text:
How can I find out the concentration of chlorine in an acid bath?
A professor I have gave the class this question, can anyone help? This is the first part of the question. (Also, my professor is from Bangladesh...so yea...)
It's the year 2395, and the world is in chaos. Two nations have conquered the world, USA and Bangladesh. You have been hired by Bangladesh's government to monitor acid baths used to remove oxidation on aerospace metallic material desperately needed for the new   F-95 fighter jet engine.
The acid bath is made up of three components:
25-30% HNO3
4-5% HF
Balance H2O
Unfortunately, a spy has broken into your facility and added large amounts of Chlorine to the acid bath. As the lead chemist, it is your job to determine how much Chlorine was added to the acid bath.
During sample collection, it is uncertain what any of the concentrations are, however once a sample is obtained you may manipulate it with any method you wish. Devise a method to determine the concentration of chlorine in the solution.

Creative methods save time and money!

The addition of chlorine to the bath will produce hydrochloric acid.  If the chloride ion concentration in the bath sample could be determined, then you could back-calculate to the amount of hydrochloric acid added.  The standard method to determine the amount of chloride ion in an aqueous sample is to precipitate it by adding silver nitrate solution followed by collecting the precipitate by filtration and then drying and weighing it.  You could use that approach here because silver nitrate and silver fluoride (which would be created from silver nitrate and the hydrofluoric acid) are soluble* in water, so only the silver chloride would precipitate if silver nitrate were added to your acid bath sample.  Here's a link to an actual gravimetric procedure for determining chloride ion content.

Note: silver fluoride is soluble to the extent of 1.8 kg/L, therefor silver fluoride should not precipitate upon addition of silver nitrate to your chlorine-contaminated sample because the hydrofluoric acid content of the bath was only 4-5%.

Nice answer however I was wondering isn't this question off topic? Since it's a homework question and no work on part of the OP is shown

Just saw the year it was asked I saw it today while searching for something maybe the rules have changed

I would suggest some form of multi-step redox titration. You'll need values for electrode potentials though.


text:
When will peroxide form?
A very typical example of a compound of peroxide is $\ce{H2O2}$. When like the title says, will such compounds form, despite usually oxygen will have a oxidation number of 2-?

I'm pretty sure such compounds can form because I can go to the drug store and very inexpensively purchase a bottle of hydrogen peroxide.

From Greenwood and Earnshaw's excellent book Chemistry of the Elements:

When the metals are burned in a free supply of air the predominant product depends on the mtal: $\ce{Li}$ forms the oxide $\ce{Li2O}$ (plus some $\ce{Li2O2}$), $\ce{Na}$ forms the peroxide $\ce{Na2O2}$ (plus some $\ce{Na2O}$) [...]

In general, formation of inorganic peroxide salts in this fashion doesn't appear to be especially favorable for other metals, at least by comparison to the variety of other oxides possible. Sodium's peculiar preference seems anomalous to me, especially given that the lattice energy for the oxide is higher (albeit slightly). This may be a function of the availability of oxygen, which is roughly $21\%$ in air, and is certainly temperature-dependent as well.
As for organic substances, formation of peroxide linkages is possible from a variety of functional groups and using various reagents. Ordinary $\ce{O2}$ will form peroxide bonds with various substrates via radical mechanisms, especially under photochemical conditions. Carboxylic acids react with $\ce{H2O2}$ to form peroxyacids. These are just two examples that come to mind, and numerous others exist.

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Can redox reactions involve only non metals?
When non metal react they form molecule, and whether the bond is polar or not, the elements in the molecular compound will not have charges what so ever, because they are not ionic. But can non metals be the only that is involved Ina redox reaction ?

A redox reaction is one in which oxidation state of some atom changes, and such reactions need not involve metals, nor any ionic species. Oxidation state is the idealized charge that a given atom would have if all its bonds are treated as completely ionic. Thus, strictly for the purpose of calculating oxidation state, the electrons of a covalent bond are typically assigned to the more electronegative atom. (I'd note, however, that IUPAC provides a different set of rules for calculating oxidation number.) Hence, in order for a redox reaction to occur, there does not need to be a complete transfer of electrons (in the sense of ionization). Examples of redox reactions involving molecular compounds are numerous.
For example, consider the reaction $\ce{N2 + 3H2 <=> 2NH3}$. This is the Haber process for the production of ammonia. Here nitrogen is reduced from an oxidation state of $0$ to $-3$, while hydrogen is oxidized from $0$ to $+1$. Again, nitrogen being the more electronegative atom, is assigned the electrons of the bonds. This idealization is useful in that it provides some rough idea of electron distribution, and hence has some utility in predicting reactivity.
Edit: Also, the statement "[...] the elements in the molecular compound will not have charges what so ever [...]" is inaccurate. There is a concept of formal charge that applies to neutral molecules as well as ions.

Related: http://chemistry.stackexchange.com/q/10944/4945

@Martin, thanks, and +1 for your answer to that question. The IUPAC choice of seemingly arbitrary rules is indeed bizarre.

Thanks. I find it a kind of reassuring, that the IUPAC sometimes makes mistakes too/ does not choose the most convenient pattern.

@Martin, yeah, I can understand that. I do feel that this is one of their more egregiously bad decisions, however. Maybe their objection is to the fact that electronegativity has many definitions. That said, I'm fairly sure the various different major scales preserve the same relative ordering the overwhelming majority of the time.

A redox reaction is any reaction in which the oxidation states of the reactants change. There are certainly reactions involving metals which fit that classification. A relatable example would be the rusting of iron (one of the possible reactions involved in rusting):
$$\ce{2Fe + O_2 + 2H_2O -> 2Fe(OH)_2}$$
Here iron is oxidized and its oxidation state changes from 0 to +2.

Well, iron is a metal. So...

All imma asking is could there only exist $non metals$?

If that doesn't answer your question, I don't know what you're asking...

If elements in a molecule do not have charges, how will the redox reaction go on? Since there is no way to know the increase or decrease in charges.

Yes. Apart from Greg's answer there are many more examples.
But since you asked specifically about a redox reaction that may produce ions, I am glad to provide an example. This reaction is very common chemistry, as it is used in public pools. It is the redox dissolution of chlorine in water (coupled with an acid-base reaction). It is also a disproportionation reaction.
\begin{array}{lcll}
\ce{1/2 Cl2 + e-  &-> & Cl-}                &\text{reduction}\\
\ce{1/2 Cl2 + H2O &-> & HClO + H+ + e-}     &\text{oxidation}\\\hline
\ce{Cl2 + H2O     &-> & HClO + HCl}         &\text{redox}\\\hline
\ce{HClO + H2O    &<=>& H3+O + ClO-}        &\text{acid-base}\\
\ce{HCl + H2O     &<=>& H3+O + Cl-}         &\text{acid-base}\\\hline
\ce{\overset{\color{\green}{0}}{Cl2} + 3H2O
                  &<=>& 2H3+O + \overset{\color{\green}{+I}}{Cl}O-
                              + \overset{\color{\green}{-I}}{Cl- } }
                                            &\text{total}\\\hline
\end{array}
Depending on the concentration and the $\ce{pH}$ of the solution, the extent of ionisation will vary.

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Synthesizing monodisperse PbS quantum dots
For a research project, I am trying to synthesize lead (II) sulfide quantum dots using the method of Li et al..  I am injecting sulfur dissolved in 1-octadecene into a mixture of $\ce{PbO}$, oleic acid, oleylamine, and 1-octadecene at $210^{\circ}\ce{C}$, waiting 5 minutes, and quenching the reaction in an ice bath. The oleic acid : $\ce{PbO}$ molar ratio is 10 : 1 which, I think (roughly extrapolating from the data presented in the paper), should result in uniform sized quantum dots with absorption peak of about 900 nm. As described in the paper, the solution I'm working with turns black almost instantly, suggesting nanocrystal nucleation.
The problem is, when I characterize them with UV-Vis-NIR (after precipitation, centrifugation, and dispersion in hexane), instead of getting a clear absorption peak, I get a weird wavelength-vs-absorbance graph that shoots up abruptly at 300 nm and then gradually curves downward, with no indication of a peak anywhere near 900 nm. Why is this, and what are possible solutions?
I think it might have something to do with how I'm extrapolating the trend for oleic acid : $\ce{PbO}$ ratio, or perhaps how I'm precipitating/centrifuging/redissolving the QDs (though I'm not sure how this would affect the size or spectra). Or maybe some other factor I haven't thought of that affects the nucleation rate. Any ideas?

Could you include an image of the spectrum? Perhaps give us more details about your UV-Vis setup? What kind of cuvette are you using?


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Is sulfuric acid or hydrochloric acid stronger?
Of my belief, sulfuric acid might be a little stronger than hydrochloric acid. Because even though they both are strong acids, one mole of sulfuric acid produces two times as much $\ce{H+}$ as one mole of hydrochloric acid.
Am I wrong?

Specifically, you might want to take a quick look at our help center page about dealing with answers to your question, if you haven't already.

When using $\mathrm{p}K_\mathrm{a}$, one typically does not consider multiple dissociations for polyprotic acids, as the acidity of the conjugate base ($\ce{HSO4-}$ in the case of sulfuric acid) can and should be measured (or calculated) separately. Using that metric, $\ce{HCl}$ is stronger (at least per Wikipedia's acid strength page). One needs to keep in mind that $\mathrm{p}K_\mathrm{a}$ is solvent-dependent, and that the values typically given are for relatively dilute solutions. Acidity can sometimes vary widely with concentration ($\ce{HF}$ being a notorious example). It should also be noted that experimentally measuring acidity for very strong acids is actually quite difficult for numerous reasons.
On the other hand, if you evaluate acid strength by, say, the $\mathrm{pH}$ of the resulting solution, then you'd need to take subsequent dissociations into account. Using the crude approximation that an aqueous "strong acid" dissociates completely, then any diprotic "strong acid" is going to be stronger than a monoprotic one for equal concentrations (assuming a non-zero $K_\mathrm{a}$ for the second dissociation). Of course, this is not accurate, and a proper calculation would take into account not only $K_\mathrm{a}$ values for all dissociations, but also water's auto-ionization. I suspect that $\ce{HCl}$ would still be stronger, given that the $K_\mathrm{a}$ for the dissociation of $\ce{HSO4-}$ is fairly small, but one would need to actually crunch the numbers to be sure.

Fun fact: Speaking of autoprotolysis and dependence on concentration, sulfuric acid is very good at removing water. $$\ce{2H2SO4 <=> H3+O + HS2O7-}$$ But in this case, there are properties that outweigh acidity.

