text:
What is the acid dissociation constant of cacodylic acid?
What is the $\mathrm pK_\mathrm a$ of cacodylic acid? I cann't seem to find one answer on the internet. I have read 1.67, 6.37, and 4.8 in my $\mathrm pK_\mathrm a$ table. What is the reason for this? Also, the problem I just did had it as 0.34.

I found this article (TW. Shin, K. Kim, IJ. Lee, J. Solution Chem. 1997, 26, 379. doi: 10.1007/BF02767677) with precise analysis of the pKa of cacodylic acid - it says that in 25 °C it's estimated to be 6.27. Values lower than 6 are wrong, but found often. Sometimes it's said to be a diprotic (which it isn't) with first pKa 1.57 and second 6.27 - pubchem says 1.57 citing Lide, D.R. (ed.). CRC Handbook of Chemistry and Physics. 76th ed. Boca Raton, FL: CRC Press Inc., 1995-1996., p. 8-45 - it might be one of the sources of problem.

Cacodylic acid is amphoteric, so it has a $K_\mathrm a$ and $K_\mathrm b$.
See table 1 of this EPA report on cacodylic acid
The $\mathrm pK_\mathrm a$  is 6.2.
The $\mathrm pK_\mathrm b$  is 12.4, which corresponds to the $\mathrm pK_\mathrm a$  of the conjugate acid being 1.6.
Also, according to "Weed Control" by the US National Research Council at page 167:

Cacodylic acid is normally monobasic, but in strong sodium hydroxide it forms the disodium salt, $\ce{(CH3)2AsO3HNa2}$, of a tribasic acid. Cacodylic acid is somewhat amphoteric, forming a hydrochloride $\ce{(CH3)2AsO2H.HCl}$ by direct action with hydrogen chloride gas.

Also, the value for the $\mathrm pK_\mathrm a$  of the conjugate acid being 1.6 was confirmed by electrophorisis is Arsoniumphospholipid in algae Proc. Natl. Acad. Sci. USA vol. 75 pages 4262–4264.


text:
Can plastic leak into water from a HDPE "2" plastic container at room temperature?
Here in USA supermarkets often sell water in plastic 1 gallon jugs. These jugs seem to all be stamped with the text "HDPE" (meaning that the material is High-density polyethylene) and a resin number of "2" on the bottom.
I had one of these jugs, containing distilled water, on a counter in the kitchen, and another one in the refrigerator. For some reason the water on the counter had a light plastic taste, whilst the water in the fridge did not.
Could this taste be because of plastic leaking into the water? Or what else could cause the plastic taste?

Polymer itself is insoluble, but additives not completely.

Nice question but do we really have an answer yet?

HDPE
has a molecular weight more than several hundreds of thousands,
"leaking" is plainly impossible.
This taste is the tase of stale water, nobody knows
what that taste is actually.
Maybe its just the taste of not cold enough water,
lacking some dissolved air, or some bacteriae
develop after contact with air and light.
(google for pseudomonas)
The only substance which ever got into the jug prior
to the water is the air (or some other gas?) used to
inflate the jug in production.
But that cause would impair the taste of the water in
the fridge as well.
BTW why do You drink such botteled water?
What about the tap water in your town/city/village?

Thank you for the explanation, Georg. I use tap water for coffee and such, the distilled water was just for other purpose.

Nah, I can tell the difference between plastic and stale.  And the water in the HDPE tasted like plastic.  It's not bacteria; it's not air; it's not temperature; it's plasticizers.  @burak-ulgut's answer is a better answer.

This should've been a comment as it only backs up Georg's answer, but it just got too long for that.
Plastic doesn't "leak", scientifically. However, it can go through the degradation process.
You don't provide enough info for anyone to decide, but if you're certain that was the "plastic" taste (I don't have much idea how you do) it's possible it may have gone through degradation, though I can tell you 95 percent of the time it isn't. And that's simply because if the plastics were too unstable to resist minor chemical or energetic "attacks" (read the wikipedia article for degradation to see what I mean), they wouldn't have produced it like that.
There is the urban hoax that doing stuff like pressurizing a mineral-water bottle by freezing the water inside releases dioxins that are carcinogenic and which cause the change of the taste of water, but this, is not true. (See here.)
Georg's answer pretty much sums it up, but I think it's better to add: It may also be possible that you (or anyone else, or mistakenly the company itself) put the "tasty" water in contact with soluble mineral salts. Thus, you drink hard water and swear for its taste!

Thank you for your comments, MARamezani. To explore your last paragraph further, could the soluble mineral salts lead to a plastic taste? Otherwise could it be that the water is not as distilled as it is marketed to be(?).

I wouldn't say it's to be blamed on the company too much. It's not only HDPE that's in contact with water. To "optimize" plastics there are many things done. (As an easy example the polymer is PE but you "HD" it) One of these things is adding material like phthalates (No phthalates, not in this case) to plastics. As mithoron points out, the minute solubility of these in the water could've caused what we know as plastic taste.

degrade? The article that is linked says 58 - 1200 years. Should we scratch that off?

I agree with the other answers, it is definitely not HDPE, however, it could be one of the chemicals known as plasticizers that are used in the process. There is always leftover plasticizer in the bottle and these are known to leach into water.
The NIH sponsored multiple studies on the levels of this, one of them published here:
http://www.ncbi.nlm.nih.gov/pubmed/21400322

Ha, why do you say you agree?  Your answer is in direct contradiction to those of the others.  It's not bacteria; it's not air; it's not temperature; it's plasticizers, like you state!  Or possibly degradation, like M.A.R. states.

@zylstra .....I will post. How do we know HDPE is degraded. The I will post again.

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text:
Should the relatively low solubility of a base be considered in the calculation of the pH?

Hydrochloric acid, $V=75.0~\mathrm{mL}$, $c=0.250~\mathrm{M}$, is added to $225.0~\mathrm{mL}$ of $0.0550~\mathrm{M}$ $\ce{Ba(OH)2}$ solution. What is the concentration of the excess $\ce{H+}$ or $\ce{OH-}$ ions left in this solution? (source)

I know the balanced equation but when considering $\ce{Ba(OH)2}$ is 'marginally soluble' which
I guess could be considered insoluble, $\ce{OH-}$ wouldn't really be in the result solution.
Am I right or at where am I wrong? thanks for reading

Welcome to the site. This question is simpler than you think. "0.0550 M Ba(OH)2 solution" means that the barium hydroxide is already dissolved. This is an acid-base question, not a solubility question.

The $K_\mathrm{sp}$ of  $\ce{Ba(OH)2}$ is $0.005$.
Therefore, the solubility is great enough that 0.055M  $\ce{Ba(OH)2}$ can all be in solution.
Secondly, you are diluting the $0.055~\mathrm{M}$ solution, so the concentration will be even less.
Finally, even if there was too much  $\ce{Ba(OH)2}$ to dissolve in water, the $\ce{HCl}$ would react with the $\ce{OH-}$ that is in solution, causing more  $\ce{Ba(OH)2}$ to dissolve.

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text:
Which base is stronger methylamine or propylamine?
Which base is stronger methylamine or propylamine? I know that it depends on the strength of inductive effect of alkyl group. Am I right?
But how can I determine the difference between methyl and propyl substituent?

Propyl has a little stronger inductive effect, but makes also some steric hinderence for protontion.

Different sources quote different values.
Wikipedia: Methylamine $\mathrm{pK_a~= 10.64}$ and Propylamine $\mathrm{pK_a~= 10.71}$
Pubchem: Methylamine $\mathrm{pK_a~= 10.6}$ and Propylamine $\mathrm{pK_a~= 10.71}$
$\mathrm{pK_a}$ is also temperature dependent and this effect may be larger than the difference in the inductive effect between the two amines in question.
http://www.zirchrom.com/organic.htm: Methylamine $\mathrm{pK_a~= 10.63~, 25^\circ C}$ and Propylamine $\mathrm{pK_a~= 10.60~, 20^\circ C}$
Overall there seems to be a consensus that propylamine is slightly more basic than methylamine at the same temperature but there is very little difference. Theoretically this can be explained by the slightly greater inductive effect of the propyl group versus the methyl group. In an experimental context I suspect that there is virtually no difference between the behaviours of the two amines.

This differences are ridiculous! Accuracy of such measurements (esp over history and whodunit) is less than 0.1 digits!

It does not make much of a difference. Both are short chain primary alkyl amines, so chemically much alike ($\ce{pKa} \approx 10.7$ due to the $\ce{-NH2}$ group). In what context do you want to know the answer to this question? That could help me direct my answer better.


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Why is aluminium carbide bent?
The geometry of aluminium carbide is attached. Why isn't it completely linear (in accordance to VSEPR), but bent on the Al atoms?

@Marko Where did you get such a structure? Aluminium carbide isn't molecular and in its structure there are no multiple bonds.

That structure (originally from ChemSpider) is incorrect.

It was from ChemSpider, I am not so imaginative.

That structure is incorrect. Several online chemical distributors have similar structures posted in their entries for aluminum carbide; so does PubChem! Some sources show the structure as linear; some show it as bent.
That structure basically assumes that aluminum forms primarily covalent bonds with an "incomplete octet" in VSEPR structures, rather like boron. But aluminum atoms are a lot larger and more metallic than boron atoms, and its compounds will have comparatively more ionic character. $\rm Al_4C_3$ isn't a molecular compound.
Cotton & Wilkinson's Advanced Inorganic Chemistry says that aluminum carbide "reacts instantly with water to produce methane, and X-ray studies have shown it to contain discrete carbon atoms (C-C = 3.16 Å); for these reasons it is sometimes considered to be a "methanide", that is, a salt containing $\rm C^{4-}$, but this is probably an oversimplification."
This is the structure that's posted on Wikipedia:

It says that "Aluminium carbide has an unusual crystal structure that consists of two types of layers. It is based on AlC$_4$ tetrahedra of two types and thus two types of carbon atoms. One is surrounded by a deformed octahedron of 6 Al atoms at a distance of 217 pm. The other is surrounded by 4 Al atoms at 190–194 pm and a fifth Al atom at 221 pm." This is attributed to Greenwood's Chemistry of the Elements, but I don't see it in the edition I have.


text:
Formation of sulfur colloids
What happens when we pass sulfur vapors through cold water? What is the reaction for it? Which type is colloid is this? And why doesn't it form in hot or lukewarm water

What happens when we pass sulfur vapors through cold water?

You get more or less coarse sulfur powder-slurry.

