Is the total entropy change of all isothermal processes 0? We define  total entropy as: $$\Delta S_\mathrm{total}=S_\mathrm{system}+S_\mathrm{surroundings}$$

I know that for an isothermal AND reversible expansion: $S_\mathrm{total}= 0$

But what about an isothermal expansion at a constant pressure? Does $S_\mathrm{total}$ also equal 0? If not, then would the equation look something like this:
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$

where $q_\mathrm{surroundings}$ is the heat lost or gained by the surroundings by a process and $q_\mathrm{rev}$ is the heat lost or gained by the system IF that process was to become reversible.

Is the total entropy change of all isothermal processes 0?

No. Only for reversible processes.
However your equation is correct.
For an irreversible process in which an ideal gas expands isothermally against a constant external pressure
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$
becomes
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{w}{T}$$
where $w=-p \Delta V$, since $q_\mathrm{surroundings}=-q_\mathrm{system}$ and for the isothermal process the first law stipulates that $w=-q_\mathrm{system}$.
For a free expansion $p_\textrm{ext}=0$ and no work is done. No heat is exchanged and the entropy of the surroundings does not change. The change in entropy of the system then drives the process.


What is the stability order of the carbocation and carboanion? So I came up with examples as below:

1
2

Now I am confused about the +I and -I effect in these structures since to my knowledge it can only be applied when there is a substituent (I might be wrong). I understand the role of resonance but do not get how to apply the +I and -I effect in these so I need help. I have basic knowledge about the substituents which give +I and -I effect but still can't get it.

Alkanes or $\mathrm{sp^3}$ hybridized carbons show +I effect.

Alkenes contain $\mathrm{sp^2}$ hybridized carbons. Due to greater $\mathrm{s}$  character they have higher electron withdrawing tendency. So they show -I effect.

Coming to the first question:

In the given figures we have $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized carbons in the given rings. Notice that in the second figure there are greater number of $\mathrm{sp^2}$ hybridized carbons. Hence, the net withdrawal of electron density is more, and accordingly, stability goes down. In the third figure there is no resonance, which is the more dominating effect than the inductive effect. So it is the least stable.

In the second question:

In figures 1 and 2, we have equal number of $\mathrm{sp^2}$ hybridized carbons. But notice in the first figure, the negative charge is on the ring and is closer to the $\mathrm{sp^3}$ carbons which are electron donating (Remember, inductive effects are distance dependent). In the third figure greater +I is being imposed on the anion, hence it has the least stability.


How to create a simple do-it-yourself procedure to measure diffusion? While I am not asking the question here, a recent question concerning osmosis at the Bio SE caught my interest. In a nutshell that question was, given isotonic solutions of sugar inside and outside an immersed semipermeable bag, what will happen if a second solute--too big to escape through the membrane-- is added to the bag?

  The question has an SAT flavor but I think it should be possible to answer it with a fairly straightforward DIY procedure. So my question here is whether the outline I have in mind would work, or whether something better can be devised.

Procedure: Make an n-molar solution of dextrose. Make an (n+x)-molar solution of dextrose [n] and a second [x] solute S (sucrose? starch? urea?--something that will not pass through the membrane). Fill the bag with the dextrose/S solution. Immerse the bag in the dextrose solution.

Osmosis alone will draw water into the bag. The question to be answered by the procedure is whether there will also be a net diffusion of dextrose out of the bag so that the final concentration of dextrose is higher outside the bag than inside, i.e., higher than the initial n-molarity.

According to one internet source, dextrose will rotate plane-polarized light about $30^o$ clockwise. If my null hypothesis is that the n-molar solution and the post-osmosis/diffusion solution outside the bag have the same molarity (i.e., that there is no net movement of dextrose out of the bag), and if I am able to construct an apparatus to measure optical rotation consistently in my kitchen, could such an apparatus be sensitive enough to reject $H_o$ if sugar in fact diffuses out of the bag?

Can anyone suggest a good practical way to set this up? Suggestions for types of membranes, necessary polarizing/light equipment, second solute S, etc. all welcome. Also, if there is a simpler/better way of going about this that would be great.

Thanks for any advice.

Notes

Sucrose reportedly rotates light more (perhaps $40^o$) but it may be too large to pass through some commercially available membranes. Fructose only rotates light about $10^o$ so I don't think it can be used here. Could sucrose be used as second solute?

The thermodynamics tag is added since I think diffusion is properly an aspect of thermodynamics (see Fermi, Thermodynamics, ch. VII). Please feel free to edit if I am mistaken.

There are way more sensitive ways to measure dextrose (glucose) concentration than polarimetry. For example, there are reducing sugar tests like the Dinitrosalicylic Acid test or (better) the 2,2'-Biconchoninic acid test. Just measure the dextrose concentrations inside and outside of the bag. I have protocols for these tests if you want; it involves getting ahold of the reagents and a spectrophotometer (unless you just want to do it qualitatively, look at the color and see if they're different), but presumably that would be no more difficult than getting hold of a polarimeter.

As far as the second solute S goes, you want something that's a lot larger than glucose. I'm only familiar with dialysis tubing that separates small molecules from proteins, so I would go with that. Cheap regenerated cellulose dialysis tubing has a molecular weight cutoff of 10,000, so use glucose (MW 180) and cheap (and soluble) Bovine Serum Albumin (MW 66,000). You always want something at least five times less than or five times greater than the cutoff for these tubes.

Be careful, there are reducing tests for sugars and reducing tests for proteins, they often use the same techniques but in different conditions, so with one protocol they'll be very sensitive for one and much less sensitive for the other. As long as you have blanks, it shouldn't be a problem, but make sure you're following the right protocol.

Or if you really want to be fancy, there's a fluorimetric test that is extremely specific to glucose, using glucose oxidase.

Or if you want to be DIY, just use a diabetes glucose meter and strips.


Is Nitroglycerine a Nitro Compound or an Ester? I looked up nitroglycerine in a science dictionary (Oxford Dictionary of Science, 2003), and it explicitly mentions that nitroglycerine is an ester and not a nitro compound.

However my teacher told us in class, that nitroglycerine is a nitro compound (She didn't say it wasn't an ester though …).

The definition for a nitro compound (courtesy Wikipedia) is:

  Nitro compounds are organic compounds that contain one or more nitro functional groups ($\ce{−NO2}$).

Assuming that definition is correct, as well and looking at the structure of nitroglycerine, it appears (to me) that nitroglycerine is both an ester and a nitro compound.

So who's right here?

The Oxford dictionary is correct.

A nitro compound is defined by the $\ce{-NO2}$ functional group, but only if that group is attached to a carbon atom. Once an additional oxygen is added, the molecule is basically inorganic nitrate and thus belongs to a wildly different compound class.

You think that is confusing? Well, an alcohol is defined by a hydroxy group — but only if that hydroxy group is attached to a carbon, which in turn is not attached to a different oxygen, nitrogen or sulphur atom, and which is not part of an aromatic system. Even more constraints in there.


Why is ethanol not an oxidizer? I am attempting the 2016 Canadian Chemistry Contest Part A, and one of the questions concerns the WHMIS symbols and whether a compound is an oxidizer or not.

The correct answer is that ethanol would not require a WHMIS oxidizer symbol. However, I am confused as to why ethanol is not an oxidizer when compounds such as potassium permanganate and nitric acid are when all of them contain oxygen.

For something to be an oxidiser. It must be capable of being reduced. The presence of oxygen is no particular indicator of the oxidising ability of a given species.

In potassium permanganate, manganese is in its +7 oxidation state (a very high oxidation state), consequently it is a good oxidant agent as it is able to gain electrons, causing oxidation of another species and itself being reduced.

In ethanol, there isn't the same possibility for reduction as the molecule is already in its lowest oxidation state. Even ethanal  (an oxidation level up) isn't particularly oxidising (despite being able to accept electrons and become formally reduced).


Why do the melting and boiling points of the noble gases increase when the atomic number increases? What causes the melting and boiling points of noble gases to rise when the atomic number increases? What role do the valence electrons play in this?

The melting and boiling points of noble gases are very low in comparison to those of other substances of comparable atomic and molecular masses. This indicates that only weak van der Waals forces or weak London dispersion forces are present between the atoms of the noble gases in the liquid or the solid state.

The van der Waals force increases with the increase in the size of the atom, and therefore, in general, the boiling and melting points increase from $\ce{He}$ to $\ce{Rn}$.

Helium boils at $-269\ \mathrm{^\circ C}$. Argon has larger mass than helium and have larger dispersion forces. Because of larger size the outer electrons are less tightly held in the larger atoms so that instantaneous dipoles are more easily induced resulting in greater interaction between argon atoms. Therefore, its boiling point ($-186\ \mathrm{^\circ C}$) is more than that of $\ce{He}$.

Similarly, because of increased dispersion forces, the boiling and melting points of monoatomic noble gases increase from helium to radon.

For more data of melting and boiling points of noble gas compounds, read this page


What is the difference between low-density and high-density Polymethyl methacrylate (PMMA) on a molecular level? I know why Polymethyl methacrylate can be made to have a low or high density but i am unable to illustrate how they look like on the molecular scale. Can someone help me with a sketch please?

Low density and high density PMMA differ by density, which is given as the ratio of mass and volume.

Given that they are prepared from the same monomer, two polymers with the same number of monomers will have the same mass.

If the polymers differ in density, this must be due to the fact that they occupy different volumes.

Note that we are talking about polymers here and not the ordered packing of molecules in a crystal. Nevertheless, inter- and intramolecular interactions (and thus "packing") is important here as well.

As a rule of thumb, polymers with higher branching will "pack" less tightly and thus occupy a higher volume.

This results in a lower density.


Volume of air required to oxidize glucose In the body, the oxidation of glucose is according to the reaction:

  $$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}$$

  Knowing that glucose consumption is $\pu{2.47 mol/d}$, What is the volume of air that must be breathed at $\pu{25 °C}$ to accomplish this oxidation for $24\ \mathrm h$ at the pressure of $1\ \mathrm{atm}$?

We are given:

  The atmosphere is composed of $21\,\%$ $\ce{O2}$ and $79\,\%$ $\ce{N2}$ (% volume).
  $R = \pu{8.314 J mol-1 K-1}$; $\pu{1 atm} = \pu{760 mmHg} = \pu{1.014e5 Pa}$.
  The molar volume of a gas at $\pu{298 K}$ is $\pu{24.45 l mol-1}$.

