Are these molecular structures of cyclodecapentaene identical? I think that the above structures are identical compounds as I doesn't exist in the shown form but it would exist naturally in the II form, so these should be identical.
But these are mentioned to be geometrical isomers in a book that I was reading. Even if I assume that I existed in it's given form, then how can they be geometrical isomers? In I, there are two trans alkenes, three cis alkenes and in II, all five are cis alkenes.
So, are these compounds isomers or identical?

If you make spring-and-ball models of both compounds, you can see that the two compounds are entirely different, and both are contorted.

The all-cis model on the left has all its hydrogens pointing outward, so the contortions of the carbon chain involve just carbon-carbon twisting to come close to 120 degree angles. The model of the cis-trans molecule on the right has two inward-pointing hydrogens (all the rest point out) which are sterically interacting so much that they introduce strain that is not evident from the carbon-only diagram in the original post. (If I put in all the hydrogens, the image is too cluttered.)
The fact that one is all cis and the other cis-trans certifies that the molecules are different, but the 2-D drawing can seem vague. The NMR spectra of these two compounds would be different. I can imagine the molecule on the left flipping all around (the model sure did!), but the molecule on the right seems pretty stuck.
Thinking in 3-D is easy after you've done it many times. But when you are starting out, it is helpful to make molecular models. And even if you've done it a million times, there are situations where your mental image is so fluid that you need to solidify it with a model, no matter how inadequate the model is. And then you mentally add onto the solid model.


What caused the flashes in the video footage of Chernobyl? There is video footage of nearby villages in Pripyat, in the first week after the Chernobyl disaster. When I watch it there are villagers leading normal lives in normal clothing. Some Russian nuclear scientists pass by in full radioactive suits. Then there these clearly visible "flashes" in the air; I have looked on youtube but haven't found the video that I watched a long time ago. My layperson understanding is that these flashes are caused by an extreme amount of radiation.

Here's footage that shows the flashes in the air. They are subtle orangish flashes that are quick. But there's one for instance just after 3 minutes.
This isn't the original one that showed the flashes better. Anyway:
Chernobyl

From Wikipedia "Ionized-air glow":

  Within minutes after the steam explosion that caused the Chernobyl
  accident at 01:23 local time, a number of employees at the power
  station went outside to get a clearer view of the extent of the
  damage. One such survivor, Alexander Yuvchenko, recounts that once he
  stopped outside and looked up towards the reactor hall he saw a "very
  beautiful" LASER-like beam of light bluish light, caused by the
  ionization of air, that appeared to flood up into infinity.

My questions are:

What amount of radioactive material is needed to make those flashes in the air happen?
Is radiation alone enough?
Are other factors such as weather, temperature or other prior chemical elements in the air needed to allow this flashing to happen?

The white flashes seen in the video are the result of radioactive contamination of the original film. Airborne radioactive particles were deposited onto the film and caused a local overexposure of the negative, quite similar to the historic experiment of Antoine Henri Becquerel. That's why the flashes only appear in individual spots in individual frames of the film. (This is also explained in the documentary The Battle of Chernobyl, which also contains these film excerpts.)

There was no corresponding visual phenomenon in the air that could be directly observed by people in Pripyat.

The ambient dose rates measured in 26 locations in Pripyat on 26 April 1986 at noon (i.e. about the time when the film was recorded) were about 0.038–1.9 R/h (i.e. about 0.38–19 mGy/h; with a geometric mean of 2.4 mGy/h). Such dose rates cannot cause any directly visible effects in air.

The flashes on the film are not related to any visual phenomenon that was caused by nuclear reactions, radiation, fire or other chemical reactions at the open reactor during the night of the accident.


Why don't decomposition reactions form pure elements as products? For example, if I had a reaction like this:

$$
\ce{NaHCO3 -> Na2CO3 + H2O + CO2}
$$

Why does it not break down all the way down to its elements?
What makes it form such "intermediate" products, stopping at water, carbon dioxide, etc.?

Using your example, you simply have not given the system enough energy to do so. Water and $\ce{CO2}$ are very stable compounds under ordinary conditions of temperature and pressure, and thus are the preferred products under standard conditions for that reaction.

If you were to heat the system further, further decomposition would occur. For example, at higher temperatures you will convert $\ce{Na2CO3}$ into $\ce{NaO}$ and more $\ce{CO2}$.  Under extreme conditions, i.e. thousands of degrees K, you will ultimately have enough energy to break all of the chemical bonds and then you will be left with the elements.


Why is water wet and fire hot? This may be a silly question but why is water wet?
All the previous questions on water do not explain the reason why is water wet. They assume that its a natural property. However we need to understand why is it wet in the first place. This could apply to other liquids such as milk, juices but the essential element is water.

Similarly another basic property which we take for granted is heat caused by fire. What causes fire to be hot as an inherent property? For the purpose of discussion consider simple fire by burning wood or paper.

Why is water wet?
Chemically speaking saying that water is wet has not much sense.We could, however, say that "Wetting process" is caused by the presence of a thin liquid particles layer over a material. In our specific case, this depends on the chemistry of water and the chemistry of skin. So the main causes of the fact that water is wet, is its wetting properties determined by a force balance between adhesive and cohesive forces over the skin. Mercury, although is a liquid, will not "wet" you because it has a strong cohesive force that doesn't permit the molecules to get in touch with the sensory cells.  When you touch water or a water wetted material part of the water molecules are transfer to your hands (a hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be caused however in some case of parasthesias even if there are no liquid particles. In fact, there is no "wet receptor" on your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in your skin.
Try to play with plasticine after a while handling it your hands will become hydrophobic you will no more fell the wet sensation when you wash them even if water is still "wet"!
What causes fire to be hot as an inherent property?
Fire is an exothermic process. So energy stored in a chemical bond is released as kinetic energy and electromagnetic wave. The fire has a high temperature because the molecules have great kinetic energies when you get in touch with fire (of course wormed air could be a medium as well for short distance) this kinetic energy is transfer to your skin and so to your thermoreceptors . And you feel what human call hot. In fact, you can feel the sensation of hot even without touching the fire because it even irradiates energy as electromagnetic waves that can be absorbed by your skin and so be feel for his receptors.


Is scandium considered a transition metal? I have seen so many conflicting answers to this question in various places so I wanted to ask it again.
IUPAC defines a transition element as

  an element whose atom has a partially filled d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

By this definition, scandium would be a transition element as its atom has an incomplete d shell.

However, I have seen many other definitions which require an incomplete d shell as a stable ion. This also adds confusion as the common scandium ion is $\ce{Sc^3+}$ which has no d electrons, so wouldn’t be a transition element, yet it’s my understanding scandium can display a +2 oxidation state also (albeit less commonly).

What’s the correct definition? Is scandium considered a transition metal or not?

It may depend on who you ask, and maybe even when you ask.  The tl, dr of what is below is that we are moving towards counting scandium as a transition metal.
Wikipedia follows the IUPAC definition and accepts the label for scandium because it forms atoms with partially filled $d$ orbitals, and also copper because it forms ions with partially filled $d$ orbitals in common settings.  Either possibility is admitted by IUPAC and thus by Wikipedia.
On the other hand, Britannica does not use the label "transition metal" for Group 3 elements, because they do not retain $d$ electrons in their ions.  Instead they render a separate classification, the "rare earth metals", in this group.
Britannica is based in the UK, of course, and we may be seeing a difference based on different countries.  Let's look at a couple other sources:
https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/trans.php, based in the United States, says yes, again following the IUPAC definition.  https://www.chemguide.co.uk/inorganic/transition/features.html, based in the UK, reports that most UK syllabuses use a definition that excludes scandium from the transition metals.  The latter source does state, however, that "the most recent IUPAC definition" would include scandium because of its neutral atom configuration.
The development of organometallic chemistry may have something to do with this evolution.  Many elements, including magnesium and more recently calcium, show lower oxidation states in some organometallic environments.  In the case of scandium this shows up as an atom retaining its $d$ electron while using the $s$ electrons and/or vacant $d$ orbitals for bonding to organic ligands.  Going back to the Wikipedia article on scandium, this gives references to several organometallic examples in which scandium has reduced oxidation states, from +2 all the way down to 0.  For example, Cloke et al. [1], show a single scandium atom in a low oxidation state in a setting similar to how a later transition metal might appear:

Cocondensation of scandium vapour with 1,3,5-tri-tert-butylbenzene affords the first example of a scandium(0) complex, the sandwich compound bis(η-1,3,5-tri-tert-butylbenzene)scandium; ESR and UV studies also the reveal the presence of a second compound from this reaction, formulated as a similarly unique example of scandium(II) arising from activation of one of the methyl groups of a tert-butyl ring substituent.

Such developments favor, and may have motivated, the broader, scandium (and yttrium) inclusive definition of a transition metal now used by IUPAC.
Reference
1.
F. Geoffrey N. Cloke,   Karl Khan  and  Robin N. Perutz, "η-Arene complexes of scandium(0) and scandium(II)", J. Chem. Soc., Chem. Commun. 1991, 1372-1373.


Is following graph is of a dibasic or monobasic acid? Is the above graph is of a dibasic or monobasic acid?
There is slight hint of it being of dibasic around 15. But is it enough to conclude ? I am confused.

As I understand it, if the data is absolutely correct, the curve should not change direction of curvature in that manner if it is monibasic. Each time it's double differentiation changes signs 2 times, I think it passes an equivalence point no matter how shallow it is. It should be a dibasic acid. You can get a function of [H+] using equilibrium constant (assume it to be monobasic) and taking x as moles of naoh added in other words, the "initial" moles of salt x is and "initial" moles of acid is a-x, a is the moles of acid we started with.By "initial" I mean the moles before equilibrium. You would get a quadratic equation, see at how many times sign of double differentiation of -log([H+]) wrt x changes for any value of x. Here is oxalic acid's curve.


What does the salt actually do during electrolysis? Pure water does not conduct electricity.
When you dissolve a salt like $\ce{Na2SO4}$ in water the solution becomes a better conductor.
If electrolysis is done on this solution, oxygen and hydrogen gases form at the electrodes.
The ions $\ce{Na+}$ and $\ce{SO^{2-}4}$ don't seem to be doing anything further.
The electrons from the cathode don't involve the ions:
$$
\begin{align}
\ce{4 H+ + 4 e- &-> 2 H2} \\
\ce{4 OH- &-> 2 H2O + O2 + 4 e-}
\end{align}
$$
How are the salt ions helping here or what are they doing?

