Structure of Ca2B6O11 What is its structure? I found an image of $\ce{B6O11}$ unit, but it looks incorrect:

 The top left corner one.

The oxygen can make max 2 bonds but atom $\mathrm{O8}$ has three!
There are six oxygen with only one bond, that is the unit has $-6$ net charge, but $\ce{2Ca^2+} = +4$ positive charge.

1) Oxygen CAN form 3 or 4 bonds. The most common example is $\ce{H3O+}$ ion This form of bonding is critical to understand basic nature of ammonia solutions, so for exact description of it look that part of the book.

2)Boron also can form more than 3 bonds if at least one its partner provides an electron pair for bonding, not one electron.

As a general rule I prefer to stay away from the chemistry of borates and alumina-silicates as it is ruled largely by ionic interactions and non-intuitive hydration/dehydration.

===========

In structural diagrams of polymeric salts 'unit' often refers to some recurring fragment... however, it is not nessesarily has same stochiometry than formal anion. In alumosilicates the anionic frame consists of $\ce{AlO4-}$ and $\ce{SiO4}$ tetrahedral units. However, they are usually connected by shared vertexes, so the exact formula of formal anion is different. Borates are a bit more complex system, since they may include both $\ce{BO3}$ and $\ce{BO4-}$ units, and oxygen is often connected to hydrogen.

==========

I looked into open cristallographic database. Unsurprisingly, it does NOT feature a compound $\ce{Ca2B6O11}$. However, salts are often written in 'dehydrated' form, so I looked for any borate with $\ce{Ca:B} = 1:3$ stochiometry. Several with rather complex crystal structure were found, so I left only two: one with rather simple structure of typical 'hidrated' borate anion and one with rather complex continous 'dehydrated' boron-oxygen layers. Note: the structures are refined from X-Ray , so they does not reveal hydrogen positions. Isolated oxygens of water molecules and cations are omitted.

$\ce{B3O8}$ unit from mineral Inyoite, $\ce{CaB3O3(OH)5·4(H2O)}$

$\ce{[B3O6]_{\infty}}$ layer from mineral Fabianite $\ce{CaB3O5(OH)}$

and its main building block


Is SO3 an ion or a molecule? Earlier I thought that $\ce{SO3^{2-}}$ (sulfite) is an ion with -2 charge but today I read on internet $\ce{SO3}$ (sulfur trioxide) is a molecule with no charge. How can this be possible? Please explain knowing that I am only a student of 10th standard (use normal chemistry concepts as far as possible).

I'll give you the structures. Then you'll understand everything:

Resonance structures of $\ce{SO3^{2-}}$ ion:

Structure of Sulfur Trioxide:

The absence of a double bond between two oxygen atoms in sulfite ion shows why it is bivalent.


Can ascorbic acid be neutralized and keep it's elastic properties in dough? Ascorbic acid (C6H8O6) is commonly used in dough to give it strength and elasticity. However, I need this dough to be alkaline, or at the very least neutral. I've used sodium carbonate (Na2CO3) before for this purpose but never at the same time with ascorbic acid.

If I use vitamin C in dough to make it elastic, as well as sodium carbonate to it neutralize it, will the dough retain the elastic properties caused by the ascorbic acid? What ratio of ascorbic acid to sodium carbonate creates a neutral solution? Thanks!

There is high probability the desired properties of ascorbic acid are directly related to acid being acidic and in the free acid form, i.e. not in the form of the ascorbate.

It is hard to advise, not being familiar with this part of the applied food chemistry, it the effect of ascorbic acid happens in the cold dough conditioning stage and/or in the stage of the later thermal processing, in case of noodles.

If the sodium carbonate is added later to the dough, it is unclear, how well would be the dough mixed to neutralized the acid, and what the sideeffect would be.

Bakerpedia.com  explicitly advices against using ascorbic salts.

I highly advice to experiment, as theory will probably not provide clear advice.
I am not sure, if the acid is applicable to the noodle dough as much as for bread dough.


When amino acids react with each other, how do you know which one reacts using it's carboxylic group and which one with it's amine group? For example the dipeptide val-ser, between the amino acids valine and serine, does valine's amino group react or does it's carboxylic group with serine's respective group?

Under physiological conditions (aqueous solution, pH near neutral), amino acids don't react with each other directly, and the equilibrium lies on the side of peptide bond hydrolysis, not formation.

  When amino acids react with each other, how do you know which one reacts using it's carboxylic group and which one with it's amine group?

In the cell, amino acids first react with ATP, then form an ester with RNA, and then their amino group forms a peptide bond with the growing amino acid chain. The last step is catalyzed by ribosomes that select the next amino acid to be added using messenger RNA as a template.

  For example the dipeptide val-ser, between the amino acids valine and serine, does valine's amino group react or does it's carboxylic group with serine's respective group?

In your example, valine and serine could form either valyl-serine or seryl-valine. As Tyberius writes in the comments, peptides are written so that in each peptide link, the amino acid on the left of the link provides the carboxylic acid group and the one of the right of the link provides the amino group. The very first amino acid (the one at the amino terminus) carries an amino groups that has not undergone peptide bond formation, and the last amino acid (the one at the carboxy terminus) carries a carboxylate group that has not (yet) undergone peptide bond formation.


Removing aluminum "corrosion?" I have some aluminum alloy, possibly 6063 or 7050 (no clue really) which has corroded. The finish might be silver-anodized but I doubt it due to the high polish, could be urethane. The item is a fishing rod holder, used in a freshwater marine environment. Hardware is stainless and steel. Several screws and bolts need to be removed, and I was wondering if there was an acceptable reducing agent for the corrosion rather than busting driver bits and/or mangling threads.

Aluminum corrosion in most environment produces $\ce{Al2O3}$, corundum, which is extremely hard (9 on the Mohs scale, where 10 is diamond). Corundum is also quite resistant to most chemicals.

There are various products called "Aluminum Jelly", such as that from Loctite and from Henkel, which can remove some of the surface corrosion. They probably attack the aluminum at its point of contact with the oxide, rather than reacting with $\ce{Al2O3}$ itself. Regrettably, this probably means that though it might brighten the surface, the jelly will not substantially free the bolts.

The drill bits break when they encounter a mass of corrosion and suddenly seize, and this breakage can be minimized by running at slow speed with lots of lubricant. Applying gentle heat and cooling to the parts immersed in lubricant (e.g. Alumaslick or Vibra-TITE) allows it to penetrate better.

In any event, any threads in the aluminum have likely been ruined, so consider drilling oversize holes and re-tapping them (aluminum is easily tapped, but shavings tend to stick in the tool, so use lubricant and clear the tap frequently).


Why does tomato soup "explode" when boiled? When you boil tomato soup it very easily "explodes" when reaching the boiling point. Why is that?

A related question has been answered here before: Why do some solutions eventually stop boiling and start "popping" as they get more concentrated?
However, the answer to that question does not fully answer the current question since it does not address what properties in a solution causes superheating/bumping. That answer describes what we see but does not describe WHY it happens (except for the words "heated too rapidly", but many solutions can be heated with the same speed as tomato soup without "exploding").

For example, to my knowledge tomato soup only suffers the risk of "exploding" THE FIRST TIME it reaches the boiling point. Therefore I have thought the reason for the explosion could be that it contains gases that leave the solution violently. (I have read this is the reason that milk also may "explode" at the boiling point.)

Friends if mine suggest that the reason for the explosive boiling is the viscosity of tomato soup. But this answer would not explain why tomato soup only risks "exploding" the FIRST time it boils. (Without extensive experience I would also say that NOT ALL viscous solutions boil explosively.)

Tha tomatoes are naturally peeled so their elastic peel/membrane does not come into account.

The previous question also relates to gradual concentration which is not the issue of the current question. The current question relates to reaching the boiling point.

So this question is not answered. To rephrase it slightly:
What properties in tomato soup causes it to boil "explosively"?

I think it's probably a combination of things you mention. First, tomato soup is a suspension of solids (tomato tissue) in water. So it's the water that's boiling. That water is turning into steam, which bubbles out of the water. The solid suspension likely traps some of that steam until the pressure builds and you get a big explosive bubble.

I happen to cook, and make a fair bit of tomato sauce, which is very similar to tomato soup. I disagree with your statement that tomato soup explodes only once upon boiling. I see bubbles popping quite continuously, and you have to stir often to prevent this. Which happens to also be the way you prevent bumping in laboratory conditions. Cooking is probably the most comparable chemistry application most people are exposed to.


Calculating mass of krypton from mole fraction There are two gases in a container: krypton and carbon dioxide. If the mass of the gases is 35 grams, and total pressure of the container is 0.708 atm, and the pressure of krypton is 0.250 atm. What is the mass of the krypton?

I've found the mole fraction of krypton is 0.3531. First, I think if I multiply mole fraction of krypton and mass of the gases, I'd get the mass of krypton. But my teacher said if the answer is false. So what is the correct answer?

Let $m_{\ce{CO2}}$ be the mass of carbon dioxide and $m_{\ce{Kr}}$ be the mass of krypton in the vessel. These two are related by:

$$m_{\ce{CO2}}+m_{\ce{Kr}}=35\ \text{g}\ \ \ \ \ (1)$$

If we can get a second equation in both variables, you can solved the system for each variable.

You were able to determine the mole fractions $\chi$:

$$\chi_{\ce{Kr}}=\frac{n_{\ce{Kr}}}{n_T}=\frac{P_{\ce{Kr}}}{P_T}=\frac{0.250\ \text{atm}}{0.708 \ \text{atm}}= 0.353\ \ \ \ \ (2)$$

$$\chi_{\ce{CO2}}=1-\chi_{\ce{Kr}}=0.647 \ \ \ \ \ (3)$$

Let $n_{\ce{CO2}}$ be the number of moles of carbon dioxide and $n_{\ce{Kr}}$ be the the number of moles of krypton in the vessel (its a little easier to work in moles). We can generate two relationships between the numbers of moles using 1) the masses and 2) the mole fractions.

  From the masses, since the mass of a sample of substance is equal to the number of moles times the molar mass (or formula weight).

$$m_x=n_x\cdot FW_x\ \ \ \ \ (4)$$

$$m_\ce{Kr}=n_\ce{Kr}\cdot FW_\ce{Kr}=n_\ce{Kr}\cdot (83.798\ \text{g/mol})\ \ \ \ \ (5)$$
$$m_{\ce{CO2}}=n_{\ce{CO2}}\cdot FW_{\ce{CO2}}=n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})\ \ \ \ \ (6)$$
$$n_{\ce{CO2}}\cdot (44.01\ \text{g/mol})+n_\ce{Kr}\cdot (83.798\ \text{g/mol})=35\ \text{g}\ \ \ \ \ (7)$$

  From mole fractions

$$\dfrac{\chi_{\ce{CO2}}}{\chi_{\ce{Kr}}}=\dfrac{\dfrac{n_{\ce{CO2}}}{n_T}}{\dfrac{n_{\ce{Kr}}}{n_T}}=\dfrac{n_{\ce{CO2}}}{n_{\ce{Kr}}}=\dfrac{0.647}{0.353}=1.832\ \ \ \ \ (8)$$
$$n_{\ce{CO2}}=1.832\cdot n_{\ce{Kr}} \ \ \ \ \ (9)$$

  Substitute equation (9) into equation (7) and solve for $n_{\ce{Kr}}$ and convert to $m_{\ce{Kr}}$.


