Cold-sensitive magic pen Yesterday, I bought some "magic" rollerball pens (those with water-based ink), available in red, blue and black.

They look and write like a normal pen, but have an eraser at the upper end, just like a pencil. You can erase the ink with this eraser, but the real miracle is: A few seconds in the freezer at −18 °C, and the ink becomes visible again.

Some additional facts:

The eraser of the pen is translucent and feels wax-like, but is quite hard for an eraser.
It does not rub off anything, you don't get that typical crumbs.
Normal erasers do not have any effects on the ink, just as expected.
It's the perfect ink eraser: You can erase a letter and directly write a new letter on it with the same pen, like a pencil.
The process can be repeated. You can erase the ink and recover it in the freezer several times.

I know those ink erasers from school or those "magic pens" which change the color of the ink of other pens, and also know, how it works.

But this pens are new to me. Has anyone an explanation of their principle of work?

The ink you describe is an example of a leuco dye, which are molecules that exist in colored or non-colored states depending on what they're interacting with and under what conditions (temperature, pH, electronic environment, etc.).

For the ink system, you need 3 components: the electron-donating leuco dye molecules; the electron-accepting molecules (the developer); and a matrix of compounds that stabilize the reaction of the first 2 components for a range of temperatures (meaning the reaction is favored one way above a certain temperature, and another way below a certain temperature).

There is a simple graphical explanation from the notable pen company Pilot, part of which is below. Note that their system uses microcapsules that interact with other molecules in the ink in certain temperature ranges that result in either colored or transparent ink. In the diagram, A is the leuco dye, B is the developer, and C is the stuff that stabilizes the reaction of A and B over a temperature range.

This detailed patent application for "Thermosensitive decolorable ink composition" details the physical and chemical processes involving formulations of leuco dyes. When you add heat to the ink by erasing, the non-colored form is present. When you cool it down, you are letting the system revert to the colored form and the ink reappears.


How to calculate the final temperature of a gas when it undergoes adiabatic expansion? The question is as follows:

  A sample of $\pu{4.0 moles}$ of a gas ($C_{v,m} = \pu{21 J mol-1 K-1}$) has an initial pressure of $\pu{304.4 kPa}$ and an initial volume of $\pu{20 dm3}$ at $\pu{270 K}$. It undergoes an adiabatic expansion against a constant external pressure of $\pu{79.8 kPa}$, until the volume has increased by a factor of $3$. Calculate the final temperature.

I have calculated:  $Q = 0$  and $w = \pu{-3192 J}$ and change in internal energy $=\Delta U = \pu{-31922 J}$.

The next question is to calculate the final temperature of the gas. However, I am unsure on how to calculate this. Originally I thought as $Q=0$, then the final temperature would be the same as the initial temperature. However, I have realized this is incorrect.

Rather than answer the question numerically I have outlined the four different cases, reversible / irreversible and isothermal / adiabatic.

In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work done must therefore be at the expense of the internal energy. If the ‘system’ is a gas then  its temperature will not remain constant during any expansion or compression.

In expansion the work done is $dw=-pdV$ and the change in internal energy $ dU=C_vdT$.

The heat change is zero then $dq=0$ which means from the First Law
$$dU = dw$$
and so
$$ C_vdT = -pdV$$

Dividing both sides by T and for one mole of an perfect gas $p=RT/V$ thus
$$C_v\frac{dT}{T}=-R\frac{dV}{V}$$
If the gas starts at $T_1,V_1$ and ends up at $T_2,V_2$ the last equation can be integrated (and rearranged) to give
$$\ln\left (\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
or
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{R/C_v}$$

using the relationship $C_p = C_v+R$
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{(C_p-C_v)/C_v}$$
Using the gas law this van be rewritten in a more useful form as
$$ p_1V_1^{\gamma } = p_2V_2^{\gamma }$$
where $\gamma = C_p/C_v$. This is also written as $pV^\gamma = const$.

The change in internal energy in an adiabatic process is
$$\Delta U=C_v(T_2-T_1)$$

Adiabatic reversible

In a reversible adiabatic change we use the formulas above to work out what happens. If n moles of a gas fill a container of volume $V_1$ at $p_1$ atm. and is expanded reversibly and adiabatically until it is in equilibrium at a final pressure $p_2$ we can calculate the final volume and temperature. The values of $C_p$ and $C_v$ are assumed to be known and are constant with temperature.

The equation to use is
$$ p_1V_1^{\gamma} = p_2V_2^{\gamma}$$
and as only $V_2$ is unknown this can be calculated.  The work done is
$$w=nC_v(T_2-T_1)$$

Adiabatic irreversible

In an irreversible adiabatic change if n moles of an perfect gas expands irreversibly from a pressure of $p_1$ against a constant external pressure $p_2$ the temperature drops from $T_1$ to $T_2$. We can calculate how much work is done and the final volume.
The internal energy change is $\Delta U= nC_v(T_2-T_1)$
and the work done is $w=p(V_2-V_1)$ and as $\Delta U = w$ (as the system is adiabatic) the final volume can be obtained.  Similarly if the volumes are known the final temperature can be obtained.

Isothermal irreversible

In an isothermal irreversible change the work done on suddenly allowing a perfect gas to expanding from $V_1$ to $V_2$ is determined by the final external pressure $p_2$ and is
$$ q=-w =\int_{V_1}^{V_2} pdV = p_2(V_2-V_1)$$
thus expansion into a vacuum does no work.

Isothermal reversible

A reversible isothermal change performs the maximum possible amount of work, and assuming a perfect gas,
$$ q=-w_{rev} =\int_{V_1}^{V_2} pdV=nRT \int_{V_1}^{V_2} \frac{1}{V}dV$$

$$w_{rev} =- nRT\ln\left(\frac{V_2}{V_1}\right)= - nRT\ln\left(\frac{p_1}{p_2}\right)$$


Cyclopentadienyl radical geometry and MO considerations Does the cyclopentadienyl radical attain a regular pentagon geometry or is there a deviation caused by the Jahn–Teller effect?
Without the Jahn-Teller effect the MO diagram would look like this:

But I guess that a more stable state would be obtained by the Jahn-Teller effect: the orbital marked with number 2 would drop in energy while 3 would rise. What is the true picture?

A typical definition outlining the Jahn-Teller effect is that for any non-linear molecule in a degenerate electronic state, there exists a displacement of the nuclei along at least one non-totally symmetric normal coordinate, that gives rise to a distortion of the molecular geometry with a concomitant lowering of the energy

The underlying cause of the Jahn–Teller effect is the presence of molecular orbitals which are both degenerate and open shell.

The cyclopentadienyl radical, has degenerate and unequally occupied molecular orbitals, thus one can expect to see JT distortions; and indeed there is plenty of evidence to support it, earliest of which comes from the work done by Thrush et. al in 1956.

Experimental data can be found here (along with a list of references to the original works).

A publication I found particularly instructive was this The potential energy surface given below has been taken from there.

This image presents a downward looking view of linear Jahn–Teller distorted lower PES appropriate for $\ce{C5J5}$ . The coordinates $Q_a$ and $Q_b$ denote the two unique dimensions along which motion resolves the degeneracy of the conical intersection corresponding to the $^2E^"_1$ state. These coordinates lie in the space spanned by the four pair of $e^"_2$
normal coordinates, $Q_{ia}$ and $Q_{ib}$. Around the perimeter of the figure are pictures of the distorted $\ce{C5H5}$ geometries for the corresponding value of the phase angle $\phi$.

The work, later goes on to summarise the geometric distortions in the table given below.
The authors also compute $\epsilon_i$ for each of the four modes, and confirm that their sum corresponds to the experimental value of $\epsilon_T$ (this is done for both $\ce{C5H5}$ and $\ce{C5D5}$)


What is the first ionization energy of the hydride ion? What is the first ionization energy for hydride?

Since there are two electrons in the s orbital, I suppose it should be more stable than hydrogen, despite the extra electric charge, so the ionization energy should be greater than the 13 eV needed to ionize hydrogen. However, I could not find any table with the energies for hydride.

You probably don't look at the "ionization energy" for hydride. Instead, you have the reverse reaction:

$\ce{H + e- -> H-}$

You'd measure the enthalpy change here as the electron affinity (EA) of the hydrogen atom. The Wikipedia data page says $72.8\ \mathrm{kJ}/\mathrm{mol}$.

You're basically asking for the opposite of that, so the enthalpy change has the same magnitude but opposite sign. Note that this is not an ionization energy because you're not creating an ion.


Can aqua regia/royal water be produced with sources of chloride and nitrate other than hydrochloric acid and nitric acid? According to Professor Poliakov and a question on Chemistry Stackexchange here, gold dissolves in aqua regia because the nitrate ion oxidizes the gold into $\ce{Au^3+}$, and the chloride bonds to the gold to form tetrachloroaurate III ions ($\ce{AuCl4−}$). This means that the actual acidity of the compound plays no role in the actual oxidization and dissolution of gold. Therefore, if I just provide nitrate ions and chloride ions, the reaction should still proceed (ex: use $\ce{NaNO_3}$ for the source of nitrate ions and $\ce{NaCl}$ for the source of chloride ions). According to the equations given by an answer to the Stackexchange question earlier, aqua regia can react with gold in two ways: $$\ce{Au + 3HNO3 + 4HCl <=>> AuCl4- +  3NO2 + H3O+ + 2H2O}$$

$$\ce{Au + HNO3 + 4HCl <=>> AuCl4- + NO + H3O+ + H2O}$$
And so therefore, by my supposed substitute for the dangerous acids, they can react with gold with the following reactions:
$$\ce{Au + 3NaNO3 + 4NaCl +3H2O<=>> AuCl4- +  3NO2 + 6NaOH + Na+}$$

$$\ce{Au + NaNO3 + 4NaCl + 2H2O<=>> AuCl4- + NO + 4 NaOH + Na+}$$

The sodium hydroxide that is produced by the reaction could be a problem, but we can then add a weak buffer like $\ce{HCO3-}$ or $\ce{H2CO3}$ afterward to deal with it.

Am I correct, or is something else at play in this reaction?

Not even zinc would react with neutral nitrate + chloride, why should gold ?

Aqua regia must be strongly acidic for nitrates to have oxidative properties for oxidation of chlorides to chlorine and to oxidize gold to trace amounts of $\ce{Au^3+}$(what redox potential allows)

Mixing of acids leads to these reactions:

$$\ce{HNO3 + 3 HCl -> NOCl + Cl2 + 2 H2O}$$
$$\ce{2 NOCl -> 2 NO + Cl2}$$
$$\ce{2 NO + O2 -> NO2 }$$

That also means aqua regia dissolution ability decreases with time.