What is the reaction for it?

Sulfur vapor ( S2 mostly) reacts to form S8
depending on speed of cooling, some S6 might
form temporarily

Which type is colloid is this?

Sometimes this might result in a "colloid",
but I recommend not to use this outdated word,
because so much nonsense was written in the classical
days of "colloids".

And why doesn't it form in hot or lukewarm water

Who says/writes this?

Thanks. I had the doubt on the last one. In the question I read, it was specified COLD water. Just wanted to know what difference would HOT water make?


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Does the probability of fluorescence increase when energy of incident X-rays match the electron's binding energy?
When analyzing manganese-rich samples with X-ray powder diffraction, sample fluorescence becomes a significant effect.
Is this because the incident X-rays' energy - which, if Cu K$\alpha_1$, is approximately 8.048 keV - is getting close to the K$\alpha_1$ (approximate binding energy) of the inner-most electrons of Mn (5.899 keV)?
Sample fluorescence from Fe and Co is even stronger, which fits with the trend mentioned above. So, is this indeed a trend? If not, how do I generalize the increase in probability for sample fluorescence?

With organic molecules (which is what I'm familiar with), the fluorescence of the molecule is a property. The amount of fluorescence is used to determine the concentration as you would use UV-Vis absorbences. So, other than using a brighter source, I don't think you can increase fluorescence.

I'm voting to close this question as off-topic because it's more physics than chemistry.

Like @LDC3, I'm familiar with fluorescence of organic compounds, rather than with XRF. If I remember correctly, XRF is not described by excitation and radiative deactivation. Instead, it results from the ionization of the sample from low-lying orbitals and subsequent electron transfer, upon which excess energy is released as photons. Consequently, a good match of incoming radiation with the relevant orbitals (spectral overlap) should result in fluorescence and more ionization events yield stronger emission.

MARamezani: The line between physics and chemistry is floating. I am a chemist, and the question is in an analytical chemical context. Anyway, I understand your reaction. @KlausWarzecha, yes, you are correct that fluorescence occurs when the inner-most electrons are ejected by EM irradiation. My understanding of the increase in probability is similar to yours, but I was not sure, and could not find any references on it earlier. I did find an article by M. Fransen (2004) which actually says the same thing (page 224, fourth bullet point). http://www.icdd.com/resources/axa/vol47/V47_32.pdf

I did XRF chemical analysis for years, so this is a chemistry question. // What you really need to solve the problem is a different x-ray tube. The K-alpha line of Cr (5.4 KeV) would not excite K lines of Mn (6.5374 KeV)
or Fe (7.1109 KeV).

Short answer, yes.
As you note, the fluorescence signal is strongest when the exciting photon has an energy close to the excitation energy for the element absorbing the photon.  There are actually two factors coming into play:
1) The closer the resonance of the incident photon with the electronic orbital it is exciting, the higher the cross-section to absorb the photon.  (See here.)
2) The higher-Z the absorbing atom, the more likely it is that once excited, the atom will emit a fluorescence photon (as opposed to an Auger electron).  That is, it has a higher fluorescence yield.  (See here.)
In your case, the Z of the atoms are pretty close, but also in portion of the periodic table where the fluorescence yield is rapidly changing.  The chance of a fluorescence photon being emitted once excited is about 20% greater for Co than Mn (and you can look it up in the literature).
The cross-section for absorption as a function of energy for these three elements looks like this (code for the plot here):

I also drew a vertical line for the exciting Cu-K$\alpha$ photon at 8048 eV, and you can see that the Co curve has the highest cross-section at about 316 cm$^2$/g, and Mn has the lowest at about 270 cm$^2$/g, which means the Co cross-section is about 17% higher than the Mn.  The primary reason for the reduction in the sensitivity for the Mn edge is that the absorption cross-section drops off as $\approx\Delta E^3$.  These data were plotted using the data files described by B. L. Henke, E. M. Gullikson, and J. C. Davis, Atomic Data and Nuclear Data Tables Vol. 54 No. 2 (July 1993)
So both these effects are colluding to increase your fluorescence for Co vs Mn.
As a note, if you want to reduce this effect in your X-ray diffraction patterns, you can change your source target material with a heavier element such as Zr which has it's K$\alpha$ up at 15.7 keV, yet the L lines are still around 2 keV (below the transition metal K-edges) or you can use a lighter target, such as Mn.  Mn, of course, would not be able to excite its K-edge using its own K$\alpha$ fluorescence photon.

I note a duplicate with http://physics.stackexchange.com/questions/162224/probability-of-fluorescence-matching-of-binding-energy-and-incoming-radiation-e  If closing in favor of the physics.SE question, then I can transfer the answer there.

I run into this effect in my research and I get around it by putting a thin piece of nickel foil (~50um thick) in front of the detector.  It doesn't eliminate it, but does reduce it enough to make the pattern usable.

Nice answer, but I'd point out a couple of additional things. (1) the question is out of whack. The "probability of fluorescence" is normally taken to mean the fluorescence yield not the resultant x-ray intensity.  (2) Once an atom is excited, it is excited. So the energy of the incident x-ray which created the excited state does not change the fluorescence yield itself.


text:
How quickly does ethanol evaporate?
Suppose I open the drinks cabinet, and pour some of the strong stuff into a saucepan.
Now the boiling point of ethanol is lower than water, so if I boil the mixture for a while, all the alcohol will surely evaporate long before the water does.
But how quickly does the alcohol evaporate? Is it all gone by the time the water starts to boil? Or does some remain unless you boil the mixture for a long time? (How long?) Or maybe the alcohol is gone by the time the mixture starts to get slightly warm?

Certainly water and ethanol and all mixtures thereof will completely evaporate without boiling given enough time.
Water and ethanol form a postive azeotrope that is 95.6% ethanol and 4.4% water and boils at 78.2 degrees C.
If you boil a mixture of water-ethanol that has less than 95.6% ethanol, the concentration of water in the liquid phase will increase, but it will only asymptotically approach pure water as all the liquid boils away.  There will not be a time when all the ethanol is gone and you are left with a finite amount of pure water.
If you boil a mixture of water-ethanol that has more than 95.6% ethanol, the concentration of ethanol is the liquid phase will increase.

...so I think what you're saying is that there will always be some alcohol left. (?) Are we talking amount "enough alcohol that you could detect it with sensitive analytical equipment" or "enough alcohol that you can still taste it"? (Bearing in mind that any mixture intended to be drunk will usually have an initial concentration far, far lower than 95%.)

The remaining alcohol content in food after cooking has been previously investigated. While some of the alcohol evaporates quickly, a macroscopic amount can remain even after hours of simmering. You can find more information in this question at Seasoned Advice, the cooking-oriented Stack Exchange. Feels good to link something from there!

@NicolauSakerNeto Interesting. So it seems both water and alcohol evaporate simultaneously (though at very different speeds) as you heat. Also, it seems my concoction probably contains A LOT of alcohol still...


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Does the speed of an ion affect the current that can flow through a solution?
Typically, when a cathode and anode are inserted into a solution that allows current to flow, the ions are moving with a certain kinetic energy.
If the ions were to gain kinetic energy (through transfer of thermal energy for example), will more current be able to flow, and will the potential difference increase?
My assumption for this is that increased ion kinetic energy will speed up the rate of ions being attracted to each electrode, thus causing more charge to arrive in a given period of time. As $I = Q/t$ where $I$ = current, $Q$ = charge and $t$ = time, would this mean an increased current?
Edit: I've performed a basic experiment in which I got a Tupperware container, stuck in some multimeter probes, and tested the difference in potential (voltage) for $\ce{NaCl}$ (table salt) in solution when cold and hot, then for water cold and hot. I appreciate there are many issues with reliability, but the general results were:
Water cold = 0.082 V
Water hot = 0.000 V (actually went into the negative then stabilised at 0)
NaCl cold = 0.065 V
NaCl hot = 0.130 V
Now, my real experiment involved $\ce{AgNO3}$ being titrated into $\ce{KCl}$ (mixed with deionised water and a little $\ce{HNO3}$ to acidify) – assuming the temperature remained constant, what effect could increasing the ions' velocity have on conductivity, without heating it up (increasing kinetic energy) to alter viscosity?

There is a strong dependence on temperature in conductivity of ionic solutions; see http://www.eutechinst.com/techtips/tech-tips44.htm, for example. However, thermal energy, i.e. molecular motion, is random, and does not favor current flow in the forward (or backward) direction. There are a number of factors that temperature influences:

Viscosity usually changes  inversely with temperature, so ions move more freely at higher temperature.
Ionic solutions have other species than OH- and H+; for example OH3+ [water + hydroxyl radical] is the hydronium ion; see http://en.wikipedia.org/wiki/Hydronium. The amount of each species varies with temperature, as shown in that reference. See also http://en.wikipedia.org/wiki/Self-ionization_of_water. Depending on mass, hydrogen bonding etc. different species have different conductivity.

Thanks. I've just done a mini-experiment at home, and I'll update with some fresh information if you could help explain that too?

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Second order reaction half life equation?
For the reaction:
$$2A\to p$$
The rate, $v$ may be given as:
$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$
Correct?
Integrating this gives:
$$-\frac12\int_{[A]_0}^{[A]}\frac{d[A]}{[A]^2}=k\int_0^tdt$$
$$\frac1{2[A]}-\frac1{2[A]_0}=kt$$
Rearranging...
$$\frac1{[A]}=\frac1{[A]_0}+2kt$$
Now substituting $t=t_{1/2}$ and $[A]=\frac{[A]_0}2$...
$$\frac2{[A]_0}=\frac1{[A]_0}+2kt_{1/2}$$
Minus $1/[A]_0$ from each side...
$$\frac1{[A]_0}=2kt_{1/2}$$
This then gives:
$$t_{1/2}=\frac1{2k[A]_0}$$
However, I have seen in textbooks that this should be written:
$$t_{1/2}=\frac1{k[A]_0}$$
Have I gone wrong somewhere? If so where?
I need to know which equation is correct because when figuring out the rate constant, would the gradient be equal to $\frac{1}{2k}$ or $\frac{1}{k}$?

What you did isn't wrong, but instead of:
$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$
some write:
$$v=-\frac{d[A]}{dt}=k[A]^2$$
which is also true.
Twice a constant is still as constant (a different constant of course).
See http://www.chem.arizona.edu/~salzmanr/480a/480ants/int2ord/int2ord.html

So if I wanted the rate constant of the reaction from the gradient of the graph (explained at the bottom of my question) which equation should I use?

