Why are there 3 columns in Group VIII of the Periodic Table? I'm currently reading in-depth about the layout of the Periodic Table, and I wondered why the table has 3 columns in its Group VIII:

As I understand, this is an old notation of the table, now deprecated. But why was Group VIII a triple-sized group?

The answer might be trivial, but it's interesting.

In the beginning of the $20$th century, the periodic table had $8$ columns, and not $18$ as today. Electrons and of course $spdf$ electrons were unknown. The numerous transition elements were included in this table by doubling the line between Argon and Krypton, then between Krypton and Xenon. It was going this way.
The $4$th period was made of two lines. So each square (each box) contained two elements written on two lines. The first seven atoms $\ce{_{19}K - _{25}Mn}$ occupied the columns $\ce{I - VII}$, and their oxidation state may always have been equal to the number of the column. Then the three atoms $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ did not belong to any column. We will come back to them later on. Then the seven next atoms $\ce{_{29}Cu - _{35}Br}$ had to be put in columns $1$ to $\ce{VII}$, under the corresponding elements : $\ce{_{29}Cu}$ in the same box and under $\ce{_{19}K}$, $\ce{_{30}Zn}$ in the same box and under $\ce{_{20}Ca}$, etc. up to $\ce{_{25}Mn}$ above $\ce{_{35}Br}$. With this system all atoms had at least one oxidation state equal to the number of their column. The same procedure was adopted for $5$th period, with $\ce{_{37}Rb}$ and $\ce{_{47}Ag}$ in the first column,$\ce{_{38}Sr}$ above $\ce{_{48}Cd}$, etc.
This table was useful, except nobody knew what to do with the three atoms  $\ce{_{26}Fe}$, $\ce{_{27}Co}$ and $\ce{_{28}Ni}$ having all valence $2$, and of course of the six corresponding atoms of the platinum group, immediately under them : $\ce{_{44}Ru}$ to $\ce{_{46}Pd}$ and $\ce{_{76}Os}$ to $\ce{_{78}Pt}$. There was plenty of discussion at that time about what to do with them. Usually these elements were "rejected" in the 8th column, with the noble gases. or they were put in a separate column made specially for them. This problem was finally solved with the discovery of the $spdf$ electrons.


What is "Woods Metal" I have a piece of what I have been told is "Woods Metal" it melts at 168F.  I am "curious" about how they make it, how low can the melting temperature be?

Here is some additional info from web

Wood's metal, also known as Lipowitz's alloy or by the commercial names Cerrobend, Bendalloy, Pewtalloy and MCP 158, is a eutectic, fusible alloy with a melting point of approximately 70 °C (158 °F). It is a eutectic alloy of 50% bismuth, 26.7% lead, 13.3% tin, and 10% cadmium by weight.

I have experience using the material for fixturing of difficult to secure /  shaped parts for machining. It can also help in fixturing/holding - then later melted away.

The eutectic composition will privide lowest Tm and get the info from a phase diagram.

Do not add mercury for safety reasons.


$P 4_{2} 22$ tetragonal group choice of origin for symmetry of special projections I was looking at the description of the tetragonal group $$P 4_{2} 2 2$$ (No 93 in the International Tables of Crystallography) and there is one aspect that I do not understand. Namely, what is the convention and implications of choosing the origin point? For example, for the two fold rotation along the [110] plane, the origin is choosen to be $$(0, 0, \frac{1}{4})$$ and thus we should have the following transformation:
$$x \rightarrow y, y \rightarrow x, z \rightarrow -z + \frac{1}{2}.$$
Why the origin is not set at $(0, 0, 0)$ and then one would have:
$$x \rightarrow y, y \rightarrow x, z \rightarrow -z?$$

As stated in the International Tables of Crystallography vol A, as a rule the origin is locatedat a centre of symmetry or at a point of high site symmetry. Anyway, the choice of such a coordinate system is not mandatory since in principle a crystal structure can be referred to any coordinate system. In particular, all centrosymmetric space groups are described with an inversion centre as origin, whereas for noncentrosymmetric space groups, the origin is at a point of highest site symmetry.
In the space group n°93 the origin is at the intersection of 3 different 2-fold rotation axes (site 2a: 0,0,0), not at 0,0,1/4. I think this you're confused on this point, since the origin is always at 0,0,0 by definition.
From the trasnformation that you have assumed (y,x,-z+1/2), I guess you are referring to the operation number (7), that is a 2-fold rotation along the crystallographic [x,x,1/4] axis (not a plane! rotations are along an axis). As stated above, you can choose another coordinate system with a new origin shifted in respect the standard one, but any case the new origin will have again 0,0,0 coordinates in the new coordinate system. If you shift the origin, the coordinates of all the symmetry operations will be obviously redefined in the new coordinate system, since they are not independently distributed in space.


Spontaneous combustion of sulfur in a steel barrel Got some free sulfur.  Went to the site, and they put a pallet with 3 barrels on it.  One of the barrels was badly corroded, but looked like I could get it home.  I had plastic barrels at home that I could transfer the material to.

About 10 km from home I noticed I was leaving a trail of smoke.  Got in back with a shovel, and found that I had spots of burning sulfur. I smothered as best I could, got home, and spent the next two hours unloading and putting out blobs of burning sulfur.  Did a number on my truck box lining in a few spots.

The lids continued to smoke -- it was a very slow motion fire, with some form of corrosion that would burn across the surface at around an inch an hour.

The smoke smelled strong of sulfur dioxide.

The bulk of the sulfur was unchanged by all this, and I was able to get it into plastic pails and barrels with no problem.

all the barrels had been stored outside, and in the non-corroded barrel had standing water on top of the sulfur.

What happened in storage?  Since sulfur always seems to have an SO2 odour, I assume that either some bacteria makes a living processing S to SO2, or there is a very slow reaction at normal temperatures.
What corrosion product would be flammable?
Why would it start to burn when transported, instead of in the former owner's parking lot?

Sulfur combustion on barrel:

A different lid:

Iron sulfide FeS is pyrophoric: flammable, and while it can start just on exposure to air, the presence of UV light will help.  As is usual, one of the key components is the surface to volume ratio.  A massive lump is unlikely to catch fire.  A crusty scale of FeS, such as forms in tanks used to store sour oil (H2S content) is more likely to combust when air is let into the tank.

Deactivation of Pyrophoric Iron Sulfides

From the descriptions in the abstract above, I suspect that I had one of the other iron sulfides.  Whatever I had did not burn fast.


What are the implications of having different conformations? I have seen videos explaining what conformations are. I am curious about whether different conformations can give rise to slightly different properties. Particularly of interest is whether you can modify the smell of something by just rotating some bonds.

That's an interesting question. To start off, what Mith said in the comments is absolutely right. Let's say that two conformations of a molecule have different smells (which is in fact not an absurd proposition by any means). The problem is that, because there is a very low barrier to the interconversion of the conformations, the individual conformations do not really hang around long enough to be individually detected; you can only really detect a weighted average of the conformations.

To give you an example, you can devise di-substituted cyclohexanes that have two predominant chair conformations. These can be investigated by proton NMR spectroscopy, since axial-axial coupling constants are larger than equatorial-axial or equatorial-equatorial couplings. From the integrals of the relevant peaks you can determine the relative proportions of both conformations.

However, if you somehow managed to synthesise a sample of the pure diequatorial conformer (the one in blue) and put it into a NMR tube, some of it would have converted into the diaxial conformer (the one in red) before you could even run your NMR. In fact, there's probably no way you can synthesise a pure conformer to begin with.

Likewise, if you replace the NMR machine with your nose, there's no way you can open a bottle of a perfume X, with conformation X1, and smell something, then open a bottle of the same perfume X with conformation X2 and smell another thing. They would both smell the same because conformers aren't isolable. (Well, not unless you are working in an argon matrix at temperatures near absolute zero or something like that!)

Mith then goes on to say that you can bind to a molecule to manipulate its conformation. That's the more interesting part, in my opinion at least! If you think about it, it's going to be very hard for us to devise some kind of surface that particularly favours one conformation over another. You'd need a very complex binding partner, with a very specific 3D shape, that "fits" with one conformation but not another conformation.

Does that ring a bell? It might, because nature has made those binding partners for us already. That previous paragraph is a perfect description of an enzyme: a protein that catalyses a biological reaction. Lysozyme is one of the most famous enzymes; it was the first enzyme whose structure was successfully solved by X-ray diffraction. It is also a great example of an enzyme that works by manipulating conformations.

Lysozyme binds to a series of carbohydrates in the peptidoglycan cell walls of some bacteria. Together with water, it hydrolyses the glycosidic bonds (a fancy name for an acetal) in these carbohydrates:

(image from Wikipedia)

I'm not going to go into the full mechanism of this, but the striking thing is that upon binding to lysozyme, the "D ring" (which is the NAM residue on the left-hand side of the diagram above) is distorted from its usual chair conformation into a half-chair conformation. This is because it "fits" better into the exact 3D conformation of the enzyme (not shown in the diagrams):

(image modified from Wikipedia)

This makes the ring more unstable and prone to hydrolysis (strictly speaking, more prone to nucleophilic attack by the aspartate-52 residue). Without this manipulation of the conformation, it's likely that the hydrolysis cannot proceed.

So, there you have it! One way in which the conformation of a molecule, upon binding to another molecule, affects its properties.


Should the relatively low solubility of a base be considered in the calculation of the pH? Hydrochloric acid, $V=75.0~\mathrm{mL}$, $c=0.250~\mathrm{M}$, is added to $225.0~\mathrm{mL}$ of $0.0550~\mathrm{M}$ $\ce{Ba(OH)2}$ solution. What is the concentration of the excess $\ce{H+}$ or $\ce{OH-}$ ions left in this solution? (source)

I know the balanced equation but when considering $\ce{Ba(OH)2}$ is 'marginally soluble' which
I guess could be considered insoluble, $\ce{OH-}$ wouldn't really be in the result solution.
Am I right or at where am I wrong? thanks for reading

The $K_\mathrm{sp}$ of  $\ce{Ba(OH)2}$ is $0.005$.

Therefore, the solubility is great enough that 0.055M  $\ce{Ba(OH)2}$ can all be in solution.

Secondly, you are diluting the $0.055~\mathrm{M}$ solution, so the concentration will be even less.

Finally, even if there was too much  $\ce{Ba(OH)2}$ to dissolve in water, the $\ce{HCl}$ would react with the $\ce{OH-}$ that is in solution, causing more  $\ce{Ba(OH)2}$ to dissolve.


What is chemistry considered the study of? I am asking this because I have a review question in my physics class which gives me a multiple-choice question.  I have heard and seen elsewhere on the internet (such as wikipedia) that chemistry is the study of reactions and the properties of matter.  That is usually a satisfactory answer, but for whatever reason this question makes me choose between reactions, properties, and matter.  Now I find that the general definition I accepted contains basically all of those, as far as I can tell.  While I don't necessarily need the answer, my curiosity has been aroused and I want to know why the question makes me choose between those answers.