So, I have been a little stuck on this question. I have been trying to use the formula:

$$pV = nRT,$$

but it seems that I am missing something and I am not getting the right answer. It's possible that it might include molar fraction or partial pressure. I am confused.

The answer is $1\,724\ \mathrm l$. I would like to know how to get to this point.

Following the advice of @matt_black:

Here are my calculations that doesn't give me the right answer of $1\,724$ liters.

$$V_\mathrm m = 24.45\ \mathrm{l/mol}$$

$$n(\ce{C6H12O6}) = 2.47\ \mathrm{mol}$$

$$n(\ce{O2})= 6\times2.47\ \mathrm{mol} = 14.82\ \mathrm{mol}$$

$$V =\ ?$$

By the ideal gas law:

$$V= n\times24.45\ \mathrm{l/mol}$$

$$= 14.82\ \mathrm{mol} \times 24.45\ \mathrm{l/mol}$$

$$= 362.349\ \mathrm l \neq 1\,724\ \mathrm l$$

Tried using $pV=nRT$

$$V= 14.82\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 293\ \mathrm K / 760\ \mathrm{mmHg}$$

$$= 47.5\ \mathrm l \neq 1\,724\ \mathrm l $$

I don't know how to implicate the fact that the atmosphere is composed of $21\,\%$ of $\ce{O2}$ and $79\,\%$ of $\ce{N2}$. Any ideas?

As matt_black commented, there is no need to use ideal gas law as you are given the molar volume already. Volume of the air $V_\mathrm{air}$ can be found from the volume fraction $\varphi(\ce{O2})$:

$$V_\mathrm{air} = \frac{V(\ce{O2})}{\varphi(\ce{O2})}$$

Volume of the oxygen can be found from the molar volume $V_\mathrm{m}(\ce{O2})$ and the amount of oxygen $n(\ce{O2})$ required to oxidize given amount of glucose:

$$V(\ce{O2}) = n(\ce{O2})\cdot V_\mathrm{m}(\ce{O2})$$

$$n(\ce{O2}) = 6n(\ce{C6H12O6})$$

Getting things together:

$$V_\mathrm{air} = \frac{6n(\ce{C6H12O6})\cdot V_\mathrm{m}(\ce{O2})}{\varphi(\ce{O2})} = \frac{6\cdot\pu{2.47 mol}\cdot\pu{24.45 L mol-1}}{0.21} = \pu{1725 L}$$

Note that you have to use appropriate units, e.g. you cannot use all base SI units in the deal gas law and put the pressure in mm Hg! Also, never drop any units — they do participate in calculations as well.


Identifying a quadruply charged anion containing three carbon atoms This linear polyatomic ion containing three atoms of carbon has a negative four charge and is only found bonded with lithium and magnesium.

Could anyone identify this for me? It's from a quiz bowl clue that doesn't have the answer with it.

Sesquicarbide

A quick Google search suggests that $\ce{Li4C3}$ is a real compound, but based on what I saw in the links, it probably is not ionic in the same way that lithium acetylide $\ce{Li2C2}$ is.

$\ce{Mg2C3}$ appears to be the formula for magnesium carbide, or at least a magnesium carbide.

A carbide is a binary compound of carbon and a less electronegative element, usually a metal or metalloid. Carbide is also a generic name for mono- and polyatomic anions containing only carbon. The following ions would all carbide ions: $\ce{C^{4-}, C2^2-, C2^{4-}, C2^{6-}, C3^{4-}, C3^{6-1}, C3^{8-}}$, etc.

The special name of three of these carbide ions are:

$\ce{C^{4-}}$ - Methide
$\ce{C2^{2-}}$ - Acetylide
$\ce{C3^{4-}}$ - Sesquicarbide


Lost track of the hydrogen in step 3 of the Krebs cycle I'm trying to follow the Krebs cycle step by step, accounting for the number of hydrogens in
each molecule before and after reactions.
I've run into a problem starting from step 3.

Isocitrate has five hydrogens.
It undergoes oxidisation to Alpha-Ketoglutarate, with NAD+ being reduced to NADH/H+.
I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+.
What I don't understand (and this lack of understanding cascades through steps 4 and 5) is that Alpha Ketoglutarate has four hydrogens.
My arithmetic says that there should be only three.
Where does the fourth hydrogen come from?
Note:
If I start from  Step 6, my accounting method works, all the way round to Step 3.
Step 6: Succinate (four hydrogens) is oxidised to Fumarate (two hydrogens), reducing FAD to FADH2.
Step 7: Fumarate (two hydrogens) is reduced to Malate (four hydrogens), by the addition of H2O, gaining
two hydrogens.
Step 8: Malate (four hydrogens) is oxidised to Oxaloacetate (two hydrogens), reducing NAD+ to NADH/H.
Etc.
I came across a similar question at https://biology.stackexchange.com/questions/93775/what-exactly-happens-to-hydrogen-atoms-in-step-4-of-citric-acid-cycle but as this was about step 4, and for me the problem occurs one step earlier.

Your problem stems from the statement
"I would say this process entails the loss of two hydrogens from Isocitrate, and the gain of those hydrogens in NADH/H+."
This statement is incorrect. In isocitrate, the carboxylic acid functional groups are deprotonated, as would be the case at neutral pH. The oxidation reaction thus involves loss of only $\ce{H-}$ to NAD. There is no loss of $\ce{H+}$, because that has already been removed.
If you start with isocitric acid, which is isocitrate in its neutral form with $\ce{-CO2H}$ instead of $\ce{-CO2-}$, then you will need to remove $\ce{H-}$ and $\ce{H+}$ to accomplish the reaction and your above statement will be correct.
The textbook image you showed takes a confusing approach of implicitly associating an $\ce{H+}$ with isocitrate even though it is not actually on the molecule, so that "NADH/H+" is indicated as a product even though only NADH has actually been produced in the reaction shown.


Oxidation state of metal ion in unit cell If a cubic unit cell for a metal nitride has nitride ions at all face and corner positions, and metal ions at all edge positions, what is the oxidation state for the metal ion in the nitride?

I think that if something is in a face position, there is half of that something present in the unit cell. So if there are nitride ions at all face positions (6) then there must be 3 total nitride ions. And at the corners, there are eighths of nitride ions. So given 8 corner positions, there must be an additional nitride ion formed by the fragments at the corners.

This gives us a total of 4 nitride ions, or a negative 12 charge that must be balanced by the metal ions at the "edge" positions. But what are edge positions? If I know what it is, then I can visualize how many metal ions there are in the unit cell.

Are the edge positions directly halfway between the corner positions and butted up against the face positions? So is there one edge position per edge of the unit cell? This gives us 12 edge positions, and each fragment is a fourth of an ion, so we have 3 metal ions total. So the oxidation state of the metal ion is +4?

Edge positions are those at the centre of an edge of the cube; i.e. in the middle between two closest corner positions. As such, there are 12 edge positions and each belongs to 4 unit cells giving us 3 metal ions.

The way the nitride positions are described is basically a verbose way of saying cubic close-packed or face-centred cubic. Therefore, we can redescribe the cation positions as ‘all octahedral voids except for the one in the centre of the unit cell’. Since there are 4 atoms per unit cell in ccp and 4 octahedral voids, that again gives us a $3:4$ ratio and thus $\ce{M3^{IV}N4}$.


Aldehyde and ketone reactions Reagents to carry out above conversion, P, Q, R respectively are:
(a) $\ce{H2C=CH-CH2-Br}$, $(\ce{HO-})$, $[\ce{HO-}, Δ]$, Wacker process.
(b) $\ce{H2C=CH-CH2-Br(HO^-)}$, Wacker process, $\ce{HO-}$, $Δ$.
(c) Wacker process, $\ce{H2C=CH-CH2-Br(HO^-)}$, $\ce{HO-} (Δ)$.
(d) Wacker process, $\ce{HO-} (Δ)$, $\ce{H2C=CH-CH2-Br(HO^-)}$.

Could someone please explain the R part of this reaction? I can't understand the five-membered cyclization step. Is it aldol condensation or there's something else?

Step P is std base-driven alpha alkylation of ketone. Step Q Wacker details here takes the terminal alkene to methyl ketone then step R is base-catalysed internal aldol mechanism here gives the product shown. So the answer is (b).


Why do some dentists claim mercury "leaks out" of amalgam fillings? From a chemistry point of view, if dental amalgam is a compound of mercury and other metals, and compounds are elements which are chemically bonded, how would it be possible for mercury to "leak out" of fillings if no further chemical reaction would be taking place?

You are labouring under a misconception about the exact meaning of the word "compound".

Normally when we describe, for example, ethanol as a compound of hydrogen, oxygen and carbon we don't mean that the component elements can "leak" out of it without a chemical reaction taking place.

Mercury amalgams are casually described as "compounds" but not in the same sense as ethanol. Amalgams are mercury alloys with varying compositions but they don't, usually, consist of strongly bonded molecules like ethanol with strong chemical bonds preventing the molecules falling apart. They are closer to mixtures than what we normally think of as compounds. Having said that they are not just mixtures as they can have very different bulk properties than their components.

The issue with mercury amalgams used in fillings is that they are mostly, but not completely, inert. Some mercury will evaporate from them and some chemical and physical processes can increase the amount that evaporates. But good estimates suggest only micrograms/day which is less than typical workplace exposure limits so probably not a source of legitimate worry. The dentists who still use mercury fillings are likely exposed to more than this and this is a better reason to limit their use. The subject is controversial with a lot of disagreement among studies. This is well summarised by the Wikipedia article on the controversy.


What's the monomer in a polycarbonate? I know what the repeating unit is, but what's the monomer in a polycarbonate?

Like Dr J said there is no single polycarbonate polymer, the most common one is formed from phosgene (or another carbon dioxide reagent) and bisphenol A. The repeating unit is $\ce{C6H4C(CH3)2C6H4-O-C(=O)-O -}$

In Nagahata, R.; Sugiyama, J.; Goyal, M.; Goto, M.; Asai, M.; Ueda, M.; Takeuchi, K. X-ray analysis of the bisphenol-A-type macrocyclic carbonate trimer. Polym. Adv. Technol. 2000, 11 (6), 294–300. DOI: 10.1002/1099-1581(200006)11:6<294::AID-PAT979>3.0.CO;2-C. (OCAMUA) a tetramer of the repeat unit of this polymer has been reported. This is a small cyclic polymer (oligomer).