I don't understand that part of the question.  Explain more exactly what you are plotting. Product concentration vs. time? Reactant concentration versus time?

$t_{1/2}$ against $1/[A]_0$

both ways would be equally correct, you just have to explain how you are defining the rate constant.

How can it be correct to get two different rate constants for one reaction?

Because you can define the rate constant to mean what you wrote (your differential equation) or what the reference wrote.

Thanks. Please may you take a look at this: http://chemistry.stackexchange.com/questions/24468/predicting-rate-equations-from-data

This is purely a mathematical answer, but shouldn't the integration produce
1/[A] - 1/[A]0 = kt
which means you don't have your 2kt being taken through your working, and gives you the same as the textbooks' answer(s)?

I have a factor of minus one half outside the integral from the rate equation

Where did your -1/2 actually come from?

Stoichiometry of the reaction

Ah, DavePhD beat me to it, but basically your value is fixed anyway so half of it is the same thing so it's not 2kt.


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The Workings of Pulse-Height Discrimination
Pulse-height discrimination is a common technique to filter out unwanted background signals from the analytical signals. But how exactly does it work?
The analytical signals are not continuously detected, but detected rather in pulses. This is because the detector has a "dead time", and there is a pile-up of ion-pairs (or whatever the principle of detection is) in this dead-time. The intensity of the built up signal - the pulse - is then detected.
Pulse-height discrimination is just that; a discrimination of the detected pulse-heights. However, to be able to discriminate the pulse-heights, surely the system must know what a "desirable" pulse-height is, or, in other words, when a detected pulse deviates from the desirable? How exactly is this done?
In the context of X-ray diffraction, we know the desirable energy of the detected radiation: in theory, this energy should be equal to the K$\alpha_1$ of the X-ray tube's anode material (since diffraction is an elastic interaction, and hence no energy lose occurs). However, the intensity of the signal has nothing to do with the energy. How do we (that is, the ones developing the analytical system) know which pulse-heights are relevant and which are irrelevant to our analysis?
I hope have presented my question clearly, and I appreciate all help. Finding free references online was not very easy, in my experience.

Pule Height Analaysis (PHA) is done on the current pulse from the detector. It only works for detectors which have a current pulse which is proportional to the energy of the x-ray (scintillation counters and gas proportional counters). It does not work for Geiger counters where the current pulse doesn't depend on the x-ray energy.
Think of dropping the current through a resistor. The current pulse then has a voltage. Basically you set a low and a high voltage (a window) for the detector pulses which you will count. Any detector signals within the window are counted, those outside the window (too low or too high) are rejected.
Given that the detector current (or voltage really) should be proportional to the energy of the x-ray detected this improves the S/N ratio. So "low" current pulses don't have enough energy to be a Cu K-alpha x-ray, and high energy pulses have too much energy. High pulses could be for example an x-ray with twice the energy.
It is important to note that the rejected pulses are not "background" counts per se. But there is a twist. The rejected pulses do create detector deadtime. So you have to be careful that the overall count rate of the detector is low enough to still be linear, or do a correction for the total count rate.


text:
Predicting rate equations from data
I am having trouble wit the following question on my problem sheet:
"The reaction $\ce{Br- + ClO^- \to BrO^- + Cl^-}$ occurs with a rate constant of $\text{0.35 dm}^3 \text{mol}^{-1}
\text{s}^{-1}$
at
298 K. For the following initial conditions calculate the time taken for the Br^-
concentration to fall to half its original value:
i) $[\ce{Br-}]=[\ce{ClO-}]=0.0017$
ii) $[\ce{Br-}]=0.0017$ and $[\ce{ClO-}]=0.084$"
I am sure I could do this question if I knew the rate equation but it isn't given. From the units of the rate constant the reaction must be second order but how am I supposed to know whether it's second order with respect to one of the reactants (and thus zeroth order w.r.t the other) or first order with respect to both? Also, what's the significance of the concentrations being equal (there's another question that stresses that the concentrations of both the reactants are equal).

I don't see a reasonable mechanism by whom the rate equation would have second order with respect to one reactant.

Expanding on @Marko 's point, if there were a couple of potential mechanisms, one for first order in both reactants and one for zeroth/second order, then you wouldn't know.  There's no particular reason the concentrations should be equal, or reason why it's important (unless perhaps there is a method of initial rates question elsewhere in your problem set.)

You have to assume it is first order in each reactant.
Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out.
The reaction can't really be zero-order in a reactant, otherwise it would imply the reaction continues after there is no reactant remaining.

how can any reactant be zeroth order then?

it can't, that's what the link is explaining

@RobChem Rewrote my comment to better explain what I meant. Suppose that the kinetics of a reaction are usually decided by a particular slow step yielding rate = k1[A], when [A] and [B] are roughly equal to one another.  If we set up the reaction with very, very small amounts of [B] instead, the rate of another reaction in the mechanism that involves B will eventually become slow enough that it becomes the slow/rate determining step.

Being the concentrations equal makes it easier to integrate the second order law (first in each reagents). Concentrations are the same at every t and therefore you have only one variable to take into account. The resulting integrated law will be analogous to that for a second order kinetics in one reagent.


text:
The anode and cathode when corrosion happens
Let's say $\ce{Fe}$ reacts with $\ce{Cu^{2+}}$ ions. $\ce{Fe}$ would oxidize and therefore give electrons to $\ce{Cu^{2+}}$so that:
$$\ce{Fe-> Fe^{2+}  +2e-}$$
$$\ce{Cu^{2+} +2e^- ->Cu}$$
The overall reaction:
$$\ce{Cu^{2+} +Fe ->Cu +Fe^{2+}}$$
Now this is an example of corrosion, right? And when this type of corrosion happens, the anode is the electrode where oxidation happens and the cathode is the electrode where reduction happens, right?
Therefore iron ($\ce{Fe}$) is the anode where oxidation happens and Copper 2+ ions ($\ce{Cu^{2+}}$) is the cathode where the reduction happens.
My question is, is this all correct? Have I successfully described corrosion (of this type) and given an example of it? Or is there something wrong? I'm 15 and soon I'm having a test in chemistry, and because of my age there's no point in making it "super-advanced", but if there's something I missed or did wrong please point it out, I would be grateful for that!

There are some problems with charges in your equation, and this type of reaction isn't what is typically called corrosion.

@Mithoron: Weird, I think my teacher said this qualified as corrosion.. Also I know that $Fe$ + $Cu$$^+$$^2$ $>$ $Cu + Fe$$^+$$^2$, I did a mistake. EDIT: Is this not qualified as corrosion as the metal doesn't change according to natural circumstances?

I checked and indeed something like this can happen in galvanic corrosion

@Mithoron: So does that mean that this example is a correct example of galvanic corrosion? I'll ask my teacher too of course if this is acceptable. EDIT: And is this description good enough when explaining galvanic corrosion?

Fe is probably oxidized to +3, otherwise I see no problem.

@Mithoron: Well I've heard about Fe oxidizing to +3, but we specifically did this experiment in my school where the iron dissolved and a layer of copper covered the solution. It was my teacher who showed me this reaction, but either way I'm thankful for your help.

You are right! What you have written is an example of Galvanic corrosion (Galvanic corrosion occurs when two different metals have physical or electrical contact with each other and are immersed in a common electrolyte).
You have  successfully described Galvanic corrosion and given an example of it


text:
Why does ionization violate the stable or lowest energy rule in atoms?
Electrons that fill orbitals always do so in such a way that the resulting structure has the lowest energy state possible, though there are anomalies like chromium and so on.
But ionization no matter which element tends to violate this rule. When electrons are ionized they are from the outermost (valence) shell. But this doesn't always produce the lowest energy state possible like the case with 3d and 4s.
To clarify, all I am asking is why in that specific case, is the electron taken out of the 4s orbital first instead of the 3d orbital, since the 3d orbital is a higher energy level?

I'm not sure what you mean by your question, but anomalies in e.config. are again because of reaching stability. The problem is for example the closeness of the energy of an empty 3d and an empty 4s. Still I was wondering if you would mind clarifying what you've written a bit, maybe some examples.

Textbooks won't bother to dig into details. Most of them are satisfied with saying that such thing exists. They're right I guess. I'll do a chemguide approach to the answer since I find it most understandable.

We draw this diagram, and then assume that it works for all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram.
If you stop and think about it, that has got to be wrong. As you move from element to element across the Periodic Table, you are adding extra protons to the nucleus, and extra electrons around the nucleus.
The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals.

Going through the Aufbau principle, when you pass Ca (Z = 20) you will face Sc, the first element in which a d orbital is going to be filled. But, you must make a really important change in your train of thoughts.
As chemguide explained, with change in the number of protons and electrons due to the change of Z, the way their orbitals' energies are defined changes. Till Ca, the behavior insisted that 4s has a lower energy state than 3d and thus, fills sooner. But, starting from Sc, the pattern changes: In Scandium, 3d has a lower energy than 4s. The behavior changed.
This change in behavior results in the current confusion of yours. This much answers the problem you have, and you can visit the page at chemguide to learn more. If you want more info, I can freely add. Hope I've helped!


text:
What ratio of bicarb to vinegar do I need in order for the result to be pH neutral?
This question stems from one I just asked over at Seasoned Advice, where my recipe tasted decidedly off because the pH was wrong.
The recipe called for $\pu{16 g}$ bicarb of soda (sodium bicarbonate, $\ce{NaHCO3}$) and $\pu{15 g}$ of vinegar (mine was 5% acidity, and I assumed a density of $\pu{1 g ml-1}$). I just googled the corresponding pHs of these two, and apparently vinegar is more acidic than bicarb is basic: vinegar has a pH of 2.4–3.4 depending on concentration, and bicarb is pH 9.5.
How can I calculate whether I put too much vinegar or too much bicarb into my recipe? The ratio in the recipe is almost $1:1$.
What ratio in grams do I need of bicarb vs vinegar in order for the result to be neutral? I know there are other contributing factors to the pH in the recipe but I just want to focus on these two first. Assuming the vinegar is 5% which mine was.
Edit:
So I found the explanation for the chemical reaction between bicarb and vinegar here. Relevant bit:

$$\ce{CH3COOH + NaHCO3 → CH3COONa + CO2 + H2O}$$
$$\frac{\pu{1 g}\;\ce{NaHCO3}}{\pu{84.0068 g mol-1}} \times \frac{1}{1} \times \pu{60.0522 g mol-1} = \pu{0.715 g}\;\ce{CH3COOH}$$
What this formula tells you is that every gram of baking soda requires \pu{0.715 g} of acetic acid. But the remaining problem is that vinegar is not pure acetic acid.  In fact, most vinegars vary from 4% to 18%.
Using those numbers, you would need somewhere between $\pu{0.715 g}/0.04 = \pu{17.9 g}$ and $\pu{0.715 g}/0.18 = \pu{3.97 g}$ of vinegar, for each gram of baking soda.
If you're lucky you can get some vinegar with the percent acetic acid listed on the label, then you can substitute that number into the equation above to determine how much vinegar you'll need.