The question is, what is the very basic idea behind chemistry?  Would it be classified as the study of reactions, properties, or matter (and you can only select one)?

While the answers so far are not wrong, I'd like to add a refinement. Chemistry is the study of matter at the level of electron interaction. The basic question in chemistry is about chemical bonding, and as far as we know this is a function of electron interaction. In contrast, radioactivity is also about matter, but is not at all about electron interaction and is not generally considered chemistry (it's considered a part of physics). Also, the chemistry of isotopes (i.e. differing neutron count) is not generally different across different isotopes. This is good, because we need solvents with are deuterium based ($\ce{D=^2H}$) as opposed to hydrogen based for NMR spectroscopy. If we go lower than electrons, we are into quantum mechanics, which is again physics. So electron interaction is a good level at which to characterize chemistry.

EDIT: I realize the original question was for a one-word answer. Frankly, that's not the greatest question, but it's generating answers that make it more interesting, and it's why I answered.

Let me give a counterexample to my own answer to show the difficulty: gases. We learn a lot about gases in chemistry which has nothing to do with chemical bonding; e.g. all the gas equations. I'd argue that most of this is not really chemistry, it's physics. Understanding how temperature, pressure, and volume interact dynamically is a very physical sort of study, not all that different from studying light, or magnetism, or force.  So why do we study gas physics in chemistry? I suspect it's because we use gas physics a lot in chemistry! You need to understand the gas laws to do any real experimentation involving actual gas chemistry; i.e. how would you measure gas as a reactant or product without knowing the gas laws?

So to be clear, if you want one word, I agree with "matter". But I think if you want to get at what chemistry is about, it's not so simple.


Why are van der Waals' forces between gas molecules strong near its critical temperature? My textbook says

  …easily liquefiable gases i.e. with higher critical temperatures are readily adsorbed as van der Waals' forces are stronger near the critical temperatures.

What I do understand is that at a temperature higher than the critical temperature $(T_\mathrm c)$ the gas molecules have higher kinetic energy which is why they're unable to form strong bonds.

But, what if the temperature is lower than the critical temperature? Or is my reasoning incorrect?

Critical temperature is  kind of measure of the strength of intermolecular van der Waals force.

Comparison of actual and critical temperature is kind of comparison of average kinetic energy of molecules with energy needed to break the intermolecular bonds.

So temperature itself does not say anything about the bonding strength. It only indirectly determine probability these bonds get broken by thermal molecular motion.


High errors in frontier orbital energy calculation using DFT In a recent paper by Faber et . al., Mean Absolute Error and DFT (B3LYP) error (relative to the experiments) for 9 properties of QM9 dataset molecules has been reported.
\begin{equation}
\begin{array}{l|ccccccccc}
{} & {\mu} & {\alpha} & {\varepsilon_{\text {HOMO }}}& {\varepsilon_{\text {LUMO }}} & {\Delta \varepsilon} & {\left\langle\mathrm{R}^{2}\right\rangle} & {\text { ZPVE }} & { U_{0}} & {C_{v}} \\
 \hline & {\text { Debye }} & {\text { Bohr }^{3}} & {\text { eV }} & {\text { eV }} &{\text { eV }} & {\text { Bohr }^{2}} & {\text { eV }} & {\text { eV }} &{\text {cal/molK}}\\ \hline
 {\mathrm{MAD}} & {1.17} & {6.29}& {0.439} & {1.05} & {1.07} & {203} & {0.717} & {8.19} & {3.21} \\
 {\mathrm{DFT}} & {0.10} & {0.4}& {2.0} & {2.6} & {1.2} &{-} & {0.0097} & {0.10} & {0.34} \\ \hline\end{array}
\end{equation}

I was wondering why the DFT results for frontier orbital energies
have such high errors in comparison with atomization energy $U_0$.
For instance, the error reported for $\varepsilon_{\text {HOMO }}$
is $2.0$ while the breadth of values is $0.439$. Now compare this
with the error $0.1$ for $U_0$ with breadth of values for the
dataset reported as $8.19$, which is much more diverse.
A study on different DFT methods has been conducted by Zhang and
    Musgrave, which discusses inaccurate computation of frontier
    orbital energies. Some works have proposed the use ML methods to
    predict these values with very low errors. Since DFT is Ab initio,
    would suggesting to use ML methods even make sense?

Edits:

In the first paper (by Faber et al.) we read

  Frontier molecular orbital energies (HOMO, LUMO and HOMO-LUMO gap) can
  not be measured directly. However, for the exact (yet unknown)
  exchange-correlation potential, HOMO and LUMO eigenvalues correspond
  to the negative of the vertical ionization potential (IP) and
  electron affinity (EA), respectively.

Based on this, I was wondering if DFT is the main approach to compute frontier orbital energies or this paper is just trying to find it using DFT in order to include the errors for DFT method.

IMHO, this particular comparison is a strange one.

You reference "Comparison of DFT Methods for Molecular Orbital Eigenvalue Calculations" by Zhang and Musgrave, and there are other similar comparisons between density functional orbital methods and ionization potentials and electron affinities.

In general, these Kohn-Sham DFT benchmark papers show that most functionals don't compute orbital energies that directly compare with ionization potentials and electron affinities. That is, in Hartree Fock, Koopmans' theorem suggests that the orbital energy of the HOMO should correspond to the negative of the ionization potential (-IP). In principal, for the exact density functional, a similar property should hold.

What does hold true (e.g., the Zhang and Musgrave paper) is that you can easily calibrate orbital eigenvalues from Kohn-Sham DFT methods to counteract systematic issues. Usually a linear regression is sufficient. (Computing the actual vertical ionization potential using two DFT calculations is generally even more accurate.)

Coming back to your question, I wouldn't be surprised if the absolute errors in orbital eigenvalues are mixed from DFT methods. But that's not how most research uses them - they either perform a calibration (e.g., linear regression) or are comparing multiple different molecules. In the latter case, the relative errors are much smaller (usually ~0.2 eV, IIRC - I'll look for some sources).

I believe the overall argument in the Faber et. al paper is that modern ML methods are very good. In some properties that's probably correct.


Which has higher lattice energy: MgO or LiF? Which has higher lattice energy: MgO or LiF ?

A compound has higher lattice energy if it's ions have smaller size and greater charge

Li cation is smaller than Mg cation => +1 for (LiF)
Fluoride anion is smaller than oxide anion => +1 for (LiF)
Mg cation has greater charge than Li cation => +1 for (MgO)
Oxide anion has greater charge than fluoride anion => +1 for (MgO)

Therefore, 'charge' factor favour MgO while 'size' factor favour LiF.

So, which one will have higher lattice energy?

$\ce {MgO}$ ($\approx 3800$ $\pu{kJ mol^{-1}}$) has higher lattice energy than $\ce {LiF}$ ($\approx 1045$ $\pu{kJ mol^{-1}}$) mainly because of the greater charge on  $\ce{Mg^2^{+}}$ ion and $\ce {O^{2-}}$ as lattice energy is directly proportional to the charges of the combing atoms.

Your reasoning isn't incorrect but remember that $\ce{Li}$ and $\ce{Mg}$ show diagonal relationship so the size of their ions has only slight difference in magnitude. Similarly, oxygen and fluorine are placed next to each other in the periodic table so their anions don't have a large difference in size.

Thus, the "charge factor" dominates.


Why can alcohol undergo elimination just by sulphuric acid? In both $\mathrm{E1}$ and $\mathrm{E2}$ reactions with alcohol, loss of a leaving group and proton transfer are required. And proton transfer requires a base to get that hydrogen in alcohol.
I know that $\ce{H_2SO_4}$ can make a bad $\ce{ROH}$ group into a good leaving group ($\ce{ROH_2^+}$) by protonation, but how can it (strong acid) serve as a base for proton transfer to happen (in this 1 step process)?
What I think:

$\ce{H_2SO_4}$ is a dehydrating agent, so it can 'suck out' the $\ce{OH}$ together with heat, like dehydration with sugar into pure carbon (from my high school textbook)? However, is this mechanism even called $\mathrm{E1/E2}$?

In E1, the way I understand it is that the resultant carbocation's protons become reasonably acidic as the conjugate base (an alkene) is very stable. They only require a weak base to be removed, so practically anything can remove them. It looks to me as if the likely weak base candidates are the water produced from the leaving group, or even another alcohol in the reaction mixture.

Primary alcohols will be less susceptible to dehydration by this method, as they follow an E2 mechanism which progresses much more slowly in acidic conditions due to the absence of a strong base (which would usually act to 'shove' the leaving group off, a weaker base gives a weaker 'push').

I have seen a few schemes for E1 and E2 in which the hydrogen sulphate ion acts as the weak base; the usual assumption is that there is a weak base that is able to capture the proton, but we don't necessarily care what it is because anything will do. As long as there isn't another half decent nucleophile in the reaction mixture, elimination will likely prevail.

As you can see from this image I found in Clayden Organic Chemistry, the proton in the reaction mechanism is often drawn as just 'falling off', emphasising that any weak base is capable, and abstraction of the proton is very likely by a large number of species.


Why isn't the ionic radius of Mg2+ bigger than F-? They will both have the same number of valence electrons, but $\ce{F-}$ is more electronegative. Why is it not the case that $\ce{Mg^2+}$ has a bigger radius than $\ce{F-}$?

These two species are isoelectronic (have the same number of electrons). In fact, they're electronically identical with the except that the magnesium ion has 3 more protons. Since all other electronic factors are equal, the net effect is that the three extra protons pull all of the electrons in closer to the nucleus. I think you understand this part.

Electronegativity is not correctly used in this context. One way to think about it is that you should only be using that concept on a neutral atom. The fact the fluoride anion is negative and magnesium cation is positive means you've already applied electronegativity to the system in order to ionize them.


Urea-Formaldehyde cleanup I work a lot with Urea-Formaldehyde resin/plastic, both in carpentry and textiles. Cleanup while still uncured is simple enough. Cleanup before even mixed is easy enough. What I don't know, is how to cleanup plastic that has set and cured. If it was reasonably safe to guess and try, I would, but in this environment/conditions acetaldehyde formation is very possible, which makes release formaldehyde a problem. Grinding or sanding the plastic is not an option. I haven't seen a MSDS say anything about cleaning up cured plastic. From accidental spills it doesn't seem to disolve in xylol, acetol, or toluol. Anyone have any ideas? As long as there is a preditable outcome I have the resources to test just about anything.

It appears it can be decomposed in the presence of Hydrogen Peroxide, with decomposition being noticeable even at 3% (extremely slowly), but working well around 12%. I didn't do anything beyond basic tests, but a white salt (likely Urea?) will precipitate out, embedding in the pores, if any, of the plastic; this can interfere with this clean up method. It also produces Aldehydes, seemingly mostly Formaldehyde from the odor, but with reactions based on the environment. Some Acetaldehyde was produced in my case, but the rate was slow enough that only small short lived flames were produced.

