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for a polydisperse polymer, why is its weight average molecular weight always bigger than its number average molecular weight?
The statement comes from my lecture slides on polymers. I hope someone can show me the maths proof for the inequality.
I will try come up with the proof, too. It is actually a maths question:D Here is the basic definition for the two average weights: https://en.wikipedia.org/wiki/Molar_mass_distribution

Surely it's axiomatic that for a positive number (in this case the molecular mass), that the squared value is larger? By definition if a polymer is described as polydisperse, the PDI is >1, so Mw must be greater than Mn

Mathematics has many examples of identities and concepts which are extremely powerful due to their simplicity, such as the pigeonhole principle and the triangle inequality. One such identity will guide us through the problem.
We start with the very innocent identity, that for a real number $z$, the following is true:
$$z^2 \geq 0$$
Now let $z = x-y$. We have:
$$(x-y)^2 \geq 0$$
$$x^2 -2xy + y^2 \geq 0$$
$$x^2 + y^2 \geq 2xy $$
$$\frac{x^2 + y^2}{2} \geq xy $$
Now this last inequality has a very powerful interpretation. To show it more clearly, we can replace $x^2$ with $a$ and $y^2$ with $b$, yielding:
$$\frac{a + b}{2} \geq \sqrt{ab} $$
In words, the arithmetic mean of two positive real numbers will always be greater than or equal to their geometric mean. This is called the AM-GM inequality, and it often shows up in the most surprising of places.

Let's get started. We define the number average molecular weight ($M_{n}$) and the weight average molecular weight ($M_{w}$) of a polydisperse polymer mixture as the following, respectively:
$$M_{n}=\frac  {\sum\limits_{i} N_{i}M_{i}}{\sum\limits_{i} N_{i}} \quad , \quad M_{w}=\frac  {\sum\limits_{i} N_{i} M_{i}^2 }{\sum\limits_{i} N_{i}M_{i}}$$
We then define the dispersity index $Đ$ (or $PDI$) as:
$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2}$$
To make things easier, let us start by analyzing the first non-trivial case ($i=2$):
$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2)(N_{1} + N_{2}) }{(N_{1} M_{1} + N_{2} M_{2})^2}$$
We multiply the bracketed terms and expand the binomial to obtain:
$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{1}N_{2}M_{1}^2 + N_{1}N_{2}M_{2}^2}{N_{1}^2 M_{1}^2 + 2N_{1}N_{2}M_{1}M_{2} + N_{2}^2 M_{2}^2}$$
Rearranging the terms slightly:
$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}$$
Notice that the terms in the numerator and denominator match except for the coloured parts.
Here is where the magic happens. From the AM-GM inequality, we must have:
$$x^2 + y^2 \geq 2xy \quad \Rightarrow \quad M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2}$$
We then replace the term $(M_{1}^2 + M_{2}^2)$ in the numerator with the term $2M_{1}M_{2}$. By doing so, the new numerator is smaller than or equal to the old numerator. Thus:
$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}} \geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2}}}=1$$
$$Đ = \frac{M_w}{M_n} \geq 1$$

What about the next non-trivial case ($i=3$)? We have:
$$Đ = \frac{M_w}{M_n} = \frac{(N_{1} M_{1}^2 + N_{2} M_{2}^2 + N_{3} M_{3}^2)(N_{1} + N_{2} + N_{3}) }{(N_{1} M_{1} + N_{2} M_{2} + N_{3} M_{3})^2}$$
Multiplying everything out and performing the appropriate arrangements yields:
$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}$$
This time, we can use the AM-GM inequality thrice:
$$M_{1}^2 + M_{2}^2 \geq 2M_{1}M_{2} \quad , \quad M_{1}^2 + M_{3}^2 \geq 2M_{1}M_{3} \quad , \quad M_{2}^2 + M_{3}^2 \geq 2M_{2}M_{3}$$
The end result is the same:
$$Đ = \frac{M_w}{M_n} = \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#ff0000}{N_{1}N_{2}(M_{1}^2 + M_{2}^2) + N_{1}N_{3}(M_{1}^2 + M_{3}^2) + N_{2}N_{3}(M_{2}^2 + M_{3}^2)}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} $$
$$\geq \frac{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3} M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}}{N_{1}^2 M_{1}^2 + N_{2}^2 M_{2}^2 + N_{3}^2M_{3}^2 + \color{#0000ff}{2N_{1}N_{2}M_{1}M_{2} + 2N_{1}N_{3}M_{1}M_{3} + 2N_{2}N_{3}M_{2}M_{3}}} = 1$$
$$Đ = \frac{M_w}{M_n} \geq 1$$

Now stating the general case:
$$Đ = \frac{M_w}{M_n} = \frac{\sum\limits_{i} N_{i} M_{i}^2 \sum\limits_{i} N_{i}}{(\sum\limits_{i} N_{i}M_{i})^2} = \frac{\sum\limits_{i} N_{i}^2 M_{i}^2  + \color{#ff0000}{\sum\limits_{i,j\ (j>i)} N_{i}N_{j}(M_{i}^2 + M_{j}^2)}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}} \geq \frac{\sum\limits_{i} N_{i}^2 M_{i}^2  + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}{\sum\limits_{i} N_{i}^2 M_{i}^2 + \color{#0000ff}{2\sum\limits_{i,j\ (j>i)} N_{i}N_{j}M_{i}M_{j}}}=1$$
$$Đ = \frac{M_w}{M_n} \geq 1$$

I was thinking about explaining it in terms of the Cauchy-Schwarz inequality, but your way is more convincing.


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Why is the curve of an isothermal process above the that of a adiabatic process during compression?
For an adiabatic process $pV^{\gamma}=k$ and after differentiating the equation we get that the slope of an adiabatic process is $$\frac{\mathrm dp}{\mathrm dV}=-{\gamma}\frac{\mathrm p}V$$
For an isothermal process $pV=k$ and after differentiating the equation we get that the slope of an isothermal process is $$\frac{\mathrm dp}{\mathrm dV}=-\frac pV$$
It says in my book that during compression adiabatic curve is above the isothermal curve and that during expansion the vice versa happens. My question is that since the slope of the adiabatic process is $\gamma$ times more than the slope of the isothermal process, shouldn't the curve of the former process always be above the latter.

The two cases you refer to seem to be  for the reversible adiabatic and reversible isothermal change in an ideal gas.
In an adiabatic expansion as there is no heat loss or gain to or from the surroundings ($q=0$) the energy needed for the work of expansion against the external pressure must come from the internal energy of the gas. This results in a lowering of the temperature.
A curve of reversible adiabatic p vs. T change must therfore be below that for a reversible isothermal change. In addition as $\gamma \gt 1$, for example $5/3$ for an monoatomic gas, the adiabatic curve becomes even further below the isothermal one. The slope of the curves is not relevent in determining which is greater.

Sorry but I believe the question was about the slope of a P-V curve (not p-T curve). (Context about me: My book also states the same results as the OP, and I too end up at the same doubt as he did.)

I think you have the answer in front of you already.
For an adiabatic process, $$PV^{\gamma}=k$$
and indeed,  $$\frac{\mathrm{d}P}{\mathrm{d}V}=-{\gamma}\frac{P}{V}$$
For an isothermal process, $$PV=k$$
and again, $$\frac{\mathrm{d}P}{\mathrm{d}V}=-\frac{P}{V}$$
So the adiabatic curve is always steeper than an isothermal curve. (Note, $\gamma > 1$)

It says in my book that during compression adiabatic curve is above the isothermal curve and that during expansion the vice versa happens.

Okay, this is very easy to see. Just consider a toy problem.
Say, you have an arbitrary system that starts of at,
$$P_i = 2\ \text{and}\ V_i = 2$$
For the isothermal case, we can write $P$ as a function of $V$, $P(V) = \frac{4}{V}$ (here $k= P_i V_i = 4$)
Similarly, we can write $P(V) = \frac{2 \times 2^{5/3}}{V}$
I have used $\gamma = 5/3$. Now, say the gas expands do a final volume $V_f = 8$, you obtain a curve like the one shown below by plotting the functions we have just written down.
Now, say we start compressing the gas, starting at an initial volume and pressure,  $V_i = 8, P_i = 0.5$ Isothermally, we trace the same curve back. Adiabatically, we obtain the green curve, since now $P(V) = \frac{0.5\ \times \ 8^{5/3}}{V}$

Your calculations are perfect.
It's just the starting point that matters... I believe that this diagram is self-explanatory:

(Sorry for the hand drawn diagram, I couldn't find the right graph online.)

An interpation of the implicit derivative by the marginal rate of substiution makes this clear

Marginal rate of substitution
In economics, when the level set R(x, y) = 0 is an indifference curve for the quantities x and y consumed of two goods, the absolute value of the implicit derivative
dy
/
dx
is interpreted as the marginal rate of substitution of the two goods: how much more of y one must receive in order to be indifferent to a loss of one unit of x.- From wiki

Going back to our expression, $$PV^{\gamma} - C = 0$$This cast as an implicit function of the from $F(V,P)=0$, now $ \frac{dP}{dV}$ is how much more of pressure should increase to be indifferent to a loss of a unit volume (say on liter is the unit of volume).
In an adiabatic curve, we find the dP/dV is greater than that of isothermal case, this means we need more drop of pressure to account for a drop in unit volume.
Also check from 3:45 of this video

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Photoelectric Effect
So If i were to define what the photoelectric effect is,
is it when light is shined on a metal and electrons just bounce off??
when do the electrons take the energy that was shined and jump excitedly to another state?

Related: http://chemistry.stackexchange.com/questions/18935/how-does-the-photoelectric-effect-depend-on-the-frequency-of-the-light, http://chemistry.stackexchange.com/questions/8678/photoelectric-effect-and-electron-loss

First, electrons are not photons... to "bounce off" implies that the incident particle is an electron, and the electron bounces off. It is better to say the photon "knocks off" an electron.
Second, "do the electrons take the energy [from the light] and jump excitedly to another state?" Yes, that is correct. It is simplistic to say that electrons are simply knocked off; it is better to say raised to a higher energy state. However, for all practical purposes, beyond a few levels higher, it is effectively as if the electron were moved to "infinity". The photoelectric effect in metals is usually observed by the removal of electrons through a circuit, moving them far beyond their source.
In some cases it is easy to observe the electron being boosted to a higher energy level, and then relaxing back with emission of a photon, as in photoluminescence. Zinc sulfide and calcium sulfide, for example, clearly demonstrate photoluminescence.

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How can I safely handle a concentrated (fuming) nitric acid spill?
I will shortly be using $90~\%$ or higher concentration nitric acid in a home environment for the purpose of decapping ICs. I understand the corrosive dangers of nitric acid, and most of the important dangers of its reactivity (e.g. exothermic with water, don't make nitrobenzene, explosive nitrations are bad) and will be performing the decapping in a well-ventilated clean area. I'll also be wearing the best thick acid-resistant gloves I could find, since it seems nothing is really rated for RFNA handling.
My primary concern is handling a spillage. I know a common disposal technique for liquid waste is to feed the $\ce{HNO3}$ very slowly into water to dilute it, then add a base such as sodium bicarbonate to level out the pH, using an indicator mixture to monitor it. However, that's not so easy if the spillage is a few splashes on the workbench — I need a reasonably safe way of wiping it up and disposing of it. Someone told me that acetone wipes are an option for cleanup, but they couldn't remember where they'd read it, so I'm wary of accepting that advice without having it verified.
So, what're my options? How can I clean up a splash of RFNA off a work surface without doing myself (or the environment) any real harm?

