text:
Does sulfur and phosphorus expand their octet?
There are many compounds in which the stability of a molecule is not governed by the presence of octet configuration in central atom. In most of the cases the central atom is generally sulfur or phosphorus. In these cases either the central atom has 6 electrons, or sometimes the electron count exceeds 8 and 10 or 12 electrons in its valence electron shell.
This makes the concept of octet somewhat incorrect. My question is how can they increase their octet to such an extent?
Could someone explain this in relation with sulfur trioxide?

Whether sulfur or phosphorous actually expand their octet is contested within the chemistry community. Another term for this octet expansion is "hypervalency." You can find many works of research regarding hypervalency. The consensus, according to Wikipedia, is that both can expand their octets, but not to a significant extent. In other words, d-orbital utilization is low at best.
Regarding sulfur trioxide, this article may be of interest.

We find no evidence to support notions of p π –d π back-donation from oxygen to sulfur.

Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: $\ce{C, N, O}$, and $\ce{F}$. And note that for carbon, there are well known exceptions in organic chemistry - carbocations (which have less than an octet of electrons). Carbocations however tend to exist mainly as reactive intermediates. Therefore, a more nuanced statement might be $\ce{C, N, O}$, and $\ce{F}$, when involved in neutral and thermodynamically stable compounds, generally have 8 valence electrons.

this makes the concept of octet somewhat incorrect.

If anyone can think of exceptions, please share || $C_7H_7^+$ , $C_8H_8^{2-}$ are well-known examples =).

Are you sure those are exceptions? I don't see how they could be exceptions.

http://www.chemtube3d.com/gallery/structurepages/C7H7_.html

The examples given above (carbocations, carbanions) are exceptions to the octet rule.  However they also belong to a class of compounds known as reactive intermediates.  It is the strong "desire" of these non-octet compounds to regain their octet that makes them so reactive.  The list of these "offenders" is endless.  If you put enough energy into methane you would generate methyl radicals, they too "violate" the octet rule, but are extremely reactive species.  I think your statement, "These elements can be expected to follow the octet rule throughout most of general and even organic

chemistry is as accurate as most general statements in chemistry. Perhaps modifying it to read, These elements, when involved in neutral, thermodynamically stable compounds, can be expected to follow the octet rule throughout most of general and even organic chemistry, would make it a bit more "correct".


text:
What copper oxide is being created upon electrolysis?
Whenever using copper as anode, I end up with cyan dust that doesn't dissolve in water.

What's that (presumably) oxide? Is it poisonous? Can I dissolve it in something to perform electroplating with it?

It appears to be copper hydroxide or copper carbonate.  Copper oxide tends to be black/grey/brown.  Copper hydroxide is blue and copper carbonate is green.  Copper carbonate forms when the hydroxide is exposed to CO2.
If you would like to dissolve it, try an acidic solution.

I would like to dissolve it in order to perform electroplating with it. Would it work?

I find that copper chloride solutions tend to be the most reactive.  If I were you I'd try dissolving it in HCl.  It will dissolve in acid and should work for electroplating.

I have dissolved it in sulphuric acid since hydrochloric was harder to obtain at the moment. The copper won't stick to most surfaces however. It just forms dust around them.

It should stick to a metallic surface.  Insure that the metal surface is free of corrosion by pickling.

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Can I use harvested plants to gain electric power?
I was just thinking - plants are mostly acid. Even grass is quite acid. Could you create battery cell from squashed plants (after cutting them in pieces)?
How much effective would it be? Could you compare it to solar panels? Would you need to replace electrodes often?

Are you thinking about a battery (i.e. rechargable) or a simple galvanic cell?

Rather a "refillable" battery - when the power is low, you remove old plants and put in a new ones. Imagine a electric device that feeds on grass...

A company in the USA called Voltree Power has gone one step further to harvest electricity generated directly from live trees and other plants!
It seems trees are able to provide somewhere between 20 and a few hundred millivolts as well as some current, derived from a difference in pH between the trees and the soil, maintained by the trees metabolic processes.
The science behind this technology was published in 2008 by researches at MIT.
The company claims to have demonstrated the technology by developing the first wireless sensor network powered exclusively by trees to detect forest fires! They are also developing the technology to monitor humidity and temperatures in the forest using electrical power harvested from the trees themselves and transmit the data using low-power wiremess mesh technology.
Another collaborative group from University of Michigan, MIT & UC Berkeley published their research in 2009 into micro-power electricity scavenged from evaporative flows within the microchannels in trees.

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Information depth in x-ray photoelectron spectroscopy (XPS)
What is 'information depth' in XPS and what does determine it? Also how does it depend on the 'take-off angle'?

Do you understand the basics of XPS?  The Wikipedia article is pretty clear. In summary, a (usually monochromatic) x-ray source irradiates a sample.  The x-ray can be absorbed by electrons in the sample, and can provide enough energy for a core electron to be set free with an energy equal to the incident x-ray energy minus its binding energy. Collecting the electrons and measuring their energy gives a plot of counts vs electron energy.  Lower electron energy indicates an electron coming from deeper in the atomic shell. Given knowledge of atomic binding levels, you can associate specific peaks in the spectrum with elements (with chemical effects on binding energies giving additional information).
So - what is the "information depth"? A slightly odd term, I've generally heard/used 'detection depth'.  Two questions to ponder - (1) how far will x-rays of a few keV penetrate into a solid (if an atom doesn't see any x-rays, it won't emit any electrons), and (2) how far can a ~100eV-1000eV electron travel in a solid before being scattered (and thus losing the useful info).  Not surprisingly, (2) dominates for XPS, and is often called the 'escape depth'.
As for the 'take-off angle' - again, the interesting part is where your electron detector is placed relative to the sample, and how far the electrons have to travel through the sample to head out in the direction of the detector - the 'escape depth' will vary with exit angle (take-off angle) since the amount of material the electron has to go through is a simple trig function of the exit angle.
I hope that helps.


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How to calculate the potential via Hess's law? Is my textbook wrong about this?

Given the two reduction half-reactions for the following fuel cell, determine the standard cell potential $E^\circ_\mathrm{cell}$,
  $$\ce{CH4 (g) + 2 O2 (g) + 2 OH- (aq) -> CO3^2- (aq) + 3 H2O}.$$
  Reduction half-reactions:
  \begin{align}
\ce{CO3^2- (aq) + 7 H2O (l) + 8 e- &-> CH4 (g) + 10 OH- (aq)}&
E^\circ &= \pu{-0.1 V}\\
\ce{O2 (g) + 2 H2O (l) + 4 e- &-> 4 OH- (aq)}&
E^\circ &= \pu{+0.55 V}
\end{align}
  Answers: (a) $\pu{+0.45 V}$ (b) $\pu{-0.45 V}$ (c) $\pu{-0.65 V}$ (d) $\pu{+0.65 V}$

It is my understanding that to apply Hess' Law to this equation the top equation sign would be reversed and the outcome would be $\pu{+0.65 V}$. Often the system produces incorrect answers. I was looking through my textbook to double check this, and there is an instance where it says
$$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode},$$
but I believe that rule does not fit this instance. Could somebody validate that this answer provided is indeed correct or not?
I would like to clarify that my answer was (d) and the system provided that the correct answer was (a).

I sent an email to my teacher to contest the question using an example from the textbook. Given that Martin has provided me with an answer concurrent with the $\pu{+0.45 V}$, then there must be a mistake in the text or I am misinterpreting something. I would really like to understand this.
At first my teacher told me that after flipping the equation you still need to subtract the anode from the cathode. So I responded like this:

I would like to reference the E Cell example from the text. In the text it explains that $$E_\mathrm{cell} = E_\mathrm{anode} - E_\mathrm{cathode}.$$
The example given in the book is identical to the one provided in the exam:
\begin{aligned}
 (1): &&\ce{3 Ag+(aq) + Fe (s)   &-> 3 Ag (s) + Fe^{3+} (aq)}\\
 (2): &&\ce{3 Ag+(aq) + 3 e-     &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\
 (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\
\end{aligned}
By using the same logic that I have applied to the exam question the $E_\mathrm{cell}$ was found to be $\pu{0.84 V}$. This is because the sign on the $E$ value of the $\ce{Fe}$ half reaction was reversed like I did to the top equation (which consequently is the cathode).
As the given equations line up with the primary equation and signs are flipped whenever the equations are flipped, then it is not important to decipher which is the anode and which is the cathode. This is also in accordance with Hess' Law.

That being said, I am confused whether the textbook example was incorrect, the exam question was incorrect, or my understanding is incorrect.

You have to take into account, that the given reactions are only reductions, while you need to have an oxidation, too. So one of the equations has to be reversed.
\begin{aligned}
(A)E^\circ = -0.10~\mathrm{V}&&\ce{CO3^{2-} + 7H2O + 8e- &-> CH4 + 10 {}^{-}OH}  \\
(B)E^\circ = +0.55~\mathrm{V}&&\ce{O2 + 2H2O + 4e- &-> 4 {}^{-}OH}  \\\hline
\color{\red}{-1}\cdot(A) + 2\cdot(B)
   &&\ce{CH4 + 2O2 + 10{}^{-}OH + 4H2O&-> CO3^{2-} + 7H2O + 4 {}^{-}OH}\\
\implies
   &&\ce{CH4 + 2O2 + 2{}^{-}OH &-> CO3^{2-} + 3H2O}
\end{aligned}
Equation $(A)$ gives you the reduction of the carbonate ions, hence the oxidation of methane. Equation $(B)$ is the reduction of oxygen. You have a galvanic cell of the form $\ce{CH4|CO3^{2-}||O2|{}^{-}OH}$, hence the standard potential difference is
\begin{aligned}
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &= E^\circ(\ce{O2|{}^{-}OH}) -E^\circ(\ce{CH4|CO3^{2-}})\\
               &= E^\circ(\ce{O2|{}^{-}OH})
                  -\left[\color{\red}{-1}\cdot E^\circ(\ce{CO3^{2-}|CH4})\right]\\
               &= +0.55~\mathrm{V}
                  -\left[\color{\red}{-1}\cdot (-0.10~\mathrm{V})\right]\\
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &= +0.45~\mathrm{V}
\end{aligned}

Just to address the second edit of the OP and to clarify:
\begin{aligned}
 (1): &&\ce{3 Ag+(aq) + Fe (s)   &-> 3 Ag (s) + Fe^{3+} (aq)}\\
 (2): &&\ce{3 Ag+(aq) + 3 e-     &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\
 (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\
\end{aligned}
The sign of the second equation has to be flipped and then subsequently subtracted from the first equation, hence the potential that should be found is $\Delta E^\circ = 0.74~\mathrm{V}$. Therefore, I believe your textbook is wrong.