This was able to be sped up with heating or HCl. I would shy away from adding HCl, as it produced a strong Chlorine smell, without any obvious yellow gas.


Volume Element in the Integral for Variational Method I'm learning how to use variational method to calculate the energy of ground state helium. I understand how we can set up a trial wavefunction that is a product of two hydrogen-like wavefunctions; however, when finding the integral $\int_{a}^{b} \phi H \phi dr_1dr_2$ (for $\phi=\phi_1(r_1)\phi_2(r_2)$), what would the volume element be? Would it be $16\pi^2 r_1^2 r_2^2$? I'm assuming the denominator is one for normalized wavefunction of hydrogen-like atom.

If you are working in spherical coordinates, and if the integrand has no dependence on the angular coordinates, then that volume element would be correct ($4\pi$ from each integration over $\theta_i \phi_i$ and $r_i^2$ for each spherical shell); however, the interelectronic potential $\frac{1}{r_{12}}$ depends not only on the radial coordinates, but also on the angular coordinates of each electron, so you can't just integrate those out.


Determine volume of methane used to heat x water to x degrees? I have 126.39 L of water. What volume of methane gas would I need to combust to heat it (at room temp, 21 degrees Celsius) to 38 degrees Celsius? How much $\ce{CO2}$ gas would be produced?
I calculated the molar enthalpy of combustion of methane and got −890.27 kJ/mol, and I know that factors into my calculation. I'm just confused as to what I actually have to do to determine exactly how much methane I need to heat the water.

A great place to start would be the heat capacity $C$ of water. It is defined to be the "amount" of heat change divided by the resulting temperature change. In symbols,

$$C=\frac{q}{\Delta T}.$$

Heat capacity is an extensive property; the size of $C$ depends on how much matter there is. We take this into account by dividing the equation by mass $m$.

$$C_s=\frac{q}{m\Delta T}$$

$C_s$ is called specific heat (capacity). Taking $\rho_{\ce{(H2O)}}\approx1$,

$$C_s \approx \frac{q}{V\Delta T}.$$

Can you take it from here? (Be careful with units.)

Note: It is probably assumed that

the combustion of $\ce{CH4}$ is complete, yielding only $\ce{CO2}$ and $\ce{H2O}$;
density of water does not change over the temperature range;
heat capacity is constant as $\Delta T$ is rather small;
vaporisation is neglible;
no heat lost to surroundings;
decreasing solubility of $\ce{CO2}$ does not contribute to the overall amount of gaseous carbon dioxide (joking...)


Is H2CO3 (carbonic acid) soluble or insoluble in water? I am asking this because I know that acids are normally aqueous but according to the solubility rules, all carbonates are insoluble.

Yes, it is soluble.

For example, the oceans are constantly getting $\ce{CO2}$ from the atmosphere that gets converted into carbonic acid and thus decreasing pH on a global scale. Or the soft drinks we drink contain carbonic acid in them in dissolved form only due to mixing of $\ce{CO2}$ gas.

Also, not all carbonates are insoluble as Ben Norris has commented.

Hard and fast 'rules' in chemistry tend to be rare.  Consider them guidelines.


Mixing alcohol with aluminum sulphate I am trying to make a mixture of the following ingredients: distilled water, alcohol, aluminum sulphate and glycerin.

The problem is that when I add alcohol to the water and aluminum sulphate mixture (the aluminum sulphate is completely dissolved), the alcohol solidifies the aluminum sulphate again, so I am left with even more solid aluminum sulphate then in the beginning.

What I need is for the mixture to be completely mixed, clear and without solids.

Thank you everyone for your time!

It looks like you added the alcohol before the glycerin.  Possibly the aluminum sulfate would be more soluble with the glycerin in there, so try adding that before the alcohol.

Wikipedia calls aluminum sulfate only "slightly soluble" in alcohol, so the alcohol addition would be expected to lower the solubility of the salt.  Nothing is reported there about the impact of glycerin.

Upon further review

Nope, we now find this did not work.  Hence as other answers indicate, we need a more careful approach.


Why is zinc oxide reduced at a temperature of 1400 degrees Celsius? I am attaching a graph of the Ellingham diagram on which my question is based.

If we interpret the graph, $\ce{ZnO}$ is reduced by $\ce{C}$ at a temperature of $\pu{800 ^\circ C}$ to $\pu{1000 ^\circ C}$. However, the actual temperature required is about $\pu{1400 ^\circ C}$. Why is this the case?

If the answer is $1400\ \mathrm{^\circ C}$ it must imply that the user does not want $\ce{CO}$ in their product. Based on the diagram, the intersection of $\ce{ZnO + C -> Zn + CO}$ occurs around $1000\ \mathrm{^\circ C}$. Full conversion of $\ce{CO}$ to $\ce{CO2}$ requires a temperature of around $1400\ \mathrm{^\circ C}$ as shown in the diagram.

As to why the reaction does not follow the $\ce{C -> CO2}$ curve (no $\ce{CO}$ involved), this is a kinetic reason. Subtract the $\Delta G$ values of each line at a specific temperature (where the lines intersect, the value is 0, meaning the reductions of the two species are in equilibrium). You will find that at the temperature of intersection between $\ce{2ZnO -> 2Zn + O2}$ and $\ce{C + O2 -> CO2}$ $(\Delta G = 0)$, there is sizeable energy driving force for $\ce{ZnO + C -> Zn + CO}$ $(G = -300\ \mathrm{kJ/mol}\ \ce{ZnO})$. The difference of the Gibbs energies is related to the ratios of the $\ce{CO}$ and $\ce{CO2}$ products by

$$\Delta G_\mathrm r(\ce{ZnO + C -> Zn + CO})-\Delta G_\mathrm r(\ce{2ZnO + C -> 2Zn + CO2}) =-RT\ln\frac{[\ce{CO}]}{[\ce{CO2}]}$$

Through plugging in the number it is a very high proportion that goes through the $\ce{CO}$ pathway. Again, the only way any of this is important is if it mentions that $\ce{CO}$ must be eliminated, else the reaction proceeds at low temperature as you say.


Thermochemical data of methanol I am trying to find the standard entropy and enthalpy of methanol as functions of temperature, but they are not available on the NIST-JANAF tables: http://kinetics.nist.gov/janaf/

Does anyone know where I could possibly find the data?

Thank you in advance!

NIST Webbook Chemistry is a reference which I would recommend as a first (as in initial) resource to consult for enthalpies of small organic molecular and inorganic materials.  The entry about methanol lists for one temperature the value in question, the experimental method of acquisition, and the primary literature reference.  The later may contain more data than the one tabulated here, too.

The references might be old, but in a scientific library you will find assistance to figure out who cited these publications which in turn may lead you to more recent determinations.


Why iron(II) spectrometry is not measured in distilled water and it requires a complexing agent like o-Phen? We know that iron(II) ions in water have a green colour. Why we cannot draw the spectrum of iron(II) directly in distilled water to measure the absorbance then determine the concentration of iron(II)? We generally use ortho-phenanthroline as a complex agent to form a red-orange coloured complex with iron(II), then measuring the absorbance and the concentration. Why we Cannot use just distilled water as complex agent knowing that it gives green colour with iron(II)?

I'm sure OP get the answer by reading all comments and the answer elsewhere. However, I'd like to point out few things on this spectrophotometric determination of $\ce{Fe^2+}$ in drinking water (it is not distilled water):

Iron can exist in natural waters where it originates from the dissolution of minerals. Even though most iron containing minerals are not very soluble in water as evident by the solubility products of iron-containing salts. However, a trace amount of iron can still dissolve in water. The dissolve iron is predominantly found in the form of $\ce{Fe^3+}$, because dissolved oxygen in water oxidizes $\ce{Fe^2+ -> Fe^3+}$.
Based on above fact, water cant be green, but probably yellow. But, it can be still determined by UV-vis spectroscopy, only if you have a lot of $\ce{Fe^3+}$ ions in your unknown drinking water sample to do so (see Ivan Neretin's comment and Maurice's answer), which is not possible.
On the other hand, 3 molecules of 1,10-phenanthroline (ortho-phenanthroline, a bidentate) chelate with one ion of $\ce{Fe^2+}$ to give highly intense orange-red $\ce{[Fe(o-phen)3]^2+}$ complex ion in water (Ref.1). The equilibrium constant for this reaction $\left(\ce{Fe^2+ + 3 o-phen -> [Fe(o-phen)3]^2+}\right)$ is $2.5 \times 10^6$ at $\pu{25 ^\circ C}$  in the $\mathrm{pH}$ range between  $\mathrm{pH} = 3$ and $\mathrm{pH} = 9$. Thus, a $\mathrm{pH}$ of about $3.5$ (using acetate buffer) is recommended to prevent precipitation of iron salts.
The molar absorption coefficient (ε) of this ferrous complex, $\ce{[Fe(o-phen)3]^2+}$ is $\pu{11,100 dm3 mol-1 cm-1}$ at the wavelength of maximum absorbance, $\lambda_{max} = \pu{508 nm}$. Thus, spectrophotometric absorbance of this complex can be easily measured in $\pu{10^{-4} M}$ to $\pu{10^{-5} M}$ or even lower levels at $\pu{508 nm}$, which is compatible with $[\ce{Fe^3+}]$ levels in drinking water (usually well water).
However, there is a problem: 1,10-Phenanthroline complexes quantitatively only with $\ce{Fe^2+}$ ions. But that can be solve using a reducing agent such as hydroxyl amine ($\ce{HCl}$ salt, which is soluble in water):
$$\ce{4Fe^3+ + 2HO-NH2.HCl -> 4Fe^2+ + N2O + 6H+ + 2Cl- + H2O}.$$ The reducing agent should be used in excess to prevent $\ce{Fe^2+}$ oxidizes back to $\ce{Fe^3+}$ by dissolve oxygen.

References:

R. Belcher, "The application of chelate compounds in analytical chemistry," Pure Appl. Chem. 1973, 34(1), 13-28 (DOI: http://dx.doi.org/10.1351/pac197334010013).


Definition of water Plasticization Can someone explain what is water plasticization? I tried to look for the definition but could not find one that explains it.

Quoting Wikipedia,

  Plasticizers (UK: plasticisers) or dispersants are additives that increase the plasticity or decrease the viscosity of a material.

As far as most research papers say, water is not plasticized, but rather it acts as a plasticizing agent for other substances such as starch films as mentioned in this article. Water makes these films permanently deformed upon external stress and resists reverting back to its original shape.


Why isn't the American nickel magnetic? I never thought that modern American nickels actually contained nickel anymore. However, according to this wiki article, the coins actually do contain 25% nickel, the rest being copper. And yet, no US coin produced today is officially magnetic. Why is this alloy of nickel not attracted to a magnet?

(and yes, I tried time and again to find and answer elsewhere online.)

There are many types of magnetic properties, including ferromagnetism, paramagnetism, diamagnetism, antiferromagnetism, ferrimagnetism, superparamagnetism, metamagnetism, spin glasses, and helimagnetism.  Many of these are too weak to cause any noticeable interaction with a magnet. The type of everyday magnetism you're thinking of, which nickel has, is ferromagnetism.

