text:
How does repeatedly freezing and thawing antibiotics decrease their antimicrobial effectiveness?
This is something I have read on forums repeatedly, but I don't understand the mechanism behind how antibiotics lose their effectiveness. I understand how repeatedly thawing and freezing cells would damage them, but not how repeatedly thawing and freezing antibiotics make them lose effectiveness.

do you have any sources for that?

http://www.protocol-online.org/biology-forums-2/posts/9077.html - It does not answer if repeated freezing and thawing actually affect antibiotics performances.

If you are talking about dry pills, I doubt if keeping them in the freezer would do any harm. Liquid formulations might use emulsifiers to keep antibiotics in a stable suspension, and freezing/thawing could break up the colloid and cause the ingredients to separate.

As indicated in the comments, freeze-thaw cycles should not affect molecules of size of typical antibiotics. Freeeze-thawing may affect colloids by disrupting their structure. Also, the effect would depend on how fast a sample was frozen/thawed etc (slow freezing would cause rearrangement of molecules and crystallization of water whereas rapid freezing would result in vitrification of water.)
The popular lab practice, in my opinion, would have perhaps arisen because it is better to keep the stock stored at the appropriate temperature (frozen) for longer life. Repeated thawing of the stock would definitely increase its chance of degradation because of mishandling and even otherwise. Therefore, it is better to aliquot smaller volumes; even if you have forgotten to keep it back into the freezer you'll end up losing only a small amount.

Not related to your question but this is an interesting fact: freeze-thaw cycles promote molecular assembly of RNA into higher order structures (ref).

Nice job here WYSIWYG. The premise of the question is just another one of those hoary molecular biology / microbiology lab myths that refuse to die...


text:
Synthesizing 2-methyl-2-butanol
I have recently been trying to come up with a method to synthesize 2-methyl-2-butanol or tert-amyl alcohol. I wish to avoid complex/expensive steps since as a home chemist/student, I don't have access to much money or equipment.
What I have done so far: I noticed that methyl ethyl ketone has pretty much the correct structure, minus a methyl group.
I then realized this meant I could use a methylating agent to get the correct molecule. What I needed was to break the $\ce{C=O}$ bond from the carbonyl group and add a methyl group.
Long story short, I found that either Grignard reagents or something like methyl lithium would do it since it would donate an electron pair to the carbon, and form a bond with the methyl. I would then have the 2-methyl-2-butanol as a conjugate base with $\ce{Li}$ ions present. I could then react them out with $\ce{HCl}$ to precipitate out the $\ce{Li}$ salt and add a hydrogen to the negatively charged oxygen.
My problem: I don't know if this would work at all. I didn't find any documentation. Also, methyl lithium is very expensive and difficult to make so I can't use it.
Does anyone have any other route suggestions? Would it be possible to make methyl sodium and if so, would it work the same? Is it possible to make Grignard reagents easily at home?

I'm a hobby chemist, so please be detailed with how you predicted the organic reaction.
Also, I don't necessarily care about what chemicals I use for the reaction, so if anyone can suggest a different starting chemical, I'd be glad to hear it. I was even thinking of using an alkene group of some kind and methylating the $\ce{C=C}$ bond, but I don't know for sure, as I'm pretty new to organic chem.

Yes it would work excellently in theory with either Grignard reagents or alkyllithiums. I can't say about in practice although both reagents can be dangerous to handle so this is something to think about. However, if you're only interested in the product then buying it is the easiest option.

Grignards are not the kind of thing that should be worked with without a hood.

Acid-catalyzed hydration of 2-methyl-2-butene should also work.

Isopropyl lithium and ethyl bromide would be my go-to if I were to try to prepare this stuff (which I will call TAA). This requires a solid fume hood, and the resulting mixture of salt, TAA, and isopropyl lithium solvent (usually n-hexane) should be easy to separate. The hexane can be stripped from the TAA via fractional distillation, with more distillation passes resulting in a more pure product. TAA boils at $\pu{102 ^\circ C}$ (Sigma Aldrich), and n-hexane boils at around $\pu{70 ^\circ C}$ - this $\pu{30 ^\circ C}$ gap in BP passes the rule-of-thumb "can I separate with distillation?" question.
All that said, even pricey Sigma Aldrich can get you 99% for $115 a liter.
I am aware that TAA is an experimental recreational drug, and a possible replacement for ethanol. I advise caution in sourcing this material for that purpose, as the impurities present are likely to be more toxic than TAA itself. I believe the industrial synth is from 2-methyl-2-butene (2m2b) with the addition of water (this would be cheapest); TAA prepared this way can be purified further by refluxing with an open condenser, enabling any unreacted alkene to evaporate (Boiling point = $\pu{39 ^\circ C}$). Acetone and ethyl Grignard would also form TAA, but with much more hassle and hazard, so the remnant ethyl chloride and acetone from such a synthesis are not likely impurities in any cheap sample.

Or, cheaper alternative: I suspect that the acid dehydration of  isoamyl alcohol would yield a carbocation with a subsequent 1,2 hydride shift, thus giving (to a yield that I don't know, and with reaction conditions that I don't know) 2-Methyl-2-butene (caution: flammable and with a bp of 40°C).
Further treatment with aqueous acid could yield, via Markovnikov hydration of the alkene, the desired product.
Any trace of a secondary alcohol product (derived from a elimination without hydride shift), if isolable, could be added to a reaction environment described in the first step, as it would easily convert to a trisubstituted alkene, that via Markovnikov hydration, again, gives 2-Methyl-2-butanol.
Please don't perform these reactions without a proper fume hood, anyway.

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text:
Do (simple, inorganic) salts ionize completely in water when they dissolve?
Specifically: Suppose I add equal parts (i.e. the same number of moles of each) sodium bicarbonate and potassium sulfate into a container of distilled water large enough to dissolve them completely. In a second container of water, I add potassium bicarbonate and sodium sulfate. Are these two containers now physically identical? (That is, is there any experiment that could tell them apart?) My understanding is that each now contains free-floating Na+, K+, S-, and HCO3- ions. And if I evaporate away all the water, what will be left behind in each container?
If I replace the potassium and/or sodium with, say, magnesium or calcium, and the bicarbonate and/or sulfate with chloride or citrate, do the same principles hold? (Assuming the resulting salts are water-soluble - some, like calcium carbonate, won't be.)
(I'm not sure if the list of cations and anions above have much in common chemically. My interest relates to the fact that they can be added to water to simulate the chemical makeup and flavor of natural mineral spring water.)
(PS: What's a more elegant way to say "the same number of moles of each?" If I say to mix things in equal parts, or in some ratio, I would normally specify whether I mean by mass, by volume or... how do I say "by number of moles?")

When a salt, no matter which salt, or any other polar compound is exposed to enough water, then its polarized molecule parts, depending on its polarity, become attracted to the oxygen or hydrogens of the water, which will result, starting with the outermost molecule layer, in the complete dissolution of the compound. Within this process both the positively and the negatively polarized sections will get surrounded by water molecules (= hydration shell) and thus shielded from one another. In this state the original compound(s) have (theoretically) disappeared completely and the two containers will indeed be identical. Possible back reactions will (mainly) result in the most energy efficient compound for the particular experimental conditions and have the same chance in both containers.
When you now evaporate all the water you may get various compounds and mixtures depending on what was dissolved.
PS: I'd say "of equal molarity" or "they are equimolar", but my mother tongue isn't English, so there might be a more elegant way.

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text:
How to remove tap water hardness?
I am looking for ways of removing water hardness from tap water, specifically permanent hardness.
I have tried working on some methods, e.g., to perform precipitates using sodium hydroxide and calculated calcium hydroxide, but it seems that either my methods or something else is wrong.
Please recommend me some methods (or resources) on how to remove the hardness from tap water.

What are the methods that you've been working on?

I tried using sodium hydroxide and calculated calcium hydroxide to form precipitates

An ion switcher is the usual way to go.

the only sure ways are distillation and reverse osmosys (de-ionisation) .  Ion-exchange resins also may be of use.

Without knowing what dissolved species are causing the hardness, it's a little difficult to know what to advise. Performing qualitative analysis would likely help you a lot.
You describe using lime softening (i.e., using hydroxides) which can help in some cases. Usually the resulting hydroxide precipitates are pretty hard to filter, requiring a membrane or a very good filtration system.
As Wikipedia notes, there are a few other reliable techniques:

reverse osmosis
distillation
ion exchange resins
chelating ligands

While the first three are definitely the best in terms of ultimate water quality, an easy thing to supplement lime softening would be to add various chelating ligands. For example, after you carefully filter the water from lime softening, maybe add citric acid?
In any case, I'd suggest you do some sort of qualitative analysis (and pH) testing as you go. This is the best way to know what steps you'll still need.


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1H (proton) NMR spectra for alkanes
I do not fully understand the idea of coupling and why chemically equivalent hydrogens do not couple.
I wanted to work with the alkanes, so let's take propane first:

There would be a triplet signal due to the $\ce{CH3}$ groups in blue and green, due to the $n+1$ rule for neighbouring protons, so the neigbouring carbon for the two $\ce{CH3}$ groups would be the $\ce{CH2}$ group circle in red.
Then the $\ce{CH2}$ group in red would have a septet signal due to the two surrounding $\ce{CH3}$ groups is what I would predict.
Moving on to butane, this is where I get more confused:

For the $\ce{CH3}$ group in blue, there would be a triplet signal due to the neighbouring $\ce{CH2}$ in red, and the same can be said for the $\ce{CH3}$ in black.
Then for the $\ce{CH2}$ groups, I am not exactly sure what would happen but here is what I think:
The $\ce{CH2}$ group in red will have a quartet signal on the NMR spectrum because equivalent hydrogens do not couple and so the $\ce{CH2}$ will not couple with the green $\ce{CH2}$ group. This means that only the $\ce{CH3}$ group will couple. Using the $n+1$ rule, a quartet will be present and the same can be said for the $\ce{CH2}$ group in green. However, I think that the red $\ce{CH2}$ will have a triplet of quartets or a multiplet if that is correct, with the same being said for the green $\ce{CH2}$ group.
I would like to extend this further to pentane, which I am also confused about. Can this be clarified with the two examples above so that I can attempt to do pentane myself?

