text:
How to tell how far equilibrium is?
I am trying to solve a question that asks to compare Q vs. K. K = 4.2, and Q = 1.0.
Problem Details:

2.0 mol of Samples of A, B, and C
$\ce{A + B \rightleftharpoons 2C}$

I know that the reaction equilibrium is to the right, but, from this information, how can I tell if it is far to the right or slightly to the right?
Thanks.

Try writing out the K expression. Remember that in general,
${K = \frac{[product]}{[reactant]}}$
In your case, ${K = \frac{[C]^2}{[A][B]}}$
Or in other words,
$K = \frac{[product]^2}{[reacant][reactant]} = \frac{[product]^2}{[reacant]^2}$
I combined the $[reactant]$ terms because the equation you have, there is 1:1 ratio of both reactants.
We can take the square root of both sides:
$\sqrt{K} = \sqrt\frac{[product]^2}{[reactant]^2} = \frac{[product]}{[reactant]}$
So now just plug in your numbers to determine how far to the right the equilibrium lies.
$\sqrt{K} = \sqrt{4.2} = \frac{[product]}{[reactant]} = 2.0$
Therefore (upon consideration of reaction stoichiometry), there is a 2:1:1 ratio of products to reactants at equilibrium.
We can also plug in the Q (reaction quotient) value; remember that K is just a special Q value (K is the Q value when the system is at equilibrium):
$\sqrt{Q} = \sqrt{1.0} = \frac{[product]}{[reactant]} = 1.0$
The Q value indicates that the system currently lies away from the equilibrium mix of a 2:1:1 ratio between products and reactant gases. At Q = 1.0 we have a 1:1:1 mix of A, B, and C.

Thanks for your reply. I have written out the expressions, but am unable to determine if the rx will shift far to the right, or only slightly to the right. Is there any way to tell this from the data?

Yes. Sorry, I had not completed my answer.

So is 2:1 a large or small ratio? (meaning, will it be only slightly shifted, or shifted far?)

I would say large. 2 is after all twice one.

So why is taking the square root of both sides necessary, if you already have the Q and K values?

If we don't take the square root then we are comparing the ratio of the products squared to the reactants squared. That doesn't make much sense.

Oh, that makes since. Does this also have anything to do with the molar ratio between C and A is 2?

No. That is irrelevant. Also if you feel that my answer helped, feel free to mark it accepted.

Ok, so lastly: if they are not perfect squares, what would happen? could you still take the square root of it to get comparable values?

Ah, in that case, you would have to take a different root. Say that you had $\frac{[C]^3}{[B]^2}$. Take the two-thirds root to get rid of the exponents.

Ok. That makes sense. Thanks for all of your help.

No problem! Glad it helped! K-expression inspection is a valuable skill.


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How to tell the pH of HC2H3O2?

Find the $\mathrm{pH}$ of a $\pu{0.500 M}$ solution of $\ce{KC2H3O2}$.

I am preparing for an exam on equilibrium and I have two questions:

How would I go about calculating this?
Is the $\ce{H3O}$ in the middle of $\ce{KC2H3O2}$ one hydroxide ion?

Here is what I have attempted to do: find the conjugate acid/bases, write out a balanced equation, then take the $-\log{[\ce{H+}]}$ of the conjugate acid. But I am a little stuck on the best course of action to find the conjugate acid/base.

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

Thanks for the reply. I have added more info and context to the question.

First of all let us introduce the given compound in a more bonding descriptive way: $\ce{H3C-COOK}$. This compound is commonly referred to as potassium acetate:

It reacts with water in the following fashion:
$$\ce{H3C-COOK_{(aq)} + H2O <=> H3C-COOH_{(aq)} + K+_{(aq)} + {}^{-}OH_{(aq)}}$$
Then the mass action law can be applied in the following way:
$$K_c=\frac{\ce{[H3C-COOH]}\ce{[K+]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}\ce{[H2O]}}$$
And you can then rearrange to form the base dissociation constant:
$$K_b=\frac{\ce{[H3C-COOH]}\ce{[{}^{-}OH]}}{\ce{[H3C-COOK]}}$$
The $\mathrm{p}K_b$ for potassium should be around $4.7$.
As dissenter pointed out, you do not need the Henderson-Hasselbalch equation to  calculate $\mathrm{p}\ce{H}$ values for a pure solution of potassium acetate. You can have
a look at wikipedia for an example calculation.

For your second question:

Hydroxide Ion: $\ce{{}^{-}OH}$ or $\ce{OH-}$ or $\ce{HO-}$ ...
Hydronium Ion: $\ce{H3+O}$ or $\ce{H3O+}$ or $\ce{H+OH2}$ ...

I like how you wrote out all the different ways of expressing hydroxide ion or hydronium ion (some more correct than others). However, I do not see why the OP needs to use the Henderson-Hasselbalch equation to solve this problem. There is essentially no conjugate acid/base pair in this system.

@Dissenter You are right, for this problem you don't need it - fixed that. However, there is a conjugate acid-base pair in this system. (Incomplete dissolution - you have all species in the first equation)

You need the $K_\mathrm{a}$ value of the solute of interest. That would be $\ce{[C2H3O2]-}$ as $\ce{K+}$ has highly limited Brønsted-Lowry acid/base properties in water.

This is more like a comment than an answer. Please consider elaborating a little more. Also I disagree on the acid-base properties. It is one of the most commonly used buffer systems.

Potassium as a buffer? Please elaborate. You are right however I should have probably expanded on my answer but when I posted it appeared that the OP had not given much thought to the question so I merely wished to nudge the OP in the right direction.

Potassium acetate has the same buffer applicability as sodium acetate just add some acid and your done.

Right, I meant potassium by itself.

I don't understand what you are talking about. Are you talking about elemental potassium, then I agree not much of  acid base properties - strung reducing agent though. But the question was about something entirely different: the weak base potassium acetate. (I also corrected your charges.)

As I wrote in my post, potassium ion has almost no acid/base properties in water.

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What is a good way for a laymen to purify alcohol to near 200 proof?
I'm making herbal and medicinal mushroom tinctures and extracts.  Normally I'll use everclear but it isn't cheap and 95% alcohol.  Also I don't know what that other 5% is.  Is it just water?
Is there a way I can reduce alcohol to near 100% without a still?  I've heard poring it over epsom salts through a few layers of coffee filters will remove some of the non alcohol contents.  Is this true?

Distillation of ethanol yields 95% ethanol and 5% water. That is the best one can do unless you don't want to drink it afterwards (it will have a little benzene, a known carcinogen). The Epsom salts may remove some water, but I don't think it will remove all the water, not to mention that some of the Epson salt will dissolve into solution.

If you have very, [very]$_n$ dry $\ce{MgSO4}$ and labgrade absolute alcohol, then it does significantly slow down the hydration process (given inert atmosphere). It is often used since $\ce{MgSO4}$ usually does not disturb (in)organic reactions.

Compared to anhydrous ethanol with about 100 US\$/L -everclear seems to be a very nice and cheap substitute (30 US\$/L). It is perfectly safe to drink diluted so the remaining stuff should be almost only water. (Almost undetectable) Trace amounts of essential oils might be present, but as it is made from grain, very unlikely.
Anhydrous Magnesium Sulphate is about 6 US\$ for a quarter pound (about 110 g) and you might need to add all of it. As LDC3 already mentioned you might solve some of it, not sure if you want that.
Coffee is also kind of expensive and in my humble opinion far too precious to use as a filter. Yet again you will solve some of its contents in your solution. (In response to LCD3's comment)
If you use your potions for outside application only, you could try and search for perfume base alcohol, which could be a little cheaper. It is denatured, so it is absolutely not safe to drink it. There is also industrial cleaning alcohol on the market - but here again you cannot be sure what exactly is in there. It is very hard and expensive to get rid of the additives in denatured alcohol.
The main price of alcohol comes from taxes anyway. If it is intended for drinking, the in most countries spirits are heavily taxed (luxury goods). Working around that is no good idea. When it comes to your own safety and your potions users, do not by too cheap - consequences could be far more expensive.

hortstu means to use the coffee filters as one would use Watman filters, not the coffee.

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Effect of temperature on Hydrogen Bonding?
We had this question in a test and I am not sure of the answer

Q. Temperature changes don't effect the extent of inter and intra molecular hydrogen bonding. State true or false.

I think that raising temperature would increase the kinetic energy of molecules and would thus lead to weakening of inter molecular hydrogen bonds. But am not sure what will happen to intra molecular hydrogen bonds...
For e.g. if ortho-nitrophenol is heated, will the intra molecular hydrogen bonds between hydrogen of the hydroxyl group and oxygen of the nitro group break?
Any ideas/suggestions?

Yes. Temperature disrupts bond of all kinds. Heat up a protein hot enough and you can even disrupt its primary structure - the linear sequence of amino acids, and amino acids are held together through covalent bonds. This is why prion contaminated organisms and instruments must be heated to extremely, extremely high temperatures. The question is whether or not the bond will break; the question is simply what temperature.
Take for example nitrous acid, which exhibits two unique $\ce{O-N-O}$ bond angles. One of these bond angles (the smaller one) results from intramolecular hydrogen bonding between the hydrogen and the oxygen on the other end. This occurs because there free-rotation is permitted around the single bonds in the nitrous acid molecule. The energy difference between the two structures is calculated to be $0.45~\mathrm{kcal~mol^{-1}}$ or $1880~\mathrm{J~mol^{-1}}$.

Do you mean that when heated, the bond rotation becomes possible and the energy decreases?

I mean heat will increase rotation and disrupt the intramolecular hydrogen bond. I expect bond rotation to be possible at all temperatures. The non-hydrogen bonded nitrous acid molecule is also actually the lower energy one.


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How to convert an organic carbonate to its formate ester?
I'm currently working on a chemistry project in which I need to convert ethyl methyl carbonate to ethyl formate.

I only have a simple background in organic chemistry (and I haven't seen anything about reaction mechanisms yet), but I was guessing water could be used in this reaction, to produce $\ce{CH3OH}$ and replace the $\ce{-OCH3}$ group by an $\ce{-OH}$ group. Then the next step would be to remove the oxygen from the  $\ce{-OH}$ group. I've seen that compounds such as $\ce{LiAlH4}$ can do this, but I'm looking for a catalyst to make the process more economically viable, in case the proposed synthetic path is correct.
Does anyone know how to perform the conversion, or even by following another synthetic path? I've got no idea whether this process is correct. I'm also trying to keep the process as low energy-consuming as possible.
I agree it would be better to start from ethanol and formic acid, but I have to start with this molecule.

This is a carbonate derivative, using water will fragment your molecule:$$\ce{EtO-CO-OMe + 2H2O -> EtOH + MeOH + "H2CO3"\ <=> EtOH + MeOH + CO2 ^ + H2O}$$ You need to reduce the central carbon with "$\ce{H-}$" and you also need a soft way to reduce. LAH is the best choice here.

