Ammonia Reaction with O2 I asked a similar question earlier that lead me to ask this question.
NH3 and O2 react in the following way $$\ce{4NH3 + 3O2 -> 2N2 + 6H2O}$$  From my understanding Ammonia will explode(react quickly/release a lot of heat quickly is how I understand the term explode) when there is 15%-25% ammonia vapor in Air.

If I injected ammonia vapor into the air (above autoignition temperature of 651°C so that it's fuel to air ratio was more like 30% ammonia to air.  Would it follow this reaction slowly until it reached the 25% mark then explode/react quickly until it reached the 15% ammonia balance air, and then react slowly from 15% ammonia to lower?

Here is the link to my similar question. Meaning of Flammability Limits and reaction in Air of Ammonia

No.  Because remember, the oxygen is being consumed too.  If we assume it's reacting via that reaction (realistically, you're going to get a complex mix of products) then we start with something like: 30% NH3 56% N2 14% O2 (the mix is 70% air, which is an 80/20 mix of N2 and O2).  For every 4% reduction in Ammonia, you're getting a 3% decrease in oxygen.  So O2/NH3 reactions will be rarer (slower), and slow down heat generation.  Also, the amount of inert gas in the vessel will be increasing, which also need to be heated up, so less of the enthalpy of reaction will go into increasing the temperature.

That said, above the autoignition temperature, it's going to be a pretty fast reaction anyway...  What we can be sure of is it will stop reacting at about 10% NH3 when all the oxygen is gone.


The Principle of Internal Standardization (in the context of ion chromatography) I am trying to grasp how internal standardization works in principle, and I am having some trouble wrapping my head around it. It is not the choosing of an IS I think about, orwhen it might benefit the analysis, but just the underlying principle.

I add a known amount of the IS to every sample: unknown samples, blank samples, calibration samples, and quality control samples. Each of these samples contain equal amounts of the internal standard (or as "equal" as our pipettes may give). Considering the calibration, when the calibration samples are analyzed, I see two peaks in the chromatogram: the analyte and the IS.

Let's say I ran calibration samples of 0, 1, 2, 5, 10, and 20 ppm, each one consiting 20 ppm internal standard. The analyte peaks will grow for each calibration sample, but the IS peaks remain more or less stationary. I construct the calibration curve by plotting the ratio of analyte peak area to IS peak area (or height), as a function of the analyte conentration in the calibration samples.

I have been told that there is a constant ratio involved, one that is used to correct analyte concentrations for loss of analyte, or varying sample introduction system. If the IS peak is lower, then the analyte concentration will be corrected upwards. This apparently is because this ratio I mentioned is constant, so when reducing the denominator, the nominator must increase. Mathematically, this is trivial; however, what is this mystical constant, and what is its value? Is that arbitrary? Can I choose this constant's value myself?

I am quite confused. Hopefully someone here can help my out!

This ratio has nothing mystical, and will not work in every case.

So let's first comment on the reasons to use an internal standard. As you mention, when you inject multiple times solutions containing the same concentration of IS:

  but the IS peaks remain more or less stationary

More or less is a major point here: this means that repeatability of the measurement process is an issue, when multiple samples with the same concentration lead to a variation in the response. If the accuracy required for the measurement is such that these small variations might not be an issue, there is no need for an IS. In chrmatography measurement, sources of variability include repeatability in the volume injected, detector sensitivity. If there is some sample preparation steps involved, loss of analyte (in transfer from one tube to the other, in the filtration steps etc...) can also lead to variations from one analysis to the other.

So how will an IS solve this variability issue? The basic point is the assulmption that these external variability issues will affect both the analyte and the IS in a similar way. So one has to look at the sources of errors. For instance, let's consider an error in the measurement of a sample volume (let's call it $\epsilon$ and assume it is close to 1) . We have:

$$m_{analyte} = V_{real} c_{analyte} = \epsilon V_{set} c_{analyte}$$
$$m_{IS} = V_{real} c_{IS} = \epsilon V_{set} c_{IS}$$

If the measurement system accurately measures $m_{analyte}$ and $m_{IS}$, it is quite obvious that by dividing both equations we can remove this variability parameter and we get:

$$R_{analyte/IS} = {m_{analyte} \over{m_{IS}}} = {c_{analyte} \over c_{IS}}$$

This is the point were the ratio comes out: taking a ratio allows to remove the variability parameter ($\epsilon$) in this case.

Now this is a bit too simplistic, and a measurement procedure is not as simple. The general concept will still be to try to factor out the variability parameter as a proportion of the measured value, and then to take a ratio which will remove this variability from the measured values.

If the IS is correctly chosen so that it behaves as the analyte through the entire analytical procedure (both in terms of physico-chemical properties, as well as in the response of the detector), and if the chromatographic properties as well as the detector are such that peak area is proportional to concentration for the analyte as for the IS, then one can write:

$$A_{analyte} = \epsilon(analyte) R_{method}(analyte) c_{analyte}$$
$$A_{IS} =  \epsilon(IS) R_{method}(IS) c_{IS}$$

where $A_{analyte}$ and $A_{IS}$ are the measured peak areas for both analyte ans IS, $R_{method}(x)$ is the method response factor, which is dependent on each compound. In general the variability factor $\epsilon$ should depend on each compound, but under the above assumption of similar properties, one can assume that $\epsilon(analyte) \approx \epsilon(IS) \approx \epsilon$. Without IS, one would determine $R_{method}(analyte)$ through a calibration procedure. With an IS, one takes the ratio and the above equation can be written as:

$$R_{area} = {A_{analyte} \over A_{IS}} = {R_{method}(analyte) \over R_{method}(IS)} {c_{analyte} \over c_{IS}}$$

The ratio:

$$R_{analyte/IS} = {R_{method}(analyte) \over R_{method}}$$

is the ratio of the response factors of the method for analyte and IS. This is probably the ratio that is referred to. And it is not mystical at all, it is obtained through a standard calibration procedure with the IS.

One caution word to conclude this answer: I have tried to be quite precise in the answer. I have often times heard some people state that the use of an, internal standard is a general solution for analytical issues. Thus I want to insist on a warning: the solution is good as long as the assumptions stated above hold. If any one of these fails, then the whole analytical procedure will also fail. And of course, the use of an IS will only protect from errors after addition of the IS, and not from variability in the addition of the IS (you mention pipetting it in the analyte solution, which will limit the repeatability to that achieved by the pipettes used).


Which two substances react spontaneously? Which two substances react spontaneously?
(A) $\ce{Ag}$ and $\ce{Cu}$
(B) $\ce{Ag+}$ and $\ce{Cu}$
(C) $\ce{Ag}$ and $\ce{Cu^2+}$
(D) $\ce{Ag+}$ and $\ce{Cu^2+}$
$\tiny{\text{Taken from 2012 Chem Olympiad}}$

The given answer is B, yet I am unclear as to why. I believe it has something to do with reduction potentials/redox or the gibbs free energy. Can someone please explain?

From the faint outline of electrochemical series I remember $Cu$ is above $Ag$. The substance which is above is able to displace the substance below from its aqueous solution(obviously lower substance is in ionic form).

Hence $(B)$ is true.

We were made to learn this outline in lower class :

K Na Ca Mg Al Zn Fe Pb H Cu Hg Ag Au Pt


Specific Heats for Non-Constant Processes Why can the specific heat at constant Pressure ($C_p$) be used to calculate the change in enthalpy as $\text{d}H = C_p\text{d}T\newcommand{\d}{\text{d}}$ for processes that are not even at constant pressure?
I have $\d H = C_p\d T + V\d p$. Why do we neglect the $V\d p$ term? For gases, $V$ the specific volume, can be very large, so an increase in pressure could have a non-negligible effect on the enthalpy. Is it so? Or not?

Cp of an ideal gas doesn't depend on pressure.  Cp of any substance is defined as the partial derivative of enthalpy respect to temperature at constant pressure.  But, if the enthalpy of the substance is independent of pressure, then it doesn't matter if the pressure is constant.  However, determining Cp by measuring the amount of heat required to change the temperature by a certain amount of the substance at constant pressure is still a standard method of measuring Cp.


How to breakdown wax, preferably beeswax I have a project that I'm working on at the moment, but I'm stuck on this detail. I'd like to find out what can I supplement beeswax with that is organic, non toxic, and safe if accidentally consumed, in order to speed up decomposition rate of beeswax to a time frame of about 1-6months. Any ideas?

As far as I know beeswax does not decompose. If you just want to melt it, use heat. If you want to break it down using a solvent, try turpentine (paint thinner).


Reverse Friedel-Crafts mechanism I need to find a mechanism for the above reaction but I cannot make much progress. I do have something in mind but it doesn't seem entirely plausible: First the tert-butyl group leaves (this seems wrong but tert-butyl cation is stable and thus a reasonable leaving group maybe? Also the negatively charged ring that it would leave behind would be stabilised by the EWG Bromines). The cation is then attacked by Cl and the ring is attacked by a proton (one product formed). Then a Friedel-Craft alkylation to make the other product (hence the excess of benzene to promote only mono-substitution). I doubt that is right, purely because of the first step. Can you help?

This reaction proceeds through a reverse Friedel-Crafts alkylation followed by a forward Friedel-Crafts alkylation. The last step of Friedel-Crafts alkylation is deprotonation, which means that the first step of the reverse process will be protonation. Next, the carbon-carbon bond breaks, with 1,2,4,5,-tetrabromobenzene acting as a leaving group. The resulting t-butyl cation can undergo the typical electrophilic aromatic substitution steps, attack by an aromatic ring (benzene here) and deprotonation to regain aromaticity.

One might wonder why aluminum chloride is present in the reaction mixture, since we can write a reasonable mechanism that excludes it. Likely it is present in case chloride (from HCl) traps it to form t-butyl chloride. Aluminum chloride will catalyze the reformation of t-butyl cation.


Using drip extraction, how much caffeine is extracted from coffee as a function of the amount of water used? When using a drip coffee maker, there are basically two choices we make: the amount of water and the amount of coffee grounds.

If we use the same amount of coffee grounds but double the water, how much extra caffeine is in the resulting brew? Does all the available caffeine in the grounds get quickly liberated, so that there's not much of a difference between brewing with more water and diluting the brew with water afterwards? Or is it closer to the other extreme, where twice the water means twice the caffeine?

I saw on wikipedia that only about 20% of the caffeine is extracted for an 'optimal brew,' but I wasn't sure if that means we can extrapolate and say that the remaining 80% can be extracted just as easily.

I have the same question about the amount of coffee grounds. I only care to consider the 'normal brewing regime', that is with water at ~80-90C, 3-6 cups of water, and 1/4-1/2 a cup of grounds.

Thanks!

Caffeine has a solubility in water of 20% at 80 $^\circ$C, while the coffee has between 1 and 4% caffeine.
That means that you have more than enough water to extract all the caffeine in the coffee.
With the quantities you have given, I estimate that you are using from 12 to 24 mL of water per gram of coffee (bulk density 0.25 g/mL).