The major oxidative reagent is free chlorine.  Hydrochloric acid saturated by chlorine dissolves gold as well.   Chlorides additionally help gold dissolution  by reaction $\ce{Au^3+ + 4 Cl- <=>> AuCl4^-}$, this decreases the redox potential $\ce{Au/Au^3+}$, making gold dissolution by chlorine and even by nitric acid easier.

$$\ce{2 Au + 3 Cl2 + 2 Cl- -> 2 AuCl4-}$$
$$\ce{Au + HNO3 + 4 HCl -> HAuCl4 + NO + 2 H2O}$$
$$\ce{Au + 3 HNO3 + 4 HCl -> HAuCl4 + 3 NO2 + 3 H2O}$$

See also https://en.wikipedia.org/wiki/Aqua_regia


Is the voltmeter in a galvanic cell in parallel? Often, a galvanic cell is illustrated as having two electrodes, two electrolytes, a salt bridge, and wires linking the electrodes to a voltmeter. However, isn’t a voltmeter connected in a circuit in parallel? This set up makes it seem as if it is being set up in series. How would it work, then?

I would much appreciate it if a circuit diagram (much like the ones you get in physics) could be included in the explanation.

Short answer, to the question title:

Yes, always.

Long answer:

  Often, a galvanic cell is illustrated as having two electrodes, two electrolytes, a salt bridge, and wires bridging the electrodes to a voltmeter.

The piece missing from your list of components, in terms of a galvanic cell being put to practical use, is the load.  You're correct, if a load were attached to the cell, it would be wired in parallel with the voltmeter.

A quick search turns up the following schematic along the lines of your description, from the UC Davis ChemWiki:

From a brief scan, the construction of the half-cells and salt bridge look okay to me. I have not double-checked the half-cell potentials for accuracy.

If I read the schematic correctly, there is no load attached, and so the only thing completing the circuit with the galvanic cell is the voltmeter.  In this case, the appropriate cell voltage would be read on the voltmeter. However, the current through the system would be negligibly small, due to the extremely high resistance of the voltmeter. The $\leftarrow e^-$ notation here indicates only the direction that current would flow through a lower-resistive load that were connected (again, yes, in parallel) with the voltmeter.


Since aqueous acids release Hydrogen cations into the solution, will an electric current create Hydrogen gas? I was wondering that since acids dissociate in water to make a lot of  $\ce{H+}$ cations. Is it possible that introducing a supply of electrons via an electric current, would create $\ce{H2}$ gas?

Yes.  In fact the first electrolysis was done with acidified water.  Pure water is only slightly ionized and is a poor conductor.  Adding ions to water increases its conductivity and speeds up the electrolysis process.  So by adding a source of protons to pure water the following reaction will occur at the cathode $$\ce{2H+ (aq) + 2e^{-}->H2(g)}$$  Of course a parallel reaction will occur at the anode $$\ce{4OH- ->2H2O + O2(g) + 4e-}$$ otherwise a current wouldn't flow and the electrolysis would not continue.


NMR spectrum of 3-hexylthiophene: why is the methyl group not a triplet? I have an example spectrum, and it shows which peaks correspond to which H, but I don't understand why the spectrum looks like that

e.g. I would expect the peak at 0.88 to be a triplet based on the (n+1) rule, why is it such a large multiplet?

The reason why the terminal methyl group is not a simple triplet is because of strong coupling, which is the scenario whereby the coupling constant between two nuclei $J$ is comparable to the difference in resonance frequency $\Delta \nu$.

As labelled in your question, protons B and C have almost the same resonance frequency (0.01 ppm difference corresponds to a 3 Hz difference on a 300 MHz spectrometer). The coupling constant between these protons, though, is probably on the order of 7 Hz or so. Hence, protons B and C are considered to be strongly coupled.

What does this have to do with protons A, which your question is asking about? Well, by virtue of the fact that protons A are coupled to protons B, which are in turn strongly coupled to protons C, there will be some "extra" coupling observed in the peak corresponding to proton A. The term for this is virtual coupling.

It turns out that this is the case no matter what nuclei they are or exactly how they are connected in the molecule - as long as X and Y are strongly coupled, any nucleus A that is coupled to either X or Y will have some degree of virtual coupling. The reason for this is fundamentally quantum mechanical in nature. Suffice it to say, though, that the n+1 rule will not predict multiplet shapes accurately under such conditions.

Further reading on Hans Reich's excellent website. In particular you may want to take a look at the "Methyl Region" section which is pretty similar to your example (although in your case it is actually even more complicated).


Chemical Reaction Alternative to O2 Respiration in H2/N2 Atmosphere Note: If this question does not belong on this Stack Exchange, please direct me to a more appropriate Stack Exchange.

This is a "creative chemistry" question as much or more-so than a "give me the facts" type of question. Should be of interest to anyone who finds the topic of plausible alternative biochemistries intriguing.

I recently came across several papers outlining a hypothetical type of photosynthesis that would be highly efficient in a Hydrogen ($\ce{H2}$) dominant atmosphere containing Methane ($\ce{CH4}$). The photosynthetic reaction is:

$$\ce{CH4 + H2O + $y$ -> CH2O + 2H2}$$

"$y$" represents energy imparted by photons. "$\ce{CH2O}$" is the simple organic compound formaldehyde, which could be used to build biomass.

This made me wonder how "animals" in such an environment would obtain the energy needed to live. In an "$\ce{O2}$" heavy atmosphere the answer is simple: use "$\ce{O2}$" to oxidize the available organic compounds, getting a ton of energy. This is aerobic respiration. What if there is essentially zero available "$\ce{O2}$"? Which brings me to...

My question in detail: What is a plausible chemical reaction (or series of reactions) that could take the place of $\ce{O2}$ respiration in a "Hydrogen Dominant" world with the following molecules and compounds to work with:

ATMOSPHERE

$\ce{H2}$ (10-90%)
$\ce{N2}$ (10-90%)
$\ce{H2O}$ Vapor (≈1%)
$\ce{CH4}$ (0.01 - 5%)
$\ce{NH3}$ Vapor (0.01 - 5%)
Other trace to minimally-present compounds may include $\ce{CO2}$, $\ce{Ar}$, etc.

BIOMASS

Built off of $\ce{CH2O}$ (the organic product of hydrogenetic
photosynthesis)

WATER & LAND

Bodies of water containing a mix of  $\ce{H2O}$ + $\ce{NH3}$, and plausibly available compounds in the crust (like iron).

By "take the place of" I mean provide enough energy to run complex life-processes. I do not mean superficially resemble aerobic respiration (i.e. The process need not require the "breathing in" of atmospheric gases. It could, but need not).

The answer should try to meet the spirit of the following criteria:

The reaction(s) could make use of compounds in the air (breathing) or could be obtained through other methods (eating, drinking) in sufficient quantity to run biological processes.
Impart significant amounts of "free" energy, more so than other plausible reactions.
Make use of available compounds (see details above).
Plausibly occur at temperatures and pressures not inimical to carbon life. $\pu{500\\K}$ temperatures or pressures of $\pu{100\\bar}$ are no-goes.
Be "plausibly likely" to evolve. This requires a measure of judgement. If the chemical reaction or series of reactions is wildly complicated, far-less efficient than other alternative reactions, or needs to make use of rare materials, then it's probably not plausible.

I'd love the form of the answer to contain the equation, a brief discussion on why it's plausible, and a discussion of the amount of energy released, and/or other "good to know" things.

Thanks!

References:

Photosynthesis in Hydrogen-Dominated Atmospheres – https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4284464/

BIOSIGNATURE GASES IN H2-DOMINATED ATMOSPHERES ON ROCKY EXOPLANETS – https://iopscience.iop.org/article/10.1088/0004-637X/777/2/95/meta

A BIOMASS-BASED MODEL TO ESTIMATE THE PLAUSIBILITY OF EXOPLANET BIOSIGNATURE GASES – https://iopscience.iop.org/article/10.1088/0004-637X/775/2/104#apj480437s4

The obvious approach to is to just do the direct analog of what Earthling metabolism does: exactly reverse the photosynthetic reaction (on net, anyway; the intermediate processes may be rather different, as they are on Earth).

In this case, that means reacting hydrogen with carbohydrates to get back methane and water. Based on this paper on photosynthesis in hydrogen-dominated atmospheres, building carbohydrates from methane and water requires considerably less energy than building them from carbon dioxide and water, so we'll equally get less energy out by decomposing them again--I'd estimate somewhere around a quarter to a third as much power density as Earthling animals relying on aerobic respiration, if we assume organisms in this environment will be building biomolecules optimized for this environment, rather than just copying ours. Given how much time I spend every day not eating, and not hyperventilating, I'm pretty sure that would still be enough to power complex animals; they just might need to eat more / breathe more / be more lazy than equivalent Earth animals. Even if you only get one fifth the power density (based on the Bains paper's estimate that hydrogenic photosynthesis can proceed with 20% of the energy required by oxygenic photosynthesis), that's still 7-8 alien-equivalent-of-ATPs per glucose, as opposed to the 2 ATPs per glucose provided by anaerobic respiration (cf. 38 ATPs-per-glucose for aerobic respiration).

As Andrew noted, however, land animals will have an extra energy supply available in the form of reducing N2 into ammonia--i.e., they can supplement their energy supply simply by breathing, and only really need to eat for the materials. Incidentally, autotrophs could also make use of excess ammonia production to power chemosynthesis in the absence of light.

Despite the relatively low abundance of ammonia in the atmosphere, however, it has much higher solubility in water than hydrogen does, and thus may be more easily available to aquatic organisms than free hydrogen is. That gives you a slightly modified "respiration" reaction that takes 3 formaldehyde units and 4 ammonia molecules to produce methane, water, and free nitrogen. Decomposing ammonia into hydrogen and nitrogen is endothermic, however, so this would be less energetic than hydrogen respiration. Not sure how well that would be balanced by the reduced energy needs for pumping water over the gills compared to processing enough water to get the equivalent quantity of free hydrogen. This is a Good Thing for the long-term stability of the atmosphere, as otherwise you'd expect life to just keep producing ammonia until either nitrogen or hydrogen is exhausted.

Now, that is not the most energetic set of reactions imaginable. Obviously, if they had access to oxygen, that would be way better. But, it's the only option that is consistently renewable. Other potential oxidizers, which are more oxidized than the plant and animal biomass that serves as food in this scenario, won't be produced by plant life (by definition--anything that is produced by plantlife is biomass available as food, and thus not more oxidized than itself!) and thus can only be regenerated in small quantities by geochemical processes, and will have very short environmental half-lives. E.g., if creature could eat lithium metal, they could react that with nitrogen, ammonia, water, or complex organic biomolecules and get energy out of any of those reactions--but you just aren't going to find large quantities of free lithium to power your biosphere!

For some back-of-the-envelope estimated specifics, let's initially suppose that plants are hydrogenically producing glucose and animals are hydrogenating glucose. That leads to the following metabolic reaction:

${C_6H_{12}O_6}^{174 kJ} + 12{H_2}^{0 kJ} \rightarrow 6{CH_4}^{-74.85 kJ} + 6H_2O^{-285.83 kJ}$

Which provides approximately 2338 kJ/mol of glucose, based on the heats of formation for each reaction component as listed in the NIST Chemistry WebBook; the net energetics for aerobic respiration are found just by replacing methane with ${CO_2}^{-393.51kJ}$ in that equation (diatomic oxygen, like diatomic hydrogen, conventionally has a heat of formation of 0, so that doesn't make any difference to the calculation), which gives us a theoretical output of 4250kJ/mol. (Looking up the actual empirical energy yield of aerobic respiration of glucose, we get 2880 kJ/mol--so, assume that all other energy calculations here are only good to within about a power of two!) So, if hydrogen-breathing animals can just manage to evolve a slightly more efficient energy system than we have (which seems plausible, especially if they are living at lower temperatures which permit better thermodynamic efficiency), they may even get up to a full half the power density of oxygen-breathing Earthlings. Given the seriously speculative and highly imprecise nature of these calculations, though, I'm more comfortable sticking with my earlier 1/3 estimate as a plausible ceiling.

