial in conformer 2a, is changed into an axial methyl group in conformer 2b. Likewise, the blue methyl group goes from axial to equatorial. The 1,2-cis relationship between the two methyl groups is retained in both conformers.

Each individual conformer can be said to be chiral, but just like how the amine is considered achiral, cis-1,2-dimethylcyclohexane as a whole is considered achiral.

Is the compound meso? According to the IUPAC Gold Book, a meso-compound is defined as:

  A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

1,2-Dimethylcyclohexane possesses two diastereomers, one cis and one trans form. The trans form is chiral, but the cis form is achiral, as explained above. Therefore, the cis form satisfies the above definition and is considered a meso-compound.


What do X-ray diffraction and NMR tell about the Proteins? I have gone through a simple idea that proteins have different conformations; it is a dynamic system. Moreover, whenever we do ensemble experiments, so we get an average picture. There are some questions-:

Q.1 Often protein structures are reported in PDB (protein data bank), so what do they signify? The most stable conformation or some averaged picture of all conformation or both?

Q.2 In the case of the NMR, proteins have the freedom to move/vibrate through space as they are in the solution phase, but in XRD, they are rigid. So what will be the difference if I get one structure from XRD and one from NMR?

Q.3  Why structure of some proteins has not been resolved yet? I get it that a large protein cannot be solved using NMR; it will be much complex. However, why we cannot use XRD?
a) Why is the crystallisation of some protein hard?
b) Will there be any other challenges even if we get the crystals of a protein?

Q.4 Investigators often use different buffers to crystalise the given protein, but how they make sure that structure that they are getting in that buffer is the "actual" structure? How they make sure that crystalisation do not affect the "actual" structure of the protein.

From "Actual" Structure, I mean the structure or set of conformations that are retained in the physiological condition (in the working conditions).

Dynamics can mean many things, and not all proteins undergo extensive conformational changes during their lifetime. Some are fairly rigid in solution.

XRD or NMR data represents an average over the conformations of a protein construct sampled under experimental conditions. A protein construct is a variant of the protein, selected to allow expression and purification, and possibly also to ensure stability and crystallization (as necessary for XRD).

NMR or XRD data is processed to derive an ensemble of geometries that are consistent with the data and satisfy other criteria, usually imposed by a protein force field or prior statistics about the frequency of geometric features, which allows scoring the quality of a geometry. Before being deposited, the ensemble undergoes quality control by the PDB, amongst other things to verify that there are not too many or extreme violations of expected protein geometric parameters (bond lengths, angles, dihedrals, vdw contacts). Deposited NMR geometries are intended as representative, not exhaustive.

Whether experimental geometries are relevant depends both on the protein and what structural (or dynamical) questions are of concern. Approximating physiological conditions improves the likelihood that the data is relevant.

Proteins under XRD have less conformational freedom, they are more "rigid" (having looked up the definition in Meriam-Webster's dictionary) than those in solution. Different XRD samples (crystals) may still capture views of significantly different conformations, and complementary techniques such as NMR can be used to fill in gaps.

Yes, bigger proteins are more difficult generally to study, but progress has been made. Some of the limits are instrumental, others with sample prep. There is no single answer (too broad). Q4 is a standard problem. You get the structure and then worry whether there are inconsistencies with what is known/expected. You can try different media and see if they generate different structures.

Why is crystallizing proteins hard? Crystallization has to happen spontaneously! The protein must be kept stable (folded) as it crystallizes, which means being restricted to a narrow range of solvent and temperature conditions. There's a lot more that can go wrong versus what needs to go right.  Even as you want the protein to associate and form an ordered crystal, you don't want it to form the wrong kind of interactions resulting in a disorderly aggregate.


Why is the 2-position of pyridine electron richer than the 4 position I am wondering about the electronic structure of pyridine.

If we look at nucleophilic substitution, which favored on electron poor centers, for 2-chloro and 4-chloro pyridine we see that the substitution in 4 position is about 5 times faster (Ethanol, 20°C, taken from The Chemistry of Heterocycles, page 277

This SE question and answer explain why this might be the case and the explanation there is that the lone pair of the nitrogen is repelling the nucleophile. This might be the case to certain extend, but from geometry of the attack I doubt that it has much of an effect. In addition if we look at the electron distribution in the pi system we get this picture (also taken from The Chemistry of Heterocycles, page 271)

So the 4 position seems to be electron poorer than the 2 position. Usually we explain electron poor and rich centers in heterocycles by using mesomeric forms:

And the usual comment is that the 2 position is closer to the nitrogen so the electron withdrawing effect is stronger than in 4 position and it should be electron poorer, which isn't the case.

I also did some DFT calculations using 2- and 4-pyridyl substituents and 4-pyridyl is more electron withdrawing than 2-pyridyl in that case.

Is there an understandable explanation why the 4-position is electron poorer despite beeing furtehr away from the nitrogen?

When you deal with an aromatic ring, pairwise covalent bond structures miss a lot of key features.  You need to consider the entire conjugated ring as a unit and take account of symmetry.

Consider almost a benzene ring.  We may describe its $\pi$ orbitals as linear combinations of the out-of-plane atomic $p$ orbital on the carbon atoms.  Let $p_k$ represent the wave function of one such atomic, with the positive lobe on a specified side of the ring plane. The same orbital but with the positive lobe on the"wrong" side of the ring plane is rendered as $-p_k$.  The subscript $k$ refers to atom number $k$ as determined in the standard IUPAC scheme.

We then describe the canonical, occupied $\pi$ orbital as the familiar linear combinations:

$\psi_1=(1/\sqrt{6})\times (p_1+p_2+p_3+p_4+p_5+p_6)$

$\psi_2=(1/\sqrt{12})\times (2p_1-p_2-p_3+2p_4-p_5-p_6)$

$\psi_3=(1/2)\times(p_2+p_3-p_5-p_6)$

Now suppose that atom $1$ is changed from carbon to nitrogen to make pyridine.  Now that atom, being more electronegative, acts an an electron sink on occupied orbitals having an interaction with that atom -- meaning $\psi_1$ and $\psi_2$ are vulberable to thus $\pi$-electron withdrawal but $\psi_3$ which has a node at atom $1$ is "protected" by its symmetry.  Then positions $2$ and $3$, with some of their $\pi$- electron density on the protected $\psi_3$ orbital, are less susceptible than atom 4 whose $\pi$ electrons are all on the vulnerable $\psi_1$ and $\psi_2$ orbitals.

Now, going a little beyond the question, suppose we were to N-oxidize the pyridine, as with a peroxy acid.  In pyridine N-oxide the oxygen acts as a $\pi$ donor so that the nitrogen (or more accurately, the N-oxide functional group) is now effectively an electron source instead of a sink.  The "vulnerable" orbitals are now electron enriched and this characteristic is most strongly imparted, by conjugation and symmetry as argued above, to position 4.  Then electrophilic substitution, enhanced (or enabled?) by the N-oxidation, is most strongly so at position 4.  So, the primary electrophilic substitution product, after subsequently reducing the N-oxide, is the 4-substituted pyridine.  Which is what we see empirically with a wide variety of pyridine N-oxide electrophilic substitutions.


Acetone Enthalpy & Observations I was looking at the enthalpy of various liquids using WolframAlpha and noticed that at 30 °C, while water and ethanol have a positive enthalpy, acetone has a negative enthalpy.

If I'm understanding this correctly, this means that as acetone evaporates, the temperature of the acetone will increase (because the evaporation will release energy, thereby raising the temperature). I started searching up in temperatures to find the zero-crossing, and that appears to be around 56.074 °C.

This suggests that if a sufficient quantity of acetone is left around, it will eventually heat up to 56.074 °C (which interestingly happens to be when its vapor pressure is really close to the STP pressure).

However, if I get acetone on my hand, I get a distinctive "cool" feel — it, like other liquids, appears to be endothermic and cools down. I may be warm-blooded, but I'm not above 56 °C.

Is there something wrong with my understanding? If not, why does the theory seem to contradict my observations?

The absolute values of enthalpy H at a single state point are meaningless. It is only the difference between two different state points that matter. Thus, the value for a single state point can be set to any arbitrary value. Many handbooks set the arbitrary state point so that the values of these properties are positive for most liquid or gas states.

Apparently, WolframAlpha is setting enthalpy H to zero for the saturated liquid at the normal boiling point (NBP) for acetone but not for water and ethanol.

For ethanol, WolframAlpha is setting specific enthalpy to h = 200 kJ/kg for the saturated liquid at 0 °C (IIR reference state).

For water, the triple point is selected as the reference state, where the internal energy U of saturated liquid is assigned a zero value. Accordingly, the enthalpy H = U + pV of liquid water at the triple point is slightly larger than zero, at a specific enthalpy of about h = 0.6115 J/kg.


Has anyone even taken a picture of a molecule to confirm the geometry predicted by theory? Have the geometries of molecules been proven, or are we operating strictly on mathematical theory? To put it over-simplistically: has anyone ever taken a picture of a molecule to compare against their math?

Yes. Researchers have been using atomic force microscopy (AFM) and scanning tunneling microscopy (STM) for some time for this purpose. Do note that these images are not photographs in the sense that we usually think of "pictures" and are indirect measurements of constituents of the molecule. However, they do yield "pictures" that show the geometry of the molecule(s) in question.

See, for instance, this piece from UC Berkeley News published in 2013 which describes the techniques in a manner that is appropriate for a non-chemist. The same article includes images of molecules before and after a chemical reaction occurs.

I also encourage you (and others) to have a look at images in the IBM STM Image Gallery, which are quite spectacular.