I'm not sure LAH is the right choice here; it's an aggressive reagent and would probably reduce the carbonyl moiety all the way to an alcohol. A more selective reagent (e.g., sterically hindered, electrophilic hydride donors such as DIBAH) is probably called for. There's still the problem of selecting which alkoxy group will leave, however.

There's also the brute force approach: reduce the carbonate ester all the way down to methanol with excess LAH, oxidize the methanol to formic acid, then esterify with ethanol. The nice thing is that the molecule you have already contains within its skeleton two molecules of methanol and one molecule of ethanol (if reduced with LAH and then quenched), which could be recovered, separated, and reused.

Economically, the conversion of ethylmethylcarbonate (98%, 100mL: 212.50 Euro) to ethylformate (97%, 500mL: 44.30 Euro) - prices from Sigma Aldrich, 2014-05-17 - is pointless.

Most likely any reduction of the molecule will result in a mixture of different reduction products, unless total reduction to alcohols is done, so, it is extremely impractical. Moreover, formate will reduce more readily than the original molecule. This said, direct approach is the last thing to consider.

First use $\ce{EtO-/EtOH}$, then lithium-tri-tert-butoxy-aluminium hydride. This will convert it to diethyl carbonate and as the reducing agent is hindered you can manipulate reaction conditions to get the desired product.
Or another way is to hydrolyse it in basic medium to get $\ce{H2CO3}$ and then evaporate $\ce{H2O}$ and $\ce{CO2}$ then after separating it will open a large number if ways as you have got ethyl alcohol and methyl alcohol.

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Potential term in the hydrogen atom Hamiltonian
Here is how I saw the Hamiltonian being written in one Quantum Mechanics book:
$$\hat{H} = -\frac{\hbar^2}{2m_\mathrm{e}r^2} \left[\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{\sin \theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right] - \frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$
Well, that's all very nice and complicated, but looking at the last term, and seeing the use of $r$ in there,
$$-\frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}$$
creates a small doubt in me - what is this $r$ there?
In electrostatics, this was simply the distance between the two charged particles, and that would mean that this was the distance between the proton and the electron, but isn't the position of the electron non-deterministic? Shouldn't that thing be a probability distribution of the distance instead? And why isn't it squared?

What is the $r$ there?

$r$ is the same in the potential energy term of the Hamiltonian as in the kinetic energy term of the Hamiltonian.
$r$ is the distance from electron to the proton.

But isn't the position of the electron non-deterministic?

In the generally accepted Copenhagen interpretation of quantum mechanics, the position of the electron is not deterministic. However, expressing the kinetic and potential energy of the the electron at each position in space does not contradict this interpretation. It is only saying, for each point, if the electron is at this point, the energy is _.

Shouldn't that thing be a probability distribution of the distance instead?

The wave function solutions represent probability amplitude in the Copenhagen interpretation.  The square of the absolute value of the wavefunction represents a probability distribution.

Why isn't it squared?

Force would be inversely proportion to $r^2$, but potential is inversely proportional to $r$.

The Copenhagen Interpretation is not "generally accepted", it's merely the interpretation that most physicists are least uncomfortable with. But that doesn't compromise your answer, +1.

I agree, maybe "generally presented in introductory texts" would have been better


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Best way to clean a gold-plated yoyo bearing?
I have a gold-plated yoyo bearing which needs cleaning. Most people clean regular non-gold-plated bearings with mineral spirits (naphtha), lighter fluid (hexane/heptane), and acetone, but these gold bearings are somewhat new and we don't have a consensus on a cleaning method. There's debate as to whether the gold plating influences spin time or if it's the precision manufacturing, but they do spin forever and are quite silent. Anyway, these bearings are $22, so constant replacement is an enormous barrier to using these outside of competitions.
I've researched gold cleaning techniques, and all the results say to rinse with water. That's not a very good option since all yoyo bearings have tight spaces where dust gets trapped. To understand this particular bearing's sensitivity: I did try rinsing with tap water last night, and now it barely spins. It seems even the contaminants in the tap water were enough to block up this bearing. Also, while most bearings can be used with a very thin lubricant to prolong life and decrease noise, this one cannot. I could try distilled water, but we really need a precision solution.

My question is: Why are you using gold metal for moving parts? Is it just for looks?

Sorry I kinda disappeared forever... It spins for a long time. As I mentioned above, I don't know if the gold is helping at all, or if it's a marketing gimmick, but either way, they are good bearings.

In most of the case distilled water is enough, but you should be sure that there are no salts inside your object if you can perform a Conductometry test on the washing water follow this procedure: Wash the item inside a buck for 3 minutes with distilled water measure Conductometry, take another bucket with the same amount of water wash the item for 3 minutes, measure Conductometry repeat these steps until conducibliy drop down.
This is a Conductometry test I've made while restoring an ancient metal object. You can see that I needed 5 washing to get good result.

If you have high conductivity you should identify the salts with some spots test to understand where they came from.
If you have grease you can use a surfactant like Plysorbate 20 2% in distilled water not more!
Eventually if there to much salts and grease you can use: Thiourea solution 8-10% Sodium thiosulphate 30% EDTA 10 % (reference here).
Remember to dry very carefully the object with a hair dryer, NEVER leave the object wet for more than a couple of minutes!
Eventually you can use a protective like Soter or Zapon these normally cause in terms of colour change a $\Delta E$ of 2-4 so ask to your self if you really need it for a gold object.
In my opinion avoid ultrasonic cleaning for composite materials! If you really care about the object ask for a metal restorer!

It's mostly dust, maybe some pocket lint. I doubt grease and salts are getting in there. The reason harsh chemicals are usually used is because the spaces are so tight, and dust gets trapped. I'll try distilled water next time, if I end up using the bearing again. Drying is an issue; a brush wouldn't be able to dry this. http://www.yoyoyo.com.au/image/cache/data/magicyoyo/magic-yoyo-10-ball-bearing-front-900x900.jpg

I wont be so sure, grease and salts are everywhere.Sorry I meant an Hairdryer!

You could use the same method they use to clean gold rings, gold-plated  watchbands, etc.  Get an ultrasonic cleaner and put in distilled water with a few drops (depending on the tank size) of dishwasher detergent.  This will remove oil, grease, contaminants, etc.  If you don't have an ultrasonic bath, just stop in at the jewelry store at the mall and 9 times out of 10 they'll be nice enough to clean your bearing in less than 5 minutes.

The problem here is that it's not the same as a ring (a "one-piece" object)

As the structure of the gold bearing is the same as other 'metal' bearings then the wash cycle will also be the same.
I am pretty sure the gold is a cosmetic finish to fleece the engineering challenged.  The gold finish on the larger parts if plastic or aluminium will likely be paint or anodizing dye and not gold either.  Any gold in the bearings will be MUCH softer than the typical hardened steel bearing balls and races and will make then fail if it gets (or is put) in there requiring regular re-purchase.  If someone is selling ceramic bearings they might be interesting, they are sometimes used in industry where lubrication is not practical (heat, speed, process contamination, viscous drag, dirt collection) but there are pretty much no common uses of gold in bearings (I had never heard of it before now)
There are very few household/hardware store chemical that can attack gold and hydrocarbon solvents are not on the list. Keep chloride salts and compounds away or the gold plating may be damaged.
EDIT:
As the gold plated, hardened steel and ceramic bearings in this application are usually unlubricated to prevent dirt pickup and drag and unsealed for the same reason so any volatile solvent will wash out dirt and evaporate quickly.  Kerosene may leave an oily film that helps with lubrication but will then get dirty quickly.
The gold plating on the balls will be smooth and any high spots will get smoothed out quickly. It will wear the gold away unevenly and no longer run quite as smoothly as when new and result in lots of repeat sales.

It could easily be a cosmetic thing, but maybe the gold's malleability is good for wearing down small bumps in the balls and the surface. Probably not though. There isn't a lot of pressure put on yoyo bearings compared to other uses. Low friction is a big concern, and ceramic bearings are the best I've found. The gold bearing spun for a long time when I got it, but it's degraded a bit. I'm using dry cleaning fluid right now, and it works as well as hexane. They probably aren't good for the gold bearings though, what do you think?


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Equation for the reaction of iodide to iodine using hydrogen peroxide under acidic conditions?
I'm building a home lab around All Lab, No Lecture Illustrated Guide to Home Chemistry Experiments and I'd like to understand the chemistry that I'm doing a bit better. For example, here's an iodine experiment.

Weigh out 2.0 g of potassium iodide and transfer it to a test tube.
Add about 1.5 mL of distilled water to the test tube and swirl to dissolve the potassium iodide.
Add 1.5 mL of concentrated hydrochloric acid (or about 1.8 mL of hardware store muriatic acid) to the test tube and swirl to mix the
  solutions.
Add about 10 mL of drugstore 3% hydrogen peroxide. The solution immediately turns dark brown as the iodide ions are oxidized to
  elemental iodine, which precipitates out.
Swirl the test tube to suspend the iodine and pour the liquid through a funnel with a folded piece of filter paper to capture the
  iodine crystals.
Rinse the iodine crystals on the filter paper several times with a few mL of distilled water. The rinse solution appears brown from
  dissolved iodine, but iodine is not very soluble in water, so you're
  not losing much of your yield.
Spread out the filter paper on a watch glass or saucer and allow the crystals to dry thoroughly. Iodine gradually sublimates (passes
  directly from solid to gaseous form) at room temperature, so don't
  leave the crystals exposed to air any longer than necessary to dry
  them.
Once the crystals are dry, transfer them to a sealed storage bottle or vial.

It's easy enough to follow and I know I have $$\ce{KI, H2O, HCl, H2O2}$$ with a result of $$\ce{I2\, +\, ???}$$ but then what?
I've had HS Chemistry, skipped Chem in college :(, so I know how to balance equations, convert from g or ml to moles, etc, but where can I learn to translate an experiment into equations without having to ask others (Textbooks or online resources are ideal, going and taking chem at a local uni. is a distant second, since I work fulltime)?
Currently I'm sticking to inorganic chem, so that should simplify things a bit.
I'm currently studying EdX's Solid State Chemistry, so I hope that'll help some.

This is the reaction equation that you're looking for: $\ce{2 HCl + 2 KI + H2O2 -> I2 + 2 KCl + 2H2O}$ (source:http://www.sciencemadness.org/talk/viewthread.php?tid=10202 , 2nd post)

Thank you. That was my "bonus" question. I suspected that the results were $$KCl,H_2O$$, but I wasn't sure. My main question still stands. Where can I learn to make that leap instead of always looking it up? Also, how do you do subscripts without using LaTeX?