Of course, you’ll want to prevent a spillage as well as possible (making sure the bottle doesn’t fall over etc.) but sometimes you just run into bad luck. Let’s hope that your work bench is at least capable of withstanding the acid for a minute or two — if you have a tiled working surface that would be preferred since they will almost certainly withstand.
If the spill is small — a few drops or so — I would take care of it by quickly adding large amounts of water (more than ten times the volume of spilt acid). I would suggest a wash bottle or something similar. Once you have added water to dilute the acid, you can wipe it up with standard kitchen wipes.
Yes, people are always told to slowly add acid to water and not the other way around. However, if it’s only a small amount of acid and using a large amount of water (and if the acid is not sulfuric acid) you can ‘get away’ with doing it the wrong way; the water will absorb the heat quicker than it can boil.
In case you spilt larger amounts, I would also prepare some baking soda to have around. Carefully (really carefully: it will bubble vigorously) sprinkle some of the baking soda onto the spill until the gas evolution is markedly reduced. Then continue as above, i.e. add water and wipe up with kitchen wipes. This intermediate baking soda step would mainly be to reduce the acid concentration by evolving $\ce{CO2}$ — it would be an overkill to use it on small spills, too.
Have all the required material ready, before you unscrew the cap of the acid bottle.
Finally, I would not use acetone wipes. Acetone can react with nitric acid in may unpredictable ways that you really don’t want to run into. Use water and simple, dry kitchen wipes.
By the way: Having eye protection (safety goggles) is greatly recommended.

Thanks for the answer. My primary work surface will be glass (I'm using an old chopping board) on concrete, outdoors but sheltered from weather -- I don't fancy explaining chemical burns on the kitchen worktop to my landlord. Eye protection is a great shout; I would've remembered it at the time but I didn't explicitly think about it until you mentioned it. I have a few pairs of old airsoft goggles that should do the trick. Water and bicarb seems like the safest and easiest option for cleanup, so I'll make sure to have those around in abundance.

@Polynomial Glass is also great since it can withstand acid as well as tiles can. I’m not used to glass surfaces from the lab,though ;)

Storage: Also note that fuming acid does just that it fumes. So storage inside is a bit tricky. A real chem lab would have a vented cabinet.

Personal Safety - Eye protection is not just recommended, it is mandatory. You'll also need leather shoes that are full covered, and an apron or lab coat (full sleeves / long pants). Some sort of rubber gloves. If you spill the acid on yourself then you'll need a method to wash yourself off. (make do emergency shower...). You need to strip with no clothes over the head. So you may need bandage scissors to cut off clothes. You should also have some method to wash your eyes (some sort of eye wash).

I've got an old lab coat, full-body Tyvek suit (not acid-resistant but should at least suffice as an extra layer), acid-resistant hazard gloves as highly spec'ed as I can get hold of, old leather boots, and ballistic grade goggles (designed to take 6mm pellets at up to 600ft/s). Our first aid kit is well-stocked and I have saline capsules and an eye bath, although I suspect if I get any RFNA in my eyes the saline will be less than pathetic at resolving the situation!

As far as storage goes, I'll not be storing more than 50ml or so at any one time, and it'll be stored in the original bottle that it arrived in. I will be using a pipette to extract the acid from the bottle and will be careful not to wet the screw threads.

Sounds good. I just wanted to point out that your safety is a lot more important than the lab bench. ;-)

As another hobbyist (with some lab safety training) decapping ICs: 1) eye protection is extremely important, and 2) respiratory protection also. Personally, I refuse to handle HNO3(conc), not even RFNA, without a full face shield and respirator, with an appropriate (acid rated) canister; being outdoors can serve you even worse than indoors if you get a sudden gust of wind in your face. I use a 3M 6700 series full mask, but anything similar should do.

We don't talk enough about safety on this site. A lot becomes second nature after a few years in the lab, but it is always a good idea to review the safety plan for all steps that you would do in an experiment. The best thing is also to plan for minor problems and not to turn small mistakes into major STUPIDS. // Long sleeves, long pants, and non-absorbing  shoes are to protect your skin. After six months of use all my lab coats had small holes here and there. Better to sacrifice the lab coat than my hide.

Since we're talking safety - you should also make sure whoever else lives with you knows what you are working with, and if they hear horrible screams coming from the shed what they should do about it.

Should you say something about the "dry kitchen wipes".  As a non-chemist, I would be worried about accidentally making "explosive nitrations": what should one avoid making contact with the $H,NO_3$?

@WetSavannaAnimalakaRodVance Which is why he should dilute the acid first. Diluted $\ce{HNO3}$ is too weak to ‘accidentally make explosive nitrations’.


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Why don’t all objects melt and liquefy when heated sufficiently?
Certain objects such as metals, glass, plastic, sugar etc. melt when heated to a certain temperature but other materials such as a piece of wood, paper, a piece of rock etc. do not. Why is that?

Half answer from a non-chemist: burning certain things will cause a reaction with oxygen that produces other substances, many of which will also be solids and gases. Some substances sublimate when heated. Although your example of rock not melting is incorrect: a large portion of our planet is moltern rock!

http://chemistry.stackexchange.com/questions/31050/can-you-melt-wood

And you can melt rock. See "Volcano"

@Polynomial actually, a large portion of our planet is no molten rock. A large portion of our planet is molten metal (mostly iron) residing in the outer core. The rocky part of our planet (mantle are crust) is overwhelmingly solid and is molten only in isolated cases, that in a minority of them we see their surface expression as volcanoes.

You can melt rock. How do you think we get pure metal and minerals (from ore)?

Usually organic substances (hydrocarbons or carbohydrates) such as paper, wood, oil and sugar combust (burn) and most other substances melt.

If you manage to isolate wood and rock then yes it will melt regularly like any other substance.

I just added an answer to https://chemistry.stackexchange.com/questions/31050/can-you-melt-wood?noredirect=1&lq=1 indicating that some melting is possible.

The problem with melting most substances is that you would be doing it in a non-innocent atmosphere and that sometimes even the functional groups inside a compound are non-innocent.
In a perfectly innocent world, everything could be molten given sufficient temperature. Given real world paper, remember that reaction $(1)$ is exothermic and the only reason why it does not spontaneously occur is the lack of activiation energy.
$$\ce{C_nH_mO_l + \frac{$2n + m - l$}{2} O2 -> n CO2 + \frac{$m$}{2} H2O}\tag{1}$$
Once you supply sufficient heat, the activation energy will be overcome and the reaction suddenly becomes spontaneous; for paper, I am told that this happens at around $451~\mathrm{^\circ F}$, which is $233~\mathrm{^\circ C}$ in proper units.
Now you could suggest melting things in an inert argon atmosphere — or maybe even take helium for additional inertness. But then still you have potential reactions that can occur with organic materials. What then will happen might be called smouldering. Technically, this is no more than burning without oxygen.

Many compounds will decompose when heated even in an inert atmosphere but I wouldn't describe that as smouldering. That would be burning slowly (perhaps because there isn't enough oxygen to create a vigorous flame).

There are multiple reasons why things don't melt
The biggest chemical explanation of why some substances don't melt is that many decompose instead. This is true for many complex organics like paper. IF you add enough energy to (theoretically) melt them you have already added enough to shake the component molecules apart. This is true whether or not they are in an inert atmosphere which they can't react with (in air, paper burns before you have added enough energy to decompose it). Some substance are so unstable they fall apart pretty quickly with any added energy (nitrogen tri iodide, for example, which explodes violently with small amounts of heat or, more commonly, just touching it).
Another reason is physical. Some compounds don't have a liquid state at ambient pressure. Carbon dioxide goes directly from solid to vapour at any pressure below about 5 atmospheres. This is a basic physical property.
Mostly, though, what happens to many substances is a reaction with the environment that happens before you have added enough heat to liquefy it. But many substances not normally seen melting will melt with enough heat. Many rocks melt (some decompose). All metals melt, even tungsten, if you heat them enough, though many require special techniques and inert atmospheres for this to be seen.

Everything has a liquid state, so your initial assumption is technically incorrect. But your question is a good one, so let me list the things needed to clarify.
Why does some compounds/solids not turn into liquids at atmospheric pressure? You have to factor in to your question that the states of matter are governed by both temperature AND pressure. Some compounds will not become liquids at room temperature no matter what you do to the pressure, and some compounds will not become liquids at 1 atmosphere, no matter what you do to the temperature. (They will certainly become "fluids" at some point though, but that is an unnecessary complication). If you plot the temperature and the pressure on y and x scale of a chart, the graph explaining the behaviour of a material will look like a slightly right leaning lowercase "y" for most compunds.
Check that out @ Wikipedia
Now, some of the things you have listed do become liquids, but at higher temperatures than what people generally see. Stones for instance, melt.
Also, as others have pointed out, some states are more reactive than others. In general reaction kinetics make reactions speed up from solids->liquids->gases so something that is idle and stable as a solid can vanish as a liquid, due to reaction. Solid states of matter, for instance, generally react so slowly that you can safely assume that solids do not burn*). Which begs alot of questions, with interesting answers. =)
*) Naturally excepting some of the more interesting reactions revolving around Cl3F and the likes

Your first sentence seems quite inaccurate. There are materials that react instead of melting. It is not about Oxygen presence, pyrolysis is a common, real and applied example.

We have to disagree on it then. Everything has a liquid state. Certainly, you can engineer temperature and pressure combinations that thermodynamically hinder liquid formation, but that does not at all prove that the liquid does not exist, only that it does not form for that certain p-T combination. That is like saying liquid argon does not exist, because the world is hotter than its boiling point.

What about CaCO3, you will never get it melt, it will loose CO2 and then CaCO will melt, this fact is used in Thermo Gravimetry. How will you get wood melt, it is complex matrix essentially based on Cellulose and Lignin, you will get reactions before it reaches a liquid state provided such state it exists for such a material. It really seems your sentence is too general to be true in real applied Chemistry.

About you example on Argon, I never said such a thing. And by the way you are only speaking in term of pure substance. I gave you an example of pure substance that cannot melt before decomposing.

Of course CaCO3 can and will melt, if the temperature and the (partial) pressure (of CO2) is high enough. For every substance there is a temperature and a pressure where a liquid exists. This is starting to sound like a really nitpicking debate, which does not answer the OPs question at all, so either lets take it to chat or agree to disagree.

Every substance is really different that every material or everything (will you melt at the correct pressure and temperature?). Your first sentence is inaccurate. The OP is asking about material (real material). Material are too far complex to be modelled as you say it could. Chemistry does not have simple rules with no exception.