While nickel is ferromagnetic,  copper is not. As you said, the American nickel is currently 25% nickel and 75% copper.  According to this paper (from 1931!), in order for a nickel-copper alloy to be ferrogmagnetic, it must contain at least 56% nickel:

  ... 56 percent nickel is required before the alloy shows ferromagnetic properties at ordinary temperatures.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.38.828

E. H. Williams, Magnetic Properties of Copper-Nickel Alloys. Phys. Rev. 38, 828 (1931).


Why do exothermic dissolution reactions occur? I've always been taught that an exothermic dissolution means that the forming of solute-solvent bonds releases more energy than the energy consumed to break solute-solute bonds and solvent-solvent bonds. However, if the attraction that occurs between solute and solvent particles are mainly based off of intermolecular forces, why would the energy released when it is formed manage to exceed the breaking of the comparably stronger intramolecular bonds between solute-solute particles?

Obviously the intramolecular bonds are not stronger than the intermolecular bonds if the majority of the solute dissociates when placed in water. There are some salts ($\ce{AgCl}$, $\ce{BaSO4}$, $\ce{ZnS}$) that have extremely low $\ce{K_{sp}}$ values, meaning the majority of the salt placed in the solution does not dissociate, and are therefore considered insoluble salts. In these salts, the intramolecular bonds are stronger than the intermolecular bonds formed with the water, which explains why only extremely small amounts of these salts dissociate in water.

There are, on the other hand, salts that do dissociate in water. Dissolution requires energy, and this energy is supplied by the kinetic motion of the water. This energy is used in breaking apart lattice structure of the salts, thereby lowering the temperature of the water.

The difference between salts with a negative enthalpy of solution and a positive enthalpy of solution, as you stated, is in formation of bonds between the water and the ions of the salt. The formation of a hydrogen bond is an extremely exothermic process, and accounts for the extremely negative enthalpies of solution of $\ce{HF}$, $\ce{NaOH}$, and $\ce{CsOH}$. Weaker intermolecular bonds also form in the dissolution of these salts and many others as a result of the interaction of the water with the electron clouds of the ions. Interactions between the water and the ions always occur, though to a greater extent in some salts than others.


Molar conductivity of transition metal complexes solutions Which of the following has the highest molar conductivity?
(A) Diaminedichloroplatinum(I)
(B) Tetraamminedichlorocobalt(III) chloride
(C) Potassium hexacyanoferrate(II)
(D) Hexaaquochromium(II) bromide

Since mobility depends on the size, charge and the number of particles, how can we determine the extent of molar conductivity from the given data?
The solution provided to me considered only the number of ions that each gives. But is this justified? Because they all have different sizes, charges and thus different mobility.

Charge transport depends on the presence of charge carriers, and an increase of the concentration of charge carriers may increase the electric conductivity.  This is the basis of the answer given to you which (implicitly) assumes each of these complexes dissociate, e.g.

$\ce{K4[Fe(CN)6] <=> 4 K+ + [Fe(CN)6]^{4-}}$

in your solvent completely.  In this view, charge transport happenz without moving these carriers on a macroscopic level (like  the salt bridge used to connect electrochemical cells). You may imagine a bucket brigade, instead people exchanging buckets of water, ions shuttle electrons.

On the other hand, there are examples where charge transport is coupled to the transport of matter (e.g., ions) on a macroscopic scale.  Proteins and nucleic acids, for example, may be separated by this (e.g., gel electrophoresis); their different size (and shape, and charge) quite noticeably influence how quickly they travel in an electric field.


What is the relationship between the values of n and ml? Example question: ml=+5 what is the lowest value of n?
I can't seem to remember the easy trick to finding the relationship if someone could explain that would be great. -Thank you in advance.

Just work it out the proper way, in my opinion you shouldn't have to resort to shortcuts for such a question.

The magnetic quantum number $m_l$ takes values between $-l, -l+1, \cdots, l-1, l$. As such, as long as $l \geq 5$ then $m_l = 5$ will be valid.

The orbital angular momentum quantum number $l$ takes values between $0, 1, 2, \cdots, n-1$. Consequently if $n \geq 6$, then $l \geq 5$ and $m_l = 5$ is permissible. The smallest possible value of $n$ is therefore $6$.


Titanium in HF vs HCl So we tried the reaction between Titanium metal with hydrochloric acid and hydrofluoric acid at a similar concentration. While boiling temperatures and constant heat over a long period of time are required to etch $\ce{Ti}$ using $\ce{HCl}$ even cold $\ce{HF}$ is enough to immediately break the protective oxide layer and the metal vigorousely reacts in the solution afterwards.

My question is now why $\ce{HF}$ is so much better at etching and dissolving $\ce{Ti}$ compared to aqua regia or $\ce{HCl}$?

Of course there is a protective oxide layer on the metal but boiling it in $\ce{HCl}$ will also form a purple solution after a while and the color intensifies if heat is applied over a long period of time. So even after the layer is destroyed the metal itself continues to react with the $\ce{HCl}$. For $\ce{HF}$ it reacts way faster so I assume it's not an issue with the protective layer itself but rather the reaction with the pure metal.

I know for glass and $\ce{HF}$ it's a combination where $\ce{H2F+}$ reacts with $\ce{Si-OH}$ endgroups and increases the electron density on the bridging oxygen which makes them more basic and the $\ce{Si-O-Si}$ bond breaks to form $\ce{Si-OH}$ bonds. But this is all for the oxides and I assume the titanium dioxide will be dissolved in a similar manner.

For metals I couldn't find much information on the topic. I know there are some similar elements, Niobium, Silicon, etc. they all require $\ce{HF}$ to be etched or dissolved.

Maybe it's the case because a $\ce{Ti-F}$ bond, much like a $\ce{Ti-O}$ "bond" is way stronger than a $\ce{Ti-Cl}$ bond? Which would explain why $\ce{HF}$ starts the reaction faster as this unique $\ce{HF/H2F+}$ mechanism seems to work pretty well for tough oxides.

Maybe it's also just the fact that more energy is produced when a $\ce{Ti-F}$ bond forms but the fast reaction starts immediately so it shouldn't just be the heat that is produced enhancing the reaction.

The process itself is a redox-reaction but the essential part here is the $\ce{H+}$ not the counter-ion. So from a redox perspective it should be pretty much the same, unless the redox potential and the overall redox reaction is influenced by the $\ce{Ti-F}$ bond formation, like the hydration enthalphy adds to the oxidative potential of oxygen and suddenly $\ce{Cu(I)}$ turns into $\ce{Cu(II)}$ once the insoluble $\ce{CuCl}$ is added to water.

(Also I'm sorry, but I cannot create a suitable tag with my score at the moment)

TL;DR. Are you asking a question about energy (thermodynamics) or a question about rate? What possible relation is there between boiling HCl(aq) and HF at room temperature as far as rate? I see no value in comparing those two reactions, but I'm probably missing something. So, if I understand you correctly, you are claiming that once you see some "purple" color, that therefore the Ti° is "bare", right? That, of course, doesn't (necessarily) follow. I'm not familiar with the reaction of Ti° + H+ → Ti(IV) + H2. (I'm assuming the +4 oxidation state) but almost certainly (I'd guess) it is not a single step reaction. Where the intermediate steps occur and how the presence of F- or Cl- ions impact one or more of those steps is the question. They could be absorbed on the metal surface, form complexes with intermediates, compete with oxide or hydroxide or oxyhalide species. It seems to me that the slow reaction rate of HCl is telling you that there continues to be something (some barrier) on the surface of the Ti° (by "continues" I don't mean to imply that it was there prior to the oxide films dissolution). About the only thing the energy can tell you about a reaction is that if its exothermic then you'd expect hotter and if hotter then faster (but there are (fairly rare) exceptions to this where hotter means slower). Since the value of comparing reaction A at 25°C to reaction B at 100°C isn't particularly informative (except if B is slower at 100° than A at RT, then we'd predict A will be even faster that B at 100° and B be even slower at RT) - especially if the reaction conditions are so different that we can't assume the mixing is the same (viscosity, acid dissociation, diffusion (turbulence?, H2 behavior?) all will be substantially different). Again, I'm not familiar with this reaction, but since it is a redox reaction, you can't necessarily even assume that the reduction is occurring near (on the molecular scale) to the oxidation, electrochemical cell formation is another possible complication... Could be complicated, IDK.


How to calculate the hybridization of ethane with H=½[V+M-C+A]? I wanted to calculate the hybridization of this molecule $\ce{CH3-CH3}$. I noticed a shortcut formula given in my class notes for calculating the hybridization. It was

$$H=\frac{1}{2}×[V+M-C+A].$$

With

$H$: Hybridization
$V$: Number of valence electrons of central atom.
$M$: Number of monovalent atoms (i.e hydrogen or halogen) around central atom.
$C$: Cationic Charge.
$A$: Anionic Charge.

When I tried to calculate the hybridisation of $\ce{CH3-CH3}$ by using above formula, where for any one of the carbon atom $V=4$, $M=3$, $C=0$, $A=0$ then $H=\frac{7}{2}$. But as we all know hybridization of every carbon in ethane is $\ce{sp^3}$, i.e. $H$ should be $4$. I know other methods of calculating $H$ like $H= \text{no. of sigma bonds} + \text{no.of lone pairs}$, which gives me the right answer, but why does the above formula give something different? And also for any molecule if using that formula gives fractional answer then what does that mean?

The mistake that you have made is incorrect understanding of M. From what I remember, M was number of single bonds, which would be only with a monovalent atom, in the context you learnt tbe formula. This was taught in chemical bonding where only atoms are bonded to central atom.
Ex- PCl5, PCl3
You should modify this definition to get the answer  here. Any monovalent group (-CH3) should be there in M.


Group Theory and Chemistry for Undergraduate Thesis I am advising a student on her senior math thesis.  She would like to involve chemistry somehow (her other major).  I know no chemistry.  I have read in several places that representation theory features prominently in some areas of chemistry.  When I try to read anything online I am not sure how much chemistry one needs to assume.  So, if my student is in her senior year of a chemistry major, are there any places where she can start to look for a project in group theory and chemistry?

The first 4 chapters of Chemical Applications of Group Theory by F A Cotton should be enough for your student. However there are numerous books available.

Again chemwiki has a good documentation on group theory.


What is the favoured geometry of chlorine trifluoride? Why is the first structure on the left more stable than the second one on the right?

In the first one the angle between the two pair of nonbonding electrons is about 120° which is a lot less than that in the second structure. And it is also very well known that lone pair-lone pair repulsions are worse than bonding pair-bonding pair repulsions. So why is the first structure more stable?

Short Answer

The structure on the left is "preferred" because the structure on the right cannot exist.  You cannot put 4 electrons in a p-orbital.

Detailed Explanation

In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized.  This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons  or bond to a ligand.  Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.

The drawing on the right has two lone pairs placed in one p-orbital.  A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.

