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Is azide paramagnetic?
Is the azide ion ($\ce{N3-}$) paramagnetic?
How do I use molecular orbital theory to show that it is or isn't?
I drew a sine wave diagram and filled up the $1\pi$ and $2\pi$ MOs, and since there are no unpaired electrons, can I conclude that the ion is not paramagnetic?

Well, it isn't  paramagnetic and even drawing few mesomeric structures looks enough for me

@Mithoron Drawing mesomeric structures would not help, since Lewis structures do not contain information about spin. That's why we conveniently write $\ce{O=O}$ even though we know it is a paramagnetic triplett.

Since $\ce{N3^-}$ is isoelectronic to $\ce{CO2}$ you might find some answers in this related question. I am not absolutely certain if we have the same idea about the sine wave diagram (could you post a picture), but I assume you are on the right track. Your conclusion is sound.

Deriving a proper MO digram for N$_3^-$ is not completely trivial a task (e.g. a random google hit, which seems to me well done: www.chem.uci.edu/~lawm/Azide%20MO%20Diagram.pdf). In the end you will have to show that the degeneracies of the orbitals together with the electron count will not require open shells. Beware of Lewis-structures and mesomery, think of O$_2$, it looks perfectly o.k. to draw O=O, only the MOs and their degeneracies will reveal that its paramagnetic.


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Water exchange at Al (III)
Why is the rate of water exchange at Al(III) centres so slow ?
According to this medical book (p.5) it is $10^{5}$ times faster at Mg(II).

I actually tend to agree with the answer from @JanJensen based on electrostatic effects, rather than with the "covalency" argument.
In particular, it has been mentioned in the comments that the $\ce{Al^{III}-OH2}$ is a "far more covalent" bond than one $\ce{Mg^{II}-OH2}$ one. This remark, however, does not actually explain anything, but rather it shifts the question to a new one, i.e. why is it, then, that the $\ce{Al-O}$ bond is more covalent?
Water exchange reactions at the Mg and Al centers have actually been studied extensively, but unfortunately not to the same extent as those for transition metal ions, or for selected lanthanides. The latter in particular are a subject of continuous and very intense interest, because water exchange at e.g. $\ce{Gd^{III}}$ or $\ce{Eu^{II}}$ ions play a pivotal role in the effectiveness of complexes of these ions as MRI contrast agents.
Nonetheless, here are a few very relevant references that will certainly offer some more insight, but probably not the final answer that you are looking for.

An $\ce{^{17}O}$-NMR Study of the Exchange of Water on $\ce{AlOH(H2O)5^{2+}(aq)}$, Inorganic Chemistry, 1998, 37 (19), 4760–4763;
Water exchange on metal ions: experiments and simulations, Coordination Chemistry Reviews, 1999, 187 (1), 151–181.

In fact, when studying exchange reactions at metal ions, the authors of these and other studies tend to consider ions with equal charge as homogeneous series, because the effect of the charge will swamp out any other effect, some times even the relatively large effects of electronic configuration in d-transition metals!
That notwithstanding, the papers above agree that the mechanism for the water exchange reactions for both Al and Mg metal complexes is dissociative in nature, which means that the rate determining step involves the loss of a $\ce{H2O}$ ligand from the inner sphere of the complex.
We can combine this piece of information with the fact that, as mentioned in the second paper above, there is a significant increase in charge density when going from $\ce{Mg^2+}$ to $\ce{Al^3+}$. First, the aluminum cation is smaller (reported values from that reference are: 72 pm for $\ce{Mg^2+}$ and 53 pm for $\ce{Al^3+}$); secondly, it has higher charge. This rather staggering difference can again be explained using the usual electrostatic arguments (i.e. the two ions are isoelectronic, but the Al nucleus contains more protons, so electrons in the latter are held more closely, leading to smaller cations).
Trying to put this all together:

removing a water molecule determines the rate of the overall exchange process according to the experimentally determined dissociative mechanism;
the smaller, more highly charged $\ce{Al^3+}$ exerts a much stronger attraction on each of its bound $\ce{H2O}$ ligands, making said removal much more costly;
in macroscopic terms, the $\ce{Al^3+}$ has a much higher activation energy, which ultimately manifests itself in the much lower rates of exchange that the OP referred to.

$\ce{Al(III)}$ has a +3 charge while $\ce{Mg(II)}$ has a +2 charge.  So the ion-dipole interaction that binds water to the ion is stronger for $\ce{Al(III)}$.
Based on transition state theory the $10^5$ increase in water exchange rate translates to a difference in activation free energy of about 30 kJ/mol. The difference in interaction energy between a 2 Debye dipole that is 3 Å from a +3 and +2 charge is about 65 kJ/mol (assuming the dipole is pointed directly at the charge).

Water is bound covalently!

This doesn't answer the question at the level of detail I need; obviously you would expect water exchange to be slower for Al(III) but the order of the magnitude of the change does not seem to be obvious from charge density arguments.

Actually, Mithoron's comment seems rather relevant: Al-water bond is far more covalent.

I agree with Greg and I encourage @Mithoron to write up an answer.

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Calculation of optimal current density
I'm trying to figure out what my optimal current density would be in respect to the limiting current density.
How do you calculate concentration polarization and limiting current density in an electrolytic cell. There are two copper electrodes and a $2\:\mathrm{M}$ copper nitrate solution. The average current density was $0.375\:\mathrm{A\:cm^{-2}}$.
I know the Tafel expression ($\eta a=a+b \log i$) can calculate the voltage drop but can I find the limiting current density just like that in the image?

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Is helium-4 a liquid at zero temperature and zero pressure?
As far as I know, for all substances other than helium, if you extrapolate the solid-gas line on the phase diagram, it passes through the origin. That is, no matter how low the temperature is, you can still vaporize the substance by lowering the pressure, and no matter how low the pressure is, you can still freeze the substance by lowering the temperature.
This makes sense to me on thermodynamic grounds. If we keep lowering the pressure at a given temperature, the entropy of vaporization increases without bound (although only logarithmically) and at some point the gas phase is favoured, whereas if we keep lowering the temperature at a given pressure, then the entropy term in free energy becomes negligible and the solid phase is favoured by enthalpy.
The phase diagrams I've found online for helium-4, however, don't follow this rule. They seem to show that helium-4 is exclusively a liquid in a neighbourhood of the origin. Example here: http://ltl.tkk.fi/research/theory/helium.html
(Edit: The linked web page seems to be unavailable. A cached version of the referenced image is available here.)
I wonder whether this is true, and if so, how it can be possible. How can it be that no matter how low the pressure is, helium-4 will not boil? I understand that superfluidity is a quantum phenomenon but I can't see how my previous reasoning could not apply.
Edit since some people think this is a duplicate: Again, my question is whether helium-4 will not boil at zero pressure, and if so, how that can be; it concerns the liquid-gas transition and not the solid-liquid one.

possible duplicate of Why can't helium be solidified at 'ordinary' pressures?

@santiago I don't think this is a duplicate. My question asks why helium-4 will not boil at zero pressure, whereas every other liquid apparently will. Not why it will not freeze.

Interesting question and preamble. I'd never considered the content in your second paragraph.

https://en.wikipedia.org/wiki/Lambda_point

It will boil all right.

You wrote:  "no matter how low the temperature is, you can still vaporize the substance by lowering the pressure, and no matter how low the pressure is, you can still freeze the substance by lowering the temperature."  These are contradictory.  The former says that, even at 0K, you can vaporize it by lowering p (=> phase at 0K & 0 bar is gas), while the latter says that, even at 0 bar, you can freeze it by lowering T (=> phase at 0K & 0 bar is solid).

@theorist It's only a contradiction if you assume that zero temperature and zero pressure are achievable, which of course they are not.

Except that you're asking specifically about what would happen at 0K and 0 bar. Even though those conditions aren't physically achievable, we can make physical statements about them, and the statements you're making are contradictory: In the first case you're saying that, in the limit of 0 bars, everything is a gas, even in the limit of 0K. And in the second case you're saying that, in the limit of 0K, everything is a solid, even in the limit of 0 bars.

@theorist No, what I'm saying is this: for every T > 0, there exists a P > 0 that is sufficiently low that you can vaporize the substance; and for every P > 0, there exists a T > 0 that is sufficiently low that you can freeze the substance. But helium-4 seems to violate this rule.

Your presentation is confusing, because in your comment you say you're referring to finite temperatures and pressures, but in your title you are referring to zero temperature and pressure.  And when your title refers to zero T and p, the most reasonable interpretation of, e.g., "no matter how low the temperature is" would be "in the limit of 0K".

See He vapor  pressure calculator for T=0.65 K ..  5.1 K.

To make a gas, He atoms must have translational kinetic energy, what they would not have at T 0K. Additionally, you cannot make any gas at p=0, as then p<>0.

It could be that the liquid-vapor line does not exactly hit zero pressure at any temperature above absolute zero, the pressure just becomes too small to distinguish on the graph. You might see the same thing, for instance, with water if you tried to plot the solid/vapor equilibrium line all the way down from the triple point to 100 Kelvins using a reasonable plotting scale. The zero pressure line, if we could achieve it exactly, would then be in vapor phase at all positive temperatures.

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Why does the value of Δo increase with increased charge on the metal?
A greater positive charge means the ligand is strongly attracted and pulled closer, and thus it creates a stronger crystal field. The potential energy value of the barycentre should thus be higher, since repulsion experienced by the d electrons is stronger. Why is though the value of the Δo increased - the potential energy difference between the various differently oriented d orbitals? Each orbital experiences stronger repulsion if charge on metal ion increases and the ligand is pulled closer, but how can anything be said of the energy difference Δo? Both t2g and eg are at a higher level, that's all.


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Why does HCN boil at a higher temperature than NH3?
The boiling point of ammonia is −33 °C while that of $\ce{HCN}$ is 25 °C. In a recent AP (Advanced Placement) Chemistry test, a free response question asked why this is the case. Can someone shine light on this?
Based on Jan's answer:

Although the $\ce{C-H}$ bond does not usually exhibit good hydrogen bond, $\ce{H-CN}$ is a special case in which the bond is polar enough to provide better hydrogen bonding than $\ce{NH3}$. I agree with Jan on this point.
However, Jan's explanation is for the polarity of the $\ce{H-CN}$ bond is based on $\mathrm pK_\mathrm a$ (lower $\mathrm pK_\mathrm a$ suggests higher acidity and good hydrogen bonding). As one user pointed out, $\mathrm pK_\mathrm a$ is only defined in aqueous solution, so stating pK$_a$ as an evidence is not suitable for the system I have which is a pure $\ce{HCN}$ liquid. In such system, $\ce{HCN}$ does not significantly dissociate. A suggestion was made on using proton affinity as the evidence for the acidity of $\ce{HCN}$.

