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What does the ‘−1’ superscript mean in units?
For example, the rate of a chemical reaction can be expressed in $\mathrm{mol}/\mathrm{L}^{-1}/\mathrm{sec}^{-1}$. Why is it ‘−1’ and not, say, ‘−2’? Does it change the meaning if the minus is removed and we simply express the rate in $\mathrm{mol}/\mathrm{L}/\mathrm{sec}$?

The answers below are correct but none seem to mention that in math $x^{-1}$ equals $\dfrac{1}{x}$ for some variable $x$. The same thing applies here.

@Calvin'sHobbies while my answer doesn't say that explicitly, it does say it implicitly with the depiction of the example as a fraction.

Please note that a solidus (/) shall not be followed by a multiplication sign or a division sign on the same line unless parentheses are inserted to avoid any ambiguity. Besides, the unit symbol for ‘second’ is ‘s’ (not ‘sec’).

@Calvin'sHobbies More precisely, given a multiplicative Group $(G, *)$ we denote for $x \in G$ an element  $x^{-1} \in G$ as an inverse of $x$. Depending on the Group it can mean different things and not necessarily $\frac{1}{x}$. In our case it is $\mathbb R$ Field we're probably talking about, but a Field encapsulates additive and multiplicative groups.

The -1 means "per" unit. So your first example mol/L-1/s-1 is not correct - it would actually be written as mol L-1 s-1, OR mol/(L s). It is also sometimes written as mol/L/s, but the double division is ambiguous and should be avoided unless parentheses are used.
If it were mol L-1 s-2, this would mean moles per litre per second per second.
This is really just a question of notation, and is not chemistry-specific at all. Yes, all the minus/plus signs and the value of numbers are important. Good examples of units can include:

area, measured in m2, or metres squared
volume, measured in m3, or metres cubed
pressure, measured in N m-2, or Newtons per metre squared
velocity, measured in m s-1, or metres per second
acceleration, measured in m s-2, or metres per second per second

The $^{-1}$ superscript can be thought of as saying "per" or as being the denominator of the fraction.
So in your example $\mathrm{mol \cdot L^{-1} sec^{-1}}$ can be thought of as saying moles per liter per second.
This is easier than writing $\mathrm{\frac{mol}{(L \cdot sec)}}$
Changing the super script from $1$ to $2$ or $3$ would change the meaning of the value.
Ex
$$1 \mathrm{cm^{3}\ is\ 1mL}$$
So, $\mathrm{cm}^{-1}$ is per centimeter, which would be a measurement of something per distance, but $\mathrm{cm^{-3}}$ would be talking about something in a given volume.

Generally correct, but fails to mention that the unit abbreviation for the second is simply s, not sec.

It may have its roots even earlier than that, but this was mainly due to people using typewriters to write scientific papers, etc.
Now we have the ability to format things like $\mathrm{\frac{mol}{L}}$, both on-screen and in print, but adjusting the carriage and line feed knob every time you had to type a complicated formula was tedious, so it was easier to type "mol-L-1" instead.  Even when the -1s became superscripts, as John points out in his answer, it was still used in typesetting to keep formulas, etc. all on the same line in books.

Even if we do not use typewriters anymore, an inline fraction  just looks plain awful and makes a manuscript very hard to read, since there will be different spacing between lines in a single paragraph.

First off: your suggestion $\require{cancel}\cancel{\mathrm{mol/L^{-1}/sec^{-1}}}$ is very wrong for three principal reasons:

the unit symbol for seconds is $\pu{s}$, not $\pu{sec}$ or anything else
you should never include two slashes for division. Does $\mathrm{mol/l/s}$ equal to $\mathrm{mol/(l/s)}$ or to $\mathrm{(mol/l)/s}$? This is ambiguous. One should always indicate with brackets which units are ‘per’ and which are not; in your example it should be $\pu{mol/(l\cdot s)}$.
your suggestion does not mean what you think it means; more on that below.

Mathematically, a negative exponent has the same effect placing the expression associated with it into the denominator.
$$\begin{align}x^{-1} &= \frac 1x\\[0.3em]
2^{-2} &= \frac1{2^2}\\[0.3em]
e^{-i\phi} &= \frac1{e^{i\phi}}\end{align}$$
Units in the natural sciences are treated much like variables in general mathematics, i.e. they can be multiplied and thereby raised to powers (e.g. $\mathrm{m^2}$) or divided by each other (e.g. $\mathrm{m/s^2}$). Only if the unit is identical, two numeric values can be added or subtracted; so $\pu{2m}+\pu{3m}=\pu{5m}$ makes sense as does $2a + 3a=5a$, but $\pu{2m}+\pu{3s}$ cannot be added akin to $2a+3b$.
The combination of units usually means what common sense would read them as. So $\pu{1m^2}$ is equivalent to a square area with the side length being $\pu{1m}$. $\pu{1 N\cdot m}$ is equivalent to a force of one newton applied over the distance of 1 meter (with a lever). And $\pu{1m/s}$ means travelling one meter per second. While more complex expressions such as $\mathrm{kg \cdot m^2 / s^2}$ do not always immediately make intuitive sense, they can usually be broken down into fragments that would make intuitive sense.
After this excursion, it becomes clear that an expression such as $\pu{mol\cdot l^-1\cdot s^-1}$ is equivalent to a fractional unit of $\mathrm{\frac{mol}{l\cdot s}}$, meaning that the concentration is increased by $\pu{1 mol/l}$ in one second. This also means that:

it does not make sense to replace the exponent of $-1$ with e.g. $-2$ as that would result in a different unit (e.g: $\mathrm{kg\cdot m^2\cdot s^{-2}}$ is joule, the unit of energy, while $\mathrm{kg\cdot m^2\cdot s^{-3}}$ is watt, the unit of power).
it does not make sense to remove the negative sign from the exponent as that would result in a different unit (e.g. $\pu{10Hz} = \pu{10s-1}$ corresponds to a frequency — ten times per second — while $\pu{10s}$ obviously corresponds to a duration).
one has to choose between either the slash or the negative exponent as both would cancel each other out.

This last one is implied by the general laws of mathematics:
$$\begin{align}\frac1{x^-1} &= \frac1{\frac1x}\\[0.5em]
&= \left(\frac11\right) / \left(\frac1x\right)\\[0.5em]
&= \left(\frac11\right) \times \left(\frac x1\right)\\[0.5em]
&= x\end{align}$$
which is also the third wrong factor in your suggestion.
In general, I would give preference to the negative exponents ($\pu{mol l-1 s-1}$) except in cases where there is only a single unit raised to a power of $-1$ and no other powers exist; in these cases, e.g. $\pu{mol/l}$ usually integrates itself better into the flow of text.


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Phenol protection
Would an acceptable way to protect a phenolic alcohol be through a Williamson ether synthesis? As in:
1) Deprotonating phenol with sodium hydroxide.
2) Addition of methyl iodide.
3) Deprotection of ether with strong, hot, concentrated acid such as HI.
I know there are more sophisticated ways of protecting alcohols, but what's wrong with this method (if any)? It certainly seems relatively simple and effective.

Protecting a phenol by using the Williamson ether synthesis to make the methyl ether is an acceptable method.  However your deprotection step using strong, hot acid is undesirable since the molecule might contain acid or thermally sensitive groups.  A milder way to deprotect a phenolic methyl ether involves the use of boron tribromide.
Here's a nice link to a page listing a variety of methods to protect and deprotect phenols, with examples for each route.  Depending on your actual molecule, the dimethyl sulfate protection route might be preferred.

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Enthalpy of formation and reaction equations
I'm disagreeing with my chemistry teacher over this enthalpy of formation problem, so can someone tell me if I am right or why I am wrong?

The enthalpy of formation of gaseous sulfur trioxide is $-396~\mathrm{kJ/mol}$. What is the enthalpy of the reaction represented by the following balanced equation?
$$\ce {2 SO_3 (g) -> 2 S (s) + 3 O2 (g) }$$
(A) $-396~\mathrm{kJ}$   (B) $396~\mathrm{kJ}$   (C) $792~\mathrm{kJ}$   (D) $-792~\mathrm{kJ}$   (E) $198~\mathrm{kJ}$

I got (C) by flipping the sign of the enthalpy of formation given (because the equation given proceeds in the opposite direction from the formation of sulfur trioxide) and multiplying it by two (because there are 2 moles of sulfur trioxide).
My teacher and most of my other classmates said that it was (B) instead. They said that I shouldn't multiply by 2 because the equation given is at it's lowest realistic coefficient values.
I don't understand this for a few reasons:

the enthalpy of formation is defined as the enthalpy change per mole of substance
in previous problems, we would change the chemical equation to make the product only have one mole

For example, here's one problem:

For which of the following reactions is the heat of reaction equal to the heat of formation of the product?
  \begin{align}
\text{(A)} &&\ce{N_2 (g) + 3 H_2 (g) &-> 2 NH_3 (g)}\\
\text{(B)} &&\ce{1/2 N_2 (g) + O_2 (g) &-> NO_2 (g) }\\
\end{align}

We narrowed it already down to (A) and (B).
My teacher said the answer was (B), because while both equations had the elements in their elemental form, the first reaction produces 2 moles of product while the second produces 1 mole of product. Thus the fraction coefficients were necessary.

I agree with you, but it has been some time since I needed to have the correct answer for this type of question.

Wait wait hold up a second. The standard enthalpy of formation is defined as the enthalpy of formation of one mole of whatever. You simply said enthalpy of formation. That could refer to 1 mole, 2 moles, 1000 moles, or 1/10000 moles. I ctrl+F'ed your post and I don't see the word "standard" anywhere, suggesting that you aren't actually referring to the std. enthalpy of formation.

I assumed it was standard because it came from the chart "Thermodynamic Quantities for  Selected  Substances at 298.15 K and 1 atm" in the appendix of our book. But even if I can't assume that, the enthalpy value given is in the format of -396 kJ for every 1 mole of sulfur trioxide.

If your teacher asks a similar question in the test you should ask whether the given value is the enthalpy of formation or the standard enthalpy of formation. If it is the former, then the question is ambiguous because without giving you the reaction equation for the formation process you can only guess the stoichiometric coefficients. If it is the latter then your reasoning is exactly right.

@2mia Yes, you may have been given standard thermodynamic data, but that doesn't mean you can't calculate non-standard thermodynamic information.

As far as I can tell, your reasoning is sound, though the question is actually not written spectacularly well. By "enthalpy of the reaction" it seems to be implicitly assuming that the stoichiometric numbers are referring to moles (making all the answers have units of kJ instead of kJ/mol). The question doesn't specify standard enthalpy change of reaction, but even if it did, the most common convention for balancing equations is to use the smallest whole numbers for stoichiometric coefficients, which is what is written there and would give the answer you got. You can calculate an enthalpy change of reaction for any reaction conditions you like and the value you get will change depending on the quantities of the substances involved, but if the question is implicitly assuming moles for coefficients, you still get the same answer.
Incidentally, there's no reason you couldn't use some other definition of "standard" in which you set the coefficients such that the smallest ones are equal to 1, though I can't remember seeing anyone do it this way. IUPAC convention is that whenever a quantity referring to some standard state is used, the conditions of the standard should be specified (though chemists like to ignore IUPAC).

