text:
Change in Gibbs free energy vs. change in enthalpy
When learning about other thermodynamic quantities, like $\Delta H$, I was taught that this represents the energy absorbed or released by a system after the reaction has gone 100% to completion. For example, if a reaction has $\Delta H = -100 \text{kJ}$, it means that the final state of the system after the reaction is complete has $100 \text{kJ}$ less energy than it did before the reaction began. As such, $\Delta H$ is a statement about the overall reaction, and is not dependent on where the reaction is at the present moment. (Is this understanding correct? Or does $\Delta H$ change throughout a reaction like $\Delta G$?)
However, when learning about Gibbs Free Energy, I have read that $\Delta G$ changes throughout a reaction, as can be seen from $\Delta G = \Delta G^\circ + RT\ln Q$, because Q varies through the course of a chemical reaction. If $\Delta G$ gives information about a specific point in the reaction, then why is there a "$\Delta$" at all? In other words, $\Delta G$ is the difference between what two values?
Thanks for the help.

To start off, $\Delta G$ is concentration-dependent and $\Delta H$ is not. You already showed the concentration-dependence of $\Delta G$. There is no equivalent relationship for $\Delta H$ because it does not depend on concentration. $\Delta H$ reflects how the bond energies change from reactants to products. Bond energies don't depend on concentration. (It also reflects other more subtle things like solvation and intermolecular interactions, but these don't change in an ideal solution).  In contrast, $\Delta G$ has an entropy term, which has to do with the entropy of mixing, resulting in a significant concentration-dependence even in ideal solutions.

Is this understanding correct? Or does ΔH change throughout a reaction like ΔG?

The enthalpy H of the system changes throughout the reaction. Because the enthalpy is not concentration-dependent, it changes in a linear way with the amount that reacts. There are two versions of $\Delta H$ with different dimensions and units. One does what you say: compare H between two states. It has dimensions of energy (typical units: kJ). The other asks what the change of enthalpy is per amount reacted. It has dimensions of energy per amount (typical units: kJ/mol) and is called molar change in enthalpy.
An example is the molar enthalpy of formation. If you want to know how big the enthalpy change is to make a certain amount of a compound, you multiply the amount by the molar enthalpy of formation. The dimensions of the result will be energy again (units: kJ / mol * mol = kJ). Introductory textbooks sometimes gloss over the difference between molar (intensive) and extensive enthalpies.

If ΔG gives information about a specific point in the reaction, then why is there a "Δ" at all?

The Gibbs energy G of the system changes throughout the reaction. The change per amount of product formed decreases as the reaction approaches equilibrium. Again, ΔG can mean two things, either the difference in Gibbs energy between an initial and a final state, or the molar change in Gibbs energy with given concentrations.
You might say concentrations change while the reaction is going on, so it is impossible to make a statement about specific concentrations. You solve this problem by either looking at a very small amount of change (infinitesimal, if you want) or by imagining very large volumes of reaction mix, where concentrations change very little even if - say - one mole of reactants turn into one mole of products. If you want to communicate that this is what you are talking about, you can either use $\Delta_r G$ or $\frac{dG}{d\xi}$. Both have dimensions of energy per amount of substance (typical units kJ/mol).
Again, introductory textbooks tend to gloss over this difference, not giving $\Delta G$ and $\Delta_r G$ different symbols even though they have different dimensions (one is extensive and the other intensive) and mean different things.


text:
Precipitation of Cu2+ and Pb2+ ions
Hi chemistry enthusiasts!
I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge although the standard may be varied across the world. And I'm not quite sure what I did wrong. Could you please have a look?

There is an aqueous solution containing Cu2+ and Pb2+ ions. The most
  suitable reagent to precipitate one of the two ions from the solution
  is (1) nitric acid (2) sodium carbonate (3) sulfuric acid (4) hydrogen
  sulfide

I picked answer (4) hydrogen sulfide because I think that it will react with copper (II) ions and will precipitate out. However, the correct answer is (3) sulfuric acid. How is this answer correct and what did I do wrong for hydrogen sulfide. Your advice will be much appreciated!

If you read the question carefully, they wish to precipitate only one ion, either copper or lead. The facts that must be in your mind are:
(i) Lead sulfate is insoluble in water but copper sulfate is.
(ii) Both lead sulfide and copper sulfide are water insoluble.

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text:
Would burning of around 200 kg of animal fat in a tunnel cause an explosion?
As shown in the movie Ironclad, they burnt a group of fat pigs together, under a tunnel, beneath the fort base and it resulted in a massive explosion.
From chemistry's perspective, and considering the tunnel environment — less oxygen, less space for energy to dissipate,  would this really cause an explosion so massive that it will dismantle the structure above the tunnel? Or would there  be no explosion, just burning?
Apologies for cruelty involved in the question.

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text:
Calorimetry: Temperature Increase due to stirring of reactants
I am not yet in university, but I hear a common first-year chemistry lab/practical is to do calorimetry taking into consideration the small (and usually negligible) rise in temperature due to the reaction container being stirred.
What is the standard method for estimating/calculating this increase in temperature, given the rate at which the stirring occurs and all the other necessary information? Is this possible to obtain an answer using molecular kinetics or are these calculations done solely on guesswork?

There is a difference between swirling the solution in a blender and giving it an "inconsequential" stir. Chemistry is often highly dependent on significant figures, not mathematical exactness. So yes a stir will raise the temperature, but if that stir only raises the temperature by a millionth of a degree when the reaction raises the temperature by 10 degrees, then the stir doesn't really matter.

I suppose you are referring to a modern reincarnation of one of Joule's experiments demonstrating the energy equivalence of heat and work? The idea is that if the container is well insulated, constant stirring will lead to an increase in temperature of the solution, even when reactions are absent. The computation is based on the work done by the stirrer (which should be relatively easy to determine), the heat capacity of the solution and/or the change in temperature (2 of these 3 are required).

This is addressed experimentally, not through theory or computation. You run one experiment starting with the reactants, measuring the change in temperature over time. Then, you run a control with a similar solution that does not react (e.g. the product mixture), setting the mixer to the same speed. This will allow you to correct for any effect not due to the reaction. The stirring will be a minor contribution (depending on the tools used, not measurable at all). Other effects are large and always have to be corrected for (for example, if the reaction needs to be started with a spark and an ignition charge).

What is the standard method for estimating/calculating this increase in temperature, given the rate at which the stirring occurs and all the other necessary information?

The work done to stir is based on the viscosity of the fluid. The faster you spin, the more work is necessary. In the classic experiment by Joule mentioned in the comments, a special stirring mechanism was used to have a large effect on the temperature. The experiment served to figure out the specific heat capacity of water (what energy is required to raise the temperature of a certain mass of water by a certain amount). You can read more about the experiment here.


text:
Storing hydrofluoric acid before the invention of plastics
The first person to synthesize hydrofluoric acid in large quantities was Carl Wilhelm Scheele in 1771. This acid is known for its ability to corrode glass and metals.
What materials were the containers in which it was stored before the invention of plastics?

Iron is resistant do HF due to the formation of a dense flouride surface layer. Not sure if he used it.

If I remember well, also nickel and copper.

Industrially they store hydrofluoric acid in steel with a few percent nickel and in magnesium boxes on rail cars because steel is too heavy.  Pure nickel stores are for when you need very low contamination.

The name gutta-percha comes to mind, which I think is a type of rubber, but with a quick look can't find any reference to back that up.

Based on research inspired by andselisk's answer, chemists stored it in glass vessels coated in wax (similar to the receiver setup Scheele used to prove the silicon dioxide precipitate was from the glassware itself.
The fourth paragraph down in this blog post on The Chronicle Flask touches on it (emphasis mine):

Where do you put something that eats through its container? Well, these days it’s stored in special plastic bottles, but in the 17th century when it was first discovered chemists had to use glass bottles coated inside with wax, and hope the coating was a good one.

I do remember seeing a wax container of hydrofluoric acid in the chemistry storage area at my high school in ~ 1974-75. The label was a mess. Hope they managed to dispose of it properly.

In the original 1771 experiment, Scheele used a very simple setup consisting of a glass retort with a glass receiver (round-bottom flask).
Yes, the glass was etched to some degree by the fumes, but it was not drastic enough to destroy the apparatus.
From Anders Lennartson's The Chemical Works of Carl Wilhelm Scheele [1, p. 22]:

3.1 Publication 1. Investigation of Fluorite and Hydroﬂuoric Acid
Original publication: Kongl. Vetenskaps Academiens Handlingar. 1771, 32,
  120–138.
Original title translated to English: Investigation of ﬂuorspar and its acid
…
The second paragraph of the paper deals with the action of acids on ﬂuorite, and
  this is where the important discovery is made. Scheele mixed powdered ﬂuorite
  with sulphuric acid (oil of vitriol) and heated it in a retort (Fig. 3.3) whereupon corrosive fumes were emitted. He repeated the experiment with some water in the receiver. He found that a white crust was formed on the water surface, and that the water contained a dissolved acid. The inner surface of the retort was etched by the fumes. Scheele identiﬁed the white crust as silicon dioxide (siliceous earth) and the residue in the retort as calcium sulphate (selenite or gypsum). This lead Scheele to the conclusion that “Fluor [ﬂuorite], therefore, consists principally of calcareous earth [calcium oxide], saturated with a peculiar acid; the nature of which we shall investigate hereafter”.

Fig. 3.3 Schematic view of the apparatus Scheele used for preparing “acid of ﬂuorspar”

In response to critics, to prove that the glassware itself was the source of the white silicon dioxide powder, Scheele improved the setup, protecting the inner walls with tin and wax.
Continuing from [1, p. 98]:

3.60 Publication 60. Experiments with Hydroﬂuoric Acid
Original publication: Chemische Annalen für die Freunde der Naturlehre.
  1786, part 1, 3–17.
Original title translated to English: New evidence for the characteristic nature of the acid of ﬂuor spar
…
Scheele powdered ﬂuorite in a brass mortar, mixed the powder with sulphuric
  acid (oil of vitriol) and transferred it to a retort lined with tin. The receiver was lined with wax, and partly ﬁlled with water. This time Scheele obtained a distillate free from precipitated silicon dioxide (siliceous earth), and addition of potassium hydroxide gave no precipitate (i.e. no $\ce{K2SiF6}$). Thus, this was pure hydroﬂuoric acid, $\ce{ HF_{(aq)} }$; similar results had been reported by Scheele’s friend, J.C.F. Meyer in Stettin.

References

Lennartson, A. The Chemical Works of Carl Wilhelm Scheele; Springer Berlin Heidelberg: New York, NY, 2017. ISBN 978-3-319-58180-4.

@CodyGray Thank you for the edit and corrections, I appreciate it!

This answer is as good as it can be and I'll be glad to accept it in a couple of days. Thank you very much.

Great answer about the original experiment, but it doesn't quite answer the storage part - since Scheele didn't actually store the hydrofluoric acid. Maybe you could explicitly challenge Ginasius' assumption that just because people could do experiments with hydrofluoric acid didn't mean they had to store it. Or add a paragraph or two about actual methods of storing hydrofluoric acid without using plastics.


text:
Is it safe to boil carbonated water in a kettle?

