text:
Perplexing claim in Chang and Overby's "General Chemistry"
I'm reading the chapter on gases in Chang and Overby's book "General Chemistry: The Essential Concepts" [1, p. 145], and the following passage is causing me some confusion:

The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle, Charles, and Gay-Lussac. In 1811 he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas must be proportional to the number of moles of molecules present; that is,
$$V \propto n$$
$$V = k_4 n \tag{5.7}$$

I don't quite understand the emphasized sentence; in particular, I don't understand how the emphasized claim "follows" from the preceding statement. In fact, I don't actually think it does.
For example, suppose instead that the volume of any gas at a given temperature and pressure was inversely proportional to the number of moles of the gas present; that is, suppose the relationship between volume and amount was instead
$$V \stackrel{?!}{=} \frac{k}{n},$$
for some constant $k$. Then it would certainly still be true that equal volumes of different gases contain the same amount of gas. For example, if I had $\pu{5 L}$ of $\ce{H2}$, then, according to the above "false law", I would have $k/5$ moles of $\ce{H2}$. Similarly, if I had $\pu{5 L}$ of $\ce{N2}$, I would have $k/5$ moles of $\ce{N2}$ — the same amount!
So the claim that the proportionality follows from the fact that equal volumes of different gases contain the same number of molecules seems illogical. In fact, it seems to me that the only thing one could logically deduce from the first hypothesis is that there is some relationship between volume and amount which holds for all gases (that is, the relationship doesn't depend on the gas itself).
Am I missing something, or is the statement in the book an error?

Added much later:
To clarify what I'm asking here (though I thought it was pretty clear), consider the following two statements:
Statement A: at a given temperature and pressure, equal volumes of different gases contain the same number of molecules.
Statement B: at a given temperature and pressure, the volume $V$ of a gas is related to the number of molecules by $V = kN$ (where $k$ depends on temperature and pressure).
My point is simply that these two statements are not logically equivalent. B implies A, but not the other way around.
So why is it that the two statements are (seemingly quite often) presented as logically equivalent?
References

Chang, R.; Overby, J. General Chemistry: The Essential Concepts, 6th ed.; McGraw-Hill: New York, NY, 2011. ISBN 978-0-07-337563-2.

I would think you are correct in saying that it doesn't necessarily follow. It could just as easily be $P=kN^a$ where "a" could be any real number just from the information provided. I think the hypothesis only emerged from Avogadro's observation that they were in fact directly proportional.

@Tyberius Thanks, but as I added above, it seems to be very common (in Chemistry textbooks etc.) to say that the law $V = kN$ "follows" from the observation that equal volumes of gas contain equal numbers of molecules. Do you have any idea why this seemingly false claim is so common? You can also see it in Oxtoby et al. and Wikipedia...

it could just be they are taking Avogadro at his word with out checking him. In the English translation of the original paper where he proposed the hypothesis, he makes very much the same statement as this textbook. I still agree that, on its own, equal molecules for equal volume doesn't imply direct proportionality. @Jesse

"The first hypothesis to present itself in this connection, and apparently even the only admissible one, is the supposition that the number of integral molecules in any gases [sic] is always the same for equal volumes, or always proportional to the volumes."

Interesting. So, arguably, Avogadro himself made a minor logical error in viewing the two hypotheses as equivalent (hence his "or" in the quote you give). But in hindsight we could say that the more fundamental hypothesis is "$V \propto N$", from which the other one follows.

Mathematically speaking the two concepts are not equivalent. So you are correct to point out the logical problem. But, for the idea to make physical sense using the simplest possible theory, the connection is much tighter. Plus, actual observations are entirely compatible with the simple formula. The law wasn't derived as a mathematical exercise but as the result of observations and physical reasoning.

The  other answer is, I think excellent.
Your question is basically "but what if there isn't a direct proportion between number of gas molecules and volume?" Instead, that was Avogadro's hypothesis in a nutshell: the volume of the gas is determined exactly proportionate to the number of molecules in it.
You are right in principal that you could have such a "false law" where the number of molecules is inversely proportionate to the volume and still have equal number of molecules for $\ce{H2}$ and $\ce{N2}$.
Let's follow your course of logic. That would mean if I get 10 L of $\ce{H2}$, I would then have $k/10$ molecules. Let's imagine I add more and more volume, to get 100 L then 1000 L of gas. At some point, I must necessarily divide the molecules in fractions in your false law. Certainly a molecule must occupy some indivisible volume, since Avogadro believed in the atomistic theory.
Instead, the obvious solution is that the number of molecules must be a measure of the amount of "stuff." So if we have more "stuff," we need more volume for it to occupy. The expression must be directly proportional between the volume and the number of molecules.

I agree that the direct proportionality is the most obvious relationship between amount and volume which would account for the fact that equal volumes of different gases contain the same number of molecules. But it seems logically dubious to claim that the proportionality follows from that fact.

@Jesse - as Geoff said, Avagadro believed in the atomistic theory. That is not explicitly mentioned in that excerpt of your book but I'm sure that in the introduction there is a discussion of the Greeks and how they believed that there is an "atom" - something that cannot be further divided - or the building block of matter.

If you believe in atomistic theory, given Avogadro's hypothesis, the relationship must follow. As illustrated above, inverse proportion ends up dividing atoms at some point.

Not sure if it's kosher to "resurrect" such an old comment thread (feel free to delete if so), but: I agree that Avogadro + atomic hypothesis rules out the law I gave. But it still doesn't imply direct proportionality (e.g. take $V = kn^2$).

@Jesse - for whatever reason, I didn't see this comment until now. The problem with .. say $V = kn^2$ is that it implies $n = \sqrt{V / k}$. If I add a bit more "stuff" ($n$) the volume goes up quadratically. But we know that we can increase volume pretty precisely. I can put just a little bit more volume into a balloon. What does that mean for the amount of "stuff" I put in? In your quadratic formula, I'll eventually get cases with fractions of molecules. Not so with direct proportion. I can add one molecule at a time or any combination thereof.

You must be familiar with the ideal gas equation: PV = nRT. Thus, an ideal gas can be completely characterised by 3 state variables (if you believe, in the ideal gas equation)
Say, you fix pressure and temperature (say, at STP). You get V = cn (where c is some constant, and n is no. of moles).
Thus it is indeed true that equal moles of gas, have equal volumes (under a given set of conditions). In fact, one calculates this volume as nearly 22.4 L (for one mole, at STP)
In fact Avogadro was the first person to make this observation. (and even though, I started my explanation with the ideal gas equation, it was formulated afterwards)
Of course, this is an idealisation. A picture that is closer to reality is given by the van der waal's equation of state, which takes into account inter-molecular interactions.
 $[P + a(\frac{n}{V})^2][\frac{V}{n}-b]= RT$
But this departure from ideal behaviour is usually not significant at low pressure, high volume, dilute conditions.

I don't quite think this addresses my question. I am familiar with the ideal gas equation, and clearly if one assumes that equation then Avogadro's hypothesis follows. My question, however, is about the converse statement; that is, how does the direct proportionality follow from the hypothesis that equal volumes of gas contain the same number of molecules?

There is a big difference between mathematical possibility and physical plausibility
You are right to point out that one claim does not lead to other logically or mathematically. But the claim makes sense physically and makes sense of the previous observations chemists had made.
The way these concepts are taught now makes it looks as though chemists leapt from one idea to another using just some form of logic. But that is not what happened. Possible mathematical relations could have been different, but chemists were not choosing between different mathematically derived ideas: they were trying to explain already observed phenomena.
They already know that twice as much stuff would give a gas with twice as much volume. They were not completely sure why. Nor did they understand that different gases did not all consist of the same components (we know that oxygen is diatomic and carbon dioxide triatomic, for example: they did not). But they knew that certain masses of reactants would combine in fixed mass ratios. They also know that gasses weighed different amounts and the volume ratios of certain gas reactions. They didn't have a unifying theory to tie these observations together.
Sure, the relationship between the composition of gases and the volume could have been more complex to a mathematician. But the idea that volume and the number of things making up the gas is simple and proportional has two benefits. It is the simplest possible theory; and it explains previous observations (not least that using twice as much of a solid in a gas producing reaction produces twice the volume of the gas). More complex theories are possible, but they won't explain that observation.
So the simple story of how the leap was made misses out some other facts relevant to the basic idea. The real theory was not a logical finger-in-the-air but a deduction based on other well-known physical observations. And one that didn't emerge from mathematical speculation but from its ability to explain many phenomena. Some modern explanations leave that story out and, as a result, sound like an imperfect logical leap.


text:
basic chemistry alkaline test for flavonoids!!! with reference please
What is flavanoid physical and chemical properties, and how alkaline reagent test for flavonoids gives yellow coulour?

This is a homework question. Please share your thoughts towards a solution otherwise it will be closed in accordance with our homework policy. A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

I don't mean to be rude, but seriously? This really does look like homework to me. You can do a Google search with the exact words of your title to get highly relevant Google results coming on top (even despite the misspelling) I'm flagging for closure.

this is not  homework question. I have to submit a report for my lab work and i was tying to find this information but i couldn't.

From this reference:

Treatment of a flavonoid with a strong base will result in ionization of its hydroxyl groups with resulting large bathochromic shifts of both absorption maxima.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


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electron passing nodes?
When one say that the electron goes from 1 orbit , say 2S, to another (higher) orbital, say 5P, it mean that the electron is excited to that orbit by some mean. In doing so it also it will also cross the nodes of 2P etc. But doing so does the electron actually cross the node or what? Cause doing so the, there is probability, although very less, of finding electron in node for a very short time.

My understanding is that the electron instantly jumps from one energy level to another. Since it isn't a discrete object the idea of it moving through space across the node is meaningless. However, I will wait for someone else to provide a more detailed answer than me.

@bon, one better says that electrons "jump" from one quantum state to another. Besides, an electron transition does not happen instantly, but rather in a course of a few nanoseconds (or less).

@Wildcat I suspect electron excitation must take at most a few hundred femtoseconds in order to explain, for example, ultrafast photoexcited electron transfer in systems with fullerenes. The duration of an electronic transition is probably given by a number close to the inverse of the absorbed photon frequency (e.g., $\mathrm{400\ nm \rightarrow 750\ THz \rightarrow 1.333\ fs}$).

@NicolauSakerNeto, no, you could not just invert the frequency. See, for instance, this question at Physics.SE.

@Wildcat That's interesting, but now I'm confused as to how an electron can be photoexcited and transferred four orders of magnitude more quickly than it would take to simply be excited.

@NicolauSakerNeto, picture it in the following way. Say, for the case of spontaneous emission, the emitted light  of some frequency can be thought of as a result of electron "oscillations" at this frequency. To produce the wave an electron "oscillates" back and forth many times (say, few million times at average) which is why transition time is nowhere close to the inverse of the frequency.

@NicolauSakerNeto, and since emission is probabilistic processes, you never know when a particular electron will stop "oscillating" and emit the light quantum. Thus, the transition time is related to the width of the spectral line, not to its frequency.

