Electronic and geometric configuration of Fe(OH)₂ I'm having serious problems in the electron configuration of iron dioxide $\ce{Fe(OH)2}$ because all I could come up with was $\ce{Fe}$ double bonded with each oxygen and single bonded with each hydrogen.

Everything seems good except for the iron that's bonded with 5 atoms (which I think is weird). Could someone you give me the right electron configuration (or geometry)?

I apologize if there's something to this that I don't know about, but my understanding of it is...

We have $\ce{Fe(OH)2}$, also called iron(II)hydroxide (iron-two-hydroxide). The reason it's not called dihydroxide, is that the number (II) describes the oxidation state of the iron, so you can easily calculate how many hydroxyl-groups (which are always -1) it needs to have an overall charge of 0.

In iron(II)hydroxide you don't have covalent bonds (double bonds, single bonds)... You have ionic bonds. This particular compound consists of a $\ce{1Fe}^{2+}\ce{+2OH-}$. They don't SHARE electrons, they are simply held together by their charges!

Ionic compounds group together in lattices. They don't have "geometric" configurations like sulfur dioxide for example. Instead, they are clustered together like in the image below:

The negative dots would be the $\ce{OH-}$ and the positive dots would be $\ce{Fe}^{2+}$. Then, of course, you would have two negatives for each positive. Ion bonds are the strongest type of bonds there is.


Why do halogen substituents make molecules more lipophilic? According to my medicinal chemistry text book, halogens increase a drug’s lipophilicty. This makes no sense to me for two reasons:

Halogens are all quite electronegative and they will form a $\ce{C-X}$ bond with a large dipole moment. Large dipole moments are associated with increased polarity.
As opposed to a methyl group, halogens are capable of donating their lone pair electrons to participate in hydrogen bonding. We see this when looking at water solubility of methyl-halogen compounds:

$$\begin{array}{cc}\hline
\text{Compound} & \text{Solubility} / (\mathrm{g \ L^{-1}})\\ \hline
\ce{CH4} & 0.055\\
\ce{CH3CH3} & 0.0568\\
\ce{CH3F} & 2.295\\
\ce{CH3Cl} & 5.325\\
\ce{CH3Br} & 15.22\\
\ce{CH3I} & 14\\ \hline \end{array}$$

Halogens, except for fluorine, aren’t all that electronegative. Chlorine is just slightly less electronegative than nitrogen and bromine is even less again. Iodine almost has carbon’s electronegativity. Thus, there won’t be so much of a dipole moment created by substituting a hydrogen with a halogen. This applies even more if a hydroxy group is replaced with a halogen.

To illustrate this, take dichloromethane. Of all the chlorinated methanes it has the highest permanent dipole moment. Yet it is still almost immiscible with water but fully miscible with hexane. Note by the way that your data somewhat reinforces that: the molecule with the highest dipole moment of all your halocarbons, fluoromethane, has the lowest water solubility.

A second point is that halogens come with free lone pairs. Lone pairs are not polarised per se but can easily be polarised in London interactions. Thus, halogens actually increase molecule’s ability to interact with lipophilic, unpolar media by allowing stronger London forces.

Note also that while nitrogen is capable of forming hydrogen bonds with surrounding aquaeous media, chlorine — in spite of its similar electronegativity — cannot since its lone pairs are much more diffuse. Only fluorine is in theory able to accept hydrogen bonds. In principle, this is another consequence of the larger size of higher-period halogens.

In all this, fluorine is somewhat separate and special. But fluorine is special in itself since perfluorinated molecules are typically insoluble in anything but fluorocarbons. Fluorous extraction is actually a technique used by some green chem labs.

To sum it up, just looking at the simple variable polarity cannot explain the solubility of halocarbons adequately. Their size needs to be considered.


Do you get a new chemical compound if you mix sodium hydroxide and water? Me and dad started fighting over this a couple of days ago and wagered on this.

He said mixing these two gives a chemical compound, I said that it doesn't.

According to Oxford's dictionary of chemistry "A compound is a mixture of two or more elements in a fixed ratio".
So if $\ce{NaOH}$ (in solid state) is added to water, it just dissociates into its constituent ions that are $\ce{Na+}$ and $\ce{OH-}$. You can add any ratios of two until you cannot add any more.


Is the mixture of alkyl dimethyl benzyl ammonium chloride and vinegar dangerous? So I was trying to clean cat pee out of my carpet and I scrubbed it with a Lysol wipe that says the active ingredient is alkyl dimethyl benzyl ammonium chloride and without thinking then sprayed white vinegar on the area. Now I'm afraid that made a bad chemical mixture? Do you know if those 2 ingredients are dangerous?

Please note that I worked in medical research, but I am not a medical doctor!

I think that you're safe, and I cannot imagine a reaction between acetic acid, the main component in vinegar, and a quarternary ammonium salt that would release harmful products.

Benzalkonium chloride is a long known disinfectant and has been used as a biocide and preservant in medical products. However, it may cause problems. The European Medicines Acency has published a Q&A on Benzalkonium chloride.

In general, it is not the worst idea to have a look at MSDS (Material Safety Data Sheets) for compounds and products, in order to be aware of what to do if something spills on your skin, gets into your eyes, is swallowed, or inhaled. Here is an example of a MSDS for benzalkonium chloride.

To sum it up:

Mixing household chemicals can release harmful products, as known for the reaction of bleach and acid, from which chlorine gas is released.
I seriously doubt that you have created a harmful mixture in your case.
Manufacturers are obligated to provide MSDS, wich have to include first aid measures! Take a look at this vital information.


What are the order of the reaction under the following conditions? The experimental rate of a reaction
$$\ce{A + 2B -> C + D}$$
$$\frac{\mathrm dx}{\mathrm dt} = K[\ce{A}][\ce{B}]^2$$
Calculate the order of the reaction when (i) $[\ce{A}]\gg[\ce{B}]$  (ii) $[\ce{B}]\gg[\ce{A}]$

My efforts :
Should we ignore the concentration of the reactant that is present negligibly? And thus the orders will be:
(i) Order = 1
(ii) Order =2
Please guide.

Let's compare the rate when $A$ and $B$ are present at their initial amounts, say $A_0$ and $B_0$, to when the reaction is 99% complete.

When $[\ce{A}]\gg[\ce{B}]$, the reaction will be 99% complete when 99% of $B$ is gone, i.e. when $B = 0.01B_0$.

At that point, $$\frac{dx}{dt} = k[A][B]^2 = k(A_0 - \frac{0.99}{2}B_0)(0.01B_0)^2 $$

Now since $A\gg B$ also means $A_0\gg B_0$, then it follows that $(A_0 - \frac{0.99}{2}B_0) \approx A_0$.  Thus, when the reaction is 99% complete and $B$ is limiting,

$$\frac{dx}{dt} \approx kA_0(0.01B_0)^2 = kA_0 (0.0001)B_0^2$$

If we lump $k$ and $A_0$ into a single parameter, $k_{app}$, then we can write $\frac{dx}{dt} = k_{app}B^2$, which shows that when $B$ is limiting, the reaction is 2nd order.

The case for $A$ being limiting shows the reverse.


Electrolysis not running, maybe dependant on distance of electrodes? So I created a DC power supply of 15 V (special transformator connected to the net) and I have a graphite anode and stainless steel cathode. When I put them in a beaker next to each other in water, the electrolysis takes place as I see gas bubbles. However, when i put them in another container that consists of two beakers connected with a tube to allow liquid to flow through but to separate the gases, the reaction doesn't run while it is the same setup more or less. How come? It is true that the tube connecting the beakers is relatively thin and the distance is also quite large (10 cm or sth), so does this lead to a higher required overpotential? I really don't know why it doesn't work in the larger container. Any help is appreciated!

Solution resistance in a two electrode cell can make a big difference as it is totally uncompensated. You may be applying 15 V between the electrodes, but the redox reaction only occurs at a very close proximity to the surface (<10 nm), so it's the interfacial potential we're concerned with, which can be significantly different than the applied voltage if the solution is highly resistive. A tight constriction in the cell can certainly increase the resistance and you should make sure you're using an electrolyte as plain water is very resistive.


How to obtain calcium carbonate? I read calcium carbonate is the main component of seashells and other animals' shells (such as snails), and also limestone. How can I prepare pure $\ce{CaCO_3}$?

Could someone describe the process and the chemical equations involved? I saw this video and apparently that's one way. Are there others? For example, is there a way to obtain calcium carbonate from bones?

Please do tell me about the risks, but don't worry about my safety, since this is not going to be done in real life, but in a Minecraft mod instead. It is important to know the risks though, because I would like to make those risks part of the game, to make very clear for my players that, if its risky, the same stuff that happens to Steve/Alex will happen to them. (I intend to make it realistic gruesome, meaning I won't exagerate a bit, but either will I minimize the risks because it's "just a game". I'm 14, so I know better, but I know Minecraft is played for kids as young as 7, so I really don't wan't little toddlers thinking Chemistry is all fun and games).

Most of the $\ce{CaCO3}$ in commercial use today is mined not synthesized. The one exception I know a bit about is the ultra-small particle sized materials. However, few of those are pure, they have surface treatments which constitute a significant portion of their total weight. Pharmaceutical and perhaps food grades are also synthesized.

The major synthetic pathway is circular: limestone is mined, then it is heated which produces lime and carbon dioxide. The lime is then mixed with water and $\ce{CO2}$ is added (bubbled in/through) and $\ce{CaCO3}$ precipitates. It is then dried.

Purification depends on the source; how impure it is to start with. It is generally difficult to separate magnesium carbonate from calcium carbonate, and given that limestone and aragonite as well as dolomite ($\ce{CaMg(CO3)2}$) are taken out of the ground (i.e. are 'dirt' cheap) there's usually no need.

As far as the risks of making it, well you could burn yourself, adding lime too fast to water could splash and it's infamous for dissolving bodies in murder mysteries, as well as being used in real life to help in outhouses and gardens where organic material is desired to rapidly decompose. Meaning you don't want to get it in your eyes, nor do you want it left on your skin for very long. It can, theoretically, fume, while being decomposed and that wouldn't be good to breathe. And if a centrifuge or mechanical press is used to squeeze the water out of it (first step in drying), then you have the assorted mechanical dangers. Its synthesis is probably one of the safest chemical syntheses you could undertake.

$\ce{CaCO3}$ decomposes with fairly mild heating (I think around 840 °C), so any material which is predominantly that (like seashells, bones) could be used to make the lime.


Cathode region in corrosion The anode region is on the iron where the following reaction occurs:
$$\ce{Fe(s) -> Fe^2+(aq) + 2e-}$$

and at the cathode region, the following reaction occurs
$$\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}$$

However, the textbook does not explicitly state what the cathode region is. Can someone provide an answer?