By the way, if you control CO2 pressure, CaCO3 will melt and decompose, both phenomena will occur provided it is technically feasible. There are reaction in materials because they are complex and they cannot be modelled the same way you argue for pure substance.

This is not a nitpicking debate, that are comments asking you to improve your post. Will you miss this opportunity? If your first sentence is accurate, could you provide us a reference?

Allright, let the nitpicking begin then... I don't mind I just think we are way beyond the question at this point =) The statement I think is imprecise (at best) is that some materials do not have liquid states. Every material has a liquid state, just that for some materials under "normal" conditions, the phase change is directly from solid to gas. Reactions aside, they are not phase changes (but the argument still holds) - Everything has a liquid state, under certain combinations of pessure and temperature. https://en.wikipedia.org/wiki/Triple_point is one source

Aaand the next source is the fact that we can dig CaCO3 out of the earth, which at some point was molten. There is ample evidence of CaCO3 minerals being deposited as lava, under pressure. If the pressure is released, however: " Explosive eruptions usually involve thick, highly viscous, silicic or felsic magma, high in volatiles like water vapor and carbon dioxide." -(wikipedia).

Could you provide us the correct conditions for metling wood (or even cite some experiment that says it is possible), or even simpler a hair fiber? Do all proteins melt? I am pretty sure this question is far away simple than you pretend. By using the term everything, you make your statement very general and this is the pitfall in the first sentence of your answer. I do not disagree with the whole post, only with the first sentence. Sorry if you think than making correct statement is nitpicking. Sorry you do not have a better reference than your argument and wikipedia. Everything is very broad!

Sure, here are the conditions. Put it in orbit around Jupiter, and slowly decrease your orbital velocity. Somewhere before you hit metallic, liquid hydrogen, your piece of wood will have liquefied and possibly solidified again. We can always discuss wether it will be a supercritical fluid and so not a liquid at all, but in order to have a critical point, it needs a triple point, so there we go. it can be a liquid,

Well I now understand your point, and I know why I disagree with. The key point is Gibbs phase rule, you are assuming that wood could exist in a gas phase (or even could have a critical phase) and this is why you assume it must have a liquid phase between those states. I really doubt it is possible (and I have no way to prove it) to reach such a phase without breaking a covalent bound or reacting. I am glad to understand why such a noise occurred. Good night.

"I really doubt it is possible (and I have no way to prove it) to reach such a phase without breaking a covalent bound" Yes, but that is what melting does. It breaks bonds.

Do you agree that wood is neither a pure substance, nor a binary mixture, not even a mixture, but essentially a complex matrix of polysaccharide polymers (cellulose) blended in cross-linked phenolic polymer (lignin)?

Let us continue this discussion in chat.

We could do that, but please first answer my question

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text:
NMR chemical shift range of different elements
A typical $\ce{^1H}$ NMR runs from approximately 0 to 10 ppm, give or take a bit. $\ce{^13C}$ NMR runs from 0 to 200. And $\ce{^59Co}$ NMR runs from -5000 to 15000 ppm!
There seems to be some correlation with atomic number, although three points hardly make a trend. Is there an explanation for this? I'm not interested in the exact numbers, but rather the relative orders of magnitude.

But $\ce{Xe}$ has only a range of 7000 ppm

Negative proton ppm's are not uncommon. They are measured relative to tms (usually) which  is why negative values can occur. In  annulenes the proton shifts are from $\approx -5$ to $\approx +20$ ppm. So negative values are because of the choice of reference, absolute values due to shielding as described by @long

That's true, 0 to 10 ppm was quite a conservative range indeed. And, cobalt just has an incredibly large range, probably due to small excitation energies of d electrons. So Xe's 7000 ppm is pretty big considering it's a p-block element.

The are a number of important factors that contribute to the shielding of a nucleus. Chemical shifts arise due to differences in the local magnetic field in the different environments within a molecule, driven by overall electron density about that nucleus. That is to say, the effective magnetic field experienced by a nucleus is a function of the shielding constant, σ. There are a number of components of σ.
$$B_\mathrm{eff} = B_\mathrm{o}(1- \sigma)$$

σdia is the local diamagnetic contribution
σpara is the local paramagnetic contribution
σm is the neighbour anisotropy effect
σrc is the ring current contribution
σef is the electric field effect
σef is the solvent effect

The two most important factors for the chemical shift range are the diamagnetic and paramagnetic contribution. All other contributions are quite small and operate within the effects of these fist two.
The diamagnetic contribution comes predominantly from electrons in the s-orbital; for most nuclei this is almost constant for all arrangements. Its effect diminishes with increasing distance from the nucleus, which is why s-orbitals play a more important role than p-orbitals. Only for those nuclei that have their outer orbitals as s-orbitals ($\ce{^1H, ^2H, ^7Li, ^{23}Na}$ for example) will the diamagnetic contribution be the dominant contributing factor, and this effect is generally small, providing only modest chemical shift ranges of up to tens of ppm.
The paramagnetic contribution is the dominating factor for all nuclei that have electrons in outer shell p, d and f orbitals. Paramagnetic shielding comes from the imbalance of electron density of the valence electrons in the outer orbitals. The way I get my brain around this is by thinking that the larger the non-spherical orbitals, the harder it is to get uniform distribution. The effect of paramagnetic contribution is large; up to many thousands of ppm.
To address your question, there is a general relationship that describes the expected chemical shift range being proportional to the inverse cube of the valence shell radius and also to the populations of the p and d orbitals. So, you are correct in your observation of chemical shift range being basically proportional to atomic number.
A useful reference for you is:
W. von Philipsborn, Chem. Soc. Rev. 1999, 28, 95. DOI: 10.1039/A706424A

I only want to add some data and quotes to supplement long's answer. I am also indebted to long since the reference he provided was what helped me to find these pieces of information.
As already mentioned in long's answer, the chemical shift $\sigma$ arises from several components. Chief amongst them are the (local) diamagnetic term $\sigma_\mathrm d$ and the (local) paramagnetic term $\sigma_\mathrm p$. The diamagnetic shielding is generally proportional to charge density and is given by the Lamb formula. The paramagnetic shielding is generally proportional to $\left<r^{-3}\right>$ (the expectation value of $r^{-3}$ for valence p or d electrons) and is inversely proportional to $\Delta E$, which is an averaged electronic excitation energy. The origins of these terms and formulae to (approximately) evaluate them are discussed by Ramsey in a seminal paper.1 Due to several issues which made the terms difficult to calculate, modifications to this model were made by adding several other terms.2–5 However, the subdivision still conceptually stands.
The variations in chemical shifts of nuclei do not depend directly on the absolute magnitude of either $\sigma_\mathrm d$ or $\sigma_\mathrm p$. They depend, instead, on the extent to which these quantities vary. Of course, there is the issue that if $\sigma_\mathrm d$ is small, then it cannot vary much. This is the case for hydrogen, in which $\sigma_\mathrm d$ is small (low electron density) and $\sigma_\mathrm p$ is small (high $\mathrm{1s}\to\mathrm{2p}$ excitation energy). However, it turns out that for the large majority of nuclei, the variation in paramagnetic shielding controls the variation in chemical shifts.
The remainder will just be quotes. Günther's NMR Spectroscopy states the trend:6

12. Selected Heteronuclei (p 434)
Existing experimental data show that chemical shift ranges tend to increase with the atomic number $Z$ across a given row of the Periodic Table and down a given group. As discussed earlier for carbon-13, the significant contribution to the shielding of heavier nuclei comes from the paramagnetic term, $\sigma_\mathrm p$, in the general equation for the shielding constant, $\sigma$.

More detailed discussion is to be found in older books, some of which are probably out of print. In NMR and the Periodic Table, edited by Harris and Mann, there is a lot of information.7

3B.5 Chemical Shift Ranges (pp 61–62)
A summary of some of the chemical shifts of $\ce{^13C}$, $\ce{^14N}$, $\ce{^17O}$ and $\ce{^19F}$, relative to the bare nuclei, is presented in Fig. 3.3. The ranges of observed shifts for these nuclei are about 650, 930, 700 and 800 ppm respectively, whilst that for boron is about 200 ppm. A somewhat similar situation arises amongst the second row nuclei, where the approximate ranges for $\ce{^27Al}$, $\ce{^29Si}$, $\ce{^31P}$, $\ce{^33S}$ and $\ce{^35Cl}$ are 270, 400, 700, 600 and 820 ppm. Thus although the nuclei with the lower values of $\left<r^{-3}\right>_{np}$ tend to have the smaller chemical shift ranges in each row, the ranges do not follow the trends of $\left<r^{-3}\right>_{np}$ shown in Fig 3.1. The greater ranges shown by $\ce{^14N}$, $\ce{^17O}$ and $\ce{^19F}$, compared with $\ce{^11B}$ and $\ce{^13C}$, reflect the importance of lone-pair electrons. In general these electrons give rise to low-energy $n \to \pi^*$ transitions and thus to large values of $\sigma_\mathrm p$. If the lone pair becomes involved in bonding their contribution to $\sigma_\mathrm p$ is removed resulting in increased nuclear screening. [...]
The much smaller ranges observed for $\ce{^7Li}$ and $\ce{^23Na}$, about 10 and 15 ppm respectively, imply that only $\sigma_\mathrm d$ is of importance for these nuclei, since $\sigma_\mathrm p$ requires the presence of p valence electrons.
Some of the heavier nuclei have very large chemical shift ranges; for example $\ce{^59Co}$ and $\ce{^205Tl}$ have ranges of about 12000 and 34000 ppm respectively. These undoubtedly reflect the greater polarisability of the higher p and d valence orbitals and their varied commitments to chemical bonding.