@solarmist - we have the mhchem package for MathJax. Learn more here and here

@LievenB - that equation is wrong. There isn't any unionized hydrochloric acid because it is a strong acid. It would be more reflective of chemical fact to write instead $H_3^+O$.

@Dissenter That's the essence of my question where can I learn to make that kind of call or determine the configuration of the product of a reaction?

Strong acids ionize completely in solution by definition. So what's really reacting isn't HCl but either $H_3^+O~or~Cl^-$.

@Dissenter but however the equation looks good is not a net ionic equation you can write it as you want...

True you can write the equation hiwever you want but what's happening in solution is independent of how you write the equation.

We know that this is a redox situation, where iodide ion is oxidized to iodine, and where hydrogen peroxide is reduced to water.
The oxidation equation is as such: $\ce{2I- -> I2 + 2e-}$
The reduction equation is as such: $\ce{2H+ + H2O2 + 2e- -> 2H2O}$
Therefore, the overal equation is as such: $\ce{2H+ + H2O2 + 2I- -> 2H2O + I2}$
You can read more about redox reactions here.


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Why is silicon used in computer chips?
Most computer chips contain or are made of silicon. Why silicon? Why not copper or tellurium or some such?

You should consider asking the exact same question on Physics.SE, a very different answer would come up. :)

Most electron components of computer chips and boards are semiconductor devices like transistors and diodes.
Silicon is, from the Wikipedia article on semiconductor devices:

By far, silicon (Si) is the most widely used material in semiconductor devices. Its combination of low raw material cost, relatively simple processing, and a useful temperature range make it currently the best compromise among the various competing materials

Silicon is the second most common element in the earth's crust after oxygen, and present in a variety of silicate minerals.
Electronics grade silicon can be produced easily by molten salt electrolysis of these minerals followed by purification by zone refining.
Copper is a conductor, and therefor not suitable for semiconductor devices. Tellurium is one of the rarest elements in the earth's crust.

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Migratory aptitude in pinacol-pinacolone rearrangement
I am confused about the migratory aptitude of various groups, as there are many different orders for the same given in different places, especially about -Ph and -H. I would like to know if someone could reliably tell the order.

Partly resolved here

According to my FIITJEE (a coaching institute in India) textbook, the relative order of migratory aptitude of groups in pinacol-pinacolone rearrangement is:
$$\ce{H > Ph > Me_3C > MeCH_2 > Me}$$
There does not seem to be a direct correlation between the mass and the migratory aptitude of these groups.

Your answer would have benefited if it also gave the reasoning of the stated order.

This is not a citable reference. Does your textbook have citations as to where it got the order from?

In the pinacol rearrangement, a 1,2-diol is treated with acid and rearranges to a carbonyl compound. Here is a reaction scheme showing a mechanism for the rearrangement.

In this case the molecule is symmetric and methyl migration is the only reaction pathway available.  Methyl migration can 1) help stabilize the developing carbocation center and 2) once fully migrated, produces a resonance stabilized hydroxyl-carbocation.
If one or more of the methyl groups is replaced with some other substituent (for example, alkyl, hydrogen, aryl), then our reaction possibilities become more complex; which hydroxyl group will protonate, which group will migrate?  Generally, both hydroxyls will protonate and lead to separate products.  In terms of which group will migrate, the only thing agreed upon is that a tertiary carbon will migrate in preference to a secondary carbon which will migrate in preference to a primary carbon.  Clearly, the ability of the migrating group to stabilize positive charge plays a role.
The reaction is reversible and product distribution can change as reaction conditions are changed (thermodynamic control vs. kinetic control).  All of these complexities make it difficult to sort things out in detail, thereby making it difficult to determine the true relative migratory preference of other groups like H and phenyl.  This is why the migratory aptitudes of these groups (H and phenyl) can change from textbook to textbook.
See section 2. Pinacol Rearrangement in this reference for a very nice, straightforward discussion of this reaction mechanism, and all its subtleties.

what do you say about -H and -Ph?

@Aditya  Can't say.  It changes when you change the other substituents, or even with the same substituents it can change if you change the reaction conditions.

In general, the group whose migration stabilizes the carbocation formed will migrate first. Although there is an accepted trend to migratory aptitude, it is ultimately arbitrary. We know that Tertiary carbocations are more stable than Secondary Carbocations which in turn are more stable than Primary Carbocations. When a conflict arises, the nature of the carbocation formed by the migration of the group is considered. If a tertiary carbocation is formed, then said group will migrate. The ability of the group to stabilize the carbocation formed is key here.


text:
Nomenclature - Common vs IUPAC Names and Alphabetical Priorities
I got a problem with this:

What if we stuck with the IUPAC name and called the isopropyl group a 1-methylethyl group? That would definitely reverse the alphabetical priorities; because the substituents here on the hydrocarbon are on equivalent positions, we must give the lower number to the substituent that comes first in the alphabet.
But that's the rub! (No isopropyl alcohol pun!)
Isopropyl - "I" comes before "M" as in methyl.
But if we name the isopropyl group systematically,
Methylethyl comes after just methyl in the alphabet.
So could the name of this also be:
2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane?

No, 2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane is not an acceptable name.  Like the rule you quoted says, start numbering at a bridgehead carbon and first pass through the largest ring (in this case the 5-membered ring) and number that ring so that a substituent will have the lowest number.  So this molecule would be named 2-isopropyl-7,7-dimethyl-bicyclo[3.2.0]heptane.  Usage of "1-methylethyl" is not acceptable.  If you weren't going to use the term "isopropyl", you could call the fragment "2-propyl". So you could write 2-(2-propyl)-7,7-dimethyl-bicyclo[3.2.0]heptane, but the term "isopropyl" is much more common usage.
Edit: After reading Greg E's comment, I looked around the internet and found that "1-methylethyl" is both correct and the official IUPAC name.  Here is a paraphrasing of the relevant IUPAC rule...
These groups (those attached off the main chain) are always numbered so that the point of attachment to the rest of the molecule is C1, with the longest continuous chain beginning from the point of attachment taken as the parent.
Then all sites I visited would add something like this...
"For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic (or common, or trivial) names."
Most of these sites seemed to (strongly) favor the trivial name.  As one site put it, "Common names are used because they are easier to say, shorter, and save paper
and ink in scientific publications when they have to be used repeatedly in manuscripts."

Are you sure the fragment is 2-propyl? It looks like a 1-methylethyl to me.

I think it is a usage thing.  Compare the possibilities: 2-isopropyl-7,7- or 2-(2-propyl)-7,7- or 2-(2-methylethyl)-7,7- .  The third item is not "wrong", it is just the longest, most cumbersome and least common usage of the 3.

@JannisAndreska, I'm fairly certain that names like "1-methylethyl" are acceptable. Q.v., the section from the 1979 IUPAC "Blue Book" on acyclic branched alkanes, wherein names such as, e.g., "1-methylbutyl" appear.

@GregE.thank you for that comment; I've too seen 1-methyl ethyl used even by textbooks. So that changes the premise; 1-methylethyl should be correct.

@ron so would my alternate naming be correct? And in general, if one decides to stick with the systematic naming, should one assign alphabetical priorities accordingly?

@Dissenter  2-(2-methylethyl)-7,7-dimethyl-bicyclo[3.2.0]heptane would be an equally correct alternative to 2-isopropyl-7,7-dimethyl-bicyclo[3.2.0]heptane.  I'm not 100% certain what you mean by "alphabetical priorities".  That said, order the naming of substituents based on alpha (3-butyl-2-octyl...).  Ignore prefixes such as di-, tri- tetra-, etc. when naming compounds (5-hexyl-3,4-dimethyl...), (3-isopropyl-6,6-dimethyl...)

@ron are you sure that 2-(2-methylethyl)-7,7-dimethy... is a correct name? We are supposed to alphabetize substituents in the names, and methyl definitely comes before methylethyl. This is what I mean by "alphabetical priorities."

Sorry, you're right.  2-isopropyl-7,7-dimethyl-bicyclo[3.2.0]heptane or 7,7-dimethyl-2-(2-methylethyl)-bicyclo[3.2.0]heptane

According to current IUPAC recommendations, the preffered name is 7,7-dimethyl-2-(propan-2-yl)bicyclo[3.2.0]heptane.

@user7951: I think 2-(propan-2-yl) should be the correct name. 2-(2-methylethyl) is not acceptable. It should have been  2-(1-methylethyl) insteard.


text:
Do undissolved salts in saturated solvents enter equilibrium with the dissolved salts?
For instance, if there is cup of water saturated with a salt to the point where any more added salt will just float around in the solution, are those undissolved salt molecules in equilibrium with the dissolved ones? For instance, do the undissolved ones become partially dissolved for a moment, while the dissolved ones swap places with them?
Or does this behave in such a manner that whatever salts were dissolved first to reach saturation are the ones that will stay dissolved, in spite of other added salts?

Yes, they enter into a constant equilibrium. Just as we write equilibrium expressions for insoluble salts as equilibrium expressions, like $\ce{CaF2} \leftrightharpoons \ce{Ca}^{2+} + \ce{2F-}$, when you reach the saturation point, solute particles are in dynamic equilibrium between aqueous and solid phase.

Yes. At the solution saturation point there is a dynamic equilibrium between the solid and solvated solutes.
From the IUPAC Gold Book:
saturated solution:
A solution which has the same concentration of a solute as one that is in equilibrium with undissolved solute at specified values of the temperature and pressure.

As others have said, they are in equilibrium. However there is a physical/mechanical (mass transfer)  barrier to this equilibrium. When a crystal is dissolving/being precipitated, all shrinkage/growth occurs at the surface of the crystal.
Once the equilibrium state is reached, the exchange occurs only at the surface of the crystal. There is no change for the material in the body of the crystal (except by diffusion, which in crystals is an extremely slow process.) For a radioactive tracer to reach the centre of the crystals, complete crystals would have to dissolve and new ones be precipitated, which is only likely to happen if there is a gradient of conditions (for example a temperature gradient) between one part of the vessel and another.
Also, if you have a layer of salt slurry with clear solution above it, the solid phase of the slurry part is only in equilibrium with the liquid phase in the slurry part. If you add a radioactive tracer to the clear solution, it will be quickly incorporated into the surface of the crystals in the top layer of the slurry, but take a long time to reach (by liquid diffusion) the bottom part of the slurry. And it may never make it into the middle of any of the crystals.
PS I am a chemist turned chemical engineer, and for quantitative design of dissolution/precipitation process, this distinction is important.


text:
Why unsaturated fats are usually cis?
I've been studying organic chemistry and I think trans fats are actually more stable than cis fats. This is because the steric hindrance (for trans) would be lower and the molecule energy would be lower too (more stable).
Why are unsaturated fats usually cis at room temperature if the steric hindrance is higher?
Is that because the room temperature provides enough energy to break the pi bond in the double covalent bond (for example) making the molecule rotate and have a isomer of higher energy?