I perceive that the answers (reasons and evidence) are mostly targeted at the chemical process of dissociation of $\ce{HCN}$. Should this question be addressed more in terms of a physical process? I mean there is little dissociation of pure-liquid $\ce{HCN}$.
What I have been thinking about is a MO picture in which the cyano group withdraws electron density from the hydrogen atom, in a similar way to how acetate group withdraws electron from the hydrogen in acetic acid. What is your thought on this?

Your MO attempt is certainly not wrong, but it would only justify a low pKa, or gas phase proton affinity, respectively. It's for one molecule only. You can extend the description to molecular clusters, but treating them with a MO only scheme will certainly give you the wrong answer.

You made a good point about one molecule. I didn't realize this sooner. However, could you tell me what you think about Jan's explanation?

I think it raises an interesting point, but I would not consider it complete. I guess in liquid phase, there are a couple of more equilibria a play in HCN compared to NH3. I really don't like the premise of the whole question, the molecules are not really similar.

The enthalpy of vaporization of $\ce{HCN}$ is higher than for $\ce{NH3}$, which suggests that $\ce{HCN}$ molecules interact more strongly than $\ce{NH3}$ molecules. $\ce{C-H}$ bonds are not usually considered good hydrogen bond donors, but $\ce{HCN}$ is unusual. For example $\ce{HCN}$ has a $\mathrm pK_\mathrm a$ value of 9.2, indicating that the $\ce{CN}$ group is electron withdrawing and that it is a reasonably good hydrogen (bond) donor. This is likely due to the electronegativity of nitrogen, and also the high "s-content" of the sp-hybridized $\ce{CH}$ bond, which keeps the electron pair close to the nucleus.

Thanks for the answer. I think you're correct that C-H of $\ce{HCN}$ is a good hydrogen bond donor. As for the reason, I'd like to think that because the electron withdrawing ability of $\ce{CN-}$ is high, the hydrogen is electronically deficient, leading to an excellent electron acceptor. I would prefer your explanation if we're speaking of a reaction (where $\ce{CN-}$ is formed)

There is a reasonably good and generally accepted (negative) correlation between hydrogen bond strength and pK$_a$ of the hydrogen donor

You're correct on the correlation between pKa and the strength of hydrogen bonding. I, however, think the polarity of $\ce{HCN}$ should be explained by MO theory and unlikely by the stabilization of the conjugate base which doesn't get produced anyway. Your argument is based on thermodynamic stabilization of a dissociation reaction, which occurs in the presence of a base. Note that there is purely $\ce{HCN}$ in the container here.

@Huy If you wanted to have an explanation in terms of MO theory, you probably better include these points (and your thoughts and reasoning) into your question.

Please don't use MathJax like NH$_3$, use the mhchem package instead: $\ce{NH3}$ It might look okay for you, but it might break in other browsers, especially at line breaks.  If you want to know more about mhchem, please have a look here and here. Please do not use markup in the title field, see here for details.

Supporting Jan's answer, actually: Consider the points below from the book Hydrogen Bonding: A Theoretical Perspective, p. 102.
"Moreover, the hydrogen in HCN is acidic enough that the molecule may act as an effective proton donor... When paired with NH$_3$, HCN acts as a proton donor..."
This agrees with Jan's note, on the pK$_a$ of HCN.
Your comment that "the hydrogen is electronically deficient, leading to an excellent electron acceptor" seems to be about hydrogen bonding, whereas there is evidence that HCN is more ionic in nature.

Besides the existing answers, which focus on the acidity of HCN, note that HCN is also a considerably larger molecule than NH3.  Thus, even if the interactions between the molecules were qualitatively identical, one would still expect a higher boiling point for HCN on the basis of the size difference (and resulting stronger dispersion interactions; see comments) alone.
For an illustrative example, we can look at methylamine, CH3NH2, which resembles ammonia in most respects, except for having one of its hydrogens replaced by a bulky methyl group, making it similar in size to HCN.  Its boiling point is −6.6 °C, well above the −33 °C for NH3.
The remaining ~32 K difference between the boiling points of HCN and CH3NH2 is then presumably explained by the stronger acidity of the HCN hydrogen, and thus the stronger hydrogen bonding between HCN molecules than for NH3 and CH3NH2.

I hear this molar mass argument a lot, and I've used it myself plenty, but I have some doubts about it. Is there some formula which shows how mass can factor in? How can we separate the mass contribution from something like molecular surface area?

@NicolauSakerNeto This is just a spontaneous thought by maybe by comparing molecules like n-pentane to neopentane? (26.5 K difference)

@Jan Indeed, those have the same mass, but rather different boiling points. It's a good example, though it doesn't approach my question from the angle I had in mind, so forgive me for not being clear. The reason I'm a bit wary of the mass argument is this: how do we guarantee that, when comparing two molecules with different masses, the change in boiling point is not dominated by the accompanying change in surface area for interactions, rather than the change in mass? Is there a pair of molecules we could compare with different masses, but same polarity and same surface area?

@NicolauSakerNeto Then let me suggest diethyl ether and d10-diethyl ether, same difference in mass as between ammonia and HCN. The other properties, especially polarity should be very similar.

@Jan That's a good idea. Do you have a reliable boiling point figure for the deuterated diethyl ether? From what I've found on a quick search, it's indistinguishable from regular diethyl ether.

@NicolauSakerNeto Unfortunately not =C But it should be measureable.

@Nicolau: Thanks for the feedback. I've edited my answer to hopefully make it clearer that, while there's a general trend for larger molecules to have higher boiling points, all else being (approximately) equal, it's not just (and perhaps not even primarily) a function of mass, but that (as Jan's neopentane example shows) surface area and geometry also play a role.

@Jan After a lot of thinking, I don't know why I didn't realize this sooner: the best case to compare isotopologue boiling point changes is dihydrogen and dideuterium! The latter is fully twice the molecular mass of the former, yet according to Air Liquide, at ambient pressure dihydrogen boils at $\mathrm{-252.78°C}$, while dideuterium boils at $\mathrm{-249.83°C}$, which is a very small difference. This suggests mass plays a limited role.

Furthermore, if I'm reading the article "Vapor pressure of the methanes" (1955) correctly, liquid $\ce{CD4}$ actually has a higher vapor pressure than liquid $\ce{CH4}$, which means the perdeuterated isotopologue seems to have a lower boiling point!

@NicolauSakerNeto And thinking about it more, the mass argument at school was only a proton argument, because more protons means more electrons (in a neutral molecule) which in turn means better polariseability and stronger van der Waals forces (or London interactions, as I’m told they’re called outside of Germany).

Just to clarify: gravitational attraction between molecules is much too small to have any measurable effect on the boiling point or any other chemical property.  The correlation between molecular mass and boiling point is primarily due to an increase in dispersion interactions (the attractive part of the van der Waals or London forces). This force exist between each electron pair, so more electrons = more dispersion.  The effect of isotopic substitution on intermolecular interactions comes mainly from change in vibrational frequencies. The sign is hard predict a priori.

@JanJensen: Naïvely, I would've expected mass to also contribute via the molecular velocity distribution (bigger molecules move slower, and so, intuitively, should have a harder time escaping the solution). But of course, thinking about it a bit more, the kinetic energy distribution is what really matters here, and that's independent of the molecular mass. Thanks, I learned something (or, at least, unlearned a wrong assumption) here.

Although $\ce{NH3}$ consist of hydrogen bonding $\ce{HCN}$ has a very stronger dipole-dipole interaction which makes its boiling point equivalent to alcohols.
See the diagram:

"Acidity" can't be responsible, since acidity refers to what happens in water solution. (I assume the user is referring to the $\mathrm{p}K_\mathrm{a}$ data.) I don't know how great the tendency of $\ce{HCN}$ is to ionize in the pure liquid, but I doubt it's significant. But if you have data, please cite it. Even for water, $K_\mathrm{w} = 10^{-14}$, its tendency to ionize does not contribute to its high boiling point.

Please think of a final answer. Include this answer. So far this is more like a comment. But thank you for saying what I wanted to say but no one tries to argue me wrong

Though $pK_a$ is usually a measure of acidity in aqueous solution, there are measures which are independent of solvent, such as proton affinity. In that table, more positive means more exothermic, which means a stronger base and thus a weaker conjugate acid. Look at a few anions formed from $\ce{C-H}$ ionization, including cyanide, cyclopentadienide and trichloromethanide; cyanide tops them. Note also that the proton affinity of cyanide is almost the same as for phenlote. That is already remarkably acidic for a $\ce{C-H}$ bond

Although Jan's explanation is based on a popularly accepted correlation between acidity and C-H bond's polarity, this is NOT a causation. IMHO, the causation would be the MO picture in which the cyano group withdraws the electron cloud from the hydrogen, causing the bond to be polar and the hydrogen bond to be stronger. If you think I am wrong, please contribute to my knowledge. Otherwise, I will not learn anything from asking this question.

@Huy If you are only stating your knowledge in the comments, I can understand why nobody takes the time to "argue" with you.

I think the crux of the question is not in hydrogen bonding. Recently, I did a question in the USA Chemistry Olympiad (Local Section) on boiling point of substances. I discovered that self-ionisation played a key role in determining boiling point and melting point. The answer to the question I attempted asking for the substance with the highest boiling point was pure sulfuric acid, which is apparently capable of self-ionisation to a large extent.
The reason why self-ionisation is able to raise the boiling point of the substance is because of the ionic interactions in the liquid of the resultant ions produced by the process. This strong ionic interactions increases the boiling point of the liquid, like in the case of sulfuric acid.
The same concept can be applied here. The hydrogen cyanide molecule being capable of self-ionisation, although not to such a significant extent, can produce ions in the liquid phase. This ionic interaction, although it may be very litte compared to sulfuric acid, does increase the boiling point of the substance.
In comparison, ammonia's self-ionisation is probably negligible and thus, the strongest interactions are only hydrogen bonding.
This post is a bit late but I hope this provides a new perspective.