Isn't it dangerous to boil carbonated water in a kettle?

Whenever I boil water in my kettle, I close its lid. Does this close lid constitute 'under pressure' (and so danger) emboldened beneath?

There was no boiling water explosion, although I more than half expected one.
The only way it could conceivably cause an explosion is if you boiled it under pressure.

Closure discussed in Meta: https://chemistry.meta.stackexchange.com/q/4456/23561

No danger!
If you are worried about flames, carbon dioxide is not flammable.
If you are worried about explosions, gas is not released so fast to cause an explosion.
For comparison, you can free much more energy per time by overheating plain water.
So nope, no danger, even if you place a lid on it (assuming that the lid is free to bump and release CO2/aqueous vapor, but even if it wasn't I doubt that it would be much more dangerous than doing the same with plain water).


text:
Borane as a reducing agent
I understand that adding borane to an alkene will produce a hydroborane. Addition of -OOH will form an alcohol.
My lecture notes state that addition of borane to:

A carboxylic acid will form an alcohol
An amide will form an amine
An aldehyde/ketone will form an alcohol
A nitrile will form an amine
An epoxide will form an alcohol
An alkene will form an alkane
An alkyne will form an alkene

I can't find the mechanisms for these anywhere. I'd appreciate seeing a few of them, namely carboxylic acid to alcohol, amide to amine, epoxide to alcohol and alkene to alkane. I should then be able to have a good crack at working out the rest.

Henry Rzepa gives a mechanism in which two borane molecules attach to the carboxyl group of a carboxylic acid.  The boron atoms are attached to the electronegative but electron-rich portions of the carboxyl group (meaning the oxygen atoms) and then transfer hydride ion moieties to the carbon while taking away one of the oxygens.  This exchange of hydrogen for oxygen reduces the carboxyl carbon.  Other carbonyl compounds may be assumed to react similarly.

I highly recommend Organic Chemistry by Clayden, Greeves and Warren. They do an unusually good job of drawing correct arrow-pushing mechanisms.

Ch 24, p.619 has the mechanism for acid and amide reductions and a table comparing reactivity of various carbonyl groups.
Ch 47, p. 1281 has the mechanism for hydroboration of alkenes and alkynes

The un-catalyzed reduction ketones, aldehydes, and epoxides with simple diborane (B2H6) are not synthetically useful. However, catalyzed versions like the Corey-Bakshi-Shibata Reduction serve to illustrate the mechanistic concept.


text:
Cp vs Cv for liquid water @ < 4 °C
I'm looking at a textbook question (Engel & Reid, Thermodynamics, Statistical Thermodynamics & Kinetics, 4 ed, Q3.1) which states the following:

The heat capacity $C_{\mathrm m,p}$ is less than $C_{\mathrm m,V}$ for $\ce{H2O(l)}$ near $\pu{4 °C}$. Explain this result.

Their answer is that water contracts as you heat it in this regime, thus (at constant $p$) the surroundings are doing work on the system, hence $C_{\mathrm m,p}\lt C_{\mathrm m,V}$.
But the textbook also contains the following equation (which is correct, and whose only restrictions are that the system is of fixed phase and composition, and $\text{đ}w = –pdV$, i.e., pV-work only):
$C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \beta^2/\kappa$
where $\beta$ is the isobaric thermal expansivity, and $\kappa$ is the isothermal compressibility.
Based on this, for a substance of fixed phase and composition, $C_{\mathrm m,p} \ge C_{\mathrm m,V}$, always, because, while $\beta$ can be $\le0$* (water near $\pu{4 °C}$ being a notable example), $\beta^2$, as well as $T, V_\mathrm m$ and  $\kappa$, are always positive. [*Edited from  $\lt 0$ to $\le 0$ based on Night Writer's answer.]
So is the statement in the textbook's question just a flat-out mistake, or am I missing something here?
It appears their error is in assuming the difference between Cp and Cv is due to pV expansion alone, when in fact there are two terms:
$$C_p - C_V = p \left(\partial V\over\partial T\right)_p+ \left(\partial U\over\partial V\right)_T \left(\partial V\over\partial T\right)_p,$$
where the first term on the RHS is the PV work per unit change in T, while the second term is the change in internal energy with respect to volume (which results from changing the intermolecular distance between interacting particles) times the rate at which the volume changes with temperature.
It would be nice to have direct reference-quality experimental values for $C_{\mathrm m,p}$ and $C_{\mathrm m,V}$ for $\ce{H2O(l)}$ near $\pu{4 °C}$, but I've not been able to locate them, nor do I expect to: because of the difficulty of accurately measuring $C_{\mathrm m,V}$ for a liquid, measurements are typically made at constant pressure, giving $C_{\mathrm m,p}$, and then $C_{\mathrm m,V}$ is calculated using the above equation.

In addition to the answers below, the data do show that $C_p \gt C_V$, albeit by a very small amount near zero $^\circ$C ($\approx 0.05$ J/mol/K) at temperatures from just above zero to 10  $^\circ$C. So the question is factually wrong.

@porphyrin Can you provide a reference for that data?

I just looked on the web: https://www.engineeringtoolbox.com/specific-heat-capacity-water-d_660.html .

@porphyrin I suspect the only experimental data there is for Cp, and that Cv was calculated from the Cp values using the above formula.  There's no way to tell without a reference to the original source.

Ok, I had'n't realised that. You will have to search for some original data, but I can see no reason why $C_v$ should be greater than $C_p$. That $\beta \to 0$ seems to be the best explanation as shown by @ Night Writer in that answer.

Repeating your line of reasoning, if $C_{\mathrm m,p}<C_{\mathrm m,V}$ you would conclude based on the expression you provide ($C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \alpha^2/\kappa$) that $\kappa<0$ which in turn would mean that
$$\left(\frac{\partial V_m}{\partial P}\right)_T > 0,\label{eqn:1}\tag{1}$$
which is not physically reasonable except presumably in some very unusual circumstances. Therefore, either the question statement contains a fallacy, or the above analysis contains a fallacy, i.e. $\eqref{eqn:1}$ is reasonable. My bet is on the first scenario, as you suggest.
What seems to happen is that near $\pu{4 °C}$
$$C_{\mathrm m,p} = C_{\mathrm m,V}\label{eqn:2}\tag{2}$$
because the change in the density or molar volume with $T$ (and therefore also the thermal expansion coefficient $\beta$ - usually designated with the symbol $\alpha$) becomes zero, that is there is a (well known) minimum in the dependence of the density or molar volume on $T$ at constant $P$, illustrated in the following figure (source: wikipedia)

In agreement with this, if you inspect available heat capacity data you will see that the isochoric and isobaric heat capacities converge as the temperature of liquid water nears $\pu{0 °C}$.
The International Association for the Properties of Water and Steam has released a document (IAPWS R7-97(2012)) containing expressions for the heat capacities of water parameterized based on experimental data (see Table 3).

That's a nice point about $\beta$ for water going to zero at 4C.

@theorist I think the answer is a combination of my explanation and the alternate answer: that answer explains some of the flaws in the book's reasoning, mine explains others (well, I echo your explanation - good catch - and follow it to it's logical conclusion based on the available data). I did find some standard data but have not crunched the numbers.

The textbook's question is flawed because, as you state, its assertion that $c_p < c_v$ is false.
The textbook's answer messes up the math a little bit. Their logic is based on the First Law:
\begin{align}
\Delta u &= q - P \Delta v
\\
q &= \Delta u + P \Delta v
\end{align}
Considering an isobaric process, we have
\begin{align}
\overbrace{c_p \Delta T}^q &= \overbrace{\left(\frac{\partial u}{\partial T}\right)_P \Delta T}^{\Delta u} + P \overbrace{\left(\frac{\partial v}{\partial T}\right)_P \Delta T}^{\Delta v}
\\
c_p &= \underbrace{\left(\frac{\partial u}{\partial T}\right)_P}_X + \underbrace{\left(\frac{\partial v}{\partial T}\right)_P}_{<0}
\end{align}
If term $X$ were the same as $c_v$ then it would follow that $c_p < c_v$, but the wrong property is held constant: $c_v$ would be $\left(\frac{\partial u}{\partial T}\right)_v$. Evidently $\left(\frac{\partial u}{\partial T}\right)_P - \left(\frac{\partial u}{\partial T}\right)_v$ is more positive than $P \left(\frac{\partial v}{\partial T}\right)_P$ is negative. The textbook seems to be incorrectly assuming that $X = c_v$.
Note that the corner case $c_p = c_v$ is allowed; this would occur when $\beta = 0$, $T = 0$, $v = 0$, or $\kappa = \infty$ (the first case being the one that can actually occur in everyday life).
Original take-down of the textbook answer:
Based on your summary, the textbook's answer fails to account for the fact that switching from an isobaric path to an isothermal path changes the end state. The (flawed) logic seems to be: "if I go from A to B isochorically then I need to provide some amount of heat, but if I go from A to B isobarically then the surroundings provide some work so I don't need to provide as much heat." The flaw is that if A-B is an isochor then an isobar beginning at A doesn't go through B. An apples-to-oranges comparison would consider the heat required to reach a third point C which is at the same pressure as A and temperature as B, and would find that this heat is never lower than that required to reach B.

You seem to be saying the textbook's answer is flawed because you can't compare Cp and Cv, the reason being that the difference in constraints implicit to Cp vs. Cv brings you to different final states (making it apples to oranges).  However, even with this, you certainly can compare Cp to Cv (their difference is $TV\beta^2/\kappa$).  Thus (and forgive me if I'm not understanding you correctly) I don't think the specific issue you raise (the difference in final states) is relevant here.  I.e., I think the textbook is wrong, but not for the reason you give.

I'm saying that their argument that the isobaric process involves less work than the isochoric process is overly simplistic (because it assumes the two processes have the same endpoint). You can certainly compare $c_p$ and $c_v$, for example using the formula you provided, but their attempt to provide an explanation based directly on the First Law/intuition is flawed.

I think their argument is flawed, but not for the reason you give.  The problem isn't that they don't go to the same endpoint; that doesn't preclude the comparison.  For instance, it's not a problem to say the reason Cp = Cv +nR for an ideal gas is that, during constant-P heating, you have to input additional thermal energy for the PV work associated with expansion.  By your argument, you couldn't make a universal statement like that for ideal gases simply because they don't go to the same final state in a constant-V vs a constant-P heating.  But that's simply not the case. [continued....]

...Hence the disparity of final states is not the logical issue per se.  Rather, I think the reason their argument is simplistic/flawed is that it doesn't account for the difference in internal energy due to changes in volume in real materials (which have attractive and repulsive interactions). I.e., it accounts only for the difference in the work done on the surroundings. I.e., you're saying that the problem is that they're comparing processes with different final states.  I'm saying it's OK for the final states to be different, so long as you account for the effect of those differences.