Electronic excitations happen in the attosecond time-scale.  See for example http://bromine.cchem.berkeley.edu/atto.htm

with regard to the question I found this very helpful: http://antoine.frostburg.edu/chem/senese/101/electrons/faq/how-electrons-cross-nodes.shtml

When one say that the electron goes from 1 orbit , say 2S, to another
  (higher) orbital, say 5P, it mean that the electron is excited to that
  orbit by some mean.

First off, there are no orbits. Electrons do not circle the nucleus. Rather they are in particular quantum states, or, in other words, they "occupy" particular orbitals. Note that when we say that an electron "occupies" an orbital, we just mean that it is in a particular quantum state. There is no spatial occupation in a classical sense out there: eletrons do not "sit" or "circle" anywhere, they are just in their quanatum states.
Consequently, when we say that electron "goes" from one orbital to another, what we actually mean is that the quantum state of the electrons changes. And again, an electron transition is not a mechanical motion, thus, electrons upon transitions do not "move" through nodes.

Can you  please explain what this image says https://en.wikipedia.org/wiki/Electron#/media/File:Hydrogen_Density_Plots.png

@ChinmayChandak, different quantum states of an electron feature (among other things) different patterns of spatial probabilities (probabilities of finding the electron in different regions of space). Upon electron transition this probability pattern changes from one which correspondes to the initial quantum state to one which corresponds to the final quantum state.

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"Crystal radius" vs "Ionic radius"
What is the difference between crystal radius and ionic radius? For $\ce{NaCl}$ for example the ions combine to a crystal, and I would have guessed they are equal. But I then found an example of data, where they are not.
I am aware that ionic radius is the radius of an atom's ion. Quite simple. So what is crystal radius?

I have found two different uses of the term "Crystal Radius." First is Linus Pauling's use of Crystal Radaii, which refers to ions that are free and not part of a crystal. The second is used to describe some parameters of the flame fusion crystal growth this paper, which would seem to be very much incorporated into the crystalling structure. This seems to be a rarely encountered term. Where did you encounter it?

Similar question: https://www.researchgate.net/post/Difference_between_Crystal_Ionic_Radius_and_Effective_Ionic_Radius

There is a table of "Pauling Empirical Crystal Radiuses", which cites the following sources: Galasso, Francis S. Structure and Properties of Inorganic Solids. Oxford: Pergamon Press, 1970.
Pauling, Linus. The Nature of the Chemical Bond, 3rd edition. Ithaca, NY: Cornell University Press, 1960.
Rohrer, Gregory S. Structure and Bonding in Crystalline Materials. Cambridge: Cambridge University Press, 2001.

To quote Erik Rakovský · 30.73  · 54.96 · Comenius University in Bratislava (from this discussion on researchgate)
Shannon published simultaneously two sets of ionic radii. So called "effective ionic radii" were made consistent with Pauling's r($\mathrm{O_2^-}$)=140 pm. "Crystal radii" are using Shannon's value r($\mathrm{O_2^-}$)=126 pm.
https://dx.doi.org/10.1107%2FS0567739476001551

Could you please summarise the content of these links?

@Gerhard The description is already above each link respectively.

This answer would have been more helpful if you would have explained what all there is in those links and then provided them as source. Just saying.

@Quark If you edit it, I will approve it.

@Dale  Only If I understood what the correct answer is. Haha.

People come to SE for answers, not for links.


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Why are electron configurations more stable when there are 8 electrons in the outer shell?
In this video on Khan Academy, Sal remarks that 'all atoms envy the noble gases, which are the most stable because they have eight electrons in their outer shell'.
I can't think of much of a reason that the number of electrons in the outer shell would affect the stability of the electron configuration. Additionally, I read that more electrons in the outer shell slightly increases the atomic radius. A larger atomic radius usually indicates a less stable electron configuration.
However, perhaps I've been thinking about this the wrong way. I've been wondering whether some property obtains when there are eight electrons in the outer shell that causes the configuration to be more stable. Perhaps, I ought to think about thus: atoms tend toward having eight electrons in their outer shell (i.e. eight electrons in the outer shell coincides with configuration stability, but doesn't cause it). The fact that the unscreened Coulomb force increases from left to right along a period is what explains why atoms tend toward having eight electrons in the outer shell: The atoms near the left of a period exhibit relatively little unscreened Coulomb force, and so are prone to giving up electrons and, so are prone to giving up the few electrons that constitute their valence shell. Whereas, those atoms near the right of the period exhibit much Coulomb force, relatively. Accordingly, those atoms are prone to attracting more electrons, and thereby prone to filling their valence shells.
Why does the stability of the electron configuration increase when the outer electron shell is full?
NOTE: I have reviewed the question Can an atom have more than 8 valence electrons? If not, why is 8 the limit? That question asks about the maximum number of valence electrons. This question asks why eight electrons in the valence shell are more stable than 1, 2, 3, ... 6, or 7 electrons in the valence shell.

@santiago, Thanks for the reference. I read it, this is definitely a different question.

http://chemistry.stackexchange.com/questions/8598/maximum-number-of-electrons-each-shell?rq=1, http://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule, http://chemistry.stackexchange.com/questions/14485/why-does-a-fluoride-ion-only-have-a-1-charge-and-not-a-2-charge-or-more

It's not the same question, but check out my answer to this one:
http://chemistry.stackexchange.com/questions/14485/why-does-a-fluoride-ion-only-have-a-1-charge-and-not-a-2-charge-or-more/14503#14503
which Mithoron also referenced above.
In it, I talk about effective nuclear charge and how it is responsible for the octet rule, which is essentially what you are saying when you talk about the Coulomb forces being unscreened as you move toward the noble gases.


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Why are compounds of V5+ and Cr6+ colored?
As far as my knowledge goes, transition metal complex ions show a color due to a transfer of electrons between the levels of split $\mathrm d$-orbitals. In the case of $\ce{V^5+}$ and $\ce{Cr^6+}$ however, there are no electrons in the $\mathrm d$-orbitals, and the electronic configuration of both is similar to $\ce{Ar}$, yet $\ce{V^5+}$ and $\ce{Cr^6+}$ show color(yellow/brown and orange respectively).

"In the case of V5+ and Cr6+ however, there are no electrons in the d-orbitals" - maybe you could get such ions using gamma rays or sth but no compound has naked cation with such high charge

http://chemistry.stackexchange.com/questions/28069/manganese-vii-oxidation-state

http://chemistry.stackexchange.com/questions/39829/how-can-the-intense-color-of-potassium-permanganate-be-explained-with-molecular

Firstly, a note about terminology. As Mithoron noted, the naked ions $\ce{V^5+}$ and $\ce{Cr^6+}$ do not exist because they require a prohibitively large amount of energy to make (just take the sum of the first 5 or 6 ionisation energies of V/Cr to see why). As such your question should really be asking about the vanadium(V) and chromium(VI) oxidation states, which can exist in acidic solution as the oxoanions $\ce{VO2+}$ and $\ce{Cr2O7^2-}$ respectively. The same idea can be extended to the manganese(VII) oxidation state, which exists as $\ce{MnO4^-}$.
Now, on to the question itself. As you rightly note, the central atom in these oxidation states have the same electronic configuration as argon; we usually call it a $\mathrm{d^0}$ configuration to indicate that there are no d electrons. The colours of these species therefore cannot arise from d-d transitions, which are the transfer of electrons from one metal orbital to another metal orbital. They are actually associated with electron transfer from the ligand (in this case, $\ce{O^2-}$) to the central metal atom. The name for this is ligand-metal charge transfer (LMCT).

Vanadium in the +5 oxidation state also exists as $\ce{VF5}$.


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How can an inexperienced chemist determine the chemical structure of a molecule?
I'm a fairly inexperienced, but nonetheless interested chemist, currently planning a small project for the coming summer: I want to isolate an unknown chemical from a plant and determine it's chemical structure.
I have ample access to Carex grayi, a sedge plant, and have chosen this plant for a few reasons.

Of the plants growing near where I live, I can definitively identify Carex grayi. I am relatively inexperienced in taxonomy and plant identification, and I want to know which species I am working with.
As far as I can tell, the plant is relatively unresearched, and may or may not contain alkaloids.

This brings me to my next point. I know it may seem immature or disheartening, but my interest in chemistry stemmed from my interest in neuroscience and neurotransmitters. From countless hours of reading, I have learned much about the pharmacological (and sometimes psychoactive) tendencies of alkaloids. I realize that ingesting unknown alkaloids is extremely dangerous, and I do not intend to ingest anything. However, due to their numerous applications in medicine and recreational use, there is a massive amount of information on the internet regarding their extraction and isolation. I am entirely open to the idea of isolating a chemical from another group, but as I am inexperienced and my interest stems from neuroscience, I only really know about alkaloids. I am only somewhat familiar with terpenoids, and know little about any other group.
So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure? The Chemical structure Wikipedia page lists the following methods for determing chemical structure:

X-ray diffraction
Proton NMR
Carbon-13 NMR
Mass spectrometry
Infrared spectroscopy

I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars. Is it too much or too difficult for a student to determine a chemical's structure over the summer? Should I do anything differently (as in, is there a certain group of chemicals that would make this project possible or easier)?
Also, is there a "classical" chemical extraction similar to this that most fledgling chemistry students would conduct? This is obviously a ridiculous made-up example: extract dandylionine from a dandy lion flower.

A nice and quite easy exercise for practising is the extraction of limonene from orange peel. Maybe you would like to try that first to get a hold on things...