Some values of $\sigma_\mathrm d$ and $\sigma_\mathrm p$ in compounds of second-row elements are also mentioned (p 55). Notice how small the variations in $\sigma_\mathrm d$ for a given nucleus are.

$$\begin{array}{cccccc} \hline
\text{Molecule} & \text{Nucleus} & \sigma_{\mathrm{d}}(\text{loc}) & \sigma_{\mathrm{d}}(\text{non-loc}) & \sigma_{\mathrm{p}}(\text{loc}) & \sigma_{\mathrm{p}}(\text{non-loc}) \\ \hline
\ce{CO} & \ce{C} & 259.36 & 0.03 & -206.24 & -4.92 \\
 & \ce{O} & 395.39 & 0.64 & -367.39 & -7.25 \\
\ce{(CH3)2CO} & \ce{C} & 257.60 & -0.12 & -212.14 & 4.99 \\
 & \ce{O} & 398.47 & 0.10 & -580.07 & 2.42 \\
\ce{H2CO} & \ce{C} & 257.77 & -0.10 & -208.24 & 5.61 \\
 & \ce{O} & 397.80 & 0.10 & -651.47 & 1.85 \\
\ce{CO2} & \ce{C} & 255.16 & -0.25 & -126.79 & -6.94 \\
 & \ce{O} & 397.32 & 0.12 & -269.51 & -3.55 \\
\ce{CH3CN} & \ce{C} & 258.94 & 0.07 & -159.96 & -5.91 \\
 & \ce{N} & 326.75 & 0.07 & -300.57 & -5.27 \\
\ce{HCN} & \ce{C} & 259.08 & 0.11 & -157.53 & -6.10 \\
 & \ce{N} & 326.70 & 0.06 & -301.30 & -5.04 \\
\ce{N2} & \ce{N} & 324.39 & 0.24 & 0382.84 & -8.62 \\ \hline
\end{array}$$

Multinuclear NMR, edited by Joan Mason, also has a wealth of information.8 A graph of the chemical shift ranges against $\left<a_0^3/r^3\right>$ is presented on p 61. I was too lazy to scan the book, so here's a photograph of it (which is actually pretty decent):

The periodic variation with atomic number of the ranges of chemical shifts observed for the nuclei of main-group and post-transition elements was pointed out in 1964 by Jameson and Gutowsky.9 The tendency of the ranges to increase across a given row of the Periodic Table, and down a given group, was compared with the similar variation of the radial term $\left<r^{-3}\right>$ for the valence p electrons in the free atom, since the paramagnetic shielding term $\sigma_\mathrm p$ has this radial dependence. Figure 1 shows the periodicity of the ranges as now known and the corresponding periodicity of $\left<r^{-3}\right>$ values. The relatively large shift ranges for the first row elements compared with heavier elements can be linked with well-known "first row anomalies" such as the readiness to form pπ bonds (C=O, C=C, N=N, etc.) and relatively high electronegativity.

One might rightly ask, "what is the physical basis for this?" The reason is that $\sigma_\mathrm d$ is a sum over all orbitals, and orbitals close to the nucleus contribute much more than orbitals distant from the nucleus; in particular, the 1s orbital is a very major contributor. However, since changes in diamagnetic shielding in molecules arise primarily due to the movement of valence electrons, the values of $\sigma_\mathrm d$ do not change much. This argument was already put forth extremely early, by Dickinson in 1950.10 $H'(0)/H$ in this paper is equal to what we now know as $\sigma_\mathrm d$.

From Table III it is seen that the outer or valence electrons give a very small contribution to $H'(0)/H$. [...] Thus it would be expected, except perhaps for the lightest atoms, that molecular perturbations of the outer electrons would have little effect on $H'(0)/H$ [...] The observed shifts of nuclear resonances from one molecule to another are often as large as, or greater than, the entire atomic diamagnetic correction for the atom concerned. The diamagnetic field at the nucleus for both molecules would be closely the same and hence the shift must be attributed chiefly to a difference in the second-order paramagnetic field in the two molecules.

References
(1) Ramsey, N. F. Magnetic Shielding of Nuclei in Molecules. Phys. Rev. 1950, 78 (6), 600–703. DOI: 10.1103/PhysRev.78.699.
(2) Saika, A.; Slichter, C. P. A Note on the Fluorine Resonance Shifts. J. Chem. Phys. 1954, 22 (1), 26–28. DOI: 10.1063/1.1739849.
(3) Pople, J. A. The Theory of Chemical Shifts in Nuclear Magnetic Resonance. I. Induced Current Densities. Proc. R. Soc. London, Ser. A 1957, 239 (1219), 541–549. DOI: 10.1098/rspa.1957.0060.
(4) Pople, J. A. The Theory of Chemical Shifts in Nuclear Magnetic Resonance. II. Interpretation of Proton Shifts. Proc. R. Soc. London, Ser. A 1957, 239 (1219), 550–556. DOI: 10.1098/rspa.1957.0061.
(5) Flygare, W. H. Magnetic Interactions in Molecules and An Analysis of Molecular Electronic Charge Distribution From Magnetic Parameters. Chem. Rev. 1974, 74 (6), 653–687. DOI: 10.1021/cr60292a003
(6) Günther, H. NMR Spectroscopy, 3rd ed.; Wiley-VCH: Weinheim, Germany, 2013.
(7) NMR and the Periodic Table; Harris, R. K.; Mann, B. E., Eds.; Academic Press, London, 1978.
(8) Multinuclear NMR; Mason, J., Ed.; Plenum, New York, 1987.
(9) Jameson, C. J.; Gutowsky, H. S. Calculation of Chemical Shifts. I. General Formulation and the Z Dependence. J. Chem. Phys. 1964, 40 (6), 1714–1724. DOI: 10.1063/1.1725387.
(10) Dickinson, W. C. Hartree Computation of the Internal Diamagnetic Field for Atoms. Phys. Rev. 1950, 80 (4), 563–566. DOI: 10.1103/PhysRev.80.563.


text:
Why can curcumin cross the blood-brain barrier, but not congo red?
I've read that both congo red and curcumin are able to inhibit the clumping together of amyloid $\mathrm{β_{42}}$ in the brain, which would otherwise result in Alzheimer's disease.
Apparently, congo red cannot cross the blood-brain barrier (BBB) while curcumin can.
From what I learned at my biology classes, the BBB is lined by squamous, endothelial cells. Those cells are, as far as I know, the most permeable kind of cells you could come across in the human body, and if congo red doesn't pass through that, then it must be one stubborn molecule.
Now the cell membrane, itself being composed of a phospholipid bilayer, normally shouldn't have a problem transporting lipophilic molecules across the membrane. From an earlier conversation with Jan, I believe congo red is a lipohilic molecule. Yet inspite of it being lipophilic, it still isn't able to cross the BBB.
Comparing the structures wasn't particularly enlightening.

But the possibility congo red could act as a bidentate ligand, does seem to make it stand out from  curcumin. Am I onto something here? Apart from that, and a somewhat larger size, I don't see why congo red shouldn't be able to cross the BBB.
Now, I'd like to know two things,
1) What enables curcumin to cross the BBB?
2) Why isn't congo red able to cross the BBB, even though it has a structure similar to that of curcumin?

Congo red is charged. That's ten times worse than merely being hydrophilic.

Preface: both molecules could act as bidentate ligands but the β-diketo structure found in curcumin is typically a much better bidentate ligands — as the 2-oxophenol moieties would be. So I would expect curcumin to be a much better ligand than congo red.

The key difference between the two molecules is comparing the extent of their lipophilicity and their hydrophilicity. Curcumin has a rather low water solubility of $3.2~\mathrm{mg/l}$, in line with the fact that it has rather little features that can enhance solubility. Basically, it boils down to the two keto-groups and the phenol moiety.
Congo red has two sulphonic acid groups and two aromatic amino groups. This means that there is a much greater relative charge present, and indeed its solubility in water is reported to be $1.16 \times 10^{2}~\mathrm{g/l}$ — notice the different unit. So congo red is already much more hydrophilic than curcuminis.
In terms of lipophilicity, the sulphonic acid groups present in congo red reduce its lipophilicity while the amino groups themselves don’t have too strong an effect. On curcumin’s side, the phenol groups can comfortably hydrogen bond to the neighbouring methoxy residues and the β-diketo fragment can tautomerise to a keto-enol form and form another intramolecular hydrogen bond. Thus, curcumin is a lot more capable of coping with low-hydrogen bond solvents and is thus expected to have a high lipophilicity.
This is the edge that curcumin has over congo red to cross the blood-brain barrier.

Congo Red is quite hydrophilic and not very lipophilic. Those sulfonates are extremly polar and are commonly used to make large aromatic molecules water soluble. Like for example bathophenanthroline, which would never be water soluble on it's own but if sulfonated twice you can dissolve 100mg/ml, which is a lot. Ofc the diazo groups and the anilines are polar too. Curcumin on the other hand is pretty apolar and lipophilic, those phenols and methoxy groups don't really have much influence.


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Is there a difference between anti-anxiety and anti-depression drugs?
[PREFACE- I've already asked this on Health.SE, but I've drawn a blank there. The place is practically deserted. Also, this is NOT a personal medical question; I'm simply trying to understand the difference between the two classes of drugs.]
I've recently come across two terms "Anxiolytic drugs" and "Anti-depressant drugs". I wanted to know how they differ from each other; if they do.
I mean, sure, "anxiety" isn't the same as "depression" but do the effects of these drugs differ widely?
For example [And this is just an example];
Chlorodiazepoxide is classified as an anxiolytic drug, while Phenelzine is classified as an anti-depressant. Say I had a patient A who's been prescribed Chlorodiazepoxide; patient B has been prescribed Phenelzine by a different physician. Barring possible allergic reactions, or some kind of known sensitivity to either drug, could they possibly take each other's medication and hope for more or less, the same effect? As in; A takes Phenelzine, but B takes Chlorodiazepoxide. [I just cooked up that example, only to see the difference between the two classes of drugs]
To make myself clear, I reiterate:
I would like to know if the two classes of drugs (anti-anxiety and anti-depression) are the same or not. If not, mentioning how the chemistry behind the action of each of the two types of drugs differ would be appreciated.

Unless you're interested in the specific chemistry of how the drugs act and whether or not they are the same then I would say this is off topic. We are not a site which deals with medical issues like which drugs to prescribe.

@bon I am, which is why I found it frustrating to have the question closed. I'll see if I can edit the question to make my intent clearer.

I don't think this is personal medical, but may be better on biology.

[Basic]

Receptor: a molecule in a cell membrane that responds specifically to a neurotransmitter, hormone, antigen or other substance.
Agonist: Substance which initiates a physiological response when combined with a receptor.
Antagonist: Substance which inhibits the physiological response when combined with a receptor.
Allosteric: Substance which alters the activity of an enzyme by means of a conformational change induced by a different molecule.

[Response]

GABA receptors are related with anxiety [1]. Anxiety occurs because of low concentration of GABA in the brain. Therefore, Chordiazepoxide is an allosteric agonist.
Dopamine receptors are related with depression [2]. Depression occurs because of low concentration of Dopamine in the brain. Therefore, Phenelzine is a non-selective and irreversible monoamine oxidase inhibitor, which means that the brain does not recycle/degrade the monoamine neurotransmitters (Which dopamine is one of them).


text:
Feasibility of using Magnesium, Beryllium, Lithium as light sources?
First off, I'm asking this for worldbuilding purposes, so don't worry, nobody will get hurt ;)
Magnesium, Beryllium and Lithium burn relatively bright. If they are somehow thinned out in some other powder so they burn weaker and last longer, would it not be possible to make torches or lamps out of them?

Berillium is toxic and quite hard to obtain, so it's off.
Lithium is more common but still quite rare. It still is used in some pyrotechnic compositions, but rarely in form of metal.
Magnesium IS used in some composition that are intended is light sources. At the dawn of photography photoflashes used magnesium-based compositions. But it is rarely used nowdays, both thanks to electrical flashes and because it generates too much UV light, so it may harm eyes.
BTW more recent photoflash compositions use aluminium to avoid avoid the bane of magnesium-based composition: extremely fine smoke.
BTW, to obtain any metal above a lot of electricity is required, and in a setting with such metals Yablochkov candle should be available and it has quite vintage/steampunk look.