So, it is not very likely that a change in volume will affect the total quantity of extracted caffeine much. Only the concentration per unit volume will be lower if more water is used.
But then, solid-liquid extraction is not that perfect and possibly you will extract some more caffeine with more water. The main property that influences the extraction is the size of the milled coffee. The smaller the better is the extraction.

Many other components of coffee are extracted fairly fast.
Therefore, here we do a percolation (expresso) which extracts everything with very less water. The concentration of caffeine per unit volume seems to be higher, but the dose of coffee is much smaller.


Choice of preferred IUPAC names When looking up names for simple organic molecules, the idea of a preferred IUPAC name (PIN) often comes up.

My understanding was that the PIN was meant to provide a unique and unambiguous name for a given molecule based upon strict applications of the IUPAC guidelines for nomenclature.

Whilst some PINs are clearly applications of these formal IUPAC guidelines, other PINs are trivial names. To give some examples:

Acetic acid (PIN, but trivial name) rather than ethanoic acid (systematic
IUPAC name)
Propan-2-one (PIN and systematic IUPAC name) rather than acetone (accepted but trivial)
Phenol (PIN, but trivial name) rather than benzene (systematic IUPAC name)

Clearly most people still refer to these molecules as acetic acid, acetone, and phenol; but it seems arbitrary that some trivial names are also the PIN whilst others are not (that is not to say that the other names are forbidden, as pointed out in the comments, with reference to P-10 of the recommendations).

  Why do we need the PIN, and why does it not always conform to the strict application of the IUPAC guidelines?

Quoting the Introduction (P-10) of the IUPAC Blue Book (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013):

[...] We have developed rules for assigning 'preferred IUPAC names', while continuing to allow alternative names in order to preserve the diversity and adaptability of the nomenclature to daily activities in chemistry [...]
Thus the existence of preferred IUPAC names does not prevent the use of other names to take into account a specific context or to emphasize structural features common to a series of compounds. Preferred IUPAC names (PINs) belong to a 'preferred IUPAC nomenclature'. Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name', in the context of a 'general IUPAC nomenclature'.

Therefore, there may be multiple names for a compound which are allowable in IUPAC nomenclature, one or more of which may be a trivial name (or derived from one). Note that there is no stipulation that the PIN must be the systematically obtained name. So, for a compound with a trivial name, either the systematic name or the trivial name may be chosen as the PIN. How does IUPAC choose which one is the PIN?
It might not be the answer we are looking for, but the end of the day, I strongly suspect it is simply arbitrary. The examples you have cited are very illustrative. Here is one more inconsistency which, to the best of my knowledge, is arbitrary.
"Acetic acid" is a PIN that may be modified by substitution (P-65-1.1.1), but only to a limited degree (P-15.1.8.2.1). Therefore "bromoacetic acid" is also a PIN. On the other hand, acetylene is also a PIN; however, substitution of any kind of not allowed (P-31.1.2.1). So, instead of bromoacetylene, the PIN is bromoethyne. (Bromoacetylene, however, remains acceptable within general use. All the names listed below that aren't followed by "(PIN)" are acceptable within general use.)

What makes IUPAC decide that - for PIN purposes - one trivial name is substitutable, whereas another is not? Or in the examples you have cited, why is one trivial name the PIN whereas another isn't? Of course, the rules state that it must be so, but to the best of my knowledge, there isn't any explanation of the philosophy behind the choice of PIN. Maybe I'll be proven wrong, though, and I'd be glad to see it.
To be honest, if you could come up with one, I'd be surprised. It's very tough to draw a line. Maybe we can agree that "acetone" should be a PIN, but should "neopentane" be a PIN as well? How about "estrogen"? Someone has to just arbitrarily choose where to draw a line, and it just so happens that IUPAC has gone with something stricter that doesn't use "acetone" as a PIN.
Anyway, just to reiterate - many of these names are perfectly valid IUPAC names. It is just that, for one reason or another, IUPAC has not selected them as preferred IUPAC names.


What is the difference between the units 'torr' and 'mm of Hg'? According to present definition torr and mm of Hg differ though slightly.
I am still confused over it. Can someone write an explicit answer differentiating between them and the reason for differentiation as earlier they were considered same. The explanation is probably related to the changes is pressure at different heights so I want the exact definitions of these units that was used earlier and what modifications have been made?

The two units torr, and mm of Hg were the same until they were redefined.

The torr was named after the Italian Evangelista Torricelli. 1 atmosphere is $101325\ \mathrm{Pa}$. The torr is defined as $1/760$ of an atmosphere. This is equal to $133.322\overline{368421052631578947}~\mathrm{Pa}$, which periodically infinitely repeats.

The mm of Hg was defined as the pressure generated by a column of mercury one millimeter high. Since the pressure of mercury depends on temperature and gravity, it is now redefined as

$$\mathrm{0.001\ mL \cdot 13595.1~mg \cdot 9.80665\ m/s^2 = 133.322387415 ~Pa}$$

This has caused a difference between the units of $0.000015~\%$, so unless you require extremely high precision, you can assume that $1\ \mathrm{torr} = 1\ \mathrm{mmHg}$


Measuring Thermodynamic Stability We can predict whether a reaction can occur spontaneously by calculating the change in Gibbs' free energy.

I was just wondering whether there is any general measure of the thermodynamic stability of any substance?

Can I look at a random substance and using a table of thermodynamic values predict whether this substance is stable or not within a range of given conditions (pressure, temperature, etc.).

Thermodynamics isn't quite that general.  It will tell you about the possibility of going from one well-defined state to another well-defined state, but as has been said, there is no way to do it in general.  What you are asking is akin to "is it hard to get there from here?" without specifying where both here and there are.

So you need to specify enough variables such as temperature or composition to unambiguously determine both the initial and final states.  How many variables do you need?  The Gibbs phase rule provides the answer.  Once that has been done, your question can be answered.


How does photosynthesis work? I wonder how photosynthesis works. I read the Wikipedia page, but I'm still confused.

My main gap of understanding is (I assume):

I know a photon changes the velocity of an electron. But I do not (fully) understand how a photon creates chemical reactions. Does the atom lose its electron due to this velocity increase and thereby attracting the electrons from the other atoms?

Is that all? Does this imply there is always an electron lost in photosynthesis or is that electron absorbed elsewhere (in another atom)?

A better way of thinking about it, without getting too bogged down in details, is that the plant absorbs the energy of photons and uses it to drive chemical processes - ultimately the synthesis of sugar out of carbon dioxide and water - that are energetically "uphill", and would not happen spontaneously without the energy from the photons.  The energy from the photons is used by the plant to move electrons from one place to another, but no electrons are actually generated or consumed in the process ... photosynthesis is a cycle, so you always end up back where you started, with some molecules of $\ce{CO2}$ and $\ce{H2O}$ having been consumed (along with a lot of light energy), and some extra molecules of sugar and $\ce{O2}$ having been produced.

The places where the photonic energy is used to mobilize the electrons are protein complexes called photosystem I and photosystem II, which are concentrated in the chloroplasts of green plants.  The details of how this happen are still not completely understood, but a LOT of progress has been made fairly recently using fancy equipment like laser spectrometers that can measure physicals processes that occur on timescales shorter than a nanosecond.  This has allowed researchers to map the progress of the electron as it moves from one part of the protein complex to another, following absorption of the photonic energy.

Does that help at all?


Arrange the following compounds in the increasing order of equilibrium constant for hydration Arrange in their increasing order of equilibrium constants for hydration:

(a) A < B < C < D < E
(b) A < C < B < E < D
(c) A < C < E < B < D
(d) C < A < B < E < D

How to compare figures E and D? Formaldehyde (figure D) doesn't have steric repulsion. Figure E has steric repulsion along with an electron withdrawing group. Given answer is (b). Does this mean steric hindrance is dominant over inductive effect while comparing? Please explain!

To my knowledge, comparing the steric hindrance of the carbonyl group is not the path you want to follow to answering this question, because the trigonal planar configuration of the carbonyl carbon provides plenty of room in order for the inductive effect to outweigh the steric hindrance. We have to examine the partial positive charge of the carbonyl carbon resulted from the C=O and the other substituents. The C=O bond has a great dipole moment resulting in a partially positive carbonyl carbon. This charge can be stabilized or increased by alkyl substituents (hyperconjugation) or electronegative atoms (inductive effect) respectively. The greater the positive charge on the carbonyl carbon the higher the equilibrium constant for hydration (due to the mechanism). Therefore, we can easily say that (A) < all, because of the two alkyl substituents. C > A, because the chlorine reduces the hyperconjugation effect. B > C (only 1 alkyl substituent). E > B (because of the same reason as the C > A scenario. And lastly D > E, because of the abscense of any alkyl groups (same reason as the B > C scenario). Now, why is B > C and D > E? Why is less alkyl substitution increasing the charge to a greater extent than 1 more alkyl group with an electronegative atom? If the chlorine was attached to the carbonyl carbon itself (attach the chlorine in E to the C=O), it would greatly icrease the charge and in fact this molecule would have the highest Constant of all. However, we have to keep in mind that the inductive effect greatly deminshes the further away you go from the chlorine atom. The carbonyl carbon is 2 σ bonds away from the chlorine (as in C and E), so yes it will feel the induction but the alkyl group will still donate elenctron density via hyperconjugation to a lower extent. In conclusion, the chlorine in C and E is reducing the hyperconjugation effect but it doesn't neutralize it completely. Therefore, B > C and D > E


Is it safe to hold silicon nitride ceramic without protection? If I have a complete silicon nitride ball bearing unit, is it safe to hold without any protection? If it cracks, or disintegrates is it also safe?

I have a few silicon nitride balls in my drawer. They look pretty cool, that's why I keep them and touch them all of the time.

It is a very hard, glasslike material. It won't be easy to crack. There is nothing coming off the surface, ball bearings have extremely smooth surfaces (nm roughness), if anything would rub off, they would not be usable as ball bearings. Besides that, I don't think it is toxic. It is an extremely inert material, much more inert than steel, does not dissolve in much. It is even used in prosthetic parts that go into the body. I assume it needs a lot of certification for that.

If you break it (which won't be easy), you'll have some sharp pieces that might cut you, that's all.

The only risk would be from a ball bearing that was ground to a very fine dust.


How to predict the color of a pH indicator using Le Chatelier's principle The question I have is above. Could someone explain why the answer is (A)?

What I think so far:

Does the sodium hydroxide turn the indicator blue because it is a strong base and hence it will react completely to form water,
shifting equilibrium really far to the right?
Does the HCl turn the indicator red for a similar reason?
I am unsure of the sodium acetate effect

You're absolutely on the right track!