So if I plug in my 5%, I get $\pu{0.715 g}/0.05 = \pu{14.3 g}$ vinegar for $\pu{1 g}$ bicarb.
Yikes! This is way more than $1:1$. But I'm not sure if it answers my question. Is the result, $\ce{CH3COONa + CO2 + H2O}$, pH neutral? If yes, then my vinegar does hardly anything to cancel out the base, is that right?
Edit 2:
Okay so according to Wikipedia, $\ce{CH3COONa}$ is a base. So even though my reaction formula shows what happens when all the bicarb/vinegar is used up, the result is not neutral (how much does the water and carbon dioxide contribute to bringing the pH down?). I need more acid to make my sodium acetate neutral (or at least neutral enough to be palatable?). This means that the needed ratio is higher than 14:1!? So almost certainly way too much bicarb?

Your calculation is correct (14.3 grams of 5% acetic acid completely reacts with 1 g sodium bicarbonate). But you're not interested really in "completely reacts"; you want something like pH 7.
You've got it just about right with your mix; you're at the endpoint, and this is a point where the pH will drop very rapidly with even a small addition of vinegar. So it would only take a few additional drops to make the pH go down to 7... but that may not be what you really want.
Sodium acetate has a mild salty taste by itself. So I'm not sure adding extra vinegar will be worthwhile, it'll make it a bit more vinegary.
Here's what I'd do: Mix vinegar and baking soda in the ratio you've figured out, and then see how that tastes. Make adjustments at that point according to taste, rather than by "titrating" to an arbitrary pH.

Why does the ph drop very rapidly with even a small addition of vinegar?

When both acetic acid and acetate are present, the equilibrium between the two can shift when you add any more acid or base to the mix so the pH doesn't change much---this is called "buffering". When you only have one form or the other present, the mix can't buffer so any added acid or base can change the pH a lot.

I am a professional baker. A good ration in volume is for every 1/4 tsp Baking Soda, use 1 Tablespoon white vinegar.   But you can try this for yourself.
in a small bowl, measure 1/4 tsp Baking soda, pinch of salt, and a TBS sugar and 2 TBS water (because your baking formula will have similar ingredients.  At this point there will be NO reaction (bubbles). Very slowly and incrementally add the 1 TBS white vinegar while stirring.  As you add the vinegar, you will observe a reaction.  You should also notice the reaction begins to weaken toward the last drops of vinegar.  At this point you can taste the liquid.  It will taste like flat soda like sprite or 7 Up but not have any objectionable acid or base taste.  in all reality, the mix will still react with more vinegar,but this is good because you don't want to burn out all your leavener while mixing.  You may increase these amounts in your recipe as required... hope this helps

I tsp a tea spoon, or a table spoon, or is that TBS? It would be preferable if you could use units of measurement that are also know outside the English-speaking world.


text:
Reaction mechanisms for organohalides from alcohols
For the syntheses of organohalides from alcohols, what would plausible reaction mechanisms be for the three following reactions?
$$
\ce{R-OH + NaBr + H2SO4 -> R-Br + NaHSO4 + H2O}
$$
$$
\ce{R-OH + PCl5 -> R-Cl + POCl3 + HCl}
$$
$$
\ce{R-OH ->[red P][Br2 or I2] R-X}
$$

Explain what exactly? Also, could you give a reference of where you found them? They look wierd to me.

@MARamezani I found those reactions in organic chemistry textbook

Oh, you edited your question and removed the typo. That was what was giving me the creeps. Still, this remains unanswered: What should be explained?

@MARamezani Explain the mechanism involved in those reactions

I may be overreacting (and sorry for removing the tag) but I think this counts as homework. You should give some try or explain a little bit about the basics of finding reaction mechanisms.

@MARamezani this is not a homework question I posted this question to understand the reaction mechanisms involved in synthesis of ogranohalides from alcohol.

I think this counts as homework, not is literal homework. I'm letting it to play out. Hopefully it won't get closed. sigh

Nucleophilic substitutions, regardless whether proceeding via $S_N1$ or $S_N2$ require a decent leaving group. $\ce{HO-}$, however, is a lousy leaving group!
To overcome this, several methods exist:

Perform the reaction under acidic conditions, where the alkanol is protonated. Your leaving group now is $\ce{H2O}$. In addition, the corresponding anion of sulfuric acid as a bad nucleophile.
React the alkanol with an acid chloride to form an ester and replace the ester group with a halide. Consider $\ce{PCl5}$ to be an acid chloride of phosphoric acid.
Form the acid halide in situ by reaction of phosphorous and iodine.


text:
How to purify water with activated carbon?
If I am not mistaken, distilling water does not remove all possible impurities. Chemicals with boiling points less than $100\ \mathrm{^\circ C}$ will be transferred as well. Does the usual tap water contain such (I think there is at least fluoride.)? I read that activated carbon can be used to purify water as well. How is it done in practice? Suppose I first use activated carbon and then distill the water (or in the reverse order). How much impurities there would be still left?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Don't confuse the fluoride anion, $\ce{F-}$ with fluorine $\ce{F2}$, which is a gas under standard conditions.

Correct. Fluoride ion is what that's in tap water and is helpful for maintaining teeth health. Plus, it's REALLY hard (in an uppercase sense) to make something 100 percent pure.

The main reason activated carbon is used in water treatment is that its large surface area is useful for adsorbing volatile organic contaminants and chlorine. It is not effective at removing inorganic ions (dissolved minerals, including fluoride), but these are removed quite well by distillation. I think distillation first would make more sense—distillation will remove less volatile contaminants so that less has to be adsorbed by the carbon.
Tap water has to meet certain standards (that depend on your area) for dissolved minerals and volatile organic compounds, so depending on the end use of the water, removing the small amount of remaining contaminants may not be necessary. Formerly, double-distilled water was the standard for making highly purified water, but certain applications are more or less tolerant of certain types of contaminants, and double-distillation requires a lot of energy, so it has largely been replaced for many purposes.
Consumer water filters usually contain filters to remove particulates, ion exchange resin to remove inorganic ions, and activated carbon to improve the taste of the water, and as far as purification for laboratory use, combinations of filtration, deionization, reverse osmosis, ultrafiltration, etc. are common these days. Different processes are better at removing some types of contaminants than others, so combinations are used. Distillation of tap water followed by treatment by activated carbon will produce quite pure water by consumer standards, but might not be pure enough for say semiconductor microfabrication or biomedical applications.

Activated Carbon is used in water filter purifiers because activated carbon removes from the water most toxic organic compounds in water like pesticides and heavy metal organic compounds.


text:
On the molecular level, how do catalysts work?
I already know what catalysts do (lower activation energy requirement for a reaction), but how exactly are they able to do that on a molecular level?

In general terms, some catalysts serve as a "meeting place" where the reacting species can get together more easily, e.g. $\ce{H2}$ and $\ce{O2}$ adsorbed onto a $\ce{Pt}$ catalyst; see this.
Others may actually participate in a reaction but are regenerated afterwards, e.g. according to Maggs and Sutton, "argentous hydroperoxide" is produced as an intermediate in the catalytic decomposition of $\ce{H2O2}$ by $\ce{Ag}$, and regenerates the silver ions as it decomposes.


text:
Why is beryllium transparent to x-rays?

Beryllium has other fancy applications. It is transparent to x-rays, so it's used in the windows of x-ray tubes, which need to be strong enough to hold a perfect vacuum, yet thin enough to let the delicate x-rays out.

The quote is from the book The Elements by Theodore Gray. Why is beryllium transparent to x-rays?

Graphene has now also been used for x-ray windows, with better transmission in the range 0.1 - 3 keV: http://arxiv.org/abs/1503.03327

The amount of X-rays absorbed by an element depends on the size of its atoms (its absorption cross-section, specifically, as affected by the size of core orbitals that contain electrons that can be excited by X-ray absorption, and the number of electrons in those orbitals) and how many atoms are packed into a given volume. Big atoms that are close together can absorb more X-rays than small atoms that are far apart.
Beryllium has very small atoms (it has a very small electronic core, with only two electrons in it), and very low density, so it will absorb relatively few X-rays compared to other materials.
According to this page, 37 % of X-rays with a wavelength of 0.56 nm will be absorbed by beryllium window that is 23.5 mm thick. If the window were made of aluminum, it would only have to be 1.4 mm thick to absorb the same amount of X-rays, and if it were made of lead it would only have to be 13 μm thick.

The notion of "size" here is hooey. Beryllium has a very low Z number so its electrons are relatively loosely bound. X-rays will in general have an energy great enough to cause some sort of electronic excited state, but as the  photons increase in  energy beyond what is needed, then the probability of an interaction between the photon and the electron goes down. The other factor here of course is that atoms of beryllium have relatively few electrons per atom of course.


text:
food chemistry maltodextrin vs organic maltodextrin
what is the difference between 'normal' food additive maltodextrin and "organic maltodextrin"? Is it only the primary source (i.e.: corn, wheat etc) that is organic or is it a different process to create maltodextrin? And is maltodextrin the same as food additive 1405?

"Food additive 1405" refers to the process for the modification of starch; in this case the process uses enzymes to break up the starch molecules into oligosaccharides between 3 and 20 saccharides.
The difference between regular maltodextrin and 'organic maltodextrin' would be that the source was grown without synthetic pesticides and synthetic fertilizers. 'Organic maltodextrin' may also need to use natural enzymes.


text:
How to rationalise differences in halogen bonding?
Please see the problem in this link. I understand the explanation in (c), but it appears to me, that accepting that would contradict the answers to parts (a), (b), and (c).
Is the reason for part (a) because the first molecule does not have resonance, and for (b) the $\ce{F3C-I}$ molecule has a larger inductive effect? If so, that would not make sense for (c) to be correct.
Can anyone explain the difference in halogen bonding to me?