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's see what's going on at each electrode.
At the cathode (negative pole), water is decomposed according to : $\ce{2 H2O + 2 e- -> H2 + 2 OH-}$. This is only possible if there are some positive ions available near the cathode to compensate the negative charges appearing in the solution with these newly created $\ce{OH-}$ ions. Without positive ions like $\ce{Na+}$ near the cathode, these new $\ce{OH-}$ ions would repel further electrons coming from the outer circuit and prevent them from arriving to the electrode and from reacting with water. This would stop the electrolysis.
At the anode (positive pole), water is decomposed according to : $\ce{4 H2O -> O2 + 4 H+ + 4 e-}$. This is only possible if there are negative ions arriving near the anode to compensate the positive charges appearing in solution because of these newly created $\ce{H+}$ ions. Without negative ions like $\ce{SO4^{2-}}$ in the solution near the anode, these new $\ce{H+}$ ions would make a positively charged solution. This would prevent further electrons to be ejected from water. This would stop the electrolysis.
At the end, the composition of the solution is modified around the electrodes. Near the cathode, the solution contains more and more $\ce{NaOH}$. Near the anode, the solution contains more and more $\ce{H2SO_{4}}$ Of ourse if you stir and mix anodic and cathodic solutions, the acid and the base neutralize each other, which regenerate the solution of the original salt.


Hierarchy of electronic wavefunctions The previous question contained too much unnecessary information and was edited.

I am wondering about the "hierarchy" of wavefunctions. If one can combine atomic orbitals (AO) into molecular orbitals (MO) through the LCAO method, could one combine MOs to form a supra wavefunction, or an LCMO?

I've done some research and found a paper on an approach to finding the MO of molecules by effectively taking a LCMO of smaller molecules or constituents. While interesting, this is not what I'm referring to as a "supra" wavefunction, if there's such a thing.

I've been reading up on this, and will take a shot at the answer. It is going to be an interesting answer, and probably not what you expect. There has been an interesting conversation going on in the Journal of Chemical Education in response to this letter advocating the retirement of the localized hybrid orbitals because they "aren't real" and cannot be used to predict "all properties" of molecules. Rebuttals included comments that delocalized molecular orbitals also fail to predict "all properties". However, the final word in the discussion came from this letter, which reminds us what orbitals are.

Orbitals are not real. The electrons that populate orbitals are real. Orbitals are mathematical constructs created to describe the curious wave-like probabilistic quantum mechanical properties of electrons. When we talk about atomic orbitals, we generally mean the 1-electron hydrogenic atomic orbitals (HAOs), which are mathematically represented as a wavefunction ($\Psi$) and graphically represented by the probability density function ($\Psi ^ * \Psi$) at either 90% or 95% level. Despite valiant and massive efforts, wavefunctions for multi-electron atoms cannot be solved exactly. They are approximated as a parameterized HAOs. When we talk about the atomic orbitals of carbon, we usually mean the HAOs that closely approximate what we think the AOs of carbon should be.

Molecules and their orbitals are even more troublesome. The LCAO approach creates a bunch of nice 1-electron wavefunctions out of the set of HAOs. These methods give rise to both the canonical delocalized molecules orbitals (CMOs), and the localized valence bond orbitals (VBOs). As the authors of that letter point out:

  The term canonical implies that CMOs are not the only possible MOs, and there are many orbitals that generate an identical “chemical reality”.

The approximate polyelectron wavefunction of a molecule (at least in HF theory) is the Slater determinant of a series of 1-electron CMOs $\psi _i$:
$$  \psi _1 \overline{\psi} _1 \psi _2 \overline{\psi} _2 \psi _3 \overline{\psi} _3 \psi _4 \overline{\psi} _4 ... $$

An interesting property of Slater determinants is that unitary operations (like rotations) on the set of CMOs return a different set of MOs, which produce the same expectation values for the polyelectron wave function as original set:

$$ \psi _1 \overline{\psi} _1 \psi _2 \overline{\psi} _2 \psi _3 \overline{\psi} _3 \psi _4 \overline{\psi} _4 ...~~ \underrightarrow{unitary~~transformation} ~~\psi _a \overline{\psi} _a \psi _b \overline{\psi} _b \psi _c \overline{\psi} _c \psi _d \overline{\psi} _d ...$$

Mathematically, it seems to me, that a linear combination of MOs that are themselves linear combinations of AOs would still be a linear combination of AOs, just a different one than you started with. If that is true, then all you have is a different set of MOs than you started with, and according to HF theory, that should give you the same expectation values.

Thus a set of "LCMOs" are no more or less valid than the original set of LCAOs. They likely do not produce any additional insights into the structure of the molecule. However, the approach might be computationally more efficient when trying to determine the MOs of very large molecules and polymers.


What does the phase diagram mean by "liquid" at NTP? (The liquid and vapor coexist even when the situation is not on the curve) [Solved. For anyone who is also confused, see the comments under Poutnik's answer. As Poutnik reminds me, remember there is no air involved. I successfully picture that with the piston experiment in Karl's answer.]

When I read the phase diagram, I think it suggests that "at certain temperature and pressure, the phase of water would be (solid, liquid, gas) or two/three of them coexisting."
But when I look up NTP on the diagram, it says "liquid", while the saturation pressure of water at 25°C is 0.03atm. So the situation can't exist since the vapor is going to condense until the pressure drops to 0.03atm.

Then what actually is the phase diagram suggesting by "liquid"?
What am I missing here?

Edited: Poutnik reminds me that the pressure is the total pressure, not the vapor pressure. Then my question should be "If the vapor and liquid coexist not only at the curve, then what is the meaning of 'liquid' on the diagram?"

The phase equilibrium diagram is for a system consisting of just water.

The diagram pressure is the total system pressure.

It may be equal to the saturated vapor pressure, if no external pressure is imposed on the system.

Or, it can be ( much ) higher, if some external pressure higher than the saturated vapour pressure exists.  In such a case all vapor is at equilibrium transformed to liquid or solid phase.

Really high pressures can be seen in the water phase diagram for very high pressures, displaying rare ice phases.


Find number grams of the solute in a dissolution given its density and Molarity Have a dissolution of $\ce{HNO_3}$ in water. The density is
  $1.42\ce{g/mL}$ with a concentration of $\ce{16M}$. What is the
  concentration in % mass/mass?

Alright. So the % mass/mass is expressed as

$$\frac{\text{grams of } \ce{HNO_3}}{\text{grams of the dissolution}}\cdot100$$

So I need to find the grams of $\ce{HNO_3}$ first. They tell me that the Molarity is

$$16 = \frac{\text{moles of } \ce{HNO_3}}{\text{liters of dissolution}}$$

$\ce{HNO_3}$ has a molar mass of $63$. So to find out how many moles of $\ce{HNO_3}$ are in the dissolution I need to divide the grams of $\ce{HNO_3}$ by $63$... But how am I supposed to find out the grams of $\ce{HNO_3}$?

I was told that the numerator is $63$. But... $63$ is the molar mass of $\ce{HNO_3}$, not the number of moles of $\ce{HNO_3}$ in the dissolution.

Anyway, let's suppose that it is indeed $63$. Therefore

$$16 = \frac{63}{\text{liters of dissolution}} \implies \text{liters of dissolution} = 3.9$$

So there are $3.9$ liters of dissolution? I'm not sure - I feel like I got this all wrong. How do you know there are $63$ moles of $\ce{HNO_3}$?

Using 1L of solution, you have 16M of HNO3, each mole of which weighs 63g. Therefore, in 1L you'd have 63 X 16 = 1008g. Mass of 1L is given as 1420g from the density.

%mass/mass = 1008/1420 = 70% HNO3


How can we recreate old pharaos' perfumes? There is ongoing research where efforts were launched to find out the original constituents of a perfume for the female pharaoh Hatshepsut:

(Source)

In a perfume flask that was found in her tomb, there are apparently some remnants of the (assumed once-liquid) perfume that she used to wear. The perfume flask was sealed, so there was probably a very low amount of gas exchange taking place during the millennia that the flask took to wait for being discovered with its treasure.

However, there should still be a lot of reactions that can happen in this timescale, and since most of the reactions that just happen by themselves tend to lower energies, at some point different organic molecules would lead to the same product (e.g. acetic acid, or similar).

It seems to me, that then the only reasonable approach is to make a list of possible constituents and then start to play with the amounts thereof and model the decay reaction pathways. Then you could check if your theory predicts the experimental results.

Is this really how you find out what an ancient perfume was made of?

References:
Here, this, this one and that (talking about Tutankhamun, but the same principles would apply).

The analysis won't be done before lunch, but neither will it be an extremely difficult process.  Certainly some of the researchers will have a fragrance industry background  and access to necessary equipment like  a tandem gas chromatograph - mass spectrometer.  Also, Pharoah's perfumes were undoubtedly less complex than todays, and all oils used in perfumes are available as "known" compounds for comparison.  Additionally, all fragrances have been "fingerprinted" making identification of the compound in a sample fairly routine and straightforward.  By "fingerprinted" I mean that if a certain pattern of compounds can be detected than they will suggest the presence of a certain oil.  For example, look at lavender oil, there are about 30 chemical components in it - just like a fingerprint.

Perhaps the base solvent has evaporated through the wall of the container and oxygen has flowed into the container.  Certainly some of the ingredients will have degraded, but the container won't be full of acetic acid.  Some amount of an alcohol may have oxidized to the corresponding ketone, aldehyde or carboxylic acid, perhaps some amount of a conjugated diene may have oxidized to the corresponding dienol; but that shouldn't make the process that much more difficult.


Remove empty space from structures with R and S groups We receive structures from a server within our company. Those structures are transferred as MOL files. Some of these structures contain R and/or S groups. Some of the creators of these structures create a really large empty space between the actual structure and the groups. We have currently ~30.000 structures in our database. Around 2.000 of these structures are flawed.

I have attached how such a structure might look at the bottom, as the image is quite large.

What I am trying to do now is to remove that empty space.

The process is currently receiving the MOL files wrapped within an XML file that does have additional information to store along with the structure. That process is written in Java. Up to now we do not have any library within the project to process and manipulate structures, as we have just parsed the MOL as String from the XML file and written it into a CLOB field in the database.

I know that libraries like the CDK exist, but right now I do not know if it can help me with that task.

I have pasted the actual mol file here.

If we open it with a viewer it looks like this:

Are you sure that the R and S are in the mol file and not part of the xml envelope?

Stereochemical R and S descriptors would not make any sense for a molecule like anthracene!

Is it conceivable that the letters come from Risk and Safety statements?

It would be helpful if you could upload the original mol file somewhere.

EDIT 1

The file that you kindly uploaded is supposed to be a V2000 mol file, but it is not standards-compliant.

(The full specifications can be obtained from Accelrys upon registration.)

The first flaw is already in the header! The header is supposed to have a length of three lines.

The following CTAB should start with the counts line (in line 4 - note that we're not using zero-based indexing here).

With this flawed format, OpenBabel, a standard tool for the conversion of chemical data formats, casts an error.