Thank you! So, ignoring the amount available, if one was to create a Lithium or Magnesium lamp, can you give me a very rough description how that could work? Can you just mix it with lots of other powder to make it burn really slowly once it's ignited, so it could burn for days and weeks?

@Krateng For litium - none practical (sensitive to air). For magnesium - like this https://en.wikipedia.org/wiki/Flare , but definitely no in-door: a lot of smoke.


text:
What does "totally reduced" mean?
My textbook mentions that:

Iron is unable to be totally reduced... the most common non-zero oxidation states of iron is (o for oxidation, r for reduction) are: +2r, +3, +4o, +5o, +6o.

I thought it maybe meant going to $\ce{Fe^4+}$ (as the most reduced I commonly see is $\ce{Fe^3+}$), but then upon googling there is also $\ce{Fe^5+}$ ions, so how far does it have to go until it is totally reduced (if this is even what it meant by totally reduced)? Edit: Sorry, got reduction and oxidation mixed up (thanks for pointing this out DVSA). However, my question still stands, though replacing the above with more to e.g.  $\ce{Fe^4-}$ than  $\ce{Fe^3-}$  .
The question, for context, is below (I have already typed up what I think are the relevant parts above). FYI the answer is D, but I want to know what B means.

Sounds fishy to me - iron can be reduced to $\ce{Fe^0}$, even though it is reactive is aqueous/oxygen-rich environments. From Wikipedia: "Elemental iron occurs in meteoroids and other low oxygen environments, but is reactive to oxygen and water. Fresh iron surfaces appear lustrous silvery-gray, but oxidize in normal air to give hydrated iron oxides, commonly known as rust. "

Thanks Todd. You were right it was a fishy statement and a misleading question from me. See comment on DSVA's answers.

Reduction means that electrons are added, you are talking about oxidation. I haven't heared the term "fully reduced" in this context before, but I guess it means a full shell. In case of iron this would mean a oxidation state of -10, while as far as I know the highest we can get is -4.
And don't confuse oxidation states with charge of ions.
Well, now that I see the actual question I think there's nothing to really understand here. Just a wrong, made up answer.

Oh. I assumed that the wrong part was that "Scandium is similiar too", rather than "Iron being unable to be totally reduced". Thank you for spotting my error.


text:
Calculating concentration of dissolved gas from its mole fraction
I came across the following (solved) problem in my textbook:

The Henry's law constant for $\ce{O2}$ dissolved in water is $\pu{4.34 \times 10^4 atm}$ at $\pu{25 ^\circ C}$. If the partial pressure of oxygen in air is $\pu{0.4 atm}$, calculate the concentration (in terms of molarity, $\pu M$) of dissolved oxygen in water, in equilibrium with air at $\pu{25 ^\circ C}$.

The following is my attempt at a solution to the above problem:
From Henry's law, we have that
$$p{\ce{O2}} = K_\mathrm H \cdot x_\ce{O2}$$
where $x_\ce{O2}$ is the mole fraction of oxygen dissolved in the solution, $K_\mathrm H$ is the Henry's law constant, and $p_{\ce{O2}}$ is the partial pressure of oxygen over the liquid.
Given that the Henry's law constant for a mixture of water and some amount of dissolved oxygen is $\pu{4.34 \times 10^4 atm}$, this gives us
$$x_\ce{O2} = p_{\ce{O2}}/K_\mathrm H = 0.4/(4.34 \times 10^4) = 9.2 \times 10^{-6}$$
To calculate the concentration of oxygen in this mixture in terms of its molarity, I need to know the number of moles of oxygen present in the solution and also the number liters of water in which the oxygen is dissolved.
To calculate the number of moles of dissolved oxygen, I can use the following equation:
$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\ce{O2} + n_\ce{H2O}}$$
However, no information regarding the number of moles of the solvent, water, has been provided.
Hence, I checked the solutions provided by my textbook and found that they have arbitrarily assumed that we are dealing with $\pu{1 L}$ of water.
My question is this: are we allowed to arbitrarily take the volume of water as $\pu{1 L}$ even when absolutely nothing has been mentioned about this in the problem?

The gist is that as long as you have a specified mole fraction, the concentration of oxygen does not depend on the amount of water present. This is what is called an "intensive property". Intensive properties don't depend on how much stuff there is, and therefore, you can arbitrarily specify any volume you like. In order to make the calculations simpler, this specified volume is usually some unit quantity, like $\pu{1 L}$.
It is somewhat analogous to determination of an empirical formula, in that it doesn't matter how many grams of the compound is present, you will always end up with the same empirical formula. A common tactic is to assume $\pu{100 g}$ of compound is present, such that if there is $x\%$ of carbon by mass, there is $x\ \pu{g}$ and therefore $x/\pu{12 mol}$ of carbon.

Here, if you have $\pu{2 L}$ of water, then $n_{\ce{H2O}}$ is double of that if you only have $\pu{1 L}$. But since you have twice as much water, to maintain the mole fraction (that you determined via Henry's law), you have to dissolve twice as much oxygen - otherwise the mole fraction won't be the same. So $n_{\ce{O2}}$ is double of that if you only have $\pu{1 L}$. The factor of 2 will cancel out in the equation you gave:
$$\begin{align}
x_{\ce{O2}} &= \frac{2n_{\ce{O2}}}{2n_{\ce{O2}} + 2n_{\ce{H2O}}} \\
&= \frac{n_{\ce{O2}}}{n_{\ce{O2}} + n_{\ce{H2O}}}
\end{align}$$
If you are really uncomfortable with assuming the volume, then simply assign an unknown value: $k$ litres, for example. You should find that $k$ cancels out and you are left with the exact same value for the molar concentration of $\ce{O2}$.

There is, perhaps, a more "elegant" way to prove this. Note that
$$\begin{align}
\frac{1}{x_{\ce{O2}}} &= \frac{n_{\ce{O2}} + n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{V_{\ce{H2O}} \times [(\pu{1000 g L-1})/(\pu{18 g mol-1})]}{n_{\ce{O2}}} \\
&= 1 + \frac{\pu{55.56 mol L-1}}{c_{\ce{O2}}}
\end{align}$$
where $V$ is the volume and $c = n/V$ the molar concentration. You can see that the molar concentration is uniquely determined by the mass fraction $x$ and the exact value of $V$ doesn't matter at all.


text:
When was the amu renamed the dalton?
I taught chemistry for 30 years and never encountered the unit until I saw it on a juice bottle!

Using "u", "amu" or "Da" is mostly a matter of taste. In chemistry, I'd stick to SI and use g/mol. Biochemists love Daltons, I don't know why.

@Karl $\mathrm{g/mol} \ne \mathrm{Da}$.

The atomic mass unit (amu) was not renamed to dalton (Da). These are different, albeit related, units.
The atomic mass unit (amu) referred to the relative atomic mass of oxygen, which was taken as 16. However, physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. This unit became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.
The unit dalton (Da) and the unified atomic mass unit (u) are non-SI units that are accepted for use with the SI. Actually, ‘dalton’ and ‘unified atomic mass unit’ are alternative names for the same unit, equal to 1/12 times the mass of a free carbon-12 atom, at rest and in its ground state, i.e.
$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$
Therefore, using the old definition of the atomic mass unit (amu) in physics
$$1\ \mathrm{Da} = 1\ \mathrm{u} \approx 1.0003\ \mathrm{amu}$$
and using the old definition of the atomic mass unit (amu) in chemistry
$$1\ \mathrm{Da} = 1\ \mathrm{u} \approx 1.00006\ \mathrm{amu}$$


text:
Drying hydrogen chloride gas
Why do we use concentrated sulphuric acid to dry hydrogen chloride gas instead of anhydrous calcium chloride? Both of them are drying agents, what's the difference between them?


text:
Is (cycloprop-2-en-1-ylidene)propanedinitrile aromatic or not?
Is (cycloprop-2-en-1-ylidene)propanedinitrile ((cycloprop-2-en-1-ylidene)malononitrile) aromatic or not? If yes? How?

At least formally, the target compound is the product of a Knoevenagel reaction of cyclopropenone with malononitrile.
For cyclopropenone, a zwitterionic mesomeric structure is conceivable. This species, a cyclopropenyl cation with a negative charge on the oxygen atom attached, is aromatic. Think that this is the little brother of tropone.
For the compound in question, charge separation is possible as well. The resulting species would again be an aromatic cyclopropenyl cation with a dicyanomethyl substituent over which the negative charge can be delocalized.

Are you basing that on experience or can you back that up with some more evidence? How would a zwitterionic tautomer differ from the non-zwitterionic form in structure, and are you implying that the resonance interpretation of cyclopropenone is wrong. The NMR shifts of these protons are more deshielded than just aromatic protons (Org. Synth. 1977, 57, 41; DOI: 10.15227/orgsyn.057.0041).

@Martin-マーチン I do have published experimental experience on spin and charge stabilization through dicyanomethyl substituents. Apparently, the old NMR findings were confusing, but there seems to be quite some evidence from later studies by Arthur Greenberg and coworkers at NJIT, cf. Chem. Eng. News, 1983, 61 (38), 33
(DOI), J. Am. Chem. Soc., 1983, 105 (23), 6855-6858 (DOI, Chem. Rev., 1989, 89, 1225-1246 (DOI).

I suggest adding those to your answer.\ The c&en article is like a little newspaper clipping somewhat summarising the JACS `83 paper. They conclude that resonance stabilisation is present in this compound, but from my point of view they fail to prove that this is due to aromaticity. As stated there, the NMR experiment cunducted until that point showed no aromatic ring current. I personally find heats of formation are not enough (at least not conclusive enough) evidence of aromaticity for this heatedly discussed topic to be settled.


text:
Is there any difference between Meprobamate and Equanil?
My textbook mentions Meprobamate and Equanil separately, under Anxiolytic drugs and even provides two (not-so) different structures to back their claim.

However, Wikipedia would beg to differ:

Meprobamate — marketed as Miltown by Wallace Laboratories and Equanil by Wyeth...

Which would naturally imply, that Meprobamate and Equanil are...the same thing. ಠ_ಠ
A quick search with Google Images got me this:

Now this particular textbook I use has, over the years, gained notoriety among several generations of us students as the Mater Omnium Errorum. So in all likelihood, my textbook's the one that's wrong here.
But since I'm not a Pharmacologist, I'd appreciate it if someone could verify if Meprobamate and Equanil are the exact, same, chemical.
I tried Google, but that endeavor wasn't particularly helpful.

The meprobamate structure appears to be the correct one and the pharmacopeias agree that the active ingredient in Equanil is meprobamate.

Pharmaceuticals with meprobamate as the active compound are marketed by different companies under different names.
Meprobamate is the International Nonproprietary Name (INN) for the drug.
Whether two pharmeceuticals with the same active components acutally have the same kinetics does depend on additional factors, such as the type of the filler.


text:
In my homemade electrolysis setup, only the negative end bubbles?
I've created an electrolysis setup by connecting a $6~\mathrm{V}$ battery to a cup filled with saline water via pencils; I am confused as to why only the negative pencil bubbles though. After running the setup for longer than five minutes, the water turns brown but so far that information has not been useful. Why is the negative pole the only one bubbling and if possible, why is the water turning brown?