Nature produces cis fats exclusively (someone may prove me wrong with some weird example.) This is why unsaturated fats such as vegetable oil have lower melting point than saturated fats of the same MW (the cis double bonds lead to a more compact molecule which gets less tangled with its neighbours.)
The reason nature produces cis bonds is due to the enzymatic mechanism in which these bonds are produced. Therefore this is down to reaction kinetics, and as often happens the product which is not the most thermodynamically stable of all possible products is produced. It just has to be more stable than the reactants.
When fat is artificially hydrogenated with a metal catalyst (for example nickel) to increase its viscosity/raise its melting point (to turn vegetable oil into margarine) some of the cis double bonds that escape being hydrogenated are able to change from cis to the more thermodynamically stable trans. These trans fats are unnatural and are therefore widely claimed to be unhealthy.

I don't see any chemical reason why evolution could not have lead to enzymes that produce trans unsaturated fats. I'd suspect that the underlying advantage for cis are the lower melting points - producing cis unsaturated fatty acids allows to have liquid fats/oils (including solving of unsaturated fats) without the need to be very warm-blooded.

I agree @cbeleites, trans unsaturated fats would probably be solid at room temperature and the mobility of this molecular compounds in a organism would be affected. That's an interesting point of view

Milk contains trans fat.  http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2596709/

Of course there is a reason why nature prefers cis: It's more reactive, the double bond can more easily be transformed to something else later, form a complex with a catalytic metal ion, addition reacion, etc. trans olefins are dead, mostly.

But in biological systems, polyunsaturated fatty acids are cis, and for an animal to oxidize them, they first have to be converted to trans, which takes energy. So I think it depends on whether you are discussing in vitro or in vivo polyunsaturated fats.


text:
Comparing formula for enthalpy change with bond dissociation energy and formation enthalpy
I learnt that given a reaction: $$\ce{A -> B}$$ the enthalpy change is given by:
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$
By convention, we keep the minus sign in the equation, and keep all bond enthalpies positive, regardless of whether bonds are made or broken. In this case, the bonds in the reactants are broken, and the bonds in the products are formed. I was also given the equation,
$$\Delta H = \sum \Delta H_\mathrm{F} \left( \text{products}\right)-\Delta H_\mathrm{F} \left( \text{reactants}\right)$$
Are both $\Delta H$'s given by the above formulas equivalent?

Yes. The two methods are equivalent.
You can try this with the dimerization of $\ce{NO_2}$ into $\ce{N_2O_4}$. In this reaction we form one $\ce{N-N}$ bond. Every other bond stays the same.
$\ce{2NO_2 <=> N_2O_4}$
If we calculate the enthalpy of the reaction through bond energies, we get the negative of the $\ce{N-N}$ bond energy. Why? Because we only formed one bond in this entire reaction. Every other bond stays the same - we start with four resonance stabilized $\ce{N-O}$ bonds and end with four resonance stabilized $\ce{N-O}$ bonds. The net change in bonds is the formation of a $\ce{N-N}$ bond. Here is a picture of the process (note that this picture doesn't show any resonance stabilization but there is; the "pi" bonding electrons are delocalized despite what the picture implies).

Bond energy is a positive value; the negative of this value implies that the dimerization reaction is exothermic - which is obviously true! We are going from a radical species - nitrogen dioxide - into a species with no unpaired electrons. Radicals are reactive!
We can also calculate this using heats of formation for both nitrogen tetraoxide and nitrogen dioxide and we'd get the same enthalpy of reaction.

Given the reaction $\ce{A -> B}$ you know that $\ce{A}$ and $\ce{B}$ must have the same atomic composition, i.e. they consist of the same number of atoms for each element, because if this wasn't the case there would be other reactants or products in the reaction equation (the reaction is some sort of rearrangement). So, $\ce{A}$ and $\ce{B}$ only differ in the way the atoms are linked together, i.e. some bonds that are present in $\ce{A}$ might not be present in $\ce{B}$ and vice versa.
Now, Hess' law, which states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps, leads you to the equation
$$\Delta H = \sum \Delta H_F \left( \text{products}\right)-\Delta H_F \left( \text{reactants}\right)$$
where $\Delta H_F$ is the enthalpy of formation. This $\Delta H_F$ is simply the energy difference between the target compound and the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature.
\begin{equation}
  \Delta H_F = H_{\text{molecule}} - H_{\text{elements}}
\end{equation}
So, $\Delta H_F$ (mostly) describes how much energy is gained by forming bonds in the molecule. Now, since $\ce{A}$ and $\ce{B}$ have the same atomic composition, i.e. they have the same $H_{\text{elements}}$, you see that the energy difference between them must come from the energy difference of the bonds in $\ce{A}$ and $\ce{B}$. The part of bonding energy that comes from bonds that both $\ce{A}$ and $\ce{B}$ have in common cancels out and what remains is
$$\Delta H = \left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds broken}\end{array}\right)-\left( \begin{array}{c} \text{total enthalpy of}\\ \text{bonds made}\end{array}\right)$$
So, yes, both methods are equivalent.


text:
How were the elements discovered before modern technology?
I recently was in need of zinc for an experiment. So I resorted to obtain it from some dry cell. But I wondered if it was pure zinc or was it $\ce{ZnO2}$ or any other compound of $\ce{Zn}$. It then occurred to me that I had knowledge of existence of pure $\ce{Zn}$ and its compounds, but how would its discovery have happened without the prior knowledge of its existence?
How did scientists discover the elements without knowing of their existence and unique properties?
Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?

Modern day scientists rely on technology and advanced nuclear physics. But in absence of these advances, how were ancient scientists able to discover them?

In many cases, they weren't - the ancient... maybe we can't call them scientists. Let's call them "natural philosophers." The ancient natural philosophers (up until about 1700) only knew about a handful of elements. This is part of the reason that it took so long to put the periodic table together.
At the time, an element was something that was not reducible to smaller components. The ancient philosophers tended to conflate properties of things with this idea of fundamental components, and as a result, things like earth, air, fire, and water were thought of as elements. When we say they "knew about" modern elements, what we mean is that they knew of the substances - they did not necessarily know that they were chemical elements in the sense that we define them today.
That modern concept of an element defined by the number of protons in the nucleus wasn't developed until around 1913. The wikipedia article on the history of chemical element definitions gives a good overview of how the definition has evolved over time.
The time period you are probably most interested in is that between the "ancient" era (pre-history up until the mid-1660's or so) to the beginnings of the modern era (say early 1900's). During this period, chemistry was evolving from alchemy into a modern science. The systematic investigation and communication of results was occurring, and a large body of known but poorly understood chemical "recipes" was available to chemists. Using a range of reactions and isolation techniques, they basically just randomly tried things (maybe not completely randomly) until they reached a point where it seemed like something could not be decomposed further, or until they identified a new substance. In many cases, they would discover the new element, but not be aware that it was an element - in other words, they could isolate and identify a substance, but because our collective understanding of chemistry was still very limited, they had no way of knowing that it was an elemental substance.
Wikipedia's list of elemental discoveries gives details about the discovery of each element - although you should keep in mind that "discovery" in this case means "identified as a new substance," not necessarily "identified as a chemical element."

Essentially, a continuous process of trying to find the purest forms of different materials, and then find patterns and distinctions in how they behave.
For example: say you pick up a random rock. You can try to measure as many qualities as possible about it: colour, hardness, density, magnetic susceptibility, and so on. Then, try to make it into something else. Heat it, freeze it, attack it with acids, water, alkali, and see if it seems to behave like anything you already know about, or whether you can convert it into something that does.
The more methods you have, the more sure you can be that two different objects are composed of the same material, or different materials. Particular similarities can cause problems, for example it took a long time for people to manage to separate the lanthanides because they behave very similarly under many tests.
In some cases, elements have also discovered after finding unexpected results -- for example, helium was discovered in the spectral lines of the sun's radiation before it was ever isolated or identified on Earth.
It's a common problem of trying to find the irreducible complexity behind an opaque emergent system -- in this case, trying to work back from the materials we have to their fundamental components. The more information you have, the easier it is to discern more elemental qualities and patterns, and the more other bits you can find.

Zinc doesn't naturally occur in elemental form.  Instead, it occurs as zinc sulfide.
It is thought that zinc was first produced in India and China by roasting the zinc sulfide with a carbon reducing agent, possible dung.
Zinc is a component of brass.  Zinc was first produced as this alloy with copper.
Purer forms of zinc were first produced in Rajasthan by distillation.

Many metals are unstable in their elemental form and naturally form oxides. Scientists from the 1600s and 1700s believed these metal oxides to be a substance called “phlogiston” until oxygen was discovered by Antoine Lavoisier ( and Joseph Priestly). After that more elements were rapidly discovered.


text:
What condenser can be used for simple distillation?
I think I need some distilling (boiling) and collection flasks and a 3-way adapter with a thermometer.  I was thinking of using an open vacuum adaptor to join the lower end of the condenser to the receiving flask.  Does this sound ok?  And if so, what kind of condenser should I use?

You generally don't need a vacuum to distill unless the BP is fairly high. If it is a fairly volatile liquid then a lot can get sucked down to the pump oil. Using water to pull the vacuum has other problems.

That sounds good.  Here's a drawing of the set up.  I'd use a water-cooled condenser.

That sounds fine. You only need the open vacuum adaptor if you're using a sealed distillation setup and you have cause to worry about the pressure while distilling.
I use a basic distillation apparatus with a Graham condenser and a Kjeldahl distillation bulb.  Currently, I skip the three-way adapter with the thermometer because I'm just doing basic distillations and while I can seal mine I generally don't because I'm not worried about my distillate evaporating .

This setup looks early 19th century and will blow in your face at the first boiling retardation.


text:
Are molecules with bond orders less than one stable?
Are molecules with bond orders less than one stable? My professor commented that they were "barely alive" but what does he mean, scientifically?
I know that molecules with fractional bond orders greater than one can exist - i.e. nitrogen oxide has a bond order of 2.5 (through MO calculations) and it exists just fine.
So are molecules with fractional bond orders below one just very unstable?

Xenon Fluoride and the most noble gas compounds  would have a bond order of 1/2 and by themselves they are quite stable though they are some of strongest oxidising agents.