EDIT 2

The rest of the file, including the stereochemical descriptors, seemed to be fine.

Once I had inserted the missing blank line in the header, a PNG could be generated using

obabel -imol 498268e9.txt -opng -O stuff.png

OpenBabel comes with bindings for Python, Java, C++ and whatnot. You might want to talk to your IT about some refactoring ;)


What is the difference between hexanes and petroleum ether I'm trying to use hexanes as the nonsolubilizing recrystallization solvent in a multisolvent recrystallization (benzene is the solubilizing solvent) that requires an oxygen free atmosphere and hexanes is present in the glovebox. I am using it in place of light petroleum ether (not in glovebox) and was wondering what the differences between hexanes and light petroleum ether were to determine if I can adequately substitute the two.

Petroleum ether:
This is a mixture of a lot of different compounds (aliphatic hydrocarbons in this case) that as a mixture have a boiling point within a specified range. The contents are dealt with as having low amounts of aromatic compounds and high amounts of this and that, not one specific formula. It's like a blackbox to me and can be used for crystallisations that are usually easy to do.

Hexane:
In the synthesis laboratory I worked in, hexane was dealt with as we would deal with petroleum ether, since it was a mixture the many possible constitutional isomers and not pure n-hexane. We bought the technical grade hexane in our chemical store and purified it with a solvent purification system to have it dry and compatible for glove box use. Since they appeared so clean in our Schlenk flasks I thought in the beginning that it was pure n-hexane but a simple 1H NMR spectrum taught me otherwise.
The hexanes mixture was used for medium difficulty crystallisations.

Pentane:
The pentane we bought and purified, however, was n-pentane and was used for difficult crystallisations that were not easy to reproduce. Already among organometallic compounds there are examples in which obtaining crystals at all is an enormous struggle and having single crystal quality is simply a matter of hope.

Therefore, by going from petroleum ether to hexane to n-hexane (or n-pentane) is a matter of going from a wide mixture to a compound to a pure chemical.


Help balancing reaction between Iron (II) chloride and potassium phosphate Very rusty with chemistry and looking to review some basics. Came across the following:

  Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

What I have so far (guessing, basically):

$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$   [still need to balance but correct right hand side?]

I can't remember the exact rules for forming the right hand side. I just put this in place to match ion charges. What prevents me from writing the right hand side as something else entirely (not sure what but I'm asking, "well, couldn't I figure something out if I break everything apart and re-arrange into another option?").

EDIT: Additional Question Below

Ok, so assume the above reaction was balanced (it's not) and valid. Then, my question is does the reaction go both ways ($\ce{<-->}$)? If so, when/why/how? I'm trying to think back many years ago and it has something to do with "reactivity" and the row of the elements on the periodic table, correct? I'm going to look into it more myself but if someone has a quick "30 second" answer I'm sure it'd be of help to me!

[Assuming this is OK and valid in the forward direction. Again, needs to be balanced.]

$$\ce{FeCl2 + K(H2PO4) -> KCl + Fe(H2PO4)2}$$

[Is this ever ok/possible?]

$$\ce{KCl + Fe(H2PO4)2 -> FeCl2 + K(H2PO4)}$$

This question was the continuation of your question and since both question are related,  I am answering the question by emphasizing the mistake you are doing and adding a relevant portion to explain it.

Mistake:

Iron(II) chloride and potassium phosphate react. Write and balance the molecular, ionic, and net ionic equations.

You need to consider potassium phosphate to be $\ce{K3PO4}$ not $\ce{KH2PO4}$ according to IUPAC's Red book (see sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.)

If $\ce{KH2PO4}$ is meant to be added, the correct term should be potassium dihydrogenphosphate.

Extra info.:

Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, only iron (II and  III) phosphate and potassium chloride forms from the reaction.

$$\ce{3FeCl2 + 2K3PO4 ⟶ Fe3(PO4)2 + 6KCl}$$

$$\ce{Fe3(PO4)2 ->[air oxidn⋅,400C] FePO4}$$

or actually

$$\ce{4Fe3(PO4)2 + 3O2 ->[400 C] 8FePO4 + 2Fe2O3}$$

Iron hydrogen phosphate can be made by reacting iron(II) phosphate with dilute hydrochloric acid.

$$\ce{Fe3(PO4)2 + 2HCl → 2FeHPO4 + FeCl2}$$


How to remove tap water hardness? I am looking for ways of removing water hardness from tap water, specifically permanent hardness.

I have tried working on some methods, e.g., to perform precipitates using sodium hydroxide and calculated calcium hydroxide, but it seems that either my methods or something else is wrong.

Please recommend me some methods (or resources) on how to remove the hardness from tap water.

Without knowing what dissolved species are causing the hardness, it's a little difficult to know what to advise. Performing qualitative analysis would likely help you a lot.

You describe using lime softening (i.e., using hydroxides) which can help in some cases. Usually the resulting hydroxide precipitates are pretty hard to filter, requiring a membrane or a very good filtration system.

As Wikipedia notes, there are a few other reliable techniques:

reverse osmosis
distillation
ion exchange resins
chelating ligands

While the first three are definitely the best in terms of ultimate water quality, an easy thing to supplement lime softening would be to add various chelating ligands. For example, after you carefully filter the water from lime softening, maybe add citric acid?

In any case, I'd suggest you do some sort of qualitative analysis (and pH) testing as you go. This is the best way to know what steps you'll still need.


Predicting major product of a reaction with nitrous acid $\hspace{4.5cm}$

For the above reaction I am asked to predict $\ce{X}$, the major product.

Of the options given, I thought product (1) would be correct, seeing as how it has the most stable intermediate cation (tertiary) and a highly substituted double bond.

$\hspace{3.9cm}$

The answer however states that product (2) is the major product.

I can't understand why the second product would be favored over the first. There's no obvious stabilizing factor here in product (2).

I think (2) is correct, but either your or your book erred in writing it. This is what the mechanism (and major product) should be:


What is the opposite of a surfactant? Surfactants decrease surface tension between two fluids.  What do you call a substance that increases the surface tension between two fluids?  How is this effect acheived?

Briefly, the cause for surface tension is the difference in the energetic state between a molecule at the surface compared to a molecule inside the liquid phase: the molecule inside has lower energetic state as it has more inermolecular interactions (or, to get to the surface, a molecule needs to break intermolecular bonds, which needs energy). This is more pronounced with higher polarity or when ions (charges) are around as the interactions are then stronger.

You can increase the polarity of a (polar) solvent to get higher surface tension, by dissolving salts.


Do polycarbonate and/or polyurethane have latex? I have a severe latex allergy which hinders my health every day via hives, inflammation, and breathing trouble. I am trying to find a cell phone that is not going to cause anaphylaxis or other reactions. Neither smartphones nor their companies are smart enough to know what their products are made of.

I know from wheelchair experience that polyurethane bothers me. I believe it has latex or a latex-like component in its make-up, but I would like to clarify this.

It appears by touch that polycarbonate is not as bad, but it just depends on the item in question...and usually I have already reacted to something else before trying to touch this.

Could you verify/explain if polycarbonate or polyurethane have latex? For that matter does polyethylene?

We have a general rule against "medical questions," but I think yours is different.

First off, latex is a form of poly-isoprene. The "isoprene" part here is the key chemical identity - the letter (so to speak) that repeats in the polymer.

So latex is chemically quite distinct from polyurethane and polycarbonate. Similarly, polyethylene and polypropylene are also distinct.

In short, from a chemical perspective, all of these materials are all different.

Now, that's not to say that:

You may have allergies to multiple polymers / plastics.
Manufacturers may blend multiple chemical components into one object.
Some latex residue may end up on various objects.

For example, many people in healthcare (as you know) use latex gloves, often with latex powder. Small amounts of the latex powder can remain on a wheelchair - causing problems for sensitive individuals like yourself.

If you're looking for a smartphone, you should be able to find one with a metal / glass case (e.g., iPhone, but also others). In that case, the amount of plastic (polymer) should be minimal and you (or a friend) can thoroughly wipe down the device before you use it (e.g, the wrapping plastic might cause sensitivity).


Negative ions and health: pseudo science or something more? Over the last year or so I've heard from friends, who subscribe to holistic medicine, talk about negative ions, saying that they are good for your mental/physical health. Being the only one among them with a chemistry background, I was suspicious when i heard this. Below are examples I've seen that use negative ions in a medicinal context:

Negative Ions Purify Air

  Virtually all particles in the air have a positive charge, while negative ions have a negative charge. In which case, negative ions and particles magnetically attract to one another. When there is a high enough concentration of negative ions in the air, they will attract to floating particles in large numbers. This causes the particle to become too heavy to remain airborne. As a result, the particle will fall out of the air, and will then be collected by normal cleaning activities, such as vacuuming or dusting. - Ion-Imbalance

Negative Ions from the Beach

  You may have experienced the power of negative ions when you last set foot on the beach or walked beneath a waterfall. While part of the euphoria is simply being around these wondrous settings and away from the normal pressures of home and work, the air circulating in the mountains and the beach is said to contain tens of thousands of negative ions -- Much more than the average home or office building, which contain dozens or hundreds, and many register a flat zero. -WebMD

Apparently they Help You Breath

  People travel from all over the world to visit salt mines like the ones found in the Himalayas, Austria, and Germany. In Germany, the hot salt springs are so revered, companies often send overworked employees there to rejuvenate.
  Why are the salt mines so beneficial?
  Quite simply... the very dry, negative ion environment of a salt mine is key.
  Well... even if you're not able to travel to Asia or Europe to take advantage of salt mines, you can still take advantage of these negative ions to help improve the air you breathe. The solution involves pure Himalayan crystal salt...- Mercola

Well it's not surprising that two thirds of the above sources are trying to sell products (salt lamps, air purifiers).

Since we're dealing with ions here, I figure the chemistry world must have some sort of input. What's the chemical or biochemical support, or the evidence against, for the above claims? If there are negative ions present from waterfalls/oceans/salt rocks, what ions are we talking about?

To address @dissenter, one of these sites claims tourmaline is a source of negative  ions...

Charge separation at waterfalls with airborne ions, resulting in a potential between the base of the waterfall and the surrounding air, is a phenomenon examined by Philipp Lenard (Über die Elektrizität der Wasserfälle, Ann Phys (Leipz), 1892, 46, 584–636).

The effect is real and more recent field studies, such as Characterization of ions at Alpine waterfalls, as well as studies on the formation of charged droplets, such as Charge Separation in the Aerodynamic Breakup of Micrometer-Sized Water Droplets should be taken seriously.

However, the sources that you cited are pseudoscientific marketing bullshit.

The first quote links to a web shop that offers marvels like magnetic bands and ion-balance bands. While the silicone band probably isn't hamful to the skin, it will not emit anything. Nada, nothing, nix!