The keys to the question are

to  order $\ce{HCl}$, $\ce{NH4Cl}$, $\ce{NaOAc}$ and $\ce{NaOH}$ in terms of strong acid, weak acid, weak base and strong base
to realize that in the presence of $\ce{HCl}$, the indicator exists as $\ce{H2Z}$, while it is fully deprotonated to $\ce{Z^{2-}}$ in the presence of $\ce{NaOH}$
in the presence of weak acids or weak bases, the indicator exists in form of each two species. Consequently, a mixed colour is observed.

In the presence of $\ce{NaOAc}$, a weak base, $\ce{H2Y}$ is completely deprotonated, thus no red colour is observed. In addition, a part of $\ce{HY-}$ is deprotonated too: the second equilibrium isn't completely on the left side. Some amount of $\ce{Y^{2-}}$ is present too. As a result, the green colour (yellow + blue) is observed.

It is all about the position of the equilibria.


Why soak filter paper in K2SO4? From a description of my experiment:

  Potassium iodide crystals are place in the middle of filter paper soaked with potassium sulphate and phenolphthalein. Two electrodes are placed on the filter on two opposite sides. The electrodes are connected to a battery. The objective of this experiment is to see the movement of ions.

Why soak the filter paper in potassium sulphate and not just soak it in potassium iodide?

For this experiment, you'll be seeing an interesting color change when Iodine is oxidized at the anode to form the brownish $\ce{I2}$ on the electrode and yellow $\ce{I3-}$ in solution.

The reduction of $\ce{H+}$ to form $\ce{H2~(g)}$ and $\ce{OH-}$ will turn the indicator pink at the cathode.

The cool thing about your experiment is that you will not see any redox reaction at all (at either electrode) until iodide ions diffuse to the anode! That is, unless you use a high voltage battery, $\gtrapprox2.0~\mathrm{V}$, you will not have sufficient energy to oxidize $\ce{SO4^2-}$ to $\ce{S2O8^2-}$.

This makes the experiment an exercise in patience, and then sudden color display! Measuring the time it takes can yield data that can help you calculate the diffusion coefficient of $\ce{I-}$ if you wanted to take it another step!

In summary: Soaking the filter paper with KI would result in the electrodes producing color changes immediately. I suspect the sulfate was used in this experiment to illustrate the diffusion of dissolved ions, because the redox reaction would not take place until the iodide ions reached the anode.


Bonding in 1,4-dihydro-1,4-diborinine In most aromatic compounds (e.g. benzene), their $\pi$-bonding molecular orbitals are completely occupied and their antibonding molecular orbitals are unoccupied. I'm wondering if there are any compounds where this is the case, but where the highest energy bonding MOs are only half-filled, so they contain unpaired electrons. My best guess at a molecule with that property is 1,4-dihydro-1,4-diborinine, as it has the same geometry as benzene (I think), including the adjacent p-orbitals, but it has two fewer electrons. Does this molecule still exhibit ring currents like benzene? Does it show the same type of delocalization?

The given dihydroborinine is indeed anti-aromatic if is it fully conjugated and planar, but counting pi electrons in a conjugated ring is not as simple as it seems when you have boron in the ring.  If the boron is not joined to a ligand, as in (mono)boronine or unhydrogenated diborinine, it could pass its electron deficiency into the sigma bonding framework to preserve pi electron aromaticity.
The basic (or Lewis acidic) case of borinine
Borinine or borabenzene, $\ce{C5H5B}$, is two electrons short of a well-known isostructural aromatic compound, pyridine.  We might suppose that as in pyridine, the sigma bonding and non-bonding orbitals including the outward-directed orbital from the boron must be filled before pi electrons can be introduced, which would leave borinine with only four pi electrons.  And we might be wrong.
Raabe et al. [1] perform an extensive quantum mechanical study of borinine molecules.  They find that in fact there are six pi electrons, the ring is aromatic, and it is the sigma electron framework that is electron-deficient because the outward-pointing boron-based orbital is empty.
To understand this, we look at the structure of borinine given in the above reference, from which the figure below is taken:

The above figure is one of several calculation results, but the all show similar results.  Not only is the ring highly distorted, it is distorted oppositely from what one might expect from the boron atom being inherently larger than the carbon atoms.  The boron is displaced inwards from where it "should be".
This inward displacement means the lobes from the adjacent carbon atoms are overlapping the "empty" outward boron-based orbital.  This overlap creates delocalized bonding around the boron and makes the empty orbital sigma antibonding, whereupon it becomes the LUMO.  The low-energy sigma orbitals are filled after all, but they do not include that outward-pointing boron orbital.  Thus borinine retains six pi electrons.
We should not conclude that borinine is therefore stable.  The pi electron component is stabilized by aromaticity, but with the overall electron deficiency in borinine this comes with an electron-deficient and highly distorted sigma-electron framework.  The antibonding orbital referenced above is too high in energy to be occupied in borinine as such, but it is low enough to combine with an electron pair from a Lewis base, such an an amine.  The resulting adduct (called a boratabenzene when the base is an anion) is the typically isolated product.  With the outward-pointing boron lobe now occupied, boratabenzenes and neutral Lewis-base adducts resume a more typical benzene-ring structure.  They still have six pi electrons, and thus remain aromatic, because the base contributes both electrons to its bond with the boron.
Going for two
I do not have a similar reference for diborinines or their hydrogenared derivatives.  However, Wahab et al. [2] provide nucleus-independent chemical shift (NICS) values for several monocyclic rings which they use in the simplified computational scheme they discuss; these are shown in the figure below, taken from this reference.

As many readers know, NICS models the inherent magnetic shielding (negative value) or deshielding (positive value) within a conjugated ring, these opposing signs then being interpreted as showing aromaticity or anti-aromaticity respectively.  Thus the unhydrogenated 1,4-diborinine is aromatic and even beats several more familiar aromatic heterocycles.  The delocalized bonding around the boron noted above apparently applies at each boron atom here, enabling the aromatic ring structure to be retained.
However, the dihydro compound shown at the right of the figure has become antiaromatic.  As was also noted earlier, when the boron is bound to a ligand the delocalized bonding around the boron is lost and we can go back to counting electrons as we would in an ordinary benzene ring.  However, whereas in Lewis base adducts both electrons needed to bond the boron to its ligand come from the ligand, in dihydrodiborinenes the hydrogen ligands provide only one electron apiece and thus the remaining two electrons to form these bonds have to come from "robbing" the pi system.  Thereby aromaticity in the pi system of dihydrodiborinines is lost, as seen in the NICS becoming positive.
Nobody else like you
Ultimately, we have to be careful rendering molecular-orbital structures in boron compounds because boron, uniquely among nonmetallic elements, lacks sufficient valence electrons to fill its orbitals through pairwise electron sharing alone. The unique bonding characteristics boron often adopts to compensate for this deficiency must be taken into account in order to achieve a proper rendering.
References
1.
Gerhard Raabe, Wolfgang Schleker, Eberhard Heyne, and Jörg Fleischhauer , "Borinine (Borabenzene): Its Structure and Vibrational Spectrum. A Quantum chemical Study", Z. Naturforsch. 42a, 352-360 (1987).  Link.
2.
Alexandra Wahab, Felix Fleckenstein, Stefan Feusi and Renana Gershoni-Poranne, "Predi-XY: a python program for automated generation of NICS-XY-scans based on an additivity scheme" Electronic Structure 2, 47002 (2020).  Link


What is the temperature dependency of the van-der-Waals constants? What is the effect of the temperature on the van-der-Waals constants $a$ and $b$?

I have checked on the effect of pressure and volume on the van-der-Waals constants, but I've not been able to get the effect of temperature.

$a$ and $b$ are supposed to be constants. But, they in fact aren't, so your question is in principle valid.

vdW is not a very good model to describe $V_m(p,T)$ of any actual non-ideal gas over a larger pressure or temperature range, and if you take a really good model (those have 6 or 8 parameters), you can just mathematically derive $a(p,T)$ and $b(p,T)$.

However I doubt there is a general trend for a and b with increasing temperature or pressure. I'm sure you could find one for similar (e.g. mono- or diatomic, elementar) gases.

vdW is useful (just like virial coefficents and other models with only two or three parameters) to accurately model $p,V,T$ in a limited range without a large computational effort, e.g. for process control.

(The value of the vdW theory is imo an educational one: It gives a convincing reason why real gases are non-ideal, and correctly describes the trends.)


Balancing the oxidative decomposition of FeS2 Doing some practice for the GRE next year.

Please could you look at my working for this and tell me why I'm wrong?

$$\ce{? FeS2 + ? O2 + ? H2O = ? Fe(OH)3 + ? H2SO4}$$

My working so far is:

There must be the same number of of $\ce{FeS2}$ as $\ce{Fe(OH)3}$
There must be twice as many $\ce{H2SO4}$ as $\ce{FeS2}$
If there is one mole of $\ce{FeS2}$ for example, there must be 1 mole of $\ce{Fe(OH)3}$ and two moles of $\ce{H2SO4}$

Tot up the $\ce{H}$ on the right: $1\times 3 + 2\times 2 = 7$. $7$ on the right so $\ce{3.5 H2O}$ on the left. ($7/2 = 3.5$)

$\ce{O}$ on the right = $1\times 3 + 2\times 4 = 11$.

This mean $11$ on the left, too. I already have $\ce{3.5 H2O}$, so we must have $11-3.5 = 7.5$ $\ce{O}$'s from the $\ce{O2}$. So $\ce{3.75 O2}$ because $7.5/2 = 3.75$.

Following the rules I've written down I arrive at:

$$\ce{1.75 FeS2 + 3.75 O2 + 3.5 H2O -> 1.75 Fe(OH)3 + 3.5 H2SO4}$$

I've multiplied these by $4$ to give integers, because that's what the question asks of me, so $7, 15, 14, 7$ and $14$.

Why am I wrong?

I found a solution which works,
$$\ce{4FeS_2 + 15O_2 +14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

This is how I found it. Do you know linear algebra ? Because that is very useful. Anyway without to know this is linear algebra you'll be able to understand.

So we are looking about the stoichiometric coefficients $\{\ce{a, b, c, d, e}\}$ of this reaction :

$$\ce{aFeS_2 + bO_2 + cH_2O + d Fe(OH)_3 + eH_2SO_4 \rightarrow 0}$$

Now what you need to do is first to find the number of different elements which are involved in your reaction, here they're four : $\{\ce{\color{\green}{H}, \color{\red}{O}, \color{\purple}{S}, \color{\orange}{Fe}}\}$. I'm not that good in the use of $\LaTeX$ but now the idea is to make column vectors under each of the reagents of your reactions and in these vectors you put in the same order for each the number of each elements there are in and if there is not just put a zero. It will gives you this :

$$a\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{0} \\ \color{\purple}2 \\ \color{\orange}{1} \end{pmatrix}+b\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{2} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+c\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}{1} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+d\cdot\begin{pmatrix} \color{\green}3 \\ \color{\red}{3} \\ \color{\purple}{0} \\ \color{\orange}1 \end{pmatrix}+e\cdot\begin{pmatrix} \color{\green}2 \\  \color{\red}4 \\ \color{\purple}1 \\ \color{\orange}0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

It can be easily transform into a system like this :

$$\begin{cases}
2c+3d+2e=0 \\
2b+c+3d+4e=0 \\
2a+e=0 \\
a+d=0
\end{cases}$$

So here we are in a good way because we have more unknowns than equation so if the system converge it has an infinity of solutions. And we can multiply by the number we want each stoichiometric coefficients and still have the same reaction.