Plants in a hydrogen-dominated atmosphere aren't necessarily going to be producing as much glucose as our plants do in our highly oxidized environment, so let's look at some other options for biomolecules that might be available for animals to eat. Suppose, for example, that plants produce acetylene as an energy storage molecule; hydrogenating that provides

$C_2H_2^{226.73 kJ} + 3H_2^{0 kJ} \rightarrow 2CH_4^{-74.85 kJ}$

376 kJ/mol of acetylene -- or about 14.5 kJ per gram. About half as much per gram as glucose.

And if we look at the above-mentioned ammonosynthesis reaction, using hydrogen and nitrogen directly from the atmosphere, we get

$3H_2 + N_2 \rightarrow 2{NH_3}^{-45.9 kJ}$

about 91 kJ/mol of nitrogen, or 5.3 kJ/gram of inhaled hydrogen/nitrogen mix. That's not a ton compared to glucose (or acetylene), but there's just so much more of it available, for cheap! After all, it takes quite a lot energy just to consume and digest food--and practically none to just breathe it in.

Also consider that, largely because of the above reaction, nitrogen is far more easily bioavailable in this world than on Earth, so maybe they use a lot more cyanide groups or even azides. Hydrogenating hydrogen cyanide gets you

${HCN}^{135.14 kJ} + 3{H_2}^{0 kJ} \rightarrow {CH_4}^{-74.85 kJ} + {NH_3}^{-45.9 kJ}$

255.75 kJ/mol, or about 9.5 kJ / gram.

Simply decomposing hydrogen azide, even before you consider the energy gains from hydrogenation, gets you

$2{HN_3}^{264 kJ} \rightarrow {H_2}^{0 kJ} + 3{N_2}^{0 kJ}$

about 6 kJ/gram.

Those last few reactions (somewhat surprisingly, given the notorious explosivity of azides) would tend to drive the average energy density of food down a smidge from the optimistic third-to-half estimates, so once again about a quarter of the total energy density of aerobically-metabolized Earthling food seems reasonable.


Does CH3CHNH exist, and what is it called? I am drawing different Lewis Structures of small compounds containing nitrogen, hydrogen, and carbon. One permutation is $\ce{CH_3CHNH}$, but I have never heard of this compound. Initially, and falsely, I thought this compound was acetonitrile. It isn't coming up in Wolfram Alpha, Pubchem, or Wikipedia.

Does this compound exist? Also, what would this compound be called regardless of its existence value?

The -CH=NH functional group is called an imine (in this case a primary imine) your molecule would be called Ethan-1-imine


What is happening to my chlorine dioxide when I add too much bleach or citric acid? I have some sodium chlorite and Chinese mystery powder which are meant to be mixed with water, then mixed together to form chlorine dioxide. I wanted to try a more food safe (and less mysterious) route for making small amounts of chlorine dioxide, so I diluted the sodium chlorite, then mixed in some bleach and a tiny bit of citric acid. I don't have a firm understanding of the reactions, but know $\ce{Cl2}$ and hypochlorous acid will react with sodium chlorite, but plain bleach may not, so the acid is to acidify the bleach and shift its equilibrium away from being hypochlorite. It worked, as evidenced by a pale yellow color. (It was very dilute, which is not a bad thing, for this chemical).
What I don't understand is that when I add more citric acid or more bleach, the color turns clear again. Either the chlorine dioxide is reacting or coming out of solution. What are the reactions with excess citric acid or bleach that might eliminate $\ce{ClO2}$?

The "mystery powder" was proposed to be sodium bisulfate, in a recent answer to another question. Sulfuric acid is used in one process to turn sodium chlorite into $\ce{ClO2}$; bisulfite would be slower, but should do the same.
Citric acid will acidify the sodium chlorite and sodium hypochlorite, but will itself be oxidized, perhaps all the way to $\ce{CO2}$. $\ce{HCl}$ is used to convert $\ce{NaClO2}$  and $\ce{NaOCl}$ to $\ce{ClO2}$ and citric acid could well do the same (be oxidized at the same time).
$$\ce{2 HCl + NaClO + 2 NaClO2 -> 2 ClO2 + 3 NaCl + H2O}$$
$$\ce{4HCl + 5NaClO2 -> 4ClO2 + 5NaCl + 2H2O}$$
Chlorite ion and hypochlorite ion are known to react rapidly to form chloride and chlorate (Lister, Can. J. Chem., 30, 879-889, 1952).
The unknown amounts of oxidizing agents and citric acid (a reducing agent) and the complex reactions make the pale yellow color a poor indicator of what might be going on. The presence of $\ce{ClO2}$ does not seem to be positively identified and might not even be needed to explain the observations.


How can x-rays show how many protons are in a nucleus? My textbook states that: Henry Moseley found using x-rays that the atomic number of each element has a characteristic positive charge.
I don't know much about x-rays and can't find an answer online or elsewhere. How does this technique work ?
My thoughts are that maybe all protons emit x-ray photons at the same frequency and rate. If this is true then some kind of x-ray detector could be used to detect the energy at that x-ray frequency that a nucleus emits. The number of protons could then be calculated.
I could be way off here.
Can someone explain how x-rays can be used to count the number of protons in a nucleus?

X-rays are not emitted by protons. They are emitted by atomic electrons when they fall down from an upper orbit (or level) to another one near the nucleus. Of course the electrons have first to be excited (= sent to this highly excited level). But whatever the way of exciting the electron, the energy of the transition corresponding to the return to the lowest level is proportional to the number of protons in the nucleus. If you are able to measure the energy of the emitted X-ray produced during this transition (or fall), you can calculate the number of protons of the nucleus. I am sorry. I have used a rather simple language. Pure theorists will prefer using a better vocabulary, which may not have been understood by Kantura. I may be wrong.


Propan-1-ol Heats Objects Faster than Propan-2-ol. Is this an anomaly? After conducting a practical revolving around burning alcohols, I found that propan-1-ol heated the 50 mL of water to 60 °C about 1 min faster than propan-2-ol.

The experiment was controlled as much as possible, including the temperature escaping the setup.

I know that propan-2-ol is more volatile than propan-1-ol, so I have a reason to believe that this is tied to the slower heating. On the other hand, I have yet to find any sources to verify this.

If I do not find any more about this, I will have to call this out as an anomaly.

If anyone knows anything about the relationship between the position of the $\ce{OH}$ bond in an alcohol and the heat output/speed of heating, please explain this to me.

The enthalpy of combustion $\Delta_\mathrm cH$ of a substance can be calculated from the enthalpy of formation $\Delta_\mathrm fH$. For a compound containing only carbon, hydrogen, and oxygen, the general combustion reaction is

$$\ce{C_$a$H_$b$O_$c$ + ($a$ + 1/4 $b$ – 1/2 $c$) O2 -> $a$ CO2(g) + 1/2 $b$ H2O(l)}$$

The corresponding standard enthalpy of combustion is

$$\Delta_\mathrm cH^\circ=-a\Delta_\mathrm fH^\circ(\ce{CO2,g})-\frac12b\Delta_\mathrm fH^\circ(\ce{H2O,l})+\Delta_\mathrm fH^\circ(\ce{C_$a$H_$b$O_$c$})$$

This equation applies if the reactants start in their standard states and the products return to the same conditions. The standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$). The definition of standard state makes no reference to fixed temperature; thus, the standard enthalpy of formation and the standard enthalpy of combustion remain functions of temperature. The most widely used reference temperature is $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$.

The following values for the standard molar enthalpy of formation at $p=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ are taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align}
\Delta_\mathrm fH^\circ(\text{propan-1-ol},\ \mathrm l)&=-302.6\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\text{propan-2-ol},\ \mathrm l)&=-318.1\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{H2O, l})&=-285.8\ \mathrm{kJ\ mol^{-1}}\\
\Delta_\mathrm fH^\circ(\ce{CO2, g})&=-393.5\ \mathrm{kJ\ mol^{-1}}
\end{align}$$

During your experiment, the products certainly did not completely return to the initial temperature. Anyway, for the difference in enthalpy of combustion of propan-1-ol and propan-2-ol, the exact state of the products is not relevant (provided that the conditions are reproduced) since the produced amounts of $\ce{CO2}$ and $\ce{H2O}$ are identical. The remaining difference is caused by the difference in enthalpy of formation of propan-1-ol and propan-2-ol.


Boat conformation: axial hydrogens In my exam, my teacher asked me if the following is true:
"The chair cyclohexane has more axial hydrogens"
I did not have time to think, so I just put false.
And apparently I was right.
But, at home, I thought to myself: Does this question even make sense? Axial hydrogens are by definition only seen on cyclohexanes according to what I know.

It says fully: TRUE OR FALSE: Boat cyclohexane is a slightly higher
  energy conformer than chair cyclohexane because: (a).... (a sub
  question) T OR F (b).... (a sub question) T OR F (c)chair cyclohexane
  has more axial hydrogens. T OR F

The answer to part (c) of that question is "False."  It turns out that the chair and boat conformations of cyclohexane both have 6 axial hydrogens - so the chair does not have more axial hydrogens.  Here's a picture of the chair (on the left) and boat (in the middle), you should be able to see the 6 axial hydrogens in both the chair and the boat.


How to perform molecular dynamics simulations of charged systems? The GROMACS official documentation (see here) states that a system with non-zero total charge will yield an error:

System has non-zero total charge
Notifies you that counter-ions may be required for the system to neutralize the charge or there may be problems with the topology.
If the charge is a non-integer, then this indicates that there is a problem with the topology. If pdb2gmx has been used, then look at the right hand comment column of the atom listing, which lists the cumulative charge. This should be an integer after every residue (and/or charge group where applicable). This will assist in finding the residue where things start departing from integer values. Also check the capping groups that have been used.
If the charge is already close to an integer, then the difference is caused by rounding errors and not a major problem.
Note for PME users: It is possible to use a uniform neutralizing background charge in PME to compensate for a system with a net background charge. There is probably nothing wrong with this in principle, because the uniform charge will not perturb the dynamics. Nevertheless, it is standard practice to actually add counter-ions to make the system net neutral.

If I understand this correctly, it means that a uniform background charge will be added to a net-charged system. What if I wanted to study a charged system, e.g. a peptide with basic groups in the gas phase, where counter ions are absent. Is it still possible to perform a gromacs calculation on such a system and how will the introduction of the compensating background charge affect the molecular dynamics?
I then looked into other packages and noted that net-charged systems seem to be generally problematic in mm calculations. Can anyone point me into a direction like literature on this?