The webshop behind the third link sells Himalayan salt lamps. It's a matter of taste to have a lamp housing made from unrefined sodium chloride with traces of iron in it (hence the colour), but I wouldn't expect any health effects through the emission of whatever from these lifestyle items. Himalayan salt is being sold for quite a while, even for nutrition but if at all, the stuff has seen the Himalaya mountains only from a distance.

There are no halite mines in the Himalaya mountains.

Instead, the material is likely to come from the Salt Range in the Punjab province in Pakistan, or from Poland.

In summary, it's business as usual: Scientific facts are mashed up with esoteric mumbo jumbo to sell overpriced and useless stuff to well-meaning but naive clients.

Oh what a tangled web we weave,
When first we practise to deceive!
-- Sir Walter Scott


Why is dinitrogen tetroxide a planar molecule? I was wondering why $\ce{N2O4}$ is a planar species in spite of having a σ-bond in between the two $\ce{N}$ atoms for free rotation. Does it not form a conformer?

I think that one would expect that a stable form of the molecule might be planar and have D2h symmetry.

Some support for the claim:

Early crystal structures suggested that the ONO bonds on either side
of the N-N bond are easily modeled by low level DFT
(doi.org/10.1039/C19660000082).
More recently, simulations of the trans and cis N2O4 structure show
planarity when optimized at B3LYP/ aug-cc-pVTZ (aka, pretty good not
amazing) See here.

The simulations also suggest that the barrier to isomerization is low, meaning that while the stable structures are planar, there are low barrier (low temperature) out of plane rotations possible.

Summary:

Just because something has sigma bonds doesn't imply that it will immediately be non-planar
yes the molecule can form a conformer


Electrode potentials in a potentiostat I am not a chemist but for a university project i am dealing with simulations concerning a potentiostat using a 3-electrode setup.

For example in the book Electrochemical Methods (Bard & Faulkner, Second edition 2001) on page 642 they give this example of a (adder) potentiostat which inputs three different potentials $e_1, e_2$ and $e_3$:

While the working electrode is grounded, point $S$ is on virtual ground and by using Kirchhoff's rule they obtain

$$-i_{ref} = i_1 + i_2 + i_3.$$

By using Ohm's law they rewrite this equation into

$$-e_{ref} = e_1 \cdot \frac{R_{ref}}{R_1} + e_2 \cdot \frac{R_{ref}}{R_2} + e_3 \cdot \frac{R_{ref}}{R_3},$$

with $e_{ref}$ being the potential of the working electrode with respect to the reference electrode.

With all resistors having the same value this can be expressed as

$$-e_{ref} = e_1 + e_2 + e_3$$

They state:

  Thus the circuit maintains the working electrode at a potential equal
  to the weighted sum of the inputs

As far as I understand, the electrode of interest is the working electrode which potential should be varied (with the reference electrode providing a stable reference potential) to study its properties. But from the description above the potential of the working electrode does not change and instead the potential at the reference electrode changes.

Bard and Faulkner are outstanding electrochemists, but that potentiostat figure always annoys me because it makes it seem as though current flows through the reference electrode and that the working electrode is literally grounded. In reality, the reference electrode should conduct extremely low currents (pA, really) and the current through the working electrode is what you care about, as a function of the electroactive species in th potentiostat's solution.

Therefore, consider Fig. 15.24 and surrounding text from from Horowitz and Hill 1:

In this three wire potentiostat, the bottom operational amplifier (op amp) simply works as a transimpedance amplifier, converting current at the working electrode to voltage. In this regard, it is basically a very low input impedance ammeter, which approximates an ideal ammeter. So the working electrode is held at virtual ground, which typically means less than a millivolt from true ground.

The potential at the inverting input of the upper op amp must equal the input reference voltage, $\mathrm V_{ref}$, to within less than a millivolt. Note that $\mathrm V_{ref}$ can be from a ramp generator, a function generator or from another op amp, as in the inverting summing op amp in the figure from Bard and Faulkner. Importantly, note that virually no current flows through the reference electrode, because of the extremely high input impedance of the op amp inputs, so it will function properly and not become polarized or damaged.

The potential between the working and reference electrodes is $\mathrm V_{ref}$ because the working electrode is clamped to virtual ground and the potential of the reference electrode is clamped at $\mathrm V_{ref}$. The solution in the cell has an impedance, which is generally complex, and the solution spans the volume between the counter electrode (also called the auxilliary electrode) and the working electrode. The reference electrode is physically placed as close to the working electrode as is feasible.

Now let $\mathrm Z_{c}$ denote the solution impedance between the counter and reference electrodes and let $\mathrm Z_{u}$ denote the solution impedance between the reference and working electrodes. Then we have a non-inverting voltage follower with gain, as shown is this figure:

The output voltage, $\mathrm V_{o}$, is the voltage at the counter electrode. Therefore, $$\mathrm V_{o} = [(Z_{c} + Z_{u})/Z_{u}] \times V_{ref} \tag{1}$$

But the voltage at the junction of $\mathrm Z_{c}$ and $\mathrm Z_{u}$ , i.e., in the solution where the reference electrode is located, is $\mathrm V_{ref}$. Therefore $$\mathrm V_{ref} = [Z_{u}/(Z_{c} + Z_{u})] \times V_{o} \tag{2}$$

If solution impedances change, the upper op amp will source or sink current, as necessary, by changing $\mathrm V_{o}$, thereby maintaining the solution potential, at the reference electrode, equal to $\mathrm V_{ref}$.

1 P. Horowitz, W. Hill, The Art of Electronics, 2nd Ed., Cambridge University Press, Cambridge, ©1989, pp. 1015-1016.


Alginate nanoparticles I want to produce alginate nanoparticles from NaAlg. I choosed nebulizing NaAlg solution into $\ce{CaCl2}$ solution but 2% (m/v) NaAlg solution is quite viscous. What should I do to make it inviscid?

Dilute the feed solution tenfold, etc.  Inelegant.  Try solution engineering,

http://www1.lsbu.ac.uk/water/kosmos.html
Chaotropes and kosmotropes.

http://en.wikipedia.org/wiki/Theta_solvent
Add a little "poor" solvent like ethanol or acetone to ball the polymer, disrupting hydrogen bond crosslinking and reducing chain entanglement.

You want a novel solution not obvious to one skilled in the art.  That qualifies it for patent protection.  A large fraction of the hard answers you need already exist.  They were done by other people doing other things.  Accumulate bone piles of knowledge.  The process for creating ceramic dislocation-pinned nickel superalloys works for Plexiglass, which "dissolves" 30 wt-% copper phthalocyanine as a dye masterbatch.  So it can't be done, so what?  Do it the "other" way.


IUPAC nomenclature of 1,2,3,4-tetramethylcyclobutane I'm having a hard time naming these molecule myself.

For the first molecule I start with one of the stereocenters, give priority "4" hydrogen and "3" to the attached CH3. How are the other two attached nodes given priority "1" and "2" in an anticlockwise fashion, like in the graph below?
For the second one I have the analogous problem, who is "1" and "2" in each stereocenter?

The Cahn-Ingold-Prelog (CIP) assignments of the configurations of the two tetramethylcyclobutanes above are accomplished as follows:
1)  Tetramethylcyclobutane 1: This isomer has all of the methyl groups on the same face of the ring. There are two mirror planes passing through C1-C3 and C2-C4. Numbering the cis-isomer is arbitrary in that there is a 4-fold axis of symmetry present. The common methods for R/S-configurations do not apply. Because the ring carbons lie in symmetry planes, they are stereogenic and achirotopic requiring r/s-assignments. When an assignment is made to one carbon, the same assignment may be applied to the remaining three.
  Temporary assignments for 1a are made to C2, C3 and C4 (r0/s0). Priorities are assigned at each of these carbons such that C1 is the highest priority group.  Returning to C1 the priorities are r0>s0>CH3>H. C1 is of the s-configuration. Digraph 1b illustrates the assignment to C1 in a different way. The horizontal line represents the plane of the ring with C2, C3 and C4 above the plane of the ring and the dotted line (C1) signifies the carbon under assignment. Since r0 precedes s0, C4 precedes C2. C1 is of the s-configuration.

2) Tetramethylcyclobutane 2: This stereoisomer has ring carbons that are both stereogenic and chirotopic. Here R/S assignments apply. There is a plane of symmetry passing through the C1-4 and C2-3 bonds while a two-fold axis of rotation exists between the C1-2 and C3-4 bonds. Applying the methodology of 1a to 2a, C1 is assigned the R-configuration. Referring to digraph 2b, R0R0 precedes R0S0 [like(lk)>unlike(ul)]. Thus, C2 precedes C4. C1 is of the R-configuration.
One may apply this method to the remaining three ring carbons but there is a shortcut. Because C1 mirrors C4, C4 is of the S-configuration. The rotational axis shows that both C1 and C2 are of the R-configuration while C3 and C4 are both of the S-configuration. The numbering system requires that the lowest numbered atoms, if possible, be labeled R.


Do filled Orbitals also hybridize? Recently I was watching a video on Valence Bond Theory, on the MIT Open-Course youtube channel. The teacher there said that

  For hybridization to occur, electron promotion from fully filled orbitals to empty ones is a must.

But here in a book it is specifically mentioned that

  Electron promotion, though common, but is not 'a must' for hybridization.

Also, it is clearly written that sometimes even the filled orbitals take part in hybridization.

I tried finding some examples of such a case on the web, but could not. It would be a great help if someone tells me even one example where the fully filled orbitals also take part in hybridization, with an explanation why it is so.

The link for the 41 min long video is here.

First off, hybridization is a concept chemists developed to help explain reality (their observations).  Just like resonance theory and Huckel MO theory, it is often (but not always) a useful way to explain the world around us. A "rule" on hybridization: hybridization occurs in response to a bonding interaction.  Further, hybridization involves the mixing of filled (or partially filled) atomic orbitals to produce new atomic orbitals that can be used to form "directed" bonds (molecular orbitals) that produce more stable molecular systems than an unhybridized atom might produce.  Finally, energetically speaking, hybridization is a net neutral process.  When low energy AOs are mixed with higher energy AOs, the resultant new AOs have the average energy of the initial AOs.  So if some electrons are "promoted", then some are lowered in energy as well

For example, in the formation of methane, carbon mixes the electrons in the 2s and three 2p atomic orbitals to produce 4 $\ce{sp^3}$ hybridized atomic orbitals.  The energy of these 4 equivalent $\ce{sp^3}$ orbitals is in between that of the starting 2s and 2p orbitals. In the case of methane this use of $\ce{sp^3}$ orbitals to produce a tetrahedrally shaped molecule, produces a molecule that is lower in energy than the alternative molecule formed from unhybridized s and p orbitals on carbon.  In the case of phosphine ($\ce{PH_3}$) the situation is reversed.  The $\ce{P-H}$ bonds in phosphine are directed along the x, y and z-axes.  The central phosphorous atom remains unhybridized.  In the case of phosphine, hybridization does not produce a lower energy molecule, so the central phosphorous atom in phosphine remains unhybridized.