A halogen bond can, quite analogous to a hydrogen bond, be considered a mostly electrostatic interaction with some covalent interaction. Hence the halogen bond acceptor is the electron donating molecule, i.e. a Lewis base. The halogen (bond) takes the role as the Lewis acid.
In (b) you compare $\ce{H3C-I}$ and $\ce{F3C-I}$. Fluorine is more electronegative than hydrogen, and the statement $$\angle(\ce{H-C-H})>\angle(\ce{F-C-F})$$ holds. The carbon directs mor $\ce{p}$ orbital character into the carbon fluorine bonds than in the carbon hydrogen bonds. As a consequence the $\ce{s}$ orbital character in the carbon iodine bond (from the carbon) is higher in $\ce{F3C-I}$ compared to $\ce{H3C-I}$. That causes the bond to be slightly shorter and having more electron density at the carbon. This is also making the iodine a little more electron deficient and a better electron acceptor, i.e. a stronger acid (see also Bent's Rule). The electronic interaction will therefore also be slightly stronger.
This observed effect is also often called inductive effect, which like you correctly stated is larger in $\ce{F3C-I}$.
In (c) you are comparing the Lewis bases $\ce{NMe3}$ and $\ce{OMe2}$ acting towards $\ce{F3C-I}$ (or $\ce{HC3-I}$, the reasoning would remain the same). Oxygen is more electronegative than nitrogen, hence the lone pair orbitals of oxygen are more compact, i.e. the maximum electron density is closer to the oxygen nucleus. This causes dimethylether a weaker electron donor. Hence the the oxygen iodine interaction in $\ce{F3C-I\bond{~}OMe2}$ can be expected to be weaker than the corresponding nitrogen iodine interaction in $\ce{F3C-I\bond{~}NMe3}$. (Just like it is stated in your sheet.)
Additionally, since methyl groups are usually slightly electron donating, and trimethylamine has one more methyl group than dimethylether, it can be expected to be an even better electron donor, stronger base.
Resonance does not contribute to these bonding situations in a significant way. (Looking at this from a valence bond perspective would certainly exceed this format and it is not necessary for a good model explanation.)
Last a word of caution. Oxygen (in neutral molecules) can in most cases not be described as $\ce{sp^{3}}$ hybridised. Symmetry considerations and it's electronegativity are usually the reason for this. It is better described as $\ce{sp}$ hybridised in terminal position (like in ketones), or $\ce{sp^{2}}$ hybridised in bridging positions (like in ethers).

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


text:
Is there a field of chemistry that doesn't do experiments?
In other words, a field of chemistry that is all theoretical? I have heard of theoretical chemistry, but my chemistry professor said you can't really major in theoretical chemistry. He said the closest thing is physical chemistry.
I like predicting in Chemistry, but I hate labs or experiments.

Here's what you do: drudge through the experiments until you have enough experience to have your own lab, then have some graduate student do all the experiments for you, thus completing the cycle.

Of course you can major in theoretical and/or computational chemistry. There are a lot of groups out there, who do exactly only that. However, you still need the groundwork from basic inorganic/organic lab work. The experience counts here and you should be familiar with what is going on in the lab. After that all possible kinds of research directions are open to you. If you want to dodge all experiments, I suggest the route, studying physics, specialise in physical chemistry/ theoretical physics and later apply (in research) your knowledge to chemical reactions.

There is computational chemistry, but IMO it's not really chemistry.

Just a tip : if you are planning to go into theoretical or computational chemistry, make sure that your maths and computer skills are good enough - most of these are usually not taught in classical chemistry programmes, but you should try to get this knowledge elsewhere early on if you want to succeed.

@G-man I hope you increase your knowledge about chemistry and revise your opinion, after all theoretical and computational chemistry helps understand experiments a lot nowadays. Hopefully, when you continue your studies you will appreciate the helping hand from the computer gals and guys.

You are correct that it is possible to study theoretical or computational chemistry, however these typically only represent a handful of courses in a normal undergraduate chemistry curriculum. Additionally, most undergraduate chemistry curriculum are designed to instruct students broadly in the core disciplines of chemistry, hence it's usually not possible to specialize.
Non-chemisty degrees may be an alternative. However any science undergraduate program will require laboratory-based coursework. Hands-on experience teaches us how scientific concepts work in the real works, giving us an appreciation of how difficult experimental work can be. It also functions to reinforce content from books and lectures; to teach technical skills (which are usually necessary for future employment); and to develop lab safety habits. So there really is no means to escape it. Moreover, in order to take advanced chemistry electives, prior courses which include lab sessions are generally necessary.
That said, there are other routes to get into chemistry. I've know individuals whom have majored in mathematics, engineering, physics, and biology who either did research in chemistry or who did graduate degrees in chemistry subsequent to obtaining an undergraduate degree in those fields. However in order to obtain undergraduate degrees in those fields and to be able to transition to chemistry programs, lab experience and some senior-level chemistry courses are typically required.
However, once in graduate school, it is possible to avoid experiments altogether: theoretical and computational chemistry graduate degrees typically do not require that the candidate engage in any hands-on experiments. Many however do require that the graduate student work as a teaching assistant in undergraduate chemistry laboratories as a condition of funding. In such cases, laboratory experience is either necessary or a significant advantage.
So while it may be possible to minimize the amount of lab work that you engage in, some will always be required. Additionally I would not recommend avoiding lab work, as it greatly enriches the study of chemistry overall.

It is possible to not perform experiments at all and only concentrate on theoretical chemistry (at least after your undergraduate studies). What your professor said might be related to your university and his group of friends.
University:
The three names of professors in theoretical chemsitry I know from our university are Hättig, Marx and Kutzelnigg: http://www.theochem.ruhr-uni-bochum.de/home.en.html . The scientists working for them did not enter a laboratory after the 6th semester (final semester of bachelor studies) as far as I know. Their major is theoretical chemistry. This is what they can have highlighted on their master certificate.
(You might want to look at this website and check out the member lists. Then you have an idea how many people are working in this field.)
Industry:
There is work for theoretical chemistry at BASF and Bayer. We heard presentations from several people of that field.
Lab work is not required. There are people who specialise in lab work and master it to a fascinating level. The people in theoretical chemistry receive the results from those in the laboratories and work their magic of interpretation, explanation and prediction. The lab skills of a theoretician are not needed if a professional is working on the experiments. The computer skills of an experimentalist are not needed if a professional is working on the calculations.
[To my understanding, theoretical chemistry does not make experiments obsolete but experiments cannot give a full explanation about the multitude of properties and comprehension of its scientific context. Furthermore, theoretical chemistry gives valuable guidance. Nowadays, theoretical and experimental chemistry are roughly on eye level (even though each thinks of himself as more important).]

Many researchers do no traditional lab experiments. Computational chemistry is important and a huge subject. The mathematical theory of chemistry is the Schrödinger equation, known since over 70 years, but unfortunately too difficult to solve for most molecules with more than a few atoms. Thus, an ever-expanding set of approximative methods have been developed and keep developing. One cool example would be Anatole von Lilienfeld who uses neural networks to predict molecule properties by pattern recognition, extrapolating from known molecules.
This is not to say that lab work can be avoided for a chemistry student. I suppose an alternate route would be to become a physicist then switch subject. I did that, although I actually do lab work and enjoy it now.

Theoretical chemistry is the branch of chemistry that focus on non-experimental side of chemistry
https://www.nature.com/subjects/theoretical-chemistry
simons.hec.utah.edu/TheoryPage/WhatisTheoreticalChem.ppt
Your professor seems to be confounding branches. Physical chemistry can be both experimental or theoretical. The same happens with other disciplines, e.g. theoretical organic chemistry.

Surprisingly, you would do a Ph.D. in chemistry, not in physical chemistry, not organic chemistry, etc. And some institutions don't even give you that; then you're just a doctor of natural sciences. The sentiment that theoretical chemistry is not a discipline is not only misleading, there is a plethora of universities and related institutions offering courses in that. Surely they cannot all be completely wrong...

@Martin-マーチン You are right, theoretical chemistry is a discipline. I edited the answer.


text:
Does the "Bat Gen" claim to produce more energy than it uses?
The following links claim that inventor Arturo Solis Herrera has created an "everlasting battery".

http://www.mexiconewsnetwork.com/news/bat-gen/
http://www.faqs.org/patents/app/20090134007#ixzz10Sh9iLQM

The article claims that:

Bat gen is the name of this ever-lasting battery. It consists of a biochemical process triggered by mixing water and melanin, which results in a substance capable of separating oxygen from hydrogen (the components of a water molecule), therefore liberating energy.
The process continues as the same molecule brings together both elements turning them into water again; as a result one more energy load is triggered.
According to Solis’ research, once this process is reached, it can continue for 100 years! Moreover, there are several ways of artificially producing the melanin found in human nails, hair and retina; two of them, which he already patented, are based on vegetables and oil.

Although they do not claim that the battery is perpetual, they seem to imply that mixing water and melanin can give an (oscillating?) reaction that can last for 100 years, providing cheap power.
Does this have any chemical basis? What matter is consumed?
Would that work without the user needing to add anything to the battery or is the article just describing the process in a misleading way?

The article is not clear (I have not read the patent), but it doesn't look like this is claimed to be a perpetual motion (or infinite energy or free energy or other thermodynamics-breaking) device; instead it seems like it's supposed to be a (rechargeable) battery with a finite capacity but very long life.

And glancing at the abstract, it seems to use a melanin-like derivative which is able to both 1) use solar power or other light source to drive electrolysis of water and produce hydrogen and oxygen (which may then be stored), and 2) recombine hydrogen and oxygen and release the energy as electricity. Basically a solar-powered rechargable hydrogen cell, not an infinite-energy claim. If the question were edited to be about whether the described technology works and would be feasible (not whether it's perpetual motion), I think it should be re-opened.

The patent says that it's using ambient light as the source of energy.

Ok that was important

hmmmm... honestly reading the article it does not seem like they say either water or melanin has to be provided (at least not for 100 years).

Oh boy is that patent painful to read. They didn't produce meaningful amounts of hydrogen, they claim it works with like 40 different substances, from 1 µL to litres, a huge list of electrodes… If it was this easy to hydrolyse water, it's a wonder our bodies don't just explode from all the melanin. Despite the craziness, it doesn't seem like they're claiming to make energy out of nothing.

I'm not sufficiently familiar with the chemistry involved to state authoritatively that this guy is full of bologna, but there are some pretty fantastic claims made in the second link.  Also, if that's a US patent application, it's a disaster.