Where do the electrons that are released in the sun go? I've heard that electrons are released from the clutches of nuclei in the plasma state that is present in the sun. My question is: Where do these electrons go?  Do they go to the product nuclei of the nuclear fusion reaction in sun (ie. He)? Or do they remain on the surface of sun?

Some of the electrons will react with positrons formed during the fusion processes; they annihilate each other and give off high-energy radiation (also called gamma rays):

Source

But this process  happens at the core, and there is no way that those electrons get out to the surface. So my best guess is that the electrons which don't get annihilated just stay there in the core plasma. I do not think that they combine with the $\ce{^{4}_2 He^{2+}}$ particles formed by this chain, because the temperature is still too high. So you start out with 6 protons and 6 electrons (for electrical neutrality reasons), two electrons get annihilated in the process and you end up with a doubly positively charged helium particle and two protons and four electrons. So the electrical neutrality is not violated.

There is not only the core of the sun but also the outer layers and the corona, from where we observe the solar wind, which is a stream of particles, mainly electrons and protons. So there you have some electrons that escape the sun. But these are not the electrons from the core, they come from the surface of the sun.


Molecular differences in the HB graphite hardness scale Today, I bought some graphite for my pencil. It is rated HB on the HB graphite scale, a measurement of the hardness of graphite. My question is how can there be different hardnesses of the same substance? Is there a difference between 4B graphite and 6H graphite on the molecular level? Are the different hardnesses not all arranged in the large, hexagonal carbon sheets that I had thought all graphite existed?

Pencil leads are not comprised of pure graphite: they're made from graphite and a varying amount of clay, which in turn is what determines the hardness of the lead. Thus, you're not actually comparing graphite to graphite but rather different mixtures of graphite and clay.

According to one article published by Jetpens.com (a vendor of writing implements) entitled Picking the Perfect Pencil Hardness Grade:

  Graphite is the key component of modern day pencil lead—in fact, lead is a misnomer as it's actually made up of a mixture of graphite and clay. The formulation of this mixture determines its lead grade. The higher the proportion of graphite content relative to clay the lead has, the softer and darker the lead will be, and vice versa.

And Wikipedia tells us:

  The various graphite pencil grades are achieved by altering the proportion of graphite to clay: the more clay the harder the pencil.

The difference in hardness is attributable to the amount of clay mixed in with graphite in your different pencil leads. There is no difference in the graphite at the molecular level.


How many significant figures should I report calculating volume concentrations? Let's say a $\pu{300mL}$ solution of $\ce{CuCl_2}$ was diluted to $\pu{900mL}$. I am then asked to calculate the concentration of the new solution. My solution to the problem is as follows:

$$\begin{align}M_1 \times V_1 &= M_2 \times V_2\\[0.5em]
\pu{300mL} \times \pu{4M} &= M_2 \times \pu{900mL}\\[0.5em]
M_2 &= \pu{1.33M}\end{align}$$

Now I'm wondering how many significant figures my final answer should have. From what I know, I have to refer to the least significant measurement, which happens to have one significant figure. Should my answer be 1 instead?

We cannot say how many significant figures you should be reporting because:

We cannot be sure about the precision of your volume measurements; and
We cannot be sure about the precision of the original solution’s concentration.

Concerning volumes: most of the time when working with such volumes in the lab, I would use measuring cylinders. These come with a rather poor precision but are adequate for most purposes. However, considering we need $\pu{300ml}$ and $\pu{900ml}$, we would probably want to work with very large cylinders so the error would be in the range of $5{-}\pu{10ml}$. It is possible to use much more precise equipment: in analytical lab courses, we would use pipettes that measure exactly $\pu{25.0 \pm 0.1 ml}$ and dilute using measuring flasks at a similar level of precision.

If, as it seems from the values you plugged, the initial solution was $\pu{4M}$ (1 significant figure), I would argue the result should be given to 1 significant figure too – introducing a large error. Arguably, a proper error propagation calculation would probably give you a better precision in the final result.


How to calculate the boiling point of a solid substance? Say that I have a solid substance; I know its density, mass, volume, melting point, specific heat, the pressure which it is under, and nothing else. (for now.)

How do I find the heat at which it boils? (So it goes solid -> liquid -> gas)

you can't...it must be found experimentally (it has) P.S Even if you know the BP at 1 atm, you can never calculate the real boiling point at any given BP, you may come close


Comparing the basicity of benzamide and acetamide What if we were asked to compare the basicity of benzamide and acetamide?

Which one of them would be more basic? I read somewhere on the internet that benzamide must be more basic than acetamide due to cross conjugation (I am not aware of this term). Is this true? Please provide an explanation.

Where do amides accept protons?

As Mithoron aptly pointed out in the comments, it is first important to establish which atom in the amide molecule receives the proton. Due to the strong delocalisation present in the amide, the nitrogen lone pair becomes very weakly basic and it is the lone pair on the carbonyl oxygen which is donated to form bonds with $\ce {H^+}$. Not only does the delocalisation diminish the basicity of the nitrogen lone pair, the donation of the lone pair from the nitrogen atom towards the carbonyl group effectively raises the energy of the lone pairs present on the oxygen atom.

This is more clearly illustrated in the molecular orbital picture, where we consider the interactions between the lone pair orbital on $\ce {N}$ and the $\ce {C=O}$ $\pi$ and $\pi^*$ MOs. The interaction with the $\pi$ MO helps to increase the energies of the electrons on the $\ce {O}$ atom while the interaction with the $\pi^*$ MO helps to lower the energy of the lone pair on the $\ce {N}$.

Let's look at some data

I personally find that a theoretical approach to this problem without looking at any data is non-trivial and may potentially give a false conclusion since many factors are at play. In fact, my initial thoughts without looking at any data was that benzamide would be more basic since the benzene ring can donate into the $\ce {C=O}$ group via resonance (in the same way as how a benzyl carbocation is stabilised by the donation of $\pi$ electrons from the ring) and this would certainly enhance the basicity of the carbonyl oxygen since the electron density at the oxygen atom would have increased significantly.

Cox et al. (1980) have determined via empirical measurements the $\ce {pk_a}$s of the conjugate acids of acetamide and benzamide to be $\ce {-0.73}$ and $\ce {-1.54}$ respectively, with an uncertainty of $\ce {0.02}$. This tells us that acetamide is actually more basic than benzamide. In the discussion of their results, the authors considered the factors which they thought underlied this basicity difference. These included the following:

Effect 1: inductive withdrawal due to the more electronegative $\ce {sp^2}$ hybridised carbon atoms of the aromatic ring, decreasing basicity of benzamide
Effect 2: resonance stabilisation of the positive charge in the conjugate acid of the benzamide molecule, increasing the basicity of benzamide
Effect 3: more negative entropy change of ionisation for benzamide, decreasing basicity of benzamide

The authors suggested that perhaps, effects 1 and 3 were more significant combined than effect 2, resulting in the observed difference. Effect 3 was not well-explained by the authors. I suppose effect 3 arises due to the difference in size of the methyl and phenyl substituents and the consequent difference in the extent of their solvation. The conjugate acid of benzamide may require more solvation and hence, a greater ordering of water molecules, giving rise to a more negative entropy change. This is my rationalisation of effect 3.

Further insights provided

In addition to determining the acidity constants of the conjugate acids, the authors also calculated the deviation from coplanarity of the the benzene ring of the various N-substituted benzamides. They observed that as the deviation from coplanarity increased due to the presence of large and bulky substituents on the $\ce {N}$ atom, the $\ce {pk_a}$ value of the N-substituted benzamides became more "acetamide-like". They then went on to say:

  This shows that whatever the nature of the effect which causes benzamide to be less basic than acetamide, the degree of coplanarity in the molecule is important
  to it.

Perhaps, there is intramolecular hydrogen bonding?

I believe that the effect at play here that was not mentioned by Cox et al. may be the presence of intramolecular hydrogen bonding in the benzamide molecule. To highlight the significance of this effect, I will now discuss the computational findings of Muthuraja et al. (2017). The following image (the grahical abstract of their study) shows the subject of the computation study and it also shows the computed Mulliken charges on the different atoms. Notice that most of the aromatic $\ce {H}$s have much more signficant positive charge (i.e. in excess of $\ce {+0.1}$ compared to the other protons not bound to the aromatic ring. Additionally, note the large negative charge value on the carbonyl oxygens.

The authors posit high possibility of intramolecular hydrogen bonding between the carbonyl oxygen of the lactam with the neighbouring aromatic proton. They cite the following evidence justifying their claim:

Firstly, the $\ce {O...H}$ distance ($\ce {2.18}$ angstroms) is shorter than their van der Waal distance ($\ce {2.72}$ angstroms)
Secondly, the carbonyl stretching frequency is red-shifted from the normal stretching frequency

From this computation study, I simply wish to point out that there is a significant chance that intramolecular hydrogen bonding may be present in benzamide due to the proximity of the carbonyl oxygen to the aromatic proton. This interaction decreases the availability of the lone pair on the carbonyl oxygen for donation to acids. Thus, benzamide is less basic. Furthermore, this agrees with the additional insight provided by Cox et al. (1980): As deviation from coplanarity increases, there is less and less intramolecular $\ce {H}$ bonding and thus, greater and greater availability of the lone pairs on the $\ce {O}$ for donation to acids.

References

Cox, R. A.; Druet, L. M.; Klausner, A. E.; Modro, T. A.; Wan, P.; Yates, K. Protonation acidity constants for some benzamides, acetamides, and lactams. Can. J. Chem 1980, 59, 1568-1573.
Muthuraja, P.; Shanmugavadivua, T.; Joselin  Beaulab, T.; Bena Jothy, V.; Dhandapani, M. Influence of intramolecular hydrogen bonding interaction on the molecular properties of N-p-tolyl-5-oxo pyrrolidine-3-carboxylic acid: A theoretical and experimental study. Chem. Phys. Lett. 2017, 691, 114-121.


How to calculate the number of protons in a sphere given the volume and the pH? The spherical radius of a nano-droplet of a solution ($\mathrm{pH}$ of which is $6$) is given as $\pu{20nm}$.

I can find the molar concentration of $\ce{H+}$ and subsequently the number of $\ce{H+}$ ions, but I am getting a fractional number ($\approx 2 \times 10^{-2}$). This is obviously wrong as there should be at least $1$ proton in a compartment which has an acidic $\mathrm{pH}$. What am I doing wrong?