Personally, I like the concept of hybridization and find it quite useful.  But you have to be careful where to apply it.  Hybridization is best when it is used to explain reality after the fact.


Something denser than lead but with lead like properties? I read that tungsten is 1.8 times more dense than lead. What other elements are denser than lead but safe to handle.  I want something that will fit under the keys of my computer keyboard so my pinky doesn't have to press so hard on the key.

Here's a link to a listing of the elements in the periodic table sorted by density.  The density of lead is 11.35 g/cm^3.  If you wanted higher density, no toxicity and to really let your pinky live in style, then gold with a density of 19.32 would be a possibility.  Other relatively available elements that are denser than lead and without toxicity concerns include palladium (12.02), rhodium (12.41), tungsten (19.35) and platinum (21.45). If you're willing to be a bit less dense than lead, then you can reduce price by using copper (8.96), molybdenum (10.22) or silver (10.5).  Alloys of some of these elements are also available at high density, but the cost (due to fabrication costs) is likely to be as high or higher than that the element itself.


Making N,N-dimethylaniline from aniline Can I make N,N-dimethylaniline from aniline just by reacting it with $\ce{CH3X}$ ?

According to me, the lone pair of nitrogen in aniline is not nucleophilic enough (since it is in conjugation with the whole benzene ring) to be able to displace the $\ce{-X}$ group.

Yes this reaction will definitely work. Use a nice leaving group like iodide and the reaction should go easily. Although the nitrogen lone pair is delocalised into the benzene ring it is still quite nucleophilic. Additionally, the delocalisation is restored when the nitrogen loses a proton to get the substituted amine. As such there probably won't be a big problem with forming a quaternary ammonium salt (which is what usually happens if you try to alkylate amines) because doesn't have any opportunity to maintain delocalisation in the product and so will be disfavoured compared to the tertiary amine.


Products of electrolysis of Ni(NO3)2 During the electrolysis of $\ce{Ni(NO3)2}$ using nickel electrodes, it's said that the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure.
But water has an oxidation potential of 0.828 V, as compared to nickel's 0.24. So while the cathodic deposition seems fine, shouldn't it be oxygen that's liberated at the anode?

Some electrolysis experimets are best first performed and then an attempt  at explanation.
For example, the products of the electrolysis of aqueous nickel nitrate with graphite electrodes are reportedly per a source (with video) as:

Electrolysis of a nickel nitrate solution produces oxygen at the anode, and hydrogen and nickel at the cathode.

Looking at related metal electrolysis, for example, Cu(NO3)2, per this source with again graphite electrodes:

Electrolysis of a copper(II) nitrate solution produces oxygen at the anode and copper at the cathode.

Lastly, with the transition metal manganese, a source on the electrolysis of Mn(NO3)2, reports, to quote:

$\ce{Mn(NO3)2 + 2H2O ± 2e- -> H2 + MnO2 + 2HNO3}$

Based on all of the above, and that hydrogen gas is forming per the reactions:
$\ce{H+ + e- -> .H }$
$\ce{.H + .H  -> H2 (g)}$
I would expect similarly some transient nitric acid formation with Ni(NO3)2 per its potential interaction between the nitrate ion and the hydrogen atom radical as follows:
$\ce{ NO3-  + .H  -> .NO2  + OH-}$
$\ce{.NO2 + .NO2 <=> N2O4 }$
$\ce{N2O4 + H2O -> HNO2 + HNO3 }$
Note, my claimed introduction of HNO3 (and its likely subsequent removal by nickel metal) can explain the comment "the molarity of the solution remains constant as nickel's active, and dissolves and deposits in equal measure", and is also consistent with the electrolysis of Mn(NO3)2 resulting in the creation of nitric acid.
And lastly, you are correct, oxygen is produced at the anode.


Does the probability of fluorescence increase when energy of incident X-rays match the electron's binding energy? When analyzing manganese-rich samples with X-ray powder diffraction, sample fluorescence becomes a significant effect.
Is this because the incident X-rays' energy - which, if Cu K$\alpha_1$, is approximately 8.048 keV - is getting close to the K$\alpha_1$ (approximate binding energy) of the inner-most electrons of Mn (5.899 keV)?

Sample fluorescence from Fe and Co is even stronger, which fits with the trend mentioned above. So, is this indeed a trend? If not, how do I generalize the increase in probability for sample fluorescence?

Short answer, yes.

As you note, the fluorescence signal is strongest when the exciting photon has an energy close to the excitation energy for the element absorbing the photon.  There are actually two factors coming into play:

1) The closer the resonance of the incident photon with the electronic orbital it is exciting, the higher the cross-section to absorb the photon.  (See here.)

2) The higher-Z the absorbing atom, the more likely it is that once excited, the atom will emit a fluorescence photon (as opposed to an Auger electron).  That is, it has a higher fluorescence yield.  (See here.)

In your case, the Z of the atoms are pretty close, but also in portion of the periodic table where the fluorescence yield is rapidly changing.  The chance of a fluorescence photon being emitted once excited is about 20% greater for Co than Mn (and you can look it up in the literature).

The cross-section for absorption as a function of energy for these three elements looks like this (code for the plot here):

I also drew a vertical line for the exciting Cu-K$\alpha$ photon at 8048 eV, and you can see that the Co curve has the highest cross-section at about 316 cm$^2$/g, and Mn has the lowest at about 270 cm$^2$/g, which means the Co cross-section is about 17% higher than the Mn.  The primary reason for the reduction in the sensitivity for the Mn edge is that the absorption cross-section drops off as $\approx\Delta E^3$.  These data were plotted using the data files described by B. L. Henke, E. M. Gullikson, and J. C. Davis, Atomic Data and Nuclear Data Tables Vol. 54 No. 2 (July 1993)

So both these effects are colluding to increase your fluorescence for Co vs Mn.

As a note, if you want to reduce this effect in your X-ray diffraction patterns, you can change your source target material with a heavier element such as Zr which has it's K$\alpha$ up at 15.7 keV, yet the L lines are still around 2 keV (below the transition metal K-edges) or you can use a lighter target, such as Mn.  Mn, of course, would not be able to excite its K-edge using its own K$\alpha$ fluorescence photon.


In the bicarbonate ion, why can the hydrogen not bond to the carbon? I've been studying electrophiles and nucleophiles recently, including how the carbonate ion ($\ce{CO3^2-}$) is stabilised through resonance. While drawing the model for bicarbonate, however, I suddenly asked myself why a different design of this ion would not be possible. The design I am proposing features the hydrogen atom attached to the carbon, with single bonds for all of the three oxygens. This is shown below:

Now, I know that this is not what we usually see, which is:

The question is, why is my own design not feasible? If it is, is it not energetically favourable due to the numerous charges that would have to be created?

First of all I appreciate your effort to think out of the box.

Now, as you are thinking of a new structure for the bicarbonate ion, you just have to compare the stability factors between the two different structures and you'll get your answer!

Here let's compare the structures:

Stability factors for the conventional structure

Octet saturation in case of each of the element.
Resonance stability (moreover, two equivalent canonical structures give more stability) in the molecule.
The negative charge resides on the more electronegative $\ce{O}$ atom.

Unstability factors for your proposed structure

Octet unsaturated in case of the positively charged $\ce{O}$ atom.
No resonance stabilisation in the molecule.
(your point) Energetically disfavoured due to numerous charge accumulation.
The positive charge resides on the more electronegative $\ce{O}$ atom.

That's why your proposed structure isn't the actual structure of bicarbonate ion.


Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene? Why is the boiling point of m-dichlorobenzene less than that of p-dichlorobenzene?  Shouldn't the former have a higher boiling point because it possesses a dipole moment? I think the van der Waals forces should be equal in both cases.

The difference between the melting points is small: ortho 180 °C, meta 172 °C, para 174 °C, so they are basically the same. It is just not possible to argue bigger or smaller with hand-waving arguments about dipoles or polarity etc. with something so complicated. However, when the thermal energy is increased, whatever intermolecular interactions exist become less important relative to the average thermal energy, and which now has a dominating and disruptive effect. Thus I would expect all isomers to have a similar boiling point.


Calculating the enthalpy of polymerisation of ethylene given the bond strengths Given the average bond dissociation enthalpies of a $\ce{C-C}$ bond (say $x$) and a $\ce{C=C}$ bond (say $y$), find the enthalpy of the following polymerisation reaction (per mole of ethylene):
$$\ce{nCH_2=CH_2 -> [-CH2-CH2 -]_n}$$
where $n$ is a large integer and $x, y$ are in $\pu{kJ/mol}$.

According to me, this should simply be $y-x$ as for every double bond broken, a single bond is formed. But the source book of this problem states the answer to be $y-2x$ claiming that for every double bond broken, 2 single bonds are formed.
How is this possible?

The mistake I was making was not noticing that every time a double bond is broken, i.e.; completely broken, 2 single bonds are formed in its place.

One bond between the 2 carbons of the same molecule and another between two molecules, the former of which I forgot to consider.

Thus, the enthalpy is in fact $y-2x$


Conceptual tussle with kinetics of equilibrium and non-equilibrium processes Prologue:
Recently started with Chemical Kinetics at class. Came across a couple of points in Levine's Physical Chemistry (the book was recommended on Chem.SE; so I thought I'd give it a read :D) that I'm having serious issues wrapping my head around. This query is about the kinetics of equilibrium (and non-equilibrium) processes, and the fact that this has puzzled me for quite a while... I attribute to my (seriously questionable) understanding of equilibrium processes (courtesy: High-school education). It would be prudent to keep my shortcomings in Kinetics (newbie) and Equilibrium (possibly flawed understanding) in mind while addressing my query in your answer or the comments section. Thanks!

According to Physical Chemistry (Levine,I.N), Chapter 15, "Kinetics":

Processes in systems in equilibrium are reversible and are comparatively easy to treat. This chapter and the next deal with non-equilibrium processes, which are irreversible and hard to treat. The rate of a reversible process is infinitesimal. Irreversible processes occur at nonzero rates.

Which, if I understood correctly, can be paraphrased as:

Equilibrium processes are reversible and have infinitesimal (practically "non-existent") rates.
Non-equilibrium process are irreversible and have non-zero ("finite" or "significant"?) rates.

From my "understanding" of Equilibrium; chemical equilibrium is an example of dynamic equilibrium. When a chemical reaction, $$\ce{A + B -> C + D}$$
has proceeded to equilibrium. The "forward" rate of reaction (rate at which $C$ + $D$ is spit out) must equal the "backward" rate of reaction (rate at which $A$ + $B$ is spit out). There is no net formation of products/reactants, but both the forward and backward rates are positive and finite.