Hydrochloric acid is the stronger of the two.  It has a $\ce{pK_{a}}$ around -6.3 while the $\ce{pK_{a}}$ of sulfuric acid is only around -3.  Here are the chemical equations for the dissociations of the two acids.  You can see that the second dissociation in the case of sulfuric acid is not very extensive and doesn't contribute much to the acidity of sulfuric acid.
\begin{aligned}
\ce{H2SO4 + H2O ~&<=>~ H3O+ + HSO4^{−}&&pK_{a}~=~-3}\\
\ce{HSO4- + H2O ~&<=>~ H3O+ + SO4^{−2}&&pK_{a}~=~2}\\
\ce{HCl + H2O ~&<=>~ H3O^+ + Cl^{-}&&pK_{a}~=~$-6.3$}\\
\end{aligned}

While the other questions cover the issue fairly well, they omit one fairly important pragmatic reason for the strength of sulfuric acid over hydrochloric:
Sulfuric acid is a liquid, and may be almost pure.  Hydrochloric acid is hydrogen chloride gas dissolved in water, with a maximum concentration of about 38%.
The extra water weakens the strength available for common use.  For example, pure sulfuric acid will char sugar, while concentrated $\ce{HCl}$ will not.

Although sulfuric acid has two acidic protons, only one of those protons dissociates completely in solution.  After the first proton has dissociated, leaving $\ce{HSO4-}$, that species is now a weak acid as well as a weak base.  As weak acids do not dissociate completely in solution, the remaining acidic proton is now "less" acidic.
A strong acid has a weak conjugate base. The $\ce{Cl-}$ ion created in solution by the dissociation of $\ce{HCl}$ is a terrible base, making hydrochloric acid very strong. The hydrogen sulfate anion is a weak base also, but it is a better base then a $\ce{Cl-}$ ion, adding to its weakened acidity.
While $K_\mathrm{a}$ values are a nice measure of acidity, only weak acids have a $K_\mathrm{a}$ value associated with them.  Both hydrochloric acid and sulfuric acid are strong acids. In fact, because the $K_\mathrm{a}$ of sulfuric acid is so high, the eleventh edition (newest printed) textbook, Chang's "Chemistry", used in College chemistry, simply states it as "extremely high", assigning no numerical value to it. Likewise, the hydrogen sulfate anion is listed as having a $K_\mathrm{b}$ as "extremely low".


text:
How does neutralization of free radicals by $\beta$-carotene work?

In my understanding, by bonding to the free radical species, say by one of the double bonds, a new free radical on the $\beta$-carotene develops.

$\beta$-Carotene is a highly delocalised $\pi$ system.
According to George Britton there are numerous mechanisms to achieve this.

Oxidation
Oxidising radicals can remove an electron from the carotene molecule.
$$\ce{CAR + .R+ -> .CAR+ + :\!R}$$
Reduction
Analogous it can accept an additional electron.
$$\ce{CAR + .R- -> .CAR- + R}$$
Hydrogen Abstraction
It is possible to abstract a hydrogen atom from a saturated carbon in allylic position.
$$\ce{H-CAR + .R -> .CAR + HR}$$
Addition
You are correct in assuming, that a free radical may form a bond with that system. This is most common for hydroxyl radicals.
$$\ce{CAR + .OH -> .CAR-OH}$$

You are also correct in assuming, that another radical will always be created. However, the single electron will be very delocalised (resonance stabilised) throughout the whole system and not nearly as reactive as the free radical before.
The remaining molecule can then of course attract more free radicals to form other singlet species, which the body can digest or excrete.

Good point, there most likely are more than just one radical species.

The extended, conjugated series of alternating double-bonds in $\beta$-carotene comprises a system of highly delocalized $\pi$ electrons. If a radical forms by hydrogen abstraction at one of the two terminal allylic positions, that lone electron is delocalized over the entire $\pi$ electron system, and thereby stabilized. Similarly, radical addition across any of the double bonds will result in a highly stabilized radical (though addition to one of the internal double bonds will actually disrupt the conjugation, but is nevertheless quite possibile for reasons of sterics and the greater lability of $\pi$-bonds generally). The stability of the resulting radical drastically lowers the energy of activation for the reaction, hence making $\beta$-carotene so liable to react. In essence, its antioxidant activity is (at least in large measure) a direct consequence of the ease with which it is itself sacrificially oxidized. The same factors are responsible for, e.g., the propensity of unsaturated fats to spoil more readily than saturated ones.
That said, as is typical of radical reactions, the resulting $\beta$-carotene radical, while reasonably stable, can still go on to propagate the radical chain reaction. There are various other antioxidants which form highly stable end products after initial radical formation, and thereby terminate the chain reaction.


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How is it thermodynamically possible to obtain energy from a piece of coal?
If we put a piece of coal in open air we don't get any energy of it. But if we put fire to it in open air we get energy. Explain it with help of Gibbs free energy.

Coal in open air can and does sponstaneously combust.
See for example Spontaneous Heating and Fire in Coal Mines and Coal’s Spontaneous Combustion Problem.
At low temperature the change in Gibbs free energy of coal oxidation is still strongly negative, just like at high temperature.
But at low temperature, the reaction rate is low.
Still, if the rate at which heat is dissipated is less that the rate at which heat is produced, temperature increases and the reaction rate increases until there is appreciable combusiton.

One of the features of a solid is that there is very little movement of the molecules. This prevents most reactions from occurring due to the lack of energy to achieve the activation energy for the reaction. If the coal is already burning, then some surface molecules could absorb the energy (of the fire) to break away from the solid so the reaction with the oxygen is far easier.

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text:
Loading nickel metal hydride battery for the first time
I just bought a new hair clipper and I was about to start shaving my head when I noticed that it was almost at 0%. It has a nickel metal hydride battery.
The manual says I have to load it for 24h hours before using it the first time or whenever it has been unused for a long time.
I started calculating. 24 hours are 1440 minutes. Dividing it by 100 makes it 14.4 minutes to gain 1%. To gain 20% it'll take 288 minutes (more than 4 hours).
Well I started to load it like an hour ago and it has almost 40% so far. That's not 4 hours. So it'll be at 100% in a few hours. I think in the evening it's done.
But the manual emphasized that it should be 24 hours ! So I wondered why do they want me to load it that long even when the percent-bar shows me it's fully loaded and ready to use? I also looked up some reviews where the customers wrote that the battery's lifetime is very short.
Would it cause a shorter battery-lifetime if I use it for like 10 up to 20 minutes in the evening (in about 6 hours)? Or more in general, is there something about the intrinsic chemistry of the battery which requires (in the most cases) so long before we use it the first time? Is that even necessary?

One thing is, that the battery power is usually measured using it's voltage. However, the voltage/charge is not linear. My notebook Li-on battery spends half of it's cycle at 5%. This all also means, that 100% doesn't need to mean 100%/

Battery on 0% usually still has some voltage, and battery on 100% usually can be overcharged. However, for many types of batteries it is not healthy, leading to premature degradation, the exact 0% and 100% charges as well as initial load procedure and storage requirements are marked at point that happens to prevent battery degradation and increase its capacity.
While I'm not aware of what exact process is intended to happen in Ni-hydride battery, I'm pretty aware that preparation procedure for battery MAY be required. For example, during initial cycles of lead accumulator, its capacity increases. The reason is in its chemistry, as it actually works on lead oxides transformations and the cycles increase layer of oxides on electrodes.
Since Ni-hydride cell includes electrode with half-reaction
$\ce{Ni(OH)2 + OH- = NiO(OH) + H2O + e-}$
I guess, that during initial charge the manufacturer expects that layer of nickel hydroxide is formed on positive electrode.  But this is just a guess, you'll have to search specific info on this particular type of battery to be sure. Ni-hydride batteries are resistant to overcharge at low currents, so if you use appropriate smart charger, say, shipped with the batteries, the overcharge is safe, so, since the manufacturer requires it, you should follow the manual.


text:
Can I perform electrolysis without electrolyte? (with distilled water)
When performing electrolysis at home, I usually used salt water as electrolyte. The problem is, that this method also produces chlorine and chlorine is aggressive to both humans and electrical equipment.
I was thinking of creating pure hydrogen and oxygen by using distilled water. I was taught that water is not conductive.
But since water has "positive" and "negative" part (which allows for it to dissolve polar bonds, doesn't it?). Couldn't I just tear the molecules apart if I use electrical field strong enough? If I can, how high the voltage must be?

I have Buder HS-72 Hydrogen Water Generator at home. I put distilled water in it and it still works! When I set it to cleaning mode it produces ozone and oxygen but not chlorine gas since I use distilled water only. As ozone has a very distinct smell and I could sense it I have therefore empirical evidence that yes, you can perform electrolysis without electrolyte. I can't exactly say how the machine technically works but the mechanism probably relies on having a bunch of conductive metal plates very close to each

This Buder HS-72 apparatus does not produce hydrogen gas. It produces a hydrogen solution in pure water. It means that the concentration of $\ce{H2}$ in water is extremely low. The maximum solubility of $\ce{H2}$ is  $1.4$ mg per liter. Such a concentration is not far from the concentration of $\ce{H+}$ ions in aerated pure water (pH = $5.6$). So it could be admitted that water is electrolyzed as such a low amount. If some electrolyte had been added, the electrolysis could produce more $\ce{H2}$, but this would be included in a mixture of $\ce{H2 + O2}$, which is explosive !

You could try using a different salt to make the water conductive. sodium carbonate, for example: no possibility of chlorine emission there.

@matt_black That's actually what I eventually figured out. I would put baking soda in an oven to turn it in sodium carbonate and than use that.

Interfacial water is more conductive than bulk due to microcrystal structure from surface nucleation which is why it works on "plates very close to each other"

If I can, how high the voltage must be?

Not possible at home.
For obtaining pure hydrogen and oxygen it is still required to use an electrolyte.  However, there is some amount of them that are not electrolyzed, allowing to electrolyze only water. I'd try sodium/potassium hydroxide or sulfate.

Even if it's not possible at home, it would be interesting. I was thinking, however, that at certain voltage a spark will go through water instead of elecrolysis.

Does sodium hydroxide get electrolyzed?

@TomášZato 1) yes, it would. However, it occurs in different way, see 'Electrical breakdown' part of 'Insulator (electricity)' on the wiki. 2) Electrolysis of $NaOH$ solution will produce hydrogen and oxygen and may be a trace amount of $H_2O_2$ will be produced as well.  If it traps some amount of $CO_2$ forming carbonates, some amount of peroxocarbonate may be produced as well. Good thing, both will stay in solution, not producing any harmful gases. Theoretically speaking, sodium may be obtained by electrolysis of water solutions, but it requires mercury electrode.

Under any low voltage, sufficient to decompose water in conducting solution, the electrolysis of pure water is going as well, as it's allowed thermodynamically. It's just very slow because the amount of product per time depends on amount of charge per time, i.e., current, not voltage.
In the case of water, you can't significantly raise the current by raising the voltage, because you sooner reach dielectric breakdown than any sufficient current.
Instead, you better invent some nanostructure to minimize space between electrodes and maximize their area :)

You mean, for example, two copper plates very close to each other? Or do you mean something that is really "nano", that is, too small to design at home?