As pointed out in the comments, the concept of pH as a thermodynamic property of nanodroplets does not quite make sense unless the nanodroplets are in equilibrium (stable). If they are in equilibrium you can still apply the equation
$$\textrm{pH}=-\log_{10}(a_{\ce{H+}})$$ but now you have to be aware that the activity will be influenced by surface (and other) effects and the approximation $$\textrm{pH}=-\log_{10}(c_{\ce{H+}})$$ may not hold. But assume it does. Then as you point out, the average number of hydronium ions per nanodroplet is

$$\begin{align} n_{\ce{H+}} &=\pu{10^{-6} M}\cdot(\frac{4}{3}\pi\times\pu{20^3 nm^3}\times\pu{10^{-27} {(m/nm)^3}}\times\pu{10^3 L/m^3})\times \pu{6.022\times10^{23} mol^{-1}}\\&=0.020 \end{align}$$

Which is ok. That just means that on average only 1 out of 50 nanodroplets contains a solvated proton. Note by the way that one nanodroplet also contains about a million water molecules.

This is a real "rabbit hole" of a question, given all that is not specified, which is why I used sleight-of-hand and wrote that I am ignoring surface effects and assuming the particles are at equilibrium. It doesn't address how to prepare nanodroplets with the desired average proton concentration or what equilibrium means (are nanodroplets exchanging matter or are they isolated?). As noted in comments, composition (buffering molecules if any, salts and other ions including hydroxyl), droplet charge etc, are being ignored and you can expect statistical fluctuations in concentrations between different droplets. However if I stick to the assumption of equilibrium then I can say (for what it's worth) that ~2% of droplets have ~pH 4.3, and most of the rest are at neutral pH.


Does rice protein have a high concentration of arsenic? According to recent reports rice is high in arsenic. What about rice protein?

I am taking a protein powder that has "yellow pea protein isolate, sprouted whole-grain rice protein concentrate, organic instantized quinoa seed protein, organic Chia Sprout protein".

Would protein concentrate from Rice have high levels of arsenic?

http://www.consumerreports.org/cro/magazine/2015/01/how-much-arsenic-is-in-your-rice/index.htm

The US FDA reported arsenic testing of 12 rice protein powders as follows:

  333863609,
  Dietary Supplements,
  Rice Protein Powders,
  Plant-Based Energizer:
  0.1

  333511892,
  Dietary Supplements,
  Rice Protein Powders,
  Protein Energizer:
  0.8

  333511937,
  Dietary Supplements,
  Rice Protein Powders,
  Protein Energizer:
  0.9

  334620956,
  Dietary Supplements,
  Rice Protein Powders,
  Raw Vegan Chocolate:
  0.9

  334094485,
  Dietary Supplements,
  Rice Protein Powders
  Plant-Based Drink Mix:
  1.0

  333511880,
  Dietary Supplements,
  Rice Protein Powders,
  Raw Protein:
  1.1

  333687971,
  Dietary Supplements,
  Rice Protein Powders,
  High Protein Energy Meal:
  1.4

  333687931,
  Dietary Supplements,
  Rice Protein Powders,
  High Protein Energy Meal:
  1.6

  334093871,
  Dietary Supplements,
  Rice Protein Powders,
  Blend of protein, fruits & veggies:
  1.6

  334094067,
  Dietary Supplements,
  Rice Protein Powders,
  Chocolate Plant Protein Shake:
  3.7

  334093956,
  Dietary Supplements,
  Rice Protein Powders,
  Blend of protein, fruits & veggies:
  4.4

  333688312,
  Dietary Supplements,
  Rice Protein Powders,
  Organic Meal Replacement:
  5.7

Where the units are micrograms of inorganic arsenic per serving.


Theoretical Drying capacity of CaSO4 What is the Theoretical Drying capacity of CaSO4 when used to dry solvents? e.g. How much water can 1 gm of CaSO4 remove?

Now since CaSO4 itself comes in several forms (anhydrous, hemihydrate, dihydrate) I am wondering which is the form best suited for water absorption?

If I assume that the drying transition is from CaSO4 (anhydr.) to CaSO4.2H2O (dihydrate) then I get a drying capacity of approx. 0.26 gm H2O per gm CaSO4 used.

Is this the right capacity? Or am I using an incorrect reaction?

The actual drying agent is calcium sulfate semihydrate. It has low drying capacity, which decreases already after the uptake of $6.6\%$ water (theoretical maximum is $15.7\%$). Within these limits, however, it is possible, for example, to dry air at room temperature down to $0.0004\ mg$ of water per liter of air. It is suitable for drying almost all liquids and gases. It can be regenerated by heating to $190-230^{\circ}C$. Above $300^{\circ}C$ it loses its drying properties. The data comes from an old Merck's handbook on drying agents.


Is there a practical re-use value gained by grinding up activated carbon that is spent/saturated? I want to buy bulk activated carbon for use in a DIY air-filtration set-up, like this, to improve air quality in my bedroom.
From what I understand, the relatively larger sizes of carbon media on the market are optimal for air filtration to maximize the amount of time the air is in contact with the carbon.
My question is, once my activated large-granule carbon is used up, say after a few months in my air filter, am I throwing away any untapped potential? Specifically, is there value in further crushing/grinding the carbon and then putting it into another application optimized for the new, more fine-grained size-- perhaps a water filter, or an air filter where the carbon is in a long tube, and air is moving slowly through it.
The same FAQ suggests it is impractical and messy to grind up the AC, but it seems to me that mess can be contained with a bit of planning. I'm also happy to be pointed to general resources that are helpful, especially if I'm barking up the wrong intuition trees, here. Thanks!

Activated carbon is called "activated" when its grains contain a large amount of large cavities. These cavities are big enough to hold large molecules of undesired substances (oils, smokes) which are present in the air. Only small molecules like oxygen and nitrogen from the air can go through without being retained. So the activated carbon removes pollution from the air. After a long time, there are more and more cavities occupied by these unwanted substances. As a consequence, the effectiveness of the activated carbon slowly decreases. It has to be replaced. But it is no use grinding it. If one wants to reuse this carbon again, one should heat it up to such a high temperature that the substances fixed in the cavities are vaporized. Unfortunately some substances like soot will never get vaporized. So the effectivity of the activated carbon will never be entirely restored


Electrophilic addition of HX to 2,3-dimethylbut-2-ene Generally, for the addition of dry HX to an alkene, it follows Markovnikov's rule. The hydrogen atom adds to the carbon with the most hydrogen atoms.

However, in the case of an alkene which does not have any hydrogen atom attached to its carbon, such as 2,3-dimethyl2-butene, would the electrophilic addition still take place?

Electrophilic addition would likely still occur. Remember that Markovnikov's rule is an observational heuristic for "in electrophilic addition reactions of alkenes in which carbocations form, the more/most stable carbocation forms preferentially, and thus the product originating from that cation is the major product".

In this case, a carbocation can form. Nothing about the tetrasubstituted pattern prevents addition. The molecule is still planar and can be aproached by the electrophile. The methyl groups are electron donating by induction. The methyl groups 1) increase the electron density of the alkene and make it more nucleophilic, and 2) stabilize the carbocation intermediate.

Thus, 2,3-dimethyl-2-butene is likely more reactive than other alkenes toward electrophiles. For example:

$$\ce{(CH3)2C=C(CH3)2 + HBr -> (CH3)2BrC-CH(CH3)2}$$


IUPAC nomenclature of alicyclic compound with an equal number of cyclic and acyclic carbons In this compound there are three cyclic and three acyclic carbons. Even the substituent comes on the 2nd position in both cases. So, should the main chain be the cyclic one, since there is a rule that cyclic has a priority over acyclic?

According to current IUPAC recommendations (2013), for the preferred IUPAC name, a ring has seniority over a chain. Therefore, the given structure is a propylcyclopropane. (However, the IUPAC recommendations allow that the context may favour the chain.)

The two prefixes of cyclopropane (chloro- and 2-bromopropyl-) are arranged alphabetically. Thus, the name is 1-(2-bromopropyl)-2-chlorocyclopropane.


Do nitrogen and oxygen react to form nitric oxide or nitrous oxide? In many books it's written that:

$\ce{N2 + O2 -> 2NO}$

and in many others it is written that:

$\ce{2N2 + O2 -> 2N2O}$

Which one of these is actually formed?

Or

Is it that both are formed in variable amounts, but depending on certain reaction conditions one is formed more than the other?

Oxygen and nitrogen actually react to form nitric oxide.

  $$\ce{O2 + N2 ⇄ 2NO}$$

  Oxygen react with nitrogen to produce nitrogen(II) oxide. This
  reaction takes place at the high temperature[about 2000°C(without catalyst) ;
  1200-1300°C(with catalyst)), an overpressure and in the presence of a
    catalyst.
  In this reaction, the catalyst is can be platinum, manganese(IV)
  oxide.(Chemiday)

In wikipedia article of nitric oxide, it is written that:

  The uncatalyzed endothermic reaction of oxygen ($\ce{O2}$) and nitrogen ($\ce{N2}$), which is performed at high temperature (>2000 °C) by lightning
  has not been developed into a practical commercial synthesis (see
  Birkeland–Eyde process):

  $$\ce{N2 + O2 → 2 ·NO}$$

  From a thermodynamic perspective, $\ce{·NO}$ is unstable with respect to $\ce{O2}$ and $\ce{N2}$, although this conversion is very slow at ambient temperatures
  in the absence of a catalyst. Because the heat of formation of $\ce{·NO}$ is
  endothermic, its synthesis from molecular nitrogen and oxygen
  requires elevated temperatures above 1000 °C.

Why nitrous oxide is not formed can be explained through thermodynamics. From Quora:

  For any chemical reaction to take place spontaneously, we look at a
  factor known as free energy available for that reaction. Gibbs's Free
  Energy is given by the equation:

  $$\ce{\Delta G = \Delta H − T\Delta S}$$     This is one of the most
  important equations in all of chemistry
  because it provides information as to whether a particular reaction is
  spontaneously possible or not.

  The quantity $\ce{\Delta G}$ must be negative for a reaction to take
  place spontaneously. For the reaction $\ce{2N2 + O2 −>2N2O}$,
  calculating the values of change in enthalpy and temperature times
  change in entropy for the atmosphere, we see that this reaction can't
  take place.