So here we will determine four unknowns in terms of an other, I chose $e$. And I got :
$$$$\begin{cases}
a=-e/2 \\
b=-15e/8 \\
c=-7e/4 \\
d=e/2 \\
e=e
\end{cases}$$$$

I multiply then by $8$ to have only integers which give me with $e=-1$ :

$$\ce{4FeS_2 + 15O_2 + 14H_2O - 4Fe(OH)_3 - 8H_2SO_4 \rightarrow 0}$$

Then $$\ce{4FeS_2 + 15O_2 + 14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$


What temperature is really required to make fused quartz? Glass transition temperature of fused quartz is 1200 °C. Does this mean it can be made by heating sand to 1200 °C, or does the melting point of $\ce{SiO2}$ has to be reached (~1700 °C) for that to happen?

Glass is not necessarily made of pure silica $\ce{SiO2}.$ Ordinary glass is a mixture of at least $50\,\%$ $\ce{SiO2},$ plus smaller proportions of $\ce{CaO},$ $\ce{Na2O},$ $\ce{K2O},$ and $\ce{Al2O3},$ with impurities like $\ce{Fe2O3}.$ Pyrex has also some $\ce{B2O3}.$
Glass has no defined melting point. It becomes like a very viscous paste at around 800 °C. Mixtures have melting points that are always much lower than their components.


entropy calculation in an irreversible process I found out that if we want to find the entropy change in an irreversible process let's say $\text{state A} \to \text{state B}$ , we find or make an reversible path for the same process and find the entropy change for that irreversible process. The thing I don't get is that, even if entropy is a state function, does it mean that the change in entropy for the irreversible process we needed to find is the same as the change in entropy for the reversible path we constructed? Or am I being mistaken? If its the latter, please provide me a solution for it. Thanks.

Quantities such as heat and work can vary depending on the path (how you get from an initial to a final state). In contrast a defining aspect of a state function is that it is independent of the path.
Entropy is a state function, as a corollary therefore a path-independent property. The experimental determination of entropy involves measuring the heat transferred as a function of temperature during a reversible process and computing $\int \frac{dq}{T}$. That is another defining property of the entropy. It is not trivial to wrap your head around why it should be so, but entropy can be thought of as helping quantify a limiting or ideal property, in particular the maximum work that might be obtained from a process, or the least amount of work required to bring about a non-spontaneous change. It defines processes in which there are no losses (losses being consistent with irreversibility).
As an example, at constant T and p the Gibbs free energy change (which is related to the limiting work) is related to the entropy as follows:
$$-\frac{\Delta G}{T}=\Delta S_{system}+\Delta S_{surrounding}$$


Positive electrode potential of copper My book (NCERT India) says that the electrode potential of metals is decided by the following factors:

Enthalpy of atomisation
Ionisation enthalpy
Hydration enthalpy

It says that the first two require energy while the latter releases energy. That's all good and I can follow till there. But then it provided this table of thermochemical data for 3d series metals (Table 8.4 page 218):

It says that the electrode potential for $\ce{Cu^2+/Cu^1+}$ is positive because the hydration enthalpy does not balance the enthalpy of atomisation and the ionisation enthalpy - which requires energy. By the same logic, simply summing up 1 & 2 (sum of the first and second ionisation enthalpies) & 3 would give a positive value. How is it feasible for any of the there elements to have a negative electrode potential?

The potential as it is given there is vs. standard hydrogen electrode(SHE) which has an arbitrary assigned value of 0.0000V. Through the choice of another reference, it is very easy to make all of those numbers positive or negative. The absolute value of the half cell potential can only be useful in comparison to others. So you could make arguments like Cu has a more positive potential compared to Ni(by 0.59V based on the above numbers), but that is it.


How do I estimate ΔG⁰f for Fe2+? How would one use the equation ΔG° = −nFE° to estimate the ΔG⁰f of Fe2+(aq)? Since Fe2+ appears in multiple entries in the following table, I'm not sure which value to use.

This is not possible. The potentials are given only as references to the standard reduction potential for hydrogen. As such, they are not absolute.

You can convert between potential and free energy for a full redox reaction.

Actually, your statement has a deeper conceptual flaw. The standard formation free energy of the Fe(II) ion does not make any sense. The standard free energy of formation requires making something from its constituent elements in their standard states, but you cannot make an ion from its constituent neutral atoms. You'll need to account for the energy of the electrons and their binding energy to the ion. Therefore, the cation alone cannot have a standard free energy of formation.


Concept of enthalpy at a given temperature In thermodynamics, we always quote a fixed temperature, whenever we mention enthalpy of a reaction. For instance, one can determine the enthalpy of combustion of methane at 25 °C. Now almost all reactions involve a change in temperature, i.e. they are either endothermic or exothermic.

How should we physically interpret the meaning of enthalpy change at a fixed temperature? I recall from one textbook on experimental physical chemistry (perhaps it was Shoemaker's Experiments in Physical Chemistry stating that these state functions; still how can physically interpret enthalpy change at a given arbitrary temperature? I have not seen any textbook explicitly addressing this point.

The enthalpy change at a fixed temperature is equal to the amount of heat you need to add in order for the final temperature to be equal to the initial temperature.  If the enthalpy change is negative, this means that the reaction is exothermic, and you need to remove heat for the final temperature to equal the initial temperature.

If the reaction is carried out adiabatically (no heat addition or removal), the temperature of the products will be either higher or lower than the temperature of the reactants, depending on whether the reaction is exothermic or endothermic, respectively.  To determine the temperature change, you can follow a two-step procedure:  1.  allow the reaction to take place isothermally, by adding or removing the heat of reaction as necessary; then, 2.  reverse the heat addition or removal that you did in step 1, by applying it to the products of the reaction from step 1.  The net effect will be the adiabatic temperature rise (or fall) produced by the chemical reaction.


Does freezing point depression depend on the charge of the solvated ions? Assuming an ideal solution, which of these 0.10 M solutions would have the least freezing point depression?
$\ce{HCl}$
$\ce{NaBr}$
$\ce{KNO3}$
$\ce{MgSO4}$

They all produce solutions with two ions per formula unit, but the correct answer is somehow supposed to be $\ce{MgSO_4}$. How is this? I don't think it can be due to the weak basicity of sulfate; that should in fact increase the ions present and have an even greater freezing point depression.

First let's gather the data from the CRC Handbook of Chemistry and Physics, 92nd edition (mol/L is moles per liter). The data below is reported as molarity rather than molality since the problem is posed in molarity. However in a solution as dilute as 0.1 M the difference shouldn't be large.

         wt%      mol/L      Delta-F
  HCl     0       0        0.00
          0.5     0.137   -0.49
          1       0.275   -0.99
          2       0.553   -2.08
          3       0.833   -3.28
          4       1.117   -4.58
          5       1.403   -5.98
  least squares fit Delta-f = = -3.4683*[HCL] - 0.5624*[HCL]^2
  [HCL] =0.1 M, Delta-f = -0.35

         wt%      mol/L      Delta-F
  NaBr    0       0.00     0.00
          0.5     0.049   -0.17
          1       0.098   -0.34
          2       0.198   -0.69
          3       0.301   -1.04
          4       0.405   -1.39
          5       0.512   -1.76
  least squares fit Delta-f = = -3.4823*[NaBr] + 0.0930*[NaBr]^2
  [NaBr] =0.1 M, Delta-f = -0.35

         wt%      mol/L      Delta-F
  KNO3   0        0        0.00
         0.5      0.05    -0.17
         1        0.099   -0.33
         2        0.2     -0.64
         3        0.302   -0.94
         4        0.405   -1.22
         5        0.509   -1.50
  least squares fit Delta-f = = -3.4264*[KNO3] + 0.9923*[KNO3]^2
  [KNO3] =0.1 M, Delta-f = -0.33

         wt%      mol/L      Delta-F
  MgSO4  0        0        0.00
         0.5      0.042    -0.10
         1        0.084    -0.19
         2        0.169    -0.36
         3        0.257    -0.52
         4        0.346    -0.69
         5        0.437    -0.87
  least squares fit Delta-f = = -2.3466*[MgSO4] + 0.9726*[MgSO4]^2
  [MgSO4] =0.1 M, Delta-f = -0.22

So there is a difference. 0.1 molar $\ce{MgSO4}$ has a freezing point depression of 0.22 degrees Celsius whereas the other 0.1 molar salts are about 0.35 degrees.

As far as the "answer" goes the problem is very poorly worded. First the problem states to "assume an ideal solution," then poses the problem in such a manor that some sort of non-ideal behavior must be assumed to obtain an answer. "Ideal behavior" for electrolytes only occurs in very dilute solutions. A 0.1 molar solution is much too concentrated for ideal electrolyte behavior.

Without any "lookup" of data the only solution seems to be some hand-waving about $\ce{MgSO4}$ being the only salt with divalent ions so the effective dissociation is smaller because of forming ion-pairs in solution.


Is there pure iron in nature? Is there pure iron in nature, would that be not become Iron Oxide (rust) immediately because of the humidity? How do we produce iron ore/dust on the lab/industry?

Fairly pure iron (or iron/nickel alloys) are found in meteoric iron. Some of these are surprisingly corrosion resistant... close to stainless steels. One well-known example is a dagger from Tutankhamun's tomb. See JSTOR for some more uses of meteoric iron.

As for iron smelting, there is much information on the web, and I suggest you try some research.


Synthesis Golf XI: Hyperielliptone HA It has been a little over a year since the last synthesis-golf was posted. I have talked to orthocresol and would like to help revive the tag!
A full FAQ post has been written on chemistry.meta.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

Hyperielliptone HA is a phloroglucinol extracted by Wu et al1 in 2008. The compound is of particular interest because of its highly unique structure and demonstration of  DNA oxidation inhibition.

Rules:

Your synthesis must involve a method for the preparation of the phloroglucinol core.
Your synthesis must be stereoselective; however, the stereocenter attached to the phloroglucinol core can be racemic. Bonus points if you can develop a stereoselective synthesis for this moiety!
All reagents must be available from Sigma Aldrich.
If you do not have time to complete the entire synthesis but have a clever idea for a specific moiety, feel free to share your answer.