Not only close, but the surface must be maximized to get as much current as possible. Don't think that "nano" is too small to design at home. I don't know how to arrange pure water decomposition at home, but there are very porous substances like medical charcoal or silicagel and the ability (at least theoretical) to coat or impregnate them.

I've ran a 240v supply through a container of de-ionised water and even rain water to connect up to a light bulb. The light bulb never worked. I even placed my finger in the waters and that didn't get a shock or trip my rcd. So, imho, water without impurities does not conduct electricity and it's only the impurities that do (don't put your finger in water that has impurities!!) So if water isn't conductive how can it be broken down into the gases by electrolysis? Well, put simply, it can't, people always add an electrolyte to the water but it is the electrolyte that is being broken down, not the water. Hope that clarifies your question.

The electrical conductivity of pure water at 25 °C is 0.055 µS/cm. This value may be very low, but it is not zero.

Try with interfacial water, that is water adsorbed onto strongly charged surfaces like a plate, works better on anode

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text:
Does washing soda (sodium carbonate) turn into baking soda (sodium bicarbonate) over time?
I have some sodium carbonate. But it's over 20 years old and was not kept in tight bottle. I wonder if this compound absorbs $\ce{H2O}$ and $\ce{CO2}$ from air.
I guess I'd need to burn it a little to turn it back in case the answer was yes.

The hydroxide tag would have been much too localised (therefore pretty soon be deleted again) and I do not see any fit for this question anyways.

Hmm...The "burn it a little" link covers conversion of sodium bicarbonate to sodium carbonate.  The current question, as I read it, is starting with the carbonate.  Over time at ambient, the hygroscopic sodium carbonate just absorbs water, no carbon dioxide, and forms different hydrates.  So it can be "burned back" (dry it) to the anhydrous carbonate.

I thought that absorbing water turns it into the bicarbonate, rather than just wet powder.

Industrially, sodium bicarbonate is produced by reacting sodium carbonate with water and carbon dioxide.  However, under normal household conditions sodium carbonate simply reacts with water or moisture in the air (it's hygroscopic) to form a hydrate - it doesn not form sodium bicarbonate. The water of hydration can be removed by mild heating (~100 C).  At high temperatures (>850 C) sodium carbonate can decompose according to the following equation.$$\ce{Na2CO3(s) -> CO2(g) + Na2O(s)}$$

I fear you've understood my question inversely. I asked if $\ce{Na2CO3 + CO2 + H2O -> NaHCO3}$ over time. Your answer only tells me that sodium bicarbone releases it's water. I thought however, that this only happens in heat over 150°C.

Oh, I see that I've stated wrongly what's in the bottle.

See my sentence beginning with "The sodium carbonate", it says that the sodium carbonate can NOT be converted back to sodium bicarbonate.  The bicarbonate can release its water and carbon dioxide (slowly) at 25 C, especially at high humidity.

Yes, I've read that and I've understood that. I just wanted the question and answer to make sense according to the title. Otherwise I thank you for the answer.

I'm aware of the SE system. I was hoping that you'll edit the answer so that it's not confusing. Though it's my fault that I had a mistake in the text of the question, the title is explanatory. The edit would be too radical for me to do it myself. But to give you a guide, After 20 years I'm sure some of it has degraded does not match question does sodium carbonate degrade? unless it indeed degrades to $\ce{Na2O}$ at room temperature.


text:
Which electrodes do not corrode at all?
I had used spare pieces of metal to perform electrolysis. They all had the disadvantage that they corrode when used as anode - and some of the oxides are toxic (copper, chromium).
I've found that carbon, which can be harvested from pencils and old batteries does not corrode contrary to my expectations (I'd expect it to produce $\ce{CO2}$). But soldering on carbon is completely impossible. Also, I need some bigger electrodes.
So are there any metals that do not corrode during electrolysis?
I'm using sodium carbonate as electrolyte.

Use Platinum coated razor blades. Old style razor. Walmart or Walgreen sells them. $5 for 10 pieces.

This highly depends on what you are electrolyzing. When using solutions that are not very acidic, oxide-coated metals sometimes work, like lead-oxide electrodes, that can be harvested from old lead accumulators (attention, lead is toxic, wear lab coat and gloves when working with them. Lead accumulators contain ~20% sulfuric acid, so wear gloves as well and unassemble them in an acid-proof working space)
If you search the web for recipes of perchlorate preparation by electrolysis at home, you will find extensive discussions on the topic of anode materials and making oxidation-resistant electrodes.
Platinum is really hard to corrode, but even it corrodes in some cases.

I'm starting with just water - but that's enough to turn most electrodes in junk and sometimes very toxic junk. Last piece of metal I used turned the gave the water colour of our profile picture.

you do not need to solder a graphite electrode to connect it. It is possible to connect mechanically (screws, wires etc.). It is possible to buy some bigger pieces here: Penta trading
To answer your question: Platinum is really hard to corrode. It is commonly used platinum coated titanium mesh, but it is quite expensive. link: Alibaba

Other than platinum, you can use Gold electrodes, but those are pretty expensive. So unless you use Aqua Regia as electrolyte, the corroding is negligible.

As mentioned before, you can use Platinum and that works great, though it is super expensive.  I have used graphite electrodes in the past with either metal fittings or alligator clips. Could that work?
Alternatively, you can do galvanic anodic protection on steel if you are up to it.

Could you expand what the last idea means and how it works?

Connect a sacrificial anode that will certainly corrode before the one you really care about. This technique is often used on boats and cars. https://en.wikipedia.org/wiki/Galvanic_anode


text:
How can we calculate enthalpy change in a closed tank containing methanol adsorbed by activated carbon?
Here's the problem in the simplest way:
I have a closed box containing methanol adsorbed by activated carbon. The box is made from copper. A pipe is connected to that box, which has condenser on the other end to return the methanol back to the liquid state.
The box is painted black and put in the sun to absorb heat.
The data I have collected are:

initial and final pressure
initial and final temperature
0.26 kg of methanol was condensed in the other end
0.3744 kg of methanol remained in the tank

How can we calculate enthalpy change in the box (heat absorbed by the box from the sun)?

Enthalpy ($H$) is defined as, $H=U+PV$ where $U$ is the internal energy and $P$ is pressure and $V$ is volume. The change in enthalpy is given by
$$\Delta H = \Delta U + \Delta(PV)$$
And $\Delta U$ is equal to the sum of the heat transferred $q$ and the work done $w=-P\Delta V$.
At constant $V$ (which is what you have in a closed system), $\Delta V =0$ (and thus $w=0$) and $\Delta(PV)=V\Delta P$.
Thus...
$$\Delta H = q+V\Delta P$$
You can determine $q$ based on the amount of methanol that vaporized using the heat of vaporization of methanol. We can assume that the majority of heat absorbed was used to distil methanol.
$$q_{vap}=m\cdot c_{vap}$$
How can we determine the volume of the system? Or... if the change in pressure is very small, perhaps that term can be ignored.


text:
How to determine amount of CO₂ in soda water
I've got a problem in determining the amount of $\ce{CO2}$ in soda water. Basically, I have boiled some purified water to remove the dissolved gas.  By using a SodaStream machine, I made carbonated water.
After that, I have tried to determine the amount of carbon dioxide through titration with phenolphthalein and also $c(\ce{NaOH})=0.1~\mathrm{mol/L}$ sodium hydroxide.
The problem is I am not sure if this is the correct method to determine the amount of dissolved $\ce{CO2}$ in water.

I'm leaving this as a comment since it is not directly answering your question, but you could check your titration (which should work fine as described) by the difference in mass of the bottle.  Weigh before carbonation and after, and account for any splashing by finding the difference in mass of a dry paper towel and the mass of the same paper towel after drying off every drop of water you can find.  It should be fairly accurate and will give you a better idea of the mass of $\ce{CO2}$ in the system immediately after carbonation.

First of all, after degasing by heating you have to be sure that no $\ce{CaCO_3}$ is formed, so you can minimise this problem by transfering the water in another flask.

More accurate would be to start directly with demineralised water which you can buy cheap at filling stations.

Secondly, you shouldn't wait to much while titrating, otherwise the water is taking up $\ce{CO_2}$ from the air.

You should use a bulb which doesn't contain much air (which contains $\ce{CO_2}$ and can interact as well. You can use a round flask and directly connect it with a stopper.

Third, Phenolphtalein is a good indicator because you titrate a weak acid with a strong base.

Make sure, you don't use too much Phenolphtalein.

Last but not least be sure that the $\ce{CO_2}$ from flask doesn't go away in the air or into the NaoH of the burette.

To cool will help, because $\ce{CO_2}$ is more soluble in cold water.

If you want to be a super-purist, do a potentiometric measurement and plot the point of change using a simple voltmeter.

In the brewing industry this problem is solved using a device made by Zahm and Nagel. Basically, carbonated beer in admitted into a chamber of known volume. The chamber is shaken which causes the CO2 to come out of solution moving a piston. The temperature, pressure and volume of gas are measured and a table consulted.
Don't use NaOH for titration here. It grabs CO2 forming sodium carbonate. If you want to titrate I guess you could add an excess (but known) amount of NaOH and then back titrate.


text:
Can lead(II) oxidize tin metal in aqueous solution?
According to my textbook, element $\ce{X}$ can oxidize element $\ce{Y}$ if $\ce{X}$ was lower in the activity series.
Lead is lower than Tin in the activity series table in the textbook, but the following question contradict that rule and I can't figure out why:

Will the reaction $\ce{Sn(s) + Pb(NO3)2(aq) -> Sn(NO3)2(aq) + Pb(s)}$ occur?

The book reveals at the end that the answer is no, but doesn't explain why.

Are you certain that your statement from the textbook is correct? The cation of X will oxidize element Y if element Y is more active than X, which is usually labeled (may be towards the top or the bottom of the list). For reference, see the activity series at Wikipedia, where the more active metals are above the less active metals in the list.

How to use the activity series:

For example, what happens when iron metal is placed into an aqueous solution of copper (II) nitrate?
Let's write the potential equation:
$$\ce{2Fe(s) + 3Cu^{2+} -> 2Fe^{3+} + 3Cu(s)}$$
Since iron is more active than copper, iron will reduce copper cations (and copper cations will oxidize iron metal). The iron will dissolve.
What happens if copper metal is placed into an aqueous solution of iron (III) nitrate?
Since copper is less active than ion, copper metal will not reduce iron cations. No reaction occurs and the copper will not dissolve.

Your question

Lead is less active than tin in all activity series that I have seen. Lead metal should reduce tin cations. This reaction should work. It is possible that the textbook/solutions manual has a typo. I find them in my books all the time.