The device is able to use any light from 900-200nm, mechanical agitation, sound, or magnetic fields (and unspecified "other") as sources of power.
He has also made all of his cells inside Tupperware containers sealed by silicone caulking, so the 100 year claim is a bit sketchy.
The inventor's testing methodology is pretty dubious as well.  The device was placed in sunlight for 30 minutes and it experienced a change of pH from 7.3 to 7.1, therefore it works.  Also, in his own words:

It is worth saying that such variation from 0.2 to 1.0 pH unit; and the reverse reaction that occurred once the containers were placed in a dark place, was anticipated by our theoric background, when we performed the experiments, we known already what was going to happen, as the matter of fact, we didn't do many experiments, we only made two or three times, obtaining the results we expected.  {All grammar and spelling mistakes are those of the original author(s).}

That said, the idea of a cell that can catalytically break down water into hydrogen and oxygen using sunlight and then recombine them into water, extracting the energy, is not impossible by any stretch of the imagination.  Designing a cell that is economically viable compared with silicon based devices has yet to be accomplished, to the best of my knowledge.

I believe he has applied for a Russian patent. Thanks


text:
Reaction Kinetics and Stoichiometry -- Mass conservation
This question has been bothering me for some time, and I can't seem to find a good answer online.
Say I have four chemical species $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, and these four react in the following ways:
\begin{align}\ce{
A + A &-> B\\
A + B &-> C\\
A + C &-> D\\
A + D &-> B + C\\
}\end{align}
The kinetic reaction equations for these four species should be:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2  -k_2[\ce{A}][\ce{B}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}
where $k_i$ is the Arrhenius coefficient for that reaction.
In my mind, the stoichiometry does not seem to add up right. It seems that when the system undergoes $\ce{A + A -> B}$, the concentration of $\ce{A}$ should decrease by twice as much as a reaction like $\ce{A + B -> C}$. Similarly, the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.
Therefore, stoichiometrically, I want to write:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= -2k_1[\ce{A}]^2-k_2[\ce{A}][\ce{B}]-k_3[\ce{A}][\ce{C}]-k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= +k_1[\ce{A}]^2  -k_2[\ce{A}][\ce{B}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= +k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +\tfrac{1}{2}k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= +k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}
Is this wrong-headed? What conceptual issue am I missing?

Reaction rates should be defined in terms of the Extent of reaction ($\xi$) that corresponds to the number of moles (or the molarity for reactions in solution) of specie $i$ divided by the the stoichiometric number, $\nu_i$:
$$\xi = \frac{[i]}{\nu_i}$$
Therefore, considering that the first reaction should be:
$$\ce{A + A -> B ~=~ 2A -> B}$$
The corresponding reaction rates are:
$$\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2} \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}$$
In this way, the reaction rate is always the same (positive) number, independently on the specie we are referring to.
The global rate equations are:
\begin{align}
\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} &= −2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]−k_3[\ce{A}][\ce{C}]−k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= 2k_1[\ce{A}]^2−k_2[\ce{A}][\ce{B}]+k_4[\ce{A}][D]\\
\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} &= k_2[\ce{A}][\ce{B}] -k_3[\ce{A}][\ce{C}] +k_4[\ce{A}][\ce{D}]\\
\frac{\mathrm{d}[\ce{D}]}{\mathrm{d}t} &= k_3[\ce{A}][\ce{C}] -k_4[\ce{A}][\ce{D}]\\
\end{align}
It is conceptually wrong to assume that

the products of $\ce{A + D}$ should be split evenly between $\ce{B}$ and $\ce{C}$.

The reaction scheme you described is the "parallel reactions" model. The amount of $\ce{A}$ consumed by each reaction depends on the specific rate. The stoichiometry of the first step says that every time 1 molecule of $\ce{B}$ is produced 2 molecules of $\ce{A}$ are consumed. But maybe $k_1$ is very small and so all the other process will consume $\ce{A}$ much more rapidly than this one.
You cannot predict what will happen in a reaction system just looking at the stoichiometry of each step. If you have a large amount of $\ce{C}$, it will probably consume a lot of $\ce{A}$ in the third step, producing a lot of $\ce{D}$ that, in turn, will produce a lot of $\ce{B}$, etc. Stoichiometry is just one piece of information. But kinetics is much more that this.

Should your third equation read $-\tfrac{1}{2}\tfrac{d[\mathrm{A}]}{dt}$? I can see how this works with one reaction, but A reacts with three other species as well; where does the factor of 1/2 fit in to the full differential equation?

I don't understand your comment. Third equation states that $\frac{d\xi}{dt} corresponds to $\frac 1 \nu_i \frac{d[i]} {dt}$ for every specie $i$.
The full differential equation is just the combination of the expressions for all the reagents in term of $\xi$ so that the stoichiometric number is automatically take into account.
I edited the answer adding the global expressions for each concentration variatiation.

I have updated your post with some more chemistry markup. If you want to know more, please have a look here and here. Even if it is tempting, please do not use markup in the title field, see here for details.
You might also want to revisit your comment, that got a little messed up.

Thank you for the more expansive explanation. What I talked about the third equation, you wrote: $\frac{\mathrm{d}\xi}{\mathrm{d}t} = -\frac{1}{2}\frac{[\ce{A}]}{\mathrm{d}t}$. The right hand side should be a derivative, $-\frac{1}{2}\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}$.

Yes, you are absolutely right. I missed a "d". I edited my answer to correct the problem.


text:
Can magnetic fields affect a chemical reaction?
This question was asked recently in an interview and I just said that "Yeah, like if the reaction involves ions or paramagnetic species". But the interviewer went on and asked me to elaborate on HOW will the reaction change, and I was stumped.
So what I am asking here is what will happen if I allow a chemical reaction to take place in a live MRI machine?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

Here is a nice review paper on the subject: Rodgers, C. T. Magnetic field effects in chemical systems. Pure Appl. Chem. 2009, 81 (1), 19–43. DOI: 10.1351/PAC-CON-08-10-18.
From the abstract:

Chemical reactions that involve radical intermediates can be influenced by magnetic
  fields, which act to alter their rate, yield, or product distribution. These effects have
  been studied extensively in liquids, solids, and constrained media such as micelles. They  may be interpreted using the radical pair mechanism (RPM). Such effects are central to the field
  of spin chemistry of which there have been several detailed and extensive reviews.

Yes, magnetic fields can affect chemical reactions. These are usually reactions in which two radicals are produced. The reaction of a molecule R-R can form two radicals and is written generically as R-R $\rightarrow$ R. + R.. The two radicals each have an unpaired electron. As the radicals are produced in close proximity, because a chemical bond is broken to produce them, the pair is called geminate. These two radicals can then recombine and so form the chemical bond again. In this case the spins are paired, conventionally written as $\uparrow \downarrow$.  However, if the spin of one is changed then instead of recombining as a singlet a triplet is formed, again conventionally $\uparrow \uparrow$ and the triplet state may undergo different reactions and not reform the starting compound.
The triplet state, as the name suggest has three substates which are of equal energy (degenerate) in the absence of a magnetic field. However, two of these states are sensitive to a magnetic field, one rises in energy, the other falls by an equal amount so that the total energy is constant, (see sketch). The triplet has a higher energy than the singlet, which, as it is spin paired, is insensitive to magnetic fields.
Normally the difference in energy is quite large compared to thermal energy and this makes the S-T crossing very slow and is unimportant in the absence of a magnetic field. (The energy gap at zero magnetic field is 2*J* where J is the exchange energy); see sketch. As can be seen there is a magnetic field at which S and T cross (dotted circle) and this will facilitate rapid crossing to triplets and thus to the possibility of different chemistry. As the crossing is quite narrow, in terms of the range of magnetic field strength, only if the magnetic field is of just the right value will S-T crossing occur.

The field strength needed to cause S-T crossing is typically of the order of a few hundred Gauss ($10^4$ gauss = 1 Tesla), far stronger than the earth's magnetic field, (approx $5.510^{-5}$ Tesla) thus there is still debate as to whether smaller fields can cause a change in chemistry in people. This may be possible as many reactions occur in enzymes and these are far more complex than simple solution phase chemistry.
(For the sake of completeness note that there are other types of reactions that are sensitive to magnetic fields such as electron transfer and triplet - triplet annihilation )

thank you, in essence if the field is too low or too big, this will not occur?

yes just in a small range of values. What that is depends on each particular molecule.

is this due to Zeeman effect?

yes, shift if energy levels due to magnetic field is Zeeman effect, as in nmr also.

@porpyrin, nice, so why researcher and science dont exploit that in tuning chemical reaction with magnetic field.

I think that it has been done, but you have to have a special type of reaction involving radicals and so is a limited range of reactions. It has been proposed that animal navigation may be influenced by the effect of magnetic fields  on radical reactions.

@porphyrin is the figure your original work?  Somebody wants to reproduce it.  https://chemistry.meta.stackexchange.com/questions/4354/i-need-permission-to-reuse-photo-from-user-porphyrin-answer-on-question-about

yes my original but I'm happy for it to be used elsewhere as long as its not 'for profit'


text:
Why is there a need for resonance?
Why are we able to show structures of compounds with different position of electron, but with same position of them? Shouldn't the structure become unstable due to this?
Let's take benzene for the example of resonance.We know that in reality benzene has no pure single or double bond. How can this be possible?
The bond length for $\ce{C-C}$ and $\ce{C=C}$ bonds in benzene is same which is $1.39~\mathring{\mathrm{A}}$. This is due to partial single and double bond between carbon atom, which is due to resonance. Why is there a resonance? Why can't benzene be happy with three alternate single and double bond between each carbon atom?

I could not say it any better than the corresponding Wiki article.

As I mentioned in my comment, the Wikipedia article on resonance provides a rather nice description of the subject where the Resonance in quantum mechanics section is of particular interest for the question asked.
However, I decided to somewhat clarify the matter emphasising few important points. And I will start by quoting the OP.

We know that in reality benzene has no pure single or double bond.

Now please, take a deep breathe and repeat with me: there are no bonds. In reality we do have electrons and nuclei that constitutes a molecule, but a chemical bond is merely a theoretical concept introduced in the framework of a scientific theory. In other words, chemical bonds do not exist in the same sense as electrons and nuclei do.
For instance, the concept of resonance was introduced by Linus Pauling in the framework of the so-called valence bond theory, according to which

a covalent bond is formed between the two atoms by the overlap of half
  filled valence atomic orbitals of each atom containing one unpaired
  electron.