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Does the term 'silver nitrate' express all of its constituent chemical elements?
My professor gave my class a set of practice questions for the mid-term. One of those questions asks for the molar mass of a number of chemicals, such as lithium bromide and silver nitrate. We haven't studied those chemicals, so there's little reason for the professor to expect us to have memorized the elements that constitute those chemicals. In the case of lithium bromide, the term informs us of its constitution. Whereas,'silver nitrate' informs us that the chemical $ \mathrm{AgNO_3}$, includes silver and nitrogen, but I don't see anything in the term that would inform the reader that the chemical includes $ \mathrm{O_3}$.
Is there anything in the term 'silver nitrate' that would inform a reader that the chemical includes $ \mathrm{O_3}$?
More generally, is there a set of rules that relates the chemical constituents of chemical substances to the linguistic constituents of the terms that designate them?

Nitrate ions are $NO_3^{-}.$
Nitrite ions are $NO_2^{-}-.$

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch2/names.html#poly

Chemists will create nomenclature wherever necessary to avoid ambiguity, unfortunately sometimes to the point that systematic IUPAC names can get very unwieldy. But there's nothing we can do about it, chemical space is enormous and extremely varied.

You should know that nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$. For example:
Sulfate is a polyatomic ion with the molecular formula $\ce{SO4^{2-}}$;
Sulfite is a polyatomic ion with the molecular formula $\ce{SO3^{2-}}$;
Nitrate is a polyatomic ion with the molecular formula $\ce{NO3^-}$;
 Nitrite is a polyatomic ion with the molecular formula $\ce{NO2^{-}}$
At first glance, the nomenclature of the polyatomic negative ions seems difficult.
When the name of the ion  ends in either -ite or -ate:

The -ite ending indicates a low oxidation state.
The -ate ending indicates a high oxidation state.

I agree with you that at the beginning, it seems difficult. But by practicing, you find this quite easy.

For $\ce{AgNO3}$, as Gokul said in the comment, ‘nitrate’ stands for $\ce{NO3-}$. Sodium nitrate is $\ce{NaNO3}$, ammonium nitrate is $\ce{NH4NO3}$, and so on. It is important to know that $\ce{AgNO3}$ is an ionic compound with ionic bond between $\ce{Ag+}$ and $\ce{NO3-}$, while nitrogen and oxygen in $\ce{NO3-}$ are bonded with covalent bond.
There are conventions for names of compounds. Some convention such as IUPAC name for organic molecule is strict and not so confusing. For ionic compound, there are rules but there are also irregularities. In this case for anions with oxygen, here are some examples:
nitrite $\ce{NO2-}$, nitrate: $\ce{NO3-}$
phosphite $\ce{PO3^3-}$, phosphate: $\ce{PO4^3-}$
sulfite $\ce{SO3^2-}$, sulfate: $\ce{SO4^2-}$
Roughly speaking, ion with the highest oxidation state of central atom got -ate name, while the second highest got -ite.

However, if central atom is in group VII, the naming convention strangely change to:
chlorite $\ce{ClO2-}$, chlorate $\ce{ClO3-}$, perchlorate $\ce{ClO4-}$
bromite $\ce{BrO2-}$, bromate $\ce{BrO3-}$, perbromate $\ce{BrO4-}$

There are also confusing things such as nitride ($\ce{N-}$), nitrile ($\ce{-CN}$), and nitrite ($\ce{NO2-}$).
In practical, it is easier to memorize the name of common ion and search the internet for the name of uncommon ion, rather than messing with confusing rules. I do this all the time when I do chemistry researches. However, this tactics might not work if you have to take a chemistry exam.

A small correction: the nitride ion is a trianion, $\ce{N^{3-}}$, which itself is not to be confused with the azide ion,  $\ce{N_3^{-}}$


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How would one construct a 3D model of a molecule based on the crystal structure of its salts?
Often crystal structures found online from past published crystallography experiments are of the salt of a molecule of biologic interest, as opposed to the freebase form of the molecule itself. For example, I have a CIF file for mescaline hydrobromide (CCDC 1211316) at the moment that I would like to create a 3D model of mescaline from. Would removing the extra hydrogen atom on the amine group (i.e., changing the $\ce{NH_{3}^+}$ to $\ce{NH_2}$) and the free bromide ion suffice or will I also have to optimize the structure's geometry? If I do need to optimize it, or otherwise modify it beyond the procedure I just pointed out, please tell me how to do it with free software (Avogadro, for instance).

In general, the crystal structure of a molecule is a highly accurate geometry. So yes, you can take a salt and modify as needed. (The caveat in the comment is useful - keep in mind that different groups may be protonated or deprotonated depending on solvent and pH.)
There are two other caveats:
1) The conformation in the crystal structure is not necessarily the lowest energy conformer in solution or the gas phase. You may wish to perform a conformer search to find a low-energy minima.
2) If you are performing further calculations (e.g., DFT for electronic structure, force fields for MD, etc.) the geometry of the crystal structure will not likely be a minima of the computational method, so geometry optimization or equilibration (for MD) is needed before further simulation.
That said, the crystal geometry is a good place to start for many uses. Certainly if a crystal structure is available, I'd suggest you use it rather than build a geometry by hand.

How would I do a conformer search for a low-energy minima & those further calculations in Avogadro? I'm a newbie when it comes to Avogadro, so step-by-step instructions in Avogadro would be much appreciated.

http://avogadro.cc/wiki/Extensions_Menu - There's Extensions > Molecular Mechanics > Conformer Search

So which type of conformer search should I use? Systematic rotor, random rotor, weighted rotor or genetic algorithm? What parameters should I use? Like I realize some of these parameters will be up to myself to decide but which ones would likely give the best results? How do I do DFT for electronic structure and MD? It might help if you could give their full names too so I might look them up for myself. Thanks for your time, though, I do appreciate it.


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Choosing fish-oil that won't oxidize in long travel?
Fish oil, is susposed   to suffer from rancidity(i.e. oxidation) when stored improperly , for example in hot temperature.
I'm interested in buying fish oil from abroad(usa) it will be shipped by air and won't be kept in a cold chain,i live in a mediternean nation(climate wise).
How should i choose my fish oil , such that when i recieve it ,it will be OK and not rancid ? is it even possible ?

Encapsulated fish oil in softgels inhibits oxidation so that storage on the shelf (even in warm climates) is possible. Some companies will ship it on ice  if you request it (it may add $50 to the order).

The recent fish oil capsules that i've bought , locally ,are rancid(fishy smell even without opening the capsule) - and they use gelatin capsule even include topotocrels (which are anti-oxidants). How come?

Unfortunately, the gelatin shells are not impervious. I know that if they are heated, the oil inside will leak out. It might be that some of the oil leaked out and oxidized, that is causing the smell (it doesn't take a lot). Also the amount of tocopherols and tocotrienols added is small (~50 mg in to  a gram of oil).

Ah ok. reading more about it - seems that they don't work well below 15C or above 30C. http://www.drugs.com/cdi/cyclosporine-soft-gelatin-capsules.html

First off, a "fishy" smell and rancidity are two different things.  A fishy smell is likely due to impurities, primarily small volatile amines such as trimethylamine.  These impurities are unrelated to oxidative degradation of the fish oil.
Rancidity in fish oil is the result of oxidation.  Oxidation leads to the formation of lipid hydroperoxides and short-chain aldehydes, which can have "off" tastes and give rise to unpleasant odors, but in a very different way than the classic fishy smell.
Fish oil's primary components are docosahexaenoic acid and eicosapentaenoic acid.  They are more sensitive to oxidation than many other fats because they contain many double bonds.
The best bet is to look for oil that is freshly prepared and highly purified.  Some fish oils are esterified with ethanol to form ethyl esters right "on the boat".  These ethyl esters can be distilled to remove most impurities.  Optionally, the purified esters can be converted back to triacylglycerols to give "molecularly distilled" fish oil in the triglyceride form.


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Dissolving metal reductions
Could reduction of an organic compound by reacting it with sodium dissolved in $\ce{NH3}$ still proceed if a metal such as aluminium were used instead of sodium? I'm not sure about the aluminium amine or amide structure that would be formed.

Aluminium does not react with anhydrous ammonia even at 500 degrees C. However I don't have a particularly good explanation why Al doesn't dissolve. It may be due to Al forming a thin protecting film on the surface; but from here it seems that the films only form when you have some water (so it's presumably Al2O3)
You can dissolve Ca, Sr, Ba, Eu, and Yb in liquid ammonia, as well the alkali metals. All of these elements have relatively low ionization energies.

Thanks I wasn't sure some things work in theory but not in reality!

To me it seems, that aluminium is the odd one out. According to Yu Qiu, Lian Gao it seems that AlN is formed and ammonia is split. I only read the abstract though. I remember vaguely from when I tried the birch reaction, my lab colleague almost dissolved a whole iron/copper spatula by accident.| @J.LS Care to write up an answer.

What if aluminium amalgamate was used? Or if naoh and a nitrate salt were used with the regular aluminium?

Is there a goal here? If you wanted to react aluminium with concentrated aq ammonia, you would get hydrogen but the rate would be proportional to temperature.  However, this would be quite a poor way to create a reductive environment and as someone has already alluded, aluminium and sodium hydroxide or a strong mineral acid is better. From what I've seen, if you use alkali, a catalyst is needed.
Ethylamine can be more conveniently used in place of ammonia.
Calcium can also be used in reductions.I think I've also seen Indium, but that seems a bit expensive.

The goal is to reduce a carboxylic acid or alcohol to the alkane using aluminium :)

Sodium in anhydrous ammonia liberate a solvated electron for reductions and produce sodium amide.
If you do the same in the lab with aluminium then the chance is big that nothing would happen because the oxide layer on the aluminium surface and the fact that ammonia is a toxic gas that you don't smell at high concentrations.
Nothing will happen if you just put aluminium in aq. ammonia. You need to transform aluminium in another form with a strong acid to make it reactive.
The answer is no.

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I don't see how the toxicity of ammonia is relevant. Perhaps you could add a bit more detail about why the reaction doesn't take place and how you can make it take place as you seem to be implying in the third paragraph.