To clarify, I'm not saying all the coefficients should be divided by two for the equation she gave. I'm saying that the enthalpy change she gave applies to an equation where the whole numbers aren't coefficents, specifically the equation $\ce{ S(s) + 3/2O2(g) -> SO3(g)}$. I infer that that enthalpy value applies to that equation because it is given as -396kJ/mol $\ce{SO3}$

If the enthalpy of formation of $\ce{SO3}$ is given in kJ/mol, i.e. as the standard enthalpy of formation, then the only way for the answer to be (B) is to divide all the coefficients. Hess' law says that the difference of the enthalpies of formation multiplied by the coefficient on each side of the equation is equal to the reaction enthalpy change. Since the right side is 0, the only way to get (B) is to halve all the coefficients.

I found an enthalpy problem set PDF file:
$$\ce {2 S (s) + 3 O2 (g) -> 2 SO3 (g)}\qquad \Delta H = -794.45 \, \mathrm{kJ}$$
So the best answer is C.
After reading all the answers and comments, I have noticed that the question asks for the enthalpy of the reaction. Since the reverse reaction has a value of $\pu{-396 kJ/mol}$, and the current reaction has 2 moles of $\ce{SO3}$, then the answer is $\pu{792 kJ}$ (the energy for the total reaction). The answer depends on the number of moles in the reaction

While ambiguously worded, if by the "enthalpy of reaction" the teacher is implying the standard enthalpy of reaction, this being the energy to form one mole of product, then by the most common usage of the words used in the problem, "B" is a better solution than "C", as it utilizes the more precise definition of energy per mole of formation.  However as the word "standard" was never used you certainly have a valid argument as to why "C" should also be considered correct.

Energy per mole of formation of what? There are two products with different coefficients. You cannot calculate a single value that would work for both. Without being specifically instructed to calculate the enthalpy of reaction for one mole of one of the two products, it seems illogical to arbitrarily divide all the coefficients by 2 rather than use the balanced, whole number equation as written.

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magnetism in metallic alloys
i am working with a multicomponent alloy system for soft magnetic application,it is iron based and have composition of Fe Mn Si B Mo,how can i calculate the net magnetic moment or saturation magnetization of the system,consider proper mixing of alloys in melt takes place.
kindly tell me the theoritical way of calculating,and individual contribution


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Precautions and tips for Qualitative Inorganic Analysis of Coins
I'm a highschool student and as part of an investigatory project, I have decided to analyse which metals are present in our everyday coins.
I chose this because in my previous academic year, I had learnt how to analyze the cations present in inorganic salts and I employed the same systematic analysis method that I had learnt to test the presence of metal cations in some coins.
Obviously, I cannot expect the coin analysis to go exactly like the salt analysis. I would have to ignore certain tests like the 'flame test'.
Could you please tell me what precaustions I should take for this experiment? I wish to write up a clear and systematic process by which any student / coin collector can easily do in a laboratory with only common chemicals and salts.
All advice, resources and tips will be most appreciated.

Just for completeness I'll add that any sort scrapping or acid test would damage the coin which would hurt its numismatic vale. XRF is nondestructive.

In addition to purely chemical aspects of the problem, be aware that some jurisdictions frown upon the destruction of money. This may be related to the stability of the monetary system (similar to counterfeiting money) or to the state symbols present on the coin.

1) The coins must be dissolved (at least partially). There is a good amount of metals that won't dissolve in common  acids (hydrochloric acids), some will require nitric acid, some will require aqua regia and some will require even worse, though that's unlikely. So, most likely you will have to deal with concentrated nitric acid, that produce nitric oxide (fox's tail) when react with metals. Given that you must have fume hood. In case something undissolved remains, you can try aqua regia, and if even that fail, you have something rare, like nioubium, with no common method of wet analisys.
2) Most metals used in coins have toxic salts. So, be cautios, and avoid skin contact. Latex gloves and protective goggles are not vital, but certainly recommended unless you believe you can be accurate enough. Usual safety requirements for work with conc. acids applies as well.
3) In case of non-destructive analisys, X-ray fluorescent spectrometry may be employed. It will give accurate analysis in almost any realistic case, but requires quite costly equipment. It is not extremely costly per analysis, though.

I only have access to fundamental chemicals and salts, it's hard enough to get purely conc. HCl + we have no fume hood. Should I just state in my project summary that I was not  able to find all the metals due to the coin not dissolving properly. I can only do cheap wet analysis.

@Nick conc. HCl is usually available via some shops, it's not super-rare staff. You will need conc. nitric acid for sure, though, there is no way around it, as I see it.

Ok, I am absolutely fine with doing destructive analysis :D Also, the coins I'll be analysing will be made of the following alloys: Ferritic Stainless Steel and Cupro-Nickel. Any special advice to verify the presence of all the metals?

As for the dissolving, if you cannot use strong acids, it might be worth trying anodic oxidation. As for the detection, go ahead with your cation analysis, just the sample preparation is more involved.

I am of the opinion, that protective goggles are vital, always, any experiment. You never know what might go wrong and losing eyesight is just terrible.

@ssavec: That's a great idea, but my teacher has asked me to verify if I should chuck the flame test because it wouldn't be systematic analysis if I don't check for Group 5 (Ba, Ca, Sr) which isn't something you'll see in coins. Should I scrap it or is there a way to do it?

@Nick the only common coin metal identifiable with flame test is copper, and honestly, flame test is not the most reliable way to check for copper. I would scrap the flame test.

@Nick: If you mean normal flame test, just scrape the coin and put the dust into flame, you are about to use the worst color and intensity detector around, i.e. your eye. Maybe using camera and then analyze the spectrum could be slightly better (see Atomic absorption spectroscopy).
But still, you are about to analyze the mixture of metals, not to identify unknown pure metal. This is much more difficult task and some metal may mask others, some do not show at all, etc...

@ssavec: How do you scrape stainless steel?

@Nick: file or sandpaper? Both in diamond version, if you want to have your sample cleaner.

I know this does not answer what you specifically asked for, but it may be a good way to get quick analyses of coins.
Try to get a quick scan with an X-Ray Fluorescence gun (XRF gun), if you want to use a modern, non-destructive analytical technique. It is common for firms and certain types of stores to have one such gun, for example jeweller stores keep such guns to check the gold or silver purity, or toy stores may keep a gun to check for dangerous elements in imported toys.
Maybe you could talk yourself into getting a free scan for your school project. They may have pre-installed metal calibration programs, and a scan takes 30 seconds. if your coin is plated with some metal, and the rest of the coin is some other metal-alloy, you can see this with an XRF gun if you scratch your coin on one side before analysis.
XRF is a nice technique, and you should be able to have a good idea of how it works (in principle) with theory from high school chemistry or physics. Modern chemical analysis is not "wet chemistry", or flame tests; it is instrumental analysis (although wet techniques are still used).
To conclude, see if you can find a local store or university or someone who has an XRF gun. That would give you the proportions (%) of the major components and a standard deviation of the analysis (error or uncertainty).
Good luck!

Aww, I appreciate this answer very much. thanks. This is a good tip (and hence in the scope of this question). Maybe an XRF gun will be able to find the chromium that I couldn't.

I don't know where you are from (and hence not what kind of coins you're analyzing), but the US Government does not mention any chromium in their coins (link at bottom). I have no idea how they're reporting the results, and maybe the coins have trace amounts of chromium. If so, XRF may not detect it; at least not an XRF gun (there are stationary XRF instruments used for laboratory analyses - the guns are meant for quick in-field analyses). Then again, maybe you would find something. I am sure the coins are not as pure as reported here.
http://tinyurl.com/67mo73b

If you are going to use a XRF gun, make sure that you do not hold the coin in your hand when you make the measurement. I suspect that some users of those XRF gadgets have some very bad radiation safety habits. I have heard of people holding the sample thus giving their hand a dose as they make the measurement.

I am thinking of three methods, you could put the coin in a scanning electron microscope if it has a X-ray spectrometer attached and then get a spectrum of the X-rays which come off the coin.
If you had access to a TRIGA nuclear reactor then you could try neutron activation combined with gamma spectroscopy. If you use a short irradation time in the reactor and count quickly after activation then you can measure things like copper. It is generally non destructive but it is possible to induce long lived radioactivity in the sample. I suspect these days that very few people will have access to a TRIGA reactor.
I think that a good option would be to digest the coin in aqua regia and then dilute and filter it into a volumetric flask which has been soaked in acid to make sure that the metal contamination level is low. I would then suggest ICPOES as a method of making a measurement of the metal concentrations in the digested coin.
In some cases such as copper it might be possible to take the digested coin and use reagents such as dimethyl phen to make strongly coloured complexes, you can then measure the concentration of the complex with a UV/vis machine and then calculate the concentration of the metal in the flask. From the mass of the coin you can then estimate the metal content in the original coin.

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Orbital overlap and size "mismatch"
Why is atomic size mismatch a sign of poor bond strength? Wouldn't orbital overlap be proportional whether the two atoms are small or the two atoms are big?
For example I heard someone comment that an argument that the C-I bond is weaker than the C-F bond based on orbital overlap made no sense to them, because why would the size of the individual orbitals matter in overlap? Small atom, big atom and small atom, small atom - shouldn't their overlaps be proportional? Or something along the lines of that. I'll have to pull up the post again ...

Atomic orbitals of similar energy lead to a better overlap than atomic orbitals of dissimilar energy.  The valence 5p orbital involved in bonding for iodine is much higher in energy than the valence 2p orbital.

Small atom, big atom and small atom, small atom - shouldn't their
  overlaps be proportional

No.  Look at the figure below and let's consider s-orbital overlap, just to keep it simple.  Remember that these are spheres with the electron density distributed throughout the entire volume.  In the "small-small" case there is effective overlap of the electron density in both 1s orbitals; we would expect the resultant bond to be strong.  In the "small-big" case, the small 1s orbital involves a significant fraction of its electron density in overlap, but not much of the electron density in the large s orbital is involved in overlap.  In the "small-big" case the overlap of electron clouds is ineffective and any resultant bond would be weak.


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Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof)
Why is the C-S bond in $\ce{CH_2SH^-}$ longer than in $\ce{CH_3SH}$? Wouldn't there be resonance in the former, forcing the C-S bond to contract?
http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431

Also while we're at it are there any non ab initio calculations/empirical studies on this interesting thiol ion?

Out of curiosity, what is this paper - can you give a DOI link I'm curious what level of theory was done. A naive thought is that there might be some triplet character to the anion.