Molecules possessing a bond of order below 1 can be perfectly stable, in the sense that their resulting molecular structure lies in an energetic potential well. Strictly speaking it is enough that at $T=\pu{0 K}$ and in the absence of any interactions with matter or fields, the molecule will not spontaneously disassemble. However, there is no need to go quite so far to protect such a molecule from decomposing; there are examples of species which are chemically important in regular laboratory conditions.
All other things being equal, it is true that species with bond orders below 1 are relatively unstable. This is mainly because the fractional bond is comparatively weak (requiring comparatively little activation energy to break, i.e. a smaller $E_\mathrm{a}$), and because in most cases the molecule can react with other substances in such a way as to form products with all covalent bonds of bond order 1 or higher (increasing the exergonicity of most reactions, i.e. a more negative $\Delta_\mathrm{r}G$).
Since the kinetic barrier for reaction is comparatively low, and the thermodynamic drive for reaction is comparatively high, species with bond order below 1 tend to need some extra protection against the world to stay in one piece. For example, diborane and trimethylaluminium are compounds possessing bonds of order 0.5, and while being indefinitely stable when pure, they spontaneously ignite in air from exposure to oxygen and moisture. As rightly pointed out in the comments, noble gas compounds require fractional bond orders, yet several compounds can be obtained and stored (especially xenon compounds), though they tend to be sensitive to moisture and heating. Boron is also responsible for a very interesting class of compounds in which many boron atoms attach to  each other in cage-like structures with very complicated bonding, in which fractional order bonds are involved. Some of the larger and more symmetric structures can be relatively stable, especially with adequate substituents.
In space, there's really not much of anything around, so you might expect to find some molecules with fractional bond orders floating about. Indeed, one can find the trihydrogen cation, which is actually one of the most common ions in the Universe!

Thank you. I suppose my teacher is wrong. I'm still trying to find his exact statement but I will be verifying with classmates in the meantime.

@Nicolau: What about resonance hybrids? They too have fractional bond orders but aren't they supposed to be more stable?

@Kaumudi Resonance is fine. There is nothing special about fractional bond orders in general (even the concept of bond order is up for debate). All I'm saying is that molecules containing especially weak bonds (which is usually the case for bond orders lower than 1) tend to be unstable with respect to a multitude of reactions capable of forming products which have overall stronger bonding (covalent, ionic or otherwise). No mystery here.


text:
How do you dry a Graham condenser?
I have a Graham condenser and tested it out by distilling water from salt water then I washed it and now I can't get either the coils or the jacket completely dry.
Any advice?  If there isn't a better way I was thinking of popping it in the oven for a while to see if that'd dry it out the rest of the way.

If you are referring to the area where the cooling water goes, I wouldn't worry about it. It will eventually dry, but why do you need it dry since you are only going to have water in that area. If you are talking about inside the coils, then water clinging to the glass would indicate that the glass is dirty. I would flush it with soapy water, rinse with water and then ethanol to clean it.

Thanks. I didn't know about flushing it with ethanol, I just washed and dried it.  How would water inside indicate that it was dirty though?

Also, why did you comment instead of answering?

Water doesn't stick to clean glass, it will roll off the glass.

Well, in this case the water doesn't have anywhere else to go, since it's in the coils. I guess I wouldn't have this problem if I dried my condenser vertically then.

There are two ways I can think of to get rid of the remaining water:

As per LDC3's answer, flush the system with ethanol which will hopefully not have the problem of slow dry time.
Put the glassware in an oven, which will also nicely sterilize the system.


text:
Incorrect Ring Flip?
Is the flip of the cis-isomer wrong? Specifically the image on the bottom right.

What I see is that on the cis-isomer in the first chair, we have an axial upward methyl group and an equatorial upward methyl group. When the ring flips we now have an equatorial upward methyl group (good) and a downward equatorial methyl group (bad). Shouldn't the second one be axial upward? From what I understand, ring flips shouldn't change the orientations of groups; just their axial/equatorial configurations.
Also, in general, would these two rings be equivalent?

Yes, the cis isomer on the right should have the methyl group on C2 in the axial (up) position.  The two rings you've drawn at the bottom of your post are equivalent, they can be interconverted by a 180 degree rotation about an axis that bisects the C1-C2 bond and the C4-C5 bond.


text:
Why does sp2 hybridization imply adjacent sp2 hybridization?
EDIT:  I misquoted my teacher; he said something more along the lines of: in conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

My teacher said that if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized.
1) Why is this true?
2) Is this even true?
I'm guessing this has to do with physical limitations. For example, consider a carbonyl oxygen. Oxygen is $\ce{sp^2}$ hybridized. The carbon the oxygen is bound to has already "used up" two of its four bonds. The carbon can only form two more sigma bonds in addition to the one sigma bond it already has in the $\ce{C=O}$ double bond.
But what about "hypervalent" (if there are such beasts) molecules? Here we come to a self-consistency issue. My teacher still believes in hypervalent molecules - i.e. he draws sulfuric acid with six bonds.
So let's consider sulfuric acid in its hypervalent form, with two double bonded $\ce{sp^2}$ oxygens. There are a total of six bonds and four sigma bonds, implying $\ce{sp^3}$ hybridization.
So my revised set of questions is as listed:
1) Is my teacher being inconsistent with himself here?
2) Does $\ce{sp^2}$ hybridization still imply $\ce{sp^2}$ hybridization for linked atoms? Note that sulfuric acid does not disprove my teacher's statement if we take sulfuric acid to have four bonds, not six (i.e. sulfuric acid is properly represented as charge separated due to the high ionic character of its $\ce{S-O}$ bonds. If I remember correctly, on a time-average basis, 1.7 electrons are in oxygen's valence and only 0.3 electrons are in sulfur's valence per bond.

To dispute the claim your teacher made, my example would be propene. There you have two $sp^2$ carbons, but bound to one of them is an $sp^3$ carbon.

and any alkyl carbonium ion where the carbonium ion carbon is $\ce{sp^2}$ hybridized as in t-butyl carbonium ion; and allene being the all carbon analogue of the $\ce{CO2}$ discussed in LDC3's answer

Acetone has the oxygen and the carbon $sp^2$ hybridized. In carbon dioxide, the oxygens are $sp^2$, but the carbon is $sp$ hybridized.
Added:
For $SO_4^{2-}$, the sulfur needs to utilize 2 of the d orbitals for the sigma bonds, so it would be spd^2 hybridized, and it uses the 2 p orbitals for the pi bonds. The sulfur is $sp^3$ hybridized. (According to Wikipedia).

Good example! I'm gonna rewatch his lecture videos just to make sure I'm not misquoting him.

About sulfate ion .. http://upload.wikimedia.org/wikipedia/commons/b/bf/Sulfate_covalent-ionic.svg
http://en.wikipedia.org/wiki/Sulfate

Looks like we both learned something!

I strongly disagree. In acetone and carbon dioxide the oxygen is better described being $\ce{sp}$ hybridised - as this would apply to all singly bound oxygens. Even though I love Pauling, his suggestions about the involvement of the d-orbitals was clarified as being overestimated. The structure of sulphate is better described as all single bonds with a +2 charge at S and -1 charges at each O. (And I do not see the relevance to the question.)

If an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

Absolutely not, here are a few examples

But only the middle molecule has a conjugated system, as his instructor mentioned.

Right ... that's what he mentioned I think. I don't recall and I haven't found the relevant 10 seconds in the many lecture videos he has made since. But in the middle atom that looks about right; we have a conjugated system and we have two sp2 atoms. Thinking about this further, this makes sense; with resonance and conjugation, we need free p-orbitals. This only happens with sp2 hybridization, not sp3 hybridization. Also we need parallel pi orbitals - again - sp2 hybridization enables this.

In conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

It can be seen as a general rule of thumb, but it is not accurate. The defining feature for conjugated systems is that they have (local) planarity. This is necessary in order to form delocalised $\pi$ orbitals. The $\pi$ orbitals are antisymmetric to that defining plane, hence have to be formed at least by $\ce{p}$ orbitals.
Since all $\ce{sp^2}$ orbitals are in a plane, there are available $\ce{p}$ orbitals perpendicular to that plane for $\pi$ orbitals. The best example for this is benzene and on a larger scale graphite.
The same concept can be applied to $\ce{sp}$ orbitals. The best example might be $\ce{HC#C-C#CH}$.
And therefore a combination of both. A fantastic example for this is phenylacetylene (BP86/def-SVP only orbitals of out of plane $\pi$ symmetry no 27, 26, 24, 21).

An even much more fantastic example is the $D_{2h}$ symmetric $\ce{HC#C-CH=CH-C#CH}$ (BP86/cc-pVTZ only orbitals of out of plane $\pi$ symmetry no 20, 17, 16)


text:
Calculate the standard oxidation potential from given potentials

The standard electrode potentials for the following redox couples are given as: >$E_0(\ce{Fe^{2+}/Fe}) = x~\mathrm{V}$ and $ E_0(\ce{Fe^{3+}/Fe^{2+}}) = y~\mathrm{V}$
  What will be potential $E_0(\ce{Fe^{3+}/Fe})$ (in $\mathrm{V}$)?

My approach: $\ce{Fe^{3+} + e- + Fe^{2+} + 2e- -> Fe^{2+} + Fe}$
So $E_0(\ce{Fe^{3+}/Fe})=(x+y)~\mathrm{V}$

You can write down the three half reactions:
\begin{aligned}
 \ce{Fe^{2+} + 2e- &-> Fe}      & E_1&= x\,\mathrm{V}\\
 \ce{Fe^{3+} + e-  &-> Fe^{2+}} & E_2&= y\,\mathrm{V}\\
 \ce{Fe^{3+} + 3e- &-> Fe}      & E_3&= z\,\mathrm{V}
\end{aligned}
The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.
An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).
Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry

$x$ joules per coulomb, or
$96500x$ joules per mole of electrons, or
$2\times 96500x$ joules per mole of reaction.

In the same fashion, the second half reaction has the electrons carrying

$y$ joules per coulomb,
$96500y$ joules per mole of electrons,
the same $96500y$ joules per mole of reaction.

The third half reaction is the sum of the first two, so its electrons carry

$2\times 96500x + 96500y$ joules per mole of reaction, which is
$3\times 96500 z$ joules per mole of reaction, or
$96500z$ joules per mole of electrons, or
$z$ joules per coulomb.

Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.

You should add that Electrode potentials of two reactions cannot be added because they are not state functions, i.e. dependent on the path. On the other hand $\ce{\Delta G}$ i.e. the energy carried by electrons, is a state function, and can be added.