My question?
Why does Levine say that equilibrium processes have "infinitesimal" (almost non-existent) reaction rates? I thought the forward and backward reaction rates in a reaction at equilibrium are finite...

OK, I think I am correct about this. If not then I suppose somebody will come along and correct me anyway.

The author here is referring to the net rate being infinitesimal. To understand where the infinitesimal comes from, we need to figure out what is being meant by an "equilibrium process", or a "reversible process". On page 43, Levine describes a reversible process as

  one where the system is always infinitesimally close to equilibrium, and an infinitesimal change in conditions can reverse the process to restore both system and surroundings to their initial states.

Often these are exemplified using thermodynamics processes, e.g. expansion/compression of a gas, but since we're talking about chemical kinetics, we could use an example where the temperature of the system is being increased in a reversible fashion. This means that the temperature is being increased in infinitesimal steps of $\mathrm{d}T$. The total change in temperature of the process is calculated via integration:

$$\Delta T = \int_{T_1}^{T_2}\mathrm{d}T = T_2 - T_1$$

and likewise for other properties. If something (e.g. the pressure $p$) depends on temperature, i.e. $p = p(T)$, then we can find the overall change in pressure

$$\Delta p = \int_{p(T_1)}^{p(T_2)} p(T)\,\mathrm{d}T$$

If the system is at equilibrium ($Q = K$), then an infinitesimal increase in temperature leads to an infinitesimal change in the equilibrium constant $K$, and hence an infinitesimal change in $Q$ (essentially this is due to Le Chatelier's principle, but more properly it is explained by the system moving in the direction of decreasing Gibbs free energy). This leads to an infinitesimal change in the concentrations of the reacting species, and hence the overall rate of reaction is infinitesimal.

If nothing is changing, the system would still be considered as being at equilibrium. The net rate in this case is, of course, zero. However, such a system is not particularly interesting, as there is no process actually occurring.

So, the bottom line is that for a system in equilibrium, its properties must be changing at most infinitesimally.


How can I determine the boiling point of a compound using simple distillation? I've graphed a distillation curve (temperature against volume of distillate) from the data obtained from a simple distillation experiment. The substance can be any of methanol, ethanol, 1-propanol or a  1-propanol and water azeotrope).

How can I find the boiling point from this curve to determine the identity of the compound?

I should also mention that there was leftover nonvolatile dye in the round bottom flask.

Do you mean that the substance could be pure methanol, ethanol or 1-propanol?

In that case there is a simple way to distinguish between the four options. A pure substance will have a T-V curve that consists of 2 straight lines, 1 running from $T=T_{start}$ to $T=T_{b}$ at $V=0$ and one running from $V=0$ to $V=V_{total}$ at $T=T_{b}$. Probably you will have some impurities which cause some small deviations from this perfect hook, but that will be the general shape. Determining the boiling temperature is just a matter of checking where the curve goes straight up. Depending on the temperature at which this happens you can identify methanol ($T_{b}\approx65$), ethanol ($T_{b}\approx78$) or 1-propanol ($T_{b}\approx97$).

The azeotropic mixture makes it a bit tricky, because the boiling points of 1-propanol and water are so close together, but the curve should show a distinct difference not going vertically up at a certain temperature.


Why the "Diverse Ion Effect" increases Ksp? I've read in chemwiki about the diverse ion effect. Assuming I have an aqueous solution of $\ce{NaCl}$ and an associated $K_\mathrm{sp}$. If I add $\ce{KI}$ there is an associated $K_\mathrm{sp}'$ for $\ce{NaCl}$ where $K_\mathrm{sp}' > K_\mathrm{sp}$. This actually means that I can dissolve more $\ce{NaCl}$ than I could before. Why does this happen? Shoudln't the new solution actually be closer to saturation and $K_\mathrm{sp}' < K_\mathrm{sp}$?

To illustrate the "diverse ion effect", we will study how the solubility of $\ce{AgCl}$ changes in absence and presence of $\ce{NaNO3}$, given that the molar solubility product of $\ce{AgCl}$ is $1.76 \times 10^{-10}$.

The exact definition of the solubility constant is: $$K_{sp}= a(\ce{Ag+}).a(\ce{Cl-})=[\ce{Ag+}].\gamma(\ce{Ag+}).[\ce{Cl-}].\gamma(\ce{Cl-})$$
Where $a(\ce{Ag+})$ and  $a(\ce{Cl-})$ are the activities of these ions, and $\gamma(\ce{Ag+})$ and $\gamma(\ce{Cl-})$

In dilute solutions, i.e. absence of $\ce{NaNO3}$:  $$\gamma(\ce{Ag+})=\gamma(\ce{Cl-})=1$$
So, the solubility of $\ce{AgCl}$ is equal to $$[\ce{Ag+}]= \sqrt{K_{sp}}= 1.33 \times 10^{-5}$$
In presence of $\ce{NaNO3}$ of a concentration of $0.1 \mathrm{mol/kg}$, then
$$[\ce{Ag+}]= \sqrt{K_{sp}}/ \sqrt{(\gamma(\ce{Na+}).\gamma(\ce{NO3-})}
= 1.33 \times 10^{-5}/ (0.769)= 1.73 \times 10^{-5} $$
i.e. in the presence of a soluble salt, the molar solubility of a slightly soluble salt increases.


Enthalpy of the reaction between hydrogen and oxygen The reaction between hydrogen and oxygen to yield water vapor has
  $\Delta H^\circ = \pu{- 484 kJ}$. How much $pV$ work is done, and what is the value of $\Delta E$ in kilojoules for the reaction of $\pu{0.50 mol}$ of $\ce{H2}$ with $\pu{0.25 mol}$ of $\ce{O2}$ at atmospheric pressure if the volume change is $-\pu{5.6L}?$

  $$\ce{2H2(g) + O2(g) -> 2H2O(g)} \qquad \Delta H^\circ = \pu{- 484 kJ}$$

I use the formula $\Delta E=\Delta H - p\Delta V$ to determine $\Delta E.$ However, when determining the enthalpy, the solutions manual does this:

  $$\Delta H = \frac{\pu{-121 kJ}}{\pu{0.50 mol}~\ce{H2}}$$

Where does the $-121$ come from? From my understanding, since there are two moles $\ce{H2}$, $\Delta H$ should be $-\pu{242 kJ}$. Or do we take into account all four hydrogen atoms? That would give us $\pu{-484 kJ}/4 = -\pu{121 kJ}.$

The confusion is partly caused by careless use of quantities and units. The value $H = -484\ \mathrm{kJ}$ denotes an enthalpy. However, what is actually meant is the enthalpy per one mole; i.e. the molar enthalpy $H_{\mathrm m} = -484\ \mathrm{kJ/mol}$.

In the definition of the molar reaction enthalpy, the ‘per mole’ does not refer to any particular substance in the equation. Instead it refers to the entire reaction as a whole. Therefore, the reaction must be specified for which this quantity applies. In this case, the enthalpy of $484\ \mathrm{kJ}$ is released when $2\ \mathrm{mol}$ of hydrogen gas react with $1\ \mathrm{mol}$ of oxygen gas to form $2\ \mathrm{mol}$ of gaseous water:
$$\ce{2H2(g) + O2(g) -> 2H2O(g)}\qquad\Delta H^\circ = -484\ \mathrm{kJ}$$
(By way of comparison, the corresponding value for liquid water is about $-572\ \mathrm{kJ}$.)

Thus, the molar enthalpy relating to the amount of hydrogen is
$$\frac{\Delta H^\circ}{n(\ce{H2})}=\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}=\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}=-242\ \mathrm{kJ/mol}$$
i.e. the released enthalpy per $1\ \mathrm{mol}$ of hydrogen gas is $242\ \mathrm{kJ}$.

However, the given question is asking about $n(\ce{H2}) = 0.50\ \mathrm{mol}$ of hydrogen (and accordingly $n(\ce{O2}) = 0.25\ \mathrm{mol}$ of oxygen). The corresponding enthalpy is
$$\begin{align}
0.5\ \mathrm{mol}\times\frac{\Delta H^\circ}{n(\ce{H2})}&= 0.5\ \mathrm{mol}\times\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}\\[6pt]
&= 0.5\ \mathrm{mol}\times\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}\\[6pt]
&= 0.5\ \mathrm{mol}\times -242\ \mathrm{kJ/mol}\\[6pt]
&= -121\ \mathrm{kJ}
\end{align}$$


Alkyl chain substituent has higher preference than benzyl substituent? The compound in question, and the name generated by ChemSketch:

I have looked at relevant (outdated) documentation for naming of substituted amines. I understand tertiary amines are named as: $\text{N-alkyl}_1\text{N-alkyl}_2\text{alkyl}_3\text{amine}$.

Now, I am confused at the first point: when we have to select the parent carbon chain. For this compound, I would have selected "benzylamine" as the parent chain, as it obviously has more number of carbon atoms and more unsaturation than both methyl and propyl. Thus, my preferred IUPAC name would have been "N-methyl-N-propylbenzylamine".

But, for some reason "propane" is the main carbon chain here instead. So, what is the logical reasoning for the naming these tertiary amines?

The longest carbon chain is the base name. Benzyl has a "one carbon chain". Compare the two structures below. In the top example, the butane chain, even though it bears a phenyl group, is the main chain. On the bottom example (yours), the phenyl is attached to a "one carbon" chain. therefore, propane is the main chain.


Reverse of E1cB reaction As we know that the first step of E1cB step is second of E1 and vice versa. So can we say that by reversing E1cB reaction we will get E1 reaction.

In this the main problem which I am getting is in E1cB reaction carbanion is formed while in E1 carbocation. By reversing we will convert carbanion into neutral not into carbocation. So is this possible?

I asked my teacher about it and he told me that no it is not possible as these two are totally different mechanism.

Sorry for english or improper answer( if it is so) as my teacher told me this much only so I can't say much clearly about it.


Is it possible to export a rotating .gif format picture from chemapps.stolaf.edu? I mainly use chemapps.stolaf.edu to get the 3D structure of a molecule from the SMILES. It renders the molecules as:

I have the option to rotate the image by specifying the $x$, $y$, and $z$ spin rates:

However I can't seem to find a way to save a spinning image as such. I can only save .png files. I need spinning images to demonstrate 3D images. Can anyone help me get a spinning image?

As of now, Java applets only work with Firefox ESR 32-bit. Assuming that OP is using Chrome and Windows 10, in my opinion the best solution would be to use a screen capture software, like ScreenToGif, which is FOSS, has nice interface and is also quite intuitive:

And here is the result:

I must remind though that the major limitation of GIF is that you are stuck with 256 colors. Structures displayed with GIF will never look as slick as those demonstrated in browser with JavaScript.