Also there will be syn- / anti- (or cis-/trans-) geometry issues with $\ce{CH2SH-}$ that may not have been considered (i.e., I suspect each form will have slightly different bond lengths .. essentially due to differences in the double-bond resonance picture)

@GeoffHutchison isn't that what you have TAs for? Sorry here's the correct link: http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-431

I'm most interested in empirical calculations lol. Because if I show my prof this paper I will get an earful about computational chemists, ahem, I mean "chemists." That's just his POV though.

How would you be able to measure this empirically? That is, how would you be able to make a carbon anion in the presence of a free thiol?

@jerepierre gas phase? And keep the concentration very low. I don't think the carboanion will form a three-membered ring and eat its tail.

Starting from what though?

On a different topic, can you explain what you find interesting about this compound?

Start from $\ce{H3CSH}$? You're right though I can't see what would selectively remove a methyl proton rather than a thiol proton. So are you saying it's impossible?

Why is the C-S bond length of the anion longer than the C-S bond length in the neutral thiol? Despite the possibility of resonance.

Let us continue this discussion in chat.

In any case @GeoffHutchison I don't mean to offend you, if that wasn't clear in my previous post about what my prof thinks. That's what he thinks, not me. Personally I'd like to prove my prof wrong about sulfur's bonding tendencies (hence this question ...)

Quick Summary
The linked paper (from 1975) is probably wrong. A slightly more recent paper (J. Am. Chem. Soc., 1989, 111 (21), pp 8112–8115) suggests the bond lengths are about equal. Some quick calculations (below) using multiple modern methods suggest that, as expected, the C-S bond length in $\ce{CH2SH-}$ is shorter than in $\ce{CH3SH}$.
For both older papers, the computational results (using basic Hartree Fock) would be considered highly suspect under modern peer review, since Hartree Fock is known to have problems with anions. Certainly the 1975 paper doesn't even mention the basis set.
Seen below, it's clear that the results are method dependent, but the best current methods, with reasonable basis sets indicate the bond length is at least slightly shorter due to some partial double-bond character.
More Details
The first thing to consider in this is that there are two forms of $\ce{CH2SH-}$, syn and anti, based on the orientation of the formally C-bound anion and the SH group:
syn form

anti form

These will have different bond lengths, since the anti form will have contribution of resonance structures II and III from Ron's answer. This gives some double-bond character and should likely have a shorter bond, although electrostatic repulsion could also be invoked.
The syn form won't have a clear effect, since the C anion lone pair and the S lone pairs are on the opposite side of the bond. My guess is that the bond length is about the same as $\ce{CH3SH}$.
I think the clear message is that we can explain a lot of things, but it's better to know if our results are accurate before trying to explain them.
What I see with more modern computational methods is that the syn form has similar C-S bond length to $\ce{CH3SH}$ and the anti form is slightly shorter. I suspect the previous result is simply wrong. Calculations involving anions are tricky, since the extra electron is usually highly diffuse. I expect the basis set was simply insufficient.
The difference in energy between the two forms is ~1 kcal/mol, so experimentally, these will interconvert readily, and an average bond length would be observed. As mentioned in the comments, this species would probably be hard to observe, even in the gas phase, since $\ce{CH3S-}$ is more stable.
Computed C-S bond lengths (in Å):
For all methods, full geometry optimization was performed using the methods indicated. I would consider the CCSD calculations to be high quality.

Is it possible to make -H2CSH

Bleh. I'll make a better table later. Suffice to say that the results are method dependent, and the syn form is about the same as CH3SH. I'll give better results later.

@GeoffHutchison Geoff, in water the 2 lone pairs are hybridized differently, one in a mostly p-orbital, the other in a roughly sp-orbital.  How are you treating the lone pairs in your calculations?

@ron These are MO calculations with full $C_s$ symmetry for the $\ce{CH2SH-}$ anion, and thus there's no notion of hybridization of lone pairs in these calculations.

I wonder about the energy difference between syn and anti. It can't be much and most likely might be considered as a free rotation, hence averaging would be appropriate.

The energy difference seems to be ~1 kcal/mol regardless of method, and so averaging would certainly be appropriate. However, it's interesting that the C-S bond length difference is sometimes fairly significant.

First, the difference we are talking about is only 0.074 A, it's always difficult to be sure what causes such small changes.  That said...
The amount of negative charge on the carbon atom in $\ce{CH3-SH}$ (-0.639) surprises me; it's actually more than in the corresponding anion $\ce{CH2^{-}-SH}$ (-0.620). Note too the significantly greater positive charge on the carbon hydrogens in $\ce{CH3-SH}$.  Perhaps this suggests that hyperconjugated resonance structures like I are important in $\ce{CH3-SH}$.  This would result in the $\ce{C-S}$ bond being shortened to begin with.

Then, in the case of the anion, resonance structures II and III suggest that there is appreciable negative charge on both carbon and sulfur, consistent with the high $\ce{q_{b}}$ and $\ce{q_{c}}$ values reported in Table I (-0.620, -0.328, respectively).  The appreciable negative charge on both carbon and sulphur, in the anion, results in coulombic repulsion and a slight elongation of the carbon-sulphur bond, from the somewhat shortened bond length observed in the neutral compound.

Can you think of anyway we can study this thiol ion empirically?

I would run calculations on the corresponding thioethers and see if the same pattern holds.  If it does, the anion should be accessible from the thioether.

How would you go from the thioether to the thio-anion? I'd imagine having to cleave the S-R bond ... but then ... we'd have to stick a proton in place of the R to make S-H. How can we selectively put the H on the S rather than on the CH2(-)? @ron

We're not putting a proton on S.  We'd simply pull a proton off of the methyl in the thioether, that's all, and then we would have a carbanion alpha to sulfur.

but then we have a inductively donating methyl stuck on the sulfur's other side too. The thing I was thinking about earlier with someone was how do we make -CH2SH so we can study it empirically? Any protecting groups to protect the thiol's proton? And any way to selectively put back the proton on the S?

@Dissenter  Correct, that's why I said I'd do a calculation first to make sure the shortening effect still occurs in the thioether series.

can we though make -CH2SH?

@Dissenter  It's CH3SCH3 that I'm referring to.

I understand, I'm just curious if synthesis of the anion is possible

Specifically the H2CSH anion, since you clearly know more about protecting groups. Or are the realities of acid/base chemistry inescapable? I'm thinking that a smart-ass answer would be yes, of course we can make it, just ionize a million moles of CH3SH and hope that probability is in your favor!

@Dissenter Yes, I suspect it could be generated in the gas phase, but would rearrange very quickly (it is sooo downhill from CH2- to RS-), perhaps too quickly to make whatever measurements are needed.

Can we zap one molecule in isolation and hope for the best?

There's a paper suggesting it might be stable, but I find no manuscripts on this experimentally, except in mass spec fragmentation.

Dissenter asked, "Can we zap one molecule in isolation and hope for the best?"  That's what the gas-phase experiment is.  But like I said, I would have guessed that intramolecular 1,2 hydrogen shift from S to C would be too fast to allow enough time to record a microwave spectrum (to determine bond lengths) of the initially generated carbon-based anion.  But @GeoffHutchison 's reference suggests an appreciable barrier (46 kcal/m) has been calculated for the hydrogen shift in the  corresponding oxygen case.  I guess that's why we do experiments.

@ron I'd even guess there's a significant H tunneling rate in the syn form (or an inversion from anti to syn followed by H shift). So I agree it would at least be tough. Unfortunately, my microwave expert retired a few years ago, so I can't ask.


text:
Are all anomers diastereomers?
I am told that all anomers are diastereomers. I think this is BS. If there is a carbohydrate with only one chiral center, then changing the configuration of the anomeric carbon gives us enantiomers.
Or is it impossible to have carbohydrates with only one chiral center?

It could be possible to have a pair of enantiomeric anomers if you started from an achiral sugar. But this is not possible. Another way would be to extend the definition of anomers. Then, using jerepierre's theoretical example, DL-glyceraldehyde would give a mixture of 4 stereoisomers including 2 pairs of enantiomeric "anomers"

Sugars are by definition chiral then @K_P?

I don't think the definition of sugars prohibits chirality. Obviously aldoses can not be achiral because of their inherent asymmetry. But after your comment I thought of one possible achiral sugar: a symmetric ketopentose. This has 2 stereocentres but is a meso compound and thus achiral. IF it exists and IF it can form anomers then it should give enantiomeric anomers.

The only saccharides I know of that have only a single chiral centre are the trioses L- and D-glyceraldehyde, but, at least according to the Gold Book definition, anomers are specifically diastereomers of cyclic forms of sugars. Trioses are too short for a cyclic form to normally arise and therefore can't anomerize.

Adding a bit to Michael's answer. Although it would be highly strained, let's look at what happens if glyceraldehyde cyclizes. Since glyceraldehyde has one stereocenter, the anomeric position becomes a second, giving diastereomers.

So in order for the anomers to be enantiomers, the compound actually must have zero stereocenters in the open form. If you consider glycoaldehyde a carbohydrate and you consider the three-membered ring cyclic hemiacetal a reasonable structure, then you have found a set of anomers that are enantiomers.

Ah, you're quite right. I hadn't considered that. Since the modern biochemical definition of saccharide seems to preclude anything shorter than trioses (and those would be even harder to cyclize), I guess we're okay.


text:
D/L Configuration
Why are D and L-glucose enantiomers???! I thought D/L refers to the "configuration" of the last chiral carbon on the open chain Fischer projection. In the D form, the higher priority group is on the right in the Fischer projection; in the L form the higher priority group is on the left. So why are D and L glucose enantiomers? Shouldn't they be diastereomers??!

Diastereomers aren't mirror images.

Simply by inspecting the two figures of the molecules, you can see that they are in fact enantiomers—non-superimposable mirror images and were they diastereomers instead of enantomers, they wouldn't both be named "Glucose". The D-/L- system is simply an antiquated naming system for enantiomers based upon the relation of the structures to glyceraldehyde. D-glucose refers to the enantiomer of glucose in which the C5 stereocentre has the same configuration as (R)-glyceraldehyde and L-glucose that of (S)-glyceraldehyde. Because this notation refers specifically to enantiomers, it is assumed that the configuration at all stereocentres is inverted and the configuration at the last stereocentre was chosen to assign unambiguous names.

Okay, thanks! I didn't know that D/L applied specifically to enantiomers .... I was just told to look at the penultimate carbon.