I would assume that the cathode region is a source of $\ce{O2(g)}$, which is probably the air in contact with the iron. Therefore the reacting should occur at the interface iron/air.


Phase equilibrium in mixtures How should I determine the total number of phases that are at equilibrium with each other in a multi-component mixture?
For example, this problem is from the 2007 GRE published by ETS (Problem 24):

What is the maximum number of phases that can be at equilibrium with each other in a three component mixture?

I initially thought the answer should be 3 since no matter what kind of mixture it is there should always be solid, liquid, and gas. But then I understood that there can be situations like immiscible liquids which increases the total number of phases. The actual answer is 5. What does this number entail?

The solution is a direct application of the phase rule
$$F=C+2-P$$
with the number of components C=3 and F the number of degrees of freedom (the number of intensive variables that can be varied while phases continue to coexist). The phase rule can be derived by considering all the conditions of balance (chemical, pressure, temperature) that must be satisfied by the phases in mutual equilibrium.
Solving for P (the number of phases).
$$P=5-F$$
The maximum number of coexisting phases is observed when there are no degrees of freedom left (a coexistence point), that is when
$$P=5$$


Converting methanol into methane I know that methane to methanol is a common practice, but is it possible to convert methanol back into methane, and would it be a relatively easy process to achieve?

Methanation is a process whereby synthesis gas is converted to methane.

$\ce{CO + 3H2 -> CH4 + H2O}$

Methanol, like nearly all organic molecules, can be gasified to form synthesis gas at elevated temperatures.

$$\ce{CH3OH -> CO + 2 H2}$$

(Note: the gasification reaction is actually a complex balance of many types of reactions.  Often times mixtures of $\ce{CO}$, $\ce{H2}$, $\ce{H2O}$, and $\ce{CO2}$ are obtained.  The reaction above is the ideal case where complete conversion is achieved.)

Thus a process for converting methanol to methane could proceed via gasification followed by methanation.

However, since methanol is made commercially from methane by essentially the reverse process, this route is unlikely to be economically viable!


Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured? Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule. In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule. Does this mean an actual electron was transferred given that oxidation numbers are fictitious? I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

Oxidation numbers are fictitious charges that pretend the entire molecule is an ion i.e. it artificially localizes electrons onto atoms within a single molecule.

Sort of. In calculating oxidation numbers we pretend that covalent bonds are ionic. Perhaps it is valid to say we are pretending the entire molecule is an ionic compound.

In a redox reaction, we find that an oxidation state of one molecule changes compared to a different molecule.

More accurate, we find the that oxidation states of atoms change going from reactants to products. Charge conservation, coupled with the way we assign oxidation states, guarantees that the sum of oxidation states is constant. So if one atom shows an increase in oxidation state, at least one other has to show a decrease in oxidation state.

Does this mean an actual electron was transferred given that oxidation numbers are fictitious?

It is not clear what "an actual electron" means. Electrons are indistinguishable, you can't paint one red and follow its path during a reaction. If the electron transfer (according to oxidation numbers) is intramolecular, it becomes a bit of a philosophical question whether there was an electron transfer. For example, is there an electron transfer during keto-enol tautomerism?

I'm trying to rationalize this, my guess is that the answer is yes because oxidation numbers are a formal way to localize electrons within a molecule, but differences in oxidation number across molecules only happen with physical electron jumps.

It is not clear what "physical electron jumps" are.
For some reactions, it is clear that you have an electron transfer because the only thing that distinguishes reactants from products (pairwise) is the number of electrons:
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
If this reaction is run as two half-cells, this is even more clear-cut.
For other reactions, the changes are more subtle. If a iodine radical combines with a methyl radical (as part of a chain termination reaction), it is formally a redox process:
$$\ce{I. + .CH3 -> I-CH3}$$
However, the electronegativities of iodine and carbon are quite similar, so it is not intuitive that an electron transfer happened. As Maurice points out in the comments for a similar case, you would have trouble writing this in terms of half-reactions.
In the reaction of methyl lithium with a carbonyl, the oxidation numbers change. You might not see it as a electron transfer and would rather think of it as methyl transfer. However, because carbon is more electronegative than lithium, the oxidation numbers of the carbonyl carbon and the methyl anion change when a bond is formed between them.

Does a change in oxidation number in a redox reaction mean that an actual electron transfer occured?

If I answer this like a certain politician: it depends upon what the meaning of the phrase "actual electron transfer" is. So it is mostly a question about how we talk about redox reactions rather than a question about the underlying chemistry.


Rate of change and kinetics order of reaction question The rate of reaction is given by the following equation, where [Ker-S-S-Ker] is just the molecule being broken down and [RSH] is an abbreviation for thioglycolic acid
The thioglycolate is in large excess and so essentially remains unchanged.

Can somebody explain why the answer is C?
I thought the answer was E (because the rate of change, i.e. the derivative, is equal to the concentration of the reactant itself and so, I thought this signifies a decaying curve?)
Many thanks.

Indeed, the only thing you'll need is solving the first order differential equation (but before that, let me define $R_0 := \ce{[RSH]_0}$ and $C :=\ce{[Ker-S-S-Ker]}$.
Now, let's solve it assuming (that's what is given) that $R_0$ can be considered a constant.
Then, we have : $$\frac{\operatorname{d}C}{\operatorname{d}t} = -kR_0C \implies \frac{\operatorname{d}C}{\operatorname{d}t} + kR_0C = 0 \implies C(t) = \exp\left(-\frac{t}{kR_0}\right) + C_0$$ (integration constants can be determined using initial values, I won't be giving further details here on how to solve this...)
Hence, since the concentration of your molecule is an exponential function of time, it is inevitably a convex function (so only curves C and E can work).
Finally, a simple thought on the values of $k$ and $R_0$ show that curve E cannot be the answer as the decay is to "fast".
Hope this helps, if needed I can also provide a "variables separation" method of solving the differential equation.


Why are the calculated NMR values on the DFT & GIAO level of cyclopentane-1,3-dione so far off? I'm trying to simulate the 13C NMR spectrum of cyclopentane-1,3-dione (PubChem CID: 77466; CAS 3859-41-4; ChemSpider ID: 69875; SDBS No: 15258; SMILES: C1CC(=O)CC1=O). There are three equivalence classes of carbons: The two directly attached to the oxygens (≈197 ppm, call this class 1), the one carbon between those two (≈105 ppm, call this class 2) and the "other two" (≈31.3 ppm, class 3).

I've been using Gaussian 16 via the guidance provided by Dean Tantillo's group at UC Davis, the chemical shift repository (CHESHIRE) in particular first performing molecular structure optimization at some level (e.g. B3LYP/6-31+G(d,p)) and then calculating shieldings (at mPW1PW91/6-311+G(2d,p) with solvent corrections, scrf=(solvent=chloroform, smd)). But I keep getting shielding values that are radically off.

In particular, regardless of the settings I arrive at isotropic shielding values for classes 1, 2, 3, of ≈-34,  ≈136, ≈144. Now of course observed chemical shifts and shieldings are often related by a linear scaling, per the Tantillo group's work (and others), which means that it's very odd that classes 2 & 3 here (the non-Oxygen-attached carbons) have nearly identical shielding values.

I am new to chemistry and DFT methods, so I worry I might be something wrong, but it might also be the case that this is just a "hard" molecule for DFT methods to get right. I wanted to know:

Are there any higher levels of theory that I should try / better methods for shielding calculation that would be worth assessing?
Is there any way of analyzing the output from Gaussian16 to get a hint that things might be going awry?
Is there something that just makes this a "hard" molecule to calculate? I get the impression that the contributions in liquid state from various rotamers, etc. is relatively minor as it seems quite rigid (although I have not quantified this).

Brief Summary

The computational results given below indicate that there is something inherently problematic with this seemingly simple system.

Computational Methodology

Full optimizations and corresponding harmonic vibrational frequency computations of cyclopentane-1,3-dione was were at two levels of theory:

B3LYP/6-31+G(d,p)
MP2/aug-cc-pVTZ

Each optimized geometry was subjected to GIAO-NMR computations at the following levels of theory:

mPW1PW91/6-311+G(2d,p)
mPW1PW91/IGLO-II
PBE0/IGLO-II
mPW1PW91-SMD/6-311+G(2d,p)
mPW1PW91-SMD/IGLO-II
PBE0-SMD/IGLO-II

The NMR levels of theory were chosen specifically to look at basis set effects as well as implicit solvation (SMD) effects.  The solvent chosen was chloroform.

All computations were carried out using the Gaussian 09 software package.  All DFT computations employed an ultrafine integration grid.  The frozen-core approximation was invoked for the MP2 computations.

Results

Table 1 presents the absolute isotropic shielding constants ($\sigma\ce{C}$) as well as the computed chemical shifts ($\delta\ce{C}$; determined using Equation 1 given by Martin in the comments) with all values given in ppm.  The carbons are labeled in figure 1 for cyclopentane-1,3-dione.  Despite the level of theory used for the optimization  or NMR computation, the $\delta\mathrm{C}^2$ values are consistently off by ≈50 ppm with respect to the 'experimental' values obtained from equation 1.

Figure 1: Cyclopentane-1,3-dione with labeled carbons

Equation 1:
$$\delta=\frac{\text{intersection}-\sigma}{-\text{slope}}= \frac{186.5242-\sigma}{1.0533}$$

I am actually quite surprised by these results as I did not expect this system to be difficult to characterize with straightforward NMR computations.


why does adding a non volatile solute not decrease the boiling point? If I add a solute that decreases the bond strength of the pure solvent then wouldn't the solvent molecules evaporate more easily as it has lesser force holding it down now?But in my textbook, it's written that BP always increases when we add a non-volatile solute.

Decrease of the solvent activity due its decreased molar fraction is bigger than the solvent activity increase due eventual solvent solute interaction.

So the pressure versus solute molar fraction curve may not be linear as by the Raoult law, but is monotonous.


Smelly matters in urine - do they have lower or higher boiling point than water? I don't know what exactly are the compounds that causes urine to be smelly, but does it (or do they) have higher or lower boiling points than water? If I have both liquid and exposed them to open air, will the water dry up first or will the smell disappear first?

There are several reasons that a smell vanishes.

One obvious reason is that a smelly component has a very low boiling point and simply evaporates. Another reason might be a chemical reaction of the "stinker" when exposed to air, which converts it to something untraceable.

On the other hand, other reactions might result in quite the opposite but I rather doubt that you want to perform a test run with open bowls of urine over the next week at home ;)

Here is an older article from 1971 on the gaschromatographic analysis of volatile organic compounds in human urine after extraction with diethylether.