Unfortunately, although the answer given by bon provides a very simplistic answer to a fairly common NMR-101 problem, it is not quite correct. It is fine for the propane case, but falls short for butane. While a very simple molecule, butane has a complicated spectrum because, though the two methylene groups are chemically equivalent, they are in fact magnetically non-equivalent. It serves as an example of how all too often NMR is poorly taught at the introductory level.
First, let's clarify a minor point of confusion that is widely taught at introductory levels - that equivalent protons do not couple to each other. They do; the reason that the coupling is not seen is that the transitions are degenerate and so sit directly on top of each other (ie not observable). Remember that the observed lines in a spectrum are the energy transitions between stable spin states - if the energy difference between two states is equivalent, then the lines in the spectrum are degenerate.
So, a better generalisation for interactions giving rise to splitting in a spectrum is that:

Coupling between magnetically equivalent nuclei cannot be detected
and does not affect the appearance of the spectrum
Coupling between
chemically equivalent, but magnetically non-equivalent nuclei will
usually affect the appearance of the NMR spectrum; commonly seen in
second order effects like that of AA'BB' systems.

Second, then, to see what a 1H spectrum of butane really looks like (well, a more realistic simulation at least):
Below is a more realistic simulated spectrum of butane (at 400MHz), which takes into account the fact that the two methylene groups couple to each other, with an average coupling of about 6.6Hz. Complicated isn't it? Have a read of Glenn Facey's page on his NMR blog, (an excellent NMR resource) where he discusses this very problem.

So why is the case? Well, we need to look at what gives rise to magnetic non-equivalence in molecules. There is an excellent on-line resource to explain about symmetry and non-equivalence at Hans Reich's NMR page. But for butane, let us consider the Newman projections:

At no time is the coupling between H2 and H3 the same as H2' and H3. Either it is psuedo-trans or pseudo-cis, but never the same. So, H2 and H2' are non-equivalent, and this will result in non-degenerate transitions, giving rise to complicated splitting patterns in the NMR spectrum. This is the same for other straight chain alkanes, with decreasing complexity as the chain length grows.
And just to show what a real spectrum of butane looks like: reproduced from Tynkkynen et al. (Magn. Reson. Chem. 2012, 50, 598–607), showing the real and calculated (at 500MHz):

long wrote, "At no time is the coupling between H2 and H3 the same as H2' and H3" (that last H3 should be H3' right?).  Aren't those couplings equivalent (or enantiotopic) in the anti-conformation pictured in the center of your figure?

For the anti conformer (centre), J2,3 != J2',3, so H2 and H2' are not equivalent in this environment. For the gauche conformers (left and right), the couple pairs J2,3 != J2',3 OR J2,3' != J2',3', again introducing non-degeneracy. Anti/gauche equilibrium is temperature dependent for these molecules, and the observed J2,3, J2,3' couplings can be used to calculate the equilibrium populations of conformers. At 300K, J2,2=-13.3, J2,3=6.12, J2,3'=8.66Hz

Very interesting - I too was never taught this.

I know that this may be a little off topic from what you have just said but what would a multiplet be and how would that form?

Also, I tried using the simplified spectra predictor from 'bon' and came up with this for pentane. [http://www.nmrdb.org/new_predictor/index.shtml?v=v2.17.5]
Why does it show 6 peaks for the CH2 group attatched to the CH3's when we just said that equivalent groups do not couple, surely the CH2 groups attatched to the CH3's should produce a quartet if we use the simplified way of looking at the problem?
I also noticed the central CH2 group had singlets only, which is what would be predicted as equivalent hydrogens do not couple. Have I gone wrong here?

Also important to note that NMR is slow compared to the time scale of molecular rotations, so you're going to be time averaging the various signals together...

@Zhe Are you sure about that? Aren't the hydrogens of the methylene group in propane homotopic and the hydrogens on the methylene in butane enantiotopic?

@TanYongBoon You're 100% correct.

Your explanation using the conformational analysis seems to suggest that each of the 2 enantiontopic hydrogens of any one of the 2 methylene groups of the butane molecule are non-equivalent and hence give rise to 2 different peaks in the NMR spectra. That means to say H2 and H2' are not equivalent? Does your explanation also suggest that H2' is magnetically equivalent to H3' and H2 is magnetically equivalent to H3?

Your analysis is mostly correct.

The blue and green $\ce{-CH3}$ groups are chemically equivalent and so will produce a single triplet signal because they couple to the two hydrogens in the red $\ce{-CH2 -}$ group. The red $\ce{-CH2 -}$ group will produce a septet (split into 7) signal due to the six equivalent hydrogens adjacent to it.

Note: This is actually a simplified view of butane's spectrum. See @long's answer for a more detailed analysis.
The black and blue $\ce{-CH3}$ groups are chemically equivalent and each couple to two equivalent protons, meaning that the two $\ce{-CH3}$ groups as a whole produce only one triplet signal. The red and green $\ce{-CH2 -}$ groups are chemically equivalent and so will not couple with each other, despite being adjacent, but each one will couple with the $\ce{-CH3}$ group next to it, producing a single quartet signal which represents both groups.

Spectra predicted using the free tool found here.

Unfortunately had to give this a -1 because the case for butane is actually incorrect. Butane and short chain alkanes should not really be used in teaching NMR at an introductory level, because they actually have far more complicated spectra than this. Nice answer for propane, though.

Indeed. The simulated spectrum is not correct. Because to simulate it was necessary to split the system and then the secondary effect of coupling constants is not considered. You need to run a simulation of the whole system. In this case you get the same spectrum as posted before:
I'm using the same tool as you, but I set the cluster size to 10
http://www.nmrdb.org/new_predictor/index.shtml?v=v2.17.5


text:
How long does it take to change states of matter?
I'm interested to know how long it takes to change states of matter and why it takes the time that it does.  For example, what is the very quickest that we could freeze a packet of chicken breasts?  In the food industry, flash freezing is often used to prevent damage from occurring by freezing meat very quickly.  But why does it take the time that it does?  Would it be possible to freeze my packet of chicken in say, a nano second instead of a few minutes?

It's rather for Physis SE

I think it's fine for Chem SE. States of matter and interrelations thereof are perfectly cromulent area of chemistry.

Heat transfer takes time.  The greater the temperature difference the faster heat transfer but it still takes time.  Even at absolute zero I doubt you could freeze a packet of chicken in a nano second.  Also it is expensive to create very low temperatures.

I understand that but why does it take the time it does?  What dictates the time it takes for the change of state to occur?  Is there some absolute constant at work here which cannot be got around or will human ingenuity continue to find new ways of freezing faster?

A change of state has set amount of energy.   You must transfer that amount of heat.    A bigger temperature difference and bigger contact area will speed the heat transfer.  Heat is real and a bit broad.  Humans are not going to be able to trick heat into finding  faster path.  With a microwave you can blast it with radiant energy for a fast heat add.  But there is not a negative radiant to suck the heat out as radiant.  You are stuck with conduction and convection for cooling.

Thanks Frisbee, most enlightening.  Can I ask the question the other way round. Using a microwave over as an example, why does it take the time it does take to bring a frozen packet of chicken to room temperature?  Obviously, throwing more energy at it would make the change of states occur faster but is there a ceiling?  If we were to throw a very large amount of energy at the frozen chicken, in the form of microwaves, would we be able to make the change from frozen to room temp occur in say, a nano second?

@SeaCat The nature of heat, temperature, and entropy is bit broad.    I am not seeing the piratical value of the question.

Let's say I was working in the food industry and am interested in finding ways to freeze and unfreeze quicker.  Traditional forms of freezing are pretty slow and during the freezing process, ice crystals form both inside each cell in the meat and between cells.  The ice crystals burst the cells open spilling their contents.  So traditionally frozen meat can never taste as good as fresh meat which has not been frozen.

But what if the meat were frozen so quickly that ice crystals did not have time to grow.  Then they could not cause any damage and, when defrosted, the meat ought to be as tasty as fresh meat.  But how to reduce the freezing time to the extent that ice crystals cannot form?  That's the question.  Now you know the context, does that help?

Flash freezing is established and the objective is not nanosecond or even as fast as possible.   Flash freeze is simply fast enough not to form ice crystals.  It is well established that liquid nitrogen is fast enough.    Flash freezing does not require nanosecond so no I do not get the context of your question.

That's true but there is no harm in trying to advance the process further.  If we were all content with the status quo nothing would ever move forward.  So the goal in this hyperthetical situation, is to freeze my packet of chicken in a few nano seconds and bring it back to room temperature from frozen, again in just a few nano seconds.  But why stop there.  What I'm interested in knowing is why things like this take the amount of time that they do.  I mean, why couldn't the process be speeded up even more so that my chicken is frozen in say, a few pico seconds?

It's not really the chicken I'm concerned with here - just the reasons why the time it takes is the time it takes.  What's the underlying rule that makes things happen in the time we experience them happening in our world?

Part of the answer is that the rate of freezing depends on the heat of fusion of the material being frozen (relatively large for water, the chief ingredient of uncooked chicken), but smaller for other substances).
Another piece of the answer is that an object can lose heat by convection, conduction, and/or radiation. You would increase the rate of heat loss by:
(1) increasing the surface area of the chicken as much as possible (flattening it),
(2) placing it against a surface held as cold as possible and made of a material such as metal with high heat capacity;
(3) hitting it with a continuous blast of extremely cold air or gas at the highest possible wind speed (remember wind chill factor?) and maximizing radiant heat loss by previously having painted the chicken with flat, black paint and conducting the experiment in a mirrored, insulated chamber. Some unusual materials can be induced to cool by irradiation with a laser (this is how we have obtained the lowest recorded temperatures ever) but this will probably not work with chicken.
I do not think there is an absolute limit on cooling rate, just diminishing returns for increasing energy expenditure. But I may be wrong. I hope you pursue this until you get an answer that satisfies you. That is how new discoveries are made.

Just spaced out the paragraphs a bit - makes this good answer clearer to read.


text:
How do I store HCl safely?
I'm in a textile unit, where we use different chemicals. My question is what type of material should I make my tank out of to store my hydrochloric acid?

Well, it's typically stored in glass bottles :) You want bigger container?

You should always have secondary containment (e.g., a deep tray) so that any spill will be contained. Redundancy is important in safe chemical storage.

How strong is your HCl?