Generally, reaction between nitrogen and oxygen produces $\ce{NO_x}$ which causes air pollution.

  $\ce{NO_x}$ gases are usually produced from the reaction among nitrogen and oxygen during combustion of fuels, such as
  hydrocarbons, in air; especially at high temperatures, such as occur
  in car engines.The term $\ce{NO_x}$ is chemistry shorthand for
  molecules containing one nitrogen and one or more oxygen atom. It is
  generally not meant to include nitrous oxide ($\ce{N2O}$).


MDLCT (Connection Table) Format specification Is there a specification for MDL Connection Table format somewhere. At some places it is mentioned that the format writes the line length as a character then the actual line characters. If I save a structure using ChemDraw, I see no such thing and the lines are separated using CarriageReturn and LineFeed characters. Can someone please point me to the correct format? Is it that the format is different while SAVING TO FILE and while COPYING TO CLIPBOARD?

I know this is an old question but like to provide some detail since this does not seem to be documented officially anywhere.
The MDL connection table format used in MOL files is described in the CT File Formats document referenced in the first answer. This is simple ASCII text with hard linebreaks (LF or CR+LF, depending on platform).

But there is also a special clipboard format MDL_CT created by ISIS/Draw, Accelrys/Draw etc. on Windows. This is not simply a MOL file put as text onto the clipboard. Instead each line of the molfile data is prefixed by a byte holding the number of characters, followed by the ASCII (one byte) characters of the line. The lines are not terminated by line breaks! This is basically the good old TurboPascal string format.

In addition to a single molecule the MDL_CT clipboard format also supports reactions. In that case the data comes from a RXN file, which is converted to the clipboard format the same way a molfile is: prefix each line with a length byte, remove hard linebreaks.


Removing sodium borate for the test proving thiocyanate ions I've just learned that when carrying out the identification test for thiocyanate ions borax can induce a false positive and therefore needs to be removed. I couldn't find the answer on google so I'm asking a question here.

I would think you first carry out the test to see if borax is present; but if it is, how would you remove it?

Lehrbuch der analytischen und präparativen anorganischen Chemie by Jander et al. [1] suggests to precipitate borate with $\ce{Ba^2+}$ in neutral solution beforehand as barium(II) thiocyanate is one of the few barium(II) salts actually soluble in water, or use an excess of $\ce{Fe^3+}$ to make sure all the interfering anions are precipitated.
From [1, p. 358]:

6 . Nachweis als $\ce{Fe(SCN)3}$
[…]
Störungen: Um den störenden Einfluss von $\ce{F-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{H3BO3}$, $\ce{C4H4O6^2-}$, $\ce{C2O4^2-}$ usw., die mit $\ce{Fe3+}$-Ionen Komplexe bilden, auszuschalten, wird $\ce{Fe^3+}$ im Überschuss zugegeben.

From [1, p. 418–419] (emphasis mine):

9. Nachweis von Fe(III) als $\ce{Fe(SCN)3}$
$$\ce{Fe^3+ + 3 SCN- <=> Fe(SCN)3}$$
[…]
Störungen: […] Ferner beeinträchtigen
$\ce{Hg^2+}$-Ionen durch Bildung von wenig dissoziiertem $\ce{Hg(SCN)2}$, $\ce{F-}$ durch $\ce{[FeF6]^3-}$-Komplexbildung, die Anionen organischer Säuren ebenfalls durch Komplexbildung und auch $\ce{PO4^3-}$, $\ce{AsO4^3-}$, Borat-Ionen sowie ein größerer Mineralsäureüberschuss die Reaktion. Es ist
daher ratsam, vor der Prüfung […] die störenden Anionen in neutraler Lösung mit $\ce{Ba^2+}$ abzutrennen.

References

Jander, G.; Blasius, E.; Strähle, J.; Schweda, E.; Rossi, R. Lehrbuch der analytischen und präparativen anorganischen Chemie, 16. Auflage.; Hirzel: Stuttgart, 2006. ISBN 978-3-7776-1388-8. (in German)


Highly conjugated compounds and the effects of substiuents TLDR, what is the effect of electron donating and withdrawing groups on the UV-Vis absorption spectra of pi-conjugated system?

For a conjugated system, my understanding is that as the extent of conjugation increases, the difference in energy between the highest unoccupied molecular orbital and the lowest unoccupied molecular orbital decreases. The delocalization of electrons in a highly conjugated molecule provides the molecule with a greater number of molecular orbitals, decreasing their spacing. Conjugated compounds which absorb visible light do so when the difference in energy of the homo and lumo correspond to a frequency in the visible spectrum.

My question is in regards to the affects of electron donating and withdrawing groups and the absorption of light by these compounds in the visible spectrum.

My intuition is that electron donating groups will cause greater delocalization of electrons within the compound, creating more relevant MO's, decreasing spacing and reducing the HOMO- LUMO energy gap, causing the compound to absorb lower energy (longer wavelength light). Is my intuition correct?

Yes, your intuition is correct.

Among others properties, systems with pi-electrons delocalisation are actually chromophore whose UV-Vis absorption spectrum is mainly dictated by the HOMO-LUMO transition.

An extension of the delocalised system results in a narrowing of the gap and thus shifts the spectrum to red. The effect is dramatic when extending, e.g., the length of a polyene chain.

However, although with different impact, the same is true if a parent system is substituted with group leading to additional resonance, both because of electron donating or withdrawing effects.

This is fundamentally no different from considering a C=C double bond, which is itself a chromophore, and functionalize it with the various auxochromes that are commonly listed when introducing UV-Vis electronic absorption spectroscopy.

An auxochrome as listed won't necessarily delocalise over the conjugated system in complicated structure, or it can marginally do so. Thus, don't assume that if there is a methoxy - just to mention an auxochrome - somewhere in the molecule or name, the spectrum of the latter is necessarily red shifted as compared to the parent molecule lacking that auxochrome.

If the auxochrome can indeed partecipate in delocalisation, then its effect is bathochromic (red shift) and generally hyperchromic (higher absorbivity).

Rules for predicting the wavelength in the absorption spectrum of conjugated systems do exist, though they fail quite early with increasing the conjugation length.

Consider that the conjugation length of a system is better described by an effective conjugation length, as for the system is susceptible of various effects, first of all departure from planarity due to hindrance. The same is true for the effect of the substituent. If you think of a very long polyene, an end substitution doesn't necessarily extend its effect all along the formally conjugated system.

Similarly the concomitant presence of electron withdrawing and donating groups can have negligible or no effect, as their influences can somehow cancel each others.

Again, is very important to consider that the effective delocalisation is not only influenced by the size of the system, but by an ensemble of parameters, such as the already mentioned steryc hindrance, structural motifs that might pin electrons (think of a phenylene in the middle of an otherwise polyene chain), solvent, etc.

With a bit of caution, you can certainly take that substitution of a conjugated system with groups having mesomeric effects results in absorption maximum at longer wavelength as well as in a higher absorptivity.


Reverse electron demand Diels-Alder Why is the reverse electron demand Diels-Alder reaction much rarer than the normal one? By reverse electron demand, I mean one with an electron-rich dienophile and an electron deficient diene.

In the book by Clayden they have drawn a diagram with the electron deficient dienophile lower in energy and saying that it would prefer to give its LUMO to the reaction.

But, I don't get what's wrong with an electron-rich dienophile lending its HOMO.

Inverse electron demand reactions are inefficient and slow. Even the prototypical Diels-Alder reaction of 1,3-butadiene and ethylene is quite sluggish. From here you can see the reaction only proceeds to 93.5% yield when ran with ethylene in excess by 50:1 at 200ºC and 350 atm for 4 hours. Yikes!
Now consider that some normal electron demand D-A reactions can occur at room temperature. In fact, 1,3-butadiene spontaneously dimerizes to 4-vinylcyclohexene while in storage, meaning the dimer must be cracked by heating immediately before use. This should give you an idea of the sensitivity of D-A reactions to substituent groups. By lowering the HOMO-LUMO gap, orbital overlap is improved, lowering the activation energy and resulting in a faster reaction.
To supplement and support my answer, here is a paraphrased excerpt from Modern Physical Organic Chemistry by Anslyn and Dougherty (2006):

Electron-withdrawing groups on the dienophile lower its LUMO. This occurs because electronegative substituents have correspondingly lower energy MOs, and lower the energies of all MOs in which they participate. Inversely, electron-donating groups raise the diene MOs.

In support of this analysis, the log($k$) for Diels-Alder reactions correlate quite well with the inverse of the difference in energy between the ionization potential of the diene (related to the energy of the HOMO) and the electron affinity of the dienophile (related to the energy of the LUMO). To further enhance the electrophilicity of the dienophile, it is common to add Lewis acids that can complex the electron withdrawing groups on the dienophile, further lowering the LUMO.


Ozonolysis mechanism question Is this mechanism will be similar to ozonolysis ?

My try :

My Question is will B be the answer ? And what does Me2S do ?

The first part of your question has been answered in the comment, as for the part about what does $\ce{Me2S}$ do:
The mechanism used here is reductive ozonolysis and $\ce{Me2S}$ acts a reducing reducing agent itself getting oxidised to dimethysulphoxide $\ce{(Me)2SO}$ (DMSO).
As for the mechanism of reductive ozonolysis with $\ce{Me2S}$ here's a helpful link.


The MO and unpaired electrons I am learning molecular orbital in general chemistry. I have come across this question that asks me to show that the Lewis structure of $\ce{O2}$ is actually an excited state when viewed in terms of the molecular orbital energy.(Zumdahl 9th edition, chapter 9, Excercise number 50) The solution book tells me that the Lewis structure of $\ce{O2}$ is an excited state because the two electrons in the $\pi ^*2p$ does not follow Hund's rule(the two electrons there exist in the same $\pi ^*2p$ orbital while leaving the other $\pi ^*2p$ orbital empty.) And it further states that the reason why the two electrons in $\ce{O2}$ Lewis structure are paired is because the Lewis structure of $\ce{O2}$ shows every electron to be paired. But my question here is, why does being paired in the Lewis structure equate to being paired in MO? I mean, when MO is formed, aren't all orbitals reconfigured, and therefore all electrons are also reconfigured?