Percent error and significant figures? For an elementary chemistry class percent error is calculated as:
 [actual-experimental]/actual x 100%  edit:bracket represents absolute value

So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?
EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then,
0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs. -> there are no significant figures so this would mean the calculated 0.0171% error would be zero correct.  If I remember my analytical correctly we will do away with sig. figs. and work with the propagation of error within each measurement but I have not thought about how this would work in a problem like this.

Thank you in advance.

Actually the % error is calculated as:
$\frac{\text{experimental} - \text{actual}}{\text{actual}}\cdot100\%$
That way

if experimental < actual then the error is negative
if experimental > actual then the error is positive

So my question is if the experimental is close enough to the actual that the subtraction would give zero significant figures, would then the percent error then be zero?

Not exactly zero...

EX: measured density of 0.997g/ml and actual density of 0.997171g/ml then, 0.997171-0.997=0.000171 which would be rounded to 0.000 due to sig. figs

Correct for subtraction you have used. Addition and subtraction round to the the last common significant decimal place of all the measurements. That would be the 8 in 10.98 in the following example.
So $10.9\bar8 + 0.6754 + 0.4795 = 12.1349 \approx 12.13 $
In actually you should have used the reverse 0.997 - 0.99717 = -0.000171

there are no significant figures so this would mean the calculated 0.0171% error would be zero correct.

Now this gets sticky. You could write that the error was 0% but that would indicate only one part in 100 precision. If you write 0.00% it isn't entirely clear if that is two or three significant figures. But you have claimed that the hundredth place in the % is significant, which it is not.
Remember that the minimum nonzero error possible was +/- 0.001 which yields an error of 0.1%.
So the error would be best written as 0.0% indicating that the error is 0 to 1 part in a thousand.

Reference: Significant Figures at Wikipedia
I'll leave the propagation of errors for a different exercise.


Is there a way to find the mixture type with just the molecular formulas and masses of the solute and solvent? If you had the molecular formula and molar mass of a solvent and a solute and no other specific information about the two, could one deduce the type of mixture (suspension, colloid, solution) they would create?

Formula alone does not give you the structure. But if you have the structure of the molecules, there are molecular/thermodynamic models which take into account group contributions for every section of a molecule. These can help calculate and predict phase behaviour and mixture type. See PC-SAFT as an example.


"Crystal radius" vs "Ionic radius" What is the difference between crystal radius and ionic radius? For $\ce{NaCl}$ for example the ions combine to a crystal, and I would have guessed they are equal. But I then found an example of data, where they are not.

I am aware that ionic radius is the radius of an atom's ion. Quite simple. So what is crystal radius?

To quote Erik Rakovský · 30.73  · 54.96 · Comenius University in Bratislava (from this discussion on researchgate)

Shannon published simultaneously two sets of ionic radii. So called "effective ionic radii" were made consistent with Pauling's r($\mathrm{O_2^-}$)=140 pm. "Crystal radii" are using Shannon's value r($\mathrm{O_2^-}$)=126 pm.

https://dx.doi.org/10.1107%2FS0567739476001551


Why doesn't H₄O²⁺ exist? I know this question has been asked previously but I cannot find a satisfactory explanation as to why is it so difficult for $\ce{H4O^2+}$ to exist. There are explanations that it is so because of $+2$ charge, but if only that was the reason then existence of species like $\ce{SO4^2-}$ should not have been possible.

So, what is exactly the reason that makes $\ce{H4O^2+}$ so unstable?

I myself was always confused why $\ce{H3O^+}$ is so well-known and yet almost nobody talks of $\ce{H4O^2+}$. I mean, $\ce{H3O^+}$ still has a lone pair, right? Why can't another proton just latch onto that? Adding to the confusion, $\ce{H4O^2+}$ is very similar to $\ce{NH4+}$, which again is extremely well-known. Even further, the methanium cation $\ce{CH5+}$ exists (admittedly not something you'll find on a shelf), and that doesn't even have an available lone pair!

It is very useful to rephrase the question "why is $\ce{H4O^2+}$ so rare?" into "why won't $\ce{H3O^+}$ accept another proton?". Now we can think of this in terms of an acid-base reaction:

$$\ce{H3O^+ + H+ -> H4O^2+}$$

Yes, that's right. In this reaction $\ce{H3O^+}$ is the base, and $\ce{H^+}$ is the acid. Because solvents can strongly influence the acidity of basicity of dissolved compounds, and because inclusion of solvent makes calculations tremendously more complicated, we will restrict ourselves to the gas phase (hence $\ce{(g)}$ next to all the formulas). This means we will be talking about proton affinities.

Before we get to business, though, let's start with something more familiar:

$$\ce{H2O(g) + H+(g) -> H3O^+(g)}$$

Because this is in the gas phase, we can visualise the process very simply. We start with a lone water molecule in a perfect vacuum. Then, from a very large distance away, a lone proton begins its approach. We can calculate the potential energy of the whole system as a function of the distance between the oxygen atom and the distant proton. We get a graph that looks something like this:

For convenience, we can set the potential energy of the system at 0 when the distance is infinite. At very large distances, the lone proton only very slightly tugs the electrons of the $\ce{H2O}$ molecule, but they attract and the system is slightly stabilised. The attraction gets stronger as the lone proton approaches. However, there is also a repulsive interaction, between the lone proton and the nuclei of the other atoms in the $\ce{H2O}$ molecule. At large distances, the attraction is stronger than the repulsion, but this flips around if the distance is too short. The happy medium is where the extra proton is close enough to dive into the molecule's electron cloud, but not close enough to experience severe repulsions with the other nuclei.

In short, a lone proton from infinity is attracted to a water molecule, and the potential energy decreases up to a critical value, the bond length. The amount of energy lost is the proton affinity: in this scenario, a mole of water molecules reacting with a mole of protons would release approximately $\mathrm{697\ kJ\ mol^{-1}}$ (values from this table). This reaction is highly exothermic

Alright, now for the next step:

$$\ce{H3O^+(g) + H+(g) -> H4O^2+(g)}$$

This should be similar, right? Actually, no. There is a very important difference between this reaction and the previous one; the reagents now both have a net positive charge. This means there is now a strong additional repulsive force between the two. In fact, the graph above changes completely. Starting from zero potential at infinity, instead of a slow decrease in potential energy, the lone proton has to climb uphill, fighting a net electrostatic repulsion. However, even more interestingly, if the proton does manage to get close enough, the electron cloud can abruptly envelop the additional proton and create a net attraction. The resulting graph now looks more like this:

Very interestingly, the bottom of the "pocket" on the left of the graph (the potential well) can have a higher potential energy than if the lone proton was infinitely far away. This means the reaction is endothermic, but with enough effort, an extra proton can be pushed into the molecule, and it gets trapped in the pocket. Indeed, according to Olah et al., J. Am. Chem. Soc. 1986,  108 (5), pp 1032-1035, the formation of $\ce{H4O^2+}$ in the gas phase was calculated to be endothermic by $\mathrm{248\ kJ\ mol^{-1}}$ (that is, the proton affinity of $\ce{H3O^+}$ is $\mathrm{-248\ kJ\ mol^{-1}}$), but once formed, it has a barrier towards decomposition (the activation energy towards release of a proton) of $\mathrm{184\ kJ\ mol^{-1}}$ (the potential well has a maximum depth of $\mathrm{184\ kJ\ mol^{-1}}$).

Due to the fact that $\ce{H4O^2+}$ was calculated to form a potential well, it can in principle exist. However, since it is the product of a highly endothermic reaction, unsurprisingly it is very hard to find. The reality in solution phase is more complicated, but its existence has been physically verified (if indirectly).

But why stop here? What about $\ce{H5O^3+}$?

$$\ce{H4O^2+(g) + H+(g) -> H5O^3+(g)}$$

I've run a rough calculation myself using computational chemistry software, and here it seems we really do reach a wall. It appears that $\ce{H5O^3+}$ is an unbound system, which is to say that its potential energy curve has no pocket like the ones above. $\ce{H5O^3+}$ could only ever be made transiently, and it would immediately spit out at least one proton. The reason here really is the massive amount of electrical repulsion, combined with the fact that the electron cloud can't reach out to the distance necessary to accommodate another atom.

You can make your own potential energy graphs here. Note how depending on the combination of parameters, the potential well can lie at negative potential energies (an exothermic reaction) or positive potential energies (an endothermic reaction). Alternatively, the pocket may not exist at all - these are the unbound systems.

EDIT: I've done some calculations of proton affinities/stabilities on several other simple molecules, for comparison. I do not claim the results to be quantitatively correct.

$$
\begin{array}{lllll}
\text{Species} & \ce{CH4}  & \ce{CH5+}    & \ce{CH6^2+}     & \ce{CH7^3+}     & \ce{CH8^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 556 & -246 & -1020   & N/A   & N/A  \\
\end{array}
$$

Notes:

Even without a lone pair, methane ($\ce{CH4}$) protonates very exothermically in the gas phase. This is a testament to the enormous reactivity of a bare proton, and the huge difference it makes to not have push a proton into an already positively-charged ion.
For most of the seemingly hypercoordinate species in these tables (more than four bonds), the excess hydrogen atoms "pair up" such that it can be viewed as a $\ce{H2}$ molecule binding sideways to the central atom. See the methanium link at the start.

$$
\begin{array}{lllll}
\text{Species} & \ce{NH3}  & \ce{NH4+}    & \ce{NH5^2+}     & \ce{NH6^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 896 & -410 & N/A   & N/A    \\
\end{array}
$$

Notes:

Even though the first protonation is easier relative to $\ce{CH4}$, the second one is harder. This is likely because increasing the electronegativity of the central atom makes the electron cloud "stiffer", and less accommodating to all those extra protons.
The $\ce{NH5^{2+}}$ ion, unlike other ions listed here with more than four hydrogens, appears to be a true hypercoordinate species.  Del Bene et al. indicate a five-coordinate square pyramidal structure with delocalized nitrogen-hydrogen bonds.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2O}  & \ce{H3O+}    & \ce{H4O^2+}     & \ce{H5O^3+}      \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 722 & -236 & N/A   & N/A    \\
\end{array}
$$

Notes:

The first series which does not accommodate proton hypercoordination.
$\ce{H3O+}$ is easier to protonate than $\ce{NH4+}$, even though oxygen is more electronegative. This is because the $\ce{H4O^2+}$ nicely accommodates all protons, while one of the protons in  $\ce{NH5^2+}$ has to fight for its space.

$$
\begin{array}{lllll}
\text{Species} & \ce{HF}  & \ce{H2F+}    & \ce{H3F^2+}     & \ce{H4F^3+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 501 & -459 & N/A   & N/A    \\
\end{array}
$$

Notes:

Even though $\ce{H3F^2+}$ still formally has a lone pair, its electron cloud is now so stiff that it cannot reach out to another proton even at normal bonding distance.

$$
\begin{array}{lllll}
\text{Species} & \ce{Ne}  & \ce{NeH+}    & \ce{NeH2^2+}    \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{No}   \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 204 & N/A   & N/A    \\
\end{array}
$$

Notes:

$\ce{Ne}$ is a notoriously unreactive noble gas, but it too will react exothermically with a bare proton in the gas phase.
Depending on the definition of electronegativity used, it is possible to determine an electronegativity for $\ce{Ne}$, which turns out to be even higher than $\ce{F}$. Accordingly, its electron cloud is even stiffer.

$$
\begin{array}{lllll}
\text{Species} & \ce{H2S}  & \ce{H3S+}    & \ce{H4S^2+}     & \ce{H5S^3+}     & \ce{H6S^4+}     \\
    \text{Stable in gas phase?} & \text{Yes} &  \text{Yes}   & \text{Yes} & \text{Yes}  & \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  & 752 & -121 & -1080   & N/A   & N/A  \\
\end{array}
$$

Notes:

The lower electronegativity and larger size of $\ce{S}$ means its electrons can reach out further and accommodate protons at a larger distance, while reducing repulsions between the nuclei. Thus, in the gas phase, $\ce{H2S}$ is a stronger base than $\ce{H2O}$. The situation is inverted in aqueous solution due to uniquely strong intermolecular interactions (hydrogen bonding) which are much more important for $\ce{H2O}$.
$\ce{H3S+}$ also has an endothermic proton affinity, but it is lower than for $\ce{H3O+}$, and therefore $\ce{H4S^2+}$ is easier to make. Accordingly, $\ce{H4S^2+}$ has been detected in milder (though still superacidic!) conditions than $\ce{H4O^2+}$.
The larger size and lower electronegativity of $\ce{S}$ once again are shown to be important; the hypercoodinate $\ce{H5S^3+}$ appears to exist, while the oxygen analogue doesn't.