This question is related: Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol
In the industries the handling of chemicals maybe periodic and, with routine, one tends to forget the hazardous and risks related to this. There's a lot of information on the web (http://water.usgs.gov/admin/memo/policy/wrdpolicy94.006.html):
STORAGE: Hydrochloric acid should be kept in a tightly closed container and
stored in chemical area that is compatible with other chemicals.
Store in a secure, well-ventilated area, that is well marked, and
away from the general work population.  Do not store near oxidizing
materials.
This manual should be helpfull too: http://www.jsia.gr.jp/data/handling_02e.pdf
1. Types of Containers:
The shipping containers used for hydrochloric acid are those rail tanks be tightly sealed, such as acid-resistant bottles (including polyethylene containers, etc.), and steel drums tank trucks, tank cars, and tankers with corrosion resistant linings
If you keep reading the manual there is valuable information about handling, transportation, and rules to keep the human safety.


text:
Why does decreasing the distance between the electrodes in an electrolytic cell increase the rate of electroplating?
I did an experiment on the effect of the distance between the electrodes on the rate of electroplating in an electrolytic cell (copper electrodes with copper sulfate electrolyte).
As I expected, when the electrodes were closer, the current was greater and the rate of copper plating was also faster. But why does less distance increase the current? My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?). I'm probably overthinking this, but any help is appreciated.

If the applied voltage is constant, lowering distance between electrodes leads to lowering the resistivity according to the:
$R=\frac{l}{A}\rho$
Lowering the rezistivity increases electric current (at constant voltage) defined by Ohm's law:
$I=\frac{V}{R}$
As Michael D. M. Dryden pointed out, this answer is simplified and do not describe a lot of mechanisms that occur in a real deposition.

It's a bit of an oversimplification to explain this with a simple invocation of Ohm's law. The thing to remember about electrochemistry is that electron transfer only happens at every close distances to the surface (<10 nm) and it's the potential difference between the electrode and this tiny region at the surface that is important. You might be applying 10 V between the two electrodes, but if you have a lot of solution resistance between them, the voltage near the surface may only be 1 V as the other 9 V are dissipated as heat in the electrolyte. Thus, by moving the electrodes closer, the resistance goes down and the voltage the electrode "sees" is higher.
The trouble with this type of two-electrode cell is that many factors are being conflated. The current that is measured (the rate of electroplating) is dependant on what is happening at both electrodes as well as the resistance of the bulk solution. As the electrode voltage increases, the current usually increases too as increasing the voltage makes it easier for electrons to transfer between the electrode and the redox-active species in solution. Eventually, however, the rate will not be limited by electron transfer, but rather by how fast copper ions can move from the bulk of the solution to the active region near the electrode and a "limiting current" is reached. You can see this below as the voltage at an electrode is swept to higher values:
 (Source: Own work)
At low voltage, there isn't enough voltage for any electron transfer to occur, but as it increases along the x-axis, we see an increase in current until it reaches a peak and the current is limited by diffusion to the electrode. (the reason it goes down after this point rather than levelling out has to do with the time-scale of the experiment, but it's not important for this)
Because the system is so complicated, when we're interested in examining the reactions that are happening at just one electrode without worrying about the other electrode or the solution resistance, an alternate setup is used. If we use a feedback amplifier (known as a potentiostat) to control the voltage applied between the two active electrodes using a third (reference) electrode to measure the voltage close to the surface of the electrode of interest, we can decouple the effects of the solution resistance and other electrode.
A sidenote:

My initial thought was that less distance to travel=ions arriving at cathode quicker=greater current. That doesn't seem right though, because different ions are depositing and liberating electrons and hence not really flowing through (right?).

You were actually onto something interesting here. You're right in that it takes a while for copper ions from one electrode to make their way to the other so most of the copper being deposited is coming from what was already in the solution (at least on relatively short time scales), but what would happen if we moved the electrodes very close to one another? Eventually we would reach a situation where copper could dissolve from one electrode and be almost immediately deposited on the other and we would not see the limiting current behaviour as above.


text:
Correct flask for reacting and distilling on hotplate?
I'm running a distillation reaction at under 100 °C with the heat coming from a hotplate (combined with magnetic stirrer).  Is a Florence flask the correct vessel to use on the hotplate?
Given the small contact area of the flask relative to the surface area of the hotplate I'm wondering if it wouldn't be better to use a fully flat-bottomed flask on the hotplate, and if so what best practices are?  Regardless, I need to keep a stirrer running during the process.
(I intend to have a Kjeldahl bulb on top of whatever flask before feeding the condenser.)

Any chance you can use a sand (or other, as appropriate) bath?

@ToddMinehardt: Maybe I could set a small layer with a dam on the hotplate?  Key is I need to keep the stirrer running during the process.

I recommend not heating directly off the hot plate. Use some kind of bath. It will be much easier to monitor the temperature, and the distillation pot will heat more uniformly.

I'm absolutely with @jerepierre here and I'd use a silicone oil bath. In addition I'd use a round bottom flask with a ground glass joint and a destillation bridge with cooler.

At under $100^\circ \mathrm{C}$ The best will be to use water bath and round-bottom flask. Most magnetic stirrers have enough range (magnetic forces do not stop working if vessel is in couple of centimeters from hot plate, though you should use non-magnetic dish for bath itself). Even if magnet in hot plate is weak I would suggest against flasks with flat bottom, because bottom may fall off under heating.

As others already said, for reaction under 100∘C, water bath with round bottom flask would be ideal and that would also provide greater temp control, distribution and prevents heating over.
You can either use big enough borosilicate evap dish, pyrex glass often found in most local supermarkets or aluminum pot for bath itself.
You could also just use flat bottom erlenmeyer flasks which do not have perfect contact with hotplate but can be used directly and from personal experience I would say that it is decent enough and not that crucial especially when you stir your reaction to prevent any hotspot formation.
In addition you can swap water for oil bath to greatly reduce evaporation rate, especially useful for really long reactions, although your limit of transferred heat increases so overheating with poor temperature control is possible.

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text:
How do I calculate how much sodium hydroxide I need to add to get a new pH?

A 100 ml $\ce{HCl}$ solution has a pH of $3.7$. You want the solution to be of pH 4.5. You have a solution of $10\ \mathrm M$ $\ce{NaOH}$. How much $\ce{NaOH}$ do you need to add to to the $100\ \mathrm{ml}$ solution of $\ce{HCl}$ to get a pH of 4.5?

I firstly calculated the concentration $\ce{HCl}$ in $100\ \mathrm{ml}$ solution = $10^{-3.7}$ = $2.0\times10^{-5}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$. After this I calculated the $\left[\ce{HCl}\right]$ of the new solution: $10^{-4.5}=3.16\times10^{-6}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$ of new solution. Now I'm stuck.

At the beginning, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-3.7}  \times 100 \times10^{-3}=10^{-4.7} \mathrm {mol}$$
After adding sodium hydroxide, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-4.5}  \times 100 \times10^{-3}=10^{-5.5} \mathrm {mol}$$
The difference represents the amount of substance of hydronium ions that have reacted with hydroxide ions.
$$n(\ce{H3O+})=10^{-4.7}-10^{-5.5}= n(\ce{OH-})$$
The volume of sodium hydroxide to add is:
$$v(\ce{OH-})=\frac {10^{-4.7}-10^{-5.5}}{10}= 1.68\times10^{-6}\mathrm{L}=1.68\,\mu \mathrm{L} $$
Assuming the volume of a droplet to be about 60 micro liter, it is much less than a droplet!

To measure a volume this small (if anybody were to actually perform that) would be quite a challenge.

$$\pu{pH}_{(\text{before adding}~\ce{NaOH})}  =-\log[\ce{H3O+}]_{(\text{before adding}~\ce{NaOH})}=3.7$$
$$[\ce{H3O+}]_{(\text{before adding}~\ce{NaOH})} =\pu{1\times{10^{-3.7}}~M}= [\ce{HCl}]_\text{(initial)}$$
$$\pu{pH}_{(\text{after adding}~\ce{NaOH})} =-\log[\ce{H3O+}]_{(\text{after adding}~\ce{NaOH})}=4.5$$
$$[\ce{H3O+}]_{(\text{after adding}~\ce{NaOH})} =\pu{1\times{10^{-4.5}}~M}$$
$$\mathrm{V}_\ce{(HCl)} = \pu{100ml}=\pu{0.1L}$$
$$\color{red}{ [\ce{H3O+}]_{(\text{after adding}~\ce{NaOH})} =\frac{\mathrm{V}_\ce{(HCl)}\times[\ce{HCl}]_\text{(initial)}-\mathrm{V}_\ce{(NaOH)_\text{added}}\times[\ce{NaOH}]}{\mathrm{V}_\ce{(HCl)}+\mathrm{V}_\ce{(NaOH)_\text{added}}}}$$
$$1\times{10^{-4.5}} =\frac{0.1\times{10^{-3.7}}-\mathrm{V}_\ce{(NaOH)_\text{added}}\times{10}}{0.1+\mathrm{V}_\ce{(NaOH)_\text{added}}}$$
$$\mathrm{V}_\ce{(NaOH)_\text{added}}=\pu{1.679\times{10^{-6}}~L}$$

I appreciate to provide me feedback about the downvote in order to improve my answer.

A bit late, but one reason for downvote might be erroneous formatting using "textual" formulas and \mathrm where it makes no sense.

authors: فيزياء وكيمياء


text:
Checking if it is a redox reaction
Ethanoic acid dissociates according to the following equation:
$$\ce{CH3COOH + H2 O -> CH3COO- + H3O+}$$
I want to know if it is a redox reaction or not. In the ethanoic acid I obtained that the oxidation number of carbon in it is 0 by calculation, but it is (-2). What is the error in my answer?

Redox involves movement of electrons. No electrons are moving in the dissociation reaction, only a proton is transferred. Draw the Lewis structures to see the bonds to the carbons. That should help you see the oxidation state better.

Isn't carbon oxidized?

No and I have no idea why you'd think so.

Perhaps you could explain how you calculated the oxidation states. This would help us to answer your question better.

I got it wrong, but by my new calculations:

in CH3COOH: 2o.nC+4o.nH+2o.nO=0, so 2o.nC+4-4=0, thus o.nC=0

In CH3COO^-: 2o.nC+3o.nH+2o.nO=-1, so 2o.nC+3-2=-1,thus 2o.nc=-1+1, then o.nC=0

So for carbon no oxidation or reduction have obtained here...