Finally, as others have commented, X-ray diffraction merits mention from both a historical and scientific perspective. As one person notes, the helical secondary structure of DNA was determined by Watson and Crick after examining the fiber X-ray diffraction plates obtained by Franklin's group.

The importance of Franklin's contribution and that of Watson and Crick are monumental achievements. That said, the structure that was elucidated was the secondary structure of a large biomolecule and as such might not fit your criteria for how "complete" the "picture" needs to be to satisfy your question.

Here is the famous Photo 51 result obtained by Rosalind Franklin's doctoral student Raymond Gosling that was subsequently used by Watson and Crick.

Recently, researchers have imaged single atoms using a combination of STM and MRI - see Nature Physics. P. Willke et al. Magnetic resonance imaging of single atoms on a surface. 1 July 2019.


C4v point group complexes and the group orbitals (MO Theory) Out of curiosity, what would the group orbitals be like in a C4v complex (square planar complexes or octahedral complexes with 5 identical ligands and 1 other) look like? I get a irreducible representation of 2A1 + B1 + E, which raises the question of how the 2 different A1 group orbitals look like.

Can someone explain this to me?

I don't have time for a full answer or derivation right now, but here is a rough explanation. We can just consider s-type orbitals on each of the five ligands.

If the group orbital is to transform under the TSIR, then each of the four "equatorial" orbitals must have the same phase as each other. This is because these orbitals can be interconverted by symmetry operations, and if the character under every operation is to be $1$, then every orbital has to have exactly the same contribution to the group orbital. However, the equatorial ones don't need to have the same phase/magnitude as the "axial" orbital, because the equatorial/axial positions are not symmetry-related.

Therefore, you end up with two group orbitals that look somewhat like this:

Obviously, the exact relative contribution will depend on the system, but this should be the correct form.

For nearly any question of this nature, an excellent reference is Albright, Burdett, and Whangbo's Orbital Interactions in Chemistry, 2nd ed. Unfortunately, I can't find a diagram for C4v MOs, but in general most things can be found there.


How to choose which reactants to have in excess? In a precipitation reaction, how do I know which soluble reactant I should put in excess?
E.G.
$$\ce{Na2SO4 +Pb(NO3)2 -> PbSO4 + 2NaNO3}$$
In my book, $\ce{Na2SO4}$ is added in excess, but can we add $\ce{Pb(NO3)2}$ in excess?

Sometimes the excess is the cheap/unlimited reagent you use to get a high yield on your expensive/limited reagent, and sometimes you use the one that you don't mind/can easily extract from the products.

In this case, if you wanted to collect and record the lead(II) sulphate as a precipitate, using an excess of the lead(II) nitrate would be a waste.  If your intention was to produce a large quantity of sodium nitrate solution, and you don't mind having a load of the lead nitrate contaminating it, then you could use an excess of the lead nitrate.


Do dipole dipole interactions need to be added explicitly to vdW forces while doing DFT simulations for organic molecules? I  am modelling the adsorption of organic molecules on metal surfaces using plane-wave dft. While there is literature on how important van der Waals forces are for organic systems, and there are also some corrections to incorporate vdW forces into DFT (DFT-D, DFT-TS, and so on...); I was wondering if these corrections incorporate electrostatic dipole-dipole interactions within them (or do the dipole-dipole terms need to be added separately)?

(Permanent dipole)-(permanent dipole) interactions are covered by the electrostatics of DFT (and Hartree-Fock, for that matter, because no [Coulomb] electron correlation is involved). In my opinion, interactions between a permanent dipole (such as the one of HCl) and an instantaneous dipole (methane) are covered by the parametrization of dispersion-accounting methods in an averaging fashion.


Does the absolute value of enthalpy have any physical significance? I know it is pretty hard to measure the change in volume in open systems. Accordingly, calculating the change in internal energy using the first law of thermodynamics is pretty hard also.
Is this the only reason for introducing enthalpy ($H$) as a new thermodynamic potential to make use of the constant atmospheric pressure, and to calculate the change in enthalpy ($\Delta{H}$) as it is equal to the in/out quantity of heat ($Q$) in this case?
If no, does the absolute value of enthalpy have any physical significance?

The absolute value of enthalpy has direct relation to the mass of the system:

$$H = U + pV = mc^2 + pV$$

with the second term mostly negligible. If the system accepts thermal energy Q, its mass change is $\Delta m = \frac {Q}{c^2}$

For most of enthalpy applications, the absolute $H$ value is irrelevant.

The absolute value interpretation is the total energy stored in the system at the constant pressure, consisting of

the equivalent energy to the mass of contained matter
work done by reaching the system volume against the external pressure.

This energy is formally extractable from the system by annihilation and shrinking to zero volume.


How can I calculate the boiling temperature of a pure substance as a function of pressure? I want to make an algorithm that will give me the boiling temperature of a substance as a function of pressure, the critical temperature of a substance, and the critical pressure of a substance. I believe this can be done using a combination of the Peng-Robinson Equation of state and the Clayperon or Clausius-Clayperon equation of state. The issue is that I believe this requires additional input criteria, specifically the latent heat but this function is supposed to be more general. Is there any way to work this method without the latent heat, or to calculate the latent heat from the other known values?

If you know the latent heat of vaporization at one temperature, you should be able to determine the latent heat of vaporization at any other temperature from Hess' law and knowledge of the heat capacities of the liquid and vapor as a function of temperature.  You may also have to include a pressure contribution as the critical point is approached.


Is solubility in Qsp affected by coefficient? Related to my previous question: Is solubility coefficient affected if ion data is given in Ksp?

  $200\,\mathrm{mL}$ solution of $0.02\,\mathrm{M}\,\ce{AgNO3}$  is added to $200\,\mathrm{mL}$ $\ce{CrO4^{2-}}$ and $\ce{PO4^3-}$ ions. Find out both of $Q_{\mathrm{sp}}$.

Actually the question is asking "Will it precipitate", but I skip it.

Here is my half approach:

$$200\,\mathrm{mL}=2\times 10^{-1}\,\mathrm{L}$$
$$V_{\mathrm{total}}=400\,\mathrm{mL}=4\times  10^{-1}\,\mathrm{L}$$
$$[\ce{Ag+}]=[\ce{CrO4^2-}]=[\ce{PO4^3-}]=\frac{0.02\,\mathrm{M}\cdot 2\times 10^{-1}\,\mathrm{L}}{4\times 10^{-1}\,\mathrm{L}}=0.01\,\mathrm{M}$$
$$\ce{Ag2CrO4 -> 2Ag^{+} +CrO4^{2-}}$$
$$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[\ce{Ag+}]^{2} \cdot [\ce{CrO4^2-}]$$
$$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[2s]^{2}[0.01\,\mathrm{M}]$$

I know that to find $K_{\mathrm{sp}}$, the $s$ needs to be multiplied (and to the power) with the coefficient from the reaction, but how about $Q_{\mathrm{sp}}$? Does the $s$ also need to be multiplied?

I asked this because my teacher said that only to the power is affecting the $Q_{\mathrm{sp}}$.

You don't need to multiply s, since $Q_\text{sp}$ is the result of multiplication of ions concentration (and you don't need to multiply with the coefficient in any $Q$). So the right way to answer $Q_\text{sp}$ is

$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[\ce{Ag^{+}}]^{2}[\ce{CrO_{4}^{2-}}]$$
$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[0.01]^{2}[0.01] = 10^{-6}$$


Is the activation barrier of a reaction related to the free energy or heat of formation of its reactants? The energy barrier for activation appears in the derivation of a bimolecular reaction through collision or transition state theory, but it seems that from looking at diagram such as this one  it seems like the barrier to activation would be related to the energy required to break bonds?

Thanks in advance.

They are certainly related, but not necessarily the same. If all bond-breaking and forming would be sequential than the activation energy would equal the bond-breaking energy.

However, it is possible in some reaction mechanisms that the forming bonds 'donate' energy to the break-up and therefore the overall activation energy will be lower. An example is the reaction of hydrogeniodide to hydrogen and iodine show in the figure below. Instead of a full dissocation, two molecules of $\ce{HI}$ line up and the breakup occurs simultaneously with the formation of the new bonds thus allowing a lower energy level.

The only thing you can safely say is that the sum of breaking energies is an upper bound for the activation energy, because single atom level is 'as bad as it gets' (excluding nuclear reactions).


Why is copper(II) carbonate not selected as an unknown compound for back titration? (Whereas another unknown carbonate is used instead?) Why is copper(II) carbonate not selected as an unknown compound for back titration? (Whereas another carbonate is used instead)

I know that the carbonate to be provided on the day (not Copper (II) obviously) will be reacting with excess HCl. The excess HCl will then react with NaOH to determine the concentration of HCl and eventually Molar Mass of the provided carbonate substance.
NOTE: phenolphthalein is the indicator being used

Copper is a transition metal. As such, its ion will be a Lewis acid. The effect will, as with iron, be stronger with increased charge--so Fe(III) salts of strong acids give acidic solutions, more so than Fe(II) salts of the same acids.

This is because lone pairs can be donated into the empty d or hybrid orbitals from water. The structure produced in this way tends to lose hydrogen ions. This means that the acidity in your experiment has two sources for hydrogen ions.


Specific Heat Capacity of Heavy Water Water has an unusually high specific heat capacity due to it's hydrogen bonds.

Why is it that the change in isotope causes a 10% difference in the heat capacity? How does the added neutron in the nucleus make any difference to the bonds?

I posted an answer here the other day which was completely wrong! I apologise and have left the incorrect text below for posterity to record how wrong people who should know better can be!