I suppose we are arguing over what is the "right" way is to do something wrong, haha. You have the advantage of actually having read the textbook's answer in its entirety ;) I have updated my answer to reflect the flaw that you see but also retained the flaw that I saw so that our comment chain still makes sense.

I've added something to my answer that I think says the same thing as your edit, though I present it differently.


text:
IUPAC Recommendations for Concentration Units (ppb, ppm)
In a recent interesting post, (Is 1 ppb equal to 1 μg/kg?)
it was pointed out that IUPAC advises to abandon parts per million, parts per billion quantities and instead it suggests to employ micromole/ mole for ppm and nanomole/mole for ppb. An example is given on page 98 of the Green Book
"The volume fraction of helium is 20 ppm".
May be I am missing a point or I see a problem here with the moles in the denominator of these suggested replacement definitions.
a) If we analyze Zr in a rock and find 5 mg Zr in a 20 gram rock, how would one apply the mole definition because one cannot define the moles of a rock. We have to stick to the classical definition:
ppm =10^6 * (mass of analyte / mass of sample).
b) How can one express ppm of Ar in air, and how should we define the moles of air? If we wish to generalize this case, what if our solvent system consists of several components.
The ambiguity I feel in micromole/mole definition is the mole in the denominator. Are we referring to the moles of the solvent or  moles of solvent plus solute.
If this replacement is analogous to molal concentration, where the denominator is clearly definied to be kg of solvent not the kg of solvent plus kg of solute, then IUPAC should clearly define it.

Who said it's all about moles? Have you actually seen other examples from that table you are quoting (I only listed one to illustrate the replacement for ppb)? The idea is not to convert everything to the amount of substance, but to use mg/g, μg/g, pg/g etc. instead when appropriate! With this in mind: a. Use mass fraction (mg/g) instead of ppm; b Use volume fraction (μmol/mol) instead of ppm. It's all in the table you are quoting, really.

Yes, I have a copy of the Green book. They do mention above the table that mg/g and other similar quantities can be formulated which are used by analytical chemists. (a) I am still not sure how to calculate "The volume fraction of helium is 20 ppm" in terms of (μmol/mol). (b) They don't specify "moles" of "what" in the denominator? An example would have helped.

They talk about "volume fraction of helium". It means that it is 20 ul per liter of sampled gas. They are not talking about moles.

You are right, this "It means that it is 20 ul per liter of sampled gas." is a classical definition. Have a look at the Table on pg 98 of the Green Book, they write μmol/mol as "Replacement". My point is that IUPAC should have provided better examples.

There are not this kind of problems with the IUPAC recommendation. The only point in it is to specify the parts of  what we refer to, nothing more.
In such a sense it seems clearer than expression as molality or molarity.
Take an amount of atmosphere at given conditions. The concentration of Ar or whatever component can surely be expressed in terms of mole number in a million or whatever number of moles that amount contains.
As for the rock analysis case in the question, note that IUPAC umol/mol is an example, it doesn't mean it should be preferred to mg/kg (or ul/l etc.). The measured amount remains the most convenient both practically and mathematically.
I add that following recommendations would - perhaps - also help clear another practical con of unspecified ppm and percentage concentration values.  Often people forget that 1 l --> 1 kg of solution is acceptable only with very dilute solutions and when the solvent is water and go around telling and even publishing % concentration in various solvent so that the reader must assume a wrong procedure in order to follow up. Explicitly saying mg/kg instead of ppm should evidence that when mg starts to be hundred, than one has to switch concentration units or being more careful when reporting.

"The concentration of Ar or whatever component can surely be expressed in terms of mole number in a million or whatever number of moles that amount contains." calculating the mole of Ar is easy, but how would be calculate moles of "air" which is the solvent for Ar. Can we work this example out?

Yes, for a start I would just consider an ideal gas. Note that this is indipendent of the unit you choose, using ppm would be very much the same (at end it is!). The iupac recommendation just tell you not to use ppm, or, if you really want, specify ppm of what. You can keep part in volume, if you want to skip the average molecular mass of that air. It will be anyway how many molecules of Argon are in a million molecules of air.

@M. Farooq ↑   isn't conceptually any different than thinking of partial Pressure, tough I perfectly understand now your concern.  There is not,  strictly, a mole of air, if not "engineeringly" speaking.

Thanks. The micromole/mol definition is similar to molal concentration not molar concentration because the denominator does not include the mass of the substance in the numerator. For example if I add 30 g NaCl to 1 kg of water, it is possible to determine molal concentration but impossible to determine its molarity (because we don't know how much volume changed after adding 30 g NaCl to 1 kg of water).

This is why I dislike in general operational dictated concentration values for which density is required. But at the end, this is pretty much used when we refer to washing/neutralizing/working up solution for which precision is the last concern. However, the umol/mol way of expressing concentration does include the stuff dissolved, as ppm does: 1 ppm = 1 part in 1 million total parts. Again, this is usually neglected when part <<< parts, which in turns lead to rounding 1 ppm = 1 mg/ml for very diluted solutions in water (d = 1). I hope this clarify.  @M. Farooq


text:
Diluting a sample of weak base before titration. What happens to equivalence point pH of diluted vs undiluted?

A student titrated a 10.00 mL of 0.20 M methylamine ($\ce{CH3NH2}$) and determined that 10.00 mL of $\ce{HCl}$ was needed to reach the equivalence point.
If you diluted your original sample of methylamine by adding 10.00 mL of water to your sample before you started adding $\ce{HCl}$, which of the following statements would be true?
a) You now only need to add 5.00 mL of $\ce{HCl}$ to reach equivalence.
b) The equivalence point pH would be the same as in the undiluted titration.
c) Both (a) and (b) are true.
d) Neither (a) nor (b) are true.

The answer is d) Neither (a) nor (b) are true.
I always thought that both the $\mathrm{p}K_\mathrm{a}$ and equivalence $\mathrm{pH}$ of an acid stay the same no matter what the concentration. Except where it is so dilute, ionization from water would have an effect. But I don't understand why or how it would change. I can't find anything in the textbook or online, can someone please explain this?

a) is certainly not true. I think the book is playing with the concept of pH. Although the acid and base have been neutralized but the pH must be different bc of dilution. This is very weak base but dilution does not change the number of moles however it changes concentration. The equivalence point should be the same. See the simulation here https://chemistry.stackexchange.com/questions/95304/how-would-a-ph-curve-look-like-for-titration-of-diluted-weak-base-compared-to-co

You could add the water after the titration, and it would also change the pH, see https://chemistry.stackexchange.com/a/33369

At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$.
If we neglect $\ce{H+}$ from water dissociation and consider $c_\ce{A-}\ll c_\ce{HA}$, then $c_\ce{A-} = c_\ce{H+}$
Therefore
$$K_\mathrm{a} = \frac{c_\ce{H+} c_\ce{A-}}{c_\ce{HA}} = \frac{c_\ce{H+} c_\ce{H+}}{c_\ce{HA} + c_\ce{A-}}$$
$$\begin{align}
\mathrm{pH} &= \frac12(\mathrm{p}K_\mathrm{a} - \log c) \\
  &= \frac12(14 - \mathrm{p}K_\mathrm{b} - \log c) \\
  &= 7 - \frac12(\mathrm{p}K_\mathrm{b} + \log c) \\
\end{align}$$
where $\mathrm{p}K_\mathrm{b}$ is the methylamine basicity constant;
$c$ is summary concentration of methylamine and it's a conjugate acid.
Therefore, if there is final volume 30 instead of 20 ml, the final
$$\Delta\mathrm{pH} = -0.5\log \left(\frac{\pu{20 ml}}{\pu{30 ml}}\right) = +0.088$$

Since the OP was confused, I think you should explicitly define what the equivalence point is.


text:
Why is reduction potential of water ( 2H₂O + 2e⁻ → H₂ + 2OH⁻ ) is -0.83 insted of 0, as in the case of (2H⁺ → H₂)?
I came across this value of reduction potential of water while solving a question. Also, I noticed that, for example
$$\ce{Ag+ → Ag}$$
has a fixed standard potential, and the same is used when we do it for a reaction like
$$\ce{AgCl → Ag + Cl-}$$
Then why here in the reduction of water the value differs from the $\ce{H+ → 1/2 H2}$ reaction?
Also, while doing electrolysis of water, which reaction actually happens at the cathode:
$$\ce{2 H2O  + 2 e- → H2 + 2 OH-}\quad\text{or}\quad\ce{2 H+ → H2}~?$$

The potential of $\ce{Ag+ -> Ag}$ is not actually fixed. It depends on the concentration of $\ce{Ag^+}$, and the form of this dependency is well known. Same thing here.

Related: Electrolysis in aqueous solution – which equations to use to predict product at each electrode?

$$\ce{Ag+(aq) + e- ->Ag(s)};\ E^\circ_{red} = \pu{0.80 V}$$
$$\ce{AgCl(s) + e- ->Ag(s) + Cl-(aq)};\ E^\circ_{red} =\pu{ 0.22 V}$$

Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1,
If we consider reactions
$$\begin{align}\ce{
2 H+ + 2e-    &<=> H2 \\
2 H2O + 2 e-     &<=>  2 OH- + H2 \\
}\end{align}$$
their standard redox potentials imply
$\rm{p}H=0$ for the former, $\rm{p}H=14$ for the latter.
For the former reaction and $\rm{p}H=14$, $E=-0.83\ \rm{V}$. As for $\ce{H+/H2}$ potential, at $\pu{p_{\ce{H_2}} = 1 atm}$
$$E_\ce{H2\ /H+}  = E^{\circ}_\ce{H2\ /H+} + 0.059 \log [\ce{H+}] = E^{\circ} - 0.059\  \rm{p}H$$
Thermodynamically, the equations they are equivalent, but kinetically, the latter reaction has in strongly alkaline environment the advantage of many concentration orders. And vice versa.
Elements can have multiple standard redox potentials, as such a potential can be expressed for any redox reaction (even if for some only formally from the thermodynamic data).
For example, some standard redox potentials for $\ce{Ag}$ are:
$$
\begin{array}{lcll}
\ce{Ag+ + e- &<=>& Ag(s)};\ & E^\circ_{red} = \pu{0.80 V}\\
\ce{AgCl(s) + e- &<=>& Ag(s) + Cl-(aq)};\ & E^\circ_{red} =\pu{ 0.22 V}
\end{array}
$$
(There are more; see the Standard electrode potential data table).
The latter can be calculated from the former using
$$E_{\ce{Ag/Ag+}} = E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+}$$
For the $\ce{Ag / AgCl / Cl-}$  system, we have to include into potential calculations the $\ce{AgCl}$ solubility constant.
$$\begin{align}
K_{\rm s, AgCl} &= c_{\ce{Ag+}}\cdot c_{\ce{Cl-}} \\[1em]
E_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+} =\\
  &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} - 0.059 \log {c_\ce{Cl-}}= \\
  &= E^{\circ}_{\ce{Ag/AgCl/Cl-}} - 0.059 \log {c_\ce{Cl-}} \\
E^{\circ}_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} =\\
 &= 0.80 + 0.059\log(1.77\cdot10^{−10}) = \\
 & = 0.22
\end{align}$$
Note that the $\ce{Ag/AgCl/Cl-}$ electrode is often used as the secondary standard of the electrode potential, aside of the calomel electrode $\ce{Hg/Hg2Cl2/Cl-}$, because maintaining of the primary potential standard electrode $\ce{Pt-H2/H+}$ is not convenient. The $\ce{Ag/AgCl/Cl-}$ electrode is used also as the electrode to measure $c_{\ce{Cl-}}$.