The question really becomes, does $\ce{ClF3}$ prefer to exist with

2 axial fluorines and 1 equatorial fluorine

or

1 axial fluorine and 2 equatorial fluorines?

We can use Bent's rule to answer this question.  Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital).  An electronegative substituent will draw electron density away from the central atom.  Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.

Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals.  Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.

A final note on the bonding in $\ce{ClF3}$, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a hypercoordinated or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons.  See these earlier answers for more detail on hypercoordinate bonding.

Why does F replace an axial bond in $\ce{PCl5}$?
How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
Hypervalency and the octet rule


Ea is the term for activation energy, what is the mean of a in this term? Our doctor said it is not stand for activation. it stands for something else, any help will be appreciated.

The subscript $\mathrm a$ in $E_\mathrm a$ may refer to Arrhenius.
The IUPAC gold book states that:

  activation energy (Arrhenius activation energy)
  An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$, $E_\mathrm a = RT^2 \frac{\mathrm d(\ln k)}{\mathrm dt}$,
  where $R$ is the gas constant and $T$ the thermodynamic temperature.


Physical properties of geometric isomers I've learnt that cis isomers have higher boiling point than the trans ones.
Why does not that rule apply on Fumaric and maleic acids?

In the case of fumaric acid, intermolecular hydrogen bonds are the reason beyond its higher melting point compared with the melting point of maleic acid ( where intramolecular hydrogen bonds dominate).


By using the magnetic susceptibility, calculate the number of unpaired elections in the complex I'm having trouble making sense of this answer. I believe I am doing all the steps right, but my main concern is that I am not arriving at a whole number. If my answer is correct, should I just round up or down?

Question

  By using the known values of $\chi$ for $\ce{HgCo(NCS)4}$ calculate the number of unpaired electrons in each of these complexes, and interpret the results in terms of the electronic structures of the complexes.

My Attempt

$$\ce{HgCo(NCS)4}\ 's \   \chi = 16.44 x10^{-6}\  \mathrm{CGS (g^{-1}\cdot cm^3) \ at \ 20°C}$$

$$\mathrm{\chi_M =16.44 \cdot 10^{-6}g^{-1}\cdot cm^3 * \frac{491.8528 g}{1 mol} =8.09\cdot 10^{-3} mol^{-1}cm^3}$$
$$\chi ' _M= \chi_M (\text{metal core electrons}) + \chi_M (\text{ligands}) + \chi_M (\text{other}) - \chi_M$$

Diamagnetic corrections include:

$$\begin{align} \ce{Hg^{2+}}&=40.0\cdot 10^{-6}\\ \ce{Co^{2+}}&=12.8\cdot 10^{-6} \\ \ce{NCS^-} &=26.2\cdot 10^{-6}\end{align}$$
Source

$$\chi ' _M=  (40\cdot 10^{-6})+(12.8\cdot 10^{-6})+(26.2\cdot 10^{-6})(3)-(16.44\cdot 10^{-6}) =-0.02418$$

then using the equation

$$\mathrm{\chi '_MT=\frac{1}{8}n(n+2)}$$
Assuming 298K and setting to zero
$$\mathrm{n^2 +2n-57.6722=0}$$

quadradic formula and i get
$$\mathrm{n=6.65}$$

I figure with all of that arithmetic it would be difficult to arrive at a whole number. So, my intuition is to round up to 7.

Is this approach correct? Which way should I round if so?

And I don't know how I would determine the electronic structure because there's two metal ions in the complex?

Let's start with "what is the species?"

While you write it $\ce{HgCo(NCS)4}$ it's probably better written as $\ce{Hg^{+2} [Co(NCS)4]^{2-}}$.

You can take the two metal ions separately as far as electronic structure.

$\ce{Hg^{+2}: d^{10}}$ So we don't expect any unpaired electrons.

$\ce{[Co(NCS)4]^{2-}}$ Well, $\ce{Co}$ must be $\ce{Co^{+2}}$, so it's $\ce{d^7}$ on paper. Next you'd need to consider the shape of the complex and possibly where the ligand is in the spectrochemical series and the likely orbital occupations.

   Here, $\ce{Co^{+2}}$ is tetrahedral, so it doesn't matter where NCS falls on the series. You expect a high spin complex with 3 unpaired spins.

Note: (NCS) can be a tricky ligand, since it can coordinate with either N or S atoms, but the crystal structure is clearly N-coordination.

I mention this, because while the Hg ion is, on paper, separated from the Co complex, in the crystal structure, there's clearly some Hg-S interactions.

So let's get to your question about the calculation. Yes, if you need an integer, you'd round up. But since you have many, many more than one molecule, your answer implies there are multiple magnetic species.

   This is clear from the electronic structure. Since the calculation above gives three unpaired spins in an isolated $\ce{Co^{+2}}$ complex, there must be some magnetic interactions between multiple species in the solid state. Indeed, this is observed at different temperatures. I'm hardly an expert in magnetic coupling here, but it certainly implies Hg must be involved as well.


Copper sulphate, Benedict's Reagent and albumin When my student's add Benedict's Reagent to albumin and heat it, the solution turns purple (as if testing for proteins).  Is it possible that the copper sulphate in the Benedict's is reacting with the protein in the albumin?

Yes, it's most likely the same reaction as in the Biuret test that detects peptide bonds.


Hyperconjugation - concept clarification As I currently understand it, Hyperconjugation occurs in carbocations and alkenes via the interaction of the electrons in the MO of a C-H bond at an adjacent carbon to bring stability to the system.The interaction requires for in phase overlap and occurs when the C-H bond is in the same plane as the vacant p orbital. (or the pi bond in case of alkenes) It is thus not seen if the beta hydrogens are in the plane perpendicular to them.

If this is true, then how does hyperconjugation work, if at all in cyclic systems where free bond rotation is impossible and thus the orbitals will never align perfectly? What conceptual bridge am i missing here?

Cyclic systems aren't planar. The system orients itself in such a way that the vacant p orbital is in plane in order to gain the ability to engage in hyperconjugation which stabilizes it very well.


Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? I am trying to calculate the relationship between the octahedral field splitting parameter ($\Delta_\mathrm{o}$) and the square planar field splitting parameter ($\Delta_\mathrm{sp}$) and thought a good way to start would be to understand why $\Delta_\mathrm{t}=4/9\Delta_\mathrm{o}$ for tetrahedral fields but couldn't find any information on it.

How does one arrive at this particular ratio mathematically?

The ratio is derived in The angular overlap model. How to use it and why J. Chem. Educ., vol. 51, page 633-640.

First, relative expressions (Shaffer Angular Overlap Factors) are derived for overlap integrals between metal and ligand orbitals as a function of angles.

For ligand sigma orbitals, the expressions are a functions of 2 angles, specifying the direction of the ligands in the coordinate system of the metal.

For pi-x and pi-y ligand orbits, there is a third angle for orientation about the ligand-metal axis.

Then, assuming the ligand-metal distances are the same in both the tetrahedral and octahedral cases, and considering the coordinate geometry of the ligands in the two cases, the relationship:

$$\Delta_{T} = -4/9 \Delta_{O}$$

is obtained.

For another discussion of this ratio in the framework of the angular overlap model, as well as extension to other geometries such as square planar, see A New Look at Structure and Bonding in Transition Metal Complexes, Advances in Inorganic Chemistry and Radiochemistry, Volume 21, pages 113-146.


Methane seems to dissolve more readily in cold water. Why? When you are diluting air in the system and/or properly evacuating it, you must consider water temperature. Methane is completely dissolved in water at 42 degrees Fahrenheit (5.5 C) but can be completely released (as a gas) at 58 degrees F (14.5 C). (source)

This would seem to imply that the enthalpy of solution for methane in water is negative (i.e. the process is exothermic). But, intuitively, it seems that you would have to input energy to force non-polar and polar to mingle.

Can anyone explain why this might be the case?

The enthalpy of solution going from the ideal gas state to the solution at infinite dilution is exothermic.  However, more importantly, the Gibbs energy of solution is positive meaning that dissolution of methane into water is not a favored process.  Most of this comes from the relatively large negative entropy change associated with the dissolution at room temperature.  This positive Gibbs energy is manifested in the very small solubilities observed and gives rise to the whole concept of something being "hydrophobic."  The fact that any methane dissolves in water is due to equilibrium effects based on a finite positive Gibbs energy.  Since methane is dissolved to a greater extent at lower temperatures must mean that the Gibbs energy is less positive at the lower temperatures.


How can I reverse engineer a commercial cleaning solution's "other ingredients"? I have a commercial lens cleaning solution. The manufacturer of the solution claims it is not only distilled water and that the formula is confidential. I suspected it was only or mostly distilled water after testing distilled water's cleaning abilities on different types of lenses with different coatings.

I have heard that you can ship a sample to a lab and they will conduct a quantitative analysis of the solution using a mass spectrometer, but there are so many different kinds of mass spectrometry that I am not sure what I would want to order.

What kind of mass spectrometry (or other lab analysis service) would be best suited to discover the formula for a non-volatile, non-corrosive, liquid sample of lens cleaner?

If that's not possible, I just want to know how much of the solution is essentially water, and would like to know what kind of service would be best to determine that.

Reverse engineering, or deformulation, is a common industry practice. This is a very common type of request for commercial analytical laboratories, and is generally not cheap. You can discover just how widespread it is, and how many commercial providers there are with a quick Google search.

To get yourself a reasonable starting point, you should request a product MSDS from your suppplier. All suppliers should be able to provide a full MSDS, which will give an approximate ingredient listing and component concentration. Sometimes, you will find ingredients listed as "Trade secret" or "Proprietary", but many times you will get more information. It is very likley that your lens cleaner solution is, in fact, predominantly distilled water; perhaps up to 90-95%. However it is the other minor ingredients that are the active ingredients, and may include alcohols or glycols, anti-fog and anti-static agents, anti-baceterials and preservatives. Examples of lens cleaner MSDS can be found via a google search, and this may give you a useful bit of background information.

Note:
1. all suppliers are legally bound to providing MSDSs upon request in the local idiom (official language).
2. this document is safety-based which means it will (a) provide (useful) information on possibly hazardous materials, (b) assess the hazardous component(s), their type, level and correlate to concentration.
3. websites such as state Chemical authorities (ECHA, FDA, ...) publish ratings, studies and lists (of registered/authorized/forbidden) of chemicals for use in particular applications such as food contact, medical, pharmaceutical devices/products.

The next step is to determine what information you really are after, as this will determine the budget you will need to outlay. If you want to just determine whether the solution is predominantly water, you could try a few things yourself (if you are experiment-savvy). Try smelling it, as a first step; check for evaporation if you run a thin film on a glass slide, test density or melting point. These will all tell you much the same thing, which is, yes, the solution is predominantly water. A simple commercial 1H NMR analysis will show you this also, and confirm water concentration up to about 99.9%. This means minor ingredients will show up, but as long as they make up about 0.1% of the ingredient composition, and they contain hydrogen (sodium bicarbonate will not show up for instance). This method is commonly used when you have a reasonable idea about the identity of the minor ingredients, but can involve extensive analysis if you really are shooting in the dark.