This is intrinsic to Ewald summation methods, not software implementations. The uniform charge arises from neglect of a reciprocal sum term. It does not directly affect the dynamics and may be a reasonable model of a spatially homogeneous system. See https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2015-October/101544.html for further details and https://www.mpibpc.mpg.de/14063977/Hub_2014_JCTC.pdf for a description and explanation of the likely artefacts.


Brass Instrument: Dezincification or just scrubbed off? I do own a brass instrument (a trombone). The lacquer is damaged.
I asked how to deal with it here.

Now I learned that the brass contains zinc, and there is a corrosion process called "dezincification". Now they told me to ask here how to identify real dezincificated parts. Here are the pictures:

Main slide handle:

Bell "middle" section:

Bell inner rim:

The issue is: On trombones (other brass instruments) red rot (dezincification) usually only appears on parts where saliva touches the instrument, i.e. in the front parts. On a trombone that would be the slide. So it is kind of unlikely but not unheard of that this might be "red rot" or dezincified.

It might also just be lacquer which went off due to improper handling (like not cleaning water spots, showering/cleaning the trombone with way too hot water, etc.).

Actual questions

How do I identify dezincifiaction correctly? Can it be seen on those pictures? (most important question)
What do you think of chemical cleaning? It sounds dangerous to me. Isn’t it going to damage the instrument?
What do you think of cleaning such an instrument soap? Is this going to damage the lacquer or material?

Brass Instrument: Dezincification or just scrubbed off?

    How do I identify dezincifiaction correctly? Can it be seen on those pictures? (most important question)

Instruments are made from solid brass, this is required as a stiffer metal like brass is required to have good sound quality. Your brass was not scraped off and the spots are indeed a result of dezincification. Water could get here for a number of reasons from fog mist to moist hands, it does not have to be saliva only.

  What do you think of chemical cleaning? It sounds dangerous to me. Isn’t it going to damage the instrument?

From what I can find chemical cleaning looks like they use a mild acid and chelatior (like citric acid or EDTA) to etch away some of the red rot. However the brass still has to be polished and the lacquer needs to be reapplied.

  What do you think of cleaning such an instrument soap? Is this going to damage the lacquer or material?

Soap won't damage the lacquer, but scrubbing might. That said your red rot shows that the lacquer is already damaged to you might as well scrub/polish the spots and reapply lacquer.


What are the factors behind making ice faster? Recently I saw a video in YouTube about how to make ice in seconds.

In this video, salt is added to a glass of water and then a straw is closed from one end and the other side is lit using a lighter and after which it is stirred for 5 seconds and removed.

As soon as it is removed the ice is formed.

What is the chemical reaction that forms ice so quickly?

The simplest explanation would be that it isn't (pure) water at all. It is most likely a supersaturated solution of sodium acetate - $\ce{CH3COONa}$ (or sometimes sodium thiosulfate - $\ce{Na2S2O3}$) in water.
A supersaturated solution is a solution that contains more solute than normally soluble at those conditions. So as soon as a rough surface is provided for nucleation (formation of crystals), the solute starts crystallizing and comes out of solution as a big crystal.
The rough surface used over here is most probably the rough end of the burnt straw. This produces enough disturbance to cause crystallization. Usually an already prepared crystal of pure sodium acetate is used for nucleation but only if you want high purity crystals.
This is a pretty common trick (also the original video is from trick life -- as mentioned on the youtube description --) and most people who don't know about this phenomenon are usually easily tricked (then again, the "ice" formed would actually be hot since it is an exothermic reaction but you can't feel it through a video right :) ).
See this page and this page if you want more info.


Why Gaussian soft (STO-3G) causes error for Helium anion? Software: G09.Rev.D.01

=======gjf-file======

%chk=F:\Hanion.chk
# hf/sto-3g geom=connectivity

Title Card Required

-1 2
 He                 0.20408162    0.07288005   -0.00095405

 1

Causes error message #2070
Log file explanation:

 There are   2 occupied orbitals but only   1 basis functions!

But the same time for the Lithium which is the same electron structure, method STO-3G work fine.

Why Gaussian soft (STO-3G) causes error for Helium anion?

An orbital of a closed-shell Hartree-Fock calculation can only contain two elctrons, one with alpha ("up") and one with beta ("down") spin.

You are calculating the Helium-anion. In a minimal STO-3G basis, it comes with one(!) basis function (or, loosly speaking, "orbital"), as it has two electrons in its neutral form. The anion however has three electrons. So, one closed-shell orbital is certainly not enough to host three electrons. The calculation works for Lithium a.) because gaussian automatically assumes that you want to perform an unrestricted-calculation as the number of electrons is uneven (this is also the case for the helium-anion, but I am not sure if you are aware of this). Furthermore,  Lithium has three electrons "by nature" (in its neutral state) and therefore, two basis-functions are enough to host four electrons.

You are using the STO-3G minimal basis, which adds one basis-function per two electrons of the system in the closed-shell case and one per electron in the open-shell case. Of importance is not the actual number of electrons in your system but the number of electrons the neutral atoms composing your system have. For a quick fix, substitute "hf/sto-3g" with "uhf/cc-pVDZ" and your calculation will run without issues.

Your system has an uneven number of electrons, so you cannot run a restricted (closed-shell) HF calculation. Either perform an unrestricted Hartree-Fock (UHF) or a restricted open-shell Hartree-Fock (ROHF) calculation. Both are available in gaussian09. As said, gaussian is already automatically doing this as a restricted HF calculation of a system with an uneven number of electrons is not possible.

Furthermore, as the calculation of the Helium-anion is quite fast, you should considering using a larger basis. You can start with "cc-pVDZ", a Dunning correlation-consistent basis-set, and move on from there.

For a comprehensive introduction to electronic-structure theory I suggest you read the book by Szabo and Ostlund. It assumes no previous mathematical or chemical knowledge above high-school-calculus und explains all the conceptual issues I have just mentioned in a detailed and understandible manner. Also, this book is quite affordable.
http://store.doverpublications.com/0486691861.html

If this is too much, I suggest you at least look up the Roothan-Hall and the Pople-Nesbet equations for running an electronic structure calculation in a basis, as Gaussian builds on them.


Is 2a¹H-cyclopenta[cd]indene aromatic? In this question, 2a1H-cyclopenta[cd]indene is said to be aromatic.
My doubt is that the central C atom (connecting the three cyclic rings) is sp³ hybridized. So I think it should be non planar thus making compound non aromatic

Can someone please explain this compound’s aromaticity?

It's a modified form of [10] annulene (https://en.wikipedia.org/wiki/Cyclodecapentaene) where internal bridges are used to hold the outer, conjugated ring in a planar geometry and thus with the $4n+2$ rule that outer ring is aromatic.

You may be more familiar with the methylene bridged version of [10] annulene that works the same way, but with a simpler bridging structure.  See the above mentioned Wikipedia link.

Addendum:

As mentioned in comments, it is overly simplistic to use the $4n+2$ rule.  IUPAC justifies it only for simple annulene systems.  When you have internal bonds in the ring like the compound here, you no longer have the symmetry and degeneracy structure on which the rule is based.

When you have saturated interior atoms surrounded by an outer conjugated ring, the $\pi$ orbitals are only slightly altered and that is why we are tempted to still apply the $4n+2$ rule.  But beware: if we unsaturate any internal atoms, in this case by removing the hydrogen from the central carbon, the $\pi$ orbitals are altered more considerably and the $4n+2$ rule is well past its use-by date.

Maybe in the end we can use the rule as a guide, but ultimately we need to back our conclusions up with hard evidence.  In this case the conclusion of aromaticity is supported by https://doi.org/10.1039/C39800000691, in which the authors studied substituted versions of the compound involved here.  They found proton n m r shifts consistent with an aromatic outer ring.


Difference between 'Preferred IUPAC Name' and 'Systematic IUPAC Name' When looking at different chemicals, I come across the 'preferred IUPAC name' and the 'systematic IUPAC name'. I've noticed that when I name chemicals, I name them by their 'systematic IUPAC name', as I've been taught to name them using this system.

When is the 'preferred IUPAC name' used? I sometimes come across it when looking at aldehydes, ketones and carboxylic acids. However, the fact that it's 'preferred' makes me wonder whether chemists use this method more often than the other system.

The problem with systematic IUPAC names is that many compounds can have more than one systematic IUPAC name. A compound may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC. Thus, asking for the systematic IUPAC name does not make sense for such compounds.

Therefore, current IUPAC recommendations (currently only for organic chemistry) include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including also the many retained names.

Whether the preferred IUPAC name is also actually preferred by chemists cannot be generalized and depends strongly on the context.


Is a hue indistinguishable from its equivalent frequency of light? A hue is the color characteristic of a pigment, a particular chemical molecule, when illuminated by sunlight.

Of course such a molecule will not emit a single frequency of light, but will emit a spectrum of light at different frequencies and intensities. This depends on the absorption characteristics of the molecule and its emission characteristics. In some cases, for example, light energy can be absorbed at one frequency and re-emitted at different frequency. In any case, what results is a spectrum of color.

The eye perceives this spectrum as being of a single color, called the pigment's hue.
https://physics.stackexchange.com/questions/ask
My question is whether this hue is truly indistinguishable from energy emitted purely at that frequency. For example, the pigment vermillion, mercuric sulfide, has a nominal wavelength of 607nm. If we create a pure light source of that frequency only, is it indistinguishable from the light of the pigment to the eye?

Certainly not

The problem is not, strictly, one of physics but one of perception. The eye doesn't detect individual wavelengths of light but the relative amount of red, green and blue light (but the sensors in the eye cover a relatively wide and overlapping range of wavelengths centred on what we might measure as "pure" blue, green and red. If a "pure" yellow light (say a wavelength of around 580nm) is shone into the eye it will excite both the green and red sensors and the brain will interpret the result as yellow. So the actual colours we perceive will match a pure light of that colour in that case. But we could get the same effect by shining two lights into the eye of pure green and pure red. The brain can't tell the difference and you would still perceive yellow.

But this system is no perfect and we can perceive colours that don't appear anywhere on the spectrum of "pure" colours. An example of this is magenta which is colour that does not appear in an spectrum of pure, single wavelength, light (or on the rainbow). The brain manufactures this colour when red and blue sensors are activated but green sensors are not. A very simple illustration and explanation of this is given in this short Royal Institution video.

This simple examples shows that colour perception does not relate to or match the perceived colour of a "pure" wavelength of light. Perceived colour is more complex that simply matching the wavelength in a light source and there are colours that we perceive that simply don't have a matching wavelength.