For D-glucose, it has the configuration (2R,3S,4R,5R). For L-glucose, it has the configuration (2S,3R,4S,5S).
As a general rule, for an enantiomer, all S will change to R and vice versa at each respective carbon centres. All the other configurations are diastereomers.
For example, (R,R) and (S,S) are enantiomers, (R,S) and (S,R) are enantiomers, but (R,R) and (R,S) are diastereomers.


text:
Assigning Carbon NMR
I have a molecule and I need to assign the carbon NMR to each of the carbons.
My molecule is

The peaks I have for the carbon NMR are shown below. I have also used the DEPT to work out if it's a $\ce{CH3}$, $\ce{CH2}$, $\ce{CH}$ or quaternary C.
17.7ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)
19.5ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)
25.5ppm $\ce{CH2}$
25.7ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)
32.1ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
36.7ppm $\ce{CH2}$
39.7ppm $\ce{CH2}$
66.0ppm $\ce{CH2}$
122.1ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
124.4ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
128.2ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
128.3ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
128.6ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
131.5ppm quaternary carbon
136.2ppm quaternary carbon
149.0ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
166.4ppm quaternary carbon
I think I can assign the quaternary carbon with the $\ce{C=O}$ in the ester as the 166.4ppm.
Out of the three methyl groups, would the lowest ppm of 17.7ppm be the one at the top of the molecule by itself and then would the other two at the bottom of the molecule on the alkene be 19.5ppm and 25.7ppm? I don't know which one is which, I know it is to do with cis/trans but I'm not sure which way around it is.
I also know that on the benzene ring, only three $\ce{CH}$ peaks show up as the two ortho and the two meta carbons are equivalent.

You should be able to narrow down the peaks with chemical shift over 100 ppm pretty easily based on hybridization.

If this is a real sample, and not a provided coursework spectrum, then the best recommendation is to run a HSQC and/or HMBC. This will provide complete data to allow this assignment fully, and in much the same time as it takes to run a DEPT135. I don't understand the urge to run DEPTS any more - I haven't run one for years. A phase edited HSQC will give you all of the information that a DEPT can provide, as well as connectivities between proton and carbons. Anyway, off my soapbox.
Below is a simulation of this molecule, which has reasonable alignment with your peaks.

A methyl centre trans to a alkyl chain will be downfield shifted to an equivalent methyl group trans to a proton. So 19 will be downfield shifted from 20, and be fairly similar in chemical shift to 16. Simulations have their limitations, and carbon shifts are solvent dependent to +/- a few ppm at least. To avoid any uncertainty, run HSQC and HMBC. These should be part of your normal procedure for assigning structures.

Have you tried a C13 NMR predictor?
Highlights according to the predictor:
166.4ppm quaternary carbon - carbonyl
136.2ppm quaternary carbon - benzene ring
131.5ppm quaternary carbon - aliphatic carbon at far end
149.0ppm CH carbon beta to carbonyl
66.0ppm CH2 between ring and ester
The chemical shifts are slightly different (as much as 5 ppm), but I think the order will be correct.


text:
Will lye damage vinyl pool liners?
I have a nutty neighbor that said someone told them to put liquid lye into their above ground pool.  I said it would eat away the lining, their answer was to only use a little bit.

Lye (Sodium Hydroxide) is sometimes added to pool water to bring about increase in pH, without making significant increases to total alkalinity. It will bring about quite rapid changes in pH. I assume the intention is to bring the overall pH back within a sensible range of 7.4-7.6.
Adding high concentrations of sodium hydroxide (liquid lye is about 50% in water) might degrade the vinyl over long periods, but so too will high levels of acid, bleach, exposure to sun, and just about every other chemical used in pools. Vinyl liner can be damaged by low pH (below 7.0), but there will be other symptoms of high pH (above 7.8) (cloudiness, calcification, precipitation) long before damage to the vinyl liner kicks in, I think.
I think the nutty neighbour may be not too far from the mark to say "only use a little bit". All things in moderation.


text:
XRD-Shorter distance between planes than actual radius of atoms/or cell parameter
I'm solving problems related to X-ray diffraction, and I get these results which apparently are correct. My question may be dumb, but I just can't see how the interplanar distance might be shorter than the atoms radius.
What's up with this?
I think it might make sense that its shorter than the cells parameter, but it's not immediately obvious to me, so if someone could prove it to me, i'd be quite glad.
Edit:
For example:
The atomic radius of chromium is 124pm. THe distance between (310) planes is .0912nm.
See problem 6

Does this help?

The distance between planes will be less than the distance between atoms in many cases because the atoms don't have to stack directly on top of one another in each plane.

First of all, diffraction is best thought of in reciprocal space rather than from planes of atoms, but let me answer your question directly.
You may not understand the meaning of a reflection index. If one imagines a simple cubic lattice made up of atoms of diameter a, then the lattice dimension of the cube is also a. The spacing between (100) planes is also a. The spacing between (200) planes is a/2, as these planes have double the density of (100) planes. As we keep increasing h,k or l the planes get closer.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

Now, that the formalities are out of the way... :) Good answer, +1 for that. However, I fear it won't help the OP very much. He seemed to look for a more visual sort of answer. If you could add any pictures that illustrate your point, I think it would greatly help to make the answer more intuitively understandable.


text:
Which has the higher chemical shift E/Z alkene?
If you have a simple alkene for example 1-methyl-But-2-ene
then you will have two methyl groups on one side of the double bond, which one of these will have the highest. Will the one on the same side as the rest of the chain of butene have the higher ppm or will it be the one in trans with the butene.
By the way, on my spectra, these two methyl groups are not equivalent. I need to know if the E or Z will have a higher chemical shift and why, thanks.

“1-Methyl-but-2-ene” is not a valid name. Do you mean “pent-2-ene”?

Are you talking about 1H NMR? In the general case, it is not so easy to say - it depends on all substituents, with geminal substitution usually contributing most to change in chemical shift. trans and cis substitution will affect chemical shifts differently, but will depend on the electronegativities of all groups concerned.
For your example, there will be very little difference between chemical shift for these two methyl groups, but the methyl trans to the alkyl chain is probably slightly downfield shifted. Coupling constants for the two methyl groups would be slightly different. In acyclic systems, 4J couplings from a methyl to a cis or trans proton can range from about -3 to +2.5 (but usually 0.8 to 2), and usually |Jtrans|>|Jcis|. Of course, nOe information would also be useful.
If you are talking about 13C NMR, a methyl peak trans to a carbon chain will be downfield shifted from a methyl group trans to a proton.

cis-butene: $\delta$(H2): 5.37 ppm [1]
trans-butene: $\delta$(H2): 5.55 ppm [2]
There are lists of increments (Schoolery, Tobey-Simon...) out there to predict chemical shifts in systems such as there, but for methyl groups in vinylic these are very similar.  In those cases it really doesn't bear to assign isomers through chemical shifts when you can get the correct assignment through coupling constants.


text:
Finding initial pressure in a chemical equilibrium
Consider the equilibrium below. If $K = 4$ and initial pressures of $\ce{CO}$ and $\ce{H2O}$ are equal, giving a total pressure of $1.5~\mathrm{bar}$ at equilibrium. What was the initial pressure of $\ce{CO}$ or $\ce{H2O}$?
$$\ce{CO(g) + H2O (g) <=> CO2 (g) + H2 (g)}$$
I don't know how to get the initial pressures in this situation. Do I have to set up an ICE chart? I'm not sure how to start, any help would be appreciated. The answer is apparently $0.75~\mathrm{bar}$.

Are we meant to assume that the system initially has no $\ce{H_2} \text{ or } \ce{CO_2}$?

If the initial pressure of $\ce{CO}$ and $\ce{H_2O}$ are equal, then let $p_{\ce{H_2O}}=p_{\ce{CO}}=x$.
Since their total is 1.5 bar, i.e. $\ce{p_{H_2O} + p_{CO}}=1.5$
Can you solve this equation?

Edit: This is under the assumption that there was originally no carbon dioxide or hydrogen, I have not yet managed to generalise it to any case.
It is given that $K=4$. From the equation for the equilibrium constant, $K=\frac{\ce{[CO2][H2]}}{\ce{[CO][H2O]}}$.
It is also given that initially, $\ce{p_{H2O} = p_{\ce{CO}}}$; from the ideal gas law,
$PV=nRT \implies P=\frac{nRT}{V}=kC  \tag{1}$
where $k$ is some constant and C is concentration.
So initially, $[\ce{H2O}]=[\ce{CO}]=y$, where $y$ is a constant. Let the concentration consumed by the equilibrium shifting to the right be x.
At equilibrium, $\frac{\ce{[CO2][H_2]}}{[\ce{CO}][\ce{H2O}]}=\frac{x^2}{(y-x)^2}=4$.
Solving this equation gives $x=\frac{2y}{3}$.
Now then, if you look back at (1), pressure is proportional to concentration. Try to use that relationship to solve the problem.

 Now, from $(1)$, we see that pressure is directly proportional to concentration. We have worked out from our algebra that: $[\ce{CO2}]=[\ce{H2}]=\frac{2y}{3}$, therefore $[\ce{CO}]=[\ce{H2O}]=\frac{y}{3}$.

So then $p_{\ce{H2}}= p_{\ce{CO2}} = 2p_{\ce{CO}} = 2p_{\ce{H2O}}$ at equilibrium.
So from substitution into Martin's equation, $\ce{p_{H_2O} + p_{CO} + p_{\ce{CO2}} + p_{\ce{H2}}}=1.5$, we find that $p_{\ce{H2O}} = p_{\ce{CO}} = \pu{0.25 bar}$.
But they are now only a third of their original concentration, and so their pressure has also decreased by a factor of 3. So their original pressure is $\pu{0.75 bar}$ each.

I am not so sure, this is the right (generic) way to solve this problem, since it is given as $$p_e(\ce{CO}) + p_e(\ce{H2O}) + p_e(\ce{CO2}) + p_e(\ce{H2}) = 1.5~\mathrm{bar}$$

@Martin Is it possible to solve this for the case where there is some $\ce{CO_2}$ or $\ce{H_2}$ present in varying amounts? I can't seem to do the algebra..

I don't understand what you mean, do you refer to $p(\ce{CO2})\neq p(\ce{H2})$ at equilibrium?|| I highly recommend you use a clearer way of presenting your calculations. I was initially not able to follow your thoughts, until I did the calculation myself. Using $K$ in two instances is never a good idea (substitution of physical properties with a nameless variable $x$ is usually not a good idea.)

Yes, that and my assumption that there was no $\ce{CO_2}$ or $\ce{H_2}$ initially. And thanks for pointing that out, I shall endeavour to clean it up and make it easier to follow in future.

Here's what I found using an ICE table. We let the initial partial pressures of $\ce{CO}$ and $\ce{H2O}$ be $x$, and assign an arbitrary value $y$ to the change.
$\begin{array}{|c|c|c|c|c|}
\hline
 &\ce{CO}  & \ce{H2O}  & \ce{CO2}  & \ce{H2}   \\
\hline
 p_0      & x  & x   & 0    & 0    \\
\hline
 \text{change}  & -y & -y  & +y   & +y  \\
\hline
 p_{\text{eq}}   & x-y & x-y  & y   & y   \\
\hline
\end{array}$
So we find that $2(x-y+y)=1.5 \Rightarrow x = 0.75$ bar, as required.

If you want to use the ICE matrix algorithm you should know how to derive the equations and understand the physics behind it.


text:
How effective is sulfur in detecting mercury vapors
I've read that powdered elemental sulfur will react to the presence of mercury vapor in the air creating red mercury sulfide, and that an open container of sulfur can be used as a means to verify leaks in a container of mercury that's being kept in a storage cabinet. But is the sulfur reaction/color change sensitive enough relative to the permissible levels of mercury vapor that might exist?