I'd write the two different half-cell reactions as reductions since that is the standard for tabular values. -- LOL, Now I'm the Standard Police.

Equilibrium reactions are neither reductions neither oxidations.  They are both.   The orientation is purely formal. Similarly, the electrode potentials do not flip signs.:-)

The point as you well know is that you can't look up the half cell reaction in a table as $$\ce{H2 <=> 2H+ + 2e-}$$ rather it will be listed as $$\ce{2H+ + 2e- <=> H2}$$

Yes, the lookup point is the valid point indeed. :-) But formal one. The multimeter does not flip it's reading when I switch sides. I will change them, additionally to be complient with $\Delta G =-nFE$

Btw, about flipping potential signs, should not it to be rather flipping equation signs ? As the flipped potential does not have meaning of a measurable potential.

Huh? re: "The multimeter does not flip it's reading when I switch sides." // When you measure a battery  don't you get a positive reading one way and when you flip the voltmeter's leads you get a negative voltage of the same magnitude. // You can't measure just one half cell of course.

When I switch sides of the reaction. :-) Additionally, the potential of the electrode does not flip the sign with the flipping the multimeter orientation.

I'd love to see your answer stressing even more prominently that -0.83 can be simply calculated via concentration using Nernst equation, and that H+'s half-reaction is no different from any other half reaction in this regard, and that it is OK for an element to have multiple half-reactions (say, there is one for Ag+ and another for AgCl, and they are related in the same manner). Oh, and please include the link: https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)

Damnnit , thanks a lot poutnik . You really explained that well..... Lol , i was struggling with this from quite a long time.

@Ivan Neretin  OK,  I will modify it after the work.

@Ivan Neretin Can you edit your edit ? I see 2 errors there. :-)

Feel free to fix them. It is your answer, after all.

Sure, I just meant it as as matter of professional honour, not to bring factual errors by editing answers of others, looking like their own errors. :-) I am on phone, so postponing editation.

The above post answers your questions, but let me add that
$$\ce{2 H+ + 2 e- <=> H2}\label{rxn:1}\tag{1}$$
and
$$\ce{2 H2O + 2 e- <=> H2 + 2 OH-}\label{rxn:2}\tag{2}$$
are two different experimental procedures. Hence these are two distinct half cell reactions. It happens that you get $\ce{H2}$ as a product in $\eqref{rxn:1}$ and $\eqref{rxn:2}$.
In $\eqref{rxn:1}$ you are reducing hydrogen ions and the experimental set-up is very specific. You have $\ce{Pt}$ electrode (or $\ce{Pt}$ black) and you bubble hydrogen over it in $\pu{1 M}$ $\ce{H+}$ ion. It is impossible to determine its absolute potential and we set this half cell to $\pu{0 V}$ (by definition). The electrode is being exposed to $\ce{H+}$ and $\ce{H2}$. This system is called NHE and SHE (normal hydrogen electrode and standard hydrogen electrode, respectively).
In $\eqref{rxn:2}$ instead you have $\ce{H2O}$ which is being reduced here in alkaline medium in the presence of $\ce{H2}$. This reaction is not $\pu{0 V}$ with reference to SHE.


text:
Disproportionation of Au(II)
$\ce{Cu(II)}$ and $\ce{Ag(II)}$ complexes are known to exist while $\ce{Au(II)}$ is unstable and disproportionates to $\ce{Au(I)}$ and $\ce{Au(III)}$.
My guess as to possible reasons to this effect is that $\ce{Au(II)}$ like $\ce{Cu(II)}$ and $\ce{Ag(II)}$ cannot undergo Jahn-Teller Distortion, the cause of their stability, because of the extremely diffuse nature of $\mathrm{d}$ orbitals of $\ce{Au(II)}$ owing to relativistic effects. Could this be the right reasoning?
Also, what is the cause of the stability of $\ce{Au(I)}$ and $\ce{Au(III)}$ over $\ce{Au(II)}$?

Somewhat related: https://chemistry.stackexchange.com/questions/87830/were-goldii-complexes-synthesized-for-the-first-time-in-2017?r=SearchResults

My suspicion is that it has nothing to do with the +1 oxidation state, but rather the inaccessibility of the +3 oxidation state for Cu and Ag (whereas for Au, it is quite easy and common to reach +3). This can partly be attributed to the increase in principal quantum number (in general heavier congeners are easier to oxidise than lighter ones), but there is maybe also a relativistic consideration (Au 5d orbitals are destabilised, so it's unusually easy to remove electrons from them).

Actually Ag(II) also disproportionates - AgO is actually  $\ce{Ag^{I}Ag^{III}O2}$

@orthocresol, I would be thankful if your comment could be made an answer.

@Mithoron true of the oxide. Silver(II) fluoride does seem to be actually silver(II).

I think this could be a counter answer. In contrast, I found a literature evidence that disproportionation of $\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ has had happens Ref.1). The abstract states that:

Interaction of $\ce{Ag+}$ salts in anhydrous liquid hydrogen fluoride, $\ce{aHF}$, with $\ce{AgF4-}$ salts gives amorphous red-brown diamagnetic $\ce{Ag^IAg^{III}F4}$, which transforms exothermally to brown, paramagnetic, microcrystalline $\ce{Ag^{II}F2}$ below $\pu{0 ^\circ C}$. $\ce{Ag^IAu^{III}F4}$ prepared from $\ce{Ag+}$ and $\ce{AuF4-}$ in $\ce{aHF}$ has a tetragonal unit cell and a $\ce{KBrF4}$ type lattice, with $a = \pu{5.788(1) Å}$, $c = \pu{10.806(2) Å}$, and $Z = 4$. Blue-green $\ce{Ag^{II}FAsF6}$ disproportionates in $\ce{aHF}$ (in the absence of $\ce{F-}$ acceptors) to colorless $\ce{Ag^IAsF6}$ and a black pseudotrifluoride, $\ce{(Ag^{II}F+)2Ag^{III}F4^-AsF6-}$. The latter and other $\ce{(AgF)2AgF4MF6}$ salts are also generated by oxidation of $\ce{AgF2}$ or $\ce{AgF+}$ salts in $\ce{aHF}$ with $\ce{F2}$ or in solutions of $\ce{O2+MF6-}$ salts ($\ce{M = As, Sb, Pt, Au, Ru}$). Single crystals of $\ce{(AgF)2AgF4AsF6}$ were grown from an $\ce{AgFAsF6/AsF5}$ solution in $\ce{aHF}$ standing over $\ce{AgF2}$ or $\ce{AgFBF4}$, with $\ce{F2}$ as the oxidant. They are monoclinic, $P2/c$, at $\pu{20 ^\circ C}$, with $a = \pu{5.6045(6) Å}$, $b = \pu{5.2567(6) Å}$, $c = \pu{7.8061(8) Å}$, $β = 96.594(9)°$, and $Z = 1$. The structure consists of $\ce{(AgF)_n^{n+}}$ chains ($\ce{F−Ag−F} = 180^\circ $, $\ce{Ag−F−Ag} = 153.9(11)^\circ $, $\ce{Ag−F} = \pu{2.003(4) Å}$), parallel to $c$, that enclose stacks of alternating $\ce{AgF4-}$ and $\ce{AsF6-}$, each anion making bridging contact with four $\ce{Ag(II)}$ cations of the four surrounding chains “caging” them. There is no registry between the ordered array in one “cage” and that in any neighboring “cage”. The $\ce{F-}$ ligand anion bridges between the anions and, with the $\ce{Ag(II)}$ of the chains, generates a trifluoride-like structure. $\ce{(AgF)_2AgF4AsF6}$ [like other $\ce{(AgF)_n^{n+}}$ salts] is a temperature-independent paramagnet except for a Curie “tail” below $\pu{50 K}$.

However, I cannot find a counter example for $\ce{Cu(II)}$ complexes. Yet, $\ce{Cu(I)}$ is known to disproportionate to $\ce{Cu(II)}$ and $\ce{Cu(0)}$, which has limited $\ce{Cu(I)}$ chemistry. For example, if you react $\ce{Cu(I)}$ oxide ($\ce{Cu2O}$) with dilute sulfuric acid (hot), you'd get a brown precipitate of $\ce{Cu(0)}$ and a blue solution of $\ce{CuSO4}$ (a $\ce{Cu(II)}$ salt), because the disproportionation reaction dominates. You would not get the expected copper(I) sulfate solution from this reaction.
$$\ce{Cu2O + H2SO4 -> Cu + CuSO4 + H2O}$$
However, if you use certain ligands to form $\ce{Cu(I)}$ complexes, they would stabilizes the $\ce{Cu(I)}$ oxidation state. For example, both $\ce{[Cu(NH3)2]+}$ and $\ce{[CuCl2]-}$ are $\ce{Cu(I)}$ complexes, which don't disproportionate.
References:

Ciping Shen, Boris Žemva, George M. Lucier, Oliver Graudejus, John A. Allman, Neil Bartlett, "Disproportionation of$\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ in Fluoride Systems and Syntheses and Structures of $\ce{(AgF+)_2AgF4^-MF6-}$ Salts ($\ce{M = As, Sb, Pt, Au, Ru}$)," Inorg. Chem. 1999, 38(20), 4570-4577 (https://doi.org/10.1021/ic9905603).
Also read for the Disproportionation of$\ce{Au(II)}$: Khaldoon A. Barakat, Thomas R. Cundari, Hassan Rabaâ, Mohammad A. Omary, "Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects," J. Phys. Chem. B 2006, 110(30), 14645-14651 (https://doi.org/10.1021/jp062501y).

I can’t find any Au(II) in your quoted abstract. If this is a counter-answer, can you highlight it? Or if you’re not countering the tendency of Au(II) to disproportionate, what are you countering?

I want to show that at least one of two metals ($\ce{Ag(II)}$) is behaving against what is claimed for. Also, I hope that for $\ce{Cu(I)}$ (instead of $\ce{Cu(II)}$) as well. I did not counter $\ce{Au(II)}$, because it is the comparison. That's why I included ref.2 to agree with OP.

The claim in the question’s first line is that silver(II) complexes exist, which your abstract proves outright by stating that $\ce{Ag^IAg^{III}F4}$ transforms exothermically to $\ce{AgF2}$ (a silver(II) complex) below $\pu{0^\circ C}$.

The question ask for Au(II) but you described for Ag(II) with no mention of Au(II). Is there something I am missing?

My answer is a counter answer. See my comment for Jan's comment above.

Ok got it. But maybe if you get some info. on Au(II) that would be good. I tried to close this question as dupe but OP also mentions that this post didn't explain the underlying mechanism/reason on why Au(II) is unstable, so you can take this opportunity.

