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Does radioactivity affect chemical reactions?
Do compounds of radioactive elements show a different behaviour? What happens when the radioactive part of the given compound decays? e.g. Radium chloride; what happens to it when radium decays?

As pure speculation I assume the compound falls apart. The chemical bonding is related to the number of electrons around an atom, and number of electrons is related to number of protons. Let's say an atom goes through alpha decay and loses 2 protons, this changes the charge balance and might disrupt the chemical bonding. If I had sources I'd try to make an answer.

@tschoppi Radioactive decay can be used to intentionally trigger reactions.

During the nuclear reaction, the atomic number Z (and therefore chemical identity) changes (transmutation).
In your example, $\ce{RaCl_2 -> \alpha^{2+} + RnCl_2^{2-} -> He + RnCl_2}$.
Radon being noble gas would react further, e.g. $\ce{RnCl_2 -> Rn + Cl_2}$
If the decaying atom would be a transition metal with ligands, it's properties  should not change so drastically so that the ligands will stay attached.

poor answer, the recoil energy when a decay occurs can break bonds cause other changes to occur.

I'd like to add to the other answers. Although not directly an answer to your question, I do feel it's relevant. Radioactive decay can cause damage to crystalline solids.
Take the example of apatite $\left(\ce{Ca5[PO4]3[F,Cl,OH]}\right)$, which can host a bit of uranium in the crystal structure. When this uranium decays, the resulting particle damages the host crystal. The crystal can then be polished and etched to reveal these so called "fission tracks":

(source: Stanford Fission Track Thermochronology Laboratory)
Heating of a damaged crystal will cause the damage to "heal" and the tracks to disappear. Thus, radioactive decay can indirectly induce chemical reactions. This routinely occurs in the deep Earth. It also has some very useful applications in geochronology, which you can read about in the Wikipedia article about fission tracks.

I think we need to avoid concetrating too much on the atom at which the decay occured, it is true that many radioactive decays will cause bonds to break as a result of the recoil energy. But alpha, beta and gamma radiations coming from the radioactive atom can deposit energy in the solution, liquid, gas or solid where the radioactivity is present and thus cause chemical reactions to occur.
If you irradate water then the first products are the solvated electrons and H2O cations, these can then react to form a range of reactive species such as HO and H radicals. These can then react further to give things like hydrogen gas and hydrogen peroxide. All of these reactive species can cause chemical reactions to occur.
It is interesting that a solution of a radioactive chemical will undergo autoradiolysis which can change the chemical form of the radionuclide. For example if I was to start with a concentrated plutonium solution in the +6 oxidation state in an aqueous solution. Then if I was to seal it up then the radiation could cause the formation of other oxidation states of Pu, so I would soon end up with a mixture of Pu(III), Pu(IV), Pu(V) and Pu(VI). It is possible with additves to try to control the oxidation state of the plutonium (or other metal), so in some systems the chemistry can be tamed.

In addition to radioactive decay, which would, in the case of radium, give radon and an alpha particle (helium nucleus) that will strip 2 electrons from chloride ions or anything else nearby, the radioisotope will have a different mass from the stable isotope. This in turn may affect its reaction rate . The so-called isotope effect is largest in the case of hydrogen (tritium being the radioactive and three times heavier isotope, and it reacts slower in reactions involving the C-H bond.

What about the fundamentally different chemical reactivities of the decay products? I fail to see how this question has anything to do with isotope effects.

One part of the question was do radioactive compounds show a bit different behavior. Isotope effects are part (but not all) of the answer.

I am not asking about reactions they show, but their properties How would example of NaCl (sodium-24, half life 15 hours) follow.

Until the moment a radioactive isotope decays, it behaves like a stable isotope. You do not seem to be asking about how the radioisotope differs up until the moment of decay, and yet after it decays, it is now a different element. Looks like some of the other forum members have covered it pretty well by now.


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Why is the maximal percent concentration of acids similar to their molar mass
Why is the usual percent concentration of a concentrated acid's solution similar to its molar mass?
For example:
                FW (amu)     mass %
sulfuric        98.08        98%
nitric          63.00        68%
hydrochloric    36.46        36%

Is there some reason for this similarity or is it coincidence?

Because of its solubility.
More concentrated solutions of, for example, nitric acid or hydrochloric acid, will result in fuming solutions where acid concentration evaporate out until the solution reaches an equilibrium.
Do not get confused. You can actually dissolve more than 370g of hydrochloirc acid in 630 mL of water.
In effect:
Solubility in water (Wikipedia)
823 g/L (0 °C)
720 g/L (20 °C)
561 g/L (60 °C)
But if we talk about a commercial solution, which should stay at the same concentration in time, you need to find a compromise between most quantity of acid possible in solution and less lose of acid due to vaporization.
Remember that pure hydrogen chloride is a gas at room temperature. If you continue increasing its concentration, you end up with man equilibrium shifted towards gas phase.
Also, as you can see, if you increase the temperature, solubility decreases, so if you heat a 50% aqueous solution of hydrochloric acid, you would end up not knowing what quantity of the reactant are you actually using.
Nitric acid is less volatile, and sulphuric acid is much less volatile, so you can have the latter almost in a pure form (98% is not an aqueous solution anymore! is just sulphuric acid with some water and other trace compounds)

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Is +6 really the most stable form of sulfur?
I think everyone might be aware that +6 is considered the most stable form of sulfur. But I think why this reaction happens very slowly and at such a high temperature and even the help of a catalyst is required, but in almost all redox reactions sulfur goes to +6:
(controversial where +6 is forms very forcefully):
$$\ce{SO2 +O2->[\ce{Pt/V2O5}][\text{high temp.}]SO3}$$
And also does it reduces easily (controversial where +6 is reduced easily):
$$\ce{P +SO3->P4O10 +SO2}$$
The reactions in my favour are(where stable +6 is formed):
$$\ce{SO2 +Cr2O7^2- ->SO4^2- +Cr^3+}$$
$$\ce{BaSO3- ->[\text{warming/$\ce{Br2-}$water}][\ce{/HNO3/H2O2/$\text{dil.}$HCl}]BaSO4 }$$
$$\ce{[AgSO3]- ->[\text{boil}]Ag + SO4^2- +SO2}$$
$$\ce{Ag2SO3+H2O ->Ag + SO4^2- +H+}$$
$$\ce{SO3^2- +MnO4- ->Mn^2+ +SO4^2- }$$
$$\ce{PbSO3->[\text{boil}]PbSO4}$$
$$\ce{S +HNO3->[\text{boil}]SO4^2-}$$
etc.

IDK what is happening with the silver compounds, but all the other reactions oxidizing sulfur use strong oxidation reagents. The energy released from the reduction may be used to created less stable oxidized products.

I would clarify the title. :)

I think it's a matter of activation energy and reaction kinetics. Although the product is much more thermodynamically stable than the reagent(s), the reaction path could feature a high activation energy which requires those harsh conditions and even a catalyst to make it happen. As shown in the graph below, despite having a favourable difference of free energy ($\Delta G < 0$), the activation energy $E_a$ required to go from reactants to products could be not negligible and hard to overcome.

The catalysts allows the reaction to happen more easily by lowering the free energy of the transition state (by stabilising it). The activation energy is also reduced by the high temperature, according to the Arrhenius equation:
$$k = A e^{{-E_a}/{RT}}$$

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What is the structure of Cl₂O₆?
I couldn't find which of these is correct and even someone I asked said there's no general covincement on it, but I'm unsure whether he's correct?

I reproduced the second structure a little wrong maybe, but arrangement and bonds are correct(I think there should be no charge on oxygen).

Dichlorine hexoxide has different structures depending upon whether it is in the gas or solid phase. In the gas phase, the molecule adopts structure (a) (reference 1, reference 2).  It is a relatively unstable molecule having a half-life measured in minutes.

In the solid phase it exists as the dark-red ionic material as shown in (b) and is often referred to as chloryl perchlorate.  Both forms are very strong oxidizing agents and react explosively with organic compunds.

the first refernce is only the reliable source but it also lists only the possible structure(NB i can read only the first page)

The second reference verifies the first reference; if you'd like to see the first reference in its entirety you can visit an academic library or purchase the article on line - your choice.  I provided the answer to your question.


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Why do Electrons leave the Zinc in a Galvanic Cell
What induces  electrons to leave the atoms around the part of the zinc metal in a solution to travel up along that zinc metal and above the solution, through the wire and to the copper metal and reduce the copper ions in that solution.
Is it a result of the positive metal ions in a 'sea of electrons' being attracted to the negative ions of the solution making the electrons surrounding them leave, are the electrons repelled by the electrons in the solution or do these electrons 'sense' the copper ions on the other side waiting to be reduced and flows towards them?
I'm just not sure what prompts them to leave in the first place if the positive copper ions are not directly in contact with the zinc.

First, let's think about what happens if you dip a single zinc electrode into some electrolyte solution (for now, it's not going to have copper or anything in it). This situation is subject to this equilibrium: $$\ce{Zn_{(s)}<=>Zn^2+_{(aq)} +2e-}$$
At first, there is no barrier to some of the zinc at the surface dissolving. The standard reduction potential is −0.7618 V (this is for the reverse of the reaction written above) so the oxidation process is energetically favourable. However, though the zinc ion can diffuse through the solution, there is nowhere for the two electrons to go, so they are trapped within the electrode. This excess charge opposes further oxidation—it becomes more and more difficult to force more charge into the electrode. For this reason, zinc electrodes do basically nothing in neutral solutions. N.B. It is possible for these electrons to be used to reduce the hydronium ion into hydrogen gas, but zinc adsorbs hydronium very poorly so this process is rather slow except in concentrated acid.
But what happens if we give these electrons somewhere to go? If we connect the zinc electrode to another electrode that is less easily oxidized than zinc (copper is often used because it's cheap, but it could be siver, gold, etc.), these electrons can flow into the other electrode and participate in a reduction reaction. In the case of a copper electrode, because there are no copper ions in our electrolyte, the reduction reaction that occurs (unless there is a more easily reduced substance in the electrolyte) is: $$\ce{2H3O+ + 2e- <=> H_{2(g)} +2H2O}$$
The two reactions balance each other out so we have no charge buildup to oppose the dissolution of the zinc.
To go back to the core of your question,
I'm just not sure what prompts them to leave in the first place if the positive copper
ions are not directly in contact with the zinc.