Would the change in enthalpy change of chemical reaction in space be considered as change in internel enrgy? The formula of enthalpy change is

change in H = U + P*(change in V)

In space since there is no air, thus atmospheric pressure = 0 , P= 0 , Pv = 0 ,

So will H = U  ?

Well, in principle yes, for example if you had a blob of liquid floating in empty space and wanted  to consider the whole blob as a thermodynamic system.

But it may be worth mentioning that wouldn't be a typical use of thermodynamics in astrophysics.  Usually you'd consider your thermodynamic system to be some small part of a larger system and the larger system would be a thermal and volume reservoir, so T and P would both be nonzero.

For example, it's common to consider the thermodynamics of stars, but what you do is consider a small packet of gas to be your system, and the rest of the star to be surroundings that establish a constant T and P for your packet.


What chemicals are in tap water that gives it a slightly basic pH level In chemistry, we recently did a lab where we observed the different ways of reading pH of several different substances. We used things such as red cabbage juice, litmus paper, pH paper and a pH probe. One of the substances we tested was water. When we used the pH probe, it came back with a reading around 8.0. When we tested the distilled water, it came back with a reading very close to 7.0. So what is different between distilled and tap water that causes the tap water to be basic?

The tap water is likely "hard," i.e., contains some dissolved mineral salts, most likely (primarily) calcium and magnesium carbonates and bicarbonates. The anions of these salts are slightly basic, consuming $\ce{H^+}$ ions and thereby raising the pH. Carbonates (with the obvious exception of alkali metal salts) are only sparingly soluble in water, while bicarbonates are typically very soluble. A couple of other subtle factors may affect the pH:

Lower atmospheric pressure encourages the evaporation of dissolved carbon dioxide from water. Since dissolved carbon dioxide exists in equilibrium with carbonic acid, it would tend to lower water pH. If you live in an area with comparatively low atmospheric pressure, your tap water might have a lower concentration of $\ce{CO_2}$, which would favor higher pH.
At higher temperatures, bicarbonate decomposes to water, carbonate, and carbon dioxide. Carbonates are relatively insoluble, and the consumption of bicarbonate should drive forward the dissociation of any dissolved carbonic acid, so a higher temperature should have the effect of lowering pH.


Is it possible to hold and rotate molecules to relativistic speeds around axis in observable way? Also, is it possible to see frame-shortening effects on electrostatic forces/van der waals at the edges, or are molecules too weak or necessarily neutral to be manipulated this way?

Rotation speed cannot be made arbitrarily close to the speed of light. Molecular rotation is quantized, and every molecule has a finite number of rotational levels, just like it has a finite number of vibrational levels based on the shape of its potential energy surface (the same can be said of vibrations: the molecule cannot vibrate arbitrarily quickly without dissociating).

However forces like the van der Waals force do get noticeably affected at the highest supported molecular energy levels. Specifically the $C_6/r^6$ van der Waals force becomes $C_7/r^7$ at certain distances, due to the Casimir-Polder effect in quantum electrodynamics (an intrinsically relativistic theory of the quantum world). The reason for this is that the derivation of the $C_6/r^6$ van der Waals force does not take into account the fact that information cannot travel at infinite speed, so it actually takes a finite amount of time (based on the finite speed of light) for one atom to "feel" the force of a distant atom. This "slowing down" of the energy transfer, is called "relativistic retardation" because "retardation" means slowing down. This causes the force to become weaker for larger distances ($1/r^7$ is much smaller than $1/r^6$ for larger $r$).

This is usually not taught at the undergraduate level, or even graduate level courses, so there is no easy "textbook" example to demonstrate it, so unfortunately we have to resort to publications in scientific journals, where the audience is far more advanced: But in the image below, I hope you can at least see that "with retardation" the experimental values here are much closer to the theoretical values of Tang (2009), versus the experimental values derived from a model "without retardation":

This shows evidence that the theoretical idea of "relativistic retardation" is in fact a much better description of the experiment than the non-relativistic theory, so it is indeed fair to say that the effect of light (electromagnetic radiation) having finite (rather than infinite) speed is observed in a molecule. The molecule here was Li$_2$ and the above image is from Figure 6 of this paper which deals with deriving information from experimental spectra of a molecule:

http://scienide2.uwaterloo.ca/~rleroy/Pubn/09JCP-Li2AX.pdf


Derivation of relationship between equilibrium constant and Gibbs free energy change Why is $\Delta G=\Delta G^o+RT\ln Q?$

It feels like all online sources were written for introductory Chemistry students! Where do I find a rigorous proof of this identity? Greatly appreciate it!

Using the fundamental equations for the state function (and its natural variables):
\begin{align}
  \mathrm{d}G &= -S\mathrm{d}T + V\mathrm{d}P\\
  V &= \left(\frac{\partial G}{\partial P}\right)_T\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p
\end{align}
Here $\bar{x}$ represents molar $x$, i.e. $x$ per mole
\begin{align}
  \bar{V} &= \frac{RT}{P}\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1}
\end{align}
Defining standard state as $P = \pu{1 bar}$ and $\bar{G}=\mu$
$$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$
consider the general gaseous reaction $\ce{a A + b B -> c C + d D}$
$$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$
for "unit progress" in reaction. Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$
\begin{align}
\Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\
\Delta G &= \Delta G^\circ + RT\ln Q
\end{align}


Calculating pH in titration of monoprotic weak acid with strong base $25.0\:\mathrm{ml}$ of $0.10\:\mathrm{M}$ $\ce{CH3COOH}$ is titrated with $0.10\:\mathrm{M}$ $\ce{NaOH}$. What is the $\mathrm{pH}$ after the addition of $30.0\:\mathrm{ml}$ $\ce{NaOH}$?

In broad strokes, as I understand it, the sodium hydroxide in this case would completely consume the acetic acid, leaving a solution of acetate ions, $\ce{Na+}$ and $\ce{OH-}$.

What makes sense to me then is to figure that the $\mathrm{pH}$ is now dominated by the concentration of hydroxide ions, so the $\mathrm{pH}$ would be given by

$\mathrm{pH} = 14-(-\log([\ce{OH-}])= 11.96$

where $[\ce{OH-}] = \frac{[(30.0 - 25.0) \times 10^{-4}]\:\mathrm{mol}}{(30.0+25.0) \times 10^{-3}\:\mathrm{l}}$

i.e., the remaining moles of hydroxide ($0.0005$) per liter of total volume ($55\:\mathrm{ml}$)

but is this correct? Do I need to take into account the conjugate base’s contribution to deprotonation of water? If so, how is that done?

I’d like to be clever and use the lack of $K_\mathrm{a}$ or related constants in the problem to indicate that I’m on the right track, but that would be giving too much credit to my professor’s design of these questions (and the prior question in this series does give $\mathrm{p}K_\mathrm{a}$ for acetic acid)

Your calculation is right.  Regarding your question to take into account the conjugate base's contribution to deprotonation of water, I'd say:

If you calculate the equilibrium constant of the neutralization reaction $$\ce{CH3COOH + OH- <=> CH3COO- + H2O}$$
$$K= \frac{K_a}{K_w}=\frac{10^{-4.75}}{10^{-14}}= 10^{9.25}$$

Now, if you write  of reaction of acetate ion with water:$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
It's the opposite of the neutralization reaction. Its  equilibrium constant  is  $10^{-9.25}$.

You can see clearly that the neutralization reaction is the "predominant reaction". So, the pH is determined by this reaction as if it's the only one happening in the aqueous solution. So, you can neglect the reaction of of acetate ion with water.


Calculating molecular weight of an acid based on titration data If $\pu{2.600 g}$ of a weak diprotic acid were dissolved in $\pu{100 mL}$ of distilled water and a $\pu{10 mL}$ aliquot of this solution required $\mathrm{21.60\ mL}$ of $\mathrm{0.1000\ M\ NaOH}$ to reach the first endpoint, what are the equivalent and formula weights of $\ce{H2A}$?

I assumed to do this: I calculate the grams of acid in solution which is

$$\mathrm{\frac{2.600\ g}{100\ mL} \cdot 10\ mL = 0.2600\ g}$$

After that I would calculate the moles of NaOH used

$$\mathrm{0.1000\ mol/L \cdot 21.60\ mL \cdot 1\ L/1000\ mL = 0.002160\ mol}$$

Then I use this to find the moles of acid present in solution; this is the part I am unsure about. I am assuming: because at the equivalence point half of the acid has been converted to it's conjugate base and so ($\mathrm{2\ mol\ NaOH = 1\ mol\ H_2A}$).

If this is correct then

$$\mathrm{0.002160\ mol\ NaOH \cdot 2\ mol\ H_2A/mol\ NaOH = 0.004320\ mol\ H_2A}$$

Which I then divide the g/mol

$$\mathrm{\frac{0.2600\ g}{0.00432\ mol} = 60.19\ g/mol}$$

Because the acid is diprotic the equivalence weight would be

$$\mathrm{\frac{60.19\ g/mol}{2} = 30.09\ g/mol}$$

I don't think this is the correct answer because I am trying to do this using real titration data I took in lab and the MW seems to be at $\approx 1/2$ the value it should be. If I am not supposed to double the NaOH moles, is there a reason for it?

Your error comes from your misunderstanding of what the equivalent points indicate, or as Ivan mystically said:

  The reason is in the words "...to reach the first endpoint".

An n-protic acid, has exactly n equivalent points. This means for your titration, that you need one base equivalent to reach the first equivalent point (EP) and two base equivalents to reach the second EP. As you are given the volume that is needed to reach the first EP, everything is quite easy, as the ratio of acid to base is exactly 1:1.

This being said, your first EP can be described by $$n_A = n_B$$
(and not as you did by assuming $n_A = 2\ n_B$, which also would not be the description of the second EP, which is $2\ n_A = n_B$ instead.)

You have to find the molecular mass of your acid, why the upper equation needs to be expanded to $$\frac{m_A}{M_A} = c_B V_B$$

Now you simply have to solve for $M_A$ and put in your numbers (0.26 g, 0.1 mol/L, 21.6 mL).


Why do ligands have such a small effect on overall absorption of a complexed ion? When a metal cation is complexed, there is strong UV-Vis absorption due to the splitting of its $d$ orbitals, thereby allowing electronic transitions.

My understanding is that ligands contribute very little to overall observed absorption. This is evident, for example, in copper solutions. Copper(II) chloride, copper(II) sulfate, and copper(II) nitrate – all solutions of complexed ions – are all very similar in color and have similar absorption spectra.

I think the simple answer is that transition metals have the capacity to allow for electronic transitions while most ligands do not, but I don't believe it is this simple. Also, it is curious that sulfate, chloride, and nitrate are all colorless when dissolved in solution.