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What do the numerals on the top right corner of the cells in the periodic table represent?
On the table published here (link) each cell has one to five numerals arranged vertically in its top right corner. The sum of those numbers is always the relative atomic mass of the element. I suspect that, in most cases, two of those numerals represent the number of protons and the number of neutrons, respectively. However, I have no idea what the other numerals represent.
What do the numerals  on the top right corner of the cells in the periodic table represent?

The numerals usually don’t add up to the atoms’ relative masses, but rather to the atomic numbers.

The numbers give you the basic ground state electron configuration of the atoms by telling you how many electrons are in each shell.
An electron's 'position' in an atom can be specified by four quantum numbers:

Principal quantum number ($n$). This tells you which shell the electron is in and can take values $1,~2,~3,~4~...$. The maximum number of electrons in each shell is given by $2n^2$ so the first shell can hold 2 electrons, the second 8, the third 18, the fourth 32 etc. The numbers given in your periodic table are the total number of electrons in each shell in the ground state (lowest energy state) of that element.
Azimuthal quantum number ($l$).
Magnetic quantum number ($m_e$).
Spin quantum number ($m_s$).

See this wikipedia article for more detail on the meaning of quantum numbers.
Electron configurations can be written in more detail using $s,p,d,f$ notation. See this question for more detail.

The numbers represent which shell the electrons occupy in the ground configuration.


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What makes pentane, hexane, heptane, and octane dangerous, while butane and nonane aren't?
In particular, how do a methylene more (in the case of the pentane) and a methylene less (in the case of the octane) make the difference in harmfulness to the environment in comparison with, respectively, the butane and the nonane?
I saw that butane and nonane are not classified as dangerous, but I have just done a research and it turns out nonane is harmful to fish, just like octane, but the difference is that the nonane becomes aqueous more difficult because of its low solubility. But I read that its soluble part is still dangerous for aquatic life. Then the question is: Shouldn't it be classified as dangerous anyway?
The propane, as well, appears to be harmful to the fish, while I can find nothing but "not classified as dangerous", as for the butane. Is LDC3's answer the explanation?

Propane and butane are gases at normal temperatures so they will evaporate into the atmosphere before getting into the water supply. The water solubility goes down with hydrocarbon size so many higher hydrocarbons will also not be available in large enough quantities in the water to cause harm. Being a hazard is a product of availability as well as the inherent toxicity if ingested. Hence the results.

Pentane, hexane, heptane, and octane are labelled with the ‘environment’ hazard pictogram (GHS09) since they are considered toxic or very toxic to aquatic life with long-lasting effects (H411 or H410) because of their toxicity to the water flea Daphnia magna in standardized tests.
Nevertheless, nonane is not completely harmless; it still may cause long-lasting harmful effects to aquatic life (H413). However, a pictogram is not required for this hazard category.
According to a comparative study, the toxicity of these substances is mainly determined by their solubility in water and their organism–water partition coefficient.

I see, thank you. Indeed, I had just found out that the soluble part of the nonane still harms aquatic life. Then, shouldn't it be classified as dangerous anyway? I think this could cause misconceptions.
How about the butane? Is it indeed because it wouldn't accumulate?

@VincenzoOliva Concerning chronic aquatic toxicity, only substances of Category 1 (H410) and Category 2 (H411) show the ‘environment’ pictogram. No pictogram is used for Category 3 (H412) and Category 4 (H413, e.g. nonane). With regard to butane, I am inclined to follow LDC3's explanation.

Butane is a gas at STP, so it would not accumulate in one location.


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Entropy increase when heating a gas
What is the entropy change when the temperature of 1.00 mol of ideal gas is raised reversibly from 10 °C to 50 °C at constant pressure, constant volume, and assuming the heat capacity is independent of temperature?
This is all the information given. I am assuming that the formula $\Delta S = C \ln (T_2/T_1)$ is used. However, I do not know how the heat capacities are supposed to be obtained from the given information. I do not have the enthalpy or internal energy of the process. I cannot calculate the heat capacity from pressure or volume.
Please say if my reasoning is incorrect.
(These aren’t the values of the original question but this is based on Atkins Chemical Principles 5th ed. Ch 8 # 5 and 6)

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 hope that my reasoning is clear now.

Do you have some idea about the atomicity of the gas? That can give you some information about the value of C.

The gas is monoatomic

Can you write the algebraic equation for the change in entropy per mole as a function of the end state temperatures and pressures?  Can you write the algebraic equation for the change in entropy per mole as a function of the end state temperatures and volumes?  What is the molar heat capacity of a monoatomic gas in terms of the universal gas constant?

With constant volume heat capacity, it is defined as, for an ideal gas, that has the internal energy equation
$$dU=TdS-\rho dV= \frac{\nu}{2} Nk_{\beta} T\ (1)$$ which is the Ideal Gas Law, obtained from the first law of thermodynamics. Using moles, (1) is equivalent to
$$ \frac{\nu}{2}nRT $$
Now, onto heat capacities.
$C_V$ is defined as,
$$C_V=(\frac{\partial U}{\partial T})_V= \frac{\partial}{\partial T}\left(\frac{\nu nRT}{2}\right)=\frac{\nu}{2} nR $$
Constant pressure heat capacity is defined as
$$C_{\rho}=C_V + nR=\frac{\nu+2}{2} nR$$, where $n$ is the amount of substance and $R$ is the universal gas constant. $\nu$ is the degrees of freedom of the molecule, which is easily obtained by thinking about a monoatomic ideal gas and the ways it can move.
Below is some fun theory surrounding the heat capacity for constant volume.
Note that constant-pressure specific heat is also defined as
$$C_{\rho}=(\frac{\partial U}{\partial T})_{\rho}=dU+\rho dV+Vd\rho=
(\frac{\partial H}{\partial T})  $$
Where $H$ is the enthalpy of heating the gas, which is
$$\mathrm{d}H=\mathrm{d}U+p\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S-\rho\mathrm{d}V+\rho\mathrm{d}V+V\mathrm{d}\rho=T\mathrm{d}S+V\mathrm{d}\rho$$
Enthalpy is commonly defined with a constant pressure process, thus $\mathrm{d}\rho=0$.
So, for a constant pressure, we have $$\mathrm{d}H=T\mathrm{d}S$$,
exploting the definition of entropy for a reversible change,
$$\mathrm{d}S=\frac{\mathrm{d}Q}{T} \rightarrow \mathrm{d}Q=T\mathrm{d}S $$
so, we finally get $$\mathrm{d}H=\mathrm{d}Q$$
which is a standard thermodynamic identity, and can also be used to get to the answer at the beginning of the post (though it is admittedly longer and more complicated to do so.)
Hope these hints help you!


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What is the meaning of signs in electrode potentials?
Question
Suppose you have a cell set up between a copper metal/copper(II) ion electrode and a reference electrode.
Under standard conditions, the emf of this cell was −0.07 V. The standard electrode potential of the copper metal / copper(II) ion electrode is +0.34 V. Hence the standard electrode potential of the reference electrode is:
A: -0.41 V
B: -0.27 V
C: +0.27 V
D: +0.41 V

My attempt:
I was taught this equation: $\mathscr{E} _{cell} = \mathscr{E} _{a} - \mathscr{E} _{b}$
where:
$\mathscr{E} _{a}$ is the emf of the more positive electrode potential
$\mathscr{E} _{b}$ is the emf of the less positive electrode potential
But using this equation, none of the combinations of electrode potentials give -0.07V as the emf.
Thanks

Chapter 14 of Harris' Quantitative Chemical Analysis has a helpful section entitled, an intuitive way to think about cell potentials.  Here's the corresponding graphic:

In this case, the author is describing a Galvanic cell with cadmium as the anode and silver as the cathode.  The potentials of the two half-cells are known and marked on the line-graph.  Electrons always flow to the more positive electrode, which in this figure is towards the right.  The difference between the two half-cells is the cell potential.
In your case, you know the separation (-0.07 V) and one of two points (+0.34 V).  If we ignore the sign for the moment, we can use the line-graph to narrow the possibilities down to C or D, since each of these differ from +0.34 V by 0.07.  The question now is, which direction are the electrons flowing?  When determining the cell potential using this line graph, we subtract the right-most number from the left-most number in order to obtain a positive potential.  Since the cell potential in your case is negative, we must be doing the opposite.  Therefore, the reference electrode potential is +0.41 V (D).

I understand it slightly better now, thanks, but I'm still unsure on why we do the opposite for a negative potential. Is it because the electrons are flowing to the less positive electrode in my case instead of to the more positive one so the calculation is the opposite?

@Cobbles it is a "frame of reference".  If you are 2 meters away from something, you may say that it is 2 m in front of you.  If you turn around, it is 2 m behind you.  The object hasn't moved, just your perception of "in front" and "behind".  We measure potentials versus a reference electrode, so regardless of where the reference electrode potential lies, we determine the cell potential as working minus reference.


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Stoichiometry reactions for water clarification
If ferric chloride was added to a water sample which had calcium bicarbonate alkalinity present, would the reaction be a ‘stoichiometry reaction’ or a ‘limiting reagent reaction’?
For example, if 1 g of ferric chloride was added, how much precipitate would form [$\ce{Fe(OH)3}$] and what type of reaction (mentioned above) would happen?
$$\ce{2FeCl3(aq) + 3Ca(HCO3)2(aq) ->2Fe(OH)3(s) + 3CaCl2(aq) + 6CO2(g)}$$

Welcome to Chemistry! What have you tried so far? Perhaps thinking in terms of moles rather than grams would be helpful. If you have 2 moles of ferric chloride, what happens per your equation?

In order to determine the reaction is "stoichiometric" or has a limiting reactant, we need to know the volume and the concentration of the calcium bicarbonate solution.

@Ben Norris: Thanks for correcting my bad balancing!

our test was to see how much precipitate formed. we did not know how much calcium bicarbonate was in the reaction


text:
Gattermann-Koch Reaction with Phenol Substrates
The Gattermann-Koch Reaction involves the formation of an aryl aldehyde using carbon monoxide and hydrochloric acid as reactants. The reaction is catalyzed by aluminium trichloride in the presence of trace amounts of cuprous chloride.