Beyond this, if you actually want to determine the ingredients, and have no idea what they might be, you will need to head down the avenue of GC-MS and/or LC-MS; a combination of chromatography to separate all of the ingredients, and then Mass Spectrometery to try to determine the mass of each component. This type of analysis is generally charged by the hour, and the more variables you are trying to evaluate, the more expensive the procedure. Generally, when analysing a competitor's products, a reasonable knowledge of the types of raw materials exists, and fingerprinting chromatography and NMR traces against raw material libraries saves a lot of time (and money). Sometimes, other techniques such as FTIR, AAS, UV-Vis or ICP may also be used, but these would really only be if the MS methods failed to give you suitable leads.

Commercial analytical labs will be able to give you a very good summary of the best techniques for your sample, but be prepared; the very first question you are likely to be asked is "what is your budget?". Good luck.


Using covalent or ionic radius when calculating dipole moment When the ionic character of a bond is above 50% should I use ionic radius instead of covalent one when calculating the dipole moment?

The formula I have for dipole moment is $\mu = \text{(% ionic character)} \cdot e \cdot d$, where $e$ is the elementary charge and $d$ the sum of the two radii.

Personally, I think "50%" as a cut-off is very arbitrary to begin with. It's not like I can suggest anything better, since proper dipole moment calculations rely on quantum mechanics.

But, to answer the question:

The whole premise of my answer is that you want to get the sum of your radii to match the actual bond length in the compound. And to do that, you have to look at the context. For example, for $\ce{BF3}$, you might want to use covalent radii since the atoms are covalently bonded. And for a ionic compound like $\ce{KF}$, you'd better use ionic radii or you might run into some problems.

Now, note that the formula you're using is already extremely inaccurate in itself - if you take the covalent radii for B and F, you get $d = (82+71) \text{ pm} = 153 \text{ pm}$. And... the actual bond length is very short - it's $130 \text{ pm}$. Clearly, there is something about that compound that the crude model of adding covalent radii cannot account for. Still, it's better than if you were to use the ionic radii: $(41 + 119)\text{ pm} = 160 \text{ pm} > 153 \text{ pm}$ (sum of covalent radii) $> 130 \text{ pm}$ (actual bond length).

  "Hey @orthocresol, using $\ce{BF3}$ isn't fair! That compound has crazy strong $\pi$ bonding between B and F!"

Okay. Let's look at $\ce{BCl3}$, $\ce{BBr3}$ and $\ce{BI3}$, where the $\pi$ bonding is weaker.

$$\begin{array}{|c|c|c|c|}
\hline
\ce{X} & \text{Actual B-X bond length / pm} & \text{Sum of covalent radii / pm} & \text{Sum of ionic radii / pm} \\
\hline
\ce{Cl} & 175 & 82 + 99 = 181 & 41 + 167 = 208\\
\ce{Br} & 187 & 82 + 114 = 196 & 41 + 182 = 223 \\
\ce{I} & 210 & 82 + 133 = 215 & 41 + 206 = 247\\
\hline
\end{array}$$

The covalent radii are closer every time.

  "That's still not fair... Can't we just use a compound without $\pi$ bonding?"

Fine. Let's talk about the tetrahedral $\ce{BF4-}$ anion. There's no $\pi$ bonding here and as a result, the B-F bond length is increases to $145 \text{ pm}$. But the covalent radii still win.

You can extend the discussion to something like $\ce{SiX4}$. But, honestly, I think that representing $\ce{SiX4}$ as $\ce{Si^4+}$ and $\ce{X-}$ is not that a great an idea, to say the least.

Also, I'm lazy to dig up bond lengths in $\ce{AlCl3}$, so I'll stop here.

  "Okay, now what about ionic compounds?"

You'd better use those ionic radii, because guess how those ionic radii are tabulated? They're calculated by measuring the distance between cations and anions in an ionic lattice and assigning a certain part of that to the cation and another certain part to the anion. So if you add them back up, they'll of course give you the internuclear separation that you want.

Let me use the simple example of $\ce{NaCl}$. You'll have to trust that I'm doing the right things here; if you're interested, you can read up on the rock salt structure. The unit cell parameter of $\ce{NaCl}$, denoted by $a$, is $564 \text{ pm}$. This corresponds to two times the distance between $\ce{Na+}$ and $\ce{Cl-}$ ions, so we could say that the actual "bond length" is $564 \text{ pm} / 2 = 282 \text{ pm}$.

Now, the ionic radii of 6-coordinate $\ce{Na+}$ and $\ce{Cl-}$ are $102\text{ pm}$ and $181\text{ pm}$ respectively. That sums up to $283\text{ pm}$ - a beautiful match. And the atomic radii? They're $191\text{ pm}$ and $199\text{ pm}$ for $\ce{Na}$ and $\ce{Cl}$ respectively. That sums up to $290\text{ pm}$ - pretty close but not quite as close as the sum of the ionic radii.

Data is taken from this website, Wikipedia:Ionic radius, Shriver & Atkins, Inorganic Chemistry, 6th ed., and Greenwood & Earnshaw, Chemistry of the Elements, 2nd ed. Apparently, the numbers in that Wikipedia figure are taken from R. D. Shannon, Acta Cryst., 1976, A32, 751, which is in fact one of the most widely-used sources of ionic radii (all inorganic textbooks reference it) - so it shouldn't be that bad.


Why is concentration used as a proxy for activity instead of mole fraction? I am trying to understand how the Gibbs energy change of a chemical reaction is computed and I do not understand why concentrations (mol/L) are used as a proxy for activity rather than mole fractions (mol/total mol).

From the Wikipedia article on the chemical potential, in an ideal solution, the chemical potential of the chemical species $i$ is expressed as

$$\mu_i \approx \mu_i^\circ + RT \ln x_i,$$

where $\mu_i^\circ$ is the chemical potential of species $i$ in standard state, $R$ is the gas constant, $T$ is the temperature and $x_i$ is the mole fraction of $i$ in the system. Because real solutions are not ideal, the activity of $i$ ($a_i = \gamma_i x_i$) is used rather than its mole fraction $x_i$ in the calculation of $\mu_i$.

When $T$ and $P$ are constant, the Gibbs energy change of a reaction is formulated as

$$\Delta_r G = \sum_i \mu_i dn_i,$$

where $dn_i$ is the change in the number of particles of $i$ caused by the reaction (its stoichiometric coefficient). Consequently;

$$\Delta_r G = \sum_i \mu_i^\circ \cdot dn_i + \sum_i dn_i \cdot RT \ln \gamma_i x_i.$$

$\sum_i \mu_i^\circ \cdot dn_i$ is equivalent to $\Delta_r G^\circ$, while $\sum_i dn_i \cdot RT \ln \gamma_i x_i$ is equivalent to $RT \ln Q$ in the more usual formulation $\Delta_r G = \Delta_rG^\circ + RT \ln Q$.

Molar concentration (mol of species per volume of solution) are often considered as a proxy for the activity of solute species, while partial pressures are used for gas species (which gives the "mass action ratio" $\Gamma$ instead of the reaction quotient $Q$). While partial pressures are a concept similar to mol fractions (they are fractions of a total), concentrations and mol fractions are not the same quantity and I do not think they scale the same way. So why are $\Delta_r G$ quantities often computed using concentrations instead of mole ratios?

EDIT: Thinking about it, I think I found a simple justification for the use of concentration as a proxy for activity.

Let's consider a fictive reaction $A \leftrightharpoons B$. The chemical potential change associated with this reaction is

$$\Delta_r \mu = \Delta \mu_B - \Delta \mu_A$$

$$\Delta_r \mu = \Delta \mu^\circ_B - \Delta \mu^\circ_A + RT \ln x_B - RT \ln x_A$$

$$\Delta_r \mu = \Delta \mu^\circ + RT \ln \frac{x_B}{x_A}$$

$$\Delta_r \mu = \Delta \mu^\circ + RT \ln Q$$

mole fractions $x_A$ and $x_B$ are respectively $\frac{n_A}{n_{total}}$ and $\frac{n_A}{n_{total}}$ with $n_{total} = n_A + n_B$, so $Q = \frac{n_B}{n_A}$ ($n_{total}$ cancels). Obviously $Q$ has the same value if concentrations ($\frac{n}{V}$) are used since the volume $V$ is the same for both species and then cancels itself. This justification is obvious for a 1:1 stoichiometry. However, with other stoichiometries, either $n_{total}$ or $V$ remains as a factor in $Q$ (to the power $\sum_i \nu_i$, where $\nu_i$ is the signed stoichiometric coefficient of species $i$). I am pretty sure that $n_{total}^{\sum_i \nu_i}$ is not always equal to $V^{\sum_i \nu_i}$; this is where the difference may happen.

The posted notation seems unorthodox if not incorrect.
For the Gibbs free energy of reaction one usually writes
$$\Delta_r G = \left(\frac{d \Delta G}{d \xi}\right) = \sum_i \nu_i\mu_i^\circ  + RT\sum_i  \nu_i \ln \gamma_i x_i$$
$$ = \Delta _r G^\circ + RT \ln Q$$
where $ \xi$ is the reaction progress, $Q=\sum_i  \nu_i \ln \gamma_i x_i$ is the reaction quotient, and one can equate $\Delta _r \mu= \sum_i  \nu_i\mu_i $ with $\Delta_r G$. For the reaction $A \leftrightharpoons B$ one would write $\Delta \mu = \mu_B - \mu_A$.
Changing concentration units is ok but a set of units usually implies a choice of thermodynamic reference state. The reference state generally defines the limit in which $\gamma \rightarrow 1$ (the limit of ideal behavior). If you decide to express the reaction quotient Q in terms of molar concentration rather than mole fraction but want to retain the same general form of the above equations, then it becomes necessary to modify both Q and $\Delta _r G^\circ$. Some reference states are better choices than others under certain circumstances, and it is not accurate to say that mole fraction is not used, it is, just not always. For solids or solvents a mole fraction unit is often convenient. When dealing with solutions, and particularly with solutes as a minor component, it is generally more convenient to deal in molal and molar units (the prior being favored), in combination with Henry's law. In addition, at small concentrations of solute the mole fraction, molarity and molality are linearly proportional, differing by a multiplicative constant.  For instance, at low concentration we can write the mole fraction of solute as
$$\chi _2 = \frac{n_2}{n_1 + \sum_i n_i} \approx \frac{n_2}{n_1} = \frac{c_2}{c_1} = \frac{c_2}{1000\delta_1/M_1} \approx \frac{M_1}{1000 \delta_1 ^\circ}c_2$$
where 1 and 2 label solvent and solute and $\delta_1 ^\circ$ is the mass density of the pure solvent, $M_1$ its molar mass.


Is Iodoform test restricted for only methyl ketones? Do only methyl ketones give iodoform test? Where can a keto methyl group be present in?