Interestingly, I have a link for a video of that exact reaction.

This is a nice example: the standard reduction potentials are close, so the Nernst equation ultimately reveals what is favored to happen spontaneously, ignoring all real world complications.
Consider a galvanic cell having a tin $(\ce{Sn})$ metal electrode in $x~\pu{M}$ aqueous $\ce{Sn(NO3)2}$ and a lead $(\ce{Pb})$ metal electrode in $y~\pu{M}$ aqueous $\ce{Pb(NO3)2}.$ Assume the salt bridge is an aqueous solution (very roughly $\pu{1 M})$ of $\ce{KNO3}.$ Under standard conditions, $x = y = \pu{1 M},$ and the standard reduction potentials are
$$
\begin{align}
\ce{Pb^2+(aq) + 2 e- &<=> Pb(s)} &\quad E^\circ &= \pu{-0.126 V} \\
\ce{Sn^2+(aq) + 2 e- &<=> Sn(s)} &\quad E^\circ &= \pu{-0.14 V}
\end{align}
$$
So $\ce{Pb^2+}$ is spontaneously reduced to $\ce{Pb(s)}$ and $\ce{Sn(s)}$ is spontaneously oxidised to $\ce{Sn^2+}$. Hence, $\ce{Sn(s)}$ is the anode (where oxidation occurs) and $\ce{Pb(s)}$ is the cathode (where reduction occurs).
The galvanic cell, in standard notation, is
$$\ce{Sn(s)|Sn^2+(aq;\pu{1 M})||Pb^2+(aq;\pu{1 M})|Pb(s)}$$
and the standard cell potential is
$$E_\mathrm{cell}^\circ = E_\mathrm{cathode}^\circ – E_\mathrm{anode}^\circ = \pu{-0.126 V} – (\pu{-0.14 V}) = \pu{+0.014 V}$$
Note that the anode is at the left and it is $\ce{Sn(s)},$ so tin metal gets oxidized.
If $x = \pu{0.6 M}$ and $y = \pu{0.1 M},$ then the Nernst equation is used in the obvious way and the non-standard reduction potentials are computed to be  $E_\ce{Pb} = \pu{-0.156 V}$ and $E_\ce{Sn} = \pu{-0.147 V}$, so
$$E_\mathrm{cell} = E_\mathrm{cathode} – E_\mathrm{anode} = \pu{-0.147 V} – (\pu{-0.156 V}) = \pu{+0.009 V}$$
and the galvanic cell notation is
$$\ce{Pb(s)|Pb^2+(aq;\pu{0.1 M})||Sn^2+(aq;\pu{0.6 M})|Sn(s)}$$
Again, the anode is at the left, but now it is $\ce{Pb(s)}$ that gets oxidized.


text:
How exactly do you measure the free energy of a reaction?
It is to my understanding that ΔG can be positive or negative given whether the reactant is exothermic or endothermic.  How would you go about measuring the energy (in joules) of a reaction using the heat of the reaction?

How would you go about measuring the energy (in joules) of a reaction
  using the heat of the reaction?

You can measure the heat of a reaction (the enthalpy of reaction, $\ce{\Delta H}$) using a calorimeter, but the enthalpy of the reaction is only one part of the free energy of the reaction ($\ce{\Delta G}$). $\ce{\Delta G}$ is related to $\ce{\Delta H}$ through the following equation $$\ce{\Delta G = \Delta H} - T \Delta S$$ where $\ce{T}$ is the temperature and $\ce{\Delta S}$ is the change in entropy for the reaction.
To use a calorimeter you put your reactants into the calorimeter and let the reaction proceed.  The calorimeter is thermally isolated from its surroundings, so any heat given off (exothermic) or absorbed (endothermic) during the course of the reaction can be accurately measured.  Knowing the moles of reactant you put into the calorimeter and the heat change in joules (using the measured $\ce{\Delta T}$ and the known heat capacity of the calorimeter) you can determine $\ce{\Delta H}$ in joules/mole.  Here is a link that provides more detail about calorimeters along with some problems that are worked out in detail.

That explains how to measure $\ce{\Delta H}$, but we still need to know $\ce{\Delta S}$ to obtain $\ce{\Delta G}$. Do you know how to measure $\ce{\Delta S}$ ?


text:
Basic chemistry of the Shinoda test for flavonoids?
In this test, four pieces of magnesium fillings (ribbon) are added to the ethanolic extract followed by a few drops of concentrated hydrochloric acid. A reddish colour indicates the presence of flavonoid.
What is the basic chemistry of the test?  What are the chemical reactions taking place in the test that lead to the formation of red precipitate, and also, why have we used magnesium and concentrated HCl in the test?

Here is a diagram of the Clemmensen reduction.  The metal supplies the electrons and the hydrochloric acid supplies the protons to achieve the carbonyl reduction.  The mechanism of reactions that occur on metal surfaces is poorly understood, but here is a link to the current thinking on the mechanism for the Clemmensen reduction.

The Shinoda test is similar to the Clemmensen reduction in that a carbonyl group is reduced.  However, in this case, a reductive-elimination occurs.  Due to the stability (extended conjugation) of the anthocyanidin product, instead of the reduction going all the way the to the corresponding methylene compound, the hydroxyl-like intermediate undergoes elimination to create the stable anthocyanidin.  In the Shinoda test, magnesium is used in place of the zinc-mercury amalgam.  The Shinoda test is a test for the presence of flavones.  If they are present in the test sample then they are reduced to anthocyanidins under the Shinoda reaction conditions as shown below.  The conjugation in flavinoid compounds produces a yellow color, while the extended conjugation in the resultant anthocyanidin shifts the color further out to the red region of the visible spectrum.  The dramatic change in color makes this a simple visual test for the presence of flavones.

Note also that the Shinoda test can provide an identification of the Flavonoids type. In instance, Flavones give orange anthocyanidin and Flavonols give red color, whereas flavanones give purplish color.

I found one source indicating that the reaction involves the conversion of the flavonoid to the corresponding anthocyanidin. The structure of that resulting anthocyanidin contains an extended system of conjugated $\pi$-electrons, which constitute a chromophore. As it happens, the electrons of molecules containing conjugated $\pi$ systems have energy level transitions corresponding to absorbance in the visible (and also ultraviolet) region of the electromagnetic spectrum. This accounts for their perceived color.
As for the precise mechanism involved, I was not able to find any sources whatsoever. It's obvious that the flavonoid is reduced, with the magnesium metal being the reducing agent. It could be the process occurs by single electron transfer (SET) with radical and anionic intermediates, or via a cyclic transition state and carbenoid intermediates with coordination of the magnesium, or something else entirely. In any case, the purpose of the acid is likely to activate the carbonyl of the flavonoid to reduction by protonating it, and providing a source of $\ce{H+}$ ions for the subsequent formation of water.
(I find it unlikely that a definitive mechanism for this reduction has been determined, if only because the mechanisms of many similar reductions involving metals and mineral acids widely used in synthetic organic chemistry have yet to be entirely resolved.)


text:
Fermentation in beer (reaction schemes)?
I've been struggling with this for a while, and I can't seem to solve the problem. Alcohol is made in the fermentation process in beer:

Sugars formed during mashing (maltose and maltotriose) will undergo hydrolysis and form multiple molecules of the simple sugar glucose. This glucose will then be fermented by the yeast under anaerobic conditions into ethyl alcohol and carbon dioxide

I understand that perfectly. My problem is to figure out the reaction schemes for the reactions, for each step. Can anyone help me with that?

Are you asking for the reaction mechanisms?

This looks more like a biology question than strictly chemistry, as yeast is definitely involved in that process. Maybe you have more luck asking your question there?

We can break the entire process into three parts:

breakdown of maltose and maltotriose into glucose
glycolysis to give pyruvate
reductive decarboxylation to give the alcohol.

Writing out full equations for each actual step should hopefully not be required; especially glycolysis is explained in detail in many places across the internet and practically every metabolism textbook. Hence I will restrict myself to generic equations for each part of the process. Furthermore, each set of reactions is catalysed by a specific set of enzymes; I will be ignoring these.

Maltose breakdown
$$\ce{\underset{maltose}{C12H22O11} + H2O -> 2 \underset{glucose}{C6H12O6}}\tag{1}$$
Glycolysis
$$\begin{multline}\ce{\underset{glucose}{C6H12O6} + 2 ADP^3- + 2 HPO4^2- + 2NAD+ -> } \\\ce{2 \underset{pyruvate}{C3H3O3-} + 2 ATP^4- + 2 H2O + 2 NADH + 2 H+}\end{multline}\tag{2}$$
Alcoholic fermentation
$$\begin{align}\ce{\underset{pyruvate}{C3H3O3-} + 2 H+ &-> CO2 + \underset{acetaldehyde}{C2H4O}}\tag{3}\\
\ce{\underset{acetaldehyde}{C2H4O} + NADH + H+ &-> \underset{ethanol}{C2H6O} + 2 NAD+}\tag{4}\end{align}$$


text:
Rusting of an Iron nail at oil water boundary surface
Recently at school, I did some experiments on corrosion. Everything mostly went as expected, but the result in one of the test was quite baffling.
This test had an iron nail in water with some of the nail protruding the surface. The tip of the nail was then covered by oil so that it was completely submerged. It was left as is for 4 weeks. At the end of this time, there was a bit of rusting (as I had expected), but most of it was right at the surface of the water, right where it and the oil meet. Is there any reason why this would be?
Edit
I had a beaker with the nail in it, leaning upright against the wall of the beaker. I put about 200 ml of water in, then about 50–100 ml of oil. The water didn't completely cover it, but the oil did. After 4 weeks, there was rust, mostly situated around (but below) the oil/water border. As for why it was done, I don't know. My lab partner decided to test this. I think it was kind of testing how much it would rust when starved of oxygen.

I think that's because at the oil/atmosphere interface, there is oxygen, and it is oxygen doing the oxidation. Try looking at this website: http://www.wiley.com/college/boyer/0470003790/reviews/redox/redox.htm

You added oil, to make the iron nail heavier, is that right? How much oil you added?

I think this is an example of localized corrosion. The oxygen reduction reaction preferably takes place on the nail surface in the medium with higher concentration of dissolved $\ce{O2}$ (water). Metal dissolution (oxidation) is then limited to the - presumably more oxygen-starved - area right at the phase border where consequently the deposit of corrosion products is formed.

@Martin I have expanded my comment into a more detailed answer.