Your intuition is correct. The answer should be +0.65V as in accordance to Hess' law.  However, your textbook contains an error on the Ecell formula as it should be $$E_\mathrm{cell} = E_\mathrm{cathode} - E_\mathrm{anode}$$
Note that the values used in this formula are the reduction potential values, hence the reason why Martin's answer leads to the incorrect +0.45V answer as it uses the oxidation potential for the anode instead. This formula also confirms our answer to be correct, with the reduction of O2 being the cathode and the other being the anode.
$$E_\mathrm{cell} = +0.55~\mathrm V - (-0.1~\mathrm V)= +0.65~\mathrm V$$
The same applies to the textbook example. The answer derived from Hess' law is the same as the answer derived from the formula $$E_\mathrm{cell} = +0.80~\mathrm V - (-0.04~\mathrm V)= +0.84~\mathrm V$$

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Block on the Periodic Table?
What are the blocks of the periodic table? What is the purpose of naming the elements per block group? What do they signify?

Below is a picture of the 4 blocks in the Periodic Table.  The elements in a block all use the same orbital to hold their valence or outer shell electrons.  For example, all of the elements in block 1 have their valence electrons in an $\ce{s}$ orbital, those in block 2 have their valence electrons in a $\ce{p}$ orbital, block 3 the $\ce{d}$ orbitals and block 4 the $\ce{f}$ orbitals.

Thanks! Now I have a better understanding of the periodic table!


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1,6-dimethylcyclohexene and 2,3-dimethylcyclohexene
Is there a difference between these two molecules?
If so how does one tell? What is the correct nomenclature for cycloalkenes? I know that the double bond is assumed to be between C1 and C2 but why can't we number a cyclohexene to get 2,3? Is it because a lower sum is not preferred but simply lower individual numbers if possible?

The IUPAC rule is to have the first branch to have the lowest number, even if it gives you 1,6-dimethylcyclohexene.

IUPAC rules for alkenes gives the two carbon atoms in the alkene functional group the lowest numbers. So, in cyclohexene, carbon atoms 1 and 2 are always in the akene.

@BenNorris Yes, but the 1,6 refers to the dimethyl groups.

So 2,3-dimethylcyclohexene is wrong?

@BenNorris If you want to assign numbers to all the items, you would have 1,6-dimethylcyclohex-1-ene.

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Note that Rule e takes precedence over Rule f.
For example, a preferred IUPAC name is 3-bromocyclohex-1-ene (not 1-bromocyclohex-2-ene): (This example is given in the Blue Book.)

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

Therefore, the structure given in the question is named as 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.


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Why is the water reduction of oxygen favored in acidic conditions?
The standard reduction potential of diatomic oxygen in acidic conditions is +1.23 volts. However, the standard reduction potential of diatomic oxygen in basic conditions is only +0.40 volts.

Why is the reduction of diatomic oxygen by acidic media more favored than the reduction of oxygen by water that is basic?
I can think of a few reasons:
1) We are likely going to be forming negative two oxidation state oxygen anions when we oxidize diatomic oxygen. Oxygen adopts an oxidation state of negative two except in peroxides, which aren't very stable, and I don't think the oxygen radical will be very stable in solution.
Can someone provide a less fluffy reason that we're going to be forming negative two oxidation state oxygen atoms? I'd imagine that the electronegativity and electron affinity of oxygen are two reasons it might favor the negative two oxidation state over the negative one oxidation state in solution. I.e. why doesn't this occur to a significant extent?
$\ce{O2 + 2e^- -> 2O^-}$
Or this?
$\ce{O2 + 2e^- -> HOOH}$
2) From the above we can deduce that water is the product; water is pretty stable in either acidic or basic medium - i.e. it won't react to a large extent with either hydronium or hydroxide ion. Plus the oxygen in water has a -2 oxidation state. So if we look at our skeleton half reaction:
$\ce{O2 + 4e^- -> 2H2O}$
We realize that we need a source of protons. And the proton source in acidic medium is likely hydronium ion.
However, in basic medium, the proton source is water. I'd imagine that water is a poorer proton source than hydronium ion.
So would this account for the difference in standard reduction potential of oxygen in media that vary in pH?
I understand that voltage is an extension of K through the Nernst equation; is what I'm seeing here basically that the protonation of a base ($\ce{O^2-}$) is more favored in acidic rather than basic media? Could there also be kinetic factors? In other words, why is this statement true?

Also:
1) What's the mechanism for the reduction of diatomic oxygen?
2) Why is the reduction of protons at high pH less favored than the reduction of protons at low pH? My first guess: you don't find as many protons in basic solutions. But then, protons don't exist by themselves anyway. So what can the proton source be? I'm guessing that in basic pH, there doesn't exist much hydronium ion. So the only two sources of protons are water and hydroxide anion. Heterolytically cleaving the hydroxide anion to form the highly basic oxide anion doesn't seem thermodynamically favorable. Cleaving water to form hydroxide anions ... are we running into a Le Chatlier's Principle effect?
ETA: Or should I be considering proton activities? Is the activity of the proton less in basic solution than in acidic solution?

1+2) standard acidic conditions: pH=0, standard basic conditions: pH=14, 1.23-0.40=0.83, 0.059*14=0.826, suspicious. Applying Nernst equation correctly may give an answer 1) It depends on the material of the electrode, but many metals and metal oxides can (reversibly) adsorb/lose oxygen on the surface. For details you'll have to study surface chemistry, which is relatively unexplored. 2)Nernst equation properly applied and learning what's behind it should give the answer

Thank you. I obtained the numbers from some presentation I found on Google. The exact numbers might be off but I'm interested in why they might differ in the way they do. What exactly about the Nernst equation should I be studying?

Derivation process. Basically, Nernst equation is an equation for equilibrium constant including electron in external electric chain.

May I know which book that information is from?

If you look at the half equations, you will see when you add them together, you get 1.23 V, 1.22 V, and 1.23 V.
As the mixture becomes more basic, it is more difficult for the oxygen to grab $H^+$ ions and easier for $H_2$ to lose electrons to generate $H^+$ ions.
The mechanism probably involves these combinations: $O_2, H_2O_2, H^+, HO_2^-, H_3O_2^-, H_2O, and\ HO^-$.

Which half reactions are you referring to?

Also why would H+ be formed in basic solution? That seems counterproductive.

@Dissenter 1. You have 3 pairs of half reactions in the table. 2. $H^+$ would immediately combine with $OH^-$ in basic solutions.

How does that affect the voltage?


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Is storing diluted (~1M) sulfuric acid, HCl, NaOH, etc. in plastic PET bottles a bad idea?
Do I really need glass bottles if the acids and NaOH is so diluted? I thought plastics dont really react with acids. And I already have them in plastic bottles (probably a different kind).
Thanks

Concentrated $HCl$ is roughly 10 M, so its pH should be -1 (in practice it is higher as at such concentration the dissociation of HCl is suppressed). Roughly same is of 1M NaOH. I would not store this liquids in PET. However, storing this liquids in polypropylene or polyethylene is OK.

Actually, it is a good idea to store $NaOH$ in polyethylene/polypropylene/teflon. Strong bases slowly eat glass away. As my buddy with great practice said, even concentrated ammonia solutions eat glass.

In any case, do not use drink or food containers to store chemicals.

Here you can find thorough and clear instructions about using plastic bottles for acids and bases:
http://www.corning.com/lifesciences/emea/en/technical_resources/product_guid/bottles/material.aspx
A common rule is to not store strong acids and bases (like sulfuric and hydrochloric acids, and NaOH) in plastic bottles if you are not sure about the type of plastics.
Polystyrene bottles are OK for strong alkali.
Here is a PDF document on this topic:
http://csmedia2.corning.com/lifesciences/media/pdf/PSUG_Storage_Bottles_CLS_GL_025.pdf

The best bet for storage of acids and bases is ACIDS: glass, teflon, and polyethylene (LDPE or HDPE); BASES: Teflon or polyethylene


text:
Interpreting Infrared Spectroscopy (IR) Spectra
I have already attempted to interpret some of the peaks. Please tell me if they are correct/incorrect and anything that I have missed.

i think you have made those interpretations very cautiously and will full effort because the trend followed by them is clearly visible . i mean as the position increases the intensity decreases i.e. they are inversely proportional to each other i cant detect any wrong or miscalleneous point in your graph

In your first spectrum, I would be skeptical about the alkynic $\ce{#C-H}$ bond you've identified at approx. $3400\mathrm{cm^{-1}}$, as there is no corresponding alkynic $\ce{C#C}$ signal (typically roughly in the $2100-2300\mathrm{cm^{-1}}$ region).

In your first spectrum, the band at 3517 $\mathrm{cm^{-1}}$ indeed fits best to the $\ce{N-H}$ stretch of a primary amine (source). However, I would not assign the shoulder of this band to a $\ce{#C-H}$ stretch. Such a band would be sharper and perhaps more intense, and as Greg E. already pointed out in his comment to your question, there is also no corresponding $\ce{C#C}$ stretch band present (reference). The absence of bands in the range between 2500-2000 $\mathrm{cm^{-1}}$ allows you to rule out some other functional groups with double or triple bonds, like nitriles, thiocyanates, isocyanates, carbodiimides or azides. The band at 1715 $\mathrm{cm^{-1}}$ is a typical $\ce{C=O}$ stretch, and while it could be a ketone, the presence of the broad band at 2932 $\mathrm{cm^{-1}}$ (which could actually be a double band) makes it likely to be an aldehyde, since the double band can then be assigned to the $\ce{=C-H}$ stretch of the CHO group (it could also include other $\ce{C-H}$ stretches due to its width).
Your second spectrum appears a bit more complex. The broad, intense band at 3298 $\mathrm{cm^{-1}}$ in combination with the band at approximately 1680 $\mathrm{cm^{-1}}$ could be assigned to an amide, but I would expect a stronger carbonyl band in this case. Nevertheless, the intensity and width of the 3298 $\mathrm{cm^{-1}}$ band suggest the presence of a functional group capable of H-bonding. The bands between 2900-2800 $\mathrm{cm^{-1}}$ could be assigned to alkyl $\ce{C-H}$ stretches (source). For the compound to be aromatic, there should be aromatic $\ce{C-H}$ stretch bands present, slightly above 3000 $\mathrm{cm^{-1}}$ (source).