Looking on the NIST fluid properties database I see that the specific heat capacity (i.e. the heat capacity per unit mass) of D2O is just 1.5% higher than the specific heat capacity of H2O. So at 298 K this makes the molar heat capacity of D2O (84.963 J/K/mol) an astonishing 12.7% higher than H2O (75.38 J/K/mol). I am astounded and have no explanation.

Everything below this line is WRONG!

The molar heat capacity is almost exactly the same. i.e. it takes the same energy to heat the same number of molecules by one degree Celsius.
The difference in the *specific* heat is just because of the difference in density. D2O molecules have a relative molar mass of about 20 compared with 18 for H2O. Since the bond lengths are similar, the density is higher by a factor 20/18 - about 11% higher. So a given mass of D2O has about 11% fewer molecules in it.


The reaction of iron with sulfur to produce Iron(II) sulfide is inconvenient (for me) I am taking a lot of iron properties this semester, but I have found some contradicting piece of information about the reaction of iron with nonmetals illustrated with two cases
\begin{align}\ce{ **First Case** }\\\end{align}
If we react the iron with chlorine, only FeCl3 will result as following:

$$\begin{align}\ce{2Fe +3Cl2 &->[\space \Delta ]2FeCl3}\tag{1}\\\end{align}$$

No FeCl2 has produced because iron prefers to have oxidation state at +3 as sublevel 3d will be half-filled.

One case that FeCl2 will result when H2 exists as
$$\begin{align}\ce{Fe +2HCl &->FeCl2 +H2}\tag{2}\\\end{align}$$
In this situatuon if any Fe2Cl3 formed, H2 will reduce it to FeCl2
\begin{align}\ce{ **Second Case** }\\\end{align}
Although, Iron favor +3 oxidation state (as illustrated in 1), If we react iron with sulfur; it will actually favor +2 oxidation state as follows:
$$\begin{align}\ce{Fe +S &->[\space \Delta ]FeS}\tag{3}\\\end{align}$$
I tried to find a convenient way that explains how this happen (as the way illustrated at 2), but

I failed to do so. I am really confused:

Does the iron actually favor +3 state or not?

If yes, Why 3 happens? Why there isn't any Fe2S3 in the products?

If no, Why 1 happens?

Is this has to do anything with some hidden conditions or unobvious reaction mechanism?

Not an answer per se (I lack sufficient reputation to comment): are these reactions in aqueous solution, or dry gases? Just a suggestion, but by omitting the charges on these species and not writing them as half-reactions you are obscuring the electron exchange in each. Thus in your first example (iron reacting with chlorine gas), you take metallic Fe, whose oxidation half-reaction can be written as
$$\ce{Fe^0 -> Fe^3+ + 3e-}$$
The equivalent for reduction of chlorine would be
$$\ce{2e- + Cl2_{(g)} -> 2Cl-}$$
Normalizing half reactions to electrons transferred ($6$), adding and simplifying stoichiometry gives

$$\ce{2Fe^0 + 3Cl2 -> 2FeCl3}$$

Thus chlorine is the electron acceptor (strong oxidizer) and iron the donor. In your reaction (2), iron is again a donor (this time of only $2$ electrons), hydrogen is the acceptor, with chloride ion just along for the ride:

$$\ce{Fe^0 -> Fe^2+ + 2e-}$$
$$\ce{e- + H+ -> \tfrac{1}{2}H2} $$

In contrast to these simple reactions, those involving sulfur are exceeding complex because of the diversity of species that can form. In your reaction to form iron monosulfide, one possible way for Fe to be oxidized but native sulfur (zero-valent S actually present as stable ring, $\mathrm{S}_8$) to be both reduced and oxidized is via a disproportionation reaction, with water reduction to hydrogen consuming additional electrons in an anoxic environment, thus yielding both FeS and iron sulfate:
$$\ce{Fe^0 + \tfrac{1}{8} S8 + 2H2O -> \tfrac{1}{2}FeSO4 + \tfrac{1}{2}FeS + 2H2}$$

Amended answer:
If you're heating metallic iron and S to significant temperatures, the elemental sulfur will form a gas, $\mathrm{S}_{n}(g)$ ($n$ indicates which T-dependent allotrope is present). To synthesize FeS however, I think you would need an oxygen-free system. If $\mathrm{O_2}$ is initially present, consumption could be accomplished by sacrificing some of the reactants via in situ oxidation of the metal, or production of $\mathrm{SO_2}$. Then this can proceed directly, using the remaining $\mathrm{S}_{n}(g)$, e.g.:
$$ \tfrac{1}{2}\mathrm{S_2}(g) + \mathrm{Fe}(s) = \mathrm{FeS}(s) $$
I believe the ratio of S/Fe of the product will increase with temperature, and at sufficiently high temperature reach 2 ($\mathrm{FeS_2}$, pyrite).


Molar absorption coefficient for potassium hydrogen phthalate I would like to draw a comparison of the sensitivity between several spectrometers. Therefor I measure the absorption of a serial dilution of potassium hydrogen phthalate (KHP). The idea is to compare the theoretical absorption with the measured absorption.

To calculate the theoretical absorption I need the molar absorption coefficient of KHP at 254 nm wavelength. Unfortunately I haven't found anything till now.

Does anyone know where I can find this coefficient?

I would imagine that you could just use one of the spectrophotometers you have available to determine the molar absorptivity constant experimentally. After all, the molar absortivity constant is simply the constant which equates the terms (those being Absorbance and Concentration) which are already proportional. So, using something like the beer-lambert law...

$$A = \epsilon b C$$

...make a solution of a known concentration, measure its absorbance at your wavelength of interest, and solve for $\epsilon$.

$$\epsilon = \frac{A}{b C}$$

You might consider doing this a number of times to get an average value and thus minimize the random error associated with your constant.


Electrolysis of copper sulphate using copper electrodes: Mass lost from anode is greater than mass deposited on cathode In short, the aim of my experiment is to determine how the concentration of sulphuric acid in solution affects the mass liberated and deposited at copper electrodes, when copper sulphate is electrolysed.
Over all data points, what I observed is that the mass deposited at the cathode was lower than the mass liberated at the anode. Theoretically, the two should be equal, and even experimentally I expected the opposite to be true, due to oxygen production taking place at the anode which could disrupt the liberation of copper. The voltage was kept constant at 3V, so I monitored the current over the duration of the experiment. I observed that the current increased, which further suggested that more ions were released at the anode than deposited on the cathode, as  that represents an increase of positively charged particles in solution and thus an increase of current density.
Now, perhaps the liberation of oxygen facilitates the oxidation of the copper anode? Is this my misunderstanding?
Additionally, I was unable to find conditions which would stop the production of hydrogen at the cathode. The voltage required to electrolyse water is 1.23V, but if I set my voltage to that low, no plating took place either. Is hydrogen also expected to be produced alongside the plating process? Maybe, the hydrogen production was disrupting the plating? In that case, what is the explanation for why hydrogen was produced at the cathode in the first place? Since copper is lower in the reactivity series than hydrogen,  no hydrogen should be produced.
Can someone point out the flaws in my experiment or where I've misunderstood things. I don't have much experience with electrochemistry. Thank you!

$\pu{3 V}$ is too high. Kinetics of deposition of $\ce{Cu}$ is saturated, so another electroactive system kicks on, creation of hydrogen. Similarly for anode. Try to evaluate volt-ampere characteristic of the system and stay on the current plateau before gases start to be created.
Generally, is is more advantageous to use a power source keeping constant current rather than constant voltage.
If gases are being evolved, you cannot expect the dissolved and deposited mass of copper would match, as various portions of the current are spent on gases, instead of copper. The current density that Maurice suggests is a good one.
For low enough voltage, copper deposition + dissolution can absorb the current. At particular current density, one of these reactions is saturated and cannot take more current any more, being diffusion controlled. Then either hydrogen, either oxygen starts to be created, depending on the electrode where it occurs first. When it happens for the other reaction as well, the other gas gets created as well.

Each electrode has a voltampere characteristic given by I=f(E) function, that has a shape of irregular rounded step-wise slope. When a cathodic potential gets low enough value for a given compound/ion, current starts to raise. When reaction gets saturated, i becomes being controlled by ion diffusion and there is zero concentration at the electrode. Then the current reaches a plateou. The plateou ends at a potential where another redox system starts to interact. Similarly for anode.
When an external voltage is applied, electrodes gain such potentials to meet 2 conditions:

The potential difference equals to the external voltage.
Currents of both electrodes, according to the above function, are equal but the sign.

Lower current density means more positive cathode potential, where $\ce{Cu^2+}$ gets selectively reduced. Higher density means lower cathode potential where both $\ce{Cu^2+}$ and $\ce{H3O+}$ may get reduced. It is like if you used mildly reducing reagent, reducing just copper ions versus strongly reducing reagent, reducing also hydronium ions. Similarly for the anode.
The only necessery thing is the cathodic current matches the anodic current. Processes on both electrodes are independent. There is no "quantum entanglement" between these processes. If $\ce{Cu^2+}$ diffusion and $\ce{Cu}$ deposition can absorb all the cathodic current, there is no need of hydrogen creation.
Try to measure the steady current as a function of the applied voltage. You can use a small resistor is a serie with a cell, measuring voltage on it, which is equivalent to the passing current as $I = U_\mathrm{R} / R.$ Find the first current plateau. Or the "like to be a plateau, just if the 2 voltage steps were not so close together."
Choose the voltage from this plateau, but be aware it may change with the change of the cell geometry or electrolyte composition.