What is the product of the reaction of diol with tosyl chloride followed by lithium aluminum hydride? This question is about the structure of products X and Y.
According to me, the answer should be only A (as OTs will substitute on benzylic carbon).

But in the answer key, they have given B and D.

Please help.

There are two replaceable $\ce{H}$s in the starting material, but only 1 equiv of $\ce{TsCl}$ is given. Thus, primary $\ce{-OH}$ will win the battle due to steric reasons, and therefore, B represent the intermediate product X. The second reaction is firstly acid-base reaction between secondary $\ce{-OH}$ and $\ce{H-}$, and then, nucleophilic substitution ($\mathrm{S_N}$2) reaction to replace $\ce{-OTs}$ group by $\ce{H-}$. That leaves D as the product Y, after acid workup.


What is the difference between an assay and an extemporaneous preparation? What is the difference between an assay and an extemporaneous preparation?

Can an extemporaneous preparation be labelled as the BP mixture?

An assay is an investigative (analytic) procedure for qualitatively
  assessing or quantitatively measuring the presence, amount, or
  functional activity of a target entity (the analyte).

E.g percentage purity of a given substance, viscosity, bulk density, total organic content etc.

The analyte can be a drug, a biochemical substance, or a cell in an organism or organic sample.

Extemporaneous preparation refers to the preparation (mixing, assembling, packaging and labelling) of a medicinal product by a pharmacist (or licensed practitioner) according to prescription requirements for a specific patient . Extemporaneous preparations  are made in  relatively small quantities  for short-term supply and use.

For example: Magesium trisilicate mixture can be extemporaneously prepared from the following:

Magnesium trisilicate
Light magnesium carbonate
Sodium Carbonate
Concentrated Peppermint emulsion
Double Strength Chloroform water

(Magnesium trisilicate mixture is an oral suspension containing 5% w/v of these excepients in suitable vehicle.)

Guidelines of extemporaneous preparations can be found in the Pharmacopoeia.

Where you want to determine the content of say one of the excepients , is assaying. For example you may wish to experimentally find the actual content of magnesium in the mixture, you may perform titration with a suitable standard solution, then you can actually be able to calculate its percentage weight in volume


Isotopes and their percentage existence in atmosphere When we talk about percentages of isotopes of an element in the atmosphere, is it percent by mass or volume or weight or something else?

Short answer: it's the percent by number of particles.

Longer answer: It's the proportion of particles of a particular mass for that element.  To chemists, a per particle basis is also commonly referred to as a per mol basis or a molar basis.

A way that it can be measured is by mass spectrometry - you pass the atoms through the apparatus which shows the mass to charge ratio. E.g. if you could pass $\ce{Cl}$ molecules, you'd see 2 peaks in the resulting graph - one is at 35 and the other at 37 marks which shows the mass/charge. First peak would be about 3 times larger (thus there were 3 times more particles). Thus chlorine would be about 75% isotope 35 and 25% isotope 37. So while this approach measures mass of each isotope, it also determines the relative molar amounts of each isotope.


Calculate the valence of Fe2O3 $$\ce{FeO}$$

Is called "Iron(II) oxide ".

$$\ce{Fe_2O_3}$$

Is called "Iron(III) oxide ".

The number in the parenthesis is "Valence" of the substance. Frankly, I don't know how to calculate such number. Can you explain to me, with this example, how come $\ce{FeO}$ has valence $2$ and the other one has valence $3$?

The only thing I know how to do is calculate the electron configurations of each element.

For Iron, it is $1s^22s^22p^63s^23p^64s^23d^6$.

For Oxygen, it is $1s^22s^22p^4$.

I am told that based on the electron configuration you can get the valence of each element. And then, you can calculate the valence of $\ce{FeO}$ and $\ce{Fe_2O_3}$. But how?

I made a question before Calculating valence of oxides, but it was wrongly formulated (they corrected me on the definition of Valence, but turns out what I actually need was how to calculate it), so I'm flagging it to be closed.

Edit The question How to get the valencies of elements? doesn't explain how to calculate the valence of a combination of two elements.

@Voldemort you learn the valencies rather than calculate them.  Valency is a fact of experiment whilst electron configuration gives an underlying explanation for the observations from experiment.

Iron has two different valencies (and oxidation states) which are a consequence of the stability of the electron configuration that remains when electrons are lost upon reaction.  So, iron can loose 2 electrons to leave a stable $\ce{Fe++}$ species; in your reaction those 2 electrons are gained by oxygen to give an anion that has 8 electrons in its outer shell which is a stable configuration.

Iron can also loose 3 electrons to leave a stable system and again, in your system, the electrons are picked up by the appropriate number of oxygen atoms to give oxygen anions that have 8 electrons in their outer shell.

If you look at the either side of the periodic table, say sodium reacting with a halogen, say chlorine (to give sodium chloride); you can see the 8 electron rule working.  But you're looking at a system where the iron is found in the middle of the periodic table and is a lot more difficult to predict when compared to the 8 electron rule.


Can an emulsion be emulsified? Can clusters of an emulsion be emulsified into another emulsion?
If so is there any common product or natural substance that would have such a structure?

Finally is there a name for this? I looked around but the question gets drowned in all emulsion topics.

Yes, they're called multiple emulsions (and more specifically: double, triple emulsions, etc.). Usually they're made from alternating polarity phases—water inside oil inside water, for example. There are bulk ways of making them though there isn't great control over how many droplets end up in each encompassing droplet without using microfluidic techniques. David Weitz's lab at Harvard has some interesting work in building microfluidic devices to make multiple emulsions.

I don't know if there are any common natural occurrences but the pharmaceutical and cosmetics industries seem to be the the main parties interested in this stuff as they can be used for encapsulating drugs for time release and the like.


Checking if it is a redox reaction Ethanoic acid dissociates according to the following equation:
$$\ce{CH3COOH + H2 O -> CH3COO- + H3O+}$$
I want to know if it is a redox reaction or not. In the ethanoic acid I obtained that the oxidation number of carbon in it is 0 by calculation, but it is (-2). What is the error in my answer?

You're getting 0 ox. state for carbon because of treating both carbons equally - 0 is mean average value from oxidation numbers of both carbons. The one in carboxylic group  has three bonds with more electronegative oxygen therefore has +3. The one in methyl group has -3 because it's connected with three hydrogens which have lower electronegativity.

See https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure - there's example with acetic acid!

In organic chemistry trying to get average value of ox. state usually doesn't make much sense as it's rare to have all carbons equivalent. @Shafter gave you good advice - without knowing the structure you were making mistake.


Can HBr add to a non-sp2 hybridized carbon? Does this reaction make sense? Mechanistically speaking, it does not for me. Is this a special case, or just a typo in the book?

There is a carbocation rearrangement. Basically, the H can produce two secondary carbocations. One of those secondary carbocations can do a 1,2-hydride shift, which produces a tertiary carbocation. This carbocation, being tertiary, will react much faster, resulting in being in the major product. So, yes, HBr can add to a non-sp2 carbon.


How to find density of element metals I was wondering, how do I determine what metal (element) has the highest density by using the periodic table? Is it possible?

One way you can do this is by looking at the packing structure of the metal.

As an example, if you look at Wikipedia, you see that Tungsten has a body-centered cubic crystal structure. This means that in each unit cell, there are going to be two Tungsten atoms. We can then predict the density of a perfect Tungsten crystal lattice using some geometry and unit conversion.

First off, I'll give you an equation which you can prove to yourself quite easily so I won't go into that. Density of a crystal is: $$\rho=\frac{n*M}{N_A*V}$$

Where, $n$ is the number of atoms in the unit cell, $M$ is the molar mass of the atom, $N_A$ is Avogadro's number, $V$ is the volume of the unit cell.

So, for Tungsten this comes out to be $$\rho=\frac{2*183.83 g*mol^{-1}}{6.022*10^{23}*(\frac{4*139*10^{-10}cm}{\sqrt{3}})^3}=18.45\frac{g}{cm^3}$$

The experimental density of Tungsten is $19.33 \frac{g}{cm^3}$.

The answer is usually a little better than that, but still quite close.

The only information you need to do this calculation which isn't on a periodic table is the packing structure and the atomic radius.

Something which is noteworthy is the atomic-packing-factor, $APF$, which comes from finding the ratio of volume of the atoms to volume of the unit cell and represents how much space the atoms fill in the cube, or how efficient the structure is at packing.

For the body-centered cubic (BCC), $$APF=\frac{2*\frac{4}{3}\pi r^3}{(\frac{4r}{\sqrt{3}})^3}=0.68$$

That means BCC, takes op 68% of the total available space per unit cell for equal sized spheres.

Check out this link if you want more info on that.

So, to answer the actual question which how do we find a trend with all this, we now know that density depends on radius, which we already have a trend for, molar mass, which also has a very simple trend, and packing structure, which is the real unknown.

There is this from this page,

  In the resonating valence bond theory, the factors that determine the choice of one from among alternative crystal structures of a metal or intermetallic compound revolve around the energy of resonance of bonds among interatomic positions. It is clear that some modes of resonance would make larger contributions (be more mechanically stable than others), and that in particular a simple ratio of number of bonds to number of positions would be exceptional. The resulting principle is that a special stability is associated with the simplest ratios or "bond numbers": 1/2, 1/3, 2/3, 1/4, 3/4, etc. The choice of structure and the value of the axial ratio (which determines the relative bond lengths) are thus a result of the effort of an atom to use its valency in the formation of stable bonds with simple fractional bond numbers.
  which I don't actually understand but seems to explain why certain lattices are chosen.

Basically, using the fact that radius decreases going right and molecular weight increases going right, we would predict that density would increase uniformly across the periodic table for elemental metals, except that various metals pack in different ways. Hexagonal Close Packed is the most efficient packing system, so I wouldn't be surprised to find that being associated with many high density metals.

I hope that gives a good idea of how there is sort of a trend, but also why no trend is truly there.

EDIT:

To figure out which has the highest density, I would start by figuring out which pack in a Hexagonal Close-Packed structure as that is the most efficient packing structure with an $APF$=.74


Bracketing Standards and dilutions You are running an ICP-MS (Inductively coupled plasma mass spectrometry), with bracketing standards of 10 ppm(mg/L) K+ and 100 ppm K+ and sample of 500 ppm, prepared 5 mL diluted to 100 mL. Shouldn’t you adjust your upper limit bracketing standard, instead of just diluting your sample? Your original sample concentration of K+ (500 ppm) should not change adjusted for the dilution factor. What am I not understanding mathematically?