So, assuming I am able to extract an unknown chemical of some specified group from a plant, is there any way for me to determine the chemical's structure?
Yes.
I don't really have access to these tools as student over the summer. I am willing to spend a small amount of money on this project, but no more than about one hundred dollars.
Then probably no (at least not with the same accuracy you would if there weren't these constraints).
Mass spectrometry and nuclear magnetic resonance are the two most powerful analytical tools to identify completely unknown substances. This doesn't mean it's as trivial as putting your purified substance in the equipment and finding its structure. For reliable results you need a purified substance, an optimized experimental routine and of course the instrument properly set up.
In fact, before trying to determine the structure of the substance, you must purify it, and that alone is a challenge. As I'm sure you are aware, the plant contains thousands, millions of different molecules. How are you going to extract something you haven't defined? The first step would have to be defining what is it that you're looking for.
Let's consider you're looking for whatever makes plants green:
You could macerate the plant, add some water, filter the mixture, check the colors of the filtrate and of the residue, and you would notice most of the green wasn't extracted. You would find different results with different solvents though, and a better option would be ethanol. Even then, it wouldn't mean you have a pure chlorophyll solution, so several other steps would probably be required to purify the substance*. Generally speaking, you would use some kind of chromatography.
For your project, I suggest you try to first precisely define what is it you're looking for. The substance that gives the plant its color? Its smell? Its pharmacological effects? You must use this property to analyze your experimental results. You can visually inspect the color, you can smell extracts, and perhaps test physiological effects (this is obviously dangerous and shouldn't be done unless you really know what you are dealing with!). Use the target substance properties' to decide what's the best way of extracting and purifying it.
Then, considering you did purify something, without the high-tech analytical tools mentioned, there are some experiments which can indicate what you have, but they require a reference substance and usually can't be used to say that a substance is the same as another, but rather that a substance isn't the same as the reference. For example, you can compare boiling/melting points and/or thin layer chromatography retardation factors, both which can be done at home home (to some extent), and also UV/Visible light absorption/emission, which might not be trivial to do at home but require instruments that are a lot easier to find in chemistry teaching labs.
For precise tips on the extraction and identifications, it would be necessary to know what you're looking for.
It's also important to note that there's a lot you can do besides determining a chemical structure. There are several properties of a substance you can study in order to characterize it. Some examples are solubility (in different solvents, at different conditions (temperature, pH, salinity)), density, reactivity (in face of pretty much anything you find at home) and physical properties such as texture, color, smell, etc.
Finally, do not think I'm saying you should give up, your wish to discover what something is made of is what made chemistry a science, but you need to have a clearer problem to be solved. Decide precisely what it is you're looking for (in terms of macroscopic properties, chemical group or something like that), and then research (and ask!) what could you do about that without a full lab.
*Chlorophyl is actually a group of substances which have a critical role in photosynthesis, and are, of course, green.
In summary:

You already have your sample. Now choose your analyte, the molecule(s) you want to study.
Do research. Look for information on how people usually study this (group of) substances. You'll have better luck looking in education-related sources, which usually have simpler experiments.
Once you've found something that fits your interests and available material, do experiments. Observe the results. Change parameters, do them again, try to understand what's happening.
If you can't explain the results, do research, check your references, and of course, ask here. We'll be glad to help.

Nicely done! But "your wish to discover what something is made of  is what made chemistry a science." In fact, it was what made physics a science. Chemistry used to be a branch of physics, and it was till the eighteenth century. But you might go on, I'm being overly pedantic here! :P

@M.A.Ramezani I actually thought I wasn't being very precise when I added that, but I just wanted to point how curiosity is important to science as a whole!

Indeed, the usual thing to do with a plant that has interesting properties is isolate the substance with interesting properties (colour, smell, psychoactivity, whatever) confirm that you have indeed isolate it (pretty easy with colour or smell, less so with psychoactivity) and only then invest the considerable effort required to elucidate its structure. Finally, spend several months synthesizing it from scratch to prove (hopefully!) thatyou got the structure right.


text:
Units of mass on the atomic scale
what are the systems of recording atomic masses and their units?
I know that the nucleon number is the number of protons and neutrons
I also know that the mass number on the periodic table is the masses of all the isotopes and their relative abundance (I believe this is called relative atomic mass and is equal in numerical value to the molar mass of that element)
I am not sure what units relative atomic mass is measured in. some say there is no units as it is a relative scale, others say that it is measured in atomic mass units (AMU), unified mass units (u) or daltons (Da).
I believe (but would like it to be confirmed) that the atomic mass unit (AMU) is an obsolete unit based on the mass relative to oxygen.
I also believe (but would like it to be confirmed) that the unified mass units (u) and dalton (Da) are equivalent units (u=Da=1/12 mass of a carbon-12 atom)
what is the difference between isotopic mass, relative atomic mass and atomic mass and what units do they all have?

related: Quick and simple explanation of molar mass, molecular mass and atomic mass

This question is partially answered in this excellent quora post here.

The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).
The quantity relative atomic mass (quantity symbol: $A_\mathrm{r}$) is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative atomic mass is the ratio of the mass of an atom to the unified atomic mass constant:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.
The value of the unified atomic mass constant (symbol: $m_\mathrm{u}$) is defined as 1/12 of the mass of a neutral atom of the nuclide $\ce{^12C}$ in the ground state at rest.
$$m_\mathrm{u} = \frac{m_\mathrm{a}(\ce{^12C})}{12}$$
The recommended value is
$$m_\mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$ (source)
Note that the value in SI units is obtained experimentally.
The unit dalton (unit symbol: $\mathrm{Da}$) and the unified atomic
mass unit (unit symbol: $\mathrm{u}$) 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.
$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$
Thus, the value of the unified atomic mass constant is, by definition, equal to one dalton or one unified atomic mass unit:
$$m_\mathrm{u} = 1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$
The old relative atomic masses (unfortunately called ‘atomic weights’) and the corresponding atomic mass unit (amu) were originally 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 $\ce{^12C}$.


text:
Oxygen-based explosives
I know that oxidanes are oxygen chains and rings. For example trioxidane is $\ce{HO-O-OH}$. I can see this turning into $\ce{O2}$ and $\ce{H2O}$ like hydrogen peroxide does. Since the double bonds in $\ce{O2}$ are strong do oxygen compounds without nitrogen ever explode?

Important examples of explosives with no nitrogen are organic peroxides, which are commonly formed upon ageing of ether solvents. Organic peroxides are highly sensitive to detonation when concentrated, and many accidents and injuries have happened in labs while evaporating ether solvents too far.

Explosion doesn't even have to produce gas - https://en.wikipedia.org/wiki/Copper%28I%29_acetylide

They do indeed explode. Oxyliquits are probably the most comprehensive example of liquid-oxygen-based primary explosives (which are considered obsolete today). Theoretically liquid oxygen can also be substituted with ozone-oxygen mixture, which in reality has never been utilized due to technical difficulties -- oxyliquits were popular in the beginning of 20th century.

Yes, famously acetone peroxide (which can be made by mixing hydrogen peroxide with acetone) is an explosive that does not contain nitrogen.  It is believed to be an entropic explosive, in that the decomposition isn't driven by oxidation, instead it's driven by the decomposition of acetone peroxide into ozone and acetone.
Dioxygen difluoride will probably also decompose explosively at room temperature, but if you are doing anything with dioxygen difluoride, you have other things to worry about; for instance, it will react with basically anything.


text:
Why do cyclic ethers have higher boiling points than their acyclic isomers?
TL;DR version is the question title. Some context and data follow.
I was creating an assignment for my organic chemistry students in which they would need to draw as many isomers as they could from a simple formula, for example $\ce{C3H6O}$ (which fits 7 structures - 9 if you count minor enol tautomers and 11 if your count stereo). In this set (ignoring stereo and enols) there are:

Two alcohols

Cyclopropanol - boiling point $=101-102\ ^\circ\text{C}$
Allyl alcohol - boiling point $=97\ ^\circ\text{C}$

Two carbonyl compounds

Propanal - boiling point $=46\ ^\circ\text{C}$
Acetone - boiling point $=56\ ^\circ\text{C}$

Three ethers

Propylene oxide - boiling point $=34\ ^\circ\text{C}$
Oxetane - boiling point
$=49-50\ ^\circ\text{C}$
Methyl vinyl ether - boiling point $=6\ ^\circ\text{C}$

The boiling point of methyl vinyl ether immediately stood out as anomalous. The fully saturated ethyl methyl ether has a boiling point of $7.4\ ^\circ\text{C}$.
Here are some more data comparing cyclic ethers to their acyclic isomers (and some analogs). The gap appears to be closing as the number of carbons increases.

Four carbons

Tetrahydrofuran - boiling point $=66\ ^\circ\text{C}$
1-Butene oxide - boiling point $=61-65\ ^\circ\text{C}$
Ethyl vinyl ether - boiling point $=33\ ^\circ\text{C}$
Allyl methyl ether - boiling point $=42-43\ ^\circ\text{C}$
Diethyl ether - boiling point $=34.6\ ^\circ\text{C}$
Methyl propyl ether - boiling point $=38.8\ ^\circ\text{C}$

Five carbons

Tetrahydropyran - boiling point $=88\ ^\circ\text{C}$
2-Methyltetrahydrofuran - boiling point $=80.3\ ^\circ\text{C}$
Allyl ethyl ether - boiling point $=65-66\ ^\circ\text{C}$
Butyl methyl ether - boiling point $=70-71\ ^\circ\text{C}$

I've wondered myself about this. It seems to me that it 's connected with  bigger accessibility of oxygen atom (sterical effects).

Apparently this behaviour is not limited to ethers; n-hexane ($\mathrm{68.5\ °C}$), 1-hexene ($\mathrm{63\ °C}$) and cyclohexane ($\mathrm{80.7\ °C}$). Did you check any other functional groups?

No, I only thought about ethers - there this effect seems to be biggest.

Ah, so penicillin isn't the only accidental finding now... :)

For anyone reading this question, I highly recommended also taking your time to go through its spinoff and the answers therein.

The data I could find suggests that cyclic ethers have higher dipole moments than their acyclic counterparts.
$$\mathbf{Four~carbons}$$
\begin{array}{c @{} c} \hline
\text{THF} & \mathrm{1.63~ D ^1, 1.75~ D ^2} \\
\text{1-butene oxide} & \mathrm{1.89~ D^2} \\
\text{ethyl vinyl ether} & \mathrm{1.26~ D ^2} \\
\text{diethyl ether} & \mathrm{1.15 ~D ^3, 1.15 D ^2} \\
\text{methyl propyl ether} & \mathrm{1.11 ~D ^2} \\\hline
\end{array}
$$\mathbf{Five~carbons}$$
\begin{array}{c @{} c} \hline
\text{Tetrahydropyran} & \mathrm{1.58~D ^2, 1.87~D ^4} \\
\text{2-Methyltetrahydrofuran} & \mathrm{1.38~ D ^5} \\
\text{dipropyl ether} & \mathrm{1.21~D^2, 1.00~D^6} \\\hline
\end{array}
This seems consistent with intuitive expectations based on conformational models where, in the cyclic ethers where the molecular geometry is constrained by the ring, the lone pairs are pointing in the opposite direction from the carbon skeleton. Whereas the lone pairs and carbon chains in the conformationally more mobile acyclic isomers are less geometrically fixed and therefore less "directed" in space.
It seems likely that the higher dipole moments in the cyclic compounds would lead to greater alignment / ordering in the liquid phase, which in turn would lead to higher boiling points.
References:

  1: http://en.wikipedia.org/wiki/Tetrahydrofuran
  2: https://physicalchemistryrosamonte.wordpress.com/material-balances/material-balances-on-a-crystallizer/physical-properties-of-pure-methanol/dipole-moment/
  3: http://en.wikipedia.org/wiki/Diethyl_ether
  4: http://www.drugfuture.com/chemdata/tetrahydropyran.html
  5: http://www.stenutz.eu/chem/solv28.php?s=1&p=20
  6: https://books.google.com/books?id=G6jaBwAAQBAJ&pg=PA50&lpg=PA50&dq=propyl%20ethyl%20ether%20dipole%20moment&source=bl&ots=VA52gqJ0kn&sig=GM8tua_QpXccFW3VXpy9qiNf7rk&hl=en&sa=X&ei=rzRvVZ__Cc32yQT9gIPACQ&ved=0CDsQ6AEwBQ#v=onepage&q=propyl%20ethyl%20ether%20dipole%20moment&f=false

The conformationally-influenced dipole explanation seems pretty consistent with observations like the similarity in boiling point between pentane $(35.9\ ^\circ\text{C})$ and diethyl ether $(34.6\ ^\circ\text{C})$, while cyclopentane $(49.2\ ^\circ\text{C})$ has a lower boiling point than THF $(66\ ^\circ\text{C})$.