My interpretation of the first spectrum:

I would say that the 3517 band belongs to OH group. This is a typical shape OH group. Amines look more like two bands with a shoulder. However secondary amine is also a possibility. Other bands of OH should be the broad band between 1400 and 1300 from which the sharp bands arise. Than probably 1166, which seems to be tertiary OH group.
≡C−H stretch is not right. Is is an overtone of the 1715 vibration.
The four bands around 2932 are CH3 and CH2 stretches.
1461 is an antisymetric deformation of CH3, 1411 is scisorring of CH2 and 1359 is an umbrella of CH3.
The rest is a fingerprint region.
I assume this is an aliphatic ester molecule.

The second spectrum I am not sure, however:

An aromatic vibrations are unlikely, there should be vibrations just over 3000, which are missing and the 2000 region also looks different.
There are aliphatic vibration just like in the first spectrum.
3298 is not typical for amine. However I cannot rule it out. It might be an OH-group as well.

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text:
why is potassium the most free element in most of plants
Potassium is known to be the most freely available mobile element.
I am wondering why that is the case?  Is the answer hidden in its atomic structure or its monovalent cationic existance?

There are many functions related to potassium some of which are shown below
Potassium regulates the opening and closing of the stomata by a potassium ion pump. Since stomata are important in water regulation, potassium reduces water loss from the leaves and increases drought tolerance. Potassium deficiency may cause necrosis or interveinal chlorosis. K+ is highly mobile and can aid in balancing the anion charges within the plant. Potassium helps in fruit colouration, shape and also increases its brix. Hence, quality fruits are produced in Potassium rich soils. It also has high solubility in water and leaches out of rocky or sandy soils. This water solubility can result in potassium deficiency. Potassium serves as an activator of enzymes used in photosynthesis and respiration[4] Potassium is used to build cellulose and aids in photosynthesis by the formation of a chlorophyll precursor. Potassium deficiency may result in higher risk of pathogens, wilting, chlorosis, brown spotting, and higher chances of damage from frost and heat.
and the answer to your question is the small size of potassium when compared to all the other cations in plants
another reason for this is its association with the chloride ion .
chloride and potassium ions will be playing follow the master game in which if any cation takes chloride ion near it then potassium will also follow chlorine and the same is the case with potassium whenever any other anion attracts it this makes potassium highly mobile.  source wikipedia
hope you got your answer. if any doubts left feel free to comment

As on the other answer, please cite your source.  Consider this a formal warning.

@jonsca i have edited my answer with the reference of source


text:
in which way calcium promotes cell division
I have read in bio chemistry that calcium plays an important role in the cell division but the in depth information is'nt available in the textbook and I cannot memorize any fact without a supporting fact I just want to know what is its role in the process of cell division

see its simple all you need to know is that calcium is responsible for the formation of middle lamellum which is a key structure in the cell wall formation if only karyokinesis occurs and cytokinesis fails to happen then the condition of resultant cell would become multinucleated which most plants do not have


text:
Heating of Enamel coated cooking utensils and health hazards
I've tried to research about the issue of health concerns about enamel coating of cooking utensils (i.e. dutch ovens) and failed. I should prepare a small report about the health concerns if they exist.
Could you please help to answer that question? Any reliable sources would be greatly appreciated, especially scientific articles.
Thanks.

Out of curiosity, what do you mean by "failed?"

@chipbuster I did not find any source about enamel-related health hazards

I unfortunately don't have time to dig through these articles in great detail, but they should help get you started. Feel free to comment with more questions.
The overall impression I got from my brief search is that there is definitely a potential health hazard depending on how the coloring was made, but in most modern countries, there are  regulations in place to prevent those kinds of problems. Enough about what I think though, let's get into the references.
According to Wikipedia, the coating on most cookware is vitreous enamel, which is made by heating a mixture of glass powder and pigment to a very high temperature and letting it melt onto a surface (much like glazing in pottery).
The glass itself is pretty much harmless, so the main danger comes from the added pigments. Here, we're stuck, because it really does depend on what was used to make the pigment. If it's a zinc pigment, you're probably fine. If it was cadmium based, you might be kinda screwed.
I've found references to several articles which describe experiments trying to find out what the leach rate of metals from the glass into a solution is (usually by boiling in vinegar), but I can't hunt down the fulltexts right now, aside from the one open-access article from the NIH. Here's what I have:
Full text:Lead, Cadmium and Cobalt (Pb, Cd, and Co) Leaching of Glass-Clay Containers by pH Effect of Food
While this study was on clay containers, it should theoretically apply to glass-on-metal as well. Keep in mind that the containers seem to have been acquired from local craftsmen in Mexico, so the results won't necessarily apply to stuff you find on the supermarket shelves.
Abstract: Hot leaching of ceramic and enameled cookware: collaborative study.
This is what you're looking for, but I can't find the full article, and the authors seem to have been more interested in establishing a method than finding the health effects. If you can find the full paper and cross-reference their results with acceptable limits of exposure (maybe from the EPA), that'd probably be your best bet.
The second link also has other articles on similar problems. If you can't find your information from these two articles, maybe you can find it by following the links on PubMed. Good luck!

great answer and the most adverse effect of glass SiO2 is silicon atoms can replace carbon atoms in many organic compounds which our body has so that must also be ranked at the top while discussing about the health hazard of enamel coatings on kitchen ware

That's true, but the binding energy of SiO2 is high enough that the danger of the pigments is probably far higher in this case. To put it in perspective, at the temperature where glass gets the tiniest bit soft (gains some atomic mobility), most food will have already burnt to a crisp.

yeah your point seems to be correct


text:
Kinetic validity; redox reactions

What's the kinetic validity of the following way of balancing the above redox reaction?
I've done some searching of the literature; I've never come across oxide anions being mentioned in redox reaction mechanisms. Also, I read something about how two-electron reduction is impossible in a single step - something about spin pairing.
I'm fairly sure the following method is a good thermodynamic method of thinking about the reaction, but what about its kinetic validity?
I'm thinking that the below method isn't kinetically valid, but neither was it meant to be kinetically valid, and neither are the traditional methods of balancing redox reactions kinetically valid. Traditional methods call for balancing "O" and "H." But we aren't dealing with 0 oxidation state oxygens and hydrogens, are we?
Here is the thought process that my professor would take in solving this problem:
1) Skeleton reaction. Note that sulfur is being reduced from a +6 oxidation state to a +4 oxidation state. This suggests that the solid zinc loses two electrons.
$\ce{Zn + HSO4^- ->SO_2 +Zn^2+}$
2) Note that two negative two oxidation state oxygens are "disappearing" from left to right. In other words, what happened to the oxygens in the hydrogen sulfate ion?
$\ce{Zn + HSO4^- ->SO_2 +Zn^2+}$
$\ce{4~ negative ~two~ oxidation~ state ~oxygens -> 2 ~negative ~2~ oxidation~ state~ oxygens }$
3) The special thing about these oxygens is that they're negative two oxidation state oxygens. This suggests that these oxygens are the oxide anion. The oxide anion is highly basic in solution. Therefore, the oxide anions must have reacted with some system component, which caused them to "disappear." This would be the strongest acid in the system.
$\ce{HSO4^- ->SO_2 + 2O^2- (unstable)}$
$\ce{2O^2- + 4H3O^+ ->6H2O}$
4) Also note that a hydrogen proton disappears from left to right in the skeleton equation. Because the proton is highly acidic, it is attacked by the strongest base in the solution - water (since we're in an acidic solution).
$\ce{HSO4^- ->SO_2 + 2O^2- + H^+}$
$\ce{H^+ + H2O -> H3O^+}$
5) Final balanced equation:
$\ce{Zn + HSO4^- + 4H3O+ + H2O ->SO_2 +Zn^2+ +6H2O + H3O+}$

whats left then you've got your answer but i too think that kinetic validation should be considered

So it's not kinetically valid?

cant say any thing about that i am also confused at this situation

I highly doubt it's kinetically valid and even if it were so I doubt one could prove it. Studying kinetics and verifying mechanisms is a time-consuming task.

i too think the same it is really a time consuming process

When you balance the equation for a redox reaction, the process of balancing it or the balanced equation itself say nothing about kinetics.  That said, redox reactions, as the name implies, deal with electron transfer.  Unless you are dealing with unusual molecules (bulky ligands for example) or unusual conditions (solids, glasses, very low temperature, etc.) electron transfer reactions are close to diffusion controlled, so kinetics is usually not an issue and is often glossed over in such reactions.

Are you saying that electrons diffuse around?

The term "diffusion controlled" describes a process that happens so quickly (electron transfer in this case) that it is only limited by the rate at which the electron donor can diffuse through the solvent and encounter the electron acceptor.  The rate constant for such a process is usually quite high.

Isn't electron transfer followed by the transfer of other elements? Is that the whole idea behind redox? Electrons move ... but so can/will other elements?

Sure, but the electron transfer is the key step.  Once this happens the die is cast.


text:
Why does the Wurtz reaction form such an unstable compound with torsional strain?
What would be the product formed when 1-bromo-3-chlorocyclobutane with two equivalents of metallic sodium in ether?
I know that there would be a free radical formation and then it will form a compound whose bond line structure would look like a square with a diagonal.
But why is such an unstable compound being formed as there is a lot of torsional strain?

In the Wurtz coupling reaction, an alkyl halide first reacts with a metal in a one-electron transfer step to produce an alkyl radical which quickly reacts in a second electron transfer with the metal to form a carbanion.  In the final step of the reaction, the carbanion displaces a halogen from a carbon-halogen bond by a $\ce{S_{N}2}$ process to form a carbon-carbon bond.  In the case of 1-bromo-3-chlorocyclobutane two reaction pathways are possible, the carbanion could react intramolecularly to form bicyclobutane or intermolecularly to from a cyclobutane dimer.  Considering the equation $$\ce{\Delta G = \Delta H - T \Delta S}$$ we would say that even though the $\ce{\Delta H}$ term is less favorable for bicyclobutane formation due to the increased ring strain, the change in entropy required to bring two separate molecules together to form the cyclobutane dimer overwhelms the enthalpic term and drives the reaction to proceed by the entropically more favorable intramolecular process.
Look at the cyclobutyl carbanion drawn in the figure below.  The carbanion is ideally set up for an intramolecular $\ce{S_{N}2}$ reaction (backside displacement) on the carbon containing the chlorine diagonally situated across from it.  Entropically, this is extremely favorable.
In summary, because 1) the carbanion and the chloride to be displaced are in the same molecule, 2) because they are so well positioned for the $\ce{S_{N}2}$ displacement reaction to occur and 3) because entropy strongly favors the unimolecular process (bicyclobutane formation) over the bimolecular process (cyclobutane dimer formation), bicyclobutane formation readily occurs.