While I was thinking about this, another question popped out of my mind. Why are the name of MOs like $\sigma$ and $\pi$? Doesn't $\sigma$ and $\pi$ get used when talking about $\sigma$ and $\pi$ bonds? Are $\sigma$ orbitals and $\sigma$ bonds correlated in anyway?

Oxygen can exist in two states, the singlet and triplet state. The triplet state is the ground state whereas the singlet state is the higher energy state. In the triplet state the highest energy electrons are unpaired whereas in the singlet state the highest energy electrons are paired.
Now, take a look at the MO diagrams of both states in Wikipedia. You'll be able to tell that the singlet state does not obey Hund's first rule.
Now lets look at the Lewis diagram for a oxygen molecule:

As you can see, all of the electrons in this diagram are paired. The oxygen represented in the diagram above is an oxygen that has two bonds and two lone pairs. The only MO state that has lone pairs is the singlet state.
Now on to "why does being paired in the Lewis structure equate to being paired in MO?" The reason is that paired electrons in Lewis diagrams represent paired electrons. The concept of paired electrons does not depend whether you are using MO theory or Lewis diagrams. Paired electrons are electrons that are in the same orbital.
Now, the next question. "Why are the name of MOs like σ and π? Doesn't σ and π get used when talking about σ and π bonds? Are σ orbitals and σ bonds correlated in anyway?" Sigma bonds and sigma orbitals refer to the same thing. Molecular orbitals are a way to represent the bonds formed by two atoms.
Sidenote: Mulliken first proposed the sigma, pi, delta notation according to this journal article..


Do acids react with neutral salts? I am new to the concept of neutralization.
My teacher told me that acids do not react with neutral salts.
I want to know why not.

If both are dissolved in water and both completely dissociate, why can't the ions of acid and salt exchange and perform a double displacement reaction.

Moreover I have seen the following reaction in a book: $$\ce{2 NaCl + H2SO4 -> 2 HCl + Na2SO4}$$

Neutralization is a reaction between acid and base which results in the formation of a salt and water.
When you have any ionic compound and dissolve it into water, dissociation occurs.
This means that you get a solution of ions.
So if you dissolve acids, bases or salts into water, you end up with a mixture of ions.
Then, more things may happen or not.

In the case of the given reaction, you will get a solution with:

$\ce{Na+ + Cl- + H+ + SO4^{2-}}$

If the system is closed, nothing will happen.
But if the system is open, then $\ce{HCl}$ will evolve because it is a gas.
Since the sulphuric acid is very polar and a liquid, water is not even required.
This is, in fact, the standard procedure to prepare anhydrous HCL in the lab. Just drop $\ce{H2SO4}$ on $\ce{NaCl}$. It must be said that $\ce{NH4Cl}$ works better.


Why Balmer series exist? I was reading about Bohr's atomic model in our textbook when came up with this question:
Since the electron in hydrogen atom is most stable in the first orbit, why should it descend from higher orbits to any orbit except the first orbit(that means balmer series for example)?

The entire process has two parts: Excitation and relaxation. The first is clear enough, we excite an electron into a higher energy level.

Now this electron wants to fall back down. There is no roadsign or intrinsic selection rule that requires the electron to immediately go back to the ground state. Any state in a lower shell will be less energetic and by chance electron 1 happens to stop in state A while electron 2 of a different atoms stops in state B. Any electron that is not in a ground state will remain there — probably for a duration of about a few femtoseconds or less — and then relax further. The universe is not deterministic so nothing is telling the electron it has to take the fastest path.

Note that this gedankenexperiment does not make any prediction which path is most likely. You would need more theory (that DavePhD is likely to give) to explain that.


Mechanism for interconversion of spin isomers of hydrogen What is the mechanism by which the ortho- and para- spin isomers of hydrogen interconvert?

If such a mechanism exists, does this mean that ortho-hydrogen increases in concentration on increasing temperature?

The first question is:

  What is the mechanism by which spin isomers of hydrogen switch between the ortho and para forms?

There is some explanation in ChemPhysChem 2006, 7 (3), 551–554 (non-paywall version here):

  One can define three situations. In the first, a magnetic conversion occurs without bond elongation or breaking. For example, in solid dihydrogen dipolar magnetic nuclear interactions are responsible for an extremely slow spin conversion.[4] In the presence of unpaired electrons the conversion is accelerated by the magnetic hyperfine interaction.[5] The second situation also involves a magnetic conversion mechanism, but is assisted by an intermediate H–H bond elongation, for example, by binding to a transition metal center. Finally, in the third situation, a H–H bond splitting and re-formation with other hydrogen atoms occurs, which corresponds to a chemical spin conversion.

The second question is:

  If such a mechanism exists, why does ortho-hydrogen not increase in concentration on increasing temperature?

The equilibrium fraction of ortho-hydrogen does increase from 0% at absolute zero to 75% at high temperature.

para-Hydrogen is the lowest energy, most stable form, since it is capable of accessing the $J=0$ rotational level (whereas ortho-hydrogen cannot). There are three microstates corresponding to ortho-hydrogen:

$$|\!\uparrow\uparrow\rangle, |\!\downarrow\downarrow\rangle, |\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle,$$

but only one microstate for para-hydrogen:

$$|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle.$$

So, according to the Boltzmann distribution, the equilibrium fraction of ortho-hydrogen goes from 0% at 0 K to 75% at infinite temperature (in practice, a temperature at which the effects of rotational quantisation are not seen).


How does sodium bicarbonate act as a deodorizing agent? I was reading this other question on the site: Deodorizing using baking powder instead of baking soda.

Apparently, baking soda is used quite often as a cleaning and deodorizing agent. From what I know, it's cleaning effects are because it can denature proteins and saponify oils (convert oils to soap) since it is a weak base.

But when you take a look the Wikipedia page on aroma compounds, the only compounds that could possibly be affected by a weakly basic substance are esters (which would undergo hydrolysis). Most of the compounds are polyunsaturated aldehydes, ketones and alcohols which are not affected by (weak) bases.

Note: I assume here that compounds which cause bad odors would have a similar structure to compounds that are fragrant since they both have to activate whatever receptors there are in the nose. Also I have read that usually the same compound can cause both a bad and a good smell depending on the concentration of the compound, such as indole.

So I ask: Does $\ce{NaHCO3}$ really have any deodorizing effect and if so, what is the mechanism (reactions involved) behind it. Also would it be possible to achieve the same effect with any random alkaline substance ($\ce{NaOH}$ maybe, although not very practical).

The key factor is that baking soda is amphoteric. This means that it can react both with strong acids and with strong bases which are typically smelly compounds (sour milk, rotten fish, fatty acids in sweat).

The explanation on the third page of article by Lamb [1] gives a nice summary:

  The chemical explanation of why bicarbonate of soda should act as a deodorant
  is of interest. Sweat (3) has an acid reaction, $\ce{pH}$ of $5.2\text{–}6.75$ which is due to the fatty acids or acid phosphates of sodium and potassium which it contains. It also contains $\ce{NaCl}$, $\ce{KCl}$, alkaline salts, organic acids and urea. Minute amounts of unstable fats and oils are secreted in sweat and hydrolysed to their corresponding fatty acids and glycerin. Free fatty acids may be secreted. Many of these acids are volatile and have a characteristic disagreeable odor. These include butyric acid, formic acid, caproic acid and valerie acid. The following are among the possible explanations of the manner in which sodium bicarbonate acts as a deodorant in the axilla.

  It forms a sodium salt with butyric, caproic and valerie acids. These sodium salts are comparatively mild in odor in comparison to the volatile "rancid-smelling" fatty acids.
  By changing the $\ce{pH}$ of the axillae, the bacteria which decompose the fats are unable to act. However, as a result of this the staphylococcic flora which prefer higher alkalinity might thrive and axillary furunculosis might result.

To give an example. The reaction of baking soda with butyric acid looks like this:

$$\ce{C3H7COOH + NaHCO3 -> C3H7COONa +H2O + CO2} $$

while butyric acid is extremely smelly (an intense vomit smell), the salt hardly smells at all.

Given this explanation, I think $\ce{NaOH}$ should obviously also work well for deodorizing in case the smell is caused by acids. The problem would be that smells caused by basic compounds are left untouched.

References

Lamb, J. H. Sodium Bicarbonate: An Excellent Deodorant. Journal of Investigative Dermatology 1946, 7 (3), 131–133. https://doi.org/10.1038/jid.1946.13


Confusion about the charge of electrons while orbiting through bohr's energy shells I came across the following part in my text book. Bohr's atomic theory explained the stability of an atom by the idea of energy shells. He said that while orbiting the nucleus through a particular orbit or energy shell the energy of electrons neither increases nor decreases. The electron does not carry energies while orbiting. The shells are charged and the energy of electrons orbiting through the shell equals to the energy of the shell and as the shells are stable therefore the energy of electrons are also stable . Now, the energy increases from shell to shell like(K,L,M,N,O,P,Q)[ascending order]. So,the energy of an electron in K shell should be less than that of an electron in Q shell. But we know that the energy of an electron is constant and is equal to (4.8*10^-10) esu. Then how can this is possible that an electron with a certain energy does not carry any energy while orbiting and it gains the same energy of its orbit while orbiting?
Note that there may be some mistakes in my informations as there may be many false informations printed in my book but still I am curious about the question. Thank you!

The main confusion I can see in your question is charge vs. energy. Energy is the capacity to do work. Energy comes in many forms. Kinetic energy is the energy of motion. Potential energy is the energy of the system due to its arrangement. Electrons in Bohr shells have energies dependent on the mass of the electron, which is constant, the charge of the electron (constant), the distance from the nucleus (fixed at certain allowable values), and some fundamental constants. Thus, an electron in a shell has unchanging energy. The only way for the electron to change its energy is to move to a different shell. $\Delta E=0$ for an electron in a Bohr shell; $E\neq0$.