While this is an interesting suggestion, I think the finding may be more general than for ethers. In particular, alkanes/alkenes also seem to show a similar boiling point increase upon cyclization (for example, this link says "Cycloalkanes have boiling points which are about 10 - 20 K higher than the corresponding straight chain alkane."). What would be the reason in this case? Dipoles may not be the whole explanation. I think it may also have to do with vaporization entropy, but have not been able to find data to back it up.

Another complication I see is 1,4-dioxane ($0.45\ \text{D}$, $101 \text{°C}$) and dimethoxyethane ($1.71\ \text{D}$, $85 \text{°C}$, which anti-correlate somewhat strongly.

In 1,4-dioxane the dipole moment may be "artificially" lowered since the local dipoles in the two halves will somewhat cancel.  Still, the local dipole moments should still facilitate ordering even though "overall" the molecule has a low dipole moment.  In the case of dimethoxyethane, perhaps the conformational mobility is somewhat restricted due to the electrostatic repulsive effect the adjacent lone pairs exert on "preferred" conformations.

@NicolauSakerNeto "What would be the reason in this case?"  I was wondering if entropic effects were in play?  Smaller ${\Delta S}$ for rings -  but wouldn't that suggest that rings would vaporize at lower temperatures than their non-ring analogues?

In my mind, molecules in general are more conformationally restricted in liquids due to mutual proximity. Cyclic molecules naturally have fewer conformational degrees of freedom in both phases, and so would not gain quite as much entropy upon vaporization. In an acyclic molecule, the chain would be much more free in the gas phase. This could make it entropically more favourable for acyclic molecules to evaporate, thus decreasing their boiling point relative to cyclic ones. This effect has been suggested elsewhere, with no confirmation.

There are many, sometime competing, some-time reinforcing, factors that affect the boiling point of these molecules:

the polarity (e.g. approximated by the dipole moment)
dispersive interactions
increase in entropy upon evaporation

Some illustrative examples (data taken from the NIST Webbook unless otherwise noted). The boiling point of tetrahydropyran is 361 K while that of  ethyl propyl ether is 337 K.  The corresponding standard enthalpies of vaporization ($\Delta H^\circ_{\text{vap}}$) are 35 and 31 kJ/mol.  This is consistent with the higher dipole moment of cyclic structures mentioned in @ron's answer.
Unfortunately I could not find the dipole moment of ethyl propyl ether, but it is probably similar to dipropyl ether and this is backed up by MolCalc calculations.
However, while the dipole moment of dipropyl ether (1.0-1.2 D) is lower than tetrahydropyran (1.6-1.8 D; taken from @ron's answer) the boiling point is 363 K - slightly higher than that of tetrahydropyran, as is the $\Delta H^\circ_{\text{vap}}$ (36 kJ/mol).  So polarity cannot be the whole story in general.  Dipropyl ether is slightly larger so larger dispersion interaction is the most likely explanation for the high boiling point here.
Finally, the boiling point of allyl ethyl ether is lower than tetrahydropyran (340 K) while the $\Delta H^\circ_{\text{vap}}$ is the same (35 kJ/mol).  (The dipole is smaller than for tetrahydropyran but bigger than for ethyl propyl ether - according to MolCalc - but the double bond is more polarizable and will lead to an increase in dispersion interactions compared to tetrahydropyran).  Since the enthalpy change is the same, allyl ethyl ether must have a higher $\Delta S^\circ_{\text{vap}}$. It's a more "floppy" molecule than tetrahydropyran so the gas phase vibrational and conformational entropy will be higher.  (According to MolCalc the vibrational entropy of ethyl propyl ether  is 54 J/molK higher than for tetrahydropyran).
So, the higher boiling point of tetrahydropyran compared to ethyl propyl ether is likely due to a combination of factors:

stronger interactions between tetrahydropyran molecules due to it being more polar
a lower increase in entropy upon vaporization, due to tetrahydropyran being less flexible.

The fact that allyl ethyl ether has roughly the same boiling point as ethyl propyl ether but the roughly the same $\Delta H^\circ_{\text{vap}}$ as tetrahydropyran suggest that point 2 contributed the most to the difference in boiling point between tetrahydropyran and ethyl propyl ether.

Great work helping with a numerical analysis and bringing everything together! I just think it might be useful to point out that yet another factor may be simple molecular packing in the liquid, as discussed in the spinoff thread. For example, ethyl propyl ether has a density of up to $0.75\ \mathrm{g\ ml^{-1}}$, while tetrahydropyran is significantly denser at $0.88\ \mathrm{g\ ml^{-1}}$.

Thanks! I am not sure how predictive the density is.  For example, n-hexane has a density of 0.65 g/ml but a higher boiling point than ethyl propyl ether, while the enthalpies of vaporization is the same.

Interesting pair for comparison. I wonder what the explanation in that case would be. Polarity would seem to favour a higher boiling point for the ether, while molecular mass, dispersion interactions, vaporization entropy and liquid packing seem similar. This is complicated stuff to predict...

If the enthalpy of vaporization is the same, then n-hexane must have a higher entropy of vaporization.  F.eks. according to MolCalc the vibrational entropy is 4 J/molK higher than ethyl propyl ether.  Also, the conformations of n-hexane will be closer in energy so the conformational entropy will be higher


text:
Are there substances people can put on themselves to prevent catching fire?
This can be very useful for people who are afraid of spontaneous human combustion! But seriously, I heard there are non-harmful chemicals used in some fire extinguishers, so can they maybe be put on our skin, clothes, etc. in advance to prevent ever catching fire?

The general class of chemicals you're interested in are called flame retardants, substances which are used on clothing already.  I'm not sure that it would be practical to smear them on your skin unless you're planning on doing something unwise near a fire.  Firefighters already use substantial protective equipment, and smoke inhalation is typically a bigger danger for them.

It depends how long one is exposed to heat, and how hot the source is, but yes, there are "fire retardants" for skin:

A mix of alcohol (ethanol or isopropanol) and water or butane in soap sud can be set afire on your palm without burning you, if done very carefully. The water soaks up heat energy both because of its high specific heat and through evaporation.
Insulation applied to the skin or between skin and flame can protect (momentarily) from a torch.
A wet hand is temporarily protected from molten lead by the leidenfrost effect. As water on the skin boils, it acts, briefly, barrier to the heat.

NB: Do not try these experiments yourself (or on others). You get no points for being a crispy critter.


text:
How to find the standard enthalpy and entropy for the dissolution of silver chloride?

A maximum of $1.89~\mathrm{mg}$ of $\ce{AgCl}$ will dissolve in $1.00~\mathrm{L}$ of solution at $25.00~^\circ\mathrm{C}$.  A maximum of $5.26~\mathrm{mg}$ of $\ce{AgCl}$ will dissolve in $1.00~\mathrm{L}$ of solution at $50.00~^\circ\mathrm{C}$.  Use this information to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for the following reaction.
  $$\ce{AgCl~(s) -> Ag+ (aq)~+ Cl- ~(aq)}$$
  Assume that ΔH and ΔS are independent of temperature.

I started off by converting the given masses into moles using molar mass of $\ce{AgCl}$ and since we're given the reaction, we can assume that $K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]$, thus

At $25~^\circ\mathrm{C}$, after doing conversion I got $1.32\cdot 10^{-5}~\mathrm{mol}$ of $\ce{AgCl}$
At $50~^\circ\mathrm{C}$ I got $3.67\cdot 10^{-5}~\mathrm{mol}$ $\ce{AgCl}$

Since, $K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]$ and mole ratio is 1:1, I calculated the two $K_\mathrm{sp}$ values:

At $25~^\circ\mathrm{C}$: $K_\mathrm{sp1} = 1.74 \cdot 10^{-10}$
At $50~^\circ\mathrm{C}$: $K_\mathrm{sp2}= 1.35\cdot 10^{-9}$

Now, we do have two temperatures and two equilibrium constants which empowers me to use this formula:
$$\ln\left(\frac{K_1}{K_2}\right)=-\frac{\Delta H}{R}\cdot \frac{T_2-T_1}{T_1T_2}$$
After solving for $H^\circ$  I get $65.6~\mathrm{kJ/mol}$ which is the correct answer.
How do I find $\Delta S$? I tried calculating it via standard $$\Delta G= \Delta H - T\Delta S$$ and averaging but it does not work.

actually I figured it out, false alarm . If anybody stumbles upon a problem like this you should use this formula : Ln(K)=-(dH/RT)+(dS/R) which gives the correct answer and ties in K, dH all together

Would you care to answer your own question? It is highly encouraged in the Stack Exchange network!


text:
How to calculate the final concentration of calcium ions from known starting masses of carbonate and calcium ions?

If $\pu{50.0 mg}$ of $\ce{CO3^2-}$ and $\pu{50.0 mg}$ of $\ce{Ca^2+}$ are present  in one liter of water, what will be the final (equilibrium) concentration of $\ce{Ca^2+}$ expressed in moles per liter?

This is a weird question. It's not like you can just add 50 mg of a cation and 50 mg of an anion. They have to come with counterions. If $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ are the only ions in solution, then they need to be in equimolar amounts. Now, if we added 50 mg of $\ce{Na2CO3}$ as a source of carbonate and 50 mg of $\ce{Ca(NO3)2}$  as a source of calcium, then we have a more viable question. Also, have you cover solubility product constants yet?

The premise of your question is weird. $\pu{50 mg}$ of calcium ion and $\pu{50 mg}$ of carbonate ion are not equimolar. However, let's assume that they have appropriate counterions somewhere. This problem can be approached two ways.
Limiting reactant
We have a precipitation reaction between the calcium and the carboncate ions,  as calcium carbonate is nearly insoluble in water:
$$\ce{Ca^2+ (aq) + CO3^2- (aq) -> CaCO3 (s) v}$$
Convert your masses of ions to amount of substance and figure out which ion is in excess. If calcium ion is in excess, then use the amount of substance in excess to determine the remaining calcium concentration.
For example: Let's say you determined that your solution has $\pu{0.1663 mol}$ of $\ce{Ca^2+}$ and $\pu{0.1221 mol}$ of $\ce{CO3^2-}$. The maximum amount of $\ce{CaCO3}$ that could precipitate is $\pu{0.1221 mol}$, because we would run out of carbonate ion. Then there would be $\pu{0.0442 mol}$ of calcium ion left in $\pu{1 L}$ of solution for a concentration of $\pu{0.0442 M}$.
Solubility product
Calcium carbonate is not completely insoluble. It's solubility product constant is $4.8\times 10^{-9}$. Thus we can set this up as an ICE equilibrium problem. I'll use the same fake amount of substances from the previous example:
\begin{align}
\ce{CaCO3 (s) &<=> Ca^2+ (aq) + CO3^2- (aq)}\\
K_\mathrm{sp}&=[\ce{Ca^2+}][\ce{CO3^2-}]
\end{align}
\begin{array}{lcc}
\hline
                   & \text{calcium}/\pu{mol}  & \text{carbonate}/\pu{mol}\\
\hline
\text{Initial}     & 0.1663                   & 0.1221     \\
\text{Change}      & -x                       & -x         \\
\text{Equilibrium} & 0.1663 - x               & 0.1221 - x \\
\hline
\end{array}
\begin{align}
4.8\times 10^{-9}&=(0.1663-x)(0.1221-x)
\end{align}
Solving for $x$ gives you $x=0.1221,0.1663$ and only the first number makes sense. In this case the final calcium amount in solution would still be $\pu{0.0442 mol}$. The carbonate amount in solution is not actually zero. It is $\pu{1.09E-7 M}$, but that is for another day.