Considering the torsional strain; the maximum relief will be in the bottom most molecule[Note that there may be other possible products from various combinations]; but from top to bottom as you can see, you require more and more molecules and thus there will be a huge loss of entropy; but as$$\Delta G=\Delta H-T\Delta S$$
;If the tempertaure is not low, so as to reduce effect of $\Delta S$, we can thus expect without a loss of entropy and by formation of an intramolecular product; it is favorable.Note that generally this is the case with intramolecular reaction favourability
Also what you mention "I know that there would be a free radical formation and ..."; this is not true, or I must say partially true, because wurtz reaction can be explained both on the basis of radical and bimolecular elimination mechanisms.


text:
Which alpha Hydrogens are more acidic?
Why is the alpha hydrogen of a acetoacetic ester less acidic than the alpha hydrogen of 1,3 diketone ? I was under the impression that acetoacetic ester has more resonance structures so it more acidic. Please explain.

I was under the impression that acetoacetic ester has more resonance structures so it more acidic.

That's actually wrong. Deprotonated acetoacetic ester has structures where electrons are donated from ester oxygen into conjugated system. This, while stabilizing system as a whole, lessens effect of electron donation from deprotonated carbon. Think about it this way: carbonyl oxygen wants electrons. While in acetylacetone electrons may be harvested from deprotonated carbon only and in acetylacetic ester they may be harvested from both deprotonated carbon and ester oxygen, stabilization effect in the first case is higher.
It is well known, that stabilization effect on acidity of nearby $CH$ grows in row
$COO^- < -CO(NH_2) < -COOR < -COR < -CHO < -NO_2$
The more electron deficient the conjugated system is, the more stabilized carbanion becomes, thus the less resonance structures with transfer of the charge onto carbonyl oxygen the group has, the more it stabilize carbanion. Thus, $\alpha-CH$-acidity in carbonyl compounds correlates with electrophilic activity.
The reason is that negative charge from deprotonated carbon must be delocalized as much as possible, which is especially easy with strongly electron-accepting groups. In case the compound already has an electron pair conjugated with carbonyl group,  this is more problematic.

Here is a picture of the basic reaction we are talking about.

We can see that we wind up placing a negative charge alpha to the carbonyl carbon.  When we examine how substituents in various positions will affect the acidity of protons alpha to a carbonyl group, we must look at how both the inductive and resonance effects exerted by the substituent will stabilize or destabilize a nearby negative charge.

In the above chart we see that by attaching a $\ce{CF3}$ to a carbonyl we make the alpha-hydrogens very acidic. This is because the $\ce{CF3}$ group is inductively extremely electron withdrawing and this effect acts to stabilize the negative charge alpha to the carbonyl.  A methyl group is electron donating both inductively ($\ce{sp^2 - sp^3}$ bond) and via resonance (hyperconjugation), so the alpha-protons in this case become less acidic.  Attaching an $\ce{OR}$ group to the carbonyl will stabilize the negative charge inductively since the oxygen is electron withdrawing, but we can now also draw resonance structures involving the $\ce{OR}$ and carbonyl groups that are electron donating to the carbonyl.

The resonance effect is larger and outweighs the inductive effect.  The net result of donating electrons and moving them closer to the negative charge destabilizes the adjacent negative charge.  Hence the ester is less acidic than the previous compounds.  Applying the same reasoning to an amino group leads us to conclude that here the stabilizing inductive effect is even further decreased and the destabilizing resonance effect increased, both because nitrogen is less electronegative than oxygen.  These same arguments can be applied to simple carbonyl compounds as well as the 1,3-dicarbonyl compounds used here as examples.

When you remove the α-hydrogen, you get the following structures:

I believe the carbonyl of the ester doesn't contribute (as much) to the stabilization since it already has some stabilization with the ester oxygen. Also the dicarbonyl tends to be flat due to Van der Waals forces. The ester group has the ability to rotate, which would inhibit stabilization.

the carbonyl of the ester doesn't contribute  | It does. Malonic ester synthesis is a very common tool to introduce acetic acid fragment, and malonic ester is quite easily deprotonated exactly because of ester groups. In fact, even deprotonated carboxyl group stabilizes anion a little.

@permeakra It is not an absolute white and black situation. Yes there is some contribution, but it is not as much as with a ketone. If you have a malonic ester with an equal amount of diketone, what would be the most prevalent product?

depends on the other substrate. It is true that 1,3-ketones are usually more acidic, but it also means that their carbanions are less nucleophilic and less reactive.


text:
Identification of unknowns with varying solubilities and states of matter

White solid A water insoluble cleaved hot to solid gas B and C. B is insoluble in water and in $\ce{HCl}$ but soluble to $\ce{HNO3}$ the react gives solution D and E colorless gas which becomes red on contact with air. The solution D gave white precipitate F with $\ce{HCl}$. The gas C is reacted by heating with magnesium G yielding a white solid which reacts with water yielding a white solid and the gas J characteristic suffocating odor. J is a basic , H is soluble in dilute $\ce{H2SO4}$. The gas J oxidized to C while the gas under specific conditions ($\ce{Pt}$ catalyst and $800^\circ\mathrm{C}$) is oxidized to the gas E.

My attempt:
C: $\ce{N2}$,
G: $\ce{Mg3N2}$,
H: $\ce{Mg(OH)2}$,
J: $\ce{NH3}$,
E: $\ce{NO}$
\begin{aligned}\ce{
N2 + 3Mg &-> Mg3N2\\
 Mg3N2 + 6H2O &-> 3Mg(OH)2 + 2NH3\\
 Mg(OH)2 + H2SO4 &<=> 2 H2O + MgSO4\\
 4 NH3 + 5 O2 &->[\ce{Pt~$800^\circ\mathrm{C}$}] 6 H2O + 4 NO\\
 4 NH3 + 3 O2 &-> 6 H2O + 2 N2\\
}\end{aligned}
I cannot really find anything else.

The solutions which you have already found look good so far. For solid A, we need to find a compound which thermally decomposes into $\ce{N2}$ and an element which is only soluble in oxidizing acids like $\ce{HNO3}$, and whose chloride is insoluble in water or acids. Silver azide could possibly be a good fit, as it can be cleaved into $\ce{Ag}$ and $\ce{N2}$ by heating, and silver chloride has a very low solubility in water. Care would have to be taken with the temperature when actually performing the decomposition of $\ce{AgN3}$, as it is explosive.
$$\ce{2AgN3 -> 3N2\uparrow + 2Ag}$$
Dissolving silver (solid B) in nitric acid gives a silver nitrate solution D and $\ce{NO}$ as colorless gas E.
$$\ce{3Ag + 4HNO3 -> 3AgNO3 + 2H2O + NO\uparrow}$$
$\ce{NO}$ turns red on contact with air when it is oxidized to nitrogen dioxide by atmospheric oxygen.
$$\ce{2NO + O2 <=> 2NO2}$$
From solution D, insoluble $\ce{AgCl}$ (F) precipitates when $\ce{HCl}$ is added.
$$\ce{Ag+ + Cl- -> AgCl\downarrow}$$


text:
Final amount of Hg present after $\beta$ decay
I have been trying to solve an exercise on nuclear chemistry, but I was not sure whether I should ask this on physics or chemistry but I think that if chemistry had an exercise based on this then chemistry experts might be having the way to overcome this problem. So here is what my nuclear chemistry question says:

One gram atom of $ _{79}^{198}\mathrm {Au} $ (having a half life of 65
  hours) decays by $\beta$-emission to produce stable nuclide of Hg.
  How much Hg will be present after 260 hours?

The thing which is making me confused is $\beta$ decay. I think I have a formula for this question but that $\beta$ thing is creating a problem. I would like to make that equation public. Here it is:
$$N_t=N_0\cdot\exp(-\lambda\cdot t)$$
where,

$\lambda$ is the radioactive decay constant
$N_0$ is initial number of nuclides present
$N_t$ is the co. of parent nuclides present at time $t$

Clarify, what part of the exercise you have problems with

I had already mentioned that what does question mean by $\beta$ decay. Means I has $N_0$, $t_{\frac{1}{2}}$ and $N_t$. With all this I can even find $\lambda$. But does $\beta$ or any other kind of particle gonna make difference by emission or they are having no use in the question.

Absolutely no. This is decay, not chain reaction or breeder reactor. For the purpose of the question you can ignore it completely.

$$\ce{^198Au -> ^198Hg + e^{-}  ~~~~~~~~ \beta~ decay}$$
260 hours corresponds to 4 half-lifes (4 x 65 = 260).  After each half-life, one-half of the $\ce{^198Au}$ will have been transformed into $\ce{^198Hg}$.  After 4 half-lifes ((1/2)^4 = 1/16) 1/16 of the gold will remain and 15/16 of the gold will have been transformed into mercury. Therefor, after 260 hours 15/16 gram atom of mercury will be present.  15/16 gram atom of mercury corresponds to (15/16 x 198 = 185.63 gm) 185.63 gm of mercury.
Using your equation we (of course) get the same result.
$$\ce{N_t=N_o * e^{-kt}~~(I)}$$ where k is the rate constant and t the elapsed time.  For radioactive decay (a first order reaction)
$$\ce{k=\frac{ln(2)}{t_{1/2}}~~(2)}$$ where $\ce{t_{1/2}}$ is the decay half-life.  Substituting into equation (I) we find
$$\ce{\frac{N_t}{N_o}= e^{-\frac{ln(2)}{t_{1/2}}*4}=1/16}$$

This is what I also got but in my book the answer is given as $\frac{15}{16}$. This is quite confusing.

The question asks how much Hg will be formed and the correct answer is 15/16.  The "equation" shows how much Au remains $\ce{(N_t/N_o)}$ which is 1/16.

Oh gosh! I didn't paid attention on that. Ok alright. Thank You.

Your welcome, glad to help.


text:
How do I find a concentration from a sample tested on a spectrophotometer?
I have an assignment to calculate the concentration from a sample. The assignment is:

You have 100 µL of RNA. You take 10 µL from the sample and put it in 990 µL of water in the cuvette. The OD is 0.05. Calculate the concentration of RNA in the sample. Calculate how many µL you have to take to make a solution of 100 ng.