What I don't understand about this reaction is that according to Wikipedia along with other online texts I've used, the Gattermann-Koch reaction is inapplicable to phenol substrates. The drawback has been left unexplained by all my resources (online and textbooks) and even my professor was oblivious to it.
I am unable to understand why the reaction doesn't apply to phenol and what exact side reactions might occur.
Additionally, why is the explanation so obscure?
[EDIT] I found the explanation. The Gattermann-Koch Reaction is not applicable to phenol and phenol ethers because they "could not be successfully formylated at atmospheric pressure in benzene as a solvent."
"This failure to react was attributed to the insolubility of the cuprous chloride in the reaction mixture."
Can someone please explain this in simple terms?

This question might be of interest to you as it is closely related.

You should try to consult it from books

You have outlined that the formyl chloride is generated as the reactive, electrophilic intermediate under these conditions.
You are planning to perform an electrophilic aromatic substitution, but in a phenol, there's high electron density on the $\ce{O}$ atom. Isn't it conceivable that your main reaction product is phenylformate here, resulting from $\ce{O}$-acylation?

I didn't consider that, but then why would that only apply for phenol substrates? Wouldn't amines and ethers also be reactive in the same manner?

@tennispro1213 Primary and secondary amines will undergo N-acylation as well, Ethers, such as anisole, on the other hand, can be formylated in a Gattermann-Koch reaction.

can we simply say that the Gattermann-Koch Reaction is generally not used in compounds containing strongly basic groups?

@tennispro1213 We could say that the Gattermann-Koch or the Vilsmeier-Haack reaction can lead to heteroatom acylation. In the case of phenols, we'd use the Reimer-Tiemann or the Duff reaction instead.

How sure can we be that we actually have formyl chloride as a reactive intermediate? I could imagine to have only protonated carbonmonoxide, according to $$\ce{CO + HCl + AlCl3 <=> H+CO + [AlCl4]^-}$$. Also cuprous chloride seems to play an important role in the mechanism.

Phenol can act as Lewis base and form adduct with AlCl3, using a lone pair on the O atom, hence deactivating the ring for further reaction.

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I suspect that, in the presence of the Lewis Acid catalyst, even if a formyl ester is formed, it will rearrange the the acylbenzene, which is irreversible. This may have consequences for the o,p-product ratio,
And as far as the cuprous chloride, use aluminum trichloride. And you could make the Scylla bromide first, with Formica acid an bromine. Much more reactive will likely form the sculpture ester first, in catalyze. Then catalyst to rearrange it

Are you simply trying to put obscure words in last sentences? I know what Scylla is and it's not some wacky mythology chemistry here...

I have to agree, this looks like an autocorrect nightmare.  Can you polish it up a bit?


text:
Why do ion cores not move in a metal?
The delocalized electrons in a metal move about because of the electrostatic attractions between them and many different positive ion cores. But why does the same not occur with the ion cores themselves? Why are electrons, which bear the same magnitude of charge as nuclei, mobile, while the nuclei themselves are not?

Yes, I apologize. I have redacted my question.

No apology necessary! good question.

The crystal structure of the metal holds the atom in place. BTW, since electrons are shared in a "sea" of electrons or conductivity band, no one atom is necessarily ionized. The crystal structure is also important for nuclear physics, e.g. using Mössbauer spectroscopy to measure gravitational effects.
Individual electrons don't actually move around the whole circuit, even in DC applications. Instead, think of it as a large stack of electrons; as one is pushed on the stack at one end, electromagnetic forces pop one off the other end. Though electrons do move a bit, the drift velocity is much less than the speed of light (in fact, it is somewhat slower than the last-placed snail in a race).
Compare the mass of a single electron and the mass of an atom, or the mass of an ion in a plasma. Note that acceleration of a particle is inversely dependent on its mass. This is important in fusion research, where confining electrons and ions is a bit tricky.

See this PPT presentation on metallic conduction.

One thing to add might be that it is only the valence electrons that "move" - the electrons in closed shells stay with their respective nuclei and hence the nuclei cannot move either.


text:
Fertiliser analysis - Why doesn't it show ureic nitrogen when it does have urea?
I have had two types of urea fertiliser analysed, the results are as follows:
$\mathrm{46\%N}$ (46% nitrogen) - which is sold as a urea fertiliser, tests at 46.1% ureic nitrogen.
$\mathrm{38\%N~ 19\%S}$ (38% nitrogen, 19% sulphur) - which I understand to contain ammonium sulphate and urea, tests at 33% nitric nitrogen and 7.4% ammoniacal nitrogen
My question is-  why does the $\mathrm{38\%N~ 19\%S}$ show no ureic nitrogen when it is supposed to contain urea - the same as the $\mathrm{46\%N}$?
edit - the analysis company has admitted a mistake- the test  result should have read 33% ureic nitrogen and not 33% nitric nitrogen.

There is no question; the OP has withdrawn the initial data.


text:
Does water ionically bond to chloride ion?
This user absolutely insists that when a chloride ion is present in water (for example, when $\ce{NaCl}$ dissolves in water) that $\ce{Cl-}$ ion is ionically bound to the hydrogen atom:
https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55324_34449
I then insist that this would form $\ce{H2OCl}$, but that's impossible due to the lack of valence electrons of $\ce{H}$:
https://biology.stackexchange.com/questions/34449/why-do-i-smell-chlorine-when-my-sinuses-are-clogged?noredirect=1#comment55386_34449
Who is right?

The most common mechanism for when $\ce{NaCl}$ dissolves is as folows:
As water is a polar molecule, as per the UC Davis ChemWiki page All About Water, shown below:

The caption reads:

the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule.

So, when $\ce{NaCl}$ salt is dissolved in water, the sodium and chloride ions are pulled apart by the water molecules with the result explained in the image below:

Image source
You'll note, that there is no ionic (nor covalent) bonding present, rather an ion-dipole interaction between the water molecules and the sodium and chloride ions (University of Waterloo).
The UCLA page Illustrated Glossary of Organic Chemistry has a very nice definition of the ion-dipole interaction:

A noncovalent attraction between one pole of a bond dipole and an oppositely-charged ion. The δ+ end of the bond dipole is attracted to an anion, and the δ- end of the bond dipole is attracted to a cation.

There is a neat YouTube clip showing the process here.
The ion $\ce{H2OCl+}$, does actually exist according to NIST, but I could not find any references that link it to the dissolution of $\ce{NaCl}$ salt.


text:
Supersaturation and Crystallization
I had a fairly simple question regarding supersaturation and crystallization. When a supersaturated solution is crystallized, is all of the solute crystallized or just the excess solute (such that the solution remains saturated)? After crystallization, would you expect to find some quantity of the solute still dissolved?

Yes, only the excess solute crystallises - the solution is still saturated (but not supersaturated) afterwards.

Think of it this way: A saturated solution is the equilibrium. Say you had a supersaturated solution, formed a crystal out of it and somehow managed to get your solution below saturation.
But once that happened, your solution is no longer saturated and there is undissolved material on the ground (your crystal). So, always striving towards equilibrium, some of your crystal would get redissolved to re-reach the saturation.
In fact, crystallisation and equilibrium are not static but rather dynamic processes. So even if there is a crystal sitting at the bottom and it looks like nothing is happening, there is a constant flux of compounds redissolving from the crystal surface or recrystallising onto the surface out of the solution. Which is also why everything seems to strive towards the equilibrium: It is the state, where the speed of one direction matches that of the other direction.
So short answer: The solution above a crystal will be saturated, but no longer supersaturated.


text:
Grubbs, Wittig & hydroboration reactions and Woodward-Hoffmann rules
Several organic mechanisms I know involve [2+2] cycloadditions and retro-[2+2] cycloadditions, namely:

The Wittig reaction
Hydroboration of an alkene
Olefin metathesis (using Grubbs' catalyst)

How are these in accordance with the Woodward–Hoffmann rules? I think that there are two $(4q+2)_s$ components and that's not an odd number, which is the criterion for a reaction to be thermally allowed.

https://en.wikipedia.org/wiki/Olefin_metathesis#Reaction_mechanism

So it happens, regardless of the forbidden symmetry.

The Woodward-Hoffmann rules for pericylic reactions assumes essentially that you can describe the orbitals in terms of Huckel approximations, which is definitely not the case for a Ru catalyst.

Could you elaborate as an answer?

I found an answer in Ahn.N.T "Frontier Orbitals" for hydroboration: "The transition state for a Woodward–Hoffmann-
allowed reaction is aromatic in character, whereas a forbidden one is antiaromatic.
However, the rules of aromaticity only apply when each atom of the annulene uses
one, and only one, AO to bind to its neighbors."

Could someone expand this into an answer?

I think you can try - don't be afraid :)

I'm afraid I can't. Would be good if someone could expand the answer to account for the Wittig and Grubbs reaction. Wittig may go through a betaine intermediate, but what about Grubbs?

Same question but specifically on the Wittig


text:
Name of these parts in CoQ10?
Am giving presentation over Coenzyme Q10 Wednesday for my final project, but we haven't studied any biochem this year.  So questions have arisen.
CoQ10 has a Quinone head, an isoprenyl tail, and these additional oxygens on the left and three methyl groups not a part of the Quinone head or tail.
Is there a name for these two oxygens and three methyl groups?  Or a category?  Like the benzene ring and two oxygens is the Quinone head, what are the additional oxygens and methyl groups called, if anything?
CoQ10 diagram: http://www.mysupplementstore.com/media/wysiwyg/structure-of-CoQ10.jpg

The additional groups are referred as methyl and methoxy groups. There is no category for these groups. It appears that the methoxy groups stabilizes the intermediates of the oxidation or reduction reactions.


text:
E2 reaction - strong base
In E2 reaction, there is more than one product. According to Zaitsev’s rule – “The alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents.” (Wikipedia)
Now, my question is, how can the base effect the percentage of the minor and major products? Does a stronger base give us more major product and vice versa?
The particular question I’m trying to solve is:
In the E2 reaction of 2-bromo-2,3-dimethyl butane, which base would give more of the minor product and which of the major product?

Could you provide the base strength for the compounds?

What do you mean? that's all the data I've got

It will be more difficult for sterically hindered bases to remove the proton of the β-carbon having the fewest hydrogen substituents, since this carbon atom has more bulky groups around it compared to the the others. So base 3, being the less hindered base, will provide a higher percentage of the major Zaitsev’s product and base 2 will provide a higher percentage of the minor product compared to the other bases.


text:
E2 reaction — which one is faster?
Will an E2 reaction that has 2 potential products go faster than one that has 3 potential products? The specific question involves both molecules below reacting with methanol. I figured it's an E2 reaction (I hope I'm right) and the question is, which one will go faster?