So till now I didn't see any oxidation or reduction reaction

Self-answers are welcome on StackExchange! Just so you'd know.

You're getting 0 ox. state for carbon because of treating both carbons equally - 0 is mean average value from oxidation numbers of both carbons. The one in carboxylic group  has three bonds with more electronegative oxygen therefore has +3. The one in methyl group has -3 because it's connected with three hydrogens which have lower electronegativity.
See https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure - there's example with acetic acid!
In organic chemistry trying to get average value of ox. state usually doesn't make much sense as it's rare to have all carbons equivalent. @Shafter gave you good advice - without knowing the structure you were making mistake.

Redox reactions involve increase in oxidation state of one chemical species and decrease in the oxidation state of the other through loss and gain of electrons respectively but since this does not occur here then it's not a redox reaction but rather water acts as a weak base to accept the proton donated by ethanoic acid

My point is not knowing if it is a redox or not, but I am lost in the oxidation numbers since I don't know why I am having the oxidation number of C as 0.


text:
pKa values for protonated nitrogens in ketoconazole
Everywhere on the internet I see that the  two $\mathrm pK_\mathrm a$'s of ketoconazole are 6.4 and 2.3, but I don't think that is correct.
The $\mathrm pK_\mathrm a$ of imidazole is around 6.4 . But ketoconazole also has a piperazine ring and normally this $\mathrm pK_\mathrm a$ is around 9.8. There are electron acceptors but I still find 2.3 to be very low. Is this correct and which of the two nitrogens will be protonated?

Can you provide references for the values you're seeing everywhere on the internet? Here is a 2-year old presentation from UCB Celltech in which the p$K_{\rm a}$ values for ketoconazole are given as 6.43 and 3.64.

The 6.4 pKa is close to the reported pKa of imidazole (6.95), so 6.4 is at the imidazole ring.
That leaves pKa 2.3 (or 3.64) for the other basic nitrogen. While that nitrogen is part of a piperazine, piperazine itself (pKa 9.8) isn't a good comparison. The nitrogen in question is also conjugated to an aromatic ring, so it would be better to look at aniline, with a pKa of 4.6. That's much closer to ketoconazole. Add in the inductive effect of the proximal amide, and a pKa in the 2-3 range makes sense.

Many common pKa's can be found on Prof. Dave Evans's pKa list.


text:
What is the correct structure for sulfur dioxide?
A question from the National Chemistry Olympiad (Germany-2008) goes as follows:

Besides the composition of a substance the structure of its molecule or ion is of interest. To predict the structure you may use the electron repulsion theory
  (VSEPR).
  The following sulphur compounds are choosen as examples
  $\ce{SO2}$, $\ce{SO3(g)}$, $\ce{SOF2}$, $\ce{SO4^2-}$, $\ce{SF4}$, $\ce{SOF4}$, $\ce{SF6}$.
b) Draw the structure of these species. Insert the free electron pairs of $\ce{S}$.
  Give the name of the geometrical shape of these compounds or ions.

I'm only interested in the structure of $\ce{SO2}$, that is represented in the answer sheet as $\ce{O=S-O}$ instead of the one I believe is the correct one $\ce{O=S=O}$.
What do you think?

See this.
It's because it can be thought as two interchangeable resonant structures. What happens in reality is that the electron is delocalized over the three atoms.


text:
Steps in mechanism of pyruvate dehydrogenase
I am having some trouble with certain steps in the mechanism below (from Mike Lu on Proteopedia) detailing the mechanism of action of pyruvate dehydrogenase converting pyruvate to acetyl-CoA. This reaction happens under the physiological conditions present in the human body ($\mathrm{p}H = 7.4, 37^{\circ}C$, 1 atm, various intracellular ionic concentrations: http://www.ld99.com/reference/old/text/2878909-566.html).

In the first step, how does the ylide of TPP form?
Why is the deprotonation of the hydroxyl group to form a ketone in the 6th step favoured, especially given that the $\ce{-OH}$ group normally has a very high $\mathrm{p}K_\mathrm{a}$?

TPP = thiamine pyrophosphate? Your first structure is not an anion, it is a neutral molecular species, depending on E1 being neutral, too. In this resonance form it is an ylide, there is another form without partial charges. It is a persistent carbene and an aromatic system. It would be good to know the conditions under which this forms. I cannot get my head around the fact that the anion of the alpha-keto acid would act as an electrophile.

I would assume it happens under the physiological conditions present in the human body.

Unfortunately I am not educated enough in that area to know what these physiological conditions are. I can only try to understand what you have posted here and therefore I might be missing information that you can better define - or that are defined in some standard way, which you could probably link to. That would possibly enhance your question and some non biochemists might give it a chance, too.

The ylide of thiamine pyrophosphate does not ‘form’; rather it is there. There are two major mesomeric structures for the five-membered ring: One including a double bond from nitrogen to the carbene carbon, the other one with a single bond (lone pair on nitrogen, sextet on the carbene carbon). As with mesomeric forms of benzene, the truth is somewhere between these two extremes.
In the carbene elimination step, you seem to assume just by the existence of the arrows that the deprotonation of the hydroxide must come first. That is not the case. In fact, I would assume the first arrow to be the recreation of the keto unit, followed by the removal of the thiamine. Only then would the proton be lost, since we are now dealing with a protonated ketone which is a very strong acid.


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What is a food-safe level of exposure of plastic containers to heat?
There are many articles about how heat can trigger the release of potentially harmful carcinogens from plastic containers into the food or liquid they contain.  Specifics they fail to mention, though, are 1) the safe level of exposure and 2) the rate of BPA release per measurement of time at a constant level of heat.  While I will never drink from a water bottle that has been in my car exposed to summer heat over the course of days or weeks, I have been known to forget my water in my car as I head into work where I'll only end up retrieving it hours later when I go to lunch.  I also store bottled water in my garage which may reach temps of 90 degrees Fahrenheit or slightly higher in the summer.
So I am basically inquiring not only about the safe level of exposure to chemicals by quantity, but also the rate of chemical release over time based on heat level.  This would allow me to produce a practical ballpark idea as to whether or not the water is safe to drink.

You ask:

I'd like to know if the information passed around the internet regarding the release of BPAs into the contents of plastic containers is grossly over-exaggerated or if there is a grain of truth to it that I should be concerned about to a degree.

In the United States, the Food and Drug Administration (FDA) has published  Bisphenol A (BPA): Use in Food Contact Application, in which they state:

FDA’s current perspective, based on its most recent safety assessment, is that BPA is safe at the current levels occurring in foods. Based on FDA’s ongoing safety review of scientific evidence, the available information continues to support the safety of BPA for the currently approved uses in food containers and packaging.

From the link (above), you can go to the 2014 Updated safety assessment of Bisphenol A (BPA) for use in food contact applications, where they conclude that the "no observed adverse effect level (NOAEL)" for BPA is 5000 $\mu\mathrm{g}$ per kg of body weight per day ($\mu\mathrm{g}$/kg-bw/day) and that exposure estimates are "0.2 $\mu\mathrm{g}$/kg-bw/day (mean) and 0.5
$\mu\mathrm{g}$/kg-bw/day (90$^{\rm th}$ percentile) for the adult US population aged 2 years and older." The estimated dietary intake (EDI) is given as 1.1 $\mu\mathrm{g}$/kg-bw/day for population under 2 years old and 0.5 $\mu\mathrm{g}$/kg-bw/day for population greater than 2 years old. Thus, the exposure levels are on the order of a thousand times less than the NOAEL.
The European Food Safety Authority (EFSA) has recently (January, 2015) revised the tolerable daily intake (TDI) of bisphenol-A (BPA) from 50 $\mu\mathrm{g}$/kg-bw/day to 4 $\mu\mathrm{g}$/kg-bw/day. See this press release and this FAQ for more details.
The summary from the press release is:

EFSA’s comprehensive re-evaluation of bisphenol A (BPA) exposure and toxicity concludes that BPA poses no health risk to consumers of any age group (including unborn children, infants and adolescents) at current exposure levels. Exposure from the diet or from a combination of sources (diet, dust, cosmetics and thermal paper) is considerably under the safe level (the “tolerable daily intake” or TDI).

The scientific opinion on the risks to public health related to the presence of bisphenol A (BPA) in foodstuffs, used by the EFSA to bolster the down-revision of the TDI for BPA, has been summarized in a very nice form suitable for the lay public, available here.
What about heating?
In one publication (Assessment of Bisphenol A Released from Reusable Plastic, Aluminium and Stainless Steel Water Bottles. J. E. Cooper, E. L. Kendig, and S. M. Belcher. Chemosphere 85(6), 943-947 (2011)), the authors state:

At room temperature the concentration of BPA migrating from polycarbonate bottles ranged from 0.2–0.3 mg/L. Under identical conditions BPA migration from aluminium bottles lined with epoxy-based resins was variable depending on manufacturer ranging from 0.08 to 1.9 mg/L. Boiling water significantly increased migration of BPA from the epoxy lined bottles. No detectable BPA contamination was observed in water stored in bottles made from Tritan™ copolyester plastic, uncoated stainless steel, or aluminium lined with EcoCare™. The results from this study demonstrate that when used according to manufactures’ recommendations reusable water bottles constructed from “BPA-free” alternative materials are suitable for consumption of beverages free of BPA contamination.

and they conclude

As shown previously by our results and those of many others, BPA migrates into water stored in polycarbonate plastic, especially when heated to high temperatures. Relatively high levels of BPA are found to be released into drinking water from reusable aluminium bottles with an epoxy-based liner. The amount of BPA released from the epoxy coating was greatly increased by heating. No detectable BPA was observed in water to migrate from unlined stainless steel, EcoCare™ copolyester lined aluminium, or copolyester Tritan™ plastic water bottles. To avoid exposures to BPA unlined stainless steel, copolyester lined aluminium or copolyester plastic drinking bottles should be used.

So, should you be worried? In my opinion, that's largely a function of whether or not you use plastic that contains BPA. A few years ago, it might have been difficult to find BPA-free plastics, but there is no shortage of BPA-free items available now. It's easy enough these days to remove any doubt from the equation by simply using containers that don't contain BPA.