I remember reading that the reaction of sulfur and mercury was much too slow to be of much use in absorbing any spilled mercury ...

OK - thanks - but what about detecting vapors?

I'd imagine it'd be too slow; i.e. you'd be poisoned before the sulfur reacted.

AFAIK, poor man's test is to use white copper(I) iodide, see wiki. Sulfur is not effective enough.

The vapour pressure of mercury is appreciable, but the evaporation kinetics are slow.  In a well-ventilated lab there shouldn't be concern about mercury poisoning from broken thermometers, Schlenk lines &c.
That said, there is Alfred Stock's famous review (Angewandte Chemie, 1926) on mercury poisoning.  He suffered from a chronic case of mercury poisoning, which IIRC really took off after the war, when they had to save on ventilation.  In old laboratory buildings the quantities found under the wood flooring during renovation are at times staggering.  Our safety guys liked to show pictures of puddles.
To detect mercury there are commercially available detectors, usually build around the resistance of a vacuum-deposited gold film (a few 100 atoms thick).  A low-pressure mercury lamp might also work.


text:
Determine what the unknown element is

A piece of electronic equipment originally contained $\pu{1.250 g}$ of a valuable metal. Corrosion over time has given a compound of formula $\ce{M(OH)3}$, which is isolated but not weighed. This compound is treated with sulfuric acid to give a compound that has beautiful red crystals and is found to have a molecular formula of $\ce{M2(SO4)2}$ and a mass of $\pu{1.860 g}$. What is this valuable metal?

My attempt:
$$ 1.860 ~\mathrm{g} - 1.250~\mathrm{g}= 0.61~\mathrm{g}\, \ce{SO4} $$
I took the $\pu{0.61 g}$ and converted them to moles of $\ce{SO4}$, then I converted 2 moles $\ce{SO4}$ to 2 moles $\ce{M}$ and got $0.0064~\mathrm{M}$.
Then I divided that by $\pu{1.250 g}$.
I got 195.31, so is platinum the correct answer?

I got Bromine for my answer? I used conversion table from SO4 in grams to moles all the way to the metal in grams?

Bromine is not a metal, so this is no possible answer. However, it would be very nice if you could [edit] your question to contain information in the scope of the aforementioned homework policy. Thank you.

Oops thats embarressing its not even a metal lol i edited with more of my thinking. How about Platinum?

Now, if you could format everything so it would be easier on the eyes, I'd give you an upvote. Check out the comment by Philipp.

The mass of the sulfate is $m_{\ce{SO4}} = 0.610~\mathrm{g}$, the mass of the metal is $m_\text{M} = 1.250~\mathrm{g}$.
With the molar mass of the sulfate anion $M_{\ce{SO4}} = 96.061~\mathrm{g\, mol^{-1}}$ we can calculate the amount of sulfate present in the final compound:
$$ n_{\ce{SO4}} = \frac{m_{\ce{SO4}}}{M_{\ce{SO4}}} = 6.35~\mathrm{mmol} $$
Since the sulfate has the formula $\ce{M2(SO4)2}$ we know that we have the same amount of metal as sulfate: $n_\text{M} = n_{\ce{SO4}}$.
We can now calculate the molar mass of the metal using the information above:
$$ M_\text{M} = \frac{m_\text{M}}{n_\text{M}} = 196.85~\mathrm{g\, mol^{-1}} $$
This number resembles most closely gold with a molar mass of $M_\text{Au} = 196.97~\mathrm{g\, mol^{-1}}$.

Note: It is important that you don't round your numbers while doing the calculations, but round them at the end. So you should keep as much precision as possible in between to minimize rounding errors along the way.
If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.35\ mmol}$ you will get $\mathrm{196.85\ g/mol}$ (gold).
If you divide $\mathrm{1.250\ g}$ by $\mathrm{6.4\ mmol}$ you will get $\mathrm{195.31 g/mol}$ (platinum).


text:
Mass number of isotopes of copernicium denoted with 'm'
On this website:
http://en.wikipedia.org/wiki/Isotopes_of_copernicium
there is a sentence "This isomer is unconfirmed" as a footnote explaining two isotopes.
What do they mean by 'isomer' in this context?
Is it possible that an element may have different spatial distribution of neutrons and protons that could have an impact on its stability?

Follow the hyperlinked word of 'isomer' on the Wiki. Clearly in a chemical context this makes no sense, but science isn't always clearly defined between disciplines. In this instance there's something called a nuclear isomer. The 'm' more than likely refers to the word 'metastable', meaning that something is locally stable, but not globally. I'd recommend moving this to the Physics.SE.

Since Glenn Seaborg won his Nobel prize in chemistry, perhaps one shouldn't be so hasty.  That being said this is more nuclear physics.  The $m$ does indicate a metastable excited nuclear state which will decay by $\gamma$ emission.  This is analogous to an excited electronic state decaying by emitting a photon.

I'll add my intuition where it's necessary, but, as a chemist, I believe it doesn't come under the remit of my expertise. I'm still not convinced that this is actually a chemistry question, in the politest way possible of course.

The 'm' does indeed stand for metastable. The word 'isomer' in this sense refers to nuclear isomers, in which at least one of the nucleons is in an excited state. Think of it as somewhat analogous to an electron in an excited state, but occurring within the nucleus. Just like electrons release photons when an excited state decays to the ground state, when nuclear isomers decay to the ground state (which is different than 'normal' nuclear decay like $\alpha$ or $\beta$ decay), a gamma ray is release.


text:
Given the mass of precipitate, how can I calculate the relative amount of the original dissolved compounds?

A simple mixture of $\ce{NaCl}$ and $\ce{NaBr}$ weighing $0.180~ \mathrm{g}$ is treated with $\ce{AgNO3}$ solution to give $0.3715~ \mathrm{g}$ of precipitate. Calculate the content of $\ce{NaCl}$ and $\ce{NaBr}$ in the mixture.

The answer given in book are $\ce{NaCl}$ = $0.0682~ \mathrm{g}$ and $\ce{NaBr}$ = $0.1118~ \mathrm{g}$.
The answers I got are $0.1519~ \mathrm{g}$ $\ce{NaCl}$ and $0.0281~ \mathrm{g}$ $\ce{NaBr}$. Please tell me where I made mistake.
Working:
$$\ce{NaCl + AgNO3 -> NaNO3 + AgCl}$$
$58.5~ \mathrm{g}$ $\ce{NaCl}$ give $143.5~ \mathrm{g}$ $\ce{AgCl}$
$0.3715~ \mathrm{g}$ $\ce{AgCl}$ give $58.5/143.5 \cdot 0.3715~ \mathrm{g}$ $\ce{NaCl}$ = $0.1519~ \mathrm{g}$ $\ce{NaCl}$
$\ce{NaBr}$ = $0.180 ~ \mathrm{g}$ - $0.1519~ \mathrm{g}$ = $0.0281~ \mathrm{g}$ $\ce{NaBr}$

Silver bromide is insoluble as well.

I think that NaCl only react with AgNO3 and NaBr does not react. If you think I am wrong please correct me

No, that's not correct.  Ag+ and Br- give an insoluble precipitate as well; all silver halides (with the exception of AgF) are poorly soluble: NaCl + AgNO3 -> NaNO3 + AgCl and NaBr + AgNO3 -> NaNO3 + AgBr.

Then please tell me how to solve this problem.

@pcforgeek you to construct a system of equations and solve it.

@permeakra Can you tell me those equations?

@pcforgeek You should figure it yourself, otherwise it would have no educational value

@permeakra Where you able to solve it and get the answer given along with question?

x = number of moles of chloride
y = number of moles of bromide
Since you know the mass of the original $\ce{NaCl}$/$\ce{NaBr }$mixture and the molar masses of $\ce{NaCl}$ and $\ce{NaBr}$, you can write an equation of the form:
$58.44 \pu{\frac{g}{mol}}~ x + 102.89 \pu{\frac{g}{mol}}~ y = 0.180~ \pu{g}$
Based on this idea, you can write a similar equation for the final products $\ce{AgCl}$ and $\ce{AgBr}$. With that in hand you will then have a system of equations you can solve (for x and y) that can be converted back into the masses of those salts.


text:
Why can't unimolecular reactions be 2nd order?
Why can't unimolecular reactions be second order? For example, doubling the concentration of a single reactant would result in four times the original rate of reaction.

A first order reaction involves the behavior of individual particles, usually their decomposition into smaller, more reactive species.  The rate of such a reaction relies solely on the number of particles present, so if we double the number of particles then the rate doubles simply because there are twice as many particles available to decompose.
A second order reaction involves the interaction of two particles with one another. (Even if both particles are the same it is still a bimolecular reaction.)  If we double the concentration of either species we double the likelihood of a collision between those species.  If both species are the same (ex: $\ce{2A -> A2}$) then the result of doubling the concentration is a quadrupling of the rate because we simultaneously double the number of particles that can interact and the number of interactions that can occur.
An metaphorical example involving people:
Unimolecular:
A couple can spontaneously break up into two separate people because of its internal properties.  (Assume there are no intercouple issues involved here.)
$\ce{C -> M + W}$
If we double the number of couples we would expect breakups to occur twice as often because there are twice as many couples to be broken up.  However, since there are no intercouple issues in this example, adding additional couples to our consideration does not increase the likelihood of existing couples breaking up.
Bimolecular with Two Different Species:
A man and a woman can join together to make a couple if they collide in the right orientation with enough energy. :-)
$\ce{M + W -> C}$
If I increase the number of either men or women, it is more likely that I will form a couple simply because there are more interactions between men and women occurring.  Each species' increase is first order though, since they require a partner of the other gender to couple with.  i.e. Doubling the number of women only doubles the rate at which couples are formed because they still have to connect with a man, and the number of men has not changed.
Bimolecular with a Single Species:
Homosexual men can join together to form a couple.
$\ce{M + M -> C'}$
If I double the number of men, then each additional man can interact with every other man in the room, greatly increasing the likelihood of a couple forming.  This means that not only are there more men available to make couples (the first doubling of the rate), but that the existing men are also more likely to be involved in a connection that results in a new couple being formed (the second doubling of the rate.)

A man and a woman can join together to make a couple if they collide in the right orientation with enough energy. :-) Finally a practical tip to find a partner. I'll try it out immediately and tell you how it worked out :) Awesome answer.