If there were any copper ions in the solution, electrons could be transferred directly from the zinc to the copper ions. If you dip a piece of zinc into a copper solution, this is exactly what happens—the zinc dissolves and small beads of copper are reduced on its surface, no second electrode required. Trying to make a zinc-copper cell with copper ion in the solution just makes the cell work worse as a good part of the zinc is lost through direct reduction of copper ions at the surface. The way to prevent this is to separate the two half-reactions with a salt bridge.
The reason the electrons leave in the first place (why the oxidation reaction above occurs) is that it is more energetically favourable for the zinc to be oxidized in the solution than to remain in its metallic state. Another way to think about is that zinc is more easily oxidized than copper so it must have electrons that are more easily lost (have a greater electrochemical potential) and these electrons will flow towards a lower electrochemical potential. Therefore, when you connect the two electrodes together, electrons flow from zinc to copper until an equilibrium is reached. Copper's electrons are not normally high enough energy to reduce the hydronium ion, but because of the electrons coming from the zinc, its electrochemical potential is raised enough for the reaction to occur.
This paper is a great reference for learning more about the lemon battery and other single solution cells (if you have ACS journal access, unfortunately).

I think the big remaining question is why does "zinc adsorbs hydronium very poorly"? The paper seems to skip any explanations and just states that as a fact. What prevents the hydronium from stealing electrons from zinc the way it steals electrons from copper?

@akhmed Here is a good primer on how it works. In short, some metals can easily form an M-H intermediate, greatly lowering the energy barrier to making H2 versus briefly having a lone H atom in the vicinity of the electrode.

This is awesome! Basically, proton needs to detach from hydronium and attach to metal first or it can't steal electrons efficiently. So for zinc electrode to work, we rely crucially on the fact that $\ce{Zn-H}$ bond is weak. If it were not so weak, hydroniums would be stealing excess electrons and we would not be able to accumulate electrons inside the zinc electrode!

I'm actually a little confused regarding your last paragraph's statement: "Copper's electrons are not normally high enough [...] ". It seems like you are describing a scenario where the electrolyte solution being used in the two half-reactions with a salt bridge is, in fact, just water. If the electrolyte solution is just water, why would electrons travel to the copper electrode (from the zinc electrode) to ultimately bind to hydronium...couldn't they have just left the zinc electrode directly to access nearby hydronium? Or does hydronium have better adsorption properties on copper than zinc?

@S.Cramer pure water is not a very good electrolyte; you still need more ions in there to make the cell work because otherwise the ionic conductivity will be very low between the two electrodes. You are on to something with your last question though: zinc can react directly with the water, making Zn2+ and H2, but hydronium adsorbs very poorly to zinc, so this happens rather slowly unless the pH is low.

@MichaelDMDryden does hydronium adsorb better to copper than it does zinc?

If you have studied semiconductors and the method of conductivity by them, mainly extrinsic semiconductors, you would recall that the movement of holes is relative and equivalently considered as the hole moving but actually the electrons are moving, in the opposite direction.
Consider the solution, copper reduces via the neighboring electrons, a very rare but favorable probability that happens similarly with zinc but with the difference that is looses some electrons in its surroundings.slowly the process starts involving electrons at far distances and suddenly when both processes approach each other, a kind of link is created which is mutually favorable and the process proceeds further more easily and now involving almost all atoms which was earlier limited to a very few atoms.
The whole process is also a particular cause of the random motion of solvent and other particles like Brownian motion. Note that random always means something that involves very small factors (do you know the butterfly effect? ) so that it is almost impossible but not exactly impossible to predict the outcome, not does it means something that is luck.

First of all, both electrodes (the copper and the zinc, in this case) are dissolving in the electrolyte. However, normally the dissolved electrode material comes out of solution just as fast. It's like trying to dig a hole in the dirt when someone else is filling it up at the exact same speed; nothing happens. Nothing happening is also called an equilibrium.
Second of all, when it dissolves, it becomes an ion, that is, it either gains or loses an electron. Now, some materials want to ionize different amounts than others. This "want" is measured in volts.
Taking into account these two factors, when you put any two electrodes together in an electrolyte, you can tell from a "Standard Electrode Potential Table" what will happen, if anything. In the case of a lemon battery, the ionic zinc wants to lose two electrons more than the copper wants to gain them. Note that one electrode must gain the electrons and one must lose them. So we have these two electrodes giving off ions into solution and also taking them back in. However, when a copper ion by chance touches the zinc metal, it transfers two electrons to the copper making metallic copper and ionic zinc. After the copper adheres to the zinc metal, it can't dissolve anymore because if it, by chance, does dissolve, the zinc metal around it instantly changes it back to metallic copper by giving it two more electrons and dissolving another zinc atom in the copper's place. Likewise, the zinc can't adhere to the copper because it would have to steal electrons from the copper, but the copper doesn't want to.
Some things to keep in mind:

This transfer of electrons is what generates the electricity in the first place.
Not all electrode pairs react.
If you want more in depth information, look up "standard reduction potential table" in Google, as well as "how to calculate the voltage of a galvanic cell".

This isn't really right because of some common misconceptions of this type of cell. The half-reaction at the copper electrode has nothing to do with copper, in a lemon battery, because there is no significant quantity of copper ions to reduce in the solution. The reduction reaction at the copper electrode is simply the production of hydrogen from the water. What happens if there is copper in the solution is that some electrons will flow through the circuit to reduce copper, but a lot of copper will simply plate onto the zinc electrode, doing no useful work in the process.

Welcome to chemistry.SE! You seem to be well aware of the rules, but I anyhow, guide you to the help center or taking a tour of the website.

@MichaelDMDryden Do zinc ions oxidate, because of the acid? You said that in neutral solutions, zinc would do nothing


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Controversy over products of iodate and bisulfite?
We know that when we are trying to recognize iodate, we use sodium bisulphite to the solution and with the help of starch paper recognize the evolved $\ce{I2}$.But when I studied the books I found actually Iodide is formed? What's the solution to this controversy?

Not much controversy here, IMO. There is more than one reaction involved, with slightly more complex kinetics.
As so often, Wikipedia already knows, so I will just quote the relevant part of the following article: http://en.wikipedia.org/wiki/Iodine_clock_reaction

Iodate variation
An alternative protocol uses a solution of iodate ion (for instance potassium iodate) to which an acidified solution (again with sulfuric acid) of sodium bisulfite is added.
In this protocol, iodide ion is generated by the following slow reaction between the iodate and bisulfite:
$$\ce{   IO3^{−} + 3 HSO3^{−} -> I^{−} + 3 HSO4^{−}}$$
This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine:
$$\ce{   IO3^{−} + 5 I^{−} + 6 H+ -> 3 I2 + 3 H2O}$$
However, the iodine is reduced immediately back to iodide by the bisulfite:
$$\ce{   I2 + HSO3^{−} + H2O → 2 I^{−} + HSO4^{−} + 2 H+}$$
When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch.

This tutorial shows you how math and chemical formulae can be nicely formatted on this site, which makes questions and answers more easily readable.

@Philipp: I appreciate your tip - but have you ever tried typing this way on a tablet? This is just not going to happen - some formatting shortcuts/buttons would be quite useful...


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Why don't unstable odd electron species dimerize so that they become stable?
Why doesn't unstable odd electron compound $\ce{NO}$ dimerize to $\ce{N2O2}$? Why doesn't this structure of ozonide dimerize? But why then does $\ce{BH3}$ dimerizes to $\ce{B2H6}$, and $\ce{AlCl3}$ to $\ce{Al2Cl6}$, both with something like bridged bonds?

Your structure of ozone is incorrect; the positive formal charge should be on the central oxygen.

Does this answer your question? Why doesn't nitrogen monoxide dimerize?

Odd electron species do dimerize.

$\ce{NO}$

Nitric oxide dimerizes, but only at low temperature (and probably high pressure}.

$\ce{NO2}$

Nitrogen dioxide does dimerize. In fact, this is a well known property of $\ce{NO2}$. $\ce{NO2}$ (orange-brown) is in equilibrium with $\ce{N2O4}$ (colorless).
$$\ce{2NO2 <=> N2O4}$$
The position of this equilibrium is dependent on temperature and pressure (like all gas equilibria), and so it makes a nice demonstration of gas properties.

$\ce{O3}$

Ozone does not dimerize. Dimerization would lead to $\ce{O6}$, which likely is a 6-membered ring of all oxygen atoms. While this structure violates no rules, the oxygen-oxygen single bond is not stable. This instability of the peroxide bond makes it both useful and dangerous.

So why does $\ce{NO2}$ dimerize readily, $\ce{NO}$ dimerize under duress, and $\ce{O3}$ not dimerize at all?

The extra electron in the nitrogen oxides is found in an antibonding orbital. Conversion to the dimer gets that electron into a bonding orbital. There are no antibonding electrons in $\ce{O3}$.

$\ce{BH3}$

Borane dimerizes because the boron atom is electron deficient - it has an empty p orbital. This species is so electron deficient that it will form bonding interactions with nearly any electron pairs, including those already in sigma bonds. (More properly the B-H-B bonds are three-center-two-electron bonds). Aluminum compounds dimerize the same way.

So why does $\ce{BH3}$ dimerize and $\ce{O3}$ does not?

Ozone has no electron deficient atoms. Count the electrons. All atoms have an octet, and while one has a positive formal charge, the compound as a whole is not electron deficient.

Molecular oxygen is pretty stable despite its single bond.

Molecular oxygen in its ground state has a bond order of 2.

http://chemwiki.ucdavis.edu/@api/deki/files/4228/image017.png?revision=1

http://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dioxygen - The Lewis dot structure of $\ce{O2}$ is insufficient to represent the molecule as a diradical with bond order of 2. https://www.youtube.com/watch?v=PFuYWFWC7-w

will the edit help explain more, i edited the question too

i'm not talking about ozone, it's ozonide!

Apologies. Your structure for ozonide looks like it has a positive charge on the rightmost oxygen atom: $\ce{^{-}O-O-O+ <-> ^{-}O-O^{+}=O}$ and not a radical: $\ce{^{-}O-O-O.}$

You might include  http://pipeline.corante.com/archives/2014/10/10/things_i_wont_work_with_peroxide_peroxides.php in your useful and dangerous list.

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A dehydrating agent acting as an hydrating agent?
We know that $\ce{P4O10}$ is a dehydrating reagent and I might list on 100's of reaction of it's dehydrating nature, mainly orgainc elimination reaction and few inorganic oxoacid dehydrations, but it is acting like a hydrating agent, cf. these:
$$\ce{PCl5 +P4O10->POCl3}$$

"Phosphorus pentachloride also forms $\ce{POCl3}$ by reaction with water, but this reaction is less easily controlled..."