As Ed said you are following the correct protocol by diluting the sample. The key reason is for doing so is to avoid a non-linear region in the calibration curve. This is why one would dilute the sample rather than make more concentrated standards. Several interesting effects happen with alkali metals at high concentrations in the flame or plasma. One of them is the concept of self-absorption.

Imagine you are monitoring the resonance line of K emission. Once you introduce lot of potassium atoms in the plasma (by using a concentrated standard) what may happen is that the ground state K atoms may absorb the light emitted by the excited potassium atoms? What is the result? You get less intensity than expected.

Try watching a sodium street lamp with a diffraction grating, or take a CD outside at night. You will see a prominent dark line in the emission spectrum of the lamp like this at 589 nm. This is the same self absorption phenomenon. There are so many Na atoms in the ground state that that they are self absorbing the light emitted by excited sodium atoms in the electrical discharge.


Gibbs free energy-minimum or zero? A reaction proceeds towards the direction of lesser Gibbs free energy (at constant $T$ (temperature) and $P$ (pressure)). So, we could say that Gibbs free energy at equilibrium is minimum.

On the other hand, we have $$\Delta G=\Delta G^\circ + RT\ln Q$$, where $Q$ is the reaction quotient. At equilibrium, $Q=K_\text{eq}$, and we already know that $\Delta G^\circ =-RT\ln K_\text{eq}$.

Substituting, we get $\Delta G=0$ at equilibrium. But, we know that $G$ minimized itself—thus there was a change in $G$ and $\Delta G < 0$.

What am I missing here?

I think your question really arises from some confusion about what $\Delta G$ represents. In general, $\Delta X$ for a thermodynamic quantity $X$ is the change of $X$ along some process. You could make it clear by actually writing $\Delta G(\text{A}\rightarrow\text{B})$ where A and B are before and after states. (We'll note that, in the general case, $\Delta X$ depends on the path take from A to B, making this notation improper. If $X$ is a function of state, though, you're good to go.)
However, in the equation you quote:
$$\Delta G = \Delta G^0 + RT \ln Q$$
the $\Delta G$ is a free energy of reaction and should thus be denoted $\Delta_\mathrm r G$, with the correct equation being:
$$\Delta_\mathrm r G = \Delta_\mathrm r G^0 + RT \ln Q$$
The free energy of reaction is defined as $\Delta_\mathrm r G = G_{\text{products}} - G_{\text{reactants}}$.

Thus, this $\Delta_\mathrm r G$ is not the variation of $G$ over the entire reaction, which would be the $\Delta G$ of the system between the start of the reaction and the equilibrium.

PS: I think this link is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!


LCAO (Linear Combination of Atomic Orbitals) and Phases So when combining atomic orbitals to form molecular orbitals, you can either add the wave functions or subtract them. But at the same time, orbitals can exist in opposite phases (say one lobe of the p orbital is '+' and the other '-'). So what happens when you subtract two 1s orbitals of opposite phase, is that the same thing as constructively interference of the wave functions?

The concept of linear combination of atomic orbitals (LCAO) to form molecular orbitals (MO) is probably best understood, while digging a little deeper into quantum chemistry. The method is an approximation that was introduced for ab initio methods like Hartree Fock. I don not want to go into too much detail, but there a some points that need to be considered before understanding what LCAO actually does.

The time independent Schrödinger equation can was postulated as $$\mathbf{H}\Psi=E\Psi,$$
 with the Hamilton operator $\mathbf{H}$, the wave function $\Psi$ and the corresponding energy eigenvalue(s) $E$. I will just note that we are working in the framework of the Born-Oppenheimer approximation and refer to many textbooks for more details.

There is a set of rules, the wave function has to obey.

It is a scalar, that can be real or complex, but the product of itself with its complex conjugated version is always positive and real. $$0\leq \Psi^*\Psi=|\Psi|^2$$
The probability of finding all $N$ electrons in all space $\mathbb{V}$ is one, hence to function is normed. $$N=\int_\mathbb{V} |\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)|^2 \mathrm{d}(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N)$$
The value of the wave function has to vanish at infinity. $$0 = \lim_{\mathbf{x}\to\infty}|\Psi(\mathbf{x})|$$
The wave function has to be continuous and continuously differentiable, due to the second order differential operator for the kinetic energy $\mathbf{T}_c$ included in $\mathbf{H}$.
The Pauli Principle has to be obeyed. $$\Psi(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_N) = -\Psi(\mathbf{x}_2,\mathbf{x}_1,\dots,\mathbf{x}_N)$$

Also the variational principle should hold for this approximation, stating that the expectation value for the energy of any trial wave function is larger that the energy eigenvalue of the true ground state. One of the most basic methods to approximately solve this problem is Hartree Fock. In it the trial wave function $\Phi$ is set up as a Slater determinant.
$$
\Phi(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_N) =
\frac{1}{\sqrt{N!}}
\left|
   \begin{matrix}
      \phi_1(\mathbf{x}_1) & \phi_2(\mathbf{x}_1) & \cdots & \phi_N(\mathbf{x}_1) \\
      \phi_1(\mathbf{x}_2) & \phi_2(\mathbf{x}_2) & \cdots & \phi_N(\mathbf{x}_2) \\
      \vdots               & \vdots               & \ddots & \vdots \\
      \phi_1(\mathbf{x}_N) & \phi_2(\mathbf{x}_N) & \cdots & \phi_N(\mathbf{x}_N)
   \end{matrix} \right|
$$

The expectation value of the energy in the Hartree Fock formalism is set up, using bra-ket notation as $$E =\langle\Phi|\mathbf{H}|\Phi\rangle.$$
Skipping through some major parts of the deviation of HF, well end up at an expression for the energy.
$$E=\sum_i^N \langle\phi_i|\mathbf{H}^c|\phi_i\rangle +\frac12 \sum_i^N\sum_j^N \langle\phi_i|\mathbf{J}_j-\mathbf{K}_j|\phi_i\rangle$$

To find the best one-electron wave functions $\phi_i$ we introduce Lagrange multiplicators $\lambda$ minimising the energy with respect to our chosen conditions. These conditions include that the molecular orbitals are ortho normal. $$\langle\phi_i|\phi_j\rangle =\delta_{ij} =\left\{
\begin{matrix}  0 & , \text{for}~i \neq j\\ 1 & , \text{for}~i = j\\\end{matrix}\right.$$
I will again skip through most of it and just show you the end expression.
$$\sum_j\lambda_{ij}|\phi_j\rangle = \mathbf{F}_i|\phi_i\rangle$$
with the Fock operator set up as
$$\mathbf{F}_i = \mathbf{H}^c +\sum_j (\mathbf{J}_j − \mathbf{K}_j)$$
with $i\in1\cdots{}N$, the total number of electrons.
We can transform these trial wavefunctions $\phi_i$ to canonical orbitals $\phi_i'$ (molecular orbitals) and obtain the pseudo eigenwertproblem $$\varepsilon_i\phi_i'=\mathbf{F}_i\phi_i'.\tag{Fock}$$
This equation is actually only well defined for occupied orbitals and these are the orbitals that give the lowest energy. In practice this formalism can be extended to include virtual (unoccupied) molecular orbitals as well.

Until now we did not use any atomic orbitals at all. This is the next step to find an approximation to actually solve these still pretty complicated systems.
LCAO a superposition method. In this approach we map a finite set of $k$ atomic (spin) orbitals $\chi_a$ onto another finite set of $l$ molecular (spin) orbitals $\phi_i'$. They are related towards each other via the expression
\begin{align}
\phi_i'(\mathbf{x}) &= c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) + \cdots + c_{k,i}\chi_k(\mathbf{x})\\
\phi_i'(\mathbf{x}) &= \sum_a^k c_{a,i}\chi_a(\mathbf{x})\\\tag{LCAO}
\end{align}

From $\text{(Fock)}$ you can see, that there will be $l$ possible equations depending on the chosen set of orbitals, in the form of
$$\sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\mathbf{F}_i|\chi_b\rangle = \varepsilon_i \sum_a^k\sum_b^k c_{ia}^* c_{ib} \langle\chi_a|\chi_b\rangle$$
or for short there are $l$ equations $$C_{ab}F_{ab}=\varepsilon_i C_{ab}S_{ab}.$$ Since $a,b \in [1,k]$ you can gather $C_{ab}$ in a $k\times k$ matrix $\mathbb{C}$. Since $i\in[1,l]$, there will only be $l$ $\varepsilon_i$ and therefore the matrix of the Fock elements $F_{ab}$ has to be a $l\times l$ matrix $\mathbb{F}$. The whole problem reduces a matrix equation $$\mathbb{F}\mathbb{C}=\mathbb{S}\mathbb{C}\epsilon\!\!\varepsilon,\tag{work}$$
from which it is obvious, that the dimension of the involved matrices have to be the same. Hence $k=l$.

Too long, didn't read

The total number of elements in the finite set of atomic orbitals is equal to the total number of elements in the finite set of molecular orbitals. Any linear combination is possible, but only the orbitals that minimise the energy will be occupied.

The coefficients $c$ from $\text{(LCAO)}$ will be chosen to minimise the energy of the wave function. This will always be constructive interference. This is also independent of the "original" phase of the orbital that is combined with another orbital. In other words, from $\text{(LCAO)}$
$$\phi_i'(\mathbf{x}) = c_{1,i}\chi_1(\mathbf{x}) + c_{2,i}\chi_2(\mathbf{x}) \equiv c_{1,i}\chi_1(\mathbf{x}) - c_{2,i}[-\chi_2(\mathbf{x})].$$


Do molecules with a hexagonal planar geometry exist? I recently read an article on Wikipedia about pentagonal planar geometry.

There are only two molecules with this kind of geometry $\ce{XeF5-}$ and  $\ce{IF5^2-}$  and it has two electron pairs at the axial positions.

As soon as I looked at this, I thought that hexagonal molecular geometry can also exist much the same way i.e. 6 substituents bonded to the central atom and two lone pairs at the axial positions opposite to each other.

On the page, its given that arrangement of 6 atoms in a plane is found in higher coordination states but that does not help either as they have not given any examples of that.

Do hexagonal planar molecules exist, and if they do, what are they?

I think it's nearly impossible to find or synthesize a "canonical" complex with hexagonal molecular geometry, but in the field of host-guest supramolecular chemistry there are numerous examples of "unusual" geometry. Probably the most well-established class of such compounds are torands ("hosts") incorporating alkali metal cations ("guests").

Check out, for example: (1) Bell, T. W.; Cragg, P. J.; Drew, M. G. B.; Firestone, A.; Kwok, D.-I. A. Angew. Chem. Int. Ed. Engl. 1992, 31 (3), 345–347, DOI 10.1002/anie.199203451.

Here is an example of the structure with potassium ion from (1), Tri-n-butyltorand-potassium picrate clathrate, which I quickly sketched in Olex2:

Top view:

Side view:

Unit cell and packing:


Hanging a spoon on the face Well, I have seen many comical acts where people balance a spoon (or a fork) on different parts of the face like on the tip of the nose, forehead etc.
The do this by blowing moist air (see below) and immediately putting it on the face

So i wanted to know is what is the force that temporarily binds the skin to the fork. I doubt it's friction or Van Der Walls force.
Can someone please clarify this for me? Is it some sort of chemical bond?