I agree that the premise of the question is weird, but I think the question means that you have 50 mg of $\ce{CO3^{2-}}$ from an appropriate carbonate source (e.g. 88.33 mg of anhydrous $\ce{Na2CO3}$) and 50 mg of $\ce{Ca^{2+}}$ from maybe 273.3 mg of $\ce{CaCl2.6H2O}$. On a side note, if we were very particular about accuracy then we should probably worry about the $\ce{HCO3^-}$/$\ce{CO3^{2-}}$ equilibrium, but that is also for another day :)

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text:
Reversible process in thermodynamics
How can we look at reversible processes in an intuitive manner?
How is it possible, that at any moment, it can be reversed by an infinitesimal change?

I sometimes explain it this way: imagine a staircase where each step is $h$ high.  If $h$ is small then it is quite easy to take a single step up or down, i.e. the process is relatively easily reversed. However, as $h$ increases it starts to become increasingly difficult to go up (and hazardous to go down).
Even for small steps there is a difference between going up and down since you expend more energy going up (against gravity) than going down, so it's not truly reversible, but as $h$ decreases the difference becomes minuscule (but never 0).
Similarly, there is no truly reversible thermodynamic process but the more slowly you let a process proceed (and equilibrate) the more it resembles the reversible ideal process

For the process to be truly reversible, you need to be able to return both the system and the surroundings to their original states, without any significant change in anything else.
I find it useful to think of the surroundings as having a limited tool kit for accomplishing this:

A set of very small weights that can be added or removed from a massless piston at different elevations
A set of constant temperature baths featuring a broad sequence of slightly different temperatures.

If you are considering the reversible compression or expansion of a gas inside of a cylinder in the transition between two different equilibrium states, you can do this by adding or removing the weights at different elevations and contacting the cylinder with a sequence of baths at slightly higher- or slightly lower temperatures than the gas temperature at each stage of the transition.  Then, if you wish to reverse the process, you just do the opposite sequence with the weights and the reservoirs.
Hope this helps.

Are all equilibrium processes reversible like for eg change in state of solid to liquid at the melting temperature

Processes like this can be carried out reversibly.

You said "can"... is there a way for these processes to be carried irreversibly

Sure, if the process involved significant temperature gradients within the system.


text:
Derivation of van 't Hoff equation for temperature dependence of equilibrium constant
While I was reading about the usefulness of the quantity $\Delta H$, I found that it can be used to calculate the how the equilibrium constant varies with temperature. How can this be done?
Does it agree with the predictions of Le Chatelier's principle (that for an exothermic reaction, increasing the temperature disfavours product formation, and vice versa)?

In this answer of mine you can find a derivation of the formula for the equilibrium constant which gives you its temperature dependence.

The equation linking $\Delta H^\circ$ and $K$ is called the van 't Hoff equation. Since Philipp's comment on your question already links to a very thorough discussion of where the equation $\Delta G^\circ = -RT\ln{K}$ comes from, I won't repeat it.
The definition of the Gibbs free energy, $G$, is $G = H - TS$. Using $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$ we obtain the Maxwell relation
$$\left(\frac{\partial G}{\partial T}\right) = -S$$
and hence the Gibbs–Helmholtz equation (derivation here)
$$\left(\frac{\partial (G/T)}{\partial T}\right) = -\frac{H}{T^2} \quad \Leftrightarrow \quad \left(\frac{\partial (\Delta G^\circ/T)}{\partial T}\right) = -\frac{\Delta H^\circ}{T^2}$$
Since $\ln K = -\Delta G^\circ/RT$, we have
$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = -\frac{1}{R}\frac{\mathrm d}{\mathrm dT}\left(\frac{\Delta G^\circ}{T}\right) = \frac{\Delta H^\circ}{RT^2}$$
This is the differential form of the van 't Hoff equation; it's not the most useful thing to us though because it only tells you the slope of a plot of $\ln{K}$ against $T$ at one given point. We usually separate the variables and integrate with respect to both sides:
$$\int_{\ln{K_1}}^{\ln{K_2}}\!\mathrm d(\ln{K}) = \int_{T_1}^{T_2}\!\frac{\Delta H^\circ}{RT^2}\,\mathrm dT$$
$$\ln{K_2} - \ln{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2} \right) $$
So, if you know the equilibrium constant $K_1$ at a certain temperature $T_1$ and you want to find the equilibrium constant $K_2$ at a different temperature $T_2$, you can just plug in your values into the equation and solve for $K_2$.
Note that this equation supports what you know of Le Chatelier's principle; if the reaction is exothermic, $\Delta H^\circ < 0$, and if you increase the temperature from $T_1$ to $T_2 > T_1$ then $(1/T_1 - 1/T_2) > 0$. The RHS of the equation is therefore negative, and that means that $\ln{K_2} < \ln{K_1} \Rightarrow K_2 < K_1$ which implies that the equilibrium position has shifted to the left.
Note that the last step (the integration) makes the assumption that $\Delta H^\circ$ is a constant over the temperature range $T_1$ to $T_2$. Note that this is, in general, not true but if the temperature range isn't too huge you will get pretty accurate results from the use of this equation.

The change in enthalpy $\Delta H^\circ$ refers to a standard state (a specific pressure). So $\Delta H^\circ$ depends also on temperature. How do we know that if a reaction is endothermic at specific conditions $(T_1,p^\circ)$ would also be endothermic in different conditions $(T_2,p^\circ)$, so we can apply Le Chatelier's principle?

@adosar you need to find the dependence of $\Delta H$ on temperature. It depends on the heat capacities of products and reactants. A full explanation is far too long for a comment, but look up Kirchhoff's law. Atkins' textbook will have a section on it. There is a brief mention of it at https://chemistry.stackexchange.com/questions/39620/temperature-dependence-of-reaction-enthalpy

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text:
Trans-hydroxylation of alkenes
In the trans-hydroxylation of alkenes to give alcohols, there is a formation of epoxide.
Does the reaction stop here if we don't add water?
Even though if we add water, why does the product depends on which medium the reaction is carried whether acidic or basic?
My text says the ring opens differently for basic and acidic medium, but both finally gives us a trans-vicinal-diol.

Oxiranes are stable in neutral water solutions and can be easly isolated.

You can stop the reaction at the epoxide. It even has its own name: the Prilezhaev Reaction (the spelling may vary).

Concerning the ring opening have a look here.

There are many natural products out there that contain an epoxide ring (in cellular, i.e. aquatic environments) and many total syntheses of these that create the epoxide at a rather early stage and keep it through the remainder of the synthesis.

This answer has technically been given in the comments, but it bears repeating.
Yes, you can stop the reaction at the epoxide stage through careful choice of conditions. If you prevent both acidic and basic aquaeous environments, then epoxide rings are stable. They are present in a wide range of molecules and even in natural products such as epothilone.
You could also open epoxides with other nucleophiles to create different 1,2-substitution patterns — with predefined stereochemistry, if the epoxide was stereoselective.
And finally, there are even methods out there to turn vicinal diols into epoxides (usually via sulphonylation, e.g. to a tosylate followed by intramolecular $\mathrm{S_N}2$ reaction).


text:
Should I consider the quantity of each atom when estimating the relative lattice energies of two compounds?
In this video, at 1:30, the presenter remarks that the charge of $\mathrm{Na}$ is $1^{+}$ and that the charge of $\mathrm{Cl}$ is $1^{-}$. Then, at 2:00, he remarks that the charge of $\mathrm{Fe}$ is $3^{+}$ and that the charge of $\mathrm{O}$ is $2^{-}$. Finally, he remarks that the result of substituting $3^{+}$ and $2^{-}$ for $q_1$ and $q_2$ in $ E_{\mathrm{latice}} =\frac{Kq_1q_2}{r}$ is six times the result of substituting $1^{+}$ and $1^{-}$ in the same equation. Hence he assumes that the value of $r$ is the same for both compounds, and concludes that the lattice energy of $\mathrm{Fe_2O_3}$ is six tiems greater than the lattice energy of $\mathrm{NaCl}$.
I thought that the fact that $\mathrm{Fe_2O_3}$ comprises more charged atoms than $\mathrm{NaCl}$ comprises would affect the differences between their lattice energies. Accordingly, I thought that the presenter should have multiplied the charge of $\mathrm{Fe}$ by $2$ and multiplied the charge of $\mathrm{O}$ by $3$, and then substituted those two products into $q_\mathrm{1}q_\mathrm{2}$ respectively? Evidently, I was wrong about that.
In estimating the relative lattice energies, should the estimator consider the number of each kind of atom in the compounds?

Experimentally some people found out before us that multiplying by the number of ions in one formula unit (i.e multiplying by 2 + 3 = 5) gave more accurate results. See: http://en.wikipedia.org/wiki/Kapustinskii_equation In this case you have $\nu = 5$, $z_+ = 2$, $z_- = 3$, making the lattice energy a grand total of 30 times larger. (I have seen different forms of this equation, but they are all proportional to $\nu$.)

Sorry, make that 15 times larger, because $\ce{NaCl}$ has $\nu = 2$. A quick Google search shows that the lattice enthalpy of $\ce{Fe2O3}$ is around 14774 kJ/mol, whereas that of $\ce{NaCl}$ is 787 kJ/mol. 14774/787 = 18.77. Close enough, I guess, considering that we are ignoring changes in the ionic radii here.


text:
Balancing reduction equations for galvanic cell
Problem: Compute the potential of the following galvanic cell: $$(\ce{Mg (s)} \mid \ce{Mg^2+ (aq)}\ (0.30\ \mathrm{M}) \mid \mid \ce{Sn(OH)4^2- (aq)}\ (0.10\ \mathrm{M}); \mathrm{pH} = 9 \mid \ce{Sn (s)})$$
Attempt at solution: I know we have to use the formula $$E_\text{cell} = E - \frac{0.0592}{n} \log Q$$ The anode is situated on the left; this means electrons are going away there. So we have for the oxidation half-reaction: $$\ce{Mg -> Mg^2+ + 2e-}$$ The potential for this is $E_1 = 2.36\ \mathrm{V}$. Then I’m not sure how to write my reduction half-reaction. I just wrote: $$\ce{Sn(OH)4^2- + 2e- -> Sn(s)}$$ and $E_2 = -0.91\ \mathrm{V}$. But this doesn’t seem correct, because if I add both reactions to get the sum, my charges aren’t balanced anymore. Also, I wonder what happens with the $\ce{(OH)4^2-}$ ? It’s not on the right side of the arrow anymore.