Thus far I have succeeded to calculate the concentration from the OD. It is 2 µg/µL. I used the formula that I found on the internet :
$$OD\times \text{Standard coeff.}(\pu{40 µg/ml})\times \text{Sample dilution}$$
but I cannot figure out how from that I can calculate the solution with 100 ng, because there is not mentioned the final volume for 100 ng, or how to calculate how much I have to take µL from the cuvette. I do not understand where to begin from.
I tried with the formula $C_1V_1=C_2V_2$, but I could not figure that out.

The question being asked is, how many uL of your solution do you need to withdraw in order for that volume to contain 100 ng of the RNA.  If there are 2 ug/uL in the RNA solution, that is equivalent to 2000 ngm/uL.  So if you want 100 ng of RNA, you will need to withdraw
$$\ce{\frac{2000 ngm}{1 uL}~=~ \frac{100 ngm}{x ~uL}}$$
solving for x yields an answer of 1/20 uL or 0.05 uL.  0.05 uL of the RNA solution will contain 100 ng of the RNA.

Are you sure? I think its to small, you cant really find a  pipette on that scale of qL.
I tried to cacluate it myself this way:
I did this : 2uq/uL* 100ng, and got 20uL. Do you think that is right and if its wrong, then why? I am really confused with all these concetration problems.

No, that is wrong for two reasons.  First, you cannot multiply 2 ug by 100 ng.  You must use the same units (ug or ng) for both.  Don't let the units confuse you, 2 ugm = 2000ngm.  Second, you must set up the ratio like I did.  When you do and solve for "x", you find that the answer is given by 100 ng divided by (2000 ng/1 uL), not multiplied as you did.

You make at least one mistake in your original calculation: the dilution factor should be $\times 100$ not $\times 1000$, which importantly reduces the concentration in the original sample to $\pu{0.2 \mu g / \mu L}$  or $\pu{200 ng / \mu L}$ RNA and makes it rather trivial to see what volume is required to obtain $\pu{100 ng}$, clearly this must be $\frac12 \pu{\mu L}$, which is reasonable (as opposed to $\pu{0.05 \mu L}$ which isn't, as noted in comments, because pipetting such a small amount is very difficult). When you obtain an unreasonable result it should make you suspect that you made a mistake somewhere.
Another point is that you can conceptualize what volume of stock contains enough of the substance (encloses the mass desired), such that if the solution is more dilute you need more of the solution, that is $V \propto c^{-1}$. If you want a sample with $\pu{100 ng}$ of RNA you just need an aliquot of the stock such that $c_{stock} \times V_\text{aliquot} = \pu{100 ng}$. This should of course provide the same result as the equation written in the other answer.


text:
Find ratio of number of atoms of $A$ and $B$
However I am thinking of this question to be damn easy but still I'll not say as I am not getting it,

There are two radio nuclei $A$ and $B$. $A$ is $\alpha$ emitter and
  $B$ is $\beta$ emitter and their decay constant are in ratio $1:2$.
  What should be the number of atoms of $A$ and $B$ at time $t=0$, so
  that probability of getting $\alpha$ and $\beta$ particles are same at
  $t=0$.

So here how can the decay constant be dependent on number of atoms of $A$ and $B$ so that $\alpha$ and $\beta$ particle are in same ratio? I really need a right approach for this one.

There are two radio nuclei A and B. A is α emitter and B is β
  emitter and their decay constant are in ratio 1:2

So B decays twice as fast as A.  If we want the actual number of decay events from A to equal the actual number of decay events from B at t=0, then we must have twice as much A present as B at t=0.
We are given that $\ce{k_{B} = 2k_{A}}$. For the decay events at T=0 to be equal for A and B, then $$\ce{[B]_{o} * k_{B}~=~[A]_{o} * k_{A}}$$
$$\ce{[B]_{o} * 2k_{A}~=~[A]_{o} * k_{A}}$$
or $$\ce{\frac{2}{1}~=~\frac{[A]_{o}}{[B]_{o}}}$$
At t=0, the concentration of A must be twice the concentration of B in order for equal decay events to occur from A and B


text:
Can nuclear fission or fusion occur together with bond formation in a reaction?
Is it possible for a nuclear reaction to occur simultaneously with a chemical reaction involving bond formation?
Can anyone suggest an example of this, please?

The question is badly worded- there are three general categories of nuclear reactions- fusion, fission and decay/recombination.

I can't think of any examples of a nuclear reaction leading to a chemical reaction with bond formation.  But you got me curious and I started wondering if the reverse process, i.e. chemical bond formation leading to a nuclear reaction might be possible. I couldn't think of any traditional bond formation reactions that would fall into this category.  The closest I could come is the following reaction.
$$\ce{^{2}_{1}H + ^{3}_{1}H -> ^{5}_{2}He -> ^{4}_{2}He + ^{1}_{0}n + Energy}$$
Once formed, the $\ce{^5He}$ is consumed by a nuclear decay process - there's half of what we were looking for. When we produce $\ce{^5He}$ from deuterium and trtium atoms, the two electrons that were once orbiting those separate atoms are now both orbiting a common atom - not a traditional chemical bond, but those two electrons are now occupying the same $\ce{1s}$ orbital and obeying the Pauli exclusion principle just as electrons in molecular orbitals do; and their once independent trajectories in space has been dramatically altered to where they are now orbiting a common nucleus.

It has been done, not with bond formation but rather with bond breaking, and is called the Szilard–Chalmers reaction (credit to Loong for pointing out the name).
When a biphasic mixture of ethyl iodide (with stable $\ce{^{127}I}$) and water is irradiated with neutrons, all the radioactivity from $\ce{^{128}I}$ is found as iodide in the aqueous phase; there is none in the organic phase.  This is because $\ce{^{128}I}$ is initially formed in an excited state that emits a γ-photon, and the recoil from that is more than enough to break the C–I bond.
The energy of a γ-photon is > 10 keV; in comparison, the strength of a typical chemical bond is ~ 1 to 10 eV. The C–I bond strength is roughly 2.5 eV.

To start with, fusion can be eliminated because of the intense pressure and high temperature needed. Chemicals would not exist since the electrons have been liberated from the nucleus.
When an atom undergoes fission, the release of energy would probably force the other nearby atoms away. Pitchblende is composed mostly of $\ce{UO2}$, and some $\ce{U3O8}$. $\ce{^238U}$ eventually decays to $\ce{^226Ra}$. The oxide formed would be $\ce{RaO}$. I don't know what happens to the freed electrons and oxygen ions, though.

You are correct about fusion. Fission is also too energetic.

This moves into the realm of particle physics rather than chemistry, but if you replace an electron in $\ce{H2}$ with a muon, the bond length shortens appreciably to the point where the hydrogen atoms are brought close enough together where they might fuse at room temperature. This process is known as Muon-catalyzed fusion and it would make hydrogen fusion much easier...if it weren't for the fact that it isn't easy to generate a lot of muons and that they seem to get "stuck" on a particular hydrogen and can't continue to catalyze the process.

authors: امیرحسین نخعی


text:
How to create a simple do-it-yourself procedure to measure diffusion?
While I am not asking the question here, a recent question concerning osmosis at the Bio SE caught my interest. In a nutshell that question was, given isotonic solutions of sugar inside and outside an immersed semipermeable bag, what will happen if a second solute--too big to escape through the membrane-- is added to the bag?

The question has an SAT flavor but I think it should be possible to answer it with a fairly straightforward DIY procedure. So my question here is whether the outline I have in mind would work, or whether something better can be devised.

Procedure: Make an n-molar solution of dextrose. Make an (n+x)-molar solution of dextrose [n] and a second [x] solute S (sucrose? starch? urea?--something that will not pass through the membrane). Fill the bag with the dextrose/S solution. Immerse the bag in the dextrose solution.
Osmosis alone will draw water into the bag. The question to be answered by the procedure is whether there will also be a net diffusion of dextrose out of the bag so that the final concentration of dextrose is higher outside the bag than inside, i.e., higher than the initial n-molarity.
According to one internet source, dextrose will rotate plane-polarized light about $30^o$ clockwise. If my null hypothesis is that the n-molar solution and the post-osmosis/diffusion solution outside the bag have the same molarity (i.e., that there is no net movement of dextrose out of the bag), and if I am able to construct an apparatus to measure optical rotation consistently in my kitchen, could such an apparatus be sensitive enough to reject $H_o$ if sugar in fact diffuses out of the bag?
Can anyone suggest a good practical way to set this up? Suggestions for types of membranes, necessary polarizing/light equipment, second solute S, etc. all welcome. Also, if there is a simpler/better way of going about this that would be great.
Thanks for any advice.
Notes
Sucrose reportedly rotates light more (perhaps $40^o$) but it may be too large to pass through some commercially available membranes. Fructose only rotates light about $10^o$ so I don't think it can be used here. Could sucrose be used as second solute?
The thermodynamics tag is added since I think diffusion is properly an aspect of thermodynamics (see Fermi, Thermodynamics, ch. VII). Please feel free to edit if I am mistaken.

There are way more sensitive ways to measure dextrose (glucose) concentration than polarimetry. For example, there are reducing sugar tests like the Dinitrosalicylic Acid test or (better) the 2,2'-Biconchoninic acid test. Just measure the dextrose concentrations inside and outside of the bag. I have protocols for these tests if you want; it involves getting ahold of the reagents and a spectrophotometer (unless you just want to do it qualitatively, look at the color and see if they're different), but presumably that would be no more difficult than getting hold of a polarimeter.
As far as the second solute S goes, you want something that's a lot larger than glucose. I'm only familiar with dialysis tubing that separates small molecules from proteins, so I would go with that. Cheap regenerated cellulose dialysis tubing has a molecular weight cutoff of 10,000, so use glucose (MW 180) and cheap (and soluble) Bovine Serum Albumin (MW 66,000). You always want something at least five times less than or five times greater than the cutoff for these tubes.
Be careful, there are reducing tests for sugars and reducing tests for proteins, they often use the same techniques but in different conditions, so with one protocol they'll be very sensitive for one and much less sensitive for the other. As long as you have blanks, it shouldn't be a problem, but make sure you're following the right protocol.
Or if you really want to be fancy, there's a fluorimetric test that is extremely specific to glucose, using glucose oxidase.
Or if you want to be DIY, just use a diabetes glucose meter and strips.

Very DIY. The strips sound like a good idea. They test in the range of 100 mg/dl so there might be some dilution to do. The protein is a good idea and I wasn't aware of the 5x rule for the tubes, thanks.