An electron in a Bohr shell certainly has kinetic energy because it has orbital angular motion. An electron in a Bohr shell certainly has potential energy because it is being held apart from the oppositely charged nucleus. The total energy of the electron in a Bohr shell ($E_n$) is a function of the principal quantum number, $n$. $Z$ is the number of protons (1 for hydrogen, the only atom for which the Bohr model works) and $R_E$ is the Rydberg energy (related to the Rydberg constant by $R_H=R_E/hc$).

$$E_n=-\frac{Z^2 R_E}{n^2}$$

The Rydberg energy is the smash up of some other more fundamental constants, like the mass of an electron ($m_e = 9.109\times10^{-34} \text{ kg}$), the fundamental unit of charge ($e=1.602\times 10^{-19} \ \text{C}$), Coulomb's constant ($k_e = 8.987 \times 10^9 \ \ce{N m^2 C^{-2}}$ also $k_e=(4\pi\varepsilon_0)^{-1}$, where $\varepsilon_0=8.854\times 10^{-12} \ \text{F/m}$ is the permittivity of a vacuum), and the Bohr radius ($a_0 = 5.29\times 10^{-11} \ \text{m}$). The Bohr radius is itself equal to a mashup of other constants  $a_0=\hbar/(m_e c \alpha)$, where $c=2.998\times10^8 \ \text{m/s}$ is the speed of light in a vacuum, $\hbar=1.054\times10^{-34} \ \text{J s}$ is the reduced Planck constant, and $\alpha=7.297\times^{-3}$ is the fine-structure constant. Eventually, $R_E$ simplifies down to a relativistic energy:

$$R_E = \frac{k_e e^2}{2a_o}= \frac{1}{2}m_e c^2 \alpha$$

Electric charge is something else. It is a unit of electrostatic interaction with other charged particles. The electric charge of an object influences the energy of that object in the presence of an electric field, see Coulomb's Law. The (magnitude of) charge on an electron ($e$) is part of the Rydberg energy, so it influences the total energy of the electron.

However, charge$\ \neq\ $energy.


Measuring mole fraction in lab using Raoult's law I was reading in my textbook of chemistry on the vapour pressure in liquid compounds, and came across a doubt regarding vapour pressure. If I have two liquids, one is A and another is B, and if I know their vapour pressure in a pure state and their initial mole fraction, would their mole fractions be the same in the vapour phase (after mixing them)? If the answer is yes, how can we prove it? Secondly, if it is practically not possible what other factors should I take into account while solving their value using Raoult's law?

The mole fractions in the gas phase also depend on the relative volatility of the two liquids, so they will not in general be equal to the mole fractions in the liquid mixture. For instance, if you have components A and B, then their vapor pressures will be

$$ \begin{align} p_A &= \chi_A p^\circ _A \\ p_B &= \chi_B p^\circ _B \end{align}$$

where $\chi_i$ is the mole fraction of component i in the liquid and $p^\circ_i$ is the vapor pressure of the component when pure. According to Dalton's law the mole fraction of a given component in the ideal gas is proportional to its contribution to the total pressure so that

$$\begin{align}  p&=p_A+p_B \tag{1}\\ &= \chi_A p^\circ _A + \chi_B p^\circ _B \\ &= \chi_A (p^\circ _A -p^\circ _B) + p^\circ _B \end{align}$$

From this last expression you can conclude that the total pressure above the mixture will vary linearly with the mole fraction of either component in the liquid.

You can also show that the mole fraction of either component in the liquid is not equal to that in the gas. For instance for component A you can write by combining Raoult's and Dalton's laws that

$$\chi_{A,g}=\frac{p_A}{p}=\chi_{A,l}\frac{p_A^\circ}{p} \tag{2}$$

where the mole fractions in the liquid and gas are labelled explicitly with subscripts $l$ and $g$. It should be clear that $\chi_{A,g} \neq \chi_{A,l}$ unless $p=p_A^\circ$, but the latter condition is only satisfied for the pure substance A.


Naming a synthetic molecule from QM9 dataset I was trying to write a name for a synthetic molecule in QM9 dataset. The molecule is:

is 3-aminooxolane-2-carbonitrile a correct name for this?

(note: The structure representation is suspicious and confusing. Linear nitrile $\ce{-C#N}$ group is "broken". But I see some software products generate such representation automatically.)

If the structure is really meant saturated, no stereochemistry is indicated.

3-aminooxolane-2-carbonitrile

(Watch the typo in your proposal …oxalane.) But it's not present at PubChem. (SMILES: N#CC1OCCC1N)

Maybe, if unsatured, furan derivative is meant:

3-aminofuran-2-carbonitrile


Why does entropy increase when the difference in temperatures is decreased? From the Wikipedia article about Entropy (Source view via The Internet Archive):

In irreversible heat transfer, heat energy is irreversibly transferred from the higher temperature system to the lower temperature system, and the combined entropy of the systems increases.

If entropy were a measure of disorder, or randomness, or the amount of information needed to describe the microstates of a system, then it seems that when the difference in temperatures is high, entropy should be high. A temperature gradient in the systems is a form of ‘disorder’ hence should be a sign of higher entropy.
So why does entropy increase after the heat transfer, as the combined system approaches a uniform temperature? It seems to me like a state with less randomness or disorder.

Consider two tanks of water. One hot, one cold.

You open a valve between the two. The temperature between the two bleeds, and the hot tank loses entropy, and the cold tank gains entropy. However, entropy is also gained just by this process happening, as it would be impossible to force the 'cold' tank to retransfer the heat back to the 'hot' tank.

Also, you must consider something else. The two conjoined systems now have more molecules than either system alone, increasing the 'randomness' as a whole. Considering dice, it's like having two sets of two, then one set of four. The set of four has more possibilities than just the set of two.


Why does nitration of aromatic amines give meta directed nitro compounds? Amines are activating groups and donate electrons to the resonance structure using the lone pair of electrons on Nitrogen.
But when aromatic amines are nitrated, for example when benzyl amine is nitrated, it gives a meta nitro benzyl amine.
Given the activating nature of amine groups, shouldn't the compound be an ortho-para directing variant?
I tried figuring it out. One explanation I've come up with is that when it's nitrated, the proton from the nitrous acid forms a coordinate covalent bond with the amine group closing the lone pair electron to the bond. As a result it becomes an $\ce{NH3+}$ group. That would mean that it has a deactivating effect or electron withdrawing effect. That's why the incoming nitro groups are attached to the meta positions. But then why doesn't the same happen for nitration of hydroxide groups, for instance phenol?

A couple things here.  First, it's nitric acid, not nitrous acid, that is the nitrating reagent.  Otherwise, your own explanation is correct.  Orthocresol points out that phenols are sufficiently inert to protons, to avoid this situation.

Second, you can make ortho/para directed nitrated aromatic amines, if you are a little more clever and if you have a tertiary amine (http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch22/ch22-3-6.html).  Use nitrous (not nitric) acid, which is weaker and will react only with the unprotonated amine.  That gives you a nitroso $\ce {-NO}$ group which you can then oxidize to the nitro group.


Oxidation Half Reaction Chemistry I was always confused as to how electrons where donated from the anode in a galvanic cell. But after doing research, I found that there are four fundamental ways as to where electrons in anodic transformations can come from:

1) Surface metal atoms

2) Solvent molecules

3) Electrolyte anions

4)Base ligands

My question is how do I know from these four options, which is providing the electron?

The only way you can know for sure is to analyze the reaction using oxidation numbers, and figuring out which elements lose/gain electrons. Here is a link explaining how to assign oxidation numbers in a reaction.

As an example, consider the reaction, what is the oxidation number of sulfur in this particular molecule, $S_2O_3$. According to these rules, which are in order of priority, the oxidation number of oxygen is $-2$, unless it is bonded with hydrogen or fluorine, which in this case, it is not. Given that we have three oxygen atoms in this molecule, the total oxidation of oxygen is $-6$, which must be offset by two sulfur atoms so the total oxidation number is $0$. Obviously, the oxidation number of a single sulfur must then be $+3$, so that the molecule is neutral.

In any reaction, you would essentially have to do this for every molecule on both the products and reactants side in order to determine which atoms are losing/gaining electrons.


Can ATP be synthesized and consumed? I do not understand why you could not fill a bowl full of adenosine triphosphate (ATP) and eat it. Is there a way to make ATP other than inside a cell and apply it to your body by injection, ingestion, topical absorption, inhalation, or another method?

To answer the question in the title:

Yes, ATP can be synthesised, isolated and you can even eat it. It would be very expensive to do so but considering others put gold on most of their food that’s not a reason in itself not to. While eating too high a dose of ATP is not beneficial as per Paracelsus’ law, small amounts are certainly not harmful.

However, ATP is rarely shuffled into or out of cells. I’m not sure whether the human body provides mechanisms to shuffle ATP across cell membranes. I highly doubt it, since ATP is typically produced cell-internally by mitochondria and what gets shuffled around via the blood stream is primarily glucose. Most transport mechanisms are adopted to that.

Furthermore, even assuming that ATP does get shuffled across cell membranes in significant amounts, eating a crumb of it would first place it into the stomach. There, in the strongly acidic environment there is practically no way that ATP won’t get hydrolysed into first adenotine and diphosphate and then further broken down. Even if some of it magically manages to get through the stomach unharmed, the slightly alkline pH of the intestines are no better and the digestive tract is swarming with enzymes designed to break down practically everything. Thus, you can be sure that ATP has been fully metabolised before the body has a chance of taking it up into the blood stream.

Some of these problems can be overcome by injecting ATP subcutanously or intravenously but the problem with getting it into the cell remains.


How to choose the correct volumes when preparing a dilution series? Original Question

A solution of dye, Amido Black 10B dye ($10~\mathrm{mM}$, $6.165~\mathrm{g\over L}$), as-provided absorbs light much too strongly for measurement of an absorbance value. It is therefore necessary to dilute it before proceeding with an experiment to construct a standard curve of absorbance vs. concentration or amount of dye. The stock $10~\mathrm{mM}$ Amido Black solution needs to be diluted $200\times$ using purified (RO) water. To achieve the dilution, it is recommended that serial dilutions be used.
How would you dilute the Amido Black 10B $200\times$ to a total volume of $3~\mathrm{mL}$? Assume you only have pipettes that cannot dispense less than $20~\mu\mathrm L$.

Comments
We've already been through this question together in class but despite asking a lot of questions I still didn't understand what my tutor was trying to explain. I already have the answers, but I want to know how to figure it out for myself. My tutor's way was this:

Adding $50~\mu\mathrm L$ of dye into $50~\mu\mathrm L~\ce{H2O}$ gives a $1$ in $2$ dilution.
Take $30~\mu\mathrm L$ of the new solution and add $2970~\mu\mathrm L~\ce{H2O}$, giving a further $1$ in $100$ dilution.

I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$. I know she split $1\!:\!200$ into $1\!:\!2$ first because the pipettes can't dispense less than $20~\mu\mathrm L$.  I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

To your questions:

  I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$.