Some addition
Also another method of doing this was to rub the concave part of the spoon on the tip of the nose. This article states how to do it properly. In the 2nd point it states

  Eventually you will feel a slight tendency for the spoon to resist on the downstroke.

What is this resisting tendency?

I would say the answer on this website is pretty near perfect.

To summarize, you need 2 things:

Enough friction between your nose and the spoon
The center of mass of the spoon should be on a line vertically downwards THROUGH the tip of your nose as illustrated in this image from the same website:

The friction is highest if both your skin and the spoon are free of greasy `lubricants'. Breathing on the spoon will add more friction by depositing a very thin (only a few atomic layers) film of water which, on this scale, behaves solid-like due to ordering effects as shown in this paper.


Determine from pKa if acid or base I'm studying medicine, not chemistry, but I hope you can help me anyway.

I just had to solve a question where it is given that the drug Propranolol has a $\text{p}K_{\text{a}}$ of 9.5 and the un-ionized form is non-polar.

Is there any way to say if propranolol is an acid or base just from the $\text{p}K_{\text{a}}$?

Hope to hear from you, and sorry if it is a stupid question.

Based on the structure, this drug is weakly basic. It has an amine moiety, which if protonated, would likely have a $\mathrm{p}K_{\mathrm{a}}$ value of around 9.5.


Why does elemental Iodine form a brown solution with Ethanol but not with Isopropyl alcohol? Although it is not the standard procedure for making iodine solution, I have dissolves pure Iodine granules in Ethanol to produce a brown solution which will produce a positive test for starch by creating a purple color.

I tried to create the same solution with isopropyl alcohol but it doesn't work.  The Iodine does not dissolve noticeably, and whatever small amount may be dissolved does not complex with starch.

How does Ethanol dissolve elemental Iodine so well especially compared to Isopropyl alcohol?

While iodine is readily soluble in ethanol and isopropanol, it is only soluble to about 300 ppm in water. I think it's pretty clear that the 30% water in your isopropanol solution was plenty to thoroughly inhibit the iodine's solubility.


Convert pg/mL to/from pmol/L? In this article, 148 pmol/L is said to be equivalent to 200 pg/mL.

But this unit conversion site says that 200 pg/mL are equivalent to ~734 pmol/L, and this one gives 44 pmol/L for the same value.

What am I missing here? What is the correct way to perform this conversion, and why are different sites giving back different values?

Your question is how to do the conversion of the units of mole/liter to gram/liter.
Mole/liter, written as mol/L, is an amount (in units of mole) per liter.

To do this conversion, we need to know the convert mole to gram or vice versa.

To this end, we need to use the physical property of molar mass in units of gram/mole. This is the mass per mole of substance, in your case the substance is Vitamin B-12.
According to Wikipedia, the molar mass of Vitamin B-12 is 1355.37 g/mol.

$$\frac{200\times 10^{-12} \frac{g}{mL}}{1355.37 \frac{g}{mol}}=0.146\times10^{-12}\frac{mol}{mL}=146\times10^{-12}\frac{mol}{L}=146\frac{pmol}{L}$$


Solubility of gold salt solution? I was wondering what salts of gold are soluble and which are insoluble. From what I understand, silver is soluble as silver nitrate but insoluble as most other common salts like chlorides and sulphates. How does this work with gold? I was also wondering if there are any rules for solubility of salts, or whether there is some explanation for the solubility that would help in guessing the solubility of a salt?

I started thinking about this because of a question I came across in which there was a mixture of gold, silver and copper to which hot concentrated nitric acid was added, and an undissolved residue of given mass remained. Then $\ce{HCl}$ was added and more precipitate was formed. The question asked you to find the percentage by mass of gold, silver and copper. Judging by the general solubility trend of these metals, I'm guessing the first precipitate was gold and the second silver, but I was also wondering if there are some solubility rules...

Old question, but I thought I'd add to it:

In your example, the first undissolved solid is indeed gold. It's not easily dissolved by $\ce{HNO3}$ alone, but requires a mixture of $\ce{HNO3}$ and $\ce{HCl}$ (aqua regia). $\ce{Ag}$ and $\ce{Cu}$ would dissolve by hot $\ce{HNO3}$. The precipitate formed by addition of $\ce{HCl}$ is indeed $\ce{AgCl}$, so you're on the right track.

A brief list from experience:
[At room temperature in aqueous solution]

$\bullet \ce{Au2S}$: insoluble (quite sure this is the precipitate formed in basic solution when sodium metabisulfite is added to $\ce{HAuCl4}$. Either that or $\ce{Au2S3}$).

$\bullet \ce{AuCl3}$: highly soluble (can be formed by dehydrating and evaporating $\ce{HCl}$ from $\ce{HAuCl4}$, which is the product of $\ce{Au}$ dissolved in aqua regia).

$\bullet \ce{KAu(CN)4}$: soluble

$\bullet \ce{KAu(CN)2}$: soluble

$\bullet \ce{AgCl}$: insoluble

$\bullet \ce{CuCl2}$: highly soluble

$\bullet \ce{CuCl}$: soluble, but harder to synthesize than $\ce{CuCl2}$

$\bullet \ce{(Au/Ag/Cu)(NO3)_x}$: soluble ($x = 1,2$)

I think the overall result here is that gold, silver, and copper salts generally follow one of the primary trends of solubility: that is, that $\ce{NO3-}$ salts are always soluble regardless of cation, and $\ce{Cl-}$ are usually soluble except in cases $\ce{Ag+}$, $\ce{Pb^{2+}}$ and $\ce{Hg+}$.


Effect of periodic acid on cyclohexane derivatives Periodic acid (per-iodic acid, $\ce{HIO4}$ ) cleaves vicinal diols(to carbonyls) via a cyclic intermediate.

Due to the cyclic intermediate, it needs a syn configuration for the $\ce{-OH}$ groups for the reaction.

Exactly which configurations are allowed on cyclohexane derivatives?

Let's take 1,2-dihydroxycyclohexane(boat conformer). If the two hydroxy groups are both on the axial positions (trans- ), then obviously the reaction will not take place.

What about when the two are cis- (one axial, one equatorial), and when they are both equatorial (trans- )?

(My original question had to do with diastereomers of cyclic glucose--aldohexoses--but I realised that the extra hydroxy groups make the reaction more complicated)

I think your problem with this reaction lies in the easily confused term syn and cis. Both appear to mean that two functional groups are oriented in the same direction. On a cyclic molecule, syn and cis are the same. However, cis is a stereochemical descriptor and syn is a conformational descriptor. Simplifying this reaction to an acyclic case may help. Consider the meso isomer of  2,3-butanediol shown.

This isomer is drawn in the anti conformation, which is probably the lowest energy for steric reasons. This isomer has a syn conformation, which looks like the following. Note that the two -OH groups are not both oriented forward with solid wedges. That drawing implies a high energy eclipsed conformation.

(source: nist.gov)
Now back to the reaction with HIO4. In order to form the cyclic intermediate, the -OH groups need to be close in space. On an acyclic molecule, this requires a syn conformation, which most acyclic molecules can achieve even if it is not the lowest energy conformation. In cyclic molecules, particularly cyclohexanes, the syn conformation might not have the -OH groups close together. For cis-1,2-cyclohexanediol, which is also syn respective to the "plane" of the ring, to move the two -OH groups into the correct arrangement, it would need to wiggle into the higher energy boat or twist-boat conformations. However, trans-1,2-cyclohexanediol, which is anti has the two -OH groups oriented close together. In a sense, the trans isomer has the two -OH groups oriented syn with respect to the C-C bond that they share. Thus, trans-1,2-cyclohexanediol should react more readily with HIO4 that cis-1,2-cyclohexanediol, because the lowest energy conformer of trans-1,2-cyclohexanediol has the close arrangement of two -OH groups. The cis isomer would have to change conformation, which increases the energy input required, increasing the activation energy and slowing the reaction rate.
The diaxial case (whether boat or chair), would definitely be unreactive.


Identify structure of compounds from their reactions and molecular formula Compound P($\ce{C7H6O}$) on refluxing with aqueous ethanolic $\ce{KCN}$ solution for 1 hour, forms Q($\ce{C14H12O2}$). Q on oxidation gives R($\ce{C14H10O2}$). R on refluxing with aqueous $\ce{NaOH}$ solution followed with acidification forms an $\alpha-$hydroxyacid S ($\ce{C14H12O3}$). What are the structures of the compounds?

My effort:
P has DBE (double bond equivalent) equal to 8, so 2 benzene rings (not possible with 7 carbons) or some cyclic or many double bonds might be necessary. Now some enolate reaction might be happening with $\ce{KCN}$, like  Aldol or Perkin or Knoevnagel or Stobbe or Claisen or Benzion etc. Oxidation doesn't affect the formula so it might be becoming an isomer (that too a functional one). After that a Cannizaro might be the thing.

You weren't far from resolution. The substrate is benzaldehyde, first reaction benzoin condensation, so Q is benzoin. It is later oxidated to benzil - R ($\ce{C14H10O2}$) (not ($\ce{H12}$) - in oxidation formula must change). Finally benzil undergoes benzilic acid rearrangement, so S is benzilic acid.

As for how one could get to the solution... Compounds having 6-8 carbons and highly unsaturated are usually benzene derivatives. In this case benzaldehyde is only common thing with this formula (it was unlikely to be for example tropone). First reaction must be dimerisation, next oxidation, and hydration.


Concentration of a species outside buffer region (polyprotic acid titration) I'm wondering if you titrate a polyprotic acid, say, $\ce{H3PO4}$, with a strong base, and you successfully pass the first equivalence point and are maybe at the $\mathrm{pH}=\mathrm{pK}_{a2}$ point now, how would you find the concentration of $\ce{H3PO4}$?

Is there none left? I don't have a specific homework question, I'm just wondering.

Attempt at a solution:
I thought maybe you would find the equilibrium concentration of $\ce{H2PO4-}$, find the $\mathrm{pH}$, and use the $\mathrm{pH}$ to find $\ce{OH-}$. Then, you could write the equilibrium, $$\ce{H2PO4- + H2O <=> OH- + H3PO4},$$ and without making an ICE table, substitute your equilibrium values into the $\mathrm K_a$ expression.

At the one side, you can always calculate the percentages of the acid or it's ions in solution:
$$\ce{\alpha(H_{n-m}A^{m-})} = \frac{x^{n-m} \prod_{j=1}^{m}k_{a,j}}{\sum_{i=0}^{n}x^{n-i} \prod_{j=1}^i k_{a,j}}$$
with $x = [\ce{H+}]$, $n>0$ as the number of protons and $0 \le m \le n$ as the number of dissociated protons.

But on the other hand there are some special points during titrations. One of those points did you choose, which is where $\mathrm{pH} = \mathrm{pK}_{a,i}$.

For $\ce{H3PO4}$ there exist three different acidity constants, which describe the three possible dissociation steps. For the first step, the reaction is as you described: $$\ce{H3PO4 + H2O \xrightleftharpoons{k_{a,1}} H2PO4- + H3O+}$$

A little work on it gives you the Henderson-Hasselbalch-equation, which usually is good approximation to buffer regions: $$\mathrm{pH} = \mathrm{pK}_a - \lg\left(\frac{[HA]}{[A^-]}\right)$$

If you enter the same values for pH and pKa you have to solve $$-\lg(x)=0$$ which is true for $x=1$. So you know that you have a 1:1 ratio at this point.