I guess what I was also looking for -- and I probably should have been more specific -- is the approximate rate of BPA release per measure of time by static heat level.  I realize that heat levels fluctuate, but it would at least give me an idea as to whether or not exposure within a hot vehicle for mere hours is anything to worry about.  I will also check whether or not Ice Mountain bottles contain BPA for good measure.

You might want to edit your question and title to narrow the specifics of your query in order to get a more useful answer.

I'll probably just keep this one and ask a new one altogether since there is plenty of good info in your answer

Actually, being BPA-free is not necessarily a better situation. http://www.scientificamerican.com/article/bpa-free-plastic-containers-may-be-just-as-hazardous/

The BPA free plastics are using a chemical cousin of BPA.  Might not be any better.  http://www.motherjones.com/environment/2014/03/tritan-certichem-eastman-bpa-free-plastic-safe/


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How do scientists count the exact number of electrons, protons, neutrons in the atom?
An atom is very small so it is probably difficult to focus exactly on a single atom. I wonder how scientists count the exact number of electrons, protons, neutrons in an atom of a certain element.
Could you explain it?
In other words, how do we know that for example, $\ce{^{19}_9F}$ has nine protons and ten neutrons and nine electrons?

@inɒzɘmɒЯ.A.M I'm not sure if OP didn't maybe ask about single atoms not atoms in general? This meaning could stem from OP's misconception, but such question would be actually much more interesting.

@Mith at first read, I didn't understand what they ask. I was gonna VTC, but then it struck me that they want to know how scientists find out how many protons are there in an atom. They could verify or correct me though. And yes, I think they're saying generally, but I figured an example would help.

Excellent question : what has been affirmed without proof can also be denied without proof

For charge neutrality, the number of electrons has to equal the number of protons.  So, the question really is how to know how many protons and how many electrons. This is actually a very interesting question dating back to the early days (and even before) of nuclear physics and chemistry.
Now, one can do it fairly easily with an ion accelerator and a magnet. With the right set up, one can get a beam of mono-energetic ions out of the accelerator, send them through a stripper to remove electrons, and then send them through a magnet to separate the components out by velocity (which depends on the ion mass through the energy). From the magnet deflections one can get a charge-to-mass ratios, and get back to number of protons (charge on a fully stripped ion) and number of neutrons (from number of protons and the mass). This is the heart of mass spectrometry (although one can use time-of-flight techniques rather than, or in addition to, magnetic separation).

So, this means we only know the ratio between the elements, but not exactly  the amount of electrons/protons? It's sounds to me that scientist all over the world agreed that H has 1 positive and 1 negative particle and based on ratio we can say that other elements like Na, has x times more mass than H and so it should have x times more electrons/protons. Am I right?

@AndaluZ - Basically, but it is a little more subtle than that - every ratio yields an integer number of protons. So, the proton could be a composite particle, which it in fact is at high enough energies. But no matter how you parse an atom, you get an integer number of units of +1 charge.

@AndaluZ Thank you for your comment !


text:
Work Done Upon Molecules

In which of the following processes is energy transferred into the substance by work ($w > 0$)?
a) Expansion of a gas against the surroundings
  b) Expansion of a gas into a vacuum
  c) Vaporization of one mole of water at 50˚C in an open container
  d) Melting of 100 g of ice on a laboratory benchtop
  e) Combustion of methane in a constant volume container.

The correct answer is given as (d), but I needed to find a physicist to explain it to me.  This problem can being thought of entirely in terms of thermodynamic work.
So, we know that
$$w = P \Delta V.$$
From this, we see that no work is done in (b) and (e).  Also, since the gas expanded against its surroundings, it did work on its surroundings, which isn't what the question asked, thus (a) is not the correct answer.
Now, although (c) appears to be a heat transfer, and it is, there is a little more going on.  The heat that was given to the water is converted into work as the water vaporizes, this causes a volume expansion (the gas rising).  Intuitively, one can think of the surroundings as a tube with a movable piston.  When vaporizing, the water vapor will push against the piston, increasing the pressure and thus the piston will move, causing the water to do work.
The last choice is (d), the correct answer, but there is also an intuitive way to think about this.  As the ice is heated, it melts (this process is also, as we correctly deduced, a heat transfer), however, this heat also causes the surrounding atmosphere to do work on the ice.
Again, think of a tube with a movable piston with a cube of ice that is slowly melting in the presence of normal air.  As the ice melts, the volume will decrease, and thus the nitrogen and oxygen molecules in the air have more space to move about, decreasing the pressure, which causes the piston to move down, thus doing work upon the ice.
Hopefully this makes sense to everyone who has been either watching or contributing to this discussion.
Unfortunately, I don't know what the authors of this question had in mind. This came up as I was working through an old Chemistry GRE exam. It wouldn't have occurred to me to consider this problem through mechanical work. As a side note, only 18% of the people who answered this question got it right, which essentially means that people just guessed the answer.

This is a bad question.  But the answer is probably supposed to be (d) because:

Water is more dense than ice.
Thus when ice melts, the total volume of "substance" goes down.
This means that atmospheric pressure did PV work on the melting ice.

Of course, heat transfer would also be required to melt the ice but the question isn't asking about that.

In (c+d) you clearly need to transfer energy into the water. But that'd be a heat flux, normally abbreviated $Q$, and not work, $W$, as was pointed out in the disk below.
In d) the surroundings of the ice block must be at least a tiny bit above $0\:\mathrm{^\circ C}$, and in c) the container gets colder.
In case d) the surrounding atmosphere does a tiny bit of work on the melting ice block, because its volume shrinks slightly (water has a larger density). It sort of gets "compressed". Not sure if the teacher thought about that when he set the question.
Aside: Ice won't melt unless you transfer a substancial amount of energy into it. Melting ice at zero degrees Celsius takes up as much heat as heating the molten water to around 75 degrees Celsius afterwards. $300\:\mathrm{kJ/kg}$ heat of melting divided by $4\:\mathrm{kJ/kg\,K}$ heat capacity $\approx 75\:\mathrm{K})$
(The exact numbers are a bit different, also the heat capacity is not constant.)

In answer (c), though, heat is also being transferred to the system, and is causing vaporization so wouldn't that also be a case where work is being transferred into a substance?

Disagree. The thermodynamic definition of work only includes certain specific modes of energy transfer. Heat interactions are, by definition, not work.

When the water evaporates, it doesn't stay an isolated system against the atmosphere -- the $\ce{H2O}$ molecules diffuse into the air. This isn't expansion work against the atmosphere.

@Brian If i heat something from the outside, by whatever means, i definitely do work. Heat transfer in a system is, you're right, not work.

Not in formal thermodynamic terms. "Thermodynamic work does not account for any energy transferred between systems as heat." https://en.wikipedia.org/wiki/Work_(thermodynamics)

Well, strictly thermodynamically speaking you're right. I'm lost as to what the original question was pointing at, though. (What happens with this disk. if i delete my answer?)

Believe it would eventually be scrubbed. To retract, I'd recommend instead striking out the original answer content and adding a note, something like "Based on discussion in comments, I'm retracting the answer." I'd gladly reverse downvote at that point.

That works! Nicely done.

You say In case d) the surrounding atmosphere does a tiny bit of work on the melting ice block, because its volume shrinks slightly (water has a larger density) and I think this the effect the question was asking about.  In other words, you got it right (+1) but you buried the lede!

@CurtF. Multiple-choice questions can only lead to a bad end. ;-)

Do you mean $4:\mathrm{kJ/(kg,K)}$?

TL;DR: If the question is asking about thermodynamic work, then none of the answers is strictly correct.
As you note, only (a) comes close to being correct: $Pv$ work is done by the gas on the surroundings as it expands.  In this case, however, energy is transferred out of the substance as it expands against its surroundings, so it is not a correct answer per the question wording.
As well, you're right that in (b), since no force is required to push the gas against vacuum (there is no resisting pressure afforded by the vacuum at the gas boundary), the $Pv$ work is zero.
I also concur that (c) and (d) are examples of heat transfer, and thus are not work interactions.  There's no electricity or shaft work involved, and no system boundary against which volume expansion/compression (i.e., $Pv$ work)
Agree also, the fixed volume of (e) means that no work interaction is possible.

I agree it's a bad question but since ice occupies a greater volume than water, when it melts, the system shrinks.  That means that the atmosphere did some PV work on the system.  See my answer below.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Cooking with vinegar and lemon juice (Carpaccio)
I know the Maillard reaction to cook at 140–160 °C. I know the low temperature cooking reaction to cook at 70 °C. But which is the reaction that allow us cooking meat or fish with vinegar or lemon juice?

See How does lemon juice "cook" meat?.

thanx. Now the question is, why we need denatured proteins to feed us? Does this question fit in chemistry or biology node of stack exchange?

You might want to edit your question to incorporate the other information you're interested in.

If you put the comment on form of answer I will give you a +1.

Apparently this question has been covered on the Cooking SE site - in this post, it is reported that:

The Oxford English Dictionary defines to cook as:
To prepare or make ready (food); to make fit for eating by due application of heat, as by boiling, baking, roasting, broiling, etc.
However, from a chemical point of view, what happens when you are cooking meat is that you are using heat, amongst other things, to denature the proteins of the meat. Denaturation can also be achieved by other means, such as marinating the meat in an acid, such as citric acid (contain in lemon juice) or acetic acid (contained in vinegar).
You can also use other agents to induce denaturation, but they are not always fit for eating.
However, it is important to understand that from a purely cooking perspective the two things are not the same: "cooking" with lemon (I guess it is fine to use the verb to cook as long as you put it in quotes), gives a completely different result than cooking with heat. This is because other chemical reactions happen during application of heat and because, obviously, you do not end up with an hot product!

which indicates that the application of acidic media to meat accomplishes the act of denaturing proteins therein in a way mechanistically similar to what cooking with applied heat does.
From Kitchen Daily:

Acids like lemon juice, lime juice and vinegar break down raw meat, allowing the marinade's oil and spices to deeply penetrate and infuse the meat with flavor and moisture. They dig out little pathways in the meat and allow the marinade to seep in. The scientific term for this breakdown is "denaturation," which is the process in which protein loses its structure by application of a strong acid, base, salt, alcohol or heat. Basically, the scientific process that cooks meat with heat in the oven is the same one that cooks it with acid. The results may taste a little different -- fish cooked in the oven is flaky, whereas fish cooked by lemon juice (a.k.a. ceviche) has a more raw-tasting consistency -- but the process is the same.


text:
What type of hybridization does phosphorus have in P2O5?
I'm inclined to think that P has a $dsp$ hybridization in that compound, because it makes 5 bonds and 2 of them are $\pi$ bonds - with oxygen. However, I found in some websites that it may have $sp^2$ hybridization, although I have no idea how could it be possible - since it has to make two $\pi$ and 3 sigma bonds.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || It's better to ask two separate questions in two separate threads.