References

Wu, C.; Yen, M.; Yang, S.; Lin, C. Phloroglucinols with Antioxidant Activity and Xanthonolignoids from the Heartwood of Hypericum geminiflorum. J. Nat. Prod. (2008) 71 (6), 1027-1031. DOI:10.1021/np8001145.

I'm sure there are a lot of issues  with my attempt I haven't spotted (or maybe some of the ones I have are in fact lethal), but I saw the question and thought I'd give it a go.
Retro:
Went straight for the Diels-Alder with something resembling Danishefsky's diene. Disconnection to reveal the species I named A and B comes from a Baylis-Hillman reaction, which works reliably with α,β-unsaturated esters (unsure whether the attached enol-looking oxygen would have anything to say about this). Asymmetry in A is provided by a Sharpless epoxidation, and in B by a bit of a longshot centred on MacMillan imidazolidinones.
Forward synthesis:

Notes:
TMS groups are used throughout but could always be beefed up to TBS as they're removed with TBAF at the end anyway.
I'd welcome suggestions of a better way of achieving step IIIa (formation of the enol-ether-like structure in A).
MacMillan catalyst sequence (steps VI to VII):
I'm aware that these catalysts (R = Bn, BnIndole etc.) are designed with one-pot systems using poor nucleophiles and electrophiles in mind, and hence expect this to be a bit of a stretch. I'm wondering if a more robust method might be to go with stoichiometric addition of the imidazolidinone before addition to THF containing the lithium alkyl cuprate shown to the left of step IV, rather than simultaneous addition of a Grignard and copper(I). The cyclisation seems like it would follow naturally and the correct stereochemistry would be provided by the imidazolidinone in accordance with MacMillan's models.
Not sure how big of a concern rearrangement during the Barton McCombie reduction step (IXa) is, but perhaps it would be better to conduct after the Diels-Alder; I don't know.
My hope is that the Diels-Alder step would get some stereocontrol from the 'B' section of the molecule folding down underneath the diene section, but expect this to be the sort of thing that has to be tried out in the lab.


Do bond energies differ for surface and bulk atoms in nanoparticles? In nano-particles, surface atoms have fewer neighboring atoms in compare to bulk atoms.
Is there any difference between bond energy of surface atoms and bulk atoms?

Yes, there is, as surface atoms and bulk atoms have different bond energies in nanoparticles. This is a reason why nano particles coagulate easily and special measures have to be taken to prevent coagulation.


How can I make an acidic pen to burn paper on writing on it? If I have a pen with refillable ink, can I just pour in concentrated Hydrochloric acid or concentrated sulfuric acid in the ink chamber to make it an acidic pen?

Would it be better to use the acid of suitable molarity instead of being concentrated?

Then if I write with it on a paper, will it burn?

Is this a good idea to play with?

Is it dangerous in any way?

Is it illegal to make such things for experimenting and fun?

Is there any other safe method to achieve the same?

What are the odds that the poured acid will burn the ink chamber itself and start leaking?

Can it be used in any constructive way? Like cutting many sheets of paper at once. Or using it in industries suitably to make smooth cuts in materials that can be burned with the used acid.

Will it be fun to you?

Thanks to everyone.

Edit:

I would like to add a few details.

Assume that the acid is not concentrated but of suitable molarity to just cut the paper. I mean, to pierce through it and burning its edges to just make a nice cut. Like a glowing incense stick makes when made to touch a paper.
Also, the nip of the pen and the ink chamber are strongly inert; enough to survive the concentrated acid's attack and prevent leaking. Like, the chamber can be made of glass (in which acids are usually kept) and the nip can be made of galvanized/electroplated iron.

Or if there is still more, we can just make this into a thought experiment.

You can pour anything into your pen. Is it a good idea? Most likely not. Is it dangerous? You bet! Is it a good idea to play with? NO! Whether it is legal or not to create such an object, you'll have to check with legislation in your country.

If you want paper to ignite on touch with this pen, you would need some truly strong acid (and probably undiluted). I'm certain that if this substance has the capability to burn paper, it would also corrode away the feed and nib of the pen. In the worst case, it will even eat through the ink container and leaking everywhere. In any way, this is no viable option to cut paper in a controlled fashion.

Is there a safer method? Well, there would be laser-assisted cutting of paper, but the good ol' knife or saw would still be cheaper.


Odd Acidity Trend in Hexaaqua Metal Complexes In the following link (three quarters down the page) it states that theoretically there should be a proportional relationship between the ionic radius and pKa of a hexaaqua metal complex (provided that the charge of each metal is the same).

The website's reasoning for this is that due to the smaller ionic radius, there is a smaller volume meaning that there is a greater charge density. Therefore this will cause a greater distorting effect on the electrons in the $\ce{O-H}$ bond. Hence this will enable the $\ce{H}$ to be deprotonated more easily, resulting in a lower pKa value.

However it later says that experimental data don't fit this relationship shown by one of the graphs in the website:

Obviously $\ce{Cu^2+, Cr^2+}$ and $\ce{V^2+}$ don't follow this relationship at all. The author of the website states that he has no idea why this happens.

I was wondering if anyone here could provide a more in depth explanation on how ionic radius effect acidity which perhaps can explain why these metals don't follow this linear relationship.

To start off with, there are two graphs on the site you linked, the second one being this one:

This, as the author of your site also states, is a much nicer graph in that it explains all but two ions ($\ce{Cr^{II}}$ and $\ce{V^{II}}$) well.

I also went around and checked different crystal structures of hexaaquametal(II) complexes for their $\ce{M-O}$ bond length to see if that may have any influence. The results are:

Vanadium: $212~\mathrm{pm}$ (octahedral)
Chromium $203~\mathrm{pm}$ (axial) and $245~\mathrm{pm}$ (equatorial; Jahn–Teller distorted octahedron)
Manganese: $214$ to $217~\mathrm{pm}$
Iron: $211~\mathrm{pm}$ (almost octahedral)
Cobalt: $208$ to $213~\mathrm{pm}$ (distorted octahedron)
Nickel: Did not find data
Copper: $200~\mathrm{pm}$ (axial) $220~\mathrm{pm}$ (equatorial)

On the lower end of this spectrum we find copper and chromium which indeed have rather low $\mathrm{p}K_\mathrm{a}$ values. This could be due to a shorter $\ce{M-O}$ bond meaning that the water molecules experience the charge stronger. (This is also why I favour the graph I put in this answer: It puts copper closer to ‘where it should be’ in ionic radii terms.)

A second thing to consider is the ligand field stabilisation enthalpy LFSE. This is strongest ($-12~\mathrm{Dq}$) for vanadium since that ion only has three electrons which all occupy the lower energy levels. This would mean that ligand exchange for $\ce{[V(H2O)6]^2+}$ is kinetically strongly inhibited especially for vanadium, since any loss of a ligand would need to overcome the stabilisation of $-12~\mathrm{Dq}$. Hence, the water molecules are strongly bound to vanadium and experience its charge all the time. On the other hand, copper has rather weakly bound ligands exemplified in how simple it is to create $\ce{[Cu(NH3)4(H2O)2]^2+}$, the deep blue tetraammincopper(II) complex.
(Note: I did not find any papers confirming or disputing this hypothesis, but I only spent a short time looking.)

Right after vanadium we encounter chromium. It only has an LFSE of $-6~\mathrm{Dq}$ — which is still pretty good, considering it all derives from unpaired electrons. Overall, I think this ion can be considered to be almost like vanadium(II) albeit a little smaller, with an extra electron and a shorter metal-water bond length which would explain its position.

Next up is manganese which looks as if it were the exception. But it actually isn’t. First off, its LFSE is $0$, thus water molecules can diffuse on and off as they see fit, therefore overall weakening the effect the ion’s charge has on them.

From iron onwards, we can again consider the water ligands bound slightly more tightly since there is again LFSE to be gained. However, remember that while the $\mathrm{t_{2g}}$ orbitals are non-bonding with respect to metal–ligand σ interactions, the $\mathrm{e_g}$ level is antibonding. We have a half-filled $\mathrm{e_g}$ level which will weaken the overall $\ce{M-O}$ bonds. This could not only explain the larger $\ce{M-O}$ distances of manganese and iron when compared to chromium but also their weaker acidity. (A weaker bond would also mean weaker charge transfer.) I suspect that this does not yet really matter for chromium due to its Jahn–Teller distortion being somehow favourable overall, and that chromium’s acidity stems from the closer water molecules.

From then on, the explanation seems rather simple, since the ion radii and the $\mathrm{p}K_\mathrm{a}$ are almost in a linear correlation. The question is how the ionic radii were deduced (which is none-trivial) and why they differ so strongly with the two sources the author of your linked site used.


Can we compare the vapour pressure of solutions whose concentrations and temperature is known? If we have some solutions, in which we know the temperature and concentration of the solute particles, can we compare their vapour pressures?

Original question:

Which solution has the highest vapour pressure among the following?
a) $0.02 ~\pu M ~\ce{NaCl}$ at $50^{\circ}$C b) $0.03 ~\pu{M}$ sucrose at $15^{\circ}$C  c) $0.005 ~\pu{M} ~\ce{CaCl_2}$ at $50^{\circ}$C d)$0.005~ \pu{M} ~ \ce{CaCl_2}$ at $25^{\circ}$C

I answered this question by noting that at higher temperature and lower concentration, vapour pressure of solution should be the highest because more temperature implies more kinetic energy which means more volatility and low concentration means that the solute is less solvated so it forms less bonds with the solvent and therefore has more vapour pressure, so answer is c). Is this reasoning correct?
Now I think that these options were set so that it becomes easy to compare.
But what if I had options like $0.02 ~\pu M ~\ce{NaCl}$ at $50^{\circ}$C ,$0.03 ~\pu M~ \ce{NaCl}$ at $60^{\circ}$C,  $0.005 ~\pu M
~\ce{CaCl_2}$ at $20^{\circ}$ C, how shall we compare their vapour pressures without doing experiments?
Now, if there existed a relation between vapour pressure and concentratin or temperature, it would become easier.
For example, relation between vapour pressure and temperature can be given by Antoine equation, but I want something that also incorporates the concentration in the equation also, so that by plugging in the values, it is easy to compare.
I am also not sure, whether van't hoff factor will have a role in it or not.