This means for phosphoric acid:
At $\mathrm{pH=pK}_{a1}$, you've got exactly $50~\%~\ce{H3PO4}$ and $50~\%~\ce{H2PO4-}$ (blue, yellow). The same procedure can be applied to the other dissociation steps. Therefor at $\mathrm{pH=pK}_{a2}$ you've got $50~\%~\ce{H2PO4-}$ and $50~\%~\ce{HPO4^2-}$ (yellow, green) and at the last possible $\mathrm{pH=pK}_{a3}$ there are $50~\%~\ce{HPO4^2-}$ and $50~\%~\ce{PO4^3-}$ (green, red).

This can be applied to most acid-base dissociation steps but you have to pay attention for close successive $\mathrm{pK}_a$ values, as if the gap gets too small, the 1:1 ratio doesn't hold.


Hypervalency and the octet rule I realize that the octet rule is more a suggestion than a rule, and that it applies mainly to non-transition metal compounds. Still, compounds that don't have an octet, like $\ce{BH3}$, tend to dimerize or disproportionate in order to achieve an octet.

In recent SE Chem posts discussing the bonding in phosphorous pentachloride and iodine heptafluoride, hypervalent bonding was used to describe the bonding in these molecules.

If we count the electrons around phosphorous in $\ce{PCl5}$ there are 6 in the three equatorial bonds (2x3) and 4 in the hypervalent axial bonds; a total of 10. However 2 of the electrons in the hypervalent bond are in a non-bonding orbital, leaving 8 bonding electrons, an octet.

If we count the electrons around iodine in $\ce{IF7}$ there are 4 in the two axial bonds (2x2) and 10 in the hypervalent equatorial bonds; a total of 14. However, 4 electrons are in a hypervalent non-bonding orbital and 2 more electrons can be viewed as being associated with a fluoride anion, leaving 8 bonding electrons, again, an octet.

Are these compounds considered to be exceptions to the octet "rule"? Do all compounds involving hypervalent bonding violate the octet "rule"? Or do these compounds comply with the octet "rule"?

Out there in the real world, university students and school pupils alike favour strict rules that are true as often as possible (or at least have clear, easily remembered exceptions), while their professors and teachers on the other hand try to tell them that these rules are merely to be considered ‘guidelines’ rather than actual rules (and attempt to welcome them aboard the Black Pearl).

And from this point on is where the question turns philosophical very quickly. One side will say

  Ah, so the rule is still valid in this surrounding; I just need to remember to do A and B when applying it!

while the other side will reply

  Ah, but this shows us that the so-called ‘rule’ can no longer be strictly applied because you need to consider data C and D that point towards the real truth.

The thing is, both sides are somehow correct while parts of their descriptions remain wrong. Maybe we can think of these as two mesomeric descriptions of the same principle?

As we all (should) know, all our concepts of describing bonding and electron location are mere approximations of the actual underlying truth. Hence, all the rules that were ever laid out — like a bond consists of two electrons between two atoms, oxidation states or the octet rule — are again just mere approximations. They cannot be correct nor can they be properly ‘valid’. Thus, what isn't properly true can't be fulfilled or invalidated.

I think that is the most theoretical answer to the question.

From a practical point of view, it mostly depends on how strict you postulated the octet rule, and how quickly you would consider it violated.

Traditionally, the octet rule wasn't valued highly for elements that weren't carbon, nitrogen, oxygen or fluorine. As such, anything that violated the octet rule in its strictest sense was considered a violation (and proof for its invalidity).

Nowadays, my observations imply that the octet rule is gaining support again, and some of the simpler cases that were hither considered violations are now something more like special cases or maybe just octet rule with additional mesomery. In the eyes of those supporting this side, these compounds comply with the octet rule.

And of course, there is the large grey space between the black and the white.

I for my part tend to favour rules like the octet rule very much — although I tend to use a not-as-standard defintion for it:

  s- or p-block elements will always favour surroundings where the number of valence electrons is 8 or less. For the two elements of the first period, this reduces itself to 2 or less.
  Mesomeric structures that allow an all-octet are to be favoured over those that do not.

Under this definition, which relies heavily on mesomery for cases such as $\ce{IF7}$ or $\ce{PCl5}$ while it allows for a single, unambiguous description of $\ce{SO4^{2-}}$, these cases comply to the octet rule and actually further prove its validity. Others will surely disagree.


Where can I find CAD blocks for lab glassware? I want to draw a sketch of my apparatus for my report. I am conversant with the AutoCAD drafting software. Where can I download CAD blocks for commonly used labware items (like round bottom flask, allihn condensor, drying tube, heating mantle, retort stand, etc.)?

ChemDraw has a couple of templates with all kinds of laboratory glasswere.
Here is a screenshot of some of the possibilities:


If aliens lived in a hydrogen (or any flammable gas) based atmosphere, would they perceive oxygen to be flammable? Energy, fuel, and oxygen are needed for a fire to burn. In a hydrogen chamber, if there was a pipe spewing small amounts of oxygen, could you light the oxygen on fire as it began to mix with the hydrogen?

Absolutely yes.

Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if you had lit a hydrogen stream in an oxygen-containing atmosphere:

$$\ce{2H2 + O2 -> 2H2O + energy}$$

The only thing we've done is reversed the sources of the $\ce{H2}$ and $\ce{O2}$, so the hydrogen-breathing aliens would perceive oxygen to be explosively flammable, just as we perceive hydrogen gas to be so.  The bottom line is that it is the two of the gases together that are flammable.


Show the acid nature of hydrogen sulfide We are studying the chalcogens at school. I should prove (show) the acid nature of $\ce{H2S}$ by writing two chemical interactions.

I am really sorry if the terms I am using aren't the right ones. It's the first time I've ever used this part of the language. Which reactions should I write, and why? I think the first one must be neutralization:

$$\ce{H2S + 2 NaOH -> Na2S + 2 H2O}$$

As orthocresol suggested, to illustrate "classic" acidity as it's understood on a school level you unquestionably want to start with Arrhenius definition, which attributes acidic properties to  any substance which increase concentration of hydronium $\ce{H3O+}$ ions in water.

That's said, the very first reaction you probably want to write is dissociation of $\ce{H2S}$ in water and demonstrating that it is a weak acid by showing the corresponding acid dissociation constants $K_\mathrm{a}$ (the larger the $\mathrm{p}K_\mathrm{a},$ the weaker the acid; data from [1, p. 5-87]):

$$
\begin{align}
\ce{H2S + H2O &<=> HS- + H3O+} &\quad \mathrm{p}K_\mathrm{a1} &= 7.05\\
\ce{HS- + H2O &<=> S^2- + H3O+} &\quad \mathrm{p}K_\mathrm{a2} &= 19
\end{align}
$$

Some reactions illustrating acidic properties of $\ce{H2S}$:

Reaction with sodium hydroxide

Note that the reaction

$$\ce{H2S(aq) + 2 NaOH(aq) -> Na2S(aq) + 2 H2O(l)}$$

is for complete neutralization (concentrated $\ce{NaOH}$ solution).
If the reactants are taken in 1:1 ratio (diluted $\ce{NaOH}$ solution), sodium hydrosulfide is formed:

$$\ce{H2S(aq) + NaOH(aq) -> NaHS(aq) + H2O(l)}$$

Reaction with sodium carbonate

Sodium hydrosulfide is also formed when saturated $\ce{H2S}$ solution reacts with $\ce{Na2CO3}$:

$$\ce{H2S(aq) + Na2CO3(aq) -> NaHS(aq) + NaHCO3(aq)}$$

Reaction with ammonia

Reaction with aqueous ammonia solution yields exclusively in ammonium hydrogen sulfide $\ce{NH4HS},$ even when concentrated ammonia solution is used:

$$\ce{H2S(aq) + NH4OH(aq) -> NH4HS(aq) + H2O(l)}$$

Ammonium sulfide $\ce{(NH4)2S}$ doesn't form as ammonium $\ce{NH4+}$ $(\mathrm{p}K_\mathrm{a} = 9.25)$ is a much stronger acid than hydrosulfide $\ce{HS-}$ by a factor of about $10^{19 - 9.25}\approx\pu{6E9}.$

So far the reactions between well-soluble compounds that were carried in aqueous solutions where Arrhenius definition works fine.
However, it cannot be used to explain acidity in non-aqueous media, where a Brønsted–Lowry definition of an acid comes in handy: the acid is a proton donor.
This can be illustrated by the reaction with liquid ammonia:

$$\ce{H2S(g) + 2 NH3(l) ->[\pu{-40 °C}] (NH4)2S(s)}$$

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J., Eds, CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data; 97th Edn.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016-2017. ISBN 978-1-4987-5429-3.


What is insoluble in drinking alcohol, but edible? What substances are edible for human consumption, but will not mix with alcohol? What is a substance that would block alcohol from mixing with its counterparts in a product?
For example: You make a cupcake and want an alcohol shot in that cupcake. What substance could the alcohol shot be contained in so it wouldn't disperse into the cupcake, but still be edible?

Ethanol does not dissolve chocolate, which consists mostly of butter and sugar. These mini alcohol filled chocolates are very common party candies. You could try inserting one of these into your cupcakes after you have baked them.

You could also try making a mold for a gelatin-based shot cup, as gelatin is also insoluble in ethanol. Gelatin is capable of absorbing alcohol, however, so I do not recommend making these too far in advance, lest they turn soggy.


Torsion angle symbol I'm reading up on molecular modelling and have come across some terms I should probably have learnt a long time ago but always scared me off a bit with the Greek letters. I guess I'm a big boy now so it's probably time to clarify...

The textbook uses the lowercase symbol omega ($\omega$) for torsion (a.k.a. dihedral) angle while Wikipedia gives the uppercase, $\Omega$, plus both upper and lower case phi ($\Phi$, $\phi$) and psi ($\Psi$, $\psi$) in its description which from reading I'm starting to understand designate particular geometrical cases from the article, though it looks like it might be a little difficult to remember at first.

My textbook also gives in the list of symbols and physical constants [lower case] tau, $\tau$, but I can't see any mention of this on the Wiki page.

Firstly I wanted to ask is the use of upper/lower case totally arbitrary, and secondly is tau just a non-specific way of expressing torsion angle compared to the other 3 (which is the only use I can imagine for it?) Or is this use totally confined to molecular modelling?

It is a habit to assign greek letters (most often lowercase) to angles. In case of "normal" angles, it's mostly $\alpha, \beta, \gamma$, whereas for dihedrals some higher letters, as you mentioned, $\phi, \psi, \omega$. As noted in comments, whatever you use, clearly state what you mean.

The difference between dihedral and torsional angle is more conceptual, then formal. In torsion you discuss rotation around a bond, dihedral is general angle of two planes, as mentioned in the $\ce{CH_4}$ example. The formal expression for both of them is the same, indeed.

To be noted, there are some well known widely used dihedral angles, where you should stick to the common notation. The most prominent example is the protein backbone arrangement, see Ramachandran plot. In this case you should not invent your own notation.


Difference between black and brown eumelanin / red and pink pheomelanin Wikipedia says the following:

  There are two types of eumelanin—brown eumelanin and black eumelanin—which chemically differ from each other in their pattern of polymeric bonds.

What are those differences in polymeric bonds and which bond produces what color? Are the bonding differences also the reason for the different colorations in pheomelanin?

First of all, melanines are mixtures of complex polymers, without highly ordered structures and precisely defined composition. It means that unlike e.g. polypropylene or PET, melanines are not composed of regularly repeating identical units (monomers).