As correct as eflaschuks approach is, I found it rather confusing, because much text and few mathematics. So I am adding them here.
First of all, let us write down the Nernst Equation for an arbitrary reaction $\ce{Ox + z\cdot e- <=> Red}$:
\begin{aligned}
  E=E^\ominus(\ce{Red/Ox}) - z^{-1}\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Red}}{\ce{Ox}}
\end{aligned}
For our given system that boils down to the following reaction and their potentials:
\begin{aligned}
  \ce{Fe^{2+} + $2\cdot$ e- &-> Fe}& E_1 &= E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}& \text{(1)}\\
  \ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &= E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}& \text{(2)}\\
  \ce{Fe^{3+} + $3\cdot$ e- &-> Fe}& E_3 &= E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}}& \text{(3)}\\
\end{aligned}
Now you are looking for the standard potential $E^\ominus(\ce{Fe/Fe^{3+}})$, so you have to consider the potential $E_3$, which can be build from $E_1$ and $E_2$:
$$E_3 = a\cdot E_1 + b\cdot E_2$$
Substituting $(1)$, $(2)$ and $(3)$ yields:
$$\begin{multline}
E^\ominus(\ce{Fe/Fe^{3+}}) - \frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} = \\
a\cdot\left(E^\ominus(\ce{Fe/Fe^{2+}}) - \frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) \\+
b\cdot\left(E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) - \frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)
\end{multline}$$
In order to obtain the standard potential the logarithmic terms have to cancel. Therfore we can split these equations:
\begin{aligned}
&&E^\ominus(\ce{Fe/Fe^{3+}}) &=
a\cdot E^\ominus(\ce{Fe/Fe^{2+}}) +
b\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}) \\
&&\frac13\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &=
a\cdot\left(\frac12\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}}\right) +
b\cdot\left(\frac11\cdot \mathcal{R}T\mathrm{F}^{-1} \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}\right)\\
\therefore&& \ln\frac{\ce{Fe}}{\ce{Fe^{3+}}} &=
\frac{3a}{2} \ln\frac{\ce{Fe}}{\ce{Fe^{2+}}} +
      3b     \ln\frac{\ce{Fe^{2+}}}{\ce{Fe^{3+}}}
\end{aligned}
In order for the $\ce{Fe^{2+}}$ terms to cancel it is fairly obvious, that $$a=\frac23,\\ b=\frac13,$$ therfore resulting in
\begin{aligned}
&&E^\ominus(\ce{Fe/Fe^{3+}}) &=
\frac{2}{3}\cdot E^\ominus(\ce{Fe/Fe^{2+}}) +
\frac{1}{3}\cdot E^\ominus(\ce{Fe^{2+}/Fe^{3+}}).
\end{aligned}
You can also check that against experimental data (from wikipedia):
\begin{array}{llrlr}\hline
\text{halfreaction} &&&& \text{potential}\\\hline
  \ce{Fe^{2+} + $2\cdot$ e- &-> Fe}     & E_1 &=& -0.44\mathrm{V} \\
  \ce{Fe^{3+} + $1\cdot$ e- &-> Fe^{2+}}& E_2 &=& 0.77\mathrm{V}\\
  \ce{Fe^{3+} + $3\cdot$ e- &-> Fe}     & E_3 &=& -0.04\mathrm{V}\\\hline
\end{array}

Your approach is a correct one.
Explanation: (I call the given two reduction half cell reactions as 1 and 2 respectively)
The required half cell reaction $\ce{Fe^{3+} -> Fe}$ could be obtained by adding reaction 1 and 2 .Thus the net $\ce{E0}$ value would be the sum of $\ce{E0}$ values of reaction 1 and 2 (as both are reduction half cell reactions.)
Hence net $\ce{E0(Fe^{3+}/Fe^{2+})} = (x + y) \, \text{V}$

Are you sure? I am quite sceptical about my approach.

Why should not the standard half reaction potentials for $\ce{Fe^{3+} -> Fe^0}$ be $\frac{2x+y}{3}$, as per calculating potentials between oxidation numbers in a Latimer diagram?


text:
Why is only one lone pair in imidazole delocalised?
I was wondering why only one of the nitrogen lone pairs in imidazole is involved in the delocalized π-system. Specifically, I was wondering why it is that lone pair B in particular is delocalized, instead of lone pair A:

Here is my conjecture about why B's lone pair is delocalized while A's lone pair cannot be.

If we delocalize A, we'd get extreme ring strain; we'd make the geometry around that nitrogen $\ce{sp}$. This implies a $120^\circ$ bond angle. That's problematic.
Effective $\unicode[Times]{x3C0}$-orbital overlap (conjugation) only occurs when the p-lobes are planar — i.e. parallel with each other. If the nitrogen with the B lone pair kept its B lone pair, then that nitrogen would be tetrahedral and that would distort the planarity of the molecule; everything else in the molecule appears to be $\ce{sp^2}$ hybridized.
We also run into formal charge issues; we get a positive formal charge on the "A" nitrogen and a negative formal charge on the B nitrogen. While this isn't a dealbreaker, it's not preferable, as there is separation of like formal charges.
Anything I miss? Any points I mangled?

The correct answer to this question is more direct, and not listed in your items. The non-bonding electron pair in nitrogen 3 is in an orbital perpendicular to the π-bonding p orbitals of all other atoms in the imidazole ring. Thus, it does not have the appropriate geometry to overlap with other orbitals forming π-bonds, and does not participate in any resonance effects.
This effect of orbital geometry is often invoked to explain the difference in basicity between pyridine and pyrrole. In the former, the nitrogen lone pair is of type 3 and does not participate in resonance, and so it has a basicity comparable to regular amines and imines. However, in the latter, the lone pair is of type 1 and does participate in the aromaticity of the ring. Protonation of the nitrogen atom would require localization of the electron pair on the nitrogen, breaking most of the resonance effect. This is energetically unfavourable, so pyrrole displays a much lower basicity than expected for amines or imines.
This section on pyridine has a little more information on why the type A lone pair does not delocalize, including a helpful picture. Notice that the lone pair is in an $\mathrm{sp^2}$ orbital parallel to the plane of the ring, while the π-bonded system is perpendicular.
I can make the following observations on your conjectures:
1) I don't think this applies, actually. The fact that orbital overlap is disallowed in the first place precludes the possibility of even thinking what would happen to the geometry of the molecule if the electron pair in nitrogen 3 formed a π-bond; it just doesn't.
However, if the molecule had a boron atom instead of carbon in position 2 (with an empty $\mathrm{sp^2}$ orbital pointing in the same direction as the full $\mathrm{sp^2}$ orbital in nitrogen 3), then I'm not sure what would happen. A more simple example of this type of structure would be 1,2-azaborine (of which only the 1,2-dihydro derivative seems to have been isolated as of yet).
I could imagine canonical structures in which the electron pair on nitrogen links with the boron atom, forming an additional bond (analogous to benzyne, which is isoelectronic with 1,2-azaborine). However, to form an extra N–B π-bond, then either the bond will be a $\pi(\mathrm{sp^2}\text{–}\mathrm{sp^2})$ bond (which is unusual as far as I know), or both the nitrogen and boron atoms would have to partially rehybridize into a mix of $\mathrm{sp^2}$ and $\mathrm{sp}$ configurations to allow a $\pi(\text{p–p})$ bond, which would likely imply an increase in bond angle and a huge increase in ring strain. Neither option seems particularly good, so I suspect such canonical structures would contribute little to the actual bonding picture in the boron-substituted ring.
2) The nature of the nitrogen atom which participates in resonance is actually something between $\mathrm{sp^2}$ and $\mathrm{sp^3}$. The concept of hybridization isn't very adept at explaining resonance naturally.
I'm also not happy with how much my arguments rely on hybridization, but I don't know other effective means to explain what's going on.

You have a problem with:

1) If we conjugate A, we'd get extreme ring strain; we'd make the geometry around that nitrogen $sp^2$. This implies a 180° bond angle.

If the nitrogen is $sp^2$, you get 120° bond angles, not 180°. For 180° bond angles, you would need $sp$ hybridization.
Actually, both A and B are conjugated. Both atoms are $sp^2$ hybridized and both have a p orbital used for pi bonding in the ring.
Added:
Here is the bonding in pyrrole. As you will notice, the nitrogen is $sp^2$ hybridized and the lone pair occupies the p orbital for pi bonding in the ring. In pyridine, the nitrogen is again $sp^2$ hybridized but the lone pair sticks out from the ring. These are the 2 nitrogen configurations in imidazole.

Edited. I meant sp and that's what the nitrogen would be. Also looks like I confused delocalization of electrons with conjugation.


text:
What is the precipitate of this reaction of magnesium?
$\ce{MgSO4}$, on reaction with $\ce{NH4OH}$ and $\ce{Na2HPO4}$, forms a white crystalline precipitate.
Identify the precipitate.
My approach: I guess the white crystalline precipitate is $\ce{Mg(OH)2}$, since it is white and insoluble in water.

Well, if we look at the possible compounds that can form here:
Mg(OH)2 = Insoluble,
MgHPO4 = Soluble,
(NH4)2SO4 = Soluble,
(NH4)2HPO4 = Soluble,
Na2SO4 = Soluble,
NaOH = Soluble
So, it appears your guess is correct.

With $\ce{(NH4)2HPO4}$, $\ce{Mg^2+}$ forms a white, crystalline precipitate of magnesium ammonium phosphate hexahydrate:
$$\ce{Mg^2+ + HPO4^2- + NH4^+ + OH- + 5H2O \rightleftharpoons Mg(NH4)PO4 \cdot 6H2O\downarrow}$$
This is a very sensitive test for $\ce{Mg^2+}$; however, calcium, strontium, barium and other heavy metal ions also form precipitates of phosphates under these conditions. $\ce{Mg(NH4)PO4 \cdot 6H2O}$ can be distinguished by the form of its crystals when they are examined under a microscope. It forms rhombic crystals with prismatic shape, and is less microcrystalline than other earth alkali metal phosphates (sources: 1, 2).


text:
How to calculate the Gibbs free energy from the equilibrium constant for the decomposition of sodium bicarbonate?

Solid $\ce{NaHCO3}$ is heated to $90~^\circ\mathrm{C}$. At equilibrium the total pressure of the gases produced is $0.545~\mathrm{atm}$. Calculate $\Delta G^\circ$ at $90~^\circ\mathrm{C}$ for the reaction.
$$\ce{2 NaHCO3(s) <=> Na2CO3(s) + H2O(g) + CO2(g)}$$

Using the $\Delta G^\circ = -\mathcal{R}T\cdot\ln(K)$ formula, I did the work below:
$$\Delta G^\circ = -(8.31)(273.15 + 90)\cdot\ln(0.545)$$
This gave me $1.83~\mathrm{kJ/mol}$ and therefore $3.66~\mathrm{kJ}$ for the entire reaction, $2~\mathrm{mol}$. Unfortunately, the answer is marked as $7.85~\mathrm{kJ}$, suggesting that $K=1$, why is this?
Shouldn't $K=(0.545)/1$ because of the gases present?