Even in cased hydronium ions do not participate on electrode reactions directly, they can participate indirectly by influencing I=f(U) characteristics of the cell via:

Keeping or affecting chemical state of involved ions. ( E.g. pushing equilibrium $\ce{[Cu(H2O)4]^2+ + H2O <=> [Cu(H2O)3(OH)]+ + H3O+}$ to the left.
Balancing the local gradient due charge disbalance
Increasing conductivity and therefore the solution potential difference, that subtracts from the applied voltage.


Why is gum so sticky and what chemicals make for easy removal of gum? To give this question context, I work in a restaurant and sometimes I'm in charge of getting all of the gum off of the tables at work. Don't mind the work, but I'd be interested in hearing what makes gum so sticky and if there is any chemical I can use to make the cleaning a bit easier and why that chemical works.

I guess another backwards question is: Is there any chemical I can apply to the underside of tables that will prevent the gum from sticking? Would be awesome if I could fix the problem at the source and would be super interested in the chemistry behind it.

The International Chewing Gum Association tells us:

Gum base is what gives chewing gum its “chew.” It is made of a combination of food-grade polymers, waxes, and softeners that give it the texture desired by consumers and enable it to effectively deliver sweetness, flavor, and various other benefits, including dental benefits.

A more complete description of what makes chewing gum sticky states:

Terence Cosgrove, Professor of Chemistry at the University of Bristol and Chief Scientific Officer of Revolymer, explains that the chemical bonds between the molecules in a polymer-based chewing gum make it difficult to remove from a surface. When you pull a piece of used gum off a surface, most of the energy goes into stretching the polymer bonds in the gum, rather than actually breaking the bonds between the surface and the gum. Since polymers are long chains of covalently bonded molecules that give the gum its elasticity and "chewiness," the attractive forces between the atoms are formed by the sharing of electrons [8]. Thus, the bonds between the repeating units tend to remain intact as they lengthen and contract from an applied external force.
The temperature at which the polymer base is exposed to the air also affects both the elasticity and adhesiveness of the chewing gum. When a piece of gum is heated by saliva and deformed by the grinding of human teeth, its polymer chains align in the direction of these forces. The degree of alignment is a function of the magnitude of applied stress, which explains why the gum becomes tougher and less elastic the longer or more rigorously you chew [8]. After the gum is removed from your mouth and placed in a cooler environment, the drop in temperature causes the orientation of the polymer chain to freeze, resulting in a hardened piece of used gum.
Most commercial polymer gum bases are hydrophobic (water-insoluble), which is the reason they stick easily to oily surfaces and are difficult to remove, even with the help of household cleaning solutions. A hydrophobic substance tends to repel polarized molecules like water and attract non-polar compounds, such as the grease and grime on streets. A hydrophilic substance, on the other hand, behaves in the opposite manner by attracting water and repelling fats and oils [8]. A compound that exhibits both hydrophobic and hydrophilic traits is known as "amphiphilic" [9]. These substances, which attract both water and oil to some degree, can be used to create synthetic polymers that could help eliminate the sticky mess of chewing gum pollution.

Note: The numbered references in the above block quote can be pulled from the linked article. The article also details efforts aimed at producing a new type of chewing gum that doesn't stick to things as much as normal/currently-manufactured ones seem to. It's a good read.
But to your main point, I've linked the MSDS for a popular product for removing gum and related compounds (it's called Goof Off), which consists almost entirely of acetone (also commonly found in nail polish remover). Given the deleterious health effects of acetone (and its high vapor pressure), it's not a good candidate for your needs.
Something non-stick and more permanent, such as a Teflon spray might be ideal for your needs.


Why is tantalum so unreactive? What is the fundamental argument behind explaining why Ta metal is so inert and does not usually participate in corrosion? I know that it forms a protective oxide, but nothing explains why in an acidic solution it is resistant to nearly everything (aside from HF). I want to know if it has to do with a bond strength of the metal or the molecular orbitals.
An in-depth citation is required.

I think the answer to your question can be described based primarily on thermodynamic considerations.  Ta has very high fusion (25 kJ/mol), vaporization (760 kJ/mol) and atomization (780 kJ/mol) enthalpies which are a result of strong metal-metal bonding in the solid (see Greenwood and Earnshaw, Chemistry of the Elements).

To further understand the protective nature of the native oxide that forms on Ta, it is fruitful to look at an Ellingham diagram, which describes the stability of a metal oxide as a function of temperature. For more authoritative sources than the Wikipedia page, see a comprehensive Inorganic Chemistry textbook such as those by Shriver and Atkins or Housecroft and Sharpe.  I found this site particularly interesting for purposes of visualizing Ellingham diagrams and used it to make the following plot:

(Apologies for the color scheme, which was not adjustable)

Here I've plotted the free energy for the oxide formation of tantalum oxide along with oxides of iron and aluminum.  The x-axis is temperature.  The large negative values for each of these oxides indicates that they are all thermodynamically favorable at the temperatures plotted.  Room temperature would be very slightly to the left of the intersection of the x and y-axes.  A complete analysis of the stability of the metal oxides would also require consideration of kinetics, but we can use thermodynamics to postulate that the very large negative values for the free energy of formation of the tantalum oxide are one factor in the stability of the oxide film.  I'll leave it to the OP to play around with thermochemical cycles, which should demonstrate why the oxide films are resistant to many types of acids.


Are PTFE stir bars resistant to all lab solvents? It seems like glass-coated stir bars are a specialty item.  Standard stir bars are coated in PTFE.  Is PTFE resistant to all typical lab solvents?  Or is there a reference to check to ensure it is inert to the chemicals being stirred?

This may not be the answer that you are looking for, but I remember a project, for which we had to replace PTFE stirring bars with their glass-coated counterparts.

It is true that PTFE is resistant to most of the solvents used, but it is not really hard and resistant to abrasion.

In a study where aqueous suspensions of metal silicides were stirred, we recognized very soon that the PTFE covers didn't last very long. The stirring bars were polished down to the metal core, which spoiled the experiment. Glass-covered stirring bars turned out to be a viable alternative here.


How to find the average half life of radioactive nuclide which undergoes two different decays? Find the average life of a radio nuclide which decays by parallel paths,
  \begin{align}
A  &\rightarrow B\\
2A &\rightarrow B,
\end{align}
  where the decay constants are $\lambda_1 = \pu{0.018 s-1}$ and $\lambda_2 = \pu{0.001 s-1}$, respectively.

I used the formula
$$\frac{1}{t} = \frac{t_1 t_2}{t_1 + t_2,}$$
where $t$ represents the mean life. The equation can be written as:
$$\frac{1}{t} = \lambda_1 + \lambda_2$$
But I am getting the correct answer only if I take $\lambda_2$ as 2 times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?

But I am getting the correct answer only if I take $\lambda_2$ as $2$ times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?

Yes.

The way I tend to approach half-life problems is to recast them as the relevant kinetic differential equations.  Assuming the first-order kinetics as stated in the problem, the contribution of reaction 1 is
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_1 = -\lambda_1[A],
$$
and that of reaction 2 is†
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_2 = -2\lambda_2[A].
$$

Implicit in the above is the assumption that the first-order decay constants apply to the stoichiometry of the reactions as-written. As you correctly note, that extra factor of $2$ in the differential equation for reaction 2 is because two $A$ nuclei are involved per "unit" of reaction progression.

The total rate of loss of $A$ is the sum of the above:
$$
\frac{\mathrm{d}[A]}{\mathrm{d}t}
= -\lambda_1[A] -2\lambda_2[A]
= -\left(\lambda_1 +  2\lambda_2\right)[A].
$$

This is a straightforward first-order ODE, readily solved for the overall half-life by standard methods.

† While it seems quite unusual to me to have a second-order reaction ($2A\longrightarrow B$) with a first-order rate constant (units of $\mathrm{s}^{-1}$), for the sake of the problem I'm content to roll with it. I'm sure they cast the problem this way so that the calculus was straightforward.

As well, simultaneously having $A \longrightarrow B$ and $2A \longrightarrow B$ processes would seem to violate all kinds of conservation laws--the problem makes no physical sense as a result. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. $2A \longrightarrow C$ would have been much less preposterous, even if implausible in the context of low-energy radioactive decay.


How actually do we decide acidity and basicity order? In a book, I read two quite contradictory statements.:

$\ce{HIO > HBrO > HClO}$ wrt acidity while

$\ce{HF > HCl > HBr}$.

Aren't they contradictory? If not can anyone please explain the accurate reasoning?

I'm not sure about the first order, but I'll try to explain the reason for the second one. The acidic strength increases as $$\ce{HF<HCl<HBr}$$

One reason is the bond strength between the hydrogen and the halogen atom. Since the bond between $\ce F$ and $\ce H$ is very strong, the molecule does not easily lose $\ce H^+$. On the other hand, as the size of the halogen atom increases, the overlapping of the orbitals between halogen and hydrogen becomes less efficient. Thus, $\ce {Cl}$ and $\ce {Br}$ can more easily lose $\ce {H^+}$ than $\ce {F}$. Hence, the acidic strength increases as size of halogen atom increases.

Second reason is the stability of formation of conjugate base. When an acid loses $\ce {H^+}$ ion, it forms what is called as conjugate base. Similarly, when a base loses $\ce {OH^-}$ ion, it forms a conjugate acid. In case of $\ce {HF}$, we have $$\ce{HF <=> H^+ +F- }$$
Here, $\ce {F^-}$ is the conjugate base. Now, due to small size of $\ce {F}$, it is difficult for $\ce {F-}$ to hold negative charge. There are strong repulsions.