Your setup, using table salt ($\ce{NaCl}$ plus additives) will initially electrolyse according to the following two reactions, separated by cathodic reduction and aniodic oxidation:
$$\ce{2 H+ + 2 e- -> H2 ^}\tag{CatRed1}$$
$$\ce{2 Cl- -> Cl2 + 2 e-}\tag{AnOx1}$$
These reactions assume no counterions. However, you don’t have dissolved protons in your solution because you aren’t electrolysing hydrochloric acid. Instead, the proton is taken from water molecules and if we include them (and sodium counterions) we get the following improved cathodic reduction equation:
$$\ce{2 Na+ + 2 H2O + 2 e- -> H2 ^ + 2 NaOH}\tag{CatRed2}$$
This still displays the hydrogen gas bubbling away at the cathode but chlorine gas at the anode should also bubble after some time (or at least evolve noticeably). However, with the increase of $\mathrm{pH}$ due to the generated hydroxide, the chlorine produced at the anode will react further:
$$\ce{Cl2 + 2 OH- -> Cl- + OCl- + H2O}\tag{AnOx2}$$
And both of these are soluble ions. The overall reaction of your electrolysis setup is:
$$\ce{NaCl + H2O -> H2 ^ + NaOCl}\tag{RedOx}$$
There will be hydrogen evolution at the cathode but no gas evolution at the anode. The browning will likely be due to a number of side reactions induced by the oxidant hypochlorite anion.
In case you’re wondering why water’s oxygen is not oxidised, this is due to its higher standard electrode potential.

are there any books you would recommend reading to learn about this sort of stuff?

@JakeWickham I generally don’t recommend books because I never use them, but there is a question here on chem.SE about good books for certain topics.

I've done this experiment using steel clips (from a car battery charger) and got a brown colour due to the steel being reduced to Fe3+ salts. After the experiment you could clearly see where the clip had been corroded.

Depends a bit on your voltage, but you might just be making chlorine at the anode. It's pretty soluble in water. I've always used sodium sulfate as the electrolyte.

Try collecting the gas at the negative electrode and testing it. Hold a match to the gas and my guess is that it will go pop. (Positive test for hydrogen)
If there is a group 1 metal salt in your solution like $\ce{NaCl}$ the ions would be Na+ and $\ce{Cl^-}$.
When the Na+ is reduced to Na by accepting the electron from the cathode (negative terminal) it becomes the metal atom.
$$\ce{Na+ + 1e^- \rightarrow Na}$$
This then reacts with the water in the solution to release hydrogen bubbles at the negative electrode.
$$\ce{2Na + 2H2O \rightarrow 2NaOH + H2}$$
This is probably why you have bubbles at the negative electrode.
If the cup vessel you are doing the experiment in has any imperfections in the ceramics or has any metals in the surface then there is probably a displacement reaction from the Na ions or the Halide ions floating around.
There could also be displacement of other ions from inside the saline (saline is a mix of different salts)
That's my best argument for your observations

This is incorrect. Sodium shall not even touch the cathode as long as species exist that can be reduced more easily (such as hydrogen in water)!

It's saline. Salts (ionic substances)would be reduced far more easily than water (simple covalent structure).

Add indicator the negative electrode would show the presence of an alkali.

Re first comment: Wrong. Please go back to your books!

I’ll agree with you on the presence of a base near the cathode; see my answer as to how this is formed.


text:
Alternative software for processing Molden files?
I am trying to salvage old data, and one of my output files is in Molden format.

I tried downloading Molden at its homesite but the link does not seem to work.
Is there another way to access Molden format files?

Thank you very much.

This is probably off-topic - rather informatics than chemistry.

Related: http://chemistry.stackexchange.com/q/33932/4945  I had no problems downloading molden from the linked homepage.  try this: ftp://ftp.cmbi.ru.nl/pub/molgraph/molden/molden5.7.tar.gz

@Mithoron no one in informatics will have an answer to this besides computational chemists there. This is the best SE to get an answer to such a question.

@Mithoron In principle questions of this type are the reason for my newest meta discussion: What's our (maybe updated) take on questions about (quantum) chemical software?.

Maybe avogadro, or VMD (Visual Molecular Dynamics)

I am fairly sure Molden can be downloaded. However, programs like JMol, and Chemcraft are usually cross-compatiblr with molden.

As far as I know, ChemCraft is not able to read molden files. I don't use JMol, so maybe that works. It would be best if you could support your claims with more data, and/ or links to documentation of the programs.

Even though ChemCraft does not open .molden files, it is able to open .wfn files.
So if using ChemCraft (or any other .wfn-compatible software that can't read .molden files) is imperative, my answer might be helpful.

In case you're OK with converting, Molden2AIM may be a good alternative.
It is able to convert .molden files into .wfn files which are easier to handle in my opinion (the software actually does a bit more than just that, such as generating .47 files for NBO analysis).
If your .molden files were generated by ORCA, you may also try an automated script that handles the details of using Molden2AIM for you.


text:
Elimination reaction of 1-bromo-2-phenylcyclopentane

In the following reaction, treating the trans-1-bromo-2-phenylcyclopentane with ethanolic KOH gives an alkene. But in this case, the double bond is on the left side (less substituted) instead of being the more substituted one.

But why does this happen? Zaitsev's rule favours the highly substituted alkene. Even if a carbocation is formed, a hydride shift would have landed it on the carbon with the phenyl group, as the carbocation would be stabilised by resonance. Does the trans configuration forbid such reaction?

There are three common elimination mechanisms: $\mathrm{E1, E2}$ and $\mathrm{E1_{cb}}$. $\mathrm{E1}$ and $\mathrm{E1_{cb}}$ both imply that an ionic intermediate is formed first; $\mathrm{E1}$ implies the same carbocationic intermediate as in an $\mathrm{S_N1}$ reaction. The $\mathrm{cb}$ in $\mathrm{E1_{cb}}$ means conjugate base, i.e. the first step is deprotonation of the compound to give an carbanionic intermediate.
$\mathrm{E1}$ reactions require good leaving groups and a moderately stabilised carbocation; this is not the case in your reactant.
$\mathrm{E1_{cb}}$ requires a very strong base and an at least mildly stabilised anion; this is not the case with your reagent and reactant.
Thus, the mechanism must be according to $\mathrm{E2}$, where the abstraction of a proton and the displacement of the leaving group happen at the same time. $\mathrm{E2}$ mechanisms require a $180^\circ$ anti-periplanar configuration of the $\ce{Br-C-C-H}$ bonds for easier attack and favour those double bonds where the hydrogen is less stericly hindered. Additionally, your reactant is trans-configured, thus meaning that there is no available proton on the higher-substituted carbon that can participate in elimination. Altogether, this leads to $\mathrm{E2}$ eliminations generally following Hofmann’s rule rather than Zaitsev’s rule. Hofmann initially observed what later became the Hofmann rule in the Hofmann elimination, which follows an $\mathrm{E2}$ mechanism.
As the ethanolic $\ce{KOH}$ is not strong enough to isomerise the double bond, the Hofmann product will be observed.

The carbocation formed by releasing $\ce{Br^-}$ ion will be a secondary carbocation. So how can we say that it will not be a stable one, as the relative order of stability is tertiary>secondary>primary (ignoring resonance effects, if any, of the benzene ring)?

@ShoubhikRajMaiti Assume only tertiary and benzylic (maybe allylic) carbocations to be stable enough to be formed. Everything else would immediately undergo Wagner-Meerwein rearrangements.

Well, if there is a carbocation formed, and there is a rearrangement, then, the charge will be shifted to the attaching point of the benzene and cyclopentane ring, as it gives maximum stability(tertiary and also resonance). Then the products will also change.

@ShoubhikRajMaiti But that is not observed since the reaction is much slower than the one by E2.

The product structure is wrong, the Br should be gone.
There's no carbokation here, this is a reaction with E2-mechanism, which works best if the abstracted hydrogen is at a dihedral angle of 180° to the leaving group.

Then the choice of product between 3-Phenylcyclopentene and 1-Phenylcyclopentene would depend upon the diastereomer, right? I was thinking about an E1-mechanism here...

Yes, if you use the cis compound you will get the double bond on the other position.

The update in the figure in the question clears all doubts.

@DSVA A more elaborate answer would have been helpful.


text:
Why would this rearrangement happen?

So, I am a first year chemistry student and I saw this chemistry reaction on a previous year's test. I attempted to figure out the mechanism, but I'm rather confused as to how the cyclopentane somehow donates a carbon to the cyclobutane? Why would such a rearrangement occur?

That is how far I get to before I am lost. Is methanol even a strong enough reagent for such a reaction?

Is this an SN1 reaction where a bromide ion is formed and then a rearrangement occurs to form a tertiary carbocation instead of a secondary carbocation, and then methanol can attack?

Is methanol even a strong enough reagent for such a reaction?

Methanol is a very poor nucleophile, so the rate of the $\mathrm{S_N2}$ pathway will be extremely slow. This is much likelier to be a $\mathrm{S_N1}$ reaction. The starting material is a benzylic bromide from the point of view of the alkyl halide, both pathways are quite possible, but the lack of a good nucleophile suggests the predominance of the $\mathrm{S_N1}$ route.
So, you go through a carbocation intermediate, which undergoes an alkyl shift to generate a more stable carbocation.

Both carbocations are benzylic and secondary, so hyperconjugation isn't the driving force for the rearrangement, as your comment hints at. Instead, it's simply related to the relief of ring strain as you expand a four-membered ring to a five-membered ring. Four-membered rings are pretty unstable to begin with and fusing one to a benzene ring increases the strain even further since you impose more geometrical constraints on the ring.
Interception by methanol and deprotonation gives the desired product.

Thank you so much!! Your comment was very helpful, as was the diagram. Does it matter which c-c bond in the cyclobutane is broken to form it? And is ring expansion a more important factor than, say, a hydride shift?

Yes, it matters. You can try migrating the other one, like this: https://i.stack.imgur.com/5BT6m.png You'll find that you end up with a non-benzylic carbocation. While this shift helps to open up the ring size, you lose a lot of electronic stabilisation (delocalisation of the positive charge into the ring). 2) As you can see from the product, yes, the ring expansion is quite important. In general cyclobutanes fused to benzene rings have a very strong tendency to open if they can. There are many more examples (e.g. electrocyclic reactions, retro-aldols).


text:
Sodium orthovanadate synthesis
My father* wants to synthesize Sodium orthovanadate, and has found out it is prepared from Vanadium pentoxide and a strongly alkaline solution of NaOH, or by fusion with NaOH but cannot find a detailed method, e.g. about quantities, separation from excess alkali, drying, etc.
I** haven't been able to find out much by googling or searching Pubmed: most methods I've found deal with stabilizing the solution, not with the complete synthesis. Could you point us to a complete protocol?
*chemist, but his English is not so great.
**not a chemist anymore

From Supniewski, J. "Inorganic Preparative Chemistry", 1958 (in Polish):

Melt $\pu{36.4 g}$ of vanadium pentoxide $\ce{(V2O5)}$ and $\pu{63.6 g}$ of sodium carbonate $\ce{(Na2CO3)}$ a nickel crucible until you get a uniform, clear melt with no bubbles of $\ce{CO2}$ coming out. Pour the melt on a cold plate and wait until it solidifies to a colorless mass (during the cooling process it should turn dark green, then yellow, then colorless). Pulverize the melt and dissolve it in a minimum amount of cold water, filter, and layer with several volumes of ethyl alcohol. After several hours, colorless crystals of $\ce{Na3VO4 * 16 H2O}$ should separate. Wash the product with ethanol and dry in a desiccator.