There is only one pi bond to oxygen and four sigma bonds per a phosphorus atom: https://en.wikipedia.org/wiki/Phosphorus_pentoxide#/media/File:Phosphorus-pentoxide-2D-dimensions.png

@Marko and even that $\pi$-bond is not neccessarily there...

It has a significant pi character as can be seen by the different bond length between the C=O and C-O bonds.

It's normal sp3 like in phosphoric - there should dative bond  to describe it properly not double.


text:
Can rocket candy explode during the preparation?
I would like to follow the steps in this video: https://www.youtube.com/watch?v=hyPysthIRss
After I mixed the Potassium nitrate KNO3 and the sugar, I start to cook it in a pot, and stir it very often. After 25-30 min, the KNO3 and the sugar will melt together, and become a brown mass.
My question is: Can this mixture explode during the heating somehow? Or it can just burn, not explode? How to prevent it, and do it the most secure way?

If it's brown, it may be too hot. Do you use a temperature probe? in my case it didn't get significantly dark. You are caramelizing the sugar... that's too hot.

See James Yawn's website ${\it Recrystalized\; Rocketry}$, which is all about solid propellant rocketry of the type you're interested in. It's really an exceptionally comprehensive site.

Can this mixture explode during the heating somehow?

It appears that it's more likely to catch fire, and do so rather vigorously, as in cases where it's exposed to a spark or an open flame. Because the system (pan and mixture of $\ce{KNO3}$ and sugar) would be open to the atmosphere (and not confined to a fixed volume), there's more of a risk of fire than of an explosion. As reported by James Yawn in the subsection on "Skillet Rcandy":

This stuff looks innocent enough in the pan.  But if you have ever seen a quantity this size burn you will have more respect for it.  This batch could send a tongue of flame 20 feet in the air and engulf the person stirring it, if only briefly.  Burns to the body would likely be second degree, and I am willing to tolerate that level of risk.  But eyeballs don't recover from this kind of trauma easily.

See this safety page from James Yawn's site.

Or it can just burn, not explode?

Yes, it can just burn: that's essentially what's going on in the rocket engine. The exhaust from the burn is channeled in one direction (${\it via}$ the hollow core in the engine itself), resulting in thrust in the other direction.

How to prevent it, and do it the most secure way?

From James Yawn's site:

Basic Safety Precautions
At present, I do NOT recommend making this propellant in a household kitchen.
Make small batches only!  No more than one standard recipe, which yields about 160 grams of propellant.
Wear protective gear. You should have a full-face shield, long sleeved heavy cotton shirt, heavy apron and heat-resistant gloves.  These must be worn at appropriate points in the process.
Do not allow children or rambunctious pets in the room where you are working.
Do not allow distracting, inattentive, inebriated or otherwise incompetent adults in the room, especially if they smoke.
Be vigilant of possible sources of sparks, flame, or high temperature that could ignite propellant on contact.
If you smoke, place your cigarettes and lighter in another room so you will not light up unconsciously.
Clear your work-area of any flammable materials.
Have a good fire extinguisher available. Also a 5-gallon bucket of water nearby and a garden hose outside.
Make sure there is good ventilation, and a quick escape route to fresh air.
Avoid stupid rocket tricks.
Avoid excessive exposure to Potassium Nitrate.

All of the above points are elaborated upon at the linked page.


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The conversion of energy during electroplating
I've made a test and i've faced a problem with a part of a question which says: "What is the conversion of energy occuring in electroplating? " I answered that it is conversion from electric energy to chemical energy, since in electroplating, the anode is reduced to X^n+ and n+, where n+ ions are transferred through DC power supply into the cathode and then combines with X^n+ to give again X and plates the cathode, so the electric energy(transfer of electrons through the DC power supply) is converted into chemical one(change  X^n+ to solid). I want to know if my answer is correct or not and why? Helping is appreciated

That's a good start, but there is also ohmic loss, as much of the energy is also converted to heat in the electrolyte, electrodes or even wiring.
BTW, during WWII, silver was used to replace copper bus-bars in stationary machinery such as electrolysis apparatus, freeing the copper for use in war materiel and slightly improving the efficiency of the apparatus due to its better conductivity then commercial copper.


text:
How do things glow in the dark?
What makes something glow in the dark? The only explanation that I can come up with for what makes things glow in the dark is that there is probably a chemical reaction slowly releasing the light that was previously absorbed. Am I correct?
How does this phenomenon work?

Related: Fluorescence versus phosphorescence

Related Why does zinc sulfide glow?

Related: Is there anything funny I can do with chemicals in glowsticks?

It depends on exactly what you're looking for. The mechanism that powers, say, chemical glowsticks is different from the one that powers children's glow-in-the-dark toys, which is in turn different from the one that powers the glow on old-school watches.

When a glow stick breaks, it breaks a thin glass barrier on the inside that allows two different substances to come into contact, mix and react. I am not sure of the exact reaction that occurs, but when chemical bonds are broken they can emit photons (light). In this case, the photons emitted are in the visible spectrum.
The case for glow in the dark stickers is a bit different. All things are absorbing and releasing photons constantly, but in most cases, the photons aren't in the visible spectrum. Glow in the dark things absorb visible light and then release the photons as visible light. This is unlike most other objects which may release them as a lower frequency instead and thus not be visible.

so is it creating the chemical bonds when they absorb photons and destroying them again when it releases them?

In this case, the photons are absorbed by the electrons of the atoms and it excites them into a higher orbit. However, electrons have a certain orbit that they are "most comfortable" and will try to stay there, also called the ground state. Soon after the electron absorbs the photon, it will release it to return to its ground state. Depending on the substance, it will release it at various wavelengths, not necessarily the same wavelength it absorbed it at. Hope this helped :)

So the electrons release the light when they return to the valence shells? thanks!

exactly so @vincentScalia

"in most cases, the photons aren't in the visible spectrum": When you shine light on objects, they do reflect light - that is how we can see their colors. The key to glow in the dark stickers is that they absorb visible light and somehow store that energy, releasing even after the lights are turned off.

The spontaneous emission of light by a substance not resulting from heat is luminescence (a type of cold-body radiation).  $\ce{^1}$
A few types of luminescence are:

Bioluminescence: Made by living creatures such as fireflies, glow-worms, and many marine creatures.
Chemoluminescence: made by a chemical reaction. Glow sticks work this way.
Electroluminescence: made by passing electricity through something like in a light emitting diode (LED).
Photoluminescence: made by shining light at "luminous" (phosphorescent) paints.
Röntgenoluminescence: made by shining X-rays at things. (The curious name comes from Wilhelm Röntgen (1845–1923), the discoverer of X-rays.)
Sonoluminescence: made by passing energetic sound waves through liquids.
Thermoluminescence: made when photons are emitted from hot materials.
Riboluminescence: made by rubbing, scratching, or physically deforming crystals.
Phosphorescence: photoluminescence as a result of triplet-singlet electronic relaxation (typical lifetime: milliseconds to hours)
A chemical absorbs and then re-emits a photon of light, where the energy of the photons matches the available energy states and allowed transitions of the chemical takes place. In phosphorescence, the electron that absorbs the photon undergoes intersystem crossing $\ce{^2}$  into a triplet state.
The excited in triplet state returns to the lower energy singlet state. Such transitions, are kinetically unfavored and progress at a slow rate. Most phosphorescent compounds are fast emitters with triplet lifetimes in milliseconds. If triplet lifetimes are minutes or even hours, very slow degradation occurs.$\ce{^3}$ $\ce{^4}$ In phosphorescent materials, there's a delay between them absorbing energy and giving out light.These substances are so-called "glow-in-the-dark" materials.

To make glow-in-the-dark toys manufacturers mix zinc sulfide or strontium aluminate with plastic giving in glow-in-the-dark toys.$\ce{^5}$ $\ce{^6}$
Fluorescence
Fluorescent materials produce light instantly, when the excited atoms return to normal, in as little as a hundred thousandth of a second.$\ce{^4}$ $\ce{^7}$
Chemoluminescence
In the case of chemiluminescence, two chemicals interact which giving energy as a byproduct in the form of light( heatless glow).
When glow sticks are bent, an inner glass tube is broken releasing hydrogen peroxide. Hydrogen peroxide reacts with diphenyl oxalate giving 1,2dioxetanedione, which is unstable. It decomposes to carbon dioxide releasing energy. The energy is absorbed by electrons in dye molecule, that falls down to ground state releasing energy in the form of light.$\ce{^8}$ $\ce{^9}$

Chemiluminescence can happen naturally as well. In such instances — like fireflies — it is bioluminescence.
Some creatures, such as fireflies and jellyfish, contain chemicals within them that cause them to glow. Some of these creatures glow for protection, camouflage or to attract mates.
Ex 1.Bobtail squid $\ce{^12}$ $\ce{^13}$

Ex.2 Mycena lampadis ( mushrooms )$\ce{^14}$

The principal chemical reaction in bioluminescence involves some light-emitting molecule and an enzyme, generally called the luciferin and the luciferase.$\ce{^10}$ $\ce{^11}$ $\ce{^15}$
Electrochemiluminescence $\ce{^16}$is a special form of chemiluminescence in which the light-emitting chemiluminescent reaction is preceded by an electrochemical reaction.$\ce{^17}$
References
1 https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Map%3A_Principles_of_Instrumental_Analysis_(Skoog_et_al.)/15%3A_Molecular_Luminescence_Spectrometry
2 https://www.shsu.edu/chm_tgc/chemilumdir/JABLONSKI.html
3 https://en.wikipedia.org/wiki/Intersystem_crossing
4 https://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/lumin1.htm
5 http://www.phys.ubbcluj.ro/~dana.maniu/OS/BS_7.pdf
6 https://science.howstuffworks.com/innovation/science-questions/question388.htm
6 https://www.wonderopolis.org/wonder/how-do-things-glow-in-the-dark
[7] https://en.wikipedia.org/wiki/Fluorescence#Lifetime
[8] https://www.compoundchem.com/2014/10/14/glowsticks/
[9] https://en.wikipedia.org/wiki/Glow_stick#Fluorophores_used
[10] https://en.wikipedia.org/wiki/Bioluminescence#Chemical_mechanism
[11] https://www.nationalgeographic.org/encyclopedia/bioluminescence/
[12] https://www.nationalgeographic.org/encyclopedia/bioluminescence/
[13] https://en.wikipedia.org/wiki/File:Sepiola_atlantica.jpg
[14]https://www.timeforkids.com/g34/bioluminescence-2/
[15] http://photobiology.info/Branchini2.html
[16] https://www.sciencedirect.com/topics/chemistry/electrochemiluminescence
[17] https://www.sciencedirect.com/science/article/pii/S0008622317304943

just wanted to mention dioxetane should have a bond between the cyclic oxygens.