This question is provocative, but there appears to be a contradiction in terms. If a reaction is unimolecular, its rate must not depend on the concentration of any other molecule including molecules of the same identity. Certainly in dilute solution, a reacting molecule in a unimolecular reaction cannot be influenced by its neighbors. At the level of stochastics there appears to be no information in the electronic (or nuclear) structure of the reacting molecule that could impart some "knowledge" of the presence of more or fewer neighboring molecules of the same kind. The rate law for a single molecule becomes just a probability; the only reason the macroscopic rate law is proportional to concentration is that to get the overall flux you have to apply that probability to all the molecules present. For example, the classic unimolecular, first order process is radioactive decay. Here the reaction is entirely nuclear (unless we are talking chain reactions - but those are bi-"molecular"), completely insulated from its neighbors by electrons, and its observed rate is first order over many orders of magnitude in initial conditions.
We could consider extremely crowded molecular environments such as some investigators think exists in the cell cytosol, but I think we could go all the way to solid state chemistry and not really address the question being asked because if you get the neighboring molecules close enough to influence the reaction rate in any way, you have, by definition, a bimolecular reaction. If we read the question as "What prevents the reaction kinetics of a unimolecular reaction from depending on the square of the concentration?", I think the answer must be that there is no physical force that could influence the local electronic chemical interactions based on the macroscopic concentration of reacting molecules.
One situation that almost meets the requirements of the question is the case where the unimolecular reaction is enzyme-catalyzed. In this case it's possible that the molecule itself is an allosteric activator and the kinetic law would have a term in S^2. A typical kinetic law for a reaction where S is the substrate and A is the activator is Flux = (VmS/Ks + bVmAS/(KaKs)) / (1 + S/Ks + (AS)/(KaKs)). You can see that if A=S then the rate law has both numerator and denominator terms in S^2.


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Structures of Thiosulfate, Dithionate & Tetrathionate Ions
I am trying to find the structures of thiosulfate, Dithionate & Tetrathionate Ions but I am not getting an exact answer. Please help.

Can you expand on your question?  Where are you getting stuck?  What have you figured out so far?  As it is, this is likely to get closed because it is a homework question that does not show your input or ask about a specific concept.

Have you tried a common web-based search engine? Thiosulfate Dithionite Tetrathionate

This question appears to be off-topic because it can be easily found using Google and/or Wikipedia.

Here you go...
Thiosulfate $\ce{[S2O3]^{-2}}$

There ia a negative charge on the apical sulfur and another negative charge distributed across the 3 oxygens by resonance
Dithionate $\ce{[S2O6]^{-2}}$

There is one negative charge distributed by resonance over each group of 3 oxygens.
Dithionite $\ce{[S2O4]^{-2}}$

There is one negative charge distributed by resonance over each group of 2 oxygens.
Tetrathionate $\ce{[S4O6]^{-2}}$

There is one negative charge distributed by resonance over each group of 3 oxygens.

You have written "Dithionate", but the formula and molecular structure shown are those of dithionite.

@JannisAndreska  Thanks for catching that, I'll edit the answer.


text:
What is a dl pair in stereochemistry?
Is it two isomers with at least one stereogenic center each but the corresponding center on the other isomer will rotate the plane of polarized light in the opposite direction. Is this applicable to meso compounds which aren't chiral overall but have stereogenic centers?

d- l- Nomenclature as well as (+), (-) refer to a physical property of the molecule, its optical rotation. Optical rotation is a result of chirality, a topological property of the molecule. So d-, l- is something that you can generally measure with a polarimeter. If you have a 1:1 mixture of the 2 enantiomers then you will see zero rotation and this mixture is called racemic and named with the dl- prefix. A meso compound also has zero optical rotation since it is achiral but as it is not a (racemic) mixture the dl- notation doesn't make sense but you might see it used sometimes.
The problem now is if you are given a pair of enantiomeric molecules that are mirror images how do you name them so you can tell them apart every time? This is why the D-, L- and R-, S- notations were introduced. D- L- is the older and more naive approach that was later improved by the R-, S- (CIP) notation. Unfortunately as chemistry is a science with long history that is also used by other non chemistry-trained professionals, several times older naming conventions have survived. So you have to know all of them!
So very simply, d-, l- reflect a real physical property of the molecule and D-, L- (and R-, S-) is a way to distinguish between molecules that contain stereogenic centres (and might be or not be chiral). Actually even the R-, S- system has been criticized for not being rigorous enough. I saw a paper where a new naming system was suggested but it is not possible to read this without knowledge of advanced maths (group theory, algebraic topology)

The d- and l- nomenclature is exactly the same as the (+)- and (-)- nomenclature. It is used to indicate the optical rotation of enantiomers. As such it is not applicable to meso- compounds (or diastereomers in general), because it is not unambiguous.
Diastereomers still have chiral centers, and as such they exhibit optical rotation, but it is absolutely useless to specify the direction of it in the name.

Note that the d- and l- nomenclature is discouraged by IUPAC. (+)- and (-)- prefixes should be used, so as to minimize confusion with the D- and L- nomenclature.

Thank you very much. Just one question: what is the difference between d-,l- and D-,L-?

See here for explanations about D and L.

Thank you so much, Just one more thing is the (+) and (-) nomenclature used exclusively for molecules as a whole or can it be used for each stereogenic center individually like R and S?

Since you can't measure the optical rotation of a single stereogenic center, it is assigned to the whole molecule. Only (R)- and (S)- are assigned to each center as far as I know, because it is the only unambiguous system for indicating the chirality.

Thank you very much - you've been very helpful. I'm a first year chemist at oxford uni and my organic chemistry tutor has set a mammoth problem sheet involving all sorts of stereochemistry questions; this has helped a lot. Thank you.


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What factors might affect the reaction between potassium permanganate and ethylene glycol?
Our chemistry club has, for the last few years, been showing off some reactions to prospective students. One of the most popular tricks we have is a bit of an explosive one: we mix some potassium permanganate with ethylene glycol in a small bottle. The permanganate oxidizes the glycol, and we usually end up with an intense, three foot flame that shoots out of the bottle, followed by some popping, and finally, the whole thing burning to the ground.
Usually.
Recently, we've been using the exact same mixtures with the same ratios and getting wildly differing results. This entire week, all we've gotten is the glycol bubbling a little and maybe spitting out some of the permanganate, and not much more. The ratios we use are almost the same, so I'm not sure what could be causing this.
Is this reaction sensitive to age of the reactants? Ambient temperature? Humidity? What could cause the reaction to not work normally?
P.S. In spite of the home-experiment tag, the chemicals used here are lab-grade (from Sigma-Aldrich), though I'd be interested in how the purity might affect the reaction as well.

What setting are you carrying out this reaction? In a hood? This sounds like a disaster waiting to happen.

@jerepierre These reactions are usually done over concrete outside of a building, with spectators kept at least 15 feet away. Also, we usually don't have any larger than a 20mL container, which limits how big we can make the boom.

Capsules of potassium permanganate injected with ethylene glycol are used as aerial ignition devices in bush and forest firefighting. They are considered a reasonably safe option because of the delayed reponse prior to ignition. They are sometimes known as dragon eggs or fireballs. The rate of chemical reaction is very much dependent upon

the particle size of the potassium permanganate. Powdered permanganate (greater surface area) works much better than larger crystals.
the concentration of the ethylene glycol. Dilute glycol may not lead to ignition at all. Perhaps your glycol solution is very wet.
ambient temperature. Unless you are trying this in the middle of winter outside in the snow, this is unlikely to be the cause of disappointment.


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How does a limiting reactant relate to the concentration of ions in solution?
Given the question

A sample of $\pu{1.50 g}$ of lead(II) nitrate is mixed with $\pu{125 mL}$ of $\pu{0.100 M}$ sodium sulfate solution. What is the limiting reactant in the reaction? Calculate the concentrations of all ions that remain in the solution after reaction.

I first found the chemical equation for this problem to be:
$$
\ce{Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq)}
$$
I found the limiting reactant to be $\ce{Pb(NO3)2}$ because there are fewer moles of that substance than the $\ce{Na2SO4}$, with there being  $0.00453 \,\mathrm{mol}\, \ce{Pb(NO3)2}$.
I then used the limiting reactant to find the concentration of each ion. I multiplied 0.00453 mol by the number of moles of each ion in the equation, and put that over 0.125 L to get the molarity. However, the correct answers are

no $\ce{Pb^{2+}}$ ions are left in solution
$[\ce{SO4^{2-}}] = 0.0638 \,\mathrm{M}$
$[\ce{Na^+}] = 0.200 \,\mathrm{M}$
$[\ce{NO3^-}] = 0.0725 \,\mathrm{M}$

I only got the concentration of $\ce{NO3^-}$ correct. What have I done wrong? I think my mistake has something to do with using the limiting reactant in the equation. Also, I understand why there is no $\ce{Pb^{2+}}$ in the solution, because it is in solid form, but doesn't that mean the $\ce{SO4^{2-}}$ should also produce no ions since it is bonded to the $\ce{Pb^{2+}}$?

Of course, the concentration of $\ce{Na^+}$ is $0.2\ce{M}$, because it is a spectator ion. It doesn't participate in the  main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$.
The concentration of $\ce{Na^+}$ is $0.1 \times 2=0.2 \ce{M}$ as there are two ions of sodium in sodium sulfate.
The number of mols  of ion nitrate, as a spectator ion, is $2\times1.5/331.2= 0.00906 \ce{mol}$.
The concentration of ion nitrate is  $ 0.00906/0.125=0.0725 \ce{M}$.
Let's go back to the main reaction between  $\ce{Pb^2+}$ and $\ce{SO4^2-}$: The limiting ion is $\ce{Pb^2+}$ as the number of moles of lead nitrate  $1.5/331.2=0.00453 \ce{mol}$ is smaller than the number of moles of sodium sulfate  $0.0125 \ce{mol}$.
So, after the reaction, there  are no  $\ce{Pb^2+}$ ions left in solution.
As for the remaining $\ce{SO4^2-}$; its concentration is:  $(0.0125-0.00453)/0.125=0.06376 \ce{M}$

Thank you, this makes so much more sense now! One question: does that mean limiting reactants only apply to solids, liquids, and gases in the products? If so, is this because aqueous substances are dissociated in the reaction?


text:
Are bond angles constant?
I'd imagine the answer to the question posed in the title is no - bond angles are not constant. Covalent bonds are not infinitely rigid; hence IR as a spectroscopic technique.
If so, how can we accurately determine bond angles? What margin of error might we expect from an experimentally measured bond angle?

No, bond angles are not constant. For example, the low-energy IR absorption of water (at $1595 \:\mathrm{cm^{-1}}$) is from the H-O-H bend
There are a few techniques to measure bond angles. For polar compounds, microwave rotational spectroscopy can often be used on small molecules in the gas phase.
I'd guess though, that most bond angles are measured from crystal structures using X-ray diffraction.
Keep in mind that spectroscopic techniques always have a timescale associated with them. So any experimental measurement is time-averaged. In the case of bond angles, while the angle will change, the average value is the equilibrium - at some points, the vibration will create a larger bond angle, and at some points in time, the bond angle will be smaller.

X-ray crystallography is useful but then again many bond angles differ in the solid and gaseous phases; HOF and HOOH's dihedral angle come to mind. How's bond angle measurement done in the gas phase?

Gas phase measurements, as I mention above, are done with microwave rotational spectroscopy.