Some clarifications as requested in the comments:
"Minimum amount" means that you need to add only so much water to dissolve the salt completely, no more. Just add water in small portions and stir well - if the salt is still visible add some more water, etc. Suppose it took you about $\pu{50 mL}$ of water to dissolve your substance. Now you need to add twice or thrice (i.e., $\pu{100 .. 150 mL}$) the amount of ethanol ("several volumes"). Add the alcohol slowly so the solutions do not mix immediately but form two separate layers, and leave it undisturbed for several hours (I think it may take a day or two).

Thanks! If someone could clarify what does "a minimum amount" and "several volumes" meant...

Could you edit this into the answer? Also if you have any idea of the expected yield for those quantities, that'd be grrrreeat.

@Adriano Varoli Piazza Unfortunately the book does not give the yield for this particular procedure, but I estimate it to be over $90 %$. I did not perform this synthesis myself. Feel free to report your yield.

According to Wikipedia, sodium orthovanadate is prepared by mixing vanadium(V) oxide in a solution of sodium hydroxide:
$$\ce{V2O5 + 6 NaOH → 2 Na3VO4 + 3 H2O}$$
But, there is a different reaction scheme as mentioned in this site:

Sodium Orthovanadate, $\ce{Na3VO4.12H2O}$, is the orthovanadate most
  frequently met with. It is readily obtained by adding excess of
  caustic soda to a solution of sodium pyrovanadate:
$$\ce{Na4V2O7 + 2NaOH <=> 2Na3VO4 + H2O}$$
It can be conveniently crystallised from caustic soda solutions, as it
  is less soluble than in water. It forms hexagonal prisms isomorphous
  with the corresponding phosphate and arsenate [...] The anhydrous
  salt, $\ce{Na3VO4}$, melts at 850°-866° C. It can be prepared by
  fusing sodium carbonate and vanadium pentoxide in the required
  molecular proportions. Extraction of the product with water and
  precipitation with alcohol gives colourless needle-shaped crystals of
  a hydrate containing sixteen molecules of water, $\ce{Na3VO4.16H2O}$.
  The hepta- and octa-hydrates, $\ce{Na3VO4.7H2O}$ and
  $\ce{Na3VO4.8H2O}$, have also been prepared.
By dissolving vanadium pentoxide in a large excess of caustic soda,
  Ditte obtained two crystalline vanadates which contained a larger
  proportion of the basic oxide than is present in the orthovanadate,
  and to which he gave the formulae $\ce{4Na2O.V2O5.30H2O}$ and
  $\ce{4Na2O.V2O5.26H2O}$.


text:
What are the resonance structures of methyl azide?
I am supposed to construct the major resonance structures of methyl azide, $\ce{CH3N3}$. I know that in compounds such as the carbonate anion $\ce{CO3^2-}$, resonance can be depicted by interconverting single and double bonds:

Is something similar happening with methyl azide but with nitrogens instead of oxygens? e.g.

You have the wrong structural isomer.

Think about the azide again. Hint: Sodium azide ($\ce{NaN3}$ hydrolyzes to $\ce{Na+}$ and $\ce{N3-}$. At the moment, your methyl azide looks like a disfigured guanidine ;-)

Your Y-shaped structures somewhat resemble guanidine, $\ce{HN=C(NH2)2}$. For an exercise on resonance structures, its protonated form, the guanidinium cation, would be an interesting case and you might want to give it a try too.
Methyl azide, however, looks completely different.

As the name suggests, there should be a methyl group, $\ce{CH3}$, and an azide group.
Draw the linear azide anion, $\ce{N3-}$.

Where would a "$\ce{CH3+}$" go?
Is methyl azide, $\ce{CH3N3}$, a linear molecule?
Draw the resonance structures for methyl azide.

Structure was determined in 1933 by Brockway and Pauling. The bond distance of the terminal two nitrogen atoms is very close to that in dinitrogen.


text:
What is "Cresol Soap"?
I recently came across a suggestion in an old (first published 1893) book that a solution of ~1% cresol soap in water could be used as a way to store a particular kind of root without allowing it to dry out or rot.
I've tried searching, but have been unable to determine what cresol soap might be. The closest I can find are several shady poorly-translated websites selling Lysol as cresol soap, or possibly a mixture of the two.
Does anyone have a clue what this book could be referring to?

Best guess is that it's just cresol (not sure which isomer). Phenol itself is considered an antiseptic agent that was the substance put forward by Joseph Lister for sterilization before surgery: https://en.wikipedia.org/wiki/Phenol#History

@orthocresol just felt like tagging you here, no reason ;)

According to an 1894 tract with the title Merck's Market Report and Pharmaceutical Journal: An Independent Monthly Magazine Devoted to the Professional and Commercial Interests of the Druggist, Volume 3, "cresol soap" was also called "crelium" or "crelium soap".
The google n-gram viewer suggests that "cresol soap" is a historical term, whose usage began in the 1890s, peaked dramatically during World War I and its immediate aftermath, and declined substantially afterwards.
A 1915 British medical report argues for using cresol soap as a vermin control agent in troop populations.  It gives a recipe as:

10 gallons of water
1.5 pounds of soft soap
Jeyes' fluid 1.5 ounces

This raises the question of what "Jeyes' fluid" is. Amazingly, Jeyes' Fluid is apparently still a commercial product, and contains a mixture of 4-chloro-m-cresol, isopropanol, terpineol, and "tar acids".
"Cresol soaps" are apparently still an item of commerce somewhere in the world, as a 2011 applications report form an analytical equipment vendor illustrates the analysis of a "cresol soap" sample, although a specific source is sadly not mentioned.  The primary detected constituents were m-cresol and p-cresol.

I suspect there are many other recipes that involve mixing soap, water, and crude cresol preparations of various kinds.  In the era of cresol soap's invention and popularity, coal was far more ubiquitous than now, and the cresol sources were probably crude fractions of distilled coal tar that were enriched in various isomers of cresol.  In those days awareness of potential toxicity from cresols was also not as developed as now.  The "tar acids" in Jeyes' fluid probably include a variety of cresol isomers.

Toxicity is kinda par for the course in a disinfectant -- they presumably started using it because it kills stuff, and it kills stuff because it's toxic. And, hey, in those days, they were so busy dying early of arsenic poisoning that they wouldn't have noticed the ffects of the cresols.

Excellent research! I would go ahead and accept this now, except it seems odd to me that an American writing in the 1980's would refer to a product from the early 1900's. I'm going to leave this open for a bit just to see if there is some other niche use of the term neither of us were able to find that another user could be aware of.

I'm sorry, I should have accepted this answer long ago. Not sure how it slipped my mind so long, but I suppose better late than never.


text:
Why do cations have higher ionization energies than their corresponding neutral atoms?
I know that metals have low ionization energies and non-metals have high ionization energies, but once a metal/non-metal becomes a cation, why does the cation itself have a ionization energy higher than the original neutral atom?
What is the scientific reason behind this increase in ionization energy?

A cation always have high ionisation energy whether it comes from a metal or a non metal.We can understand it in this way:
As the atom become a cation,its number of electrons decreases(but notice that number of protons are same,and hence positive charge is more as compared to negetive charge).Since the number of electrons is less now, the effective nuclear charge has increased due to decrease in electron-electron repulsion.
As the number of electron decreases (which happen in a cation),the repulsion between the remaining electrons decreases. Since electron-electron repulsion make it easier to pull out a electron hence a decrease in electron-electron repulsion will contribute towards an increase in resistance of being pulled outside.
Conclusion: As you decrease number of electrons in atom by taking them out,effective nuclear charge increases, due to which electrons are bounded to nucleus strongly, making  them hard to remove.

This is very loose terminology and is potentially misleading. The protons don't attract the electrons harder, because the nuclear charge is constant. The increased ionisation energy results from the decrease in electron-electron repulsion, not an increase in electron-nucleus attraction.

But the number of electron has decreased,shouldn't it affect the attraction??

As a pre-emptive statement I will say that there is no doubt that Zeff increases. If you want to say that that represents an increase in the net attraction of electrons to nucleus, fine by me. However the increase in Zeff is due to a decrease in σ (an e-e repulsion term) and not an increase in Z (an e-p attraction term). Which is why I said that it was potentially misleading (not outright misleading).

Right, I like this much better. I'm not the person who downvoted though, so I can't help you with that.

That's okay.I just want to give a correct answer,a loss of 2 reputations does not matter.

@orthocresol Can you define $\sigma$? I've never it used in this context.

@pentavalentcarbon inorganic chemistry wise it's just a fudge factor to account for the difference between Z and Zeff. I'm not sure there is a mathematical formula for it short of its definition. Out now so can't check but I suspect Zeff is the quantity that is determined variationally.

Consider a simple thought experiment. Your ionisation process is essentially removing a (negatively charged) electron from a remainderatom.
$$\ce{Atom^n+ -> Remainderatom^{(n+1)+} + e-}$$
Throughout this process, you are separating charges, thus there is an electrostatic force driving the electron back to its remainderatom.
If the remainderatom has a charge of $2\ (\Rightarrow n=1)$, the attractive force is twice as strong in simplistic theories than if remainderatom’s charge is $1\ (\Rightarrow n=0)$.
Thus, it is twice as hard to remove a second electron.

This thought experiment makes use of a number of simplifications. For example, it does not consider electron energy levels (orbitals), it does not consider orbital contraction upon ionisation and much more. But it does a good job at explaining the general trend.

Yes, it has a higher ionization energy. Cations are also smaller than the neutral atom, and the reason for both is the same.
The positive charge in the core stays the same, if you remove electrons the interaction between the core and your electrons increases, due to lower shielding, lower Pauli repulsion,and so on, which lowers the energy of those electrons. This directly converts to smaller orbitals and higher ionization energy.

Your question is an application of the electrostatic Coulomb force.
Removing $1$ electron from a neutral atom requires some energy. Removing a second electron from this same cation is more difficult, because it is difficult to extract a negative charge from an already  positive ion.
Removing a third electron is still more difficult, because a huge energy is needed to remove a third negative charge from a positive ion that has already lost $2$ electrons. Etc.


text:
Do the compounds in a chemical equation need a net charge of zero when predicting reaction products?
For predicting reaction products in chemical equations, I am not sure of what the subscripts would look like in the product side. I know the types of reactions, but when you predict the products, do the compounds need a net charge of zero? If the net charge of zero does not matter, would the subscripts in the products look the same in the reactants and just the coefficients are different? Thank you for your help in advance. Here are some examples as well:
EX:

__C2H2 + __O2 -> (<-Reactants)

(Products->)   -> __CO2 + __H2O (<-my guess)
EX:

__LiOH + __Fe(NO3)3 ->

                > LiNO3 + FeOH

The products that are shown in the example are my guess, so they may not be correct. I am also not sure how I would balance equations for combustion.