My question is: why do ligands have such a small effect on overall absorption of a complexed metal cation?

This is absolutely not true. Many ligands can strongly dictate the color of transition metal solutions.

Your example picks three simple salts and then questions whether the color of the solutions (dictated largely by $\ce{[Cu(H2O)6]^{2+}}$) is very different. No, because the resulting majority complex in aqueous solution is likely identical.

Even taking a simple ammonia complex, e.g. $\ce{[Cu(NH3)4(H2O)2]^{2+}}$ you can see a substantial color change.

Many copper acetonitrile compounds are weakly colored or colorless, with UV/Vis optical absorption occurring near the edge of the red to near-IR.

When you get into octahedral complexes, you can find substantial changes in color due to MLCT and LMCT absorptions, as well as significant modulation of the d-d transitions due to the ligand (i.e., high-field and low-field ligands).


What would be the mechanism of the given reaction? Problem:

  $\ce{A}$ can be?

  concentrated sulphuric acid
  alcoholic potassium hydroxide
  triethylamine
  potassium tert-butoxide

My comments:

Option (A): Conc. $\ce{H2SO4}$, is an acid, it gives $\ce{H+}$ ions which must act like electrophile and add to the double bond, which must produce a carbocation and then, $\ce{HSO4^-}$ ion must act like a nucleophile and form a bond with the carbocation. Hence, the reaction must be addition of $\ce{H+}$ and $\ce{HSO4^-}$

Options (B),(C),(D): are bases, with basicity order: (D) > (B) > (C). But what would they do to an alkene?

Answer given in the book is A. How is this so?

Bottom line: $\ce{HSO4-}$ is a really bad nucleophile, and it doesn't enter in the picture.

(I realize after scanning that I'm missing the "+ H+" alongside the final product.)

Question for extra credit: after the second step, why is the third step (carbocation migration) favorable rather than directly reforming the double bond on the right-hand side ring?


How to detect azide group on a primer? I bought azide-labeled primer for click chemistry, but it didn't work. So I want to prove the existence of azide group. How could I achieve this? I found a method to add PPh3 to Kaiser test reagents, but the concentration of the primer is too low (the protocol needs the solution be higher than 0.1M) and I cannot see the reaction.

One possibility would be to check the IR spectrum.  If there's any azide in there the azide band should show up as a strong band at around 2150 cm-1.  The trouble is of course that this isn't quantitative, you can only prove the absence of azide, not how much is present.

The Staudinger reaction might be a viable option.  PPh3 won't go up in aqueous solution, but sulfonated PPh3 will.  One should be able to distinguish starting material and product by 31P NMR.  With a known quantity of TPPTS you can even determine % azide in the starting material.


Copper is a metallic element, so it is likely to be? Copper is a metallic element, so it is likely to be?

A. Poor conductor of electricity       B. A substance with a low melting point      C. Malleable     D. Poor conductor of heat     E. An element that forms negative ions

For me, C should be the only right. But are metallic elements really likely to be malleable?

Yes, It is C.

Metals are:

-Generally good conductors of electricity
-Substances with high melting points since they are mostly in liquid/solid states. Only liquid metal I can recall offhand is probably mercury, at room temperature and pressure of course.
-They form positive ions because they usually lose electrons to gain octet configuration.

well, Wikipedia says "Metals are described as malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond."

Other examples of malleable metals are Gold, Silver and Iron.


Fructofuranose vs fructopyranose What's the difference between the two molecules? I can always read about β-D-fructofuranose in the chemistry books, but unfortunately, I've never found any info concerning β-D-fructopyranose. Is there a big difference between them (apart from their ring structure)?
And where can the furanose and the pyranose form be found in nature?

They are isomers. Both of them have the same molecular formula ($\ce{C6H12O6}$) but different structures.

Structure of β-D-fructopyranose

Structure of β-D-fructofuranose


Where to find a look-up table for mechanical properties of thermoplastics? Does anybody know of a good compilation, regarding mechanical properties of thermoplastics or in general polymers?

I suggest MatWeb:

  MatWeb's searchable database of material properties includes data sheets of thermoplastic and thermoset polymers such as ABS, nylon, polycarbonate, polyester, polyethylene and polypropylene; metals such as aluminum, cobalt, copper, lead, magnesium, nickel, steel, superalloys, titanium and zinc alloys; ceramics; plus semiconductors, fibers, and other engineering materials.

There are 84,173 entries in the Polymers tab in the physical properties search page at the time of my writing this. Note that you can select up to 3 properties at a time to narrow your search (or 10 if you pay for the service: MatWeb is a freemium deal, so be sure to look into whether or not your needs are satisfied by the free/pay versions.)


Effect of dillution on titration When we are titrating acid/base using ph meter we add distilled water to immerse the ph electrode.
Won't this affect the concentration of the acid/base: I mean isn't  this dillution.
Won't  this affect our calcultion of the concentration at the end.

What's important to note in this situation is that the number of moles of the titrand isn't changing. This matters because all acid/base reactions go to completion, so every drop of titrant that falls into the solution will react with the titrand until there are no moles of titrand left regardless of concentration, at which point you've reached the equivalence point.

Adding a little water here or there doesn't really change the pH at the equivalence point, because ideally there are no moles of acid or base present in solution.

The reason this doesn't mess up your measurement for the concentration is because you measured the volume of titrand before the titration, and then recorded the volume of the known titrant needed to reach the equivalence point. From this information, you can calculate the moles of titrant needed to reach equivalence, which in most cases equals the number of moles of titrand present, and with the original volume of titrand noted previously, you can calculate the original concentration of the solution the sample was taken from. voila!


Isomers of coordination compounds Does [Pt(CH3-CH(NH2)(COO-))2] show geometrical and optical isomerism?

According to me Pt should be dsp2 hybridised giving a square planar geometry. So it is of the form M(aa)2 type. The ligand itself has a chiral carbon. Hence it can show optical isomerism. But can it show geometrical isomerism?

First of all, the compound is not M(aa)2 type.M(aa)2 shows no kind of isomerism (with normal ligands). Since the donor atoms on a ligand are different, it would be M(ab)2. And square planar structures show only geometrical isomerism and not optical isomers, as you think. Why?

Optical isomerism is when the compounds can have non super-imposable mirror image like structure. They should not have a plane (or axis) of symmetry. However, in a square complex, it is not possible. Since it will always have a plane of symmetry or a mirror image which can be superimposed on the previous compound, even if it is of type  M(abcd) (Try this out for yourself)

But, you are right, carbon on your compound is chiral, so it can have different configuration of methyl and hydrogen groups in space so as to remove that plane of symmetry factor and make them optical isomers.

Geometrical isomers, however are variations of a compound about the central metal atom atom or ring or a double bond, in which the arrangement of groups is different. So square planar complexes can show geometrical isomers. The one is your question can be depicted as -

If you include the configuration of the chiral carbon of the ligand, there can be a few more structures made.


Is there a way to export detailed data about atom connectivity, dihedral angle, etc from GaussView? As far as I know, opening the .com files as .txt files (with a text editor like simple notepad) yields a nicely formatted table of (atom) (x) (y) (z) values that can then be worked with. The problem is that this data has no information about what atom is connected to what. But within the actual Gaussview software, there seems to be a data set with more detailed information. Is there any way to easily extract that data via code?

To my knowledge, neither .xyz nor .com format preserve connectivity information by default. Gaussview (and many other programs) makes an educated guess based on atomic covalent radii and positions of atoms. Gauss .com may also store a set of internal coordinates used for optimization that gaussview derive from connectivity information. Gaussian .com may contain connectivity information if geom=connectivity option was used, but you cannot expect it. The format of connectivity section is documented in geom=connectivity suboption part of Gaussian input docs.

That said, some other formats like .pdb allow to store that data explicitly and are well documented. Also, Gaussian can be used to perform various forms of population analysis printing a lot of nice tables, some of them might be interpreted as connectivity information (then again, some cannot).


Henry's Law with varying temperature I am trying to determine what the derived partial pressure would be for CO2 in the global atmosphere from historical inconstant temperatures and CO2 concentrations of the global atmosphere:

Is it possible to do this?  If so, how?

The short answer is no.

The long answer is:

Henry's law works for small concentrations of ideal mixtures at equilibrium. Henry's Law constant varies with temperature according to the Van't Hoff equation:

$$
k(T)=k(T^{\circ})exp\left[-C\left(\frac{1}{T}-\frac{1}{T\circ}\right)\right]
$$

Here $C$ is a constant related to the enthalpy of solvation for each gas, and $^{\circ}$ represents the standard state, T = 298 K and 1 atm.

Using standard state data, you could get an estimate of the partial pressure at a given temperature. However, two key assumptions that are implicit to Henry's law and the Van't Hoff equation do not hold for global atmospheric gas composition:

The atmosphere and the oceans are not at equilibrium (not even close when you consider atmospheric layers and convection)
$\ce{CO2}$ in seawater is not an ideal mixture (very non-ideal since $\ce{CO2}$ reacts with water).

This is one of the many reasons that climate modeling is difficult - the oceans act as huge carbon sinks with a very complex and not-very-stable relationship between dissolved $\ce{CO2}$ content, temperature, composition of rock on the sea floor, convection currents in the atmosphere and the ocean, amount of shellfish (no kidding), and probably lots of other factors (I'm not a climate expert).

The net result is that while Henry's law can get you a quick idea of how much $\ce{CO2}$ could be dissolved at a given temperature under pseudo-equilibrium conditions (or what the corresponding partial pressure would be for a given concentration in water), it isn't enough to get you anywhere close to a good prediction of what the global atmospheric composition will be, particularly for $\ce{CO2}$ (due to its reactivity with water and incorporation into biomass).

Now, if you really just want to convert $\ce{CO2}$ concentration in ppm to a partial pressure, you can do that.

$\ce{CO2}$ in air is close enough to ideal that we can use the ideal gas law. For an ideal gas, the partial pressure is the same as whatever the pressure would be if it were in a container by itself:

$$
P_1 = \frac{n_1RT}{V}
$$

You can find V for a given mass of air, find the mass of $\ce{CO2}$ from the concentration, then convert to moles and plug it into the above.


Does a benzene ring continue to oscillate at absolute zero? The carbon bonds in a benzene molecule oscillate between single and double bonds. As they do so, the distances between the carbon atoms changes (as the attached animated GIF illustrates). At absolute zero all molecular motion ceases; that is, the molecules stop bouncing into each other, but does that mean that molecules like benzene also cease their internal molecular motion?

"At absolute zero all molecular motion ceases; that is, the molecules stop bouncing into each other"

This statement is a common misconception about absolute zero and is totally false. At absolute zero, we are simply in the lowest possible energy state. That does not mean that molecular motion ceases. That would violate the uncertainty principle.