How to calculate pH of the Na2CO3 solution given ambiguous Ka values This is the exact question I faced on an exam:

Calculate the $\mathrm{pH}$ of $\pu{0.05 M}~\ce{Na2CO3}. \\(\ce{H2CO3}: K_\mathrm{a_1}= 4\times 10^{-7},~ K_\mathrm{a_2}= 4.7\times 10^{-11})$

Solution
$$\ce{Na2CO3 ->2Na+ + CO3^2-}$$
I suppose nothing that can contribute to the $\mathrm{pH}$ of a solution happens to $\ce{Na+}$ ions and we proceed with $\ce{CO3^2-}$ which has a concentration of $\pu{0.05M}$
$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$
To calculate $\mathrm{pH}$, I need to first figure out concentration of $\ce{OH-}$ ions, and to do so I have to know $K_\mathrm{b}$ dissociation constant for $\ce{CO3^2-}$
On equilibrium concentration of species are as follows: $$[\ce{CO3^2-}] = \pu{0.05 M}-x,[\ce{HCO3-}] = [\ce{OH-}] = x$$
So, we have:
$$\dfrac{x^2}{0.05-x}=K_\mathrm b$$
All I need to know is $K_\mathrm{b}$ value for $\ce{CO3^2-}$ ion. I tried to derive $K_\mathrm{b}$ from $K_\mathrm{a}$ values using $K_\mathrm{a} \times K_\mathrm{b} = 1\times 10^{-14}$ but apparently obtained the incorrect answer. What method should I use to do it the right way?

The  carbonate ion is the Conjugate base of the weak acid
$\ce{HCO_3^-}\ (K={4.7\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H_2CO_3}$ , so the solution of this problem as a solution of a monoprotic weak base:
$\ce{CO_3^{-2} + H_2O <=> HCO_3^-  + OH^-}$
$$\ce{K_b}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{Ka}=\frac{10^{-14}}{4.7\times 10^{-11}}=\ce{10^{-3.7}}$$
Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and thus
  $$\dfrac{[OH^-]^2}{0.05-{[OH^-]}}=K_b= \ce{10^{-3.7}}$$
The equilibrium expression must be solved as a quadratic and yields the root $\ce{[OH^-]}$=0.00306 Which corresponds to pOH = 2.5 or pH = 11.5


Energy in chemical reactions My question concerns how energy is used in chemical reactions. I am working with a reaction between magnesium and hydrochloric acid in a class (although this is not a homework question). What I've found is that the magnesium gives away its valence electrons to either the hydrogen or to the chloride. But in the second case, the chloride ions would first give their valence electrons to the hydrogen, and then they would accept the electrons from a magnesium atom. As I was thinking about which of the cases was more probable, I started thinking about the energy involved. For example, I would think that moving the electrons first from the chloride ions to the hydrogen, and then from the magnesium to the chloride, would use much more energy than the magnesium just giving its electrons to the hydrogen ions.

My question is, as a general rule of thumb, although I wouldn't mind a detailed answer, are reactions that, like this one (as far as I know), have multiple "ways" of reacting going to take the way that uses the minimal amount of energy?  Or are the electrons free to move to whichever atom is closest?

Magnesium cannot give one or two electrons to the chloride ion. Because Chloride ions have the maximum number of 8 electrons in their external shells. There is no possibility for a new electron to stay on a Chloride ion. And the Chloride ion will be reluctant to accept or to loose one of its electrons, because it has the ideal conformation (8 outer electrons). So there are no "multiple ways" of reacting, as you imagine. The only thing that may happen is for magnesium to loose two electrons per atom, because at the end of this departure, it will have 8 outer electrons, and this seems to be to ultimate goal for all atoms. These electrons will be caught by $H^+$ ions, which are transformed into $H$ atoms and then into $H_2$ molecules.


Can titrant and analyte be swapped? I just did a titration experiment at school, our teacher told us multiple times to add the solution of known concentration into the solution of unknown concentration. I am wondering if we can do it the other way, if not, why?

From what I know, you can add the solution of unknown to the solution of known concentration. The only thing that matters is you have a known volume of the solution to which you're adding the other solution.

For example, suppose we're titrating $\pu{0.15 M}$ $\ce{Na2CO3}$ with an unknown concentration of $\ce{HCl}$, if we need to add $\pu{say 15 mL}$ $\ce{Na2CO3}$ to a known volume of say $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can calculate the concentration of the HCl solution.

$$\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$$

$n(\ce{Na2CO3}) = \pu{0.15 mol/L} \cdot \pu{0.015 L}$

$n(\ce{Na2CO3}) = \pu{0.00225 mol}$ (i.e. the known concentration times 15 mL change to L)

We know that 2 moles of $\ce{HCl}$ will react with 1 mole of $\ce{Na2CO3}$ to form the products on the right hand side of the equation. So,

$n(\ce{HCl}) = 2 \cdot n(\ce{Na2CO3})$

which we calculated above using the relation of moles as product of concentration times volume. So

$n(\ce{HCl}) = \pu{0.0045 mol}$

Since we know that we had $\pu{30 mL}$ of a $\ce{HCl}$ solution with $\pu{0.0045 mol}$ of $\ce{HCl}$ in it,

$[\ce{HCl}] = \frac{\pu{0.0045 mol}}{\pu{0.03 L}}$

$[\ce{HCl}] = \pu{0.15 mol/L}$

So, we had a $\pu{0.15 M}$ solution of $\ce{HCl}$.

If we're now adding $\ce{HCl}$ to $\pu{15 mL}$ $\ce{Na2CO3}$ and it takes $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can use similar calculations to once again find the concentration of $\ce{HCl}$ (and get the same answer).

It doesn't matter whether you add the known solution to the unknown one, the only thing you need to know is the volume of the thing you're adding to and how much of the other one you need to add to reach the endpoint.
Think of it like this,
We have 4 variables so to speak;
- concentration of the known solution (we know)

concentration of the unknown (what we want to work out)
volume of the known solution (either adding or added to) which we measure
volume of the unknown solution (either adding or added to)which we also measure

So we know 3 things no matter what we add what to and these 3 values are enough to work out the concentration of the unknown solution. You should clarify with your teacher what she meant. She may have said always add the acid to the base because if we place the base in the burette it may react with the glass, $\ce{SiO2}$ which acts as an acid and that might degrade up the precision for future use.


Calculation of rupture force in a covalent bond Is there any way to theoretically or mathematically calculate the rupture force of a covalent bond?
One of the articles I saw said that it is the maximum derivative of the potential energy curve. Is this true?

A force is the derivative of a potential energy:
$$\vec{F} = -\vec{\nabla} U$$
If you choose to apply a continuous force to stretch a bond until rupture, the force will need to exceed the restorative force (as described say by Hooke's law for small displacements from equilibrium) due to mutual attraction of the atoms. The maximum value of that restorative force is the maximum value of the gradient of the potential, therefore:
$$\vec{F}_{\text{rupture, continuous}} = (\vec{\nabla} U )_{\text{max}}$$
Note a quadratic potential (such as leading to Hooke's law) does not predict rupture. You would require a more accurate description of the internuclear potential to compute the rupture force.


Deciding the order of heat of combustion of isomeric alkanes I'm asked which one of the alkane has the lowest heat of combustion (note: the question says heat, not enthalpy) and according to the answers, the right one is d.

I'm trying to understand why d is the right answer but with no success.
Please help me understand the concept behind it for future problems.

Branched-chain alkanes have lower values of ΔcH⊖ than straight-chain alkanes of the same number of carbon atoms, and so can be seen to be somewhat more stable. https://en.wikipedia.org/wiki/Alkane#Branched_alkanes

The heats of formation are given in https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation  and go from -40.0 kcal/mol for a to -41.8 for b to -42.5 for c to -44.5 for d. Compound d has the largest heat of formation, therefore will have the smallest heat of combustion.

The explanation is complicated. Sometimes you have to wonder why teachers spend so much time on effects that are so minor. The data are interesting, in a trivial sort of way. Show the data; it's interesting; move on.

Highly branched alkanes may stretch the bond angle. This can cause steric hindrance, and can substantially increase the reactivity. "Substantially". That's interesting. In addition, the reactivity of highly branched alkanes can be used to increase the octane rating of gasoline, because the stability of highly branched free radicals slows down the combustion and avoids explosions in the cylinder (pinging). Substantial is the key word, whether it it is reactivity or stability. The difference between 40 and 44.5 is hardly substantial. I would call that difference about half of substantial.

Wikipedia goes on: "However, in general and perhaps surprisingly, when branching is not extensive, branched alkanes are actually more thermodynamically stable than their linear (or less branched) isomers. For example, the highly branched 2,2,3,3-tetramethylbutane is about 1.9 kcal/mol more stable than its linear isomer, n-octane. Due to the subtlety of this effect, the exact reasons for this rule have been vigorously debated in the chemical literature and is yet unsettled. Several explanations, including stabilization of branched alkanes by electron correlation, destabilization of linear alkanes by steric repulsion, stabilization by neutral hyperconjugation, and/or electrostatic effects have been advanced as possibilities. The controversy is related to the question of whether the traditional explanation of hyperconjugation is the primary factor governing the stability of alkyl radicals." Surprisingly subtle and unsettled. Not really a great teaching topic.

The question regularly pops up in secondary school exams. Why? The reasoning seems unsettled. How to answer the question? The teacher must have given the answer if the teacher is at all competent and meant to provide the reasoning; so regurgitate the correct answer without reserving a special place in your memory for it. If the teacher did not provide the answer, then you must do your own research, in advance of the test - first, to discover what kind of questions your teacher is likely to ask (can you search out previous year's tests?) and second, to research the possible answers.

If you do well at this, you may be asked to teach the next class.


How to separate KCl from NaCl and MgCl₂? Having mixed powder of $\ce{KCl}$, $\ce{NaCl}$, and $\ce{MgCl2}$ (and, perhaps, some crap), how can the components be separated?

As I mentioned in the comments, a quantitative separation of the three compounds will probably not be possible outside of a proper laboratory, however a coarse separation is possible. A quick search on the internet yields this link for the separation of $\ce{NaCl}$ and $\ce{KCl}$. You didn't give us the exact composition of the table salt substitute, so I assume $\ce{MgCl2}$ is present in much smaller amounts than the first two compounds.

The simplest strategy is to use the difference between the solubilities of the salts in water or ethanol. Using water is much more convenient, so we'll stick to that. I'll outline the method to obtain the highest purity $\ce{KCl}$. First, add a very large amount of table salt substitute to a container with boiling water (you'll want to add some 200-250 g per 100 g of water) and let the mixture dissolve as much as possible (though it probably won't even come close to dissolving completely). What you have then is a boiling water solution saturated with both $\ce{NaCl}$ and $\ce{KCl}$. Now, while the water is still very hot, quickly filter off the excess salt (might be doable with a fine sieve, or a funnel with some cotton would work better) and reserve the liquid in another container. Let the liquid cool down (using an ice bath is optional but increases yield and purity), and it will start to precipitate the salts. Now you perform another filtration, but this time you keep the solid retained on the filter, not the liquid. This yields purified $\ce{KCl}$.

This strategy works because it turns out that $\ce{NaCl}$ and $\ce{KCl}$ behave rather differently in how their solubility in water is affected by temperature.