Could you provide specifics?

What are the conditions for this reaction? Only methanol is mentioned, which leads me to believe this is E1 not E2.

For the E1 pathway, molecule A will react faster since the intermediate is more stable than from molecule B. According to this web site, the major product for A is 1,2-dimethylcyclohexene; for B, it is 1-methylcyclohexene.
For the E2 pathway, both reactions will have nearly the same rate. Depending on the orientation of the α-methyl group (cis or trans to Br), the major product is the same for cis, but for trans it will be 2,3-dimethylcyclohexene for A and 2-methylcyclohexene for B.

why is the intermediate more stable in molecule A?

The intermediate is a tertiary cation; it's more stable than a secondary cation.

But in the E2 mechanism there is no cation formed.

@Martin-マーチン It's been a while since I needed to know the difference between E1 and E2. I have them switched.

I was confused, too, and had to look it up. And now I see a comment to the question, that makes me believe your answer is absolutely on target.


text:
Explain why these ions are more strongly hydrated in an aqueous solution?
I've already answered this question. All I want to know is if my reasoning is correct.
Case 1:

$\ce{Rb+}$
$\ce{Na+} \leftarrow$ This ion is more strongly favored

Case 2:

$\ce{Mg^{2+}} \leftarrow$ This ion is more strongly favored
$\ce{Na+}$

Reasoning:
For the first case, my guess would have to be because the electrons are much closer to the center of the nucleus (the atomic radius of the atom is very small) and makes it much easier to interact the positive dipole end of the water molecule (the hydrogen molecules).
For the second case, the reason why the magnesium ion is more favored is because even though both ions have roughly the same molar mass, one is more positively charged than the other and thus is more likely to attract the negative dipole end of the water molecule (the oxygen molecule).
Is my reasoning justified or is there something in particular that I'm not paying attention to?

There's a problem with the reasoning for your first case: positive ions will much more strongly attract the oxygen atom than the hydrogen atoms in water. The oxygen atom in water has a locally high electron density (partial negative charge), while the hydrogen atoms has a locally low electron density (partial positive charge). The strongest attraction happens when oxygen atoms point towards the cations, and the hydrogen atoms point away.

I believe your reasoning is correct (more or less). Solvation is influenced by charge to size ratios. Higher the charge to size ratio, the easier it is for the species to be solvated by water molecules.
Apply this idea to both your cases. (note: in the second case, I don't get your argument about molar mass. The idea is much simpler, magnesium's atomic radius is smaller than sodium's and it has a higher positive charge. What can we conclude from this?
Also, reconsider the first case: a positive ion would not be attracted to the positive end of a dipole.

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Calculating the number of ligands of complex ions in d- block
How do you know the number of ligands surrounding a metal ion?
Example: the obtained complex of $\ce{Ni^2+}$ ion in $\ce{NH3}$ solution is written as $\ce{[Ni(NH3)6]^2+}$ but $\ce{Cu^2+}$ in $\ce{NH3}$ solution is $\ce{[Cu(NH3)4]^2+}$.
So how do you find the number for different metal ions like $\ce{Ag^2+}$?

The coordination number (number of ligands) changes depending on the type of ligand. In this case ammonia.

I suspect you may have erroneously tried to use the silver dication $\ce{Ag^{2+}}$ as a standard example of transition metal ion by extrapolating the most common aqueous copper ion, $\ce{Cu^{2+}}$. In reality, $\ce{Ag^{2+}}$ is rather unstable, and silver is much more commonly found as a monocation, $\ce{Ag+}$.

@NicolauSakerNeto $\ce{Ag^{2+}}$ is perfectly stable in proper neighborhood. In fact $\ce{Ag+}$ disproportionate giving $\ce{Ag^{2+}}$ and metallic silver in some conditions.

@permeakra Quite true, I didn't mean to imply it wasn't stable ever. To my understanding $\ce{Ag^{2+}}$ is stable in a narrower range of conditions, though, or at least doesn't overlap much with "common" conditions, for some reasonable definition of common. Coordination compounds in aqueous media with $\ce{Ag^{2+}}$ ions are relatively scarce, I would believe.

@NicolauSakerNeto you are incorrect in it. $\ce{Ag^{2+}}$ is quite stable in bipiridyl complexes and again, $\ce{Ag+}$ disproportionate in presence of some ligands http://pubs.acs.org/doi/abs/10.1021/ja00775a074

@permeakra Very well then, thank you for pointing it out!

To my knowledge, there is no "rule" to predict the coordination number of a given metal ion.  There are, however, guidelines, which are nicely summarized here.

Larger ions can accommodate more ligands
Bulky ligands will reduce the overall coordination number
Highly charged ions will tend to accept a greater number of Lewis Bases

With your example, we can make an educated guess as to the coordination number of Ag2+ using periodic trends.  It is in the same group as copper and will be bigger since it is lower on the periodic table.  Based on the three points above, I would predict that Ag2+ could have a coordination number greater than 4.  It turns out that (distorted) octahedral coordination of the silver dication is not all that uncommon.

1st-row $d$-element + 2nd-row $p$-element - usually octahedron except for $\ce{Ni(II),Cu(II),Co(I)}$ (often a square or a distorted octahedron with two weakly bound ligands) and $\ce{Zn}$ (usually a tetrahedron). 1st-row $d$-element + 3+-row $p$-element - usually a tetrahedron.
2nd/3rd row $d$-element are somewhat more complicated as they allow higher coordination number up to 9 ($\ce{ReH9^{2-}}$), but the situation is so much influenced by electron configuration, ligands and metal-metal interactions here that simple rules loose a lot of meaning.
In short, the coordination number is usually a compromise between 18-electron rule and size of ligands. For example, $\ce{V(CO)6}$ should dimerize, giving c.n. of 7 to vanadium, but can't do so because of size constrains. Somewhat similar happens with bulky alcoholats of titanium, that tend to include $\ce{Ti(IV)}$ in octahedral coordination, but lower c.n. is observable in case of tritox (tris-tert-butyl) alcoholat.
Though it is generally not recommended to guess the c.n., it is always better to google for some reference compounds.


text:
Why any other element except fluorine wouldn't take electrons from oxygen?
My teacher said that fluorine is more electronegative than oxygen thus it will take electrons of oxygen.
But now, why won't other Group 17 elements, like chlorine, bromine, iodine, would take electrons from oxygen? They too have just 7 electrons in their valence shells.
Technically, they should but yet they have lesser electronegativity and this is something that I'm not able to digest.

Flourine's nucleus is less shielded from electrons than in the other halogens therefore it is a better at attracting electrons.

The only time oxygen atoms have a positive oxidation number in compounds is when fluorine is somehow involved, be it $\ce{OF2}$, $\ce{O2F2}$ or $\ce{O_2^+[PtF6]^{-}}$. There are many compounds where oxygen has a positive formal charge, however, such as in trialkyloxonium ions.

http://chemistry.stackexchange.com/questions/8762/why-is-fluorine-the-most-electronegative-atom?rq=1

You can consider thinking about charge to volume density. Fluorine is very small and has highest charge to volume ratio among other group 17 elements. For that reason only fluorine is able to snatch electrons from oxygen.

There are two simple ways to calculate charges on atoms which can be done by hand.
The first relies on the concept of oxidation numbers. To define these, draw the Lewis structure of a target molecule/ion and break all covalent bonds by transferring both electrons in the bond to the more electronegative atom (usually based on the Pauling scale of electronegativity). In this sense, by definition oxygen only presents a positive charge when either an electron is directly removed from oxygen, or when it shares a bond with fluorine. There are relatively few examples of this, as $\ce{O-F}$ bonds are weak and oxygen atoms are hard to ionize. Examples of molecules with $\ce{O-F}$ bonds and positive oxidation number on the oxygen atom are $\ce{OF2}$ (+2) and $\ce{O2F2}$ (+1). An interesting example of oxygen with positive oxidation number happens in the dioxygenyl cation, $\ce{O2^{+}}$ (+0.5 on both atoms), which is quite difficult to produce except upon reaction with highly fluorinated substances such as $\ce{PtF5}$ or $\ce{AsF5}$. These are actually avid enough for electrons that they tear one off the neutral dioxygen molecule $\ce{O2}$, producing the ionic salts $\ce{O2^+[PtF5]^{-}}$ and $\ce{O2^+[AsF5]^{-}}$, among others. The sum of all oxidation numbers in a molecule/ion must equal its net charge.
The second way to calculate charges relies on the concept of formal charges. To find these, draw the Lewis structure of a target molecule/ion and break all covalent bonds right in the middle, transferring one electron to each of the bonding atoms, regardless of electronegativity. This definition does not rely on electronegativity, so as one might suspect, it is much more common for oxygen to present a positive formal charge. Simple examples (all of the following with a positive formal charge of +1) include the hydronium cation $\ce{H3O^+}$, trialkyloxonium cations such as trimethyloxonium $\ce{(CH3)3O^+}$, the central oxygen atom in ozone $\ce{O3}$, and pyrylium aromatic rings, the last of which are found in many natural compounds. The sum of all formal charges in a molecule/ion must also equal its net charge.
(A more pictorial explanation for the ways to calculate oxidation numbers and formal charges can be found in this Wikipedia article)
So, which of these two concepts is correct? In reality, neither. They're both extremes of a scale, and reality often sits somewhere in the middle. Both are meant more as a bookkeeping mechanism to keep track of the total charge in an ion/molecule, and provide some qualitative idea of where electrons might be more concentrated or sparse. Attempts at determining precise atomic charges rely on complicated calculations in computational chemistry. These calculations need not agree (and most often do not agree) with either of the values given by oxidation numbers or formal charges, and the actual charges can be represented by any real number, rarely being an integer or a simple fraction.
In light of this, let's look at your claim. You suggest that fluorine takes electrons from oxygen while other halogens don't. Let us take the two simple examples $\ce{OF2}$ and $\ce{Cl2O}$. These are both gasses at room temperature, which suggests bonding in these substances is not predominantly ionic, so no atom abstracts electrons completely from any other; instead they share their electrons. Thus, they are well-described by considering the oxygen and halogen atoms are covalently bound into a molecule much like water, except with halogen atoms in place of hydrogen atoms. Applying the concept of oxidation numbers, the oxygen atom in $\ce{OF2}$ has an oxidation number of +2 as stated previously, while in $\ce{Cl2O}$ the oxygen has an oxidation number of -2. Applying the concept of formal charges, we find that the formal charge on the oxygen atom is exactly zero in both molecules. Neither concept is correct, so we expect reality to be something in between. Exactly what the charge is, we don't know without resorting to computer calculations, though it is probably quite safe to say at least that the oxygen atom is more deficient in electron density in $\ce{OF2}$ than in $\ce{Cl2O}$.


text:
Can an atom bond with more than 8 other atoms?
Is it possible for an atom to bond with 8 other elements (same or other type)? If yes, then please give some examples. If no, then what could be the possible reason for it?
My question is not about valence shell electrons. My question is about whether an element could combine with more than 8 other same or different elements? It could be rather covalent bonds or any bond.