Reaction with water is not always a hydration reaction.
A hydration reaction is one in which all of the atoms of the water molecule get added to the reactant to form a product:
$$\ce{MX + H2O -> MH2XO}$$
$$\ce{C2H4 + H2O}\overset{\ce{H+}}{\longrightarrow}\ce{C2H6O}$$
$$\ce{P4O10 + 6H2O -> 4H3PO4}$$
The reaction of water with phosphorous pentachloride is more properly a substitution or partial metathesis reaction. The hydrogen atoms in water are not incorporated into the product:
$$\ce{PCl5 + H2O -> POCl3 + 2HCl}$$
This reaction is difficult to control because $\ce{POCl3}$ is reactive with water also,
and under the right conditions can yield phosphoric acid or even $\ce{P4O10}$.
$$\ce{POCl3 + 3H2O -> H3PO4 + 3HCl}$$
$$\ce{4POCl3 + 6H2O -> P4O10 + 12HCl}$$
The reaction of $\ce{P4O10}$ and $\ce{PCl5}$ is thus also a partial metathesis. It is more obvious when you balance the equation. The oxygen and chlorine atoms are rearranging between the phosphorous atoms.
$$\ce{P4O10 + 6PCl5 -> 10POCl3}$$


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What happens in reaction of Lead and concentrated Nitric Acid?
I have two oppsing information, one that it forms a protective layer of $\ce{PbO}$ and doesn't reacts,other that it forms lead nitrate.Can someone clarify. NB the acid is cc.

According to the CRC Handbook of Chemistry and Physics 41st edition, lead "dissolves" in nitric acid. As is the case with other unreactive metals like copper and silver, nitric acid acts as both an acid and an oxidizing agent, and dissolves the metal with formation of the metal nitrate and evolution of nitrogen dioxide.


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Is the definition of anode and cathode failing?
We know anode=oxidation=loss of electrons and cathode=reduction=gain of electrons but in the photoelectric effect the electrons are gained at the anode and lost at the cathode of the discharge tube?
References: Anode,
Cathode, Oxidation, Reduction

entropid's answer is correct, but I'll try to express it differently. As Bard et al. state in Electrochemical Dictionary (1), the concept of anode and cathode is related only to the direction of electron flow: The anode is the electrode at which oxidation occurs (in an electrochemical cell) and electrons flow from electrolyte to electrode and the cathode is where reduction occurs and electrons flow from electrode to electrolyte. In a galvanic cell with a zinc anode, the electrons are flowing away from the interface in the electrode. Similarly, in a phototube, electrons emitted by the cathode are collected by the anode and again, electrons are moving away from the interface in the anode.
Where the electrons are actually coming from isn't important, just what direction they flow with respect to the electrode interface. Electrons moving from interface towards electrode = anode, electrons moving from electrode towards interface = cathode.
(1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008, pp 31.

I don't think this answer adequately explains the confusion. In a battery the current in the external circuit flows from the positive electrode, the cathode, to the negative electrode, the anode. The problem here is that Benjamin Franklin guessed wrongly that the charge carrier, the electron, was positively charged.

The definition of anode, according to Wikipedia, is:

An anode is an electrode through which positive electric charge flows into a polarized electrical device. The flow of charge is an electric current.

The confusion here arises from the fact that you should not look at the chamber where the photoelectric effect occurs but at the circuit instead. This way, you will notice that the anode is the one which is losing electrons (accumulated by the photoelectric effect onto it). These electrons flow through the circuit to the cathode, which is gaining electrons.

see here

The definition does not change anything. You need to consider the oxidation related to the circuit not to the chamber, which just acts as a “generator”.

but then why does anode gain electrons(wrt anode) from zinc when it oxidises to zn2+, but looses electrons(wrt zinc)?

I didn't get your comment, can you explain better?

if you start considering flow of electrons related to circuit in the galvanic cell, your argument fails, but when seen in respect of present ions it survives.

i was talking about galvanic cell

Nope it does not. In a galvanic cell the anode is the Zinc electrode, where the electrons flows away from the electrode, therefore a positive electric change flows into it, making it an anode.

no anode attracts negative charge not positive

No. Have you read the definition?

I addressed this here:
Cathode + Anode + Rechargeable Battery
There is a prevalent terminology inconsistency, where the positive terminal of a rechargeable battery is incorrectly labeled as the cathode. This is half correct, because it is correct during discharge where the positive terminal performs the reduction half reaction and receives electrons, but it is the anode during charge where it performs the oxidation half reaction and produces electrons.
A minority label the two physical terminals positive/negative instead of cathode/anode which is more correct and logical, but the more confusing notation is prevalent.


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Determine the best hydride donor
In the given question, the solution (according to the book) says that the electron withdrawing group will facilitate hydride transfer (not mentioning how so) and therefore, said the answer to be (D). But should not (C) be the answer, as the electron donating $\ce {-OCH3}$ group creates a "pushing" effect?

I would like to say that you must be knowing that the best hydride donor is almost always the entity with most partial positive charge, which is most in $\bf D$ due to strong electron withdrawing group $\bf \ce{-NO_2}$ both via inductive and resonance effects,. Remember $\bf\ce{-OCH3}$ in $\bf C$ is actually an electron donating group via resonance and electron withdrawing group via inductive effect of which former dominates largely. I think the rest two are pretty simple.

That, in fact, was my doubt- why is the compound with greatest partial positive charge the best hydride donor?

By the way -O- is an electron donating group, not electron withdrawing.

@Abhishek the reason of first point is mechanism and second point is false, consider -I

@Abhishek the O- thing was correct by you


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Non-zero dipole moment of hydroquinone
Why does hydroquinone possess a non-zero dipole moment? The $\ce{OH}$ groups present at para positions on the benzene ring should cancel the effect of each other... if there is a plane change then exactly when do atoms in a molecule change planes?

According to J. Am. Chem. Soc. 1945, 67 (2), 322–324:

Planar configurations for resorcinol and hydroquinone are more acceptable on comparison of experiment and calculation than are the structures assuming either free or partially restricted rotation of the hydroxyl groups.

It states that the dipole is present because there is restricted rotation with the hydroxyl groups. The preferred conformation for hydroquinone is a flat molecule with the hydrogens in the plane.

there are two possible planar configurations for hidroquinone, one having zero dipole moment. Why one with non-zero dipole moment dominates?

@permeakra It doesn't. The measured dipole is 1.4 D and the paper reports 2.97 D for the second conformation. So, I get a concentration of about 47.1% for the second conformation.

I don't understand why in table II, of the linked paper, the calculated value of dipole moments for all three molecules is equal. What kind of calculation would result into that?

If you look at the two possible planar structures of hydroquinone, one will have both hydroxyl hydrogens on the same side of the molecule (with a nonzero net dipole moment since the two OH dipoles are additive as vectors), whereas the other will have them on opposite sides (and a zero dipole moment since the OH dipoles will cancel each other out).

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When are the terms diphosphate and pyrophosphate used?
On the Wikipedia article about pyrophosphate, it states that:

The group (pyrophosphate) is also called diphosphate or dipolyphosphate.

But for example, why is ADP apparently called adenosine diphosphate but not adenosine pyrophosphate? The same for geranyl pyrophosphate (GPP) instead of geranyl diphosphate?  In which context the terms diphosphate and pyrophosphate are used?

Great question, but unfortunately without a great answer.  Chemistry is very inconsistent in its naming of phosphates.  In addition to the terms diphosphate and pyrophosphate, the term bisphosphate is also used.
Someone will probably put in an answer that cites "official" IUPAC or IUBMB rules.  In many cases (but not all) these rules are more sensical and well thought-out than the terms in common usage in the field, but unfortunately, they are not in universal (or even wide) use.
Here is a rough breakdown of my sense of how these three terms are used:
Inorganic chemistry

Pyrophosphate indicates a phosphate anhydride.
Diphosphate indicates a salt that contains more than one (ortho)-phosphate anion.
Bisphosphate is not widely used.

Biochemistry

Pyrophosphate: used inconsistently to name an organic derivative of a phosphate anhydride.  This is what you see in geranyl pyrophosphate (GPP).  It is also used to refer to "inorganic" pyrophosphate anion, often abbreviated as PPi.
Diphosphate: an unfortunate term that is ambiguous, although usage is shifting towards synonymity with pyrophosphate.  It is often used to mean "pyrophosphate" and but sometimes still used to mean "bisphosphate".  Biochemically, the most important phosphate anhydrides are undoubtedly ATP, ADP, etc.  These are never referred to as pyrophosphates even though that term would be more accurate.  History has chosen "triphosphate" and "diphosphate" as the terms to use for adenosine-substituted phosphate anhydrides, icky though the term may be.
Older literature will also use "diphosphate" to mean "bisphosphate".  The name "fructose diphosphate was formerly in wide usage for what is now known as "fructose-1,6-bisphosphate", for example.  That compound contains no phosphate anhydrides.
Geranyl diphosphate is still a fairly widely used synonym for geranyl pyrophosphate.  You see abbreviations for this molecule of both GPP and GDP.  I prefer the name GPP, but unfortunately not everyone uses it.
bisphosphate: Now in wide usage as the preferred term for an organic molecule with multiple ortho-phosphate esters, such as fructose 1,6-bisphosphate.  Biochemists have done a relatively good job of adopting this usage in modern times.

A common confusion that stems from this nomenclature concerns the stability of organophosphate compounds. Phosphate esters, e.g. glucose-1-phosphate or glucose-6-phosphate or adenosine monophosphate, are much more stable and less reactive than phosphate anhydrides such as acetyl phosphate (a mixed anhydride of acetic acid and phosphoric acid) or the phosphoanhydride bonds in adenosine triphosphate (ATP).

Pyrophosphate came about from heating inorganic phosphates, with the loss of water to create new compounds. Since ADP and GDP are organic compounds, diphosphate is used.
See the Dictionary definition:

Chemistry. a combining form used in the names of inorganic acids, indicating that the acid's water content is intermediate between that of the corresponding ortho- (more water) and meta- (least water) acids ...

But geranyl pyrophosphate is also an organic compound, and is not generated by heat. My guess is that the researchers studying terpenoid chemistry were different from those working on nucleotides and the different nomenclatures just stuck.

@user137 It's possible that when it was synthesized for confirmation, that they used pyrophosphate and a terpenoid to create geranyl pyrophosphate. The structure of ADP was probably determined later, and confirmed by a different team, so the diphosphate name was used. Unfortunately, a quick search finds a lot of information about geranyl pyrophosphate synthase.

Yeah, but if geranyl pyrophosphate was discovered and named before the enzyme, they would have named the enzyme after the molecule, so that's reasonable.

Pyro- is for when both phosphate are together, di- means the phosphates can occur anywhere in the molecule.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post.

@M.A.Ramezani I think it does provide an answer to the question, it just needs fleshing out a bit. Perhaps if you could expand on what you mean by 'both phosphate are together' then this could form the basis of a good answer.


text:
How to solve 1-deuterio-3-methylcyclohexene dehydrohalogenation?