It's always problematic expressing dilutions as ratios, because it's ambiguous as to what the numbers are referring to.  Here, the "$1$" is referring to the volume of the solution to be diluted, and the "$2$" is referring to the final volume after dilution.  Thus, the $50~\mu\mathrm L$ ("$1\times$") is being diluted to $100~\mu\mathrm L$ ("$2\times$"), and in order to do that you need to add $100~\mu\mathrm L-50~\mu\mathrm L=50~\mu\mathrm L$ of the RO $\ce{H2O}$.  This could just as validly be expressed as a $1\!:\!1$ dilution; thus the ambiguity.

  I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

Per the same $\mathit{Initial~Volume : Final~Volume}$ ratio convention, a $1\!:\!100$ dilution to a final volume of $3~\mathrm{mL}=3000~\mu\mathrm L$ will require starting with $3000\div 100 = 30~\mu\mathrm L$ of the intermediate solution.

More generally:

I prefer to express dilutions using an "n-fold" notation, e.g.:

Dilute $50~\mu\mathrm L$ of dye solution $2$-fold with RO $\ce{H2O}$
Dilute $30~\mu\mathrm L$ of the solution from step ($1$) $100$-fold with RO $\ce{H2O}$

Though there is still room for ambiguity in this phrasing, to me diluting $n$-fold indicates much more clearly that the final volume $V_f$ should be $n\times$ the initial volume $V_i$, and thus that the volume of diluent needed is:

$$
V_d = V_f-V_i = nV_i-V_i = V_i\left(n-1\right)
$$


Unit of the equilibrium constant: contradiction of Bridgman's theorem? The following equation is standard in thermodynamics:

$$
\Delta G^\circ=-RT\log(K)
$$

where $K$ is the equilibrium constant.  In dimensional analysis, Bridgman's theorem tells us that the argument of a transcendental function (like $\log$) must always be dimensionless.  But $K$ may have dimensions (depending on the particular equilibrium).  Why is this OK?

Note: working out dimensions explicitly gives:

\begin{align}
[R]&=EN^{-1}\Theta^{-1}\\
[T]&=\Theta\\
[\Delta G^\circ]&=EN^{-1}
\end{align}

where $\Theta$ is the dimensions of temperature, $N$ is the dimensions of number and $E$ = dimensions of energy = $MLT^{-2}$.  From these, we see that the quantity $\log(K)$ ought to be dimensionless, implying that $K$ should be dimensionless as well.  But it isn't always.

Furthermore, suppose $K$ has dimensions $NL^{-3}$ (for an equilibrium of the form $A+B\leftrightarrow AB$, say).  Suppose now that we scale the units for number by a factor $a$.  Then we get new values for $K,R,\Delta G^\circ$, given by:

\begin{align}
\hat K&=K/a\\
\hat R&=aR\\
\hat{\Delta G^\circ}&=a\Delta G^\circ
\end{align}

From our original equation:

\begin{align}
\Delta G^\circ&=-RT\log(K)\\
a\Delta G^\circ&=-aRT\log(K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(a\hat K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(\hat K)-\hat RT\log(a)
\end{align}

So the new quantities do not satisfy the old equation.  Rather, they satisfy the old equation, but with a constant factor of $-\hat RT\log(a)$ added on.  What is going on here?

The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected.

From thermodynamics it is known that the Gibbs free energy of reation is given by
\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,
\end{equation}

where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively.
Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium.
Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely
\begin{equation}
 \mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).
\end{equation}
From Maxwell's relations it is known that
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}
\end{equation}

but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to
\begin{equation}
 \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}
\end{equation}

and one gets
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,
\end{equation}

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase.
Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get
\begin{equation}
 \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p   \ ,
\end{equation}
so that, introducing the mole fraction $x_{i}$,
\begin{equation}
 \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .
\end{equation}
Please, note that there is a dimensionless quantity inside the logarithm.
Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$
\begin{equation}
a_{i} = \frac{\varphi_{i} p_{i}}{p^0}
\end{equation}
where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition.
When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises
\begin{equation}
 \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .
\end{equation}
Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets
\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}
where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure.
Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that
\begin{equation}
\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}
So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant
\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}
which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via
\begin{equation}
 K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
it is more or less proportional to $K$ and thus gives qualitatively the same information.

Edit: If the solution at hand behaves like an ideal solution then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$
\begin{equation}
 K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor.

Edit: I forgot to mention there are also "versions" of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those "versions" can be traced back to the original equilibrium constant.


Why do internal combustion engines only burn to carbon monoxide, not dioxide? Why don't diesel and regular car engines burn all the way to $\ce{CO2}$?

I think this is sort of covered by What really happens atomically in an explosion? because the burning of the fuel air mixture is basically an explosion. The point is that there isn't a nice simple answer because the reaction is fearsomely complex.

Actually most of the fuel does burn to carbon dioxide. The proportion of carbon monoxide is quite small. In a normally tuned engine there is enough oxygen to burn all the fuel to carbon dioxide but the fuel/air mixture isn't in the cylinder long enough for the reaction to go to completion. You probably know that all modern cars are fitted with catalytic convertors, and what they do is to complete the reaction as the exhaust gases pass through them.


Is the name of this chemical 3-oxopropanoic acid? Is the name of this chemical 3-oxopropanoic acid?

I included "oxo"- for the aldehyde group, and "-oic acid" for the carboxylate group.

However, I'm unsure whether this is the correct name since I cannot find it on the internet, while I can find almost all other names I've been trying to figure out.

The name ‘3-oxopropanoic acid’ for the compound $\ce{OHC-CH2-COOH}$ given in the question is correct.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

  (…)
  7. acids
  (…)
  15. aldehydes
  (…)

Thus, the carboxylic acid group has priority to be cited as a suffix.

Systematic names of carboxylic acids are formed as follows:

  P-65.1.2 Systematic names

  Carboxylic acids are named substitutively using the suffix ‘oic acid’ or ‘carboxylic acid’ and the prefix ‘carboxy’ to describe carboxy groups that cannot be included as the principal characteristic group or to express the carboxy group in the presence of a higher principal characteristic group.

  P-65.1.2.1 Carboxylic acid groups, $\ce{-COOH}$, that conceptually replace a $\ce{-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; (…). Except for (…), systematically formed names are preferred IUPAC names; (…)

Therefore, the preferred IUPAC name (PIN) of the unmodified carboxylic acid is ‘propanoic acid’. (Note that the retained name ‘propionic acid’ may be used in general nomenclature.)

Aldehydes can be systematically named in various ways. Monoaldehydes derived from alkanes are preferably named substitutively using the suffix ‘al’. In this case, however, this method cannot be used since the carboxylic acid group has priority to be cited as a suffix. Therefore, the aldehyde group has to be expressed by a prefix as follows:

  P-66.6.1.3 In the presence of a characteristic group having priority to be cited as a suffix or when present on a side chain, a $\ce{-CHO}$ group is expressed by the preferred prefix ‘oxo’ if located at an end of a carbon chain, or, otherwise, by the preferred prefix ‘formyl’.

Thus, the PIN for the complete structure is ‘3-oxopropanoic acid’.

The corresponding name that is formed by using the prefix ‘formyl’ is ‘formylacetic acid’. Such names are not acceptable as PINs but may be used in general nomenclature.

Furthermore, a special rule for aldehydic acids applies to the given compound:

  P-65.1.6.3 Aldehydic acids

  When a dicarboxylic acid has a retained name (…) and when one of its carboxy groups is replaced by a formyl group, $\ce{-CHO}$ (…), the resulting structure is called an aldehydic acid and, in general nomenclature may be named by replacing the ending ‘ic acid’ of the name of the dicarboxylic acid by the ending ‘aldehydic acid’. Preferred IUPAC names for aldehydic acids derived from all dicarboxylic acids are constructed systematically. The prefix ‘formyl’ is used in preferred IUPAC names, except for a $\ce{-CHO}$ group at the end of an acyclic chain, which is designated by the prefix ‘oxo’.

For the compound given in the question, the PIN of the corresponding dicarboxylic acid is ‘propanedioic acid’.

This dicarboxylic acid also has the retained name ‘malonic acid’, which may be used in general nomenclature. Hence, according to P-65.1.6.3, the corresponding aldehydic acid may be named ‘malonaldehydic acid’ in general nomenclature.

By way of comparison, for the analogous example $\ce{OHC-CH2-CH2-COOH}$, all three corresponding names are explicitly mentioned in the Blue Book:

4-oxobutanoic acid (PIN)
3-formylpropanoic acid
succinaldehydic acid


Entropy and total differential In an exercise, I have derived that

$$\mathrm dS_\mathrm{m} = \frac{3R}{2T}\mathrm dT + \frac{R}{V}\mathrm dV$$

using the 1st and 2nd law of thermodynamics (1 mol ideal gas).

Now, I want to write $\mathrm dS$ as a function of $T$ and $P$ instead of $T$ and $V$. How can I do that?

I have tried to do this:

$$\frac{R}{V} = \frac{P}{T}$$

and

$$\mathrm dV = \frac{\mathrm dV}{\mathrm dP}\mathrm dP = -\frac{RT}{P^2}\mathrm dP$$

This gives me

$$\frac{R}{V}\mathrm dV = -\frac{R}{P}\mathrm dP,$$

which is wrong according to my T.A.

How can I write $\mathrm dS$ as a function of $T$ and $P$?

The short way is to proceed as noted in the comments. The equivalent long way is to start from the total differential for $S_m$ in terms of $p$ and $T$:

$$dS_m = \left(\frac{\partial S_m}{\partial p}\right)_Tdp + \left(\frac{\partial S_m}{\partial T}\right)_pdT$$

The second term is given by

$$ \left(\frac{\partial S_m}{\partial T}\right)_p = \frac{C_p}{T}=\frac{5R}{2T}$$

The first can be determined using the identity

$$ \left(\frac{dS_m}{dp}\right)_T = -\left(\frac{\partial V_m}{\partial T}\right)_p=-\frac{R}{p}$$

Substituting,

$$dS_m = \left(-\frac{R}{p}\right)dp + \left(\frac{5R}{2T}\right)dT$$

It is not difficult to show that this is equivalent to the result obtained by substituting

$$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$

into the expansion in terms of $V$ and $T$ shown in the OP.