On a first view, however, the ketones that they found usually have a rather nice smell.

In order to get a hand on the components in the vapour phase above the urine, others have performed a headspace GC-MS analysis, see this application note
and this article.

Again, the ketones usually have a pleasant smell, but they also identified some of the highly volatile bad boys, such as trimethylamine, methanthiol, and dimethylsulfide. These have boiling points of ~5, ~6 and ~40 °C, respectively.


Is this reaction at equilibrium? The following equilibrium occurs:

$$\ce{2 NOCl(g) <=> 2 NO(g) + Cl2(g)}$$

A gaseous mixture of $\ce{NOCl}$, $\ce{NO}$ and $\ce{Cl2}$ is put in a container. After a few minutes it is found that two moles of $\ce{NOCl}$ react for every three moles of products which react. Is the mixture at equilibrium? Why? In other words, how does the mole ratio relate to whether or not the reaction is at equilibrium? Thanks.

The information you provided ("two moles of NOCl react for every three moles of products which react") confirms that the reaction occurs as depicted, but is insufficient to establish that the system is at equilibrium. The important missing information is not the relative proportions of these three species in the mixture, but whether those proportions are changing (system is not at equilibrium) or staying the same (system is at equilibrium).


Reaction of hypohalous acids with alkenes I have read that hypohalous acids react with alkene to give halohydrin. For example the below reaction of hypochlorous acid with 2-methyl-2-butene which I found in Wikipedia.

Now my question is: Since hypochlorous acid is an acid it should give $\ce{H+}$ and $\ce{OCl-}$ ions which it does according to Wikipedia (although it is a weak acid), so shouldn't it add as $\ce{H+}$ and $\ce{OCl-}$ like other hydrogen halides add to alkene? Why would it add as $\ce{HO-}$ and $\ce{Cl+}$?

Let us deconstruct our reagent:

Hypochlorous acids, more commonly, 'bleach', is an oxidant.  Additionally, it is not a great acid, pKa 7.40.

And let us deconstruct our product:

Additionally the chloride is on the least substituted carbon (anti-Markovnikoff) without any indication of added radical source.

In this context it is unwise to view this as an analogous to HCl addition.

Your product could be readily explained by an epoxidation-type reaction followed by a nucleophilic substitution with residual chloride ion.  The regioselectivity of Cl on the less substituted carbon would seem to support that, as well as the absence of a O-Cl bond in the product.


Energies of atomic orbitals on molecular orbital diagrams When we plot molecular orbital diagrams we use a linear combination of atomic orbitals.

Where can I find the energies of particular atomic orbitals?

In the picture below, the energies of 2s and 2p orbitals of Oxygen are lower than Carbon's 2s and 2p orbitals of Carbon, what is the logic behind this?

Ben Norris's comment is correct.

An estimate of the atomic energies for orbitals may be gleaned from ionization energies, for example here:

http://periodictable.com/Properties/A/IonizationEnergies.html

Otherwise, the energies need to be calculated quantum mechanically.


Is Na$^+$ more stable than Na? I had a question in my textbook:-
Na -> Na$^+$ + e$^-$
Cl + e$^-$ -> Cl$^-$
But Na$^+$ and Cl$^-$ have a stable electronic configuration. So, why do they react to form NaCl?
So, this is my question. Is Na$^+$ really more stable than Na?
Na$^+$ doesn't react with atoms, only with ions, like Cl$^-$. Moreover, Na$^+$ has a noble gas electronic configuration, and Na is highly reactive BECAUSE of the extra electron in its valence shell. So, in this respect, Na$^+$ should be more stable, right?
Edit: I looked this up on StackExchange, and got varying answers. That's why I'm confused.

I think you are mixing things up in your question and it would be helpful to specify what the question in your textbook was.
NaCl is a solid made up from Na$^+$ and Cl$^-$ ions, not from uncharged atoms as stated in your question,  arranged in a periodic crystal lattice. So the redox equations for oxidation and reduction in your question are not relevant for the formation of NaCl from its ions e.g. from a supersaturated solution, where the formation of the solid releases lattice energy. Although the redox equations are relevant, when looking at the formation of NaCl$_{(s)}$ from its elements Na and Cl, which you rightly pointed out yield the favorable noble gas configuration for both. (See Born-Harber cycle for more information on the formation of salts from its elements and the related energies)
As stated in Nilay Ghosh's comment, a question (and answer) comparing the stability of Na vs Na$^+$ can be found here.
Also: Sodium ions do in fact react with other (uncharged) atoms, it's just a question of a strong enough reducing agent under the right environment. Commercially, sodium was produced in the Deville process at 1100 °C with carbon as the reducing agent:
$\ce{Na2CO3 + 2 C -> 2 Na + 3 CO}$


Why don't hydride shifts occur with hydrohalogenation of 4-methyl-cyclohexene? My book (see bottom) implies that hydrohalogenation of 4-methyl-cyclohexene follows this mechanism:

Why not undergo consecutive hydride shifts to furnish a tertiary carbocation?
I would guess that the second step, nucleophilic attack by bromine, is much faster and prevents this from happening. But when I try to confirm this by looking up some examples I see this one at masterorganicchemistry.com:

This implies the following very similar mechanism:

The carbocation in the example is a smaller ring, meaning it should be less stable and even more reactive (less able to perform hydride shifting.) But then again the charge is closer to the tertiary carbon and easier accessible.
Could someone explain what is happening?

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 6th edition; W. H. Freeman: New York, 2014, p491 exercise 12-6:

Exercise 12-6
Predict the outcome of the addition of HBr to (a) 1-hexene; (b) trans-2-pentene; (c) 2-methyl-2-butene; (d) 4-methylcyclohexene. How many isomers can be formed in each case?

The reason for not having consecutive hydride shifts is that generally, a new carbocation is formed through shift only when it is more stable than the immediately previous one.

To form a tertiary carbocation, 3 consecutive hydride shifts would have to occur consecutively (when + is at c-4 from the alkyl group, your first product) in which more than one intermediates have same stability (same number of alpha-H). Hydride shifts will not occur for your second product to form tertiary carbocation as the next intermediate is infact less stable (3 alpha-H as compared to 4 alpha-H previously).

Since the ring already has 6 members (least strain) so ring expansion/contraction will not occur either.


Why is the pKa of a ketone higher than of water? The pKa of water is 16, while the pKa of a ketone, HC3(CO)CH3, is closer to 20.
The lower the pKa the stronger the acid. When a ketone loses a hydrogen, the NBE are resonance stabilized into the carbonyl group, lowering their energy, increasing the acidity. When a proton is removed from water, the NBE have no resonance structures. Yet it has a lower pKa, so it is a stronger acid.

Why is this?

As I'm sure you know, (p)Ka is a measure of the equilibrium between an acid and its conjugate base (what you're calling NBE?). Both the conjugate bases of water (hydroxide) and a ketone (enolate) have a negative charge on the oxygen. I'm summarizing your question as: If both conjugate bases have a negative charge on the same atom type, but one base (the enolate) is stabilized by resonance, shouldn't that corresponding acid (ketone) be more acidic? (Correct me if I'm misreading you.)

I think the key here is what (p)Ka really represents. Fot water, with room temp pKa 15.7, the equilibrium favors water but some dissociation is present. For acetone, pKa 19.2, the equilibrium lies further on the non-dissociated side than for water. We take this to mean that acetone is a weaker acid that water. If we only looked at the stability of the conjugate bases, we might conclude the opposite (previous paragraph). But as pKa measures a balance between a conjugate pair, we must consider the acid forms as well.

The ketone contains the very "stable" (energetically favorable) C=O carbonyl bond. When a ketone donates an alpha proton to become the enolate conjugate base, the carbonyl bond is disrupted. This is an extra energetic cost that is not present in the water equilibrium, and I propose it is the reason why the ketones tend to dissociate less than water.

By this logic, water is a better acid than a ketone because acting as an acid requires a ketone to disrupt an energetically preferable C=O bond. Thus the balance lies closer to the undissociated side for a ketone (lower Ka, higher pKa) than for water (lower pKa), so we say that water is a stronger acid and a ketone is weaker.

(One might argue that the carbonyl need not be disrupted, the lone pair could localize on the alpha carbon, but I hope we can agree such a carbanion is quite unstable, and a carbanion not preferable to breaking the carbonyl and localizing on the more electronegative oxygen.)


Is activated charcoal an air purifier? It's well known that activated charcoal is an adsorbent and due to very big surface area per unit of mass it can adsorb a lot of particles.

Let's take as an example for my question this product on Amazon.

My questions are the followings:

How much do this apply to the air purifiers? According to them it is supposed to work as an air purifier but I don't see it working unless the air you want to purify is being pushed through a activated charcoal filter. AFAIK by itself the activated charcoal is not attracting the smelly particles. Or is it?
I have read that activated charcoal adsorbs bad smells(gases from air) and when the activated charcoal is placed in a hot environment it releases the adsorbed gases. What is that temperature at which adsorbed gases are released back in the air? Is this why they advice placing the product in the sun? Taken from product description: "Placing activated bamboo charcoal bag in the sun for 2 hours once a month, naturally rejuvenates the charcoal by removing excess moisture - extending the life of these charcoal room odor absorbers for up to 2 years."

Agreed: unless air is pushed through the activated charcoal bed, few impurities would be adsorbed.
It would depend on how tightly the substance clings to the carbon, i.e. it would vary with impurity, but Trox Technik KS Filters recommends, "saturated activated carbon is reprocessed... with a gradual increase of temperature from 20 to 830 °C and a delay period of approximately 30 minutes."
Finally, how long does the air-stream remain in contact with the filter? It would seem that most air bypasses the pores of the carbon on inexpensive commercial air filters. Activated charcoal might be of more use in reducing foul odors in a closed container, such as bottles of medicines.


Why do dihaloalkanes lose both halogen atoms during mass spectrometry? I was looking through mass spectrums of haloalkanes and found the one for 1,2-dibromobutane confusing:

The base peak is at 55 m/z, and the only fragment I can think of with that ration will be $\ce{C4H7+}$. The two bromine atoms must be absent as the relative molecular mass of one bromine atom is 80. I can't think of how this will happen. Does the removal of one bromine atom force the other off? Does it form an alkene? It is present in all positional isomers (1,3-dibromobutane, 1,4-dibromobutane etc.) I initially thought that the lone pair of the bromine will be donated to the positively charged carbon, but if the carbon is 2 carbons away from the bromine, I can't imagine this happening.
I feel like this is somehow related to the addition of bromine to alkenes, maybe the bromonium ion is an intermediate and bromine gas is formed? The chemical formula might be that of an butene radical. But since the peak still shows in the spectrum of 1,4-dibromobutane, I don't think that is the case.
If anyone has any theories, please do share!