Ive been struggling to solve this dehydrohalogenataion excercise.
How does the E2 mechanism work in this case and how is it affected by the stereoselectivity and antiperiplanar position? How do I get to the transition state and how does the chair conformations for the cyclohexane ring look like?

It will help to write the cyclohexane ring in the chair configuration.  What does the transition state look like, and how to get there?

Welcome to Chemistry.se! Enjoy the features of this site by taking the [tour]. Additional information can be obtained in the [help]. In accordance with our homework policy, I would like you to add some more of your thoughts towards the solution of the problem (Maybe start with @AbelFriedman's suggestion). You can simply [edit] your question.

As you noted, the chlorine and the other group being eliminated must be antiperiplanar.  In the case of the chair form of cyclohexane, this means that the chlorine and other group being eliminated must be trans-diaxial.  In the figure below I've drawn the 2 possible conformations (interconverted by a chair flip) for your deuterated cyclohexane.  Only in the top conformation is the chlorine axial, and the hydrogen on the adjacent (non-methylated) carbon is also positioned trans and axial.  Only this hydrogen can be eliminated as it is the only one that is axial and antiperiplanar to the chlorine.  Therefore, the only product produced when this material is treated with base and an E2 occurs, is the pictured 1-deuterio, 3-methylcyclohexene.  The hydrogen on the methylated carbon cannot be eliminated since it is never axial when the chlorine is axial.

The hardest part for me is to visualize the different conformations of the cyclohexane chair, it takes time and lots of practice. Thank you so much for your answer, it really cleared up all my doubts on this topic!

Your welcome!  It is extremely helpful to build a model until you become proficient in the visualization step.  If the answer was helpful, please mark it as accepted.


text:
Electrochemistry (Spontaneous Reactions)
Determine which of the following pairs of reactants will result in a spontaneous reaction at 25 celsius?
Pb2+(aq) + Cu(s)
Ag+(aq) + Br-(aq)
Li+(aq) + Cr(s)
Fe3+(aq) + Cd(s)
None of the Above.
I've tried using using tables like this http://ch302.cm.utexas.edu/images302/Electrochemistry_Reduction_Potentials.jpg to use the equation Ecell = Ecath - Eanode.
I'm hoping anyone could answer the question, and explain why each of them could be, or could not be spontaneous. (I know that an Ecell being positive results in spontaneity).
Here's my work:
Pb2+(aq) + Cu(s)
.34-(-.13) = .47
Ag+(aq) + Br-(aq)
.80 - (.76) = .04
Li+(aq) + Cr(s)
-.90-(-3.05) = 2.15
Fe3+(aq) + Cd(s)
-.04 - (-.81) = 77
None of the Above.
As you can see, they're all spontaneous. I'm wondering if I grabbed the wrong values from the above table. The question asks about ONE spontaneous reaction, from my calculations, they're all spontaneous.
@MichaelD.M.Dryden

You seem to have the correct equation and criterion for spontaneity. What exactly are you having trouble with?

@MichaelD.M.Dryden Pb2+(aq) + Cu(s)
.34-(-.13) = .47
Ag+(aq) + Br-(aq)
.80 - (.76) = .04
Li+(aq) + Cr(s)
-.90-(-3.05) = 2.15
Fe3+(aq) + Cd(s)
-.04 - (-.81) = 77
None of the Above. ------------------------------------------All of these are spontaneous. The question is asking to identify only one. This is where I'm stuck. Did I grab the wrong information from the Electro Potential Chart? (sorry for the horrible format, whenever I try to press enter, it sends the reply)

@MichaelD.M.Dryden Sorry for the previous post. But you can look at the Original Post to see my work in a more clear format. Could you tell me where I went wrong, and why? Thanks very much!

Also, see here for details on how you can format chemical equations clearly.

For $\ce{Pb^2+_{(aq)} + Cu_{(s)}<=> Pb_{(s)} + Cu^2+_{(aq)}}$, your potentials are the wrong way around. The cathode is the electrode at which the reduction occurs so it's lead minus copper.
For $\ce{2 Ag+_{(aq)} + 2 Br^{-}_{(aq)}<=> Ag_{(s)} + Br2_{(l)}}$, I don't know where you got .76 V for bromide, but that's not the right number. Also, consider the $K_{sp}$ of $\ce{AgBr}$ and think about whether a non-redox reaction is occurring.
For $\ce{3Li^+_{(aq)} + Cr_{(s)}<=> 3Li_{(s)} + Cr^3+_{(aq)}}$, your potentials are backwards again and the potential for $\ce{Cr}$ is wrong. Consider how easily elemental lithium is oxidized; just putting it in water will evolve hydrogen and oxidize the lithium, so anything that reduces the lithium cation would have to be a phenomenal reducing agent.
For $\ce{2Fe^3+_{(aq)} + 3Cd_{(s)}<=> 2Fe_{(s)} + 3Cd^2+_{(aq)}}$, you have the right order, but the wrong potential for $\ce{Cd}$.
If you're having trouble keeping the order straight, it may help to think of it this way:
$$E_{\mathrm{cell}}=E_{\mathrm{reduction}} + E_{\mathrm{oxidation}}$$ where $E_{\mathrm{oxidation}}=-E_{\mathrm{reduction}}$. It's the same equation, but instead of keeping track of anode and cathode, the half-cell where reduction is happening will match what the table says and you use the potential as-is (since it's a table of reduction potentials), and the half-cell where oxidation is happening will be backwards from that in the table, so you negate the potential.


text:
A defining feature for acids / bases
I know of 3 different definitions for acids and bases including Arrhenius, Brønsted–Lowry and Lewis. Can anyone tell me what is the common feature among these 3, that we have to label the compounds under those definitions as acids or bases? Is there any defining physical or chemical feature that guides this?

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

tl;dr version: You can only define an acid or a base in context of a reaction. Individual compounds are not acids by themselves. They are acids in reactions in which they lose protons (or gain electron pairs).

The Lewis definition is the broadest acid-base definition. Each of the others are specific subsets.
The Arrhenius definition is the simplest:

Acid = something that creates $\ce{H+}$ ions in water.
Base = something that creates $\ce{OH-}$ ions in water.

Thus any compound that can ionize in or react with water, like $\ce{HCl}$ or $\ce{HCO2H}$ is an acid and any compound that contains hydroxide ions or reacts with water to produce them, like $\ce{NaNH2}$, is a base.
However, the Arrhenius definition breaks down in non-aqueous media.
The Brønsted-Lowry definition is broader and more inclusive.

Acid = proton donor - All Arrhenius acids are Brønsted-Lowry acids
Base = proton acceptor - All Arrhenius bases are Brønsted-Lowry bases

The Brønsted-Lowry definition allows us to extend our discussion of acids and bases to, say, solutions in methanol $\ce{CH3OH}$. Note that the Brønsted-Lowry definition requires a proton transfer reaction to be occurring to properly identify acids. For example, water is amphiprotic - it can react as either an acid or a base. Many other compounds are similar. There is no absolute "this is an acid always". Instead, you have "this compound is an acid in this reaction". For example, water is an acid in the first reaction and a base in the second.
$$\ce{H2O + H- -> H2  + OH-}$$
$$\ce{H2O + H2SO4 -> H3O+ + HSO4-}$$
The Lewis definition is broader and more inclusive still. It allows extension of acids and bases into aprotic media and situations:

Acid = electron pair acceptor - all Brønsted-Lowry acids are Lewis acids
Base = electron pair donor - all Brønsted-Lowry bases are Lewis bases

Again, the Lewis definition requires a reaction to occur in order to classify compounds.
However, in all cases, there are compounds that behave as acids much more frequently than they act as bases, and vice versa. For example, $\ce{HNO3}$ is almost always an acid, but not exclusively. The first step of the mechanism of aromatic nitration is considered to be a proton transfer step between sufluric acid and nitric acid in which nitric acid is the base.
$$\ce{HOSO3H + HONO2 <=> HOSO3- + O2NOH2+}$$


text:
How to derive the energy of solvation equation?
The work done as a ion is solvated is given by: $-z^2e^2/8\pi\epsilon r$ but since the top line can be written $Q^2$, for simplicity I will say the equation is $-Q^2/8\pi\epsilon r$. Is this derived from integrating potential with respect to charge $Q$?
i.e    $$w=\int_Q^0 VdQ$$ where $$V=Q/4 \pi \epsilon r$$
I believe this would give the correct units and equation but is this an acceptable derivation?
Also, as a, sort of, side question why can't I write:
$$\triangle w=\int_Q^0 VdQ$$.
I understand that work is an inexact differential but I fail to understand the argument regarding paths - surely if you change the charge of a particle from one value to another, you must be doing the same amount of work no matter how you do it. This argument must be wrong but I don't know why and why does this mean I cant write $\triangle w$? I know you can do this for things like entropy and enthalpy because they are state functions - I just don't understand the difference. Thanks.

Although the meaning of the symbols is rather clear in this case, it is generally advisable to include the definitions of all symbols in any question or answer you post. Furthermore, could you include a reference where you got the equation from? Your substitution $Q^2 = -z^2 e^2$ is a little fishy because this would make $Q$ a complex number and this would entail a lot of trouble with the integral for the work as complex integration is not straightforward. But maybe it is just a typo and the minus sign is not part of $Q^2$.

Sorry, yes it was a typo. I have made amendments.


text:
How is 2D spectroscopy able to show quantum coherent transport through networks?
For wave-like (quantum coherent) energy transfer in networks (eg. propagation of excitation in photosynthetic protein complexes of algae or FMO complex in plants) 2D electronic (photon echo?) spectroscopy is used (eg. in this article). I have read an article (unfortunately I can't cite it here) about how it works, but I didn't understand it. I know that the off-diagonal peaks in the spectrum mean some kind of coherence in the network but I don't know why. I also don't know what kind of coherence it means. What are meant by electronic eigenstates? What are the so called 'quantum beatings'? It would be very much appreciated if somebody would be so kind as to make these thing clear.

I don't understand why you can't cite the article here.  You don't remember the title or you don't have permission?

The second. I could cite the abstract, but I don't see much point in it. It's G. R. Fleming's "Two Dimensional Electronic Spectroscopy of Molecular Complexes" by the way.

Why can't you cite it? You can generate an academic citation for anything, even personal communications.

@GeoffHutchison I can cite it, but there's no point since it is not open acces, but since two of you have asked, here you go: M. Cho, T. Brixner, I. Stiopkin, H. Vaswani, and G. R. Fleming, “Two dimensional
electronic spectroscopy of molecular complexes”, Journal of the Chinese Chemical
Society, vol. 53, no. 1, pp. 15–24, Sep. 25, 2013. doi: 10.1002/jccs.200600002.