The point of the question is to test your understanding
i. of Raoult's law (that increasing the solute concentration decreases the vapor pressure and that the identity of the solute is not important in an ideal solution; the mole fraction of solute determines $\Delta p$: $\Delta p = p-p^\circ = -\chi_\textrm{solute}p^\circ$
ii. that vapor pressure increases with T
Other things you might want to remember are that molality and molarity are linearly proportional for dilute solutions (ie then $m \propto c_M$), and that ideality is approached for dilute solutions of small solutes (typically at low $\pu{mM}$ concentrations such as in this problem).
You can answer the question based entirely  on consideration of the van't Hoff factor to compute total concentrations, and by assuming that the vapor pressure increases with T.
First compute effective total solute concentrations $im$ (equal to either the solute concentration for a non-dissociating solute such as sucrose, or the total ion concentration for the electrolyte solutions):
$$\begin{array}{|c|c|c|} \hline \text{species} &T (\pu{^\circ C})& c_M (\pu{M})& i & ic_M(\pu{M}) \\\hline  1.~\ce{NaCl} & 50 & 0.02 & 2 & 0.04\\
2.~\text{sucrose} & 15 & 0.02 & 1 & 0.02\\
3.~\ce{CaCl2 } & 15 & 0.005 & 3 & 0.015\\
4.~\ce{CaCl2 } & 50 & 0.005 & 3 & 0.015\\\hline
\end{array}$$
Inspection should quickly eliminate all options but row 4 since they have a higher effective concentration at the same T, or since they are at a lower T at the same effective concentration.
Actual vapor pressures are not stated in the problem. In the absence of that information there is no point in attempting to compare things in more detail, and that's not the point of the problem. However, provided Raoult's law holds it can be used to determine the change in vapor pressure regardless of T. Raoult's law assumes the identity of the solute does not matter, that the effect of added solute on the activity of solvent is entirely entropic (accounted for by ideal mixing entropy) and that the gas is ideal. In general, to predict vapor pressure at any T and solute concentration you need a model of the activity of the solvent as a function of the solute concentration, and of the activity in the gas phase. You can determine the vapor pressure by finding the conditions that render the activities identical.


How to represent a 3D model of cocaine? For my AP Chemistry class we were told to make a molecular diagram that was "better than excellent" and the rest is up to us (we could choose the molecule and how we wanted to represent it).

My group chose benzoylmethylecgonine, or $\ce{C17H21NO4}$, also known as cocaine.

This is proving to be a difficult task as I now realize that cocaine is a pretty complicated molecule and I couldn't find much on the 3D molecular structure.  If anyone could explain its chemical structure, it would be very appreciated.

While I do not think that cocaine is a particularly interesting molecule, the crystal structure has been reported in Ronald J. Hrynchuk, Richard J. Barton, and Beverley E. Robertson, Can. J. Chem., 1983, 61, 481. There is also an open access paper of the XII IUCr Congress available, R. D. Hrynchuk, R. J. Barton and B. E. Robertson, Acta Cryst A, 1981, A37, C72. The crystal structure file has been deposited to ccdc.cam.ac.uk and can be downloaded free of charge. However, as far as I can see, it does only contain the information about the space group, etc, but no atomic co-ordinates.

However, the hydrochloride structure has also been reported in E. J. Gabe and W. H. Barnes, Acta Cryst., 1963, 16, 796-801. And the cif file can be also obtained at ccdc.cam.ac.uk.

I did a quick calculation on this geometry on the DF-BP86/def2-SVP level of theory, the plain xyz coordinates are attached at the end of this post. You can open them with molecular viewers such as Avogadro to create your own model.

The following, however, has been produced with the commercial ChemCraft program.

I do not know how far you want to go with your presentation, there are some pictures on wikipedia. I can also recommend checking the record of it on ChemSpider.

Geometry of cocaine hydrochloride:

symmetry c1
Cl      -2.051693000     -0.493009000      4.153090000
C       -2.451519000     -3.565091000      6.602382000
C       -6.214292000      1.164927000      8.772468000
C       -7.187531000      2.140573000      9.037661000
C       -8.160694000      1.917991000     10.029303000
C       -8.167271000      0.711328000     10.754343000
C       -7.200746000     -0.269358000     10.489848000
C       -4.535361000     -2.199809000      5.904903000
C       -4.965503000     -2.410764000      3.577976000
C       -0.472992000     -3.311748000      5.082266000
C       -3.681531000     -2.746076000      7.063993000
C       -3.208615000     -1.681350000      8.100725000
C       -1.959949000     -0.929634000      7.627840000
C       -0.886578000     -1.913758000      7.125382000
C       -0.580604000     -3.058883000      8.128726000
C       -1.633721000     -4.157723000      7.786625000
C       -5.209882000     -1.120147000      9.257751000
C       -6.215633000     -0.043947000      9.503197000
H       -2.776079000     -4.325870000      5.870251000
H       -1.141638000     -5.101415000      7.477122000
H       -5.458897000      1.318599000      7.989093000
H       -7.190344000      3.080452000      8.463274000
H       -8.921920000      2.687588000     10.235464000
H       -8.932096000      0.535630000     11.527409000
H       -7.186414000     -1.224606000     11.036739000
H       -4.915209000     -3.233215000      2.840991000
H       -4.408859000     -1.522439000      3.213266000
H       -6.015572000     -2.140911000      3.803492000
H        0.202195000     -3.979332000      5.655364000
H       -4.361351000     -3.437105000      7.610613000
H       -1.009855000     -3.900141000      4.313786000
H        0.111577000     -2.526413000      4.564425000
H       -2.986886000     -2.247300000      9.031392000
H       -2.216795000     -0.241161000      6.793930000
H       -1.551394000     -0.325210000      8.463897000
H        0.011514000     -1.355303000      6.795935000
H        0.453167000     -3.437387000      7.998115000
H       -0.655253000     -2.701279000      9.175171000
H       -2.290875000     -4.410651000      8.642544000
N       -1.471722000     -2.634200000      5.937688000
O       -4.202728000     -0.704253000      8.439562000
O       -5.256578000     -2.245544000      9.737495000
O       -5.375360000     -1.329144000      6.012186000
O       -4.299429000     -2.894414000      4.763398000
H       -1.903686000     -1.736871000      5.153339000


Iodine value calculation A $\pu{1.1g}$ sample of an oil was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with $\pu{7.3 cm3}$ of $\pu{0.1 mol/dm3}$ sodium thiosulfate solution. The blank titration required $\pu{43.5cm3}$ of sodium thiosulfate. What is the iodine value?

I've been asked this question but there are no examples of calculating iodine value in my textbook. I'm aware that the iodine value is the mass of iodine to react with $\pu{100g}$ fat/oil and the question gives a $\pu{1.1g}$ sample of an oil. Does this mean $1.1$ should be divided by $\pu{100g}$? I'm just a bit confused how to even start this question.

Assuming you are given normality of sodium thiosulphate as 0.1 N (or sometimes called standard), you could use this formula to find the iodine value:

iodine number=$$\frac{(mL 0.1N Na_2S_2O_3 blank  – mL 0.1 N Na_2S_2O_3 test) (12.7)(100)}{(1000) (1.1)}$$

Where “mL 0.1 N Na2S2O3 Blank” = the volume of sodium thiosulphate required to titrate
the blank solution,
Where “mL 0.1 N Na2S2O3 Test” = the volume of sodium thiosulphate required to titrate the
test solution that contained either solid fat or oil as the sample,
Where “12.7” is the amount of grams of iodine contained in one litre of 0.1 N iodine,
Where “1.1” is the amount of fat taken in grams  and,
Where “100” converts the sample to per 100 grams of fat.


Is an allene or a conjugated diene formed? If we dehydrate the following molecule, which hydrogen ($\ce{H1}$ or $\ce{H2}$) gets removed?

The conclusion that I have drawn from the molecule: The carbon to which $\ce{H1}$ is attached is $\ce{sp2}$ hybridized which makes $\ce{H1}$ slightly more acidic than $\ce{H2}$.

As the intermediate passes through an allylic carbocation which is fairly stable, it does not matter which H is 'more acidic' - the whole system is in equilibrium. The product you get is the thermodynamically most stable which will be a conjugated diene.


Why does chemisorption increase with rise in temperature, even though it is an exothermic process? Chemisorption is exothermic process, but still it increases with increase in temperature. Why?

I know that there needs to be a minimum activation energy to be achieved in order to initiate chemisorption. Also physiosorption can change to chemisorption in some cases on merely rising the temperature.

But why does it increase with an increase in temperature? How is this related to Le Chatelier's principle?

To answer that first think about a reaction like combustion reaction. Even though combustion reactions are exothermic you need to provide heat to the system by increasing its temperature. See, every reaction or chemical change happening requires a certain minimum amount energy called the "activation energy" for the reaction to start. This energy is required to break the bonds of the reactants so that they can react to form product. That's why we say that at low temperatures, kinetics controls the reaction rather than thermodynamics. Now, once you are at a temperature which is able to provide the sufficient activation energy to the reactants, the thermodynamics comes into play which essentially says in this case that if a reaction is exothermic, increasing the temperature will shift the reaction in backward direction.

Now coming to chemisorption, in this case too you need to increase the temperature to provide sufficient activation energy so that the reaction can start. So increasing the temperature increases the rate of reaction. After a certain temperature when thermodynamics become more important, the reaction shifts backward.

Hope it helps!


Open source for atomic spectra raw data Is anyone aware of a open source atomic spectra database? In particular I am looking for high resolution Na atomic emission which can be downloaded in csv or Excel. I am aware of NIST but I could not find a atomic spectrum. Secondly do they allow re-use of their figures? Thanks.

Based on your previous question and your comment above in lines of «NIST's policy to use published data may be fine» I question if your literature search included a visit at NIST's Handbook of Basic Atomic Spectroscopic Data.
In the case of sodium, recorded lines for either air or vacuum are listed next to the data about persistent lines, and analogue data about the single ionized form.  While not yet in a .csv format, copy-paste into an editor is eased because you may toggle between the output to be table-like formatted form, or ASCII.
Addition:
An alternative could be to digitize emission spectra like the following:

(reference, final spectrum about the hot high pressure sodium lamp)


Can two precipitates form in a precipitation reaction that starts with only two compounds in water? Can two precipitates form in a precipitation reaction that starts with only two compounds in water?

The number of examples is considerably great.
Consider the following:

$$\ce{Ba(OH)2(aq) + MgSO4(aq)  -> BaSO4(s) + Mg(OH)2(s)} $$

Above reactants are reasonably soluble in water.
From Wikipedia, $\ce{Ba(OH)2} $ has a solubility of 1.67 g / 100 mL at 20 °C and $\ce{MgSO4} $ a solubility of 25.5 g / 100 mL at 20 °C


Central Atom in EXAFS I read somewhere that EXAFS (Extended X-ray absorption fine structure) gives us the distances between central atom and its neighboring atoms. Is that true? Why can't we get the distance between two atoms, which none of them are central? How is EXAFS method distinguishing between central atom and other ones? So for example for CF4, I think we get one EXAFS peak for C...F distance and not F...F distance, why?

(Copied and edited from comments)

The central atom here refers to the center of attention (i.e. where the given peak assigned), the center of scattering which is often a center of an active site/coordination complex. I.e anything can be a central atom, if you tune your X-ray frequency to their peaks - you choose the central atom. In the given example theoretically you can choose to study C or F, and their closes neighbors. Theoretically you can see their neighbors; C-F distance is much shorter than F..F distance, though.