Now, for $\ce {HBr}$, we have $$\ce{HBr <=> H^+ +Br^-}$$

$\ce {Br}$ has a greater atomic radius. It is also less electronegative than $\ce {F}$. Hence, the repulsions are less in $\ce {Br-}$. Thus, $\ce {Br^-}$ is quite stable. The more stable is the conjugate base, the more readily the acid loses $\ce {H^+}$ ion, and so, more is its acidic strength.

Check this link. Maybe you'll find it useful.


Line structure of formaldehyde Would the line structure of formaldehyde, H2COH, simply be a straight line with a OH on the end of the line?

Formaldehyde, $\ce{H_2CO}$, has the following structure

There is no $\ce{OH}$ group, just a central carbon atom with a single bond to each of two hydrogens and a double bond to oxygen.


Why are these ions classified as isoelectronic? According to Chemistry the Central Science 13 Edition, these ions are isoelectronic:

$\ce{Ca^2+}$ (18 electrons); $\ce{Cs+}$ (54 electrons);  $\ce{Y^3+}$ (36 electrons)

Furthermore, the textbook states that the caesium ion is the largest.

Now, every definition I've looked up, including from the book, states that in an isoelectronic series, the atoms have the same number of electrons. Clearly, these don't. I have noticed, however, that they are in multiples of 18, where the Calcium ion is 18, and Caesium is 3 times 18, and Yttrium is 18 times 2.

A friend, who took chemistry 1 a few years ago, recalls that there was some formula (or something) to determine isoelectronic series.

I suspect that they have something to do with the electrons in their valence.

The book does not give an explanation. It moves on to ionization energy.

But it also states that of these three, caesium is the largest, which compounds my confusion since just a few lines up, the book states that the nuclear charge is inversely proportional to the ionic radius, so an increasing nuclear charge (atomic number) means a decreasing atomic radius. It would seem appropriate that the calcium ion has the largest ionic radius since it has the smallest nuclear charge.

Can anyone help?

Core Electrons don't count. It's the valence electrons that are counted (outermost shell). The valence shell of all 3 is empty. So they all have 0 electrons in their valence shell.


What are some examples of pure substances with no liquid state? For example,

A molecular substance that decomposes before melting, and under pressure can only assume a liquid state that doesn't contain the same molecules, for example a mixture of the elements of which it's composed
A polymer that breaks down into monomers when it melts, and doesn't have a liquid state with the same polymeric bonding regardless of the pressure
An ionic solid that sublimates upon heating, and under pressure can only assume a liquid state that clearly does not exhibit ionic bonding

What are some such substances? (Non-pure substances, such as wood, are non-examples.)

Interestingly enough, when considering not just the set of "typical" or "known" substances, but the vastly larger amount of all possible compounds, almost all pure substances do not have a thermodynamically stable liquid state or gaseous state. To become a liquid, a molecule must gain enough energy to set itself loose from the tug of its neighbours in the solid state. Typically intermolecular forces are fairly weak, and so for simple molecules this energy can be provided at relatively low temperatures.

However, there is no upper bound on how strong the intermolecular forces between two molecules can become; just take an enormous molecule. Even with relatively weak intermolecular interactions between each atom, when summed up across a huge molecule, the required energy becomes vast.

Combine this with the fact that, thanks to combinatorics, almost all possible substances are composed of massive molecules (in a chemical analogue of the frivolous theorem of arithmetic). For example, there are only five possible distinct substances with the formula $\ce{C6H14}$, but over a trillion trillion trillion trillion ($10^{48}$) distinct molecules with formula $\ce{C120H242}$.

The conclusion is that almost certainly the molecules would decompose (breakage of covalent bonds) well before getting close to the extremely high temperatures needed to produce a true liquid state.

Therefore, there should be many, many examples to choose from. A heavy-handed example would be maitotoxin (or many other natural products). Though the theoretical melting point would be very high, the presence of all those hydroxyl groups suggests it would decompose by dehydration around ~200 °C, as is typical of polyols (I'm sure it would actually decompose much sooner).

In the same polyol vein, a much more relatable example would be sucrose (sugar). In the strict thermodynamic sense, sugar doesn't actually melt when heated on a stove. The clear colourless syrup that's formed is actually the first step in caramelization, the complicated thermal decomposition of sugar.

The above cases represent your first suggestion of possibilities. Polymers also fit, with the complication that artificial polymers are almost never actually pure substances but mixtures of very similar molecular chains, varying in size. Something like  your last suggestion is interesting, but examples seem hard to provide. Perhaps dinitrogen pentoxide is close, due to its unusual transition between ionic and molecular forms.


Fundamental equation of thermodynamics for magnetocaloric materials The fundamental equation of thermodynamics, as us chemists (and chemical engineers!) are used to seeing it, is

$$ dG = - S~dT + V~dP + \sum_{i}\mu_i~dN_i$$

This gives the Gibbs free energy as a function of temperature, pressure, and composition, assuming there are no other relevant forces other than mechanical pressure.

The other day I watched a video on the magnetocaloric effect.  Obviously, there are non-pressure magnetic forces acting in such systems.  What's the proper form of the fundamental equation for magnetocaloric materials?

Suppose that the magnetocaloric material used is chemically pure and non-reactive during the magnetization process.  Then we could get rid of the $\sum_{i}\mu_i~dN_i$ term.  I suppose its also reasonable to assume that pressure is constant during magnetization / demagnetization process, and that the volume of the material is unchanged by magnetization so probably we could dispense with the $V~dP$ term as well (is that true?).

That leaves us with $dG = -S~dT + \rm{MAGNETIC~STUFF}$.  The $\rm{MAGNETIC ~STUFF}$ term probably has a $B$ or $H$ or something like that in it to represent the imposed magnetic field, but what else goes in there?

The fundamental equation for paramagnetic system is as follows
$$dU=TdS+BdM+\mu dN$$
Where B is the external magnetic field intensity and M is the magnetic moment (Here I have neglected the P-V work of the system). For your chemically pure and non-reactive system the above equation simplifies to
$$dU=TdS+BdM$$
Using Legendre Transformations:
$$y(0)=U(S,M)$$
$$dy(0)=dU=TdS+BdM$$
$$dy(1)=dA(T,M)=-SdT+BdM$$
$$dy(2)=dG(T,B)=-SdT-MdB$$
Now changing the order of the equation, we can find enthalpy,
$$y(0)=U(M,S)$$
$$dy(0)=dU=BdM+TdS$$
$$dy(1)=dH(S,B)=-MdB+TdS$$


Are there IP limitations on (M)SDS documents? Much effort is put in putting together (M)SDS documents:

  Many companies offer the service of collecting, or writing and
  revising, data sheets to ensure they are up to date and available for
  their subscribers or users. Some jurisdictions impose an explicit
  duty of care that each SDS be regularly updated, usually every three
  to five years. However, when new information becomes available, the SDS must be revised without delay.

Are there any intellectual property (IP) limitations on MSDS documents? i.e. are they copyrighted?

A really nice question. First, let's point out that specifics about these issues will vary from country to country, as they all relate to the laws of that country. Secondly, let's clarify that copyright and IP aren't exactly the same. Copyright covers the exclusive right of that owner to make copies, license and otherwise exploit their works (printed, artistic etc). Intellectual property is property that results from creative thought, and can include patents, copyrighted material and trademarks. So, IP is a bit like an umbrella that encompasses copyright material.

The specifics relating to copyright and IP differ slightly from country to country, but it is safe to say that the (M)SDS document is a copyrighted document. Most SDS documents will have this specified, usually at the very end of the document - I just checked some of the ones I have on file from commercial suppliers, and they all have a copyright statement. This means that you cannot make copies etc under the usual copyright limitations.

There are requirements about what information goes into a SDS. Again, there will be differences from country to country, but for the US, for example, I think all hazardous components above 1% must be specified, and known carcinogens above 0.1% must also be listed. The exact numbers might be a little different, but the general gist is that hazardous components must be listed.

HOWEVER, every country has a clause in their code governing SDS requirements that specifies, essentially, an exemption from listing ingredients that may be regarded as a 'trade secret', or puts at risk an identified Intellectual Property claim. How this is expressed varies; in some countries, a trade name may be used to 'hide' the true identity of a IP-protected ingredient. In other countries, a summary of functional groups can be used to represent that key ingredient. Usually, though, the nature of the hazard must be represented somehow on the SDS document. I know in some countries, even if an ingredient is protected by an IP clause, its true identity must be revealed to health practitioners if requested. I'd have to dig around for the exact laws for each country/zone if you needed more specific answers.


Why does Raoult's law work? I am trying to look for the entropy explanation but I do not understand anything I can find. Please keep it simple but talk about it from a entropy point of view. Use water and a volatile solute as an example if you need one. I do not understand how the addition of a volatile solute to water increases the vapour pressure of the solution.

Raoult's law is an approximation for ideal solutions.

Those links have more detail, but the basic idea is that when you assume that the interactions between solute and solvent molecules don't depend on the identity of the molecules, but the vapor pressure does, then it makes sense that the vapor pressure of the mixture would be a linear combination of the vapor pressure of the molecules in the mixture.

Think about it like a bunch of slightly sticky spheres bouncing around in a bucket. A certain percentage will randomly bounce out of the bucket - that's the vapor pressure. If you have a different type of sphere with less mass, more will bounce out in the same amount of time - the vapor pressure will be higher. If you mix them, the total vapor pressure is given by the total number of both types that bounce out. The more volatile ones leave more easily, so the total increases.

Since ideal gases don't interact with each other and since ideal mixtures have negligible enthalpy of solvation, all that is left is entropy terms. If you are looking for a connection to entropy, that is it.