I thought about glucose strips. I wonder though if you could not simply weigh the water inside, and evaporate the water to find the mass of glucose and get the molarity from that. Note that test strips estimate plasma glucose from whole blood glucose so the absolute number is not be accurate. I think you can find the answer if you know the volume inside and out of the bag and the end, and then evaporate a small sample known volume on a dish to get the mass and therefore the concentration of solute-I may write it up as a solution later.

In addition, glucose will spontaneously glycosylate BSA

I first considered that you could use glucose test strips to measure the concentration of glucose inside and outside of the bag after allowing the system to come to equilibrium, but there are some issues with them.
First of all, they are not a direct measure of glucose in a solution.  An enzyme in the test strip known as glucose oxidase reacts with glucose in the aqueous component of the blood, in a hypothetically fixed volume in the test strip, and allows a current to pass through the solution.  The strength of the current is used to give the readout on the glucometer.  Rather than measuring glucose, it actually directly measures electric current.  Test strips are manufactured to specifications that allow them to very by up to 20% at 100 mg/dl or above, and to vary by any amount on 5% of tests, and they are allowed to vary by 20 mg/dl in ranges below 100 mg/dl, as well as to vary by any amount up to 5% of the time.
Second, test strips do not measure glucose concentration, but rather estimate SERUM glucose concentration from whole blood glucose concentration.
(It might be interesting to test a known concentration of glucose with several test strips to find the correlation between the measurement of whole blood and serum, as well as the deviation)
It appears to me that there is a simple, direct method of finding out what happens.  Take a known volume and concentration of a solution of glucose in a "bath". (You could even make and leave this in a large graduated cylinder).  Make another solution that has the same concentration of glucose, and add your larger solute.  Cover the system to prevent evaporation.
After equilibrium has been reached, remove the bag, measure the volume of the bath, and take a very small known volume of the solution (depending on the precision of your scale).  Weigh a watch glass, or even a microscope slide if you have 0.001 precision.  Place the known volume on the watch glass.  Weigh it  and then allow the water to evaporate and weigh the dry sugar.  That will give you the concentration of glucose in the solution after the system has reached equilibrium as well as the mass of water per volume of solution (the solute concentration will effect the overall volume of the solution).
You now know everything that changed with the solution bath.
-The change in volume
-The change in concentration, therefore the change in mass/moles of sugar
-The change in the amount of water (mass of purified volume) in the bath because the weight of the known volume of solution on the watch glass minus the weight of the sugar after evaporation will be the weight of water per volume of solution.
Since you can derive how much water and glucose moved into the bath, you can also calculate the actual amount of water and sugar that exited the bag (ie everything needed to derive anything else).
I would use starch rather than Albumin in the bag because glucose will spontaneously glycosylate protein and be bound to them, preventing them from flowing through the dialysis bag and also otherwise changing the energetics.


text:
Why pipetting with a 30° angle?
After seeing this youtube video, I was left to wonder why to use a $30^\circ$ angle between the flask and the pipette when emptying it. I can imagine it is important to touch the flask with your pipette when emptying, because the liquid prefers being attached to a solid surface due to capillary forces. When just emptying it from above the flask droplets would remain at the tip of the pipette. But why the $30^\circ$ angle? Is this an important standard or just arbitrary?

I believe what you want is "capillary". This occurs when the adhesion between the water and the glass is stronger than gravity, which is surprisingly often.

The angle of 30 degrees is not so important.
You have to make sure that the pipette is straight and the liquid runs along inside of the glass. You should not hold the pipette directly into the liquid. The drop in the tip has been included in the calibration. At the end you just have to wipe the pipette at the glass edge.

The main advantage is that you don't risk sprinkling your solution outside the beaker, so choose the angle you are more comfortable with to obtain this goal. Sometimes you don't see little drops spreading around!
It is a good practice but if you make multiple pipetting using the same pipette you could contaminate your reagent touching the beaker, so maybe sometimes emptying the pipette slowly, quite near to the solution, is the best thing to do.
Some pipettes must be completely empty when you use them, while others take into account that some amount of solution will remain inside. Check what type yours is but generally touching the beaker's side does not change the result.
I think it is more important to use gloves and cut your nails (the men in the photo seems to ignore it!).

What’s wrong with long nails? D=

@Jan eh if you don't use the gloves some residues might stay in your nails, if you have the gloves you might cut them ehehe

From my personal experience of now around five years full-time in chemistry labs with long nails the answer is: the gloves survive. (Longer than those in the picture btw.)

For volumetric instruments, such as bulb pipettes and graduated measuring pipettes, used for delivery of a liquid, the volume delivered is always less than the volume contained, due to the film of liquid left on the inner walls of the volumetric instrument. Most pipettes are calibrated ‘to deliver’ (TD, Ex); i.e. the delivered (not the contained) quantity of liquid corresponds to the value printed on the pipette.
Volumetric instruments – like any other measuring instrument – have a limited accuracy. Various sources of error are inherent in calibration and use. When the greatest possible accuracy is desired, the pipette should be used as closely as possible to the manner in which it has been calibrated. Methods for the testing, calibration and use of volumetric instruments made from glass are specified in the international standard ISO 4787:2010. The shown handling of a pipette including the angle of about 30° is part of the standardized calibration methods:
The pipette shall be held in a vertical position. Delivery shall be made into another glass vessel (the receiving vessel), with the receiving vessel held inclined at an angle of about 30°, so that the tip of the jet is in contact with the inner surface of the receiving vessel, above the level of any collected liquid, but without movement of one against the other throughout the delivery period, and finally drawing the tip over a distance of about 10 mm.
The resulting capillary effects influence the time required for the pipette to deliver its nominal volume and the volume of the film of liquid that is left on the inner walls of the pipette. Therefore, the volume delivered depends on the manner in which the pipette is used. Since capillary effects depend considerably on the material on which the liquid runs down, it is important that a receiving vessel made of glass is used.


text:
Polymer Crystallinity
See the following question:

I think the answer is A because of the hydrogen bond in the side group, thus the degree of secondary bonding between parallel chain segments is higher.  Is my thinking correct?

debatable, as glycerol and sugars often do not crystallize.

That's why the question asks for which one would be crystallized better.
Five factors that influence the efficiency of packing polymer chains:
1 - Size of side groups
2 - Extent of chain branching
3 - Tacticity
4 - Complexity of the repeat unit
5 - Degree of secondary bonding between parallel chain segments

Question is, what is "better"? Faster, or higher melting enthalpy, or degree of crystallinity? The methyl group is bulkier and thus hard to get into the right backbone conformation, and the OH group is polar, as you've noted, leading to a melting point of PVA of 230°C (iPP ~175°C).
But then, it all depends on the actual tacticity, which is not specified. The isotactic species have a totally different crystal structure than the syndiotactic, and sample with low tacticity (PVA!) might still have another one. That leads to totally different crystallisation kinetics.
So while a lot speaks for "A", I say there can be no definite answer.


text:
Does hybridization take place in other atoms apart from central atom in a molecule?
I have seen my textbooks always give formula and description of hybridization of central atom of a molecule. But I wonder will other atoms present (not Hydrogen) go in hybridization or only the central atom only uses hybrid orbital and other use pure orbital?

Yes, all atoms with p or d orbitals can hybridize in a molecule, not just the central atom.  Take acetone for example, the carbonyl carbon is $\ce{sp^2}$ hybridized and the carbonyl oxygen is also $\ce{sp^2}$ hybridized.  Another example is methyl chloride, the methyl carbon is $\ce{sp^3}$ hybridized and the chlorine is also $\ce{sp^3}$ hybridized.  As a final example consider dimethyl ether, each of the carbon atoms is $\ce{sp^3}$ hybridized and the central oxygen is also $\ce{sp^3}$ hybridized.

Ok,then can I say all atoms in a molecule use hybridised orbitals instead of pure atomic orbitals?

How to find the state of hybridization of other atoms apart from central atom?

For your first question, the answer is "yes" when we are talking about orbitals that 1) contain electrons and 2) are used to form bonds.  For your second question, look at the various bond angles (including those involving hydrogen) around the atom in question.  From this information you can infer hybridization.  For example, if the angles around an atom are all 120 degrees, then the atom is $\ce{sp^2}$ hybridized.  See my answer here for a more detailed example  http://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test/10666#10666

I do partially disagree with that: Terminal atoms are almost always (lack of external field) better described as having at the most sp hybrid orbitals. In acetone the oxygen would therefore (roughly) have 1 sp hybrid orbital forming the sigma bond, 1 p orbital for the pi bond, 1 perpendicular p orbital for the lone pair (stabilised by hyperconjugation), another sp lone pair orbital.

Hybridization is the rearrangement of the orbitals predicted by the Schrodinger Equations to coincide with the observed orbital arrangements. So any atom with electrons in the S and P orbitals (lithium and heavier elements) can have hybridization rearrangements.


text:
On Spontaneity of the Redox Reactions
According to this Mc Graw-Hill link during a spontaneous redox reaction Gibbs Free Enthalpy must decrease and at the same time the change in the Gibbs Free Enthalpy is the maximum electric work that the reaction can do:
$$\Delta G = W_\text{max}=-n F E_\text{cell}$$
in which $F$ stands for the Faraday constant, the amount of charge carried by one mole of electrons, and $n$ represents the number of moles of electrons to be transferred from the reducing agent to the oxidizing agent
according to the balanced equation, and $E_\text{cell}$ is the electromotive force of the cell, which equals $V_\text{cathod} - V_\text{anode}$, or alternatively, $E_\text{oxidizer} - E_\text{reducer}$ wherein $E$ here stands for the reduction potential. As $\Delta G$ must be nagative in the spontanous direction so $E_\text{cell}$ must be positive.
However, according to the first and second law we have
\begin{align}
&\delta Q-\delta W=\mathrm dU=T\mathrm dS-p\mathrm dV+\sum_i\mu_i\mathrm dN_i\\
&\Rightarrow\;\delta W-p\mathrm dV+\sum_i\mu_i\mathrm dN_i=\delta Q-T\mathrm dS\le0\qquad\because\text{2nd law}\\
&\Rightarrow\;\delta W-p\mathrm dV+\mathrm dG\bigr|_{p,T}\le0\\
&\Rightarrow\;\delta W-p\mathrm dV\le-\mathrm dG\bigr|_{p,T}\\
\end{align}
If $\delta W=p\mathrm dV$ then $\mathrm dG\le 0$, that is, the condition $\mathrm dG\le 0$ coincide with the 2nd law only id the only work done by the system is the pressure work.
Also if $\delta W=p\mathrm dV+\text{other works}$, then $\text{other works}\le-\mathrm dG\bigr|_{p,T}$. This means the change in the "minus Gibbs function" is the maximum work attainable from the system beside the pressure work. This extra work can be positive or negative, in the form of electric work or friction work or etc.
Now here I have some problems with the content of the link provided at the beginning of my question, as follows:

First of all, the maximum work that a system can do on its surrounding equals minus the change of the Gibbs Free Enthalpy, so the equality $\Delta G=W_\text{max}$ doesn't hold!?
If $\Delta G=W_\text{max}$ is valid, that is, I mean if the work attained is the maximum possible, then the process must be assumed as reversible , but if the process is reversible thenwe will need $\Delta G=0$ and not $\Delta G<0$, so that the process should be at equilibrium and spontaneous in no direction!?
That the Gibbs Free Energy must decrease in a closed system in the constant temperature and pressure is true only when the only type of work which is done is in the form of pressure work in expansion of the system's volume, not if other works like friction or electric works are also available, so that again using at the same time $\Delta G=W_\text{max}$ and $\Delta G\le0$ is misleading!?
The electric work is an internal work for a system containing the redox reaction and shouldn't enter the above formulation at all, as by work in the 1st law formulated above we mean only those work-based energy transport that occur at the boundaries of the system. So the electric force can never enter this formulation unless we change the control system under study, isn't it right?
And the main question. if I was wrong in analyzing the content of the provided link (which is very probable) where is my mistakes and if I was correct then how can I obtain the same result (which is certainly correct), that "in a spontaneous redox reaction we should have $E_\text{oxidizer}-E_\text{reducer}\ge0$"?