@ Blaise it is a software error ,will be correted

@Melanie Shebel noted.The question is so fabulous and vast that i concentrated on the bioluminescence  in vast .My attempt was to bring that beauty that entwines chemistry to living that have developed these millions of years ago.While we try to anslyse it now. Perfection is desired but to bring beauty was my attempt.

Understood, but it's very important to give credit to the photo author and not a third party that posted the photo. It's good practice to get as close to the original source as possible. I will edit the answer shortly to give the proper citation.

@A.K. could you please explain your recent edit to this post? Why is "gas" wrong?

@WilliamR.Ebenezer because the gas example is actually incandescence.

@A.K. Alrighto, thanks for clarifying. (It'd have been great if you'd have mentioned it in the edit summary.) I think you should leave an official comment notifying the poster about the edit.

The glow in the dark phenomena is called luminescence. It can be caused by chemical reactions, electrical energy, subatomic motions, or stress on a crystal. Three of the common mechanisms are:

Fluorescence: Light is emitted during the excitation with invisible
electromagnetic radiation.
Chemoluminescence: A chemical reaction radiates energy in for of electromagnetic radiation (e.g. glow sticks, luziferase reaction,...)
Phosphorescence: You can read about the quantum mechanical processes here. There is one interesting thing I want to point out. While the most phosphorescent materials are solid, in this paper, they have reported the synthesis and afterglow of water-soluble ZnS:Cu,Co nanoparticles.

I know Wikipedia is not at all a reference, but there you can find an overview of all types of luminescence.

Reactions emits energy in different forms. Could be heat, light or electrical. If the form emited is light energy form it surely will glow irrespective of the external environmental characteristics in this case dark!


text:
Program to convert WFN-file to Molden
I have tried google but without success so here the question: does anybody know a tool that can covert WFN-file to Molden or to Molpro format (including orbital info not the coordinates only).

Which program do you use to get those wfn files?

ADF (indirectly via adf2aim), Gaussian

I found sth that does not convert wfn to molden but ADF output to molden. It is quite old (1999), but maybe it still works. ADFrom There seems to be some attention from the programmer on building a suite including adfrom but on Github it's the Fortran code, so you have to compile it and hope that it works.

I know this one and nope that will not work. This code is not supported by original programmer and is using a link to the libraries from SCM (i.e. ADF). These libraries have to be linked during compilation, however they are not working with any "relatively" current version of ADF. A year ago or so, I have even tried to contact the authors but they change a few times institutions they were working for and they do not use ADF anymore.

I am the author of ADFrom, and Kris is correct. I do not maintain anymore the ADFrom code. The main reason is that I am not using ADF since a while and I lost track for the updates. Indeed it needs ADF libraries for licensing reasons (I am not sure about today, I needed to read the binary output for some specific data). The code is in the suite (freemol) for historical reasons and in case anybody wants to fork it ;)
But the full suite still compiles even on a mac!

sorry, but molden reads gaussian data, then saves it in molden format. or?

When a wfn-file is loaded into Multiwfn and one tries to convert it into a molden file via 100 > 2 > 6, it says Error: This function works only when input file contains basis function information.
So I guess, that as the wfn file lacks important information, it’s not easily possible to convert it into a molden file.


text:
Molecules made of molecules?
When a molecule of water ($\ce{H2O}$) and a molecule of hydrochloric acid ($\ce{HCl}$) react together, would they become a molecule made of molecules? What's the name for such thing? How would it be represented - just as $\ce{H2OHCl}$?

The key to your question lies in the word "bound". Without mentioning the state of your species we can say in general, species which are joined by covalent bonds are single molecules, there are countless examples of such reactions in organic chemistry where two small molecules condense into a single molecule.
Species with intramolecular forces (hydrogen bonds, van der waals etc) can be called bound or perhaps "clusters", which is quite a broad term. Real bonding however is not so easily defined as the molecular orbitals may  generate a wide spectrum of charge distributions etc. Quantum mechanics gives a much clearer description of the bonding of a particular molecule or extended system.
The terminology you write your $\ce{H2OHCl}$ is similar to how we write water of crystallisation in a solid. If this is where your idea came from, it should be remarked that this is a distinct concept.
To conclude; species that are covalently bound (that could have strong polarisation even to bordering ionic bonding) are molecules. Intermolecular forces are precisely that, between distinct molecules. We should remark that we can also have extended solids with covalent bonding (e.g. group 14). In addition some regard metals as molecules too, molecules whose molecular orbitals are distributed over the entire structure etc (research tight binding model for more information). But I feel that wasn't the focus of your question?

In fact, such clusters are sometimes referred to as "van der Waals molecules".

@IvanNeretin T be honest I think there is a bit of a grey area in terms of the nomenclature. It is best to think of interactions in terms of MO diagrams in my opinion. When one draws the cut off between a molecule or not is vague.


text:
Combustion without hydrogen
Is there any fuels without hydrogen that will combust. My friend believes there might be and do to my experience I have never heard of one. Please explain and if u can show a balanced equation of this combustion.

Other then base element's suck as sulphate.

And carbon and iron and magnesium and... Why would elements be special? Would steel count than? Your friend is quite wrong...

Cyanogen seems to be an example ($\ce{(CN)2 + O2 -> 2CO + N2}$). Check this paper, it briefly discusses the reaction. Despite not being open access, the first page is available - and it contains only one page...

Prior to electricity, most households were heated by burning solid fuel known as coal (which is basically carbon). Does that count?

Fuels like coal are made out of carbon hydrogen chains

A combustion reaction is classified as a reaction between a fuel and oxygen to produce energy. The reaction: 2CO(g) + O2(g) = 2CO2(g) fulfills this definition, is exothermic, and does not use hydrogen.

So does the reaction C + O2 -> CO2, i.e. the combustion of coke (or graphite or diamonds).


text:
How much does a light heat the air around it?
My dad told me to turn off the lights when I'm not using them because it will keep the house cooler. I don't think he is correct that a light can heat up the room. How much of a difference in the temperature would it make? Assume a 60 Watt light bulb running for one hour in a room with approx. 3000ft^3/85m^3 of air. Also assume the air is dry and in constant thermal equilibrium at 1 atm. I think the specific heat capacity of air is about 1 J/g*K, but I am not confident.

Looks like you have all figures you need. Why wouldn't you plug them into the formula and do the math?

During a period of $\Delta t = 1\ \mathrm h =3600\ \mathrm s$, the light bulb converts electric power of $P = 60\ \mathrm W$ to light and heat. Assuming that none of this energy escapes from the room, the entire power is eventually converted to heat:
$$\begin{align}
Q &= P \cdot \Delta t \\
&= 60\ \mathrm W \times 3600\ \mathrm s \\
&= 216\ \mathrm{kJ}
\end{align}$$
According to REFPROP (NIST Standard Reference Database 23, Version 9.0), the relevant parameter values of air at the assumed initial temperature of $T_0 = 20\ \mathrm{^\circ C}$ and pressure $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$ are a density of $\rho_0 = 1.2046\ \mathrm{kg/m^3}$ and a specific heat capacity at constant pressure of $c_{p,0} = 1.0061\ \mathrm{kJ/(kg\ K)}$.
The initial volume is given as $V_0 = 85\ \mathrm{m^3}$. The corresponding mass is
$$\begin{align}
m &= V_0 \cdot \rho_0 \\
&= 85\ \mathrm{m^3} \times 1.2046\ \mathrm{kg/m^3} \\
&= 102.391\ \mathrm{kg}
\end{align}$$
Ignoring the heat capacity of the walls and inventory, the temperature of the air is approximately increased by
$$\begin{align}
\Delta T &= \frac QC \\
&= \frac Q{m \cdot c_{p,0}} \\
&= \frac{216\ \mathrm{kJ}}{102.391\ \mathrm{kg} \times 1.0061\ \mathrm{kJ/(kg\ K)}} \\
&= 2.1\ \mathrm{K}
\end{align}$$
By way of comparison, a better result may be obtained using the enthalpy of air, since the heat capacity of air is not constant. In this case, however, the difference is insignificant because the temperature change is small. The initial specific enthalpy of air is $h_0 = 419.40\ \mathrm{kJ/kg}$. Thus, the enthalpy is
$$\begin{align}
H_0 &= m \cdot h_0 \\
&= 102.391\ \mathrm{kg} \times 419.40\ \mathrm{kJ/kg} \\
&= 42\,943\ \mathrm{kJ}
\end{align}$$
At constant pressure, the change in enthalpy corresponds to the heat transferred to the air:
$$\begin{align}
H_1 &= H_0 + \Delta H \\
&= H_0 + Q \\
&= 42\,943\ \mathrm{kJ} + 216\ \mathrm{kJ} \\
&= 43\,159\ \mathrm{kJ}
\end{align}$$
Thus, the final specific enthalpy is
$$\begin{align}
h_1 &= \frac{H_1}{m} \\
&= \frac{43\,159\ \mathrm{kJ}}{102.391\ \mathrm{kg}} \\
&= 421.51\ \mathrm{kJ/kg}
\end{align}$$
At a pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, this specific enthalpy corresponds to a temperature of $T = 22.1\ \mathrm{^\circ C}$.
Note that, at the final temperature of $T = 22.1\ \mathrm{^\circ C}$ and the pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, the new density of air amounts to $\rho_1 = 1.1960\ \mathrm{kg/m^3}$. Thus, the new volume is
$$\begin{align}
V_1 &= \frac{m}{\rho_1} \\
&= \frac{102.391\ \mathrm{kg}}{1.1960\ \mathrm{kg/m^3}} \\
&= 85.6\ \mathrm{m^3}
\end{align}$$
Therefore, a small amount of air must escape from the room.
Finally, if the heat from a $60\ \mathrm W$ light bulb really is a problem, you might want to leave the room since the human body generates about $100\ \mathrm W$.