I'm going to focus on one question here. You really have several and it'd be better if you asked these as separate questions on the site.
First off, 2D spectroscopy involves using multiple pulses, just like multi-dimensional NMR experiments. There are a variety of techniques but in all cases, you're sending some sort of pump pulse which excites the system (e.g., electronic or vibrational levels) and then probing the system over time.
The resulting data is inherently two-dimensional. You know the absorption of the system along one axis, and the emission (or second absorption) along another axis. Often, you can watch this 2D data over time, giving you even more information.
Now how does this show coherence? Let's say you have some complex system. You take a UV/Vis electronic absorption spectra and you see two peaks.

Now normally, you'd think of this system as having two distinct absorptions. Maybe it's a metal complex and you see a ligand-metal or metal-ligand charge transfer band and a d-d transition at longer wavelength.
But perhaps instead, some of the intensity in a peak comes from energy transfer. This can particularly be the case in biology or materials systems, where there are multiple photosynthetic reaction centers with ever-so-slightly different energies. So you excite one center, but then energy dumps into another one (e.g., at lower energy).
By using two-dimensional spectroscopy, that will show up as an off-diagonal peak. You'll excite at one wavelength, but it fluoresces at another one!

That's an example of coupling between oscillators (either vibrational or electronic). Now there are multiple ways that energy transfer could occur. Usually, this happens essentially as a diffusive random walk - all phase information is lost.
In some cases, the wave packet transfers from one state to another, but retains its phase. That's coherent.
In 2D photon echo spectra, you'll see oscillations in the peak heights when coherent transfer occurs, thus "beats."

For applications of 2DES to photosynthesis, I highly recommend "Two-dimensional electronic spectroscopy and photosynthesis: Fundamentals and applications to photosynthetic light-harvesting" Chemical Physics 386 (2011) 1–22.


text:
Need for bubbling hydrogen gas as part of hydrogen electrode

Why the hydrogen gas should be bubbled continuously through the electrolyte in a hydrogen electrode?

To keep the solution saturated with hydrogen gas
To clear the platinum electrode
To gate the highest electrode potential
To minimize loss of current
To speed up the process
To minimize the cost

In my opinion:

This is false because the solution doesn't need to be saturated.
Maybe true because if the platinum electrode is full with hydrogen gas which isn't bubbled continuously, there will not be enough space on platinum electrode to let another $\ce{H2}$ gas dissociate.
This is false because it doesn't relate.
Maybe true because if the hydrogen gas isn't bubbled continuously, $\ce{H2}$ will not dissociate continuously and will not transfer electrons. So there will be lack of current through this electric cell.
Maybe true because if there's lack of current, it means that the process is too slow.
False because this choice doesn't relate.

But the question asks to choose just one choice. What choice is the best? Do my reasons for each choice make sense?

Is this talking about the standard hydrogen electrode? If you want a steady response and not deplete the H2 gas you need to continuously bubble hydrogen. The correct answer is the first one, you want to saturate the solution to whatever pressure you have of hydrogen gas, otherwise the concentration in solution will change over time and so will your potential.

It is the Pressure of H2 at the electrode not the dissolved H2 in solution. The answer is probably to maintain the correct [max?] potential. The question is confusing to one who actually thinks about it because current is related to potential even tho the cell is measured at zero current draw. The questioner, as most of us, has probably never used or seen a hydrogen electrode.

The hydrogen gas is the reactant of the half reaction, so it has to be present:
$$\ce{H2(g) ->[Pt(s)] 2H+(aq) + 2e-}$$
The platinum electrode is the catalyst, and water is the solvent or proton acceptor for the hydrogen ions that are generated. Once the hydrogen adsorbed on the platinum surface is depleted, the half reaction stops unless you supply more hydrogen gas.
In a standard hydrogen electrode, the pressure of the hydrogen gas is important because the voltage measured will depend on the pressure.
I am glad I don’t have to answer the multiple choice question posted by the OP. None of the answers seem appropriate.

Hydrogen gas must be bubbled into the solution so that it is able to oxidize by losing electrons which they can't just in air. Platinum electrode need not be cleaned. Potential depends upon concentration of hydrogen, by "Nernst" equation. If hydrogen was there or not, current can never be lost, speed of process is kinetics and in no way related to thermodynamics. Cost, rubbish.

oxidize H2 or reduce H+. Last[penultimate?] sentence is confusing. Pt black takes care of the reaction rates. To measure the thermodynamics accurately [voltages] the kinetics must be sufficient, so they are relevant.


text:
Why doesn't baking soda react with the solid form of an acid?
Baking soda (powder) will react when it comes into contact with a solution of citric acid crystals dissolved in a liquid medium such as water. But why doesn't baking soda react when mixed with a solid form of an acid (citric acid crystals)? Why is water needed for a reaction to occur?
Baking Soda (Solid) + Citric Acid Solution (Liquid) = Fizzy Reaction
Baking Soda (Solid) + Citric Acid Crystals (Solid) = No Reaction (??)

The solids very likely do react, actually, just to a very limited extent since the contact between the substances is small. However, if you were to grind the powders in a mortar, the reaction should proceed to completion after enough effort. An indication would be that the initially dry powders would become clumpy (maybe even soggy) from the water produced in the reaction.

How do you know a reaction is not occurring when you mix the two solids? The only cue we have that a reaction occurs in the solution state is the formation of bubbles of escaping carbon dioxide. If there is no liquid to displace, would we be able to tell if carbon dioxide was being produced? Yes, but not with the naked eye.

The real issue is that the solid state reaction is much, much slower.

To produce carbon dioxide, the bicarbonate ion needs to come in contact with an acidic proton on citric acid. This is unlikely in the solid state since there is very little kinetic energy.
In solution, there is more kinetic energy, so the molecules are more likely to encounter each other. Additionally, the proton transfer can propagate quickly through the hydrogen bond network in water.
$$\ce{HCO3- \bond{...} H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-OH\bond{...}H-Cit}$$

Is there really more kinetic energy in the liquid state compared to the solid? I thought kinetic energy depended only on temperature, not on the phase of matter.

@NicolauSakerNeto - Total internal energy perhaps is dependent only on temperature. However, which fraction is kinetic energy is going to also depend on the identity of the substance. In a "perfect" gas-like substance with no forces between particles, almost all of the energy will be kinetic. Water is not any such thing. A lot of the energy in water is potential energy stored in the intermolecular attractions.

$\ce{KE=\frac{1}{2}mv^2}$, where the velocity will depend not only on the temperature, but also on the strength of intermolecular forces. Particles with strong interactions (both attractive and repulsive) will have inelastic collisions, and thus kinetic energy and potential energy will interconvert when particles collide.


text:
Polarity of amino acids
If I understand correctly, solubility of organic compounds depends mainly on the presence of polar groups plus putting in the factor of the number of carbon atoms in the molecule.
So putting that in mind I do not understand why Glycine, an amino acid with two polar groups, the amine and the carboxyl groups, is considered non-polar and insoluble in water .

Most amino acid molecules are polar in the sense that they have polar functional groups. Even the "least polar" amino acids like leucine and phenylalanine are likely soluble in water.
However, we also classify amino acids by the polarity of their side chains. When incorporated into proteins, clusters of amino acids with nonpolar side chains can create pockets of hydrophobicity in an otherwise aqueous environment.
Glycine is the only amino acid without a side chain. As such, it is not appropriate to classify its side chain as polar or nonpolar.
Polar acidic side chain:             aspartic acid, glutamic acid
Polar basic side chain:              arginine, histidine, lysine,
Other polar side chain:              asparagine, glutamine, serine, threonine
Nonpolar weakly acidic side chain:   cysteine, tyrosine
Other nonpolar side chain:           alanine, isoleucine, leucine, methionine,
                                     phenylalanine, proline, tryptophan, valine
No side chain:                       glycine

According to Wikipedia and my experience, glycine is soluble in water and is a polar molecule. The reference you saw, uses polar and non-polar to mean if there is another group other than the carboxylic acid and the amine group.

Solubility in water   24.99 g/100 mL (25 °C)


text:
What is the difference between "reaction in both directions" and "equilibrium"?
According to the Wikipedia page on Chemical Equations:

Symbols are used to differentiate between different types of reactions. To denote the type of reaction:
  "$=$" symbol is used to denote a stoichiometric relation.
  "$\rightarrow$" symbol is used to denote a net forward reaction.
  "$\rightleftarrows$" symbol is used to denote a reaction in both directions.
  "$\rightleftharpoons$" symbol is used to denote an equilibrium.

How is a "reaction in both directions" and an "equilibrium" any different? Aren't they supposed to be identical?

Unfortunately this is the same definition as in the goldbook and no other explanations were given.

As far as I know, they are used indistinguishably and interchangeably.
One or another could be chosen to emphasize whether you are talking about chemical equilibrium or just trying to show that a reaction can take place in both directions depending on the conditions (which can be understood as completely the same...).
I saw the $⇄$ symbol used to specify some conditions for the reaction to take place in one direction and other kind of conditions for the reaction to occur in the other direction, and $⇌$ if conditions are the same for both reactions
For example:

First case is not an equilibrium, but it's true that the reaction can go both directions using the appropriated reagents in each case. But that could not be expressed using $⇌$.
The second reaction is an equilibrium, both reactions are happening continuously under the same conditions.
It also could be expressed using $⇄$.
That is the only difference I can tell. Like Martin pointed out, IUPAC does not give more hints about it, so the question cannot be answered for 100% sure.

This seems reasonable. You can have effects in both directions without them being in equilibrium and I think that this is what is being denoted here.

What about $\ce{<<=>}$ and $\ce{<=>>}$?

A reaction in both directions may be equilibrating, but is not necessarily equilibrating (like if the system is being excited), and it certainly isn't necessarily at equilibrium.
For example, if one puts pure ammonia gas in a container, it will be far from equilibrium with respect to its decomposition products (nitrogen and hydrogen gas) but it will spontaneously proceed in the following direction:
$2\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 3\mathrm{H}_2$.
Some time after that, the rate of the reverse reaction becomes non-negligible;
$\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$,
so that both reactions simultaneously occurring can be summarized as:
$2\mathrm{NH}_3 \rightleftarrows \mathrm{N}_2 + 3\mathrm{H}_2$. That doesn't mean the system's at equilibrium, just that the forward and reverse rates are non-negligible.
Finally, the equilibrium condition (so that the concentration of all species remains constant over time) applies, and the forward and reverse rates become equivalent. That's the distinction that's specified with the double harpoon arrows:
$2\mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3\mathrm{H}_2$.
I think the harpoon arrows are most often used to specify reaction rates, so the equilibrium condition, where the forward and reverse rates balance, is marked using them.


text:
number of ions present from moles
I have $0.0585$ moles of $\ce{Al^{3+}}$, and need to find the number of actual ions. I thought I just divided by $6.022\times 10^{23}$ but I get $9.7\times 10^{-26}$, and the book says it should be $3.5\times 10^{22}$.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Hint: Think about the unit of avogadro's number $6.022\times 10^{23}$; if you divide your number of moles by avogadro's number, what unit do you get and what unit should you actually get?