I was thinking in 3D; most 6 coordinate atoms have an octahedral shape. But bon's answer made me think of ferrocene.

Very relevant previous question.

Yes, there are coordination complexes of large elements which have coordination numbers greater than eight. Some examples are:

$\ce{[ReH9]^2-}$ with a tricapped trigonal prismatic structure. The nine hydride ligands are small enough to fit around the relatively large rhenium atom fairly easily. This ion can be isolated as a potassium salt $\ce{K2ReH9}$.

$\ce{[U(NO3)6]^2-}$ with an impressive coordination number of 12, consisting of six nitrate ligands each forming two coordinate bonds to the central uranium ion.

Counting only monohaptic ligands, coordination number 9 seems to be the highest currently known, in compounds such $\ce{[ReH9]^{2-}}$, though a coordination number as high as 15 has been predicted for $\ce{[PbHe_15]^{2+}}$. Allowing for multihaptic ligands, a coordination number of 20 is found in $\ce{Th}\mathrm{(\eta ^5-}\ce{C5H5}\mathrm{)_4}$.

@NicolauSakerNeto Does the last one really count as 20?

@Alizter Orbitals from all 20 carbon atoms are involved in the bond with the central thorium ion, so yes.

@NicolauSakerNeto I think the question is asking for ligands rather than atoms or else any large molecule would suffice. The Thorium forms 4 ligands.

@NicolauSakerNeto  I've seen a number of papers argue that 20-coordination in $\ce{M(C5H5)4}$ is not an accurate view.  The argument goes that each double bond should count as one 2-electron contact.  Hence there would be 3 contacts for each $\ce{C5H5}$ for a total (Werner) coordination number of 12.

@ron The x-ray data for the cyclopentadienyl ligands have equal bond distances and the entity is flat, which means that the bonding of cyclopentadienyl is distributed equally on the ring. Since there are 5 orbitals available, how could the bonding be characterized as 3?

@LDC3  Read the second paragraph in ref_2 in my answer.  Also, Wikipedia says, "There are various ways of assigning the contribution made to the coordination number of the central iron atom by each cyclopentadienide ligand. The contribution could be assigned as one since there is one ligand, or as five since there are five neighbouring atoms, or as three since there are three electron pairs involved. Normally the count of electron pairs is taken" (emphasis mine).

An isostructural to $\ce{[U(NO3)6]^2-}$ complex  $\ce{[Ce(NO3)6]^2-}$ even found its way to Shriver's Inorganic Chemistry cover:)

14 coordination is claimed in $\ce{U(BH4)4}$ (ref_1, p. 268).  The molecule exists as a polymer in the solid state.  Six hydrogens from two of the $\ce{BH4}$ groups bond between the boron and uranium (a bridge bond).  Two hydrogens from each of the two remaining $\ce{BH4}$ groups also bridge bond to uranium; the other two hydrogens bond to an adjacent uranium atom to build the polymer.  So far this takes us to 10 hydrogens (6+4) bonding to the central uranium. Two more hydrogens in two $\ce{BH4}$ groups in two adjacent uraniums also bond to the initial uranium, just as described above, as part of the polymer extension.  These 4 additional bonds to hydrogen take the coordination number to 14.
Coordination number 15 has been found in $\ce{Th(H3BNMe2BH3)4}$, and it is suggested that the molecule would have a coordination number of 16 in the gas phase (ref_2)

(image source: see ref_2 above)
There are 7 short $\ce{B-Th}$ contacts and one long $\ce{B-Th}$ contact.  Each short contact provides 2 hydrogens that bond to thorium, while the long contact provides 1 hydrogen for bonding to thorium, making for a total of 15 $\ce{Th-H}$ bonds.

There has been quite some interesting work by my former co-worker and my supervisor on metal-rich molecule with the co-ordination higher than eight.

For more see Timo Bollermann, Thomas Cadenbach, Christian Gemel, Moritz von Hopffgarten, Gernot Frenking, Roland A. Fischer, Chem. Eur. J., 2010, 16 (45), 13372-13384. Also available through researchgate.com. For the purely theoretical bonding analysis see Moritz von Hopffgarten, Gernot Frenking, J. Phys. Chem. A, 2011, 115 (45), 12758–12768.

In october 2015, 5 months after all the current answers were given, a new cluster compound witn C.N. 16, $\ce{CoB16-}$, has been reported [1]:

Here we report the observation of a large metal-doped boron cluster of $\ce{CoB16-}$, which is produced using a laser vaporization cluster source and characterized by photoelectron spectroscopy (PES). Extensive computational searches reveal that there are two nearly degenerate structures for $\ce{CoB16-}$, which are indistinguishable at the highest level of theory employed. They both possess tubular double-ring framework and give similar photoelectron spectral patterns. The structures can be viewed as two $\ce{B8}$ rings sandwiching a $\ce{Co}$ atom, reminiscent of a drum and giving rise to the highest coordination number known in chemistry thus far. [...]
The previous highest coordination number known experimentally was 15 for
  $\ce{[Th(H3BNMe2BH3)4]}$ (ref. 33), though theoretical studies have suggested the highest coordination numbers of 15 in $\ce{PbHe15^2+}$ (ref. 34) and 16 in the Friauf–Laves phases in $\ce{MgZn2}$ or $\ce{MgNi2}$ (ref. 35). Endohedral fullerenes ($\ce{M@C60}$) have been observed 36,37, but the metal atom in those cases interacts with the $\ce{C60}$ shell primarily ionically and it does not stay in the centre of $\ce{C60}$.

It needs to be pointed out, that no crystal structure has yet been determined  (as of 2017Q3).
References

Popov, I. A.; Jian, T.; Lopez, G. V.; Boldyrev, A. I.; Wang, L.-S. Nature Communications 2015, 6, ncomms9654 DOI: 10.1038/ncomms9654.
(Open Access)

"no crystal structure has yet been determined (as of 2017Q3)." do you have any updates on this now that Q1 2018 is about to end? Thanks!

Maybe it's a "cheat", but coordination to ten atoms has been known since ferrocene.  The iron is bonded symmetrically to two five-carbon rings.  Variations with larger rings and thus more atoms bonded to the metal are also known, as in uranocene.

With π-donor ligands it's not that trivial. Actual C.N. has little to do with the number of atoms in each ring, instead one has to determine the number of donated π-electron pairs. So, actual maximum estimated C.N. of ferrocene and uranocene are 6 and 10, respectively.

In metals with hexagonal or face-centered cubic packing, every atom has 12 closest neighbors. Libretext illustration


text:
Ene-yne metathesis product
In the ene-yne metathesis how to know which product will form predominately, as the mechanisms to both are plausible?

It depends on what the R groups are and what sort of catalyst you are using.

How is that do?


text:
What is the Difference between difference between Standard Electrode Potentials and Standard Reduction Potentials?
As the question suggests I am wondering the difference between standard electrode potentials and standard reduction potentials. I know that they are both a measure of the Voltage from certain reactions, and can be used to calculate the feasibility of a reaction - but when do you use which / what is the difference, or are they interchangeable?

Are they not the same thing? Do you have specific data which suggests they are different?

Well in a data booklet (such as: https://www.scribd.com/doc/16633801/GCE-Chemistry-Data-Booklet-Issue-2 , Page 14-17), they are it separate sections, with one being 'Electrode Reaction' and the other being 'Right-hand electrode system'.

They appear to be exactly the same except that one page lists reduction equations and their potentials whereas the other page just lists the species being reduced and the species formed and leaves you to work out the equation.


text:
Why is it the case that KE = 0.5mv^2
To the best of my knowledge, energy equals work, $\mathrm{E = W}$; work equals force multiplied by distance, $\mathrm{W = Fm}$ ; force equals mass multiplied by acceleration,  $\mathrm{F = MA}$; and acceleration equals distance per second squared, $\mathrm{A = m \setminus s^2}$. However, when I substitute those values in for the dimensions of $ \mathrm{W}$, I can't transform the result of those substitutions into  $ \mathrm{W= \frac{1}{2}Mv^2}$. Perhaps I've done the algebra incorrectly.
Does the result of those substitutions transform into $ \mathrm{W= \frac{1}{2}Mv^2}$?
If it doesn't transform into that equation, why doesn't it do so, and why do we use an equation that isn't compatible with our other equations?

This is more of a physics question.

@Alizter I thought about that, but it's straight from my chemistry text book. Chemistry and physics overlap in a number of places, I don't see any reason why physics.se should trump chemistry.se in those cases.

What do you get when you try it?