All compounds with $CH_3CO- $ or $CH_3CH(OH)-$ group answer iodoform test.


Neutralization of alkali solution by a mix of strong acids $\pu{0.1 M}$ $\ce{HCl}$ and $\pu{0.2 M}$ $\ce{H2SO4}$ solutions are mixed in equal volume. This solution is diluted to double the volume. Find the volume of $\pu{0.1 M}$ $\ce{NaOH}$ which can be neutralized with $\pu{20 mL}$ of the former solution.

The answer is $\pu{50 mL}$.

I have tried a bit, but the answer I get is wrong. Let volume be $V$ at the beginning, therefore, no. of moles of $\ce{HCl}$ is $0.1V~\pu{mol}$, no. of moles of $\ce{H2SO4}$ is $0.2V~\pu{mol}$.

Total volume of solution after dilution is $4V$, therefore, no. of moles of $\ce{H+}$ in the solution is $(0.1V + 0.4V)/4V$.

No. of moles of $\ce{Cl-}$ in the solution is $0.1V/4V = 1/40$.

No. of moles of $\ce{SO4^2-}$ in the solution is $1/20$.

Since $\ce{NaOH}$ reacts only with $\ce{Cl-}$ and $\ce{SO4^2-}$, therefore total moles of $\ce{NaOH}$ needed is $(1/20 + 1/40)(20/1000) = 1/100~\pu{mol}$

Therefore, $\text{volume needed} = \text{molarity/no. of moles} = 0.1/0.01 = \pu{10 L} = \pu{10000 mL}$.

I Don't Know what I am doing wrong.

I tried again using some of the suggestions given: for the first step of mixing H2SO4 and HCl, M1V+M2V=M3(2V),
which comes as M3=0.5/2.
Then for dilution,
(0.5/2)*2V=M4*4V
thus M4=0.5/4.
Taking 20ml, (0.5/4)*20=0.1*x,     [0.1M of NaOH]
thus x=25ml.
Why doesn't this work??

I think you over-complicated things a bit by focusing on the formation of sodium salts, when in reality all you need to take into account is the net neutralization reaction:

$$\ce{H3O+ + OH- -> 2 H2O}$$

From the equivalence point condition of any neutralization reaction one can find volume of the base $V_\mathrm{b}$ expressed via the volume of acid(s) $V_\mathrm{a}$:

$$V_\mathrm{b} = \frac{C(\ce{H3O+})}{C(\ce{OH-})} \cdot V_\mathrm{a} \tag{1}$$

Since there are two moderately diluted strong acids and a base, one can conclude that the dissociation of each acid and a base is independent and complete:

\begin{align}
\ce{HCl + H2O &<=>> H3O+ + Cl-} \\
\ce{H2SO4 + 2H2O &<=>> 2H3O+ + SO4^2-} \\
& \\
\ce{NaOH &<=>> Na+ + OH-}
\end{align}

Note that concentration of protons is doubled in case of sulfuric acid. Total concentration of protons in the solution is

$$C(\ce{H3O+}) = \frac{n(\ce{H3O+})}{V(\ce{H3O+})} = \frac{n(\ce{HCl}) + 2 \cdot n(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \tag{2}$$

Taking into account that $n = CV$ and that two identical volumes of acid were mixed ($V(\ce{HCl}) = V(\ce{H2SO4}) = V(\text{acid})$) (2) can be transferred into:

$$
\require{cancel}
\begin{align}
C(\ce{H3O+}) &= \frac{C(\ce{HCl}) \cdot V(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}) \cdot V(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \\
&= \frac{\cancel{V(\text{acid})} \cdot (C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}))}{2 \cdot \cancel{V(\text{acid})}} \\
&= \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2} \tag{3}
\end{align}
$$

Now let's plug (3) in (1) and finally find the volume of alkali:

$$V_\mathrm{b} = \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2 \cdot C(\ce{OH-})}  \cdot V_\mathrm{a} = \frac{\pu{0.1 M} + 2 \cdot \pu{0.2 M}}{2 \cdot \pu{0.1 M}} \cdot \pu{20 mL} = \pu{50 mL}$$


hypo residual chlorine testing using a 10ml reagent pack Why when testing a sample for chlorine residual do I get the same results for 5 ml sample of water as I do with a 10 ml sample, and a 15 ml sample of water using a reagent pack that asks for a 10 ml. The reagent container does not give the weight of the reagent in the packet (it is powder), but I think it might weigh 0.1 grams. I am just using a Hach DR 900 to test the free chlorine residual.

My thinking is if the reagent packet is asking for 10 ml sample of water shouldn't my results differ when using different sample sizes. For the three different volumes of samples, I got a free chlorine residual of 1.00 mg/L for all three sample sizes.

Sorry, the packet is a DPD free chlorine reagent for 10 ml sample. The packet contains Sodium Phosphate, Dibasic, EDTA Disodium Salt, DPD Salt, and Carboxylate Salt.

This is sort of an extended comment, not necessarily an answer.

Overall, yes I'd expect different numbers at least. How they would vary is a matter of some conjecture.

(1) The Hach DR 900 is a colorimeter. I'm not sure how large of a sample you need to fill active volume of the colorimeter cell.

(2) The ratio of water volume relative to the amount of solid reagent is changing. So the chemical equilibrium on which the calibration curve depends might shift. If you dilute 5 and 10 ml samples to 15 ml with "pure" water and take 10 ml of each of the three samples (2 dilutions plus "full strength"), then you be comparing better.

(3) Even given (2), the results seems odd. Normal drinking water should have between 0.2 and 2.0 mg/L.

Is pH of the water ok?
Does water have some unusual buffer capacity?
Reagent pack old?
Do you have some standards which you can test?


What does acidity of para-alkyl substituted phenols depend on? Problem

Arrange the following organic compounds in decreasing order of acidity.

Answer

 S, R, Q, P.

Question
According to my knowledge the answer should be P, Q, R, S. As we proceed from P to S we are increasing the number of methyl groups, and so the number of electron donating groups increases, decreasing the polarity of of the C−O bond, which is the opposite to the answer. This obviously means that whatever I'm thinking is the right reason is the wrong logic, but I'm not sure how.

The reason provided by ML cannot explain the difference in acidities since the +I effect doesn't come into play normally for the para positions.
A more likely reason is hyperconjugation (or no bond resonance) where the number of hydrogens directly attached to the benzyl carbon decreases as it goes from P to S.
This means that the number of destabilizing structures that are produced via this resonance decreases as you go down the list. This leads to greater stability in the deprotonated form compared to each other.
Therefore the correct order becomes S, R, Q, P.


carbon electrodes have potential difference even with no electrical input? What is happening in this photo? I set up a simple electrolytic cell today of copper (II) sulfate solution with carbon electrodes. It was connected to a powerpack (set at ~8.0V), and a variable resistor with the resistance cranked up to its maximum as in the image below. You can also see the ammeter and the voltmeter in the photo.

When the powerpack was switched off, the current fell immediately to zero, as expected, but the voltmeter still read a non-zero number (initially 0.80V, then later seemed to decrease to 0.18 V). I turned the electrodes upside-down so that the clean ends were in solution (rather than the end covered in copper), but reading still read the same. Removed the electrodes to a beaker of water (but not deionised water), and it still read 0.18V or close to.

Tried the exact same thing with pure copper electrodes, but it read 0.00V immediately upon turning off the power supply.

Tried completely taking the leads out of the power supply (which was off anyway), and the carbon rods still read a non-zero value even with the circuit obviously incomplete.

What's happening here? Copper electrodes did not show the same effect. Is there something about carbon electrodes in particular that can make them act as a capacitor? The current stayed at 0.00A the whole time. This effect was NOT seen when the electrodes were lifted out of the solution/water.

Edit, to respond to the answer below:

This would be the most interesting answer.
I did consider initially this, so first I tried up-ending the
electrodes so the clean ends were in solution, but the reading
remained the same. Second, it showed no such behaviour with the pure
copper electrodes.
I think it's unlikely to be a case of the voltmeter left floating.
Surely if that had been it, regardless of which electrodes (copper or
carbon) were used, random static charges would've been present? Plus,
the phenomenon was reproduced when I swapped out different electrodes
and then swapped back in the funny-acting ones, so the behaviour
wasn't exactly random.

However, I did forget to mention, that I also tested this with a
second clean set of carbon electrodes and got 0.00V! Support for 2nd
explanation (but if so, how does that work exactly?). I still would've
liked it to have been the 1st explanation.

I also note that I left the apparatus overnight, and by next morning
the reading was 0.02V. I've unfortunately have had to clear the
materials away now, but if I get a moment I may try replicating the
result.

Some possibilities:

Porous carbon electrodes do act as a capacitor, the basis for supercapacitors (aka ultracapacitors).
You are electrolytically depositing copper on the negative electrode when you run current through the solution, even though it might not be enough to see. This creates an electrochemical cell.
Simply disconnecting the power supply leads will leave a high-impedance voltmeter floating, so it might be detecting random static charges. You might need to bypass the voltmeter with a high resistance (10 or 20 megohms) to reduce static sensitivity.


Solvation effect on cyclohexane I ran molecular dynamics simulation with a molecule of methylcyclohexane soaked into a water box. On the other hand I ran Monte Carlo with the molecule alone (no soaking). I performed this for the equatorial and axial conformers.

The axial conformer with MD stayed axial (the methyl group). The equatorial with MD stayed equatorial. So far no visual change. Is there a way to compute the average energy in VMD for these systems (I have around 250 frames).

On the other hand, the axial conformer alone with MC was transformed into equatorial. While the equatorial stayed the same (visually speaking).

I am trying to understand what's happening here. What I read is that always the equatorial configuration is more stable as it has less steric hindrance (BTW I am computer scientist learning this), so this explains the interconversion in Monte Carlo simulation from axial to equatorial.

On the other hand, I would expect also that in MD the axial conformer turn into equatorial but it didn't, what I believe here is that the water molecules are having some "Solvation effect" on the methylcyclohexane that prevents the interconversion.

So I would like to know if my thought are good and if you could give some references to get more info about solvation effects.

The difference in behavior between the two methods is likely how much "wiggle" the molecule is allowed to have, or how much additional kinetic energy you allow the molecule to have when searching for the minimum energy conformation. The equatorial conformation is the global minimum conformation, but the axial conformation is a local minimum. Additionally, between the two conformations, there is another minimum called the twist-boat. You can read about them on this Wikipedia article or you could check out an introductory organic chemistry book from a library. Molecular dynamics tend to stop when a minimum is found. In other words, if a small alteration to the geometry increases the potential energy, the calculation stops. Monte Carlo calculations on molecules are designed to examine the evolution of conformation with time, and so geometry moves out of energy minima are allowed.

Unsubstituted cyclohexane has four different conformations: Chair (relative energy 0 kJ/mol), half chair (+45 kJ/mol), twist boat (+23 kJ/mol), and boat (+30 kJ/mol). A molecular dynamics calculation may not allow large enough steps to get over the activation barrier to get out of a chair conformation. The following energy diagram shows the conformational space of cyclohexane.