For your reaction:
$$\ce{2 NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g)}$$
$$K=P_{\ce{CO2}}\cdot P_{\ce{H2O}}$$
However, $K\ne0.545\text{ atm}$ because the sum of the pressures is $0.545\text{ atm}$ and the equilibrium constant is the product of the pressures.
$$P_{\ce{CO2}}+P_{\ce{H2O}}=0.545\text{ atm}$$
Since you started with no gas, and the stoichiometric coefficients of both gases are equal, then
$$P_{\ce{CO2}}=P_{\ce{H2O}}=\dfrac{0.545\text{ atm}}{2}$$

What is the value of $K$ now?

$K=\frac{{\left(\frac{0.545}{2}\right)}^2}{1}$ which means ∆G° is 7846J or 7.85kJ. Thank you for pointing out my mistake!


text:
What is the minimum pressure for existence of a liquid solution?

The minimum pressure for the existence of liquid solution is?

$0.6\;\mathrm{atm}$
$0.8\;\mathrm{atm}$
$0.77\;\mathrm{atm}$
$0.9\;\mathrm{atm}$

Although this seems to be an informative question, I want to know the process by which we get this answer.
I know that the critical pressure for a liquid solution is $p={ a\over {27b^2}}$ but what is the value of a and b in that case?
I expect an easy to understand answer for high-school level.

My answer makes an attempt at this problem, but something quickly breaks down in the results of the maths. Can you provide some information on the origin of that $p=\dfrac{a}{27b^2}$ equation?

Very concentrated aqueous brine solutions seem to flow on the surface of Mars in the summer, where the atmospheric pressure is as low as 0.01 atm. They don't last indefinitely and pool in lakes, however, because it's only just above the triple point pressure of pure water (0.006 atm), and Mars is usually quite cold, so they tend to freeze or evaporate easily. But as long as a liquid is above its triple point pressure, solutions should be possible in a range of temperatures. In fact, concentrated solutions can be stable at pressures slightly below the triple point pressure of the pure solvent

@NicolauSakerNeto Yes,you're right..But since i have to zero down on only one option..(all options are above the triple point pressure of water)I can't really understand what to do.The correct answer is $0.77 atm$ and so how we arrived that is the problem.

Either you omitted something, or you can safely prove to your exam committee that the question does not give enough information to answer it properly.

The only things I can figure for $a$ and $b$ in $p=\dfrac{a}{27b^2}$ are the coefficients from the van der Waals equation for nonideal gases:
$$\left(p+\dfrac{n^2a}{V^2}\right)\left(V-nb\right)=nRT$$
These constants have appropriate units to give you pressure: $a\ \left( \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}\right)$ and $b \ \left(\dfrac{\text{L}}{\text{mol}} \right)$.
$$\dfrac{a}{27b^2}\implies\dfrac{\dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left(\dfrac{\text{L}}{\text{mol}}\right)^2}=\dfrac{\text{atm}}{27}\implies p$$
However, these constants are empirically derived, and thus different for every substance, and so the critical pressure would be different for every substance.
For example, for argon $a=1.355 \ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}$ and $b=0.0320\ \dfrac{\text{L}}{\text{mol}}$, so $p$ for argon is
$$p=\dfrac{1.355\ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left( 0.0320\ \dfrac{\text{L}}{\text{mol}} \right)^2}=49.01\ \text{atm}$$
Were you told which substance you should be using? For all substances, $b$ appears to be less than $1$, so we need to find the minimum $a$ and maximum $b$ combination to get the minimum $p$. However, by my calculation, all values of $p$ will be greater than $1$. The value for water is $220.5\text{ atm}$ (which does not make sense given that a vast number of liquid solutions are stable at normal atmosphere pressure).

No,I wasn't told which substance, but I have been asked about it in general.I have posted the question as it was, and no other data is given to me.The answer should come to $0.77 atm$but how should I arrive at it is a problem.See,using the critical pressure relation is my assumption over it,and I'm not certain about it.The reason I posted.

The use of the equation is imo correct and it refers to the pressure of the critical point - for water this is 218 atm, so you are very close. However, I do not see how you can use this equation for the given problem. (But I do not have any clue whatsoever.)

Also it's liquid 'solution' as stated in the question, so its not pure either. Its either a really good question or as @permekra stated there might be a lack in proper information.

The minimum pressure for the existence of a liquid solution is just below 1 atm. Because at 1 atm pressure(vapour pressure) every liquid gets boiled off corresponding to the temperature at which it attains the atmospheric pressure as it's vapour pressure.


text:
Which d-block cation has the maximum hydration enthalpy?

Among $\ce{Mn^2+}$, $\ce{V^2+}$, $\ce{Ni^2+}$ and $\ce{Ti^2+}$, which cation is having the highest hydration energy to form the aqua complex $\ce{[M(H2O)6]^2+}$?

I guess $\ce{Mn^2+}$ should be the answer as it has the most stable electronic configuration.

Note that hydration energy is the energy released when water molecules are adsorbed on the surface of the cation. Hence the cation should have a high charge density, which means it should have a small size. So, the smaller, the better.
Here, $\ce{Ni}$ has the smallest atomic radii compared to all other options due to greater effective nuclear charge, so it's ionic radii would also follow the same trend. Hence, $\ce{Ni^2+}$ would have greater hydration enthalpy.


text:
Hazardous materials handled by amateur chemists
I'm a enthusiastic and devoted amateur chemist (in disguise, I'm actually a mathematician), but lately I've been thinking to what extend it is safe to pursue my hobby. I'm after all an amateur with just high-school lab skills and some unnecessary math to back me up. I really love chemical experiments, but I don't want to get seriously injured. The point is that same very basic materials, like sulfuric acid or cyclohexylamine, already have a 3 on the NFPA 704 scale of health hazards, quoting:

Short exposure could cause serious temporary or moderate residual injury (e.g. chlorine)

Of course I read the MSDS of the chemicals I'm working with, wear a protective lab coat and tightly connected safety goggles, plus safety gloves, but I don't have nearly the safety equipment of a normal lab (fumehood, eye showers, sprinklers, etc.) To to concretize my question:

Speaking for all amateur chemists out there: to what extend do you as a professional chemist consider amateurs like me to be safe. Can we handle level 3 or perhaps even level 4 chemicals, or should we stick with 2 (greatly reducing our potential experimenting terrain)?
What measure can we (and should we) take to ensure our safety?

I understand that the capabilities vary greatly between people, but there was not sufficient information regarding this topic (from reputable sources) for me to be found on the internet, so I though I post it here. I don't feel like this is a primarily opinion based question, as there is at least to some degree consensus that amateurs cannot (and are not allowed to!) safely handle, say for example tabun or potassium cyanide (just a ridiculous example to illustrate my point).
I hope you can appreciate my concerns and that if this is not the right place for this type of question (I think it is, there is even a safety tag here) you cordially redirect me to an other more suited place and close this.

to what extend do you as a professional chemist consider amateurs like me to be safe. That's a pretty good indicator that at least a part of this is opinion-based.  However, I think there are some good questions buried in here (it's honestly too broad at present as well).  I would split this up into a series of very specific questions about lab safety, not just a blanket question.

As an aside, please do seek advice from someone that has used some of these chemicals before (not just random folks on the internet) if you are unsure.  This can be an exciting hobby, but is a lot less enjoyable when missing appendages or organs.

It may further be helpful if you specify the quantities you're dealing with and give a more exact account of which chemicals you're interested in. The hazards of macroscale experiments are often distinct from microscale ones, and there are too many disparate classes of chemicals to give useful advice beyond the most obvious of generalizations.

To give a concrete example of why more details are needed: you note that you wear safety gloves, but you haven't specified the material. If you're dealing with strong acids, for example, then common materials like latex and nitrile rubber will likely be unsuitable at all but low concentrations. If you use, e.g., ethyl acetate as a solvent, latex should be fine, while nitrile will dissolve. Probably only gloves with laminate film are resistant to most common chemicals across the board, but then you have to sacrifice manual dexterity.

@jonsca, Of course there is some opinion in this question, but I don't think it is primarily opinion based. Splitting could indeed be an option, let me think about it how this could be split in the most efficient and clear way "As an aside, please do seek advice from someone that has used some of these chemicals before" whenever I can I do that, but this is not always possible.

@GregE. In the case of sulfuric acid and cyclohexylamine (just two random example) I'm talking about handling at most something under a liter (this can really range from 5ml to 1L). In fact I don't think, I even reach that except for solvents like acetone, ethanol, toluene, etc. When I say, I wear safety gloves I mean those that are (as far to my knowledge) most suited for the chemicals being used, not any specific. This could probably be a good split question: what gloves should be worn under which circumstances (when working with acid or bases or some organic compound) (to be continued...)

as far as this can be generalized (perhaps including some reference material where the amateur chemist can deduce for himself what gloves are best suited for the occasion).

@Jori, yes, I definitely concur that splitting this into separate questions of narrower scope would be preferable. That would permit you to provide more specifics and elicit more useful responses.

@GregE. Also, this question was more about to which extend amateur chemist can generally work with chemicals (like level 3 as indicated by NFPA 704 or some other well expected scale). The dangers of the specific chemicals can be learned from MSDS's. Thanks for all your input, I'll try to narrow down as best as is possible and favorable.

@Jori, fair enough, but that's a question that's both broad and subjective. I don't personally see a way to answer that which is both definitive and goes beyond generic advice/platitudes. That may be a failure on my part, however.

@Jori I don't know the legalities of where you are located, but in general, handling hazardous materials requires a license, so it's not necessarily a matter of "can work with" as "are permitted to work with".  Anyway, we've given you plenty of food for thought here, so I look forward to your edited question.

@jonsca, Yes, plenty food for thought haha. I'm located in the Netherlands. Yes we have laws here regarding chemicals (of course), but most of them are to stop people from making illicit drugs. As far as I can oversee it, it seems that most level 3 chemicals can be obtained, but most level 4 chemicals not (potassium cyanide requires registration here for example). It will cost me some time to split this, so that it will be useful to my as well as useful to a more broad (amateur) public.

Ahem. If you have place, you can buy or assemble improvised fume hood. As for storage, I'd say that anything stable below 100 C and having bp above 50 C is safe to store, if obvious safety rules are followed (do not store things with low bp open, do not store oxidizers and reducers together, keep reducers and solvents in dedicated metal cases.) With lack of fume hood anything with heavy gas production is probably out of question as most organic solvents and reagents with bp below 100-150. Other than that, just do an exhaustive research and you should be fine. But learn to perfrorm first aid.

However, for obvious reason, I'd avoid anything with toxic dose below ~ 1 g aeven having access to all required equipment just as anything involving high explosives and most explosive mixtures. Also, many things in a lab will put a skunk to a shame, and that is another reason to be picky. anything that can produce toxic vapors is a strict no-no without a fume hood or at least open air lab.