In cases where the ideal approximation is not realistic (water and a volatile solute, for example), Raoult's law doesn't really hold, but it is still a reasonable approximation when the concentration is low simply because most functions look approximately linear for small deviations. It also tends to work reasonably well because most things that will form a solution tend to have similar interaction energies - otherwise they are less likely to mix in the first place.

Of course, there are many, many practical cases where Raoult's law doesn't work, and so we have corrections for that: fugacity and activity.


Combustion and pyrolysis of sulfides and organosulfurs Last week there was a huge fire in french chemical factory Lubrizol, we now know the main products that burnt, to simplify the discussion I extracted a few compounds:

isobutyl and isobutene polysulfide 500 tons,
dithiophosphoric acid and corresponding zinc salt (R = isopropyl) 200 tons,
various hydrocarbons (mainly mineral oil) 2000 tons
as well as  calcium sulfonate and calcium salt of methylphenol sulfide (100 tons), amides/esters of phosphoric acid, polymers, aromatic amines & alcohols, various reaction products of $\ce{P2S5}$ with isopropanol 400 tons

  What kind of sulfur compounds should we expect to be emitted from pyrolysis and combustion of such organosulfur compounds?

Note the authorities claim the smoke analysis didn't show any $\ce{H2S}$ and that the $\ce{SO2}$ level wasn't higher than everyday measurements in cities.

There is a paper on the pyrolysis of polyethylenesulfide showing cyclization leads to a bunch of products, I wonder if it is applicable here  (also I would like to know why no $\ce{H2S}$ is formed?)

I'd expect $\ce{H2O, CO, CO2}$, and $\ce{SO2}$ to be created during the combustion of these compounds. I'd wager a decent amount of $\ce{SO2}$ was created, but was diluted in the atmosphere.

Regarding $\ce{H2S}$, maybe it was produced, but then reacted according to the following:

$$\ce{H2S + O2 -> SO2 + H2O}$$

Additionally, there were probably a wide variety of odd sulfur compounds, but their existence was transient as they reacted to form $\ce{SO2}$ for instance. Furthermore, the authorities may not even be looking for these more complex molecules.


Photoelectric effect and electron loss I've read that when a certain amount of UV light is shone on a metal surface, electrons are ejected (the photoelectric effect).

Are these electrons from the metal atoms themselves? and in case they are - what happens to the metal atoms seen as they have lost an electron? (Are they chemically different?)

It is NOT the amount of light (number of photons), it is the energy of each photon (its frequency).  When the energy of a photon exceeds the vacuum work function of the surface, an electron is unbound.  There obtains a charged surface and a free electron.  Unless the surface is part of a circuit (e.g., grounded or connected to the cathode), the next electron out must overcome the vacuum work function plus the electrical potential.

The electron is emitted from the conduction band of the bulk metal surface.  The positive charge is delocalized across the surface if it is of constant radius of curvature.   Surface electric field divergence is inversely proportional to radius of curvature.  Unless the surface is very smooth you can get cold cathode emission or electrical breakdown of the encapsulating medium.


Why does increasing number of salt bridges increase voltage of electrochemical battery? In an experiment, I set up a cell with lead nitrate (w/ lead electrode) and zinc sulfate (w/ zinc electrode), with a salt bridge containing potassium nitrate.

I observed that by increasing the number of salt bridges, the voltage measured increased. However, given what I know about electrochemical cells, I can't work out why. I initially thought it would have no effect on the voltage/potential difference.

The voltage you measure between the terminals of a voltaic cell will depend on two factors:

The intrinsic maximum voltage $(V_\mathrm{max} = E_\mathrm{cell})$ that the cell could produce, depending on the $E^o_{red}$ of each half cell, the ion concentrations and the temperature. This is calculated from the Nernst equation:
$$E_\mathrm{cell} = E^⦵_\mathrm{cell} - \frac{RT}{zF}\ln{Q_r}$$

Resistive terms that come from within the cell, including overpotentials at the electrodes and other factors that influence the internal resistance of the cell.

Overall, the observed voltage $(V_L)$ measured at the terminals of the cell will be lower than the theoretical $V_\mathrm{max}$ because the internal resistance of the cell involves a voltage drop, in the same way that you would see an external voltage drop across a resistor in series in the circuit.
$V_L$ can be calculated in the following way:
$$V_L = V_\mathrm{max} - V_I$$
where $V_I$ is the internal voltage drop from the internal resistance of the cell. This drop in voltage depends on the internal resistance of the cell ($R_I$) and the total current $(I)$ running through the cell (and the circuit, see farside.ph.utexas.edu):
$$V_I = I \cdot R_I$$
Your question seems to revolve around how the salt bridge affects the internal resistance of the cell. It would make sense that the more salt bridges you add, the less internal resistance would result, so the smaller the voltage drop and the larger the measured voltage from the cell.
It is possible to think of the salt bridge, which involves the flow of ions, as being analogous to a wire, which involves the flow of electrons. The resistance of a wire $(R)$ is dependent on it cross-sectional area (see http://hyperphysics.phy-astr.gsu.edu):
$$R = \frac{ρL}{A}$$
where $ρ$ is resistivity; $L$ is length; $A$ is cross sectional area.
So it follows that the same may be true for a salt bridge. By adding more salt bridges, you are increasing the cross-sectional area, decreasing the internal resistance and increasing the observed voltage, which would approach the theoretical maximum voltage as the internal resistance approaches zero.


Identification of aqueous solutions of sodium hydroxide, sodium acetate and potassium thiocyanate using nickel(II) I'm very new to chem — taking an intro to chemistry course where we are currently focused on substance identification. I have two substances, both aqueous solutions of either $\ce{NaOH},$ $\ce{KSCN},$ or $\ce{NaCH3COO}.$ Both substances are colorless with a pH higher than 7.
When mixed with $\ce{NiCl2},$ substance A forms a precipitate, and substance B does not. I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{OH-}$ ions, and I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{CH3COO-}$ ions, so if $\ce{KSCN}$ was not a potential substance, I would know that substance A was $\ce{NaOH}$ and substance B was $\ce{NaCH3COO}.$
However, I am confused as to how $\ce{KSCN}$ fits into this. It's my understanding that since $\ce{SCN-}$ is the anion of thiocyanic acid, which is a weak acid, alkali metal compounds with $\ce{SCN-}$ anions are weak bases.
I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form a precipitate. Is substance A $\ce{KSCN}$ or $\ce{NaOH}?$ Is there even a way to distinguish the two with this data?

I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are
exposed to $\ce{OH−}$ ions

Yes, you are right, they will indeed form a green precipitate, nickel(II) hydroxide:
$$\ce{NiCl2 + 2NaOH -> Ni(OH)2 + 2NaCl}$$
Nickel(II) hydroxide does not dissolve in excess $\ce{NaOH}$.

I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions
are exposed to $\ce{CH3COO−}$ ions

I haven't found any reaction like that to form nickel(II) acetate as it is generally formed from the reaction of nickel(II) carbonate and acetic acid but a little bit of searching gave me a site which shows the formation of nickel(II) acetate from various nickel(II) salts and acetic acid and various salts of acetic acid. The closest reactions I could find from that site is the reaction between acetic acid/lead acetate and nickel(II) chloride and reaction between sodium acetate and nickel(II) nitrate. Technically, the reaction can be written and balanced on pen-paper as it is a simple double displacement reaction but the reaction hasn't been performed practically. Also, I don't know the credibility of the site, so take that site as a reference in your discretion.

I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form
a precipitate

Theoretically, it can be formed from a double displacement reaction, but practically it is found to be unfeasible as the end products get contaminated with unreacted reactants and byproducts to the point where yield is somewhat negligible. From a paper1:

Compounds containing nickel(II) thiocyanate are prepared mostly by metathesis between nickel(II) chloride or nitrate with alkali metal
thiocyanates. However, in such syntheses if precautions are not
taken, the final product may be contaminated with starting
materials. Methods known for preparing this compound include:
treatment of a dilute thiocyanic acid solution with nickel hydroxide
or carbonate, followed by evaporating the mixture at $\pu{150 ^\circ C}$; mixing and boiling aqueous solutions of ammonium thiocyanate and
barium hydroxide, followed by addition of an aqueous nickel sulphate.

Conclusion
A: $\ce{NaOH}$
B: $\ce{CH3COONa}$, as it is somewhat a weak base, pH = 8.9
Reference

Sarma, K.P., Poddar, R.K. A convenient method of preparing nickel(II) thiocyanate and its use in synthesis. Transition Met Chem 9, 135–138 (1984). DOI: 10.1007/BF00935928 (PDF)


Reference value for two-electron repulsion integral over GTO's I am currently trying to implement a Full CI program from scratch. The energies I get are a bit too high, so I'm looking for the mistake.

One possibility is my implementation of the two-electron repulsion integrals,
$$\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{e^{-\alpha_1 \boldsymbol{r}_1^2} e^{-\alpha_2 \boldsymbol{r}_1^2} e^{-\beta_1 \boldsymbol{r}_2^2} e^{-\beta_2 \boldsymbol{r}_2^2}} {\left| \boldsymbol{r}_1-\boldsymbol{r}_2 \right|\ } \mathrm{d}\boldsymbol{r}_1^3 \mathrm{d}\boldsymbol{r}_2^3 $$

with $\boldsymbol{r}_1=(x_1,y_1,z_1),  \boldsymbol{r}_2=(x_2,y_2,z_2).$

However, I cannot find any values I could compare my results to. I tried (numerical) integration in Maple, but that's way too slow and/or numerically unstable because of the singularity at  $\boldsymbol{r}_1=\boldsymbol{r}_2$. I tried installing the libraries libint and Libcint, and the python module pyscf, which all should be able to do this kind of computations, but I horribly fail at installing things not made for Windows (MINGW is only half working, and I don't have a proper Linux installation available right now...).