$\ce{NaCl}$ soubility in water:

35.6 g/100 g water (0°C)
35.9 g/100 g water (20°C)
39.0 g/100 g water (100°C)

$\ce{KCl}$ soubility in water:

28.0 g/100 g water (0°C)
34.2 g/100 g water (20°C)
56.3 g/100 g water (100°C)

Evidently, as the filtered liquid cools down, $\ce{NaCl}$ remains almost as soluble in cooler water as in hot water, while $\ce{KCl}$ is only about half as soluble. This means that a large amount of $\ce{KCl}$ will precipitate as the filtered liquid cools, while only a bit of $\ce{NaCl}$ will follow. Hence, ideally, from 100 g of water saturated with both salts at 100°C (and containing nothing else), you will be able to extract 25.0 g of a mixture of salts containing 87.6% $\ce{KCl}$ if you perform the second filtration at 20°C. If you filter at 0°C, then the yield increases to 31.7 g of a mixture containing 89.3% $\ce{KCl}$. If for some reason you don't want to do the second filtration, you can just let the water evaporate completely after the first filtration, producing 95.3 g of a mixture containing 59.1% $\ce{KCl}$ from 100 g of saturated boiling water.

In practice your yield and the purity of the $\ce{KCl}$ will likely be somewhat lower due to losses in manipulation and the presence of other impurities. The procedure may be complicated by the very large initial amount of table salt mixture that you have to dissolve in water. You can reduce the amount of the mixture, but at some point the purification yield will start to go down.


Are there any strategies to determine bond order and hybridization rankings without drawing the Lewis Structure? I'm not Chemist. I'm taking a chemistry class in college, and the test review contains many questions where I'm asked to rank from least to greatest based on bond order, polarity, and bond angles. I feel like I can rate on electronegativity.

The problem is that I don't know a way to rationalize this in my mind without drawing the Lewis Structures. Many times each possible choice has like 5 or so, so it takes a lot of time. We are only given 50 minutes for the test and there are many more problems that require large amounts of time.

My question is... As a chemist or someone who is more knowledgable in this field than I, are there any techniques that would help you now or when you were first learning?

Thanks.

Although it is often taught otherwise, hybridisation is a function of geometry. Therefore you must know (or be able to predict) the geometry. On a basic level, the easiest way to do that are Lewis structures.
A different way is using the VESPR theory, a rudimentary model to predict geometries. Be warned though that this model will fail when it comes to more complex molecules. You need a good imagination of the three dimensions to be able to do it without drawing it.
However you want to spin it, you will have to know which atoms are bonded to other atoms. If you can imagine that, that is good, if not you will have to draw it out. The longer you study, the more present some common molecules are and you will be able to tell certain properties by what we like to call chemical intuition. But that takes time.

Bond order is a bit of a tricky topic, because you have to know the bonding situation. This would include all possible resonance structures for some molecules. While knowing or judging this also comes with experience, it is probably best to draw such structures for learning purposes.

I personally do not like exam questions like these, because they usually just test if a student is good at cramming facts, rather than asking about if these things are understood. But things will get better, if you don't lose interest in the mean time.


Is there a way to detect bleach residue on a surface? I was wondering if there is a way to detect for presence or residue of bleach using household items.

I am curious about either chlorine-based or peroxide-based. It can be any sort of reaction that shows the presence of it. Well, as long as the reaction doesn't produce a lot of heat or is somehow toxic.

What immediately comes to mind is something rather obvious, don't use ammonia on any surface that could possibly be contaminated with bleach.

A search revealed a video showing red cabbage juice used as an indicator.  I do not believe the limit of detection would be anything to write home about.  However, this got me thinking about a homemade pH paper.  This would allow you to dampen the surface with a very small amount of neutral water, apply the paper, and possibly get an accurate reading on whether something acidic or basic is present on the spot you tested.

Fortunately, someone else has already put together a procedure on how to make these strips, found here.  With another source showing the expected colors using this indicator.  Since bleach aka sodium hypochlorite is a base, I would expect the color of the test strips to be in the blue to green range.

This could also be a cool home science project.  The strips will be more effective at detection for lower concentrations of bleach than a liquid solution of the indicator. (pH is concentration dependent).

Good luck!

P.S. I linked many articles to fill in some relevant information but I was limited to 2 links.  If you have questions please feel free to ask!


What is the mechanism for this Horner-Wadsworth-Emmons (HWE) reaction? In this reaction the HWE reaction does not give a "normal" alkene product, but instead gives (R/S)-2. Provide a reasonable mechanism for its formation?

First dil. aq. HCl hydrolyses the starting material to the dialdehyde.
Second the diethylphosphonoacetate anion reacts with one of the aldehyde groups to give the expected intermediate alkoxide. This is in equilibrium with the carbanion shown below. This is not normally depicted in the HWE mechanism as there is usually no pathway forward from it, however in this case there is. This carbanion can, and does, undergo a second HWE with the other aldehyde which proceeds through the usual mechanism to give the products shown in the question


Is FA 18:2 undifferentiated the same as linoleic acid? On wiki, linoleic acid has the code number FA 18:2 cis,cis-9,12. But then what exactly is FA 18:2 undifferentiated?

It's part of an old nomenclature used in the fatty acid industry.  FA 18:2 translates to F atty A cid with 18 carbon atoms and 2 double bonds.  You can read more about these "Lipid numbers" here.  So to answer your question, there are number of fatty acids with 18 carbons and 2 double bonds, linoleic acid is just one of them.  FA 18:2 includes linoleic acid, but is not limited to just linoleic acid.


Lithium batteries a reality - why not sodium, pottasium batteries? Lithium batteries seem to be mainstream now, offering high energy density, high output current capability.

Stepping down the periodic table there are alkali metals with greater electronegativity than Lithium: sodium and potassium, etc.

So why not sodium and potassium batteries. Can't these offer higher energy density, higher current output? Is the reactivity just so great that the chemistry can't be engineered to provide a practical battery? Or are the safety issues just too great?

The sodium–sulfur battery is already in use for power leveling and other stationary applications, such as a 1.2 MWh installation in WV, USA. The Ford Ecostar did use a Na-S battery, but it did not have sufficient energy for distant travel (and some of the experimental batteries caught fire, but so do Li-ion cells on occasion. Oops!).

Sodium-ion batteries are also in development. In the JACS, the article Na-ion batteries get closer to replacing Li-ion batteries discusses some improvements in maximum current capacity and total energy storage.

That said, for mobile use, lithium has a higher energy/weight ratio than heavier metals, but for other use, sodium is less expensive and is readily available.


How to calculate a rough estimate for the volume of a nitric oxide molecule? I was wondering what would be the best way to calculate a rough estimate for size of a nitric oxide molecule to use in comparison to a box.

I was thinking volume would be the best method, but this gave me an answer in dm3/mol which I can't use for comparing against the volume of a box.

Take the volume per mole value (b) from a van der waals constants table (0.0289 L/mol) and divide by $6.022 \times 10^{23}$ molecules per mole.

This will give you a volume that you can compare to the box.


Cis-trans isomers internal energy Whilst reading about stereochemistry, I came to know that the stability and melting point of cis isomers is lower than trans isomers. By searching on net I've found the reason behind that. But one question arises - why is the internal energy of cis isomers bigger than of trans isomers?

I can think of two reasons:
1) Steric repulsion
2) Electrostatic interaction

Steric repulsion: cis means in same side. So, sterically it is suitable to be on opposite side. It is a fundamental nature of both human and chemical entity to define a personal boundary (My personal opinion :) )
In the image you can see the personal boundary of $\ce{-CH_3}$ is the furthest in the trans configuration and that's wh the trans configuration is more stable than the cis configuration.

Electrostatic interaction: Due to electrostatic interaction, molecules or functional group try to distance themselve as far as possible. This also results in lower energy in trans configuration.


Why is the +3 oxidation state of Cerium more stable than +4? For the 4f inner transition metals, +3 is the most common oxidation state (OS). Why is the +3 OS of cerium considered more stable than +4, at which it attains noble gas configuration?

Are you sure $\ce {Ce}$ (III) is really considered more stable than $\ce {Ce}$ (IV)? Since the most common cerium based compound is its dioxide so $\ce {Ce}$ (IV) is actually more stable.

In addition to that, $\ce{Ce}$ has 4 electrons in its outer orbital, thus leaving only one unpaired electron on the $\ce{6s}$ orbital doesn't sound really stable to me.


What's the difference between radial probability and probability density? I understand that radial probability is the probability of finding an electron at some distance from nucleus in a shell. But what does that exactly mean? What does it tell us about the atom and electrons?

Probability density is probability per volume.  Probability density at a given point means probability per volume in the limit that the volume is infinitesimally small.

Radial probability distribution at a given radius is the probability per distance that the event occurs in a infinitesimally thin spherical shell at that radius. (The distance being the thickness of the shell).


Is dissolving salt a chemical reaction or a physical change? Physical changes are limited to changes that result in a difference in display without changing the composition.
  (source)

e.g. melting water or dissolving sugar in water

  Chemical reactions are the processes by which chemicals interact to form new chemicals with different compositions. (source)

e.g. burning $\ce{H2}$ to $\ce{H2O}$

But what is dissolving salt in water? Is that a physical change (like dissolving sugar) or is it a chemical reaction?

I would think that dissolving a non-reactive salt, like $\ce{NaCl}$ in water can be considered as a physical change. This is because, you can recover the salt by evaporating the water away.

From the Wikipedia page of physical change

  In general a physical change is reversible using physical means. For example, salt dissolved in water can be recovered by allowing the water to evaporate.

But the distinction between physical and chemical change can be misleading and confusing at times. In general, anything that does not change the chemical constitution of the substance, is a physical change.

When you dissolve $\ce{NH3}$ in water, the following equilibrium is set up:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

But if you heat the water up, you will get back the original ammonia without any chemical change in its composition. However, the ammonia does change its chemical composition in water, as $\ce{NH4+}$ ions are formed. By that argument, you can say that the dissolution of an ionizable salt is a chemical change.

In some cases, dissolution of a compound can be called a true chemical reaction. For example, when you dissolve Calcium oxide ($\ce{CaO}$) in water, a vigorous chemical reaction takes place, forming $\ce{Ca(OH)2}$. You cannot get back the original $\ce{CaO}$ by just evaporating the water. In case of aqueous solution of $\ce{AlCl3}$, if you try to evaporate the water by heating the solution, you will get back only $\ce{Al(OH)3}$ as anhydrous $\ce{AlCl3}$ reacts with water as soon as it is in contact with water, to form $\ce{Al[H2O]6Cl3}$.

What I am trying to say is that some changes can be classified into one of these types, while some other changes can not be given a clear classification. Melting of water is clearly a physical change, while reaction of $\ce{H2}$ and $\ce{O2}$ is clearly a chemical change. One another fact, putting paint on a canvas is not always a physical change. In fact some paints are manufactured in such a way that during drying, some chemical reactions occur, due to which the paint fixes firmly on the surface. This is the reason many water thinnable paints can not be dissolved in water once they become dry.

To conclude, you should not hold hard classifications like that in your mind. However, my final verdict: Dissolving a salt is physical change (mostly).