It isn't just both bromine atoms that are lost, although that might be expected in mass spec given the relative weakness of the carbon-bromine bond.  You also lose a hydrogen atom.  And that is key.
We might suppose that a single bromide ion (or bromine atom plus electron) comes off first, forming a $\ce{C4H8Br^+}$ ion.  That is the peaks at 135 and 137, recalling that in mass spec the mass of a bromine atom is both 79 and 81 which are the most common isotopes.  There is nothing spectacular about such an ion per se, but now it can eliminate $\ce{HBr}$, or $\ce{H + Br}$ to give a resonance-stabilized cation:
$\ce{CH2=CH - \overset{+}{C}H - CH3}\leftrightarrow\ce{\overset{+}{C}H2 - CH=CH - CH3}$
Thus the formation of this ion is energetically favored, and it would account for $\ce{C4H7^+}$.
In the case of 1,4-dibromobutane you should expect the initial (primary) $\ce{C4H8Br^+}$ carbocation to rearrange with a hydride transfer to the end carbon that first lost the bromine, making the ion secondary.  Then the ion will match that obtained with the other isomers.


How does VSEPR theory explain the formation of different bond angles in PCl₅? We know, that in the exited state, electrons of Phosphorus could go to higher states and form $\ce{dsp^3}$ hybridization. Therefore it will create five equal energy states of electrons. But my question is that, why are the angles of the bonds in $\ce{PCl5}$  90 and 120 degree instead of equally distributed? How does VSEPR theory explain the difference in the angles?

Why do the  five equal bonding pairs of electrons have a trigonal
 bipyramidal arrangement, with two different internal bond angles -
 120° in the plane and 90° between the plane and the axial atoms?

There are only five Platonic solids:

tetrahedron,
octahedron,
cube,
dodecahedron, and
isocahedron.

Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane.
Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages Min-Energy Configurations of Electrons On A Sphere (Stable link via The Internet Archive).

Also there are not five equal energy states as demonstrated in the lecture notes by Marcel Schlaf and Kathryn Preuss from the University of Guelph:
Chapter 5.5 (continued 2) MO Theory ICl4, PCl5. (pdf) (Stable pdf link via The Internet Archive)


A liquid that forms a surface membrane that prevents evaporation but allows oxygen to pass through I'm a biologist. I have a water solution with living cells in a tiny well (10 um). I need to keep the solution from evaporating for several days in a warm incubator. I'm looking for a liquid to place on the surface of the solution that will create a surface membrane/film that prevents or slows evaporation but continues to allow oxygen exchange. I welcome any other suggestions for achieving the same result, prevent/slow evaporation while allowing oxygen exchange.

Here are some things that we've tried and some constraints:

Increasing the humidity in the incubator is not sufficient (we've tried)
Lowering the temperature in the incubator is not an option (we want the cells to keep doing what they do)
Sealing the well is not an option (we need the solution to remain oxygenated)
What can be added to the solution is limited to substance we are sure won't react with the cells

I don't think you or your cells want any strange low molecular weight substance in your experiment.

You can buy polymethylpentene (PMP) foil, this stuff has a rather low density compared to PE and other polyolefines, and therefore a high permeability for gases (10x higher than PE for oxygen, and PE is already not exactly tight against gases, x20 higher permeability than PVC for example). 50µm films are readily available, you find them online in a minute.


Regularity of electron configuration for heavier atoms Edit: Is there structure in the nucleus that might be responsible for part of the regularity of the (ground-state) Aufbau process? The periodic addition of electrons is more than one might naively predict from Coulomb forces alone--so is this an intrinsic property of electrons or does the nucleus somehow mediate or influence the Aufbau process?

This is a very naive question.

Only after making graphs of electron counts by orbital for elements in sequence and seeing the symmetry of some of the heavier (and possibly non-realizable) elements (below is 170 or so, from the Wiki electron configuration data page) did it occur to me that the picture seems to imply a lot about the nucleus that I do not recall from general chemistry.

In particular for heavier elements, despite some irregularity along the way, it seems to imply that there are mirroring forces inside the nucleus that govern the buildup of electrons and that in some sense the nucleus is at least as complex as the rest of the atom. Not only are the electrons balanced in terms of charge numerically, but there is seemingly some structural (for lack of a better word) equivalence as well.

If someone could give a qualitative synopsis of what is known about this and/or correct any misconceptions that would be great. An introductory reference would also be of interest.

Yes, the nucleus has a complex structure, too.  The most basic evidence for this comes in the existence of patterns in which nuclei are most stable, which manifests itself in the decay lifetime and to some extent in the cosmic abundance of isotopes.  I don't think there is any reflection of it in the electronic structure: from the point of view of the electrons, the nucleus is pretty much just a point positive charge.  The patterns in the electron configuration you are noticing are due to the Pauli exclusion principle and the various ways quantum mechanical particles can fit into a spherical potential well.   You can find a description in any basic general chemistry textbook, and the math will be developed in any undergraduate physical chemistry textbook.

You can get started on nuclear structure by googling "nuclear shell structure."  That will lead you to very basic introductions, e.g. on Wikipedia or Hyperphysics, and you can decide where to go from there.  I don't know any textbooks that treat it simply, because the assumption is that if you are interested in the nuclear physics of nuclei you are already very familiar with the quantum mechanics of electrons.


I need to find the Ka of a weak acid in titration with a strong base I have been working on this lab forever and have not been able to get a $K_\text{a}$ anywhere close to the accepted value. I’m sure I’m missing something really simple.

We titrated $\ce{CH3COOH}$ with $\ce{NaOH}$.

I used $10.00\ \mathrm{mL}$ of $\ce{CH3COOH}$. Volume of $\ce{NaOH}$ at the equivalence point was $10.14\ \mathrm{mL}$, and the concentration of $\ce{NaOH}$ was $0.1530\ \mathrm{mol \cdot L^{-1}}$.

First I found the concentration of $\ce{CH_3COOH}$ by using $c_\text{a} \cdot V_\text{a} = c_\text{b} \cdot V_\text{b}$, so $c_\text{a} \cdot 10.00\ \mathrm{mL} = 0.1530\ \mathrm{mol \cdot L^{-1}} \cdot 10.14\ \mathrm{mL}$
$c_\text{a} = 0.1551\ \mathrm{mol \cdot L^{-1}}$

Then I used the equation: $\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$
and the pH (at the midpoint of $5.07\ \mathrm{mL}\ \ce{NaOH}$) which was $4.87$

$$K_\text{a}= \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]
[\ce{H3O+}]}=[\ce{CH3COO-}]= 10^{-4.87}= 1.36 \times 10^5
$$
$$
K_\text{a}= \frac{(1.36 \times 10^5)^2}{ 0.1551} = 1.19 \times 10^{-9}
$$
Any help on what I am doing wrong would be greatly appreciated!

I’m not sure if I’m missing your question, but I thought that perhaps you could use the relation $\mathrm{pH}=\mathrm{p}K_\text{a}$ at the half equivalence point? The answer would be still incorrect but closer?

Otherwise, you could probably use stoichiometry to solve for $K_\text{a}$ at the beginning or the end of the titration by constructing an ICE table or otherwise. By considering the deprotonation at the start, so the concentration of protons would be equivalent to concentration of conjugate base. Then, instead of inserting the pH at the half equivalence point, insert the pH before the titration.

If you need help doing the algebra, I could try...

But I think what you did wrong was inserting the concentration at the half equivalence point, rather than before the titration.


Stability of cyclohexane type species I came across a question where we have to find out the stability of species that are in chair conformations. The species are as follows:

Comparing the 'A' values tells me that compound I, III and VI should be most stable, but the answer given is: I,IV and V.
Can anybody tell me where did I go wrong?

According to the Ref.1, 5-hydroxy-1,3-dioxane (the compound of structures V and VI) both in the gaseous state and in dilute $\ce{CCl4}$ solution exists as a chair conformer, the hydroxy group in an axial position with an intramolecular hydrogen bond of the $\ce{O—H⋯O}$ type:

Accordingly, the axial conformer (V) is more stable by $\Delta G^\circ = \pu{-1.2 kcal mol−1}$). Microwave spectroscopy and the $^3J_\ce{H(5eq),OH}$ coupling constant suggest that the $\ce{OH}$ group lies in the plane of symmetry $(\ce{C_{(2)}-C_{(5)}-O}\text{-plane})$ and is a part of a bifurcated hydrogen bond to the two ring oxygen atoms (Ref.2). Therefore, it is safe to say the conformer V is more stable than conformer VI between two possible conformers of 5-hydroxy-1,3-dioxane (marked with $\color{green}{\text{green}}$ oval).
It is well known that the most stable conformer of tert-butylcyclohexane is tert-butyl in equatorial position. In a 1,4-substituted version of 1-tert-butyl-4-methylcyclohexane, since methyl group is so small compared to tert-butyl group that I is almost exclusive in equilibrium (marked with $\color{green}{\text{green}}$ oval).
However, I wasn't so sure about stability between III and IV. There is clearly no H-bonding opportunities except theoretical chemist may argue hyper conjugation with ring oxygen by MO calculations. I leave it open for any computational chemist to prove that possibility, yet my best guess is III more stable between them (marked with $\color{orange}{\text{orange}}$ oval with question mark).
References:

J. C. Jochims, Y. Kobayashi, “Bifurcated hydrogen bonds in z-2-phenyl-1,3-dioxan-5-ols,” Tetrahedron Letters 1976, 17(24), 2065-2068 (https://doi.org/10.1016/S0040-4039(00)93819-X).
Jose L. Alonso, E. Bright Wilson, “Study of an intramolecular, bifurcated hydrogen bond in 1,3-dioxan-5-ol by microwave spectroscopy,” J. Am. Chem. Soc. 1980, 102(4), 1248–1251 (https://doi.org/10.1021/ja00524a005).


Hydrolysis of Radiolabeled Esters Hydrolysis of C$_6$H$_5$COO$^{18}$CH$_3$ with aqueous NaOH produces what final product.

(a) CH$_3$$^{18}$OH

(b) CH$_3$CO$^{18}$ONa

(c) CH$_3$C$^{18}$OONa

(d) CH$_3$$^{18}$ONa

(e) CH$_3$OH

The correct answer is (a), which is puzzling to me.  I think it should be (d), as when the radiolabeled methoxy leaving group is pushed out by the reformation of the carbonyl, it will be negatively charged.  The only cation that is floating around is Na$^+$.

Where, oh were, did this magical proton appear in basic conditions?