There is no real answer to this because light bulbs do not have a exact temperature do to the various shapes sizes. Also some are Built with different chemicals or metals. But let's say it is a basic light bulb if you touch it warms Ur hand to burning. There is also light radiation as a factor warming the surfaces it connects with so yes a light bulb warms the standard room. To prove this just take a wind mill and hold it over a lamp the windmill will spin slowly form the rise of hot air.

A 60 watt equivalent LED bulb uses about 9 watts of power, so we can conclude that an incandescent bulb produces at least 51 watts of heat energy for every 9 watts of light energy.

Actually, a 60-watt lamp produces about 60 watts of heat energy, not counting any light that escapes through windows or converted to glucose by house plants. All the light produced, as well as IR, will bounce around the room until it is absorbed and turned to heat. LED and CFL lamps produce just about as much heat as the electricity consumed.

You are right, DrMoishe.


text:
At what point does Avogadro's law not apply?
A mole of hydrogen gas takes up 22.4 L, but so does a larger molecule like $\ce{CO2}$. Methane ($\ce{CH4}$) is an even larger sized molecule, but a mole of it will still take up 22.4 L. I would assume that an even bigger molecule like decane ($\ce{C10H22}$) would still act the same as the others. A mole of sand however does not take up 22.4 L. This begs the question: is there a certain size particle at which Avogadro's law is not valid anymore? Perhaps at a molecular size larger than buckyballs ($\ce{C60}$)? (This also happens to be the point that wave-particle duality does not apply.)

https://en.wikipedia.org/wiki/Avogadro's_law - law would be really valid for ideal gas

My feeling is that the size of the molecule matters much less than you would think. For instance, from a purely ideal gas perspective (as Avogadro's law assumes), the expected value for the speed of a particle is $$v=\sqrt{\frac{8kT}{\pi m}}$$
where $k$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the particle.
So, that means that as the mass of your particle doubles, the average speed of the particle is decreased by a factor of $1/\sqrt{2}$. But, since pressure, from a kinetic molecular theory perspective treats pressure as being the average force which the particles apply to a wall (over a certain area), and force is the time derivative of momentum $p=mv$, we see that $p$ increases by a factor of $\sqrt{2}$ when the mass is doubled (because $v$ decreased by $1/\sqrt{2}$.) That means, the average force on the wall will be greater per collision, but not very much greater.
For instance, I went ahead and calculated the pressure which a mole of hexane exerts on $22.42\ \mathrm{L}$ according to the van der Waals (VDW) equation and found it was $1.05\ \mathrm{atm}$. So, as expected, the value is larger, but not much larger. That's because VDW actually accounts for a parameter (b) which is the volume excluded per mole (meaning the volume that particle actually takes up, sort of).
So, hexane, which is significantly larger than $\ce{CO2}$ or anything else mentioned, only increases pressure by 5% when confined to the same volume. That also means the volume should be 5% larger.
Be wary of scaling that to larger molecules though! As hydrocarbons get larger (and being linear helps) they have a much greater polarizable volume which results in (how about this for a name) greater van der Waals attractions. This extra attraction counteracts the increase in size of the molecule a pretty significant amount.
All that to say, the deviation from Avogadro's Law at SATP is actually quite small.
You're only going to get noticeable deviations at SATP when you have very polar molecules, even then, however, it's not too bad. Water for instance takes up $22.45\ \mathrm{L}$ at SATP which isn't even deviating in the direction we would expect, but if you increase the pressure to $1.1\ \mathrm{bar}$, then the volume occupied for water is $22.33\ \mathrm{L}$.
Intermolecular forces are more significant than the particle's size, particularly at lower temperatures and higher pressures.


text:
Why is monohydrogen phosphate an incorrect name?
In my Ap chem class I got this sheet with the names of a bunch of common polyatomic ions. One of them is named monohydrogen phosphate ($\ce{HPO4^2-}$). In naming, I learned not to use ‘mono-’ if it’s at the beginning of a name.
Is this simply a mistake? Or is there a reason?

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
(…) The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances (…).
(…)

Therefore, hydrogenphosphate ($\ce{HPO4^2-}$) may be called monohydrogenphosphate in order to emphasize the difference to dihydrogenphosphate $\ce{H2PO4-}$.
Note that, according to IUPAC Recommendations, the systematic hydrogen name of $\ce{HPO4^2-}$ actually is ‘hydrogen(tetraoxidophosphate)(2−)’; however, ‘hydrogenphosphate’ is an accepted simplified hydrogen name.

The reason you don't generally need "mono"at the beginning is that the charge of the ion implies how many hydrogens would be there if you simply said "hydrogen phosphate". That would be three hydrogens because phosphate has a charge of -3.
That being said, I think you are correct in saying mono-hydrogen phosphate because because "hydrogen phosphate" implies three hydrogen, not the one that you want to specify.
In the end everyone is going to write out the chemical symbols for small molecules like that so don't stress too much beyond getting those easy test questions.


text:
Can we have Aromatic rings in a non-aromatic structure
Is it possible to have aromatic rings inside a structure that itself is non-aromatic? Is there an example.

Also this answer: http://chemistry.stackexchange.com/a/9221/9961

Sounds like a contradiction in terms to me. If a compound has aromatic ring(s), then it is aromatic - that's the definition.
Though aromatic compounds with aliphatic side chains are possible, of course.
Your compound is azulene; look it up on Wikipedia or elsewhere. Indeed, it is aromatic, and so is each of its rings. You may be confused by the fact that cyclopentadiene (which looks just like one of these rings) as an individual compound is not aromatic, but that's another story. See, a 5-membered ring needs an extra electron to be truly aromatic. At the same time, the 7-membered ring has one electron more than it needs. So they exchange that electron and live happily ever after.

Thanks. I have added a structure. It looks like the structure is aromatic but the individual rings are not.


text:
Calculation of the lowest energy conformation with a webbased solution or freeware
I like to draw a molecule in different conformations in a) website or b) freeware and

Calculate its relative energy difference (for example to the lowest global energy conformation)
Display the lowest energy conformation

I'm particularly interested in the calculation of diaxial interactions, gauche interactions, double gauche pentane interactions (and less in ring strain, like chair / half-chair / twisted rings).
I know there are plenty of tools for modelling conformational and electrostacic properties. Most of those are not freely available. Do you an easy solution?
It does not must to be the most precise one, relative energy differences in steps of 0.1 kcal/mol between two drawn conformations would be fine for me.
May be this question is off-topic, because it's a software request, but I didn't know where else I could ask this question.

0.1 kcal/mol is very precise

I agree with @PH13 that 1 kcal/mol, much less 0.1 kcal/mol is extremely precise based on existing computational methods.

@GeoffHutchison Geoff, I started playing around a bit with Avogadro + Gaussian and had a question. Since you are one of its authors, could you please answer me: what exactly does the conformer energy obtained from Molecular Mechanics -> Calculate Energy signify? Is this enthalpy of formation or energy relative to some standard form or w/e? I tried searching the Wiki first, but it's not really helpful as most articles either don't exist or are only half written. Thanks in advance!

@Jori If you are using MMFF94, it loosely corresponds to a heat of formation, but relative comparisons are best. Using molecular mechanics for an absolute heat of formation is unreliable at best.

Avogadro can do a lot of this for you. It might not be precisely what you want but the molecular mechanics from Open Babel integrated into it can do at least some of the tasks you mentioned. It's free and has a very intuitive and smooth user interface. A real pleasure. The downside is that it's poorly documented.
To play around with conformations, go to Extensions -> Molecular Mechanics. To view the generated conformers (from the Conformer Search option) go to View -> Properties -> Conformer Properties...

I've drawn methylcyclohexane and then clicked: Extension -> Optimize Geometry. Now I get a planar molecule, whats only from symmetrical point of view optimized. I want now to see that Avogadro changes the conformation (global energy minimization) with the chair and with Me equatorially placed. How can I do this? (Afterwards I'd like to put the Me axially calculate the energy and then see if the energy calculation approximately gives the A value of Me of 1.74 kcal/mol.)

@laminin Just follow my instructions :) . I didn't say to use Extension -> Optimize Geometry!

I have to add that conformer search doesn't work in Avogadro for cyclic parts of the molecules (as I've read and experienced).

When I "draw a molecule, click Extension -> Molecular Mechanics -> Conformer Search -> Systematic rotor search -> OK", I also get a planar methyl-cyclopropane :(

Besides the draw mode there is another Energy optimization mode, available through the "E with the green arrow beneath"-Button. There you can choose a force field (choose UFF or MMFF92(s)) and then your molecule should not be planar when you optimize it. If it does, just try to move one atom by clicking and holding it and move it around.

I'll point out that there's now a manual which covers conformer searching.


text:
Regioselectivity for electrophilic aromatic substitution with multiple strongly electron withdrawing groups

At which position will electrophilic substitution on p-cyanonitrobenzene occur?

As both are -M groups, there are two possible sites of reaction, meta to each of the groups.
The answer is meta to $\ce{CN-}$. Why is this so? According to me it should be meta to $\ce{NO2}$, because $\ce{NO2}$ is the stronger -M group?

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

For such compound most eletrophilic  subst. wouldn't work

As Mithoron suggests, the correct answer is probably "no reaction." There are no examples of electrophilic aromatic substitution of p-cyanonitrobenzene in SciFinder, by the introductory organic chemistry mechanism at least.
There is one report of halogenation (with N-halosuccinimide) of this compound with palladium catalysis. The cyano group serves as a directing group to place a halogen ortho to the cyano. Chlorination and bromination are both reported.
Reference: Palladium-Catalyzed Highly Selective ortho-Halogenation (I, Br, Cl) of Arylnitriles via sp2 C–H Bond Activation Using Cyano as Directing Group

Do you mean that there is no such reaction at all ?

@arutoregni using the typical conditions. For example, reaction of p-cyanonitrobenzene with bromine and iron tribromide has never been reported in the literature. Halogenation has been accomplished by a different mechanism, which is what is described in the answer.