Yes, the compounds on the products side do need a net charge of zero.  You are correct in the products of the combustion reaction being CO2 and H2O.  I balanced it using a trial and error method, but it does take a few trials to get to the answer.  I will give you a hint, the coefficient for O2 is 5.
The second equation is a precipitation reaction, assuming it takes place in solution.  Since Iron (III) Hydroxide is insoluble then the reaction will occur.  In order to determine the subscripts for the products, you need to first determine the charges on the individual ions.  Li is a group 1 metal.  It will have a charge of +1.  OH is a hydroxide ion, it always has a charge of -1.  Iron (Fe) typically has a charge of either +2 or +3.  In order to determine its charge for the products, you have to determine its charge on the reactant side.  Since N03 always has a charge of -1 and there are 3 of them on the reactant side, than Fe has a +3 charge.  Because Fe has a +3 charge and the OH has a -1 charge you will need 3 OH ions for every 1 Fe ion =  Fe(OH)3.  The Li(NO3) is already neutral since Li is +1 and NO3 is -1.
Hope this helps!

Compounds on the product side of a reaction do not necessarily have to have zero charge.
However, the fundamental principles of conservation of mass and conservation of charge mean that you must have the same number of each atom type on both sides of a reaction arrow, and the same net charge. Net charge means, that a reaction is allowed if an overall reactant charge of $1-$ gives a set of products with an overall charge of $1-$, but not if the products’ overall charge is $2-$. An electron is permitted to be on either side of the reaction.


text:
How to balance chemical equations with polyatomic ions
I know that you can not split poly-atomic ions but I am confused on how to balance the following equation:
9PbSO4 -> 10PBSO3 + 3O2
Pb = 9    Pb = 10
S = 9     S = 10
O = 36    O = 36
Obviously the equation is not balanced and there is an oxygen outside of a poly-atomic ion which makes it harder to balance. Do I count the oxygen with the oxygen in SO?
What numbers could I use to get PB and S the same on both sides?

Start with the original unbalanced equation $\ce{PbSO4 -> PbSO3 + O2}.$
$\ce{PbSO3}$ needs only one more oxygen atom to balance $\ce{PbSO4}$; one atom of oxygen is $\frac{1}{2}\ce{O2}$.
This balances, then, as $\ce{PbSO4 -> PbSO3 + 1/2 O2}$.
Since half molecules are not permitted, double everything to:
$\ce{2 PbSO4 -> 2 PbSO3 + O2}$

I edited your equations. Have a look at http://meta.chemistry.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know/3046#3046 for details.

Welcome to Chemistry.SE! Given that you have a lot of experience in teaching chemistry, you might find the current discussion on homework questions interesting.


text:
What is the alternative to chromyl chloride test?
Chromyl chloride test is by far the best test to determine the presence of chloride ions in a salt because no analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides. This test is actually considered selective for $\ce{Cl-}$.
But this test possess certain drawbacks. This test does not give positive results to covalent chloride like $\ce{HgCl2, SbCl3, SnCl2 }$ etc. owing to their slight ionisation and only  partial conversion occurs when added to chromyl chloride solution. Thus, this test is only positive to ionic chlorides. There are also other tests like conc. sulfuric acid test, $\ce{MnO2}$ test or lead acetate test which will almost determine the presence of chloride ions though it possess some drawbacks (see comment section).
For a particular cation or anion, a selective test is chosen in such a way that it will give positive result to any salt, no matter how complex it is and has the capability to differentiate from other analogous salts. So, why Chromyl chloride is still considered as the selective test for $\ce{Cl-}$ despite its drawbacks?

You can use $\ce{AgNO_3}$ test to get white precipitate of $\ce{AgCl}$, and then identify that it is $\ce{AgCl}$, not $\ce{AgBr}$ or $\ce{AgI}$ by reacting the precipitate with $\ce{Na_3AsO_3}$ solution. If it is $\ce{AgCl}$ then yellow coloured $\ce{Ag_3AsO_3}$ will be formed. This can be used as an alternative to chromyl chloride test, as a confirmatory test of chloride.

Are you sure that $\ce{Ag3AsO3}$ will not form when $\ce{AgBr/I}$ is reacted with $\ce{Na3AsO3}$?

"If the silver chloride precipitate is filtered off, washed with distilled water, and then shaken with sodium arsenite solution it is converted into yellow silver arsenite (distinction from silver bromide and silver iodide, which are unaffected by this treatment)." - Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis, 5th edition, Longman Group Limited, 1979.

Hmm good find. Could you convert your comment into answer so that I can accept? In this, you can add any other detail information/reference.

Sorry, but I don't know the answer to your question in the last sentence: "Why Chromyl chloride is still considered as the....". As I can't answer your question fully, I will not post an answer.

The tests you have mentioned, do have drawbacks. In case of lead acetate test, bromide ion also forms a white crystalline ppt. of PbBr2 as does chloride ion. What's more, PbBr2 is soluble in hot water, just like PbCl2, and unless you can control the temperature and amounts of precipitate, there is no way to differentiate them.

And in case of conc. H2SO4 or MnO2 test, the greenish hue of chlorine is rather hard to detect without proper light in the room, if the quantity of sample is low (my own experience). Besides, if nitrate ion is present in the sample, then conc. H2SO4 will produce red vapours of NO2 which masks the colour of Cl2.

@Shoubhik Hmm. For a test to be selective, its sensitivity should be very high, way higher than other 4-5 tests. Perhaps, chromyl chloride test has the highest sensitivity than the other tests. So it is considered selective. I will try to find more information and convert it to an answer.

I should add that chromyl chloride test is not so sensitive. I have had 3 cases of failed chromyl chloride tests till today. In those cases, instead of red vapours of chromyl chloride, suffocating white fumes of HCl were produced. This happens particaulary if the sample is not anhydrous. In those cases, the salt had waters of crystallisation.

Why chromyl chloride test is used as confirmatory test for $\ce{Cl-}$:

According to this Wikipedia page:

The chromyl chloride test entails heating a sample suspected of
  containing chloride with potassium dichromate and concentrated
  sulfuric acid. If chloride is present, chromyl chloride is formed and
  red fumes of $\ce{CrO2Cl2}$ are evident. If there is no chloride
  present, no red fumes are produced. No analogous compounds are formed
  with fluorides, bromides, iodides and cyanides, so this test is
  therefore specific for chlorides. [1]

Most chemistry lab manuals mention that only red vapour is not the confirmation for presence of chloride ion. Rather, if the red vapour dissolves in a solution of $\ce{NaOH}$ and turns the solution yellow (due to $\ce{Na2CrO4}$), and that solution gives yellow precipitate with lead acetate solution, then only, presence of chloride is confirmed.
However, this paper mentions that chromyl fluoride is produced on heating calcium fluoride with chromate and sulfuric acid. The compound chromyl fluoride is a red-vapour at normal temperatures, and is also produced by reaction of dichromates and $\ce{HF}$.[2].
Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis[3] mentions that for chromyl chloride test:

Fluorides give rise to volatile chromyl fluoride, $\ce{CrO2F2}$, which
  is decomposed by water, and hence should be absent or removed.

These suggest that there is a possibility of production of chromyl fluoride by reaction of potassium dichromate, concentrated sulphuric acid and a fluoride sample. As chromyl fluoride is red vapour, this would result in a false positive. Moreover, chromyl fluoride is also hydrolysed in water, [4]  $$\ce{CrO2F2 + H2O->CrO3 + 2HF}$$ and thus it should also turn $\ce{NaOH}$ solution yellow, and produce yellow precipitate of $\ce{PbCrO4}$ with lead acetate. So, there is no way they can be distinguished without performing some other test for chloride or fluoride.
It is not possible to know whether chromyl fluoride is actually produced in large quantities like chromyl chloride without performing the actual experiment. But, I think there is a chance of error with fluoride sample.
Also, bromide salts produce red vapours of bromine when heated with $\ce{K2Cr2O7}$ and conc. $\ce{H2SO4}$(this is from my own experience). I don't know why the Wikipedia page specifically mentions that no red fumes are produced in absence of chlorine, and no analogous compound is formed with fluoride. It appears to me that the information is questionable.

Alternative to chromyl chloride test:

A feasible (and in my opinion, easier) alternative is this procedure—

Prepare the soda extract or sodium carbonate extract of the salt in question. This would produce $\ce{NaCl}$, even from covalent chlorides. For insoluble chlorides, it has to be done in melt.
Make a water solution of the extract, and filter it.
Add $\ce{AgNO3}$ soln. to this solution. A white precipitate of $\ce{AgCl}$ would form.
To confirm that it is indeed $\ce{AgCl}$ and not $\ce{AgBr}$ or $\ce{AgI}$, add sodium arsenite($\ce{Na3AsO3}$) solution, and shake. Yellow precipitate of $\ce{Ag3AsO3}$ is formed[3]. This does not occur in case of $\ce{AgBr}$ or $\ce{AgI}$. [I am not sure about $\ce{AgCN}$ though.]

Another alternative is—

Add $\ce{AgNO3}$ solution to the sodium carbonate extract.
$\ce{AgCl}$ is white and soluble in $\ce{NH4OH}$. $\ce{AgBr}$ is pale yellow and partially soluble in $\ce{NH4OH}$. $\ce{AgI}$ is yellow and insoluble in $\ce{NH4OH}$. To improve colour detection, use a grooved tile made from black granite, and perform spot test[5].
For detecting cyanide, add iron(II) sulphate to the soda extract and then add a mineral acid. If a precipitate of prussian blue does not appear, then there is no cyanide.

References:
[1] https://en.wikipedia.org/wiki/Chromyl_chloride
[2] Pure Chromyl Fluoride. Alfred Engelbrecht, Aristid V. Grosse. J. Am. Chem. Soc., 1952, 74 (21), pp 5262–5264 |http://pubs.acs.org/doi/abs/10.1021/ja01141a007?journalCode=jacsat
[3] Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis, 5th edition, Longman Group Limited, 1979.
[4] https://en.wikipedia.org/wiki/Chromyl_fluoride
[5] Greener Qualitative Analysis for Anions in a Mixture.
Indu Tucker Sidhwani, Sushmita Chowdhury, Veena Tucker, Shweta Bansal, Elipsha Sharma and Parul Johar. DU Journal of Undergraduate Research and Innovation. Volume 2, Issue 2 pp 70-79, 2016| http://journals.du.ac.in/ugresearch/DU%20e-Journal%20-%20UG%20Research_files/DUJ4%20Vol%202%20Issue1/J%201.8.pdf