Why is hydrogen sulfate put together as it is? I recently looked up hydrogen sulfate ($\ce{HSO4-}$) to see how it's put together, and found this image of methyl hydrogen sulfate:

I'd assume that hydrogen sulfate looks the same, but without the methyl group attached. However I'm confused, how can it be constructed like that? This shows the sulfur having a total of 12 valence electrons, but shouldn't the maximum be 8?

I might be asking something that'll be obvious to me later on in chemistry, but this (and other groups that act as atoms) appear early on. I probably don't need to know it yet but I'd prefer to know how it works if I can as opposed to just accepting it.

Hypervalency is an obsolete concept - it was used to explain bonding situations of molecules, that seem to exceed the octet rule. It was accomplished by the help of $\ce{spd}$ hybrid orbitals. It is now known, that for almost all molecules that were described this way the contribution of $\ce{d}$ orbitals is negligible (<1%).

Focussing on the question at hand, the bonding situation of the core $\ce{SO4^{2-}}$ anion is isoelectronic to the phosphate anion $\ce{PO4^{3-}}$.

For the methylsulfate this situation may be extended. A qualitative analysis of the natural charges (BP86/cc-pVDZ) reveals no difference. All of the oxygen atoms carry a negative charge, while sulfur is highly positive charged. Further more, NBO analysis confirms that the terminal oxygens have three lone pairs, while the other oxygens have two lone pairs. A more accurate Lewis structure would represent that fact and only use single bonds.


Is it possible to synthesize amino acids from hydrocarbons? I recently heard that during some period of time Purina pet food was being produced from crude oil via a set of complex chemical transformations. Which, I think, is mostly bluff and marketing. So is it possible, and have somebody had any real positive practical experience of getting amino acids from hydrocarbons?

Out of memory, I don't recall a method for the direct synthesis of $\alpha$-amino acids from plain hydrocarbons (including alkenes and alkynes).

The situation is different for $\beta$-amino acids. The $\ce{Pd(II)}$-catalyzed reaction of alkenes with amines in the presence of carbon monoxide, known as aminocarbonylation can furnish these amino acids or their derivates.

  […] from crude oil via a set of complex chemical transformations

however, is pretty vague and may involve anything, including the formation of aldehydes and their subsequent use in a Strecker reaction.


Find ratio of quantities of water and of argon in airtight container I am working on my EJU exam past papers (2013) and got stuck at this question. Knowledge of high school or first year university chemistry is assumed. I honestly do not know where I should start.

  When argon and water were placed in an airtight container with a
  constant volume and the gas mixture was kept at 373 K, the pressure of
  the gas mixture became P1 with a portion of water remained as liquid.
  When the whole was heated to 403 K, all the water was vaporized and
  the pressure became P2. From (1) to (6) below choose the correct
  formula representing the ratio, water/argon, of the quantities (in
  mol) of water and of argon in the container. Assume that the vapor
  pressure of water at 373 K is Pv, that argon is an ideal gas not
  dissolving in water, and that the volume of water before heating to
  403 K can be neglected.

Please note that answer key provided is (1). I tried to think that we could form an equation from total moles of gases equate to the sum of total moles of water and argon at 403 K since all are vaporized. Then use the ideal gas law to find the total amount of gases at 403 K. However, I cannot connect these pieces and information given at 373 K together. Could anyone give me a solution to this question or at least a hint so I can continue working on it?

You should use the ideal gas law to convert between $n_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write
$$\begin{align}n_{g,1}=\frac{p_1V}{RT_1}\\n_{g,2}=\frac{p_2V}{RT_2}\end{align}$$
where
$$\begin{align}n_{g,1}&=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&=n_{H_2O}+n_{Ar}\end{align}$$
You also know that for each individual gas the partial pressure is proportional to the number of moles (Dalton's law):
$$n_i=\frac{p_iV}{RT}$$
 The "trick" is that at $T_1=\pu{373K}$
$$p_{Ar} = p_1 - p_V$$
where $p_V$ is given as the partial pressure of water at that temperature.
You should be able to show that
$$\frac{n_{H_2O}}{n_{Ar}}=\frac{n_{g,2}}{n_{Ar}}-1$$
and making appropriate substitutions it should not be difficult to arrive at the answer.


Can density functional approximations including dispersion correction perform worse than without it? I am trying correlate some experimental properties of a molecule to its calculated equilibrium geometry. To do so, I calculated the geometry and energy of some conformers.
Now, I obtain mixed results when performing the same energetic calculation with dispersion-corrected (D3) and not corrected functional (B3LYP): when no correction is applied, a conformer has lower energy, while with D3 correction another conformer has lower energy.
The idea of using D3 looked reasonable, as the molecule has two aromatic and two aliphatic wings, which from the D3-uncorrected geometry seemed to promote to some degree an aliphatic-aromatic stacking interaction.

The point is: my intuition drove me to the use of the dispersion correction, and indeed the D3-corrected result would agree wonderfully with spectroscopic results. But is there a case in which a D3-uncorrected result would be more accurate than the corrected one? Or, alternatively, is there a way to understand quantitatively if the uncorrected model does not describe correctly the system?

I saw a lot of reviews talking about the "benefits" of DFT-D3, but could it do worse than DFT alone?

Edit, to answer Marvin's questions:
Yes, the molecule is organic. It all started with a "routine calculation set-up": b3lyp/6-31g*, but it later became evident that more refinement was required, so the current level might be seen as an expansion of the original calculation. Beyond the 6-31+g** level, I don't stress on Pople's basis anymore, and if more refinements will be required, Alrichs basis sets will be employed. Yes, I omitted it, but B-J damping was employed. At the moment I have optimized one structure with D3(BJ), the other one is being optimized. Of both the geometries optimized with B3LYP, the D3BJ-energy was calculated, together with the uncorrected one.
The two molecules arise indeed from a semiempirical conformational search, followed by a fast and inaccurate hf/3-21g* optimization on 100 geometries. A dozen of low energy conformers was optimized with hf/6-31g*, and the two outliers/low lying conformers were subject to further study with the methods discussed. The molecule contains 90 atoms, and the calculated energy difference is less than 0.5 kcal/mol, but as NMR experiments show a net predominance of one geometry, it's likely that the low energy difference is coupled with an high isomerization barrier, which I still haven't calculated: at the moment I am focusing on the equilibrium geometry.
As soon as the "new" geometry calculations will complete, I will probably do more calculations with a functional better suited for dispersion corrections, thank you very much!

Note: an NMR calculation is planned, but as good results require time and a strong basis set, it will probably be the last thing I will do, starting with better geometries.

tldr; I'll check the detailed results, but the overall answer is that dispersion correction is absolutely worth it

We just finished a benchmark over ~700 molecules with up to 10 conformers, so ~6700 single points at the DLPNO-CCSD(T) / def2-TZVP level. I will post a link to the preprint when it's available soon.

In the meantime, here's a quick summary for the effects of dispersion corrections:

  Comparing un-corrected PBE, B3LYP, and ωB97X single-point energies to DLPNO-CCSD(T) illustrate a significant effect. The uncorrected median R2 values drop by ~0.12, and the median Spearman correlations drop by ~0.08. For example, the median R2 of B3LYP / TZ drops from 0.920 to 0.706 without the D3BJ dispersion correction.

The metric here is the correlation between a method (i.e. B3LYP) with the DLPNO-CCSD(T) energies.

I'll look to see if there's ever a case where the dispersion correction does worse. But there are many, many papers illustrating that conformer energetics are dominated by non-bonded interactions .. which are improved with dispersion-corrected density functional methods.

Your methodology seems solid, although if you want good energy evaluations of different conformers, I'd consider either RI-MP2 or DLPNO-CCSD(T) single-points. Our timings in the paper show that RI-MP2 can be about as fast as hybrid DFT methods, but more accurate at ranking conformers.


Can a solvent be a gas? A solvent is an element which dissolves something, basically the solute. Normally, all the solvents we encounter are in liquid phase. So, I was wondering if there could be any solvent in a gaseous phase. Can there be any gaseous solvent, after all?

Solvents imply dissolving process as intermolecular interactions between solvent and solute, leading to disperging the solute from the condensed form.

"Permanent" gases near normal conditions do not provide such interactions. Volatile compounds evaporate with or without presence of gas. Gas is rather a bystander/spectator of such evaporation.

Fluids near critical conditions like supercritical CO2 share many properties of liquids and gases, so such interactions are present, so in some sense and with both eyes closed, one can consider it as solvent and gas.


Chemical Equilibrium - Le Chatelier's Principle, Change in Volume Question: Consider the following reaction at equilibrium at a total pressure that we will call $P_1$.

  $$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}$$

  Suppose the volume of the system is compressed to $\frac 12$ its initial volume and then equilibrium is reestablished. The new equilibrium total pressure will be:

  a. Twice $P_1$

  b. Three times $P_1$

  c. 3.5 times $P_1$

  d. less than twice $P_1$

  e. unchanged

My thought process: Since I knew the volume had decreased, that meant the pressure had increased and so the reaction will head towards the side with less moles, which would be the products. I thought that since the volume halved, that meant the pressure should increase two times.

However, the answer is D. Could someone please explain this to me?

Expanding on what Buck Thorn said in the comments and providing a more rigorous answer.

The partial pressure would change as follows:
$$\begin{array}{c|c|c|c}
& \ce{SO2} & \ce{O2} & \ce{SO3} \\
\hline
\text{Initial state} & P_A & P_B & P_C \\
\text{On halving volume} & 2P_A & 2P_B & 2P_c\\
\text{Final state} & 2P_A-2x & 2P_B-x & 2P_c + 2x\\
\end{array}$$

where $P_A + P_B + P_C = P_1$

Since initial and final states are in equilibrium,
$$ K_p = \frac{P_C^2}{P_A^2 \cdot P_B} = \frac{(2P_C+2x)^2}{(2P_A-2x)^2 \cdot (P_B+x)}$$

We do not need to solve this equation: looking at both the sides, we can see that the value of x would be positive (for a negative x, the numerator would decrease and denominator would comparatively increase due to the $(2P_A - 2x)^2$ term increasing more than the $(P_B + x)$ term due to the square). You could go ahead and solve the equation for a more accurate answer but looking at the options, we can make do with an approximation.

Adding the partial pressures at the final state, we have
$$P_2 = 2(P_A + P_B + P_C) - x = 2P_1 - x < 2P_1$$

Which corresponds to option (d)


Calculating the pH of a solution of Ca(OH)2 A solution prepared by dissolving $2.8 g$ of lime, $\ce{CaO}$ in enough water to make $1.00 l$ of lime water ($\ce{Ca(OH)_{2(aq)}}$). If solubility of $\ce{Ca(OH)2}$ in water is $1.48 g$. The $p\ce H$ of the solution obtained will be:

[$\log 2 = 0.3$, Atomic masses are $\ce Ca$ = $40$ , $\ce O$ = $16$, $\ce H$ = $1$]

[Note : $\ce {Ca(OH)2}$ is considered to be a weak base because  of its low solubility but it is 100% dissociated]

(A) $12.3$

(B) $12.6$

(C) $1.3$

(D) $13$

My attempt:As $2.8g$ of $\ce {CaO}$ dissolves in $1l$ water the molarity of the solution is 0.05M. So,the concentration of OH- ions in the solution is $0.1M$.So $p\ce {OH}$ is $1$ and hence $p\ce H$ is $13$. But, the answer is given to be $12.6$.What am I doing wrong?