In a great simplification, brown and black eumelanines are composed of units of DHI (5,6‐dihydroxy‐1H‐indole) and DHICA (5,6‐dihydroxy‐1H‐indole‐3‐carboxylic acid). The order of these units is random and the ratio of DHI to DHICA depends on the origin of the pigment. Darker eumelanines are enriched in DHICA, which form branched structures (each DHICA unit can be bonded to three other units). Light-brown eumelanines contain more DHI units. Pheomelanines are build of different units - alanylhydroxybenzothiazines.

Now to the color. The color of melanines results from the absorption of light by electrons of chromophores (i.e. fragments of a molecule that strongly absorb light at certain wavelengths). In eumelanines the chromophores are dihydroxyindole units, and in pheomelanines benzothiazine rings. The color of melanines does not result directly from how are these bonded with each other, but from how many chromophores are in a melanine molecule, how densely are they packed, and from a number of different factors (e.g. presence of metal ions, oxydation). Black eumelanines absorb more light than brown eumelanines, because they have more chromophores (they are more densly packed due to the branching of the polymer) and that is why they appear darker. And pheomelanines have different chromophores that absorb light of a bit different wavelength (blue region). Different bonding patterns affect the optical density of melanines, but not directly their color.


Calculate the volume of ammonium hydroxide needed to get desired pH I have an $29\%$ aqueous solution of ammonium hydroxide. How much do I need to add to $\pu{20 ml}$ of water to get $\mathrm{pH}$ $12.8$. Ammonium hydroxide $K_\mathrm{b}$ is $1.774\cdot 10^{-5}$.

What I tried:

$$K_\mathrm{b} = \frac{[\ce{OH-}][\ce{NH4+}]}{[\ce{NH4OH}]} \approx \frac{[\ce{OH-}]^2}{[\ce{NH4OH}]}$$

Could someone explain the rational for this approximation if it is correct?

Then

$$
\begin{align}
-\log{K_\mathrm{b}} &= -2\log{[\ce{OH-}]} + \log{[\ce{NH4OH}]} \\
\mathrm{p}K_\mathrm{b} &= 2\mathrm{pOH} + \log{[\ce{NH4OH}]} \\
[\ce{NH4OH}] &= 10^{\mathrm{p}K_\mathrm{b} - 2\mathrm{pOH}} \\
&= 10^{\mathrm{p}K_\mathrm{b}-2(\mathrm{p}K_\mathrm{w}-\mathrm{pH})} \\
\mathrm{p}K_\mathrm{b} &= -\log{(1.774\cdot 10^{-5})} = 4.75 \\
\mathrm{p}K_\mathrm{w} &= 14 \\
\mathrm{pH} &= 12.8
\end{align}
$$

Plugging in:

$$[\ce{NH4OH}] = 10^{4.75-2(14-12.8)} = 223.9$$

This is much too high and does not make sense.

Technically, you calculated the concentration correctly (math-wise). However, your initial hypothesis is wrong.

First, you cannot get ammonia hydroxide solution of $\mathrm{pH}$ $12.8$ at the room temperature to work with. At $\pu{20^\circ C}$ concentrated solution would be approx. $35\%$ with $\mathrm{pH} \approx 12.0$. The higher the temperature, the lower the solubility and the $\mathrm{pH}$ will be. Having $29\%$ stock ammonia solution, there is no way to reach $\mathrm{pH}$ $12.8$.

Second, when you work with borderline/concentrated solutions, you are no longer allowed to use concentrations when describing the equilibrium, you need to use activities instead. The formula you derived would give a reasonable result for the moderately concentrated solutions only. You can adapt it to establish the relationship between the mass fraction $\omega$ and $\mathrm{pH}$, knowing the density of the solution $\rho$ ($M$ is molar mass of ammonia)

$$\mathrm{pH} = 14 + 0.5\lg{\frac{\omega\rho K_\mathrm{b}}{M}}\, ,$$

but that's pretty much all to it.


Does nitrogen tetrafluoride act as an acid or a base in its reaction with water? $\ce{NF4+}$ ion reacts with water as:

$$\ce{NF4+ + H2O -> NF3 + H2F+ + {1\over 2}O2}$$

My question is, does $\ce{NF4+}$ function as an acid or base in this reaction, and why?

Neither. The reaction is NOT one of ion exchange; it is a redox reaction. Thus, $\ce{NF4+}$ works as electron acceptor here, like this:

$$\ce{NF4+ + 2e- -> NF3 + F-}$$

Since the reaction mechanism is unclear, it is possible that some of its stages may have acid-base character (for example, $\ce{F-}$ sometimes acts as a base), but $\ce{NF4+}$ certainly isn't acting primarily as base or acid here.


Could a 'polymer' be formed by hydrogen bonding? For example, given you had a molecule with a carboxyl functional group on either side, could it form a continuous chain of dimers? I understand this may not fit under the definition for polymer but I lack better terminology to describe such a thing. For Example:
Alternatively;

Since a polymer is a

"molecule of high relative molecular mass, the structure of which essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass." (IUPAC Gold Book)

joining two monomers into a dimer is just a specific wording for the more general term polymer like it is for trimer, tetramer, pentamer, hexamer, etc.  While the term polymers often is used about plastics, cellulose, starch, or chitin equally are polymers, too.  Note, this definition does constraint the binding of the monomers to covalent bonds.
The depictions shown as examples by you are valid representations when describing, e.g., the hydrogen-bond pattern in organic crystals.  Here, hydrogen bond donors and hydrogen bond acceptors may contribute significantly to the overall stabilization of the supramolecule an organic crystal may be considered.  Magaret C. Etter and Joel Bernstein popularized  formalisms to identify and systematically describe these pattern.
The one described by you matches for example the situation for $p$-terephthalic acid yielding (in principle) infinite bands hold by the motif $R^2_2(8)$ (a ring of eight atoms in total, comprising 2 donor, and 2 acceptors):

(1990AccChemRes120, includes a guide how to define these graph sets)
This approach equally may be extended to describe the interactions if the monomers are chemically different:

($R^2_4(8)$ motif about the interaction of phosphine / arsine oxides and  nitroanilines in co-crystals, reviewed in 1995ACIE1555)


What's the configuration of chromium(III) in aqueous medium according to crystal field theory? It has $\mathrm{3d^3}$ configuration, but how will we decide if they enter $\mathrm{t_{2g}}$ or $\mathrm{e_g}$ orbitals since we don't know which complex will form?

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call.

In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they are lower in energy. (Look up any introduction on crystal field theory or check out the answers on this site.) In a tetrahedral complex, $\mathrm{e}$ — not $\mathrm{e_g}$, as there is no centre of symmetry, meaning the the $\mathrm{u/g}$ designators do not make sense, even though the orbitals are still often called $\mathrm{e_g}$ at introductory level — are the orbitals of choice to begin with. Again, just look at the scheme.

Thankfully, however, we can make an educated guess based on various factors.

octahedral complexes are more common especially at introductory levels and for the ‘top middle’ of the d block elements (roughly from titanium to nickel)
the LFSE (ligand field stabilisation enthalpy) can be calculated to $\pu{-12Dq}$ in the octahedral case and $\pu{-8Dq}$ in the tetrahedral case which twice prefers octahedral
there is no compelling reason (size or charge balance) to consider tetrahedral to be preferable to octahedral.

Thus, the complex you are looking at will most likely be octahedral — and so it is: it is hexaaquachromium(III) $\ce{[Cr(H2O)6]^3+}$, which is octahedral.


Participation of Pi-bonds and lone pairs in resonance Below is a quote from Klein's Organic Chemistry 3rd edition:

  Here is the bottom line: Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

However, I am confused as a few pages prior it was shown how allylic lone pairs and pi-bonds participated in resonance:

  The first curved arrow goes from the lone pair to form a π
  bond, while the second curved arrow goes from the π bond to form a lone pair:

What have I understood wrongly?

You have probably misunderstood what the sentence actually means.

It means that only one, either $\pi$ bond or lone pair will participate in resonance if the atom has both.

For example, in pyridine, the nitrogen has lone pair as well as is attached with a $\pi$ bond.

While resonance, its $\pi$ bond participates in resonance because it (and not the lone pair) is in the plane perpendicular to the molecule.

Observe that the lone pair remains at the place it is meaning it doesn't participate.

Conclusion: The pair of electrons present in the orbital perpendicular to the plane of molecule participates in resonance if it is present.


Can an exothermic (endothermic) process increase (decrease) internal energy? We classify exothermic and endothermic processes based on enthalpy change, $\Delta H < 0$ and $\Delta H > 0$ respectively.

I assume this translates to $\Delta U < 0$ and $\Delta U > 0$, respectively, for most processes. (where $\Delta H = \Delta U + \Delta (PV)$, in general).

But not all? In which case we might have processes where $\Delta H < 0$ (exothermic) but $\Delta U > 0$? Could you give an example?

Otherwise exo and endo could just be classified based on $\Delta U$ alone instead of $\Delta U + \Delta (PV)$.

Background: I'm just trying to find a motivation for using the concept of enthalpy. I know about the motivation due to additional $PV$ work needed for  a given process (hence $U$ not enough to account for all the energy involved). Just wondering if the exo/endo classification is another motivation for using enthalpy?

Side note: We know something like this for $\Delta H$ v $\Delta G$, for classifying spontaneous v non-spontaneous processes. In that case $\Delta H$ is not enough and we need $\Delta G$. Just wondering if a similar thing happens in case of $\Delta U$ v $\Delta H$ for exo/endothermic classification.

A primary reason for using enthalpy is the fact that heat transferred ($q$) is a fairly straightforward quantity to measure, whereas change in internal energy ($\Delta U$) is often not.

Under conditions where volume is held constant and no form of work other than pressure-volume work is possible, $\Delta U = q$, so we can get direct information about the change in energy from the amount of heat transferred.

But under the much more common laboratory condition of constant pressure, $q \neq \Delta U$. For this case, it is useful to equate $q$ with the change in a state function, so we have defined enthalpy such that $\Delta H = q$ when pressure is constant. Because enthalpy is a state function, we can make general conclusions about other paths between states based on our measurements of $q$, even though $q$ is path-dependent.


How do I combine z-matrices of multiple molecules into one input stack? I'd like to run a calculation in Gaussian which requires me to provide cartesian coordinates for reactants and products respectively. My question is how do I combine the cartesian coordinates for two reactants (that I would obtain from a database)? For instance, CH3 + N?

Is there a specific program/procedure that I need to use?

To be more specific, say I would like to combine the z-matrix of $\ce{CH3}$

0 3
C
X 1 1.0
H 1 r 2 90.0
H 1 r 2 90.0 3 120.0
H 1 r 2 90.0 3 -120.0

r 1.0828

with a z-matrix of $\ce{N}$

0 3
N

And I would like to combine the z-matrix of $\ce{HCN}$

0 1
C
X 1 1.0
H 1 rh 2 90.0
N 1 rn 2 90.0 3 180.0

rh 1.0587
rn 1.1326

with that of $\ce{H2}$

0 1
H
H 1 r

r 0.741

In other words, I want to combine these z-matrices (or geometries) of the two reactant/ product species into a single coordinate, so that I can use them for my QST2 calculation. What do I do?

The Setup

The first thing you need to do is to get the difference between a Z-matrix (sometimes also called internal coordinates) and cartesian coordinates straight. They are truly different things.