I don't think it makes sense to compare gold with silver and copper. In the 6th line of the periodic table, relativistic effects become important. And one of the relativistic effect is a contraction of the lengths. This effect is proportional to the sum of the two first quantum numbers. Gold outer orbitals are 4f, 5d and 6s. Its configuration is [Xe] 4f14 5d10 6s1. These 4f, 5d and 6s orbitals are defined by n=4, l=4, then n=5, l=3, then n=6, l=1. The total n+l is equal to 8, 8, and 7. It is lot. So the 4f, 5d and 6s are compressed and disappear among the other electrons. The Au atom looks as if its outer shell has 10+14+2 = 26 electrons "lost". As Au has a total of 79 electrons, it looks as if it has only 79 - 26 = 53 electrons. The atom with 53 electrons is iodine. So The chemical properties of gold looks like iodine. Iodine is known to have -1, +1, +3 as possible oxidation states. Like gold. In fact it is even known that gold may be reduced to -1, in the compound CsAu!
I may also add that the next atom after gold is mercury, which looks like the next atom after iodine: xenon. It looks like a "heavy xenon", since it boiling point is relatively low, much lower than the usual metals. It is nearly a noble gas. Same things for the next atoms, Tl and Pb. Thallium is similar to an alkaline metal, and lead to barium: PbSO4 is as insoluble as BaSO4.

It is seen that Au(III) and Au(I) complexes are more stable than Au(II).I have an explanation but that is my observation-
Let's consider their electronic configuration.
Au → 5d10 6s1
Au+ → 5d10
Au2+ → 5d9
Au3+ → 5d8
So, if we split d-orbital into t2g and eg, we get

For Au →

For Au+ →

For Au2+ →

For Au3+ →

So, As you can see,

Au (I) - Fully Filled Configuration
Au (III)- Half-Filled Configuration

Therefore, both of them are more stable than Au(II) because of the fact that fully filled and half-filled orbitals render more stability than any other configuration.

This could happen to Ag and Cu too, but that is not generally the case in reality. I am more interested in the comparison between Cu, Ag and Au. Also yours could be only part of the complete answer.

I know in case of Cu Stability of Cu+2 is more than Cu+1 because of a higher amount of Hydration Energy of Cu+2 than Cu+1 which compensates the energy difference.(Even if Cu+1 > Cu+2 by stability if we consider electronic configuration ionly) Perhaps, you could think about your question in this angle.


text:
Will evaporation from two immiscible liquids blow up a balloon?
Consider the following setup:

Two immiscible liquids A and B are placed inside a balloon at a temperature where neither one of them boils. The inside of the balloon contains some argon, and the outside is air at atmospheric pressure. The balloon is made of mylar or some other similar material (flexible but not elastic).
I expect that the volume of the balloon would increase a bit due to evaporation of A and B into the argon. If we start without any gas in the balloon, is there a scenario (choice of liquids, choice of temperature) where the combined evaporation of A and B will be sufficient to fill up the balloon?
This question is related to Is it possible to boil a liquid by just mixing many immiscible liquids together? It addresses the question how to get both liquids into contact with the gas phase without worrying that the more dense liquid would be covered by the less dense one. It specifically does not ask about boiling, though, but rather about combined vapor pressure higher than ambient pressure, i.e. sufficiently high to displace gas at atmospheric pressure.
If this setup works, I wonder if it is possible to build a balloon pump where the balloon blows up "automatically", i.e. by heat transfer from the environment to drive evaporation of the two liquids.

The 'system' is possible and true according to what my (physical chemistry) teacher has taught us regarding Raoult's Law and corresponding Vapour pressure for a two-immiscible-liquids system. Is it correct that I understand 'automatically' does not refer to 'perpetual'? As liquids would be consumed.

It is surely possible to blow up a balloon by evaporation of one  liquid (provided it boils low enough; liquid nitrogen will do). Two liquids hardly provide any new insight.

@Ivan_Neretin Well, the liquid nitrogen would boil. I am asking about a temperature below the boiling point where the liquid would evaporate. Evaporation of a single liquid would not blow up the balloon.

@glucose No, not 'perpetual', just until the balloon is full.

If each liquid had a vapor pressure of 0.5 atmosphere at some temperature T, then combined they would blow up the balloon, and neither would be boiling. Water for instance has a vapor pressure of 0.5 atm at about 82 C.


text:
Why do some ions form tetrahedral complexes more readily than others?
Why does Co$^{2+}$ readily form tetrahedral complexes such as [CoCl$_4$]$^{2-}$, whereas Fe$^{2+}$ usually forms octahedral complexes?

Welcome to StackExchange Chemistry! If you list crystal field theory as a tag, you should mention your thoughts about how crystal field theory predicts or does not predict the geometry of these complexes.

Ligands use lone pairs of electrons to form coordinate bonds with the metal ion of the complex. The coordination number is the number of coordinate bonds from the ligands to the central ion. For instance, if a central ion has two coordinate bonds formed to it, one from each ligand, then the coordination number would be 2.(1)
According to Lumen Learning, for certain metals the maximum coordination number is related to the "electronic configuration of the metal ion (specifically, the number of empty orbitals) and to the ratio of the size of the ligands and the metal ion". Furthermore, it states that large metals with small ligands result in high coordination numbers, presumably due to there being more attraction from the positively charged central ion. Conversely, small metals with large ligands will have low coordination numbers, due to there being less electrostatic attraction towards the lone pairs on the ligands as its surface area is smaller.(2) Extrapolating this knowledge towards your question, we can parse the question to look at its components:
Co2+ is has an ionic radius of 74 pm while Fe2+ has an ionic radius of 76 pm. If we examine the example complex you gave, [CoCl4]2-, Chlorine has a large atomic radius of 175 pm relative to Co2+ ionic radius.(3) Using what we learned from Lumen Learning, this is a small central ion with relatively large ligands. Hence, the coordinate number must be smaller, such that it is 4, meaning it forms a complex with a tetrahedral geometry. Fe2+ on the other hand has an ionic radius of 76 pm and for a complex with an octahedral geometry, let us say [Fe(H2O)6]2+, it has 6 water molecules that function as ligands. The hydrogen and oxygen atoms have atomic radii of 25 and 60 pm respectively.(4) Consequently, the size of the water atoms are considerably smaller than the Chlorine atoms and thus, according to Lumen Learning, since we have smaller ligands, more of them can form coordinate bonds with the Fe2+ central ion. If we are to look at a different example, e.g. [Fe(CO)4]2−, we can see that it has ligands with large atomic radii (i.e. Carbon and Oxygen) and thus fewer ligands can bond with the central ion, resulting in a tetrahedral geometry.
References:

Brown, Catrin. "Chapter 3 – Periodicity." Higher Level Chemistry. Ed. Mike Ford. 2nd ed. N.p.: n.p., n.d. 123-33. Print.
Mott, Vallerie. "Introduction to Chemistry." Lumen. N.p., n.d. Web. 16 Apr. 2019.
"Cobalt Chemistry." Cobalt Transition Metal Chemistry Cobalt(II)Co2+ Complex Ions Stabilised Ligand Substitution Cobalt(III) Co3+ Complexes Redox Chemical Reactions +2 +3 Principal Oxidation States GCE AS A2 IB A Level Inorganic Chemistry Revision Notes. N.p., n.d. Web. 16 Apr. 2019.
Kwilkis. "Helium Atom, Radon Atom and Water Molecule Sizes and Concrete?" Helium Atom, Radon Atom and Water Molecule Size... | ACS Network. N.p., 03 Oct. 2018. Web. 16 Apr. 2019.


text:
Why do tetrahedral organic molecules show stereo-isomerism, but not tetrahedral transition metal complexes?
Both types of molecules show different arrangements of the molecule in space, so they should qualify for optical isomerism right?

Could you give an example of a tetrahedral transition metal complex with four different ligands that might have an optical isomer?

Actually they can and you didn't mix up anything. It may be hard to come up with an example that wouldn't involve organic ligands, but it's definitely possible that metal would be center of chirality.

Mithoron is right for tetrahedral metal complexes but the compounds you have shown are octahedral. But some octahedral metal complexes can also be chiral (the cis compound you drew has, if it is stable and the bi.dentate ligands don't swap around quickly, two mirror images)


text:
notation on MO diagrams
I wanted to ask a question about notations on MO diagrams.
Our first lesson about MO diagrams was today and I was taken through an example with $\ce{F2}$ as shown below:

but one thing that the lecturer did not explain, and I couldn't find over the internet with explanation, was the notation $\ce{3\sigma _g}$, $\ce{1\pi _u}$, $\ce{1\pi _{g}^{*}}$ and $\ce{3\sigma _{u}^{*}}$.
Is this notation relevant to the symmetry of the orbitals, or is there something else?
I haven't found any information that can explain this thus far.

The *'s are for antibonding versus bonding. u and g are for ungerade and gerade, which are antisymmetric and symmetric with respect to a center of symmetry.

Reference: J D Lee Concise Inorganic Chemistry

There is likely a question that addresses this on the site already, but I couldn't find it either, so I'll add an answer here.
You are correct that these symbols relate to the symmetry of the molecular orbitals that they label. In chemistry, we often label molecules by their point group symmetry, which describes how the molecule behaves with respect to particular symmetry operations (e.g. rotations, reflections, inversions).
If a molecule has a particular point group symmetry, its orbitals (and some other properties like vibrations) can be divided up into irreducible representations, which are essentially just different ways in which the overall symmetry can occur. One can find these irreducible representations (often shortened to irreps) in a character table for that point group.
$\sigma_\mathrm{g},\sigma_\mathrm{u},\pi_\mathrm{g},\pi_\mathrm{u}$ are all irreps of the $D_{\infty\mathrm{h}}$ point group, which you will most commonly encounter as the symmetry of a homonuclear diatomic. Since you are just being introduced to these, you won't likely use a character table and will just be given these MO diagrams.
At the level you will need for the moment, $\sigma$ orbitals come from the overlap of a single lobe from each atomic orbital and $\pi$ orbitals come the overlap of two lobes from each atomic orbital. The $\mathrm{g}$ and $\mathrm{u}$ labels tell you whether the orbital is symmetric or antisymmetric under inversion, that is, if you move every point $(x,y,z)$ to $(-x,-y,-z)$, does the sign of the orbital remain the same or does it become the opposite sign. The * label is used to describe whether an orbital is bonding (overlap is between atomic orbitals/lobes of the same sign) or antibonding (overlap is between atomic orbitals/lobes of opposite sign).

oh ok. fabulous! The numbers $3$ and $1$, are they anything special that we would need to know right now? I thought they meant the 3rd and 1st sigma and pi orbital respectively in the MO diagram, but I thought I was wrong.