@JannisAndreska Thank you very much, I immediately upvoted ;)

@Martin It was my pleasure :)

According to your description, this looks like localized corrosion. Due to differences in the availability of oxygen in the solution, the two half-reactions of the redox process take place on different areas of the surface of the corroding metal. The oxygen reduction reaction preferably takes place on the nail surface in the medium with higher concentration of dissolved $\ce{O2}$ (water).
$$\ce{O2 + 4e- + 2H2O -> 4 OH-}$$
The metal dissolution (oxidation) is then limited to the more oxygen-starved surface area, which is presumably right at the the phase border, where consequently the deposit of corrosion products is formed.
\begin{aligned}
\ce{Fe &->~ Fe^2+ + 2e-}\\
\ce{Fe^2+ + 2OH- &->~ Fe(OH)2 \rightleftharpoons FeO + H2O}\\
\end{aligned}
Since these two half-reactions are coupled, the amount of corrosion products formed depends on the oxygen concentration. When there is sufficiently dissolved oxygen available, $\ce{Fe^2+}$ is easily oxidized by to $\ce{Fe^3+}$, and thus the final product, which is commonly called rust, is a mixture of more or less hydrated iron(III) oxides and hydroxides.
$$\ce{Fe^3+ + 3OH- -> Fe(OH)3 \rightleftharpoons FeO(OH) + H2O}$$
$$\ce{2FeO(OH) \rightleftharpoons Fe2O3 + H2O}$$

Its local general corrosion not localised . General corrosion occurs on big areas on the microscale but it is often uneven on the macroscale on short time scales.


text:
Determine the concentration of the original NaOH solution
This was a question on the mt exam two days ago :
A chemist added 40.0 mL of an $NaOH$ solution to 90.0 mL of 0.400 M $HCl$.
The solution was then treated with an excess of $\text {nickel(II) nitrate}$, resulting in the formation of 1.06 of $Ni(OH)_2$ precipitate. Determine the concentration of the original $NaOH$ solution.
I didn't know what to do with this question
I knew that I should get the number of mol of NaOH to get the solution's concentration
Then I  wrote first the equation that describes the first part of the reaction :
$NaOH +  HCl \rightarrow NaCl + H2O $
After that I wrote the second part of the reaction :
$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$
I don't know what will be the the other products ?
I stopped here and I guessed an answer.

I am going to need a unit for the 1.06. The question needs you to figure out the amount in moles of Nickel Hydroxide. Then multiply this by two as two moles of NaOH are required to form one mole of the Nickel precipitate. Then this is the number of moles that was surplus in the reaction with HCl (NaOH is added in excess - the key to the question). Find the number of moles of HCl using the volume and concentration. Add this number to the number of moles that were used to form the Nickel precipitate and then find the concentration using the moles and the volume.

user3764899 - good answer.  But what a strange question. Why didn't  the chemist just titrate the $\ce{NaOH}$ with the nickel directly? If the concentration of $\ce{HCl}$ were unknown and that of $\ce{NaOH}$ were known, the procedure described would make more sense, but as it is, it makes me wonder whether this teacher ever did any chemistry.


text:
Can I use sodium hydroxide (NaOH) to remove oxides from metals?
Sodium hydroxide dissolved in water is a very strong base. I think this could mean it could de-oxidate the metal oxides during re-ox reaction. Basically, I ask whether the following happens:
$$\ce{NaOH + Fe2O3 -> Fe + something}$$
I have read that it does reduce carbon dioxide producing sodium carbonate and water:
$$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$
Would that work with metals? If yes, which ones?

The short answer is no. Sodium Hydroxide is a soluble base (alkali) and Iron (III) oxide is a base too. An acid will react with metal oxides forming a salt and water. For example, if a sample of Iron(III) oxide was reacted with HCl, Iron chloride and water would form. It's possible to extract the pure Iron by electrolysis. Although, I am not sure that Iron Chloride is soluble so another acid may be more suitable or electrolysis could be performed using molten iron chloride.

Same applies on copper and other common metals?

Your second reaction isn't a redox reaction. When carbon dioxide is dissolved in water, two equilibriums are forms:
\begin{eqnarray}
    \ce{H_2CO_3 + H_2O &<=> HCO_3^- + H_3O^+} \\
    \ce{HCO_3^- + H_2O &<=> CO_3^{2-} + H_3O^+}
  \end{eqnarray}
Remember that water also is in equilibrium:
$$\ce{2H_2O <=> OH^- + H_3O^+}$$
Adding $\ce{NaOH}$ will — according to Le Chatelier's principle, the equilibriums are disturbed, and more carbonate ($\ce{CO_3^{2-}}$) is formed.
If you're interested in a nice redox reaction involving iron(III) oxide, aluminium is much better choice.


text:
Can I perform electroplating with sodium carbonate and copper?
I want to avoid anything poisonous during electroplating. I can't avoid copper oxide though.
I have dissolved $\ce{Na2CO3}$ in water and I'm trying to electroplate aluminium with copper:

As you can see I'm doing it with no actual chemical equipment. Actually I'm trying to discover ways that can be done at home.
In the water, I can see copper hydroxide ($\ce{Cu(OH)2}$) and copper carbonate ($\ce{CuCO3}$), both formed from the copper anode. But the aluminium heatsink does not seem to have any copper on it. Will it take time or am I just wasting copper here?
If this is not gonna work, what is gonna work? Could I use vinegar?

What does it mean some copper oxide? What kind of substance did you use for adding copper in the solution? $CuSO_4$?

The copper oxide is generated by the oxidation of anode (which is a copper PCB). If I had $\ce{CuSO4}$ I'd use it as an electrolyte. And I was wondering if vinegar could help instead (because I don't have $\ce{H2SO4}$).

It is possible to buy copper sulfate in a shop (it was used for pool cleaning). Without copper in the solution the deposition will not occur (oxidation of the anode is in this case insufficient). I have seen some recipes where vinegar was used, but the deposition will be complicated. For direct aluminium copper plating is generally used cyanide baths (I am strongly against using them).

I was thinking that if  dissolve the copper hydroxide and copper carbonate in vinegar I could electroplate with them. But I'm not sure if it's that easy. I want to avoid any poisons. From what I have discovered, both copper compounds in the solution are not poisonous. Copper sulfate is poisonous (especially dangerous to human eye) and I guess cyanide isn't safe either.

You can try the electrodeposition in a solution prepared by mixing baking soda with vinegar and the copper sulfate. Wear glove and protection glasses. The aluminium must be cleaned before plating. Please let me know the result.
Acidic copper plating on aluminium will cause evolution of hydrogen, so less acidic solution is better. On the other hand you need to have the solution conductive and the baking soda could help.

Will not baking soda neutralise vinegar? As far as I know, this combination is used to amuse children with safe "explosions" (eg. volcanoe). Also what's the point of these two when copper sulfate is present?


text:
Reduction potential and electron affinity
Why is it that fluorine has a much greater reduction potential than chlorine, despite the fact fluorine has a lower electron affinity than chlorine?
I understand that fluorine has a lower electron affinity than chlorine due to fluorine's compact size and higher negative charge density.
So what makes fluorine a good oxidizing agent - one better than chlorine?
Could it be that the reduction potentials are referring to diatomic fluorine rather than elemental fluorine? And similarly diatomic chlorine rather than elemental chlorine?

Oxidising power , i.e the ability to accept eelectrons for any element depends on its electron gain enthalpy as also 2 other factors:
1) Hydration energy - greater the better, which basically means that water molecules easily overcome attractive forces in the solute particles to break chemical bonds and that easy separation of solvent molecules to accommodate the solute occurs in a solution and that formation of new attractive interactions between solute and solvent particles is not that common.
Rem: sum of delta H 1, 2, and 3 is the overall enthalpy of the solution process. Rem also that H1, H2 are positive, H3 are negative...........read this up if you have forgotten.
Due to its small size Fluorine's hydration energy is very high.
2) Enthalpy of dissociation (i.e. bond dissociation energy) - lower the better
Fluoride has a greater electron-electron repulsion among the lone pairs in the small sized F2 molecule where they are much closer to each other than in case of Cl2, hence the enthalpy of dissociation of F2 is lower than Cl2.
These two results in fluorine to have a greater positive electrode potential than chlorine, which results in greater oxidising power for F2.

authors: Řųhãńï Šhäŕmå


text:
How does a lone pair of a central atom affect the dipole moment?
Dipole moment is the degree of polarity, i.e. the seperation of positive and negative charges. But I am not getting the intuition why and how lone pairs affect the polarity and dipole moment. I cannot connect the  separation of positive and negative charges  definition with lone pair.

How does lone pair of a central atom affect the dipole moment?

There is no single answer to your question, let me explain.  Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around the atom that has the lone pair of electrons.  This increased electron density could lead to a more significant contribution from the lone pair electrons to the molecular dipole moment than from electrons spread out more diffusely in a covalent bond.  Next we must understand the directionality of the lone pair of electrons.  Consider the two molecules pictured below, ammonia and phosphine.  The molecules appear to be very similar, they are in the same column in the Periodic Table.

However in ammonia the $\ce{H-N-H}$ angle is around 107 degrees and the molecule is roughly $\ce{sp^3}$ hybridized, the lone pair and the 3 $\ce{N-H}$ bonds roughly pointing towards the corners of a tetrahedron.  You can see that in this case (as shown by the arrows, the "arrowhead" end representing the negative end of a dipole), the lone pair on nitrogen will make a contribution to the molecular dipole moment.  Next, let's examine phosphine.  The $\ce{H-P-H}$ angle is around 90 degrees and the molecule can be viewed as being unhybridized, the lone pair is an $\ce{s}$ orbital and the 3 $\ce{P-H}$ bonds are constructed from phosphorous $\ce{p}$ orbitals.  You can see that in this case, the lone pair on phosphorous, due to its spherical symmetry will not make a contribution to the overall molecular dipole moment.
So in summary, a lone pair of electrons can make a significant contribution to the magnitude of a molecular dipole moment due to the fact that they are more localized than bonding electrons and consequently there is a high electron density.  But, the directionality (or lack thereof) of the lone pair must also be assessed, since a lack of directionality may preclude it from making a significant contribution to the overall molecular dipole moment.

Could you elaborate a bit more on the directionality part? I don't understand why the dipole moment of the lone pair in ammonia points away from the nitrogen atom. Shouldn't more electron density be closer to the nitrogen atom (where nuclear attraction is greater)? This would imply that the dipole moment's direction (due to the lone pair) would be opposite to the one actually observed.

@Gerard  The nitrogen nucleus is positively charged.  The lone pair electrons are contained in an orbital directed away from the nucleus.  The vector describing the nitrogen nucleus - lone pair dipole would be a straight line having it's positive end around the nucleus and its negative end around the center of the lone pair orbital - as I've drawn in the figure above.

@ron , JD lee says, “ the lone pair has no contribution to he dipole moment of the molecule , if the lone pair is present in the pure s or p orbital as s is spherically symmetrical and p is objected equally in opposite directions”
Does this statement agree with your answer?

@AaryanDewan Yes, read what I say about lone pair "directionality".

@ron You say the nitrogen nucleus is positively charged. How? As far as I can see, nitrogen has no formal charge here and it is receiving electron density from the nitrogen-hydrogen bonds.