To construct reduction equations there are a set of steps you can follow.
For a balanced equation, the number of atoms of each element, and the total charge, must be the same on each side of the equation.
Firstly identify the main reactants and products which you have correctly done in this case as: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s)}$$
Under acidic conditions the only things we can add to balance the reaction are $\ce{H=}$, $\ce{H2O}$ and $\ce{e-}$. Under alkaline conditions we can add $\ce{OH-}$ instead of $\ce{H+}$. Usually, when dealing with alkaline reactions the best way of dealing with them is to treat them as acidic and then neutralize the acid later. However, in this instance, since the only reactant involves hydroxide ions, we can immediately balance the hydrogens and oxygens just by adding four hydroxide ions to the products: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4OH- (aq)}$$
Then balance the charges to get the overall equation: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$
The alternative method which works for any alkaline reactions, which are sometimes less obvious than the example above, is to balance the reaction under acidic conditions and then neutralize the acid.
Since we only have one option for balancing the oxygens we add $\ce{H2O}$ first: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4H2O(l)}$$
Now balance the hydrogens: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) -> Sn(s) + 4H2O(l)}$$
Now the charges: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) + 2e- -> Sn(s) + 4H2O(l)}$$
If this were in acidic solution then this would be the final balanced equation. However, since we are in alkaline solution, we add $\ce{OH-}$ ions to cancel out the $\ce{H+}$, forming $\ce{H2O}$. $$\ce{[Sn(OH)4]^{2-}(aq) + 4H2O(l) + 2e- -> Sn(s) + 4H2O(l) + 4OH- (aq)}$$
Finally we can cancel the excess $\ce{H2O}$ to give: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$
This is the same result as we got earlier by the quicker method.

Thanks for the reply. So if we add everything we get $\ce{Mg + Sn(OH)4^{2-} -> Mg^{2+} + Sn + 4OH-}$. I used $E_1 = -2.36$ V and $E_2 = -0.91$V. Then should I add these two? If I do $E{kathode} - E_{anode}$ I get $1.45$ V. If I then use the Nernst equation with $\log(0.30/0.10)$ I get $1.43$ V as answer. But my textbook says it should be $2.03$ V?

I agree with you. I get 1.44V as my answer (1.44/1.43 depends on what values for constants you use and your rounding). Possibly an error in the book (not uncommon for textbook questions in my experience) or maybe someone else will spot something we haven't.


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Equilibrium constant in terms of mole fractions

$$\ce{CO + H2O <=> CO2 + H2}$$
  A stoichiometric mixture of $\ce{CO(g)}$ and $\ce{H2O(g)}$ is allowed to reach equilibrium. $x$ is the mole fraction of $\ce{H2(g)}$ present.

I am meant to find the result that $$K_p=\frac{4x^2}{(1-2x)^2}$$
However, I only get: $K_p=\frac{x^2}{(1-x)^2}$
Here's what I am thinking:
$$K_p=\frac{x_{\ce{CO_2}}x_{\ce{H_2O}}}{x_{\ce{CO}}x_{\ce{H_2}}}$$
But: $x(\ce{CO})=1-\alpha=x(\ce{H_2O})$ and $x(\ce{CO_2})=\alpha=x(\ce{H_2})$.
Then substituting this in gives me my result. What am I doing wrong?

The initial mole fraction of $\ce{CO}$ present must be $y=0.5$.  Then the initial mole fraction of $\ce{H2O}$ present is also $y=0.5$, and the initial moles of both $\ce{H2}$ and $\ce{CO2}$ are 0.
After equilibration, both $\ce{H2}$ and $\ce{CO2}$ are present in the amount $x$ and so by stoichiometry, $\ce{H2O}$ and $\ce{CO}$ must be present in the amount $y-x=0.5-x=\frac{1}{2}-x$.
Since the reaction doesn't change the number of moles of gas, it doesn't matter much if we use $K_p$ or $K_c$, so $K_p=\frac{x^2}{(\frac{1}{2}-x)^2}=\frac{x^2}{\frac{1}{4}-x+x^2}=~...$
I'm sure you can figure out the rest.


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Why do cyclic hydrocarbons have higher boiling points than their acyclic isomers?
As pointed out in the comments to this question, cyclic hydrocarbons have higher boiling points than their acyclic isomers. The major attractive force for hydrocarbons should be the London forces, which scale with surface area. My intuition has me thinking the cyclic molecules have less surface area, but they might enable more productive interactions over larger parts of their surface area due to their fixed conformations. I would love to see a study or some data that supports my intuition.
Boiling point data

Three carbons

Cyclopropane - BP $=-33\ ^\circ\text{C}$
Propene - BP $=-47.6\ ^\circ\text{C}$
Propane - BP $=-42\ ^\circ\text{C}$

Four carbons

Cyclobutane - BP $=12.5\ ^\circ\text{C}$
Methylcyclopropane - BP $=4\ ^\circ\text{C}$
1-butene - BP $=-6.7\ ^\circ\text{C}$
Butane - BP $=-1\ ^\circ\text{C}$

Five carbons

Cyclopentane - BP $=49.2\ ^\circ\text{C}$
1-Pentene - BP $=30\ ^\circ\text{C}$
Pentane - BP $=35.9\ ^\circ\text{C}$

Six carbons

Cyclohexane - BP $=80.7\ ^\circ\text{C}$
1-Hexene - BP $=63\ ^\circ\text{C}$
Hexane - BP $=68.5\ ^\circ\text{C}$

Your intuition is indeed correct! Several sources provide the same answer (1, 2, 3). Perhaps the simplest and most direct evidence comes from comparing the densities of the liquid unbranched alkanes and cycloalkanes:

(Source)
The cycloalkanes have slightly lower molecular mass than their parent unbranched alkanes with the same number of carbon atoms, yet the cycloalkanes are consistently more dense in the liquid phase by a factor of ~20%. The only way this can happen is if each individual cycloalkane molecule effectively occupies less volume in the liquid phase than the related alkane. This means cycloalkane molecules are closer on average in the liquid, and as van der Waals forces are attractive (at intermolecular separations) and their strength varies inversely with distance, the intermolecular attractions are stronger in cycloalkanes, and so the boiling point of the cycloalkanes should be higher.
This can be explained by considering that the unbranched linear chains in alkanes have many possible conformations, and thus are in more geometrically irregular contact in the liquid, leaving more space on average between the molecules. Cycloalkanes do not have as many conformations available, and so their molecules will approach each other more orderly and effectively, leaving less intermolecular space.
Presumably this effect can be extended to all substances (including the ethers mentioned in the question you linked), but it need not always be the dominant factor in determining boiling points.

The relation to density is certainly compelling. I'm wondering if there is any computational data that can be present on the relative surface areas (or perhaps more importantly surface area to volume ratios).

It should be pointed out that the type of molecular volume I mention (macroscopic volume occupied by liquid divided by number of molecules present = effective molecular volume) is not equivalent to the volume most often mentioned in computational chemistry, which is the region enclosed by the van der Waals surface. The only way to obtain the effective molecular volume would be to simulate a large box filled with molecules at a temperature high enough for them to be liquid, and not just by calculating the geometry of a single molecule.

I agree with the points made above regarding attractive forces in the liquid being diminished in the gas phase.
However, I also think there may be an entropic component to all of this.  When a molecule undergoes a phase change from liquid to gas (e.g. it boils)
$$\Delta G_\mathrm{vap} = \Delta H_\mathrm{vap} - T_\mathrm{vap} \times \Delta S_\mathrm{vap} = 0$$
or
$$\frac{\Delta H_\mathrm{vap}}{T_\mathrm{vap}} = \Delta S_\mathrm{vap}$$
For a ring, torsional vibrations (rotations about the $\ce{C-C}$ single bonds) are constrained in both the liquid and gas phase compared to an acyclic analogue where there is free rotation about the single bonds.  Hence, we would expect $\Delta S_\mathrm{vap}$ to be smaller for a cyclic compound compared to a noncyclic analogue. For $\Delta S_\mathrm{vap}$ to be smaller either $T$ must be larger (higher boiling point) and/or $\Delta H_\mathrm{vap}$ must be smaller.
It's easy to see how the arguments presented in the other answers can produce an increase in the enthalpic term for rings, but this would tend to move $\Delta S_\mathrm{vap}$ in the wrong direction to completely explain the effect (e.g. $\Delta S_\mathrm{vap}$ would be larger for the rings than non-rings).
Therefore, when comparing compounds like cyclopentane and pentane, I suspect that the enthalpic term for the ring is similar to or larger than that for the non-ring.  Yet since the entropic term is smaller for the ring compared to the non-ring, this requires that $T$ must be larger for the ring compared to the non-ring.  In other words, the boiling point for the ring will be higher than the boiling point of the non-ring.  Also, this reasoning would explain why, as rings get larger and are able to rotate about their bonds more freely, the magnitude of the difference between rings and non-rings diminishes.

Also of interest would be to consider even more conformationally restricted molecules, such as polycyclic saturated hydrocarbons (decalin, norbornane, etc). Data on these are significantly sparser, though.

The best that I have been able to determine is that it has to do with the rigidity of the rings. With the rings locked into certain conformations, the molecules are able to come into closer contact, as the movement of the chains in the non ring compounds can clear a larger around the molecules. Rings, on the other hands, are locked into certain conformations, so their area of contact is consistently larger.

This is probably on the right track. My intuition tells me this also, but I would love to see some data that would back it up.

Stated slightly differently - the rings aren't able to spin (as much) about their carbon-carbon bond to dislodge a neighboring molecule.


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Finding the mM concentration of capric acid (decanoic acid) in coconut oil
Im trying to figure out the mM concentration of capric acid in coconut oil. I know that There is 8% capric acid in coconut oil, but I would like to know how much capric acid is in coconut oil in a mM ammount, not a percent. I know the molecular weight of capric acid (decanoic acid) is 172.26
The reason being: I read a study that capric acid solutions are effective against microbes and fungus at a mM concentration of 20 mM. Im wondering if coconut oil's 8% of capric acid converts to roughly around a 20mM concentration of capric acid within the oil.
Thanks very much for your answers!