You should understand the use/limitations of the method: if you have chemically nonequivalent atoms of the same type in your sample, you will have their EXAFS peaks overlap. Both will give you a signal at the same place, which are hard to separate. In practice it means that you can use this analysis for e.g. central metal atom of complexes, but not for carbon atoms of a ligand as you generally have a lot of different carbon atoms in the sample.


Stability and resonance structures In school they ask us to count the number of resonance structures in different organic compounds to determine it's stability. The one with more resonance structures is more stable. Is this notion correct?

Stability is a funny word. It gets tossed around all the time, but a molecule's stability really depends on the context you are considering it in. For example, benzene is totally unreactive towards electrophilic addition of $\ce{Br2}$ like most alkenes, but $\Delta_\mathrm{f}H^\circ(\text{benzene}) = 49.0\ \mathrm{kJ/mol}$, while $\Delta_\mathrm{f}H^\circ(\text{cyclohexane}) = -37.8\ \mathrm{kJ/mol}$.

So when we talk about resonance and stability, what are we really talking about? What this really refers to the stabilization energy associated with conjugated systems. For every p orbital that is added to $\pi$-system, it mixes in phase to create a new molecular orbital that is lower in energy than the two starting orbitals, or mixes out of phase to create an orbital that is higher in energy than the two starting orbitals, and in some instances doesn't mix at all. This is illustrated here:

You can't directly compare the relative energies of the molecular orbitals of two unlike molecules simply based on their number of resonance structures, though. Different atoms and substituents will cause deviations and make comparison meaningless, so it is really only useful to compare homologous compounds, like butane, but-1-ene, and buta-1,3-diene.


What causes methane to have 25 times higher global warming potential than CO2 What causes methane to have 25 times higher global warming potential than $\ce{CO2}$?

I think it has to do with $\ce{CH4}$ being a less massive molecule, but I'm not sure.

First we have to look how the greenhouse effect works for all gases:

Light with a wavelength not absorbed by the atmosphere gets absorbed by the soil and heats up the earth. Because of black body radiation for 300K the earth starts to emit light itself with a maximum in the IR spectrum. Light with this frequency gets readily absorbed by vibrational excitation of molecules. But one has to keep in mind that intensity of absorption between different molecules and even between different vibration in one molecule differs a lot. This can not be explained with mass, or just vibration itself you actually have to look at https://en.wikipedia.org/wiki/Transition_dipole_moment or https://en.wikipedia.org/wiki/Selection_rule

But just believe me for a moment, methane is actually very bad at absorbing infrared light. Especially its by far worse than Water or $CO_2$.

But then there is another effect. You can not absorb more light than there actually is. So if you practically absorb 100% of the light for a given frequency then there is no effect in increasing the concentration of the molecule absorbing this frequency.

Both water and $CO_2$ absorb with nearly 100% in their frequency range (compare fig. 1 which shows the IR absorption of the atmosphere).
So in order to absorb more light you have to absorb in other frequency ranges.

This is where methane gets its role. If you look at figure 2 and keep in mind to change the wavenumber into wavelengths you will see that it absorbs in a frequency range that is covered neither by water nor by $CO_2$.


Gas evolution in reaction between zinc and HCl So I was doing a chemistry lab for school, and we were analyzing how the concentration of a solution affects the rate of a reaction. In order to do this, we used the inverse burette method for the reaction between zinc and HCl, which would produce hydrogen gas. We had an Erlenmeyer flask connected to the burette by a red tube, and the reaction would occur in the flask. I noticed that the flask stayed blurred, meaning that the hydrogen gas would be trapped inside of it. I asked my teacher if that meant that the pressure was not enough for the hydrogen to displace the water, but he said that was wrong. I cannot think, however, another reason for that. I would appreciate any help.

When HCl reacts with zinc or any other active metal such as aluminium, hydrogen is produced. This is what your textbook tells you. Whats is perplexing is the appearance of "fog" in the Erlenmeyer. In the beginning there is air above the liquid. When hydrogen evolution starts, the flask begins to fill with $\ce{H2}$ gas and creates enough pressure to displace the column of water in the buret. So that part of your observation is correct and you see a colorless gas displacing water column in the buret. What is in the conical flask that is causing the fog? If you were to touch the Erlenmeyer you could have noticed that it was hot, secondly there was a very vigorous evolution of hydrogen bubbles. There are several reasons, each contributing to the fogginess in the flask. That "fog" is just not hydrogen, air but vapors and very fine droplets of acid.

(i) creation of very fine droplets of acid by vigorous bubbling process (major contribution)
(ii) steam generated by the heat of the reaction
(iii) Hydrochloric acid has the tendency to produce fumes in moist air (seen with very concentrated acid). You did not mention the concentration of the acid used.


Formation of green precipitation whilst adding potassium fluoride to aqueous solution of copper(II) sulfate Aqueous copper sulfate solution (blue) gives a green precipitation with aqueous potassium fluoride. Explain these experimental results.

I've read answers like, $\ce{H2O}$ is a weak ligand and fluorine can replace it, but isn't $\ce{H2O}$ a stronger ligand than fluorine according to the spectrochemical series?

$\ce{CuSO4}$ exists as $\ce{[Cu(H2O)4]SO4}$ in solution. It is blue in colour due to the presence of $\ce{[Cu(H2O)4]^{2+}}$ ions.

Now if aqueous $\ce{KF}$ is added, the solution turns green due to formation of complexes $\ce{[CuF4]^2-}$ and $\ce{[CuF6]^4-}$(Initially copper(II) fluoride($\ce{CuF2}$) is formed but it formed complexe in present of water). The complete reaction is given by:

$$\ce{[Cu(H2O)4]SO4 + 4KF -> K2[CuF4] + K2SO4 + 4H2O}$$


nomenclature of citric acid I named Citric acid : "3-hydroxypentan-1,3,5-trioic acid"
but when I searched , I found that the right name is "3-carboxy-3-hydroxypentane-1,5-dioic acid"

Can anyone explain to me why my nomenclature is wrong?

The "-oic" ending means to convert the carbon into a COOH group, not that there's an additional COOH group added to that carbon.

Look at your main chain. You have five carbons, of which carbons 1 and 5 are fully oxidized, hence "pentane-1,5-dioic acid". If the "COOH" group was counted as an independent entity adding an additional carbon, then it would be a three carbon chain, to which was added two COOH groups.

That's why it can't be a 3-oic acid - that would imply that the third carbon of main chain has been oxidized to a COOH group (which is preposterous), not that there's a COOH group that's been added. Instead, the COOH group (carboxy) has been appended to the three position, and the molecule is named accordingly.


Why do ionic substances with higher lattice energies tend to be less soluble in water than substances with lower lattice energies? I understand that that energy is required to overcome intermolecular forces holding solute particles together in the crystal, but I don't understand how this relates to lattice energies and respective solubilities in water.

When an ionic compound dissolves in water, the water molecules surround the individual ions. Since water is very polar, when water surrounds an ion there is a decrease in overall energy. However, to pull the ions apart in the first place, there has to be an energy increase to overcome the lattice energy. If the lattice energy (cost) is larger than the energy of solvation of the ions (gain) the compound won't dissociate, and it will be insoluble.

There is a large grey area where lattice energy and solvation energy are close, and for those compounds temperature has a big influence on solubility.


How to Convert Sucrose to Glucose and Fructose If you have some sort of sports drink that contains sucrose, how would you take that sucrose and break it down into its components, glucose and fructose? I know that sucrose + water will break it apart, but how can it still remain together (as sucrose) while in the bottle of sports drink?

I am trying to compare the levels of glucose in various sport drinks. So I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the colouring from it before placing in the spectroscope).

I will prepare standards by adding 5 grams of glucose to 100 mL of water, 10 grams of glucose to 100 mL water, and so on.

As a high school student I only have limited resources. Any information about my proposed plan or alternatives that are low-cost are appreciated.

Sounds like a great experiment!

As you state, a sucrose solution in water eventually hydrolyzes to fructose and glucose, but without a catalyst, the reaction rate is extremely slow; beverages are stable at room temperature for years. However it is hydrolyzed in hours at room temperature by adding sucrase enzyme (secreted in the small intestine), invertase enzyme (derived from yeast) or even a weak acid such as citric acid, tartaric acid, or even dilute hydrochloric acid.

You may need to test whichever of these are available for compatibility with Benedict's solution, though, and you may need to test some of the other ingredients in the drinks to see if they also cause false positive otr negative effects. You might compare the results from the Benedict's test with those from a blood-sugar glucometer (the solutions may need to be diluted in both tests to stay within a measurable range).

BTW, study of yeast fermentation of sucrose is responsible for the name enzyme, meaning leavened with yeast, so one could say invertase is the epitome of an enzyme.

Please share your experimental results here.


Can carbon and titanium form an ionic bond? Can carbon and titanium form an ionic bond because they both have four valence electrons? If titanium transfers four electrons to carbon, we get $\ce{Ti^4+}$ and $\ce{C^4-}$ which both have noble gas configurations.

Yes, carbon can form ionic bonds with Titanium. $\ce{TiC}$ is titanium carbide, an ionic compound with $\ce{C^{4-}}$ (the carbide anion) and $\ce{Ti^4+}$ (titanium cation). It has the same crystal structure as $\ce{NaCl}$ (which probably has to do with the $\ce{1\!\!:\!1}$ ratio of anion to cation).

This is a space filling model of titanium carbide:

Here is sodium chloride for a crystal comparison (click the image for a larger version):

The second model (the $\ce{NaCl}$) shows the ionic bonds as covalent bonds but still, both titanium carbide and sodium chloride have a cubic lattice structure.

However, carbon does not often form ionic bonds with metals since it prefers to be covalent (much like oxygen). In fact carbon has much more covalent character than oxygen does which is why oxides form easily (just $\ce{O2 + element}$ and usually, oxides will form) but carbides don't (a catalyst is needed to form carbides and some oxides like gold oxide).


Bond Potential Energy Lowering "Bonds will form only if the potential energy will be lower when the
atoms are bonded than when they are separate."

Is this statement true? If so, how do endothermic reactions exist? Can't you add energy to a system to make the bonds form albeit unfavorably?

Two different factors play a role when considering a reaction. The energy of the bonds etc. (enthalpy) and the entropy. Together, these two components make up what is called the 'Gibbs Free Energy'. (Wikipedia)

The change in Gibbs free energy ($\Delta{}G$) of a system must be $< 0$ for a reaction to be spontaneous. From this it is possible to see that a reaction where the entropy increases drastically, but the enthalpy increases. In this case a reaction can be endothermic.

A case of terminology: The 'more correct' term (my teacher always said) is exergonic or endergonic. What this means is that the reaction has a $\Delta{}G < 0$ or $\Delta{}G > 0$ respectively.

As @Cohorent mentioned before, it is also possible to have a system that is larger than just the chemical reaction (adding a battery for example). Meaning that the reaction is maybe endergonic, but a different energy source is used to drive the reaction. Note then, that the $\Delta{}G$ for the entire system is still below zero.