EDIT:  Since I think my comments will get longer I'll just post my question here.  I agree that the methoxy is in equilibrium with water:

CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$

But the pK$_a$ of methanol is 15.54, and the pK$_a$ of water is 15.74.  Thus this equilibrium will favor the methoxy/water side.  Thus, the slight majority of product should be a methoxy salt.

You are surely right in that the labeled oxygen will stay with the alcohol and not with the acid. Now is the time to recall what do you know about the comparative acidity of alcohol and... wait, what does the word "aqueous" stand for, really?


Stability of ortho- and para-hydrogen can the stability of ortho-hydrogen greater than para-hydrogen can be stated as this?

  The intrinsic energy of ortho is greater than para-hydrogen and also with rise in temperature %of ortho form increase while para decreases and ultimately at room temperature it's 75% ortho and 25% para-hydrogen. Plus as we know that the everything in nature tries to attain lowest possible energy level, and ortho form with opposite spins counteracts each others magnetic moments, making it energetically stable.

Why is it that pure para can be obtained at lower temperature such as 20 K but pure ortho form is difficult to obtain?

I mean when ortho is more stable it should be ortho which should be available more easily than para form?

From "Statistical Mechanics: A Concise Introduction for Chemists" by Benjamin Widom (pp. 35-36):

  For the diatom of ordinary hydrogen ($\ce{{}^{1}_{1} H}$), where each
  nucleus is just a proton of spin quantum number 1/2, there are three
  possible symmetric states of the pair of spins (nuclear spin triplet)
  and only one possible antisymmetric nuclear spin state (spin
  singlet). [...]

  At high temperatures, which generally means $T \gg
> \theta_{\mathrm{rot}}$ (where $\theta_{\mathrm{rot}}$ is the
  characteristic rotational temperature) and so for $\ce{H2}$ means $T$
  well above 88 K, the three ortho nuclear-spin states and one para
  state are equally populated, so equilibrium hydrogen at high
  temperatures is a mixture of ortho- and para-hydrogen in the ratio
  3:1. In pure hydrogen the magnetic interactions between molecules that
  could cause transitions between nuclear spin states are so weak that
  such hydrogen may be cooled to arbitrarily low temperatures with the
  ortho:para ratio remaining 3:1. That is normal hydrogen. Paramagnetic
  impurities such as $\ce{O2}$, or a catalyst such as $\ce{Pt}$ on the
  surface of which $\ce{H2}$ molecules may dissociate, allowing the
  atoms to recombine with their nuclear spins in a different relative
  orientation, catalyze the ortho-para conversion. Cooled in the
  presence of such a paramagnetic impurity or such a catalyst, then, the
  hydrogen may have its ortho:para ratio continually adjust to that
  which would correspond to thermal equilibrium at the given low
  temperature. [...]

  Cooled in such a way, then, hydrogen becomes pure para-$\ce{H2}$ at
  low temperatures, in contrast to "normal" hydrogen, which remains a
  mixture of ortho and para in the ratio 3:1. If the hydrogen is cooled
  in the presence of a catalyst that catalyzes the ortho-para conversion
  and thus allows it to become pure para-$\ce{H2}$ at low T, and if the
  catalyst is then removed and the hydrogen warmed back up to 298 K,
  say, the result is a sample of pure para-$\ce{H2}$ at normal
  laboratory temperatures. Pure ortho-$\ce{H2}$ cannot be prepared in
  this way.


What dangers/risks are there in the production of carbon subnitride? I was reading that carbon subnitride or Dicyanoacetylene has one of the hottest flames when used as a fuel, but it also is unstable and can explode easily into carbon powder (no specifics on cause).  The only thing I know about its production is that it requires extremely hot graphite to have nitrogen blown on it.  What specific risks are there in producing it or storing it, and how does it get produced and stored safely?  For example is it safe to store at room temperature? Can it be produced safely in a normal atmospheric environment?

There is no way, dicyanoacetylene can be produced at room temperature. According to the paper "Syntheses of Dicyanoacetylene", decomposing 4,5-dicy ano-1,3-dithiol-2-one at 600-800°C, a 59-76% yield of dicyanoacetylene is obtained along with side-products like carbon disulfide, carbonyl sulfide and sulfur. Another scheme is pyrolyzing solid dicyanodiazomethane at 220°C, which polymerizes. The polymerised product is decomposed at 700°C to get desired product. Further, a note on the storage of dicyanoacetylene is mentioned:

Dicyanoacetylene is a hazardous material. Because of its toxicity and
high vapor pressure at room temperature, it should be handled only in
a hood. It is potentially explosive, both in the pure state and in
concentrated solutions, but its thermal stability in dilute solutions
in inert solvents is greater than previously re-ported. Pure
dicyanoacetylene turns dark slowly at room temperature, but it can be
kept almost indefinitely at Dry Ice temperature.

Dicyanoacetylene can only be kept at very low temperatures, in dry ice or liquid nitrogen, like a dewar, away from light and oxygen, as it may polymerize. Do not store it for long. Diluted solutions of dicyanoacetylene are much more stable, and can be safely kept at standard conditions.
Reference

Syntheses of dicyanoacetylene, Engelbert Ciganek and Carl G. Krespan, The Journal of Organic Chemistry 1968 33 (2), 541-544, DOI: 10.1021/jo01266a014


Which is the strongest base among the given anilines? I am facing difficulty in the following problem.

In my view N,N-dimethylaniline 2 should be the strongest base because it has two methyl groups which increases electron density of N and hence it can donate it easily than in other cases. However I am not sure about it.

In my opinion, option 4 should be the answer. This is because too much crowding near the nitrogen atom will throw it out of the plane, as a result of the which the lone pair of electrons of nitrogen atom get localized and are better available for accepting the protons. Moreover, presence of $\ce{-CH3}$ groups also increase basicity due to +I effect.


Would electrolysis work with a conductive (but waterproof) divider between anode and cathode? Note I am not trying to "double fuel efficiency" in a car or something scamey like that. My purpose is to fill balloons with hydrogen and possibly power an HHO torch (also make it safer because pure hydrogen or pure oxygen will not explode).

I found this https://www.thingiverse.com/thing:1200458 HHO generator someone made which seems perfect for my needs except that it outputs HHO together. I cannot find a design for actually useful HHO production that actually separates H2 and O (by useful I mean large volumes of gas, not just a science experiment).

My plan for alterations is to make that design thicker so that it can hold two arrays of plates and split it in halves. I would also add in separate nozzles for each side. I would then print both halves and glue them together with a metal plate (aluminum) in the middle. Would this work? I don't see why it would not however I have not seen a design like this which probably means that I'm wrong rather than everyone else :D

A cell with two electrodes separated from each other and an aluminum divider plate in the middle is really two cells: one has an aluminum plate as a cathode, and the other cell has the same aluminum plate as the anode.

In this situation, the solution does not directly connect the external anode and cathode; the aluminum divider separates the solution into two separate cells.

When you pass current (electrons) thru this arrangement, ions travel. Protons (or other cations) travel toward the negative electrode. In the cell attached to the external positive source, protons move away, and discharge at the aluminum. Simultaneously, protons in the adjacent cell move away from the aluminum and discharge at the stainless steel cathode. Hydroxyls (or other anions) do the reverse.

You will need twice the cell voltage to get a high current, and hydrogen will be evolved from the external cathode (-) and the farther side of the aluminum divider plate. Oxygen will be evolved from the other side of this aluminum plate and from the external anode (+).

Now make a small change: let the aluminum divider go from the top of your case to below the water line, say, halfway down. Then ions can travel thru the solution. This is now just one cell. Hydrogen will be evolved at the (one) cathode and oxygen will be evolved at the (one) anode. Your separation of the gas phase by the aluminum divider allows hydrogen to be collected separately from the oxygen. BTW, you might as well use stainless steel for this divider, since it is used for the electrodes.

When collecting the gasses, hydrogen will be evolved at twice the rate of oxygen. If you build up pressure, the cell will respond by pushing liquid from the cathode compartment into the anode compartment to equalize the pressure. You may have to use a check valve or some kind of flow equalizer to keep liquid from spurting  out of one of the gas outlets.


If vibrational modes are quantised, why is infrared spectroscopy smooth looking? I have heard that the vibrational energy level in a molecule is quantised. However, when looking at the infrared spectroscopy graph, the curves look highly continuous and smooth - not something I would expect from a quantised quantity.

The sharp pointy bits in the fingerprint region (on the right) make sense, given that only certain quantised frequencies would be absorbed - but the alcohol $\ce{OH}$ group seems to be able to absorb a very large range. So, why is the graph smooth when the vibrational energy is quantised?

You should first have a look at the gas phase spectrum of water to convince yourself that vibrational levels are indeed "discrete". The textbook treatment of vibrating molecules assumes that there is no molecule nearby and each and every unit is independent. Which state of matter is closest to these assumptions? The gas phase, the lower the pressure, the further the molecules are apart (large mean free paths). Therefore if you wish to see quantitization visually, it is better to look at the spectrum of gases (rotations, vibrations and electronic transitions). Recall you study atomic spectrum of hydrogen gas not of solid hydrogen or liquid hydrogen.

Now, you may ask, that water is a triatomic molecules with 3N-6= 3 vibrational modes. What are the rest of those lines? Recall that water can also rotate in the gas phase. Those very thin lines represent rotational energy transitions.

The story changes when you deal with liquids or solids. Now the vibrating molecules are so close to each other that they are no longer independent from each other's influence. The rotational structure is completely smeared. The vibrations are still quantized but now you see a collective behavior millions of close-by water molecules which are strongly interacting by hydrogen bond. If we had a very fancy ultra ultra high resolution infrared spectrometer, we might be able to see some structure in each broad band, but the smearing is so bad in solids and liquids that we all see a overall wide peak, which is just an envelope of all possible vobrational transitions by millions of molecules in different environments on the liquid and solid state.


When measuring avidity should you use the ligand concentration or the nanoparticle concentration? Looking at avidity between a ligand-receptor, you're looking at an enhancement of the $K_d$ compared to a lone ligand.

Is it more appropriate to compare the $K_d$ using the concentration of the "free" ligand or the concentration of the free complex?

Using the ligand is odd at the lower regimes since the ligand efficency is fairly low but using the nanoparticle changes how you have to calculate the effective concentration. I've seen papers that do both and I'm curious which is most appropriate/correct

In the case where there is not cooperativity between binding events, the Scatchard Equation is the generalized solution to multiple ligand binding to a macromolecule. The Scatchard plot uses the free ligand concentration.

There are multiple models used to address cooperative binding. The Hill Equation is the most frequently encountered in biochemistry literature:
$$ \theta = \frac{[L]^n}{K_d+[L]^n} $$
Where $[L]$ is the concentration of free ligand, $\theta$ is the fraction of bound ligand-receptor complex, $K_d$ is the apparent dissociation constant, and $n$ is the Hill coefficient, describing cooperativity.