Question 1:

You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done by the system on the surroundings is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy change of the system is negative (energy is lost). Chemistry is concerned ultimately only with the system (usually the chemical reaction under investigation), and not the surroundings (apparatus, equipment, experimenter, laboratory, and the rest of the universe).
Thus, for pressure-volume work:
$$W=-p\Delta V$$
$$\delta W=-pdV$$
We need this negative sign. The system does pressure-volume work on the surroundings when the system expands (think a piston in an automotive engine) so $dV>0$. Without the negative sign, we would assume that the system gains energy and the surroundings lose energy when the system expands.
Thus, in the chemistry formulation of the first law (assuming only pressure-volume work) is:
$$dU=\delta Q + \delta W =TdS - pdV$$
In physics, some consideration is given to the surroundings (and in some cases the surroundings may be more interesting than the system, such as in the proper definition of entropy). Especially in the early days of physics, work leaving the system was useful in that it could be utilized. Thus, in physics $W=p\Delta V$ means work is being done by the system on the surroundings. Thus, the first law as formulated in physics is:
$$dU=\delta Q - \delta W = TdS -pdV$$
Note however, regardless of the sign convention in the definition of $W$ as a function of $p$ and $\Delta V$, we still end up with $$dU=TdS-pdV$$
The sign of $W_{\text{max}}$ in chemistry needs to be negative to follow the same conditions of chemistry. Thus, the equation in your chemistry textbook is correct. An equivalent definition in physics would be:
$$W_{\text{max}}=nFE_{\text{cell}}$$
$$\Delta G_\text{cell}=-W_\text{max}=-nFE_\text{cell}$$
In chemistry the negative sign goes in the definition of work. In physics the negative sign goes in front of the work term. Don't mix and match conventions, and you will get consistent answers.

Question 2:

$\Delta G = 0$ is the definition of equilibrium not the conditions for reversibility. A reaction can be reversible (and most electrochemical reactions are) and still be far from equilibrium at the initial conditions.

Question 3:

Temperature is not required to be constant here. Also, note that the term $W_\text{max}$ refers to the maximum possible electrochemical work. Usually less electrochemical work is done because heat is also generated (which changes the temperature). Also, most electrochemical systems are not closed. Even a battery is not a closed system (by the strict physics definition). It's hard to maintain constant pressure in a closed system.

Question 4:

Technically the formulation of the first law that you used is explicitly for pressure-volume work done by closed gaseous systems. As most redox reactions occur in solution in open systems, that formulation is invalid for a different reason.

Question 5:

Given that your initial equation is correct, then for a spontaneous reaction:
$$\Delta G < 0$$
$$\therefore -nFE_\text{cell} < 0$$
$$nFE_\text{cell} > 0$$
$$E_\text{cell} > 0$$
$$E_\text{red} - E_\text{ox} > 0$$

Thanks for mentioning the sign convention in chemistry about work, but that really solves no problem, since work is substituted by an expression and the minus would go inside that expression. / About the reversibility and equilibrium mention why $\Delta G=0$ is used to define the equilibrium. According to the second law we should always have $dG\le 0$ so that if $G$ reaches a local minimum that would imply an equilibrium. But a local minimum is reached whenever $dG=0$, which also occurs when the equality in the $dG\le 0$ holds. (to be continued)

That is, a reversible process is an idealistic process always being at equilibrium, a quasi equilibrium process would be still irreversible though is very close to being reversible. Therefore, if one assumes the problem as reversible then the process is always at a local equilibrium and $dG=0$ and not $dg<0$. Maybe here we should run a little bit out of precise speaking and say the process is quasi equilibrium and thus $dG<0$ although $dG\approx 0$, say, a very very slow process. That the formulation works for points far from equilibrium is another point which is interesting at its own place.

Yet I'm not convinced with the fourth part of your answer, which is the key part for me to understand where I am wrong. Actually I don't understand why you assume the system as open, in a Redox reaction there exists no source of electrons other than the reactants, and if we consider the reactants inside the system, as they have been considered by writing the term $\sum_i \mu_i dN_i$ in the internal energy change of the system, then the whole electric current and so the whole electrochemical cell is inside the system with no electric current flowing outside the cell. please explain more, thanks

I'm using the open/closed definition of open = system can exchange matter and energy with the surroundings and closed = system can exchange energy with the surroundings but not matter.

Yes, and you assume a Redox reaction involves matter transport to the environment but to me if we are considering both the reactants and the products inside the system then the system would be closed, not open, and we are indeed assuming such when $dU$ is expanded by $T dS - p dV + \sum_i \mu_i dN_i$ with $i$ counting all the reactants and products involved in the reaction. You don't agree?

Actually, I assume than an electrochemical reaction taking place in an electrochemical cell in aqueous solution almost certainly must be an open system. Just because matter can be exchanged does not mean matter must be exchanged.

Sorry I cannot follow your answer and understand why $\Delta G=W_{electric}$ and at the same time $\Delta G\le0$.


text:
How can we justify setting the affinity $\sum_i \mu_i \nu_i$ equal to zero at chemical equilibrium?
Let $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$ be a general reaction whose progress during time interval $dt$ is measured by $d\zeta$, so the amount of reactants consumed and products generated in mole would be $d N_i=\nu_i d\zeta$, wherein, $\nu_i$ would take negative values for the reactants being consumed.
From on the other hand we already know from the second law of thermodynamics that the reaction will occur in the direction in which $\sum_i \mu_i dN_i \le 0$, wherein, $\mu_i=\frac{\partial G}{\partial N_i}\bigr|_{p,T,N_j\;(j\ne i)}$ are the chemical potentials, so that we would have:
$$\left(\sum_i \mu_i \nu_i\right)\; d\zeta \le 0$$
The equality implies reversibility and should denote the equilibrium (am I right?). At the equilibrium it is clear that the progression of the reaction should vanish and we should have $d\zeta=0$, so that the term $\left(\sum_i \mu_i \nu_i\right)$ should be free to gain any positive, zero or negative finite value. However, this is not what Guggenheim has written in his Thermodynamics book. He has first assumed in a given direction the reaction progresses, so that in that direction $d\zeta>0$, then has canceled out this positive quantity from the inequality and achieved: $\left(\sum_i \mu_i \nu_i\right) \le 0$, and eventually concluded that at the equilibrium the equality holds and we should have $\left(\sum_i \mu_i \nu_i\right) = 0$.
But how is it justified to be true when we already know at the equilibrium $d\zeta>0$ doesn\t hold and we should instead write: $d\zeta=0$?
Thanks for bearing with me, I'm new to chemistry.

please specify what is $\mu_i$

@Aditya, their formal definition has been added.

Ad 1. Haven't you forgot to add a mixing entropy term? Ad 2. Maybe I misunderstand you, but I think your problem is that: It is a standard method to find a stable equilibrium point saying that the first derivative of the enthalpy etc. is zero, but even when the systems suffers a perturbation, it returns to the equilibirum. I.e. even for small, non-zero dζ the system stays in equilibrium which can only happen if  (∑iμiνi)=0.

@Greg, about the mixing entropy I'm not sure I have understood what you mean (sorry I'm new to chemistry), but the inequality mentioned is indeed derived from $dG\le 0$ at constant pressure and temperature. Actually I saw another book and yet the same reasoning was mentioned there, that $d\zeta$ is considered positive and so on. Hence, it seems no term has been forgotten here as the formulation is not mine.

@Greg, about your second point maybe you are right, I cannot judge it very easily at this point, but what you say seems more precisely addressing the question if the equilibrium achieved is a stable equilibrium. By the way, at the very equilibrium no advancement is expectable and thus we should have $d\zeta=0$. Maybe the authors are implicitly assuming that the equilibrium should also be stable, so that the term $\sum_i \mu_i \nu_i$ should be finite and preferably small, but zero is somewhat a lot idealization, or perhaps I am still wrong?

@Greg, sorry I was wrong, before equilibrium is achieved we have $d\zeta\ge 0$ and past the equilibrium point $d\zeta\le 0$, therefore, the greater be the term $\sum_i \mu_i \nu_i$ the more stable would be the equilibrium, unless it also changes in sign and honestly I don't know anything about its behavior!?

@topology I've tried giving this more of a descriptive title, but I haven't studied this stuff for [a long while], so please make any corrections to it as necessary.

@topology This equilibrium is a stable one, and from chemical experience / intuition we know that the system is going toward the equilibrium from both directions. The  dζ=0 condition is just not enough to describe the system, any combination of chemical potential can satisfy it. I think it is easier to re-order your arguments: at a given ∑iμiνi we can determine the sign of dζ, i.e. which direction the system progress. If ∑iμiνi is non-zero then there is always a dζ, which satisfies dG < 0, i.e. the reaction will minimize G. You need a vanishing  ∑iμiνi, so no dζ can further decrease G.

@jonsca, thanks for caring, I felt the question was not around setting $d\zeta$ equal to zero but about justification of $\sum \mu_i\nu_i=0$, so changed the title again, but meanwhile if you have been far from the stuff for a long time but I have never been involved in this sort of stuffs, so yet you will know better anyway.