@YUSUFHASAN  The key word is nucleus.  The nucleus contains protons and neutrons and is always positively charged.  Add the electrons and you then have a neutral atom.

@ron As a rule of Thumb, Can I say that when the lone pair rests in an hybridized orbital, it will contribute to molecules dipole moment but when it is in the pure orbital (unhybridized) it will not contribute to the molecules dipole moment ?

@Tushar I think that might be a bit too binary.  I would prefer to say that the lone pair will contribute less when in an unhybridized orbital.

Dipole moment can be defined as the product of magnitude of charges and the distance of separation between the charges. Dipole moment may refer to:

Electric dipole moment, the measure of the electrical polarity of a system of charges
Transition dipole moment, the electrical dipole moment in quantum mechanics
Molecular dipole moment, the electric dipole moment of a molecule.
Bond dipole moment, the measure of polarity of a chemical bond
Electron electric dipole moment, the measure of the charge distribution within an electron
Magnetic dipole moment, the measure of the magnetic polarity of a system of charges
Electron magnetic dipole moment
Nuclear magnetic moment, the magnetic moment of an atomic nucleus
Topological dipole moment, the measure of the topological defect charge distribution
The first order term (or the second term) of the multipole expansion of a function

(from wikipedia)

The dipole moment is actually affected by the presence of a lone pair of electrons because the electrons on the central atom can cause shielding effect as the inner orbital electrons does for the outer-orbitals.
It can cause hindrances by charge imbalances which could result in imbalance of charges and which could destabilise the atom due to decrease in dipole moment.
Consider a molecule of carbon dioxide where two oxygens are present at each side of carbon:

Here the formal charge on carbon is zero and the same formal charge is there on both the oxygens. Coming to the topic of dipole moments you should consider the lone pair of electrons on oxygen. The lone pair is generally resided on the other side of the bond [as shown in figure]. These electrons are a part of oxygen and these makes oxygen more electronegative because of which the dipole moment increases ans hence the bond becomes stronger.
In most of the cases the central atom in any compound of oxygen is more electropositive atom and electropositive atoms do not contain lone pairs. You can just imagine there are lone pairs on the central atom carbon which could generate some partial negative character in them and hence the bond between them must not be as strong as the bond between an electropositive and electronegative atom.
This all is governed by dipole moment only.

the pure shielding effect is not seen in the MOLECULES they are just seen in ATOMS in order to make you understand just used the term but it does'nt mean that it is as same as in atoms here it just hinders the central atom [make's a sort of covering ] from the attached atoms so that the complete atom is'nt exposed to the central atom and there will surely be a difference in the lenght of bond but this is not what you asked keep this topic aside

i will try to make the same same thing with ammonia

This was hard to read, edit, and I believe most of it is wrong. For example the central atom in $\ce{NH3}$ is the most electronegative element. The picture of $\ce{CO2}$ is just utter garbage and has no source given. The lone pairs of oxygen do not make oxygen more electronegative.


text:
Something denser than lead but with lead like properties?
I read that tungsten is 1.8 times more dense than lead. What other elements are denser than lead but safe to handle.  I want something that will fit under the keys of my computer keyboard so my pinky doesn't have to press so hard on the key.

Here's a link to a listing of the elements in the periodic table sorted by density.  The density of lead is 11.35 g/cm^3.  If you wanted higher density, no toxicity and to really let your pinky live in style, then gold with a density of 19.32 would be a possibility.  Other relatively available elements that are denser than lead and without toxicity concerns include palladium (12.02), rhodium (12.41), tungsten (19.35) and platinum (21.45). If you're willing to be a bit less dense than lead, then you can reduce price by using copper (8.96), molybdenum (10.22) or silver (10.5).  Alloys of some of these elements are also available at high density, but the cost (due to fabrication costs) is likely to be as high or higher than that the element itself.

I totally forgot about gold.

Just whatever you do, don't install osmium key weights.


text:
Water flow in condenser
How should the water flow in your condenser? From the flask to the top of the condenser, or from the top of the condenser back to the flask? I could imagine both: flask to top will fill the entire condenser with water, because of the resistance of gravity, enlarging cooling area, or top to flask, because this will generate counter current exchange increasing heat flow, but then it is almost impossible to get a decent cooling surface without extreme water pressure.
This picture suggests the first.

In general the cooling water is attached according to the countercurrent exchange mechanism to maximize the efficiency. Of course this doesn't work well for the Liebig condenser you pictured, but you generally wouldn't use this kind of condenser for reflux anyway. The most common condenser for reflux is the Dimroth condenser which can be used for countercurrent exchange without any problems as the entry and exit of the water are both at the top of the condenser.
Here is an image from Arbeitsmethoden in der organischen Chemie (page 43) that shows where to attach the cooling water for different condenser types.

Water should always enter from the bottom of a condenser (the end closest to the flask) and exit from the top of the condenser.  Doing it this way always ensures that your condenser will be full of cooling water.  If you set it up the opposite way, and if for some reason the rate of water flow into the condenser decreased and became slower than the rate of water exiting the condenser, then the condenser would not remain filled with water.  If the reaction were unattended, this could cause problems.  BTW, here is a link you might enjoy.  It contains a lot of lab tips, including "how to hook up the water tubing to a condenser".

Yes, I thought that would be the case, but then I remembered counter current exchange and started to doubt.

How should an empty space suddenly appear in the water? Bubbles, OK, one of several reasons why a Liebig cooler is totally unsuitable as a reflux condensor.

@Karl Gravity. If the flow come from the bottom, any flow will fill the condenser; if it flows from the top, the water will tend to flow under gravity to the bottom and low flow will mean the coils/free space isn't full as the flow will be determined by the rate water flows out under gravity not the inherent flow rate of the water.

@matt_black 1000hPa is equivalent to a 10m column of water. Once the condensor is full of water, there is no way some vacuum space will appear in there. It can indeed be difficult to fill the condenser if you hook it up the wrong way.

@Karl "How should an empty space suddenly appear in the water? "  If the top condenser connector is used as a water inlet, and if the inlet tubing falls off or otherwise fails, or if the water flow from the main is interrupted, then water will drain out the bottom condenser connector leaving the condenser jacket filled with only air.

If the inlet tubing falls off, you firstly get a flooded lab, and secondly the condensor will empty itself, no matter where it was connected. Ever heard of Archimedes' communicating vessels? ;-)

Hook from the bottom flowing upward in case the total cocondensation doesn't  happen before the vapor reached the totally liquid state. it will get the cooling first before the exit giving a positive condensation  with easy  cooling water adjustments.


text:
Making a solution with two solutes
I need to make a solution that has a 15% (wt/wt%) HCl acid concentration and a 6% (wt/wt%) NaI concentration. I'm not sure yet how much of this solution I will be required to make so a general recipie would be awesome.  I would also like to know how to make the concentrations wt/vol% as well.
Additional information:
I have a bottle of Muriatic Acid. Its label says the contents are HCl acid, 20 degree Baume (31.45% by weight). The product's MSDS sheet says its 65-75% water and 25-35% HCl, and its specific gravity is 1.16 (1.16 g/ml).
I have Sodium Iodide (a brick of it). Its label says 99+%, pure, anhydrous.

Are you sure it is vol/vol% for HCl and NaI? There are two problems with defining composition in this way. The first is that volumes are not additive, especially over such a large range of concentrations (100% to 15%). The second is that NaI is not a liquid at 100% concentration - it's an ionic solid, and HCl is a gas at room temperature.
Typically concentrations made from combining aqueous solutions are specified in either molarity, molality, wt/wt%, or wt/vol%. You can use vol/vol% for mixtures of pure substances if you ignore $\Delta V$ of mixing.

@thomij i've edited my question.

You need 15 g HCl and 6 g NaI per 100 g of the solution. Treating 31.45% (I don't know much about practice) as exact, there is 15 g HCl in 15 g / 31.45% = 47.69 g of your muriatic acid. So you need 6 g of sodium iodide, 47.69 g of your muriatic acid and the rest, i.e. 46.31 g, of pure water per 100 g of the solution.


text:
Mechanism for reaction of Tollens' reagent with aldehydes
Upon treatment with Tollens' reagent (ammoniacal silver(I) nitrate), aldehydes are oxidised to carboxylic acid, and silver(I) is reduced to silver metal.
I am trying to find a mechanism for the this reaction online, but the only thing I can find is the balanced equation. Can someone propose or help me find the mechanism?

Here are the two half reactions:
$$\begin{align}
\ce{[Ag(NH3)2]+ + e- &-> Ag^0 + 2NH3} \\
\ce{RCHO + 3OH- &-> RCO2- + 2H2O + 2e-}
\end{align}$$
which together yield the overall reaction
$$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -> 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$
Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the process and the $\ce{Ag^+}$ is reduced. The resultant oxidized aldehyde (now a radical cation) reacts with hydroxide to form a tetrahedral intermediate. A gem-diol like intermediate is formed via a hydrogen shift, which then continues on to the final carboxylate anion.

How would we explain from this mechanism whether primary alcohols give a positive Tollens' test or not?

@GaurangTandon Your question is quite different from the original question so you might want to post it as a new question. My view would be that since a typical alcohol does not have a readily available electron (e.g. high-lying HOMO as a carbonyl has), an alcohol would not undergo 1 electron oxidation and react with Tollens reagent in a normal fashion to produce the characteristic silver mirror. That said, if you heat things up, or have oxidizing impurities in the sample, then all bets are off and anything could happen.

Oh, that's interesting. Thanks! Though do have a look here, as you said, gentle heating is causing silver ppt to form, as bon says

@GaurangTandon No, he says it produces a "fine black precipitate" with the alcohol.

Do you mind posting a new question about this? I don't know how to phrase it whilst avoiding being called a duplicate :/ Or please tell me how to?

A Google search for "Tollens' test mechanism" gave me a link to this article: J. Chem. Res. 2011, 35 (12), 675–677. A quick glance at the text of the article leads to the conclusion that the currently proposed mechanism is as follows:
$$\begin{align}
\ce{R-CHO + H2O &-> R-CH(OH)2} \\
\ce{R-CH(OH)2 + Ag+ &-> R-C^.(OH)2 + H+ + Ag^0} \\
\ce{R-C^.(OH)2 + Ag+ &-> R-COOH + H+ + Ag^0}
\end{align}$$
In strongly alkaline solution (pH > 10) the mechanism changes to the following:
$$\begin{align}
\ce{R-CHO + OH &-> R-CH(OH)O-} \\
\ce{R-CH(OH)O- + Ag+ + OH- &-> R-C^.(OH)O- + H2O + Ag^0} \\
\ce{R-C^.(OH)O- + Ag+ + OH- &-> R-COO- + H2O + Ag^0}
\end{align}$$
However, it seems that there is no direct evidence for these exact mechanisms, though they seem believable. That is normal, though; proving any mechanism is a long and tedious task that can take decades of dedicated studies, so we have to live with it.