Well, to answer the question properly, we really need to know more information. Does the 8% correspond to w/w, or w/v? And if it is w/w, then we need to know the density of the coconut oil before we can calculate the molarity of the acid. To give an unfounded answer though, whether it's 8% w/w or w/v, it sounds like more than enough to make a 20 mM solution. For comparison, a 8% w/v aqueous solution of NaCl corresponds to 1.37 M, three orders of magnitude larger than the target concentration. Btw, I think you mean microbes, not microbicides. Microbicides are the things that kill microbes :)

Thank you very much for your reply and for the correction in my terminology! Unfortunately the study doesnt specify if the 20mM of capric acid was based on a w/v or a w/w calculation. My guess, since capric acid is crystal like in its pure state, then turned into a powder, they are using a w/w calculation. Since I believe most aqueous/powder mixtures would be calculated by w/w. I just measured 100ml of coconut oil and it weighs exactly 90 grams. Thank you for any additional input! Your guestamation makes sense and is good enough for me if there is still not enough info to know exactly

Also, I found the density (not the weight) of coconut oil is 924.27 kg/cu.m

Thank you very much for a well-specified question.  If 8% of coconut oil were free capric acid, then as noted in the comments the units of "%", i.e. w/w, w/v, or v/v, would affect the calculation somewhat, but since capric acid likely has a density that is very close to that of bulk coconut oil, the difference will be slight.  8% w/w, and assuming a density of 0.8 g/mL for coconut oil would yield $0.08~{\frac{\rm g~capric~acid}{\rm g~oil}}\times \frac{\rm mol~capric~acid}{172~\rm g~capric~acid}\times0.8\frac{\rm g~oil}{\rm mL}\times\frac{1000~\rm mL}{\rm L}=0.372\frac{\rm mol~capric~acid}{\rm L}=372~\rm{mM}$.
However, I suspect that the capric acid in coconut oil is not free acid.  It is part of triglycerides.  Instead of containing capric acid, the oil contains an ester of capric acid with glycerol.  Triglycerides are molecules where all of the available 3 hydroxyl groups of glycerol are esterified with a fatty acid.  Because triglycerides are large complex molecules, they were historically difficult for chemists to analyze directly.  The method of choice for analysis of plant oils was transesterification or ester hydrolysis, converting the triester triglyceride molecules into three smaller monoester molecules that were easier to detect by analysis.  The 8% figure you are using is probably the result of such an analysis.
The bottom line is that coconut oil that is "8% capric acid" actually contains very little capric acid, but contains a large amount of glycerol-esterified capric acid.  But only free capric acid is likely to be antimicrobial.  So the capric acid content of coconut oil is unlikely to make it antimicrobial.
The presence of free fatty acids in plant oils is usually a sign of degradation.  In olive oil, an acidity above 0.5% (i.e. of all the fatty acid chains in the olive oil, 1 in 200 are in the free form and 199 in 200 are in the esterified form) is a sign of lower quality.  I would imagine the for coconut oil the situation is similar, although the specific numbers used as quality thresholds may be different.


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Why can there not be more than one sigma bond in a set of bonds?
A question on an exam asked why there is exactly one sigma bond in double and triple covalent bonds. I looked in my text and online after the exam, but couldn't find an anawer to the question.
Why can there not be more than one sigma bond in a set of covalent bonds?

What atomic orbitals overlap to form a sigma bond?

@LordStryker, the s and p orbitals overlap to form the orbitals involved in sigma bonds,  if that's what you mean.

"Why can there not be more than one sigma bond in a set of covalent bonds?" Actually that is quite a profound question.  Your thinking is far ahead of your test.  The answer to your question is that while introductory texts often display double and triple bonds as one sigma  bond and the rest pi bonds, they can also be equivalently described as 2 or 3 "bent"  sigma bonds. So double bonds and triple bonds can be described using only sigma  bonds or as a mixture of pi and sigma bonds.

Strictly speaking, you can get two $\sigma$ bonds between the same two atoms, though it is rare. One example is in the gaseous dimolybdenum molecule $\ce{Mo2}$ with its sextuple bond, which has both a $s! -! s$ $\sigma$ sigma bond and a $d! -! d$ $\sigma$ sigma bond, as well as two $d! -! d$ $\pi$ pi bonds and two $d! -! d$ $\delta$ delta bonds. I've never heard of a double sigma bond for $\ce{C=C}$, though in some interpretations dicarbon $\ce{C2}$ has two $\pi$ bonds with no $\sigma$ bond.

@Nicolau Saker Neto Oh, sorry it were those pi bonds of dicarbon, I heard of.

This is where chemistry really gets interesting -- when you try to pin a concept down and, after some wild times down the rabbit hole, find out that things are far more subtle and more weird than you ever thought possible.  Nearly every concept is initially taught at a (relatively) comprehensible, modest-to-extreme level of approximation. As learning proceeds, the approximations are gradually stripped away until one finally starts to bump up against the limits of human knowledge.

Why can there not be more than one sigma bond in a set of bonds?

There can be, even in simple carbon compounds. Bent bonds, tau bonds or banana bonds; whatever you might like to call them were proposed by Linus Pauling; Erich Hückel proposed the alternative $\sigma - \pi$ bonding formalism. Hückel's description is the one commonly seen in introductory texts, but both methods produce equivalent descriptions of the electron distribution in a molecule.
In order to better understand the bent bond model let's first consider its application to cyclopropane and then move to ethylene.
In cyclopropane it has been found that significant electron density lies off the internuclear axis, rather than along the axis.

(image source)
Further, the $\ce{H-C-H}$ angle in cyclopropane has been measured and found to be 114 degrees.  From this, and using Coulson's Theorem $$\ce{1+\lambda^2cos(114)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\ce{sp^{2.46}}$ hybridized.  Now, using the equation $$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-C}^2}=1}$$ (which says that summing the "s" character in all bonds at a given carbon must total to 1) we find that $\ce{\lambda_{c-c}^2~=~}$$\mathrm{3.74}$, or the $\ce{C-C}$ bond is $\ce{sp^{3.74}}$ hybridized.
Pictorially, the bonds look as follows.  They are bent (hence the strain in cyclopropane) and concentrate their electron density off of the internuclear axis as experimentally observed.

We can apply these same concepts to the description of ethylene. Using the known $\ce{H-C-H}$ bond angle of 117 degrees, Coulson's theorem and assuming that we have one p-orbital on each carbon (Hückel's $\sigma - \pi$ formalism), we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bond are $\ce{sp^{2.2}}$ hybridized and the one carbon orbital involved in the $\ce{C-C}$ sigma bond is $\ce{sp^{1.7}}$ hybridized (plus there is the unhybridized p-orbital). This is the "$\ce{sp^2}$" description we see in most textbooks.
Alternately, if we only change our assumption of one pi bond and one sigma bond between the two carbon atoms to two equivalent sigma bonds (Pauling's bent bond formalism), we would find that the carbon orbitals involved in the $\ce{C-H}$ bond are again $\ce{sp^{2.2}}$ hybridized, but the two equivalent $\ce{C-C}$ sigma bonds are $\ce{sp^{4.3}}$ hybridized. We have constructed a two-membered ring cycloalkane analogue!

Hybridization is just a mathematical construct, another model to help us understand and describe molecular bonding.  As shown in the ethylene example above, we can mix carbon's atomic s- and p-orbitals in different ways to describe ethylene. However, the two different ways we mixed these orbitals

one s-orbital plus two p-orbitals plus one unmixed p-orbital (Hückel's  $\sigma - \pi$), or
two s-orbitals plus 2 p-orbitals (Pauling's bent bond)

must lead to equivalent electronic descriptions of ethylene. As noted in the Wikipedia article (first link at the beginning of this answer),

There is still some debate as to which of the two representations is
  better,[10] although both models are mathematically equivalent. In a
  1996 review, Kenneth B. Wiberg concluded that "although a conclusive
  statement cannot be made on the basis of the currently available
  information, it seems likely that we can continue to consider the σ/π
  and bent-bond descriptions of ethylene to be equivalent.2 Ian
  Fleming goes further in a 2010 textbook, noting that "the overall
  distribution of electrons [...] is exactly the same" in the two
  models.[11]

The bent bond model has the advantage of explaining both cyclopropane (and other strained molecules) as well as olefins.  The strain in cyclopropane and ethylene (e.g. heats of hydrogenation) also make intuitive sense with the bent bond model where the term "bent bond" conjures up an image of strain.
So, bent bonds are a mix of both sigma and pi properties and multiple bent bonds  can be used to describe the bonding between adjacent atoms.

The last statement (in bold) is technically incorrect, because a bent bond is by definition something uniquely different different from a sigma bond.

@Martin-マーチン  I've thought some more about your comment. I've always thought of a bent bond as just another sigma bond that lies along the continuum from pure s to pure p.  For me, the differentiating factor between sigma and pi bonds has been the number of areas of electron overlap. In in a sigma bond there is one such area, while there are two such areas in a pi bond.  What definition are you using that excludes a bent bond from being considered as a sigma bond?

And I thought about your comment, too. The separation in sigma, pi, ... and bent bond are just two completely different approaches to one problem. Hückel's approach preserves orbital symmetry, while tau bonds don't. I don't have a clear grasp about bonds one way or the other - both are simplifications-, that is why I prefer the delocalised picture - which does not necessarily give the correct answer either. So I can only rely on the definitions of the underlying description, which are different, hence sigma bonds and bent bonds cannot be the same. When it comes to bonds, it gets philosophical.

I would like to duscuss  a sort of not comfortable feeling that the depiction of ethylene in terms of bent bonds convey to me, at least at a first glance. It seems to me that the two equivalent bonds leave the molecule more susceptible of a total breaking, at least if I could imagine two addition reactions taking place at the some moment, or if simply matematically I deliver the necessary amount of energy to break apart the" two bonds of the double bond". This might be due to  my way of thinking of sigma bonds as skeletons along with distribution of more mobile and diffuse pi electrons occurs

Continuing :  in particular because this is a quite working pictures in my field, where extended pi-conjugated systems are the working horses. On the other hand, from a vibrational point of view, the bent bond picture is easily visualised as two springs having the same force constant, while in the pi-sigma treatment it is like one spring is there loose and waiting for the other one to break before itself starts start oscillate. As you see this is pictorial but I think it convey what I like to express (without doubting of the value and o of equivalence of both treatments).

@ron The C-C-C angle is 60 degrees. Aren't the hybridized orbitals you have pictured in the second image $\ce{sp^{3}}$? What is the right model? The hybridization is different than $\ce{sp^{3}}$ or the hybridization is  $\ce{sp^{3}}$ but the overlap is bad?

@Anton Yes, the C-C-C internuclear angle is 60° in cyclopropane, but the interorbital angle is larger (use the first equation with $\ce{\lambda^2=3.74}$ and you can solve for the interorbital angle). The $\ce{C-C}$ orbital hybridization is around $\ce{sp^{3.74}}$ as described above.  The overlap is less than ideal, hence the weak $\ce{C-C}$ bond in cyclopropane. Hope this helps.