@Greg, you are totally right that as stated in my question "any combination of chemical potential can satisfy it", which cannot be correct. So let me write $dG|_{p,T}=\frac{\partial G}{\partial\zeta} d\zeta$ and equilibrium would then be defined by $\frac{\partial G}{\partial\zeta}=0$ and not $d\zeta=0$. Thank you very much for your inspiring comments, you can prepare them as a complete answer and God-willing I'll accept it.

@Greg Would you condense your comments into an answer? Otherwise, if you don't have the time I could formulate an answer.

Chemical equilibrium is a stable equilibrium, i.e. the system returns to equilibrium to whatever perturbations. This also leads to the condition of equilibrium, as it should  contain this information.
The short answer to your question:
Due to the stability, the system stays/returns to equilibrium at any kind of small perturbation. It also means, $\left(\sum_i \mu_i \nu_i\right)\; d\zeta = 0$ not only in the equilibrium point, but also for any small $d\zeta$. Therefore the $\left(\sum_i \mu_i \nu_i\right)$ itself must be zero.
Somewhat longer answer:
The reaction goes from one direction to the other ($\sum_A \nu_A A \rightarrow \sum_B \nu_B B$)  spontaneously only till it reaches the equilibrium condition. In this direction  $\sum_i \mu_i dN_i \le 0$.
We followed the reaction from a given concentration toward equilibrium. If we follow the same $\sum_A \nu_A A \rightarrow \sum_B \nu_B B$  transformation BEYOND the equilibrium point, $ \sum_i \mu_i dN_i $ must change sign at the equilibrium point, and it must be positive should be valid (since it is not a spontaneous reaction).
Therefore the condition of equilibrium is not $d\zeta = 0 $ , but sign change: $ \sum_i \mu_i dN_i =0$.


text:
Why does the lone pair of phosphine not contribute to its dipole moment?
Unlike ammonia, the lone pair of phosphorous does not add to the dipole moment of phosphine. My textbook says this is due to the presence of lone pair of phosphorous in $\ce{PH3}$ in an s-orbital. I couldn't understand my book's explanation. How can the orbital of lone pair determine its contribution to the dipole moment?

I elaborate on that in the second half of my answer here. Also, user ron already seems to have answered that in another question you made, so you might want to reread it.

Ron has beautifully answered that ; i also understood the 1st part but couldn't decipher how the orbital of lone pair is important in determining the dipole moment. I only want to know why it happens so?

@Nikolau Saker :How does the asymmetry of an orbital give rise to dipole moment?

@user36790  Do you understand why the lone pair in ammonia contributes to the dipole moment?

@ron : Sir,yes but no! Sir, I could not understand how an asymmetric hybridised orbital can cause polarity (in case of ammonia) and why does a symmetric one cannot add to polarity (in case of phosphine).

Let's see if we can use the following picture to help clarify things.

Let's look at the molecule $\ce{H3X-Z}$.  Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$.  Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment.  Are you with me so far?
If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased.  If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$.  These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture.  The orbital and the electrons in it still "point" towards where $\ce{Z}$ used to be.  The orbital and the electrons in it still have directionality.  They still contribute to a dipole moment.  Can you see this?
Let's continue assuming you can.  Let's repeat all that we've done above with one change.  Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized.  Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital.  An $\ce{s}$ orbital is spherically symmetric.  It has no directionality, it doesn't "point" in any specific direction because it is spherically symmetric.  So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.

@ ron:sir,if instead of taking $s$ ,if I take other pure atomic orbitals,then also will I get the same result?

@user36790  No, all other orbitals, p, d and f have directionality; the distribution of electrons in them is not spherically symmetric.  Only the electrons in an s orbital are distributed in a spherically symmetric manner.

@ ron:Thanks for such an excellent answer. It is clear the orbital which has direction can be responsible for polarity.

Great, I think you've got it now!


text:
Why are amidines so much more basic than amides?
Both delocalize the added proton. Is it because the bond lengths in amidines are the same, or perhaps because oxygen is more electronegative than nitrogen and thus is more reluctant at keeping the proton? But how can that explane the huge difference in basicity between amidines and amides (resp $12$/$13$ versus $-1$/$0$ of $\mathrm{p}K_\mathrm{aH}$)?

Since amides are protonated to oxygen, the question you're asking can be rephrased as 'why oxygen is easier protonated then nitrogen' . The easiest explanation is 'electronegativity'. When protonated, oxygen has to share an electron with proton, and it is much easier for nitrogen than for oxygen.

Here is a diagram comparing what happens when we protonate an amide and an amidine.

Let's compare the protonated products in the two reactions.  In the case of the amidine we can draw two important (the fact that they are equivalent resonance structures is why they both are important) resonance structures to describe the protonated product.  In the case of the protonated amide the resonance structure with the double bond to oxygen places a positive charge on oxygen.  Therefor this resonance structure is a less significant contributor than the corresponding structure for the protonated amidine where the positive charge resides on the less electronegative nitrogen.  Since we can draw more significant resonance structures for the amidine case, the protonated amidine is more stable than the protonated amide.  This difference in protonated product stabilities explains why amidines are more basic than amides.

Uhm, the proton adds to oxygen in the amide, not to the nitrogen?

See edited version.

but can this electronegativity difference of 0.5 solely explain the guge pH difference of about 13?

I didn't say anything about electronegativity, I argued based on the relative stabilities of the protonated forms.

Yes you did. Because oxygen is a more (0.5 Pauling) electronegative than nitrogen, the positive charge is less delocalized and thus is the resonance structure less stable. Right?

Yes, that's right.  All of the textbooks that compare amides and amidines, argue that the increased basicity of the amidine is due to increased resonance stabilization of the protonated form.

Well all right, I just couldn't imagine it with such small electronegativity difference and so large pH difference. But as I dont see any other explanation, I'll accept yours until someone says we're not right haha :-)

authors: Ñąwshąđ Ãhmmęđ


text:
Database with organic reaction mechanisms (curly arrows)
Is there any database or book that has an index of organic reactions where the mechanisms are written out, complete with curly arrows? It would really help if I could look up a reaction and see a proposed mechanism with curly arrows.

As far as free online resources go, I've found the following useful:

Organic Reactions (organic-chemistry.org)
Name-reaction.com
List of organic reactions (en.wikipedia.org)

I would also strongly recommend acquiring a copy of March's Advanced Organic Chemistry at some point. I think it's fair to say that it's widely regarded as one of the most essential reference works in the field of organic chemistry. It's concise, more comprehensive than one would've thought possible, beautifully organized, and contains thousands of primary literature references. While not every reaction is presented with an accompanying mechanism, many are, and useful exposition is typically provided.
Another book you might find useful is Strategic Applications of Named Reactions in Organic Synthesis. More than a mere reference text, it includes an overview describing important applications for the reactions it covers, examples of uses in practical syntheses, as well as comprehensive stepwise mechanisms.

I second recommendation for acquiring of "March's Advanced Organic Chemistry". It is an excellent work with deep discussion of common mechanisms.

This is the closest I've seen to what you're asking for; some of the reactions have curly arrows. The Chemical Thesaurus

The answer to your question depends on what type of reactions you are looking for. If you are looking for simple 'name reactions' such as aldol condensation reaction, or Grignard reaction, then the links suggested by Greg would suffice. On the other hand, if you are looking for very specific reactions or reactions published in recent literature, then you must search in a dedicated chemistry database such as Scifinder. You may check the website here: http://www.cas.org/products/scifinder/what-s-new-in-scifinder
However, scifinder is a paid database, and your academic institution will need to have an active subscribtion to this service. Most universities in the US and the Europe have subscription to Scifinder. You will not be access this service from your home, as it is IP restricted.


text:
What makes teflon such a strong material?
I don't know much about polymers but recently teflon made me get addicted to study their properties and nature.
As far as I know, the teflon polymer contains fluorine which is one of the most reactive elements but in the polymer state it is totally stable and moreover, it is applied as a coating to kitchen-wares and many utensils which are exposed to very high temperatures.
How did a reactive element like fluorine got such an exciting stability?

Teflon (PTFE, polytetrafluoroethylene) is prepared by the polymerization of tetrafluoroethylene as shown below.  If we took polyethylene and replaced all of the hydrogens with fluorines, we would have PTFE.

Here is a Table that compares the strengths of some single bonds involving a methyl group.
\begin{array}\hline
Experimental ~Bond~ Enthalpies for~ CH_3 - X~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
I & 57.6    \\ \hline
Br & 72.1    \\ \hline
Cl & 83.7    \\ \hline
CH_{3} & 90.1    \\ \hline
H & 104.9    \\ \hline
F & 115.0    \\ \hline
\end{array}
The $\ce{C-F}$ bond is one of the strongest bonds to carbon known.  As you add additional fluorine atoms to the carbon, the bond strength increases even further.  While the $\ce{C-F}$ bond is a covalent bond, it has considerable ionic character making the carbon positively polarized and the fluorine negatively polarized.  This bond polarization results in a strong electrostatic attraction between carbon and fluorine which accounts for the high bond strength observed.  It is this very high bond strength that makes fluorocarbons so stable (unreactive).  There are not many conditions that can supply enough energy to break this very strong bond.

One condition for breaking these bonds is cooking. Cooking with teflon at even ordinary temperatures can cause the "Teflon flu," in which gases from the pyrolysis of teflon are breathed in. Birds, which have extremely high respiratory rates, have died in the kitchens of owners with teflon.

@Dissenter wrote, "One condition for breaking these bonds.." Why do you assume that the $\ce{C-F}$ bond in PTFE is involved? Is "Teflon flu" due to the PTFE breaking down or decomposition of some other material used in the manufacture and coating of PTFE that happens to be carried along with the PTFE? If it is the PTFE breaking down, is it the $\ce{C-C}$ or $\ce{C-F}$ bond that is breaking?

@ron will the same thing be applicable to pvc [poly vinyl chloride]

No, PVC has the structure $\ce{-[CH_{2}-CHCl]_{n} -}$.  Both the C-Cl and C-H bonds can react.  I don't believe that poly-(tetrachloroethylene), which would be the chlorine analogue of PTFE, exists.  I think I once read that steric hindrance from the bulky chlorines prevents polymerization - however copolymers exist.

@ron i cant understand what bulky chlorine means as chlorine and fluorine belong to the same group so there must not be such a big difference in those

@agha rehan abbas  Ah, but there is a big difference.  The atomic radius for fluorine is 42 pm, for the much larger ("bulky") chlorine it is 79 pm, almost double!