You should multiply by Avogadro's number $6.022\times 10^{23}$ instead of dividing by it.
To make sure you always use a conversion factor (Avogadro's number is a conversion factor between moles and numbers) correctly, you need to think about units.

Step 1 - What are the units of the answer?

You want to find the number of ions. The units of the answer is number.

Step 2 - What are the units of the quantities you were given?

You were given 0.0585 moles of $\ce{Al^{3+}}$. The units you are given is "moles".

Step 3 - What are the units of the conversion factor?

Avogadro's number (with units) is $6.022\times 10^{23} \text{ mol}^{-1}$. The units of Avogadro's number are "number per mole".

Step 4 - Set up the problem so that units cancel and you are left with only the units you need.

Multiplying by Avogadro's number
$$\require{cancel} \cancel{\text{mol}}\times \dfrac{\text{number}}{\cancel{\text{mol}}}=\text{number}$$
The "moles" units cancel, leaving you with "number".
Dividing by Avogadro's number
$$\dfrac{\text{mol}}{\frac{\text{number}}{\text{mol}}}=\text{mol}\times \dfrac{\text{mol}}{\text{number}}=\dfrac{\text{mol}^2}{\text{number}}$$
The units do not cancel appropriately, leaving us with units that do not match the desired outcome. This approach is an application of dimensional analysis commonly called the factor-label method. With practice and conscientious use of this method, you should not make unit conversion errors.

Step 5 - Check to see if your answer makes sense.

$0.0585$ moles is about $\frac{1}{20}$ of a mole (or one half of one tenth). Your answer should be about $\frac{1}{20}$ of Avogadro's number.


text:
Can I change viscosity of an HDPE polymer?
I want to make composite with HDPE and wood, but when I melt it I have problem to mix them, because this polymer have high viscosity.

If you use an extruder to prepare your composite material, try to use HDPE with a melt flow index: MFI of 0.1 g/10 min.


text:
Where does this equation for the activity in the gas phase equilibrium between N₂, H₂ and NH₃ come from?
I have a question with Equilibrium Constant.
I understand that for a reaction in equilibrium $aA +bB \rightleftharpoons cC +dD$
$$K_c= \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
and $K_p$ is just using the partial pressures.
I'm looking through my PChem1 tutorials and I saw that they wrote for
$$\ce{1/2 N2 + 3/2 H2 <=> NH_3}$$
$$K=\frac{a(NH_3)}{a(N_2)^\frac12a(H_2)^\frac32}$$
I believe that a represents activity. However what i don't get is what follows:
$$K=\frac{a(NH_3)}{a(N_2)^\frac12a(H_2)^\frac32}=\frac{\left(\frac{n(NH_3)RT}{p_oV}\right)}{\left(\frac{n(N_2)RT}{p_0V}\right)^\frac12\left(\frac{n(H_2)RT}{p_oV}\right)^\frac32}=\frac{n(NH_3)}{n(N_2)^\frac12n(H_2)^\frac32}\left(\frac{p_oV}{RT}\right)$$
I cant' figure out how $a(NH_3)=\frac{n(NH_3)RT}{p_oV}$
It obviously has something to do with the ideal gas equation $pV=nRT$ but I don't get how.
additional context:
The question asks for the equilibrium composition starting with 1 mol each of $\ce{N_2}$, $\ce{H_2}$ and $\ce{NH_3}$ in a 1L reaction vessel at 298K
P.s. I get that I am to use extent of reaction to solve this.

The activity of a molecule $\ce{M}$ as a function of its concentration in the solution can be written as
\begin{equation}
  a_{\ce{M}} = \frac{\gamma_{\ce{M}} c_{\ce{M}}}{c^0} \ ,
\end{equation}
where $c^0$ is the standard concentration and $\gamma_{\ce{M}}$ is the (dimensionless) activity coefficient (see also: here and Wikipedia).
If you have a sufficiently dilute solution so that non-ideal contributions become very small, the activity coefficient can be approximated by 1, i.e. $\gamma_{\ce{M}} \approx 1$.
So, what you're left with for the activity under these conditions is
\begin{equation}
  a_{\ce{M}} \approx \frac{c_{\ce{M}}}{c^0} \ ,
\end{equation}
Now, the concentration is defined by $c = \frac{n}{V}$ and with the ideal gas law $\frac{n}{V} = \frac{p}{RT}$ you get:
\begin{align}
  a_{\ce{M}} &\approx \frac{c_{\ce{M}}}{c^0} = \frac{\frac{n_{\ce{M}}}{V}}{\frac{n^0}{V}} \\
  &= \frac{\frac{n_{\ce{M}}}{V}}{\frac{p^{0}}{RT}} = \frac{n_{\ce{M}} R T}{p^{0} V}
\end{align}
where $p^{0}$ is the standard pressure.

@Danxe Oh, and welcome to Chemistry.SE by the way ;) Furthermore, you can format chemical equations more easily with mhchem on this site, see here.


text:
What makes elements radioactive?
I know that certain elements like plutonium and certain isotopes of caesium are radioactive but what gives them this ability? If it's their electron/proton/neutron arrangement, what makes that arrangement give radioactive properties and how?

Temperature has no effect on the radio activeness of a species.

You'd likely find the answer in the Physics section.

Radioactivity is not at all dependent on the atomic or molecular structure: it a purely nuclear phenomenon. The only thing that is relevant is the structure of the nucleus.
The nucleus contains only protons and neutrons and the distinguishing feature that makes nuclei radioactive is, crudely, the ratio of the two. Some arrangements are unstable for a variety of reasons a full understanding of which requires fairly advanced quantum theory. What you need to remember is that some ratios or arrangements of protons and neutrons are inherently unstable and spontaneously decay in a variety of ways to more stable nuclei. For example, 39K (which contains 19 protons and 20 neutrons) is stable but adding a neutron makes 40K which is slightly unstable and spontaneously decays in several possible ways. You can tell this isn't a simple phenomenon, though, as 41K, with two extra neutrons is stable again.
It is also important to note that radioactive decay is essentially uninfluenced by anything outside the nucleus. For any given nucleus there is a probability that it will decay but that event occurs randomly and we can't tell when it will happen (but the rate of decay is stable and not influenced by temperature, pressure or the chemical compounds the nuclei are present in).

It's not correct that "Radioactivity is not at all dependent on the atomic or molecular structure". Variations in half life have been observed as a function of chemical state, see http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/decay_rates.html for examples.

@IanBush To be fair, I did qualify that in the final paragraph "essentially uninfluenced" which I think is a fair summary of an effect that is both rare and almost always small.


text:
What is the role of POCl₃ in the reaction to form (NPCl2)3 (hexachlorophosphazene)?
I did a reaction to form $\ce{(NPCl2)3}$. The reagents I used were:

phosphorus pentachloride $\ce{PCl5}$
ammonium chloride $\ce{NH4Cl}$
chlorobenzene - the solvent
phosphorus oxychloride $\ce{POCl3}$

The reaction is $\ce{3 PCl5 + 3 NH4Cl -> (NPCl2)3 + 12HCl}$
Where the chlorobenzene is the solvent.
But what role does the phosphorus oxychloride $\ce{POCl3}$ have in this reaction?
Is it some sort of catalyst? This reaction also needs to be performed under more acidic conditions, so it is the phosphorus oxychloride that creates the acidic conditions?

If your goal was to prepare hexachlorophosphazene $\ce{(NPCl2)_3}$,

then I'm not sure why $\ce{POCl3}$ was added.  The synthesis involving $\ce{POCl3}$ certainly works, but hexachlorophosphazene can be prepared more simply by heating phosphorous pentachloride and ammonium chloride in chlorobenzene.  Careful sublimation affords the desired product, pure and in high yield.  This route is more economical and less toxic since it avoids the use of $\ce{POCl3}$.
$\ce{POCl3}$ is typically added to the reaction when one is interested in preparing higher polymers of $\ce{(NPCl2)_{n}}$.  In this case, use of $\ce{POCl3}$ allows you to prepare and isolate $\ce{Cl3P=N-POCl2}~(1)$ in high yield and purity (ref. 1).  The purity is important in obtaining high molecular weight polymers of $\ce{(NPCl2)_{n}}$.  Condensation of $(1)$ to produce hexachlorophosphazene typically occurs when reaction temperatures are maintained well below $\pu{200^\circ C}$, while high molecular weight polymers are produced at temperatures over $\pu{200^\circ  C}$.

Reference:

Synthesis and Characterizations of Poly(organophosphazenes) By Roger De Jaeger, Mario Gleria. See page 26 here


text:
How can the hydrogen bonds in solid HF be best represented?

The hydrogen bonds in solid HF can be best represented as:

The correct answer is supposed to be (c), but I don't understand why this is so. Does the strength of hydrogen bonding depend on the $\ce{F-H\bond{...}F}$ or the $\ce{H-F\bond{...}H}$ bond angles?

Of course it does depend. That's one of most important things about hydrogen bonds.

At Wikipedia you find this structure (drawn by Benjah-bmm27 on Wikimedia Commons):

In the crystal, there is a hexagonal lattice of these chains, so the packing of the fluorine (green spheres) has an impact on the exact geometry of the hydrogen bonds.

So, the answer (c) is correct. Yomen Atassi correctly stated that in such a hydrogen bond the two electronegative partners and the hydrogen prefer a linear arrangement, as this maximizes the orbital overlap for the hydrogen bond. That the configuration (c) is preferred over (a) can basically be explained via the VSEPR theory: a fluoride ion in HF is surrounded by 3 electron pairs and 1 H–F bond: those 4 "ligands" should roughly be arranged tetrahedrally – approximately, not exactly, because electron pairs need more space than bonding electrons – around the F atom, and this leads to the zigzag chains from answer (c).

Here is a link to the crystal structure: https://www.researchgate.net/profile/Ezatollah-Arzi/publication/250905639_The_crystal_structure_of_deuterium_fluoride/links/5876090a08aebf17d3b910e1/The-crystal-structure-of-deuterium-fluoride.pdf?origin=publication_detail

Generally hydrogen bonds $\ce{A-H\bond{...}B}$ can be taken to be approximately linear.
In fact, the highest bonding energy for the hydrogen bond is obtained when the two electronegative atoms (fluorine here) are approximately in line with the electropositive hydrogen atom in between ($\ce{O\bond{...}H-O}$ in water, $\ce{F\bond{...}H-F}$ in hydrogen fluoride). Deviations from linearity will reduce the bonding energy rapidly. It follows that the "best" hydrogen bond is "close" to linear.