I’m not sure if this answering attempt is correct in the light of Mithoron’s and ron’s comments on your question, but this is the way I learnt it, so if this is wrong I will at least learn something, too.
We all know what s-, p- and d-orbitals look like, but what is the significance, and why do these orbitals preferentially form $\sigma$, $\pi$ and $\delta$ bonds, respectively?
Mathematically spoken, orbitals are functions of the hydrogen atom that solve the Schrödinger equation. The model in question is a non-rigid rotor,* i.e. the rotor’s axis is not fixed in any spatial direction (the electron can rotate freely around the nucleus). For solving this equation, it is helpful to use polar coordinates $(r, \varphi, \theta)$, mainly because the solution can be split into a radial factor (dependent only on $r$) and angular factors (dependent on $\varphi$ and $\theta$).
$$\Psi (r, \varphi, \theta) = R(r) \cdot Y(\varphi, \theta)$$
$R(r)$ can be thought of giving an orbital its extension into space while $Y(\varphi, \theta)$ gives it its shape. Both functions depend heavily on quantum numbers: $R(r)$ does so for $n$ and $l$ while $Y(\varphi, \theta)$ depends on $l$ and $m_l$. For the simplest case ($l = 0; m_l = 0$, s-orbital), $Y (\varphi, \theta)$ degenerates to a simple constant, meaning that the orbital will have a totally symmetrical spherical shape. $l = 1$, (the p-orbital) while loosing spherical symmetry, still keeps total symmetry with respect to one axis, i.e. every slice you take through that orbital perpendicular to the axis of symmetry will be a circle. Higher quantum numbers lose more symmetry but it’s not always as easily visualised, so I’ll stick with these.
But you were talking about bonds, where do they come into play? Well, bonds also have a symmetry, but they also have an axis instead of a nucleus, so their symmetry will be reduced per se. The simplest symmetry along a bond axis is total rotational symmetry around the bonds axis. I hope you see the similarity between the s-orbital (total symmetry around a central point) and a  $\sigma$ bond (total symmetry around the bond’s central axis). Similarly, a $\pi$ bond will always have one degree of symmetry less, which turns out to mean ‘having a plane of symmetry that includes the bond axis’. And a $\delta$ bond will have two planes of symmetry — yet another degree of symmetry less.
According to this definition, an orbital that can take part in a $\sigma$ bond needs to have full rotational symmetry along the bond’s axis. That means, that there is only one, at most two orbitals that fulfil the criterion (but if there are two, one is going to be an unmodified s-orbital and likely not take part in bonding at all). Therefore, only one $\sigma$ bond would be possible between two atoms.

Writing this up, I remembered the ‘banana bonds’ that were introduced to us to explain the extremely small ($60°$) bond angles in $\ce{P4}$. I would need to go back, recheck and rethink what I would think of those and if I would treat them as exceptions of this ‘rule’ or simply as special cases that need additional information to be discussed. They certainly deserve consideration, as they are, de facto $\sigma$ bonds from the way they look, even though they do bend.

An interesting comment was left on the question pointing to sextuple bonds. I didn’t know bonds of that order existed; my knowledge was stuck at 4. For a quadruple bond, possible between certain transition metals such as in $\ce{[Re2Cl8]^2-}$, four of the five d-orbitals form a bond to the other metal; one being $\sigma$, two being $\pi$ and a fourth one of $\delta$ type (to planes of symmetry). Extending that to a quintuple bond by adding a second $\delta$ layer with the last remaining pair of d-orbitals isn’t hard.
The sextuple bond – eg $\ce{Mo2}$ — derives from an additional $\sigma$ bond between the s-orbitals of the higher shell. You thereby solve a problem you would otherwise have: The $4\mathrm{d}_{z^2}$ orbital can take part in $\sigma$ bonding along the $z$-axis; and the higher $5\mathrm{s}$ orbital is more diffuse, extends further into space and therefore is still able to form a contact to the neighbouring atom’s counterpart. Because it is more or less a sphere, it can only form $\sigma$ bonds.

* I don’t think this is the model’s correct name. In my German quantum chemistry class, the rigid rotor was a raumstarrer Rotator and thus the model here was a raumfreier Rotator. Somebody who might know the proper name please comment (or edit).

As for the name, what I was taught in QM was simply "3D rigid rotor".

@orthocresol So that would mean that ‘rigid’ does not refer to an axis of rotation being ‘rigid’ with respect the the surrounding space, because otherwise ‘3D rigid’ doesn’t make sense at all?

I think the definition of "rigid" is that $r$ is a constant. Which doesn't seem to make sense either...

Maybe it's because the potential in an atom is purely a function of $r$, so the angular wavefunctions are similar in form to the solutions of the rigid rotor model (where $r$ is fixed). I am actually rather out of my depth here so maybe someone else can come along and answer. BTW Atkins calls it a "particle on a sphere" - these give rise to the spherical harmonics $Y(\theta,\phi)$ which are the angular part of the wavefunctions.

Short answer
The main thing that determines the shape of orbitals is that they must have zero net overlap with all other orbitals (i.e. that they are orthogonal). It's quite hard to construct two orthogonal $\sigma$-bonding MOs connecting the same two atoms.
More details
$\sigma$-bonding MOs are typically $sp^x$ hybridized orbitals with maximum overlap along the line connecting two atomic centers

It's hard to construct another bonding MO with maximum overlap along the same line that is orthogonal to this MO. To stay orthogonal we have introduce a node and the next $\sigma$ MO becomes anti-bonding

The usual way to get another, orthogonal, bonding MO is a $\pi$-bond.

If you have valence $d$-electrons available then it is possible to construct another, orthogonal, $\sigma$-orbital

However, it is still rare to have two $\sigma$ bonding MOs because transition metals tend to loose the $s$-electrons needed for one of the $\sigma$-MOs very readily. So double $\sigma$-bonding tend to be only observed in special cases such as gas phase $\ce{Mo2}$.

It is why to my modest opinion a po-sigma is more straightforward as compared to tau approach in case of ethylene. However it is nice to remember that it is matter of what orbitals one combines.


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Entropy change when liquids at two different temperatures are mixed

Two quantities of water ($M_\mathrm{r}=18$) each of mass $216\ \mathrm{g}$ are mixed together in a
  vessel. The temperatures of the two quantities before mixing are $303\ \mathrm{K}$ and $333\ \mathrm{K}$. The entire system is perfectly insulated and the vessel has a negligible heat capacity. Calculate the change in entropy after equilibrium has been established, and comment on your result in the light of the Second Law of Thermodynamics.
  Note that $C_{\mathrm{m},p} = 75.5\ \mathrm{J\ K^{−1}\ mol^{−1}}$ for liquid water in this temperature range.

Is it correct to assume that the temperature of the water at equilibrium will be half way between the two values (i.e. $318\ \mathrm{K}$)? If so, I think the rest of the problem can be solved by using $\Delta S=\frac{C_V}T\Delta T$. Is this correct or is it more complicated?

I would say the equation is $dS = \frac{C_v }{T}dT$, and thus you'd better integrate it: $\Delta S=C_v \Delta (\ln T)$

Good point. Is the final temperature 318K?

Yeah, it would be. You don't need to assume it, and you shouldn't, because it's easily proven. Since the system is isolated, the heat gained by the water at 303 K must be equal to the heat lost by the water at 333 K, and since you are mixing equal amounts of the same substance, you can use $q = mc\Delta T$ to show that $T_f = 318$ K.

Also, your equation should read $C_p$, not $C_v$, because your process is being carried out at constant pressure: $dq = C_p,dT$ so $dS = dq/T = C_p, dT/T$.

@RobChem, not sure where we are on this question, but if the comments so far have enabled you to solve it, would you be able to write up an answer?  I'd upvote it if you did.


text:
Law of chemical equivalence
In this reaction,
$$\ce{KIO_3 + KI -> I_2}$$
According to law of chemical equivalence, shouldn't $m_{eq}$ of $\ce{KIO_3}$ = $m_{eq}$ of $\ce{KI}$ = $m_{eq}$ of $\ce{I_2}$?
However, the book that I was reading from, said:
Total $m_{eq}$. of $\ce{I_2}$ = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO_3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$).
Why is it so?

This Reaction is a Disproportionation Reaction, that is, a reaction involving both oxidation and reduction of a single species, in this case : Iodine. This is not one reaction but a combination of two reactions:
$$\ce{KIO3 -> I2}$$
And
$$\ce{KI -> I2}$$
Therefore, by applying law of chemical equivalence in both reactions and adding the equations :
$m_{eq}$ of $\ce{I_2}$ (from both reactions) = $m_{eq}$ of $\ce{I_2}$ (from $\ce{KIO3}$) + $m_{eq}$ of $\ce{I_2}$ (from $\ce{KI}$)

Welcome to chem.SE! You might also write the reactions completely to avoid future confusions from the readers of this answer.

Technically it is a comproportionation (two different oxidation states going to the same oxidation state), not a disproportionation (the same oxidation state going to different oxidation states, such as $\ce{Cl2 -> Cl- + OCl-}$. I think it's a trivial matter though.


text:
Electroless Plating of Zinc onto Copper in NaOH Solution
I’m researching how the classic “alchemy” demonstration works in which zinc is plated onto a copper penny in a solution of sodium hydroxide. Most online resources refer to the electroplating of copper with zinc, which is not what I’m interested in. The best article I’ve been able to find on the electroless plating of zinc onto copper is from the Royal Chemistry Society. However, I have several questions. The three reactions this article describes are:
$\ce{Zn(s) + 2NaOH(aq) + 2H2O(l) -> Na2Zn(OH)4(aq) + H2(g)}$, which is the “formation of sodium zincate”
$\ce{Zn(s) -> Zn^2+(aq) + 2e- }$, which is the “electrode reaction” at the “zinc electrode” and is “followed by the complexing of the zinc ions as $\ce{Zn(OH)4^2- (aq)}$”
$\ce{Zn(OH)4^2- (aq) + 2e- -> Zn(s) + 4OH- (aq)}$, which is the “electrode reaction” at the “copper electrode”
I really don’t understand how these three reactions come together to describe the plating of the zinc onto the copper. Specifically, I don’t understand:
1) What is the role of zincate, $\ce{Zn(OH)4^2- (aq)}$, in the reaction?
2) On the molecular level, how are the zinc and copper actually attached together?
3) How is this an electrochemical cell (i.e. how is electricity involved)? Where can I learn more about electrochemical cells?

An explanation is provided in this reference. The idea is that the electrode reactions are
$\ce{Zn(s) + 4OH^- -> [Zn(OH)_4]^{2-} + 2e-}$
and
$\ce{[Zn(OH)_4]^{2-} + 2e- -> Zn(alloy) + 4OH^-}$
So the zinc powder remaining will go into solution and the zincate is reduced on the copper surface to form a form of zinc rich brass. Brass is an alloy and the atoms are held together via metallic bonding.


text:
Will selenium react or bond, ionic or covalent, with vanadium?
I was recently asked if $\ce{Se}$ will bond with $\ce{V}$, creating a compound $\ce{SeV}$ . I am not sure if this is even possible but if so I'd like to find out.

Afaik, selenides are quite hard to prepare and require smelting of the elements or use of transport reaction.

http://www.sciencedirect.com/science/article/pii/0022190266800088

The ASM Alloy Phase Diagram Database (at asminternational.org if you have access) shows that $V_{2}Se_{9}$, $V_{5}Se_{8}$, $V_{3}Se_{4}$ $V_{5}Se_{4}$, and $VSe$ are all accepted stable phases in the Se-V binary system.

Brilliant. Thank you very much. This has helped me out a lot.  @joncluster