Please avoid using Latex in titles due to searching issues

The kinetic energy $E_\text{k}$ equals the work $W$ done on an object:
$$E_\text{k}=W$$
The work $W$ is the result of a force $F$ on a distance $s$. If $F$ is constant, the work $W$ is equal to the force $F$ multiplied by the distance $s$:
$$W=Fs$$
The force $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$:
$$F=ma$$
For uniform acceleration $a$, the final velocity $v$ is
$$v=at$$
and the distance $s$ is
$$s=\frac{1}{2}vt=\frac{1}{2}at^2$$
Therefore,
$$\begin{align}
E_\text{k}&=W\\
&=Fs\\
&=mas\\
&=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\
&=\frac{1}{2}mv^2
\end{align}$$

@Loong's answer is perfect but I just want to show another method which can be used even if the acceleration is not uniform.
$$a=\frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt}$$
Hence $$a=v\frac{dv}{dx}$$
Force is mass times acceleration.
Hence $$F=ma=mv \frac{dv}{dx}$$
For a small distance $dx$ moved, the work done by the force is $F \times dx$.
By the Work-Energy Theorem, this is also the gain in kinetic energy.
Hence $$\Delta KE = \int_{x_1}^{x_2} Fdx = \int_{x_1}^{x_2} (mv \frac{dv}{dx}) dx = \int_{v_1}^{v_2}mvdv$$
Hence $$\Delta KE = \frac{1}{2}m{v_2}^2 - \frac{1}{2}m{v_1}^2$$
If the body starts at rest $v_1=0$ and we get the kinetic energy as $\frac{1}{2}mv^2$


text:
When I heat up a balloon, does the air inside increase in pressure as well as volume?
When I heat up a balloon, does the air inside the balloon increase in pressure as well as volume?  I thought pressure and volume were inversely proportional?  Or does pressure and volume increase as temperature increases?

What exactly do you mean by pressure? I think your notion of pressure is quite different.

The pressure of a gas is the force that the gas exerts on the walls of its container, or in this case the balloon.

Yes, the vapor pressure increases.

What is the difference between pressure and volume in this scenario?  Is there one?  Thats where I am getting so confused.

As you said, pressure and volume are inversely proportional but I believe that you are getting confused in a different matter which I will go over in the answer.

Are you talking about a latex balloon (party balloon) or a hot air balloon?

A Latex Balloon.

If the balloon is closed, then yes, both volume and pressure will increase when the gas inside is heated. Let's look at two simpler cases first.
If the gas were completely free to expand against ambient pressure (say, inside of a container sealed with a freely moving piston, with no friction), then the heated gas would expand until it created as much force per area (gas pressure) as the force per area acting on it (ambient pressure), so that the forces cancel out and the piston stops moving. Here, a temperature increase in the gas would translate solely to a volume increase.
If the gas were confined in a perfectly rigid box, then an increase in gas temperature would cause the molecules inside to bump harder against the inner surfaces, but to no avail, as the walls do not budge and the box stays exactly the same size. Here, a temperature increase in the gas would translate solely to a pressure increase.
In a balloon, the gas is free to expand, but not completely free. In other words, it's a situation somewhere between the two described above. The balloon skin is an elastic which pulls itself, creating a force vector pointing towards the interior of the balloon, and the more the skin is stretched, the stronger the force becomes. Now the gas inside the balloon has to create enough pressure to compensate for both the ambient pressure, and the elastic force trying to pull the balloon skin inwards. This means that after heating, the gas inside a balloon will expand since the balloon is not perfectly rigid, but the equilibrium pressure of the gas inside the balloon will be higher than before because the balloon is pressing against a more tightly stretched balloon.
This should be provable experimentally without much difficulty. Take a rubber balloon, preferably one with as thin of a membrane as you can get, and open its mouth to the atmosphere. Then, clamp it shut without blowing any air inside (in reality, you might have to blow some air in to unstick the rubber walls, but then let all the excess air out). The air inside will have a pressure exactly equal to ambient pressure, because the elastic is not being stretched. Now cover the balloon completely in warm water for a few minutes, and it should inflate slightly. Finally, remove the balloon from the water and quickly perforate it with a sharp object. You might be able to hear a small pop, and feel a rush of air. Both are an indication that pressure inside the heated balloon was higher than ambient pressure.

Notes:

Here is a very related Physics.SE question - Why is the pressure inside a soap bubble higher than outside? which deals with bubbles rather than balloons. The diagram and equations are applicable in both cases.
Evidently, taking into account the elasticity of the balloon requires a more subtle treatment of the problem. Take a look at Floris' interesting input in the comments. It seems that even an idealized balloon starts off acting as a quasi-rigid container. After reaching a maximum pressure, the balloon starts expanding, and from there the walls get weaker as expansion continues. This means that for a range of temperatures higher than some critical value, the pressure of the gas actually decreases as the temperature increases. The interior pressure of the balloon will still always be greater than ambient pressure, though.

It is worth pointing out that the elasticity of the balloon material usually varies (a lot) with temperature: as the material becomes more compliant (lower modulus) the balloon will expand more and the pressure could actually drop. Note also that an expanding balloon has lower pressure as the volume increases. See for example this answer on Physics.SE

@Floris Very interesting, thank you for the link! I'm trying to wrap my head around this. Even assuming the elasticity doesn't vary with temperature, does that mean that the unstretched balloon with gas at ambient pressure, when heated, initially goes through a quick increase in pressure with little volume change, reaches a pressure maximum, and from there the pressure decreases as the balloon expands?

@NicolauSakerNeto - yes, I think that is correct when you assume perfect elasticity. The thinner balloon is less able to support high pressures.

This answer provides a great treatment of buoyant gas balloons.  Aeronautics fans:  A hot air balloon is open at the bottom - the gas is completely free to expand and escape.  Pressure and volume stays constant, density and weight decreases.

Almost six years later, I have come across something interesting. The physical description of the two-balloon experiment provides further insight relevant to this problem.

You may recall the ideal gas law: $$PV = nRT.$$
Here, $P$ is pressure, $V$ is volume, $n$ is the amount of gas present (in moles), $R$ is the ideal gas constant, and $T$ is temperature.
In an enclosed system, with no gas flowing in or out, $n$ is constant (as is also, obviously, $R$).  We can rearrange the equation above to pull all the constant terms to one side, like this: $$\frac{PV}{T} = nR = \text{const}.$$
Thus, we can see that, with $nR$ staying constant, the left-hand side of the equation, $\frac{PV}{T}$, must also stay constant.  Thus, the three quantities $P$, $V$ and $\frac1T$ are inversely related to each other: if you increase one of them, at least one of the other two must decrease so that their product remains constant.
Having noted this, it should be obvious what's happening with your balloon.  By heating it, you're increasing the temperature $T$ (and thus decreasing $\frac1T$); to keep $\frac{PV}{T}$ constant, at least one of $P$ and $V$ must therefore also increase.
As it happens, in a rubber balloon, both the pressure and the volume will typically increase, since any increase in pressure will cause the balloon skin to expand, increasing the volume (and reducing the pressure) until the inward force exerted by the stretched balloon skin is balanced by the pressure of the gas inside it.
With a mylar balloon, however, things would be different, since mylar is not particularly elastic.  In that case, starting with a partially inflated balloon, the pressure would remain approximately constant (and equal to ambient pressure) up to the point where the mylar skin was stretched taut; at that point, the volume would stop increasing, and the pressure would start to build up instead (potentially up to the point where the balloon would burst).


text:
Why is mercury liquid?
Why is mercury the only metal that is liquid at room temperature?
Note: I am a high school student and would appreciate an answer, if possible, suited for my level of understanding.
Thank you.

This question addresses it quite nicely I think. Another great post is by Frank Senese. Especially the referenced publication, J.  Chem. Ed., 1991, 68, 110-113 is interesting, but probably behind a paywall http://pubs.acs.org/doi/abs/10.1021/ed068p110

BTW, gallium melts ~30C, "room temperature" in a warm climate (it would melt in your hand, too).

An interesting paper on this was published in 2013. Take a look at this news article with an overview and a link to the paper.

There are a couple of reasons why Mercury is a liquid at STP, I will go through each of them with an explanation that won't be too complex for your understanding.
First of all understanding why most metals are metallic is a good place to start. This happens because of the sharing of valence electrons. The sharing of these electrons make metals malleable and ductile and thus not liquid. Generally the more the electrons shared means the higher the melting point and a harder metal.

OK, getting into the Mercury. Like others have probably mentioned above, Mercury doesn't like to share its electrons. It hangs onto them very tightly and thus Mercury-Mercury bonding is very weak and thus the compound melts and boils at very low temperatures. In other words, Mercury lacks the sea of electrons.
So then, why doesn't it like to share electrons? The $\ce{6s}$ electrons come very close to the nucleus and swing around it at near the speed of light. Due to this, relativistic effects occur. In other words, the $\ce{s}$ electrons behave as if they are more massive than other electrons. This lowers the energy of the electrons. To summarize, the electrons are buried deep in the atomic core and thus aren't able to participate in chemistry.
Overall, since you are a high school student, I would probably just say that Mercury doesn't readily bond well and that is due to the $\ce{6s}$ electrons being very attracted to the atomic core.

Isn't the sea of electrons what give metals their metallic look? If so, then why is mercury shiny and silvery?

There is a theory of relativistic effect. According to it, in elements with atomic number 80 or greater, the electrostatic attractive force is very high in the valence shell. This causes the 6s valence electrons of mercury to be actually strongly bonded. Now there is also a theory 'electron sea model' where metals share their valence electrons which leads to adhesive forces among metal atoms. As mentioned earlier mercury's valence shell electrons are strongly attracted to its nucleus so mercury atoms do not benefit from the adhesive forces due to very less sharing of the valence electrons.
Hope this answers your question.

At room temp at atmospheric pressure mercury will not readily share valence electrons with other atoms. The mercury atoms won't interact with each other to form a structure or lattice to become a solid.

This is quite a short answer, and does not really address the why. It might be better suited as a comment or extended to include more information.

Sorry I've got more info on the topic but it said to keep it simple.  What aspect of it should I include more info on if I were to edit the answer? Thank you for feedback :)

I think there should at least be a reason for your first statement, why does mercury not share electrons easily and why is that the reason for not forming a lattice. Right now, it seems you are just stating the obvious. Keeping it simple means here probably only, that the argumentative structure should be easy to follow, with basic explanations.

And you should not be sorry, I am sure the OP appreciates your help in any case. I just wanted to direct you a little bit, that your answer might not be very satisfactory. In the end we are all working together here.

I just noticed when I've gone to edit that others have left some answers similar to how I was going to edit mine. Should I remove my answer or just leave it a simplified response. ? Others don't mention the lattice structure ?

Just give it a go, add another aspect to the question. The part about the lattice is certainly interesting enough. With a little more information I think you are on a good way to a nice answer.

Thanks heaps, it was what I got told when I asked my Chem teacher the same thing. I'll spend a bit more time constructing the answer so not to waste anyone's time :)

It's nearly a physics question.