How to name compounds with more than one spiro atom and cyclic compounds with more than two fused rings? The nomenclature of the bicyclic and spiro (mono) compounds are easy to understand. But when the number of spiro atoms or the number of rings exceeds 2, the nomenclature turns very complex. I cannot understand the method of numbering the carbons and the method of writing those bracketed numbers with dots and superscripts.

I have read the IUPAC Recommendations 2004, but the text given there is really hard to understand (for me only, I hope). I want to know the method of writing those numbers with the superscripts inside the bracket for spiro and cyclo compounds. Diagrams with arrows or a detailed text with one particular example, will be helpful.

Bicyclic compounds

Locate the bridgehead carbons, the carbons part of two rings.
Now, identify the three connections between the 2 bridgeheads, called bridges: arrange these in decreasing order of number of carbon atoms (excluding the bridgehead). Say these three numbers are m, n, o [m > n > o].
Write the name of your compound as: bicyclo[m.n.o]"m+n+o+2"alkane.
In case there is a substituent or a functional group is present, you need to number your carbons: start from of the bridge-head, through the longest bridge to the shorter and then the shortest (if any).
Other rules of IUPAC apply as such.

Spirocyclic compounds

Locate the bridgehead carbon, the only carbon part of the two rings.
Now, spot the two ring systems having having 2 or more carbons: arrange these in increasing order of number of carbon atoms (excluding the bridgehead, again).
Say these three numbers are p,q [p < q].
Write the name of your compound as: spiro[p.q]"p+q+1"alkane.
In case there is a substituent or a functional group is present, you need to number your carbons: start from of the atom next to bridgehead in the smaller ring, through the bridgehead carbon to the bigger ring.
Other rules of IUPAC apply as such.


What effects would an electron in the nucleus have? Under the understanding of quantum chemistry:

  There is a very miniscule chance that an electron could appear inside the nucleus of an atom.

What could possibly happen if this "phenomenon" were to occur?  Would it wreak havoc upon the atom and tear it apart?  Or will it bind with the protons and create a temporary vacuum?

Under my current research, the electron would jump out of the nucleus as soon as possible because this phenomenon should not occur for more than a fraction of a second. Though it shouldn't leave without some adverse effects.

An electron being at/in the nucleus results in the Fermi Contact Interaction.

This interaction can be observed experimentally through NMR, EPR, and electron capture.

In NMR of paramagnetic compounds, chemical shifts can be far outside the usual diamagentic range. The contribution to chemical shift of the electron density at the nucleus is referred to as the "contact shift".

For more information on Fermi contact, see "Origin and meaning of the Fermi contact interaction" Theor Chim Acta (1988) 73:173-200.


What is the size of a Sulfate ion? In terms of nanometers, what would the size be of an individual sulfate ion in solution?

Would its size exceed 0.5 nm? Enough to displace something that is held together with a gap of 0.35 nm?

The radius of a sulfate ion is  0.242 +/- 0.007 nm. (Yirhak Marcus. Chem. Rev. 1988, 88, 1475-1498.)
In order to know if it would displace something could you tell us what exactly you are trying to displace?


How many stereoisomers are possible for hepta-1,6-diene? I am asked to identify the number of stereoisomers that are possible for hepta-1,6-diene $(\ce{H2C=CHCH2CH2CH2CH=CH2})$.

I drew the bond-line structure as follows, and I think that there can be cis and trans isomers, but my answer is wrong. The correct answer states that since neither of the double bonds exhibit stereoisomerism, this compound does not have any stereoisomers.

Originally, I thought that single bonds can be considered too. But I think I my error is that by looking at the single bonds, we are considering the constitutional isomers of this compound, rather than the stereoisomers (which are determined by the 'fixed' double bonds). Since double bonds lack the free rotation that single bonds have, for any compound that may exhibit stereoisomerism, we can only consider the double bonds. Also, are we considering only one double bond at a time?

Is this reasoning correct? Can someone clarify and expound more on this please?

In the case of your original drawing with the terminal double bonds, there are no E or Z (trans or cis) isomers!

E and Z isomers are only possible if both ends of the $\ce{C=C}$ double bond each carry a substituent other than a hydrogen atom.

As far as the other structure is concerned, E and Z isomers look as follows:


What is the meaning of the phase state label (cr,l) appearing in ATcT thermochemical data tables? I was examining data associated with this post on the heat of solution of ammonium nitrate when I came upon the NIST ATcT data page which lists the heat of formation of $\ce{(NH4)NO3(cr,l)}$. In this formula, what states does "(cr,l)" refer to?

The nearest I came to an explanation is an article (Ref. 1) by authors of the database explaining that, in the case of $\ce{H2O (cr,l)}$, the notation (cr,l) refers to the condensed state. Also, a link from the above data page explains that "cr=crystal" and "l=liquid" (thanks MaxW for hinting at the solution).

A further update: it finally dawned on me that the website sometimes provides enthalpy data at two different temperatures ($\pu{0 K}$ and $\pu{298.15 K}$) in one line, so evidently the lower T refers to the crystalline state, the higher to the liquid state. But it remains somewhat confusing notation. In the case of an aqueous solution of $\ce{(NH4)NO3}$, it would seem to make sense to use the notation "(aq)" for the liquid state. Evidently "(l)" is synonymous with "(aq)" in the case of solutes in aqueous solution, and "(cr)" refers to a frozen dilute aqueous solution at $\ce{0 K}$. But this remains confusing, since in this table, the last entry describes the following reaction: $$\ce{(NH4)NO3 (cr,l) → [NH4]+ (aq) + [NO3]- (aq) }$$ Here a distinction is made between the hydrated (or aqueous) ions and the species $\ce{(NH4)NO3 (cr,l)}$, whatever it may be. It seems unlikely that $\ce{(NH4)NO3}$ can exist as a molten salt. It seems (cr,l) is used to signify a condensed phase of the pure substance, whichever makes sense under the given conditions. In the above reaction this is presumably the crystalline solid, and the reaction represents solubilization.

References

Ruscic et al., J. Phys. Chem. A, Vol. 108, No. 45, 2004. See the caption to Table 1.


How is Oxygen Toxic? We talk about divers using Scuba tanks that contain a only a small concentration of oxygen diluted in a mixture of other gasses such as nitrogen or helium.

From what I've read, this is because breathing pure oxygen, or a concentration that exceeds the normal partial-pressure of oxygen at sea level, when diving can cause major health problems.

Oxygen Toxicity is a phrase that keeps getting repeated, though I've found sparse few details surrounding what it actually is/means.

I ask therefore: How does oxygen become toxic to humans when breathed-in at high concentrations or underwater or wherever — on a chemical/biolgical level — and also why this is the case?

Because really, isn't good-old $\ce{O2}$ meant to be the gas that keeps us alive?

To get one thing out of the way first: toxic does not always mean the same thing. Some things are acutely toxic (e.g. carbon monoxide), some things are toxic upon accumulation (e.g. lead). Some things can get removed from the organism given enough time (e.g. methanol), some things will stay in your body ‘forever’ (e.g. lead). Some things kill you quickly (e.g. cyanide) while others will cause cancer that will kill you in maybe decades (e.g. benzene). And the organism can cope with some things given small enough concentrations (e.g. nitrite) while others are harmful at even the smallest concentrations (e.g. dimethyl mercury). Paracelsus is typically quoted in German with the saying:

Alle Dinge sind Gift, und nichts ist ohne Gift; allein die dosis machts, daß ein Ding kein Gift sei.
(All things are poison and nothing is without poison; only the dosis determines that a thing may not be poison.)

Thus, the question is only which category oxygen falls into: both in terms of acute/chronic/instant/accumulative/… and in terms of $\mathrm{LD_{50}}$ values.

Oxygen is, in principle, a very reactive gas. It is very electronegative and often reacts from its neutral $\pm 0$ oxidation state to compounds in which it features a $\mathrm{-II}$ oxidation state.
Life and living matter builds on being able to selectively use this type of oxidation enthalpy for its own uses, i.e. to build up complex structures (proteins, DNA) or to overcome entropic problems. From its beginning, ‘life’ was basically a method of using inspecific chemical energy for its own uses. Spontaneous reactions out of life’s control are typically suppressed as well as possible, workarounds made and so on. For example, aldehydes are useful, reactive chemical species that can be derivatised in may different reactions. Living organisms tend to make use of similar chemistry (as least partially) but it also tends to avoid the generation of free aldehydes as much as possible. Indeed, one way life keeps aldehydes ‘safe’ is by masking them as Schiff bases or as acetals (e.g. sugars). Life relies a lot on in situ generation of reactive species.
Indeed, imagining an evolution of the first living being in an oxygen-rich surrounding is very hard. When the first life forms came into being, the atmosphere and the conditions on Earth were generally much more reductive, much less oxidative than today. The first living organisms probably used chemosynthetic energy sources that did not rely on oxygen. It took a while for organisms to develop photosynthesis. This had quite a number of beneficial effects for those organisms: the light energy source is much more easily accessible, carbon dioxide to build up organic molecules was plentiful and these organic molecules could then be synthesised according to the organism’s needs either as building blocks or as an energy source. Photosynthesis only has one disadvantage: It has a strongly toxic by-product which can ruin the carefully-tuned enzymes your cell has by oxidising their metal cores, their sulfur atoms and more; and it can react with other molecules in the cell in unwanted ways. It had to be removed from the cell and the first photosynthetic life forms indeed proceeded to dump it into the environment.
At first, the dumped oxygen would react with reactive species and be chemically bound e.g. in rocks at the ocean floor. Partially, it was also absorbed and diluted in the oceans. After a certain oceanic concentration was reached, oxygen was released into the atmosphere and started to accumulate in land-side oxygen sinks by similar mechanisms. After all these oxygen sinks (‘dumps’) were filled, atmospheric oxygen levels began to increase dramatically within a few million years.
The effect that this had on existing life forms — even during the early evolution of oxygen-generating organisms — was catastrophic (hence the term Oxygen Catastrophy). Most life forms — obligate anaerobic ones — would have become extinct and paved the way for an entirely new generation of species: oxygen-tolerant and finally aerobic ones. The obligate anaerobes had no method to control cellular oxygen while later forms had these mechanisms in place.
The great advantage of having molecular, atmospheric or ocean-dissolved oxygen as an energy source is that it can supply a lot more energy than non-oxygenic processes: think of the amount of heat generated by burning and compare that to most other chemical processes to understand the difference. However, any organism relying on the use of oxygen must also cope with the by-products that can occur during oxidation — both desired oxidation procedures and undesired, spontaneous ones. Aerobic organisms tend to have a large battery of peroxidases and superoxide-dismutases for the controlled degradation of peroxides $\ce{R-O-O-H}$ and superoxides $\ce{H-O-O^.}$ — even though these species autodegrade almost at diffusion-controlled rates. The reason is that these reactive species, generated by the use of oxygen, would create too much havoc to allow them to degrade themselves; organisms must degrade them specifically and with minimal danger to the cell as quickly as possible.
And this is also the question why high oxygen levels can be considered toxic. Oxygen reacts with organic molecules present in the cell in an undesired fashion to generate the more reactive peroxides and superoxides. If they cannot be captured and destroyed quickly enough, they create more havoc by randomly oxidising further parts of the cell in undesired fashions potentially causing protein malfunction, DNA damage, cell wall damage and more. The higher the exterior oxygen concentration, the higher the lung oxygen concentration and the higher the blood oxygen level and cellular oxygen level. The higher the cellular oxygen concentration, the more damage it can perform to a cell and the harder it is for the peroxidases and superoxide-dismutases to keep the reactive species under control. Thus, more damage occurs leading, finally, to that bit of damage that can no longer be controlled.
A catch-it-all term for this is oxidative stress. All the different reactions are too numerous and too diverse to be presented here (or even on a medium as Wikipedia), but they all involve the reactive peroxides, superoxides and hydroxyl radicals $\ce{HO^.}$ (which I didn’t mention yet, but which are also terribly reactive).