How could you dissolve a coating of CaCO3(s) from a metal plate? A metal plate has an unwanted coating of $\ce{CaCO3(s)}$. You must find a way to clean this metal plate without causing any scratches or abrasion. How might you dissolve the coating?

I know I should find a substance that will dissolve the coating, and then soak the plate in it. How do I determine an appropriate substance?

It might take some time, but I would use vinegar.

The acid in vinegar is strong enough to dissolve the calcium carbonate, but it is not strong enough to dissolve metals (unless the metal reacts with water).


Melting point of obsidian I have tried 4 experiments at various temperatures to melt obsidian.
At varying temperatures it just bubbles up and I don't seem to reach a melting point.

Do you know what gases obsidian might contain and would they be the cause of the bubbles?

Obsidian

This is a volcanic glass, formed when rhyolitic lavas cool too quickly for crystals to form.  From a chemistry point of view, it's a mixture of silicon dioxide, aluminium oxide, with sodium ,potassium, calcium and iron oxides in various quantities. These lavas will also contain significant volatiles (water and $\ce{CO2}$) held in solution by pressure, although they should be lost at the pressures required for glass formation (higher pressures means slower cooling due to more insulation).

Melting

Note that when molten, rhyolite lavas can have a very high viscosity. So it's possible that you have melted your sample, but just didn't give it enough time to flow. Indeed, with substances like this you will see partial melting - some components will melt at a lower temperature than others. If you are getting bubbles, this would indicate that your glass formed under enough pressure to retain some $\ce{CO2}$ and/or $\ce{H2O}$ in its lattice, and you are now allowing this to be released by heating at surface pressure.  This is all a bit speculative, without a closer inspection.


How to synthesize Boron tri-fluoride using Boron Oxide I am finding different equations for the synthesis. My lecture handout suggests a reaction with Sodium tetrafluroborate and Sulfuric acid. This is fine but I have to remember this for an exam and that reaction looks extremely hard to remember/balance in an exam. A second way seems to be reaction of boron oxide with Calcium fluoride and Sulfuric acid. However, I cannot find consistent equations for this...

Wikipedia simply suggests that Boron oxide reacts with $\ce{HF}$ which is made in situ by the reaction between sulfuric acid and calcium fluoride:

$$\ce{B2O3 + 6HF -> 2BF3 + 3H2O}$$

It's easy to remember but I am reluctant to accept this because it excludes two of the reactants (sulfuric acid and calcium fluoride). Would it be acceptable to merge the formation of $\ce{HF}$ into the above equation - thus including all reactants:

$$\ce{B2O3 + 3CaF2 + 6H2SO4 -> 2BF3 + 3H2O + 3Ca(HSO4)2}$$

It's a little tricky to balance but not too hard to remember. Is there anything wrong with the above equation? I haven't seen it anywhere but only variations on it.

The reaction of boron trioxide with sulfuric acid and calcium fluoride to obtain boron fluoride is given here.

$$\ce{B2O3 + 3CaF2 + 3H2SO4 ->[\Delta] 2BF3 + 3CaSO4 + 3H2O}$$

Conditions

  Boron trioxide react with calcium fluoride and sulfuric acid to produce boron trifluoride, calcium sulfate and water. Sulfuric acid - concentrated solution.

The reaction will be easy to understand if you break the reaction into two reaction.

Babounet says:-

  $\ce{CaF2 + H2SO4}$  will give you $\ce{HF}$ so why bother if you add directly the $\ce{HF}$ instead ?

Indeed $\ce{CaF2 + H2SO4}$  will indeed yield  $\ce{HF}$ along with calcium sulfate as indicated here.

$$\ce{CaF2 + H2SO4 ->[\Delta] 2HF + CaSO4}$$

This $\ce{HF}$ will now react with boron trioxide to yield boron trifluoride as indicated here.

$$\ce{B2O3 + 6HF → 2BF3 + 3H2O}$$

Thus the  products are boron trifluoride, calcium sulfate and water and not calcium bisulfate.


How would I correct this Hess cycle? Given
$$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \quad \Delta_\mathrm{r}H = \pu{-196.0 kJ mol^-1}$$
and $\Delta_\mathrm{f}H(\ce{SO2}) = \pu{-296.0 kJ mol^-1},$ find $\Delta_\mathrm{f}H(\ce{SO3}).$

I have chosen to solve this by constructing a Hess cycle and have come up with the following:

This method has lead me to the correct procedure of solving the equation: $$\Delta_\mathrm{r}H = -2\Delta_\mathrm{f}H(\ce{SO2}) + 2\Delta_\mathrm{f}H(\ce{SO3})$$
However, it is not entirely complete in my opinion and I am unsure on how to proceed in completing it, specifically on what I should place below the two arrows.
I am aware that it should be some combination of $\ce{S + O2}$ but am unaware on how exactly the coefficients would work.

You did the right thing. However the method you used sounds confusing sometimes, it gave me a hard time too. Hess's law says that the resultant enthalpy change in a reaction is same whether it occurs in one or several steps .
We can use that to our advantage by assuming a hypothetical situation where the given reaction proceed in a number of steps for our convinience.
In your example, we disassociated $\ce{SO_2}$ molecule to corresponding standard state constituents and the enthalpy change associated with it will be $-\Delta H_f(\ce{SO_2})$.
$$\ce{2SO_2(g) + O_2(g) ->[-2\Delta H_f(SO_2)] 2S(s) + 2O_2(g) + O_2(g)->[2\Delta H_f(SO_3)] 2SO_3(g)}$$
I took the following reaction through above mentioned 2 steps $$\ce{2SO_2(g) + O_2(g)->[\Delta_r H] 2SO_3(g)}$$
According to Hess law, the Enthalpy of reaction $\Delta_r H$ will be the summation of the all enthalpy changes occuring in the hypothetical intermediate steps.
$$\Delta_r H = -2\Delta H_f(\ce{SO_2}) + 2\Delta H_f(\ce{SO_3})$$


Are there any reactions with no activation energies? Are there any reactions with no activation energies? Our professor just told us there are no chemical cliffs. Is this true?

I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work?

Marcus-Hush theory describes electron/charge transfer rates using conventional transition state theory, e.g.:

$\ce{A+ + B -> A + B+}$

Now one might naively assume that there's no activation barrier in a simple electron transfer, but that's not true. Since the geometry of $\ce{A+}$ and $\ce{A}$ as well as $\ce{B}$ and  $\ce{B+}$ are not identical, there is a reorganization energy of both the molecular species and the solvent or environment. This reorganization energy ($\lambda$) serves as an activation barrier for the charge transfer.

Now the convenient thing about charge transfer reactions is that you can tune the $\Delta G$ of the reactants.

Marcus predicted that the maximum rate would occur when the $\Delta G^0 = -\lambda$ and thus the $\Delta G^\ddagger = 0$. If the reaction becomes more thermodynamically favorable, you enter the inverted region and the activation barrier goes up and the reaction rate goes down.

This is a long-winded answer to say there are some chemical reactions with no activation barrier, in addition to the other cases here. Usually the key phrase describing these in papers is "barrier-less reactions"


How to calculate a molar mass based on density? I have been asked to calculate the molar mass and mole fraction of a 35 w/w% HClO4 solution with a density of 1.3g per ml.

This is the equation I used to try to solve it:

$\frac{1}{M} = \frac{0.35}{100.46} + \frac{0.65}{18.02}$

Is this correct?

What we are talking here are 3 measures of concentration:

mass fraction is about $m_{solute}/m_{solution}$
mole fraction is about $n_{solute}/n_{solution}$
molar mass is about the mass of one mole solution $m_{solution}/n_{solution}$

What we know that the mass fraction is $0.35w/w$ and the density is $1.3g/mL$. You can count the molar mass of water and HClO4 I assume, let's call them $M_{solvent} (M_{H_2O})$ and $M_{solute} (M_{HClO_4})$.

I will use the following equations by the calculations:

$$m_{solvent} = m_{solution} - m_{solute}$$
$$n_{solute} = \frac{m_{solute}}{M_{solute}}$$
$$n_{solvent} = \frac{m_{solvent}}{M_{solvent}}$$
$$n_{solution} = n_{solute} + n_{solvent} = \frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}$$

From the mass fraction you can get the mole fraction this way:

$$\frac{n_{solute}}{n_{solution}} = \frac{\frac{m_{solute}}{M_{solute}}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{1}{1+\frac{\frac{m_{solvent}}{M_{solvent}}}{\frac{m_{solute}}{M_{solute}}}} = \frac{1}{1+\frac{m_{solvent} \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(m_{solution} - m_{solute}) \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(1 - \frac{m_{solute}}{m_{solution}}) \cdot M_{solute}}{M_{solvent} \cdot \frac{m_{solute}}{m_{solution}}}}$$

You can count the molar mass the following way:

$$\frac{m_{solution}}{n_{solution}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solution} - m_{solute}}{M_{solvent}}} = \frac{1}{\frac{\frac{m_{solute}}{m_{solution}}}{M_{solute}} + \frac{1 - \frac{m_{solute}}{m_{solution}}}{M_{solvent}}}$$

You don't need the density for these calculations.

$$M_{HClO_4} = 100.46g/mol$$
$$M_{H_2O} = 18.02g/mol$$
$$m_{HClO_4}/m_{solution} = 0.35$$
$$\frac{n_{HClO_4}}{n_{solution}} = \frac{1}{1+\frac{(1 - \frac{m_{HClO_4}}{m_{solution}}) \cdot M_{HClO_4}}{M_{H_2O} \cdot \frac{m_{HClO_4}}{m_{solution}}}} = \frac{1}{1+\frac{(1 - 0.35) \cdot 100.46g/mol}{18.02g/mol \cdot 0.35}} = 0.088$$
$$M_{solution} = \frac{m_{solution}}{n_{solution}} = \frac{1}{\frac{\frac{m_{HClO_4}}{m_{solution}}}{M_{HClO_4}} + \frac{1 - \frac{m_{HClO_4}}{m_{solution}}}{M_{H_2O}}} = \frac{1}{\frac{0.35}{100.46g/mol} + \frac{1 - 0.35}{18.02g/mol}} = 25.28g/mol$$


What is the meaning of n in n-octane and similar compounds? Straight-chain octane ($\ce{CH3(CH2)6CH3}$) is also called n-octane. What does the n stand for, and where did the term originate?

For reference, see the PubChem entry for octane (beside "Chemical Names:")

In this case, n means "normal", i.e. straight chain. This is somewhat of a holdover from the olden days, but in the case of octane in particular, it makes sense to specify n-octane if that is what you have. For example, the "octane" used as a standard gasoline engine fuel against which gasolines are rated (by definition, 100 octane) is actually not n-octane, but rather 2,2,4-trimethylpentane - a branched isomer of octane - in contrast to the straight chain version.