So, this is the way the concept of chemical bonding enters the scene in the valence bond theory: a chemical bond is used to describe overlapped atomic orbitals of two atoms in a molecule. Note that atomic orbitals also do not exist in the same sense as electrons and nuclei do, but an orbital has at least some strict mathematical and physical meaning, while a chemical bond, as introduced above, does not. Read the above quote carefully one more time: it is not even a definition! What we just say is that whenever half filled valence atomic orbitals of two atoms overlap a covalent bond is formed, but we do not say what exactly is a covalent bond. It is just something; something which is formed whenever orbitals overlap. It is a concept, purely theoretical construction. It is practically helpful in description of a reality, but no one will say that it is actually a part of it.
Do not be afraid of all this business, because it is the way science generally works. Scientific theories basically tell us how nature could be described, and they do so in the language of mathematics, since it is the only language nature can unambiguously "speak" with us. However, it is often so tempting to force it to speak with us in a different language, in a language of our theoretical concepts (such as, for instance, the concept of a chemical bonding), no matter how fuzzy they are. This is a very dangerous road, because, after all, who are we to tell nature which language to speak? Nature will likely refuse to "speak" with us in that language, and that would be it.
That is what actually happened with the valence bond theory as it was originally formulated. Indeed, all carbon-carbon bond lengths are equal in benzene, but the valence bond theory were not able to account for that simple fact. Nature refused to "speak" with us in the language of chemical bonding. At that point chemists had two choices: either abandon the valence bond theory or introduce another theoretical concept into it, introduce a new word in our language making it possible to communicate with nature in this language. The concept of a chemical bonding was already so tremendously useful in chemistry that the choice was actually predetermined, and in 1928 Linus Pauling introduced the concept of resonance between several valence-bond structures of a molecule.
What is resonance? Is it a physical process? No way! It is just a concept, a word, used to describe

delocalized electrons within certain molecules or polyatomic ions
  where the bonding cannot be expressed by a single Lewis formula.

The word "resonance" was probably not the best choice by Pauling, since it gives an unwanted feeling that the molecule actually do oscillate back and forth between the structures, which is, as I said, not the case. But putting aside its somewhat misleading name, the concept of resonance is as good as the concept of chemical bonding and many other concepts in chemistry and other sciences.

Delocalization is characteristic of the molecular orbital theory
  concerning the structure of atoms. Rather than the lone pair of
  electrons contained in specific bonds (as in the valence-bond theory),
  the MO (molecular-orbital theory) theorizes that electrons exist in
  orbitals that are spread over the entire molecule. The MO theory
  explains molecules such as ozone and benzene, which cannot be drawn
  satisfactorily with one Lewis structure, and are therefore described
  as resonance hybrids. Molecular orbitals solve this issue through the
  concept of delocalized electrons.
When orbitals overlap, many of the orbitals are in hybrid states. Like
  delocalization, hybridization occurs to promote symmetry and
  stability. Delocalized electrons are often found in covalently bonded
  molecules that alternate single and multiple (usually double) bonds.
  Also, they frequently occur in aromatic systems and in mesoionic
  systems which means that they cannot exist in merely one resonance
  structure (they can and do exist in many resonance structures).
Delocalization matters for several reasons. One, spreading electron
  densities over a great area creates greater stability for the
  molecule.   Spreading electrons creates charge distribution.
  Therefore, chemical reactions expected because of a hypothesized
  configuration may not occur, a stable product is formed more readily
  in a reaction. Delocalization of electrons also paves the way for
  conductivity while in spectroscopy, wavelenght absorbtion is also
  based on resonance stability and delocalization of electrons.

Source: Here

Thank you for adding the source.  Rather than quoting the entire passage, which is already available online, it's best to illuminate what is written with your own prose.

One way to look at it is the Electron Clouds.
A chemical becomes reactive when it has either excess or insufficient electron cloud. For example, in case of Ammonia $\ce {NH}_3$, Nitrogen has a lone pair of electron. Thus, it has excess of electron cloud. It can donate these electrons. Hence, Ammonia is reactive. Similarly, Aluminium is highly electropositive and has empty p-orbital. Hence, it can accept electrons and is also reactive.
In case of Benzene, the electrons are delocalised, that is, the electron cloud is spread over the entire ring of Benzene. Thus, there is no accumulation of charge in the ring at one place. This gives extra stability to Benzene. And every molecule wants to attain maximum stability.
If resonance didn't exist in Benzene, then it would react easily with many reagents. Since each of the carbons is $\ce {sp}^2$ hybridised, that is, each carbon is quite electronegative, and has the capacity to react.
Maybe there is a better answer to your question. This is what I can think as the reason.


text:
Why reduced atoms are more energetic?
For example, in glycolysis, $\ce{NAD+} $ is reduced to $\ce{NADH}$ and then it's more energetic. Why adding electrons to an atoms gives it more energy?

All electrochemistry is relative.  And any time you have a reduction or oxidation reaction, you are dealing with an electrochemistry problem.
NADH is "more energetic" than NAD+ because the hidden assumption is that energy levels are relative to reaction with the atmosphere.  NADH can react with oxygen in the atmosphere to release energy.  But on a planet with no oxygen and lots of hydrogen in the atmosphere, NAD+ would be the "more energetic" molecule.
Another less exotic example is anaerobic fermentation in yeast.  NADH builds up in glycolysis, and must be converted back to NAD+ through formation of ethanol + CO2.  That process consumes energy; thus, in anaerobic environments, often times NAD+ can be thought of as a higher-energy molecule than NADH.

NADH is more energetic because it holds more bonds, therefore more energy is required to keep this molecule intact. So when trying to break the bond, it would require more energy to break, and therefore it is also more stable then NAD+. NAD+ is energetically unfavorable because it is a cation. Adding a hydrogen will give it favorability, but it will also hold the molecule with more energy.

Energy of polyelectronic atoms is hard to calculate since you have to take care of the individual energy of each electrons bu also of interaction between them.
To simplify, you can assume the addition of an electron will bring the energy of the electron plus maybe some other interaction wich will rise the total energy of the atom.


text:
Why is the inductive effect of σ-electrons only along saturated carbon chains?
In my book the following is written:

When an electron withdrawing atom such as a halogen is attached to the end of a carbon chain, the $\sigma$-electrons of the $\ce{C-X}$ bond are attached to or displaced towards the more electronegative halogen atom, as a result $\ce{X}$ acquires a small negative charge and $\ce{C_1}$ acquires a small positive charge. The small positive charge on the $\ce{C_1}$ in turn attracts the $\sigma$-electrons of the $\ce{C_1-C_2}$ bond towards it. As a result $\ce{C_2}$ acquires a small charge, of course smaller than on $\ce{C_1}$.

Then as $\ce{C_2}$ acquires a small a charge shouldn't the positive charge of $\ce{C_1}$ become less than earlier? Shouldn't it be true?

...This type of a displacement of $\sigma$-electrons along a saturated carbon chain whenever an electron withdrawing group is present the end of the chain is called inductive effect.

Why is there displacement of only $\sigma$-electrons not $\pi$-electrons and why does it happen along a saturated carbon chain only?

Your, right in 1 case and it seems to me your over interpreting this - it's only an example and inductive effect affects all bonds, but there may be stronger mesomeric effects

The reason why the inductive effect is applicable only on sigma electrons and not pi electrons is because sigma bonds are stronger and more stable than pi bonds. Sigma bonds involve head-on overlapping of orbitals, so the electrons in these bonds are more likely to stay in place. This allows the inductive effect to propagate along a chain of atoms connected by sigma bonds3.
On the other hand, pi bonds involve side-on overlapping of orbitals and the electron density is spread out above and below the plane of the atoms. This makes the electrons in pi bonds more mobile and less likely to stay in one place, which is why the inductive effect does not apply to them


text:
Formula of entropy change
In my 12th standard book, the formula for entropy change is given as $\Delta S = \frac{q_\text{reversible}}{T}$. What is the importance of absorbing heat reversibly and not irreversibly? What does absorbing heat reversibly mean? Does it mean that the process should be reversible, and if it does, then how would we define change in entropy of a spontaneous process i.e an irreversible process?
 Would its formula also include $q_\text{reversible}$?

A process is thermodynamically reversible if it is essentially at equilibrium. Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\text{system}} = T_{\text{surroundings}}$ if heat can be exchanged reversibly; any temperature difference will result in an irreversible exchange of heat.
What's special about reversible heat transfer? When a tiny amount of heat $dq_{\text{rev}}$ is released by the system during a reversible process, the surroundings will absorb exactly that amount of heat, no matter what path the process takes.
If you have an irreversible process, the amount of heat (and work) depends on how the system and surroundings change during the process.
What's the difference between reversible and irreversible heat? The heat transferred by an irreversible process $dq_{\text{irrev}}$ is less than $dq_{\text{rev}}$, where $dq_{\text{rev}}$ is the heat that would have been transferred if the process had been performed reversibly. The difference between the two is the amount of work you'd have to do on the system to reverse the effects of the irreversible process.
How do we know that? Energy $U$ is a state function, so it'll be the same whichever way we do the process:
$$dU = dq_{\text{irrev}} + dw_{\text{irrev}} = dq_{\text{rev}} + dw_{\text{rev}}$$
so the difference between the two heats is
$$dq_{\text{rev}} - dq_{\text{irrev}} = dw_{\text{irrev}} - dw_{\text{rev}}$$
The differences on either side are positive, so $$dq_{rev} > dq_{irrev}$$
This means that an irreversible process will always absorb less heat from the surroundings and perform less work on the surroundings than it would have if the process had been done reversibly.
Why $dS = dq_{\text{rev}}/T$? With a reversible process, $$\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{system}} = -\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{surroundings}}$$
so if we define $dS \equiv dq_{\text{rev}}/T$, then we have $dS_{\text{universe}} = dS_{\text{system}} + dS_{\text{surroundings}} = 0$.
Consider what would happen in an irreversible process that took the system from state A to state B. The inequality $dq_{\text{rev}} > dq_{\text{irrev}}$ together with the definition of entropy in terms of reversible heats gives
$$\Delta S = S_B - S_A = \int_A^B\frac{dq_{\text{rev}}}{T} > \int_A^B\frac{dq_{\text{irrev}}}{T}$$ for any such process. This also implies that for any cyclic process (e. g. A to B back to A), we have $$\oint\frac{dq_{\text{irrev}}}{T} \lt 0, \ \text{and}\  \Delta S = \oint\frac{dq_{\text{rev}}}{T} = 0$$
This shows that $q_{\text{rev}}/T$ is a state function (it's zero for any cyclic process) but  $q_{\text{irrev}}/T$ is not a state function. If entropy is to be a state function, then, you can't define it as simply $q/T$; you've got to use $\Delta S = q_{\text{rev}}/T$, or for the differential form $dS = dq_{\text{rev}}/T$.
How could we relate the entropy change for an irreversible process to $dq_{\text{irrev}}$? From the difference between $q_{rev}$ and $dq_{irrev}$ above and the definition of entropy, we have
$$dS = \frac{dq_{\text{rev}}}{T} = \frac{dq_{\text{irrev}}}{T} + \frac{dw_{\text{irrev}}-dw_{\text{rev}}}{T}$$

TL;DR: There IS no TL;DR; you're asking some important questions here, and the answers need to be made carefully.