Secondly, take the Z-matrices of a sufficiently large molecule. The basic structure is:

C1 .  .  .
C2 r2 .  .
C3 r3 a3 .
C4 r4 a4 d4
C5 r5 a5 d5
(and so on from this point)

where in this simplified notation, I have omitted the anchor centers for the coordinates. One notices that there are 6 coordinates left out, marked by dots. They can be omitted because three degrees of freedom for translation and three degrees of freedom for rotation of the entire molecule are irrelevant for the calculation of properties.

Now imagine a second molecule tacked onto the other one:

C1 .  .  .
C2 r2 .  .
C3 r3 a3 .
C4 r4 a4 d4
C5 r5 a5 d5
C6 *  *  *
C7 r7 *  *
C8 r8 a8 *
C9 r9 a9 d9

This Z-matrix is invalid. One needs to fill the missing coordinates marked by the asterisks. In a sense, the second molecule does not have the 6 degrees of freedom mentioned above, because any change in them would change the orientation relative to the first one, impacting all properties.

So, what can you do?

Filling in the missing coordinates is hard, at the very least by hand or some other automated fashion. Doing it in an editor is basically the same amount of work as the solution I actually suggest: Create the system of both molecules in one Z-matrix and split that Z-matrix. Remove the superfluous coordinates in the resulting matrices and run calculations as necessary.


Osmium tetroxide reaction with cell membrane Background: $\ce{OsO4}$ is a widely used staining agent used in transmission electron microscopy (TEM) to provide contrast to the image. In the staining of the plasma membrane, osmium(VIII) oxide binds phospholipid head regions, thus creating contrast with the neighbouring protoplasm (cytoplasm).The osmium tetroxide molecule is tetrahedral and therefore non-polar. This nonpolarity helps $\ce{OsO4}$ penetrate charged cell membranes.

Question:

What would be the product formed by reaction of $\ce{OsO4}$ with polar head of phospholipid of cell membrane?

My attempt , since $\ce{OsO4}$ is non-polar and phospholipid is polar, therefore it must be some sort of covalent interaction, I couldn't find any literature regarding further at wikipedia.

According to the paper The chemical nature of osmium tetroxide fixation and staining of membranes by x-ray photoelectron spectroscopy[1] by White et al. (1976),

  The results support a scheme for the reaction of osmium tetroxide with tissues in which the initial reaction site is the double bonds of unsaturated lipids to form $\ce{Os(VI)}$ derivatives. Subsequent hydrolysis and further reduction yield complexes of $\ce{Os(IV)}$ and $\ce{Os(III)}$. A mixture of these three states is present in membrane specimens during microscopic observation. $\ce{Os(VI)}$ and $\ce{Os(IV)}$ could be present as osmate esters and osmium dioxide, respectively; $\ce{Os(III)}$ could be present as an oxo- or amino complex(es).

However, since $\ce{OsO4}$ reacts with biological material, the above list is probably incomplete.

Riemersma[2] suggests that negatively charged colloidally dispersed osmium dioxide hydrate is deposited at the polar ends of lipids that contain a positively charged (phosphoryl)choline moiety (lecithin, phosphatidylcholines, etc). In this case we would have a pure Coulomb interaction.

References

White DL, Andrews SB, Faller JW, Barrnett RJ, Biochim Biophys Acta, 436(3):577-92 (1976).
Jelle C. Riemersma, J. Histochem. Cytochem, 11, 436-442 (1963)


Distinguishing propan-2-ol from ethanol using IR spectroscopy A question requires me to distinguish between propan-2-ol and ethanol from IR spectra. To me, they both look nearly identical, even though I know both of them contain the same functional group and that IR spectra are only capable of identifying functional groups.

My question is, how do you distinguish the two, without using a mass spectrometer?

Here is the spectrum for ethanol:

Here is the spectrum for propan-2-ol:

Here are the tabulated peak positions ($\pu{cm^{-1}}$) and percent transmittance (%T) for the two. I've tried to match the appropriate peaks.

$$
\begin{array}{cccc}%
\hline%
\text{ethanol position} & \text{ethanol %T} & \text{propan-2-ol position} & \text{propan-2-ol %T} \\
\hline%
3358 & 12 & 3346 & 9  \\
     &    & 3334 & 9  \\
2974 & 13 & 2972 & 5  \\
2927 & 38 & 2933 & 19 \\
     &    & 2907 & 23 \\
2887 & 35 & 2884 & 20 \\
     &    & 2722 & 60 \\
     &    & 2669 & 60 \\
     &    & 2521 & 74 \\
     &    & 2408 & 79 \\
     &    & 2387 & 81 \\
     &    & 2198 & 84 \\
     &    & 1903 & 85 \\
     &    & 1766 & 84 \\
1455 & 59 & 1467 & 24 \\
     &    & 1409 & 29 \\
1381 & 54 & 1379 & 16 \\
     &    & 1368 & 17 \\
1330 & 68 & 1341 & 28 \\
1274 & 74 & 1309 & 23 \\
     &    & 1162 & 10 \\
1090 & 22 & 1130 & 9  \\
1050 & 7  & 1110 & 15 \\
     &    & \color{red}{954}  & 4  \\
881  & 37 & 818  & 13 \\
669  & 67 & 660  & 37 \\
     &    & 654  & 37 \\
     &    & 490  & 66 \\
     &    & 436  & 68 \\
\hline%
\end{array}%
$$

One way of differentiating between the two is that I don't believe the peak at 954 wavenumbers in propan-2-ol appears in ethanol. Another better way that doesn't require looking at the fingerprint region is to see that the hydrogen-bonding peaks (the broadest ones to the left) are red-shifted in propan-2-ol compared to ethanol.


In Graham's law, is the rate of effusion only dependent on molar mass? According to Graham's Law, the effusion rate of a gas is inversely proportional to the square root of the molar mass of the gas.

However, consider this situation: We have s balloon with 1 mole of neon gas and another with 2 moles of neon gas. The gases are in thermal equilibrium and occupy the same volume. In this scenario, would the rate of effusion be the same?

The molar mass of the neon in both balloons is the same, so according to Graham's Law the rates would be the same. But intuitively, thinking about 2 moles versus 1, the pressure is doubled if volume and temperature are kept constant. And if the pressure is doubled, more collisions per second occur, so I would think that despite Graham's Law, the rate of effusion is also affected by the number of moles.

Molar mass is just one of the factors affecting effusion. Graham's law states that everything else remaining same, the rate of effusion of a gas is inversely proportional to $\sqrt M$. In other words, Graham's law describes the relation of the rate of effusion with only one of the factors involved in determining the overall rate.

For the general rate of effusion of a gas at pressure $P$, and the hole area $A$ at a temperature $T$, using kinetic molecular theory, it turns out to be: (derivation on Physics.SE)

$$r=\frac{PA}{\sqrt{2\pi MRT}}$$

Now, in your question, the balloon with two moles will have a greater pressure (more moles in the same volume) and therefore it will have a greater rate of effusion for the same hole area and temperature and molecular mass.


Ionic or covalent bond Which bond is stronger, ionic or covalent? I have a lot of confusion about this.

Covalent and ionic bonds are best thought of as opposite ends of a spectrum, where electrons are shared evenly (as in the case of a symmetric molecule like N2) or unevenly (like HF).

That being said, bonds with more ionic character are generally considered stronger due to the significant electostatic contribution to bonding (the cationic and anionic components experience attraction).

This principle is the core concept of Pauling Electronegativities, where the degree to which electrons in a bond are "attracted" to one atom over the other (that is, the degree to which a bond is ionic) is ultimately derived from experimental bond dissociation energies (i.e. bond strengths).


What is the importance of using a pKa value instead of a pKb value when describing drug chemistry? Why do we only consider the $\mathrm{p}K_\text{a}$ value of a drug regardless of whether it is an acid or base? Why do we not use the $\mathrm{p}K_\text{b}$ value?

For easier comparability:

Comparing two pKa values is much easier than comparing a pKa with a pKb value.


Which characteristics of the compound affect the number of theoretical plates in a GC experiment? Some days ago, I tried for the first time a chromatography machine.
The data I collected are the following:
$$
\begin{array}{lrrr}
\hline
\text{Compound}                 & t_\mathrm{R}/\pu{min} & w/\pu{min} & N     \\
\hline
\text{Benzene}                  & 2.325                 & 0.047      & 39153 \\
\text{Toluene}                  & 2.712                 & 0.067      & 26214 \\
\text{Chlorobenzene}            & 3.690                  & 0.099      & 22228 \\
\textit{p}\text{-Chlorotoluene} & 4.332                 & 0.106      & 26723 \\
\hline
\end{array}
$$
I know the relationship between retention time $t_\mathrm{R},$ peak width $w$ and theoretical plates $N.$
Known the different properties, boiling points, polarities, or shapes of the molecules involved, is there a way to predict which molecule will yield an higher number of theoretical plates?
Which characteristics of the molecules affect the different number of plates for each molecule?

The number of theoretical plates can be calculated as:

  $\mathrm{N_t = \frac {H}{HETP}}$

  $\mathrm{N_t =}$ number of theoretical plates
  $\mathrm{H =}$ total height (or usually length) of the column [$\mathrm{m}$]
  $\mathrm{HETP =}$ Height Equivalent to a Theoretical Plate [$\mathrm{m}$]

Where $\mathrm{HETP =}$ can be calculated from the Van Deemter equation:

  $\mathrm{HETP = A + \frac {B}{u} + (C_s + C_m) * u}$

  $\mathrm{A =}$ Eddy-diffusion parameter, related to channeling through a non-ideal packing [$\mathrm{m}$]
  $\mathrm{B =}$ diffusion coefficient of the eluting particles in the longitudinal direction, resulting in dispersion [$\mathrm{m^2 s^{−1}}$]
  $\mathrm{C =}$ Resistance to mass transfer coefficient of the analyte between mobile and stationary phase [$\mathrm{s}$]
  $\mathrm{u =}$ Linear Velocity [$\mathrm{m s^{−1}}$]

So, the characteristics specific to the analyte molecule that determine the number of theoretical plates are B and, unfortunately from the standpoint of developing an ab initio model, C.

Various models have been developed, however.  Below is a list of references to some articles of interest in this field.

[ 1 (abstract)]"The problem of plate modeling: Theoretical and computational results", Computer Methods in Applied Mechanics and Engineering, Volume 100, Issue 2, October 1992, Pages 249–273

[ 2 ]"Plate Models in Chromatography: Analysis and Implication for Scale Up", Chromatograpy,Springer-Verlag

[ 3 (abstract)]"Neural Network Modeling of Structured Packing Height Equivalent to a Theoretical Plate", Ind. Eng. Chem. Res., 2000, 39 (5), pp 1520–1525


Relationship between electric dipole moment and polarization? $$\vec{p}_\text{ind} = \alpha \vec{E}$$
The induced dipole moment is the polarizability times the electric field vector.
$$\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}(\omega)$$
The polarization is proportional to the susceptibility times the electric field vector.

In spectroscopy we used the dipole approximation. Unfortunately, I often mix up the different formulas. Is there a relationship between both formulas, or do both equations tell us exactly the same thing?

In fact, the first equation is applicable on a microscopic scale  (for an atom or a molecule):
  $\space$ $\vec{p}_\text{ind}=\alpha \vec{E}_\text{local} $

While the second one is applicable on a macroscopic scale (for the bulk):
$\space$ $\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}_\text{ext}(\omega) $
So, the polarisability is a microscopic quantity, while the susceptibility is a macroscopic quantity.

You have to notice that the electric field in the first equation is the local field, while it's the external field in the second one.
For more details, please see this page.