The correct combination of straight and bent bond angles may be explained with the molecular orbitals description of hydrogen bonding.
In an isolated hydrogen fluoride molecule the valence orbitals combine to give an occupied bonding orbital (upper orbital in the figure below) and an empty antibonding orbital (lower orbital):

Note the positioning: the lower-energy bonding orbital is concentrated between the atoms where the bond forms and also around the more electronegative fluorine atom, while the higher-energy antibonding orbital is mostly in less stable regions outside the bond and around the hydrogen atom.
Now introduce a second hydrogen fluoride molecule. In this molecule a nonbonding electron pair on the fluorine can overlap the antibonding orbital on the first molecule, which is mostly around the hydrogen atom; this slightly weakens the bobd within the first molecule but creates a bond between the two of them. The net additional bonding energy is the net hydrogen bonding energy. Because the nonbonding orbitals on fluorine are not aligned with the bond but the antibonding orbital is so aligned, the best hydrogen-bond overlap occurs when the geometry at hydrogen is linear but the geometry at fluorine is bent:

Molecules other than hydrogen fluoride, such as water, also tend to have molecular orbitals aligned along the bonds between hydrogen atoms and electronegative atoms, so these too commonly form the strongest intermolecular hydrogen bonds if the hydrogen is linearly between the neighboring atoms.

I like the argument you make about the orbital interactions, something that is entirely missing in the other answers. However, your first image is quite confusing in this regard. It makes it seem that linear HFH would be quite likely. It also appears to only use s orbitals, which would be incorrect. There's plenty going on in this molecular framework, I wouldn't be sure how to start unraveling it. (You also might want to check the post for typos.)

I have limited options for sketching these orbitals, which I generated on a Powerpoint presentation. You may want to improve the depiction if you have a better app/software. None of this should affect the overlaps I am describing, which are necessary to account for hydrogen bonding using less electronegative elements (even carbon!).

Indeed. I think I've expressed this similarly here: https://chemistry.stackexchange.com/a/163456/4945 Unfortunately I don't have the equipment to make nice images. I'm now even having access to something like PowerPoint. I'm just on my tablet doing some commenting.


text:
What home chemical can I use to make droplet lens?
I'm very interested in making homemade droplet lens, I want to know what home element can I use to cook them.
Here's a video showing the process, with no further information:
https://www.youtube.com/watch?v=NQLWS5j8gB4
And a research paper about it:
http://www.opticsinfobase.org/boe/fulltext.cfm?uri=boe-5-5-1626&id=284161

The paper says it's just PDMS, I doubt there are any home substitutes that would work very well. Silicone caulk, perhaps? Most of those don't cure transparent enough, I don't think, though.

There are a bunch of thermoplastics that are crystal clear when solid.  Getting them to form a nice drop like this might be a challenge though.  To get the same effect you can use the lens from a laser pointer.


text:
Is the speed of gamma rays equal to the speed of radio waves?
Does a radio wave or gamma radiation have a faster speed? I know that all light travels at $\pu{3E8 m s-1}$, but does that include these forms of electromagnetic radiation? Or do radio waves travel at $700~\mathrm{nm}$ and gamma rays $400~\mathrm{nm}$?

travels the same speed in the same medium.

nm is short for nanometers which is 1x10^-9 meters. It's a unit of distance not speed

The 700nm and 400nm are wavelengths. An electromagnetic wave of wavelength $\lambda$ still travels at the speed of light $c$, but has a different frequency , $f=c/\lambda$ and hence energy $e=hf$.

The wavelengths of radio waves are actually in the range $10^{-3}$ to $10^5$ m, whereas those of gamma rays are less than $10^{-11}$ m. As mentioned in one of the answers, $700$ nm and $400$ nm are the wavelengths of red and violet visible light respectively.

In a vacuum all light travels the same speed no matter what the wavelength or frequency. Gamma radiation has a smaller wavelength and larger frequency, radio waves a smaller frequency and larger wavelength.
Wavelength $\lambda$ and frequency $f$ are inversely proportional. As $\lambda$ gets bigger, $f$ gets smaller. But if you multiply them together they always equal a constant, the speed of light $c$:
$$\lambda \cdot ν = c$$

[...] do radio waves travel at $\pu{700 nm}$ [...]

The value $\pu{700 nm}$ is the measure of wavelength; meter is a measure of distance; speed would be meters per second.

All light travels at the same speed in vacuum. What makes different kinds of light different is their wavelength. The electromagnetic spectrum gives you the range of wavelengths that the different kinds of light have. So, $\pu{400 - 700 nm}$ would be the visible light range. Basically, the light is moving in a particular direction at speed $\pu{3x10^8 m/s}$, while the electric field is moving up and down and the magnetic field is moving left and right, both perpendicular to the direction the light is moving and perpendicular to each other. If the electric field went up as high as it can go, then went as low as it can go, and then came back up, the distance the light traveled during this time will be the wavelength, and the time it took will be the period ($t$), and the amount of times it happens in one second is the frequency($1/t$ or $f$). More energy makes the electric field oscillate faster up and down, so the wavelength is shorter. The speed of propagation remains the same.

All electromagnetic waves will travel at around the same speed in a vacuum. The differences are negligible so we just refer to the speed as $c$, or the speed of light which is $\pu{3.00E8 m/s}$.
The $\pu{700 nm}$ and $\pu{400 nm}$ you are referring to is probably referring to wavelength which is usually measured in $\pu{nm}$ because it is so small. A wavelength of $\pu{700 nm}$ and $\pu{400 nm}$ will place those waves in the visible light spectrum because the visible light runs from $\pu{390 nm}$ to $\pu{700 nm}$. Frequency also plays a big role and it is usually measured in $\pu{Hz}$ or hertz which is just another name for $\pu{1/s}$, because it is measuring how many times the wave cycles per second. Wavelength is inversely related to frequency. Gamma rays have one of the highest frequencies and smallest wavelengths in the electromagnetic spectrum while radio waves are at the opposite end with low frequency large wavelengths.


text:
Stuck with derivation of the integrated rate equation for a pseudo first order equilibrium reaction
I was reading about the integrated rate law. However I have problems to follow the solution.
I have an equilibrium reaction:
$$\ce{A + B<=>[{k_{on}}][{k_{off}}]AB}$$
with forward and back reaction. I approximate the measurement to a pseudo first order reaction. The complex can be measured.
B can be calculated:
$$[\ce{B}]\overset{\text{def}}{=}[\ce{B_0}]-[\ce{AB}]$$
Therefore the equilibrium is
$$0 = k_\mathrm{on} \cdot [\ce{A_0}] \cdot ([\ce{B_0}]-[\ce{AB}])-k_\mathrm{off} \cdot [\ce{AB}]$$
which can be expressed also like this:
$$0 = k_\mathrm{on} \cdot \ce{[A_0]} \cdot \ce{[B_0]} - (k_\mathrm{on} \cdot A_0 - k_\mathrm{off})F$$
To solve the differential equation I simplified to this:
\begin{align}
\frac{\mathrm{d}F(t)}{\mathrm{d}t} &= c_1 - c_2 \cdot F(t)\\
c_1 &= k_\mathrm{on} \cdot A_0 \cdot B_0\\
c_2 &= k_\mathrm{on} \cdot A_0 - k_\mathrm{off}\\
\end{align}
So far the theory was easy and I am sure everything is correct, but now I am stuck. The solution of this differential equation would be straightforward:
$$F(t) = \frac{c_1}{c_2}+k\cdot \mathrm{e}^{-(c_2\cdot t)}$$
However, the solution to fit and simulate the kinetic should be (in the simplified writing). This equation is published in http://afm1.pharm.utah.edu/pnscourse/Anal_Biochem_1995.pdf.
$$F(t) = \frac{c_1\cdot (1-\mathrm{e}^{-(c_2\cdot t)})}{c_2}$$
I really would like to know, where is my mistake and how the derivations of this formula has to be.

Crossposted to http://physics.stackexchange.com/q/149712/2451

In your equation 0=k_onA_0B_0-(k_onA_0-k_off )F : If I understand you correctly then F = [AB], right? If so, then I think, you have a sign mistake there: it should be 0=k_onA_0B_0-(k_onA_0 + k_off )F instead.

Maybe, but this term doesn't influence the differential equation.

Looking at your equations again, aren't your solution and the solution from the paper identical? I mean the solution from the paper is $F(t) = c_1 \frac{1 - e^{-(c_2 t)}}{c_2} = \frac{c_1}{c_2} - \frac{c_1}{c_2} e^{-(c_2 t)}$ which is exactly the form your equation has, i.e $F(t) = \frac{c_1}{c_2} + k e^{-(c_2 t)}$ with $k = -\frac{c_1}{c_2}$, right?

Yes the equation is nearly okay, and k should be -c1/c2. But k came from the integration step. Unfortunately, I don't know why it is okay to define k=-c1/c2.

You have made no mistake. Your solution and the solution from the paper are practically identical. The solution from the paper is
\begin{align}
  F(t)=c_1 \frac{1−e^{−c_2 t}}{c_2}=\frac{c_1}{c_2} - \frac{c_1}{c_2} e^{−c_2 t}
\end{align}
which is exactly the form your equation has, i.e.
\begin{align}
  F(t)=\frac{c_1}{c_2} + K e^{−c_2 t} \qquad \text{with} \qquad K = - \frac{c_1}{c_2} \ .
\end{align}
The integration constant $K$ is not defined the way it is but fixed via the boundary conditions of the reaction. In this case you get $K = - \frac{c_1}{c_2}$ by requiring that at the start of the reaction, i.e. $t=0$, the reactants have not yet reacted with each other and there is no product $\ce{AB}$ present initially, i.e. $F(t\!=\!0) = 0$. This leads to the desired result:
\begin{align}
  F(t\!=\!0) \overset{!}{=} 0 &=\frac{c_1}{c_2} + K e^{0} \qquad \Rightarrow \quad K=- \frac{c_1}{c_2} \ .
\end{align}


text:
What do short-range and long-range corrections mean in DFT methods?
Currently I am looking for the most accurate calculation method for a simple non-conjugate molecule consist of C, N, O and H. Normally I would try each method for a similar and known molecule then compare them with the experimental results but I don't have anything to compare this time.
So I need to predict a method for ground state and excited state calculations. While I was searching I kept coming across the terms "long-range" and "short-range". In the original publication of CAM-B3LYP for example, the authors claim that CAM-B3LYP provides additional long-range corrections, the weak part of B3LYP. What does it suppose to mean to me? I was unable to find any definitions of these terms.
Any additional tips about choosing a method are appreciated, also.

I think "choosing a method" would be a separate question, and you'd need to provide a bit more detail on your system.