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Recrystallization and purification techniques for getting a pure sample
Could someone explain in detail as to why exactly recrystallization gives a more pure solid sample? As well as why is would be better 'technique' to allow for the crystal to form slowly?
I was thinking that it would be because being a purification technique it essentially breaks the lattice and allows the structure to reform again. As for why it needs to be done slowly, well, because rapid changes in heat would cause the structure to reform into the impure solid it was.
I'm looking for a more detailed, better expressed, answer to this.

Why recrystallization gives a more pure solid sample than what? Precipitation? Sublimation? Evaporation? Fractional crystallization?

Recrystallization is a purification method because it is a slow selective generation of a crystal framework, (mostly) free of trapped impurities. It is only effective as a purification method if done properly. Poor recrystallization techniques; rapid cooling, dramatic changes in solubility with temperature, bad choice of solvent, over evaporation of solvent, can all produce crystals of low purity which are not much better than the starting crop.
Impurities in crystals can occur via three mechanisms;

inclusion: impurities occupy a lattice site within the crystal framework; that is, they become part of a defective crystal unit themselves
occlusion: impurities become physically trapped within the crystal framework. This can be a common occurence for some solvent molecules
adsorption: impurities form a weakly bound interaction to the crystal surface.

The faster the crystallization takes place, the greater the chances these mechanisms have of occurring.  Crystals are specific odd-shaped units that need to pack in nicely to their neighbours to form a good crystal. This takes time. A new molecule may attach to the growing lattice, but if it is not the right fit (such as because it is an impurity), it will leave in favour of a better suited fit attaching to the lattice. I may be showing my age here, but we used to liken crystal growth to a game of Tetris. Heck you can still play it! Done slowly, there is time to rotate all the pieces and get the perfect fit. As you speed things up, it becomes harder to get the right fit, holes are left, and before long the whole crystal goes to the dogs. That is essentially what is happening with rapid crystal growth. It's why you get smaller crystals and greater levels of impurities.
Of course, there other methods for purifying solids which are effective also, if used correctly. Precipitation, sublimation and fractional crystallization are all methods for purifying solids, which can give high levels of purity, matching purities obtained be recrystallization.

Crystallization is an equilibrium process where we count on the pure compound interacting with itself to be a lower energy state than interacting with impurities. If impurities become trapped in the growing precipitate, we want the system to have enough time and energy to redissolve the surface of the crystal and trapped impurities in the bulk solution. Said another way, effective crystallization achieves the lowest energy state, so we should use thermodynamic conditions: long time and high temperature (i.e., without cooling if possible).

Well, you are counting on the solubility of the impurity being different in the solid and in the melt (which in general, of course, is true).  Time can certainly be a factor, because if the impurities segregated to the melt cannot escape the moving crystal front they will be kinetically trapped. Look into 'zone melt refining'.  Further on a classic binary alloy diagram, you do have to continue to cool (until you reach the eutectic) or crystal growth will not continue - that assumes you have a very large impurity concentration.


text:
What would be the most suitable solvent for a single-solvent recrystallization?
The compound in question is acetanilide, and
the solvents in question are: water, ethanol, and ethyl acetate
I was thinking all three would be suitable since upon experimentation, acetanilide dissolved in all three solvents when heated.
Is it possible to rank the suitability?
As, well could someone provide an explanation as to why one is more suitable than the other ?

The ideal properties for a recrystallization solvent are that the compound of interest is soluble at high temperature but insoluble at low temperature. You have addressed the first part, so you just have to assess the second part.
Given that acetanilide is a non-polar compound, I would expect that it would have the lowest solubility in the most polar solvent. For this reason, water is probably the best, then ethanol. I would expect acetanilide to be readily soluble in ethyl acetate at room temperature, making it unsuitable for this recrystallization.

It's not enough to be soluble. You want a solvent to have high solubility at high temperature, plus you want to have low solubility of the solute at low temperature.
Moreover, in an ideal world, you'd like the solubility of any impurities to remain higher than the target compound (i.e., they'll stay dissolved while your target crystallizes out).
So yes, it should be possible to rank the suitability, but that requires:

knowing the solubility at high temperature (i.e., very high)
knowing the solubility at low temperature (i.e., very low)
knowing potential impurities

The first two are absolutely necessary to rank different solvents. The third is useful, but sometimes hard to know completely.

Acetanilide is readily soluble in ethanol at room temperature. So we can not use the ethanol as a solvent for acetanilide recrystallization. But it can soluble in water when heating. Therefore, water is the beat solvent for acetanilide.


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What does 'preconcentration techniques' mean?
In the scope of the analysis of organic compounds in column-switching liquid chromatography, what does the following statement mean?

sample preconcentration techniques with high enrichment factor(EFs)

The word  preconcentration is not understandable for me.
Is it a technique for extracting the components?

Welcome to chemistry.se! Take the [tour] to further acquaint yourself with the page. For formatting options and further advice visit the [help]. I took the liberty to edit your question to make is a little more clearer, please check if I introduced any errors.

Preconcentration means to increase the concentration of a sample prior to analysis or detection. For example you can do preconcentration for an non-volatile organic sample with evaporation its solvent.
An operation (process) as the result of which microcomponents are transferred from the sample of larger mass into the sample of smaller mass, so that the concentration of the microcomponents is increased. Examples include the decrease in solvent volume during distillation or evaporation, and the transfer of microcomponents from an aqueous solution into a smaller volume of organic solvent by extraction.(IUPAC)

There are lots of methods for preconcentration: solvent evaporation, using a different phase (ex: a solid or liquid that could absorbs sample from the solvent) and ...


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Perovskite characterization
Does anyone know of groups working on water sorbtion testing and/or the use of inverse gas chromatography for characterizing Perovskites (as a class of materials)? Especially with regard to applications eg solar PV

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

When you say groups working on this material, do you mean research facilities?

Yes, either academic or industry


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How many liters of hydrogen gas is produced per gram of aluminium?

How many liters of $\ce{H2(g)}$ at STP is produced per gram of $\ce{Al(s)}$ consumed in the following reaction?  $$\ce{2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)}$$
  Express your answer to four significant figures and include the appropriate units.

My attempted solution is included below:
$$ 1.00~\mathrm{g~Al} \left( \dfrac{1~\mathrm{mol~Al}}{26.9815385} \right)
\left(\dfrac{3~\mathrm{mol~H_2}}{2~\mathrm{mol~Al}} \right)
\left( \dfrac{22.4~\mathrm{L}}{1~\mathrm{mol~H_2}}\right) = 1.245~\mathrm{L}$$
This is being marked as the wrong answer.  Where is my error?

Seems fine to me. I can only imagine that they expected that you use the standard molar volume constant of perfect gasses, which is slightly higher (and also wrong of course) $V_m=24.5$ L/mol at 298K and $p=p_0$.

The value $V_\mathrm m=22.4~\mathrm{l/mol}$ for the molar volume of an ideal gas corresponds to the old definition of standard temperature and pressure (STP), i.e. a temperature of $T=273.15~\mathrm K$ and a pressure of $p=1~\mathrm{atm}=101,325~\mathrm{Pa}$. Since 1982, the recommended STP correspond to a pressure of $p=1~\mathrm{bar}=100,000~\mathrm{Pa}$. At this pressure, the molar volume actually is $V_\mathrm m=22.710,947(13)~\mathrm{l/mol}$. However, many textbooks still use the old values.

Your work is correct.  I have to assume that this is being graded by a computer, so trivial differences between your answer and the answer that it is expecting matter.
Possible source of error 1: Ideal Gas Law
Instead of using $22.4\ \mathrm{L/mol}$ they may have been expecting you to use the ideal gas law.
$$PV = nRT$$
$$101.3\ \mathrm{kPa}\times V = 0.05559\ \mathrm{mol} \times 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 273.15\ \mathrm K$$
$$V = 1.246\ \mathrm L$$
Possible source of error 2: Units
The computer may be looking for units in liters per gram of aluminum, something like:
$$\frac{\mathrm L}{\mathrm g} \ \text{or}\ \frac{\mathrm L}{\mathrm g\ \ce{Al}}$$
I doubt that this is the case, but it's possible.
Possible source of error 3: Typo
Either you or the person who set up the quiz made a mistake and typed in the answer incorrectly.

I just asked my professor and he told me that "the molar volume at STP (1 bar, 0 oC) is slighly not 22.4 L" Looking in the book at can not find the value?

@user137452 The molar volume at STP ($T=0~\mathrm{^\circ C}=273.15~\mathrm K$, $p=1~\mathrm{bar}=100,000~\mathrm{Pa}$) is $V_\mathrm m=22.710,947(13)~\mathrm{l/mol}$

The answer is 1.263 One has to use 1 mol gas = 22.414 (at 0 degrees Celsius, 1 atm) = 22.711 (at STP) in the calculation.


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Why isn't organic chemistry taught in 3D?
As the title says. What's the point of learning everything in 2D and trying to visualize and interpret stereochemistry, newman structures, isomers etc, with hundreds of hours wasted, when a 3d visualization suite could be used to teach everything instead, without any ambiguities as to what the structure looks like? The 2D format seems completely irrelevant besides the communication of chemical structures on paper.
Surely a better system would teach everything in 3D first to gain a true visual and theoretical understanding of the subject, and ONLY THEN taught in 2D on how to simplify things for proper communication?

I disagree. It's easier to learn in 2D, simply because it's simple. Once you know the 2D foundations, feel free to look at it in 3D. 3D will involve too many specials effects and distract from the actual structure (IMO).

The problem is that 3d is not enough either.
Most organic molecules can have multiple conformations, and drawing them all is impossible. However, influence of groups quickly reduces with distance growth, so only local 3d neighborhood is required most of the time, and it can be adequately described in 2d using shaded and unshaded triangles, even if needed. On the other hand, 2d sketches can be done quickly, while adequate 3d structures are not that easy to come with, and in some cases the exact 3d structure is debatable at best.
Still, visualisation and 3d-models for some key molecules and transition states are useful.

Face it, a molecule is really a multinuclear atom - a dynamic entity consisting of atomic nuclei and electrons that is constantly in motion and is governed primarily by electostatic forces and quantum laws. Concepts such as hybridization, bonds, bond order, atomic radii, polarization, etc., along with Dreiding models, space filling models, MO models, etc. are all imperfect crutches we use in trying to understand this beast.

iad22agp, these are my exact thoughts. And I don't feel, as a student, visualizing everything on paper is building a good theoretical understanding. What's two protons with nuclear force next to two neutrons? Helium.. but separate those two protons until they're no longer under that nuclear force, and now it's hydrogen gas. Nothing changed only the distance and electron densities. I think the proper physics must be visualized in a 3D space, including protons and all along with an electron cloud and then the abstractions like hybridization etc taught so more complex molecules can be understood.