So, could someone who has such a program installed please give me a number this integral evaluates to, for whatever numbers $\alpha_1 \ldots \beta_2$ ?

You don't state which method you implemented in order to compute two-electron integrals, therefore I will list all the main references at first.

Cook's book [1] contains analytical formulas for overlap, kinetic, electron-nucleus attraction and electron-electron repulsion integrals. The analytical formula for electron-electron repulsion integrals is wrong in the book, but you can look at this discussion for the errata.

However, a Full CI calculation is computationally costly. For this reason I believe a better approach would be to use a more efficient scheme to compute integrals. You can look at the following:

Obara-Saika scheme
McMurchie-Davidson scheme
Rys quadrature

You can find a good explanation of these methods in Ref. [2].

Now, if you want to check what you already implemented, you can find a list of the two-electron integrals in the STO-3G basis set for $\ce{HeH+}$ in Szabo's book appendix [3]. Here I can give you a set of two-electron integrals I obtain for $\ce{H2}$ (always in the STO-3G basis set) for a bond length of $1.4$ Bohr:

 (           1           1           1           1 )   0.77460834925515787
 (           1           1           1           2 )   0.44410904384277344
 (           1           1           2           1 )   0.44410904384277350
 (           1           1           2           2 )   0.56967771733030592
 (           1           2           1           1 )   0.44410904384277361
 (           1           2           1           2 )   0.29702946944511982
 (           1           2           2           1 )   0.29702946944511982
 (           1           2           2           2 )   0.44410904384277333
 (           2           1           1           1 )   0.44410904384277333
 (           2           1           1           2 )   0.29702946944511982
 (           2           1           2           1 )   0.29702946944511982
 (           2           1           2           2 )   0.44410904384277361
 (           2           2           1           1 )   0.56967771733030592
 (           2           2           1           2 )   0.44410904384277350
 (           2           2           2           1 )   0.44410904384277344
 (           2           2           2           2 )   0.77460834925515787

Note that these results came from a program I wrote myself, but it usually match very well Szabo's [3] and Gaussian09 values for the total energy. If these results match your calculations, then the problem might be in integrals with higher angular momentum.

[1] Cook, Handbook of Computational Chemistry, Oxford University Press, 1998.

[2] T. Helgaker, P. Jørgensen and J. Olsen, Molecular Electronic-Structure Theory, Wiley, 2000.

[3]  A. Szabo and N. Ostlund, Modern Quantum Chemistry, Dover, 1996.


Butanone or butan-2-one? According to IUPAC rules, am I right in assuming that butanone is sufficient, because there's only one possible molecule? I keep noticing butan-2-one, so I want to make sure.

The preferred IUPAC name is butan-2-one. This is explicitly mentioned in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the Blue Book:

P-14.3.4.1 terminal locants
P-14.3.4.2 the locant ‘1’
P-14.3.4.3 where there is only one kind of substitutable hydrogen
P-14.3.4.4 when no isomer can be generated by moving suffixes or prefixes from their position to another or by interchanging them between two different positions
P-14.3.4.5 when all substitutable positions are completely substituted
P-14.3.4.6 when all substitutable hydrogen atoms have the same locant.

None of these cases applies to butan-2-one. Therefore, the locant cannot be omitted in the preferred IUPAC name.


Why do Magnesium and Lithium form *covalent* organometallic compounds? Lithium and magnesium are Group 1 and Group 2 elements respectively. Elements of these groups are highly ionic, and I've never heard of them forming significantly covalent inorganic compounds.

Yet these elements form a variety of organometallic compounds ($\ce{PhLi}$, the whole family of Grignard reagents, etc). Organometallic compounds have significant covalent character (i.e., the bond can be called covalent)  in the carbon–metal bond.

What's so special about carbon that makes these elements form covalent bonds?

The character of the bond is determined by the electronegativity of the atoms.

Speaking of bonds as purely ionic or covalent is not always correct - usually it is more correct to say that a bond has ionic or covalent characteristics.

So comparing the difference in electronegativities gives us the following:

$$\begin{array}{cc}\hline
\text{Difference in electronegativity} & \text{Type of bond} \\ \hline
< 0.4 & \text{Non-polar covalent} \\
0.4 \mathrm{-} 1.7 & \text{Polar covalent} \\
>1.7 & \text{Ionic} \\ \hline
\end{array}$$

At the upper end of the polar covalent spectrum, the bonds frequently have both covalent and ionic characteristics.

For organometallic compounds, the difference in electronegativity between Li and C is $1.57$.  So while this is still in the polar covalent range, it is also close to ionic.  Similarly, the difference between Mg and C is $1.24$ - again, a very polar covalent bond.

Compare this to the difference between H and C ($0.35$) - a non-polar covalent bond.

So to answer your question, the thing that is "special" about carbon is that it has a fairly low electronegativity compared to the chalcogens and halogens.  Granted, bonds with carbon are also going to be weaker than in say LiCl, but that's what makes organometallic compounds actually work to form carbon-carbon bonds.

(The electronegativity values came from wikipedia's great chart)


Why do we write NH3? We've learnt that the electropositive element is written first. Then why is ammonia written as $\ce{NH3}$ ?

According to current nomenclature rules, $\ce{H3N}$ would be correct and acceptable.  However some chemical formulas, like $\ce{NH3}$ for ammonia, that were in use long before the rules came out, are still accepted today.


Why does silver sulfate not precipitate in this double displacement reaction? Aqueous silver nitrate is mixed with aqueous sodium sulfate.  We were asked to perform this reaction in lab in my chemistry class, and we did not observe a precipitant being formed. However, the equation indicates otherwise:

$$\ce{2AgNO3 (aq) + Na2SO4 (aq) -> Ag2SO4 (s) + 2NaNO3 (aq)}$$

This shows the formation of $\ce{Ag2SO4}$, silver sulfate. I found in the Wikipedia article for silver sulfate a reference to "ruby red illumination" and thus am led to believe that, like some silver-based substances used in photographic film, the precipitant might have been immediately been affected by light in the room and thus decomposed into something soluble in water. I have not, however, been able to find references to this anywhere else; I have found here a reference to it being classified as "sparingly soluble," but I haven't the slightest idea what this means.

Since the lab is past due now, I'll give what I think is the answer.

$\ce{Ag2SO4}$ is somewhat soluble in water. It is most likely that you simply didn't get enough to cause precipitation.

The other possibility, which I don't think applies here, is that you have a supersaturated solution. There are some precipitates for which crystals are just stubborn to nucleate.


Electrolysis of molten Calcium Chloride: Identifying the product I was trying to make Calcium metal by electrolyzing molten $\ce{CaCl_2}$. I used a stainless steel spoon and a thick steel wire as electrodes. Then I attached them to a power supply. I placed some $\ce{CaCl_2}$ in the spoon and melted with a gas torch. I then placed the steel electrode in the molten pool and gradually pulled it out. A black glassy substance formed on the rod. I ran water over it and it fizzed as I would expect from Calcium metal, however, when it was exposed to air for a period of time it start to "melt" as it absorbed moisture from the air. As I was breaking it off of the electrode, it shattered and I could see that it was the glassy substance all the way through.

I am trying to figure out what I have made. Could it possibly be a mixture of unreacted $\ce{CaCl_2}$ and a small amount of Calcium?

I would hypothesize that the "black glassy stuff" is nothing more than glassified $\ce{CaCl_2}$. Metal is an excellent heat conductor and the liquid $\ce{CaCl_2}$ likely cooled rapidly to form glass when the cold wire was inserted.

Two things could be done to avoid this:

1.) heat the wire before inserting it into the mixture of molten $\ce{CaCl_2}$.

2.) heat the $\ce{CaCl_2}$ well above its melting point.

If you are determined to make Calcium metal it might be worth electrolysing an aqueous solution of $\ce{CaCl_2}$ to evolve the $\ce{Cl2(g)}$ (do this responsibly outside/in fume hood of course as $\ce{Cl2(g)}$ is poisonous.) Calcium Hydroxide should be the result. Evaporating the water will yield solid $\ce{CaOH}$, which should be much easier to electrolize using your method since it has a melting point of 580°C vs. 772°C for $\ce{CaCl_2}$.


Heating p-Bromobenzenesulfonic acid in the presence of dilute HCl How do I predict the outcome of this reaction?

I expected this to be an acid base reaction leading to the formation of the weak acid para-chlorobenzenesulfonic acid. However, the answer given in the worksheet was surprising $\ce{-SO3H}$ was eliminated and the product stated was $\ce{C6H5Br}$. I would like to be acquainted with the mechanism of this unusual  reaction. I couldn't find it in Clayden or Solomon Frhyle.

I think @ron has hit the bull's eye in the comments! I would like to elaborate upon that:

Since each step in the sulfonation of benzene is an equilbirum, sulfonation is a reversible reaction. According to Le-Chatlier's principle addition of $\ce{H+}$ ions in the solution would shift the equilibrium to the left in the last step of sulfonation where arenium is deprotonated, hence addition of $\ce{HCl(aq)}$ accompanied with heating for kinetic requirements results in the production of 4-bromobenzenesulfonic acid in the given reaction.  This process is called protodesulphonylation.