@DavidSmith Those are just labeling how many of that type you have. So $3\sigma_g$ is the third $\sigma_g$ orbital, $1\pi_u$ is the first $\pi_u$ orbital.


text:
Why can't a thermosetting plastic be stretched like an elastomer?
So far I've learnt that there are 2 'types' of thermosets: thermosetting plastics and thermosetting elastomers. Both of them consists of crosslinks that when stretched, allows for the material to spring back into its original shape. So why can't a thermosetting plastic be stretched like an elastomer? Is it something to do  with the degree of crystallinity and elastomers being amorphous?
Sorry if my question is a bit messy, I'm quite new to this topic.

Only has to do with the chain lenghts between crosslink points. There are thermoplastic elastomers (TPEs) with crystalline crosslinks instead of covalently bonded ones, and they can be relatively rigid or highly elastic. It's all about the geometric molecular structure, not so much what exactly the chemical structure of the crosslink is.


text:
What happens in this reaction! Some hints needed

I understand step A involves reductive ozonolysis. and that step B is addition involves addition of ammonia to the carbonyl site to form imines. But what does the sodium salt of trichloroacetic acid do? I'm inclined to say C or D since I cannot explain how else the extra double bonds form with a strong slant towards C because I would expect di-chlorination for D.

It doesn't look like a common reaction, indeed. I think you're right about the first step. As for the second step, OK, let's say that you formed an imine with one carbonyl. What happens then? As for the last step, I think it's like the carbene reaction described here: https://books.google.be/books?id=RH1l_VQcFDQC&pg=PA159&lpg=PA159&dq=trichloroacetate+reaction+with+pyrrole&source=bl&ots=6ril4UfpEj&sig=ACfU3U3EJlEj3mNdLjnnZ7RKsw3J4_NhMw&hl=en&sa=X&ved=2ahUKEwiC0dfqitXhAhURCuwKHfnuDb8Q6AEwC3oECAcQAQ#v=onepage&q=trichloroacetate%20reaction%20with%20pyrrole&f=false

I do know the carbene mechanism but it doesn't seem  how even the dichloro carbene is generated! I am also familiar with the abnormal reimer teimann reaction

Here is a report on decomposition of sodium trichloroacetate in the presence of amines: https://pubs.acs.org/doi/abs/10.1021/jo01050a541?journalCode=joceah

Here is another one: https://onlinelibrary.wiley.com/doi/abs/10.1002/recl.19610800707. If it does not help you directly, it might help someone else to write an answer. Both papers establish that under the right conditions, sodium trichloroacetate decomposes to form dichloro carbene, CO2 and sodium chloride. (I'm not at the level to understand them either...)

Step 1 is ozonolysis with reductive work-up giving butan-1,4-dialdehyde.
Step 2 reaction with NH3 gives pyrrole.
Step 3 is the interesting one. As the discussion in the comments states the thermal decomposition of sodium trichloroacetate gives dichlorocarbene. This adds to one of the double bonds of pyrrole to give initially the dichlorocyclopropane. This is unstable and rearranges to give 3-Chloropyridine. This gives higher yields in neutral aprotic solvents reference here. Further discussion and mechanism here

Image from reference 2


text:
Basicity of subsituted amines
In comparing hydroxylamine, hydrazine, and methylamine for basicity, the order of increasing basicity is hydroxylamine < hydrazine < methylamine.
I tried to reason this trend based on electron donating power of the -$\ce{OH}$, -$\ce{NH2}$, and -$\ce{CH3}$ groups, however that yields the order of basicity: methylamine < hydroxylamine < hydrazine, because more electron-donating power concentrates electron density on the nitrogen atom, making it more basic.
Other factors, such as steric hindrance, do not seem to be impacting anything here, because $\ce{-OH}$, $\ce{-NH2}$, and $\ce{-CH3}$ groups are all around the same size.
What is the key factor that influences the basicity here? I realize that hydrazine has two nitrogen atoms, whereas hydroxylamine and methylamine have one nitrogen atom, so that may be affecting the basicity.

Try this one out. Sort carbon, nitrogen and oxygen according to their electronegativity.

O is more electronegative than N, which is more electronegative than C. So the more electronegative substituent decreases the charge density of the nitrogen atom (thereby stabilizing it) and making it less basic, correct?

I would say yes. Physical chemists may have a much more sophisticated and articulate explanation, but for a simple organic chemist like myself this seems reasonable.

On the other hand, hydrazine is much more basic than ammonia, although H is less electronegative than N. For that one, if I remember correctly, they told us it was because of the repulsion between the electron pairs on the two adjacent nitrogen atoms, so one of the two wants to catch a proton really bad.


text:
Could there be a class of drugs beyond, similar to barbiturates and benzodiazepines?
I'm talking about group of drugs similar to them but different though.Is there anything beyond benzodiazepines?

Similar in structure or effect - and if so, which effect?

I'm talking generally sedatives/anxiolytics. If you have any knowledge about potentials in structre i'd also appreciate it.

The Serotonin/Noradrenaline reuptake inhibitors have activity against anxiety/depression

Wikipedia has a list of sedatives divided into 13 categories: barbiturates, benzodiazepines, orexin antagonists, herbal sedatives, general anesthetics, antidepressants, antihistamines, opioids, etc.
A list of anxiolytics greatly overlaps with sedatives, but additionally includes sympatholytics, carbamates and some other individual drugs.

I'm talking more about undiscovered could there be something beyond benzos

There could be, this remains an area of interest to many pharma companies. Small molecules that act at glutamate receptors for example: https://www.sciencedaily.com/releases/2013/10/131006142321.htm


text:
Hydrogen peroxide and carbon as hybrid rocket propellant?
I was thinking that a hybrid rocket could be made using hydrogen peroxide and activated charcoal or some other carbon source.
Step 1: Decompose hydrogen peroxide to steam and oxygen, as in a monopropellant H₂O₂ rocket. This step is exothermic.
2H₂O₂ → 2H₂O + O₂
Step 2: Have the superheated steam flow over the finely divided carbon, which is the "water gas" reaction and is endothermic:
H₂O + C → H₂ + CO
Step 3: The oxygen produced by the hydrogen peroxide incompletely burns the water gas, which should be exothermic.
2H₂ + 2CO + O₂ → x(H₂O + CO) + y(H₂ + CO₂), where x + y = 2
Total reaction: 2H₂O₂ + 2C → x(H₂O + CO) + y(H₂ + CO₂). This is certainly exothermic in total, and also favored entropically.
Depending on the kinetics, either reaction x or reaction y would dominate; I'm hoping y will, as having lots of diatomic hydrogen would be good for specific impulse.
According to the article on water gas, the carbon bed has to be heated red-hot. I'm hoping that having the steam be hot and the carbon be porous would be sufficient. I'm also wondering if the oxygen would react right there in the chamber, or if it'd flow out the back and need to be ignited. Small amounts of catalysts might be mixed in with the carbon as well. You might even be able to just coat the carbon with silver catalyst so the decomposition of hydrogen peroxide heats the coal bed, but replacing the catalyst every time would probably be expensive.
The primary advantage of such a rocket would be that the oxidizer is relatively nontoxic and nonreactive; the fuel is safe enough you could eat it like a biscuit; any fire caused by an oxidizer tank leak or rupture could be far more easily contained; and design of the rocket should be fairly simple, involving just placing a fuel biscuit in the exhaust of a H₂O₂ monoprop rocket; this could theoretically supercharge the peroxide-using USAF Astronaut Maneuvering Unit.
Has this idea been tried before by serious rocketers? What are the numbers (enthalpy, Gibbs free energy, calculated specific impulse, and so on)? How can I make sure the carbon biscuit is porous enough that it reacts easily and completely with the steam and oxygen, yet not so porous that it just gets shredded and blasted through the exhaust by the pressure?

I would guess that the reaction is not exergonic enough to reach a very high exhaust temperature.

The formation enthalpy of water is -240kJ/mol, of H2O2 -190kJ/mol. 50 kJ/mol is just enough to evaporate the water afterwards. Making O-H bonds releases a lot of energy, but you already have them. Your H2O2 is around 50% ballast. ;-)

You're overthinking it. Highly conc. H2O2 is a rocket  fuel oxidiser and you sure could used C dust as fuel, but both aren't particularly good in practice.

Well, yes, I could get better dV by using LOX/RP1 or nitric acid/UDMH, but both of those have major handling and safety issues. I'm looking more for something that's an evolution of the Bell Rocket Belt, and having a single, relatively safe oxidizer is part of that goal.


text:
Why is my conclusion inconsistent with the van't Hoff equation?
Let's say I hypothesize that a graph of $\ln K$ vs. $1/T$ has a slope of $-∆G^\circ/R$ and a $y$-intercept of $0$. I prove it simply:
$$∆G^\circ = -RT\ln K \quad\to\quad \ln K = -\frac{∆G^\circ}{RT}$$
This matches the linear form $y = mx + b$. Thus, plotting $\ln K$ vs. $1/T$ would have a slope $m = -∆G^\circ/R$ and a $y$-intercept $b = 0$.
However, I understand that a van't Hoff plot defines a graph of $\ln K$ vs. $1/T$ to have a slope of $-ΔH^\circ/R$ and a $y$-intercept of $∆S^\circ/R$. It is clear from the relation $∆G^\circ = ∆H^\circ - TΔS^\circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that
$$\ln K = -\frac{∆H^\circ}{RT} + \frac{∆S^\circ}{R},$$
but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^\circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^\circ/R$ instead of $-ΔG^\circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $\ln K = -∆G^\circ/(RT)$?

The fact of the matter is that the differential version of your equation
$$\frac{\mathrm{d}\ln{K}}{\mathrm{d}\left(\frac{1}{T}\right)} = -\frac{\Delta G^\circ}{R}$$
is not exact (because $\Delta G^\circ$ is a function of $T$) while the form of the van't Hoff equation involving differentials
$$\frac{\mathrm{d}\ln{K}}{\mathrm{d}\left(\frac{1}{T}\right)} = -\frac{\Delta H^\circ}{R}$$
is exact.  Moreover, the derivation of the van't Hoff equation properly takes into account the fact that, in varying temperature $T$, the initial and final states for $\Delta G^\circ$ are constrained to be at 1 bar.  So, in the van't Hoff development, the temperature derivative of $\Delta G^\circ$ is exactly given by
$$\frac{\Delta G^\circ}{\mathrm{d}T} = -\Delta S^\circ$$
This important constraint is not even addressed in your approach.
Finally, do you have a reference where it asserts that the intercept at $(1/T) \to 0$ is supposed to be $\Delta S^\circ/R$?

I took the intercept to be ∆Sº/R as 1/T approaches 0 just from the linear form of the van't Hoff equation, ln(K) = -∆Hº/RT + ∆Sº/R. If this is in y=mx+b form, b is ∆Sº/R. I guess my reference is the Wikipedia page for the van't Hoff equation, which mentions it a few times (such as in the "van't Hoff plot" section). Is this not an accurate assertion to make?

Actually, if you take the tangent line to the curve of ln(K) vs 1/T at the point (T*, K*), the intercept is $$\ln{K^}+\frac{\Delta S^0(T^)}{R}$$

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $\Delta G^\circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -\frac{\Delta H}{R} + \frac{T \Delta S}{R}$$
The $y$-intercept corresponds to an infinitely high temperature where $-\frac{\Delta H}{R} \times \frac{1}{T}$ tends to zero and $\frac{T \Delta S}{R} \times \frac{1}{T}$ cancels to be just $\frac{\Delta S}{R}$.