Tl;dr: oxygen is great in principle, but too reactive. Cells can cope well with ambient oxygen levels and keep the reactive species under control. A rise in oxygen levels will generate too many harmful, reactive species that the cell can no longer control.


Why does SO2 have a larger bond angle than H2O, H2S, and NH3 So we have to find which of the follwing compounds has the smallest bond angle:
$\ce{H2O}$, $\ce{H2S}$, $\ce{NH3}$, $\ce{SO2}$.

So L $\propto \frac{1}{BA}$
where $L$ is the number of lone pairs and $BA$ is bond angle

So the number of lone pairs is:-

$\ce{H2O}$: 2

$\ce{H2S}$: 2

$\ce{NH3}$: 1

$\ce{SO2}$: 2

Since $\ce{NH3}$ has the fewest lone pairs, it must have the highest bond angle, so it is out.

Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.
The electronegativiy of the O atom in $\ce{H2O}$ is the highest, so it is out.

Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom. The electronegativiy of the H atom in $\ce{H2S}$ is the lowest of the remaining two. The only one left is $\ce{SO2}$ which must have the lowest bond angle but when I checked on google it had the highest bond angle. How is that possible?

Here are the bond angles for each molecule (data from wikipedia):

\begin{array}{|c|c|}\hline
\mathrm{Molecule} & \mathrm{Bond \space Angle \space (^\circ)} \\ \hline
\ce{H2S} & 92.1 \\ \hline
\ce{H2O} & 104.5 \\ \hline
\ce{NH3} & 107.8 \\ \hline
\ce{SO2} & 119 \\ \hline
\end{array}

  So $L \propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle.

This is true but only in very specific situations; when dealing with molecules that have a central atom in the same period and outer atoms of the same element (e.g $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$). It breaks down as soon as you start comparing molecules with central atoms from different periods (e.g $\ce{PH3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle) or when you compare molecules with different outer atoms (e.g $\ce{NF3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle). For the reasons behind this I direct you to an excellent previous answer from @ron.

Additionally, I think you need to reconsider the number of lone pairs in $\ce{SO2}$, which can be described by these two resonance structures.

As you can see there is only one lone pair, but unfortunately this doesn't help us very much as we have no other molecule to compare it to.

  Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.

This is incorrect. In fact the reverse is true - that $BA \propto \frac{1}{ENC}$ - but only in the same situations as mentioned before. In this case the trend is only really pronounced in the second period; it is very slight in the third period and virtually non existent in the fourth. In fact, this 'rule' doesn't really have anything to do with electronegativity (that's just a coincidence) and it's essentially just a result of the first rule.

  Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom.

This is sort of true - it's known as Bent's rule and it can be very useful but it's not really applicable here. It's been discussed many times on this site but here and here are some good introductions.

So how should you answer this question?

The first thing to note is that $\ce{SO2}$ only has three 'groups' on the central atom (sometimes called 'effective electron pairs' in VSEPR theory) - two bonds intermediate between a double and a single bond and a lone pair - whereas all the other molecules have three. Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle. We can apply the hybridisation arguments given by @ron in the answer I linked earlier to determine that $\ce{H2S}$ has the smallest bond angle, and indeed we find that it is almost unhybridised with a bond angle very close to $\mathrm{90~^\circ}$.


Confusion in the name of this organic compound I think the name of the above compound should be 3-Ethyl-2-methylcyclohex-1-ene. However, according to my book (Organic Chemistry Resonance DLPD, page #43, question #5), this organic compound is 6-Ethyl-1-methylcyclohex-1-ene.

A quick Google search shows relevant results for the later name but not for the former.

As far as I know as a high school student, substituents should be given the lowest number and double bonds should be given the number 1 whenever possible.

The name I thought of first satisfies these conditions better than the name the book gives as the answer.

What did I do wrong? Which rule did I not follow?

From here:

"Number the carbon atoms of the double bond as “1” and “2” in the direction that
gives the substituent encountered first the smaller number."

The methyl group right on the double bond is deemed first, so the carbon at that point on the ring gets 1 and the other double-bonded carbon gets 2, forcing 6 for the ethyl substituted carbon.


Compound with a bright green emission when electrically excited I rolled back the edit because I really am looking for the following:

I am looking for an element or compound that would have a bright green emission when electrically excited and is a gas at between 1 torr and 4 torr. If such a compound could be nontoxic, that would be beneficial.

The pressure range is important.

I'm familiar with the emissions spectra of all the noble gases, but none of them work for me. I want to find such a compound to use in a Geissler tube, which will be excited with about 50kVac at 300 to 500kHz. I am capable of pulling a vacuum down to 0.2 torr, and I normally work between 1 torr and 4 torr.

I know of a Xenon/nitrogen mix that produces the desired color, but it's too fussy for me to achieve, and Xenon is too expensive for this purpose.

Relatedly, I was wondering why the Xenon/nitrogen mixture emits a green light. Neither Xenon or Nitrogen has much green at the appropriate pressure, but somehow the combination, within a narrow range of Nitrogen partial pressure, produces a bright green.  What causes this color when the two compounds are mixed together, and how can I predict the color of a given mixture of gases at a given pressure?

I know of the bohr model and balmer series etc, but I don't understand how to apply that to compounds or mixtures of compounds. Is there some other way to predict the spectra produced by mixtures of gases?

It's not easy gettin' green!  (Apologies to Kermit)

While it is not a gas, copper and barium compounds come to mind. If you could introduce a volatile copper compound, you could get something remotely like a mercury vapor lamp, except with copper.


Keeping an open, partially empty soda bottle bubbly longer I remember as a kid being able to get a little pump which you could screw onto a half empty soda bottle (I believe the bottles were glass at the time) to pump air into it to keep it fizzy for longer.  I surmised or was told that this worked by increasing the density of the air, thus making it harder for gas to leave the liquid and enter the air.

I now find myself with partially empty soda bottles.  Can I keep the fizz longer by squeezing the air out of the bottles and screwing on the lid?    (My guess would be that this would work if the compressed bottle is altered such that it does not try and reform back to its original round shape, but if it would pop back save for the lid being screwed on, I'm actually going to make it loose its fizz slightly faster ?)

The fizz is dissolved carbon dioxide (or carbonic acid with which it is in equilibrium) which is put into the bottle under quite some pressure (around $\pu{120 kPa}$ at $\pu{4 ^\circ C}$ or $\pu{250 kPa}$ at $\pu{20 ^\circ C}$; from a first glance on the google results).
$$\ce{CO2 (g) + H2O (aq) <=> H2CO3 (aq)}$$

Because the bottle (in first approximation) is a closed system, there is an established equilibrium between the "air" in the bottle (gaseous $\ce{CO2}$) and the dissolved $\ce{CO2}$. Once the bottle is opened the equilibrium is disrupted and the dissolved carbon dioxide is no longer as soluble as before.

If you close the bottle again and use the little pump you essentially add pressure again, shifting the equilibrium back to where more dissolved carbon dioxide is stable. It won't be as effective as the factory produced bottle, but it will slow down loosing the fizz.

The volumes also play some role here, ideally the aqueous phase should be much larger than the gaseous phase, as you need more molecules producing the pressure if you have more space. An half empty bottle therefore is not ideal to start with.

Squeezing the air out of the bottle would in the worst case result in the opposite effect that you want, in the best case it won't change much. As you correctly remarked, the bottle reforms which means that you would have created lower pressure, essentially dragging out more carbon dioxide molecules. If that was not the case, you would reduce the volume of the gaseous phase, which would probably help a little. Without adding high pressure though it will still loose most of the fizz over time, until equilibrium is reestablished.

In principle there is not really very much you can do. Keep the bottles refrigerated (as carbon dioxide is more soluble at lower temperatures) and try to finish them quickly. Alternatively, get yourself a soda maker and enjoy all the fizzy drinks in all the customised quantities whenever you want. (I guess, this is not a plug for soda maker machines, I personally prefer non fizzy drinks.)


Are there safe inks that allow one to eat a fish after painting it? Just came across a specific type of art called Gyotaku where specific types of ink are used to paint over one side of a fish and then a paper (rice paper, shoji paper, ...) is placed on top in order to retrieve the fish impression. After the impression, the ink is removed with water.
As catching the fish just to print doesn't seem worthy, I am wondering if there there safe types of ink that let's one eat the fish afterwards? Will washing it fully remove the ink?
Is it safe to eat the fish if one uses acrylic paints?
What kind of inks should one avoid?

Edit
Here's a video of an artist painting a oni, washing the ink out and then eating it. It is not clear if the ink is absorbed by the fish before he eats it or not. He states that he uses water-based acrylic and that it is water soluble and washable.

I Googled "edible ink" and found this item, which is a set of edible ink-containing pens.
The keyword is "edible." I'd stay away from anything not specifically labeled "edible."
Finally, it is safe to eat fish adorned with ink specifically certified to be safe to eat.
Disclaimer: There might still be some issues with the fish, so keep your eye/nose on that.


Relation in the sign (+/-) for gyromagnetic ratio and magnetic susceptibility Is there a relation (or formula) between the sign of the magnetic susceptibility and the gyromagnetic ratio of a nucleus in NMR?

I should preface this with the caveat that I'm a chemist and not a physicist, so it's possible that I haven't understood your question correctly.