@permeakra You mean a LD50 below 1g? Or just hazardous dose? Even ammonia is hazardous when ingesting >1g (according to Wikipedia this could kill you quite easily, but this could be an error from their side as it seems a bit extreme). All potential hazardous chemicals I'm working with are bp > 100 °C (lowest I use is 112.0 °C, keeping reaction refluxed at about 100 °C). I now a fumehood can be easily constructed, but I lack the space to place the tubing (I could hang it out a window simply, but then the neighbors will think I'm running a hemp plantation).

However, there is a small ventilating system in place (I'm using an unused bathroom) and I could open the window and adding an extra ventilator (something like this). Of course I do my research on the chemicals I'm using, but it is sometimes hard for an amateur to judge them (e.g. I can easily judge that reactions that produce phosgene are not in my league, especially without fumehood, but inspecting if something like toluene or nitroethane is safe for me is much more diffecult). I appreciate your help :-)

Ammonia is relatively safe as it stinks like hell, it is almost impossible to ingest significant amount of it. However, by 'toxic dose' I mean the minimum dose to give risks of permanent or severe injury (not even death). The problem is that working on large scale it is surprisingly easy to get a small dose. Some toxins can accumulate (most notably been organic mercury compounds), so they are ok to work once or twice but a thing to avoid for long use. I understand your concerns about PR, but if you work with anything producing toxic vapours or dust, you have to install fume hood anyway.

For compounds there are two risks other then immidiate toxity: explosive hazard and cancer risk. The first is always noted, the second is not so easy to measure as it accumulates. For example, it is generally assumed that all aromatics is potentially carcinogenic, especially polycondensed, but alkyl derivatives are less dangerous (so, toluene instead of benzene). Flammability is something to live with, unfortunately.

For reactions the main risks are dangerous by-product/intermediate compound (usually mentioned) or risk of autospeeding reaction, that can run out of control (usually also mentioned). Because of it it is highly recommended to make several tests on small scale before going large scale and make an effort to keep temperature from rising too high for exothermic reactions (keep an ice bath nearby, use low-boiling solvents etc).


text:
Calculating empirical formula of vitamin C
Combustion of 0.2000 g of vitamin C gives 0.2998 g $\ce{CO2}$ and 0.819 g of water. What is the empirical formula of vitamin C?
My work:
I calculated moles of products and found the mole ratio but it's not coming to be a whole number.

We cannot answer homework questions. Would you mind sharing more of your work and thought process though so we can better assist you?

You are using the wrong number for the grams of water, it should be 0.0819.  Go back and redo your calculations with that change.  Now you should get the correct ratio of the elements in the molecular formula for ascorbic acid.  If you don't, report back and show your math.

When you get the mole ratios but they are not whole numbers, divide through by the smallest number. Say you get for some hypothetical compound the following formula:
${C_{2.5}H_{5}O_{7.5}}$
In this case to get the empirical formula, just divide each subscript by 2.5, the smallest subscript. You will get:
${CH_{2}O_{3}}$


text:
Bond character and bond angles
Somewhere along the line of researching why the $\ce{Cl-C-H}$ bond angle in methyl chloride is less than what is predicted ($\pu{109.5^\circ}$), I ran into a book which said that the electron withdrawing effects of the chlorine atom gave the $\ce{C-Cl}$ bond more "p-character".
What does this mean? Is he simply using this term to describe the elongated nature of the bond - i.e. s-orbitals are spherical and p-orbitals not spherical but rather elongated (at least as depicted by cartoon drawings in textbooks)?
Is the author saying that the electron-withdrawing effects of chlorine cause the electrons to spend more time further away from the nucleus; hence the "p-character"? I understand that in a spherical shell, electrons are basically equidistant from the nucleus. But in an elongated, "p-character" shell, electrons can go further away from the nucleus.

S orbitals are lower in energy than P orbitals.  Electrons prefer to be in as low an energy orbital as possible.  Therefor, when we mix S and P orbitals to make hybrid orbitals, the more S character the orbital has the lower the energy of electrons occupying that orbital. Let me use chloroform, $\ce{CHCl3}$, as an example to make my point (I have data for it and the $\ce{C-Cl}$ bonds are analogous to those in methyl chloride).  The chlorine atom is electronegative, so electrons in the $\ce{C-Cl}$ bond will spend more of their time closer to chlorine - away from the carbon, then they would in say a $\ce{C-H}$ bond.  So if the electron density in the hybrid orbital contributed by carbon to the $\ce{C-Cl}$ bond is going to be reduced, why put as much S character in it?  Instead save that S character for other hybrid orbitals emanating from the carbon that have higher electron density in them.  It turns out in the case of chloroform, knowing the various bond angles and the $\ce{C_{3V}}$ symmetry of the molecule, that the carbon portion of the $\ce{C-Cl}$ has a hybridization of about $\ce{sp^4}$ (more P character and less S character just as we predicted since it has a lower electron density) and the $\ce{C-H}$ bonds are approximately $\ce{sp^{2.75}}$ (using that S character we saved from the $\ce{C-Cl}$ bond to stabilize the electrons in these 3 orbitals, just as we predicted).  The same effects play out in methyl chloride, just not as dramatically (e.g. the hybridization index of the $\ce{C-Cl}$ bond will be greater than 3 but less than 4).

Thank you. Makes sense. Reduced electron density about the carbon in the carbon-chlorine bond reduces the need for s-character as s-character helps stabilize electron density. But if there isn't that much electron density immediately around the carbon there is less of a need to stabilize said electron density.

Also speaking of point-group symmetry, what resource would you recommend as a good primer on point-group symmetry?

The first book where I came across this topic was "Intoduction to Stereochemistry" by Kurt Mislow (a brilliant thinker and chemist).  It is a paperback and in one of the early chapters he spends several pages on the point groups.  I think it is an easy and excellent introduction to the subject.  The rest of the book is a wonderful introduction to many of the other important concepts in stereochemistry.


text:
How to determine the volume of a substance produced when only given the equation at STP?

At standard temperature and pressure, what volume in $\pu{mL}$ of $\ce{H2O}$ is produced in the following reaction?
  $$\ce{2 KOH(l)  +  H3PO4(l)  ->  K2HPO4(l)  + 2 H2O(l)}$$

Most questions I've found have more information than this so I'm confused as to how to do this problem.
I've tried multiplying $\pu{2 mol}$ of $\ce{H2O}$ by $\pu{22.4 L}$ and then converting it to $\pu{mL}$, but I've a feeling that isn't right.

I think you erred in multiplying 2 moles of water by 22.4 L. There are 55.5 moles of water per liter of water. Keeping this relationship in mind, use dimensional analysis to find the moles of water produced in your reaction.
You're confusing the number 22.4 with the fact that 1 mol of gas at STP occupies approximately 22.4 L.


text:
Constitutional Isomers or Stereoisomers?
With regard to a trigonal bipyramidal molecule, such as $\ce{PIBrClF_2}$, can there be any constituional isomers? I'm thinking no, because constitutional isomers differ in the connectivity of their atoms. The molecule only has single bonds. All are bound to a single central atom. There is no way for the connectivity of the atoms to change versus something such as ethanol, $\ce{C_2H_6O}$. I can think of two constitutional isomers off the bat:
$\ce{H_3CCH_2OH}$ and $\ce{H_3COCH_3}$ in which the connectivity of the atoms differ; in the former, we have a hydroxyl group; in the latter, we are missing the hydroxyl group.
Therefore, $\ce{PIBrClF_2}$ should not have any constitutional isomers but does have stereoisomers, correct?

Correct, no constitutional isomers, but there are stereoisomers (different substituents in the apical and equatorial positions).  Just for the record, what is your fifth substituent?

Oops, good catch! Also I've always heard the apical position referred to as the axial position but both terms should work; they mean the same thing.

Yes, they mean the same thing.  I guess it's another one of those "style" things (like 1-methylethyl vs. isopropyl).  Chemists I hung around with would tend to use "apical" more for non-carbon systems, like your phosphorous example.

True; I've only seen textbooks describe carbon systems as axial or equatorial. Never really seen them use the terminology for inorganic compounds but my professor seems to have appropriated the terminology to apply to inorganic compounds as well.

Apical did remind me of biology and "apical meristems."


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Drawing Constitutional Isomers
Are these all valid isomers for the formula $\ce{C_3H_6O}$? I drew the first five; my friend came up with the last three, which have double bonds. Are these all isomers or different? More importantly, are there any more?
Also in general, how can one come up with all the isomers for a molecule? I heard of graph theory, but I think that's a little bit too advanced for me. I guess the only way is to draw out as many as you can and name them to weed out redundant structures?

The figures are for $C_3H_6O$, not $C_3H_8O$. Also, the last configuration from your friend is not very stable.

Thank you! Corrected in text. Ah, stability. Let me see if I can figure this out. I don't see any Texas carbons; I don't see any non-zero formal charges. So I'm guessing it's less about electronic factors as it is about steric/geometric factors. The rightmost carbon is tetrahedral; the center two carbons are both trigonal planar_ and we have a hydroxyl group hanging off; I'm guessing there's some kind of steric hindrance?

You can find out about enols here. http://en.wikipedia.org/wiki/Enol

Oops, back to electronic effects; the electron withdrawing hydroxyl group makes a carbon more nucleophilic right? By concentrating electron density in the C = C?

You are missing acetone-enol

They are all constitutional isomers.  If you are counting enols (your friends #3, it is the enol of your #1), then include the enol of acetone as an additional constitutional isomer.


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How could a scientist from the 1800s to the 1920s run spectroscopy of chemicals?
How would such a scientist be able to get the spectroscopy of chemicals?  Without access to modern computing or electronics how would it be possible?
Any video link or reference link are much appreciated here.

What type of spectroscopy are we talking about? Many spectroscopic techniques are only possible because of electronics.

I think you are referring to VIS spectroscopy, right?

UV, IR, Visible, xray, mass spec were all known by the 1920s.  NMR and EPR not until 1944.

@BenNorris - Is visible spectroscopy the only non-electronic method?

Victor, UV, IR and visible spectroscopy were all practiced without use of electricity.  Xrays are generated by acceleration of electrons, so "electronics" but not a computer is needed.  Similarly, for mass spec, "electonics" is needed to make ions, but electronic computation is not needed.

Light from a flame, the sun, or a star, was passed though a narrow slit, then directed through a prism (F) and observed.  The prism is rotatable about a vertical axis to adjust the wavelength range being observed.  Bright lines in the spectrum are emission transitions and dark lines are absorption transitions.
Further progress was made using diffraction gratings instead of prisms and using photography instead of naked eye observation.  Photography permitted observation of infrared and ultraviolet lines.
By these techniques, known elements were detected in the sun and other stars, and helium was discovered in the sun before it was discovered on Earth, all before 1900.
For further information see Spectroscopy History.