What is the internal energy at constant pressure? What is the internal energy when a reaction is run at constant volume and $9.19\ \mathrm{kJ}$ of heat is absorbed, when run at constant pressure $8.62\ \mathrm{kJ}$ is absorbed? (for an ideal gas)

I know that internal energy for constant volume is equal to the heat then $\Delta U=9.19\ \mathrm{kJ}$, but what is the internal energy for a reaction at constant pressure? I know that $\Delta H=\Delta U+p\Delta V$ and from this I know that $\Delta H=8.62\ \mathrm{kJ}$ but how does it relate to the internal energy?

If you are working with an ideal gas mixture, for which the internal energy of the reactants and products is not a function of pressure, the change in internal energy in going from reactants to products at constant temperature and volume is the same as the change in internal energy in going from reactants to products at constant temperature and pressure.  With regard to the changes in enthalpy, the same is true.  If the number of moles of reactants and products is the same, then the change in enthalpy for each of these processes will be the same as the change in internal energy.  However, if the number of moles changes, the change in enthalpy will differ from the change in internal energy in proportion to the change in the number of moles.


Why do molecules have to first transform to gas before doing a reaction? I have to calculate the enthalpy of the reaction
$$\ce{Cs(l) + 1/2I2(s) -> CsI(s)}$$
and the answer is
$$\Delta H_f = \Delta H_{vap}^{\ce{Cs}} + \frac{1}{2}\Delta H_{subl}^{\ce{I2}} + \frac{1}{2}E_{diss}^{\ce{I2}} + I_1^{\ce{Cs}} + E_A^{\ce{I}} + E_N.$$
I can see that the molecules are first put to gas state before creating an ionic bond.

Why does it have to do it ?
In which case do the molecules have to first go in gas state before reacting ?
Is it only in case of ionic bond ?

Notes:

$E_{diss}$ is the energy of dissociation between the to atoms $\ce{I}$.
$I_1$ is the energy of first ionisation.
$E_A$ is the energy of electron affinity.
$E_N$ is the energy of neutralisation between $\ce{Cs+}$ ans $\ce{I-}$.

Not all reactions require vaporization. This process looks like the Born-Haber cycle for cesium iodide, an exercise in Hess's Law usually used to fill in one of the missing values in the cycle that might be hard to measure (usually lattice energy, but not always). The Born-Haber cycle can also be used to drive general chemistry students crazy.

The $E_N$ "energy of neutralization" looks like the lattice energy, which is defined as the enthalpy/energy change accompanying the formation of a solid crystal lattice from the gas phase ions. In your case, the lattice energy of $\ce{CsI}$ is defined as the energy change accompanying the following reaction:

$$\ce{Cs+ (g) + I^- (g) -> CsI(s)}$$

Because the lattice energy is defined for gas phase ions, you need the vaporization enthalpies for this process. Not all reactions require this vaporization step.


How can I solve a non-cubic crystal structure? Say I want to solve the crystal structure of barium sulfate, and I want to not only determine crystal type and the lattice parameter, but also the orientation of the oxygen atoms on the sulfate anion. How do I proceed? All I have is the powder X-ray diffractogram (i.e. intensity vs $2\theta$ for the compound).

I don't want to use any existing software as this is more about learning how crystal structures are solved. I have looked at books and they seem to be incredibly difficult to follow.

The lattice constants and the symmetry of the lattice determine the position of the diffraction peaks observed along the $2\theta$-scale.  The intensity of the diffracton peaks, leaving out increase by accidental or systematic superposition of them, is due to the arrangement within the asymmetric unit, i.e. the motif which -- after application of the symmetry operators (centres of inversion, proper axes, rotoinversion axes, glide planes, mirror planes) -- builds the unit cell.

So you face the phase problem (as single crystal diffraction analysis does, too). If you do not want to rely on already existing software (like DASH, MAUD, Jana, or FOX to mention a few), you will look up the Rietveld refinement.  Beside by the comment by @porphyrin, as shown in this lecture briefly providing some background, the numerous properties of sample and experiment (including their statistics) equally lead me to the recommendation to use software already proven to work.  The International Union of Crystallography maintains a software directory, along with -- under their educational umbrella -- the notes of a dedicated workshop about powder diffraction.

Perhaps contacting a local crystallographer may be helpful, too; like via the American Crystallographic Association ACA.


Are certain substances in furniture truly toxic? I'm looking to purchase a sofa, but have been bombarded with all kinds of internet info about the toxicity of certain materials used to make sofas.  For example, polyurathane foam is labeled as toxic--is this true?  Is something like a blend of cotton/polyester and polyurethane toxic?

The polyurethane would have to be absorbed through your pores when you're in contact with the sofa. That's not going to happen easily?
Theres a few tangent articles about the safety of these so called "cheap" sofas and mattress on livestrong and other websites.
I'd say it's not a concern. Sofa on bro.


What is equivalents per mole? Which has maximum number of equivalents per mole of the oxidant?
  $$
\begin{align}
\ce{MnO4- + H2O2 &-> Mn^2+ + O2}\tag{i}\\
\ce{Cr2O7^2- + Fe^2+ &-> 2 Cr^3+ + Fe^3+}\tag{ii}
\end{align}
$$

The answer is the second option. I wanna know why

If there's another term different from equivalent mass like equivalents, then I tried doing it using the formula = No.of moles of oxidant x n factor (no of electrons gained)

The question can be understood as:
Which one of the following oxidants has a greater n-factor as there's only 1 mole each of permanganate ion and dichromate ion.

The n-factor of permanganate ion is (7-2)=5

while

The n-factor of dichromate ion is (6*2-3*2)=6

So, the answer is the 2nd reaction.


Physical meaning of (negative) pressure in molecular dynamics simulations Lately I have been simulating a system in an NPT-ensemble where $N$ is constant and $P$ and $T$ are coupled to a Berendsen barostat/thermostat. The system is cubic and periodic in all directions.

I am trying to fathom how the pressure affects my system, why it fluctuates so much and how negative pressure should be interpreted in systems such as the one described above.

I have some knowledge of classical MD, and it is my understanding that (in general three dimensional system), the total pressure of the system is computed with the virial equation

\begin{equation}
P=Nk_{b}TV^{-1}+\frac{1}{3}\bigg<\sum_{i=1}^{N}r_{i}\cdot F_{i}\bigg>V^{-1}\tag{1}
\end{equation}
where $N$ is the number of particles, $T$ is temperature and $k_{b}$ is the Boltzmann constant.

By looking at equation 1, it is not fully clear to me what negative pressure would imply, so I looked how the equation is derived.

The derivation starts from the Clausius virial function

\begin{equation}
W(r_{1},...,r_{n})=\sum_{i}^{N}r_{i}\cdot F_{i}^{TOT} \tag{2}
\end{equation}

where $F_{i}^{TOT}$ is the total force acting on particle $i$.

This can be averaged over the whole MD trajectory, and integrated to obtain

\begin{equation}
\big<W\big> = -\lim_{t\rightarrow 0}\frac{1}{t}\int^{t}_{0}d\tau\sum_{i=1}^{N}m_{i}|r'(\tau)|^{2} \tag{3}
\end{equation}

from which, by using equipartition of energy, one can obtain

\begin{equation}
\big<W\big>=-3Nk_{b}T \tag{4}
\end{equation}

where 3 is the dimensionality, $N$ is the number of particles, $T$ is temperature and $k_{b}$ is the Boltzmann constant.

The force exerted by the container walls can be taken in to account by 'splitting' the $F_{i}^{TOT}$ term in to internal and external part, and then
evaluate $\big<W^{EXT.}\big>$ using:

\begin{equation}
\big<W^{EXT.}\big> = 3L³(-P) = -3PV\tag{5}
\end{equation}

Where $L$ is the length of the cubic box (so that $L_{x}~=~L_{y}~=~L_{z}~=~L$)

Now, since $\big<W\big> = -3Nk_{b}T$, one can rewrite it as:

\begin{equation}
\bigg<\sum_{i=1}^{N}r_{i}\cdot F_{i}\bigg> - 3PV = -3Nk_{b}T\tag{6}
\end{equation}

and get the original virial equation (1).

My questions are:

Is this correct? Mathematics (unfortunately) is not yet my strong suit, and I know that my reasoning (even with great help from different MD-related tutorials and sites) is prone to errors.
Could you explain the negative pressures in MD simulation simply by arguing that the force exerted by the container walls is negative?
If so, can one just argue that because the force is negative, the volume of the container must be increasing?

Thank you for your time.

Looks fine. 2. I'd phrase that more generally as the simulation volume would tend to expand. 3. But can only do so if the ensemble permits it (pressure can be negative in NVT).

The fluctuations are a consequence of making an observation of a macroscopic quantity over small numbers of degrees of freedom, or time. See http://www.gromacs.org/Documentation/Terminology/Pressure


Why the formation of PH4+ is difficult compared to NH4+? My textbook asks this as an objective question.
What I had in mind was that the lone pair on phosporous was more distributed as compared to nitrogen(since nitrogen is more electronegative and smaller in size ) and hence the statement. But the option given in the textbook says the answer is lone pair of phosphorous resides in almost pure s orbital.
How do you know if the lone pair is placed in pure s orbital or pure p orbital

The bond angle in $\ce{PH3}$ is about $92-93^\circ$, and by the percentage $s$ character formula i.e $\cos(\theta) = s/(s-1)$ you can calculate the $s$ character and that comes out to be 6%. And that where drago rule holds true , that is the orbitals involved in bonding are pure $p_x$ ,$p_y$, $p_z$ and as a character is high the lone pair is present in pure s orbital and highly penetrated towards phosphorus. Hence acts as a poor Lewis base.