The energy difference between equatorial and axial methylcyclohexane is 7.6 kJ/mol (1.6 kcal/mol on this table), which is small compared to the difference between the chair and the half-chair conformation. The energy diagram below approximates the energy steps between equatorial and axial methylcyclohexane.

Any computational method that allows potential energy steps large enough to get to the half-chair will certainly isomerize the axial conformer to the equatorial conformer. Presence or absence of solvent molecules in the simulation may be irrelevant.


Atomic radii of Sc, Ti, Fe, and Co The atomic radius of Sc is $\pu{162pm}$, Ti is $\pu{147pm}$, $\ce{Fe}$ is $\pu{126pm}$, and that of $\ce{Co}$ is $\pu{125pm}$.

The electronic configuration of $\ce{Fe}$ is $\ce{[Ar] 3d^6 4s^2}$, and that of $\ce{Co}$ is $\ce{[Ar] 3d^7 4s^2}$.  The difference in atomic number, and hence the difference in the number of 3d electrons, is 1. So, due to screening, the extra charge is "canceled" and they have nearly the same radii.

The difference in atomic number and 3d electron count between $\ce{Sc}$ and $\ce{Ti}$ is also the same — 1.

So, I want to know why there is a considerable difference between the radii of Sc and Ti but not between the radii of Fe and Co.

There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal.  This notion is very sensitive to the number of electrons per atom involved in bonding.  Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4.  These all participate, in some extent, in the "electron soup" that holds metals together.  I haven't been able to figure out exactly to what extent, but  It's fair to say though that the 4 valence electrons of Ti bind the nuclei together significantly more tightly than the 3 of $\ce{Sc}$. As a result, the $\ce{Ti}$ atoms get significantly closer together.  (An analogous situation is the covalent radius of $\ce{F2}$, of around $\pu{70 pm}$, versus that of $\ce{O2}$, of around $\pu{60 pm}$; although the covalent radius tends to decrease across a period, it grows from $\ce{O}$ to $\ce{F}$ because $\ce{F2}$ has a single bond while $\ce{O2}$ has a double one.)

As you progress further along the transition metals the delocalization of d electrons in the metal goes down.  That is, although there are more d electrons in $\ce{Co}$ than in $\ce{Fe}$, their effectiveness in binding atoms together isn't really any larger.  As a result, the distance between neighbors (hence the metallic radius) is the same for both.


Rearrangement of anions My book says a. But i think it is wrong.

Rearrangement is not  possible in anion me, ans also
 there is group migration but here the group has been  divided .

The most fascinating thing about aldol reaction is that it is reversible. So, when the aldol product is kept in a basic medium, it can actually undergo reverse mechanistic pathway to regenerate the two carbonyl compounds between which condensation occured to form the product . Then those two carbonyl compounds can literally undergo condensation in a different way to produce some different aldol product. This phenomenon is called as Retro-aldol reaction.

In this case, also if you observe carefully, the compound (A) is an $\alpha-\beta $ unsaturated ketone (basically a dehydrated aldol product), which by hydrolysis can convert itself into an aldol and thus undergo retro-aldol reaction. Here the aldol condensation was done intramolecularly, so the two ketones produced undergoes intramolecular aldol condensation in a different way to produce another isomer (B).

The mechanism of formation of (B) is as follows,

Thus, option (a) should be the correct answer.


How to get a "rechargable" chemical reaction for a glow stick I want to create my own glow stick that lasts a while while ligting up the surroundings. How would I go about this? I know there are youtube videos and such. But the way they do it doesn't allow for any way of "recharging"

How can i make a "rechargable" glow stick?

Chemoluminescent processes based on the reaction of diaryloxalates with hydrogen peroxide in the presence of a fluorescent dye are not reversible.

While the dye remains unchanged, both the oxalates and the hydrogen peroxide are consumed. "Recharging" a device (glow stick) isn't possible. When performing the reaction in a flask, only could probably prolong the light emission by subsequently adding portions of the starting materials over time.

Photoluminescent glow-in-the-dark devices based on the emission of long-lived excited triplet states can indeed be "recharged" by holding the device in the light again. But this has nothing to do with the reactions happening in glow sticks.


How can I find the emf of a cell given the individual reduction and oxidation reactions and their respective potentials? If I am given the individual oxidation and reduction half-equations, how can I find the emf of the combination? For example:

$$
\begin{align}
E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) &= \pu{0.77 V}\\
E^\circ(\ce{Fe^2+}/\ce{Fe^0}) &= \pu{-0.44V}
\end{align}
$$

How can I find the emf of this redox pair, i.e. $E^\circ(\ce{Fe^3+}/\ce{Fe^0})?$

This is my first answer in stack exchange if some mistake comment on.
$$E^\circ_\ce{Fe^3+/Fe} \neq E^\circ_\ce{Fe^3+/Fe^2+} + E^\circ_\ce{Fe^2+/Fe}$$

but

$$\Delta_r G^\circ_1 = \Delta_r G^\circ_2 + \Delta_r G^\circ_3$$

for the three reactions:

$$\ce{Fe^3+ + 3/2 H2 -> Fe + 3H+}\tag{1}$$

$$\ce{Fe^3+ + 1/2 H2 -> Fe^2+ + H+}\tag{2}$$

$$\ce{Fe^2+ +  H2 -> Fe + 2H+}\tag{3}$$

With $$\Delta_r G^\circ = - n F E^\circ , $$ we can establish a relationship between the reduction potentials:

$$n_1 F E^\circ_1 = n_2 F E^\circ_2 + n_3 F E^\circ_3$$

Cancelling F and solving for $E^\circ_1$ allows us to calculate the reduction potential of the iron/iron(III) half reaction:

$$n_1 E^\circ_1 = n_2 E^\circ_2 + n_3 E^\circ_3$$

$$E^\circ_1 = \frac{n_2 E^\circ_2 + n_3 E^\circ_3}{n_1}$$

$$ = \frac{1 \cdot \pu{0.77 V} + 2 \cdot (\pu{-0.44 V})}{3}$$

$$ = \pu{-0.367 V}$$


What is an alloy called when it melts at the same temperature as something else? What is an alloy called when it melts at the same temperature as something else? For example, if there is an alloy that melts at the same temperature as glass then it will be susceptible to enameling and it is called an XXX-alloy.

As I recall, the term begins with the letters "eu-".

Isotectic substances are those which melt at the same temperature.

Ref.: Heycock, C. T.; Neville, F. H. "On the freezing points of triple alloys of gold, cadmium, and tin". J. Chem. Soc., Trans., 1891, 59, 936-966

But be aware that this term is very rarely used.


Posible Pair of Enantiomers What is the number of possible enantiomers in monochlorination of 2-methybutane?

My approach

When $\ce{Cl2}$ is added to any alkane it prefers $3^\circ > 2^\circ > 1^\circ$.
When it is being added to 2-methybutane, final product (major) formed will be 2-chloro-2-methybutane, which will show 0 chiral centre, hence the maximum number of enantiomers will be

$$2^n = 2^0 = 1$$

My book gave me answer $2$, but no explanation is provided.

Assuming monochlorination of 2-Methyl Butane ,the following products are possible as shown in the figure below.


What is the significance of the kink in the [A] vs T graph for a 2nd order reaction? The 2nd order graph shows a bit of a kink, whereas the 1st order graph is smooth.

What is the significance of this kink, and why does it appear in only the 2nd order graph.
What is the significance of the numerical value at this kink? The graph below show it occurs at $[A]=1.5-2$.

There's no significance to the shape, that's simply how any graph of the form $y = 1/(c+x)$ looks like ($c$ being any constant you like).
Even if the supposed "kink" has some significance, there isn't any significance to the number, especially since the number depends on your system anyway (it's dependent on the value of your rate constant as well as the initial concentration).


Does any compound exist with multiple liquid phases? I've seen some phase diagrams and I found really interesting, how many exotic solid phases of most elements exist on higher pressure and temperature.

But I've seen always only a single liquid phase and also only a single gaseous phase.

The second is understable, but the first not. I think, if there is an unconventional potential energy field between the molecules, maybe it is not impossible.

Does any chemical compounds with multiple liquid phases exist?

If not, why not?

Extension: Well, superfluid helium is one of them. Maybe I don't formulate enough well, but I think on a distinct phase border, like between ice-I and water. As I know, the superfluid helium is a mix of its superfluid and normal fluid phases.

There's no reason why chemical substances with multiple liquid phases shouldn't exist.

In fact there is evidence for the existence of a least a second form of (surprise) liquid water. If you search the Web for liquid-liquid phase transitions you will find more of these, e.g. cerium.


Phosphorous acid vs phosphonic acid? I want to know the difference b/w phosphonic acid and phosphorous acid.

The formula for both is H3PO3.

Websites give me conflicting answers.
Which one of them is dibasic and which one is tribasic?

The difference is that phosphonic acid ($\ce{HPO(OH)2}$) is an existing, shelf-stable compound while phosphorous acid ($\ce{P(OH)3}$) is extremely short-lived if at all and immediately tautomerises to phosphonic acid.

All the -ous acids are theoretically produced in the same manner: take the -ic acid and remove one unprotonated oxygen, leaving an additional electron pair with the central atom which reduces its oxidation state by two. So while sulphuric acid is $\ce{SO2(OH)2}$, sulphurous acid is $\ce{SO(OH)2}$.

Likewise, you can always theoretically generate the -onic acids by tautomerisation from the ‘parent’ -ous acid; one proton that previously was attached to an oxygen bonds to the central atom‘s newly-formed (when coming from the -ic acid) lone pair. Thus, sulphonic acid is $\ce{HSO2(OH)}$.

In case of sulphur, while both acids have been observed and crystal structures of both have been determined to the best of my knowledge,[citation needed] the two can still tautomerise. In the case of phosphorus, only the phosphonic acid is shelf-stable and all isolated salts have been determined to be phosphonate salts.[citation needed]

In both cases, however, organic esters have been generated in both ways. Thus, you can have e.g. triethyl phosphite $\ce{P(OEt)3}$ or diethyl ethylphosphonate $\ce{EtPO(OEt)2}$.


How can we write the wave function in quantum mechanics? In quantum mechanics, while writing the wave functions we can take a product of the spatial and the spin parts.But what does it means? What is the meaning of the spatial wave function and the spin wave function?

The wavefunction contains all the information about the system of interest.  This is a basic premise of quantum mechanics.

Within the Born-Oppenheimer approximation, we 'index' all the values required to describe where an electron is.  This includes the spatial coordinates, ($\textbf{r}$), and the spin coordinate, $\omega$.  Electrons are characterized by their spin ($\uparrow$ vs. $\downarrow$).  Another way to think about it is this.  The quantum numbers are used to describe everything we need to know about 'where' an electron is ($n$, $l$, $m_l$, $m_s$).  The spatial coordinates (e.g. Cartesian coordinates) take care of the first 3 quantum numbers.  We need the fourth coordinate to characterize $m_s$.