I couldn't see any mechanism  or text describing the mechanism on the link you provided, you could?  Also, your mechanism suggests that silver(I) will oxidize saturated carbons containing a hydrogen, for example, dimethoxymethane.  Is this reported in the literature?

You have to download and read the article for it. 2.a) It was reported that aldehydes forming stable hydrates, like $\ce{CCl3CHO}$, reacts much faster. 2.b) Tollen reagent does give false positive, but I am too lazy to find full list.

The balanced equation can tell you a lot. This is the balanced equation:
$\ce{R-CHO + 2[Ag(NH3)2]+ + 3HO- -> R-COO- + 2Ag + 2H2O + 4NH3}$
Let's check out what's happening:
1) First, the aldehyde is being oxidized. Specifically, the carbon in the aldehyde is being oxidized; it is losing electrons to silver.
Oxidation half reaction: $\ce{R-CHO -> R-COO- + 2e^- + H+}$
Oxidation states of C: $~~~~~~+1~~~~~~~~~~~~~+3$
2) Mass isn't balanced; we need a source of $\ce{O^2-}$ (negative 2 oxidation state oxygen atoms). Given that this reaction is happening with a solution of silver ammonia, it makes thermodynamic sense that your best source of $\ce{O^2-}$ oxygen is the hydroxide anion: $\ce{HO^-}$.
Water is a possible source but it's pricier to heterolytically cleave two $\ce{H-O}$ bonds as opposed to just one $\ce{H-O}$ bond.
Thermodynamically favorable: $\ce{H-O^- -> H^+ +O^2-}$
Not so thermodynamically favorable: $\ce{H-O-H -> 2H^+ +O^2-}$
3) The freeing of one $\ce{O^2-}$ yields the hydrogen proton, $\ce{H+}$. This is an acid; it will react immediately with the strongest base in the system, which would be the hydroxide anion.
Also, note that a hydrogen proton is freed from the aldehyde. The hydrogen atom is freed heterolytically as to yield 2 electrons, which go to the silver ammonia complex.
$\ce{R-CHO -> R-COO- + H+ + 2e^-}$
Two protons require two additional hydroxide anions to react with; this is why we use a total of three hydroxide anions and this yields two water molecules.
$\ce{2H+ + 2HO- -> 2H2O}$
4) The above explains the hard stuff. The reduction of silver is easy:
$\ce{[Ag(NH3)_2]+ + e^- -> Ag + 2NH3}$
The silver in the silver ammonia complex ion has an oxidation state of +1; gaining an electron allows solid silver to precipitate out, giving a positive Tollens test for aldehydes. The ammonia remains unchanged.

The aldehyde is not losing electrons to oxygen. It's losing electrons to silver. Note that silver goes from the +1 oxidation state to the 0 oxidation state.

I see what you're saying; I was referring to the cause of the +1 to +3 oxidation state change of the carbon from reactant to product.

You will never form oxygen dianion in an aqueous solution. It is far too basic. In this case, deprotonating hydroxide is equally thermodynamically disfavorable to doubly deprotonating water.

What's your source? Do you have a citation?

I don't know what the pKa of hydroxide is. I've never seen it in a table, presumably because it's not acidic at all. Here's a discussion though: http://www.chemicalforums.com/index.php?topic=62198.0 According to your mechanism, hydroxide is the destination of the proton that comes off the original hydroxide. So we can estimate the equilibrium constant for that reaction. The pKa of hydroxide is 50 (estimate) and the pKa of water is 16. The equilibrium constant for the deprotonation of hydroxide by hydroxide then is 10^(16-50), which is a very small number.

Check out a general chemistry book's section on equilibrium, and in particular the section on diprotic acids, which is essentially what you're doing with water.


text:
Mechanism for the Beilstein test?
What is a possible mechanism for the Beilstein test for halogens ? I can't seem to find a arrow pushing mechanism for this reaction.

You can easily find it if you dig google enough. I do know the answer, but not going to write one and recommend to search it yourself, it is easy.

Please, I tried a Google search but I could not find one. If writing the mechanism is too much work, please give me a link. Thank You !

I was under the impression that the goal of a stack exchange was so that, when you googled a question, the top result would be a stack exchange result. This was the stated goal on Stack Overflow; I'm not sure if it's extended to all sites on the stack exchange network

hmm..well its listed as the fourth result. Maybe if someone helped me figure out the mechanism, it might become the first result...

@permeakra Please share your wisdom. I tried finding it on google (for a good while) and managed nothing. And frankly, I find your attitude a little disturbing.

@permeakra I agree with Martin; I would like to add that you usually have useful contributions, and I would like to hear them even if answers are readily found online. It's always great getting a second opinion.

@PeprikaDesilva Actually, wiki article includes pretty correct description, wich is duplicated now in the answer. Beilstein test is a variation of flame test. In flame test many elements form particles with strong emission lines in spectrum, coloring the flame. Copper is somewhat pathological case as it may form several different species, giving different colors: monochloride (700-1000 C, Cl present. blue), atomic species (green), and oxide (oxidative atmosphere, golden). In reductive atmosphere of the reasonably hot flame organics breaks and copper monochloride is formed. Learn google-fu.

@Martin With all respect, google-fu is something I strongly recommend to learn. Yes, there IS much to learn using google: what words to use, how to generalize them, secondary searches (search for topics found on pages after first request), advanced search features. But in modern days it is something one really need to know.

@permeakra With all respect, sir, I very much encourage you to share your knowledge. Maybe a link to a tutorial on how to intelligent use the search engines would be not only very helpful, but also very much appreciated. I might have missed some important information and I wish to not continue living without it. Maybe I am just plain stupid, according to your tone, you must think that. (One could guess, that I am outraged, and I really have to behave myself to not use strong language here.)

@permeakra The given answer as well as your previous comment cannot be seen as an actual mechanism. Those are net reaction equations, with no information about the process, that can be referred to as a reaction path, whatsoever. And if you had read the question carefully, you would have noticed, that the mechanism was looked for.

@Martin Google has advanced search http://www.google.com/advanced_search . Two most useful options has shortcuts usable directly from the main page: "-word", to find pages without the word and "site:url", performing search on the site with specific prefix. When searching for scientific information, scholar.google.com is also very handful. Other than that it is mostly the matter of practices, I guess.

@Martin What do you want? Specific details how organic chlorides breaks?

I spent about a half hour looking this up yesterday, even going as far as to read several papers in JACS. Nowhere did I find an arrow-pushing mechanism (though I did find a fun one that claimed that CO2 can be generated from nothing). Yes, the info in prooffreader's answer can be found widely, but it doesn't explain how the chlorine is stripped from the organic, or what happens to the organic afterward (I can make several really, really good guesses, but that doesn't make the answer obvious)

@chipbuster ... and I readily got http://prr.hec.gov.pk/Thesis/3645H.pdf from 'thermal dehydrohalogenation' from google.

When you heat the copper wire in a flame, it is oxidized on the surface to copper (II) oxide:
$$\ce{2Cu(s) + O2(g) -> 2CuO(s)}$$
Then when you mix it with the halide and heat it, the higher reduction potential makes it displace the oxygen.
Here's an example with sodium chloride:
$$\ce{CuO(s) + 2NaCl(s) -> CuCl2(g) + Na2O}$$
Copper halides are volatile (except for fluorides); the gas is hot enough to push some electrons into an excited state, and as they cool a tiny bit, the electrons drop back into their rest state and emit the excess energy in the form of a photon with a wavelength corresponding to the color green.

Chlorine, however, is less electronegative than oxygen.

Thanks for pointing that out, I meant reduction potential. I'm not my best at one in the morning!

I heard that even before mixing the halide, when you are heating the copper wire, a green colour flame might be seen. So you must wait till that colour disappears before you mix the halide. This is probably due to impurities of halides in the environment or on the wire. So what do these halides convert into - considering the green colour disappears after some time?


text:
How can I determine/quantify the presence of boron nitride on a metal surface?
We use micron-grade hexagonal boron nitride (hBN) to coat the surface of bullets that are jacketed in gilding metal.  This coating is mechanically embedded/plated onto the surface in order to increase the lubricity of the bullets as they are propelled down the rifled bore of a gun.
We can capture fired bullets in a water tank.  There is some question as to whether any of the hBN survives the temperature, pressure, and friction involved in shooting the bullet through a firearm.
How can we determine whether the hBN coating survives the firing process?  Or better yet, can we characterize the quantity that remains on the bullet, or any part of it, after firing and recovery?
Destructive methods of analysis are fine.

I'd suggest looking into the following:
scanning electron microscopy (should be able to see the film and how intact it is on the surface); Inductively Coupled Plasma tests (destructive).  I believe after your sample is prepared and presented ICP can find B and N emission lines

Yep: We actually did find a lab that did SEM, which was quite effective because the boron shows up black against the metal in the micrographs.


text:
Binary Molecular Compounds of Nonmetals
How does one determine the formula for two non metal elements?
Such as Nitrogen and Oxygen.
I look at the charges with Nitrogen having -3 and Oxygen having -2. How would I get to zero?
How do you build a formula based on those two elements.
Why does it equal $\ce{N2O5}$         ?

Nonmetals form covalent bonds with other nonmetals, not ionic bonds. This means you can't use ionic bonding rules to figure out what the structure will be. In fact, there is no way to predict the structure - all you could do is propose Lewis structures for various combinations to predict what it could be.
This is also why there is a special set of nomenclature rules for molecular compounds - it's not enough to just give the names of the elements, you also have to specify the number of each atom.
In your example, if all you are given is N and O, you would have no way of predicting the formula. It could be any one of maybe half a dozen compounds (maybe more). To end up with
$\ce{N2O5}$, you would need to know the name of the compound first.
For binary molecular compounds, the naming rules are pretty easy. First, you need to know the prefixes:

1: mono-
2: di-
3: tri-
4: tetra-
5: penta-
6: hexa-
7: hepta-
8: octa-
9: nona-
10: deca-

Then, for each element (starting with the least electronegative one - the one furthest to the left on the periodic table) you just use the prefix in front of the element name. $\ce{N2O5}$ would be called dinitrogen pentoxide.
There are two small details. The first is that we don't use "mono" when there is only one atom of the first element. This means that $\ce{CO}$ is carbon monoxide rather than monocarbon monoxide. The second is that when an element starts with a vowel and the prefix ends in a vowel, we drop the prefix's ending vowel. This means that we say pentoxide instead of pentaoxide.
To find the formula given the name, you just reverse this process.
$\ce{NO}$ - nitrogen monoxide
$\ce{N2O}$ - dinitrogen monoxide
... you get the idea.