Isn't ∆Gº a constant defined at 298 K? The relation ∆Gº=-RTlnK gives the standard change in free energy (i.e. 298 K, 100 kPa, 1M), so wouldn't this value be constant for a given reaction? Plus, if we were talking about non-standard values, doesn't change in enthalpy depend on temperature as well?

@MateenKasim You are varying the temperature. The Gibbs energy and the equilibrium constant are significantly temperature-dependent. The enthalpy is temperature-dependent too, but to a much lesser degree.

If you want to see the rigorous treatment, look at the answer posted by @Chet_Miller.


text:
Final product when sp2 oxygen of acid is O18 in fischer esterification

In the question above, i was curious about option 3. If the carbonyl oxygen is O18, shouldn't half of the product we obtain contain O18 on the carbonyl since we have two possible leaving groups with equal probability of leaving?

$\rm^{18}O$ is not radioactive, which might give us some insight as to the level of this quiz and its authors. Your concern is also legitimate. Forget it. Just give them the answer they want.

@IvanNeretin I understood that and deleted my comment after that. Thank you.

@AdityaGarg I don't think its possible to isolate O18 as only the carbonyl oxygen. I maybe wrong but I think that since the Acid would be partially(although small) dissociated, i t will lead to resonance between O16 and O18 and thus at equillibrim, the carbonyl oxygen will have both the O18 and O16.


text:
Importance of substrate stereo chemistry in semi pinacol rearrangement
Consider the following two reactions (source):

I don't quite get why the products shown are the main ones. In reaction 3A., if you rotate the carbon facing us by 120 degrees clockwise, the anisole would be antiperiplanar with the leaving group and should migrate instead of the phenyl group. In reaction 3B., the same rotation should lead to a migration of the phenyl group instead of the anisole group.
Why does phenyl migrate in the erythro isomer and anisole migrate in the threo isomer?

@Esha Manideep, as a suggestion for future questions:  If you intend to substantiate your question by pictures, have a look how to crop and rescale them and eventually insert them here on ChemSE.  This will ease visitors of ChemSE to see your question in one glimpse compared to your original post relaying to a page outside ChemSE.


text:
Size of hydrocarbons and boiling point
Increasing the contract surface area between hydrocarbons raises boiling point, so hexane should have a higher boiling point than propane. This doesn't really make sense to me. What is the difference between pulling apart a pair of propane twice, versus one pair of hexane? Why would the total amount of energy needed differ? It's seems to go against the concept of pressure.

It is also about entropy. If you pull two propane molecules out of liquid propane you get two molecules with more freedom to move as opposed to just one when it is hexane. Using your argument, you would get two different boiling points for hexane, depending whether to pull apart a single hexane (lower boiling point) or two at a time (higher boiling point).

Sorry, Karsten, I don't really agree with you. Entropy has little to do with this process, which can actually be explained only in terms of enthalpy. The boiling point of a molecule is unique ,that means that you can't have two boiling points for water, for instance, or for any other substance. If you pull apart hexane, then you are dissociating the molecule into two different fragments, which is different than separating two molecules using heat.

@Pier You are using entropy implicitly in your answer as you compare the enthalpy per molecule to disrupt dispersion forces. You could have compared it per mass, in which case all hydrocarbons would have similar boiling points. There is a nice article on Wikipedia, https://en.wikipedia.org/wiki/Entropy_of_vaporization. The (molar) entropy of vaporization happens to be very similar for hydrocarbons irrespective of size, so you get good results just comparing enthalpies of vaporization.

Some data on size of hydrocarbons and boiling point
I found complete data on the butane/octane pair but not the propane/hexane pair, so I am using those as illustration (I think they are taken at a temperature near the normal boiling point, respectively).
$$
\begin{array}{|c|c|c|c|c|}
\hline
\text{substance} &  \Delta H_{vap} & \Delta S_{vap}  & T_b & \text{ref}\\
\hline
\text{butane} & \pu{22.4 kJ/mol}  &  \pu{82.30 J/mol K}& 273 K & [1]\\
\hline
\text{octane} & \pu{39.1  kJ/mol}  &  \pu{98.3  J/mol K}& 398.7 K & [2]\\
\hline
\end{array}
$$
Ref. 1: https://webbook.nist.gov/cgi/cbook.cgi?ID=C106978&Mask=4
Ref. 2: https://file.scirp.org/pdf/OALibJ_2016091314302030.pdf
The enthalpy of vaporization is determined by measuring equilibrium vapor pressure over a range of temperatures (Clausius Clapeyron equation). Then, the entropy of vaporization is calculated from the boiling point and the enthalpy of vaporization ($\Delta G_{vap} = 0$ at the boiling point):
$$ \Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}$$

What is the difference between pulling apart a pair of propane twice, versus one pair of hexane?

As far as enthalpy is concerned, there is not much difference. You have to disrupt about the same dispersion forces removing one molecule of hexane from liquid hexane or removing two molecules of propane.
However, if enthalpy were the only aspect you had to consider, liquids would never turn into gases because you lose all interactions in the gas phase. Experience shows that if you increase the temperature sufficiently, at one point you disrupt the interactions between molecules in a liquid and get a gas.

Why would the total amount of energy needed differ?

What determines whether a process happens or not is not the change in enthalpy alone, but the change in Gibbs energy. The Gibbs energy contains an enthalpy component and a highly temperature-dependent entropy component. The entropy change of a molecule from the liquid to the gas state is about the same for butane and octane (also see Trouton's rule), but the enthalpy change is different by a factor of two. That's why the two substances have different boiling points.
$$ \Delta G_{vap} = \Delta H_{vap} - T_b \Delta S_{vap} = 0$$
$$ T_b = \frac{\Delta H_{vap}}{\Delta S_{vap}}$$
If you ask for the analysis for evaporating a kilogram each of butane and octane, the enthalpy change for the entire sample would be similar, but the entropy change for the entire sample would be off by a factor of about 2. This is because the molar entropy of vaporization is similar for butane and octane, but a kilogram of butane contains about twice the number of molecules compared to a kilogram of octane.

It's seems to go against the concept of pressure.

This is a bit of a puzzling statement. There are two things that require energy when a liquid turns into a gas at constant pressure. The first is disrupting the intermolecular forces in the liquid. The second is pushing against the pressure to increase the volume of the sample as liquid turns into gas. The latter is abbreviated as P-V-work. If you look at internal energy changes, you have to consider it. If you look at enthalpy changes, however, it is taken out of the equation because enthalpy is defined as
$$ H = U + P V$$
As we are discussing Gibbs energy and enthalpy, we don't have to worry about P-V-work. ("It is not part of the bill").

I understand and agree with your explanation, but I feel that your last paragraph is a bit confusing as you use kg not mole as your comparison in the first 2 lines. You say ' ..is not the change in enthalpy'  in line 3 but 'the enthalpy change is different by a factor of 2 ' in the last line. You write also in para 1 that dispersion forces are similar 'there is not much difference' but then use this difference to explain the change in boiling point in the last para.

@porphyrin - I added some data and broke the paragraph you mentioned into two parts.

very clear now :)

To answer your question, keep in mind that for hydrocarbons the main attractive force at play is dispersion interactions. They increase as the size of the molecule increases. In other words, since the molecule of hexane is "larger" than the molecule of propane, it means that it has a larger surface capable of interacting with the surface of another molecule of the same kind. The more surface you have, the more intense is this interaction, and the more energy you need to tear the two molecules apart.
Take a look at the table listed in this website: https://crippen.education.ufl.edu/calculators/activities/BP.html
You see that the larger the mass (or surface of the molecule), the higher the boiling point.
Can you tell us a little bit more about your last question? Why should it go against the pressure?
I hope that this helps you a little.


text:
CIP Rules - How to deal with a double bonded nitrogen?

If a substituent were to exist, how will one deal with the doubly bonded nitrogen? In the tree will the nitrogen with the double bond have (C,C) or (C,C,C) or (C,C,lone pair)?
I could not find this case on the CIP Wikipedia.
Also if it were (C,C), then how would one compare the priority of this with say (C,C,C)?

It would be helpful if you could draw the complete hierarchical digraph.
Here is my shot at it:

I have taken your functionality and I have incorporated it into structure 1 that has stereochemistry. The two double bonds in 1 are shown in the CIP shorthand of structure 2. C3 and C4 are non-duplicate atoms and they are attached to duplicate carbon atoms (4) and (3), respectively. Similarly, in the C=N double bond C1 is attached to the duplicate red nitrogen (N) while the non-duplicate red nitrogen is attached to duplicate carbon (1). Both non-duplicate and duplicate atoms of the same element have the same atomic number, however, duplicate atoms are attached to three phantom atoms of atomic number zero. In addition, all atoms (C,N,O,Cl,etc.) are treated as though they are quadrivalent.
In the digraph the blue numbers are the successive spheres emanating from the stereogenic center C2. In sphere 1 hydrogen has the lowest priority while C5 has the next lowest priority owing to the 3 H's in sphere 2. C1 is attached to N,(N),N in sphere 2, having the highest priority, while C3 is connected to 4,(4),H in sphere 2. The overall priorities are C1>C3>C5>H.
Now to the gist of your question. While N=N=(N), in sphere 3, (N) is attached to 0,0,0. (N) has the lowest priority of the three nitrogens. Of the remaining two nitrogens, C,C,0 = C,(1),0 in sphere 3 but this tie is broken in sphere 4 where H,H,H/H,H,H has priority over H,H,H/0,0,0. The nitrogen priorities are black N>red N>(N).


text:
Why do metals form cations?
metals that have a low number of occupied shells such as lithium and potassium should have a stronger electrostatic attraction to their nuclei, so what causes them to always lose their electrons during ionic bonding with a non metal?

Potassium is in the 4th period and therefore has at least 4 shells occupied with electrons while lithium is in period 2 and thus only has 2 shells occupied with electrons. Perhaps you mean they both have 1 valence electron that must be ionised before they become a cation? Anyhow, ionic bonds are formed due to the strong electrostatic attraction between oppositely charged ions. All elements would like to achieve a noble gas electron arrangement due to its stability. However, to be able to achieve this, elements must either gain or lose electrons depending on how close their electron arrangement is to their closest noble gas, e.g. Fluorine would like to gain 1 electron while Magnesium is more likely to lose 2 electrons to form its closest noble gas.
For instance, NaCl: Na would like to lose an electron and Cl would like to gain an electron. Together this is possible. The nucleus of the Cl atom electrostatically attracts the valence electron from the Na atom– this attraction is strong enough the overcome the the attraction this valence electron feels from its own nucleus. As a result, the Cl gains a negative charge due to it gaining a negatively charged electron and the Na atom now has a positive charge as it has one less negatively charged electron. The Cl- and Na+ are oppositely charged and therefore attract each other. This is known as ionic bonding and results in the formation of an ionic lattice.
Now to your question. All elements in group 1 have approximately the same effective nuclear charge, i.e. the attraction the valence electrons feel from the nucleus of their atom after the shielding of the inner electrons is considered. Hence, the fact that they are in group 1 should not significantly affect the ability to lose their valence electrons. However, as one goes down periods, the number of occupied outer shells increases, which means that the valence shells experience a weaker attraction from their nucleus. This is confirmed by the lower first ionisation energy trends. To answer your question frankly (I perhaps overcomplicated with my verbose writing style), it is the strong affinity for electrons of the non-metals that result in the loss of the valence electrons from the metals. Even if the the metal has a smaller atomic radius due to occupying less shells, the affinity for the electrons from the non-metals is strong enough to take the metal's valence electrons and make it into a cation.
References:
Pearson Higher Level Chemistry Textbook, 2nd Edition. By Catrin Brown and Mike Ford.

thank you for the explanation- so to summarise, the outer electron of a metal atom is more attracted to the more densely positive non metal nucleus than its own nucleus?