Organic chemistry is often taught in 3D.  There have long been 3D molecular models used in classrooms, as well as Fischer projections.
Sometimes 2D is enough for a given situation.  Maybe curved 4-dimensional spacetime is needed for some situations, vicinity of neutron stars or black holes lets say.
Use the simplist description that includes all the relavent concepts.

My first chemistry class was in 1968. I still think that a model kit that you can hold in your hands is better than 3D graphics to get a solid understanding of how a molecule fits together in 3D. After you handle some simple molecules by hand then you are prepared to  graduate to more complex models in 3D computer graphics which are available today.

The problem it boils down to is ease of communication. When you develop a new synthetic pathway, or you find the newest cancer drug, you want to tell the whole scientific community about it, and the medium for that is scientific articles, published in scientific journals after rigorous peer-review.
Those articles were traditionally printed on dead trees (i.e. paper) and are now mostly just read on-screen in PDF format. With both of these cases, it is not easily feasible to include 3D visualisation more than what is already done in class on a blackboard. Wedges, bold lines and parallel perspective are used all over the world and they mean the same thing. We already have a system that works well enough, why change?

I too sometimes run into the problem that I cannot easily visualize the 3D structure of some molecule. Then I build it from a model by hand, or on the computer. Those few cases where I had to do that simply do not justify the overhead if you want to do this for every molecule.

Surely a better system would teach everything in 3D first to gain a true visual and theoretical understanding of the subject, and ONLY THEN taught in 2D on how to simplify things for proper communication?

You are absolutely free to teach organic chemistry the way you want to. But you will then see, that this generates a bunch of work for the set-up of that system (figuring out which software to use, writing guides for the students to use the software) and basically everyone needs a screen in the lecture hall. A blackboard is readily available, easily erasable and as such the best option.

I absolutely agree here. But I just mean as a self-study student attempting to gain a theoretical understanding of the subject, using software like chembiosuite seems more effective than the tutorials that teach everything in a 2d sense.

But then every student has to find out for himself whether he/she needs that kind of supplement. If that is what you really want to talk about, this question was rightfully closed as being primarily opinion-based.


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find the pH where is maximum the concentration of a amphiprotic substance
Considering a general amphiprotic substance NaHA. I know that the maximum concentration of the specie $HA^-$ is found at $pH = \dfrac{pKa1+pKa2}{2}$. Why? How i can confirm this statment?
The equations are:
$$ \ce{HA^- + H_2O <=> A^{2-} + H_3O^{+}} \quad Ka2$$
and
$$ \ce{HA^- + H_2O <=> H_2A + OH^{-}} \quad Kb=\dfrac{Kw}{Ka1}$$

Let's take an aqueous solution of a salt $\ce{NaHA}$ with the initial concentration $C$ when added to water. It will completely dissociate  according to the eaquation: $\ce{NaHA(s) \rightarrow Na^+ +HA^-}$.
$\ce{HA^-}$ will participate in three equilibria:
$\ce{2HA^- \leftrightarrows H2A +A^{2-}\quad \quad \quad  }$ ${K_1^0=K_{A2}/K_{A1}}$
$\ce{HA^- +H2O\leftrightarrows H3O^+ +A^{2-}\quad }$ $K_2^0=K_{A2}$
$\ce{HA^- +H2O\leftrightarrows OH^- +H_2A^\quad }$ $K_3^0=K_{B1}={K_w}/{K_{A1}}$
In most cases, $K_1^0$ is far bigger than $K_2^0$ and $K_3^0$. So, the first equilibrium is the preponderant reaction, and this reaction will impose the pH of the solution.
Let's now calculate the product $K_{A2}\times K_{A1}$:
$K_{A2}\times K_{A1}= \frac{[\ce{}A^{2-}].[\ce{H3O+}]}{\ce{[HA^-]}}.\frac{[\ce{}HA^{-}].[\ce{H3O+}]}{\ce{[H_2A]}}$
According to the stoichiometry  of the preponderant reaction, we have $\ce{[A^{2-}]=[H_2A^]}$. So the product
$K_{A2}\times K_{A1}= \ce{[H3O+}]^2}$ .
i.e. $\ce{pH}=0.5(\ce{p}K_{A2} +\ce{p} K_{A1})$

Thanks. this is the only way to solve this problem.  If you  try to make the mass and charge balances and finding the roots of the derivative $\dfrac{d[HA^-]}{d[H_3O^+]}$, this don't work because your roots are not real.

It's interesant to see (if i omit the first reaction) that:

${[HA^-]} = - \dfrac{K_{a1} ([H_3O^+]^2-K_w)}{[H_3O^+]^2- K_{a1}K_{a2}}$ is not defined at $[H_3O^+]=\sqrt{K_{a1}K_{a2}}$

The usual answer for this kind of problem is to write down all species ($HA, A^-, H_2A^+, H_3O^+, OH^-$) and all equations for the various equilibria (don't forget autoionization!) and then find the maximum concentration of $HA$.
Try it!  Try also various approximations.  The exercise will increase your understanding of that kind of equilibrium problems, and of buffers, titration curves, acid-base reactions in general.


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Why does Nitrogen form N₂ (and not N₄) while but Phosphorus forms P₄ (and not P₂)?
$\ce{N2}$ is a gas
$\ce{P4}$ is a solid
$\ce{N}$ and $\ce{P}$ are in the same group in the periodic table
The bond enthalpies are:
$\ce{N#N}$ triple bond is $949~ \mathrm{kJ/mol}$
$\ce{P#P}$ triple bond is $490~ \mathrm{kJ/mol}$
$\ce{P-P}$ single bond is $200 ~ \mathrm{kJ/mol}$
$\ce{N-N}$ single bond is $159-296 ~ \mathrm{kJ/mol}$
So why does $\ce{P}$ go to $\ce{P4}$ not $\ce{P2}$ and why does $\ce{N}$ go to $\ce{N2}$ and not $\ce{N4}$?

A related question has some good answers here.

Due to larger atomic size $\ce{P}$ is unable to form pi bonds and so it is tetra-atomic in which each $\ce{P}$ atom is linked with 3 other $\ce{P}$ atoms by 3 sigma bond. But, due to smaller atomic size $\ce{N}$ forms 1 sigma and 2 pi bonds i.e. triple bonds with other $\ce{N}$ atom and exists as diatomic molecule.

Reference: http://philmintz.tripod.com/Chemistry/page4.html


text:
Finding compounds with B, Si, Ca, and Mg in differing amounts
Not sure if my question is better suited to a software related group, but here goes:
Is there a software program that lets you add differing amounts of elements, minerals, etc., then tells you what the items added might make?  Specifically, I am looking for information about what boron, silicon, calcium and magnesium together (no idea about amounts for each) would make.

I think this question may be well suited for this forum because if such a software exists someone here may have used it.

This would be an exceedingly difficult problem.
My suggestion might be to use a database like the Crystallography Open Database and perform searches with the elements you indicate.
For example, the search interface ensures B, Ca, Mg, and Si are all in the compound. I get 141 matches.
It would be incredibly hard to predict compounds with arbitrary stoichiometry. There are several genetic algorithm search packages that let one attempt to find low-energy crystal structures given a formula, including:

XtalOpt
GASP
USPEX

This is, however, an incredibly active area of research. I'm not sure if any of these packages can handle arbitrary amounts of different elements, or how they would handle missing elements. (For example, the elements you describe could be found in various borosilicate minerals given that oxygen is also allowed.)


text:
Grignards and diatomic halogens
Can Grignards react with diatomic halogens as to perform an substitution reaction?
I'd imagine that the first part of making the reaction work - polarizing the halogen as to prepare to pull it apart - would be easy for a Grignard. Hell, even alkene pi bonds can polarize halogens sufficiently; why can't a Grignard?
So in essence, can phenylmagnesiumbromide react with $\ce{Cl2}$ to make chlorobenzene? The other product of this reaction would be $\ce{Cl-}$ ... which I could see combining with the $\ce{MgBr+}$ in some way.

I don't know the answer to this, but why would you do this? The Grignard is made from the corresponding halide and now you want to go back to the halide.

It's a theoretical exercise. The question was how to go from a bromide grignard to a chloride grignard

I am not sure how good it works but if it does it could be a nice way to exchange a halogen. For example how many ways are there to go from an aromatic/heteroaromatic bromide to its chloride?

So you know it works, just not how well?

No I am afraid. I was commenting on the practicality of the transformation after @jerepierre 's comment. It looks possible maybe even with a different Cl source like N-chlorosuccinimide. Have a look at some literature. Is it a question from school or your own question?

My own question my prof couldn't answer.

Can Grignards react with diatomic halogens as to perform an
  substitution reaction?

Yes, but it looks like a mixture of products results, including something explosive.  Here is a link to an abstract of some early work that reports,
"Chlorination of Grignard and Iotsich reagents, RMgX, (where X = Br, I) gives considerable quantities of RI and RBr together with RCl. Bromination of RMgI results in RI and RBr. Bromination of RMgCl leads to RBr, and iodination of RMgX (X =Cl, Br), to RI."
An "Iotsich" reagent is an acetylenic Grignard ($\ce{RC#CMgBr}$)
Judging by the fact that 1) the abstract suggests that product mixtures of the various possible alkyl\aryl halides are often formed and 2) no further reports claiming synthetic utlity have emerged, I suspect that the reaction is of limited synthetic value.  Nonetheless, your initial suspicion that reaction should occur seems supported.
One further note, many MSDS's and "Hazardous Laboratory Chemicals Disposal Guide, Third Edition" report that when phenylmagnesium bromide is reacted with chlorine, a solid is formed that explodes when shaken.  The reference provided is the same J. Organomet. Chem. paper cited above.  Given the lack of detail provided on the solid, my guess is that it wasn't characterized.

what does this mean? "The abstract that you found, which is an alternative to your original claim, I believe is what is known as transmetallation.   It would be helpful to see the full paper to understand yields but I don't think it is worth $30 to find out.  Transmetallation is the very reason why productive alkyl substitution does not work well with Grignards and alkyl halides."

I'm confused because why is there transmellation in the reaction of a Grignard with an alkyl halide .... there's only one metal cation (Mg2+) ... is he saying that the Mg(2+) switches places or something? Plus I think the paper is characterizing something best described as trans-halogenation (hence RI, RBr, and RCl as products).

I'm not following, 1) "what does this mean" is "this" referring to the abstract, what don't you follow, 2) what was my "original claim", 3)  I'm also confused as to what you mean by "transmetallation".

Sorry, that was my teacher's response to me. I gave him the abstract you gave me and told him it looks like chlorination should work. The last line is what I don't understand. Transmetallation occurs with Grignards and alkyl halides?

The last line in the abstract is confusing because they omit the triple bond between the 2 carbons.  However that line is not relevant to your initial question.  My understanding of transmetallation involves ligand exchange between 2 different metals.  It does not occur here.

does transmetallation occur between Grignards and alkyl halides ...?

okay I see, it doesn't occur between a Grignard and alkyl halide because there are not two different metals.

That's correct.

I guess now I gotta figure out why the prof said it was trans metallic