Many more Nitrogen Oxides than just half a dozen, allthough not a lot of them are stable at room temperature and pressure. And then there are ionic forms ...

If you're asking how can you determine a molecular formula when simply given two elements, then there are no hard and fast rules, especially with nitrogen. The only rule I can think of is that molecules, as opposed to ions, are electrically neutral. This still leaves you with plenty of possibilities.
Nitrogen is a versatile element and can adopt a variety of oxidation states. So you'll find nitrogen in a variety of compounds in various proportions.

authors: Михаил Фаузель


text:
How is chlorophyll made in plants?
I've been studying photosynthesis and something made me think about how chlorophyll itself is made in plants.How exactly is it formed ?

This might be of interest: http://en.wikipedia.org/wiki/Chlorophyll#Biosynthesis

In the first phase of chlorophyll biosynthesis, the amino acid glutamic acid is converted to 5-aminolevulinic acid (ALA) . This reaction is unusual in that it involves a covalent intermediate in which the glutamic acid is attached to a transfer RNA molecule. This is one of a very small number of examples in biochemistry in which a tRNA is utilized in a process other than protein synthesis. Two molecules of ALA are then condensed to form porphobilinogen (PBG), which ultimately form the pyrrole rings in chlorophyll. The next phase is the assembly of a porphyrin structure from four molecules of PBG. This phase consists of six distinct enzymatic steps, ending with the product protoporphyrin IX.
All the biosynthesis steps up to this point are the same for the synthesis of both chlorophyll and heme. But here the pathway branches, and the fate of the molecule depends on which metal is inserted into the center of the porphyrin. If magnesium is inserted by an enzyme called magnesium chelatase, then the additional steps needed to convert the molecule into chlorophyll take place; if iron is inserted, the species ultimately becomes heme.
The next phase of the chlorophyll biosynthetic pathway is the formation of the fifth ring (ring E) by cyclization of one of the propionic acid side chains to form protochlorophyllide. The pathway involves the reduction of one of the double bonds in ring D, using NADPH. This process is driven by light in angiosperms and is carried out by an enzyme called protochlorophyllide oxidoreductase (POR). Non-oxygen-evolving photosynthetic bacteria carry out this reaction without light, using a completely different set of enzymes. Cyanobacteria, algae, lower plants, and gymnosperms contain both the light-dependent POR pathway and the light-independent pathway. Seedlings of angiosperms grown in complete darkness lack chlorophyll, because the POR enzyme requires light. These etiolated plants very rapidly turn green when exposed to light. The final step in the chlorophyll biosynthetic pathway is the attachment of the phytol tail, which is catalyzed by an enzyme called chlorophyll synthetase (Malkin and Nyogi 2000).

Source: Plant Physiology Online: Chlorophyll Biosynthesis


text:
Finding the molarity of the combination of two solution reacting with each other
I have a question that gives two concentrations and asks for the mass of $\ce{HCl}$ formed by the reaction.
$$\ce{H2SO4 + NaCl ->Na2SO4 + HCl}$$
I have two concentrations:
$\pu{250 mL}$ of $\pu{4.00 M}$ $\ce{H2SO4}$, and $\pu{250 mL}$ of $\pu{1.00 M}$ $\ce{NaCl}$.
Here is the balanced reaction equation:
$$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$
I know how to find the mass once I find the moles, molecular weights and then grams by multiplying the two.
However, how do I add those two concentrations? I assume the mixture will be at least $\pu{500 mL}$, but how do I add the molarity?

Your working equation is correct.
$$\ce{H2SO4 + 2NaCl ->Na2SO4 + 2HCl}$$

Find the amount of substance of protons of each solution via $n=c\cdot V$.

 $n(\ce{H+})=2~\mathrm{mol}$, $n(\ce{Cl-})=0.25~\mathrm{mol}$

What is the limiting agent?

 Chlorine

How many moles of hydrogen chloride can only be formed?

 $n(\ce{HCl})=0.25~\mathrm{mol}$

Calculate the mass of hydrogen chloride via $m = n\cdot M$

 $M(\ce{HCl})=36.5~\mathrm{g/mol}$, $m(\ce{HCl})=9.1~\mathrm{g}$

Martin, you are a formatting genius.

@Dissenter As a child I loved playing in a sandbox... so I enjoy that now and again too: Check this out!

You're right; you can't directly add the two concentrations since these concentrations are specific to a certain volume of solvent.
If you want to recalculate concentrations of each system component in a combined solution, you need to find the moles of each system component in this combined solution, and then divide by the total solution volume.

So does that become 500mL of 2.500 M solution?

That is not quite correct, as you have a limiting agent present.


text:
Recrystallization of two unknown compounds?
I managed to separate a binary mixture(in solid form) of two unknown substances into two individual solid compounds. For the separation, I used an acid-base extraction with dichloromethane, NaOH, and after separation HCl.  My next step is to purify the two separated solids using recrystallization(this is the technique mandated by my instructor for solid purification) but I am not sure which solvents to use. Can anyone suggest how I should proceed ? Thanks a lot !
Edit: I was thinking since I used NaOH and HCl on my aqueous layer, NaCl would have formed. Since compound doesn't dissolve in water, is it okay to just put the compound in water, let NaCl dissolve and vacuum filter the solid(which presumably should be pure solid) ?

Choosing a recrystallization solvent is as much an art as a science, even when you know what the substance is. Without knowing, it is a matter of making educated guesses. You want a solvent where it is much more soluble in the hot solvent than in the cold. Given that this is a lab exercise, we can assume that these are common substances. So, the one that dissolves in base is likely a carboxylic acid--perhaps benzoic acid or something close. That substance dissolves about 10 times as much in hot water as in cold - so I would try water first.  If it is too insoluble, then you could go to methanol, ethanol, etc., moving toward more nonpolar solvents until you get it to dissolve when heated. Then, hopefully it will crystallize out on cooling.  For the other substance, it is obviously soluble in dichloromethane and is not an acid. Perhaps it is non-polar. I would start with ethanol, and see how that works - if you need to go to a more non-polar solvent, toluene might be a good choice. Good luck!

Washing your crude product (from acidifying the aqueous layer) to remove salt makes sense, as long as its water solubility is low. If you use water as your recrystallization solvent, that step is unnecessary.


text:
How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid?

Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists:
  \begin{align}
\ce{HCO3- &<=> H+ + CO3^{2-}}&
\frac{\ce{[CO3^{2-}]}}{\ce{[HCO3-]}}  &= 0.958
\end{align}
  When $\pu{10 cm^3}$ of lake water were titrated with  $\pu{0.2 mol/dm^3}$ $\ce{HCl}$, $\pu{22 cm^3}$ of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations:
  \begin{align}
  \ce{H+ + HCO3- &-> H2O + CO2}\\
  \ce{2H+ + CO3^{2-} &-> H2O + CO2}\\
\end{align}
  Calculate the total amount of substance of acid used, and thus, by using the ratio quoted, calculate $\ce{[CO3^{2-}]}$ and $\ce{[HCO3-]}$ in the lake.

I was able to find the amount of substance of acid $\ce{HCl}$, which is $\pu{4.4E-3 mol}.$ How do I go on?

You have correctly identified the amount of substance $\ce{HCl}$ which were necessary to neutralize all $\ce{HCO3-}$ and $\ce{CO3^2-}$ in the sample. With the given sample volume of $\pu{10 cm^3}$, we can calculate the concentration of both ions in the sample:
$$
\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}}
= \pu{4.4E-4 mol//cm^3}
= [\ce{HCO3-}]+[\ce{CO3^{2-}}]
$$
Using the given ratio of $0.958$, we can express $[\ce{CO3^{2-}}]$ in terms of $[\ce{HCO3^{-}}]$:
\begin{gather}
[\ce{CO3^2-}]
  = 0.958\times[\ce{HCO3-}]\\
\pu{4.4E-4 mol//cm^3}
  = [\ce{HCO3-}] + [\ce{CO3^2-}]
  = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}]
  = 1.958[\ce{HCO3-}]\\
[\ce{HCO3-}]
  = \pu{2.25E-4 mol//cm3}
\end{gather}
We then use the calculated hydrogen carbonate concentration and the ratio to find $[\ce{CO3^2-}]$:
$$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$

I would do it like follows: By using the ratio of 0.958 you can simply state the ratio of amount of substances. Then by using the balanced equations you can find the ratio of amount of substances of HCl required for each. Having both the ratio and total amount, you can get the the HCl for each; and then you can find the the required concentraion of those two ions from the balanced equation and total volume.


text:
What's the difference between Carvacrol and Thymol?
What is the difference when it comes to separating Carvacrol and Thymol on HPPLC with reverse phase - C18 column?
Carvacrol elutes first, is that because it is ortho?

On a reverse phase C18 column  (which itself is hydrophobic, it is an octadecyl hydrocarbon chain bound to the support material) the more hydrophobic the molecule, the more strongly it will bind to the stationary phase [like dissolves (or in this case binds to) like], and the longer it will take to elute with an organic solvent. Below are the structures of carvacrol and thymol.  The more exposed the hydroxyl group, the less hydrophobic the molecule.  The hydroxyl group in carvacrol (ortho to a methyl group) is more exposed than the hydroxyl group in thymol (ortho to an isopropyl group) because of the relative steric sizes of a methyl group vs. an isopropyl group. Therefor thymol is more hydrophobic than carvacrol.  Consequently we would expect thymol to bind to the column more strongly than carvacrol, and therfor we would expect the carvacrol to elute first which matches your observation.


text:
Relationship between Van der Waals forces and the boiling point
Choose the right answers in the following statement:
The boiling point of a liquid substance is higher as the Van der Waals forces (in/between) the molecules of the substance are (lower/higher).
I don't understand why the Van der Waals forces are lower when the boiling point is lower. Second, do you say: the forces in the molecules of a substance or the forces between the molecules of a substance?

We talk about the forces between molecules.  The Van der Waals force represents the summation of all attractive and repulsive forces between molecules, other than those due to electrostatic charges or actual bonds.  This summation most often turns out to be a weak attractive force.  Typically the magnitude of the Van der Waals force is small compared to electrostatic or bonding forces and (again typically) only apply at relatively short range.  So they usually serve to weakly hold molecules together.  The stronger the Van der Waals force, the more strongly molecules will be held together and the more energy will be needed to be put into the system to break them apart from each other.  In the case of boiling a liquid, you are really talking about breaking molecules apart from one another.  So in the case of boiling, the weaker the Van der Waals attraction between molecules, less energy will needed to break them apart from each other and allow them to boil.
Here is a link to a Wikipedia article on Van der Waals forces if you'd like to read more about them.

The van der Waals force is the sum of the attractive or repulsive forces between molecules (or between parts of the same molecule) other than those due to covalent bonds, the hydrogen bonds, or the electrostatic interaction of ions with one another or with neutral molecules or charged molecules.