How do we define system and surrounding here?

The question is "how would we determine the change in entropy of a spontaneous process i.e irreversible process."  To do this, you first focus exclusively on the initial and final thermodynamic equilibrium states of the system, resulting from the irreversible path.  Then you need to conceive of a reversible path between the exact same initial and final equilibrium states (the reversible path may or may not necessarily bear any resemblance to the actual irreversible process).  Any reversible path between the two equilibrium states will suffice.  You then calculate the integral of dq/T for this reversible path.  This will be the change in entropy.


text:
Are geminal disubstituted alkenes more stable than their cis/trans isomers?
I know for a fact that cis alkenes are less stable than trans alkenes due to steric repulsions.
But what bugs me is the fact that why systems such as 2-methylprop-1-ene, where I believe steric repulsions are greater than that in cis-but-2-ene, are more stable than its cis and trans counterparts.
I saw this being given in the stability order of alkenes in Solomons and Frhyle.
I looked up Morrison and Boyd and it said the same thing. Paula Bruice said otherwise stating trans to be more stable.
Which is the correct order, and why?
Paula Bruice

Solomons and Frhyle

For reference, free energies of formation:
$$
\begin{array}{c|c} \text{Compound} & \Delta G^{\circ}_{f}\text{ in kJ mol}^{-1} \\\hline \text{2-methylpropene} & 58.1 \\ E\text{-2-butene} & 63.0 \\ Z\text{-2-butene} & 65.9  \\\hline \end{array}
$$

This is a very good question and if popular books give conflicting answers, then it must be reasoned out. Unfortunately, Paula Bruice has given the wrong answer while the other two books have given no explanation for this comparison.
For the answer I assume that you know about hyperconjugation and the various contributing structures it involves.  This gives a basic idea of hyperconjugation:

Source: tutorcircle.com
Note that on hyperconjugation, the other carbon of the alkene gets a negative charge on it.
If the hyperconjugation structures of 2-methylprop-1-ene and trans-but-2-ene are drawn:

you will notice that:

$ \ce{C_1} $ of the former has a negative charge on it
$ \ce{C_2} $ of the latter has a negative charge on it
The latter's methyl group makes the negatve charge less stable due to +I (inductive effect)
The former has no such destablizing effect as the carbon containing the negative charge is just connected to hydrogens
As all the hyperconjugation structure have this effect applying on them, the net effect cant be ignored for the overall stability

Hence the overall stability of  2-methylprop-1-ene > trans-but-2-ene

While I agree with Del Pate's explanation, I also have an additional reason as to why 2-methylpropene is more stable than (E) but-2-ene.
If we attempt the hyperconjugation of 2-methylpropene, we quickly realize that the carbanion obtained thus has equivalent resonance structures (as drawn below). Equivalent resonance structures are degenerate, and delocalise rapidly. I have shown one such hyperconjugation below, but the lower methyl group can just as easily be invoked in hyperconjugation as the upper methyl group.

Thus, this rapid delocalisation via equivalent resonance structures in hyperconjugation is the reason why 2-methylpropene happens to be much more stable than (E) but-2-ene. The curious reader will also draw hyperconjugative structures of (E) but-2-ene, and realize that they are not equivalent to each other.


text:
Effect of inductive effect on boiling point,melting point and dipole movement
In my book the following was written:-

Inductive effect is a permanent effect operating in the ground state of the organic molecules and hence is responsible for higher melting point, boiling point and dipole movement of polar compounds.

How is inductive effect responsible for higher melting point, boiling point and dipole movement of polar compounds?
Why is inductive effect responsible for higher melting point, boiling point and dipole movement of polar compounds only?

Not quite sure about melting and boiling points but things like polarity ,dipole moment and inductive effect are kind of interconnected.Inductive effect is a consequence of electronegativity differences.More inductive effect leads to more dipole moment but that also depends from one case to another.For example each of the $\ce{C-Cl}$ bonds in $\ce{CCl4}$ is highly polar but net dipole moment is 0 due to the fact that Dipole moments are added through vector addition and $\ce{CCl4}$ is tetrahedral.


text:
How do I extract cyanide from apple seeds?
I'm working on a crime story about cyanide poisoning from apple seeds. I just would like to have an idea of what processes and extraction techniques might be involved in getting cyanide from the seeds. The character is supposed to have access to a high school chemistry lab; so I was thinking equipment and tools that can be found in a typical HS lab. I prefer a detailed answer, step by step, if possible. If not then just generalized terminologies would work.

"I'm working on a crime story" is the new "I'm asking for a friend who's trying to murder people."  :)

I have a strange feeling that this question will be used as evidence in an upcoming murder trial...

Plot twist: OP is someone you know.

There are "better" ways, even on a budget and with limited equipment ;)

Sorry, I couldn't ask my chemistry teacher - he will be the victim.

Forget about the apple seeds, they contain about 1 to 4 mg amygdalin per gramm seeds (DOI).
Instead, collect apricot seeds during the right season, the amygdalin content varies though the year and can be as high as 5% of the dry weight of the seed (DOI).
It is probably advantagenous to break the husk with a nut cracker, a plier, etc. and cut the softer inner material to smaller pieces.
Extraction of amygdalin can be performed by

immersing the material in methanol and subsequent ultrasonification
Soxhlet extraction with methanol
reflux extraction in water in the presence of citric acid

A comparison of the extraction methods is given here.
Removal of the solvent in vacuum will yield a crude material with significant amounts of amygdalin. You might want to have a look at this article from the Western Journal of Medicine on the toxicity. Here, an $\mathrm{LD_{50}}$ of 522 mg amygdalin per kg body weight was estimated for the oral application to rats. The online resource of the U.S. National Library of Medicine gives a value of 405 mg/kg.

Further information on the health risk of apricot kernels are provided by of the German Bundesinstitut für Risikobewertung (Federal Institute for Risk Assessment) and the British Committee on Toxicity.
A note in the a German medical journal, Deutsche Ärzteblatt, (PDF) describes a case where boy of four years (110 cm, 18 kg body weight) was given apricot kernels during an alternative cancer treatment. Upon additional treatment with a single dose of 500 mg amygdalin, the kid showed agitation, spasms and the eyes started to roll.

I'll leave it up to your fantasy as a writer on how to apply the poison, but spicing some marzipan with it might help ;)

Weird that the NFPA health rating is 1 if amygdalin is a potential murder toxic. How much do you need to ingest to become severely ill?

@Jori Unless I'm completely off, the blue NFPA ratings are for acute intoxications and health risks upon direct exposure. Release of cyanide from amygdalin takes place in the smaller intestine (Dünndarm). Anyway, I've added some more information to my answer.

There is bitter almonds and "normal" almonds, the same seems to be the case for apricot. Of course only the bitter variants contain amygdalin.

@Georg To my knowledge, this is true for the sweet and bitter variants of the almonds, but not for the apricots (Prunus armeniaca).

@Georg All variations of apricot contains amygdalin

The leaves of many Prunus species contain high concentrations of amygdalin and prunasin, both of which are cyanogenic glucosides capable of doing the "job". There are better common garden plants for poisoning people, but I could see that there is something poetic about apples.

Honestly, you probably wouldn't be able to get this done with just a highschool chem lab, at least not to a high purity (whether or not this is needed is debatable; in fact it could be interesting in the story because it would give a clue as to how it was produced, whereas pure cyanide wouldn't really tell the detective much).
Having said that, apples don't contain the harmful version of cyanide (hydrogen cyanide) in and of themselves. They do however contain amygdalin, which can be metabolized to hydrogen cyanide. A quick search ends up with wikipedia giving a brief overview of how it is isolated:

Amygdalin is extracted from almonds or apricot kernels by boiling in ethanol; on evaporation of the solution and the addition of diethyl ether, amygdalin is precipitated as white minute crystals.

In case your orgo isn't all that great (mine is a little rusty too), what you should read from the above quote is that ethanol is used to oxydize the primary alcohol group to an aldehyde or a carboxylic acid (depending on how long you let it boil for) making it polar and so non-soluble the diethyl ether (an organic solvent) which leads it to precipitating out as a white crystal. Once it precipitates out the easiest way to isolate it is just to use a paper filter, since it should be the only solid present. Suction filtration would be optimal but that's generally not available in high schools. Lastly you could use sublimation to get a really pure final product, but again this isn't normally available in high school labs.
If you're writing a novel I recommend you take the time to read up on it and how it is metabolized within the body. The more you understand the process the more authentic it'll feel, imo. Best of luck.
EDIT: You may want to consider switching to apricot kernels as opposed to apple seeds. The process is effectively the same but it's more realistic, since you'd need a LOT of apple seeds to get a decent amount of cyanide. Also, like I said my orgo is a bit rusty so if someone else could look over this that would be nice.

Ethanol can oxidize another alcohol?


text:
Concentration of hydrochloric acid for electrolysis?
How much hydrochloric acid (maximum) should I use to do electrolysis of water, and how much
amount of voltage should I use to get the maximum of hydrogen and oxygen?

Welcome to chemistry.SE! || I see a few problems in your question. To begin with, you can do this modification: Industrial chemistry deals with minimum, not maximum. I even have a hunch you meant "minimum".

You will not get oxygen if you use HCl, you will get Chlorine gas instead.
You could use sulfuric acid (like your title said before someone edited it) or NaOH as an electrolyte.
0.5 M would be a reasonable concentration.

You need to see the concentration of acid. If it is a concentrated acid, you will get chlorine gas. But if you use diluted acid, you will get oxygen gas.