You did not account for the fact that $\ce{Ca(OH)2}$ is partly precipitated. Firstly, to find out the moles of $\ce{Ca(OH)2}$ precipitated:-
The number of moles in solution is $\frac{1.48}{74}M$ which is $0.02M$. The total moles precipitated is just the moles of $\ce{CaO}$ minus $0.02M$ since the initial number of moles of
$\ce{Ca(OH)2}$ is equal to the moles of $\ce{CaO}$ taken and hence the amount precipitated turns out to be $0.03M$. Nevertheless, this is not required. With $0.02M$ $\ce{Ca(OH)2}$ in solution, the $\ce{OH-}$ concentration turns out to be $0.04M$ which, assuming no contribution of $\ce{OH-}$ from water and complete ionization, gives a $p\ce H$of $12.6$.


Is the Ebullioscopic constant temperature-dependent? In tables, the ebullioscopic constant is usually written as a single value.

But, when the value is not available, it can be calculated by applying the formula: $$K_{\mathrm b} = RT_{\mathrm b}^2/\Delta h_\text{vap}$$

I was wondering: since in this last definition the ebullioscopic constant does depend on temperature, being the value of $\Delta h_\text{vap}$ temperature-dependent (its value is different if calculated at room temperature or at the boiling point @1 atm) what is, in real life, the best approximation of the behaviour of $K_b$?

In other words, does it behave like a constant or its value changes at two different temperatures? (eg: boiling point elevation for a solution @atmospheric pressure vs boiling point elevation for the same solution @reduced pressure)

Short answer

As long as external pressure is constant, it does not make sense to discuss $K_b$'s temperature dependence. If external pressure is changed, boiling point of solvent changes and hence also the ebullioscopic constant.

Longer version

Condition for equilibrium (skip if familiar)

Boiling is a reversible equilibrium process. By the second law of thermodynamics,

$$\mathrm{d}S_\text{universe}=0.$$

Assuming constant pressure,

$$\mathrm{d}S_\text{universe}= \mathrm{d}S-\frac{\mathrm{d}H}{T}.$$

Provided additional constant temperature,

$$-T\mathrm{d}S_\text{universe}= \mathrm{d}H - T\mathrm{d}S = \mathrm{d}G_\text{P,T}.$$

We also know that $G=G(P,T,n_1,\ldots, n_k).$ Thus

$$\mathrm{dG}=V\mathrm{d}P-S\mathrm{d}T + \sum_{i=1}^{k}\mu_i\mathrm{d}n_i,$$

and therefore

$$\mathrm{d}G_\text{P,T} = \sum_{i=1}^{k}\mu_i\mathrm{d}n_i=0.$$
Currently, we have two phases: a gas and a liquid. Boiling is a dynamic equilibrium, the differentials are non-zero, $\mathrm{d}n_1 = -\mathrm{d}n_2$, and so

$$\mathrm{d}G_\text{P,T} = 0 \iff G_\text{liquid} = G_\text{gas} \iff \mu_\text{liquid} = \mu_\text{gas}.$$

Clausius$-$Clapeyron equation (skip if familiar)

Suppose a closed system with initial equilibrium, constant temperature and pressure. We saw that

$$G_\text{liquid} = G_\text{gas}.$$

Now change the temperature. To achieve a new equilibrium in a closed system, the pressure would also need to change. At this new equilibrium, $G'_\text{liquid} = G'_\text{gas}$, which is why

$$\mathrm{d}G_\text{liquid} = \mathrm{d}G_\text{gas},$$

or equivalently,

$$V_\text{liquid}\mathrm{d}P - S_\text{liquid}\mathrm{d}T = V_\text{gas}\mathrm{d}P - S_\text{gas}\mathrm{d}T.$$

Now

$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{S_\text{gas}-S_\text{liquid}}{V_\text{gas}-V_\text{liquid}}.$$

Note that the RHS depends again only on the initial and final conditions of phase equilibrium. Thus we are allowed to use $\Delta S = \Delta H/T$ to arrive at

$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{\Delta H}{T(V_\text{gas}-V_\text{liquid})}.$$

Because $V_\text{gas} >> V_\text{liquid}$, and due to the ideal gas law

$$\frac{1}{P}\frac{\mathrm{d}P}{\mathrm{d}T}\approx\frac{\Delta H}{RT^2}.$$

Assuming $\Delta H$ is not a function of temperature,

$$\ln \frac{P_2}{P_1} \approx -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).\tag{Clausius–Clapeyron}$$

NOTE. There is a small but important detail to keep in mind here. The assumption that $\Delta H$ is independent of temperature will only make our equations more inaccurate but it will not affect our answer. See the definition further below.

Boiling point elevation

We have

$$\ln \frac{P^\text{s}}{P^\circ} \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right)\tag1$$

where

$P^\text{s}$ – vapor pressure of solution at initial boiling temperature $T^\circ_\text{b}$,
$P^\circ$ – saturation pressure of solvent, by definition of boiling, $P^\circ = P_\text{ext}$,
$\Delta H_{vap}(T^\circ_\text{b})$ – entalphy of vaporisation at initial boiling temperature $T^\circ_\text{b}$,
$T^\text{s}_\text{b}$ – new boiling temperature at some external pressure $P_\text{ext}$,
$T^\circ_\text{b}$ – boiling temperature of solute at the same external pressure $P_\text{ext}$.

From Raoult's law

$$P^\text{s} \approx P^\circ X_\text{solvent} = P^\circ(1 - X_\text{solute}),$$

and thus from $(1)$

$$\ln (1 - X_\text{solute}) \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right).$$

From first-order Maclaurin series, and since boiling point elevation is usually small for our equations to be a good approximation,

$$\Delta T_\text{b} \approx \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}X_\text{solute}.$$

Granted molar mass $M$ is in units $\mathrm{g/mol}$, and molality $c_m$ has units of $\mathrm{mol\ /\ 1\ kg\ solvent}$, then

$$\Delta T_\text{b} \approx \underbrace{\frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}}_{K_\text{b}}c_m \equiv K_\text{b}c_m$$

or

$$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}. \tag2$$

It does not matter whether our previous equations are mere approximations, this is now a definition of the ebullioscopic constant. It is evident that it is in no manner dependent on external temperature. It is only dependent on the boiling temperature of a pure solvent at some well-defined external pressure $P_\text{ext}$. To change this temperature, we would have to change the external pressure, or pick a different solvent altogether.

If the external pressure is transformed, $T^\circ_\text{b}$ will not remain the same. There is no reason to believe that $(T^\circ_\text{b})^2$ at this new pressure $P'_\text{ext}$ will cancel with the change in vaporisation entalphy. And thus, the ebullioscopic constant is, in fact, not constant.

Conclusion

If external pressure remains the same, it makes no sense to talk about temperature dependence. This is because the ebullioscopic constant is defined as $$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}$$
which is dependent only on the original boiling temperature of the pure solvent at some well-defined external pressure.
If external pressure is changed, the pure solvent's boiling point $T^\circ_\text{b}$ will change to a new value. There is no cause to believe that the square of the boiling temperature will cancel with the altered entalphy of vaporisation. Hence, ebullioscopic constant will change, and is, in this sense, dependent on temperature. The exact function will vary with the solvent at hand but is most likely approximated by polynomials with empirical coefficients if need be.


What chemical explosives detonate when exposed to alpha particles or nuclear fission products? Wikipedia's article on nitrogen triiodide $\ce{NI3}$ claims that

  Nitrogen triiodide is also notable for being the only known chemical explosive that detonates when exposed to alpha particles and nuclear fission products

referring to the article published 60 years ago [1]. Are there any other examples of chemical explosives capable of that discovered afterwards, or is it still a prerogative of nitrogen triiodide?

References

Bowden F. P.; Young D. A. The Initiation of Explosion by Neutrons, α-Particles and Fission Products. Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences 1958, 246 (1245), 216–219. https://doi.org/10.1098/rspa.1958.0123.

(not really an answer here, but anyway ...)
I would suspect any highly sensitive explosive is theoretically capable to do that, possibly depending on how close you are to the point where it would just thermally explode. They checked  "Crystals of lead azide, silver azide, cadmium azide, silver acetylyde and nitrogen iodide" in the paper cited above. The four with heavy metal ions don't work, I guess it makes some sense that those can (better) absorb a very local energy surge like from an alpha particle.

My list of suspects would thus include organic ozonides, peroxides, azides, tetrazoles and other compounds that only contain CNOH. Number 1: solid $\ce{H2O2}$.

And basically anything that is known to sometimes explode without an obvious external triggering event.


Calculating the ionic product of water In the question I've to find out the ionic product of water using the facts that the degree of dissociation of water at $18\ \mathrm{^\circ C}$ is $1.8\times10^{-9}$.

My attempt

Let the concentration of water be $1\ \mathrm M$.

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

Initial:

$$\begin{align}
[\ce{H2O}] &= 1\ \mathrm M\\[6pt]
[\ce{H+}] &= 0\ \mathrm M\\[6pt]
[\ce{OH-}] &= 0\ \mathrm M
\end{align}$$

At equilibrium:

$$\begin{align}
[\ce{H2O}] &= (1 - 1.8×10^{-9})\ \mathrm M\\[6pt]
[\ce{H+}] &= 1.8×10^{-9}\ \mathrm M\\[6pt]
[\ce{OH-}] &= 1.8×10^{-9}\ \mathrm M
\end{align}$$

Therefore,
$$\begin{align}
K_\mathrm w &= \frac{[\ce{H+}] + [\ce{OH-}]}{[\ce{H2O}]}\\[6pt]
&= \left(1.8\times10^{-9}\right) \times \left(1.8\times10^{-9}\right)\\[6pt]
&= 3.24 \times 10^{-18}
\end{align}$$

According to my book answer is $1.0\times10^{-14}$, which I know is correct. I want to know where am I going wrong?

What you did wrong was adding the concentration of the hydron and the hydroxide ion. Also, although it hardly matters, the concentration of $\ce{H_2O}$ is considered a constant and ignored in the product.


equivalents for products? I do not understand how to use the equivalents to find the millimoles of the theoretical product. This is the part that I'm struggling with the most. My TA says that after I find the millimoles using the equivalents, then I can solve for grams and the mL of water.

First, you have to find out the limiting reactant in the reaction, in your case it's seems that cyclohexanone is limiting reactant. Theoretically, one equivalent of cyclohexanone (27mmol) react completely with 2 equivalent of benzaldehye (54mmol),but benzaldehye is used in excess (73mmol) to gave one equivalent of product (27mmol) and two equivalent of water (54mmol).
When you multiply 27mmol of product with mol. weight (274.36g/mol) which is equal to =7408mg/1000= 7.4g

Similar you can do for water as well..
I hope this would be helpful, right now i don't have pencil with me, otherwise i will post a picture to this problem.
anyway if you didn't understand let me know..
sincerely,