Not because the non-metal nucleus is more densely positive, but because it has a stronger affinity for the electron in combination with the metal wanting to lose its valence electron to attain a noble gas electron configuration.


text:
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
Totally an elementary question.
Staring at a candle, it appears that the bottom of the wick is dark whereas the top glows. However the bottom of the flame (the blue) is the hottest.
Is the reason for this that the concentration of liquid wax is greater at the bottom, offsetting the greater temperature at the bottom?

The wick temperature does not have to be the same as the flame temperature.The flame is hottest at the bottom, but the wick is hottest at the top.
For a candle, the wick burning isn't the intended purpose of the wick; light comes from burning wax (more generally: fuel), you want to burn the wax not the wick. Rather the purpose of a wick is to help fuel evaporate by soaking up wax and allowing the radiant energy from the flame to heat the wax causing it to evaporate and burn also.
As wax travels up the wick, it evaporates and less wax is in the wick the further up you go. Eventually the wax dries up and the radiant energy is heating a wick without any wax. Eventually the wick gets so hot at the tip, that it will glow due to black-body radiation.
In summary: though the blue is the hottest part of the flame, the wick can evaporate wax to cool towards the bottom. There is no wax at the top and thus as the radiant energy of the flame causes it to get hotter until it starts to glow.
Extra: I will note that the top of the wick does not burn when lit as the gas around the wick has too little oxygen to burn (unless the wick slumps out side the flame). When you blow out the candle the low oxygen environment of a flame is gone and thus the hot wick will smoulder.

The physics behind the inner workings of a candle is surprisingly interesting.

@T.Sar you should check out: Can flames be colored black?

Because fire is not the same thing as light.
Michael Faraday did a wonderful job of explaining how the candle works, and I direct you to look at it (there are also Youtube videos giving a modern take on this work) if you're interested.
In short, the candle produces light, not because it is hot, but because it is sooty. The particles of soot glow when they are hot (blackbody radiation), and that's what produces light.
If you take a flame where the fuel is very well-mixed with oxygen and the flame that's produced is not sooty, it does not glow like a candle even though it's very hot. This is precisely why Bunsen burners are used in a laboratory. They give a hot but non-illuminating flame that's great for doing analysis.

I just browsed a bit in the Faraday lecture for children: wonderful!

This answer seems to be explaining the visibility of a flame, while the question is asking about the color of the wick.

@KarstenTheis Or, if you like it per video, Bill Hammack (U of Illinois) created a series of five + one videos of the Faraday lectures itself plus five commentary videos and supplementary material: http://www.engineerguy.com/faraday/ or https://www.youtube.com/watch?v=RrHnLXMTOWM&list=PL0INsTTU1k2UCpOfRuMDR-wlvWkLan1_r is an entry to it.

@Buttonwood That is exactly what I was alluding to in my answer. :)

@Zhe This is a terrific explanation. While A.K's answer more directly addresses the question regarding the wick, I've found that your answer was enlightening and I very much appreciate your response. I'm glad how many other users have agreed by show of upvote! Have one from me too!

@ZAR Thanks! I probably should've read your question a bit closer though... :)


text:
Gibbs Free Energy and Maximum Work
I am a high school student and my professor mentioned that the Gibbs free energy is the maximum amount of work (or useful work) that a system can do, whereas entropy is a measure of the non-available enthalpy.
I am not able to understand this and it would be helpful if you could explain this or link to some resources that could explain this at a high school level (I know some multivariable calculus, but not that comfortable with it) as I have difficulty finding them.
Also, we used the fact that Gibbs free energy is the maximum amount of work to derive the equation $∆G = -nFE$ and then proceeded to derive the Nernst equation.

It is impossible to understand this without spending a lot of time on it, time which you typical only have in your third year or so in college. However, I think it is possible to come up with a better hand-waiving account of what is going on than what you summarized.

Gibb's free energy is defined as the only available energy to do work.

Entropy represents the degree of randomness. It is a function that tells us which substance has more random motion of molecules

@Divy Sancheti, Can you access the Journal of Chemical Education? It has excellent conceptual articles on Gibbs free energy. Search Gibbs free energy in the titles.

https://chemistry.stackexchange.com/questions/63967/why-does-the-gibbs-free-energy-change-determine-spontaneity-of-a-reaction/63992#63992 Note that $\Delta G$ measures the excess of enthalpy relative to a scaled entropy. Any excess is potentially available for work, but only this excess as you must have excess or nothing to conform to the second law of thermodynamics.

In order for a process to happen, it has to increase the combined entropy of the system in which it happens and of the surrounding (2nd law of thermodynamics). As we will see in a bit, the more work the process does on the surroundings, the less the entropy of the surroundings increases, putting a limit on how much work the process can do.
What happens to the energy leaving a system?
When a process brings the system from a higher to a lower energy, the energy leaves the system in the form of work done on the surroundings or heat transferred to the surroundings (1st law of thermodynamics). For a given set of initial and final states, the amounts of work and heat can vary, but the sum has to be the same. For example, if our system is a charged battery, we can either run a motor with it, lifting up a rock (work done on the surrounding). Or we can short the battery, creating thermal energy that is transferred to the surrounding in the form of heat. We can even have a situation where heat and work have opposite signs (e.g. do more work and have heat transferred into the system to get the same energy balance).
How is heat and work related to entropy changes?
The change in entropy inside the system depends solely on the initial and final state. For the surrounding (with nothing else going on in the surrounding), the change in entropy depends on the heat transferred. The more heat is transferred to the surrounding, the more the entropy of the surrounding increases. To maximize the work, you want to minimize the heat transferred to the surroundings (or even transfer heat to the system), but only to the point the that total entropy still increases (otherwise the process could not run). So if you match the entropy change in the system with the opposite entropy change in the surrounding (via heat exchange), you get the maximum work out of your process.
Can I see a picture?
Sure. The picture shows that the sum of heat and work is equal to the enthalpy change of the system. Values shown with a downward arrow are negative and with an upward arrow positive. For heat and work, negative values mean energy is transferred out of the system. For example, work done by the system on the surrounding is a downward arrow, and exothermic reactions have a downward arrow for $\Delta H$.
The picture also indirectly shows the entropy (via the arrows labeled heat and $-T \Delta S$. If the arrow for $-T \Delta S$ is downwards, entropy in the system increases (scenarios labeled "more dispersed"). If the arrow for heat is downwards, entropy in the surrounding increases. To calculate the actual entropy, you would have to divide by the temperature, but the way it is shown everything can go into a single picture.
I have three scenarios, depending on whether the process is exothermic or endothermic and whether the entropy in the system increases or decreases. One scenario (endothermic reaction with a decrease of the system's entropy) is not shown (you would have to do work on the system to drive such a process, and we are trying to make the process do work for us instead).
In scenario 1, the process is exothermic. Some of that available energy is transferred in the form of heat (to make sure that the overall entropy does not decrease) and some of it is transferred in the form of work done by the system.
In scenario 2, the process is also exothermic, but different from scenario 1 the entropy of the system increases in this scenario. The work done by the system is (surprisingly, perhaps) larger than the enthalpy change, and the extra energy comes from heat transfer from the surroundings.
In scenario 3, the process is endothermic. You might think that this process can't do any work because - where would the energy come from? It comes from the surrounding, partially to provide the energy for the endothermic process and partially to be transferred back into the surrounding in the form of work. This can happen because just as in scenario 2, the entropy of the system increases.

[...]the Gibbs free energy is the maximum amount of work (or useful work) that a system can do, whereas entropy is a measure of the non-available enthalpy.

This sort of describes scenario 1, which I pasted again below with two boxes for emphasis.

The pink box represents the 1st law of thermodynamics. There is some enthalpy available from the process, and this will leave the system either as work or heat or some combination, as long as it adds up to the enthalpy change in the system. The blue box represents the 2nd law of thermodynamics. The entropy increase of the surrounding (through transfer of heat to it) has to be larger or equal to the entropy loss in the system. If you combine the two laws, you get the maximum work the system can do. In this scenario, the change in entropy of the system times the temperature represents that portion of the enthalpy not available for work (labeled $-T \Delta S$ in the diagram). You can't, however, use that label in the other scenarios where the entropy of the system increases. In those scenarios, the entropy change in the system results in more work available than the enthalpy change would suggest.
This is all rather complex and might make you dizzy. The Gibbs energy helps us to summarize all this in a single quantity. It combines entropy and enthalpy changes in the system in a neat way that tells us about the maximum work. In the absence of work, it also tells whether the process will happen in that direction or not.
Disclaimer for those that have a physical chemistry background
This is for a closed system at constant pressure and temperature (i.e. P-V work is done against a constant pressure, and initial and final state have the same temperature). Whenever it says work, it refers to non-P-V work only. Labels $\Delta H$, $\Delta G$, $\Delta S$ should read $\Delta H_\text{sys}$, $\Delta G_\text{sys}$, $\Delta S_\text{sys}$.


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Lewis structure of benzenediazonium
In the benzenediazonium $\ce{C6H5N2+}$ the nitrogen contributes its lone pair to the carbon in benzene. So where is one electron of that carbon, since it is not used for bonding with nitrogen where has one of the electron of carbon gone?

Same place as it's hydrogen ;>

If the diazonium ion were formed in the way you described, it would be formed by N2 acting as a nucleophile and attacking the ring with a lone pair, displacing a leaving group - a nucleophilic aromatic substituion. In this case, the "one electron of carbon" would have left with the leaving group that was on the ring before. The carbon would now be neutral as it had lost an electron and essentially gained one from the dative bond from the N.
However, this isn't the usual way to form a diazonium ion. In fact, we normally start with a primary aromatic amine, and react it with a diazotizing mixture of NaNO2 and HCl. In this case, the positive charge on the N ends up, not from the N forming a dative bond whilst acting as a nucleophile towards the ring but in the final elimination of water to form the triple bond in the diazonium ion:

Image: Wikipedia

In the last second structure, why positive charge  is sent to terminal Nitrogen from charged middle nitrogen ?