There are many good review articles, but I really like the freely available "The ABC of DFT" by Kieron Burke (and friends) and a recent review article that tries to keep to understandable language.
The problem is treating the Coulomb interactions. Conventional DFT is not "asymptotically correct." That is, at long range, the 1/r behavior is not met, and dispersion interactions (e.g., van der Waals forces) are not properly handled either.
For example, in Neon atom (from ABC of DFT by Burke):

The problem of course is that most DFT methods start from the local density approximation (LDA) which only considers the electron density at a particular (local) point. LDA indicates that the long-range behavior falls off exponentially instead of 1/r:
$$v_{xc}^{\mathrm{LDA}}(\mathbf{r}) = \frac{\delta E^{\mathrm{LDA}}}{\delta\rho(\mathbf{r})} = \epsilon_{xc}(\rho(\mathbf{r})) + \rho(\mathbf{r})\frac{\partial \epsilon_{xc}(\rho(\mathbf{r}))}{\partial\rho(\mathbf{r})}\ $$
So some recent efforts have used "range separated" functionals, e.g.,
$$ \frac{1}{r} = \frac{1-g(r)}{r} + \frac{g(r)}{r}$$
The first (short-range) term can be handled by the generalized gradient approximation (GGA) used by basically all modern functionals. The second term is usually handled by Hartree-Fock (i.e., exact exchange), and there's some function g(r) which smoothly scales between the two.
I think modern range-separated hybrid functionals are quite good, but choosing a functional depends a bit on the property that you want to predict. Thermodynamics? Excitation energies? Oxidation/reduction potentials from orbital eigenvalues? etc.

Geoff basically answers your question. As the tip about choosing a method: if you are looking for an accuracy beyond 1 kcal/mol, then you should consider wave-function based quantum chemistry methods, e.g. coupled-cluster methods. CCSD(T) method would be very appropriate for ground state calculations, while you can use EOM-CCSD methods for the excited state calculations. However, these methods have a much worse scaling compared to DFT, so depending on the size of your molecule and the computational sources you have, these methods might be out of reach. You may want to check this review on coupled-cluster methods.

Welcome! There are definitely some CCSD(T) proponents on here. I'd say that many modern DFT functionals, particularly the range-separated hybrids do extremely well at a wide range of properties. I do suggest CCSD and CCSD(T) methods, and with modern hardware and -F12 implementations, these are much more widely applicable.


text:
Where can I find the infrared band positions for phosphorus compounds?
I am trying to interpret an IR spectrum where I have $\ce{(NPCl2)3}$
I have $\ce{P-N}$, $\ce{P=N}$ and $\ce{P-Cl}$ bonds and it is cyclic.
Where can I find or does anyone know what the general ranges for these stretches are?
I found a website that told me $\ce{P=N}$ is around $1170 \, \mathrm{cm}^{-1}$.

Have you checked the National Institute of Standards and Technology Atomic Spectra Database? http://www.nist.gov/pml/data/asd.cfm They also have a Handbook of Basic Atomic Spectroscopic Data which you may find useful: http://physics.nist.gov/PhysRefData/Handbook/Tables/phosphorustable2.htm

A good resource for IR and Raman bands is "Infrared and Raman Characteristic Group Frequencies: Tables and Charts" by George Socrates. There you can find a chart containing the positions and intensities of bands observed in the infrared spectra of phosphorus compounds (explanation: "w", "m", "s"  specify the observed intensity of the bands means whereby "w" stands for "weak", "m" for "medium", and "s" for "strong"; additionally the prefix "v" can be given meaning "very", e.g. "vs" means "very strong"):

More specifically you find the following numbers for the different bands you are looking for:

There is also another comment on $\ce{P-Cl}$ bands:

Compounds containing the $\ce{P-Cl}$ bond have a medium-to-strong absorption
  at $605 - 435$ $\mathrm{cm}^{-1}$ due to the $\ce{P-Cl}$ stretching vibration.
  The position of the band due to the $\ce{P-X}$ ($\ce{X}$=$\ce{F}$ or $\ce{Cl}$) stretching vibration is affected by the oxidation state of the phosphorus atom. 'In the presence of
  more than one halogen atom directly attached to the phosphorus atom, two
  peaks are observed due to the asymmetric and symmetric stretching vibrations
  respectively. Difluorides of the type $\ce{-P(O)F2}$ absorb at $930-895$ $\mathrm{cm}^{-1}$ and $890-870$ $\mathrm{cm}^{-1}$


text:
Chemistry Experiments at Home
I'm looking to set up a small chemistry lab in my garage for doing demos. I think it would be good for a freshman student I'm mentoring and also as a hobby. Does anyone know where I can buy apparatus (like flasks, pipettes, heaters) and chemicals?

All over the internet, e.g. http://www.amazon.com/b/ref=dp_brw_link?ie=UTF8&node=318049011

If you live close to a research university, I would recommend you to look up if they have a surplus store. Often times when professors retire a lot of the lab supplies get sent to surplus, where they are sold very cheap. You can also build equipment out of basic stuff you can find at any thrift shop (pressure cookers, blenders, etc).
As for chemicals, it really depends on what you are interested in doing. Metals are all around you. You can take some pennies, dilute hydrogen peroxide, any carboxylic acid (vinegar, citric acid, amino acid supplements, etc.), and stick the mixture into a microwave for a few seconds. The copper will form complexes with the carboxylic acid you add. Different acids will give you different colors. Leave them standing for a bit and you will get crystals.
If you get to know the plants around you, you will have access to a huge array of naturally occurring small molecules and enzymes. Some palm trees have a high concentration peroxidases, onions and garlic undergo some pretty cool chemistry which results in the formation of pigments (blue, green, pink), there's pH indicators, capsaicin, etc.
Buying pure chemicals for organic synthesis is usually tricky, and for many of them you need to be affiliated with a research institute.
There are several YouTube channels with chemistry experiments which mentions where they obtained the chemicals. I recommend 'NurdRage'.
Good luck!


text:
Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule?
This is a continuation of Is tropone aromatic?

I'm looking at the resonance form that does have the carbonyl double-bond, 1a. Can this resonance structure also be considered aromatic? Would the electrons from the C=O double bond also be delocalized through extended conjugation?
To me, this looks like a very similar situation to styrene, when we have an alkenyl sidechain attached to a benzene ring. Apparently, for alkenyl side-chains, Huckel's number is not affected, leaving an already aromatic compound to stay aromatic even though more electrons come into the conjugation.
In the case of tropone, would the C=O double bond interfere with electron counting for Hückel's rule?

In addition to Philipp's answer I'd like to share some orbital pictures which shall make the case rather obvious.
Let me shortly remind you of the key part of the IUPAC definition for aromaticity:

In the traditional sense, 'having a chemistry typified by benzene'.

I will therefore follow the same approach as in my answer to "Is phenoxide aromatic?" in comparing the orbital pictures.
The following orbitals are for the π system of benzene, in increasing order of energy (DF-BP86/def2SVP). The two orbitals on the right are degenerate:

Analogously, the π system for cyclohepta-2,4,6-trien-1-one is extensively delocalised and holds eight electrons. Note that the symmetry of this molecule is only $C_\mathrm{2v}$ removing a lot of constraint from the electronic structure. This is also why the Hückel rule breaks down, since there are no more degeneracies observed.

Also note, that the LUMO of this molecule (pictured below) still has considerable bonding character and that reduction would most likely not destroy aromaticity. (The same statement also applies to oxidation.)

Maybe my answer concerning why aniline is considered aromatic might help a bit. There I describe that groups cross-conjugated with the aromatic ring are not included in the "Hückel-count" and give some arguments for why this is so.
A similar argument would apply here: The keto group is cross-conjugated with the ring's π-system but in a different way than the amino group is in aniline because the carbonyl carbon is part of the ring. So, the role of the keto group's π system is only to act as a bridge between the $\ce{C=C}$ π-orbitals so that the ring is "closed". Thus, you wouldn't count the keto group's electrons but only the $\ce{C=C}$ π-electrons. And this would lead to the conclusion that your cyclic ketone will be aromatic if it really is flat (that will have to be checked by NMR or other methods) because there are 6 π-electrons in the ring.

So para-quinone is antiaromatic because not counting the conjugated pi electrons there are four in the ring.  Does that make sense?  Para-quinone is usually described as nonaromatic instead, and a look at molecular orbitals (even qualitatively) should convince you that the eight total pi electrons are in fact optimal for pi bonding in this molecule.

You can't apply Huckel's 4n+2 rule except for purely cyclic conjugation.  When there are pendant pi bonds like the one in tropone all bets are off on 4n+2.  You need to look at the MO structure to see if the total pi electron count, eight in this case,  is optimal for the whole molecule, not just the ring. (In tropone it is.)


text:
Resonance structures of lysine and cause of its basic nature
Lysine is one of the most important amino acids. IUPAC name it as 2,6 - diaminohexanoic acid . But what are its resonance forms?? And can resonance tell why it is basic??? Please help.

The group that constitutes the side chain is nothing but a primary amine. That explains the high pKa of ~ 10.5 (compare propylamine (pka ~10.9).  No need to invoke resonance.  Are you perhaps thinking of arginine?

+1. Sir, first of all, thanks for the answer. I did want to know the resonating structures. So, can you please??

There is just one Lewis structure.  The charge is confined to the side-chain nitrogen.

Sir,can you tell why there is dot-dot bond between $O$ atoms? Doesn't this mean resonance??

That's the carboxylate functionality, which explains the smallest pKa of ~ 2.2

The alpha amine is also basic, though not as much.


text:
Is it possible to construct a 3D molecular model using x-ray powder diffraction data?
I've got x-ray powder diffraction (XRPD) data on bendamustine HCl from a patent and I want to know how to construct a 3D model using this data with software.

It is possible to reconstruct molecular structure from powder X-ray data. It is called the Rietveld method:
http://home.wxs.nl/~rietv025/
http://www.aps.anl.gov/Xray_Science_Division/Powder_Diffraction_Crystallography/
However, it requires high quality, high resolution data (meaning generally synchrothrone powder data), and it is an involved process which requires some experience. Most probably you don't have that quality data at hand, you don't have time to learn and figure out yourself try-and-error and most probably you don't need the result either. If you need some general insight how the molecule look like, I would join to Abel Friedman and suggest using simple calculation techniques. Molecular mechanics and other low-cost methods can give very good structure with minimal effort.

What do you hope to achieve?  It is possible to obtain accurate interatomic distances from powder data (here is an example, the structures of several alkali metal cyclopentadienides), but the method is non-routine, and the data-parameter ratio is low, which means that you cannot refine many parameters and must put much information in through appropiate restraints.  You still need accurate peak shape data.
If you just want the general shape of the molecule to illustrate Wikipedia, you are better off with obgen from the OpenBabel project.  It usually gives good results.