Magnetic susceptibility, $\chi$, is an electronic property, not a nuclear property, so one could simply write the question off as not making sense.  That being said, let's look a little closer as I believe we can demonstrate an answer to this question.

First, magnetic susceptibilities are typically divided into diamagnetic susceptibilities (which are negative and arise from paired electrons), paramagnetic susceptibilities (which are positive and arise from unpaired electrons), and then a few others (ferromagnetic, ferrimagnetic, etc. which arise from the cooperative behavior of unpaired electrons).

Second, the shell theory of the nucleus provides a nice picture when discussing the NMR properties of different isotopes as we can picture protons and neutrons existing in discrete nuclear energy levels and being paired or unpaired, much like chemists picture electrons in atomic or molecular orbitals with their discrete energy levels.  One of the neat results from this approach is that any isotope with an even number of protons and and even number of neutrons will have all of the protons paired up and all of the neutrons paired up, resulting in a total nuclear spin = 0.  Thus $\ce{^4He}$, $\ce{^12C}$, and $\ce{^16O}$ are all expected to be spin 0.  On the other hand, an isotope with an odd number of protons will have an unpaired proton; and an odd number of neutrons gives rise to an unpaired neutron.  Having a single unpaired nucleon gives rise to a half-integer nuclear spin (1/2, 3/2, 5/2, etc.) where the unpaired nucleon provides a unit of 1/2 to the angular momentum and any remaining integer spin comes from the nuclear equivalent of orbital angular momentum ($l=0,1,2,$ etc.)  Having both an unpaired proton and neutron gives you an integer-spin nucleus (1, 2, 3, 4, etc.) where 1 unit of angular momentum comes from the two unpaired nucleons and the rest comes from the shell theory equivalent of orbital angular momentum.

OK, so why I am boring you with all of this detail?  Well, nuclei with unpaired nucleons are our nuclear analogues to atoms or molecules with unpaired electrons.  Thus, all NMR active isotopes (spin > 0) can be considered to exhibit nuclear paramagnetism.  It's the presence of the unpaired nucleon that gives rise to a non-zero spin angular momentum.  On the other hand, all NMR inactive isotopes (spin = 0) would be expected to exhibit nuclear diamagnetism.

Now, at this point I'm venturing beyond what I actually know, as I have never seen any values reported for nuclear magnetic susceptibility.  However, one would expect that, analogous to the electronic case, nuclear paramagnetism should have a positive susceptibility and nuclear diamagnetism should have a negative susceptibility.  In other words, the only nuclei that should have negative nuclear magnetic susceptibilities have a spin of 0 and no gyromagnetic ratio.  Any nucleus that is NMR active should have a positive nuclear magnetic susceptibility; but, some NMR active isotopes have positive gyromagnetic ratios and others have negative gyromagnetic ratios.

So the answer to your question (if I've understood it correctly) would be NO, there is no formula relating the sign of the gyromagnetic ratio to the magnetic susceptibility.

In the case of normal (electronic) magnetic susceptibility, the nucleus plays no role (or at least such a small role that it's completely negligible) and in the case of the extremely weak nuclear magnetic susceptibility, all NMR active nuclei have positive susceptibilities regardless of the sign of their gyromagnetic ratios and all NMR inactive nuclei have negative susceptibilities and no gyromagnetic ratios.

As a footnote that you may or may not find interesting.  Picturing integer-spin nuclei as the nuclear equivalent of di-radicals gives a nice rationalization for why they are so rare (only ~8 stable isotopes with integer spin vs. ~112 stable isotopes with half-integer spin vs. I don't know how many isotopes with spin = 0).  Just like molecules, the di-radical isotopes are quite unstable, with some exceptions (e.g., $\ce{^2H}$ is spin=1 and has an unpaired proton and neutron, but is a stable isotope; similarly, molecular oxygen is probably the most abundant di-radical (note: you need M.O. theory to see why it's a di-radical, but you can easily show that liquid $\ce{O2}$ is paramagentic - go do a youtube search) and is relatively stable, at least compared to many other di-radicals).


What changes occur when adding differing amounts of acids & bases? In an experiment, $\pu{5 cm^3}$ of $\pu{1 mol/dm^3}$ sodium
  hydroxide are gradually added to $\pu{10 cm^3}$ of $\pu{1 mol/dm^3}$ hydrochloric acid containing methyl orange. What changes occur in the mixture?

  The concentration of $\ce{H+}$ ions increase (decreases)
  The methyl orange changes colour
  More water molecules are formed (water is formed, but use of the word 'more'??)
  A precipitate is formed (no ppt formed, $\ce{NaCl}$ is soluble)

I thought the answer would be (2). However when thinking about the question the methyl orange would be red in the acid and given the proportions of the reactants and the 1:1 mole ratio, the $\ce{HCl}$ would be in excess, so would it still change colour?

It is perhaps difficult to comment on the intended semantics of the option 3, but I think that is the most appropriate answer.

Option 1 and 4 are wrong because of the reasons you cite.

To comment upon option 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\pu{\frac 13 M}$ solution of $\ce{HCl}$, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding option 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' $\ce{NaCl}$ molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water


Separating glucose and fructose If I have a mixture of glucose and fructose in aqueous solution, what is the simplest method to separate them? Heating to 110 °C to melt only the fructose would probably work theoretically, but this would cause a huge mess.

The most common industrial method is one of various types of liquid chromatographies, which is usually how glucose-fructose solutions are enriched in fructose.
I did see a couple other possible methods though:

This one from 1965—before good LC methods existed—relies on the difference in solubility in partially water-miscible solvents like n-butanol to perform a liquid-liquid extraction.
Unfortunately, the pdf seems to be incomplete, but basically the water-solvent-sugar system separates into two phases with each sugar partitioning more into a different phase.
They calculated that it would take around 40 cycles to get 90% fructose from this method, but it doesn't seem to be very efficient due to slow diffusion of the sugars (and since LC is the dominant method today, it doesn't sound too promising on any reasonable scale).
In the introduction, they also note that some of the other attempts of the era involved chemical means—specifically precipitation of the fructose as calcium fructosate, getting the fructose back out by reaction with carbon dioxide, oxalic acid, or sulfuric acid in what is known as Dubrunfaut's method.
Again, this did not result in an industrially useful process.

This one involves careful fractional crystallization of glucose-fructose solutions, as fructose is far more soluble in water than glucose.
The solution is cooled from room temperature to -20 °C fairly slowly (0.5 K/min) while being agitated and in the presence of small glucose seed crystals.
Some of the glucose crystallizes out of the solution, which is separated by centrifugation, and the process is repeated going to -35 °C.
The crystals they get are 89% glucose and 11% fructose, but a big red flag is they don't actually measure the concentrations in the liquid left behind (separating fructose is far more important industrially because pure glucose is easy to get by hydrolysis of starch).
They say that "a concentrated solution of fructose containing a small amount of glucose was obtained", but don't actually prove it, and apparently it only works properly if you can maintain a total sugar concentration near 50%, so I'm pretty sceptical of the method's utility.
You could probably do something similar at home if all you're interested in is getting some somewhat purified glucose.

Unfortunately, if there were a simple and effective way to do this, industrial production probably wouldn't bother with LC.
If you just want to play around with separations, these might be some things to try, but if you just want glucose and/or fructose, I have seen both at bulk food stores. (glucose is often labelled as dextrose and fructose is sold as a low glycemic index sweetener)


pKa and pH graphically I'm quite a newbie in chemistry and I'm trying to learn how logarithmic diagrams work. I have been told that you can calculate pKa and pH graphically:

pKa, when [HA] = [A-]
pH, when [H+] = [A-]

Why does this happen?

You should state the charge balance or the proton balance.

[H3O+] = [OH-] + [CH3COO-]

We take the logarithm of this and get:
log [H3O+] = log([OH-] + [CH3COO-]).

Now, you can plot this in your nice diagram.

log [H3O+] will follow the straight line in your diagram, i.e. log [H+].

The line for log([OH-] + [CH3COO-]) will at very high pH follow the line for log([OH-], but quite soon it will bend off and only follow the line for log [CH3COO-]. This is because [OH+] << [CH3COO-] at lower pH.

When [OH+] << [CH3COO-], you get log [H3O+] = log [CH3COO-]). When the line for log [H3O+] crosses the line for log [CH3COO-], you can calculate the pH of the solution from the diagram.
The lines are crossing each other at right angel: pH = (1.00 + 4.65)/2 = 2.82(5).
Very good, I am encouraging you to use logarithmic diagrams for solving acid-base equilibria.


Why is there no electron swapping after excitation? If we have 2 electrons in the 1st energy level named X & Y, and one electron in the 2nd energy level named Z.

If the electron X was excited and transferred to the higher energy level, why didn't Z replace it in the 1st energy level?
Why don't we say that there is a type of nuclear gravity, and it increases when the atom becomes unstable and that gravitates Z electron towards the nucleus instead of the excited one?

On the first part: it will, or won't, in a subsequent, separate process.

If you excite the system and promote an electron from the lowest orbital to the next orbital, the immediate result is that you have a lowest orbital populated by one electron and a higher orbital populated by two electrons. It is an excited state, and will tend to relax back to the fundamental state if there is an available mechanism to do so - but that will be a second distinct event after the excitation. In fact, the excited state can, depending on the circumstances, last long.

The second part makes no sense. The nature of the interaction between nucleus and electron is not gravitational (it's electromagnetic), and the lowest energy level isn't "below" in any spatial sense. The gravitational analogy is misleading.


Why is the ionic product for water $10^{-14}$? Why is the reference value of $K_w=10^{-14}\ \mathrm{M}^2$, such that $[\ce{H+}]= 10^{-7}\ \mathrm{M}$ and $[\ce{OH^-}]= 10^{-7}\ \mathrm{M}$ for pure water at $20\ ^\circ \mathrm{C}$?

For example, why isn't $K_w=10^{-18}\ \mathrm{M}^2$, and thus $[\ce{H+}]= 10^{-9}\ \mathrm{M}$ and $[\ce{OH-}]= 10^{-9}\ \mathrm{M}$?

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

$$K_w = \frac{a(\ce{H3O+})a(\ce{OH-})}{[a(\ce{H2O})]^2}$$

where $a$ is "activity of".

Then you can approximate activity of water as 1 if appropriate and activity of the ions as concentration of the ions.

Just like any other equilibrium constant, the relative chemical potential (Gibbs energy) of the products and reactant determine the constant.

Consider the relative strengths of the bonds of $\ce{H2O}$, compare to those of $\ce{OH-}$ and $\ce{H3O+}$, as well as the number and strength of intermolecular hydrogen bonds to each of these.

For $\ce{D2O}$, the equilibrium constant is 8 times small.